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http://mathhelpforum.com/advanced-algebra/92352-conflicting-defintions-explanations.html
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# Thread:
1. ## Conflicting defintions/explanations
My professor states in one instance that if $A$ is invertible, then $B=A^{-1}$ when $AB=BA=I$, leading to $AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $B=A^{-1}$ implies that $B$ is invertible when $BA=AB=I$, leading to $A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $AB=BA=I$ or $BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $B=A^{-1}$ leads to $AB=BA=I$, and then later on states that it is $A=B^{-1}$ that leads to $AB=BA=I$. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.
So I need clarification on the theorem. If $AB=BA=I$ then $A$ is invertible and $B=A^{-1}$, or does $B=A^{-1}$ imply $B$ is invertible and therefore $BA=AB=I$. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.
2. Originally Posted by Pinkk
My professor states in one instance that if $A$ is invertible, then $B=A^{-1}$ when $AB=BA=I$, leading to $AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $B=A^{-1}$ implies that $B$ is invertible when $BA=AB=I$, leading to $A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $AB=BA=I$ or $BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $B=A^{-1}$ leads to $AB=BA=I$, and then later on states that it is $A=B^{-1}$ that leads to $AB=BA=I$. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.
So I need clarification on the theorem. If $AB=BA=I$ then $A$ is invertible and $B=A^{-1}$, or does $B=A^{-1}$ imply $B$ is invertible and therefore $BA=AB=I$. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.
if $A$ is a square matrix, say $n \times n,$ then $A$ is invertible if and only if $AB=I_n$ or $BA=I_n,$ for some $n \times n$ matrix $B.$ the important point, which unfortunately is not mentioned in most elementary
linear algebra courses, is that for square matrices $A,B$ we have $AB=I$ if and only if $BA=I.$ anyway, in this case we call $B$ the inverse of $A$ and similarly $A$ would be the inverse of $B.$
so for square matrices the following four statements are equivalent, meaning any of them implies the other three:
1. $AB=I,$
2. $BA=I,$
3. $A=B^{-1},$
4. $B=A^{-1}.$
for non-square matrices things are different. an $n \times m, \ n \neq m,$ matrix $A$ could have left inverse, i.e. $BA=I_m,$ for some $m \times n$ matrix $B,$ but no right inverse, i.e. $AC \neq I_n,$ for any $m \times n$
matrix $C.$ it is also possible that $A$ has a right inverse but not a left inverse (or of course $A$ might have neither left nor right inverse). but if $A$ has both left and right inverse, then it has to be
a square matrix, i.e. $n=m$ and then the left and right inverse must be equal.
3. I understand that $A=B^{-1}$ implies $B=A^{-1}$ given the necessary conditions, but my professor used a proof to show this where he stated the definition I mentioned in my first post. But he is not consistent with that definition in the proof. The definition he first gave was $AB=BA=I$ implies $A$ is invertible and thus $B=A^{-1}$. He uses this defintion to show that $A=B^{-1}$, but as I mention in my first post, he somewhat "contradicts" himself because he put great emphasis on the order of those equations.
4. Originally Posted by Pinkk
I understand that $A=B^{-1}$ implies $B=A^{-1}$ given the necessary conditions, but my professor used a proof to show this where he stated the definition I mentioned in my first post. But he is not consistent with that definition in the proof. The definition he first gave was $AB=BA=I$ implies $A$ is invertible and thus $B=A^{-1}$. He uses this defintion to show that $A=B^{-1}$, but as I mention in my first post, he somewhat "contradicts" himself because he put great emphasis on the order of those equations.
in my previous post i actually ignored what your prof said because it's nonsense. i also said that the definition is equivalent to saying that $AB=I$ or $BA=I$ because for square matrices
$AB=I$ if and only if $BA=I.$ anyway, let's just stick to your prof's definition: then in order to show that $A=B^{-1}$ we need to show that $BA=AB=I,$ which is obviously the same as
$AB=BA=I.$ so, yes, the order is not important and you're right. another point that i mentioned is the "uniqueness", i.e. if $AB=BA=I$ and $AC=CA=I,$ then $B=C.$ now we can
5. Ah okay, thank you. And yes, I had a hunch that my professor was making it more convoluted than it has to be. The if and only if statement applies only to square matrices since performing $BA$ would result in an $m \times m$ and $AC$ results in an $n \times n$, and therefore $BA \ne AC$, and if $AB$ are $n \times n$ square matrices of equal size , then $AB=BA$ since $\sum_{r=0}^{n}a_{ir}b_{rj}=\sum_{r=0}^{n}b_{ir}a_{ rj}$ for square matrices.
6. see how an MIT prof explains the inverse of a matrix. he mentions what i said, i.e. for square matrices $AB=I$ if and only if $BA=I$ but he also says that the proof is not "easy",
(well, it's not terribly hard either! haha) and so he skips it. i like this guy! his lecture is very accurate, although he talks too much! haha watch the video from 21':30" if you like:
MIT OpenCourseWare | Mathematics | 18.06 Linear Algebra, Spring 2005 | Video Lectures | detail
7. No doubt someone can prove me wrong, but I don't think there is such a thing as a one-sided inverse for square matrices. If AB=I then BA=I.
But in a more abstract setting there are certainly monoids in which one can have AB=I but not BA=I. But this can only happen in rather restricted circumstances:
it can't happen for a finite monoid
if AB=I but BA != I then there is no C such that BC=I nor D such that DA=I. If there were then it is easy to show that AB=BA=I.
8. Originally Posted by alunw
No doubt someone can prove me wrong, but I don't think there is such a thing as a one-sided inverse for square matrices. If AB=I then BA=I.
But in a more abstract setting there are certainly monoids in which one can have AB=I but not BA=I. But this can only happen in rather restricted circumstances:
it can't happen for a finite monoid
if AB=I but BA != I then there is no C such that BC=I nor D such that DA=I. If there were then it is easy to show that AB=BA=I.
For instance, the Bicyclic Monoid: $<b,c|bc=1>$.
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http://mathoverflow.net/revisions/90129/list
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2 edited title
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# Notions of orthogonality in normed spaces which are not inner product spaces
I have come across a notion of orthogonality of two vectors in a normed space not necessarily inner product space. Two vectors $x$ and $y$ in a normed space are said to be orthogonal (represented $x\perp y$) if $||x||\leq ||x+\alpha y||,$ for every $\alpha,$ a scalar.
1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$.
2) Is this unique generalization of the concept of orthogonality from inner product spaces?
Thank you.
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http://math.stackexchange.com/questions/141407/global-minimal-model-of-elliptic-curve-over-mathbbq
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# Global minimal model of elliptic curve over $\mathbb{Q}$
I am basically trying to solve the cannonball problem using elliptic curves.
In other words I have to show that the only integer points on the "elliptic curve" $6y^2 = 2x^3 + 3x^2 + x$ are $(0,0), (1,\pm 1), (24,\pm 70)$.
Now my plan is to find the torsion group via Nagell-Lutz and then show (somehow) that there is no integral point with infinite order (maybe even find that there is no point of infinite order).
My problem is with the fact that when written in standard form the curve is defined over $\mathbb{Q}$. How do I find a global minimal model for this curve?
Also does my strategy sound about right?
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The problem is mentioned in Chapter 1 of Larry Washington's textbook, Elliptic Curves: Number Theory and Cryptography. I know that doesn't answer the question - it just gives you a place to look. – Gerry Myerson May 6 '12 at 3:56
– Gerry Myerson May 6 '12 at 4:04
Well actually Washington is where I found the problem, he translates into elliptic curves and says that the problem can then be solved using some of the theorems but only gives a reference and never returns to the problem. The reference he gives is a journal entry and solves the problem in some other way (elliptic curves are never mentioned). I have just read Silverman so reckon I should know all I need to know to solve this. – fretty May 6 '12 at 8:42
Just curious, do you need the equation to be a global minimal model for some reason, or would it be enough to have an integral model? – Álvaro Lozano-Robledo May 7 '12 at 3:54
Probably an integral model will be enough. Even better if we get an isogeny between the two curves. – fretty May 7 '12 at 8:03
show 3 more comments
## 1 Answer
Tate's algorithm for computing the conductors was shown by Michael Laska to be able to be adapted for computing the minimal Weierstrass equation.
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http://math.stackexchange.com/questions/206346/big-o-with-log-base-equivalences-and-a-question-about-sum-of-series
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# Big O with Log base equivalences and a question about sum of series.
Hi I need help figuring these out:
True or False: $\log_2 n$ is $O(\log_3 n)$
I used the definition of Big O in Dasgupta's book: ${f(n)}\over g{(n)}$ $\leq c$
So I used the base transformation rule to get: ${(\log_2 3)(\log_3 n)}\over \log_3 n$ $\leq c$
Which just gives me $\log_2 3 \leq c$
So I got false as the answer, what I'm not sure about is if I did this correctly and if it makes sense in terms of Big-O notation, because I still have trouble with Big O.
For another question: $2^{(\log_2 n)} = O(2^{(\log_3 n)})$
I simplfied this to $2^{(\log_2 n)}\over 2^{(\log_3 n)}$ to $n\over n 2^{\log_3 2}$ to $1 \over 2^{\log_3 2)}$ but I'm not sure that I did that correctly when I simplified it down.
And for this question
True or False: $(\log_2 n)^2$ is $O(\log_3 n)^2$
I'm not even sure how I would begin this problem, any hints?
One last quick question about sum of series for the series $S(n) = 1^c + 2^c + 3^c + \cdots + n^c$
I get $\sum_{i=1}^n i^c$ Does that equal the equation $n^{c+1} -1 \over n-1$?
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Forget about $n=1$, the (not so good) definition has problems there. One log is a constant multiple of another. So the big O relationship trivially holds. – André Nicolas Oct 3 '12 at 4:06
## 1 Answer
$$\log_2 n = \frac{\log_3 n}{\log_3 2} = c\log_3 n\text{ where } c = \frac{1}{\log_3 2}=\log_2 3.$$ If $\log_2 n = c\log_3 n$ then $\log_2 n\le c\log_3 n$. If $\log_2 n\le c\log_3 n$ then $\log_2 n = O(\log_3 n)$.
(A crucial point here is that the quantity called "$c$" does not depend on $n$. "Constant" always means "not depending on something". What "something" is depends on the context.)
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I'm not sure how showing c = $1\over log_3 2$ $= log_2 3$ fits into showing that it is $O(log_3 n)$. Is it just showing that in this context that is what c is? And if that is the value of c the inequalities follow and that it is $O(log_3 n)$? Or well actually they would be equal in this situation correct? not less than, and is that why the answer is true? – Kevin Oct 3 '12 at 4:28
The constant is either $1/\log_3 2$ or $\log_2 3$; it's the same number either way. – Michael Hardy Oct 3 '12 at 4:47
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http://physics.stackexchange.com/questions/tagged/renormalization+ising-model
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# Tagged Questions
1answer
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### Renormalization Group and Ising with d=1 and D=1
I have a question about the results of RG on Ising model. I know it's possible to obtain two couple of relations $K'(K)$, $q(K')$ $K(K')$, $q(K)$ between the coupling costants. My problem arise ...
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### Freedom in the Choice of a Beta Functions in RG
Assume we're given a certain statistical model, say the infinite range Ising model \begin{equation} H_{N}\{\vec\sigma_{N}\}~=~ - \frac{x_{N}}{2N} \sum_{i,j =1}^{N} \sigma_{i} \sigma_{j} ...
1answer
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### What's the meaning of the coupling change after a renormalization (in the 1-dim Ising Model)?
What does it mean that after the theory (1-dim Ising model here, but the question is general) is renormalized one time and $g_i\rightarrow g_i'$, that the couplings are weaker, even if the theory ...
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### Scaling with the Ising Model
I am stuck with one formula in the CFT book by Di Francesco and al. Chapter 3. Equation 3.46 third step, for those who don't have the book, he integrates out degrees of freedom from the Ising Model by ...
2answers
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### Identifying a critical phenomena?
I have a system with a number of measurables (in time). Some measurables are discrete some are continuous (within the measurement accuracy). How can I determine whether my system experiences ...
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http://crypto.stackexchange.com/tags/discrete-logarithm/new
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Tag Info
New answers tagged discrete-logarithm
2
Trying to better understand the failure of the Index Calculus for ECDLP
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Why is the discrete log problem easy when the exponent comes from a binomial distribution?
Here's the deal. The discrete log problem is feasible in the special case where the exponent is known to come from a small range of possibilities (e.g., the exponent is not too large). Suppose we are given $y=g^m$, and we want to find $m$. Suppose moreover we know that $m$ is small: $0 \le m < 2^{30}$, say. Then it turns out it is easy to recover $m$, ...
0
What should be the size of a Diffie-Hellman private key?
The size of the modulus for ssh implementations of Diffie-Hellman (according to RFC4419 from 2006) should be between 1024 and 8192 bits. This seems to be a bit off from what @poncho recommended (128 bits) in 2012.
Top 50 recent answers are included
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http://mathoverflow.net/questions/109528/character-degree-in-gap
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## character degree in Gap
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let S be a finite set of integers, do I can check with Gap that this set be a set of character degrees of small group?
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No one has mentioned yet: the order of the group must be a multiple of the LCM of the elements of S. – John Wiltshire-Gordon Oct 13 at 19:38
## 2 Answers
Dima is correct that the problem is much easier with multisets. Furthermore, taking direct products with Abelian groups, it is clear that for any given set of irreducible character degrees which actually occurs, there are infinitely many groups for which it occurs, if we allow multiplicities. I had a recollection that Isaacs and Passman proved a kind of converse of this: if a finite group $G$ has all its irreducible character degrees of degree less than $d,$ then $G$ has an Abelian normal subgroup $A$ whose index is bounded in terms of $d.$ I could not locate a reference at the time, though I did not have Isaacs' book to hand. Since my original post, however, Yemon Choi kindly provided a reference in the comments below (see MR0167539). The wording of Isaacs and Passman's result is perhaps mildly ambiguos: they speak of an Abelian subgroup of index bounded in terms of $d$, omitting the word "normal", but of course if $G$ has an Abelian subgroup $A$ with $[G:A] \leq h,$ then the intersection of the $G$-conjugates of $A$ is normal, and has index at most $h!$. Hence, in general, if we know the irreducible character degrees of a finite group $H,$ there is an Abelian normal subgroup $A$ of $H$ such that $|H/A|$ is bounded in terms of the largest degree. So in small cases, it may be possible to use GAP to determine all possibilities for $H/A.$
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I am out of the office right now so cannot view it, but I think the result of Isaacs and Passman that you refer to is in their paper Groups with representations of bounded degree, Canad. J. Math. 16 (1964) 299--309. ams.org/mathscinet-getitem?mr=167539 – Yemon Choi Oct 13 at 23:58
Thanks, that figures- I put a date range in Mathscinet starting at 1965! – Geoff Robinson Oct 14 at 1:24
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You most probably mean "the set of degrees of irreducible characters". Indeed, given a finite set of integers, it should not be hard to construct a finite group that has irreducible characters of that degrees (and many more other degrees).
Still, for "the set of degrees" the problem does not look easy, as degrees can repeat, and you don't have a good way to control the order of the group $G$. For "the multiset of degrees" $d_1$,...,$d_t$ things are much better, as then $G$ has order equal to $\sum_{j=1}^t d_j^2$, and so you can just look through GAP's database of small groups for such a group.
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http://mathoverflow.net/questions/32088?sort=newest
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## Motivation
One of the methods for strictly extending a theory $T$ (which is axiomatizable and consistent, and includes enough arithmetic) is adding the sentence expressing the consistency of $T$ ( $Con(T)$ ) to $T$. But this extension ( $T+Con(T)$ ) looks very artificial from the mathematical viewpoint, i.e. does not seem to have any mathematically interesting new consequences, and therefore is probably of no interest to a typical mathematician.
I would like to know if there is a natural theory (like PA, ZFC, ... ) which by adding the consistency statement we can prove new mathematically interesting statements. I don't have a definition for what is a natural theory or a mathematically interesting statement, but a theory artificially build for the sole purpose of this question would not be natural, and a purely metamathematical statement (like consistency of $T$, or a statement depending on the encoding of $T$ or its language, or ...) would not count as a mathematically interesting statement.
Questions:
1. Is there a natural theory $T$ and an mathematically interesting statement $\varphi$, such that it is not known that $T \vdash \varphi$, but $T + Con(T) \vdash \varphi$?
2. Is there a natural theory $T$ and an interesting mathematical statement $\varphi$, such that $T \nvdash \varphi$ but $T + Con(T) \vdash \varphi$?
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I've been told that the consistency of ZFC can be used to give a construction of the Rado graph (the "unique" countably infinite random graph), but there are many other constructions that do not require this. – Charles Staats Jul 16 2010 at 0:20
You may be interesed in this related question: mathoverflow.net/questions/12865/… – Joel David Hamkins Jul 16 2010 at 1:39
This question mathoverflow.net/questions/26411/… is also related, in that it inquires about adding less than Con(T) to T (but no naturality requirement). – Joel David Hamkins Jul 16 2010 at 1:44
## 3 Answers
Vitali famously constructed a set of reals that is not Lebesgue measurable by using the Axiom of Choice. Most people expect that it is not possible to carry out such a construction without the Axiom of Choice.
Solovay and Shelah, however, proved that this expectation is exactly equiconsistent with the existence of an inaccessible cardinal over ZFC. Thus, the consistency statement Con(ZFC + inaccessible) is exactly equivalent to our inability to carry out a Vitali construction without appealing to AC (beyond Dependent Choice).
Thus, if $T$ is the theory $ZFC+$inaccessible, then T+Con(T) can prove "You will not be able to perform a Vitali construction without AC", but $T$, if consistent, does not prove this.
I find both this theory and the statement to be natural (even though the statement can also be expressed itself as a consistency statement). Most mathematicians simply believe the statement to be true, and are often surprised to learn that it has large cardinal strength.
There is another general observation to be made. For any consistent theory $T$ whose axioms can be computably enumerated, and this likely includes most or all of the natural theories you might have in mind, there is a polynomial $p(\vec x)$ over the integers such that $T$ does not prove that $p(\vec x)=0$ has no solutions in the integers, but $T+Con(T)$ does prove this. So if you regard the question of whether these diophantine equations have solutions as natural, then they would be examples of the kind you seek. And the argument shows that every computable theory has such examples.
The proof of this fact is to use the MRDP solution of Hilbert's 10th problem. Namely, Con(T) is the assertion that there is no proof of a contradiction from $T$, and the MRDP methods show that such computable properties can be coded into diophantine equations. Basically, the polynomial $p(\vec x)$ has a solution exactly at a Goedel code of a proof of a contradiction from $T$, so the existence of a solution to $p(\vec x)=0$ is equivalent to $Con(T)$. If $T$ is consistent, then it will not prove $Con(T)$, and so will not prove there are no integer solutions, but $T+Con(T)$ does prove that there are no integer solutions.
By the way, it is not true in general that if $T$ is consistent, then so is $T+Con(T)$. Although it might be surprising, some consistent theories actually prove their own inconsistency! For example, if PA is consistent, then so is the theory $T=PA+\neg Con(PA)$, but this theory $T$ proves $\neg Con(T)$. Thus, there are interesting consistent theories $T$, such as the one I just gave, such that $T+Con(T)$ proves any statement at all!
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Joel, I believe the question is about $T+Con(T+\varphi)$ proving a statement $\psi$, not about $Con(T+\varphi)\leftrightarrow Con(T+\psi)$ being provable in $T$ (or in a weaker system). Your example is of the second kind. – Andres Caicedo Jul 16 2010 at 0:34
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My statement $\psi$ was "our inability to carry out a Vitali construction without using AC." While it is true that this statement can be thought of as a consistency statement, it needn't be. – Joel David Hamkins Jul 16 2010 at 0:45
It is really great answer! – Sergei Tropanets Jul 17 2010 at 1:03
This is a nice result that I was not aware of. However, Kaveh might disqualify it as a "metamathematical" statement. – Timothy Chow Jul 20 2010 at 2:44
Thank you Joel. The first half is really nice, but I have to think about it a little more. The second part refers to the theory since the polynomial from the MRDP theorem will depend on the encoding of the theory. This will make it metamathematical. – Kaveh Jul 21 2010 at 23:46
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The Paris–Harrington Theorem is equivalent (over IΣ1) to Con(PA + Tr(Π1)), where Tr(Π1) is the set of all true Π1 sentences in the language of arithmetic.
For clarity, the 1-consistency of PA, i.e. what is meant by Con(PA + Tr(Π1)), is the following statement:
If φ is a (Gödel code for a) true Π1 sentence, then PA ⊬ ¬φ.
Or the dual reflection principle:
If φ is a (Gödel code for a) Σ1 sentence and PA ⊦ φ, then φ is true.
Note that IΣ1 proves the existence of Σ1 truth predicates, so the above is expressible as a Π2 statement.
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In fact, most "combinatorial" statements (Hydra games, Kanamori-McAloon, ...) are equivalent to this theory. On the other hand, the theory is not recursive, so I am not sure it qualifies as "natural". (I find it natural, but I also think that a formal definition of naturalness ought to include that it is recursive.) – Andres Caicedo Jul 16 2010 at 0:26
By a natural theory I mean something that ordinary mathematicians use, i.e. prove theorems using it (theorems inside it, not about it), so I think being recursively enumerable is probably a necessary condition. – Kaveh Jul 23 2010 at 0:32
PA is recursively enumerable, it's just that 1-consistency is stronger than just plain consistency. – François G. Dorais♦ Jul 23 2010 at 1:00
The short answer is no. Con(T) is a very weak assumption and it is asking a lot for it to have interesting mathematical consequences. A slightly less ambitious question is whether "ZFC + the consistency of some large cardinal axiom" has any interesting mathematical consequences. Here the work of Harvey Friedman is relevant, as I explained in this answer to a related MO question. I don't think Friedman's examples are quite there yet but they're getting close.
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Note that Friedman uses not simple consistency but the rather stronger notion of Sigma-1 soundness. – Charles Stewart Jul 17 2010 at 12:34
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http://math.stackexchange.com/questions/210826/probability-of-tossing-a-coin
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Probability of tossing a coin
This question comes from an exercise in book: If I'm tossing coins. I tossed 2 heads in a row using a coin. What’s the probability now that the next coin will be heads?
Here's what I thought:
Tossing a coin three times are independent events. So,
Probability of getting a head will be 1/2 = 50 percent chance.
Probability of getting a head second time will be = 1/4 = 25 percent chance.
Probability of getting a head third time will be = 1/8 = 12.5 percent chance.
Is this correct? Because the answer says that it would be 50 percent.
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throws are assumed independant of each other, getting $2,4,40$ heads in a row doesn't change the probability of the next throw. – Jean-Sébastien Oct 11 '12 at 2:09
What you compute is the probability of having $3$ Heads in a row, what you really want is to have a Head, given that you just had $2$ – Jean-Sébastien Oct 11 '12 at 2:09
What would be the correct wording for a question according to my answer? – user121314 Oct 11 '12 at 2:18
Of course it is "50 percent" or $\frac{1}{2}$, since previous tosses do not affect current outcome. In case you want $HHH$ out of three tosses, it is $\frac{1}{2^3}$ – Alex Oct 11 '12 at 2:31
1 Answer
To answer your question in your comment, If you have seen conditionnal probabilities, denote $A$ the event getting a HeadS on the third throw, and $B$ the event getting HeadS on throw $1$ and $2$. Argue that $A$ and $B$ are independant and so $\Pr(A|B)=\Pr(A)=\frac{1}{2}$.
If you haven't seen conditional probabilities, then you can say that since each throws are independant (I at least assume so), the first run of HeadS does not change the probabilities, so it remains $1/2$
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http://mathoverflow.net/questions/59422/cyclotomic-polynomials/59426
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## Cyclotomic Polynomials
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Let $\phi_{n}(x)$ be the $n$-th cyclotomic polynomial. What are the restrictions to $n$ (if any) to have $\phi_{n}(x)$ divides $\phi_{2n}(x)$ (where division is in $\mathbb{Z}[x]$)?Or is it true that $\frac{\phi_{2n}(x)}{\phi_{n}(x)}\in\mathbb{Z}[x]$ for all integers $n$?
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Oh I see...but is it still impossible to have "$\phi_{n}(x)$ divides $\phi_{2n}(x)$" (not necessarily over \mathbb{Z}[x])? – Kikiriku Mar 24 2011 at 13:04
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It's just not possible. – Charles Matthews Mar 24 2011 at 15:05
## 3 Answers
When is a primitive *n*th root of unity also a primitive 2*n*th root of unity? Please note that the answer is never, and this can also be seen by unique factorisation.
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Clear...thx.... – Kikiriku Mar 24 2011 at 13:11
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Those polynomials are irreducible in $\mathbb Z[X]$ and have different degree... see http://en.wikipedia.org/wiki/Cyclotomic_polynomial
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well, $\phi_{2011}$ and $\phi_{4022}$ have the same degree. – Xandi Tuni Mar 24 2011 at 13:10
Maybe I should think a bit before writing... Of course those polynomials will very often have the same degree (as soon as n is odd)... Kikiriku's answer is much better and does not use the irreducibility of those polynomials... – Aurelien Mar 25 2011 at 13:35
The restrictions are $n$ nonnegative with $n \le 0$. Another characterization is $n=2n.$
I mention that mainly for the humor value. The OEIS comments:
We follow Maple in defining $\Phi_0$ to be $x$; it could equally well be taken to be $1$.
I suppose one could equally well just not define it, a number of sources don't.
$\Phi_{2n}(x)$ is $\Phi_{n}(-x)$ for odd $n$ and $\Phi_n(x^2)$ for even $n$. That would argue that $\Phi_0$ should be $1$ if it is defined.
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http://mathhelpforum.com/statistics/213194-arrangements-word-mathematics-restriction.html
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2Thanks
• 1 Post By Plato
• 1 Post By Plato
# Thread:
1. ## Arrangements of the word MATHEMATICS with restriction.
Hello all, I am trying to count the number of arrangements of the word MATHEMATICS where each consonant in adjacent in a vowel. (7 consonants, 4 vowels)
Consider the following diagram:
$\left(\;\;\;\;\right)\;\;v\;\;\left(\;\;\;\;\right )\;\;v\;\;\left(\;\;\;\;\right)\;\;v\;\;\left(\;\; \;\;\right)\;\;v\;\;\left(\;\;\;\;\right)$
where v is a vowel and $\left(\;\;\;\;\right)$ is a bubble for consonants to go in
Now in order to deal with the constraint of each consonant being next to a vowel, I believe the following is true
Two of the three middle bubbles must contain exactly two consonants
One of the end bubbles must contain exactly one consonant.
So we have the following possible unordered distributions (partitions?) of 7:
{2,2,1,1,1} and {2,2,2,1,0}
For {2,2,2,1,0}, all three twos must go in the middle three bubbles. The 0,1 can go on either end.
This leads to two possible distribution of consonants 0v2v2v2v1 and 1v2v2v2v0
For {2,2,1,1,1}, choose two of the three middle bubbles for the 2's, fill in the rest with 1's.
This leads to three possible distribution of consonants 1v2v2v1v1, 1v2v1v2v1, 1v1v2v2v1
So there are 5 possible ways we can place the vowels and consonants. Now we just need to order them
$\frac{7!}{2!2!}$ ways to order consonants
$\frac{4!}{2!}$ ways to order vowels
Final answer:
$5 * \frac{7!}{2!2!} * \frac{4!}{2!}$
Any thoughts? clarifications? glaring errors?
2. ## Re: Arrangements of the word MATHEMATICS with restriction.
Originally Posted by Jame
Hello all, I am trying to count the number of arrangements of the word MATHEMATICS where each consonant in adjacent in a vowel. (7 consonants, 4 vowels)
Consider the following diagram:
$\left(\;\;\;\;\right)\;\;v\;\;\left(\;\;\;\;\right )\;\;v\;\;\left(\;\;\;\;\right)\;\;v\;\;\left(\;\; \;\;\right)\;\;v\;\;\left(\;\;\;\;\right)$
where v is a vowel and $\left(\;\;\;\;\right)$ is a bubble for consonants to go in
Now in order to deal with the constraint of each consonant being next to a vowel, I believe the following is true
Two of the three middle bubbles must contain exactly two consonants
One of the end bubbles must contain exactly one consonant.
So we have the following possible unordered distributions (partitions?) of 7:
{2,2,1,1,1} and {2,2,2,1,0}
For {2,2,2,1,0}, all three twos must go in the middle three bubbles. The 0,1 can go on either end.
This leads to two possible distribution of consonants 0v2v2v2v1 and 1v2v2v2v0
For {2,2,1,1,1}, choose two of the three middle bubbles for the 2's, fill in the rest with 1's.
This leads to three possible distribution of consonants 1v2v2v1v1, 1v2v1v2v1, 1v1v2v2v1
So there are 5 possible ways we can place the vowels and consonants. Now we just need to order them
$\frac{7!}{2!2!}$ ways to order consonants
$\frac{4!}{2!}$ ways to order vowels
Final answer:
$5 * \frac{7!}{2!2!} * \frac{4!}{2!}$
Any thoughts? clarifications? glaring errors?
I did it an entirely different ways getting the same answer.
3. ## Re: Arrangements of the word MATHEMATICS with restriction.
Thank you very much for replying! May I ask how you arrived at your answer? I am always interested in seeing a different way to do a problem.
4. ## Re: Arrangements of the word MATHEMATICS with restriction.
Originally Posted by Jame
Thank you very much for replying! May I ask how you arrived at your answer? I am always interested in seeing a different way to do a problem.
I did it using a generating function.
Expand $\left( {1 + x} \right)^2 \left( {1 + x + x^2 } \right)^3$.
5 is the coefficient of $x^7$. Then use your arrangement.
5. ## Re: Arrangements of the word MATHEMATICS with restriction.
Interesting. So we build this function to partition a number into five summands, two of which are between 0 and 1 and three of which are between 0 and 2.
And as it turns out, at least two of the five summands must be 2 in order to get a sum of 7. (the ones mentioned eariler)
Thanks again!
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http://math.stackexchange.com/questions/53676/blowing-up-a-subvariety-what-can-happen-to-the-singular-locus
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# Blowing up a subvariety - what can happen to the singular locus?
I know that blow ups can be used for the resolution of singular points on a variety $X$.
What I need to know is - if I blow-up along some arbitrary subvariety of $X$, what are the possible outcomes for the dimension of the singular locus of the variety? If the subvariety lies outside the singular locus of $X$, then it stays the same, if it is a carefully chosen singular point, it might go down. $\textbf{Can it go up?}$
To be more specific, my variety is a high dimensional hypersurface, and the subvariety I am blowing up is a $\textbf{linear}$ space of much smaller dimension than the singular locus. I don't know if this changes the situation.
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http://www.physicsforums.com/showthread.php?p=4236974
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Physics Forums
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## Solving heat equation for heat-pulse in a point on the surface
Hi everybody, I'm trying to find a solution for the 3D heat equation for pulsed surface heating of a semi-infinte solid with insulated surface. I know the method of reflection is required, and that a point source in an infinite solid gives the following solution:
$U(x,y,z,t)= \frac{Q}{8\sqrt{(πκt)^3}}*e^{-\frac{x^2+y^2+z^2}{4κt}}$
Where κ is thermal conductivity and Q is a measure for the strength of the heat source. However, I have only found a solution for a semi-infinite solid with surface temperature zero and a heat source inside the solid.
In my case however, the heat source is on the surface, (let's say in point (0,0,0)), hence surface temperature cannot be zero, yet to make matters (a little less) complicated, let's assume a perfectly insulated surface with no heat transfer..
Thanks in advance for help, or tips for usefull literature
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
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Gold Member
Quote by Jbari Hi everybody, I'm trying to find a solution for the 3D heat equation for pulsed surface heating of a semi-infinte solid with insulated surface. I know the method of reflection is required, and that a point source in an infinite solid gives the following solution: $U(x,y,z,t)= \frac{Q}{8\sqrt{(πκt)^3}}*e^{-\frac{x^2+y^2+z^2}{4κt}}$ Where κ is thermal conductivity and Q is a measure for the strength of the heat source. However, I have only found a solution for a semi-infinite solid with surface temperature zero and a heat source inside the solid. In my case however, the heat source is on the surface, (let's say in point (0,0,0)), hence surface temperature cannot be zero, yet to make matters (a little less) complicated, let's assume a perfectly insulated surface with no heat transfer.. Thanks in advance for help, or tips for usefull literature
Your solution looks OK to me. The partial derivative of U with respect to z is zero, so the surface heat flux (normal to the x-y plane) is zero. Your semi-infinite solid lies above the x-y plane (z > = 0).
Quote by Chestermiller Your solution looks OK to me. The partial derivative of U with respect to z is zero, so the surface heat flux (normal to the x-y plane) is zero. Your semi-infinite solid lies above the x-y plane (z > = 0).
Thanks for the quick response, but are you sure this is correct? Because this is the solution for a point source inside an infinite solid, and it looks quite odd to me that a surface heat source in a semi-infinite solid has te same solution. I mean, intuitively, one would expect a different heat propagation in this semi-infinite solid since there is only one way the heat is dissipated (in the part where z>=0) and the heat is 'blocked' in the other direction (where z<0)? Or is this assumption not correct?
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## Solving heat equation for heat-pulse in a point on the surface
Think of it first as an infinite solid. You have a sudden instantaneous spherically symmetric injection of heat at the origin. Then you let the heat diffuse away. Half the previously injected heat goes up, and half the previously injected heat goes down. So there is symmetry of the temperature distribution with respect to the x-y plane (z = 0). No heat crosses this boundary (after t = 0). This is exactly what your solution tells you. The upward heat flux at z = 0 is -kdU/dz, but dU/dz is zero at all times after t = 0.
Thanks a lot for the extra information, your explanation looks correct indeed, and I understand that in this case my equation covers the problem since I cannot see a theoretical error in your explanation . However, I still find it odd that there is no effect of the insulated boundary (although I might of course be mistaken): Isn't the heat that normally goes down in the z<0 area (when the solid is infinite), 'trapped' due to thermal insulation of the surface, resulting in the fact that it is dissipated to the other side, leading to a higher heat input in the z>=0 area? And if this is the case, does this only affect the factor Q (heat input) in the equation?
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Science Advisor
Quote by Jbari However, I still find it odd that there is no effect of the insulated boundary
The point is that, because of the symmetry of the heat flow, there would be no heat flow across the "boundary" plane even if the solid was infinite, not semi-infinite.
If there is no heat flow through a surface, replacing that surface with an insulated boundary doesn't change the heat flow.
Ok, thanks a lot! I didn't realise I had the correct solution lying around all this time :).
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Quote by AlephZero The point is that, because of the symmetry of the heat flow, there would be no heat flow across the "boundary" plane even if the solid was infinite, not semi-infinite. If there is no heat flow through a surface, replacing that surface with an insulated boundary doesn't change the heat flow.
I like your explanation much better than mine. I was struggling to say just this, but was unable to articulate it as simply and concisely.
Chet
Quote by Chestermiller I like your explanation much better than mine. I was struggling to say just this, but was unable to articulate it as simply and concisely. Chet
Very counter-intuitive. So adding a thin insulating plane to an infinite solid at z=0 would have no influence at all on the T distribution?
About the insulation: in my question I assumed a perfect insulation with no heat transfer across the boundary, but what if we try to make the situation more realistic and insulation is not perfect.. So instead of a semi-infinite solid, we now actually have 'two semi-infinite solids with their surfaces in thermal contact' (or something like that). Does this change anything for the heat transfer? Again, intuitively, one would assume it does, but I cannot rely on my intuition :). Furthermore, small differences are quite important here, because I'm also intrested in the heat propagation on the boundary plane. Hope this isn't too much to ask?
Recognitions: Gold Member Let me say this back so that I understand correctly. You now have 2 semi-infinite solids with an imperfect insulating material of thickness ?? between then, and you release the heat at a point on the surface of one of the solids, but not at the mirror image point on the other solid. And the thermal properties of the insulating material is different from that of the two semi-infinite solids. Correct? Chet
Ah no, sorry for the confusion, there is 1 semi-infinite solid, and the other semi infinite solid IS the insulating material.. So 2 materials in contact with eachother, with different thermal properties, and with a point heat source on the boundary surface between them. I hope this makes it more clear to you?
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Quote by Jbari Ah no, sorry for the confusion, there is 1 semi-infinite solid, and the other semi infinite solid IS the insulating material.. So 2 materials in contact with eachother, with different thermal properties, and with a point heat source on the boundary surface between them. I hope this makes it more clear to you?
Is the other semi-infinite solid a perfect insulator?
Quote by Chestermiller Is the other semi-infinite solid a perfect insulator?
No, it is a 'realistic' insulator, so just a material with a lower thermal conductivity than the other semi-infinite solid, so from the mathematical point of view it could be any material..
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Gold Member
Quote by Jbari No, it is a 'realistic' insulator, so just a material with a lower thermal conductivity than the other semi-infinite solid, so from the mathematical point of view it could be any material..
If the thermal conductivities are not equal, you can still have zero heat flux at the interface if the thermal diffusivities are equal. The heat pulse will just initially partition between the two slabs in proportion to the thermal conductivities. After that, no heat flow will occur across the interface.
If the thermal diffusivities are unequal, there will be heat flow across the interface. I'm not sure whether this problem has an analytic solution. Of course, it can always be solved numerically.
What confuses me is how come the OP's original equation does not have a diffusion term at all? Shouldn't k, Cp and density always occur in combination (i.e. Diffusivity) in the solutions of the heat equation?
Quote by rollingstein What confuses me is how come the OP's original equation does not have a diffusion term at all? Shouldn't k, Cp and density always occur in combination (i.e. Diffusivity) in the solutions of the heat equation?
You are correct, I ment κ to be thermal diffusivity (= K/(ρ*Cp)) but I defined it incorrectly in my explanation as thermal conductivity..
And to return to the problem: the thermal diffusivities are NOT equal, but I would still like to find an analytical solution (if possible of course). Any suggestions on how/where to find it (maybe in literature, but my search has been fruitless untill now).
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heat equation, pde, semi-infinite solid
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http://unapologetic.wordpress.com/2011/05/25/identifying-vector-fields/?like=1&_wpnonce=9d779cb1c0
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# The Unapologetic Mathematician
## Identifying Vector Fields
We know what vector fields are on a region $U\subseteq M$, but to identify them in the wild we need to verify that a given function sending each $p\in U$ to a vector in $\mathcal{T}_pM$ is smooth. This might not always be so easy to check directly, so we need some equivalent conditions. First we need to define how vector fields act on functions.
If $X\in\mathfrak{X}U$ is a vector field and $f\in\mathcal{O}U$ is a smooth function then we get another function $Xf$ by defining $Xf(p)=\left[X(p)\right](f)$. Indeed, $X(p)\in\mathcal{T}_pM$, so it can take (the germ of) a smooth function at $p$ and give us a number. Essentially, at each point the vector field defines a displacement, and we ask how the function $f$ changes along this displacement. This action is key to our conditions, and to how we will actually use vector fields.
Firstly, if $X$ is a vector field — a differentiable function — and if $(V,x)$ is a chart with $V\subseteq U$, then $Xx^i$ is always smooth. Indeed, remember that $(V,x)$ gives us a coordinate patch $(\pi^{-1}(V),\bar{x})$ on the tangent bundle. Since $\bar{x}$ is smooth and $X$ is smooth, the composition
$\displaystyle\bar{x}\circ X\vert_V=(x\circ I\vert_V;X\vert_V(x^1),\dots,x\vert_V(x^n))$
is also smooth. And thus each component $Xx^i$ is smooth on $V$.
Next, we do not assume that $X$ is a vector field — it is a function but not necessarily a differentiable one — but we assume that it satisfies the conclusion of the preceding paragraph. That is, for every chart $(V,x)$ with $V\subseteq U$ each $Xx^i$ is smooth. Now we will show that $Xf$ is smooth for every smooth $f\in\mathcal{O}V$, not just those that arise as coordinate functions. To see this, we use the decomposition of $X$ into coordinate vector fields:
$\displaystyle X=\sum\limits_{i=1}^nX^i\frac{\partial}{\partial x^i}$
which didn’t assume that $X$ was smooth, except to show that the coefficient functions were smooth. We can now calculate that $X^i=Xx^i$, since
$\displaystyle Xx^i=\sum\limits_{j=1}^nX^j\frac{\partial x^i}{\partial x^j}=\sum\limits_{j=1}^nX^j\delta^i_j=X^i$
But this means we can write
$\displaystyle Xf=\sum\limits_{i=1}^nXx^i\frac{\partial f}{\partial x^i}$
which makes $Xf$ a linear combination of the smooth (by assumption) functions $Xx^i$ with the coefficients $\frac{\partial f}{\partial x^i}$, proving that it is itself smooth.
Okay, now I say that if $Xf$ is smooth for every smooth function $f\in\mathcal{O}V$ on some region $V\subseteq U$, then $X$ is smooth as a function, and thus is a vector field. In this case around any $p\in U$ we can find some coordinate patch $(V,x)$. Now we go back up to the composition above:
$\displaystyle\bar{x}\circ X\vert_V=(x\circ I\vert_V;X\vert_V(x^1),\dots,x\vert_V(x^n))$
Everything in sight on the right is smooth, and so the left is also smooth. But this is exactly what we need to check when we’re using the local coordinates $(V,x)$ and $(\pi^{-1}(V),\bar{x})$ to verify the smoothness of $X$ at $p$.
The upshot is that when we want to verify that a function $X$ really is a smooth vector field, we take an arbitrary smooth “test function” and feed it into $X$. If the result is always smooth, then $X$ is smooth. In fact, some authors take this as the definition, regarding the action of $X$ on functions as fundamental, and only later talking in terms of its “value at a point”.
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Posted by John Armstrong | Differential Topology, Topology
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2. [...] know that any vector field can act as an endomorphism on the space of smooth functions on . What happens if we act by one vector field followed by [...]
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3. [...] as for vector fields, we need a good condition to identify tensor fields in the wild. And the condition we will use is [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://physics.stackexchange.com/questions/tagged/magnetohydrodynamics
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# Tagged Questions
The magnetohydrodynamics tag has no wiki summary.
1answer
61 views
### A rod filled with water and charged particles is rotated in magnetic field
Assume we have a rod of diameter $d$ and length $l$. It is filled with a mixture of water and, say, sodium chloride. Thus there are positively charged sodium ions and negatively charged chloride ions ...
1answer
114 views
### The equation of a ferrofluid under a magnetic field?
What is the parametric equation guiding the geometry of a ferrofluid under a magnetic field? See also this Wikipedia page. From previous research, Maxwell's Equations and Navier-Stokes Equations were ...
0answers
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### Richtmyer Meshkov instability in MHD [duplicate]
Possible Duplicate: Richtmyer Meshkov instability in MHD The Richtmyer Meshkov instability in Hydrodynamics:- 1) When shock wave interacts with the contact discontinuity, it accumulates the ...
1answer
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### Richtmyer Meshkov instability in MHD
In magnetohydrodynamics, the Richtmyer Meshkov instability is found to get suppressed by application of longitudinal magnetic field. Exactly what happens at the interface? Why instability gets ...
0answers
149 views
### Solar wind and the Earth's magnetic field
I have again an old question from a comprehensive exam I took a couple of months ago. Lucky for me one could pick 5 out of 8 questions, because on some of the problems I didn't even know how to start. ...
2answers
155 views
### Dynamic ferrofluid sculptures
http://www.youtube.com/watch?v=UJJuq_pcyIQ What exactly is going on in the video example? I understand the phenomena occurs because of magnetism but I am trying to figure out the mechanics behind ...
5answers
1k views
### What sustains the rotation of earth's core (faster than surface)?
I recently read that the earth's core rotates faster than the surface. Well, firstly, it's easier to digest the concept of planetary bodies, stars, galaxies in rotation and/or orbital motion. But, ...
1answer
263 views
### Producing energy from magnetic flux loops
I've simplified this down by quite a lot but as far as I understand it, magnetic fields on the Sun's surface twist together and when it gets all a bit too much they release energy. My question is why ...
4answers
532 views
### Help fill in my understanding of the Polywell fusion reactor
Polywell is a proposed new type of fusion reactor, which is designed to use magnetic fields to overcome the problems with the Elmore-Tuck-Watson fusor. I'm trying to understand exactly how it works. ...
3answers
488 views
### How can we make an order-of-magnitude estimate of the strength of Earth's magnetic field?
The source of Earth's magnetic field is a dynamo driven by convection current in the molten core. Using some basic physics principles (Maxwell's equations, fluid mechanics equations), properties of ...
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http://medlibrary.org/medwiki/Yang-Baxter_equation
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# Yang-Baxter equation
Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below:
The Yang–Baxter equation (or star-triangle relation) is an equation which was first introduced in the field of statistical mechanics. It takes its name from independent work of C. N. Yang from 1968, and R. J. Baxter from 1971. It refers to a principle in integrable systems taking the form of local equivalence transformations which appear in a variety of contexts, such as electric networks, knot theory and braid groups, and spin systems, to name just a few.
## Parameter-dependent Yang–Baxter equation
Let $A$ be a unital associative algebra. The parameter-dependent Yang–Baxter equation is an equation for $R(u)$, a parameter-dependent invertible element of the tensor product $A \otimes A$ (here, $u$ is the parameter, which usually ranges over all real numbers in the case of an additive parameter, or over all positive real numbers in the case of a multiplicative parameter). The Yang–Baxter equation is
$R_{12}(u) \ R_{13}(u+v) \ R_{23}(v) = R_{23}(v) \ R_{13}(u+v) \ R_{12}(u),$
for all values of $u$ and $v$, in the case of an additive parameter. At some value of the parameter $R(u)$ can turn into one dimensional projector, this gives rise to quantum determinant. For multiplicative parameter Yang–Baxter equation is
$R_{12}(u) \ R_{13}(uv) \ R_{23}(v) = R_{23}(v) \ R_{13}(uv) \ R_{12}(u),$
for all values of $u$ and $v$, where $R_{12}(w) = \phi_{12}(R(w))$, $R_{13}(w) = \phi_{13}(R(w))$, and $R_{23}(w) = \phi_{23}(R(w))$, for all values of the parameter $w$, and $\phi_{12} : A \otimes A \to A \otimes A \otimes A$, $\phi_{13} : A \otimes A \to A \otimes A \otimes A$, and $\phi_{23} : A \otimes A \to A \otimes A \otimes A$ are algebra morphisms determined by
$\phi_{12}(a \otimes b) = a \otimes b \otimes 1,$
$\phi_{13}(a \otimes b) = a \otimes 1 \otimes b,$
$\phi_{23}(a \otimes b) = 1 \otimes a \otimes b.$
In some cases the determinant of $R (u)$ can vanish at specific values of the spectral parameter $u=u_{0}$. Some $R$ matrices turn into one dimensional projector at $u=u_{0}$. In this case quantum determinant can be defined.
## Parameter-independent Yang–Baxter equation
Let $A$ be a unital associative algebra. The parameter-independent Yang–Baxter equation is an equation for $R$, an invertible element of the tensor product $A \otimes A$. The Yang–Baxter equation is
$R_{12} \ R_{13} \ R_{23} = R_{23} \ R_{13} \ R_{12},$
where $R_{12} = \phi_{12}(R)$, $R_{13} = \phi_{13}(R)$, and $R_{23} = \phi_{23}(R)$.
Let $V$ be a module of $A$. Let $T : V \otimes V \to V \otimes V$ be the linear map satisfying $T(x \otimes y) = y \otimes x$ for all $x, y \in V$. Then a representation of the braid group, $B_n$, can be constructed on $V^{\otimes n}$ by $\sigma_i = 1^{\otimes i-1} \otimes \check{R} \otimes 1^{\otimes n-i-1}$ for $i = 1,\dots,n-1$, where $\check{R} = T \circ R$ on $V \otimes V$. This representation can be used to determine quasi-invariants of braids, knots and links.
## References
• H.-D. Doebner, J.-D. Hennig, eds, Quantum groups, Proceedings of the 8th International Workshop on Mathematical Physics, Arnold Sommerfeld Institute, Clausthal, FRG, 1989, Springer-Verlag Berlin, ISBN 3-540-53503-9 [Amazon-US | Amazon-UK].
• Vyjayanthi Chari and Andrew Pressley, A Guide to Quantum Groups, (1994), Cambridge University Press, Cambridge ISBN 0-521-55884-0 [Amazon-US | Amazon-UK].
• Jacques H.H. Perk and Helen Au-Yang, "Yang–Baxter Equations", (2006), arXiv:math-ph/0606053.
Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Yang-Baxter equation", available in its original form here:
http://en.wikipedia.org/w/index.php?title=Yang-Baxter_equation
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http://mathoverflow.net/questions/21599/transitivity-of-the-stone-cech-compactification/21605
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“Transitivity” of the Stone-Cech compactification
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Let $\beta \mathbb{N}$ be the Stone-Cech compactification of the natural numbers $\mathbb{N}$, and let $x, y \in \beta \mathbb{N} \backslash \mathbb{N}$ be two non-principal elements of this compactification (or equivalently, $x$ and $y$ are two non-principal ultrafilters). I am interested in ways to "model" the ultrafilter $y$ using the ultrafilter $x$. More precisely,
Q1. (Existence) Does there necessarily exist a continuous map $f: \beta \mathbb{N} \to \beta \mathbb{N}$ which maps $x$ to $y$, while mapping $\mathbb{N}$ to $\mathbb{N}$? To put it another way: does there exist a function $f: \mathbb{N} \to \mathbb{N}$ such that $\lim_{n \to x} f(n) = y$?
Q2. (Uniqueness) Suppose there are two continuous maps $f, g: \beta \mathbb{N} \to \beta\mathbb{N}$ with $f(x)=g(x)=y$, which map $\mathbb{N}$ to $\mathbb{N}$. Is it then true that $f$ and $g$ must then be equal on a neighbourhood of $x$?
I suspect the answer to both questions is either "no" or "undecidable in ZFC", though perhaps there exist "universal" ultrafilters $x$ for which the answers become yes. But I do not have enough intuition regarding the topology of $\beta \mathbb{N}$ (other than that it is somewhat pathological) to make this more precise. (The fact that $\beta\mathbb{N}$ is not first countable does seem to indicate that the answers should be negative, though.)
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2 Answers
The answer to Q1 is no. This has been well studied in set theory; you're basically asking whether any two non-principal ultrafilters on $\mathbb{N}$ are comparable under the Rudin-Keisler ordering. Variations on your question have led to many, many interesting developments in set theory, but your question Q1 is easy to answer by a cardinality argument.
First note that every $f:\mathbb{N}\to\mathbb{N}$ has a unique extension to a continuous function $\bar{f}:\beta\mathbb{N}\to\beta\mathbb{N}$. Any $x \in \beta\mathbb{N}$ has at most $2^{\aleph_0}$ images through such $\bar{f}$, but there are $2^{2^{\aleph_0}}$ ultrafilters on $\mathbb{N}$, so there are very many $y \in \beta\mathbb{N}$ which are not images of $x$ through such $\bar{f}$.
The answer to Q2 is also no. Let $y$ be a nonprincipal ultrafilter on $\mathbb{N}$. The sets $A \times A\setminus\Delta$ where $A \in y$ and `$\Delta = \{(n,n) : n \in \mathbb{N}\}$` form a filter base on $\mathbb{N}\times\mathbb{N}$. Let $x$ be an ultrafilter on $\mathbb{N}\times\mathbb{N}$ that contains all these sets. The left and right projections $\pi_1,\pi_2:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ both send $x$ to $y$, but they are not equal on any neighborhood of $x$.
However, the answer to Q2 is yes when $x$ is a selective ultrafilter. Recall that $x$ is selective if for every $h:\mathbb{N}\to\mathbb{N}$ there is a set $A \in x$ on which $h$ is either constant or one-to-one. Given $f,g:\mathbb{N}\to\mathbb{N}$ such that $\bar{f}(x) = \bar{g}(x)$ is nonprincipal, then we can find $A \in x$ on which $f$ and $g$ are both one-to-one. In that case, $f\circ g^{-1}$ must be well-defined on some $A' \in x$. Any extension of $f\circ g^{-1}$ to the complement of $A'$ must map $x$ to $x$, which means that $f \circ g^{-1}$ is the identity on some $A'' \in x$. Thus $f$ and $g$ are equal on the neighborhood of $x$ defined by $A''$.
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I don't understand your cardinality argument - there are very many ultrafilters, yes, but there are only $2^{\alpha_0}$ choices that each ultrafilter can pick for $y$, as far as I can tell. What am I not seeing? – Jacques Carette Apr 16 2010 at 20:38
(meant to type $\aleph$ but slipped and typed $\alpha$, sorry) – Jacques Carette Apr 16 2010 at 20:38
As I understand François's argument: the cardinality of $\beta\mathbb{N}$ is bigger than the cardinality of the set of continuous functions $\beta\mathbb{N}\to\beta\mathbb{N}$ taking $\mathbb{N}\to\mathbb{N}$. So given $x$ you can find $y$ so that no such function takes $x$ to $y$. – Tom Church Apr 16 2010 at 20:50
That's exactly right Tom, there are simply too few continuous functions that map $\mathbb{N}\to\mathbb{N}$. – François G. Dorais♦ Apr 16 2010 at 20:57
I think the answer to Q2 is yes when $x$ is a selective ultrafilter, which is why I had deleted my earlier remark on Q2. – François G. Dorais♦ Apr 16 2010 at 20:58
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The answer to Q1 is even more `no': Kunen showed that there are x and y such that x cannot be mapped to y and y cannot be mapped to x, see this review. This has been strengthened by Rudin and Shelah. It is still open, as far as I know, whether given x there is a y such that neither can be mapped to the other.
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I maybe miss something, as I don't see how this question can be open: if you consider the oriented labeled graph with the Stone-Cech boundary as set of vertices, and oriented edge $(x,y)$ labeled by $f:I\to J$ whenever $f(x)=y$, and consider the underlying unoriented graph, then its valency is at most continuum, and therefore its components have cardinality at most continuum. So given $x$ you can find $y$ in another component. – Yves Cornulier Oct 25 at 18:57
The valency of the graph is $2^{\mathfrak{c}}$, rather than $\mathfrak{c}$; all points of the remainder have have $2^{\mathfrak{c}}$ preimages under non-trivial finite-to-one maps from $\mathbb{N}$ to itself (those where the fibers have arbitrarily large finite cardinality). – KP Hart Oct 29 at 20:35
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http://unapologetic.wordpress.com/2007/02/28/rubiks-group-notation/?like=1&source=post_flair&_wpnonce=78d053c4d8
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# The Unapologetic Mathematician
## Rubik’s Group notation
Today I want to introduce some notation for discussing the cube.
Take a cube — either a real one, the Java version, or just one in your mind’s eye — and hold it with the center cubies fixed in place pointing up, down, left, right, front, and back. Twists of the faces will generate Rubik’s goup $G$. We pick the six generators as follows
• $U$ is a twist of the upper face by a quarter turn clockwise, looking down at the top of the cube.
• $D$ is a twist of the lower face by a quarter turn clockwise, looking up at the bottom of the cube.
• $R$ is a twist of the right face by a quarter turn clockwise, looking left at the right side of the cube.
• $L$ is a twist of the left face by a quarter turn clockwise, looking right at the left of the cube.
• $F$ is a twist of the front face by a quarter turn clockwise, looking at the front of the cube.
• $B$ is a twist of the back face by a quarter turn clockwise, looking at the back of the cube.
For instance, executing $B$ involves turning the whole cube around to look at the back, twisting that face clockwise by 90°, and turning the cube back to the original orientation.
For each twist $T$, the 180° twist of the corresponding face is $T^2$, and the anticlockwist twist is $T^{-1}$. Four quarter-twists is the same as doing nothing, so $T^4$ is the identity.
It’s also useful to label the cubicles. I’ll do this by listing the faces it touches. The face cubicle on the left side of the cube is $l$, the lower edge on the back ls $db$ or $bd$. The upper-right corner on the front of the cube is $urf$, $ufr$, $ruf$, $rfu$, $fur$, or $fru$. The order of the faces doesn’t matter so much for a single cubicle, but becomes important when we start thinking about how maneuvers affect the state of the cube.
For instance, the effect of the maneuver $R^2U^{-1}DB^2UD^{-1}$ has an effect on the cube I’ll write as $(fr\,bl\,br)$. This takes the cubie in the front-right cubicle and puts it in the back-left cubicle, with the facelet that was in the front now in the back and the facelet from the right now on the left. It takes the cubie from the back-left cubicle and puts it in the back-right cubicle with the facelet from the back still on the back and the facelet from the left now on the right. Finally it takes the cubie from the back-right and moves it to the front-right with the right facelet still on the right and the back facelet now on the front.
As another example, $(R^{-1}U^2RB^{-1}U^2B)^2$ takes the cubie from the upper-right-front cubicle and leaves it where it is, but twists it to the right, and similarly twists the cubie in the lower-left-back cubicle. We write this transformation as $(urf\,rfu\,fur)(dlb\,lbd,bdl)$. For shorthand we’ll modify this permutation notation slightly. When a cycle brings a cubie back to its starting cubicle but rotated, we add a sign. In this example we’ll write $(+urf)(-dlb)$, since if we look directly at the upper-right-front cubie it’s been rotated clockwise by 1/3 of a turn and looking directly at the lower-left-back it’s been rotated anticlockwise by 1/3 of a turn.
This notation does make sense. Let’s say that we have a maneuver whose effect contains some cycle on the corners of the cube of length $k$. If we apply that maneuver $k$ times each cubie in the cycle comes back where it started, but the cubies may have been twisted in the process. Each one will be twisted by the same amount: either 1/3 to the right, 1/3 to the left, or untwisted. The sign in the notation tells us what that twist is.
The same notation goes for edges. The maneuver $RUD^{-1}FUD^{-1}LUD^{-1}BUD^{-1}$ has effect $(+ur)(+fr)(+fl)(+bl)$, flipping those four edges. We could write this out as $(ur\,ru)(fr\,rf)(fl\,lf)(bl\,lb)$, but the “twisted permutation” notation is more compact.
From this notation it’s easy to compute the order of a maneuver — what power of the maneuver returns to the identity transformation. A corner cycle of length $k$ has order $k$ if there’s no twist, and order $3k$ if there is a twist. Similarly, an edge $k$-cycle has order $k$ if there’s no flip and order $2k$ if there is a flip. So if we write the effect of a maneuver as a twisted permutation we can find the order of each twisted cycle. The order the the whole maneuver is the least common multiple of those orders.
As an example, consider the maneuver $RU$. This has effect $(+urf)(-ufl\,ulb\,ubr\,bdr\,dfr)(fr\,uf\,ul\,ub\,ur\,br\,dr)$. The three twisted cycles have orders 3, 15, and 7, so the total order is 105. If you actually sit and perform $(RU)^{105}$ you’ll get back exactly where you started.
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Posted by John Armstrong | Group theory, Rubik\'s Cube
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http://en.wikipedia.org/wiki/Manifold_learning
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# Nonlinear dimensionality reduction
(Redirected from Manifold learning)
High-dimensional data, meaning data that requires more than two or three dimensions to represent, can be difficult to interpret. One approach to simplification is to assume that the data of interest lie on an embedded non-linear manifold within the higher-dimensional space. If the manifold is of low enough dimension, the data can be visualised in the low dimensional space.
Top-left: a 3D dataset of 1000 points in a spiraling band (a.k.a. the swiss roll) with a rectangular hole in the middle. Top-right: the original 2D manifold used to generate the 3D dataset. Bottom left and right: 2D recoveries of the manifold respectively using the LLE and Hessian LLE algorithms as implemented by the Modular Data Processing toolkit.
Below is a summary of some of the important algorithms from the history of manifold learning and nonlinear dimensionality reduction (NLDR).[1] Many of these non-linear dimensionality reduction methods are related to the linear methods listed below. Non-linear methods can be broadly classified into two groups: those that provide a mapping (either from the high dimensional space to the low dimensional embedding or vice versa), and those that just give a visualisation. In the context of machine learning, mapping methods may be viewed as a preliminary feature extraction step, after which pattern recognition algorithms are applied. Typically those that just give a visualisation are based on proximity data – that is, distance measurements.
## Linear methods
• Independent component analysis (ICA).
• Principal component analysis (PCA) (also called Karhunen–Loève transform — KLT).
• Singular value decomposition (SVD).
• Factor analysis.
## Uses for NLDR
Consider a dataset represented as a matrix (or a database table), such that each row represents a set of attributes (or features or dimensions) that describe a particular instance of something. If the number of attributes is large, then the space of unique possible rows is exponentially large. Thus, the larger the dimensionality, the more difficult it becomes to sample the space. This causes many problems. Algorithms that operate on high-dimensional data tend to have a very high time complexity. Many machine learning algorithms, for example, struggle with high-dimensional data. This has become known as the curse of dimensionality. Reducing data into fewer dimensions often makes analysis algorithms more efficient, and can help machine learning algorithms make more accurate predictions.
Humans often have difficulty comprehending data in many dimensions. Thus, reducing data to a small number of dimensions is useful for visualization purposes.
Plot of the two-dimensional points that results from using a NLDR algorithm. In this case, Manifold Sculpting used to reduce the data into just two dimensions (rotation and scale).
The reduced-dimensional representations of data are often referred to as "intrinsic variables". This description implies that these are the values from which the data was produced. For example, consider a dataset that contains images of a letter 'A', which has been scaled and rotated by varying amounts. Each image has 32x32 pixels. Each image can be represented as a vector of 1024 pixel values. Each row is a sample on a two-dimensional manifold in 1024-dimensional space (a Hamming space). The intrinsic dimensionality is two, because two variables (rotation and scale) were varied in order to produce the data. Information about the shape or look of a letter 'A' is not part of the intrinsic variables because it is the same in every instance. Nonlinear dimensionality reduction will discard the correlated information (the letter 'A') and recover only the varying information (rotation and scale). The image to the left shows sample images from this dataset (to save space, not all input images are shown), and a plot of the two-dimensional points that results from using a NLDR algorithm (in this case, Manifold Sculpting was used) to reduce the data into just two dimensions.
PCA (a linear dimensionality reduction algorithm) is used to reduce this same dataset into two dimensions, the resulting values are not so well organized.
By comparison, if PCA (a linear dimensionality reduction algorithm) is used to reduce this same dataset into two dimensions, the resulting values are not so well organized. This demonstrates that the high-dimensional vectors (each representing a letter 'A') that sample this manifold vary in a non-linear manner.
It should be apparent, therefore, that NLDR has several applications in the field of computer-vision. For example, consider a robot that uses a camera to navigate in a closed static environment. The images obtained by that camera can be considered to be samples on a manifold in high-dimensional space, and the intrinsic variables of that manifold will represent the robot's position and orientation. This utility is not limited to robots. Dynamical systems, a more general class of systems, which includes robots, are defined in terms of a manifold. Active research in NLDR seeks to unfold the observation manifolds associated dynamical systems to develop techniques for modeling such systems and enable them to operate autonomously.[2]
## Manifold learning algorithms
Some of the more prominent manifold learning algorithms are listed below (in approximately chronological order). An algorithm may learn an internal model of the data, which can be used to map points unavailable at training time into the embedding in a process often called out-of-sample extension.
### Sammon's mapping
Sammon's mapping is one of the first NLDR techniques.
Approximation of a principal curve by one-dimensional SOM (a broken line with red squares, 20 nodes). The first principal component is presented by a blue straight line. Data points are the small grey circles. For PCA, the Fraction of variance unexplained in this example is 23.23%, for SOM it is 6.86%.[3]
### Kohonen maps
Kohonen maps (also called self-organizing maps or SOM) and its probabilistic variant generative topographic mapping (GTM) use a point representation in the embedded space to form a latent variable model based on a non-linear mapping from the embedded space to the high dimensional space.[4] These techniques are related to work on density networks, which also are based around the same probabilistic model.
### Principal curves and manifolds
Application of principal curves: Nonlinear quality of life index.[5] Points represent data of the UN 171 countries in 4-dimensional space formed by the values of 4 indicators: gross product per capita, life expectancy, infant mortality, tuberculosis incidence. Different forms and colors correspond to various geographical locations. Red bold line represents the principal curve, approximating the dataset. This principal curve was produced by the method of elastic map. Software is available for free non-commercial use.[6][7]
Principal curves and manifolds give the natural geometric framework for nonlinear dimensionality reduction and extend the geometric interpretation of PCA by explicitly constructing an embedded manifold, and by encoding using standard geometric projection onto the manifold. This approach was proposed by Trevor Hastie in his thesis (1984)[8] and developed further by many authors.[9] How to define the "simplicity" of the manifold is problem-dependent, however, it is commonly measured by the intrinsic dimensionality and/or the smoothness of the manifold. Usually, the principal manifold is defined as a solution to an optimization problem. The objective function includes a quality of data approximation and some penalty terms for the bending of the manifold. The popular initial approximations are generated by linear PCA, Kohonen's SOM or autoencoders. The elastic map method provides the expectation-maximization algorithm for principal manifold learning with minimization of quadratic energy functional at the "maximization" step.
### Autoencoders
An autoencoder is a feed-forward neural network which is trained to approximate the identity function. That is, it is trained to map from a vector of values to the same vector. When used for dimensionality reduction purposes, one of the hidden layers in the network is limited to contain only a small number of network units. Thus, the network must learn to encode the vector into a small number of dimensions and then decode it back into the original space. Thus, the first half of the network is a model which maps from high to low-dimensional space, and the second half maps from low to high-dimensional space. Although the idea of autoencoders is quite old, training of deep autoencoders has only recently become possible through the use of restricted Boltzmann machines and stacked denoising autoencoders. Related to autoencoders is the NeuroScale algorithm, which uses stress functions inspired by multidimensional scaling and Sammon mappings (see below) to learn a non-linear mapping from the high dimensional to the embedded space. The mappings in NeuroScale are based on radial basis function networks.
### Gaussian process latent variable models
Gaussian process latent variable models (GPLVM)[10] are a probabilistic non-linear PCA. Like kernel PCA they use a kernel function to form the mapping (in the form of a Gaussian process). However in the GPLVM the mapping is from the embedded space to the data space (like density networks and GTM) whereas in kernel PCA it is in the opposite direction.
### Curvilinear component analysis
Curvilinear component analysis (CCA)[11] looks for the configuration of points in the output space that preserves original distances as much as possible while focusing on small distances in the output space (conversely to Sammon's mapping which focus on small distances in original space).
It should be noticed that CCA, as an iterative learning algorithm, actually starts with focus on large distances (like the Sammon algorithm), then gradually change focus to small distances. The small distance information will overwrite the large distance information, if compromises between the two have to be made.
The stress function of CCA is related to a sum of right Bregman divergences[12]
### Curvilinear distance analysis
CDA[11] trains a self-organizing neural network to fit the manifold and seeks to preserve geodesic distances in its embedding. It is based on Curvilinear Component Analysis (which extended Sammon's mapping), but uses geodesic distances instead.
### Diffeomorphic dimensionality reduction
Diffeomorphic Dimensionality Reduction or Diffeomap[13] learns a smooth diffeomorphic mapping which transports the data onto a lower dimensional linear subspace. The methods solves for a smooth time indexed vector field such that path integrals along the field which start at the data points will end at a lower dimensional linear subspace, thereby attempting to preserve pairwise differences under both the forward and inverse mapping.
### Kernel principal component analysis
Perhaps the most widely used algorithm for manifold learning is kernel PCA.[14] It is a combination of Principal component analysis and the kernel trick. PCA begins by computing the covariance matrix of the $m \times n$ matrix $\mathbf{X}$
$C = \frac{1}{m}\sum_{i=1}^m{\mathbf{x}_i\mathbf{x}_i^\mathsf{T}}.$
It then projects the data onto the first k eigenvectors of that matrix. By comparison, KPCA begins by computing the covariance matrix of the data after being transformed into a higher-dimensional space,
$C = \frac{1}{m}\sum_{i=1}^m{\Phi(\mathbf{x}_i)\Phi(\mathbf{x}_i)^\mathsf{T}}.$
It then projects the transformed data onto the first k eigenvectors of that matrix, just like PCA. It uses the kernel trick to factor away much of the computation, such that the entire process can be performed without actually computing $\Phi(\mathbf{x})$. Of course $\Phi$ must be chosen such that it has a known corresponding kernel. Unfortunately, it is not trivial to find a good kernel for a given problem, so KPCA does not yield good results with some problems. For example, it is known to perform poorly with the swiss roll manifold.
KPCA has an internal model, so it can be used to map points onto its embedding that were not available at training time.
### Isomap
Isomap[15] is a combination of the Floyd–Warshall algorithm with classic Multidimensional Scaling. Classic Multidimensional Scaling (MDS) takes a matrix of pair-wise distances between all points, and computes a position for each point. Isomap assumes that the pair-wise distances are only known between neighboring points, and uses the Floyd–Warshall algorithm to compute the pair-wise distances between all other points. This effectively estimates the full matrix of pair-wise geodesic distances between all of the points. Isomap then uses classic MDS to compute the reduced-dimensional positions of all the points.
Landmark-Isomap is a variant of this algorithm that uses landmarks to increase speed, at the cost of some accuracy.
### Locally-linear embedding
Locally-Linear Embedding (LLE)[16] was presented at approximately the same time as Isomap. It has several advantages over Isomap, including faster optimization when implemented to take advantage of sparse matrix algorithms, and better results with many problems. LLE also begins by finding a set of the nearest neighbors of each point. It then computes a set of weights for each point that best describe the point as a linear combination of its neighbors. Finally, it uses an eigenvector-based optimization technique to find the low-dimensional embedding of points, such that each point is still described with the same linear combination of its neighbors. LLE tends to handle non-uniform sample densities poorly because there is no fixed unit to prevent the weights from drifting as various regions differ in sample densities. LLE has no internal model.
LLE computes the barycentric coordinates of a point Xi based on its neighbors Xj. The original point is reconstructed by a linear combination, given by the weight matrix Wij, of its neighbors. The reconstruction error is given by the cost function E(W).
$E(W) = \sum_i |{\mathbf{X}_i - \sum_j {\mathbf{W}_{ij}\mathbf{X}_j}|}^\mathsf{2}$
The weights Wij refer to the amount of contribution the point Xj has while reconstructing the point Xi. The cost function is minimized under two constraints: (a) Each data point Xi is reconstructed only from its neighbors, thus enforcing Wij to be zero if point Xj is not a neighbor of the point Xi and (b) The sum of every row of the weight matrix equals 1.
$\sum_j {\mathbf{W}_{ij}} = 1$
The original data points are collected in a D dimensional space and the goal of the algorithm is to reduce the dimensionality to d such that D >> d. The same weights Wij that reconstructs the ith data point in the D dimensional space will be used to reconstruct the same point in the lower d dimensional space. A neighborhood preserving map is created based on this idea. Each point Xi in the D dimensional space is mapped onto a point Yi in the d dimensional space by minimizing the cost function
$C(Y) = \sum_i |{\mathbf{Y}_i - \sum_j {\mathbf{W}_{ij}\mathbf{Y}_j}|}^\mathsf{2}$
In this cost function, unlike the previous one, the weights Wij are kept fixed and the minimization is done on the points Yi to optimize the coordinates. This minimization problem can be solved by solving a sparse N X N eigen value problem (N being the number of data points), whose bottom d nonzero eigen vectors provide an orthogonal set of coordinates. Generally the data points are reconstructed from K nearest neighbors, as measured by Euclidean distance. For such an implementation the algorithm has only one free parameter K, which can be chosen by cross validation.
### Laplacian eigenmaps
Laplacian Eigenmaps[17] uses spectral techniques to perform dimensionality reduction. This technique relies on the basic assumption that the data lies in a low dimensional manifold in a high dimensional space.[18] This algorithm cannot embed out of sample points, but techniques based on Reproducing kernel Hilbert space regularization exist for adding this capability.[19] Such techniques can be applied to other nonlinear dimensionality reduction algorithms as well.
Traditional techniques like principal component analysis do not consider the intrinsic geometry of the data. Laplacian eigenmaps builds a graph from neighborhood information of the data set. Each data point serves as a node on the graph and connectivity between nodes is governed by the proximity of neighboring points (using e.g. the k-nearest neighbor algorithm). The graph thus generated can be considered as a discrete approximation of the low dimensional manifold in the high dimensional space. Minimization of a cost function based on the graph ensures that points close to each other on the manifold are mapped close to each other in the low dimensional space, preserving local distances. The eigenfunctions of the Laplace–Beltrami operator on the manifold serve as the embedding dimensions, since under mild conditions this operator has a countable spectrum that is a basis for square integrable functions on the manifold (compare to Fourier series on the unit circle manifold). Attempts to place Laplacian eigenmaps on solid theoretical ground have met with some success, as under certain nonrestrictive assumptions, the graph Laplacian matrix has been shown to converge to the Laplace–Beltrami operator as the number of points goes to infinity.[20] Matlab code for Laplacian Eigenmaps can be found in algorithms[21] and the PhD thesis of Belkin can be found at the Ohio State University.[22]
### Manifold alignment
Manifold alignment takes advantage of the assumption that disparate data sets produced by similar generating processes will share a similar underlying manifold representation. By learning projections from each original space to the shared manifold, correspondences are recovered and knowledge from one domain can be transferred to another. Most manifold alignment techniques consider only two data sets, but the concept extends to arbitrarily many initial data sets.[23]
### Diffusion maps
Diffusion maps leverages the relationship between heat diffusion and a random walk (Markov Chain); an analogy is drawn between the diffusion operator on a manifold and a Markov transition matrix operating on functions defined on the graph whose nodes were sampled from the manifold.[24] In particular let a data set be represented by $\mathbf{X} = [x_1,x_2,\ldots,x_n] \in \Omega \subset \mathbf {R^D}$. The underlying assumption of diffusion map is that the data although high-dimensional, lies on a low-dimensional manifold of dimensions $\mathbf{d}$.X represents the data set and let $\mu$ represent the distribution of the data points on X. In addition to this lets define a kernel which represents some notion of affinity of the points in X. The kernel $\mathit{k}$ has the following properties[25]
$k(x,y) = k(y,x), \,$
k is symmetric
$k(x,y) \geq 0\qquad \forall x,y, k$
k is positivity preserving
Thus one can think of the individual data points as the nodes of a graph and the kernel k defining some sort of affinity on that graph. The graph is symmetric by construction since the kernel is symmetric. It is easy to see here that from the tuple {X,k} one can construct a reversible Markov Chain. This technique is fairly popular in a variety of fields and is known as the graph laplacian.
The graph K = (X,E) can be constructed for example using a Gaussian kernel.
$K_{ij} = \begin{cases} e^{-||x_i -x_j||/\sigma ^2} & \text{if } x_i \sim x_j \\ 0 & \text{otherwise} \end{cases}$
In this above equation $x_i \sim x_j$ denotes that $x_i$ is a nearest neighbor of $x_j$. In reality Geodesic distance should be used to actually measure distances on the manifold. Since the exact structure of the manifold is not available, the geodesic distance is approximated by euclidean distances with only nearest neighbors. The choice $\sigma$ modulates our notion of proximity in the sense that if $||x_i - x_j||_2 >> \sigma$ then $K_{ij} = 0$ and if $||x_i - x_j||_2 << \sigma$ then $K_{ij} = 1$. The former means that very little diffusion has taken place while the latter implies that the diffusion process is nearly complete. Different strategies to choose $\sigma$ can be found in.[26] If $K$ has to faithfully represent a Markov matrix, then it has to be normalized by the corresponding degree matrix $D$:
$P = D^{-1}K. \,$
$P$ now represents a Markov chain. $P(x_i,x_j)$ is the probability of transitioning from $x_i$ to $x_j$ in one a time step. Similarly the probability of transitioning from $x_i$ to $x_j$ in t time steps is given by $P^t (x_i,x_j)$. Here $P^t$ is the matrix $P$ multiplied to itself t times. Now the Markov matrix $P$ constitutes some notion of local geometry of the data set X. The major difference between diffusion maps and principal component analysis is that only local features of the data is considered in diffusion maps as opposed to taking correlations of the entire data set.
$K$ defines a random walk on the data set which means that the kernel captures some local geometry of data set. The Markov chain defines fast and slow directions of propagation, based on the values taken by the kernel, and as one propagates the walk forward in time, the local geometry information aggregates in the same way as local transitions (defined by differential equations) of the dynamical system.[25] The concept of diffusion arises from the definition of a family diffusion distance {$D_t$}$_{t \in N}$
$D_t^2(x,y) = ||p_t(x,\cdot) - p_t(y,\cdot)||^2$
For a given value of t $D_t$ defines a distance between any two points of the data set. This means that the value of $D_t(x,y)$ will be small if there are many paths that connect x to y and vice versa. The quantity $D_t(x,y)$ involves summing over of all paths of length t, as a result of which $D_t$ is extremely robust to noise in the data as opposed to geodesic distance. $D_t$ takes into account all the relation between points x and y while calculating the distance and serves as a better notion of proximity than just Euclidean distance or even geodesic distance.
### Hessian LLE
Like LLE, Hessian LLE[27] is also based on sparse matrix techniques. It tends to yield results of a much higher quality than LLE. Unfortunately, it has a very costly computational complexity, so it is not well-suited for heavily-sampled manifolds. It has no internal model.
### Modified LLE
Modified LLE (MLLE)[28] is another LLE variant which uses multiple weights in each neighborhood to address the local weight matrix conditioning problem which leads to distortions in LLE maps. MLLE produces robust projections similar to Hessian LLE, but without the significant additional computational cost.
### Local tangent space alignment
Main article: Local tangent space alignment
LTSA[29] is based on the intuition that when a manifold is correctly unfolded, all of the tangent hyperplanes to the manifold will become aligned. It begins by computing the k-nearest neighbors of every point. It computes the tangent space at every point by computing the d-first principal components in each local neighborhood. It then optimizes to find an embedding that aligns the tangent spaces.
### Local multidimensional scaling
Local Multidimensional Scaling[30] performs multidimensional scaling in local regions, and then uses convex optimization to fit all the pieces together.
### Maximum variance unfolding
Maximum Variance Unfolding was formerly known as Semidefinite Embedding. The intuition for this algorithm is that when a manifold is properly unfolded, the variance over the points is maximized. This algorithm also begins by finding the k-nearest neighbors of every point. It then seeks to solve the problem of maximizing the distance between all non-neighboring points, constrained such that the distances between neighboring points are preserved. The primary contribution of this algorithm is a technique for casting this problem as a semidefinite programming problem. Unfortunately, semidefinite programming solvers have a high computational cost. The Landmark–MVU variant of this algorithm uses landmarks to increase speed with some cost to accuracy. It has no model.
### Nonlinear PCA
Nonlinear PCA[31] (NLPCA) uses backpropagation to train a multi-layer perceptron to fit to a manifold. Unlike typical MLP training, which only updates the weights, NLPCA updates both the weights and the inputs. That is, both the weights and inputs are treated as latent values. After training, the latent inputs are a low-dimensional representation of the observed vectors, and the MLP maps from that low-dimensional representation to the high-dimensional observation space.
### Data-driven high-dimensional scaling
Data-Driven High Dimensional Scaling (DD-HDS)[32] is closely related to Sammon's mapping and curvilinear component analysis except that (1) it simultaneously penalizes false neighborhoods and tears by focusing on small distances in both original and output space, and that (2) it accounts for concentration of measure phenomenon by adapting the weighting function to the distance distribution.
### Manifold sculpting
Manifold Sculpting[33] uses graduated optimization to find an embedding. Like other algorithms, it computes the k-nearest neighbors and tries to seek an embedding that preserves relationships in local neighborhoods. It slowly scales variance out of higher dimensions, while simultaneously adjusting points in lower dimensions to preserve those relationships. If the rate of scaling is small, it can find very precise embeddings. It boasts higher empirical accuracy than other algorithms with several problems. It can also be used to refine the results from other manifold learning algorithms. It struggles to unfold some manifolds, however, unless a very slow scaling rate is used. It has no model.
### RankVisu
RankVisu[34] is designed to preserve rank of neighborhood rather than distance. RankVisu is especially useful on difficult tasks (when the preservation of distance cannot be achieved satisfyingly). Indeed, the rank of neighborhood is less informative than distance (ranks can be deduced from distances but distances cannot be deduced from ranks) and its preservation is thus easier.
### Topologically constrained isometric embedding
Topologically Constrained Isometric Embedding (TCIE)[35] is an algorithm based approximating geodesic distances after filtering geodesics inconsistent with the Euclidean metric. Aimed at correcting the distortions caused when Isomap is used to map intrinsically non-convex data, TCIE uses weight least-squares MDS in order to obtain a more accurate mapping. The TCIE algorithm first detects possible boundary points in the data, and during computation of the geodesic length marks inconsistent geodesics, to be given a small weight in the weighted Stress majorization that follows.
### Relational perspective map
Relational perspective map is a multidimensional scaling algorithm. The algorithm finds a configuration of data points on a manifold by simulating a multi-particle dynamic system on a closed manifold, where data points are mapped to particles and distances (or dissimilarity) between data points represent a repulsive force. As the manifold gradually grows in size the multi-particle system cools down gradually and converges to a configuration that reflects the distance information of the data points.
Relational perspective map was inspired by a physical model in which positively charged particles move freely on the surface of a ball. Guided by the Coulomb force between particles, the minimal energy configuration of the particles will reflect the strength of repulsive forces between the particles.
The Relational perspective map was introduced in.[36] The algorithm firstly used the flat torus as the image manifold, then it has been extended (in the software VisuMap to use other types of closed manifolds, like the sphere, projective space, and Klein bottle, as image manifolds.
## Methods based on proximity matrices
A method based on proximity matrices is one where the data is presented to the algorithm in the form of a similarity matrix or a distance matrix. These methods all fall under the broader class of metric multidimensional scaling. The variations tend to be differences in how the proximity data is computed; for example, Isomap, locally linear embeddings, maximum variance unfolding, and Sammon mapping (which is not in fact a mapping) are examples of metric multidimensional scaling methods.
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26. Rosman G., Bronstein M. M., Bronstein A. M. and Kimmel R., Nonlinear Dimensionality Reduction by Topologically Constrained Isometric Embedding, International Journal of Computer Vision, Volume 89, Number 1, 56–68, 2010
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http://mathoverflow.net/questions/29104/why-are-proofs-so-valuable-although-we-do-not-know-that-our-axiom-system-is-cons/29110
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## Why are proofs so valuable, although we do not know that our axiom system is consistent? [closed]
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As a person who has been spending significant time to learn mathematics, I have to admit that I sometimes find the fact uncovered by Godel very upsetting: we never can know that our axiom system is consistent. The consistency of ZFC can only be proved in a larger system, whose consistency is unknown.
That means proofs are not like as I once used to believe: a certificate that a counterexample for a statement can not be found. For example, in spite of the proof of Wiles, it is conceivable that someday someone can come up with integers a,b and c and n>2 such that a^n + b^n = c^n, which would mean that our axiom system happened to be inconsistent.
I would like to learn about the reasons that, in spite of Godel's thoerem, mathematicians (or you) think that proofs are still very valuable. Why do they worry less and less each day about Godel's theorem (edit: or do they)?
I would also appreciate references written for non-experts addressing this question.
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Do mathematicians really "worry less and less each day about Godel's theorem"? – Robin Chapman Jun 22 2010 at 15:42
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Something to note: Supposing we could actually prove some consistency statement Con(ZFC) within ZFC, that's still no reason to believe that ZFC is consistent! Why? Well, suppose ZFC is inconsistent, then it would prove Con(ZFC) for sure! – Brendan Cordy Jun 22 2010 at 16:29
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@Tom: my impression was that if one is looking for certainty, mathematics is the least wrong field. – Pete L. Clark Jun 22 2010 at 18:02
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I can only speak for myself, but I don't worry about such things for the same reason that I still walk to work every day even though I could get hit by a car at any minute. If I spend the rest of my life trying to convince myself that what I think is a proof really is a proof and one day I actually succeed, then I will just wish I had spent all that time thinking about geometry instead. – Paul Siegel Jun 22 2010 at 18:27
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@Pete: That's a widely held view, and I'm not really denying it. Let me revise my statement: If you're looking for utter certainty, then even mathematics is not entirely the right field. – Tom Goodwillie Jun 22 2010 at 19:04
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## 9 Answers
If you like, you can view proofs of a statement in some formal system (e.g. ZFC) as a certificate that a counterexample cannot be found without demonstrating the inconsistency of ZFC, which would be a major mathematical event, and probably one of far greater significance than whether one's given statement was true or false.
In practice, a given proof is not going to be closely tied to a single formal system such as ZFC, but will be robust enough that it can follow from any number of reasonable sets of axioms, including those much weaker than ZFC. Only one of these sets of axioms then needs to be consistent in order to guarantee that no counterexample would ever be found, and this is about as close to an ironclad guarantee as one can ever hope for.
But ultimately, mathematicians are not really after proofs, despite appearances; they are after understanding. This is discussed quite well in Thurston's article "On proof and progress in mathematics".
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One fascinating case of a proof that (possibly) fails this "robustness" is Laver's proof that the periodicity of Laver tables tends to infinity from the assumption of a rank-into-rank embedding (largest cardinal assumption not known to be inconsistent). There is no known proof of this simple arithmetic statement in any weaker system, although Dougherty & Jech showed that PRA (primitive recursive arithmetic) is insufficient to prove the theorem. – Kiochi Jun 22 2010 at 22:37
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Gödel's theorems do not say that we can never know our axiom systems are consistent. Not at all. What they say is that we can never prove that certain systems are consistent within those systems themselves. This leaves open the possibility that we can prove their consistency in other axiom systems, or can convince ourselves of their consistency by methods that are not completely formal.
My recommended reference on the incompleteness theorems for a general reader is "Gödel's Theorem: An Incomplete Guide to its Use and Abuse" by Torkel Franzén. This book has the rare combination of being written to be broadly accessible while still being precise enough to be satisfying.
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Your argument explains why we might believe in the consistency of some axioms; however, we technically still can't "know" that the original system is consistent, as its consistency is dependent on whatever system proves its consistency. A real "proof" of consistency would require something like writing out all formal proofs from a given set of axioms and checking whether a contradiction is reached, which is clearly impossible. Of course, we choose axioms that we like, not because we can prove they are consistent. – Kiochi Jun 22 2010 at 18:36
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I assert that I do know the Peano axioms are consistent. Most of the time, when people say that they do not "know" this, it's because they have unnaturally limited the meaning of the word "know" specifically to exclude the consistency proofs. Chacun à son goût. – Carl Mummert Jun 22 2010 at 19:23
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OK, I think we understand each other; we are using different definitions of "know." I personally would say that I "strongly believe" that the Peano axioms, and even ZFC are consistent, which may seem overly cautious to those who realize that the likelihood that they are inconsistent is very small. Still, I feel justified on the basis of Occam's razor. I also feel compelled to point out that many people "know" that God exists (and they have proofs for it, too). – Kiochi Jun 22 2010 at 22:17
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@Kiochi: Do you "know" anything? For example, do you "know" that there infinitely many primes? On what basis? On the basis of a proof from some axioms that you do not know to be consistent? – Timothy Chow Jun 22 2010 at 23:46
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When is mathematical philosophy ever popular, convenient, or romantic? More seriously, if you don't believe there are actually infinitely many primes (insisting on qualifying the sentence), why do you strongly believe that ZFC is consistent (without qualification)? "ZFC is consistent" is also an arithmetic sentence, just like "there are infinitely any primes" or Wiles' theorem. Shouldn't you at least strongly believe the infinitude of primes? (Which incidentally is provable in elementary recursive arithmetic, something much weaker than PA.) – Daniel Mehkeri Jun 24 2010 at 21:43
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To address the issue of Fermat's Last Theorem: the reasoning behind Fermat's Last Theorem, while elaborate, in the end rests on basic intuition about the integers. (I'm not sure that it is actually proved in first order Peano arithmetic, since the proof as written certainly uses concepts outside of PA, but nevertheless, it is basically a result about numbers, proved using our fundamental notions about numbers.)
If the proof was correct, but the statement wrong (due to an inconsistency), there would be something fundamentally wrong in our conception of numbers. I don't think this would be like the crisis in set-theory: it would be much more fundamental. For example, if induction turns out to be inconsistent (and this is the kind of thing being speculated about here), this says that our basic intuition for the natural numbers, namely that non-empty subsets have least elements, is wrong. If that is true, then all mathematics goes out the window!
I think that most mathematicians (indeed, most humans who have been taught arithmetic) have a mental model of the natural numbers which says that you can always add 1 to get a new number, and that between any two natural numbers there are only finitely many more (so that any non-empty subset of the naturals has a least element). Given this, they know that PA is in fact consistent, even though PA doesn't prove this. They are proving it by exhibiting a (mental) model; they don't need formal arguments. (This falls under the class of "not completely formal" methods alluded to by Carl Mummert.)
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Wiles' proof of FLT was actually from something stronger than ZFC, however, as you say, being a statement about integers it almost certainly could be proven from much less, probably PA and maybe something more elementary. See cs.nyu.edu/pipermail/fom/2010-May/014728.html. Errett Bishop said, "A proof is any completely convincing argument." – Daniel Mehkeri Jun 23 2010 at 2:03
Dear Daniel, Thanks for the link to this very interesting article. – Emerton Jun 23 2010 at 3:03
To stray from mathematical logic to how other mathematicians might think about proofs...
I think many mathematicians go with Carl's "convince ourselves of their consistency by methods that are not completely formal." Many mathematicians use set theory simply as a language--probably similar sorts of mathematicians as do not concern themselves with categories too much, of which type there are still many. Mathematicians with a physics bent often enjoy "informal" arguments based on physical intuition, a mechanical construction, or the nonrigorous arguments of Archimedes or Appolonius or Cavalieri using a primitive version of infinitesimals to compute volumes, etc. The insight gained from less formal arguments, while less definitive, perhaps, probably outweighs the worries about set-theoretic and proof-theoretic issues for many mathematicians. (A graph theorist, for example, could be perfectly happy proving results for classes of graphs and graph properties for their whole career, knowing that the results are true for graphs the way one usually pictures them, without worrying about the consistency of ZFC).
Since your question is part mathematical logic and part psychology (whether people "worry"), may I suggest some of the literature on pedagogy for higher mathematics? Many people have thought a lot about how to treat the concept of proof and other issues in courses for various sorts of students in order to maximize the understanding and value gained. See, for example, David Henderson's work on "educational mathematics" at Cornell.
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To answer the question posed:
A proof is valuable because it helps convince oneself and others of the validity of that result from the axioms [whether those axioms are consistent or not]. Mathematicians, I believe, are not worried that the axioms are inconsistent, but rather hopeful that they are consistent; or even more precisely, optimistic that if the axioms are inconsistent they can be modified [if necessary] to be consistent and still encompass most things proved. But even if they are inconsistent, we won't figure that out without lots and lots of proofs in the meantime.
To answer the philosophical question from a personal point of view:
From my point of view, I do mathematics because I love certainty and truth. I also enjoy discovery. Godel's theorems simply tell me that there are some things I will never be certain of or discover (inside a formal system). This may be disappointing, but at some level we all have to deal with uncertainty. For example, I could be deceiving myself that I'm typing this message. But I (and most others I know) are willing to accept a few things on faith; and if shown we are wrong, modify our beliefs accordingly.
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We adopt axioms not because we can prove their consistency, but because we believe that they accurately describe something that we want to study. A proof from these axioms will have value in that it shows how the proposition (which may be surprising or complicated) follows from things that we already believe and are simple. If we someday prove an inconsistency using a given set of axioms, this shows that our possibly naïve intuition for reasonable axioms was incorrect. (e.g. Russel's paradox showing that unrestricted comprehension is a bad idea)
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Most practically-useful mathematics does not need the strength of a theory such as ZFC. The reverse mathematics programme shows that many notable theorems can be stated in $RCA_0$, which is a finitistic system. Most other systems studied in revese mathematics (such as $WKL_0$, $ACA_0$ or $ATR_0$) are subsystems of second-order arithmetic, and also enjoy conservativity properties over $RCA_0$.
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Not sure why Gödel's theorems are relevant here. It is indeed unknown whether ZFC is consistent. Somewhere at the undergraduate level students are lead to believe that a proof gives an absolute certainty. Later they learn that this is not quite true, big deal. Most working mathematicians believe in consistency of ZFC, and nobody yet proved them wrong. To date proofs have served us remarkably well.
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Perhaps you should be upset if you've been led to believe as a generalisation that "...we never can know that our axiom system is consistent", and that "...proofs are not like as I once used to believe: a certificate that a counterexample for a statement can not be found".
These beliefs may be understandable (even if not justifiable) in the case of a set-theoretical language such as ZF, which can have no finitary interpretation.
They are misleading in the case of a first-order language such as PA which, in a digitally communicating universe, would be of more consequence than ZF.
Misleading, because they reflect the subjective belief that Aristotle's particularisation holds over the structure N of the natural numbers; which is the postulation that from an assertion such as:
'It is not the case that, for any given x, P(x) does not hold in N',
usually denoted symbolically by '~(Ax)~P(x)', we may always validly infer that:
'There exists an unspecified x such that P(x) holds in N',
usually denoted symbolically by '(Ex)P(x)'.
However, as Brouwer had noted in a seminal 1908 doctoral dissertation, the presumption that Aristotle's particularisation holds over infinite domains such as that of the natural numbers does not lie (as self-evident) within a common human intuition; and such a presumption has no logical basis in the objective decidability and computability of number-theoretic relations and functions over the domain of the natural numbers.
Now, although Kurt Goedel's arguments in his seminal 1931 on undecidable arithmetical propositions avoid assuming that Aristotle's particularisation holds over N, Goedel found it necessary to assume that the Peano Arithmetic he was considering was omega-consistent.
Omega-consistency: PA is omega-consistent if, and only if, there is no PA formula [F(x)] such that [~(Ax) F(x)] is PA-provable and also that, for any PA-numeral [n], [F(n)] is PA-provable.
However, it is easily seen that PA is omega-consistent if, and only if, Aristotle's particularisation holds over N.
(Although J. Barkley Rosser claimed in a 1936 paper to prove the existence of undecidable arithmetical propositions without the assumption of omega-consistency, his argument appears to implicitly presume that Aristotle's particularisation holds over N.)
The assumption of omega-consistency has a history.
David Hilbert was one who firmly believed that an arithmetical proof is, indeed, a certificate that a counterexample for a true arithmetical statement can not be found.
Thus, as part of his program for giving mathematical reasoning a finitary foundation, Hilbert proposed an omega-Rule as a finitary means of extending a Peano Arithmetic to a possible completion (i.e. to logically showing that, given any arithmetical proposition, either the proposition or its negation is formally provable from the axioms and rules of inference of the extended Arithmetic).
Hilbert also believed that the standard interpretation of PA is sound, which implies that Aristotle's particularisation holds over N.
In a contemporary context, Hilbert's omega-rule can thus be expressed as:
Hilbert's omega-Rule: If it is proved that a formula [F(x)] of the first-order Peano Arithmetic PA interprets under the standard interpretation of PA as an arithmetical relation F(x) that is true for any given natural number n, then the PA formula [(Ax) F(x)] can be admitted as an initial formula (axiom) in PA.
Goedel's 1931 paper on formally undecidable arithmetical propositions can, not unreasonably, be seen as the outcome of a presumed attempt to validate Hilbert's omega-rule by his assumption of omega-consistency.
However, Goedel discovered a PA formula [R(x)] such that if PA is assumed omega-consistent, then both [(Ax)R(x)] and [~(Ax)R(x)] are not PA-provable (Goedel's First Incompleteness Theorem).
Further, assuming that Aristotle's particularisation holds over N, Goedel defined a number-theoretic relation Wid(PA) which holds in N if, and only if, PA is consistent.
Goedel then discovered (his Second Incompleteness Theorem) that, if PA is assumed consistent, and we assume that some PA formula [W] expresses Wid(PA) in PA, then [(Ax)R(x)] is PA-provable if [W] is PA-provable.
Ergo, if PA is omega-consistent, then we cannot express the assertion 'PA is consistent' in PA by a PA-provable formula.
Now the points to note are that:
1. PA is omega-consistent if, and only if, Aristotle's particularisation holds over the domain N of the natural numbers.
2. If the standard interpretation of PA is logically sound, then Aristotle's particularisation holds over N.
3. Aristotle's particularisation over N can be expressed in contemporary terms as:
From an assertion such as:
'It is not the case that, for any given x, any witness Witness_N of N can decide that P(x) does not hold in N',
usually denoted symbolically by '~(Ax)~P(x)', we may always validly infer that:
'There exists an unspecified x such that any witness Witness_N of N can decide that P(x) holds in N',
usually denoted symbolically by '(Ex)P(x)'.
The validity of Brouwer's objection follows since Aristotle's particularisation does not hold over N if we take the witness Witness_N as a Turing machine, and P(x) is a Halting-type of number-theoretic relation.
It follows that if PA is not omega-consistent, then we cannot conclude from Goedel's Incompleteness Theorems that Goedel's [~(Ax)R(x)] is unprovable in PA.
The significance of the above is that issues involving number-theoretic functions and relations containing quantification over N lie naturally within the domains of:
(a) First-order Peano Arithmetic PA, which attempts to capture in a formal language the objective essence of how a human intelligence intuitively reasons about number-theoretic predicates, and;
(b) Computability Theory, which attempts to capture in a formal language the objective essence of how a human intelligence intuitively computes number-theoretic functions.
Now Goedel had also shown in Theorem VII of his 1931 paper that every recursive relation can be expressed arithmetically.
This suggests that if we can, conversely, define a finitary interpretation of first-order PA over N as sought by Hilbert, then any number-theoretic problem can be expressed - and addressed - formally in PA and its solution, if any, interpreted finitarily over N.
Now, if [A(x1, x2, ..., xn)] is an atomic formula of PA then, for any given sequence of numerals [b1, b2, ..., bn], the PA formula [A(b1, b2, ..., bn)] is an atomic formula of the form [c=d], where [c] and [d] are atomic PA formulas that denote PA numerals. Since [c] and [d] are recursively defined formulas in the language of PA, it follows from a standard result that, if PA is consistent, then [c=d] is algorithmically computable as either true or false in N.
In other words, if PA is consistent, then [A(x1, x2, ..., xn)] is algorithmically decidable over N in the sense that there is a Turing machine TM_A that, for any given sequence of numerals [b1, b2, ..., bn], will accept the natural number m if, and only if, m is the Goedel number of the PA formula [A(b1, b2, ..., bn)], and halt with output 0 if [A(b1, b2, ..., bn)] interprets as true in N; and halt with output 1 if [A(b1, b2, ..., bn)] interprets as false in N.
Moreover, since Tarski has shown that the satisfaction and truth of the compound formulas of PA (i.e., the formulas involving the logical connectives and the quantifiers) under an interpretation of PA is definable inductively in terms of only the satisfaction (non-satisfaction) of the atomic formulas of PA, it follows that the satisfaction and/or truth of the formulas of PA under the usual interpretation of the PA symbols is algorithmically decidable.
This is clearly a finitary interpretation of PA over N.
So, if Aristotle's particularisation does not hold over N, and the standard interpretation of PA is not logically sound, then - instead of considering Hilbert's omega-rule - the question that one ought to consider is whether the above is a sound algorithmic interpretation of PA such that:
Algorithmic omega-Rule: If it is proved that the PA formula [F(x)] interprets as an arithmetical relation F(x) that is algorithmically decidable as true for any given natural number n, then the PA formula [(Ax)F(x)] can be admitted as an initial formula (axiom) in PA.
This question is far more amenable to finitary reasoning, and I argue that it can be answered affirmatively. See:
http://alixcomsi.com/27_Resolving_PvNP_Update.pdf
If so, it would confirm that your faith in the classical notion of a proof is not misplaced, and an arithmetical proof is, indeed, a certificate that a counterexample for a true arithmetical statement can not be found.
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http://mathhelpforum.com/advanced-algebra/58536-cyclic-group.html
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Thread:
1. Cyclic group
Prove that all subgroup of a cyclic group is cyclic.
Thanks!
2. Let $(G,.)$ be a cyclic group, $H$ a subgroup of $G$, and $\alpha \in G$ such that $<\alpha >=G$.
Let $\phi : \mathbb{Z} \rightarrow G : n \mapsto \alpha^{n}$ be the only morphism of groups such that $\phi(1)=\alpha$.
Then $\phi^{-1}(H)$ is a subgroup of $(\mathbb{Z},+)$, so it's a cyclic group.
So $\phi(\phi^{-1}(H))=H$ (because of surjectivity) is a cyclic group.
Why, if $K$ is a cyclic group, $L$ another group, and $\psi$ a morphism of groups between $K$ and $L$, then $\psi(K)$ is cyclic?
Let $a$ be a generator of $K$, and $b$ an element of $\psi(K)$. There is a $n \in \mathbb{N}$ such that $b=\psi(a^{n})=\psi(a)^{n}$. Thus $<\psi(a)>=\psi(K)$
3. Originally Posted by roporte
Prove that all subgroup of a cyclic group is cyclic.
Thanks!
Let $H$ be a subgroup. And $a$ generate the large group $G$. Choose $n\in \mathbb{Z}^{\times}$ to be the smallest such that $a^n\in H$. Now prove that $\left< a^n \right> = H$.
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http://mathhelpforum.com/advanced-statistics/204102-converge-probability-sample-variance-print.html
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# Converge in probability of sample variance
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• September 25th 2012, 10:41 PM
usagi_killer
Converge in probability of sample variance
If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?
• September 28th 2012, 11:00 PM
chiro
Re: Converge in probability of sample variance
Hey usagi_killer.
Maybe what you can do is to look at S^2 in terms of individual expectations. So typically with variance we know Var[X] = E[X^2] - {E[X]}^2 so if you can show that both expectations converge to their respective results (you already know what happens with E[X]) then basically you show that the two converge to what they are meant to and that the whole thing converges to what's it meant to.
For this you will need to correct the estimator so that it's un-biased by the idea is still the same: expand out the expectation of the sample variance and show that it approaches the parameter.
All times are GMT -8. The time now is 06:56 AM.
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http://physics.stackexchange.com/questions/30638/gauge-symmetry-description-for-phi4
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# Gauge symmetry description for $\phi^4$?
That is a follow-up to this question: Gauge symmetry is not a symmetry?
Ok, gauge symmetry is not a symmetry, but ...
... a redundancy in our description, by introducing fake degrees of freedom to facilitate calculations.
I want some simple and practical example for this.
So If I, say, take a simple $\phi^4$ theory, then I can gauge it by...
... introducing the proper fake degrees of freedom
Can I?
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## 1 Answer
A trivial example:
Take your original field $\phi$ to be a free real scalar field on $\mathbb{R}^n$. Double the number of fields by adding another free real scalar field $\chi$ to your list of fields
Now introduce a gauge symmetry by making the group of functions $g: \mathbb{R}^n \to \mathbb{R}$ act by $g: (\phi,\chi) \mapsto (\phi,\chi+ g)$. So your group of gauge transformations acts trivially on the space of $\phi$'s and freely and transitively on the space of $\chi$'s.
Now fix a gauge in your favorite way. You can grind through the BRST machinery, or you can just choose the gauge slice $\chi= 0$. Either way your original free scalar field is precisely equivalent to the new theory with two fields and the wacky gauge symmetry I described above.
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Could you explain what does gauge away mean? I came across this expression while reading about the Higgs mechanism but didn't understand what it means and what is the significance of doing it. I wanted to start a question on it but found it stated here, hence my question. (And by the way I'm trying to comprehend what you wrote here but this stuff is a bit heavy for me.) – user09876 Jun 22 '12 at 23:15
A gauge symmetry is a symmetry of the variables used to describe a physical system which isn't actually a symmetry of the physical system. Gauging away a variable means using a gauge symmetry to eliminate it from the description of the system. I was getting ahead of myself when I said 'gauge away' $\chi$, since I hadn't actually introduced the gauge symmetry or used it to eliminate $\chi$. Fixed that. – user1504 Jun 23 '12 at 2:08
When you say "not actually a symmetry" can you elaborate? Isn't U(1) symmetry of QED "due to" conservation of electric charge? I mean obviously since we're talking about a bundle over space-time the "translations" (phase changes) aren't actual translations but otherwise i'm not sure what you meant by this. – tachyonicbrane Jun 23 '12 at 2:50
– DJBunk Jun 23 '12 at 3:32
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@JerrySchirmer: The imaginary part in your example in the comments is not a gauge field, it is just zero after the constraint. It isn't arbitrary, and you don't pick out a slice in the path integral. – Ron Maimon Jun 23 '12 at 11:14
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http://mathhelpforum.com/advanced-algebra/169052-roots-unity.html
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# Thread:
1. ## Roots of Unity
Hello, I have a basic (I think) question about Roots of Unity. First I feel like I should mention that I have only the most basic understanding of complex numbers. Here is what I have been asked to prove:
An $n$th root of unity is a complex number $z$ such that $z^{n}=1$. Prove that the $n$th roots of unity form a cyclic subgroup of $\mathbb{C}^{\times}$ of order $n$.
This is my first exposure to roots of unity, but after looking through a number theory book, I feel like the first sentence should include "...where $n\in\mathbb{Z}^{+}$. Am I right about this?
If I am correct about this then how do I prove that each $n$th root of unity has an inverse?
Then, how do I express the identity element of $\mathbb{C}^{\times}$?
I am feeling pretty lost here. Thanks in advance.
2. The $n~~n^{th}$ roots of unity are $\xi _k = e^{i\frac{{2\pi k}}{n}} ,\;k = 0,1 \cdots ,n - 1$.
Now $\xi _0 =1$ is the identity.
Also $\xi _1 = e^{i\frac{{2\pi}}{n}}$ is the generator.
3. Well, if you can assume that $\mathbb{C}^{\times}$ is a group, then you only have two things you have to prove: closure and inverses.
Yes, I would definitely say that $n\in\mathbb{Z}^{+}.$
For inverses, think about a root of unity, $z^{j},$ where $z^{n}=1.$ What do you have to multiply $z^{j}$ by to get $1?$
For closure, the product of two roots of unity must be a root of unity. So, take $z^{j}$ and $z^{k}.$ The product is $z^{j+k}.$ What happens when you raise this product to the $n$th power?
4. The problem isn't just showing that each root has an inverse (all roots are nonzero and so have inverses in C), but that each inverse is also a root of unity. But that's really not too hard if you know enough about complex numbers. Specifically, I'm thinking that for any complex number $z$ with absolute value 1, $z\overline{z}=1$, combined with the fact that $z^n-1=0$ and so $\overline{z}^n-1=$...
If this is more than you know, then it's not hard to list all nth roots of unity explicitly and then figure out the inverses that way. Also, if you know how the nth roots of unity are located in the complex plane and how multiplication of complex numbers adds angles and multiplies absolute values, then finding at least one generator shouldn't be too bad.
If you're not sure about the geometric view of complex numbers and their operations I'd suggest learning that because it makes problems like this so much easier to visualize.
5. How is this? We are still in the first week of this class (abstract algebra) so I'm not sure how rigorous he expects us to be.
Let $U$ be the set of $n$th roots of unity, with elements $a$ and $b$. Then $a^{n}=1$ and $b^{n}=1$. If we multiply both sides of $a^{n}=1$ on the right by $b^{n}$ we have
$a^{n}b^{n}=b^{n}$
$(ab)^{n}=b^{n}$
But since $b^{n}=1$ we have $(ab)^{n}=1$ and therefore $ab$ is an $n$th root of unity. Hence, $ab\in U$.
Since 1 is an identity element in $\mathbb{C}^{\times}$ and $1^{n}=1$, the identity element in $\mathbb{C}^{\times}$ is also in $U$.
Finally, let the inverse of $a^{n}$ be $a^{-n}$. Since $a^{-n}=(a^{n})^{-1}=1^{-1}=1$, we have that $a^{-n}=1$. Hence, $a^{-n}\in U$.
6. You need to show that $a^{-1}\in U,$ not $a^{-n}\in U.$
7. Ah yes, well that is a glaring mistake! Thanks for your help. I've got a feeling that I'll have several more questions throughout the semester.
Finally, let the inverse of $a$ be $a^{-1}$. Since $(a^{-1})^n=a^{-n}=(a^{n})^{-1}=1^{-1}=1$, we have that $a^{-1}$ is an $n$th root of unity. Hence, $a^{-1}\in U$.
8. Right. That looks better.
9. Hello again,
I believe that I have found a problem with my proof. While I have shown that a set $U$ consisting of the $n$th roots of unity form a subgroup of $\mathbb{C}^{\times}$, I have not shown that $U$ is cyclic, or that it is of order $n$.
I am kinda stumped as to how I should go about showing this. While it seems obvious that my proof holds for any powers of $a$, the issue seems that the order of $U$ could be infinite.
Thoughts?
10. Take a closer look at Plato's post # 2. There's a generator there.
11. As for possibly being infinite, there are several ways to show that it isn't. You can argue that there are only finitely many places the root can lie on the unit circle. For example, one 6th root of unity makes an angle of 360/6 degrees with the real axis. This will "go around" the circle once on its trip to 1. But a point can also go around twice. This would correspond to an angle of 360/3. Sooner or later you start to repeat points.
Also, you can show that an infinite cyclic group is isomorphic to $\langle\mathbb{Z}, +\rangle$. In that group the equation $x^n=1$ would be written $nx=0$ using additive notation. This equation only has one solution in $\langle\mathbb{Z}, +\rangle$, whereas the corresponding equation in the group of roots of unity has at least n.
Of course the easiest way is just to invoke the fact that an nth degree polynomial in any field can have at most n zeros. If you're allowed to assume this then it's the easiest way to go.
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http://physics.stackexchange.com/questions/26856/examples-of-number-theory-showing-up-in-physics/26862
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# Examples of number theory showing up in physics
My question is very simple: Are there any interesting examples of number theory showing up unexpectedly in physics?
This probably sounds like rather strange question, or rather like one of the trivial to ask but often unhelpful questions like "give some examples of topic A occurring in relation to topic B", so let me try to motivate it.
In quantum computing one well known question is to quantify the number of mutually unbiased (orthonormal) bases (MUBs) in a $d$-dimensional Hilbert space. A set of bases is said to be mutually unbiased if $|\langle a_i | b_j \rangle|^2 = d^{-1}$ for every pair of vectors from chosen from different bases within the set. As each basis is orthonormal we also have $\langle a_i | a_j \rangle =\delta_{ij}$ for vectors within the same basis. We know the answer when $d$ is prime (it's $d+1$) or when $d$ is an exact power of a prime (still $d+1$), but have been unable to determine the number for other composite $d$ (even the case of $d=6$ is open). Further, there is a reasonable amount of evidence that for $d=6$ there are significantly less than $7$ MUBs. If correct, this strikes me as very weird. It feels (to me at least) like number theoretic properties like primality have no business showing up in physics like this. Are there other examples of this kind of thing showing up in physics in a fundamental way?
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1
MUBs is a really fascinating subjects. They are also linked with Latin squares, as e.g; shown here. I find this link with number theory more surprising than the role of prime numbers. – Frédéric Grosshans Dec 1 '11 at 19:30
1
The MUB connection is really to finite fields (and latin squares as @FrédéricGrosshans mentions) and only through that to prime numbers. I guess we could say this is a connection to number theory, but really seems like a connection to abstract algebra, which is not nearly as surprising. – Artem Kaznatcheev Dec 2 '11 at 4:22
@Artem: It's true that you can arrive at above result via finite fields, but the structure of the partial results is governed by number theoretic properties. I don't really see the way of arriving at a given result as particularly fundamental, as there are often multiple paths to the result. – Joe Fitzsimons Dec 2 '11 at 12:43
– Qmechanic♦ Aug 25 '12 at 10:26
## 9 Answers
There are many attempts for a physical proof of the Riemann hypothesis. The major work in this direction was summarized in a recent review by: Schumayer and Hutchinson.
One of these attempts was proposed by: Berry and Keating. Their suggestion is within the framework of the Hilbert–Pólya conjecture, according to which, the Hilbert–Pólya Hamiltonian, whose spectrum is the imaginary part of the zeta zeros, can be obtained by quantizing a classical Hamiltonian of a chaotic system having periodic orbits with log prime periods. They argue that the classical Hamiltonian can be $xp$ (with appropriate yet unknown boundary conditions).
Another suggestion is due to Freeman Dyson in his Birds and Frogs lecture who suggests that the Riemann hypothesis might be proved through the classification of one dimensional quasicrystals.
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This seems more like engineering a physical system to embody certain number theoretic properties, rather than them occurring unexpectedly. – Joe Fitzsimons Dec 1 '11 at 15:15
I wouldn't count it as "engineering". It's more like using physical intuition in order to make a breakthrough in maths. We know the properties the Hilbert-Polya hamiltonian should have, so we whether it might be implemented in a physical system. – Javier Rodriguez Laguna Dec 3 '11 at 19:12
Here is a toy example; I don't know how interesting this will be to physicists. The eigenvalues of the Laplacian acting on, say, smooth functions $\mathbb{R}^k/(2\pi \mathbb{Z})^k \to \mathbb{C}$ are given by $$\{ m_1^2 + ... + m_k^2 : m_i \in \mathbb{Z} \}.$$
as a multiset (that is, with multiplicities). These are the energy eigenvalues of $n$ free non-interacting quantum particles on a circle. The multiplicity of a given eigenvalue is therefore the number of ways to write it as a sum of $k$ (integer) squares.
This is a classical number-theoretic problem. For example, it is a classical result that the number of ways to write a non-negative integer $n$ as the sum of two squares is $$r_2(n) = 4 \sum_{d | n} \chi_4(d)$$
where $\chi_4(d)$ is equal to $0$ if $d \equiv 0, 2 \bmod 4$, equal to $1$ if $d \equiv 1 \bmod 4$, and equal to $-1$ if $d \equiv 3 \bmod 4$. In general, the number of ways $r_k(n)$ to write a non-negative integer $n$ as the sum of $k$ squares has generating function $$\sum r_k(n) q^n = \left( \sum_{m \in \mathbb{Z}} q^{m^2} \right)^k = \theta(q)^k.$$
The function $\theta(q)$ is a theta function. Theta functions are closely related to modular forms, an important topic in number theory, and in fact the classical proof of the closed form $$r_4(n) = 8 \sum_{d | n} [4 \nmid d]$$
(where we have used the Iverson bracket above) proceeds by showing that $\theta(q)^4$ is a modular form; see Wikipedia.
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The $\theta$ page is crowded with greeks like the acropolis. Can you specify which $\theta$ you mean? – draks ... May 14 '12 at 18:23
@draks: it's defined immediately before I use the symbol. – Qiaochu Yuan May 14 '12 at 18:26
Ah, that was too close for me, thanks. (i) How are $k$, the number of squares and $n$, the number of particles on the circle related? And (ii) are there comparable formulas for $r_k(n)$ when one sums powers other than 2? – draks ... May 14 '12 at 18:32
@draks: $k$ is the number of particles. $n$ is (up to some normalization) an energy eigenvalue. The situation when summing powers other than $2$ is considerably more complicated because the close relationship to modular forms disappears and I don't know what's known about it. – Qiaochu Yuan May 14 '12 at 18:34
Thanks a lot. – draks ... May 14 '12 at 18:35
I have an example of my own :). It appeared trying to calculate the dimension of a Hilbert space associated with rotationally invariant systems of n spins. The dimension was given in terms of the Moebius function. for details, check the appendix of Phys. Rev. E 76, 061127 (2007) or arXiv:quant-ph/0702164.
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That's weird, and certainly interesting. I'll give the paper a look. – Joe Fitzsimons Dec 2 '11 at 12:59
I've encoutered Diophantine equations (a variant of Pell's equation) in an (unpublished) attempt to turn a molecular system into a classical logical gate. The goal was to (approximately) synchronize incommensurable oscillations, and successives solution to the Diophantine equation gave me better fidelities.
I don't know if it qualifies for number theory, or even for physics, but I was surprised to find this equation as a good tool for my physics problem.
If anyone is interested, I can probably unearth my old notes and write something more detailed on the problem and the solution I found. Just ask in the comment.
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This is a general property--- diophantine approximation is related to resonance in classical systems like in KAM. – Ron Maimon May 8 '12 at 15:45
There are many theorems in quantum information which only apply to qudits of prime dimension. In particular, this seems to happen with graph states. In that case many theorems rely on the fact that multiplication modulo a prime is an invertible operation.
The Chinese Remainder Theorem can be used to show that graph states made of qudits of square-free dimension are equivalent to collections of graph states of qudits of prime dimension (the primes being the prime factorization of the original dimension).
Related to number theory is algebra. Group theory in particular tends to play an important role in quantum computing (e.g. the hidden subgroup problem).
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1
Thanks for taking the time to compose an answer. I don't really consider quantum algorithms as fundamental physics in the sense of this question, particularly given that the hidden subgroup stuff is driven by a generalization of problems from number theory (factoring/discrete logs). The graph state observation seems more related to the fact that you are looking at factoring a Hilbert space, which directly relates to primality of the dimesnionality, etc. – Joe Fitzsimons Dec 2 '11 at 12:50
For quantized cat maps, the inverse of Planck's constant is an integer N . There are various results for the special cases, where N is a power of a prime. So, the arithmetic properties of N play an important role here.
For references, see http://www.math.kth.se/~rikardo/cat2.pdf .
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Could you elaborate on this a bit? – András Bátkai Dec 2 '11 at 7:11
Added a reference. – jjcale Dec 3 '11 at 16:52
This answer is closely related to jjcale's answer. In this article, Gurevich and Hadani prove Rudnick's quantum ergodicity conjecture about the Berry-Hannay model. To do it they construct a number-theoretical description of the quantization of a torus phase space at rational values of hbar, involving l-adic sheaves on an algebraic variety of positive characteristic.
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This sounds more like physics showing up in number theory. – MBN Dec 5 '11 at 11:47
@MBN not quite. The result they prove belongs to the realm of quantum dynamical systems, not number theory. Number theory, or, more precisely, arithmetic geometry is a tool to solve the problem. – Squark Dec 5 '11 at 21:52
in general the Riemann xi function can be proved to be a functonal determinant
$\frac{\xi(s)}{\xi(0)}= \frac{det(H+1/4+s(1-s)}{det(H+1/4)}$ with $H=p^{2}+ V(x)$ and $V^{-1} (x)= 2 \sqrt \pi \frac{d^{1/2}}{dx^{1/2}}\frac{1}{\pi}arg\xi(1/2+i\sqrt x)$
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There's the Langlands program in supersymmetric quantum gauge theories, and string theory.
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http://math.stackexchange.com/questions/43562/universal-digraph?answertab=active
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# Universal digraph
The Rado graph is universal since it contains an isomorphic copy of every countable graph (infinitely often). The Rado graph can be explicitely constructed and an isomorphic copy of any countable graph can be effectively computed.
Is there also a universal directed graph, especially one with an explicit construction?
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## 1 Answer
Yes. Use the same construction as the Rado graph, but replace the base 2 by base 4. The vertices are in one-to-one correspondence with the natural numbers, with edges obeying the following rule.
For any two vertices $x,y$ with $x<y$:
1. If the $x$th bit of the base-4 expansion of $y$ is a $0$, then there are no edges between $x$ and $y$.
2. If the $x$th bit of the base-4 expansion of $y$ is a $1$, then there is a directed edge from $x$ to $y$.
3. If the $x$th bit of the base-4 expansion of $y$ is a $2$, then there is a directed edge from $y$ to $x$.
4. If the $x$th bit of the base-4 expansion of $y$ is a $3$, then there are directed edges in both directions between $x$ and $y$.
This has very similar properties to the Rado graph, and roughly the same algorithm works for finding induced subgraphs of a given isomorphism type.
Incidentally, the idea here is presumably the same as the construction of 4-edge-colored Rado graph $G_4$ mentioned in "related concepts" section of the same Wikipedia article.
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Thanks. Do you have a reference where I can learn a little bit more about the directed Rado graph? – Hans Stricker Jun 6 '11 at 7:14
@Hans: I would recommend looking up references on the 4-edge-colored Rado graph, e.g. the reference to Truss (1985) mentioned in the article. A directed graph really isn't very different from the type of 4-edge-colored graph they are considering. – Jim Belk Jun 6 '11 at 7:25
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http://mathoverflow.net/revisions/94094/list
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## Return to Answer
2 added 2 characters in body
Well, $B$ is the Hadamard product of $A$ with itself, so if $A$ is nonnegative or psd, its radius is bounded from above by the square of the radius of $A$. For real matrices, at least.
Well, $B$ is the Hadamard product of $A$ with itself, so its radius is bounded from above by the square of the radius of $A$. For real matrices, at least.
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http://mathhelpforum.com/calculus/182714-need-help-finding-critical-points-inflection-points-graphing-print.html
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# Need Help finding critical points and Inflection points. (graphing)
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• June 9th 2011, 11:06 AM
mattyc33
Need Help finding critical points and Inflection points. (graphing)
The question asks for the critical points and inflection points of:
(x^2-1)/(x^3+1)
I've tried to take the derivitave and second derivitave but it just got too complicated. If anyone could help me that would be great.
My quick commands menu doesn't seem to be working but the question is quite simple.
Thanks!
• June 9th 2011, 11:36 AM
Quacky
Quote:
Originally Posted by mattyc33
The question asks for the critical points and inflection points of:
(x^2-1)/(x^3+1)
I've tried to take the derivitave and second derivitave but it just got too complicated. If anyone could help me that would be great.
My quick commands menu doesn't seem to be working but the question is quite simple.
Thanks!
I don't see any shorthand, I think we're just going to have to use the quotient rule.
So Let $u=x^2-1$
$\frac{du}{dx}=2x$
Let $v=x^3+1$
$\frac{dv}{dx}=3x^2$
We have:
$\dfrac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
$=\dfrac{(x^3+1)(2x)-(x^2-1)(3x^2)}{(x^3+1)^2}$
$=\dfrac{2x^4+2x-3x^4+3x^2}{x^6+2x^3+1}$
$=\dfrac{-x^4+3x^2+2x}{x^6+2x^3+1}$
$=\dfrac{x(-x^3+3x+2)}{(x^3+1)^2}$
$\dfrac{-x(x-2)(x+1)(x+1)}{((x+1)(x^2-x+1))^2}$
$=\dfrac{-x(x-2)}{(x^2-x+1)^2}$
...which I suppose you could differentiate again, although there's got to be a faster approach? If there is, it's at a level that's beyond me, though.
• June 9th 2011, 11:39 AM
waqarhaider
expression can written as
y = x^5 - x^3 + x^2 - 1
then y' = 5x^4 - 3x^2 + 2x
for critical points 5x^4 - 3x^2 + 2x = 0 gives two real roots x = 0 and x = -1 now you can do it
• June 9th 2011, 12:00 PM
mattyc33
Quote:
Originally Posted by waqarhaider
expression can written as
y = x^5 - x^3 + x^2 - 1
then y' = 5x^4 - 3x^2 + 2x
for critical points 5x^4 - 3x^2 + 2x = 0 gives two real roots x = 0 and x = -1 now you can do it
Unfortunatly, I have no idea how you got that expression (Thinking).
When I plugged what Quacky said into my calculator I got two x values of x=0 and x=2.
Could anyone please explain how he got this expression (Wondering).
• June 9th 2011, 12:21 PM
Quacky
Quote:
Originally Posted by mattyc33
Unfortunatly, I have no idea how you got that expression (Thinking).
When I plugged what Quacky said into my calculator I got two x values of x=0 and x=2.
Could anyone please explain how he got this expression (Wondering).
Editted out a sign error with my original response - and to be honest I'm not entirely sure how he got there either.
• June 9th 2011, 01:42 PM
mattyc33
Thanks alot for clearing up the critical points.
I'm pretty terrible at my derivitave calculation though so once again i'm having alot of trouble finding the second derivitave in order to find the inflection points.
As Quacky pointed out the first derivitave is -x(x-2)/(x^2-x+1)^2
If anyone could help me that would be great (Worried).
• June 9th 2011, 01:44 PM
Quacky
Quote:
Originally Posted by mattyc33
Thanks alot for clearing up the critical points.
I'm pretty terrible at my derivitave calculation though so once again i'm having alot of trouble finding the second derivitave in order to find the inflection points.
As Quacky pointed out the first derivitave is -x(x-2)/(x^2-x+1)^2
If anyone could help me that would be great (Worried).
Again, I think there must be a shorter method, but what have you tried? Let $u=-x^2+2x$ and $v=(x^2-x+1)^2$
• June 9th 2011, 01:49 PM
mattyc33
Yeah, I've tried using the quotient rule, it just gets very messy and I'm 99% sure i've been making mistakes in this mess 0.0
• June 9th 2011, 01:52 PM
Quacky
Quote:
Originally Posted by mattyc33
Yeah, I've tried using the quotient rule, it just gets very messy and I'm 99% sure i've been making mistakes in this mess 0.0
Please post some effort. Doing the first one took me long enough!
• June 9th 2011, 03:14 PM
mattyc33
Well all I have to say is that there must be an easier way:
(-x^2+2x)'(x^2-x+1)^2 - (x^2+2x)(x^2-x+1)' / (x^2-x+1)^4
(-2x+2)(x^2-x+1)^2 - (x^2+2x)[2(x^2-x+1) x (2x-1)] / (x^2-x+1)^4
(-2x+2)(x^4-2x^3+3x^2-2x+1) - (x^2+2x)(4x-2)(x^2-x+1) / (x^2-x+1)^4
(-2x^5+2x^4-6x^3+4x^2-2x+2x^4-4x^3+6x^2-4x+2) - (4x^5-4x^3-4x) / (x^2-x+1)^4
And from there... I have no idea (Headbang).
I'm not saying to try it haha, I know there must be another way, thanks for the help!
• June 9th 2011, 03:43 PM
Quacky
Indeed: The roots are horrid!
http://www.wolframalpha.com/input/?i=Differentiate+-x%28x-2%29%2F%28x^2-x%2B1%29^2
• June 9th 2011, 03:51 PM
mattyc33
Quote:
Originally Posted by Quacky
Indeed: The roots are horrid!
http://www.wolframalpha.com/input/?i=Differentiate+-x%28x-2%29%2F%28x^2-x%2B1%29^2
Whew thats great thanks!
• June 9th 2011, 06:29 PM
Ackbeet
Try this:
$\frac{x^{2}-1}{x^{3}+1}=\frac{(x-1)(x+1)}{(x+1)(x^{2}-x+1)}=\frac{x-1}{x^{2}-x+1},$
for $x\not=-1.$
I think that might simplify things for you a bit.
• June 10th 2011, 06:03 AM
waqarhaider
Sorry for the first reply where I missed '/' sign and solved (x^2-1) (x^3+1)
expression can be written as
y = ( x - 1 ) / ( x^2 - x + 1 )
then y' = [ ( x^2 - x + 1 ) 1 - ( x - 1 ) ( 2x - 1 ) ] / ( x^2 - x + 1 )^2
gives y' = ( 2x - x^2 ) / ( x^2 - x + 1 )
again y" = [ ( x^2 - x + 1 )^2 ( 2 - 2x ) - ( 2x - x^2 ) 2( x^2 - x + 1 )( 2x - 1 ) ]/ ( x^2 -x + 1 )^4
gives y" = ( 2x^3 - 6x^2 + 2 ) / ( x^2 - x + 1 )^3 and we find point of inflection
• June 10th 2011, 06:11 AM
Ackbeet
Your derivatives look correct to me. So what do you get for critical points and points of inflection?
Show 40 post(s) from this thread on one page
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http://mathhelpforum.com/trigonometry/142582-finding-composite-inverse-function.html
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# Thread:
1. ## Finding a composite inverse function.
Hey guys! I need to find:
$tan(cot^{-1}(\frac{-1}{\sqrt{3}}))$
Here's how I'm going about finding the answer:
I read this as, "Tangent of the angle whose cotangent is negative 1 over the square root of 3."
As I understand it, I need to draw the triangle for the inverse cotangent in quadrant 4 since it is negative. Here's my drawing:
Now, using this drawing, I would find the tangent (and the answer to the expression) to be: $-\sqrt{3}$
Did I do this correctly? I'm having a hard time with the concept of inverse functions in general, so any advice or somewhere I could read to better understand the concept would be helpful. I've read my trig book and it's not helping me very much.
I'm doubting that it is correct because when I evaluate the expression on my TI-84, the answer doesn't match.
Thanks for any help!
2. Originally Posted by JennyFlowers
Hey guys! I need to find:
$tan(cot^{-1}(\frac{-1}{\sqrt{3}}))$
Here's how I'm going about finding the answer:
I read this as, "Tangent of the angle whose cotangent is negative 1 over the square root of 3."
As I understand it, I need to draw the triangle for the inverse cotangent in quadrant 4 since it is negative. Here's my drawing:
Now, using this drawing, I would find the tangent (and the answer to the expression) to be: $-\sqrt{3}$
Did I do this correctly? I'm having a hard time with the concept of inverse functions in general, so any advice or somewhere I could read to better understand the concept would be helpful. I've read my trig book and it's not helping me very much.
I'm doubting that it is correct because when I evaluate the expression on my TI-84, the answer doesn't match.
Thanks for any help!
Actually,
$cot^{-1}(\frac{-1}{\sqrt 3})$ returns you an angle
so, $cot^{-1}(\frac{-1}{\sqrt 3}) = \frac{-\pi}{3}$
and $tan(\frac{-\pi}{3})= -tan(\frac{\pi}{3})= -\sqrt{3}$
$\therefore tan(cot^{-1}(\frac{-1}{\sqrt 3})) = -\sqrt{3}$
3. Out of curiosity, what difference would this make?
$tan^{-1}(cot(\frac{-1}{\sqrt{3}}))$
Does switching which function is the inverse change the outcome?
Thanks for the help!
4. Hello, JennyFlowers!
With a little thought, this is an "eyeball" problem.
(You get the answer by looking at the problem.)
Find: . $\tan\left[\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right)\right]$
Here's how I'm going about finding the answer:
I read this as: "Tangent of the angle whose cotangent is $-\tfrac{1}{\sqrt{3}}$ " . Good!
$\text{Let }\theta \,=\,\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) \quad \hdots \quad \theta\text{ is an angle whose cotangent is }-\tfrac{1}{\sqrt{3}}$
Then: . $\cot\theta \:=\:-\frac{1}{\sqrt{3}} \:=\:\frac{adj}{opp}$
The right triangle looks like this:
Code:
``` *
* |
* | _
* | √3
* |
* θ |
* - - - - - *
-1```
And we see that: . $\tan\theta \:=\:\frac{\sqrt{3}}{\text{-}1} \:=\:\text{-}\sqrt{3}$
. . Hence: . $\theta \:=\:\tan^{-1}\left(\text{-}\sqrt{3}\right)$
Do you see what we've done?
We've shown that: . $\cot^{-1}\left(\text{-}\tfrac{1}{\sqrt{3}}\right) \;=\;\tan^{-1}\left(\text{-}\sqrt{3}\right)$
Give it some thought . . . The reasoning should become clear. .**
So we can replace $\cot^{-1}\left(\text{-}\tfrac{1}{\sqrt{3}}\right)$ with $\tan^{-1}\left(\text{-}\sqrt{3}\right)$
And the original problem becomes: . $\tan\left(\tan^{-1}(\text{-}\sqrt{3})\right)$
. . which equals (of course): . $-\sqrt{3}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
The same reasoning that allows us to say:
. . $\begin{array}{ccc}\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) &=& \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) \\ \\[-2mm]<br /> <br /> \csc^{-1}(\sqrt{2}) &=& \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) \end{array}$
5. Thank you, Soroban, I think I understand now. But how did you determine that the angle should be drawn in quadrant II? Was it just luck that I got the same answer by drawing in quadrant IV?
6. "Tangent" is periodic with period $\pi$. Drawing your angle in quadrant IV rather than quadrant II was just adding [itex]\pi[/itex] radians to it.
In any case, since tan(x)= 1/cot(x), $tan(cot^{-1}(-\frac{1}{\sqrt{3}})= 1/cot(cot^{-1}(\frac{1}{\sqrt{3}}))= -1/\frac{1}{\sqrt{3}}= -\sqrt{3}$.
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http://mathhelpforum.com/algebra/107347-wheel-question.html
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# Thread:
1. ## Wheel Question
Hi, Please help me solve this question. I have no idea where to begin, other than knowing that A must be larger than one, because the question asks, 'what is the probability that the fraction $\frac{a}{b}$ is greater than 1.
The question is attached as an image below.
Attached Thumbnails
2. (0+1+2+3+4+5) / 36 = 15/36 = 5/9
0: 1/1, 1/2, 1/3, 1/4, 1/5, 1/6
....
5: 6/1, 6/2, 6/3, 6/4, 6/5, 6/6
3. Originally Posted by gs.sh11
Hi, Please help me solve this question. I have no idea where to begin, other than knowing that A must be larger than one, because the question asks, 'what is the probability that the fraction $\frac{a}{b}$ is greater than 1.
The question is attached as an image below.
$\frac{a}{b} > 1$ if and only if $a > b$.
From the picture, all values of a from 1 to 6, are equally likely, and the same for b. So all 36 ordered pairs (a,b) are equally likely. List the pairs and count the ones where a > b.
4. A little more quickly that listing them all: of the 36 pairs, exactly 6 of them are pairs of equal numbers. Of the remaining 30, exactly half, 15 have a> b and the other half have b>a. The probability that, on a single roll, a> b is 15/36= 5/9.
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http://nrich.maths.org/2382/note
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nrich enriching mathematicsSkip over navigation
Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Adding All Nine
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
GOT IT Now
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
Ben's Game
Why do this problem?
This problem requires some simple knowledge of fractions and multiples and demands some strategic thinking. It may offer a good opportunity to compare methods between students - there isn't just one route to the solution. Note that there is no need to use algebra in this problem.
Possible approach
Choose three students to act out the scenario with a (real or imaginary) pot of $40$ counters, as described under support below.
Ask students to work in pairs or small groups to try and find the answers. If any groups are successful too quickly (!) ask them to change one or more of the fractions, and to adapt their strategies to the new situations.
As a group discuss the methods used.
What worked? What didn't work?
If faced with a similar problem in future, which methods would the class use?
Methods students have used include:
• trial and error, making strong use of the upper bound of 40,
• working backwards from the fact that they end up with the same amount.
• focusing on one individual child's initial share of counters
• use the fact that each child passes a certain fraction to their neighbour.
• algebraic representation
Because the problem has many variables, students will need to devise a clear recording system
Key questions
Can Ben start with $10$ counters?
Why can't they use all $40$ counters in this game?
What are the possibilities for Emma's first pile of counters?
What possible numbers of counters could each child end up with?
Will there be just one solution? No solution? Many solutions?
Possible extension
Change the numbers. What if the $40$ were $100$, or a limitless supply of counters?
What if the third, quarter and fifth were different fractions - unitary or not?
What generalisations can you find in the solutions?
Possible support
Group students in $3$s and provide sets of $40$ counters. Ask them to play this game as a group and challenge them to be the first group to find the answer. Use these rules
1. Decide who will be Emma, Jack and Ben.
2. Emma chooses any number of counters from the $40$ counters and notes this number down
3. Jack then chooses any number of counters from those left over and notes this number down
4. Ben then chooses any number of counters from those left over and notes this number down
5. Emma, Jack and Ben then find a fifth, quarter and third of their counters respectively
6. All pass the counters to their neighbour
7. If they all have the same number then the group has won.
As students try to play this game they will encounter difficulties in making the fractions and see that the number of counters taken must be quite specific in order to end up with the same amount at the end. Students should be encouraged to learn from their mistakes to try to find out the winning combination.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://quant.stackexchange.com/tags/expected-return/hot
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# Tag Info
## Hot answers tagged expected-return
10
### What methods do you use to improve expected return estimates when constructing a portfolio in a mean-variance framework?
Short of having a 'reasonable' predictive model for expected returns and the covariance matrix, there are a couple lines of attack. Shrinkage estimators (via Bayesian inference or Stein-class of estimators) Robust portfolio optimization Michaud's Resampled Efficient Frontier Imposing norm constraints on portfolio weights Naively, shrinkage methods ...
8
### How are distributions for tail risk measures estimated in practice?
Perhaps you may want to consider article by D. Levine - Modeling Tail Behavior with Extreme Value Theory who gives practicale example on how EVT can be used to calculate probabilities on returns in tails with use of the Pickands-Balkema-de Haan Theorem and generalized Pareto distribution. It also contains some criterias and points on other methods that can ...
8
### How are distributions for tail risk measures estimated in practice?
One approach is Conditional Value at Risk (CVaR) a.k.a. Expected Shortfall (ES). It does, as you suggest, take into account the whole set of returns. However, instead of traditional VaR which asks "what is the worst 1% or 5% loss I can expect" in a given time frame, conditional VaR asks "assuming I sustain losses of at least 95% or 99% (and perhaps am ...
8
### What methods do you use to improve expected return estimates when constructing a portfolio in a mean-variance framework?
Both answers from Shane and Vishal Belsare make sense and detail different models. In my experience, I have never been satisfied by a unique model since the majority of papers out there can be split in two categories: Those that predict the mean component of the problem. Those that predict the variance component of the problem. The ideal (to read ...
6
### What methods do you use to improve expected return estimates when constructing a portfolio in a mean-variance framework?
You raise a very important point, which unfortunately doesn't have a simple answer. Black-Litterman addresses the allocation problem by allowing you to provide a prior within a bayesian framework. It doesn't really tell you how to produce the prior itself. But more importantly, it doesn't address the fundamental problem: it's difficult to accurately ...
6
### How does left tail risk differ from right tail risk?
Here's a partial answer: This partly depends on the return characteristics. One way to look at this is to analyze the skewness and kurtosis of the returns. Most strategies have a negative skewness, which roughly means that they have mostly consistent small positive returns, with the occasional large negative return. Alternatively, some strategies have ...
5
### What is the expected return I should use for the momentum strategy in MV optimization framework?
If you don’t have any specific model which describes the behavior of the asset being traded, you can estimate the empirical distribution of returns by backtesting your momentum strategy. Then you can adjust this estimate during your strategy’s lifetime from your trading results. Additionally you can enhance this by accounting for different market regimes ...
5
### How to apply the Kelly criterion when expected return may be negative?
You are trying to apply the Kelly Criterion, supposedly to maximize how aggressively to bet, and you are having trouble when the Kelly Value turns negative. The naive answer to your question is that when your kelly value turns negative, then $f=\frac{bp-q}{b}$ turning negative means the instantaneous expected return is negative, which means you should not ...
4
### Expected return from a multiple linear regression?
If the equation satisfies all the assumptions of OLS, particularly homoscedasticity and no autocorrelation in the errors, then the expected return for the equation you laid out is $E[r_{future}|r_{history},x_{news}]=\alpha+\beta_1r_{history}+\beta_2x_{news}+\beta_3r_{history}*x_{news}$ If the unconditional expected return is zero (as is likely to be ...
3
### St Petersburg lottery pricing & short investing horizons
What a great question -- it touches on many issues at the core of quantitative finance. This answer might be a lot more than you bargained for, but it's too interesting to pass up. References Mostly, this subject falls somewhere at the intersection of these three highly-interrelated topics: risk-neutral valuation, rational pricing and the fundamental ...
3
### What is the expected return I should use for the momentum strategy in MV optimization framework?
Another possible approach is taking views a la Black-Litterman. There is a 2006 paper "Incorporating Trading Strategies in the Black-Litterman Framework" that discusses the methodology in more detail. There are several practical issues that one should consider when implementing a momentum strategy with optimization. I would pay careful attention to the ...
3
### How does left tail risk differ from right tail risk?
Tail risk represents the probability that the magnitude of returns on an asset/portfolio will exceed some threshold (usually three standard deviations) on the normal curve. If you visualize a normal curve on standard axes, the tail on the left side corresponds to an extreme low return and the tail on the right side corresponds to an extreme high return. In ...
2
### How does return-based analysis calculate expected return of a trading system?
Returns-based analysis cannot calculate the expected return of a trading system. It yields nonsensical results and is not suited to this particular calculation. Consider a game where every time you play, you win 25% twice and lose 40% once. There are basically three permutations of this game. Represented in R vectors: first <- c(.25, .25, -.4) second ...
2
### How does return-based analysis calculate expected return of a trading system?
You can't add returns. You must multiply them. In your example above where daily returns are 25%, 25%, and -40% To compute expected return from a return series, simply use this formula: return = product( 1+return); in the case of you example this yields: return = (1.25 * 1.25 * .6) = .9375 To get the expected daily return use the geometric mean: ...
2
### How to model the daily return using intraday data?
It seems that your real question is: is the PFP (Price Formation Process) diffusive from intraday to weekly sampling rate? It is a very good question since on intraday, some academics found some multifractal features into intraday returns, meaning that the PFP is not a Geometric Brownian Motion at small scales (even considering stochastic volatility). You ...
1
### St Petersburg lottery pricing & short investing horizons
This may or may not be helpful, since I don't have anything to point you to that specifically addresses the high skewness of the distribution you mention. However, this sounds like it is probably an idiosyncratic risk, and that certainly has bearing on whether or not it would be priced. In the standard capital asset pricing model, the marginal investor ...
1
### How to model the daily return using intraday data?
Isn't this a simple mathematical rule? $$\Delta r_{t}=r_{t} - r_{t-1} = ln(p_{t}) - ln(p_{0}) - ln(p_{t-1}) + ln(p_{0})=ln(\frac{p_{t}}{p_{t-1}})$$ i.e. logarithmic or continuously compounded return. As a result: E(\Delta r_{t})=\frac{1}{T}\sum_{t=1}^{T}\Delta r_{t} = ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://mathhelpforum.com/calculus/212516-using-integral-parts.html
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# Thread:
1. ## Using integral by parts
Hello everyone it's me again...
For this question I have to use integral by parts to find the integral of (t+5)/(sqrt 8-t)
I have:
u= t+5
v' = (8-t)^-1/2
u'= 1
v= 2(8-t)^1/2
then I plugged in the numbers
(t+5)(2(8-t))^1/2 = integral (8-t)^1/2
I don't really know what to do could someone help me out? /:
2. ## Re: Using integral by parts
Integration by Parts is not the best method to use here.
$\displaystyle \begin{align*} \int{\frac{t+5}{\sqrt{8-t}}\,dt} = -\int{-\frac{t+5}{\sqrt{8-t}}\,dt} \end{align*}$
Let $\displaystyle \begin{align*} u = 8 - t \implies du = -dt \end{align*}$ and note that $\displaystyle \begin{align*} t + 5 = 13 - u \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} -\int{-\frac{t + 5}{\sqrt{8-t}}\,dt} &= -\int{\frac{13 - u}{\sqrt{u}}\,du} \\ &= -\int{13u^{-\frac{1}{2}} - u^{\frac{1}{2}} \, du} \end{align*}$
This should now be easy to solve.
3. ## Re: Using integral by parts
Originally Posted by Prove It
Integration by Parts is not the best method to use here.
$\displaystyle \begin{align*} \int{\frac{t+5}{\sqrt{8-t}}\,dt} = -\int{-\frac{t+5}{\sqrt{8-t}}\,dt} \end{align*}$
Let $\displaystyle \begin{align*} u = 8 - t \implies du = -dt \end{align*}$ and note that $\displaystyle \begin{align*} t + 5 = 13 - u \end{align*}$ and the integral becomes
$\displaystyle \begin{align*} -\int{-\frac{t + 5}{\sqrt{8-t}}\,dt} &= -\int{\frac{13 - u}{\sqrt{u}}\,du} \\ &= -\int{13u^{-\frac{1}{2}} - u^{\frac{1}{2}} \, du} \end{align*}$
This should now be easy to solve.
Unfortunately I have to use integration by parts as instructed by my teacher but this should help me come up with the answer. Thank you.
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http://crypto.stackexchange.com/questions/2627/algorithm-design-for-only-mutual-information-sharing/2630
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# Algorithm Design for only Mutual Information Sharing
Bob and Alice each have a bit string they want to keep private. They each want to know what the bitwise AND of their two strings would be without telling the other or anyone else listening to their exchange their actual bit strings... how can they do this? Keep in mind that even once they both hold the AND of their two bit strings, they should still not be able to calculate the other person's string exactly (unless of course one of their strings was all 1s).
I asked this on Stack Overflow and got yelled at to move it here. Not really sure how to tag it either if anyone knows what it would fit under better please feel free to edit. I know that I have seen something similar before in some sort of mutual key system/voting system but I couldn't remember where. It has to be something like make a private random key, xor it and use that somehow... but I couldn't work out the details. Any clever encryption design people out there?
-
I've been wondering about similar questions for a long time. We can ignore the question of the eavesdropper, since any solution that allows $A$ and $B$ to exchange the right amount of information can be encrypted more or less unbreakably with the usual methods so that an eavesdropper gets no information at all. The tricky part is making sure that $A$ doesn't find out any of the bits of $B$'s string where the corresponding bit of $A$'s string is 0. – Mark Dominus May 16 '12 at 0:23
right because of course A will know exactly the values of B's string at any locations where A has a 1 – hackartist May 16 '12 at 0:24
My own favorite question of this type is whether $A$ and $B$ can learn how many 1 bits there are in the XOR of their strings without learning anything else about the other person's string. (I don't know.) – Mark Dominus May 16 '12 at 0:29
– Zev Chonoles May 16 '12 at 2:25
ok thank you. I wasn't aware about the multiple site policy and I will be more careful in the future. – hackartist May 16 '12 at 2:28
## 2 Answers
Aaron Roth on theoretical CS was kind enough to answer with the following answer for anyone out there who is interested.
What you want to do is called "Private Set Intersection". You can think of Alice and Bob as each holding sets (the indices for which their strings are "1"), and they want to compute the intersection (the bitwise AND) so that neither of them learns anything about the other's set except what is implied by the intersection itself.
This problem is well studied. See, for example, Freedman, Nissim, and Pinkas: http://www.pinkas.net/PAPERS/FNP04.pdf
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Garbled circuits might be a good way to do this. There are plenty of libraries that will allow you to do garbled circuits easy enough. The two that come to mind are Fairplay/FairplayMP, or a more updated system done by the University of Virginia.
The advantages of these systems over the paper you referenced is that they don't use Public-Key crypto, so they should be faster. If you go with the UVa system, I would expect your performance to be similar to their results on the hamming distance.
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http://physics.stackexchange.com/questions/36440/what-causes-an-electric-shock-current-or-voltage
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# What causes an electric shock - Current or Voltage?
Though voltage and current are two interdependent physical quantity, I would like to know what gives more "shock" to a person - Voltage or Current? In simple words, will it cause more "electric - shock" when the voltage is high or Current is more?
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1
– Qmechanic♦ Sep 14 '12 at 20:18
1
Lots of very good informed answers in that related question – Jaime Sep 14 '12 at 22:05
## 6 Answers
You'd have to define 'shock', but what kills you is enough current during enough time, not voltage. Of course you need enough voltage to keep the current going over your body's resistance, but it definitely plays a secondary role.
In a former professional life I worked developing Residual Current Circuit Breakers, and 30 mA is the usual rating for devices aiming at protecting lives. In wet environments, such as bathrooms or swimming pools, sometimes 10 mA is recommended.
According to wikipedia's RCCB article, 25-40 ms of 30 mA is enough to send your heart into fibrillation, which probably qualifies as a pretty strong shock. That would require your heart being in the path of the current, though. This link has more information on what to expect depending, again, on the current, not the voltage.
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I suggest this is self evident. You can have a high voltage between two metal plates and put your hand through safely as long as you do not touch them. Otherwise nobody would be able to repair fuses so close to live wires. If there is no current there is no damage, it is the current we feel as "shock". – anna v Sep 15 '12 at 4:22
2
@annav, this problem is in the context of a circuit which implies a closed path for a current to exist. While I appreciate the point you've made, I believe it is actually out of context as there is no circuit in your example. – Alfred Centauri Sep 15 '12 at 21:43
You see, current in one direction indicates the flow of free electrons in the other direction. Hence, current flows. But voltage does not flow. It's actually the work done per unit charge. (Joule/Coulomb). In other words, it's an energy or simply the potential difference between two points. Sometimes, a bird does not get a shock while resting on a carrier wire (Not all time the bird sits on the same wire). Whereas By touching the live wire and ground, we create a potential difference (a path which is sufficient enough for current to flow) with live wire relative to earth. Hence, we get a shock...
Electric shock is the sufficient amount of current flowing through human body which the person could feel. But, this doesn't mean that shock is not caused by Voltage. Of course, it depends upon voltage which is given by Ohm's law... $$I=\frac{V}{R}$$
The shock your body feels, depends upon the applied voltage & resistance of your body. If your body has a resistance of about 10,000 $\Omega$, and the voltage is 230 V, then $$I=\frac{230}{10000}=23 mA$$
You would get a shock, but seriously... (You can't let go off this current)
Hence, you cannot differentiate a shock based on voltage or current. It depends on both voltage and resistance of your body (which implies current).
Here's a page which shows current paths and another one which provides comparison info..!
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If we model the path for the current through the human body as a resistor, then by Ohm's law, the current and voltage are proportional.
That is, a greater current through the body will be associated with a greater voltage across the body.
Having said that, let's consider the source of the shock. It is the case that sources may produce a large voltage but are not capable or sourcing significant current. These sources are said to have high internal resistance.
The point is that a high-voltage source with high internal resistance may not give you much of a shock at all while a lower voltage source with low internal resistance may kill you.
So, there really isn't a simple answer to your question other than "it depends".
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The shock is the product of the current and the voltage, so in principle you need both. A friction spark from your shoe is high voltage by low ampage, but a shock from a power main is high ampage and comparatively low voltage.
The type of current, whether it is alternating or steady, is also important, as an alternating current will interfere with biological processes more than a steady current. But the energy deposited is the product of the two, so it is not meaningful to separate them out.
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Thanks for your quick response Ron but I would like to understand two more points on this : 1. When a bird sits on high voltage wire, nothing happens - of course the conductivity is not there so in spite of high voltage there is no "electric-shock".. Doesn't that suggest "Voltage" has no role in the intensity of the "shock" ? – Tabish Sep 14 '12 at 20:17
2. When you say "shock is the product of the current and the voltage", do you say this is how we can quantify the intensity of "shock" ? Is this the actual formula ? – Tabish Sep 14 '12 at 20:24
@Tabish: For the first thing, by voltage, one always means "voltage difference", the bird has no voltage difference between it's feet, so there is no effect. The product is the energy delivered by the shock, so it's how much you heat up, how much burning you get at the skin and flesh, but it isn't the severity of stopping your heart, which is a different thing, related to the electrical flows that regulate the heartbeat. Also, in terms of feeling the shock, setting up nerve impulses is not by power, but by the change in intermembrane voltages, but power is good rough guide to the effect. – Ron Maimon Sep 14 '12 at 20:33
Thanks Qmechanic, It helped. Thanks Ron for taking time to address this. – Tabish Sep 14 '12 at 20:36
"A friction spark from your shoe is high voltage by low ampage" No, it's several amps. It's harmless because of the short duration. – endolith May 11 at 4:27
show 3 more comments
Higher voltage causes more shock at the same wattage.
Low-voltage high-current exposure will be felt as warming up not as a shock.
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Could you please site any source to support it? – Tabish Sep 14 '12 at 20:43
low-voltage high-current will actually kill you, so you may not actually feel the shock... – Jaime Sep 14 '12 at 22:02
@Jaime it will fry you. – Anixx Sep 14 '12 at 22:13
How can a low voltage create a high current through a human being with high resistance? – endolith May 11 at 4:28
the current generates the heat.....but the voltage (potential difference) generates shock...so i think shock is generat due to high voltage!!
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## protected by Qmechanic♦Feb 15 at 18:26
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://mathhelpforum.com/algebra/102311-confused-about-how-solve-simple-problem.html
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Thread:
1. Confused about how to solve simple problem...?
The distance between two cities is 10km more than one third of the distance between them?
I understand the answer is 15 km - but how do i put the answer in an algebraic equation?
2. Originally Posted by EyesThatSparkle02
The distance between two cities is 10km more than one third of the distance between them?
I understand the answer is 15 km - but how do i put the answer in an algebraic equation?
If $d$ is the distance between the two cities then
$d = 10 + \frac{1}{3}d$
$\frac{2}{3}d = 10$
$d = \frac{3}{2}\cdot 10$
$d = 15$.
3. Let the distance between the two cities be X
X is 10km more than a third the distance between them
Hint: When they say 1/3 distance between them, they're referring to X
Algebraically
$X=10+\frac{1}{3}X$
$\frac{2}{3}X=10$
$X=15$
Edit: Beaten by Prove It, I'm slow tonight
4. Thanks guys,
New Question: (Similar)
If a 5 cm piece of wire is cut into two parts such that a square is formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part?
5. Originally Posted by EyesThatSparkle02
Thanks guys,
New Question: (Similar)
If a 5 cm piece of wire is cut into two parts such that a square is formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part?
Let $P_1 = x$ be the perimeter of one of the squares (length of one of the pieces of wire).
Its side length is therefore $L_1 = \frac{1}{4}x$.
Thus its area is $A_1 = L_1^2$
$= \left(\frac{1}{4}x\right)^2$
$= \frac{1}{16}x^2$.
Let $P_2 = 5 - x$ be the perimeter of the other piece.
Clearly $P_1 + P_2 = 5$.
Its side length is therefore $L_2 = \frac{1}{4}(5 - x)$.
Thus its area is $A_2 = \left[\frac{1}{4}(5 - x)\right]^2$
$= \frac{1}{16}(5 - x)^2$
$= \frac{1}{16}(25 - 10x - x^2)$
$= -\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16}$.
We also know that $A_2 = 4A_1$
Therefore
$-\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16} = \frac{1}{4}x^2$
$0 = \frac{5}{16}x^2 + \frac{5}{8}x - \frac{25}{16}$
$0 = x^2 + 2x - 5$
$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2(1)}$
$= \frac{-2 \pm \sqrt{24}}{2}$
$= \frac{-2 \pm 2\sqrt{6}}{2}$
$= -1 \pm \sqrt{6}$.
Since you can not have a negative length
$x = -1 + \sqrt{6}$.
Remembering that the other length $= 5 - x$
$= 5 - (-1 + \sqrt{6})$
$= 5 + 1 - \sqrt{6}$
$= 6 - \sqrt{6}$.
Put this into the calculator and you will know how many cm this length is.
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http://rjlipton.wordpress.com/2010/11/19/is-complexity-theory-on-the-brink/
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## a personal view of the theory of computation
by
A discussion of some of the great recent results of theory
Victor Klee was one of the greatest geometry experts ever, and is known for many wonderful results and problems. For example, he was the first to raise the question now called the “art gallery problem,” and also the question now called Klee’s measure problem. The latter is especially relevant, since it is a pure computational geometry problem, that is still not completely settled.
Today I want to share a feeling that I have, a feeling that is very positive, yet is just a feeling. I think that complexity theory is on the verge of major breakthroughs of all kinds. I feel this is already starting to happen, and I wish to share the reasons I feel this way.
Our field, like most areas of science, have times of huge advances and times when the advances are small. I think that we are in a time of potentially major advances across the entire spectrum of theory. It is an exciting time, and much of that excitement is in seeing problems solved or at least partially solved that were open for years and sometimes decades.
I think Klee would have liked one of the new results very much. I met him only once or twice, and never had a chance to work together. But I believe that he would be excited to see his famous lower bound on Simplex finally improved.
Three Big Results
I feel that there are many results already that are great advances, but I also feel that they point to even greater results in the near future. The next few weeks, months, and years could be a most exciting time for computational complexity.
${\bullet }$ Unique Games:
The result of Sanjeev Arora, Boaz Barak, and David Steurer on the Unique Games Conjecture is a clear example. Their beautiful result has proved that Unique Games can be approximately solved in time ${2^{n^{\epsilon}}}$ for any ${\epsilon>0}$. Their result does leave an interesting open problem: Is the approximation problem for unique games in polynomial time?
${\bullet }$ Circuit Lower Bounds:
The new paper of Ryan Williams is another clear example. His beautiful result is that ${\mathsf{NTIME}[2^n]}$ does not have non-uniform ${\mathsf{ACC}^0}$ circuits of polynomial size. His result leaves open many problems including: Can the circuit class be improved to general polynomial size circuits?
I will not say much more about either of these, since they have both been discussed at length here and elsewhere. For example, I just discussed Ryan’s proof in some detail here.
${\bullet }$ Simplex: The result of Oliver Friedmann, Thomas Hansen, and Uri Zwick is another wonderful example. They have proved strong, almost exponential, lower bounds on randomized pivoting rules for the Simplex method. See Gil Kalai, who is an expert in this area, for comments on their result. Gil also is the discoverer of one of the best methods known for running Simplex: his method requires at most
$\displaystyle 2^{ {\tilde O} ({\sqrt m})},$
for ${m}$ constraints.
Let me give a short history of the search for “good” pivoting rules for Simplex. The Simplex method is really a family of algorithms: at each step the algorithm essentially moves around the vertices of the polytope defined by the linear constraints. The key is the decision of where to move next: the pivot rule. In practice there are many simple rules that work extremely well; however, the complexity question is to find a rule that always works efficiently.
In 1977 Victor Klee and George Minty discovered a “difficult” linear program. Their program was based on a polytopes that are now called Klee-Minty cubes:
$\displaystyle \begin{array}{rcl} \min x_n : \\ 0 & \le & x_1 \le 1, \\ \epsilon x_{i-1} & \le & x_i \le 1-\epsilon x_{i-1} \end{array}$
for ${2 \le i \le n}$ and ${0 < \epsilon < 1/2}$. Here is a picture for ${\epsilon = 1/3}$, it is from the paper by Bernd Gärtner, Martin Henk, and Günter Ziegler.
Klee and Minty proved that the naive pivot rule had a worst case exponential running time on these programs. Quickly other rules were suggested and for a long time various random rules were consider potential winners: that is many believed these rules took expected polynomial time.
I am definitely not an expert in this area, but I am amazed that Friedmann, Hansen, and Zwick were able to prove that various random pivoting rules do not run in expected polynomial time. Finding a worst-case example for any algorithm can be hard. Finding one for a randomized algorithm can be quite difficult: the example must cause the algorithm to fail on paths weighted by their probability, something which is often difficult to show.
Finally, they exploit a connection between linear programming and certain games that seems quite important. We have long known that games and linear programming are related, but that the relationship could be used to prove a lower bound seems quite surprising—at least to me. Their result does leave an interesting open problem: Is there a pivot rule that even is a reasonable conjectured efficient one?
One More
${\bullet }$ Learning Theory: The result of Amir Shpilka and Avishay Tal is my last example of a great new result. They prove new bounds on the Fourier degree of symmetric boolean functions.
The truth is that I am not quite unbiased. I was part of a team that had the previous best result. Mihail Kolountzakis, Evangelos Markakis, Aranyak Mehta, Nisheeth Vishnoi, and I studied: What is the smallest ${t}$ such that every symmetric boolean function on ${k}$ variables—that is not a constant or a parity function—has a non-zero Fourier coefficient of order at least ${1}$ and at most ${t}$? Let ${\tau(k)}$ be the smallest such ${t}$. Our main result was that ${\tau(k) \le 4k/\log k}$ for symmetric boolean functions on ${k}$ variables. The question arises in learning theory and in particular in the special case of the Junta problem for symmetric boolean functions.
Our proof involves careful analysis of the structure of a symmetric boolean function. Even though we could only prove ${o(k)}$, we did conjecture that the correct upper bound was much better—perhaps even ${O(1)}$. Shpilka and Tal prove a theorem that is much stronger than our result. Much. Further their proof is quite pretty and uses some quite interesting connections between this pure complexity question and the structure of prime numbers. Very neat.
They use a deep result from the distribution of primes. Define ${\Gamma(m)}$ to be,
$\displaystyle \max \{ b-a \mid 1 \le a < b\le m \text{ and there is no prime in the interval} (a,b] \}.$
It is known that ${\Gamma(m) \le m^{.525}}$; that is it is known to those who study the deep structure of the primes. It is believed that there are better bounds on ${\Gamma(m)}$, and it has been conjectured that ${\Gamma(m)}$ is ${O(\sqrt{m})}$ and perhaps even ${O(\log^2 m)}$.
Shpilka and Tal prove the following theorem:
Theorem: The function ${\tau(k) \le O(k^{.525})}$ for symmetric boolean functions with ${k}$ inputs.
This immediately gives new bounds on learning symmetric boolean functions—see their paper for details. Their theorem does leave an interesting open problem: What is the best bound on ${\tau(k)}$? Can it really be ${O(1)}$?
Open Problems
Do you agree that there are many new exciting results being discovered in complexity theory? I think so, and I am optimistic that the trend will continue. That more and more problems will be solved or at least chipped away some.
$\displaystyle \S$
The answer to the two secrets in the last post were the digits of ${\pi}$, and each subsection’s label being a song title. Okay, I like to add some fun to the discussions.
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http://math.stackexchange.com/questions/84058/what-are-the-additional-symbols-used-in-numeral-systems-with-more-than-10-base-d
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# What are the additional symbols used in numeral systems with more than 10 base digits?
What would the last digit be in a base 11 system? Base 12?
I'm mostly wondering because I was thinking of a base 36 system with digits 0-9 immediately followed by the letters A-Z but I wasn't sure if that would be considered a proper numeral system, or if there was something else in place.
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8
Traditionally, A, B, C, D, E, F, etc are used; this is standard for base-16, for instance, which is very common in computer science. But in principle, you can use any symbol you wish; we could agree that `#` represents 10, `&` represents, etc. – Arturo Magidin Nov 20 '11 at 23:07
## 3 Answers
The extra digits can be anything you want (the symbols 0-9 themselves are arbitrary, and could be switched around or replaced by other things, though that'd be adding unnecessary confusion). As Arturo mentions above, hexadecimal notation is common in computing, and there it is conventional to use $A=10$ through $F=15$.
If you want to stick to using the 10 Hindu-Arabic numerals and 26 English letters, and want to go past base 36, the Wikipedia article on positional notation describes how one might solve this by using groups of these symbols in each position.
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In addition to the standard A-F, etc. for hexadecimal and other systems, this old Schoolhouse Rock video (which implicitly talks about duodecimal) uses symbols that look an awful lot like the Greek chi ($\chi$) and epsilon ($\epsilon$) for ten and eleven, respectively.
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3
– J. M. Nov 21 '11 at 0:55
+1 for "School House Rock" episode – Hartley Brody Nov 21 '11 at 2:51
2
– J. M. Nov 21 '11 at 14:28
IIRC, on the SWAC at UCLA in the 50's and 60's, the 6 hex digits beyond 9 were U V W X Y Z.
Also, for some reason, this pops into my mind for the hex digits: G B N J F L. These stood for "Great Big Numbers Just For Laughs".
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http://mathoverflow.net/revisions/32911/list
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## Return to Question
2 improved formatting
An important invariant of a knot in $S^3$ is its Alexander polynomial, related also to Reidemeister torsion. Is there something like that for knotted surfaces in $S^4$? If not, what are the difficulties?
1
# Alexander polynomial or Reidemeister torsion for knotted surfaces?
An important invariant of a knot in $S^3$ is its Alexander polynomial, related also to Reidemeister torsion. Is there something like that for knotted surfaces in $S^4$? If not, what are the difficulties?
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http://math.stackexchange.com/questions/218741/how-to-prove-that-estimator-is-plug-in-estimator?answertab=oldest
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# How to prove that estimator is plug-in estimator?
I need to prove that:
a) sample median is a plug-in estimator
b) sample quantile is a plug-in estimator,
but I have no idea where to start;
In a) :
$$M = \begin{cases} X_{\left(\frac{n+1}{2}:n\right)} & \text{if } n \text{ is odd} \\[6pt] \frac{X_{\left(\frac{n}{2}:n\right)} + X_{\left(\frac{n}{2}+1:n\right)} }{2} & \text{if } n \text{ is even} \end{cases}$$
but I don't know how to connect this with empirical CDF
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http://mathoverflow.net/questions/57749/laplace-derham-operator-for-1-forms-on-the-sphere/57753
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## Laplace-deRham operator for 1-forms on the sphere
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
What do the eigenforms of the 1-form Laplace-de Rham operator look like on the 2-sphere, seen as vector fields via the inner product?
For the standard Laplace-de Rham operator on 0-forms (functions) the simple answer is the spherical harmonics. What about for the 1-form operator?
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This is probably overkill, but you can take a look at Folland, "Harmonic analysis of the de Rham complex on the sphere. " Crelles 1989 – Donu Arapura Mar 8 2011 at 0:26
## 1 Answer
If $f:S^2\to R$ satisfies $\Delta f=\lambda f$, then $$\Delta(df)=(dd^*+d^*d)(df)=\lambda df$$ and similarly
$$\Delta(\ast df)=(dd^*+d^*d)(*df)=\lambda \ast df$$
Since $H^1(S^2)=0$, these are all eigenvectors on 1-forms. Here $*$ is the Hodge * operator and $d^*=-\ast d \ast$.
The vector field is the unique $X$ so that $df(v)=(X,v)$.
On 2-forms all eigenvectors are of the form $\ast f$.
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http://www.math.uah.edu/stat/applets/SecretaryGame.html
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### The Secretary Game
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|--------|--------|--------|--------|--------|--------|--------|--------|--------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|---------|
| Ball 1 | Ball 2 | Ball 3 | Ball 4 | Ball 5 | Ball 6 | Ball 7 | Ball 8 | Ball 9 | Ball 10 | Ball 11 | Ball 12 | Ball 13 | Ball 14 | Ball 15 | Ball 16 | Ball 17 | Ball 18 | Ball 19 | Ball 20 | Ball 21 | Ball 22 | Ball 23 | Ball 24 | Ball 25 |
| Ball 1 | Ball 2 | Ball 3 | Ball 4 | Ball 5 | Ball 6 | Ball 7 | Ball 8 | Ball 9 | Ball 10 | Ball 11 | Ball 12 | Ball 13 | Ball 14 | Ball 15 | Ball 16 | Ball 17 | Ball 18 | Ball 19 | Ball 20 | Ball 21 | Ball 22 | Ball 23 | Ball 24 | Ball 25 |
Distribution graph
#### Description
In the secretary problem, there are $$n$$ candidates, totally ranked from best to worst, with no ties. The candidates arrive sequentially, in random order. We can not observe the absolute ranks of the candidates as they arrive, only the relative ranks. Our goal is to choose the best candidate; any other outcome is failure.
In the secretary game, the candidates are represented as balls. The applet has buttons for starting a new game, rejecting a candidate, and accepting a candidate. The labels on the balls show the relative ranks of the candidates (1 is best). Candidates with relative ranks greater than 1 (non-optimal) are colored red; the candidate with relative rank 1 (the best so far) is colored green. Once a candidate is accepted, the game is over and the second row of balls shows the absolute ranks, again with non-optimal candidates red and the best candidate green. The data table and data graph show the relative frequency of successes and failures. The number (arrival order) of the selected candidate $$X$$, the number of the best candidate $$Y$$, and the outcome (success or failure) $$W$$ are recorded in the record table for each game. The number of candidates can be varied with the input control.
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http://math.stackexchange.com/questions/43525/solving-sec-3-beta-10-csc-beta-8
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# solving $\sec (3 \beta + 10) = \csc (\beta + 8)$
$\sec (3 \beta + 10) = \csc (\beta + 8)$ (in degrees)
I am supposed to find one solution, and the angles are acute.
I do not know the answer or how to get the answer.
It is confusing for me because I don't know what to do with the $3 \beta$.
I am guessing that maybe I get rid of the $\sec$ and $\csc$ somehow but I am not sure how I would go about doing that.
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@Adam: I wonder about how often you are asking these homework questions. So I will hold back answering for a while to allow you proper time to work on it. As far as I can tell, you have asked over 6 today, and over 10 in the last few days. – mixedmath♦ Jun 6 '11 at 2:01
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– Arturo Magidin Jun 6 '11 at 2:02
Well I don't know what to do, all I learned from the last one is the change the sign. – Adam Jun 6 '11 at 2:04
@Adam: Then you didn't learn the right thing. It's exactly the same thing; secant and cosecant are "function and cofunction", to use your parlance. There is one case in which each function and its cofunction are always equal. That gives the answer. – Arturo Magidin Jun 6 '11 at 2:05
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You seem to have a lot of attitude problems and take offense when I don't understand what you say. I don't know why this is, but I am being honest. Most of what you say does not at all further my understand of the material at all. – Adam Jun 6 '11 at 2:23
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## 2 Answers
These two functions are cofunctions. Using this fact gives you one easy solution.
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I don't quite see what to do, is this correct? So 4b+18 = 90 so b=18? – Adam Jun 6 '11 at 2:32
Yes. That yields $\csc(26) = \sec(64)$. (degrees) – ncmathsadist Jun 6 '11 at 2:35
Awesome so if it adds to 90 I know it is correct. – Adam Jun 6 '11 at 2:37
look at this site
http://en.wikipedia.org/wiki/Trigonometric_functions here is explanation what sec and csc are
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http://mathoverflow.net/questions/99728/supremum-of-continuous-functions-on-mathbbr
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## Supremum of continuous functions on $\mathbb{R}$ [closed]
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If $f(x)$ is a continuous function on all of $\mathbb R$, with the property that $\sup_{x\in\mathbb R}|f(x)|\leq 1$. If this is the case, how I can test if the sup is attained or not? (i.e., if there exists at least $x_{o}\in \mathbb R$ such that $|f(x_{o})|\geq |f(x)|, \forall x\in \mathbb R$).
Should we have something like $\lim_{x\to\pm\infty}|f(x)|=0$, or something else?
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$\lim\sup_{x\to\pm\infty} |f(x)|<1$ is sufficient but not necessary. – Will Sawin Jun 15 at 17:39
Will is incorrect. It is neither. But MathOverflow is probably not the place for this question. – Gerald Edgar Jun 15 at 18:06
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(he probably meant $\limsup |fx| < \sup |f(x)|$) @Catherine: a good place for your question is e.g. en.wikipedia.org/wiki/Wikipedia:Reference_desk/… – Pietro Majer Jun 15 at 18:22
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http://mathoverflow.net/questions/29005?sort=oldest
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## Characterisation of parabolic subalgebras: reference sought
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Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $\mathfrak{p}$ a subalgebra. As we all know, $\mathfrak{p}$ is parabolic if it contains a Borel (thus maximal solvable) subalgebra. In this case, with $\mathfrak{p}^\perp$ the orthocomplement of $\mathfrak{p}$ with respect to the Killing form of $\mathfrak{g}$, $\mathfrak{p}^\perp$ is the nilradical of $\mathfrak{p}$.
There is a handy converse to this statement which goes as follows: a subalgebra $\mathfrak{p}$ of $\mathfrak{g}$ is parabolic if $\mathfrak{p}^\perp$ is a nilpotent (thus central descending series terminates) subalgebra of $\mathfrak{g}$. Note that there is no a priori demand that $\mathfrak{p}^\perp$ is even contained in $\mathfrak{p}$ (though that is, of course. part of the conclusion).
My question: does anyone know a reference for this (not difficult to prove) fact? (I have, in the past, incorrectly attributed it to Grothendieck.)
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I think you need to be more careful about what you mean by nilpotent subalgebra here, since you are implicitly requiring that it consist of "nilpotent" elements in the sense of the abstract Jordan decomposition in `$\mathfrak{g}$`. A Cartan subalgebra is also nilpotent, for example, but consists of "semisimple" elements. An arbitrary nilpotent subalgebra could involve both types. Unless you assume $\mathfrak{n}$` consists of nilpotent elements, the discussion gets more subtle (and the orthocomplement need not even be a subalgebra of `$\mathfrak{g}$`) – Jim Humphreys Jun 22 2010 at 12:35
I think I am being careful, Jim: I only require that $\mathfrak{p}^\perp$ is nilpotent in the usual sense that the central descending series terminates. However, I am definitely requiring that the orthocomplement be a subalgebra, else, as you say, a CSA would provide a counterexample. I will edit my question to make my meaning clearer. – Fran Burstall Jun 22 2010 at 17:58
Maybe I understand better: the essential statement not already implied by the Bourbaki theorem is that the orthocomplement of a nilpotent subalgebra containing nonzero semisimple elements is never a Lie subalgebra (since if it were, it would have to be parabolic and thus its orthocomplement in turn would be a nil algebra)? This is somewhat roundabout to state though probably true. I haven't seen it in print, however. – Jim Humphreys Jun 22 2010 at 22:16
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A more "forward" way is to say that if $\mathfrak{n}$ is a nilpotent Lie subalgebra of $\mathfrak{g}$ such that its orthocomplement $\mathfrak{p}:=\mathfrak{n}^\perp$ is a Lie subalgebra then $\mathfrak{p}$ is a parabolic subalgebra (and hence $\mathfrak{n}$ is its nilradical). I can't help thinking that Ozeki and Wakimoto paper, which proves that any polarizable subalgebra is parabolic, is somehow relevant; at least, it gives the right conclusion. – Victor Protsak Jun 23 2010 at 0:09
Victor: thank you for drawing the Ozeki-Wakimoto paper to my attention. It is indeed interesting but does not, as far as I can see, prove my statement. In fact, their result seems much deeper and uses 'non-algebraic' considerations: they look at the analytic subgroup corresponding to a w-polarisable subalgebra and see that the resulting coset space is compact whence the subgroup and so, eventually, the subalgebra is parabolic. – Fran Burstall Jun 23 2010 at 18:29
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## 2 Answers
I think that this follows from Bourbaki's Éléments de Mathématique. Groupes et algèbres de Lie, Chapitre VIII, §10, Theorem 1 (see below) applied to the adjoint representation. Alas, I cannot provide the google books link because the book that Google Books claims to be this one, is actually Algèbre commutative, Chapitres 5 à 7! (And the "Feedback" link does not allow me to point this out, since in their arrogance, Google does not even allow for the possibility of such an error!)
Théorème 1. --- Soient $V$ un espace vectoriel de dimension finie, $\mathfrak{g}$ une sous-algèbre de Lie réductive dans $\mathfrak{gl}(V)$, $\mathfrak{q}$ une sous-algèbre de Lie de $\mathfrak{g}$ et $\Phi$ la forme bilinéaire $(x,y) \mapsto \mathrm{Tr}(xy)$ sur $\mathfrak{g} \times \mathfrak{g}$. On suppose que l'orthogonal $\mathfrak{n}$ de $\mathfrak{q}$ par rapport à $\Phi$ est une sous-algèbre de Lie de $\mathfrak{g}$ composée d'endomorphismes nilpotents de $V$. Alors, $\mathfrak{q}$ est une sous-algèbre parabolique de $\mathfrak{g}$.
And here's a possible translation:
Theorem 1. --- Let $V$ be a finite-dimensional vector space, $\mathfrak{g}$ a reductive Lie subalgebra of $\mathfrak{gl}(V)$, $\mathfrak{q}$ a Lie subalgebra of $\mathfrak{g}$ and $\Phi$ the bilinear form $(x,y) \mapsto \mathrm{Tr}(xy)$ on $\mathfrak{g} \times \mathfrak{g}$. If the orthogonal complement $\mathfrak{n}$ of $\mathfrak{q}$ relative to $\Phi$ is a Lie subalgebra of $\mathfrak{g}$ consisting of nilpotent endomorphisms of $V$, then $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}$.
Edit
As Fran points out in the comments below, my original translation was incorrect and had $\mathfrak{n}$ nilpotent instead of consisting of nilpotent endomorphisms. Happily, for the case of the adjoint representation, one has Engel's theorem, which says that the the two notions agree.
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Almost, but not quite. What Bourbaki has is Grothendieck's result which, for the adjoint representation, requires that each element of $\xi\in\mathfrak{n}$ have $\mathrm{ad}\xi$ nilpotent on $\mathfrak{g}$. This is a stronger condition than simply requiring $\mathfrak{n}$ to be nilpotent---a distinction which got Lost in Translation! – Fran Burstall Jun 22 2010 at 7:09
But isn't that just Engel's theorem? i.e., a Lie algebra $\mathfrak{g}$ is nilpotent if and only if the endomorphisms $\mathrm{ad}_x$ are nilpotent for every $x \in \mathfrak{g}$. – José Figueroa-O'Farrill Jun 22 2010 at 11:19
I'll edit the translation, though -- since as you point out it is not literal! – José Figueroa-O'Farrill Jun 22 2010 at 11:20
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Engel's theorem isn't directly applicable here, since you have to look at the adjoint representation of the semisimple Lie algebra and not that of its nilpotent subalgebra. The Bourbaki theorem has to be applied carefully to an abstract Lie algebra and its Killing form. – Jim Humphreys Jun 22 2010 at 11:56
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Disclaimer: I wrote that book when much younger to learn the subject better, but still haven't gotten to "expert" level. Writing a book is definitely the best way for the author to learn a subject in mathematics, though not invariably so for the reader. – Jim Humphreys Jun 22 2010 at 22:22
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
As alluded to in the comments above, there is a rather old paper by Jacques Dixmier that contains a result in this direction. The reference is
Dixmier, Jacques. Polarisations dans les algèbres de Lie. II. Bull. Soc. Math. France 104 (1976), no. 2, 145--164.
The result is Lemme 1.1, but the proof is attributed to P. Tauvel. The proof proceeds with a judicious application of the invariance of the Killing form and, for the nilpotency of $\mathfrak{p}^{\perp}$, Bourbaki's Groupes et algèbres de Lie, Chapitre I, §5, Lemme 3.
However, there is a caveat: a slightly different notion of co-isotropy is assumed. Dixmier's notion of co-isotropy is to define the orthogonal complement $\mathfrak{p}^f$ with respect to an anti-symmetric bilinear form $B_f$ derived from the Killing form.
Edit: I claimed earlier that Dixmier's notion of co-isotropy implies that $\mathfrak{p}^{\perp}$ is contained in $\mathfrak{p}$. This is not quite correct: you can prove, as is done there, that $\mathfrak{p}^{\perp} = [x,\mathfrak{p}^f]$ is an ideal of $\mathfrak{p}$ without using the assumption of co-isotropy that Dixmier made. The assumption kicks in only for the proof of nilpotency.
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Thanks for the reference! I shall have a look at this. – Fran Burstall Mar 22 2012 at 8:35
You're welcome! Your proof of the statement is also very interesting. – Rongmin Lu Mar 23 2012 at 4:30
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http://crypto.stackexchange.com/questions/2136/is-it-generally-possible-to-employ-brute-force-methods-when-the-encryption-schem
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# Is it generally possible to employ brute force methods when the encryption scheme is not known? Why or why not?
Lets say you are presented with an encrypted string of bits or text, and no other information. What would be necessary before you could apply brute force methods to decrypting the string? How would a cryptanalyst approach such a case? (lets assume the conditions are such that the encryption can be brute forced in a reasonable time frame).
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## 2 Answers
This is called ciphertext-only cryptanalysis*, and it's pretty difficult unless the cipher is quite weak. Therefore, the first priority for a cryptanalyst in such a situation is usually to try to find more information about the algorithm. Fortunately (for the cryptanalyst), as Kerckhoff's principle suggests, there are often ways to find out how the algorithm works. For example:
• If the string was encrypted (or meant to be decrypted) by a computer program, one could try to obtain a copy of the program and inspect it in a debugger to determine the algorithm.
• Similarly, if the encryption and/or decryption was done by a hardware device, one could try to obtain such a device and reverse engineer it.
• One might also try to obtain a copy of a manual explaining how the cipher is used, or design documents explaining how it works, or even look for hints in public articles written about it. Or one could even try to capture someone who knows how the cipher works and interrogate them.
That said, there are some things that can be done even without knowing the algorithm in advance, particularly if the cipher is a simple one:
• One can look at the character frequencies in the cipher, or more generally at $n$-gram frequencies or at correlations between characters $n$ positions apart. Many simple classical ciphers can reveal characteristic patterns under such an examination, allowing the algorithm and perhaps eventually the key to be guessed.
• More generally, one can look for patterns that may reveal information about the algorithm. For example, modern ciphers usually operate on bytes, or blocks of bytes, while classical ciphers typically use a smaller alphabet of letters and maybe a few symbols. Some modern operating modes may reveal the block size of the cipher; also, comparing the length of the ciphertext with that of the plaintext, if known, may reveal the presence (or absence) of an IV and/or a MAC. And sometimes, e.g. when used in ECB mode, even the output of a modern block cipher may have patterns that can reveal information about the plaintext, and, incidentally, the cipher and mode used.
• Finally, one can simply try known algorithms and variants of them and see if they work. This works particularly well if one has already narrowed the choices down somehow, and/or has additional information available (such as knowing the key but not the algorithm).
For an interesting historical example, see the cryptanalysis of the Enigma (and of the other WWII axis ciphers), which provides examples of several of these methods and their combinations.
* The term "ciphertext-only cryptanalysis" is also used for the broader situation where the algorithm is known but no information about the key or the plaintext is available. Even that can be quite challenging, although obviously easier than if even the algorithm used is uncertain.
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If you can't tell what function was applied to create the cipher text, your search space is as many bits as the message. It's the perfect secrecy achieved with a one time pad. (x + y = z, given z what are x and y?) During an exhaustive search the attacker could find as many messages as they were willing to compute, but they will never know which one was right.
That being said, the added clause "lets assume the conditions are such that the encryption can be brute forced in a reasonable time frame" really means you just confine your search to any ciphers that are insecure enough to be cracked in the chosen time frame.
Ilmari's answer is way cooler though. :-)
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http://mathoverflow.net/questions/62635/are-there-any-mathematical-objects-that-exist-but-have-no-concrete-examples/62684
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## Are there any mathematical objects that exist but have no concrete examples? [closed]
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I am curious as to whether there exists a mathematical object in any field that can be proven to exist but has no concrete examples? I.e., something completely non-constructive. The closest example I know of are ultrafilters, which only have one example that can be written down. MathOverflow user Harrison Brown mentioned to me that there are examples in Ramsey theory of objects that are proven to exist but have no known deterministic construction (but there might be), which is close to what I'm looking for. He also mentioned that the absolute Galois group of the rationals has only two elements that you can write down - the identity element and complex conjugation.
I am worried that this might be a terribly silly question, since typically there is a trivial example of an object, and a definition that specifically did not include the trivial case would be 'cheating' as far as I'm concerned. My motivation for this question is purely out of curiosity. Also, this is my first question on MO, so I probably need help with tags and such (I'm not terribly sure what this would belong to). I think that this should be a community wiki, but I do not have the reputation to make it so as far as I can tell.
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Lots of things proved to exist by Zorn's lemma are non-constructive, like a basis for R as a Q-vector space, a transcendence basis for R as a field extension of Q, a well-ordering of R, a nontrivial non-archimedean absolute value on C, a field isomorphism between the algebraic closure of Q_p and C,... – KConrad Apr 22 2011 at 16:47
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“No concrete examples” does not imply “non-constructive”. For example, in the game of hex the first player has a winning strategy, which can be constructed by marking the (finite) game graph. However, I don't think anyone has described a winning strategy for the game. en.wikipedia.org/wiki/Hex_%28board_game%29 – Harald Hanche-Olsen Apr 22 2011 at 17:46
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@Qiaochu: Pretty sure the intention was that only the principal ultrafilters can be "explicitly" described. – Andres Caicedo Apr 22 2011 at 19:03
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@Jon: You need to specify better what you mean by "concrete", or the question becomes too vague to be useful. – Andres Caicedo Apr 22 2011 at 19:04
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This question is, at time of writing, likely to be closed. Since the comments section is already long, please bring discussion to meta.mathoverflow.net/discussion/1019 . And please vote up this comment so that it appears "above the fold". – Theo Johnson-Freyd Apr 22 2011 at 21:14
show 9 more comments
## 8 Answers
If I remember correctly, there is a theorem that asserts that all but possibly zero, one or two prime numbers generate infinitely many of the (cyclic) multiplicative groups $\mathbb{Z}/q\mathbb{Z}^{\times}$ where $q$ varies among the primes. Yet not even one such prime is known, not even $2$ or $3$. Thus, among $2,3$ and $5$, at least one of them has the property, but no one knows which do.
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It is due to Heath-Brown: qjmath.oxfordjournals.org/content/37/1/… – S. Carnahan♦ May 11 2011 at 4:28
But note also that each of the numbers $2$, $3$ and $5$ is concrete individually and can be given as explicitly as anything can be in mathematics. So this is not really a case where we prove something exists but there is no concrete example; rather, it is a case where among several concrete examples, we can prove that one of them has a certain property, but we don't know which one. – Joel David Hamkins May 11 2011 at 12:21
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think the meaning of the term "exist" needs to be clarified. All of the examples you describe except the Ramsey-theoretic one depend on axioms independent of ZF (e.g. the ultrafilter lemma). On the other hand, the probabilistic method can prove, in ZF, that plenty of objects exist (e.g. efficient sphere packings, families of graphs realizing bounds on the Ramsey numbers) for which we do not have efficient deterministic constructions. I assume this is what Harrison is referring to (the use of the probabilistic method in Ramsey theory).
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I was going to mention error-correcting codes saturating the Shannon limit, but this answer implicitly covers that. – Steve Huntsman Apr 22 2011 at 21:10
"for which we do not have deterministic constructions" is an exaggeration. Whenever finite objects are concerned, one can always produce one by enumeration. You wouldn't say that we do not have a deterministic way of finding prime decompositions, would you? – Ori Gurel-Gurevich May 10 2011 at 20:49
@Ori: sorry, I guess I meant "efficient deterministic constructions." – Qiaochu Yuan May 10 2011 at 23:48
You should look at Handbook of Analysis and its Foundations by Eric Schecter. Here is an excerpt from the preface:
Students and researchers need examples; it is a basic precept of pedagogy that every abstract idea should be accompanied by one or more concrete examples. Therefore, when I began writing this book (originally a conventional analysis book), I resolved to give examples of everything. However, as I searched through the literature, I was unable to find explicit examples of several important pathological objects, which I now call intangibles:
finitely additive probabilities that are not countably additive, elements of $(l_\infty)^*- l_1$(a customary corollary of the Hahn- Banach Theorem), universal nets that are not eventually constant, free ultrafilters (used very freely in nonstandard analysis!), well orderings for R, inequivalent complete norms on a vector space, etc. In analysis books it has been customary to prove the existence of these and other pathological objects without constructing any explicit examples, without explaining the omission of examples, and without even mentioning that anything has been omitted. Typically, the student does not consciously notice the omission, but is left with a vague uneasiness about these unillustrated objects that are so difficult to visualize.
I could not understand the dearth of examples until I accidentally ventured beyond the traditional confines of analysis. I was surprised to learn that the examples of these mysterious objects are omitted from the literature because they must be omitted: Although the objects exist, it can also be proved that explicit constructions do not exist. That may sound paradoxical, but it merely reflects a peculiarity in our language: The customary requirements for an "explicit construction" are more stringent than the customary requirements for an "existence proof." In an existence proof we are permitted to postulate arbitrary choices, but in an explicit construction we are expected to make choices in an algorithmic fashion. (To make this observation more precise requires some definitions, which are given in 14.76 and 14.77.)
Though existence without examples has puzzled some analysts, the relevant concepts have been a part of logic for many years. The nonconstructive nature of the Axiom of Choice was controversial when set theory was born about a century ago, but our understanding and acceptance of it has gradually grown. An account of its history is given by Moore [1982]. It is now easy to observe that nonconstructive techniques are used in many of the classical existence proofs for pathological objects of analysis. It can also be shown, though less easily, that many of those existence theorems cannot be proved by other, constructive techniques. Thus, the pathological objects in question are inherently unconstructible.
The paradox of existence without examples has become a part of the logicians' folklore, which is not easily accessible to nonlogicians. Most modern books and papers on logic are written in a specialized, technical language that is unfamiliar and nonintuitive to outsiders: Symbols are used where other mathematicians are accustomed to seeing words, and distinctions are made which other mathematicians are accustomed to blurring -- e.g., the distinction between first-order and higher-order languages. Moreover, those books and papers of logic generally do not focus on the intangibles of analysis.
On the other hand, analysis books and papers invoke nonconstructive principles like magical incantations, without much accompanying explanation and -- in some cases -- without much understanding. One recent analysis book asserts that analysts would gain little from questioning the Axiom of Choice. I disagree. The present work was motivated in part by my feeling that students deserve a more "honest" explanation of some of the non-examples of analysis -- especially of some of the consequences of the Hahn- Banach Theorem. When we cannot construct an explicit example, we should say so. The student who cannot visualize some object should be reassured that no one else can visualize it either. Because examples are so important in the learning process, the lack of examples should be discussed at least briefly when that lack is first encountered; it should not be postponed until some more advanced course or ignored altogether.
-
Very interesting! Definitely looks like something I will jack from the library. – Jon Paprocki Apr 23 2011 at 15:53
In the game which is just like chess, except each player makes two moves in a row, the first player has a strategy that draws at least, but no explicit such strategy is known.
-
Do you have a reference for that? – Mariano Suárez-Alvarez Apr 23 2011 at 0:10
14
If the second player has a winning strategy, the first player starts by moving a knight and then undoing their move. More generally, this applies to any game in which "do nothing" is a legal move. – Eric Wofsey Apr 23 2011 at 0:16
Eric, I don't quite follow. White plays Nf3, Black plays, White plays Ng1, Black plays. Now Black has made <i>two</i> moves. – Todd Trimble May 11 2011 at 0:12
Todd, did you miss the part about each player making two moves in a row? White plays Nf3 and Ng1 before Black gets to do anything. – Gerry Myerson May 11 2011 at 0:37
3
Of course, if the best opening for the first player is to do nothing, effectively reversing roles, then the best follow-up for the second player is also to do nothing, reversing roles back. So two ideal players would just keep infinitely passing the buck back and forth; the game would stall out with no action. Which, yeah, in some sense is still a validation of this as a guaranteed non-loss strategy, but it's alas a possibility which is not very interesting. – Sridhar Ramesh May 11 2011 at 12:15
show 2 more comments
-The Robertson-Thomas-Seymour Graph Minor theorem says there exists of a polynomial time algorithm for determining if a graph has a heritable property P.
-Banach-Tarski decomposition of a ball into two balls of the same volume is another example.
-The proposition which is true but not provable in Godel's incompleteness theorem.
-The linear PDE which admits no solution (like in the last Chapter of John Fritz's book).
• Pretty much any proof that uses the axiom of choice to construct something has this problem. I'll post more if I can come up with other examples. There are tons.
-
1
Taylor, the last comment you make is a bit delicate. In set theory there is what we call "canonical structures", whose existence needs the axiom of choice, but are 'explicit' infinitary objects, not dependent on any well-ordering or choice function used in their construction. – Andres Caicedo Apr 22 2011 at 19:31
1
"-The proposition which is true but not provable in Godel's incompleteness theorem." I'll just mention that for Peanos' axioms such a proposition had been found: for any infinite sequence of binary words (the words of symbols of 0 and 1) there are two of them, s.t. the first contains the second as a subword. – zroslav Apr 22 2011 at 19:37
4
Godel's proof provided a perfectly explicit construction of a statement (in a given formal system) which is true but unprovable in the system if the system is consistent. Godel's second incompleteness theorem shows that one such statement expresses the consistency of the system. – Robert Israel Apr 22 2011 at 19:39
What I think is a nicer example from the graph minor theorem, is that there "exists" a finite set of forbidden minors for toroidal graphs (or any family of graphs which is closed under minors). – Daniel Mehkeri Apr 22 2011 at 21:17
1
It's actually Fritz John. – David Hansen Apr 23 2011 at 1:58
Eigenvalues of the Laplacian $\Delta$ acting on $L^2 (G/ \Gamma)$, where $G = SL_2 (\mathbb{R})$ and $\Gamma = SL_2 (\mathbb{Z}) < G$ (one can consider more general groups $G$ and take any lattice $\Gamma$ in $G$), or the so called Maass forms. It is known, by Selberg's trace formula and other related results, that such eigenvalues do exist, and we even have theorems describing their asymptotic count, but not a single, concrete example of a Maass form is known, even for this specific choice of $G$ and $\Gamma$. Quoting from Goldfeld's "Automorphic forms and L-functions for the group GL(n,R)":
"Up to now no one has found a single example of a Maass form for $SL_2 (\mathbb{Z})$".
-
6
This quote must be interpreted carefully. Numerical values of these functions' Laplace eigenvalues and Hecke eigenvalues have been computed to over one hundred decimal places. Goldfeld is referring to the fact that they will never be constructed explicitly "from simpler functions", just as the solutions to any moderately complicated PDE will never be constructed explicitly in this way. – David Hansen Apr 23 2011 at 2:02
I think the best known example is the subset of the plane, s.t. its intersection with any line has exactly 2 points in it. This can be proven by the axiom of choice but there are no constructions of it.
-
4
In what sense is that a better than, say, a basis of $\mathbb R$ as a rational vector space? It's not like your set obviously exists! :) – Mariano Suárez-Alvarez Apr 22 2011 at 20:04
1
This sense is my own sense of beauty :) My set is the most argued to be existing. The existing of this set is the main reason for elementary geometers to beleive that the axiom of choice is not true :) – zroslav Apr 22 2011 at 22:45
4
I'm pretty confident that if/when those "elementary geometers" become algebraic geometers, though, they'll want their commutative rings to have lots of maximal ideals! :) – Mariano Suárez-Alvarez Apr 23 2011 at 0:13
A lot of existence proofs use arguments such as Cantor's diagonal argument, Baire category etc. Unlike the Zorn's lemma arguments, they can "in principle" yield examples. For instance, we could construct a transcendental number by enumerating the algebraic numbers and picking a number that differs from the nth algebraic number in the nth decimal digit. We can compute this number to as many digits as we want. Of course, this is not a transcendental number that anyone wants to know about.
-
You mean the Baire category theorem which assumes local compactness? Because the one assuming completeness sounds very constructive to me. – darij grinberg Apr 22 2011 at 23:55
Unfortunately for much of mathematical physics, Arzela-Ascoli is unconstructive. Maybe you meant that? – darij grinberg Apr 22 2011 at 23:57
1
Why is Arzela-Ascoli unconstructive? The proof gives a construction of the convergent subsequence of functions, once we know how to get a convergent subsequence for given argument. In any case, the point of my comment was that some existence proofs may appear to be unconstructive (such as proving existence of transcendental numbers based on uncountability), but in principle allow a construction. – Michael Renardy Apr 23 2011 at 0:46
"once we know how to get a convergent subsequence for given argument" This. – darij grinberg Apr 23 2011 at 8:27
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http://physics.stackexchange.com/questions/tagged/nuclear-physics+quantum-mechanics
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# Tagged Questions
0answers
31 views
### where do courses offered for quantum mechanics(for m.tech) in India [closed]
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### Direct nuclear reactions problems [closed]
can anyone explain Multi-step nuclear reactions in terms of direct nuclear reactions .
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### Fermi's Golden Rule
It is well known that to calculate the probability of transition in the scattering processes, as a first approximation, we use the Fermi golden rule. This rule is obtained considering the initial ...
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http://lukepalmer.wordpress.com/tag/relativism/
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# Luke Palmer
Functional programming and mathematical philosophy with musical interludes
# Beliefs and Truth
By on March 3, 2012 | 9 Comments
I have now met the fourth person who has said that they don’t have beliefs.
Perhaps I am still stuck in a naive conception of truth that they have transcended. I still unconsciously assign beliefs to be axioms, as assumed truths upon which to base my inferences, and as such not having beliefs would seem impossible. Perhaps they have already achieved what I merely strive for: just living, just being the little perceptrons they are, already embodying the consequences of truth as a linguistic construction and not a fact of the world. They know that whether an idea is true is irrelevant — that there is nothing more than successful ideas being successful — and as such to “believe in” any truth is only to be enslaved by a clever, self-reinforcing idea: that ideas can be true.
This transcendence must have been achieved after many years of thought and meditation — we are perhaps even born clinging to truth as though it were unitary and absolute. Wars have been fought over is and is not, as if ignoring the evidence shining in their swords, both could not coexist. We have a deep genetic drive, because the uncertainty introduced in realizing the paradox of accessible truths is enough to delay a life-saving decision by a few milliseconds, and thus has been bred out of us. The option that there is a representational barrier between your perceptions and the world is not an option for the animal at the edge of survival. But perhaps there is a latent genetic drive toward the non-believer’s enlightened state after all — once you stop worrying about what is true, you can react faster, having closed the analytical gap between cause and effect. You are a wild animal, your thoughts having proregressed into instincts. Indeed, when time is of the essence, this idea could be more successful than the idea of truth — perhaps their meditation was to put themselves in life-threatening situations in which they needed to be lightningfast to survive.
They see the intimate connection between the words “belief” and “truth”. An idea must be able to be true in order to be believed. But they do not reject these words, for an idea must be able to be false to be rejected. The collusion of “belief” and “truth” makes them very hard to break out of: each reinforces the other. When it comes time to communicate, the non-believers see that language is built around truth, and one cannot communicate without presupposing it. So for them to communicate that they are not where you think they are, they must use a sentence which by its very utterance contradicts itself: “I do not have beliefs.”
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Posted in: Uncategorized | Tagged: beliefs, philosophy, realism, relativism, thought
# Computably Uncountable
By on January 26, 2012 | 5 Comments
We are all familiar with Cantor’s diagonal argument that proves there exist infinite sets which are “larger” than the set of natural numbers. In this post I will show that we can express this argument in the form of a program, thus showing that there are countable sets which are “computably uncountable”.
I begin with the program itself:
```type Cantor = Nat -> Bool
diagonal :: (Nat -> Cantor) -> Cantor
diagonal cs n = not (cs n n)
```
Cantor is “the cantor space”, the type of infinite sequences of booleans. We will call such an infinite sequence “a Cantor“. There are clearly infinitely many Cantors; e.g. take the range of this function which gives False at every position except the one specified:
```unit :: Nat -> Cantor
unit m n = m == n
```
diagonal is (Georg) Cantor’s diagonal argument written as a program — it takes an alleged sequence of all Cantors, and returns a Cantor which does not occur in the sequence, by construction. This function shows by contradiction that we cannot put Cantors in 1 to 1 correspondence with naturals, and thus that there are more Cantors than there are naturals.
So how many Cantors are there? Since Nat -> Bool is a Haskell type — the type of computable functions from Nat to Bool — Cantors must be representable by programs. We can encode programs as numbers by treating their source code as base-128 numbers. Hence, there are no more Cantors than naturals, and so Cantors can be put into 1 to 1 correspondence with naturals.
Wait — what? There are more Cantors than Nats, but they both have the same size? Something is wrong. Indeed, in the process of this argument we have asserted both
1. “We cannot put Cantors in 1 to 1 correspondence with naturals”
2. “Cantors can be put into 1 to 1 correspondence with naturals”
We clearly can’t have both.
### I
The erroneous statement is (2). It is undecidable whether a given program represents a Cantor. If the nth Cantor is ⊥ at n, then diagonal will fail: diagonal cs n = not (cs n n) = not ⊥ = ⊥. Because ⊥ is a fixed point of not, diagonal cannot return an element different from the one it was given. Thus for diagonal to work, we must require that Cantors be fully-defined — no infinite loops!
With this requirement, we can no longer put Cantors in 1 to 1 correspondence with the naturals, because we would have to solve the halting problem. It is not enough that the type of the term is a Cantor, it now must be fully defined for all inputs, and determining that given arbitrary source code is an undecidable problem.
### II
The erroneous statement is (1). Cantors are computable functions, so as we have argued, they have the same cardinality as the naturals. There are no more programs than numbers, so by the definition of equal cardinality we can put them in 1 to 1 correspondence with a function.
The problem with (1) occurs because diagonal takes as its first argument not an arbitrary sequence of Cantors, but a computable sequence of Cantors. If cs is not computable, then neither is diagonal cs (for we no longer have cs‘s source code with which to construct it), and Cantors are defined to be computable sequences. So diagonal fails to contradict our bijection.
### III
The erroneous statement is (2). Section II claims to put Cantors and naturals in 1 to 1 correspondence, but it is lying. Suppose Section II is formulated with respect to some axiom system A. If it were “telling the truth”, we would expect there to be some term f in the language of A such that for every fully defined Cantor program c, there is some natural number n such that we have $A \vdash f(\bar{n}) = \bar{c}$ (i.e. it is a theorem of A that f(1 + 1 + … + 1) = (source code of c)).
Let’s suppose we have written down the axioms of A into a Haskell program, and we have a (partial) function proofSearch :: Nat -> Cantor, which, given a number n, searches for theorems of the form $f(\bar{n}) = \bar{c}$ and compiles and returns the first such c it finds. In the case that there is no such statement, it just runs forever; similarly for the case that c fails to compile. Although cumbersome to write, I’m sure we agree that this is possible to write. If section II is not lying, then we expect that for every natural n, proofSearch n does in fact return a valid Cantor.
Now, let us return to familiar lands with a new program:
```evidence :: Cantor
evidence = diagonal proofSearch
```
Oh my! If section II is the truth, then proofSearch is a total, computable function of type Nat -> Cantor, which we can pass to diagonal to find a Cantor that it missed! So it must have been lying, either (1) about its function f finding every possible Cantor or (2) about it actually possessing such a function (i.e. it “proved” that there is such a function, but it couldn’t actually represent it). In either case, it did not actually create a 1 to 1 correspondence between the naturals and Cantors.
### IV
Left as an exercise for the reader.
Which one is it really?
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Posted in: Uncategorized | Tagged: constructive logic, haskell, math, relativism
# Relativism and Language
By on January 18, 2012 | 3 Comments
It is hard for me to imagine that so many people are so wrong. Sure, core beliefs go unexamined. Yes, we often unconsciously repeat taglines we have heard from those we respect instead of attempting to translate our true views. But I must admit I think of all people as essentially wanting to figure it out. Life, the universe, their meaning, how to make the world a better place. Some, who see the world as a competitive, dog-eat-dog place, want to figure it out because it will help them survive. Others, like me, who see the modern (Western — that is all I have direct experience with) world as an essentially benign place, just want to figure it out because of a innate curiosity (no doubt a result of past generations with the former motivation).
So when someone says something which strikes me as wrong, when I have the kneejerking impulse to correct them, this belief of mine kicks in and stops me. Oh my, it didn’t used to; I would happily correct the abundant wrongness in the world. After all, if people think the right way, they will do better for themselves and others. I can’t remember a time when I didn’t have this belief, however, but it has taken a while to trickle its way into my choice of actions.
All through my youth, I was told that I was smart (a pedagogically questionable practice). I didn’t buy it (I’ve always had a rebellious streak). What makes me so special? I wasn’t just born with smartness, I thought. At first this manifested as an individualistic self-righteousness: I must be smart because of the intelligent ways I chose to spend my youth (what? Trampoline, video games, and Power Rangers?). More recently it has manifested as a skepticism of the views of those who tell me I am smart: you only say that because I am articulating things you agree with, so the compliment is a way of affirming your own worldview. Those both seem naive to me now. I don’t know what I currently think about it, I will probably only be able to articulate that once I move to some other view.
I am still skeptical of any innate superiority (however not enough so to avoid writing this post in a way that comes across as advice). So when I stop myself from correcting a wrongness, what do I do? This is the relativism I’ve been talking about.
Words don’t have meaning; in a conversation, the speaker translates meaning into words, and then the listener translates the words into meaning. We have a soft social agreement about how words are used, and that gives rise to trends in our patterns of thought. But the possibility remains — and I use the word possibility only because of a timidness, I really think of it more as a high probability — that the meanings that I have assigned to the words when I hear them are different from the meanings that were used to form them. Indeed, it is unclear what is even meant by two people having the same thought. My brain is not likely to have the ability to represent the thought that produced the words, especially if I disagree with them.
The exercise, then, is this: try to represent those thoughts anyway. How can I think of these words so that the sentence becomes true? Not just slightly less false, but really true. I might have to temporarily reorient my value system; I might have to imagine I grew up in a religious family; I might have to picture the scary possible worlds that might result if the statement were false (that is, beyond the proportion of the consequences I actually predict, already thinking the statement is false). When I remember to do this, I am brought to a calm, understanding level, with few fiery arguments in sight. My contributions to these conversations are transformed into questions instead of assertions — not Socratic “let me lead you to the right answer” questions, but genuine “I want to understand you” questions.
And that is the essence of relativism to me. What you mean by your words is not what I mean by your words. Sentences are uttered with the concept of their truth in mind, and before blasting forth a correction, I first have to understand how they are true. And more often than not, my planned correction is dismantled and replaced by a connected friendship.
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Posted in: Uncategorized | Tagged: language, relativism, thought
# Perspectives on Truth and Realism
By on January 8, 2012 | 18 Comments
Lately I have been considering myself a relativist. To cast away the kneejerks, I don’t consider all belief systems equally valid (with caveats1). Wikipedia sums it up nicely:
… that truth is always relative to some particular frame of reference, such as a language or a culture.
I have noticed an increase in my opposition to what I am currently calling “scientific realism” — the belief that discoveries made by science are true, and other things are false (basically just an incarnation of absolutism). Yesterday I had an impassioned argument (still in good fun, though) with my roommate about our differences in perception. I noticed my emotions firing up around this subject, a symptom begging me to analyze its cause. Humans get very emotional when their thoughts approach a shattering of a core belief, so I am curious if one is near.
This time, instead of a philosophical persuasive essay, I’m just going to write down some of my observations.
In the conversation with my roommate Monty (who I consider quite intelligent), mostly a battle over semantics, I found the following ensemble of his ideas to leave an impression on me:
1. Newtonian gravity is false, and General Relativity is true.
2. If he lived 200 years ago, Newtonian physics would be true.
3. One thing cannot be more true than another (except in the trivial case of one thing being true and the other false, of course).
4. General Relativity and The Standard Model, which are mathematically incompatible, can both be true at the same time.
5. He hasn’t yet seen any evidence that would suggest there are things that can’t eventually be explained by our current scientific ideas.
Taken together, these ideas are fascinating to me. They indicate a different definition of truth than the one I use, and I’m fascinated because I don’t have a concept that I could substitute for it. On surface interpretation, these statements seem inconsistent to me, so I am really curious about the concept from which they arise. (I am pretty sure (5) is just a fallacy though: what would such evidence look like?)
I have met others who claim that they do not have beliefs. I find this to be common among scientific realists. I wonder what definition of “belief” they use to be able to consider themselves devoid of it; so far when I have pried I am just evaded. There are two reasons I evade inquiries: (1) I am not taking the conversation seriously, which may be because it is threatening my beliefs, or other reasons; and (2) the inquiries are using words in ways that don’t have meaning to me, so I answer in riddles that bring out the dissonance2. I usually assume they are doing it because their beliefs are being threatened3; what makes me curious is the possibility that they are evading because of (2)4. Perhaps I am using “belief” incorrectly when asking that question.
Among Skeptics, there is another possible reason to avoid the word “belief”: because it is very close to “faith”, the buzzword of the enemy. Maybe they use the word “truth” to mean what I call “belief”… but then the idea that someone’s beliefs can be false would be nonsense.
I think most of my anti-realism comes from a desire to (at least give due diligence to) respect the belief systems of others. I think I may start considering “true” to be a value judgement (which, as an experiment, I am trying to avoid). I had a debate with a young earth creationist, a belief system I typically have a hard time respecting. After a long time, I think I heard an essential difference, when he said (paraphrasing): “I believe there is a God because I don’t want to live in a world without a God.” That indicates to me a different relationship to truth — that truth and desirability are related concepts — and opened to me the possibility of respecting his belief system a little more.
Dan Piponi made a brilliant comment on twitter during a conversation about realism: “I don’t think ‘reality’ means much. It’s just a placeholder to make the sentence grammatical.”
Notes
1 What exactly does a belief system being “valid” mean?
2 This will happen, for example, if you ask me whether I believe extra-terrestrial life exists, because I get hung up on the definition of “life”. People seem to acknowledge the subtlety of that word, but then keep using the word anyway as if the inability to define it is no big thing: “you know what I mean.” No, I actually don’t.
3 Probably because it confirms my superiority.
4 Possibly because it threatens my superiority.
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http://math.stackexchange.com/questions/49023/compare-which-two-cube-is-the-same
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# compare which two cube is the same
I am solving following problem:
The problem states that on figure 1 there is shown a cube with three facets on which there is drawn three section(length). This cube was put on other facet and turned such that there is also shown these three length on one cube from list of cubes which are drawn below.our task is find which one is this cube?please help i think that D,E,and C is not, in my point of view it must be A but not sure.
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Try to rotate the cube in your mind (or get a dice and actually do it). You are right that D, E and C are not it. – Leonardo Fontoura Jul 2 '11 at 14:24
## 3 Answers
$A$ is incorrect. Pretend you are standing on the line segment which touches a corner of the cube. Orient yourself so you are facing the this corner. Then the line which touches the side is to your left, but on $A$ it is to the right. So the answer can't be $A.$
The answer is $B.$ See if once you know the answer you have an easier time about the spacial reasoning.
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B but D is not wrong because of the hidden sides
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1
Since there are three hidden sides, any could be the answer. – Carl Brannen Jul 2 '11 at 20:38
Hint: Take a paper and some glue and verify that your thoughts are correct.
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"Protip" - ??? I guess if we're not allowed to upvote our answers, that's the next best thing... – The Chaz 2.0 Jul 2 '11 at 14:36
Maybe I should have called it hint. Actually thats a method to solve also very hard problems of that kind, of course it is cheating but it can be used to control the result. – Listing Jul 2 '11 at 14:38
This is a very good method to get precisely one question right on the SAT. – jspecter Jul 2 '11 at 14:46
@jspecter: I didn't say this method is good to apply it in a math test, its rather didactical. – Listing Jul 2 '11 at 15:01
@Listing: That's not cheating; it's a legit problem-solving strategy! – The Chaz 2.0 Jul 2 '11 at 15:33
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http://theoreticalatlas.wordpress.com/2009/06/08/my-talk-at-categories-quanta-concepts/
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# Theoretical Atlas
He had bought a large map representing the sea, / Without the least vestige of land: / And the crew were much pleased when they found it to be / A map they could all understand.
June 8, 2009
## My Talk at “Categories, Quanta, Concepts”
Posted by Jeffrey Morton under category theory, conferences, groupoids, quantum mechanics, spans, talks
[7] Comments
I spent most of last week attending four of the five days of the workshop “Categories, Quanta, Concepts”, at the Perimeter Institute. In the next few days I plan to write up many of the talks, but it was quite a lot. For the moment, I’d like to do a little writeup on the talk I gave. I wasn’t originally expecting to speak, but the organizers wanted the grad students and postdocs who weren’t talking in the scheduled sessions to give little talks. So I gave a short version of this one which I gave in Ottawa but as a blackboard talk, so I have no slides for it.
Now, the workshop had about ten people from Oxford’s Comlab visiting, including Samson Abramsky and Bob Coecke, Marni Sheppard, Jamie Vicary, and about half a dozen others. Many folks in this group work in the context of dagger compact categories, which is a nice abstract setting that captures a lot of the features of the category $Hilb$ which are relevant to quantum mechanics. Jamie Vicary had, earlier that day, given a talk about n-dimensional TQFT’s and n-categories – specifically, n-Hilbert spaces. I’ll write up their talks in a later, but it was a nice context in which to give the talk.
The point of this talk is to describe, briefly, $Span(Gpd)$ – as a category and as a 2-category; to explain why it’s a good conceptual setting for quantum theory; and to show how it bridges the gap between Hilbert spaces and 2-Hilbert spaces.
History and Symmetry
In the course of an afternoon discussion session, we were talking about the various approaches people are taking in fundamentals of quantum theory, and in trying to find a “quantum theory of gravity” (whatever that ends up meaning). I raised a question about robust ideas: basically, it seems to me that if an idea shows up across many different domains, that’s probably a sign it belongs in a good theory. I was hoping people knew of a number of these notions, because there are really only two I’ve seen in this light, and really there probably should be more.
The two physical notions that motivate everything here are (1) symmetry, and (2) emphasis on histories. Both ideas are applied to states: states have symmetries; histories link starting states to ending states. Combining them suggests histories should have symmetries of their own, which ought to get along with the symmetries of the states they begin and end with.
Both concepts are rather fundamental. Hermann Weyl wrote a whole book, “Symmetry”, about the first, and wrote: As far as I can see, all a-priori statements in physics are based on symmetry. From diffeomorphism invariance in general relativity, to gauge symmetry in quantum field theory, to symmetric tensor products involved in Fock space, through classical examples like Noether’s theorem. Noether’s theorem is also about histories: it applies when a symmetry holds along an entire history of a system: in fact, Langrangian mechanics generally is all about histories, and how they’re selected to be “real” in a classical system (by having a critical value of the action functional). The Lagrangian point of view appears in quantum theory (and this was what Richard Feynman did in his thesis) as the famous “sum over histories”, or path integral. General relativity embraces histories as real – they’re spacetimes, which is what GR is all about. So these concepts seem to hold up rather well across different contexts.
I began by drawing this table:
$Sets$ $Span(Sets) \rightarrow Rel$
$Grpd$ $Span(Grpd)$
The names are all those of categories. Moving left to right moves from a category describing collections of states, to one describing states-and-histories. It so happens that it also takes a cartesian category (or 2-category) to a symmetric monoidal one. Moving from top to bottom goes from a setting with no symmetry to one with symmetry. In both cases, the key concept is naturally expressed with a category, and shows up in morphisms. Now, since groupoids are already categories, both of the bottom entries properly ought to be 2-categories, but when we choose to, we can ignore that fact.
Why Spans?
I’ve written a bunch on spans here before, but to recap, a span in a category $C$ is a diagram like: $X \stackrel{s}{\leftarrow} H \stackrel{t}{\rightarrow} Y$. Say we’re in $Sets$, so all these objects are sets: we interpret $X$ and $Y$ as sets of states. Each one describes some system by collecting all its possible (“pure”) states. (To be better, we could start with a different base category – symplectic manifolds, say – and see if the rest of the analysis goes through). For now, we just realize that $H$ is a set of histories leading the system $X$ to the system $Y$ (notice there’s no assumption the system is the same). The maps $s,t$ are source and target maps: they specify the unique state where a history $h \in H$ starts and where it ends.
If $C$ has pullbacks (or at least any we may need), we can use them to compose spans:
$X \stackrel{s_1}{\leftarrow} H_1 \stackrel{t_1}{\rightarrow} Y \stackrel{s_2}{\leftarrow} H_2 \stackrel{t_2}{\rightarrow} Z \stackrel{\circ}{\Longrightarrow} X \stackrel{S}{\leftarrow} H_1 \times_Y H_2 \stackrel{T}{\rightarrow} Z$
The pullback $H_1 \times_Y H_2$ – a fibred product if we’re in $Sets$ – picks out pairs of histories in $H_1 \times H_2$ which match at $Y$. This should be exactly the possible histories taking $X$ to $Z$.
I’ve included an arrow to the category $Rel$: this is the category whose objects are sets, and whose morphisms are relations. A number of people at CQC mentioned $Rel$ as an example of a monoidal category which supports toy models having some but not all features of quantum mechanics. It happens to be a quotient of $Span(Sets)$. A relation is an equivalence class of spans, where we only notice whether the set of histories connecting $x \in X$ to $y \in Y$ is empty or not. $Span(Sets)$ is more like quantum mechanics, because its composition is just like matrix multiplication: counting the number of histories from $x$ to $y$ turns the span into a $|X| \times |Y|$ matrix – so we can think of $X$ and $Y$ as being like vector spaces.
In fact, there’s a map $L : Span(Sets) \rightarrow Hilb$ taking an object $X$ to $\mathbb{C}^X$ and a span to the matrix I just mentioned, which faithfully represents $Span(Sets)$. A more conceptual way to say this is: a function $f : X \rightarrow \mathbb{C}$ can be transported across the span. It lifts to $H$ as $f \circ s : H \rightarrow \mathbb{C}$. Getting down the other leg, we add all the contributions of each history ending at a given $y$: $t_*(s \circ f) = \sum_{t(h)=y} f \circ s (h)$.
This “sum over histories” is what matrix multiplication actually is.
Why Groupoids?
The point of groupoids is that they represent sets with a notion of (local) symmetry. A groupoid is a category with invertible morphisms. Each such isomorphism tells us that two states are in some sense “the same”. The beginning example is the “action groupoid” that comes from a group $G$ acting on a set $X$, which we call $X /\!\!/ G$ (or the “weak quotient” of $X$ by $G$).
This suggests how groupoids come into the physical picture – the intuition is that $X$ is the set (or, in later variations, space) of states, and $G$ is a group of symmetries. For example, $G$ could be a group of coordinate transformations: states which can be transformed into each other by a rotation, say, are formally but not physically different. The Extended TQFT example comes from the case where $X$ is a set of connections, and $G$ the group of gauge transformations. Of course, not all physically interesting cases come from a single group action: for the harmonic oscillator, the states (“pure states”) are just energy levels – nonnegative integers. On each state $n$, there is an action of the permutation group $S_n$ – a “local” symmetry.
One nice thing about groupoids is that one often really only wants to think about them up to equivalence – as a result, it becomes a matter of convention whether formally different but physically indistinguishable states are really considered different. There’s a side effect, though: $Gpd$ is a 2-category. In particular, this has two consequences for $Span(Gpd)$: it ought to have 2-morphisms, so we stop thinking about spans up to isomorphism. Instead, we allow spans of span maps as 2-morphisms. Also, when composing spans (which are no longer taken up to isomorphism) we have to use a weak pullback, not an ordinary one. I didn’t have time to say much about the 2-morphism level in the CQC talk, but the slides above do.
In any case, moving into $Span(Gpd)$ means that the arrows in the spans are now functors – in particular, a symmetry of a history$h$ now has to map to a symmetry of the start and end states, $s(h)$ and $t(h)$. In particular, the functors give homomorphisms of the symmetry groups of each object.
Physics in Hilb and 2Hilb
So the point of the above is really to motivate the claim that there’s a clear physical meaning to groupoids (states and symmetries), and spans of them (putting histories on an even footing with states). There’s less obvious physical meaning to the usual setting of quantum theory, the category $Hilb$ – but it’s a slightly nicer category than $Span(Gpd)$. For one thing, there is a concept of a “dual” of a span – it’s the same span, with the roles of $s$ and $t$ interchanged. However (as Jamie Vicary pointed out to me), it’s not an “adjoint” in $Span(Gpd)$ in the technical sense. In particular, $Span(Gpd)$ is a symmetric monoidal category, like $Hilb$, but it’s not “dagger compact”, the kind of category all the folks from Oxford like so much.
Now, groupoidification lets us generalize the map $L : Span(Sets) \rightarrow Hilb$ to groupoids making as few changes as possible. We still use Hilbert space $\mathbb{C}^X$, but now $X$ is the set of isomorphism classes of objects in the groupoid. The “sum over histories” – in other words, the linear map associated to a span – is found in almost the same way, but histories now have “weights” found using groupoid cardinality (see any of the papers on groupoidification, or my slides above, for the details). This reproduces a lot of known physics (see my paper on the harmonic oscillator; TQFT’s can also be defined this way).
While this is “as much like” linearization of $Span(Set)$ as possible in some sense, it’s not exactly analogous. It also is rather violent to the structure of the groupoids: at the level of objects it treats $X /\!\!/ G$ as $X/G$. At the morphism level, it ignores everything about the structure of symmetries in the system except how many of them there are. Since a groupoid is a category, the more direct analogy for $\mathbb{C}^X$ – the set of functions (fancier versions use, say, $L^2$ functions only) from $X$ to $\mathbb{C}$ is $Hilb^G$ – the category of functors from a groupoid into $Hilb$. That is, representations of $X$.
One of the attractions here is that, because of a generalization of Tanaka-Krein duality, this category will actually be enough to reconstruct the groupoid if it’s reasonably nice. The representation of $Span(Gpd)$ in $2Hilb$, unlike in $Hilb$ is actually faithful for objects, at least for compact or finite groupoids.
Then you can “pull and push” a representation$F$ across a span to get $t_*(F \circ s)$ – using $t_*$, the adjoint functor to pulling back. This is the 1-morphism level of the 2-functor I call $\Lambda$, generalizing the functor $L$ in the world of sets. The result is still a “direct sum over histories” – but because we’re dealing with pushing representations through homomorphisms, this adjoint is a bit more complicated than in the 0-category world of $\mathbb{C}$. (See my slides or paper for the details). But it remains true that the weights and so forth used in ordinary groupoidification show up here at the level of 2-morphisms. So the representation in $2Hilb$ is not a faithful representation of the (intuitively meaningful) category $Span(Gpd)$ either. But it does capture a fair bit more than Hilbert spaces.
One point of my talk was to try to motivate the use of 2-Hilbert spaces in physics from an a-priori point of view. One thing I think is nice, for this purpose, is to see how our physical intuitions motivate $Span(Gpd)$ – a nice point itself – and then observe that there is this “higher level” span around:
$Hilb \stackrel{|\cdot |}{\leftarrow} Span(Gpd) \stackrel{\Lambda}{\rightarrow} 2Hilb$
Further Thoughts
Where can one take this? There seem to be theories whose states and symmetries naturally want to form n-groupoids: in “higher gauge theory“, a sort of gauge theory for categorical groups, one would have connections as states, gauge transformations as symmetries, and some kind of “symmetry of symmetries”, rather as 2-categories have functors, natural transformations between them, and modifications of these. Perhaps these could be organized into n-dimensional spans-of-spans-of-spans… of n-groupoids. Then representations of an n-groupoid – namely, n-functors into $(n-1)-Hilb$ – could be subjected to the kind of “pull-push” process we’ve just looked at.
Finally, part of the point here was to see how some fundamental physical notions – symmetry and histories – appear across physics, and lead to $Span(Gpd)$. Presumably these two aren’t enough. The next principle that looks appealing – because it appears across domains – is some form of an action principle.
But that would be a different talk altogether.
### 7 Responses to “My Talk at “Categories, Quanta, Concepts””
1. Kea Says:
June 8, 2009 at 8:44 am
It was an excellent talk, and it would have been nicer to hear a longer version.
2. John Baez Says:
June 9, 2009 at 12:50 am
Too bad I couldn’t be there!
Someone should whisper in my ear what they thought of Mike Stay’s talk.
3. Jamie Vicary Says:
June 9, 2009 at 1:57 pm
Great post, Jeff! I’ll reply in more detail soon.
Mike’s talk was excellent — but why ask what everyone else thought, when you can find out for yourself here!
4. bob Says:
June 9, 2009 at 4:50 pm
Sorry your talk is not there Jeff; if we would have known a bit more in advance that you were coming we would have included you in the main schedule, …
Btw Mike didn’t just give this talk but also gave a tutorial Monday afternoon on Monoidal categories.
5. June 11, 2009 at 8:31 pm
[...] online, and lots of good discussion and presentations, which unfortunately can’t. (But see Jeff Morton’s comments.) My talk was on the Rosetta Stone paper I co-authored with Dr. Baez. This entry was written by [...]
6. June 30, 2010 at 8:01 am
[...] a little further into how this fits into a more general picture. To repeat a bit of what’s in this post, 2-linearization describes a (weak) [...]
7. January 21, 2011 at 5:12 pm
[...] of groupoids: pairs of maps . Since I’ve already discussed the backgroup here before (e.g. here and to a lesser extent here), and the papers I just mentioned give plenty more detail (as does [...]
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http://math.stackexchange.com/questions/181984/complete-course-of-self-study/182068
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# Complete course of self-study
I am about 16 years old and I have just started studying some college mathematics. I may never manage to get into a proper or good university (I do not trust fate) but I want to really study mathematics.
I request people to tell me what topics an undergraduate may/must study and the books that you highly recommend (please do not ask me to define an undergraduate).
Background:
1. Single variable calculus from Apostol's book Calculus;
2. I have started IN Herstein's topics in algebra;
3. I have a limited knowledge of linear algebra: I only know what a basis is, a dimension is, a bit of transpose, inverse of a matrix, determinants defined in terms of co-factors, etc., but no more;
4. absolutely elementary point set topology. I think open and closed balls, limit points, compactness, Bolzano-Weirstrass theorem (I may have forgotten this topology bit);
5. binomial coefficients, recursions, bijections;
6. very elementary number theory: divisibility, modular arithmetic, Fermat's little theorem, Euler's phi function, etc.
I asked a similar question (covering less ground than this one) some time back which received no answers and which I deleted. Even if I do not manage to get into a good university, I wish to self-study mathematics. I thanks all those who help me and all those who give me their valuable criticism and advice.
P.S.: Thanks all of you. Time for me to start studying.
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The most important thing is to focus your study around solving problems. Improving your ability to DO mathematics is much, much more important than increasing the amount of mathematical knowledge that you know. There will be plenty of time to learn all that you need to know. So pick topics that you enjoy and seek out resources that include challenging problems. – Michael Joyce Aug 13 '12 at 14:55
1
I haven't got a clue if any of this is good material, but I found a list of free textbooks from various universities and colleges: openculture.com/free_textbooks (Press ctrl+f and type 'mathematics' to jump to the mathematics part). If you like video courses, try this: openculture.com/math_free_courses . The website also covers many other topics if you happen to be interested in them. – Simon Verbeke Aug 13 '12 at 17:58
## 21 Answers
This is a recapitulation and extension of what we talked about in chat.
Whatever you do, I recommend that you try a variety of areas in order to find out what you like best. Don’t feel obliged to stick to the most common ones, either; for instance, if you find that you’ve a taste for set theory, give it a try.
My own interests are outside the undergraduate mainstream, so in mainstream areas others can probably give better recommendations. I do know that you’re working through Herstein for algebra; although it’s a little old-fashioned, it’s still a fine book, and anyone who can do the harder problems in it is doing well.
You mentioned that you’d prefer books and notes that are freely available. The revised version of Judy Roitman’s Introduction to Modern Set Theory is pretty good and is available here as a PDF. You can also get it from Barnes & Noble for \$8.99. Introduction to Set Theory by Hrbacek & Jech is also good, but it’s not freely available (or at least not legitimately so).
I’ve not seen a freely available topology text that I like; in particular, I’m not fond of Morris, Topology Without Tears, though I’ve certainly seen worse. If you’re willing to spend a little and like the idea of a book that proves only the hardest results and leaves the rest to the reader, you could do a lot worse than John Greever’s Theory and Examples of Point-Set Topology. It’s out of print, but Amazon has several very inexpensive used copies. (This book was designed for use in a course taught using the so-called Moore method. It’s excellent for self-study if you have someone available to answer questions if you get stuck, but SE offers exactly that. In the interests of full disclosure I should probably mention that I first learned topology from this book when it was still mimeographed typescript.) If I were to pick a single undergraduate topology book to serve both as a text and a reference, it would probably be Topology, by James Munkres, but I don’t believe that it’s (legitimately) freely available. You might instead consider Stephen Willard, General Topology; it’s at a very slightly higher level than the Munkres, but it’s also well-written, and the Dover edition is very inexpensive.
I can’t speak to its quality, but Robert B. Ash has a first-year graduate algebra text available here; it includes solutions to the exercises, and it introduces some topics not touched by Herstein. He has some other texts available from this page; the algebra ones are more advanced graduate level texts, but the complex analysis text requires only a basic real analysis or advanced calculus course.
This page has links to quite a collection of freely available math books, including several real analysis texts; I’ve not looked at them, so I can’t make any very confident recommendations, but if nothing else there may be some useful ancillary texts there. I will say that this analysis text by Elias Zakon and the companion second volume look pretty decent at first glance. For that matter, the intermediate-level book on number theory by Leo Moser available here looks pretty good, too, apart from having very few exercises. Oh, come to think of it there is one real analysis book that I want to mention: DePree and Swartz, Introduction to Real Analysis, if only for its wonderful introduction to the gauge integral.
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Thank you in particular for pointing me to the set theory text by Judy Roitman. – user37450 Aug 13 '12 at 16:35
I am perhaps going to study(later) Munkres,Roitman and Rudin.It's Herstein for now before I am in a position to assess my self . – user37450 Aug 13 '12 at 16:51
Books may be available freely from public and academic libraries' own collections, and typically free or cheap via inter-library loans. WorldCat.org can help search library catalogues to find copies locally. – mctylr Aug 13 '12 at 19:22
"The principle of Mathematic Analysis" by Rudin is strongly recommended. Although it may be difficult for you, it will be very impressive if you can read through it. Although I don't know your interest, this book is critical for all students studying mathematics.
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1
"Understanding Analysis" by Steven Abott is amazing. I think it is a good bridge if one wants to read Rudin. – Peter Tamaroff Aug 13 '12 at 16:51
Does that book have good examples ? Georges Elencwajg expressed some reservations about Rudin's style?(I have not yet bought anything but Herstein, so asking) – user37450 Aug 13 '12 at 19:01
The generality, breath-stopping elegance and economy of presentation of Rudin's Real and Complex Analysis (a bit more advanced text than The Principle of Mathematical Analysis) made me almost cry when I was using it as a self-found substitute to prepare for Complex Analysis examination as a second year student. If you like general way of presenting mathematics and want to learn mathematical thinking, I think Rudin is very very good. – FooF Aug 14 '12 at 4:56
1
I'd like to join in with Peter Tamaroff in recommending Abbott's book, which I called "the best written introductory analysis book that's appeared in the past couple of decades" back in January 2003. If you do wind up working through Abbott's book, a great follow-up to Abbott's book is Charles Chapman Pugh's Real Mathematical Analysis. – Dave L. Renfro Aug 14 '12 at 16:38
Lang's Undergraduate Analysis is richly illustrated, rigorous, very geometric (vector fields on spheres are discussed) and contains some juicy calculations (look at the treatment of the Fejer or the heat kernel ).
Above all it prepares you to advanced modern analysis: Chapter XIV on the Fourier integral for example starts with a section on Schwartz space (again with some non-trivial calculations thrown in) which will ease the transition to distribution theory and partial differential equations.
Don't forget that analysis is not a sterile exercise in axiomatics, despite what some boring books would make you believe, but one of the most useful and exciting subjects in mathematics (and physics).
NB Lang was not an analyst (he was a student of the great arithmetician Emil Artin) but was capable of using quite tough analysis: just look at his book $SL_2(\mathbb R)$ and you will see Sobolev spaces, Mellin and zeta transforms, resolvants of Laplace operators,... at work.
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Something that surprisingly has not been said, i find it amazing that at 16 years old you're into math that much and from the list stating your current knowledge, i see that you already acquired some overview on different fields of mathematics, which is great.
Now since there have been some books recommended, i won't go into that anymore. But i would like to give you the following tips:
If you look at the website of a university, often you can see a curriculum of the first few years in math. I'm pretty sure that you can find a university that has detailed course descriptions, including what literature they use. If you look further there might even be course notes online. This not only provides you some of the requested literature, but also can give you a good idea of what essential mathematical knowledge is, treated in the first years at uni. For example, my uni www.uva.nl has course descriptions, and you can select the language to be English.
Also, if you look around and aren't afraid of rejections, i guess that if you email professors at some uni (anywhere around the world), there might be one enthousiastic and willing to help in some way. You might not realize it, but what you're doing is special. Maybe 1 out of 10 or 100 will actually respond, but hey, emailing is free. You might even have a better shot if you dont mail professors but graduate or post-graduate students.
Lastly, considering your enthousiasm and the knowledge you already acquired, you should definitely go for a place at a uni. I have no idea where you live, but i believe some European universities offer grants to foreign students as well (i have NO personal experience with this though), so look around, there might be more possibilities then you think..
Good luck!
Oh and by the way, Singh: Fermat's last theorem is a great popular math book!
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I do not know in which field you are interested, though it is too early for you to select any field of interest, and rather learn all the basic techniques. Apart from Herstein, you should look at Artin's Algebra. Not only it is lucid, but also it clears several concepts from a practical (applied?) point of view. In particular, you should see the linear algebra portions. Also the exercises are important and should be attempted to make any real progress.
Along with Rudin, you may try to work out Calculus I and II by Apostol. If these are too easy for you, then check differential geometry by Pressley. Some knowledge in probability theory is always useful. You can try Chung.
All the best.
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[Not a direct answer but may help in conjunction with the other answers that give topics and subjects of study]
If you've not already checked it out then take a look at MITs "OpenCourseWare" site - I used this to brush up on my linear algebra when getting back into study around 3 years ago. As well as a wealth of lectures in many subjects the site also provides many, many references, including some that are out of print. If you pick through the courses carefully you will also find a number of assignments (with answers).
http://ocw.mit.edu/courses/mathematics/
Full disclosure: I am not associated with MIT or the OCW site in any way, I am just a (very) happy user of their site!
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Here's a bunch of lesser known material I found really good. It may be a somewhat biased list though, since I tend to like reading "easy" books first to get the main idea, then solving hard problems and moving on to more difficult books later. So, some of this list is less textbook'y and more motivational.
Linear Algebra:
Numerical Linear Algebra:
General techniques for solving mathematical problems:
Abstract Algebra:
Convex analysis and optimization:
Real Analysis:
Mathematical inequalities:
Overview of mathematics from Courant:
Topology:
Overview of mathematical physics from Penrose:
Differential topology/geometry:
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+1: Strang is a master expositor and his video lectures are indeed incredibly good . – Georges Elencwajg Aug 13 '12 at 20:22
I wouldn't call the Penrose book mathematical physics. – Nick Kidman Aug 14 '12 at 8:16
Well, it's not at the graduate or research level but it takes the mathematics seriously, which is to be expected since Penrose is a mathematical physicist himself. It certainly has a much more mathematical style than most physics textbooks which tend to handwave away difficulties. – Nick Alger Aug 14 '12 at 16:10
Benedict Gross's and Francis Su's video lectures seem amazing. – user37450 Aug 14 '12 at 17:06
Linear algebra is the most indispensable subject in all of mathematics: I can easily imagine someone getting a Fields medal without knowing what the sine function is, but I don't believe one could get that prize without being familiar with linear maps.
Linear algebra is an easy subject and the main difficulty is choosing between the thousands of books on the subject and even between the tens of excellent ones.
1) My initiation was through Lang's Linear Algebra but I cannot guarantee that my genuine enthusiasm for that book is not nostalgia-tinted .
2) Lipschutz-Lipson's book in the Schaum series is very elementary, pleasant and richly illustrated, as befits the subject.
As in all books in the series, the theory is kept to a minimum and the reader is encouraged to recreate the subject by solving judicious exercises (provided with complete solutions, just in case!)
3) Another excellent classic is Hoffman's-Kunze's Linear Algebra , which is very solid albeit a little austere. It is more advanced than the preceding two.
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I would begin with the following:
1) Everyone has to learn multivariable calculus. There are many books to choose from here. The one I liked best was Calculus: A Complete Course by Adams.
2) You should be comfortable with linear algebra. I see that you know something about it, but you should really learn more. I'm not really sure which book you should use here (I used Elementary Linear Algebra by Edwards and Penney which was ok). Any suggestions?
Then you can learn more advanced mathematics like the text by Rudin which user37787 recommended. For topology I would recommend Topology by Munkres.
You could also look into applied mathematics like statistics and numerical mathematics. I don't know much about these fields so I cannot recommend anything in particular here.
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I think the following subjects are absolute minimum. IMO, you should learn those before you learn other subjects(PDE, algebraic topology, differential geometry, algebraic geometry, etc.). Perhaps other people will recommend good books on each subject.
- Linear algebra
- Calculus(single and multivariable)
- General topology
- Abstract algebra(basics)
- Naive set theory(basics)
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This is precisely what I intend to study for now. – user37450 Aug 13 '12 at 16:38
When I was your age doing what you are doing, my Analysis professor said I should read Irving Kaplansky's book Set Theory and Metric Spaces. This was good advice. This book was my introduction to set theory, and ideas like well-orderings whose ideas underlie much of modern mathematics. I found it very readable and enjoyed it; it was an excellent supplement to Rudin, which others in this thread have recommended.
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As many have said, Rudin's 'Principle of Mathematical Analysis' is a classic for Analysis. My personal recommendation for point-set topology would be Sutherland's 'Introduction to Metric and Topological Spaces'. It basically builds on from what we know in Analysis into more general spaces and the proofs inside are quite neat in my opinion.
Apart from studying textbooks alone, have you considered reading books on mathematics itself? Books such as 'What is Mathematics' by Stewart and Courant, 'A very short introduction to Mathematics' by Tim Gowers are must reads, and I would also recommend 'The mathematical experience' by Hersh and Davis for a more philosophical insight.
Happy reading!
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This question is quite interesting and possibly deserves a big list of answers, so I would add my point of view. This is just an opinion.
First of all, the goals should be set much clearer that it is done in the current version of the question. Even when studying at a proper University one needs to be very specific on what he or she wants to achieve. Examples are: "I want to prepare for a course of General Relativity", or "I want to improve my mathematics for a Computer Graphics project", or "I want to contribute to ... and this is going to be my hobby" etc.
Secondly, I would recommend to find a challenging book that is not a leisure reading but rather a masterpiece in the area that you want to make your own. For instance, you could find one of S.Tabachnikov's books here as a motivating conundrum to start with. Just a hint. no more.
Then, trying to work through the book you will encounter the parts of the story that are blurry or confusing. Thus you will get a good reason to learn more, and more and more. This is perhaps the way how to learn to ask right questions. (I believe that mathematics is all about that, and I feel that I still have to improve my question asking skills...)
Of course, you will need to learn the cornerstones. Try to learn as much Linear Algebra as you can (this includes multilinear algebra). This will help you to master multivariable calculus to the level when it is done on smooth manifolds. A good knowledge of combinatorics is often an advantage and helps to master things like representation theory. My personal preference is geometry that is an enormous area but a course on differential geometry of curves and surfaces would dramatically expand the dimensionality of one's perception of mathematical problems.
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The Foundations of Mathematics by Ian Stewart and David Tall (available second hand at Abe books). This is a great book on a variety of topics before "more serious" study, but considering what you say in you have already been reading it may be on the easy side, nevertheless I think it is worthy of a read.
Definitely recommended is Galois Theory (3rd edition) by Ian Stewart, not only is it a beautiful story but beautiful mathematics, you can "Look Inside" a large part of it on the Amazon website before you buy.
Something that is the upper end of undergraduate, but worth mentioning, is the free-online book Algebraic Topology by Allen Hatcher. Many tricky concepts, it's essentially an encyclopedia of the subject, and should be on everyone's bookshelf/stored on their computer.
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You can use this link to get Great Stuff like: free video lectures (from top Indian professors) along with Lecture Notes and Good References:
http://nptel.iitm.ac.in/ (Select "Mathematics" from the list of courses available).
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I am not a mathematician, but when I was learning mathematics at the undergraduate level, a great professor recommended A Radical Approach to Real Analysis. I found it informative, full of interesting problems, and quite enlightening (it couches real analysis in the problems it was developed to solve, and as such added rich context to a subject I found otherwise difficult to access).
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I would suggest some mathematical modeling or other practical application of mathematics. Also Finite automata and graph-theory is interesting as it is further away from "pure math" as I see it, it has given me another perspective of math.
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I highly recommend The Princeton Companion to Mathematics - an encyclopedic overview of pure math and some theoretical physics with chapters on proof, many areas of math and biographies of famous mathematicians.Math blogs and personal websites. It is great for getting motivation and an overview on most subject areas so you can pick which to study further. It lists further reading for most topics. You should be able to read at your local library.
As well as book there are some great math blogs, here are 3 that are a good start.
• Tim Gowers - lots of interesting articles and solved problems http://gowers.wordpress.com/
• Terence Tao - great discussion of topics, open problems and career advice http://terrytao.wordpress.com/
• Tom Körner - learning guides, lecture notes and more http://www.dpmms.cam.ac.uk/~twk/
Any many universities put course materials online so that you can both see what topics math undergraduates study and read the materials and problem sets for free.
• Cambridge - Pure math example sheets, lecture notes http://www.maths.cam.ac.uk/undergrad/course/ (drill down into the staff person pages https://www.dpmms.cam.ac.uk/people/ to see posted lecture notes)
• MIT - math lecture notes and example sheets. Some are full OCW Scholar courses that are designed for independent learners who have few additional resources available to them. These courses include exam solution notes, online study groups, video and simulations. http://ocw.mit.edu/courses/mathematics/
• Open University - OpenLearn free course materials on pure, applied and statistics http://openlearn.open.ac.uk/course/category.php?id=8
• Udacity - free video lectures, online tests and learning community mainly related to applied, applicable and computer science topics. http://www.udacity.com/
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On top of all helpful answers above I can add one very popular book: $\mathit{Concrete \ Mathematics}$ by Graham, Knuth and Patashnik (1995 edition). The main reasons are:
1. This book is aimed at Computer Scientists that consider (or want to consider) themselves mathematicians through better understanding of mathematics behind programming and algorithms
2. It has heaps of awesome problems in areas usually not very well covered by Discrete Math books: Generating functions, series, probability, asymptotics .
3. Complexity of problems varies from easy high-school till PhD/open problems. All have an answer or hints (not just 'odd-numbered').
4. It blends continuous and discrete (hence 'concrete') math.
5. It focuses a lot on recurrences, from very simple to very complicated.
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I applaud you for taking the initiative to study on your own. I noticed the post was more than 6 months ago, so not sure if you are still looking for help. I recently started a website, www.redhoop.org, to help self learners like yourself. Search for math courses and see if you find any of them useful to you. Feel free to tell me what you think. Again, keep up your learning! Knowledge is the most important and only long lasting asset we have.
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At your age, it is more useful for you to socialise with friends, read good literature, watch TV, play sports and generally immerse yourself in hobbies. When you get to my age (I'm 22 so younger than you may think) you will not have that much time and if you now spend all your time worrying about undergraduate mathematics you may regret it. It's great to be young so don't waste it!!
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3
If this answer is meant as a joke, it is certainly not very funny. – Alex Nov 15 '12 at 2:36
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http://tamino.wordpress.com/2011/07/06/aligning-station-records/
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Science, Politics, Life, the Universe, and Everything
# Aligning Station Records
Posted on July 6, 2011
As some of you know, I devised a method for aligning temperature data records which I believe is better than the “reference station method” used by NASA GISS. However, the difference is small and it doesn’t change the overall global result when small regions are averaged, then those regional results are area-weight-averaged to produce a global estimate. It’s an interesting, and possibly useful, refinement which doesn’t change the overall final answer.
In any case, the method computes an offset for each station which will bring it into alignment with the other stations, based on the idea that the difference between station data sets which are at nearby locations is approximately constant. RomanM modified my method by computing a separate offset for each station for each month. This compensates for possible differences in the annual cycle of different nearby stations. If your purpose is to study the annual cycle then this is a loss of information, but if your purpose is to study anomaly (the departure from the annual cycle) this is a gain. Most of the time we really want to get at the anomalies, so on the whole I’d say his method is better.
But nowadays I use yet another method. It’s based on the Berkeley method for combining station data to form a global average. They use a “weighting function” to represent the area over which a station should be considered to have an influence. The temperature estimate for a specific location is the weighted average of nearby stations, with closer stations given greater weight by the weighting function. I ignored the weighting function altogether so that all stations in a given region can be equally weighted to compute a regional average. My modification is therefore specifically tailored to produce a local estimate, whereas the Berkeley method is designed to include everything in one fell swoop and produce a global estimate.
Anyway, here are the details.
We have a number of station data records $y_j(t_k)$ for the temperature estimate at station j, at time $t_k$. These can be temperature anomalies so the seasonal cycle is already removed, or they can be raw temperature. Note that the straight Berkeley method uses anomalies, because they intend to combine stations worldwide, but I allow raw data so that nearby stations can be combined to give an estimate which includes the annual cycle.
We assume that there’s an offset $\mu_j$ for station j (one offset for each station), and the offsets together make up a vector of offsets. Therefore, while the data record for station j is $y_j(t_k)$, its contribution to the regional average will be
$x_j(t_k) = y_j(t_k) - \mu_j$.
We also assume that there’s a single regional average temperature $\gamma_k = \gamma(t_k)$ for each time $t_k$. The offsets and the temperature averages are determined simultaneously, by minimizing the sum of squared deviations of offset station records from the regional average. In other words, we minimize the quantity
$Q = \sum_{times~k} \sum_{stations~j} \Bigl [ (y_j(t_k) - \mu_j) - \gamma_k \Bigr ]^2$.
Defining equations for the offsets $\mu_j$ and average temperatures $\gamma_k$ result from taking partial derivatives of this expression and setting them equal to zero. Therefore
$\partial Q / \partial \gamma_k = -2 \sum_{stations~j} \gamma_k \Bigl [ (y_j(t_k) - \mu_j) - \gamma_k \Bigr ] = 0$,
and
$\partial Q / \partial \mu_j = - 2 \sum_{times~k} \mu_j \Bigl [ (y_j(t_k) - \mu_j) - \gamma_k \Bigr ] = 0$.
We can rearrange these two equations. The 1st says that for any time $t_k$ we have what I’ll call equation (1):
$\gamma_k = (1/N_k) \sum_{stations~j} (y_j(t_k) - \mu_j )$, (1),
where $N_k$ is the number of stations with data at time $t_k$. The 2nd says that for any station j we have what I’ll call equation (2):
$\mu_j = (1/N_j) \sum_{times~k} (y_j(t_k) - \gamma_k )$, (2)
where $N_j$ is the number of time points for which station j has data.
This system of equations is actually underdetermined. If I add a constant c to all the station offsets $\mu_j$, and subtract that same constant c from all the average temperatures $\gamma_k$, then the quantity Q to be minimized is unchanged — so the offsets and average temperatures are only determined up to an additive constant. Therefore I impose the additional condition that the offset for the very first station, $\mu_1$, is required to be zero. Other choices would do as well. In fact, there might be merit in requiring that the average of all the station offsets must be zero.
To compute the values, I begin by setting all station offsets to zero. Then I use equation (1) to compute the average temperature $\gamma_k$ at all times $t_k$. Then I use those values for $\gamma_k$ in equation (2) to compute updated values for the station offsets $\mu_j$. Then I subtract to the value of $\mu_1$ from all the offsets so that I’ll meet the requirement $\mu_1 = 0$.
These values are only approximate, so I repeat that procedure, at each step computing new estimates of offsets $\mu_j$ and temperatures $\gamma_k$. I also compute the sum of the absolute values of the differences in the offsets at each iteration. When this sum of differences is less than a particular threshold, I consider that the algorithm has converged and use the latest values of the offsets to compute final values of the average temperatures.
It’s an R function. Its inputs are a set of times t, a set of temperatures x, and a set of station identifiers part. It returns a list with two elements, the computed average temperatures and the station offsets.
The first entry in the list is actually a data frame with 5 columns. Column 1 is the time, column 2 is the average temperature for that time, column 3 is the number of stations with data for that time, column 4 is the standard error of the average, column 5 is the standard deviation of the offset temperatures. The second entry is simply a vector of the station offsets, and the 1st element of this vector will be zero.
There are a few other options. You may want the data frame to include the raw input data, in which case you can include the option “full=T” and the data frame will have more than 5 columns, it will also include a column for each raw input data set.
You can also set your own threshold for when the calculation is considered to have converged. By default, if the sum of absolute values of the changes in station offsets is less than 0.001, it is considered to have converged. If you want to insist it must be less than 0.000001, include the option “offres=.000001″.
For data to be combined at a particular time, the time values must match exactly. However, if you want to round off the times to a particular number of significant digits after the decimal point before insisting that they match, say rounding times to 3 decimal digits before computing, include the option “tround=3″.
By default the average temperatures will be rounded to 4 digits after the decimal point. If you want 6 digits instead, include the option “xround=6″.
To get the code itself, download this pdf file, then copy-and-paste it into a text editor and save that as an R function.
There’s always the possibility that I’ve made an error, or that my code is fine but my description is faulty.
And it almost certainly could be optimized — perhaps someone can modify it so that it’s more efficient either in the time of computation or in the memory required. Feel free.
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### 16 Responses to Aligning Station Records
1. Robert Way
Hey Tamino,
Interesting post and interesting approach. Do you by any chance have a script for the downloading of the data and putting into the formatting this method requires. Something like that would be really useful, particularly if there was a way to specify via lat/longs. I.e. to cover regions such as the arctic for example.
[Response: No, but maybe Mosher's programs will do some of that for you.]
2. Robert Way
Is there any particular reason to conclude that it would be wise to choose one threshold over another with respect to the absolute value of the sum of the offsets. For a publication to be submitting very soon we used this approach you outlined for a study region and found it hard to rationalize why we chose 0.0001 as our value for the sum of the offsets. I guess I should mention that it was your previous posts on the subject that led us to eventually use the adapted method we did so a little bit of a thanks to you is in order.
I should note that ours was done by manually downloading the information from gchn and environment Canada and manually doing this process in excel. It is therefore great that the R code is available and hopefully I can figure out how to use it.
[Response: There probably is a rational choice, but honestly I just pulled it out of my ***. It seems to work fine, and I expect the net error will be bounded by the threshold. In fact that's probably not too hard to show.]
3. g2-b31f1590b0e74a6d1af4639162aa7f3f
Some time ago, I whipped up a program that implements a *very crude* global-average anomaly gridding/averaging procedure.
I simply averaged all data from all stations sharing the 5-digit WMO “base” identification number to generate a single temperature time-series per 5-digit WMO base ID.
Then I computed the 1951-1980 baseline temps for each 5-digit WMO ID for each month. Threw out all station/month data for all temperature time-series for which there weren’t at least 15 out of 30 years of good data for each station/month (results were very insensitive to this number, btw).
Computed station anomalies per the above baseline values, ran it all through a very simple gridding/averaging routine (20deg x 20deg lat/long grids at the equator, crudely adjusted to give approximately equal grid areas as you go N/S).
Very crude, very amateurish, I’ll freely admit — something that high-school AP students could do.
The results? GHCN monthly-mean raw data run though this program generated results that matched NASA’s official “Meteorological Stations” pretty decently. Results here: http://img541.imageshack.us/img541/9994/landstationindexnosmoot.jpg
No offset-alignments or anything fancy like that — just simple, dumb anomaly gridding/averaging of raw GHCN temperature data.
Nothing new or exciting here — Tamino could do this in his sleep after pounding down a fifth of Jack Daniel’s (and chased by a couple of pints of pale ale).
But what this does show is how robust the global-warming temperature signal is; no matter how crudely you process the surface temperature data, you will get results not only in NASA’s ballpark, but pretty darned close to home plate — unless you completely f*@! it up, that is.
Kinda puts some perspective on all the *years* that deniers have been attacking the NASA/GISS global-temperature results, without doing so much of a lick of high-school-quality analysis of their own.
4. g2-b31f1590b0e74a6d1af4639162aa7f3f
That should be “Meteorological Stations” *temperature index*….
5. cce
I mentioned this in another thread, but it seems to me the “Berkeley Method” of treating each station discontinuity as the beginning/end of a unique series would be ideal for analyzing the radiosonde data, which is replete with problems due to the introduction of different measuring equipment.
6. yourmommycalledandsaidbehave@gmail.com
Tamino Had you ever considered using something like a Barnes(1964,1973,.1994) objective analysis to provide an area averaged value for the temperatures. A Barnes objective analysis is commonly used to determine the “correct” value for an area given a number of samples. The basis for the analysis is a Fourier series fit to the data.
7. Chad
Nice post Tamino. I was fooling around with a similar method not too long ago. I modeled each station as T(i,t) = T(t) + k_i*dz_i + e, where T(t) is the grid point temperature, T(i,t) is the ith station value, k_i is the lapse rate at the ith station and dz_i is the difference in altitude between a reference station and the ith station. I estimate the lapse rate using each station as a reference, so for N stations, I have N estimates of the lapse rate for the ith station. Then I adjust each station’s temperature by its mean local lapse rate to bring it up or down to the elevation of the desired grid point. This procedure is performed for all twelve months to allow for annual variation in the lapse rate. The purpose of this method is to create not just a grid of temperature anomalies, but a grid of temperature. I was also thinking about introducing kriging into the mix to avoid having to predefine a radius of influence for the weighting function, since that has been shown to have significant seasonal variability.
8. tamino
I’m taking a long weekend with the wife, so comment moderation will be slow and possibly nonexistent until Sunday night. Patience is a virtue.
9. Michael
“I devised a method for aligning temperature data records which I believe is better than the “reference station method” used by NASA GISS…..” – tamino
….THEREFORE ALL CLIMATE SCIENCE IS A FRAUD!! (hat tip Steve, Anthony et al.)
10. Nick Stokes
Tamino,
I wrote a program last year using a very similar least squares optimisation. It was also similar to the Berkeley idea, but I’ve used it for a lot of local temperature fittings, with anomalies. It uses density-based weightings, arguing essentially that the sums should really represent space integration formulae. I went on in V2 to represent the global temp by a spatial distribution rather than a single value.
There’s an overview of the early stages of the project here. Math description of V1 is linked here – this also links to the R codes.
The release note for the latest version is here. I’m now using a kind of Voronoi tesselation for the weighting. The Math basis for the spatial model is here.
11. Steven Mosher
I’ve completed a verification script for folks who want to play around with tamino’s algorithm.
I’ll do a posting of it along with instructions.
The verification consists of
1. A routine to create data to feed to his “align” function
2. His “align” function.
3. His function split into two smaller functions:
A. a function to create the big data.frame (zz)
B. the core math function ( which takes a data.frame as an input)
by splitting A and B, integration, testing and verification of
any attempts to speed up the code, (its already fast) will be
easier. ( we dont have to rebuild the data.frame every time)
I’ll be integrating the core math function in the Rghcnv3 package
over the next week. And then i’ll look at getting rid of the loops
in the code ( except the while loop) and see if that speeds things up.
Changes to the core math, of course, will be checked with the verification
script.
Great work tamino. Your code was clean and understandable and quick.
Now I get to give romanM and nick stokes some shit if their stuff is harder
to work with.
12. Steven Mosher
Here is the post.
I’ve split your code into two functions, the core math function should drop right into RghcnV3.
http://stevemosher.wordpress.com/2011/07/11/taminos-method-regional-temperatures/
13. steven mosher
Package built and submitted to CRAN. usually takes 2-3 days to get all the binaries tested and posted, source should be up in 24 hours or less.
RghcnV3 version1.2 now supports your method for aligning stations and calculating a mean.
The function is called regionalAverage()
a demo of it is in the demo files demo(Texas)
the documents need more work, but its there for people to play with
Its so fast I have to go work on the other methods to speed them up.
Thanks!
[Response: I'm surprised it's that fast.]
14. MP
@Steven Mosher,
Would it be difficult to add bootstrapping, e.g. resampling of stations, to your package and obtain some information about the size of the confidence intervals? I guess fast methods will help to get a decent sample.
15. Steven Mosher
Tamino: yes it’s quick relative to all the crap I have to do for a common anomaly method:
1. compute the base period means
2. subtract
3. area average and weight ( weights change per month)
etc etc, plus all the checks along the way. When I get around to removing
your three loops it’ll be even faster. Getting the data in the right structure
( you use a matrix which is faster than data.frame) is half the battle. The next thing I will add for people is a method of selecting regions.. in due course
MP:
Bootstrapping should be easy.
I assume what you want is this.
You select a region. That region has say 50 stations. You want to
run tamino’s regionalizer on multiple random samples of that 50?
That’s a snap. I dont even have to add it to the package since you can do that in R easily. Just describe what you want done clearly. It’s fun for me and keeps me from saying stupid stuff on blogs.
16. MP
@Steven Mosher,
Exactly yes, I noticed that Zeke had the same idea and posted some results. It would be a useful functionality in your package. I can imagine there are several possible sampling techniques. One catch with this method is that wit non uniform coverage the meaning of the confidence intervals is not unambiguous, but does give an impression.
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• ### mathematics
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http://crypto.stackexchange.com/questions/tagged/pairings+elliptic-curves
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# Tagged Questions
1answer
54 views
### How are Elliptic Curve Cryptography and Pairing Based Cryptography related?
I have been doing a project that uses the PBC library developed by Ben Lynn. But I am still not clear on how PBC is related to ECC. I know that this is a site for complex crypto QA, but I did not know ...
0answers
37 views
### Generating non-supersingular elliptic curves for symmetric pairings
I am looking into the application of pairings in CPABE in particular. I've notice that the scheme uses a supersingular curve as the basis of the pairing. Looking through Ben Lynn's thesis for the ...
1answer
51 views
### Discrete logs on elliptic curve with embedding degree 3 with the 'MOV' attack
The curve $E(\mathbb{F}_{47}):y^2=x^3+x+38$ has order $61$ and $61|47^3-1$ so the embedding degree of $E$ is $3$ and therefore the MOV attack, presumably using some sort of distortion map and a ...
2answers
137 views
### Modulus for elliptic curve point multiplication
I want to implement a point multiplication ($k \cdot P$) operation on FPGA. I have a BN curve $y^2=x^3+2$, and a scalar value $k$. The $x$ and $y$ coordinates of point $P$ are of 256 bits. In the ...
2answers
244 views
### Pairing-friendly curves in small characteristic fields
There are several well-known techniques to generate pairing-friendly curves of degrees 1 to 36 on prime fields GF(p): Cocks-Pinch, MNT, Brezing-Weng, and several others. In extension fields GF(p^n), ...
1answer
395 views
### Mapping points between elliptic curves and the integers
My primary question is: Is there an easy way to create a bijective mapping from points on an elliptic curve E (over a finite field) to the integers (desirably to $\mathbb{Z}^*_q$ where $q$ is the ...
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http://mathhelpforum.com/calculus/62838-functional-equations.html
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# Thread:
1. ## functional equations
(i)If $f(x)$ is a polynomial function $f<img src=$0,\infty)\rightarrow(0,\infty)" alt="f0,\infty)\rightarrow(0,\infty)" />satisfies the equation $f(f(x))=6x-f(x)$then find $f(17)$
(ii)If $f(x)$ is a polynomial function satisfying the condition such that $f(x^2+1)=(f(x))^2+1$ and that $f(0)=0$ then find $f(34).$
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http://stats.stackexchange.com/tags/mcmc/hot?filter=all
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# Tag Info
## Hot answers tagged mcmc
10
### How to sample from $\{1, 2, …, K\}$ for $n$ random variables, each with different mass functions, in R?
We can do this in a couple of simple ways. The first is easy to code, easy to understand and reasonably fast. The second is a little trickier, but much more efficient for this size of problem than the first method or other approaches mentioned here. Method 1: Quick and dirty. To get a single observation from the probability distribution of each row, we can ...
9
### Is there a standard technique to debug MCMC programs?
Standard programming practice: -when debugging run the simulation with fixed sources of randomness (i.e. same seed) so that any changes are due to code changes and not different random numbers -try your code on a model (or several models) where the answer IS known -adopt good programming habits so that you introduce fewer bugs -think very hard & long ...
9
### Examples of errors in MCMC algorithms
Darren Wilkinson on his blog gives a detailed example of a common mistake in random walk Metropolis-Hastings. I recommend reading it in full, but here is the tl;dr version. If the target distribution is positive (like Gamma distributions etc) in one dimension, it is tempting to reject proposals that have a negative value on that dimension straight away. The ...
8
### Given a 10D MCMC chain, how can I determine its posterior mode(s) in R?
Have you considered using a nearest neighbour approach ? e.g. building a list of the k nearest neighbours for each of the 100'000 points and then consider the data point with the smallest distance of the kth neighbour a mode. In other words: find the point with the 'smallest bubble' containing k other points around this point. I'm not sure how robust ...
8
### Does standardising independent variables reduce collinearity?
It doesn't change the collinearity between the main effects at all. Scaling doesn't either. Any linear transform won't do that. What it changes is the correlation between main effects and their interactions. Even if A and B are independent with a correlation of 0, the correlation between A, and A:B will be dependent upon scale factors. Try this... a ...
8
### Good summaries (reviews, books) on various applications of Markov chain Monte Carlo (MCMC)?
Yes the Handbook of MCMC is a very up-to-date collection of papers on MCMC, Also the book by Robert and Casella is a more current account than Markov Chain Monte Carlo in Practice. But I think MCMC in Practice is really a good place to start learning the subject. Here are amazon links to descriptions of the books I mentioned above. Introducing Monte Carlo ...
7
### Is it possible to apply Bayes Theorem with only samples from the prior?
The short answer is yes. Have a look at sequential MCMC/ particle filters. Essentially, your prior consists of a bunch of particles ($M$). So to sample from your prior, just select a particle with probability $1/M$. Since each particle has equal probability of being chosen, this term disappears in the M-H ratio. A big problem with particle filters is ...
7
### What is the best method for checking convergence in MCMC?
I use the Gelman-Rubin convergence diagnostic as well. A potential problem with Gelman-Rubin is that it may mis-diagnose convergence if the shrink factor happens to be close to 1 by chance, in which case you can use a Gelman-Rubin-Brooks plot. See the "General Methods for Monitoring Convergence of Iterative Simulations" paper for details. This is ...
7
### Is there a standard method to deal with label switching problem in MCMC estimation of mixture models?
There is a nice and reasonably recent discussion of this problem here: http://www.icms.org.uk/downloads/mixtures/Robert.pdf Essentially, there are several standard strategies, and each has pros and cons. The most obvious thing to do is to formulate the prior in such a way as to ensure there is only one posterior mode (eg. order the means of the mixuture ...
7
### Acceptance rates for Metropolis-Hastings with uniform candidate distribution
I believe that Weak convergence and optimal scaling of random walk Metropolis algorithms by Roberts, Gelman and Gilks is the source for the 0.234 optimal acceptance rate. What the paper shows is that, under certain assumptions, you can scale the random walk Metropolis-Hastings algorithm as the dimension of the space goes to infinity to get a limiting ...
7
### Acceptance rates for Metropolis-Hastings with uniform candidate distribution
Just to add to answer by @NRH. The general idea follows the Goldilocks principal: If the jumps are "too large", then the chain sticks; If the jumps are "too small", then the chain explores the parameter space very slower; We want the jumps to be just right. Of course the question is, what do we mean by "just right". Essentially, for a particular case ...
7
### Winbugs and other MCMC without information for prior distribution
Unfortunately, harmless seeming priors can be very dangerous (and have even fooled some seasoned Bayesians). This recent paper, provides a nice introduction along with plotting methods to visualize the prior and posterior (usually marginal priors/posterior for the parameter(s) of interest). Hidden Dangers of Specifying Noninformative Priors. John W. ...
6
### Forecasting unemployment rate
The Arellano-Bond estimator has been designed for precisely this type of problems. You will find a short non-technical paper with a examples here. In a nutshell, it combines the information embedded in the large number of cross-section to make up for the small number of points in each series. This estimator is widely used and implemented: it is avalaible in ...
6
### What MCMC algorithms/techniques are used for discrete parameters?
So the simple answer is yes: Metropolis-Hastings and its special case Gibbs sampling :) General and powerful; whether or not it scales depends on the problem at hand. I'm not sure why you think sampling an arbitrary discrete distribution is more difficult than an arbitrary continuous distribution. If you can calculate the discrete distribution and the ...
6
### Why should we care about rapid mixing in MCMC chains?
The ideal Monte Carlo algorithm uses independent successive random values. In MCMC, successive values are not independant, which makes the method converge slower than ideal Monte Carlo; however, the faster it mixes, the faster the dependence decays in successive iterations¹, and the faster it converges. ¹ I mean here that the successive values are quickly ...
6
### MCMC to handle flat likelihood issues
I find it surprising that a flat likelihood produces convergence issues: it is usually the opposite case that causes problems! The usual first check for such situations is to make sure that your posterior is proper: if not it would explain for endless excursions in the "tails". If the posterior is indeed proper, you could use fatter tail proposals like a ...
6
### Winbugs and other MCMC without information for prior distribution
Parameters in linear predictor are t-distributed. When the number of records goes to infinity, it converges to normal distribution. So yes, normally it is considered correct to assume normal distribution of parameters. Anyways, in bayesian statistics, you need not to assume parameter distribution. Normally you specify so called uninformative priors. For ...
5
### Reversible jump MCMC code (Matlab or R)
RJMCMC was introduced by Peter Green in a 1995 paper that is a citation classic. He wrote a Fortran program called AutoRJ for automatic RJMCMC; his page on this links to David Hastie's C program AutoMix. There's a list of freely available software for various RJMCMC algorithms in Table 1 of a 2005 paper by Scott Sisson. A Google search also finds some ...
5
### Is there a standard method to deal with label switching problem in MCMC estimation of mixture models?
Gilles Celeux also worked on the problem of label switching, e.g. G. Celeux, Bayesian inference for Mixture: the label switching problem. Proceedings Compstat 98, pp. 227-232, Physica-Verlag (1998). As a complement to @darrenjw's fine answer, here are two online papers that reviewed alternative strategies: Jasra et al., Markov Chain Monte Carlo ...
5
### Measures of autocorrelation in categorical values of a Markov Chain?
You can always choose one or several real valued functions of the categorical variables and look at the auto-correlation for the resulting sequence(s). You can, for instance, consider indicators of some subsets of the variables. However, if I understood your question correctly, your sequence is obtained by an MCMC algorithm on the discrete space. In that ...
5
### Which distribution to use with MCMC and empirical data?
Kolmogorov Smirnoff is always a good test to see if an arbitrary distribution fits. You can use the test cited below to see if two sets of data came from the same distribution: Li, Q. and E. Maasoumi and J.S. Racine (2009), “A Nonparametric Test for Equality of Distributions with Mixed Categorical and Continuous Data,” Journal of Econometrics, ...
5
### Why should we care about rapid mixing in MCMC chains?
In completion of both earlier answers, mixing is only one aspect of MCMC convergence. It is indeed directly connected with the speed of forgetting the initial value or distribution of the Markov chain $(X_n)$. For instance,the mathematical notion of $\alpha$-mixing is defined by the measure \alpha(n) = \sup_{A,B} \left\{\,|P(X_0\in A,X_n\in\cap B) - ...
5
### MCMC for infinite variance posteriors
There is nothing wrong with infinite variance distributions, per se... For instance, simulating a Cauchy using rcauchy(10^3) produces a sample truly from a Cauchy distribution! Hence MCMC has no specific feature to "fight" for or against infinite variance distributions. The difficulty with infinite variance distributions is at the Monte Carlo level, for ...
5
### Accept rate in Metropolis–Hastings algorithm
In order to get this, and to simplify the matters, I always think first in just one parameter with uniform (long-range) a-priori distribution, so that in this case, the MAP estimate of the parameter is the same as the MLE. However, assume that your likelihood function is complicated enough to have several local maxima. What MCMC does in this example in 1-D ...
5
### How to sample from $\{1, 2, …, K\}$ for $n$ random variables, each with different mass functions, in R?
A for loop may be terribly slow in R. How about this simple vectorization with sapply? n <- 10000 k <- 200 S <- 1:k p <- matrix(rep(1 / k, n * k), nrow = n, ncol = k) x <- numeric(n) x <- sapply(1:n, function(i) sample(S, 1, prob = p[i,])) Of course, this uniform p is just for testing.
5
### Textbook deriving Metropolis-Hastings and Gibbs Sampling
For a handbook and an extensive coverage, the following one is very moderately priced. Brooks, et al. (ed.), Handbook of Markov Chain Monte Carlo, Chapman & Hall/CRC, 2011. Robert and Casella (2010) have a good deal of theory.
5
### Textbook deriving Metropolis-Hastings and Gibbs Sampling
I'm not sure whether this is exactly what you're after, but a couple of articles I've found useful on theoretical properties of various Metropolis-Hastings algorithms are: Optimal scaling for various Metropolis-Hastings algorithms - Roberts & Rosenthal, 2001. (This summarises some earlier results for the Ransom walk Metropolis and the ...
5
### Parameters without defined priors in Stan
From (an earlier version of) the Stan reference manual: Not specifying a prior is equivalent to specifying a uniform prior. A uniform prior is only proper if the parameter is bounded[...] Improper priors are also allowed in Stan programs; they arise from unconstrained parameters without sampling statements. In some cases, an improper prior may ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://mathhelpforum.com/calculus/191017-integration-step-pde-laplace-transform-method.html
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# Thread:
1. ## An integration step in PDE by Laplace transform method
I am stuck with an integration step in solving a PDE by Laplace transform.
The integration is $\int e^{\frac{s x^2}{2}} \frac{x}{s} dx = \frac{1}{s^2} e^{\frac{s x^2}{2}}$. How is the integral obtained? I want to use integration by parts but I don't know how to integrate $e^{\frac{sx^2}{2}}$.
2. ## Re: An integration step in PDE by Laplace transform method
Let $u = \frac{sx^2}{2}$. The integration will come out fine!
3. ## Re: An integration step in PDE by Laplace transform method
Thank you. That worked.
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http://unapologetic.wordpress.com/2007/11/20/nets-part-ii/?like=1&source=post_flair&_wpnonce=671450efa2
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# The Unapologetic Mathematician
## Nets, Part II
Okay, let’s pick up our characterization of topologies with nets by, well, actually using them to characterize a topology. First we’re going to need yet another way of looking at the points in the closure of a set.
Here goes: a point $x$ is in $\mathrm{Cl}(A)$ if and only if every open neighborhood of $x$ has a nonempty intersection with $A$. To see this, remember that the closure of $A$ is the complement of the interior of the complement of $A$. And we defined the interior of the complement of $A$ as the set of points that have at least one open neighborhood completely contained in the complement of $A$. And so the closure of $A$ is the set of points that have no open neighborhoods completely contained in the complement of $A$. So they all touch $A$ somewhere. Cool?
Okay, so let’s kick it up to nets. The closure $\mathrm{Cl}(A)$ consists of exactly the accumulation points of nets in $A$. Well, since we know that every accumulation point of a net is the limit of some subnet, we can equivalently say that $\mathrm{Cl}(A)$ consists of the limit points of nets in $A$. So for every point in $\mathrm{Cl}(A)$ we need a net converging to it, and conversely we need to show that the limit of any convergent net in $A$ lands in $\mathrm{Cl}(A)$.
First, let’s take an $x\in\mathrm{Cl}(A)$. Then every open neighborhood $U$ of $x$ meets $A$ in a nonempty intersection, and so we can pick an $x_U\in U\cap A$. The collection of all open neighborhoods is partially ordered by inclusion, and we’ll write $U\geq V$ if $U\subseteq V$. Also, for any $U_1$ and $U_2$ we have $U_1\cap U_2\geq U_1$ and $U_1\cap U_2\geq U_2$. Thus the open neighborhoods themselves form a directed set. The function $U\mapsto x_U$ is then a net in $A$. And given any neighborhood $N$ of $x$ there is a neighborhood $U$ contained in $N$. And then for any $V\geq U$ we have $x_V\in V\subseteq U\subseteq N$, so our net is eventually in $N$. Thus we have a net which converges to $x$.
Now let’s say $\Phi:D\rightarrow A\subseteq X$ is a net converging to $x\in X$. Then $\Phi$ is eventually in any open neighborhood $U$ of $x$. That is, every open neighborhood of $x$ meets $A$ in at least one point, and thus $x\in\mathrm{Cl}(A)$.
So for any $A$, the closure $\mathrm{Cl}(A)$ is the collection of accumulation points of all nets in $A$. And now we can turn this around and define a closure operator by this condition. That is, we specify for each net $\Phi$ the collection of its accumulation points, and from these we derive a topology with this closure operator.
Let’s see that we really have a closure operator. First of all, clearly $U\subseteq\mathrm{Cl}(U)$ for all $U$ because we can just take constant nets. Even easier is to see that there are no nets into $\varnothing$, and so its closure is still empty. Any accumulation point of a net in $U\cup V$ is the limit of a subnet, which we can pick to lie completely within either $U$ or $V$, and so $\mathrm{Cl}(U\cup V)=\mathrm{Cl}(U)\cup\mathrm{Cl}(V)$.
To finish, we must show that this purported closure operator is idempotent. For this, I’ll use a really nifty trick. A point in $\mathrm{Cl}(\mathrm{Cl}(U))$ is the limit of some net $\Phi:D\rightarrow\mathrm{Cl}(U)$, and each of the points $\Phi(d)$ is the limit of a net $\Phi_d:D_d\rightarrow U$. So let’s build a new directed set by taking the disjoint union $\biguplus\limits_{d\in D}D_d$ and defining the order as follows: if $a\in D_c$ and $b\in D_d$ for $c$ and $d$ in $D$, then $a\geq b$ if $c\geq d$, or if $c=d$ and $a\geq b$ in $D_d$. Then combining the nets $\Phi_d$ we get a net from this new directed set, which clearly has an accumulation point at the limit point of $\Phi$, and which is completely contained within $U$. This completes the verification that $\mathrm{Cl}$ is indeed a closure operator, and thus defines a topology.
### Like this:
Posted by John Armstrong | Point-Set Topology, Topology
## 6 Comments »
1. hi, i was reading about nets recently and became quite excited about how so many thing in topology become so much more natural when thinking in terms of sets(like definitons of continuity, compactness, product topology, tychonoff theorem). it would be nice to have a definition of a topological space just in terms of nets. but all the characterizations of topological spaces in terms of nets prove that the map from topological spaces to
the class {sets with information about which nets converge to which points} is an injective map. But it would a nice to know what what subset of the above class come from topological spaces,
for instance, one obvious requirement is that the set of accumulation points of a subnet is a subset of the accumulation points of the net and that isomorphic nets have
the same accumulation points…
but these conditions aren’t enough, there could be no topological structure on the set which has the same info about accumulation points of nets.
If there were some sufficient conditions though, then one get rid entirely of the open sets approach and start with nets from the begining and that would be a very nice thing.
Comment by harsha | November 20, 2007 | Reply
2. harsha, I don’t know the answer exactly. Something similar goes on here to what happened when I talked about neighborhoods. That is, any collection of filters gives a topology, but many different collections will give rise to the same topology. In effect, there’s a sort of “closure” going on.
Here instead we have the fact that no matter what accumulation points we assign to each net, we get some closure operator, but not every such collection of data arises from a closure operator. So the process of going from data to operator to data is a sort of “closure” on the class of such data.
So here’s how I’d suggest going about it: try to figure out how this process works. That is, if you start with an arbitrary collection of accumulation points for each net, use that data to define a topology, and see what accumulation points each net now has. Then determine what the fixed points of this process are. Those will be your answers.
First, though, I’d ask around your department. I know at least one venerable-mathematician-who-knows-everything sort of topologist who hangs around there you might try. If nothing else you might be able to verify that this actually hasn’t been done, and then it would make an excellent thesis problem
Comment by | November 20, 2007 | Reply
3. [...] and Subbases We’ve defined topologies by convergence of nets, by neighborhood systems, and by closure operators. In each case, we saw some additional hypothesis [...]
Pingback by | November 22, 2007 | Reply
4. I just aksed myself the same question harsha brought up here, and hence found this webpage. Is there the notion of a basis for a space’s nets? I only have a vague notion of what it should be: A basis in terms of nets is a full subcategory of the category of nets in some set X, together with their accumulation points such that the topology that arises from this subcategory agrees with the original one.
For instance, if a space is 1st countable, then instead of considering all nets and their accumulation points it suffices to take all sequences into account. There are various other examples.
Still, the difficulty is what properties needs the accumulation point map to have to give rise to a topology?
Comment by olf | March 11, 2008 | Reply
5. Olf, I really wish I knew. But when it comes down to it I’m not a point-set topologist and I just haven’t spent a lot of time on this problem. I’d love to hear if you turn up anything, though.
Comment by | March 11, 2008 | Reply
6. [...] The first thing to note about is that it’s a closed set. That is, it should contain all its accumulation points. So let be such an accumulation point and assume that is isn’t in , so . So there must exist [...]
Pingback by | December 9, 2009 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://scienceblogs.com/principles/2013/03/12/science-kids-fictitious-forces-and-frictionless-surfaces/
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## National Geographic SocietyIn partnership with
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# Science Kids, Fictitious Forces, and Frictionless Surfaces
Posted by on March 12, 2013
Sid the Science Kid screenshot, originally from http://www.confessionsofapsychotichousewife.com/2010/10/sid-science-kid-party.html
SteelyKid has started to demand Sid the Science Kid videos, which of course we are implacably opposed to around here. One of the recent episodes available online was “Slide to the Side,” talking about friction. While this partakes a bit of the Feynman “Energy makes it go” problem, it was generally pretty good, and prompted a question from Kate that (combined with Rhett’s latest.
Kate and SteelyKid were talking about how friction keeps things from sliding, then Kate asked me whether it would be possible to keep anything at rest on a truly frictionless surface, due to the rotation of the Earth. My initial answer was that it wouldn’t be a problem, because the surface and whateveer object you were placing on it would initially be at rest with respect to the rotating surface of the Earth, and thus would continue to move together. It’s not like the Earth would spin out from under the object once you put it on a frictionless surface, or anything.
But after reading Rhett’s analysis of people flying off a spinning Earth, I want to change my answer slightly. As Rhett notes, if you’re going to talk about motion on the surface of the Earth, you need to talk about some fictitious forces that come in due to the rotation, and when you do it out carefully, I think you do get a slight force making things move, but not in the way you might expect.
The relevant force is the one Rhett discusses, the “centrifugal force.” This is a somewhat dangerous term to use in intro classes, because students want to invoke it for everything, and it’s not quite right. There’s no interaction that leads to a real force that would be measurable in a stationary lab frame; the “force” that appears occurs only when you transfer to a rotating frame, where things appear to start moving even without a specific interaction causing them to move. This is the basis of a bunch of amusement-park rides, and is what marks the rotating frame as non-inertial, that is, a frame of reference where Newton’s Laws of Motion don’t appear to work properly.
That said, there’s nothing inherently wrong about looking at the world from the perspective of someone in a non-inertial frame– we do it all the time, as beings riding on a spinning planet that’s orbiting a star. So let’s think about these forces, using the not-remotely-to-scale diagram below:
The black arc here represents a slice of the surface of the Earth, and the blue circles are spherical frictionless objects placed on the surface. If you look at the lower of the two objects, sitting on the equator, from the perspective of an observer who is also rotating with the surface of the Earth, this object experiences two forces: a gravitational force pulling the object toward the center of the Earth (represented by the blue arrow), and a “centrifugal force” which is directed outward, away from the axis of rotation (represented by the green arrow). These are exactly opposite one another, and the gravitational force is vastly bigger than the centrifugal force, so an object at the equator will fall straight down, and just sit on the ground nicely.
For the other object, at higher latitude, though, the situation is more complicated. The two forces aren’t quite in the same direction any more– the “centrifugal force” is still directly out from the axis, to the right in the figure, but the gravitational force is now down and to the left, toward the center of the Earth. This mismatch in directions makes all the difference.
To add these two forces together, we really need to break them into components. since the force of gravity is vastly bigger than the “centrifugal force,” we’ll leave it alone, and only think about what happens to the red arrow. We can break this into two pieces, one that goes along the same line as the gravitational force (the small blue arrow), and one perpendicular to it (the small green arrow). As with the object at the equator, the vertical component is more than balanced out by the gravitational force, so an object will still appear to fall straight down when dropped.
The other component, though, will be unaffected by gravity, and should look like a small force pushing the object toward the equator– south, in the figure above, assuming the usual America-centric view of the globe. If you had a truly frictionless object, and set it down on a frictionless surface in the middle latitudes, you would see it slowly drift to the south.
So, how big is this force? Well, if we call the latitude of the object (the angle measured up from the equator) $\theta$, the green arrow will have a magnitude given by:
$|F_{south}| = |F_{cent}|\sin \theta$
(a little bit of trigonometry there, but an easy bit), where $|F_{cent}|$ is the magnitude of the centrifugal force. this, in turn, is given by:
$|F_{cent}| = m \omega^2 R_{\perp} = m \omega^2 R_{Earth} \cos \theta$
where $\omega$ is the rate of the Earth’s rotation in radians/sec. The cosine creeps in there because the radius that matters is the radius away from the axis of rotation, which gets smaller as you move to higher latitudes. Putting these together, we find that:
$|F_{south}| = m \omega^2 R_{Earth} \cos \theta \sin \theta$
This will have its biggest value at a latitude of 45 degrees, coincidentally not too far from where I happen to be writing this, where the product of trig functions will be equal to exactly 1/2. Putting in numbers for everything (the Earth turns through 2 π radians in 86400s, the radius of the Earth is about 6,400,000 meters), we get a force equal to the mass multiplied by 0.017 m/s/s. Which means that an object placed on the surface of the Earth at that latitude should accelerate southward at 0.017 m/s/s, or about 2/1000ths the acceleration of gravity.
What’s that mean for your frictionless surfaces? Well, it suggests that an object placed on a truly frictionless surface ought to take about 11 seconds to slide 1m to the south. That’s actually a bit faster than I would’ve expected, which may indicate that I’ve made an arithmetic error someplace, but I’ve already spent more time on this than I should’ve, so I’ll leave that for people to point out in the comments.
Anyway, there you go. From SteelyKid to the Sid the Science Kid to Rhett the Science Blogger, your weird physics-y blog tidbit for the week.
For extra credit, calculate the oscillation frequency of a mass sliding back and forth across the equator on a smooth frictionless sphere the size of the Earth. this is, of course, anharmonic, but for a first pass you can use the small-angle approximation for the latitude.
For extra extra credit, add the Coriolis force to the problem. Send your extra credit problems for grading to Neil de Grasse Tyson, The Hayden Planetarium, 81 Central Park West, New York, NY, 10023.
(The Sid screenshot that’s the “Featured Image” for this post came from Confessions of a Psychotic Housewife.)
## Comments
1. NL
March 12, 2013
I think you mean “red”, not “green” when talking about the equator?
2. Eric Lund
March 12, 2013
You haven’t made any arithmetic errors that I can see, but you did assume a spherical earth. The actual shape is an oblate spheroid, so the normal force of the surface does not point exactly toward the center of the earth. Instead, it will have a slight poleward component, countering the equatorward component due to centrifugal force. I haven’t done the full calculation, so I don’t know if they exactly cancel, but I would expect them to be the same order of magnitude.
3. March 12, 2013
Yes, I meant “red” when talking about the equatorial point. This is what I get for spending so much time talking to the dog– my color vision gets all wonky.
I had meant to mention the shape of the Earth in this, but forgot when I started to close in on my self-imposed deadline. You’re probably right about the order of magnitude– the difference in radii is about 0.3%, and if you took the really naive approximation of saying there was an 0.3% axial component of the gravitational force, that would be bigger than the force above. It’s probably more complicated than that, but almost certainly on the same order of magnitude as the force described here.
Clearly, what we need is a perfectly frictionless surface so we can see which way, if any, an object slides on it. Somebody get on that.
4. CCPhysicist
March 12, 2013
And you forgot that the Earth’s axis is tilted.
If the earth were an inviscid liquid, it would take a shape that eliminates the force you are calculating. It isn’t, of course, but that is why the oblate shape of the earth works against the simplistic spherical argument.
And the statements above (and your analysis) ignore the moon and sun. Their gravity, normally manifest as tidal forces, act on that mass as well and the tilted axis gives a seasonal contribution.
5. Ian
March 13, 2013
Surely the oblate spheroidiness of the earth would exactly cancel the rotation.
The non-sphericity of the earth is determined by the exact same two forces as you are using to determine the object’s acceleration.
Or to put it another way. For the purposes of determining the shape of the earth, it is a fluid. A fluid in a steady-state must have no shear force. A particle of fluid (or a body floating on the surface) on the surface, feels gravity, buoyancy along the surface normal, and the centrifugal force. If the three did not cancel exactly, the surface particle would feel a sheer force. Since it is a fluid, it would deform ans so the shape of the earth would not be a steady-state. But it is, so the forces must cancel exactly.
6. Wilson
March 13, 2013
From the sublime to the ridiculous: Not having seen any episodes of Sid the Science Kid, why the implacable opposition? Bad science?
7. March 13, 2013
That was a joke. We’re happy to have her watch science-y shows– Mythbusters is one of the tv staples in our house.
I should’ve included an explicit sarcasm flag, because Internet, but I was writing in a hurry.
8. MobiusKlein
March 13, 2013
What about gyroscopic effects? An object on a frictionless surface will still be changing orientation as the Earth spins.
(except at the poles)
9. Wilson
March 14, 2013
Gotcha.
Yes, the Internet needs a sarcasm tag of some kind. Text communication in general does, but I guess the bulk of that is the Internet anyway.
Thanks for the clarification!
10. Clay B
United States
March 15, 2013
I think Ian is onto something there. The sliding particle is going “uphill” in terms of distance from the center of the Earth, so gravity is opposed to (cancels out?) the Equatorward push. At least my physics intuition says everything should cancel out.
11. thomas
NZ
March 15, 2013
I would have thought the oblateness and centrifugal components would have cancelled out exactly at the time the crust solidified. The earth was rotating faster then, so now, with a slower rotation, the oblateness should win and the object should slide away from the equator — but probably very slowly.
12. Ian
March 19, 2013
The solid crust is flexed by the moon’s gravity on a daily cycle. It is only a few 10s of km thick at most, compared to the 1000skm diameter of the earth. It will have zero rigidity over geological time periods.
Another way of thinking about the whole problem is to think of what mean down, and what means a flat, level surface. A flat level surface is one which has no net gravitational force, as resolved in the directions of the plane. When setting up this surface, assuming that you use a spirit level, you can’t tell which part of the forces are gravitational and which are centrifugal. So, almost by definition, you will put this surface down so that there is no acceleration along the surface.
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http://mathhelpforum.com/calculus/74054-least-greatest-upper-bound-set.html
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# Thread:
1. ## Least and greatest upper bound of a set
please help, i dont know how to post on latex, for example the attachment is on the link below but I dont know how to post on this message
Attached Files
• exx2.pdf (18.8 KB, 16 views)
2. Just surround your $\text\LaTeX$ code with [tex] and [/tex]. The board sets math mode automatically, so don't use dollar signs around everything.
For example:
Code:
`[tex]\sum_{n=1}^\infty\frac1n=\frac{\pi^2}6[/tex]`
Produces
$\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$
Use \text{} to get regular non-math text. There are limits on the number of characters and the size of each $\text\LaTeX$ image, so you may need to break things up into several sections. For full paragraphs of text, like in your PDF, post the non-math text outside of the [tex] tags.
3. Find the least upper bound and the greatest lower bound of the following sets. For a given $\epsilon$>0, find a number in the set that exceeds (l.u.b.A)- $\epsilon$ and a number in the set that is smaller than (g.l.b. A)+ $\epsilon$.
(a). A= $\lbrace<br />$ $\frac{4+x}{x}$ $\vert$x
$\geq$
1 $\rbrace$
4. Originally Posted by mancillaj3
Find the least upper bound and the greatest lower bound of the following sets. For a given $\epsilon$>0, find a number in the set that exceeds (l.u.b.A)- $\epsilon$ and a number in the set that is smaller than (g.l.b. A)+ $\epsilon$.
(a). A= $\lbrace<br />$ $\frac{4+x}{x}$ $\vert$x
$\geq$
1 $\rbrace$
For the last line, try
Code:
`\text A=\left\{\left.\frac{4+x}x\;\right\vert\;x\geq1\right\}`
which gives
$\text A=\left\{\left.\frac{4+x}x\;\right\vert\;x\geq1\ri ght\}$
5. Originally Posted by mancillaj3
Find the least upper bound and the greatest lower bound of the following sets. For a given $\epsilon$>0, find a number in the set that exceeds (l.u.b.A)- $\epsilon$ and a number in the set that is smaller than (g.l.b. A)+ $\epsilon$.
(a). A= $\lbrace<br />$ $\frac{4+x}{x}$ $\vert$x
$\geq$
1 $\rbrace$
So, for the record, the actual question that the OP wants an answer for is:
Find the least upper bound and the greatest lower bound of the following sets. For a given $\epsilon$>0, find a number in the set that exceeds (l.u.b.A)- $\epsilon$ and a number in the set that is smaller than (g.l.b. A)+ $\epsilon$.
$\text A=\left\{\left.\frac{4+x}x\;\right\vert\;x\geq1\ri ght\}$
l.u.b. = 5, g.l.b. = 1. Do you see why?
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http://mathoverflow.net/questions/70889/determinant-of-exterior-power
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determinant of exterior power
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose A is a n times n matrix.
what is the determinant of the i-th exterior power of A, in terms of determinant of A ?
thanks..
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1 Answer
The answer is just $\det(\Lambda^i(A))=\det(A)^m$, where $m=\left(\begin{matrix}n-1\\ i-1\end{matrix}\right)$. Indeed, by continuity it is enough to prove this when $A$ is diagonalisable, and by conjugation-invariance it suffices to prove it when $A$ is diagonal, and in that case it is straightforward. Alternatively, you can show that $SL_n(\mathbb{C})$ is the commutator subgroup of $GL_n(\mathbb{C})$ and thus that any polynomial homomorphism $GL_n(\mathbb{C})\to\mathbb{C}^\times$ is a power of the determinant. To find the relevant power, just take $A$ to be a multiple of the identity.
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1
This result is called the Sylvester--Franke theorem. – KConrad Jul 21 2011 at 13:02
1
Neil's first argument works over an arbitrary field: first observe that WLOG the field is alg closed, and then use the Zariski topology for the continuity argument. – Kevin Buzzard Jul 21 2011 at 13:32
1
In that case I'd say the result is true over any commutative ring, since the formula can be regarded as a universal polynomial identity over Z and thus it suffices to check it over the complex numbers. – KConrad Jul 21 2011 at 22:28
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http://math.stackexchange.com/questions/279929/wedge-product-vocabulary
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# Wedge Product Vocabulary
I am looking for a word or phrase that is similar in meaning to the "support of a function", but in the context of differential forms.
The "support of a function" is the subset of the domain of a function where the function is non-zero.
In my case, consider the bilinear functional $(dx \wedge dy)$ acting on vector pairs in $xyz$-space.
I would like to say something like, "The support of $(dx \wedge dy)$ is the $xy$ plane.", because the result of the calculation is zero unless both vectors have a no-zero projection on the $xy$-plane.
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The notion of support of a differential form $\omega$ already exists. Unfortunately it means the closure of the set of points on which $\omega$ is non-zero. (Recall, a differential form is a smooth assignment of alternating multilinear forms to each point of a manifold...) – Zhen Lin Jan 16 at 9:59
1
The more fundamental space associated with $dx\wedge dy$ is actually the $z$-axis, which is the set of vectors $v$ such that $dx\wedge dy(v,\cdot)=0$. This is sometimes called the kernel, or characteristic space, or annihilator of the differential form; notice that it's defined without reference to the inner product. If you're willing to rely on the inner product of $\mathbb R^3$, you can single out the $xy$-plane as the space orthogonal to the kernel. But a different inner product would result in a different such $2$-dimensional space, while the kernel wouldn't change. – Jack Lee Jan 17 at 5:15
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http://mathoverflow.net/questions/42389/applications-of-stacks/42390
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## Applications of Stacks
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I've been aware of stacks since grad school, and I can usually follow in rough lines a discussion about stacks, but I've often wondered what particular (purely!) scheme-theoretic argument or theorem is significantly simplified by the introduction of stacks. I'm sure there are many, but since I don't deal with stacks on a regular basis I don't encounter them as frequently, and I thought maybe some of you can enlighten me.
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3
Cohomology of stacks provide useful "algebraic/arithmetic" constructions superior in some respects to what can be done using coarse moduli schemes. For example, if one wants to equip spaces of modular forms with $\mathbf{Z}$-structure in a manner that handles all primes in a unified manner, it can be simpler to work with moduli stacks of (generalized) elliptic curves rather than adjoining "extra level" and using fine moduli schemes. For example "$X_0(N)$" is always a regular proper Artin stack, whereas the coarse moduli scheme cousin is neither regular nor fine (but is also useful!). – BCnrd Oct 17 2010 at 1:09
I haven't met stacks often, but I used to think that they are tautological reformulations of: a) moduli problems b) quotient problems, that allow to "speak geometrically" (e.g. a moduli stack has a tangent bundle, a "moduli problem" doesn't) – Qfwfq Oct 17 2010 at 17:15
## 2 Answers
The classic application is Deligne & Mumford's paper proving the irreducibility of the coarse moduli scheme $\overline{M}_g$ of stable genus g curves over any algebraically closed field. They proved irreducibility first for the moduli stack $\overline{\mathcal{M}}_g$ of curves, and then inferred from this result the irreducibility of the scheme.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The DeRham space is a stack $X_{DR}$ associated to a smooth variety $X$, so that modules on $X_{DR}$ are D-modules on $X$. This is accomplished by declaring the maps from $Y$ into $X_{DR}$ are the same as maps from $Y^{red}$ (the reduced scheme) into $X$. This has the effect of identifying points with their infinitesmal neighborhoods.
The DeRham space is often most useful as a conceptual tool. However, a specific application of it was by Ben-Zvi and Nevins, who used it (and other tools) to show that certain cusped versions $\widetilde{X}$ of $X$ had equivalent categories of D-modules. The idea being, these cusps were identifying some of the infinitesmal neighborhoods of some of the points, and so they should be intermediate between a variety and its DeRham space.
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http://mathoverflow.net/questions/1887?sort=votes
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## Given a Spanning Tree and an Edge Not on the Spanning Tree, How to Form a Cycle Base?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a graph with Edge `E` and Vertex `V`, I can find the spanning tree using Kruskal algorithm, now I want to find all the cycle bases that are created by utilitizing that spanning tree and the edges that are not on the tree, any algorithm that allows me to do that, besides brute force search?
I can, of course, starts from one vertex of the non-spanning tree edge, gets all the edges, explore all of them, retracts if I find dead end, until I come back to the other vertex of the edge. But this is a bit, err... brutal. Any other ideas?
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You don't need Kruskal, Ngu! You can find a spanning tree trivially -- it doesn't have to be minimum. Newcomers, see mathoverflow.net/questions/1443/… for Ngu's initial question. – TonyK Oct 22 2009 at 18:15
Sure, I can find spanning tree-- with or without Kruskal. But my question is how to form the cycle base without resorting to something what I call 'brute force'. – Graviton Oct 23 2009 at 1:34
## 2 Answers
I think you're likely to get better answers if you post this question on Stack Overflow rather than here.
But anyways, if your graph doesn't have weights on edges, you don't need Kruskal's algorithm to find a spanning tree; you can just use DFS. As you compute the tree, store for each vertex its parent and the distance to the root. Then when you encounter a non-tree edge, you can find the nearest common ancestor of its two endpoints efficiently.
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Even with weights, you can use DFS to find a spanning tree. It might not have any nice properties, though. – Michael Lugo Oct 22 2009 at 18:03
>if your graph doesn't have weights on edges I agree, but But my question is how to form the cycle base without resorting to something what I call 'brute force'. – Graviton Oct 23 2009 at 1:34
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Although this thread is long-dead, there is a very simple way to do this with linear algebra, so I will post an answer.
You are interested in a basis of the cycle space of $G$. Orient $G$ in any way. Let $\partial(G)$ be the edge-vertex incidence matrix of $G$. A basis of the cycle space is given by a basis for $\ker \partial(G)$. To get the answer to your question note that if you take any spanning tree $T$ and you consider the edges of $E-T$, there will be a unique element of the cycle space that is $1$ on a given edge of $E-T$ and $0$ on the others. Finding such an element for each edge of $E-T$ will yield a new basis for $\ker \partial(G)$.
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http://johncarlosbaez.wordpress.com/2010/12/24/this-weeks-finds-week-308/
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Azimuth
This Week’s Finds (Week 308)
Last week we met the El Niño-Southern Oscillation, or ENSO. I like to explain things as I learn about them. So, often I look back and find my explanations naive. But this time it took less than a week!
What did it was reading this:
• J. D. Neelin, D. S. Battisti, A. C. Hirst et al., ENSO theory, J. Geophys. Res. 103 (1998), 14261-14290.
I wouldn’t recommend this to the faint of heart. It’s a bit terrifying. It’s well-written, but it tells the long and tangled tale of how theories of the ENSO phenomenon evolved from 1969 to 1998 — a period that saw much progress, but did not end with a neat, clean understanding of this phenomenon. It’s packed with hundreds of references, and sprinkled with somewhat intimidating remarks like:
The Fourier-decomposed longitude and time dependence of these eigensolutions obey dispersion relations familiar to every physical oceanographer…
Nonetheless I found it fascinating — so, I’ll pick off one small idea and explain it now.
As I’m sure you’ve heard, climate science involves some extremely complicated models: some of the most complex known to science. But it also involves models of lesser complexity, like the "box model" explained by Nathan Urban in "week304". And it also involves some extremely simple models that are designed to isolate some interesting phenomena and display them in their Platonic ideal form, stripped of all distractions.
Because of their simplicity, these models are great for mathematicians to think about: we can even prove theorems about them! And simplicity goes along with generality, so the simplest models of all tend to be applicable — in a rough way — not just to the Earth’s climate, but to a vast number of systems. They are, one might say, general possibilities of behavior.
Of course, we can’t expect simple models to describe complicated real-world situations very accurately. That’s not what they’re good for. So, even calling them "models" could be a bit misleading. It might be better to call them "patterns": patterns that can help organize our thinking about complex systems.
There’s a nice mathematical theory of these patterns… indeed, several such theories. But instead of taking a top-down approach, which gets a bit abstract, I’d rather tell you about some examples, which I can illustrate using pictures. But I didn’t make these pictures. They were created by Tim van Beek as part of the Azimuth Code Project. The Azimuth Code Project is a way for programmers to help save the planet. More about that later, at the end of this article.
As we saw last time, the ENSO cycle relies crucially on interactions between the ocean and atmosphere. In some models, we can artificially adjust the strength of these interactions, and we find something interesting. If we set the interaction strength to less than a certain amount, the Pacific Ocean will settle down to a stable equilibrium state. But when we turn it up past that point, we instead see periodic oscillations! Instead of a stable equilibrium state where nothing happens, we have a stable cycle.
This pattern, or at least one pattern of this sort, is called the "Hopf bifurcation". There are various differential equations that exhibit a Hopf bifurcation, but here’s my favorite:
$\frac{d x}{d t} = -y + \beta x - x (x^2 + y^2)$
$\frac{d y}{d t} = \; x + \beta y - y (x^2 + y^2)$
Here $x$ and $y$ are functions of time, $t$, so these equations describe a point moving around on the plane. It’s easier to see what’s going on in polar coordinates:
$\frac{d r}{d t} = \beta r - r^3$
$\frac{d \theta}{d t} = 1$
The angle $\theta$ goes around at a constant rate while the radius $r$ does something more interesting. When $\beta \le 0$, you can see that any solution spirals in towards the origin! Or, if it starts at the origin, it stays there. So, we call the origin a "stable equilibrium".
Here’s a typical solution for $\beta = -1/4$, drawn as a curve in the $x y$ plane. As time passes, the solution spirals in towards the origin:
The equations are more interesting for $\beta > 0$. Then $dr/dt = 0$ whenever
$\beta r - r^3 = 0$
This has two solutions, $r = 0$ and $r = \sqrt{\beta}$. Since $r = 0$ is a solution, the origin is still an equilibrium. But now it’s not stable: if $r$ is between $0$ and $\sqrt{\beta}$, we’ll have $\beta r - r^3 > 0$, so our solution will spiral out, away from the origin and towards the circle $r = \sqrt{\beta}$. So, we say the origin is an "unstable equilibrium". On the other hand, if $r$ starts out bigger than $\sqrt{\beta}$, our solution will spiral in towards that circle.
Here’s a picture of two solutions for $\beta = 1$:
The red solution starts near the origin and spirals out towards the circle $r = \sqrt{\beta}$. The green solution starts outside this circle and spirals in towards it, soon becoming indistinguishable from the circle itself. So, this equation describes a system where $x$ and $y$ quickly settle down to a periodic oscillating behavior.
Since solutions that start anywhere near the circle $r = \sqrt{\beta}$ will keep going round and round getting closer to this circle, it’s called a "stable limit cycle".
This is what the Hopf bifurcation is all about! We’ve got a dynamical system that depends on a parameter, and as we change this parameter, a stable fixed point become unstable, and a stable limit cycle forms around it.
This isn’t quite a mathematical definition yet, but it’s close enough for now. If you want something a bit more precise, try:
• Yuri A. Kuznetsov, Andronov-Hopf bifurcation, Scholarpedia, 2006.
Now, clearly the Hopf bifurcation idea is too simple for describing real-world weather cycles like the ENSO. In the Hopf bifurcation, our system settles down into an orbit very close to the limit cycle, which is perfectly periodic. The ENSO cycle is only roughly periodic:
The time between El Niños varies between 3 and 7 years, averaging around 4 years. There can also be two El Niños without an intervening La Niña, or vice versa. One can try to explain this in various ways.
One very simple, general idea to add random noise to whatever differential equation we were using to model the ENSO cycle, obtaining a so-called stochastic differential equation: a differential equation describing a random process. Richard Kleeman discusses this idea in Tim Palmer’s book:
• Richard Kleeman, Stochastic theories for the irregularity of ENSO, in Stochastic Physics and Climate Modelling, eds. Tim Palmer and Paul Williams, Cambridge U. Press, Cambridge, 2010, pp. 248-265.
Kleeman mentions three general theories for the irregularity of the ENSO. They all involve the idea of separating the weather into "modes" — roughly speaking, different ways that things can oscillate. Some modes are slow and some are fast. The ENSO cycle is defined by the behavior of certain slow modes, but of course these interact with the fast modes. So, there are various options:
1. Perhaps the relevant slow modes interact with each other in a chaotic way.
2. Perhaps the relevant slow modes interact with each other in a non-chaotic way, but also interact with chaotic fast modes, which inject noise into what would otherwise be simple periodic behavior.
3. Perhaps the relevant slow modes interact with each other in a chaotic way, and also interact in a significant way with chaotic fast modes.
Kleeman reviews work on the first option but focuses on the second. The third option is the most complicated, so the pessimist in me suspects that’s what’s really going on. Still, it’s good to start by studying simple models!
How can we get a simple model that illustrates the second option? Simple: take the model we just saw, and add some noise! This idea is discussed in detail here:
• H. A. Dijkstra, L. M. Frankcombe and A.S von der Heydt, The Atlantic Multidecadal Oscillation: a stochastic dynamical systems view, in Stochastic Physics and Climate Modelling, eds. Tim Palmer and Paul Williams, Cambridge U. Press, Cambridge, 2010, pp. 287-306.
This paper is not about the ENSO cycle, but another one, which is often nicknamed the AMO. I would love to talk about it — but not now. Let me just show you the equations for a Hopf bifurcation with noise:
$\frac{d x}{d t} = -y + \beta x - x (x^2 + y^2) + \lambda \frac{d W_1}{d t}$
$\frac{d y}{d t} = \; x + \beta y - y (x^2 + y^2) + \lambda \frac{d W_2}{d t}$
They’re the same as before, but with some new extra terms at the end: that’s the noise.
This could easily get a bit technical, but I don’t want it to. So, I’ll just say some buzzwords and let you click on the links if you want more detail. $W_1$ and $W_2$ are two independent Wiener processes, so they describe Brownian motion in the $x$ and $y$ coordinates. When we differentiate a Wiener process we get white noise. So, we’re adding some amount of white noise to the equations we had before, and the number $\lambda$ says precisely how much. That means that $x$ and $y$ are no longer specific functions of time: they’re random functions, also known as stochastic processes.
If this were a math course, I’d feel obliged to precisely define all the terms I just dropped on you. But it’s not, so I’ll just show you some pictures!
If $\beta = 1$ and $\lambda = 0.1$, here are some typical solutions:
They look similar to the solutions we saw before for $\beta = 1$, but now they have some random wiggles added on.
(You may be wondering what this picture really shows. After all, I said the solutions were random functions of time, not specific functions. But it’s tough to draw a "random function". So, to get one of the curves shown above, what Tim did is randomly choose a function according to some rule for computing probabilities, and draw that.)
If we turn up the noise, our solutions get more wiggly. If $\beta = 1$ and $\lambda = 0.3$, they look like this:
In these examples, $\beta > 0$, so we would have a limit cycle if there weren’t any noise — and you can see that even with noise, the solutions approximately tend towards the limit cycle. So, we can use an equation of this sort to describe systems that oscillate, but in a somewhat random way.
But now comes the really interesting part! Suppose $\beta \le 0$. Then we’ve seen that without noise, there’s no limit cycle: any solution quickly spirals in towards the origin. But with noise, something a bit different happens. If $\beta = -1/4$ and $\lambda = 0.1$ we get a picture like this:
We get irregular oscillations even though there’s no limit cycle! Roughly speaking, the noise keeps knocking the solution away from the stable fixed point at $x = y = 0$, so it keeps going round and round, but in an irregular way. It may seem to be spiralling in, but if we waited a bit longer it would get kicked out again.
This is a lot easier to see if we plot just $x$ as a function of $t$. Then we can run our solution for a longer time without the picture becoming a horrible mess:
If you compare this with the ENSO cycle, you’ll see they look roughly similar:
That’s nice. Of course it doesn’t prove that a model based on a Hopf bifurcation plus noise is "right" — indeed, we don’t really have a model until we’ve chosen variables for both $x$ and $y$. But it suggests that a model of this sort could be worth studying.
If you want to see how the Hopf bifurcation plus noise is applied to climate cycles, I suggest starting with the paper by Dijkstra, Frankcombe and von der Heydt. If you want to see it applied to the El Niño-Southern Oscillation, start with Section 6.3 of the ENSO theory paper, and then dig into the many references. Here it seems a model with $\beta > 0$ may work best. If so, noise is not required to keep the ENSO cycle going, but it makes the cycle irregular.
To a mathematician like me, what’s really interesting is how the addition of noise "smooths out" the Hopf bifurcation. When there’s no noise, the qualitative behavior of solutions jumps drastically at $\beta = 0$. For $\beta \le 0$ we have a stable equilibrium, while for $\beta > 0$ we have a stable limit cycle. But in the presence of noise, we get irregular cycles not only for $\beta > 0$ but also $\beta \le 0$. This is not really surprising, but it suggests a bunch of questions. Such as: what are some quantities we can use to describe the behavior of "irregular cycles", and how do these quantities change as a function of $\lambda$ and $\beta$?
You’ll see some answers to this question in Dijkstra, Frankcombe and von der Heydt’s paper. However, if you’re a mathematician, you’ll instantly think of dozens more questions — like, how can I prove what these guys are saying?
If you make any progress, let me know. If you don’t know where to start, you might try the Dijkstra et al. paper, and then learn a bit about the Hopf bifurcation, stochastic processes, and stochastic differential equations:
• John Guckenheimer and Philip Holmes, Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields, Springer, Berlin, 1983.
• Zdzisław Brzeźniak and Tomasz Zastawniak, Basic Stochastic Processes: A Course Through Exercises, Springer, Berlin, 1999.
• Bernt Øksendal, Stochastic Differential Equations: An Introduction with Applications, 6th edition, Springer, Berlin, 2003.
Now, about the Azimuth Code Project. Tim van Beek started it just recently, but the Azimuth Project seems to be attracting people who can program, so I have high hopes for it. Tim wrote:
My main objectives to start the Azimuth Code Project were:
• to have a central repository for the code used for simulations or data analysis on the Azimuth Project,
• to have an online free access repository and make all software open source, to enable anyone to use the software, for example to reproduce the results on the Azimuth Project. Also to show by example that this can and should be done for every scientific publication.
Of less importance is:
• to implement the software with an eye to software engineering principles.
This less important because the world of numerical high performance computing differs significantly from the rest of the software industry: it has special requirements and it is not clear at all which paradigms that are useful for the rest will turn out to be useful here. Nevertheless I’m confident that parts of the scientific community will profit from a closer interaction with software engineering.
So, if you like programming, I hope you’ll chat with us and consider joining in! Our next projects involve limit cycles in predator-prey models, stochastic resonance in some theories of the ice ages, and delay differential equations in ENSO models.
And in case you’re wondering, the code used for the pictures above is a simple implementation in Java of the Euler scheme, using random number generating algorithms from Numerical Recipes. Pictures were generated with gnuplot.
There are two ways of constructing a software design. One way is to make it so simple that there are obviously no deficiencies. And the other way is to make it so complicated that there are no obvious deficiencies. – C.A.R. Hoare
This entry was posted on Friday, December 24th, 2010 at 2:32 am and is filed under azimuth, climate, mathematics, this week's finds. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
48 Responses to This Week’s Finds (Week 308)
1. John Baez says:
By the way, people reading this blog might prefer the version of This Week’s Finds on my website. You can see “week308″ here, and latest edition can always be found here.
The version on my website doesn’t suffer from the “skinny column” format of the blog version — in particular, some of the pictures are bigger than 450 pixels across!
The equations also look better — at least to me, now that I’ve started using jsMath to render TeX on my website. If the math symbols on “week308″ look smaller than the other letters, let me know — and let me know what browswer you’re using. To me, using Firefox on Windows, they look just right, but apparently that varies from browser to browser. They’re easy to adjust at my end, but I may not be able to make everyone happy.
On my website, you may see an annoying red box at the top of “week308″ saying “no jsMath TeX fonts found”. This should not impede your ability to see the math symbols! The great thing about jsMath is that you don’t need to download any math symbol fonts. But the math symbols will show up faster, and maybe better, if you download the fonts. You can do this by clicking the little grey box saying “jsMath Control Panel” and then doing some (supposedly) obvious things.
• srp says:
Great post and a very worthy software project. Best of luck with it. BTW, this blog’s graphics look fine on Safari.
• John Baez says:
Thanks!
• some guy dotdotdot says:
needs your home directory, not just “home”!
2. John Baez says:
Tim van Beek notes that Terry Tao has a blog entry that explains Brownian motion in a rigorous way. This is a good way for mathematicians to get started on stochastic differential equations.
3. Arrow says:
I find the idea of using random noise to make a model of something quite absurd. A model should capture causal relations of the modeled phenomenon. A “model” which produces a similar output for completely different reasons then the ones governing the actual phenomenon is a failure.
If one thinks that there is some noise which for some reason is impossible to model then it would make much more sense to filter this noise out and then compare such filtered data to a model without any noise.
• srp says:
I’m not sure your precept is valid in general. If you believe your model has to have exogenous variables and you are trying to fit data where those exogenous variables arrive in an irregular and uncertain way, then you neither want to filter them or try to explain them (which would threaten an infinite regress to another set of exogenous variables).
It is true that introducing noise without a clear understanding of where it’s coming from (no specification of the exogenous variables) is less convincing, especially if it plays a major role in generating the structure of interest, as in some of the pictures above. But in that less-satisfying case, perhaps the results can be interpreted as a candidate pattern that any causal variable would have to match.
• John Baez says:
If a model increases our probability of making successful predictions, it’s a useful tool, regardless of its nature.
Scientists are in love with knowledge, and engaged in an all-out war against ignorance. All is fair in love and war.
Models that incorporate randomness are widely used in physics and engineering. For a brief history, try this.
In “week306″ I interviewed Tim Palmer, who is a big proponent of stochastic (i.e., probabilistic) methods in weather and climate modelling. He explained how thanks to the “real butterfly effect”, it’s impossible to make accurate deterministic weather predictions beyond a certain short threshold. Unmeasured wind velocities, temperatures and pressures at short distance scales are examples of the ‘exogenous variables’ srp mentioned. This umeasured information at short distance scales quickly percolates up to affect larger distance scales. But we can model this unmeasured information stochastically, by treating it as random noise.
In short: just because we can’t know things for sure, doesn’t mean we should throw in the towel. It can help a lot to know the probability it’s going to rain in a week.
For more on Palmer’s ideas, try this:
• Steve Easterbrook, AGU talk: Tim Palmer on Building Probabilistic Climate Models, Serendipity, 19 December 2010.
My plan is to explain a tiny bit of this book he helped edit:
• Tim Palmer and Paul Williams, editors, Stochastic Physics and Climate Modelling, Cambridge U. Press, Cambridge, 2010.
So, after an intro to El Niño in “week307″, this week I talked about stochastic models of the El Niño and the Atlantic Multidecadal Oscillation, based on these two papers:
• Richard Kleeman, Stochastic theories for the irregularity of ENSO, in Stochastic Physics and Climate Modelling, eds. Tim Palmer and Paul Williams, Cambridge U. Press, Cambridge, 2010, pp. 248-265.
• H. A. Dijkstra, L. M. Frankcombe and A. S. von der Heydt, The Atlantic Multidecadal Oscillation: a stochastic dynamical systems view, in Stochastic Physics and Climate Modelling, eds. Tim Palmer and Paul Williams, Cambridge U. Press, Cambridge, 2010, pp. 287-306.
Of course I only scratched the surface: the Hopf bifurcation plus noise is pretty, but far from the whole story. These papers are really fun to read, so I hope folks dig into them!
In future episodes I want to talk about glacial cycles, and how the noise from short-term weather may act to amplify the effects of cycles in the Earth’s orbit
(Milankovitch cycles), via a phenomenon called stochastic resonance. Luckily for me, Tim van Beek has been studying stochastic resonance and writing programs to show how it works.
But before I get to that, I’ll probably give another example of a Hopf bifurcation plus noise, developed by Graham Jones over on the Azimuth Project. I won’t give the link because I want it to be a surprise!
If anyone reading this likes to program, I hope you join the fun: drop us a line on the Azimuth Forum!
Merry Christmas, everyone. My wife and I going to Vietnam in 2 hours. We’ll spend seven days in Hanoi and five days in Huế, which was the imperial capital of the Nguyễn Dynasty until 1945. I’ll be back on January 5th. Be good while I’m gone.
• Tim van Beek says:
Happy Holidays!
Arrow said:
A model should capture causal relations of the modeled phenomenon. A “model” which produces a similar output for completely different reasons then the ones governing the actual phenomenon is a failure.
The actual phenomenon is governed by an interaction of a slowly varying, long memory phenomenon, namely oscillations of temperature and other variables of the Pacific Ocean, and very short term, no memory influences from the atmosphere. The noise models the latter.
If one thinks that there is some noise which for some reason is impossible to model then it would make much more sense to filter this noise out and then compare such filtered data to a model without any noise.
The “noise” is used to include effects into the model which would otherwise not be included, therefore making the model more accurate. However, models can have different uses, for example:
1) make predictions,
2) explain phenomena qualitatively,
3) as a null hypothesis for other, more sophisticaed models.
The model in week 308 cannot be used to make predictions: to predict the next ENSO cycle one needs more sophisticated models that are closer to a full model using Navier-Stokes equations with real world data as initial and boundary data. But I think it can be useful for 2) and 3).
But: Note that “noise” can be and has been used very successfully for models that make very accurate predictions, examples can be found in financial mathematics (Black-Scholes formula), but also in engineering, for example in control theory.
There is a lot of literature out there, simply look for the keyword “stochastic control theory”.
• DavidTweed says:
Note that something subtle is going on: adding noise is using a model, it’s just that it’s a very weak model which where we’ve decided (due to either ignorance of the details, inability to measure inputs to the details, computational intractability of the details or some combination thereof) is that all the model tells us is the probabilistic distribution of possibilities. Using those to model small residuals after dominant contributions have been modeled more closely is typical of modeling, and use of models in science. Have you ever tried to come up with a measure of the coefficient of friction between two surfaces directly without measuring the atomic level details and computing all the interactions? That’s the same thing, except in this case — as with most cases in typical simple physics — the probability distribution is very, very tightly peaked so you can neglect the probabilistic dimension. What’s different here is that (analogies to) very complex, huge-number-of-component systems are being considered where the system behaves very non-linearly to small perturbations.
• John Baez says:
David wrote:
Note that something subtle is going on: adding noise is using a model…
Right: and this is very easy to see by noticing that there are infinitely many different kinds of noise, so building a model involving noise requires that you make assumptions about the nature of the noise involved.
In the simple “Hopf bifurcation plus noise” model described in this blog entry, we’re adding independent identically distributed white noise terms to the formulas for the time derivatives of both variables (x and y). This has the mathematical advantage of being as simple as possible, and not destroying the circular symmetry of the problem. But in any serious attempt to model a specific real-world phenomenon, one should try to make the noise realistic, not just mathematically elegant.
For example, in weather simulation, where the noise might model the effects of turbulence at length scales not included in the model, one would want to think hard about how turbulence actually works. Luckily, people have been thinking about this ever since Kolmogorov wrote his amazing paper back in 1941. So, a lot of the hard thinking has already been done.
• John F says:
Although I never worked in weather, the same issues affect any modeling w/noise. The unmeasured stuff, the residuals, the fast variables, uncertainty, the details of how the noise is designated is not very important. What is important is that what is being modeled be correspondable to what you want you want it to model! For instance, you wouldn’t want to model an audio amplifer as being only fully off (too quiet to hear) and wide open (too loud to hear). For one reason, because then what should be actual sound is instead noise in that model. In the context of control theory this modelability is controllability
http://en.wikipedia.org/wiki/Controllability
Anyway, in almost all applications (such as finance – where are those finance guys when you want them?) you have to use past measured behavior to predict future behavior, especially to make realistic noise aspects. Because you almost never have a clear idea of where the noise is coming from or what it’s properties should be, especially apart from those measurements.
What exactly do they use for the noise terms in weather forecasting? Are there trade secrets?
• Tim van Beek says:
What exactly do they use for the noise terms in weather forecasting? Are there trade secrets?
So far I have not discovered a weather model using stochastic elements. People include subgrid phenomena via so called “parameterizations”, which are heuristic deterministic influences on each grid cell based on an estimation of concrete physical effects. They then use “ensemble runs” which means running the models multiple times with different parameterizations and different initial data, to find out which predictions of the models are stable and which are not.
This is significantly different from having a stochastic effect on every time step of the model, like one has in stochastic partial or ordinary differential equations.
Stochastic equations are used with great success in engineering, for example for controlling the flow of fluids through pipes, as an approximation to the unaccessible solution of Navier-Stokes equations. So, IMHO, this could turn out to be of use for weather and climate models, too, but if I’m correct there would be the need for a minor paradigm change in the weather modeling community.
• John F says:
Tim,
I’m surprised that weather models are only studied via sensitivity analyses
http://en.wikipedia.org/wiki/Sensitivity_analysis
without stochastic parameters. It seems blind to me.
4. Tim van Beek says:
John wrote:
To a mathematician like me, what’s really interesting is how the addition of noise “smooths out” the Hopf bifurcation.
The topic of “qualitative analysis of differential equations” is now over 100 years old, it was started by Henri Poincaré, and a lot of concepts bear his name, like Poincaré map. Of course, Poincaré could not resort to computer approximations to get an idea how solutions to certain differential equations look like. But although we have powerful computers today, the topic itself is still very important to supplement and cross check numerical solutions.
The topic of qualitative analysis of stochastic differential equations however lies idle, as far as I know (I don’t know of any interesting results or methods). The example in week 308 shows that there are additional phenomena that have to be investigated and taken care of, the influence of noise clearly adds a layer of complexity. Equillibrium methods, like the evolution of probability distributions described by Fokker-Planck equations, are of limited use here, since we are interested in the qualitative behaviour of single sample paths…
• Phil Henshaw says:
Would it count for what you’re looking for, as “interesting results or methods”, to find a result that energy conservation implies that the stochastic equations of natural processes need to incorporate finite developmental periods displaying accumulative proportional change?
Somebody’s going to get it some day, why not you guys??
http://www.synapse9.com/drtheo.pdf
5. srp says:
I own a wonderful desk toy called the Randomly Oscillating Magnetic Pendulum (ROMP). It gives some intuitive feel for how a simple system can exhibit limit points and cycles and can abruptly switch from one to another “for no good reason.” You can rearrange the magnets into any configuration on the plate below the pendulum, causing a new set of complex behaviors. Entertaining, enlightening, and slightly hypnotic.
6. Giampiero Campa says:
I think Simulink is the ideal tool to do these kind of simulations, however, this is the quick and dirty Matlab code to do this, if anyone wants to try:
beta=1;lambda=0.1;
dv=@(t,v) [-v(2);v(1)]+beta*v-v’*v*v+lambda*randn(2,1);
[t,v]=ode45(dv,[0 10],[0 2]);
plot(v(:,1),v(:,2));axis([-2 2 -1.5 2])
The first line sets up the parameters, the second sets up the dynamical system (where v is the vector having x and y as components), the third solves the ode from 0 to 10 seconds, with an initial condition of [0 2], and the fourth line plots the results (use plot(t,v(:,1)) to plot the first component versus time).
The plots are smoother than the one shown above, perhaps because I have assumed that the variance of the white noise is 1, and maybe a larger one was assumed above. Multiplying lambda by 10 produces very similar plots.
• Tim van Beek says:
Interesting!
The main reason why I did not use Matlab is that it is not free and I don’t have a license – I know that most universities have one, but I’m not affiliated with any university :-)
Besides that, Matlab is not open source, so I wonder
a) what algorithm does it use to generate random numbers? Does it always use the same seed i.e. does it always reproduce the same graphs?
b) what algorithm does it use to solve the SDE?
c) do different versions of Matlab produce the same graphics? (I doubt that, to be honest).
The plots are smoother than the one shown above, perhaps because I have assumed that the variance of the white noise is 1, and maybe a larger one was assumed above.
No, the variance of white noise is not a free parameter in an Itô stochastic differential equation driven by Brownian motion, which makes me wonder why you have to specify it at all and to what effect.
The graphics above were produced with the explicit Euler scheme with a step size of 0.001. Therefore, in every timestep one has to add noise in the form of a random variable drawn from $\lambda N(0, \sqrt(0.001))$, because the increments of Brownian motion $W_{t + h} - W_{t}$ are Gaussian with mean zero and variance h.
• Tim van Beek says:
Sorry, it should be:
in every timestep one has to add noise in the form of a random variable drawn from $N(0, 0.001)$ multiplied by $\lambda$.
• Giampiero Campa says:
Tim, are you sure about this ? If the variance is 0.001 and lambda is 0.1 then the noise should be pretty much non-existent, and the plots would be really smooth.
Maybe you meant that the variance should be 1/sqrt(0.001) ? (and in fact i get plots that are really similar to yours in that case).
In any case i have checked also with forward euler and results are essentially the same, so it comes down entirely to the noise level.
Let me know if you want me to send you the time histories for comparison.
• Tim van Beek says:
Giampiero said:
If the variance is 0.001 and lambda is 0.1 then the noise should be pretty much non-existent, and the plots would be really smooth.
It is important to keep in mind that the Euler scheme is a discrete approximation with a fixed stepsize h, and when we simulate a sample path we have to add noise at every step with variance h. So if the stepsize is rather small (like 0.001 in our sample) the variance of the noise added at each step is also very small. But the effect adds up!
No, this is of course an important question! I don’t trust my intution to tell me what a certain noise level should look like – in fact I should have written some tests before the publication of week 308!
A good test for the noise level is this: It is possible to formulate the Fokker-Planck equation for our example in polar coordinates and find the stationary, rotationally invariant solution which describes the distribution of r, the radius, in the equillibrium. This can be calculated in closed form.
Then it is possible to create a histogram approximation of the distribution of r by sampling a very long run of the SDE solver, and compare this to the solution obtained from the Fokker-Planck equation.
I did this, and the results match! so, yes, know I’m pretty confident that the SDE solver for week 308 gets the noise right.
I will explain all of this in more detail on the Azimuth project once I have more time (maybe tomorrow).
Thanks for your help and for getting me to finally do these chross-checks!
• Giampiero Campa says:
Hi Tim, yes of course h multiplies everything on the right hand side if you use Euler approx. But i am talking about the variance of the noise in the equations given by John.
In other words, let’s suppose the vector [x y] is [1 1] and have a look at the derivatives:
The first term is [-1 1], the second is [1 1], and the third is -[2 2]. Now, here is the important part, we are adding a noise here and I want to be sure about the variance of the noise that we are adding. If you really add a noise of variance 0.001 and multiply it by 0.1 here in this equation (and i don’t think you meant to add it here), you are adding something that is 3 to 4 orders of magnitudes below the other terms so you will hardly see anything !
I don’t think there is a bug anywhere, i just think we are using different levels of noise. If you tell me the variance of noise that is supposed to be added to those equations (not to their Euler approx) then I’ll redo the plots and upload them to you so we will start comparing apples to apples :)
• Tim van Beek says:
Giampiero said:
I don’t think there is a bug anywhere…
I don’t think that there is a bug with the noise level, too, because the SDE solver reproduces the distribution of r pretty good :-)
In other words, let’s suppose the vector [x y] is [1 1] and have a look at the derivatives
I’m sorry, but I don’t understand you: Is [1 1] the starting point of the process and you calculate the derivatives?
…but I am talking about the variance of the noise in the equations given by John.
The only free parameters are $\beta$ and $\lambda$. As you can see on the page stochastic differential equation, the Euler scheme is a simple generalization of the classical deterministic Euler scheme, in every step the position of the process is changed according to the deterministic part + noise.
Once a step size h is fixed, the increment $W_{t+h} - W_t$ has a fixed distribution, namely \$N(0, h)\$. There is no further choice involved here.
(Higher order schemes like Runge-Kutta schemes use higher terms from the stochastic Taylor series, which involve iterated integrals of Brownian motion, which are a little bit more complicated to simulate, so it is not true that all discrete approximations use the simple increment of Brownian motion only.)
The deterministic part gets multiplied by h, where the noise has variance h. This scaling behaviour is the reason why Brownian motion has a.s. sample paths that are not of bounded variation, so that the Riemann-Stieltjes integral with respect to the sample paths does not converge a.s. This is the reason why there is a stochastic calculus at all :-)
But when we compare the influence of the deterministic path to the influence of additive noise in the Euler scheme, we have to keep in mind that the influence of the deterministic part gets multiplied by h, ergo it is not true that the noise term is negligible.
• Giampiero Campa says:
So the same code works fine in octave, (you just have to download ode45.m which you can just google for, you’ll find several open sourced versions, there is one at the university of Boston. There is also an ode78.m which implements Runge-Kutta with 7th order formulas).
http://www.mathworks.com/help/techdoc/ref/randn.html
(yes you can reset the seed to repeat the same sequence).
By the way if you specifically want an sde solver there is an apposite sde toolbox here (free):
http://sdetoolbox.sourceforge.net/
(also, i think there are specific SDE objects and solvers in the econometrics toolbox, but i have never used them).
I have tried a couple of versions before 2008 and the graphs are identical, with R2010a it is slightly different, i think something changed in the number generator around that time.
• Giampiero Campa says:
Tim, I also meant to say that if you are using a simple forward Euler integration it’s already a big difference right there, obviously.
• Tim van Beek says:
Alright, I finally took the time to look at the Matlab script, here are my thoughts:
1. Matlab does not provide any SDE solvers, at least I did not find any description of one,
2. ode45 is a ODE solver, it chooses it’s stepsize dynamically,
3. it would seem that the line
“dv=@(t,v) [-v(2);v(1)]+beta*v-v’*v*v+lambda*randn(2,1)”
gets Matlab to add random numbers at every step that are distributed according to N(0, 1).
Now, with some luck ode45 will use a constant stepsize h, in this case you’ll get sample paths that look similar to those of TWF if \$\lambda*h*N(0,1)\$ looks close enough to \$\lambda*N(0, h)\$. But this is more of an accident if the stepsize \$h\$ is much smaller than the resolution of the graphic. Although ode45 chose 1.0 as the default initial stepsize, I suspect that it will calculate a much smaller one and use this according to the default error bounds, but I don’t know how to find out which one it actually uses.
If I’m correct then this is a very good example that one needs more than simple intuition to check if an SDE solver is correct :-)
• Giampiero Campa says:
Ok, got it! dW/dt is approximated by (W(t+h)-W(t))/h, so you have a noise with variance 0.001 that _is divided by h_.
Indeed using the line dv=@(t,v) [-v(2);v(1)]+beta*v-v’*v*v+lambda*sqrt(0.001)*randn(2,1)/0.001;gives very similar plots to yours, especially when you fix the step size to 0.001.
To double check, I have also tried a simple euler approximation, using the following 4 lines, which can be copied and pasted directly in Octave or Matlab, and don’t need any ode solver (and besides execute very fast):
beta=1;lambda=0.1;t=[0:0.001:10];z=[0 2]‘*t;z(:,1)=[0 2]‘;
dv=@(t,v) 0.001*([-v(2);v(1)]+beta*v-v’*v*v)+lambda*sqrt(0.001)*randn(2,1);
for i=1:length(t)-1,z(:,i+1)=z(:,i)+dv(t(i),z(:,i));end
plot(z(1,:),z(2,:));axis([-2 2 -1.5 2])
This also gives plots that are basically identical to yours, so problem solved.
Thanks Tim !
• Tim van Beek says:
Yes, that’s right :-)
There should be also several additional packages for Matlab that are not part of the core with SDE solvers, also with adaptive step size control – at least such algorithms exist. Of course the noise level has to be adapted to the chosen step size at each step.
• John F says:
Matlab fan here. I also constantly write numerical routines in Basic, Fortran, C, and IDL, but (except for OPC – Other People’s Code) I always develop numerical analysis algorithms in Matlab first.
By the way, almost all Matlab packages are just someone’s collections of ordinary Matlab routines, so you probably don’t need them or even want to have to learn them unless you’re going to be mostly focused on something specialized, like say Computational Biology.
• Tim van Beek says:
John F said:
By the way, almost all Matlab packages are just someone’s collections of ordinary Matlab routines, so you probably don’t need them or even want to have to learn them unless you’re going to be mostly focused on something specialized, like say Computational Biology.
I’m trying to learn a little bit more about climate models. I understand that Matlab is a good tool to quickly code a simple algorithm and play with it. Climate models need an effort of several man decades. I doubt that Matlab is suitable to handle models that are comprised of millions of lines of code and are developed by hundreds of people over several years. On the other hand, this is quite usual for applications using either the Java or the .NET tool stack.
• Tim van Beek says:
This is the first bug report for the Azimuth Code Project, so I created an issue with the issue tracker over at google projects here.
Thanks to Giampiero: It’s very important that we cross-check our results and fix all bugs before they have any impact.
(I’m pretty sure that the pictures give a qualitative correct impression of the behaviour of the solutions, even if it turns out that there is some scaling error in the noise level – so such an error would not be that bad now. It would be very bad if we used the code to calculate e.g. crossing times for pricing options :-).
• John Baez says:
Hi, Tim! I’m in Vietnam, which is a fascinating place, so I don’t want to spend spend much time blogging, but if you could fix the the pictures for “week308″, that would be great. That way I can put up fixed versions when I get back.
• Tim van Beek says:
Sure, no problem :-)
Every report of unusual behaviour should result in a bug report, or an issue (these are synonymous, especially if one uses the bug tracker of google project). This does not imply that there actually is a bug in the software.
In this case I became suspicious that I maybe did something wrong in the random number generator when I translated it from C++ to Java, so I wrote a chi-square-test for it – which it passed, so now I’m pretty sure that there isn’t a bug.
Nevertheless I’ll calculate the Fokker-Planck equation in polar coordinates and compare the stationary solution to the approximation I get from my SDE solver – this is another sanity check (I’ll try to get this done before new year’s eve, since I won’t have any time for that afterwardts :-(.
But right now I’m more supicious about what Matlab actually does when fed with the lines that Giampiero posted and could some help with that, since I don’t have and don’t know Matlab.
7. Tim van Beek says:
Giampiero has updated the issue over at the Azimuth Code Project, I need to take a closer look at it – meanwhile I had enough time to publish my plausibility test of the noise level of the SDE solver over at the Azimuth project:
• Fokker-Planck equation, look for “Hopf-Bifurcation Example of TWF 308″.
The distribution of r is very sensitive to the noise level λ, so I find it hard to believe that a rather exact match of both calculated and simulated values as displayed in the graphic over there is an accident.
8. Tim van Beek says:
I just noted a typo, it should be:
Bernt Øksendal
His book contains many quite successful applications of stochastic differential equations to convince the sceptics that this topic is worthwhile to study :-)
• John Baez says:
Hi, I’m back in Singapore. I fixed this typo — thanks for catching it.
9. Chris Goddard says:
This post had me thinking about catastrophes. Interestingly, there were 7 types discussed in Thom’s investigations (including the “butterfly catastrophe”), which suggests to me 3-cat level complexity. If viewed with minimal structure & in a box model context, I believe this could be a fruitful direction of further work in this instance.
• John Baez says:
Hi, Chris! I’ve tried to fix the html in your post. To make a link in a comment on this blog, you should type something like this:
<a href = “http://math.ucr.edu/home/baez/music/”>John Baez’s music</a>
which produces this result:
John Baez’s music
You wrote:
This post had me thinking about catastrophes.
Indeed, I was wanting to say a lot more about catastrophe theory, but I restrained myself because I thought it would be too distracting. I decided it would be better to see some examples of catastrophe theory in climate science before chatting about the mathematical subject in its abstract glory!
This week we saw the Hopf bifurcation in the context of the El Niño-Southern Oscillation. Next week we’ll see it in the context of predator-prey models. Later we’ll see the fold catastrophe in the context of ‘Snowball Earth’ and the glacial cycles. But if you’re impatient, you can see the basic idea now:
• Azimuth Project, Snowball Earth.
In all cases these catastrophes display some extra interesting behavior when we add noise.
And yes, I’m very interested in thinking about them in terms of box models, because box models are really just applied category theory.
I’ll probably spend a year or two writing some papers that make the last sentence more precise. I’m eager to bring some category-theoretic tools into climate physics, systems ecology and the like. I don’t want to get too sidetracked by this theoretical desire — I doubt it will ‘help save the planet’ very much — but I do need to keep publishing math papers, and this is one obvious way.
10. Chris Goddard says:
Hi John,
Thanks for fix – this wordpress stuff is tricky business =|
Incidentally, related to your aside, procedural generation of music is something that I’d like to learn how to do at some stage (and the implementation of procedural techniques in general).
But I suspect that that would be sidetracking this discussion just a bit!
11. Joris Vankerschaver says:
Nice post! I never thought adding noise could have this effect.
Just for fun I create an interactive Sage version of this phenomenon: you can play with the model at http://www.sagenb.org/home/pub/2630.
On a side-note, I think that Sage fits in very well with the Azimuth philosophy:
It is open source (it suffices to add ?? to any method name to view the implementation) and actively maintained. It doesn’t aim to reinvent the wheel but instead incorporates GAP, Maxima, scipy, … into one consistent interface.
It’s written in Python, which is easy to learn and (thanks to cython) can be as fast as compiled C. For instance, the whole interactive demo is no more than 10 lines.
Code that goes into Sage is peer reviewed and unit tested, so you don’t need to worry about stress-testing the code (though it is always good to be cautious).
• John Baez says:
Joris wrote:
Just for fun I created an interactive Sage version of this phenomenon: you can play with the model at http://www.sagenb.org/home/pub/2630.
For some reason that doesn’t work for me. I see some sliders at the bottom, and I can slide them, but I don’t see a plot of the solution.
Also, for some mysterious reason the jsmath doesn’t work for me either.
Does this stuff work for anyone else reading this? Maybe it’s just me.
• Nathan Urban says:
I have the same problems as John (using Mac Firefox 3.6.1 or Safari 5.0.3).
• Tim van Beek says:
Unfortunately it does not work for me, either, I have the same problems as John.
Over at the Azimuth forum Staffan Liljegren already pointed out Sage but I haven’t taken a closer look at it yet. The motivation of the authors of Sage is very similar to the motivation of the Azimuth project, so I do hope that we can use and maybe extend Sage for our purposes.
• Joris Vankerschaver says:
Thanks for pointing this out! It seems that we have stumbled upon a limitation within Sage: apparently, to see the interactive elements in a worksheet one needs to be logged in. Rather than asking you all to create an account with Sage, I noticed that there are some patches lying around to enable this, and once I get back from this conference I’m attending, I’ll try to set up my own server running this worksheet.
Sorry for this unexpected setback! You can still see the niftyness by logging in though…
12. peter waaben says:
The Poincaré–Bendixson theorem states that any orbit that stays in a compact region of the state space of a 2-dimensional planar continuous dynamical system, all of whose fixed points are isolated, approaches its ω-limit set, which is either a fixed point, a periodic orbit, or is a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these. Thus chaotic behaviour can only arise in continuous dynamical systems whose phase space has three or more dimensions. However the theorem does not apply to discrete dynamical systems, where chaotic behaviour can arise in two or even one dimensional systems.
A weaker version of the theorem was originally conceived by Henri Poincaré, although he lacked a complete proof. Ivar Bendixson (1901) gave a rigorous proof of the full theorem.
I.e., this is a proof that discrete and continous systems are not-isomorphic thus care seems appropriate in the development of models etc.
13. [...] Software for investigating the Hopf bifurcation and its stochastic version: see week308 of This Week’s [...]
14. In This Weeks Finds, back in ‘week308′, we had a look at a model with ‘noise’. A lot of systems can be described by ordinary [...]
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|
http://mathematica.stackexchange.com/questions/tagged/list-manipulation?sort=active&pagesize=50
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http://mathoverflow.net/revisions/71522/list
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## Return to Answer
3 added 20 characters in body
Well, the thing that may or may not be a "real functor" (and which may even fail to exist if the limit(/colimit) does not always exist) is in any case a "profunctor" (that is, a functor into $Set^{C^{op}}$ (or $Set^C$ for colimits) rather than into $C$). The limit of a diagram will actually exist just in case the profunctor's value at that diagram is a representable presheaf (that is, one in the range (up to isomorphism) of the Yoneda embedding). If one makes a choice of such a representation at every diagram, one can factor the entire profunctor through the Yoneda embedding, into a genuine functor. This of course is precisely the choice you want to avoid, but it indicates that one can at least still treat the profunctor as an "anafunctor" in such cases (essentially, a functor whose value at an object/morphism is only determined up to isomorphism, in a coherent way). Further reading on profunctors and anafunctors (for example, at the nLab) may be precisely the sort of thing you are looking for.
In short: profunctors are the way to describe adjoints which may exist only partially, while anafunctors are the way to describe functors whose construction requires a number of arbitrary choices (with anafunctors both avoiding the "evil" in making any single choice and the need for the Axiom of Choice in making so many of them).
2 added 4 characters in body
Well, the thing that may or may not be a "real functor" (and which may even fail to exist if the limit(/colimit) does not always exist) is in any case a "profunctor" (that is, a functor into $Set^{C^{op}}$ (or $Set^C$ for colimits) rather than into $C$). The limit of a diagram will actually exist just in case the profunctor's value at that diagram is a representable presheaf (that is, one in the range of the Yoneda embedding). If one makes a choice of such a representation at every diagram, one can factor the entire profunctor through the Yoneda embedding, into a genuine functor. This of course is precisely the choice you want to avoid, but it indicates that one can at least still treat the profunctor as an "anafunctor" in such cases (essentially, a functor whose value at an object/morphism is only determined up to isomorphism, in a coherent way). Further reading on profunctors and anafunctors (for example, at the nLab) may be precisely the sort of thing you are looking for.
In short: profunctors are the way to describe adjoints which may or may not exist only partially, while anafunctors are the way to describe functors whose construction requires a number of arbitrary choices (with anafunctors both avoiding the "evil" in making any single choice and the need for the Axiom of Choice in making so many of them).
1
Well, the thing that may or may not be a "real functor" (and which may even fail to exist if the limit(/colimit) does not always exist) is in any case a "profunctor" (that is, a functor into $Set^{C^{op}}$ (or $Set^C$ for colimits) rather than into $C$). The limit of a diagram will actually exist just in case the profunctor's value at that diagram is a representable presheaf (that is, one in the range of the Yoneda embedding). If one makes a choice of such a representation at every diagram, one can factor the entire profunctor through the Yoneda embedding, into a genuine functor. This of course is precisely the choice you want to avoid, but it indicates that one can at least still treat the profunctor as an "anafunctor" in such cases (essentially, a functor whose value at an object/morphism is only determined up to isomorphism, in a coherent way). Further reading on profunctors and anafunctors (for example, at the nLab) may be precisely the sort of thing you are looking for.
In short: profunctors are the way to describe adjoints which may or may not exist, while anafunctors are the way to describe functors whose construction requires a number of arbitrary choices (with anafunctors both avoiding the "evil" in making any single choice and the need for the Axiom of Choice in making so many of them).
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http://mathoverflow.net/questions/15755?sort=newest
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## Weak partitioning vs. strong partitioning
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $U$ is a complete lattice with least element 0.
Weak partitioning is a collection $S$ of nonempty subsets of $U$ such that $\forall x\in S: x\cap\bigcup(S\setminus\{x\})=0$.
Strong partitioning is a collection $S$ of nonempty subsets of $U$ such that $\forall A,B\in PS:(A\cap B=\emptyset \Rightarrow \bigcup A\cap\bigcup B=0)$.
Easy to show that every strong partitioning is weak partitioning.
Is weak and strong partitioning the same?
If not, under which additional conditions these are the same?
-
Is U a complete lattice or a complete lattice of sets? (In the former case, changing your cups and caps with vees and wedges where appropriate would make the question easier to parse.) The answer to your question hinges on the validity of some distributive laws, these do hold in lattices of sets but not necessarily in general lattices. – François G. Dorais♦ Feb 18 2010 at 21:08
## 2 Answers
If the lattice $U$ satisfies the meet distributive law $$x \wedge \bigvee_{i \in I} y_i = \bigvee_{i \in I} x \wedge y_i$$ where $(y_i)_{i \in I}$ is an arbitrary collection of elements of $U$, then "weak partitioning" implies "strong partitioning." More precisely, you only need the above to hold when the right hand side is $0$.
An example of a complete lattice where weak and strong partitioning are inequivalent is the lattice $U$ consisting of all closed subsets of `$\{1,\frac12,\frac13,\ldots,0\}$` (as a subspace of $\mathbb{R}$) and the collection `$S = \{\{\frac1n\}: n \geq 1\}$`. The weak-partitioning property is easily verified since the points $\frac1n$ are isolated. The strong partitioning property fails for the two sets `$A = \{\{\frac1{2n}\}: n \geq 1\}$` and `$B = \{\{\frac1{2n+1}\} : n \geq 0\}$`, for example, since $\bigvee A = \overline{\bigcup A}$ and $\bigvee B = \overline{\bigcup B}$ both contain the point $0$.
PS: In your formulation of weak and strong partitioning, I interpret $S$ as a collection of nonzero elements of $U$, since "nonempty subsets" doesn't make much sense in context.
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I probably should have mentioned that complete lattices which satisfy the meet distributive law mentioned above are called frames or complete Heyting algebras. en.wikipedia.org/wiki/Pointless_topology en.wikipedia.org/wiki/Complete_Heyting_algebra – François G. Dorais♦ Feb 18 2010 at 23:54
Sorry, I misunderstand something. First: all elements of $U$ except $S$ are subsets of $\mathbb{R}$ but $S$ is not a subset of $\mathbb{R}$. How this may be? Second: What are infimuma and suprema, are these infimuma and suprema on the induced order? Are infima and suprema opposite to these defined by the induced order? – porton Feb 19 2010 at 18:19
S is a subset of U. (Please see the PS and correct your question to reflect what you wanted to say.) Thes infs and sups of the second paragraph are computed in the complete lattice of closed subsets of the given subspace of R. – François G. Dorais♦ Feb 19 2010 at 18:26
I again don't understand. Why we introduce $S$? Elements $\frac1n$ of $S$ are anyway closed subsets of $\{1,\frac12,\frac13,\ldots,0\}$ and so we have them in $U$. Why we need $S$? – porton Feb 19 2010 at 19:04
The set S is in your question. – François G. Dorais♦ Feb 19 2010 at 19:23
show 2 more comments
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I wrote more detailed proof based on the counterexample by François G. Dorais in this online article.
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http://physics.stackexchange.com/questions/tagged/uncertainty-principle+commutator
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# Tagged Questions
4answers
286 views
### Does uncertainty imply noncommutativity?
We already know that non-commutativity of observables leads to uncertainty in quantum mechanics cf. e.g. this and this Phys.SE post. What about the opposite: Does uncertainty imply noncommutativity? ...
3answers
1k views
### Proof of Canonical Commutation Relation (CCR)
I am not sure how $QP-PQ =i\hbar$ where $P$ represent momentum and $Q$ represent position. $Q$ and $P$ are matrices. The question would be, how can $Q$ and $P$ be formulated as a matrix? Also, what is ...
2answers
160 views
### Why does $i ( LK-KL )$ represent a real quantity?
According to my textbook, it says that $i( LK-KL )$ represents a real quantity when $K$ and $L$ represent a real quantity. $K$ and $L$ are matrices. It says that this is because of basic rules. ...
3answers
348 views
### Generalizing Heisenberg Uncertainty Priniciple
Writing the relationship between canonical momenta $\pi _i$ and canonical coordinates $x_i$ $$\pi _i =\text{ }\frac{\partial \mathcal{L}}{\partial \left(\frac{\partial x_i}{\partial t}\right)}$$ ...
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http://en.wikipedia.org/wiki/Dirac_delta_function
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# Dirac delta function
"Delta function" redirects here. For other uses, see Delta function (disambiguation).
Schematic representation of the Dirac delta function by a line surmounted by an arrow. The height of the arrow is usually used to specify the value of any multiplicative constant, which will give the area under the function. The other convention is to write the area next to the arrowhead.
The Dirac delta function as the limit (in the sense of distributions) of the sequence of zero-centered normal distributions $\delta_a(x) = \frac{1}{a \sqrt{\pi}} \mathrm{e}^{-x^2/a^2}$ as $a \rightarrow 0$.
In mathematics, the Dirac delta function, or δ function, is (informally) a generalized function on the real number line that is zero everywhere except at zero, with an integral of one over the entire real line.[1][2][3] The delta function is sometimes thought of as an infinitely high, infinitely thin spike at the origin, with total area one under the spike, and physically represents an idealized point mass or point charge.[4] It was introduced by theoretical physicist Paul Dirac. Dirac explicitly spoke of infinitely great values of his integrand.[5] In the context of signal processing it is often referred to as the unit impulse symbol (or function).[6] Its discrete analog is the Kronecker delta function which is usually defined on a finite domain and takes values 0 and 1.
From a purely mathematical viewpoint, the Dirac delta is not strictly a function, because any extended-real function that is equal to zero everywhere but a single point must have total integral zero.[7] The delta function only makes sense as a mathematical object when it appears inside an integral. While from this perspective the Dirac delta can usually be manipulated as though it were a function, formally it must be defined as a distribution that is also a measure. In many applications, the Dirac delta is regarded as a kind of limit (a weak limit) of a sequence of functions having a tall spike at the origin. The approximating functions of the sequence are thus "approximate" or "nascent" delta functions.
## Overview
The graph of the delta function is usually thought of as following the whole x-axis and the positive y-axis. Despite its name, the delta function is not truly a function, at least not a usual one with range in real numbers. For example, the objects f(x) = δ(x) and g(x) = 0 are equal everywhere except at x = 0 yet have integrals that are different. According to Lebesgue integration theory, if f and g are functions such that f = g almost everywhere, then f is integrable if and only if g is integrable and the integrals of f and g are identical. Rigorous treatment of the Dirac delta requires measure theory or the theory of distributions.
The Dirac delta is used to model a tall narrow spike function (an impulse), and other similar abstractions such as a point charge, point mass or electron point. For example, to calculate the dynamics of a baseball being hit by a bat, one can approximate the force of the bat hitting the baseball by a delta function. In doing so, one not only simplifies the equations, but one also is able to calculate the motion of the baseball by only considering the total impulse of the bat against the ball rather than requiring knowledge of the details of how the bat transferred energy to the ball.
In applied mathematics, the delta function is often manipulated as a kind of limit (a weak limit) of a sequence of functions, each member of which has a tall spike at the origin: for example, a sequence of Gaussian distributions centered at the origin with variance tending to zero.
## History
Joseph Fourier presented what is now called the Fourier integral theorem in his treatise Théorie analytique de la chaleur in the form:[8]
$f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty\ \ d\alpha f(\alpha) \ \int_{-\infty}^\infty dp\ \cos (px-p\alpha)\ ,$
which is tantamount to the introduction of the δ-function in the form:[9]
$\delta(x-\alpha)=\frac{1}{2\pi} \int_{-\infty}^\infty dp\ \cos (px-p\alpha) \ .$
Later, Augustin Cauchy expressed the theorem using exponentials:[10][11]
$f(x)=\frac{1}{2\pi} \int_{-\infty} ^ \infty \ e^{ipx}\left(\int_{-\infty}^\infty e^{-ip\alpha }f(\alpha)\ d \alpha \right) \ dp.$
Cauchy pointed out that in some circumstances the order of integration in this result was significant.[12][13]
As justified using the theory of distributions, the Cauchy equation can be rearranged to resemble Fourier's original formulation and expose the δ-function as:
$\begin{align} f(x)&=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ipx}\left(\int_{-\infty}^\infty e^{-ip\alpha }f(\alpha)\ d \alpha \right) \ dp \\ &=\frac{1}{2\pi} \int_{-\infty}^\infty \left(\int_{-\infty}^\infty e^{ipx} e^{-ip\alpha } \ dp \right)f(\alpha)\ d \alpha =\int_{-\infty}^\infty \delta (x-\alpha) f(\alpha) \ d \alpha, \end{align}$
where the δ-function is expressed as:
$\delta(x-\alpha)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ip(x-\alpha)}\ dp \ .$
A rigorous interpretation of the exponential form and the various limitations upon the function f necessary for its application extended over several centuries. The problems with a classical interpretation are explained as follows:[14]
"The greatest drawback of the classical Fourier transformation is a rather narrow class of functions (originals) for which it can be effectively computed. Namely, it is necessary that these functions decrease sufficiently rapidly to zero (in the neighborhood of infinity) in order to insure the existence of the Fourier integral. For example, the Fourier transform of such simple functions as polynomials does not exist in the classical sense. The extension of the classical Fourier transformation to distributions considerably enlarged the class of functions that could be transformed and this removed many obstacles."
Further developments included generalization of the Fourier integral, "beginning with Plancherel's pathbreaking L2-theory (1910), continuing with Wiener's and Bochner's works (around 1930) and culminating with the amalgamation into L. Schwartz's theory of distributions (1945)...",[15] and leading to the formal development of the Dirac delta function.
An infinitesimal formula for an infinitely tall, unit impulse delta function (infinitesimal version of Cauchy distribution) explicitly appears in an 1827 text of Augustin Louis Cauchy.[16] Siméon Denis Poisson considered the issue in connection with the study of wave propagation as did Gustav Kirchhoff somewhat later. Kirchhoff and Hermann von Helmholtz also introduced the unit impulse as a limit of Gaussians, which also corresponded to Lord Kelvin's notion of a point heat source. At the end of the 19th century, Oliver Heaviside used formal Fourier series to manipulate the unit impulse.[17] The Dirac delta function as such was introduced as a "convenient notation" by Paul Dirac in his influential 1930 book Principles of Quantum Mechanics.[18] He called it the "delta function" since he used it as a continuous analogue of the discrete Kronecker delta.
## Definitions
The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite,
$\delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 \end{cases}$
and which is also constrained to satisfy the identity
$\int_{-\infty}^\infty \delta(x) \, dx = 1.$[19]
This is merely a heuristic characterization. The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties.[18] The Dirac delta function can be rigorously defined either as a distribution or as a measure.
### As a measure
One way to rigorously define the delta function is as a measure, which accepts as an argument a subset A of the real line R, and returns δ(A) = 1 if 0 ∈ A, and δ(A) = 0 otherwise.[20] If the delta function is conceptualized as modeling an idealized point mass at 0, then δ(A) represents the mass contained in the set A. One may then define the integral against δ as the integral of a function against this mass distribution. Formally, the Lebesgue integral provides the necessary analytic device. The Lebesgue integral with respect to the measure δ satisfies
$\int_{-\infty}^\infty f(x) \, \delta\{dx\} = f(0)$
for all continuous compactly supported functions f. The measure δ is not absolutely continuous with respect to the Lebesgue measure — in fact, it is a singular measure. Consequently, the delta measure has no Radon–Nikodym derivative — no true function for which the property
$\int_{-\infty}^\infty f(x)\delta(x)\, dx = f(0)$
holds.[21] As a result, the latter notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral.
As a probability measure on R, the delta measure is characterized by its cumulative distribution function, which is the unit step function[22]
$H(x) = \begin{cases} 1 & \text{if } x\ge 0\\ 0 & \text{if } x < 0. \end{cases}$
This means that H(x) is the integral of the cumulative indicator function 1(−∞, x] with respect to the measure δ; to wit,
$H(x) = \int_{\mathbf{R}}\mathbf{1}_{(-\infty,x]}(t)\,\delta\{dt\} = \delta(-\infty,x].$
Thus in particular the integral of the delta function against a continuous function can be properly understood as a Stieltjes integral:[23]
$\int_{-\infty}^\infty f(x)\delta\{dx\} = \int_{-\infty}^\infty f(x) \, dH(x).$
All higher moments of δ are zero. In particular, characteristic function and moment generating function are both equal to one.
### As a distribution
In the theory of distributions a generalized function is thought of not as a function itself, but only in relation to how it affects other functions when it is "integrated" against them. In keeping with this philosophy, to define the delta function properly, it is enough to say what the "integral" of the delta function against a sufficiently "good" test function is. If the delta function is already understood as a measure, then the Lebesgue integral of a test function against that measure supplies the necessary integral.
A typical space of test functions consists of all smooth functions on R with compact support. As a distribution, the Dirac delta is a linear functional on the space of test functions and is defined by[24]
$\delta[\varphi] = \varphi(0)\,$
()
for every test function φ.
For δ to be properly a distribution, it must be "continuous" in a suitable sense. In general, for a linear functional S on the space of test functions to define a distribution, it is necessary and sufficient that, for every positive integer N there is an integer MN and a constant CN such that for every test function φ, one has the inequality[25]
$|S[\phi]| \le C_N \sum_{k=0}^{M_N}\sup_{x\in [-N,N]}|\phi^{(k)}(x)|.$
With the δ distribution, one has such an inequality (with CN = 1) with MN = 0 for all N. Thus δ is a distribution of order zero. It is, furthermore, a distribution with compact support (the support being {0}).
The delta distribution can also be defined in a number of equivalent ways. For instance, it is the distributional derivative of the Heaviside step function. This means that, for every test function φ, one has
$\delta[\phi] = -\int_{-\infty}^\infty \phi'(x)H(x)\, dx.$
Intuitively, if integration by parts were permitted, then the latter integral should simplify to
$\int_{-\infty}^\infty \phi(x)H'(x)\, dx = \int_{-\infty}^\infty \phi(x)\delta(x)\, dx,$
and indeed, a form of integration by parts is permitted for the Stieltjes integral, and in that case one does have
$-\int_{-\infty}^\infty \phi'(x)H(x)\, dx = \int_{-\infty}^\infty \phi(x)\,dH(x).$
In the context of measure theory, the Dirac measure gives rise to a distribution by integration. Conversely, equation (1) defines a Daniell integral on the space of all compactly supported continuous functions φ which, by the Riesz representation theorem, can be represented as the Lebesgue integral of φ with respect to some Radon measure.
### Generalizations
The delta function can be defined in n-dimensional Euclidean space Rn as the measure such that
$\int_{\mathbf{R}^n} f(\mathbf{x})\delta\{d\mathbf{x}\} = f(\mathbf{0})$
for every compactly supported continuous function f. As a measure, the n-dimensional delta function is the product measure of the 1-dimensional delta functions in each variable separately. Thus, formally, with x = (x1,x2,...,xn), one has[6]
$\delta(\mathbf{x}) = \delta(x_1)\delta(x_2)\dots\delta(x_n).$
()
The delta function can also be defined in the sense of distributions exactly as above in the one-dimensional case.[26] However, despite widespread use in engineering contexts, (2) should be manipulated with care, since the product of distributions can only be defined under quite narrow circumstances.[27]
The notion of a Dirac measure makes sense on any set whatsoever.[20] Thus if X is a set, x0 ∈ X is a marked point, and Σ is any sigma algebra of subsets of X, then the measure defined on sets A ∈ Σ by
$\delta_{x_0}(A)=\begin{cases} 1 &\rm{if\ }x_0\in A\\ 0 &\rm{if\ }x_0\notin A \end{cases}$
is the delta measure or unit mass concentrated at x0.
Another common generalization of the delta function is to a differentiable manifold where most of its properties as a distribution can also be exploited because of the differentiable structure. The delta function on a manifold M centered at the point x0 ∈ M is defined as the following distribution:
$\delta_{x_0}[\phi] = \phi(x_0)$
()
for all compactly supported smooth real-valued functions φ on M.[28] A common special case of this construction is when M is an open set in the Euclidean space Rn.
On a locally compact Hausdorff space X, the Dirac delta measure concentrated at a point x is the Radon measure associated with the Daniell integral (3) on compactly supported continuous functions φ. At this level of generality, calculus as such is no longer possible, however a variety of techniques from abstract analysis are available. For instance, the mapping $x_0\mapsto \delta_{x_0}$ is a continuous embedding of X into the space of finite Radon measures on X, equipped with its vague topology. Moreover, the convex hull of the image of X under this embedding is dense in the space of probability measures on X.[29]
## Properties
### Scaling and symmetry
The delta function satisfies the following scaling property for a non-zero scalar α:[30]
$\int_{-\infty}^\infty \delta(\alpha x)\,dx =\int_{-\infty}^\infty \delta(u)\,\frac{du}{|\alpha|} =\frac{1}{|\alpha|}$
and so
$\delta(\alpha x) = \frac{\delta(x)}{|\alpha|}.$
()
In particular, the delta function is an even distribution, in the sense that
$\delta(-x) = \delta(x)$
which is homogeneous of degree −1.
### Algebraic properties
The distributional product of δ with x is equal to zero:
$x\delta(x) = 0.$
Conversely, if xf(x) = xg(x), where f and g are distributions, then
$f(x) = g(x) +c \delta(x)$
for some constant c.[31]
### Translation
The integral of the time-delayed Dirac delta is given by:
$\int_{-\infty}^\infty f(t) \delta(t-T)\,dt = f(T).$
This is sometimes referred to as the sifting property[32] or the sampling property. The delta function is said to "sift out" the value at t = T.
It follows that the effect of convolving a function f(t) with the time-delayed Dirac delta is to time-delay f(t) by the same amount:
$(f(t) * \delta(t-T))\,$ $\ \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau) \cdot \delta(t-T-\tau) \, d\tau$
$= \int\limits_{-\infty}^\infty f(\tau) \cdot \delta(\tau-(t-T)) \, d\tau$ (using (4): $\delta(-x)=\delta(x)$)
$= f(t-T).\,$
This holds under the precise condition that f be a tempered distribution (see the discussion of the Fourier transform below). As a special case, for instance, we have the identity (understood in the distribution sense)
$\int_{-\infty}^\infty \delta (\xi-x) \delta(x-\eta) \, dx = \delta(\xi-\eta).$
### Composition with a function
More generally, the delta distribution may be composed with a smooth function g(x) in such a way that the familiar change of variables formula holds, that
$\int_{\mathbf{R}} \delta\bigl(g(x)\bigr) f\bigl(g(x)\bigr) |g'(x)|\,dx = \int_{g(\mathbf{R})} \delta(u)f(u)\, du$
provided that g is a continuously differentiable function with g′ nowhere zero.[33] That is, there is a unique way to assign meaning to the distribution $\delta\circ g$ so that this identity holds for all compactly supported test functions f. This distribution satisfies δ(g(x)) = 0 if g is nowhere zero, and otherwise if g has a real root at x0, then
$\delta(g(x)) = \frac{\delta(x-x_0)}{|g'(x_0)|}.$
It is natural therefore to define the composition δ(g(x)) for continuously differentiable functions g by
$\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$
where the sum extends over all roots of g(x), which are assumed to be simple.[33] Thus, for example
$\delta\left(x^2-\alpha^2\right) = \frac{1}{2|\alpha|}\Big[\delta\left(x+\alpha\right)+\delta\left(x-\alpha\right)\Big].$
In the integral form the generalized scaling property may be written as
$\int_{-\infty}^\infty f(x) \, \delta(g(x)) \, dx = \sum_{i}\frac{f(x_i)}{|g'(x_i)|}.$
### Properties in n dimensions
The delta distribution in an n-dimensional space satisfies the following scaling property instead:
$\delta(\alpha\mathbf{x}) = |\alpha|^{-n}\delta(\mathbf{x})$
so that δ is a homogeneous distribution of degree −n. Under any reflection or rotation ρ, the delta function is invariant:
$\delta(\rho \mathbf{x}) = \delta(\mathbf{x}).$
As in the one-variable case, it is possible to define the composition of δ with a bi-Lipschitz function[34] g: Rn → Rn uniquely so that the identity
$\int_{\mathbf{R}^n} \delta(g(\mathbf{x}))\, f(g(\mathbf{x}))\, |\det g'(\mathbf{x})|\, d\mathbf{x} = \int_{g(\mathbf{R}^n)} \delta(\mathbf{u}) f(\mathbf{u})\,d\mathbf{u}$
for all compactly supported functions f.
Using the coarea formula from geometric measure theory, one can also define the composition of the delta function with a submersion from one Euclidean space to another one of different dimension; the result is a type of current. In the special case of a continuously differentiable function g: Rn → R such that the gradient of g is nowhere zero, the following identity holds[35]
$\int_{\mathbf{R}^n} f(\mathbf{x}) \, \delta(g(\mathbf{x})) \, d\mathbf{x} = \int_{g^{-1}(0)}\frac{f(\mathbf{x})}{|\mathbf{\nabla}g|}\,d\sigma(\mathbf{x})$
where the integral on the right is over g−1(0), the n − 1 dimensional surface defined by g(x) = 0 with respect to the Minkowski content measure. This is known as a simple layer integral.
More generally, if S is a smooth hypersurface of Rn, then we can associated to S the distribution that integrates any compactly supported smooth function g over S:
$\delta_S[g] = \int_S g(\mathbf{s})\,d\sigma(\mathbf{s})$
where σ is the hypersurface measure associated to S. This generalization is associated with the potential theory of simple layer potentials on S. If D is a domain in Rn with smooth boundary S, then δS is equal to the normal derivative of the indicator function of D in the distribution sense:
$-\int_{\mathbf{R}^n}g(\mathbf{x})\,\frac{\partial 1_D(\mathbf{x})}{\partial n}\;d\mathbf{x}=\int_S\,g(\mathbf{s})\;d\sigma(\mathbf{s}),$
where n is the outward normal.[36][37]
## Fourier transform
The delta function is a tempered distribution, and therefore it has a well-defined Fourier transform. Formally, one finds[38]
$\hat{\delta}(\xi)=\int_{-\infty}^\infty e^{-2\pi i x \xi}\delta(x)\,dx = 1.$
Properly speaking, the Fourier transform of a distribution is defined by imposing self-adjointness of the Fourier transform under the duality pairing $\langle\cdot,\cdot\rangle$ of tempered distributions with Schwartz functions. Thus $\hat{\delta}$ is defined as the unique tempered distribution satisfying
$\langle\hat{\delta},\phi\rangle = \langle\delta,\hat{\phi}\rangle$
for all Schwartz functions φ. And indeed it follows from this that $\hat{\delta}=1.$
As a result of this identity, the convolution of the delta function with any other tempered distribution S is simply S:
$S*\delta = S.\,$
That is to say that δ is an identity element for the convolution on tempered distributions, and in fact the space of compactly supported distributions under convolution is an associative algebra with identity the delta function. This property is fundamental in signal processing, as convolution with a tempered distribution is a linear time-invariant system, and applying the linear time-invariant system measures its impulse response. The impulse response can be computed to any desired degree of accuracy by choosing a suitable approximation for δ, and once it is known, it characterizes the system completely. See LTI system theory:Impulse response and convolution.
The inverse Fourier transform of the tempered distribution f(ξ) = 1 is the delta function. Formally, this is expressed
$\int_{-\infty}^\infty 1 \cdot e^{2\pi i x\xi}\,d\xi = \delta(x)$
and more rigorously, it follows since
$\langle 1, f^\vee\rangle = f(0) = \langle\delta,f\rangle$
for all Schwartz functions f.
In these terms, the delta function provides a suggestive statement of the orthogonality property of the Fourier kernel on R. Formally, one has
$\int_{-\infty}^\infty e^{i 2\pi \xi_1 t} \left[e^{i 2\pi \xi_2 t}\right]^*\,dt = \int_{-\infty}^\infty e^{-i 2\pi (\xi_2 - \xi_1) t} \,dt = \delta(\xi_2 - \xi_1).$
This is, of course, shorthand for the assertion that the Fourier transform of the tempered distribution
$f(t) = e^{i2\pi\xi_1 t}$
is
$\hat{f}(\xi_2) = \delta(\xi_1-\xi_2)$
which again follows by imposing self-adjointness of the Fourier transform.
By analytic continuation of the Fourier transform, the Laplace transform of the delta function is found to be[39]
$\int_{0}^{\infty}\delta (t-a)e^{-st} \, dt=e^{-sa}.$
## Distributional derivatives
The distributional derivative of the Dirac delta distribution is the distribution δ′ defined on compactly supported smooth test functions φ by[40]
$\delta'[\varphi] = -\delta[\varphi']=-\varphi'(0).$
The first equality here is a kind of integration by parts, for if δ were a true function then
$\int_{-\infty}^\infty \delta'(x)\varphi(x)\,dx = -\int_{-\infty}^\infty \delta(x)\varphi'(x)\,dx.$
The k-th derivative of δ is defined similarly as the distribution given on test functions by
$\delta^{(k)}[\varphi] = (-1)^k \varphi^{(k)}(0).$
In particular δ is an infinitely differentiable distribution.
The first derivative of the delta function is the distributional limit of the difference quotients:[41]
$\delta'(x) = \lim_{h\to 0} \frac{\delta(x+h)-\delta(x)}{h}.$
More properly, one has
$\delta' = \lim_{h\to 0} \frac{1}{h}(\tau_h\delta - \delta)$
where τh is the translation operator, defined on functions by τhφ(x) = φ(x+h), and on a distribution S by
$(\tau_h S)[\varphi] = S[\tau_{-h}\varphi].$
In the theory of electromagnetism, the first derivative of the delta function represents a point magnetic dipole situated at the origin. Accordingly, it is referred to as a dipole or the doublet function.[42]
The derivative of the delta function satisfies a number of basic properties, including:
$\frac{d}{dx}\delta(-x) = -\frac{d}{dx}\delta(x)$
$x\delta'(x) = -\delta(x).$
Furthermore, the convolution of δ' with a compactly supported smooth function f is
$\delta'*f = \delta*f' = f',$
which follows from the properties of the distributional derivative of a convolution.
### Higher dimensions
More generally, on an open set U in the n-dimensional Euclidean space Rn, the Dirac delta distribution centered at a point a ∈ U is defined by[43]
$\delta_a[\phi]=\phi(a)$
for all φ ∈ S(U), the space of all smooth compactly supported functions on U. If α = (α1, ..., αn) is any multi-index and ∂α denotes the associated mixed partial derivative operator, then the αth derivative ∂αδa of δa is given by[43]
$\left\langle \partial^{\alpha} \delta_{a}, \varphi \right\rangle = (-1)^{| \alpha |} \left\langle \delta_{a}, \partial^{\alpha} \varphi \right\rangle = \left. (-1)^{| \alpha |} \partial^{\alpha} \varphi (x) \right|_{x = a} \mbox{ for all } \varphi \in S(U).$
That is, the αth derivative of δa is the distribution whose value on any test function φ is the αth derivative of φ at a (with the appropriate positive or negative sign).
The first partial derivatives of the delta function are thought of as double layers along the coordinate planes. More generally, the normal derivative of a simple layer supported on a surface is a double layer supported on that surface, and represents a laminar magnetic monopole. Higher derivatives of the delta function are known in physics as multipoles.
Higher derivatives enter into mathematics naturally as the building blocks for the complete structure of distributions with point support. If S is any distribution on U supported on the set {a} consisting of a single point, then there is an integer m and coefficients cα such that[44]
$S = \sum_{|\alpha|\le m} c_\alpha \partial^\alpha\delta_a.$
## Representations of the delta function
The delta function can be viewed as the limit of a sequence of functions
$\delta (x) = \lim_{\varepsilon\to 0^+} \eta_\varepsilon(x), \,$
where ηε(x) is sometimes called a nascent delta function. This limit is meant in a weak sense: either that
$\lim_{\varepsilon\to 0^+} \int_{-\infty}^{\infty}\eta_\varepsilon(x)f(x) \, dx = f(0) \$
()
for all continuous functions f having compact support, or that this limit holds for all smooth functions f with compact support. The difference between these two slightly different modes of weak convergence is often subtle: the former is convergence in the vague topology of measures, and the latter is convergence in the sense of distributions.
### Approximations to the identity
Typically a nascent delta function ηε can be constructed in the following manner. Let η be an absolutely integrable function on R of total integral 1, and define
$\eta_\varepsilon(x) = \varepsilon^{-1} \eta \left (\frac{x}{\varepsilon} \right).$
In n dimensions, one uses instead the scaling
$\eta_\varepsilon(x) = \varepsilon^{-n} \eta \left (\frac{x}{\varepsilon} \right).$
Then a simple change of variables shows that ηε also has integral 1.[45] One shows easily that (5) holds for all continuous compactly supported functions f, and so ηε converges weakly to δ in the sense of measures.
The ηε constructed in this way are known as an approximation to the identity.[46] This terminology is because the space L1(R) of absolutely integrable functions is closed under the operation of convolution of functions: f∗g ∈ L1(R) whenever f and g are in L1(R). However, there is no identity in L1(R) for the convolution product: no element h such that f∗h = f for all f. Nevertheless, the sequence ηε does approximate such an identity in the sense that
$f*\eta_\varepsilon \to f\quad\rm{as\ }\varepsilon\to 0.$
This limit holds in the sense of mean convergence (convergence in L1). Further conditions on the ηε, for instance that it be a mollifier associated to a compactly supported function,[47] are needed to ensure pointwise convergence almost everywhere.
If the initial η = η1 is itself smooth and compactly supported then the sequence is called a mollifier. The standard mollifier is obtained by choosing η to be a suitably normalized bump function, for instance
$\eta(x) = \begin{cases} e^{-\frac{1}{1-|x|^2}}& \text{ if } |x| < 1\\ 0& \text{ if } |x|\geq 1. \end{cases}$
In some situations such as numerical analysis, a piecewise linear approximation to the identity is desirable. This can be obtained by taking η1 to be a hat function. With this choice of η1, one has
$\eta_\varepsilon(x) = \varepsilon^{-1}\max \left (1-|\frac{x}{\varepsilon}|,0 \right)$
which are all continuous and compactly supported, although not smooth and so not a mollifier.
### Probabilistic considerations
In the context of probability theory, it is natural to impose the additional condition that the initial η1 in an approximation to the identity should be positive, as such a function then represents a probability distribution. Convolution with a probability distribution is sometimes favorable because it does not result in overshoot or undershoot, as the output is a convex combination of the input values, and thus falls between the maximum and minimum of the input function. Taking η1 to be any probability distribution at all, and letting ηε(x) = η1(x/ε)/ε as above will give rise to an approximation to the identity. In general this converges more rapidly to a delta function if, in addition, η has mean 0 and has small higher moments. For instance, if η1 is the uniform distribution on [−1/2, 1/2], also known as the rectangular function, then:[48]
$\eta_\varepsilon(x) = \frac{1}{\varepsilon}\ \textrm{rect}\left(\frac{x}{\varepsilon}\right)= \begin{cases} \frac{1}{\varepsilon},&-\frac{\varepsilon}{2}<x<\frac{\varepsilon}{2}\\ 0,&\text{otherwise}. \end{cases}$
Another example is with the Wigner semicircle distribution
$\eta_\varepsilon(x)= \begin{cases} \frac{2}{\pi \varepsilon^2}\sqrt{\varepsilon^2 - x^2}, & -\varepsilon < x < \varepsilon \\ 0, & \text{otherwise} \end{cases}$
This is continuous and compactly supported, but not a mollifier because it is not smooth.
### Semigroups
Nascent delta functions often arise as convolution semigroups. This amounts to the further constraint that the convolution of ηε with ηδ must satisfy
$\eta_\varepsilon * \eta_\delta = \eta_{\varepsilon+\delta}$
for all ε, δ > 0. Convolution semigroups in L1 that form a nascent delta function are always an approximation to the identity in the above sense, however the semigroup condition is quite a strong restriction.
In practice, semigroups approximating the delta function arise as fundamental solutions or Green's functions to physically motivated elliptic or parabolic partial differential equations. In the context of applied mathematics, semigroups arise as the output of a linear time-invariant system. Abstractly, if A is a linear operator acting on functions of x, then a convolution semigroup arises by solving the initial value problem
$\begin{cases} \frac{\partial}{\partial t}\eta(t,x) = A\eta(t,x), \quad t>0 \\ \displaystyle\lim_{t\to 0^+} \eta(t,x) = \delta(x) \end{cases}$
in which the limit is as usual understood in the weak sense. Setting ηε(x) = η(ε, x) gives the associated nascent delta function.
Some examples of physically important convolution semigroups arising from such a fundamental solution include the following.
The heat kernel
The heat kernel, defined by
$\eta_\varepsilon(x) = \frac{1}{\sqrt{2\pi\varepsilon}} \mathrm{e}^{-\frac{x^2}{2\varepsilon}}$
represents the temperature in an infinite wire at time t > 0, if a unit of heat energy is stored at the origin of the wire at time t = 0. This semigroup evolves according to the one-dimensional heat equation:
$\frac{\partial u}{\partial t} = \frac{1}{2}\frac{\partial^2 u}{\partial x^2}.$
In probability theory, ηε(x) is a normal distribution of variance ε and mean 0. It represents the probability density at time t = ε of the position of a particle starting at the origin following a standard Brownian motion. In this context, the semigroup condition is then an expression of the Markov property of Brownian motion.
In higher dimensional Euclidean space Rn, the heat kernel is
$\eta_\varepsilon = \frac{1}{(2\pi\varepsilon)^{n/2}}\mathrm{e}^{-\frac{x\cdot x}{2\varepsilon}},$
and has the same physical interpretation, mutatis mutandis. It also represents a nascent delta function in the sense that ηε → δ in the distribution sense as ε → 0.
The Poisson kernel
The Poisson kernel
$\eta_\varepsilon(x) = \frac{1}{\pi} \frac{\varepsilon}{\varepsilon^2 + x^2}=\int_{-\infty}^{\infty}\mathrm{e}^{2\pi\mathrm{i} \xi x-|\varepsilon \xi|}\;d\xi$
is the fundamental solution of the Laplace equation in the upper half-plane.[49] It represents the electrostatic potential in a semi-infinite plate whose potential along the edge is held at fixed at the delta function. The Poisson kernel is also closely related to the Cauchy distribution. This semigroup evolves according to the equation
$\frac{\partial u}{\partial t} = -\left (-\frac{\partial^2}{\partial x^2} \right)^{\frac{1}{2}}u(t,x)$
where the operator is rigorously defined as the Fourier multiplier
$\mathcal{F}\left[\left(-\frac{\partial^2}{\partial x^2} \right)^{\frac{1}{2}}f\right](\xi) = |2\pi\xi|\mathcal{F}f(\xi).$
### Oscillatory integrals
In areas of physics such as wave propagation and wave mechanics, the equations involved are hyperbolic and so may have more singular solutions. As a result, the nascent delta functions that arise as fundamental solutions of the associated Cauchy problems are generally oscillatory integrals. An example, which comes from a solution of the Euler–Tricomi equation of transonic gas dynamics,[50] is the rescaled Airy function
$\varepsilon^{-\frac{1}{3}}\operatorname{Ai}\left (x\varepsilon^{-\frac{1}{3}} \right).$
Although using the Fourier transform, it is easy to see that this generates a semigroup in some sense, it is not absolutely integrable and so cannot define a semigroup in the above strong sense. Many nascent delta functions constructed as oscillatory integrals only converge in the sense of distributions (an example is the Dirichlet kernel below), rather than in the sense of measures.
Another example is the Cauchy problem for the wave equation in R1+1:[51]
$\begin{align} c^{-2}\frac{\partial^2u}{\partial t^2} - \Delta u &= 0\\ u=0,\quad \frac{\partial u}{\partial t} = \delta &\qquad \text{for }t=0. \end{align}$
The solution u represents the displacement from equilibrium of an infinite elastic string, with an initial disturbance at the origin.
Other approximations to the identity of this kind include the sinc function
$\eta_\varepsilon(x)=\frac{1}{\pi x}\sin\left(\frac{x}{\varepsilon}\right)=\frac{1}{2\pi}\int_{-\frac{1}{\varepsilon}}^{\frac{1}{\varepsilon}} \cos(kx)\;dk$
and the Bessel function
$\eta_\varepsilon(x) = \frac{1}{\varepsilon}J_{\frac{1}{\varepsilon}} \left(\frac{x+1}{\varepsilon}\right).$
### Plane wave decomposition
One approach to the study of a linear partial differential equation
$L[u]=f,\,$
where L is a differential operator on Rn, is to seek first a fundamental solution, which is a solution of the equation
$L[u]=\delta.\,$
When L is particularly simple, this problem can often be resolved using the Fourier transform directly (as in the case of the Poisson kernel and heat kernel already mentioned). For more complicated operators, it is sometimes easier first to consider an equation of the form
$L[u]=h\,$
where h is a plane wave function, meaning that it has the form
$h = h(x\cdot\xi)$
for some vector ξ. Such an equation can be resolved (if the coefficients of L are analytic functions) by the Cauchy–Kovalevskaya theorem or (if the coefficients of L are constant) by quadrature. So, if the delta function can be decomposed into plane waves, then one can in principle solve linear partial differential equations.
Such a decomposition of the delta function into plane waves was part of a general technique first introduced essentially by Johann Radon, and then developed in this form by Fritz John (1955).[52] Choose k so that n + k is an even integer, and for a real number s, put
$g(s) = \operatorname{Re}\left[\frac{-s^k\log(-is)}{k!(2\pi i)^n}\right] =\begin{cases} \frac{|s|^k}{4k!(2\pi i)^{n-1}}&n \text{ odd}\\ &\\ -\frac{|s|^k\log|s|}{k!(2\pi i)^{n}}&n \text{ even.} \end{cases}$
Then δ is obtained by applying a power of the Laplacian to the integral with respect to the unit sphere measure dω of g(x · ξ) for ξ in the unit sphere Sn−1:
$\delta(x) = \Delta_x^{\frac{n+k}{2}} \int_{S^{n-1}} g(x\cdot\xi)\,d\omega_\xi.$
The Laplacian here is interpreted as a weak derivative, so that this equation is taken to mean that, for any test function φ,
$\varphi(x) = \int_{\mathbf{R}^n}\varphi(y)\,dy\,\Delta_x^{\frac{n+k}{2}} \int_{S^{n-1}} g((x-y)\cdot\xi)\,d\omega_\xi.$
The result follows from the formula for the Newtonian potential (the fundamental solution of Poisson's equation). This is essentially a form of the inversion formula for the Radon transform, because it recovers the value of φ(x) from its integrals over hyperplanes. For instance, if n is odd and k = 1, then the integral on the right hand side is
$c_n \Delta^{\frac{n+1}{2}}_x\int\int_{S^{n-1}} \varphi(y)|(y-x)\cdot\xi|\,d\omega_\xi\,dy = c_n\Delta^{\frac{n+1}{2}}_x\int_{S^{n-1}} \, d\omega_\xi \int_{-\infty}^\infty |p|R\varphi(\xi,p+x\cdot\xi)\,dp$
where Rφ(ξ, p) is the Radon transform of φ:
$R\varphi(\xi,p) = \int_{x\cdot\xi=p} f(x)\,d^{n-1}x.$
An alternative equivalent expression of the plane wave decomposition, from Gel'fand & Shilov (1966–1968, I, §3.10), is
$\delta(x) = \frac{(n-1)!}{(2\pi i)^n}\int_{S^{n-1}}(x\cdot\xi)^{-n}\,d\omega_\xi$
for n even, and
$\delta(x) = \frac{1}{2(2\pi i)^{n-1}}\int_{S^{n-1}}\delta^{(n-1)}(x\cdot\xi)\,d\omega_\xi$
for n odd.
### Fourier kernels
See also: Convergence of Fourier series
In the study of Fourier series, a major question consists of determining whether and in what sense the Fourier series associated with a periodic function converges to the function. The nth partial sum of the Fourier series of a function f of period 2π is defined by convolution (on the interval [−π,π]) with the Dirichlet kernel:
$D_N(x) = \sum_{n=-N}^N e^{inx} = \frac{\sin\left((N+\tfrac12)x\right)}{\sin(x/2)}.$
Thus,
$s_N(f)(x) = D_N*f(x) = \sum_{n=-N}^N a_n e^{inx}$
where
$a_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(y)e^{-iny}\,dy.$
A fundamental result of elementary Fourier series states that the Dirichlet kernel tends to the a multiple of the delta function as N → ∞. This is interpreted in the distribution sense, that
$s_N(f)(0) = \int_{\mathbf{R}} D_N(x)f(x)\,dx \to 2\pi f(0)$
for every compactly supported smooth function f. Thus, formally one has
$\delta(x) = \frac1{2\pi} \sum_{n=-\infty}^\infty e^{inx}$
on the interval [−π,π].
In spite of this, the result does not hold for all compactly supported continuous functions: that is DN does not converge weakly in the sense of measures. The lack of convergence of the Fourier series has led to the introduction of a variety of summability methods in order to produce convergence. The method of Cesàro summation leads to the Fejér kernel[53]
$F_N(x) = \sum_{n=0}^N D_n(x) = \frac{1}{N}\left(\frac{\sin \frac{Nx}{2}}{\sin \frac{x}{2}}\right)^2.$
The Fejér kernels tend to the delta function in a stronger sense that[54]
$\int_{\mathbf{R}} F_N(x)f(x)\,dx \to 2\pi f(0)$
for every compactly supported continuous function f. The implication is that the Fourier series of any continuous function is Cesàro summable to the value of the function at every point.
### Hilbert space theory
The Dirac delta distribution is a densely defined unbounded linear functional on the Hilbert space L2 of square integrable functions. Indeed, smooth compactly support functions are dense in L2, and the action of the delta distribution on such functions is well-defined. In many applications, it is possible to identify subspaces of L2 and to give a stronger topology on which the delta function defines a bounded linear functional.
Sobolev spaces
The Sobolev embedding theorem for Sobolev spaces on the real line R implies that any square-integrable function f such that
$\|f\|_{H^1}^2 = \int_{-\infty}^\infty |\hat{f}(\xi)|^2 (1+|\xi|^2)\,d\xi < \infty$
is automatically continuous, and satisfies in particular
$\delta[f]=|f(0)| < C \|f\|_{H^1}.$
Thus δ is a bounded linear functional on the Sobolev space H1. Equivalently δ is an element of the continuous dual space H−1 of H1. More generally, in n dimensions, one has δ ∈ H−s(Rn) provided s > n / 2.
#### Spaces of holomorphic functions
In complex analysis, the delta function enters via Cauchy's integral formula which asserts that if D is a domain in the complex plane with smooth boundary, then
$f(z) = \frac{1}{2\pi i} \oint_{\partial D} \frac{f(\zeta)\,d\zeta}{\zeta-z},\quad z\in D$
for all holomorphic functions f in D that are continuous on the closure of D. As a result, the delta function δz is represented on this class of holomorphic functions by the Cauchy integral:
$\delta_z[f] = f(z) = \frac{1}{2\pi i} \oint_{\partial D} \frac{f(\zeta)\,d\zeta}{\zeta-z}.$
More generally, let H2(∂D) be the Hardy space consisting of the closure in L2(∂D) of all holomorphic functions in D continuous up to the boundary of D. Then functions in H2(∂D) uniquely extend to holomorphic functions in D, and the Cauchy integral formula continues to hold. In particular for z ∈ D, the delta function δz is a continuous linear functional on H2(∂D). This is a special case of the situation in several complex variables in which, for smooth domains D, the Szegő kernel plays the role of the Cauchy integral.
#### Resolutions of the identity
Given a complete orthonormal basis set of functions {φn} in a separable Hilbert space, for example, the normalized eigenvectors of a compact self-adjoint operator, any vector f can be expressed as:
$f = \sum_{n=1}^\infty \alpha_n \varphi_n.$
The coefficients {αn} are found as:
$\alpha_n = \langle \varphi_n, f \rangle,$
which may be represented by the notation:
$\alpha_n = \varphi_n^\dagger f,$
a form of the bra-ket notation of Dirac.[55] Adopting this notation, the expansion of f takes the dyadic form:[56]
$f = \sum_{n=1}^\infty \varphi_n \left ( \varphi_n^\dagger f \right).$
Letting I denote the identity operator on the Hilbert space, the expression
$I = \sum_{n=1}^\infty \varphi_n \varphi_n^\dagger,$
is called a resolution of the identity. When the Hilbert space is the space L2(D) of square-integrable functions on a domain D, the quantity:
$\varphi_n \varphi_n^\dagger,$
is an integral operator, and the expression for f can be rewritten as:
$f(x) = \sum_{n=1}^\infty \int_D\, \left( \varphi_n (x) \varphi_n^*(\xi)\right) f(\xi) \, d \xi.$
The right-hand side converges to f in the L2 sense. It need not hold in a pointwise sense, even when f is a continuous function. Nevertheless, it is common to abuse notation and write
$f(x) = \int \, \delta(x-\xi) f (\xi)\, d\xi,$
resulting in the representation of the delta function:[57]
$\delta(x-\xi) = \sum_{n=1}^\infty \varphi_n (x) \varphi_n^*(\xi).$
With a suitable rigged Hilbert space (Φ, L2(D), Φ*) where Φ ⊂ L2(D) contains all compactly supported smooth functions, this summation may converge in Φ*, depending on the properties of the basis φn. In most cases of practical interest, the orthonormal basis comes from an integral or differential operator, in which case the series converges in the distribution sense.[58]
### Infinitesimal delta functions
Cauchy used an infinitesimal α to write down a unit impulse, infinitely tall and narrow Dirac-type delta function δα satisfying $\int F(x)\delta_\alpha(x) = F(0)$ in a number of articles in 1827.[59] Cauchy defined an infinitesimal in Cours d'Analyse (1827) in terms of a sequence tending to zero. Namely, such a null sequence becomes an infinitesimal in Cauchy's and Lazare Carnot's terminology.
Modern set-theoretic approaches allow one to define infinitesimals via the ultrapower construction, where a null sequence becomes an infinitesimal in the sense of an equivalence class modulo a relation defined in terms of a suitable ultrafilter. The article by Yamashita (2007) contains a bibliography on modern Dirac delta functions in the context of an infinitesimal-enriched continuum provided by the hyperreals. Here the Dirac delta can be given by an actual function, having the property that for every real function F one has $\int F(x)\delta_\alpha(x) = F(0)$ as anticipated by Fourier and Cauchy.
## Dirac comb
Main article: Dirac comb
A Dirac comb is an infinite series of Dirac delta functions spaced at intervals of T
A so-called uniform "pulse train" of Dirac delta measures, which is known as a Dirac comb, or as the Shah distribution, creates a sampling function, often used in digital signal processing (DSP) and discrete time signal analysis. The Dirac comb is given as the infinite sum, whose limit is understood in the distribution sense,
$\Delta(x) = \sum_{n=-\infty}^\infty \delta(x-n),$
which is a sequence of point masses at each of the integers.
Up to an overall normalizing constant, the Dirac comb is equal to its own Fourier transform. This is significant because if f is any Schwartz function, then the periodization of f is given by the convolution
$(f*\Delta)(x) = \sum_{n=-\infty}^\infty f(x-n).$
In particular,
$(f*\Delta)^\wedge = \hat{f}\widehat{\Delta} = \hat{f}\Delta$
is precisely the Poisson summation formula.[60]
## Sokhotski–Plemelj theorem
The Sokhotski–Plemelj theorem, important in quantum mechanics, relates the delta function to the distribution p.v.1/x, the Cauchy principal value of the function 1/x, defined by
$\left\langle\operatorname{p.v.}\frac{1}{x}, \phi\right\rangle = \lim_{\varepsilon\to 0^+}\int_{|x|>\varepsilon} \frac{\phi(x)}{x}\,dx.$
Sokhatsky's formula states that[61]
$\lim_{\varepsilon\to 0^+} \frac{1}{x\pm i\varepsilon} = \operatorname{p.v.}\frac{1}{x} \mp i\pi\delta(x),$
Here the limit is understood in the distribution sense, that for all compactly supported smooth functions f,
$\lim_{\varepsilon\to 0^+} \int_{-\infty}^\infty\frac{f(x)}{x\pm i\varepsilon}\,dx = \mp i\pi f(0) + \lim_{\varepsilon\to 0^+} \int_{|x|>\varepsilon}\frac{f(x)}{x}\,dx.$
## Relationship to the Kronecker delta
The Kronecker delta δij is the quantity defined by
$\delta_{ij} = \begin{cases} 1 & i=j\\ 0 &i\not=j \end{cases}$
for all integers i, j. This function then satisfies the following analog of the sifting property: if $(a_i)_{i \in \mathbf{Z}}$ is any doubly infinite sequence, then
$\sum_{i=-\infty}^\infty a_i \delta_{ik}=a_k.$
Similarly, for any real or complex valued continuous function f on R, the Dirac delta satisfies the sifting property
$\int_{-\infty}^\infty f(x)\delta(x-x_0)\,dx=f(x_0).$
This exhibits the Kronecker delta function as a discrete analog of the Dirac delta function.[62]
## Applications to probability theory
In probability theory and statistics, the Dirac delta function is often used to represent a discrete distribution, or a partially discrete, partially continuous distribution, using a probability density function (which is normally used to represent fully continuous distributions). For example, the probability density function f(x) of a discrete distribution consisting of points x = {x1, ..., xn}, with corresponding probabilities p1, ..., pn, can be written as
$f(x) = \sum_{i=1}^n p_i \delta(x-x_i).$
As another example, consider a distribution which 6/10 of the time returns a standard normal distribution, and 4/10 of the time returns exactly the value 3.5 (i.e. a partly continuous, partly discrete mixture distribution). The density function of this distribution can be written as
$f(x) = 0.6 \, \frac {1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} + 0.4 \, \delta(x-3.5).$
The delta function is also used in a completely different way to represent the local time of a diffusion process (like Brownian motion). The local time of a stochastic process B(t) is given by
$\ell(x,t) = \int_0^t \delta(x-B(s))\,ds$
and represents the amount of time that the process spends at the point x in the range of the process. More precisely, in one dimension this integral can be written
$\ell(x,t) = \lim_{\varepsilon\to 0^+}\frac{1}{2\varepsilon}\int_0^t \mathbf{1}_{[x-\varepsilon,x+\varepsilon]}(B(s))\,ds$
where 1[x−ε, x+ε] is the indicator function of the interval [x−ε, x+ε].
## Application to quantum mechanics
We give an example of how the delta function is expedient in quantum mechanics. The wave function of a particle gives the probability amplitude of finding a particle within a given region of space. Wave functions are assumed to be elements of the Hilbert space L2 of square-integrable functions, and the total probability of finding a particle within a given interval is the integral of the magnitude of the wave function squared over the interval. A set {φn} of wave functions is orthonormal if they are normalized by
$\langle\phi_n|\phi_m\rangle = \delta_{nm}$
where δ here refers to the Kronecker delta. A set of orthonormal wave functions is complete in the space of square-integrable functions if any wave function ψ can be expressed as a combination of the φn:
$\psi = \sum c_n \phi_n,$
with $c_n = \langle \phi_n | \psi \rangle$. Complete orthonormal systems of wave functions appear naturally as the eigenfunctions of the Hamiltonian (of a bound system) in quantum mechanics that measures the energy levels, which are called the eigenvalues. The set of eigenvalues, in this case, is known as the spectrum of the Hamiltonian. In bra-ket notation, as above, this equality implies the resolution of the identity:
$I = \sum |\phi_n\rangle\langle\phi_n|.$
Here the eigenvalues are assumed to be discrete, but the set of eigenvalues of an observable may be continuous rather than discrete. An example is the position observable, Qψ(x) = xψ(x). The spectrum of the position (in one dimension) is the entire real line, and is called a continuous spectrum. However, unlike the Hamiltonian, the position operator lacks proper eigenfunctions. The conventional way to overcome this shortcoming is to widen the class of available functions by allowing distributions as well: that is, to replace the Hilbert space of quantum mechanics by an appropriate rigged Hilbert space.[63] In this context, the position operator has a complete set of eigen-distributions, labeled by the points y of the real line, given by
$\phi_y(x) = \delta(x-y).\;$
The eigenfunctions of position are denoted by $\phi_y = |y\rangle$ in Dirac notation, and are known as position eigenstates.
Similar considerations apply to the eigenstates of the momentum operator, or indeed any other self-adjoint unbounded operator P on the Hilbert space, provided the spectrum of P is continuous and there are no degenerate eigenvalues. In that case, there is a set Ω of real numbers (the spectrum), and a collection φy of distributions indexed by the elements of Ω, such that
$P\phi_y = y\phi_y.\;$
That is, φy are the eigenvectors of P. If the eigenvectors are normalized so that
$\langle \phi_y,\phi_{y'}\rangle = \delta(y-y')$
in the distribution sense, then for any test function ψ,
$\psi(x) = \int_\Omega c(y) \phi_y(x) \, dy$
where
$c(y) = \langle \psi, \phi_y \rangle.$
That is, as in the discrete case, there is a resolution of the identity
$I = \int_\Omega |\phi_y\rangle\, \langle\phi_y|\,dy$
where the operator-valued integral is again understood in the weak sense. If the spectrum of P has both continuous and discrete parts, then the resolution of the identity involves a summation over the discrete spectrum and an integral over the continuous spectrum.
The delta function also has many more specialized applications in quantum mechanics, such as the delta potential models for a single and double potential well.
## Application to structural mechanics
The delta function can be used in structural mechanics to describe transient loads or point loads acting on structures. The governing equation of a simple mass–spring system excited by a sudden force impulse I at time t = 0 can be written
$m \frac{\mathrm{d}^2 \xi}{\mathrm{d} t^2} + k \xi = I \delta(t),$
where m is the mass, ξ the deflection and k the spring constant.
As another example, the equation governing the static deflection of a slender beam is, according to Euler-Bernoulli theory,
$EI \frac{\mathrm{d}^4 w}{\mathrm{d} x^4} = q(x),\,$
where EI is the bending stiffness of the beam, w the deflection, x the spatial coordinate and q(x) the load distribution. If a beam is loaded by a point force F at x = x0, the load distribution is written
$q(x) = F \delta(x-x_0).\,$
As integration of the delta function results in the Heaviside step function, it follows that the static deflection of a slender beam subject to multiple point loads is described by a set of piecewise polynomials.
Also a point moment acting on a beam can be described by delta functions. Consider two opposing point forces F at a distance d apart. They then produce a moment M = Fd acting on the beam. Now, let the distance d approach the limit zero, while M is kept constant. The load distribution, assuming a clockwise moment acting at x = 0, is written
$\begin{align} q(x) &= \lim_{d \to 0} \Big( F \delta(x) - F \delta(x-d) \Big) \\ &= \lim_{d \to 0} \left( \frac{M}{d} \delta(x) - \frac{M}{d} \delta(x-d) \right) \\ &= M \lim_{d \to 0} \frac{\delta(x) - \delta(x - d)}{d}\\ &= M \delta'(x). \end{align}$
Point moments can thus be represented by the derivative of the delta function. Integration of the beam equation again results in piecewise polynomial deflection.
## Notes
1. Katz, Mikhail; Tall, David (2012), "A Cauchy-Dirac delta function", , arXiv:1206.0119, doi:10.1007/s10699-012-9289-4 .
2. ^ a b
3. The original French text can be found here.
4. Hikosaburo Komatsu (2002). "Fourier's hyperfunctions and Heaviside's pseudodifferential operators". In Takahiro Kawai, Keiko Fujita, eds. Microlocal Analysis and Complex Fourier Analysis. World Scientific. p. 200. ISBN 9812381619.
5. Tyn Myint-U., Lokenath Debnath (2007). Linear Partial Differential Equations for Scientists And Engineers (4th ed.). Springer. p. 4. ISBN 0817643931.
6. Lokenath Debnath, Dambaru Bhatta (2007). Integral Transforms And Their Applications (2nd ed.). CRC Press. p. 2. ISBN 1584885750.
7. Ivor Grattan-Guinness (2009). Convolutions in French Mathematics, 1800-1840: From the Calculus and Mechanics to Mathematical Analysis and Mathematical Physics, Volume 2. Birkhäuser. p. 653. ISBN 3764322381.
8. Dragiša Mitrović, Darko Žubrinić (1998). Fundamentals of Applied Functional Analysis: Distributions, Sobolev Spaces. CRC Press. p. 62. ISBN 0582246946.
9. Manfred Kracht, Erwin Kreyszig (1989). "On singular integral operators and generalizations". In Themistocles M. Rassias, ed. Topics in Mathematical Analysis: A Volume Dedicated to the Memory of A.L. Cauchy. World Scientific. p. 553. ISBN 9971506661.
10. ^ a b
11. ^ a b
12. Driggers 2003, p. 2321. See also Bracewell 1986, Chapter 5 for a different interpretation. Other conventions for the assigning the value of the Heaviside function at zero exist, and some of these are not consistent with what follows.
13. Strichartz 1994, §2.3; Hörmander 1983, §8.2
14. ^ a b
15. In some conventions for the Fourier transform.
16. ^ a b
17. Hörmander 1983, p. 56; Rudin 1991, Theorem 6.25
18. More generally, one only needs η = η1 to have an integrable radially symmetric decreasing rearrangement.
19. The development of this section in bra-ket notation is found in (Levin 2002, Coordinate-space wave functions and completeness, pp.=109ff)
20. Córdoba 1988; Hörmander 1983, §7.2
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• Federer, Herbert (1969), Geometric measure theory, Die Grundlehren der mathematischen Wissenschaften 153, New York: Springer-Verlag, pp. xiv+676, ISBN 978-3-540-60656-7, MR 0257325 .
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• Lange, Rutger-Jan (2012), "Potential theory, path integrals and the Laplacian of the indicator", Journal of High Energy Physics (Springer) 2012 (11): 29–30, Bibcode:2012JHEP...11..032L, doi:10.1007/JHEP11(2012)032 .
• Laugwitz, D. (1989), "Definite values of infinite sums: aspects of the foundations of infinitesimal analysis around 1820", Arch. Hist. Exact Sci. 39 (3): 195–245, doi:10.1007/BF00329867 .
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• Li, Y. T.; Wong, R. (2008), "Integral and series representations of the Dirac delta function", Commun. Pure Appl. Anal. 7 (2): 229–247, doi:10.3934/cpaa.2008.7.229, MR 2373214 .
• de la Madrid, R.; Bohm, A.; Gadella, M. (2002), "Rigged Hilbert Space Treatment of Continuous Spectrum", Fortschr. Phys. 50 (2): 185–216, arXiv:quant-ph/0109154, Bibcode:2002ForPh..50..185D, doi:10.1002/1521-3978(200203)50:2<185::AID-PROP185>3.0.CO;2-S .
• McMahon, D. (2005-11-22), "An Introduction to State Space", Quantum Mechanics Demystified, A Self-Teaching Guide, Demystified Series, New York: McGraw-Hill, p. 108, doi:10.1036/0071455469, ISBN 0-07-145546-9, retrieved 2008-03-17 .
• van der Pol, Balth.; Bremmer, H. (1987), Operational calculus (3rd ed.), New York: Chelsea Publishing Co., ISBN 978-0-8284-0327-6, MR 904873 .
• Rudin, W. (1991), Functional Analysis (2nd ed.), McGraw-Hill, ISBN 0-07-054236-8 .
• Soares, Manuel; Vallée, Olivier (2004), Airy functions and applications to physics, London: Imperial College Press .
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• Schwartz, L. (1950), Théorie des distributions 1, Hermann .
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• Stein, Elias; Weiss, Guido (1971), Introduction to Fourier Analysis on Euclidean Spaces, Princeton University Press, ISBN 0-691-08078-X .
• Strichartz, R. (1994), A Guide to Distribution Theory and Fourier Transforms, CRC Press, ISBN 0-8493-8273-4 .
• Vladimirov, V. S. (1971), Equations of mathematical physics, Marcel Dekker, ISBN 0-8247-1713-9 .
• Yamashita, H. (2006), "Pointwise analysis of scalar fields: A nonstandard approach", 47 (9): 092301, Bibcode:2006JMP....47i2301Y, doi:10.1063/1.2339017
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|
http://www.physicsforums.com/showthread.php?s=333c00b4a6961bbf5c68446f81ab5b6a&p=3940382
|
Physics Forums
How would you go about solving these set of equations?
Say you have a set of equations of the form
[itex]\left\{ \begin{array}{rl}
x+y+z &=a \\
xy + xz + yz &= b \\
xyz &= c
\end{array} \right. [/itex]
(for clarity: I'm working over the regular numbers)
how would you go about solving it elegantly? (or at least rewriting it as linear equations)
I'm thinking of something analogous to how you can solve
[itex]\left\{ \begin{array}{rl}
x+y &=a \\
xy &= b
\end{array} \right. [/itex]
namely by noting that (x-y)² = (x+y)² - 4xy = a² - 4b (and after taking the square root we're left with two good ol' linear equations, i.e. x+y=... and x-y=..., a form which I regard as "being solved")
Quote by mr. vodka Say you have a set of equations of the form $\left\{ \begin{array}{rl} x+y+z &=a \\ xy + xz + yz &= b \\ xyz &= c \end{array} \right.$ (for clarity: I'm working over the regular numbers) how would you go about solving it elegantly? (or at least rewriting it as linear equations) I'm thinking of something analogous to how you can solve $\left\{ \begin{array}{rl} x+y &=a \\ xy &= b \end{array} \right.$ namely by noting that (x-y)² = (x+y)² - 4xy = a² - 4b (and after taking the square root we're left with two good ol' linear equations, i.e. x+y=... and x-y=..., a form which I regard as "being solved")
Hey mr. vodka.
Have you tried just subsituting two of the variables to get the whole thing in a third one?
So for example you could everything in terms of z by taking (1) to get x = a - y - z and taking (2) to get y = (b - xz)(x + z) and then plug in (1) and (2) and simplify to get an equation for (3) giving z = c/xy where you get an expression in terms of only z.
You could do it in more than one way (this is only one possible way), but then you would end up with some kind of equation and at the worst you can use a root-finding algorithm, and probably use a few transformations to get the thing in surd form if it exists (using your ideas in your original post).
Call the equations 1, 2 and 3 Multiply 2 by z xyz+xz2+yz2=bz Subtract 3 (x+y)z2 = bz-c Substitute into 1 (bz-c)/z2+z=a Can you take the cubic from there?
How would you go about solving these set of equations?
Thank you both for your trouble. Studiot's method was more like something I was looking for, thank you!
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Mean_Value_Theorem
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# All Science Fair Projects
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# Mean value theorem
(Redirected from Mean Value Theorem)
In calculus, the mean value theorem states, roughly, that given a section of a smooth curve, there is a point on that section at which the gradient (slope) of the curve is equal to the "average" gradient of the section. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.
This theorem was developed by Lagrange. Some mathematicians consider this theorem to be the most important theorem of calculus (see also: the fundamental theorem of calculus). The theorem is not often used to solve mathematical problems; rather, it is more commonly used to prove other theorems. The mean value theorem can be used to prove Taylor's theorem, of which it is a special case.
More precisely, the theorem states: for some continually differentiable curve; for every secant, there is some parallel tangent. In addition, the tangent runs through a point located between the intersection points of said secant.
Let f : [a, b] → R be continuous on the closed interval [a, b], and differentiable on the open interval (a, b). Then there exists some c in (a, b) such that
$f ' (c) = \frac{f(b) - f(a)}{b - a}$
Generalization: The theorem is usually stated in the form above, but it is actually valid in a slightly more general setting: We only need to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ exists or is equal to ± infinity.
Contents
## Proof
An understanding of this and the Point-Slope Formula will make it clear that the equation of a secant (which intersects (a, f(a)) and (b, f(b)) ) is: y = {[f(b) - f(a)] / [b - a]}(x - a) - f(a).
The formula ( f(b) - f(a) ) / (b - a) gives the slope of the line joining the points (a, f(a)) and (b, f(b)), which we call a chord of the curve, while f ' (x) gives the slope of the tangent to the curve at the point (x, f(x) ). Thus the Mean value theorem says that given any chord of a smooth curve, we can find a point lying between the end-points of the chord such that the tangent at that point is parallel to the chord. The following proof illustrates this idea.
Define g(x) = f(x) + rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is true of g. We choose r so that g satisfies the conditions of Rolle's theorem, which means
$g(a) = g(b) \qquad \Rightarrow \qquad f(a) + ra = f(b) + rb$
$\Rightarrow \qquad r = - \frac{ f(b) - f(a) }{ b - a}$
By Rolle's Theorem, there is some c in (a, b) for which g '(c) = 0, and it follows
$f ' (c) = g ' (c) - r = 0 - r = \frac{ f(b) - f(a) }{ b - a}$
as required. The mean value theorem in the following form is considered more useful.
f(b) - f(a) = f'(c)(b - a)
## Cauchy's mean value theorem
Cauchy's mean value is the more generalised form of mean value theorem. It states: If functions f(t) and g(t) are both continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists some c in (a, b), such that
$\frac {f'(c)} {g'(c)} = \frac {f(b) - f(a)} {g(b) - g(a)}.$
Cauchy's mean value theorem can be used to prove l'Hopital's rule.
### Proof of Cauchy's mean value theorem
The proof of Cauchy's mean value theorem is based on the same idea as the proof of mean value theorem. We aim to transform the curve defined by y = y(t) and x = x(t), so that it satisfies the conditions of Rolle's theorem.
We define a new function:
F(t) = y(t) - mx(t)
where m is a constant, so that
$F(a) = F(b) \qquad \Rightarrow \qquad m = \frac {y(b) - y(a)} {x(b) - x(a)}$
Since F is continuous and F(a) = F(b), by Rolle's theorem, there exists some c in (a, b) such that F′(c) = 0, i.e.
$F'(c) = 0 \ = \ y'(c) - \frac {y(b) - y(a)} {x(b) -x(a)} x'(c)$
$\Rightarrow \qquad \frac {y'(c)} {x'(c)}\ = \ \frac {y(b) - y(a)} {x(b) - x(a)}$
as required.
## Mean value theorems for integration
The first mean value theorem for integration states:
If f : [a, b] → R is a continuous function and φ : [a, b] → R is an integrable positive function, then there exists a number x in (a, b) such that
$\int_a^b f(t)\varphi (t) \, dt \quad = \quad f(x) \int_a^b \varphi (t) \, dt.$
In particular (φ(t) = 1), there exists x in (a, b) with
$\int_a^b f(t) \, dt \quad = \quad f(x) (b - a).$
The second mean value theorem for integration states:
If f : [a, b] → R is a positive and monotone decreasing function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b] such that
$\int_a^b f(t) \varphi (t) \, dt \quad = \quad ( \lim_{t \to a} f(t) ) \cdot \int_a^x \varphi (t) \, dt.$
## See also
Last updated: 08-20-2005 16:11:22
03-10-2013 05:06:04
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http://wikipedia.sfstate.us/Sampling_distribution
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edit
# Sampling distribution
In statistics, a sampling distribution or finite-sample distribution is the probability distribution of a given statistic based on a random sample. Sampling distributions are important in statistics because they provide a major simplification on the route to statistical inference. More specifically, they allow analytical considerations to be based on the sampling distribution of a statistic, rather than on the joint probability distribution of all the individual sample values.
## Introduction
The sampling distribution of a statistic is the distribution of that statistic, considered as a random variable, when derived from a random sample of size n. It may be considered as the distribution of the statistic for all possible samples from the same population of a given size. The sampling distribution depends on the underlying distribution of the population, the statistic being considered, the sampling procedure employed and the sample size used. There is often considerable interest in whether the sampling distribution can be approximated by an asymptotic distribution, which corresponds to the limiting case as n → ∞.
For example, consider a normal population with mean μ and variance σ². Assume we repeatedly take samples of a given size from this population and calculate the arithmetic mean $\scriptstyle \bar x$ for each sample — this statistic is called the sample mean. Each sample has its own average value, and the distribution of these averages is called the "sampling distribution of the sample mean". This distribution is normal $\scriptstyle \mathcal{N}(\mu,\, \sigma^2/n)$ since the underlying population is normal, although sampling distributions may also often be close to normal even when the population distribution is not (see central limit theorem). An alternative to the sample mean is the sample median. When calculated from the same population, it has a different sampling distribution to that of the mean and is generally not normal (but it may be close for large sample sizes).
The mean of a sample from a population having a normal distribution is an example of a simple statistic taken from one of the simplest statistical populations. For other statistics and other populations the formulas are more complicated, and often they don't exist in closed-form. In such cases the sampling distributions may be approximated through Monte-Carlo simulations, bootstrap methods, or asymptotic distribution theory.
## Standard error
The standard deviation of the sampling distribution of the statistic is referred to as the standard error of that quantity. For the case where the statistic is the sample mean, and samples are uncorrelated, the standard error is:
$\sigma_{\bar x} = \frac{\sigma}{\sqrt{n}}$
where $\sigma$ is the standard deviation of the population distribution of that quantity and n is the size (number of items) in the sample.
An important implication of this formula is that the sample size must be quadrupled (multiplied by 4) to achieve half (1/2) the measurement error. When designing statistical studies where cost is a factor, this may have a role in understanding cost-benefit tradeoffs.
## Examples
Population Statistic Sampling distribution
Normal: $\mathcal{N}(\mu, \sigma^2)$ Sample mean $\bar X$ from samples of size n $\bar X \sim \mathcal{N}\Big(\mu,\, \frac{\sigma^2}{n} \Big)$
Bernoulli: $\operatorname{Bernoulli}(p)$ Sample proportion of "successful trials" $\bar X$ $n \bar X \sim \operatorname{Binomial}(n, p)$
Two independent normal populations:
$\mathcal{N}(\mu_1, \sigma_1^2)$ and $\mathcal{N}(\mu_2, \sigma_2^2)$
Difference between sample means, $\bar X_1 - \bar X_2$ $\bar X_1 - \bar X_2 \sim \mathcal{N}\! \left(\mu_1 - \mu_2,\, \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \right)$
Any absolutely continuous distribution F with density ƒ Median $X_{(k)}$ from a sample of size n = 2k − 1, where sample is ordered $X_{(1)}$ to $X_{(n)}$ $f_{X_{(k)}}(x) = \frac{(2k-1)!}{(k-1)!^2}f(x)\Big(F(x)(1-F(x))\Big)^{k-1}$
Any distribution with distribution function F Maximum $M=\max\ X_k$ from a random sample of size n $F_M(x) = P(M\le x) = \prod P(X_k\le x)= \left(F(x)\right)^n$
## Statistical inference
In the theory of statistical inference, the idea of a sufficient statistic provides the basis of choosing a statistic (as a function of the sample data points) in such a way that no information is lost by replacing the full probabilistic description of the sample with the sampling distribution of the selected statistic.
In frequentist inference, for example in the development of a statistical hypothesis test or a confidence interval, the availability of the sampling distribution of a statistic (or an approximation to this in the form of an asymptotic distribution) can allow the ready formulation of such procedures, whereas the development of procedures starting from the joint distribution of the sample would be less straightforward.
In Bayesian inference, when the sampling distribution of a statistic is available, one can consider replacing the final outcome of such procedures, specifically the conditional distributions of any unknown quantities given the sample data, by the conditional distributions of any unknown quantities given selected sample statistics. Such a procedure would involve the sampling distribution of the statistics. The results would be identical provided the statistics chosen are jointly sufficient statistics.
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http://www.physicsforums.com/showthread.php?t=238578
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Physics Forums
## Representation Theory
I'm done a basic course on representation theory and character theory of finite groups, mainly over a complex field. When the order of the group divides the characteristic of the field clearly things are very different.
What I'd like to learn about is what happens when the field is not complex but still quite well-behaved. In particular if we have an algebraically closed field whose characteristic doesn't divide the order of the group what changes?
The reason I ask is that there doesn't seem to be a very good treatment of this in any of the books I've seen. Can anyone offer any suggestions? I guess I could start from scratch and go though all the proofs in the complex case from the bottom up checking whether they still hold, but it would be nice to have a reference.
Are there any major pitfalls when trying to transfer the theory from the complex case?
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Quote by Hello Kitty In particular if we have an algebraically closed field whose characteristic doesn't divide the order of the group what changes?
Nothing. That's why there are no treatises on the subject - the only added complication is it not being over C.
So the character degrees over $$\overline{\mathbb F_p}$$ are the same as for $$\mathbb C$$ provided $$p \not\vert \ |G|$$?
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## Representation Theory
Look at your notes - where does it use that the characteristic of C is zero? It will only use that it is co-prime to |G|.
The method of going from C to char p was given by Brauer in the 50s. Reps over C are actually realizable over the algebraic integers, A. Pick a maximal ideal containing the prime ideal (p) in A, and reduce modulo this ideal. This yields the projective modules over the field of char p, which are all the modules if p is coprime to |G|.
I have a question in representation theory. There is a result that says that if I have a linear character of a subgroup H of a group G with kernel K, then the induced character is irreducible iff (H,K) is a Shoda pair. The proof uses the fact that If, chi(ghg-1)=chi(h) for all h in H ∩ g(-1)Hg, then [H,g]∩H ⊂ K. I am not able to prove this one...can sumbody help??
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http://en.wikipedia.org/wiki/Hypotrochoid
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# Hypotrochoid
The red curve is a hypotrochoid drawn as the smaller black circle rolls around inside the larger blue circle (parameters are R = 5, r = 3, d = 5).
A hypotrochoid is a roulette traced by a point attached to a circle of radius r rolling around the inside of a fixed circle of radius R, where the point is a distance d from the center of the interior circle.
The parametric equations for a hypotrochoid are:[citation needed]
$x (\theta) = (R - r)\cos\theta + d\cos\left({R - r \over r}\theta\right)$
$y (\theta) = (R - r)\sin\theta - d\sin\left({R - r \over r}\theta\right).$
Where $\theta$ is the angle formed by the horizontal and the center of the rolling circle (note that these are not polar equations because $\theta$ is not the polar angle).
Special cases include the hypocycloid with d = r and the ellipse with R = 2r.
The ellipse (drawn in red) may be expressed as a special case of the hypotrochoid, with R = 2r; here R = 10, r = 5, d = 1.
The classic Spirograph toy traces out hypotrochoid and epitrochoid curves.
## References
• J. Dennis Lawrence (1972). A catalog of special plane curves. Dover Publications. pp. 165–168. ISBN 0-486-60288-5.
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http://mathhelpforum.com/discrete-math/127168-yet-two-more-questions-i-don-t-understand.html
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# Thread:
1. ## Yet two more questions I don't understand.
Question 1:
We have seen that congruence modulo m is an equivalence relation on Z(all integers) for any integer m >= 1. Describe the set of equivalence classes for congruence modulo 8 - how many distinct classes are there and what are they?
What I am getting out of this is that there are 7 distinct classes and they are 1-7. But then again I am completely clueless on this modulo stuff and I've already gone to the professor with no luck on understanding it any better.
Question 2:
Show that addition modulo 8 is well-defined. (Use arbitrary elements from two equivalence classes [a] and [b] and how that the result is an element in [a] + [b]).
We briefly went over this in class and I again didn't understand it because of the whole modulo thing. I think it is something to do with taking two classes that are equivalent to mod 8 (which I am not sure how to figure out) and then proving that if you add them together you get whatever the question is asking.
That's just it though, I am not understanding the question nor how to find a part of what it is asking. Can anyone clarify this for me?
Help is appreciated!
2. I'll give it a try Yaeger- and hopefully someone more senior on the board will fix up the misakes. I guess you know how the mod function works...
We'll do it with mod 3 because that's easier:
Notice how (2 mod 3) = (5 mod 3) = 2. Well, when you think about mod 3 as being a relation on the set of whole numbers, there are infinitely many numbers (every 3rd member of the Set to be exact) that are related to 2 by the mod 3 relation. So when you start to write down the members of the relationship, you get:
$<br /> \rho:= \{(0, 0), (1, 1), (2, 2), (3, 0), (4, 1), (5, 2), ..., (8, 2), ... (2 + 3n, 2), ...\}<br />$
An equivalence class is a short-hand way of naming the three partitions of the relationship, that is all the members of the relationship that have either (x, 0), (y, 1) or (z, 3) in them, so $[1]_{mod 3} = [4]_{mod 3}$ are two equivalence classes which are congruent. Both of them are elements from the same representational system, or partition.
Another example of equivalence classes are in the rational numbers, where $\frac{1}{2} = \frac{2}{4} \ldots$
Question 2 just shows you how you can do some basic math stuff with them. Try it out.
3. Thanks for the post. I understand the equivalence class thing now. But I am still unsure about the second question. But I am currently trying to figure it out now.
Thanks again!
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http://gowers.wordpress.com/2010/08/21/icm2010-ngo-laudatio/
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# Gowers's Weblog
Mathematics related discussions
## ICM2010 — Ngô laudatio
Before I continue with brief descriptions of the laudationes, let me mention that Julie Rehmeyer has written descriptions of their work for a general audience and Terence Tao has now posted about the work of the Fields medallists and the other prizewinners. And as I have already said, the ICM website has links to the full texts of the laudationes themselves. So anybody now wanting to understand the mathematics has an excellent starting point, and I am free to concentrate on the more frivolous details of the talks, perhaps slipping in the odd mathematical comment as I do so.
Jim Arthur went next. His was the terrifying task (though much less terrifying for him than for most) of explaining the work of Ngô Bảo Châu to a general mathematical audience. I’d say that he did about as well as it is possible to do, which meant that he was able to convey some of the flavour, but obviously without managing to transmit to the non-expert the sort of wisdom tht it takes the experts in this particular area years to accumulate.
Ngô’s big result is a proof of the so-called Fundamental Lemma, conjectured by Langlands. Not bad to get a Fields medal for a lemma, you might think. The story here is that Langlands, while working out his famous programme, recognised the need for this lemma, and thought that it would be reasonably straightforward to prove. However, it turned out that he had wildly underestimated its difficulty. As Arthur put it, what Langlands was seeing was the visible part of the iceberg, and to prove the lemma it was necessary to uncover and understand the iceberg in its entirety. (I can’t remember his exact words, but he did not use the phrase “tip of the iceberg” and definitely did talk about icebergs.)
Arthur then gave a general description of the Langlands programme (having apologized in advance that much of what he was going to say would be “murky”). After the laudationes I found myself having coffee with Assaf Naor and Irit Dinur. (If you haven’t heard of them, they are both fabulous mathematicians, and both speaking here. In fact, Irit has just given her talk, the theoretical computer science plenary lecture, about the famous PCP theorem, of which she has a famous new proof.) The conversation turned to the topic of how many Fields medals could in theory be given for advances in the Langlands programme. My view was I suppose the official line, which is that it is such a deep and difficult area that any major advance is huge news, though I couldn’t resist a joke comparison to pole vault records, where people who are in a position to beat them deliberately don’t beat them by much because you get big money for beating world records. (I’m not seriously suggesting that somebody who had a proof of all the Langlands conjectures would sit on it, or release it only gradually.) Assaf (and I hope he won’t mind my making his views public) was more sceptical, maintaining that all mathematicians have their icebergs to explore and that the Langlands programme was not as unusual in this respect as perhaps it is sometimes conveyed as being. He said that he likes to ask the experts whether if they could assume all the results they wanted that are currently conjectural, they would know more about any concrete Diophantine equations. Apparently they don’t particularly like this question. Whether it is an appropriate criterion to judge the area is of course a matter for debate. In fact, that is what prompted me to say that perhaps the iceberg was the true and fascinating object of study in that area. I didn’t think of saying it at the time, but after a while there is not much interest in solving more and more Diophantine equations (not that Assaf was claiming that there was), and attention must turn to more global phenomena somehow. Perhaps that is what algebraic number theory is. I’m not sure why I’m musing on this at such length, but one more thought is that the question, “What is the most general statement of which Gauss’s law of quadratic reciprocity is a particular example?” is an obviously entirely valid and interesting one, and if I understand correctly, one of the Langlands conjectures is more or less a proposed answer to it.
I don’t understand what an automorphic form is, but there are levels of non-understanding (I would be enjoying several deeper ones later in the talk) and Jim Arthur lifted me to a slightly higher one — by which I mean that I had a slightly better idea what automorphic forms were after the next section of his talk. Before, I just thought of them as particularly nice kinds of functions that number theorists liked, and often mentioned in the same breath as modular forms (which I understand slightly better but still by no means fully). Anyhow, automorphic forms are eigenforms of natural operators on arithmetic symmetric spaces. He then said that these natural operators were Hecke operators, which themselves were Laplace-Beltrami operators on … er … I can’t remember. Hang on, we’ve got some spaces around — those symmetric spaces — so that’s OK.
What does one get out of a portion of talk like that? That is, what does one get out of hearing one concept one does not understand explained in terms of others? Let me try to say in this case. I don’t know exactly what an eigenform is but I presume it’s an eigenvector (and in fact he described them as simultaneous eigenvectors, so perhaps they were simultaneous eigenvectors for all Hecke operators — hmm, not sure about that). He talks about natural operators, which is obviously not meant to be precise and can therefore be understood in a non-precise way. Then he said “arithmetic symmetric spaces”. I don’t know what those are, though I imagine they are one of those definitions that is rather simple when you finally get told it. (I had that experience with algebraic groups, objects that I was afraid of until I learned that they were just groups where the set and the group operation are defined by means of polynomials.) I happen to know that a Laplace-Beltrami operator is what you get when you ask what the right analogue of the Laplace operator should be for a function defined on a manifold. (These last two definitions I know only as a result of editing the Princeton Companion to Mathematics, which forced me to pick up quite a lot of this kind of general knowledge.) And I’ve heard Hecke operators mentioned numerous times without ever actually finding out what they are. As a result of all that I now know that automorphic forms are eigenfunctions of operators that come up in a nice natural way in a number-theoretic context and that relate to all sorts of buzzwords I’ve heard many times. That doesn’t tell me exactly what an automorphic form is but it is non-trivial information. (What are they good for? There I cannot say anything that’s worth saying.)
Anyhow, the Langlands programme is about connecting automorphic forms with representation theory and looking at objects called automorphic representations. The difficulty of the area is this. One has some nice concrete operators (the Hecke operators mentioned above) and would like to know about their eigenvalues. However, just because an operator is concrete, it doesn’t mean you can write down its eigenvalues, and in this case you can’t. However, what you can hope to do is relate automorphic representations for different groups to each other, and this, if you manage it, gives you very deep reciprocity laws.
There’s something else called the principle of functoriality, which I won’t attempt to describe here even vaguely. Jim Arthur said, “The principle of functoriality awaits the efforts of future Fields medallists.”
The slides were getting more and more difficult to get anything out of by this stage, and I think I won’t say very much more. But Arthur gave us some idea of why the Fundamental Lemma has so many interesting consequences, and then started to explain what was so remarkable about Ngô’s proof.
If you want to impress your friends, here’s how to pretend you understand the proof in detail. If someone asks what his main idea was, you can reply, “Well, his deepest insight was to show that the Hitchin fibration of the anisotropic part of the trace formula is a Deligne-Mumford stack.” If that doesn’t do the job, then try to drop the phrase “perverse sheaves” into the conversation — they are relevant apparently. If you want to show that you have a broad view, then you could also say that Ngô very remarkably used global methods such as Hitchin fibrations to prove a local theorem. If you’re looking for a single amazing idea, then probably the use of Hitchin fibrations was it. Finally, here’s a list of names to splash about: Goresky, Hales, Kottwitz, Langlands, Laumon, MacPherson, Shelstad and Waldspurger. (This is apparently a far from complete list of the people on whose work Ngo builds.) A final summary from Jim Arthur: Ngô’s work opens up automorphic forms to some wonderful applications.
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### 27 Responses to “ICM2010 — Ngô laudatio”
1. David Ben-Zvi Says:
August 21, 2010 at 12:15 pm | Reply
Dear Tim,
I recently gave what I hope was an accessible – and certainly very informal! – talk on the fundamental lemma – notes and video at http://media.cit.utexas.edu/math-grasp/David_Ben-Zvi_lecture.html
(a much better attempt by David Nadler will appear in the Bulletin of AMS I believe). It seems one needs to find amateurs like me who are willing to lie enough to make this more broadly comprehensible.. In any case, to summarize, Ngo has made a great breakthrough in our understanding of the relation between eigenvalues, characteristic polynomials and conjugacy classes of matrices, in a form suited to deepen the ancient theme of the relation between conjugacy classes and representations in groups… Hope that helps.
David
• gowers Says:
August 21, 2010 at 12:20 pm
I’ll definitely check that out. Many thanks for letting me (and others) know.
• Andrea Says:
August 21, 2010 at 4:13 pm
I tried to watch the talk, but apparently my browser is not able to embed Quicktime video. Moreover, there is no link to download the file itself to view on a local video player. Would it be possible to either make the video available in another format or making the video available for download?
Thank you
• Andrea Says:
August 21, 2010 at 4:33 pm
Never mind, I’m able both to watch it embedded and to download it using the version for iPhone. Maybe it is just a problem with embedding a video with Flash?
2. Anonymous Says:
August 21, 2010 at 1:03 pm | Reply
Highly recommended. It’s an awesome talk for aspiring mathematicians. Especially for those interested in that area -it’s a great introduction-, or in David.
3. Todd Trimble Says:
August 21, 2010 at 2:22 pm | Reply
If you want to impress your friends
One of the things I find really heartening about these and all your blog posts is that this is obviously not what you’re setting out to do: impress your friends. The absolutely uninhibited intellectual honesty, not only in confessing ignorance and confusion (as in this post), but also thinking hard and out loud about ostensibly “elementary” topics on occasion. I cannot find a trace of glibness or braggadocio in anything you write, and that’s really appreciated. Wish more people would write this way!
Hard for me to guess whether being a Fields Medalist makes it easier or harder.
• Sune Kristian Jakobsen Says:
August 21, 2010 at 9:00 pm
Couldn’t agree more!
4. Lior Says:
August 21, 2010 at 2:33 pm | Reply
Regarding Hecke operators: on an arithmetic locally symmetric space, in addition to the notion of “nearby” coming from the Riemannian metric, there is a combinatorial notion of “nearby” coming from the number-theoretic symmetries of the manifold, in fact infinitely many such notions — each for one prime number.
In the simplest cases for all (but finitely many) prime numbers p there is a naturally defined graph (with vertex degrees depending on p) whose vertex set is the manifold. The “Hecke operator” is then the combinatorial graph Laplacian. Moreover, all the notions of “nearby” commute — locally speaking moving to a neighbour in the “p – graph structure” commutes with moving to a neighbour in the “q – graph structure” where p,q are different primes, and similarly moving a small distance in the Riemannian metric commutes with moving along one of the graphs. What this means is that all these Laplacians commute.
Since the Laplace-Beltrami operator and all these combinatorial Laplacians commute (and they are all self-adjoint) there are functions on the manifold which are joint eigenvectors of all of them, and these are the functions under consideration.
The structure of the graph you get is well-understood — except for a set of vertices of measure zero (in the usual sense on the manifold), the connected components are all trees, so the graph structure is basically a forest.
In greater generality what you get is not always a disjoint union of trees but usually a disjoint union of higher-dimensional simplicial complexes, but the basic idea is the same.
• gowers Says:
August 21, 2010 at 7:43 pm
Many thanks for that — the fog is clearing further as a result.
5. Harald Says:
August 22, 2010 at 6:40 am | Reply
I was wondering whether I would be the first person to be provoked into saying that Hecke operators *commute* with Laplace-Beltrami operators (though they can themselves be seen as a discrete or p-adic analogue of Laplace operators, roughly speaking). I think I am the second person.
• gowers Says:
August 22, 2010 at 7:13 am
Even though you are the second, it is only now that the message is sinking in.
6. Harald Says:
August 22, 2010 at 6:53 am | Reply
And modular forms are just the classical case of automorphic forms: modular forms are automorphic forms on the upper half plane (= the one-dimensional complex manifold consisting of complex numbers with positive real part, endowed with the usual Riemannian metric). To confuse matters slightly, people nowadays prefer to define automorphic forms on groups and their quotients; one then has to prove the not completely trivial fact that there is a simple transformation that gives you a bijection between modular forms (with certain parameters specified, e.g., the weight k) and automorphic forms on the group SL_2(R) (with the corresponding parameters specified accordingly – e.g., the same weight k).
7. Thomas Sauvaget Says:
August 22, 2010 at 7:16 am | Reply
Unrelatedly, I would like to mention something a bit curious that I’ve noticed in the abstracts booklet located at http://www.icm2010.in/wp-content/icmfiles/abstracts/Contributed-Abstracts-5July2010.pdf
Indeed while very estimeed researchers make short communications in the number theory section, e.g. Granville/Alon/Ubis-Martinez at page 102 about sumsets in finite fields, there appears to be very possibly bogus contributions by amateurs: I’ve found no less than 4 different abstracts claiming a proof of FLT!! See pages 72-73, 86-87, 92-93 and 99-100. I’m a bit puzzled how these have been accepted at all, I hope they won’t appear in the proceedings. The same phenomenom is perhaps occuring in other sections, I haven’t had a look there yet.
8. Assaf Naor Says:
August 22, 2010 at 1:43 pm | Reply
To clarify, I asked the above question in the context of our discussion on how one could explain long-term research programs in a talk intended for the general mathematical audience. The research programs of this type with which I am closely familiar do not usually lend themselves to such an exercise, but I suspect that the Langlands program might. If it were possible to give an elementary number theory result that is currently unproven, but would follow from the assumption that all the Langlands conjectures are true, then it would be a cool way to start a general audience talk on this topic. This isn’t necessary, it is definitely not a criterion to judge an area, and I completely agree with you that in good mathematics often attention must turn to more global phenomena. I asked a couple of experts, and they didn’t know of immediate applications to Diophantine equations off the top of their head, but my hunch is that there should be such applications that are easily stated, and if so it seems worthwhile to work them out, if only for the purpose of talks.
• gowers Says:
August 22, 2010 at 5:48 pm
Thanks for that — sorry not to have consulted you before reporting on our conversation, but I hope that I didn’t misrepresent your views too much, even if what you say in your comment is more nuanced …
• Emmanuel Kowalski Says:
August 22, 2010 at 6:44 pm
Along these lines, one of my first courses on automorphic forms introduced (parts of) Langlands’s conjectures as follows: if you look, for some fixed k, at the arithmetic function r_k(n) which is the number of representations of n as sum of k squares (say k=24 for concreteness), then there is an asymptotic formula where the main term is an “elementary” sum-of-divisor-type function, and an error term. How large can this error term be as n grows? The Ramanujan-Petersson gives the answer, and the point is that this can be predicted very easily using the conjectured (by Langlands) existence of what are called “symmetric powers” of classical modular forms (it’s based on a variant of the “tensor power” trick that Tao blogged about a while ago).
Now, it turns out that — in that special case — the desired conclusion was proved without going through symmetric powers. But that was not easy either: Deligne did this using both his proof of the general Riemann Hypothesis over finite fields, and some highly non-trivial algebraic geometry. But his proof of RH over finite field uses itself some tensor-power trick, and I’ve heard it suggested that this was inspired by Langlands’s conjectural approach.
In any case there are many instances of more general cases of the Ramanujan-Petersson conjecture which remain unproved… but would follow immediately from general forms of Langlands functoriality.
P.S. Incidentally, Deligne’s result for the Ramanujan tau function, which is more or less the case of 24 squares, is on the logo of the ICM 2010…
9. observer Says:
August 22, 2010 at 8:36 pm | Reply
Emmanuel: RH is an unbelievable problem because of the near misses. These other theories building around it: do they have a similar phenomenon?
10. Matthew Emerton Says:
August 24, 2010 at 9:43 pm | Reply
Here is one consequence of Langlands’s conjectures: if \$X\$ is a quotient of the upper half-plane by a congruence subgroup of SL_2(Z) (i.e. by the kernel of the map
SL_2(Z) –> SL_2(Z/nZ) for some n), then the least eigenvalue of the Laplacian on L_2(X) is at least 1/4.
(This is Selberg’s 1/4 conjecture; it is an analogue for the Laplacian of the Ramanujan–Petersson conjecture mentioned by Emmanuel. It follows from the same tensor power technique that Emmanuel describes; indeed, in Langlands’s viewpoint, the Laplacians and the Hecke operators have exactly the same status, and functoriality for symmetric powers of GL_2 (which is currently still a conjecture!) implies both Ramanujan—Petersson and Selberg.)
11. ohisee Says:
August 31, 2010 at 2:13 pm | Reply
Ngo Bao Chau — if I knew how to do circumflex accents I’d add them
I am a Vietnamese and I’ll help you add the accents: Ngô Bảo Châu.
• gowers Says:
August 31, 2010 at 2:27 pm
That was a great help — accents now added.
12. Fundamental Lemma and Hitchin Fibration « INNOVATION WORLD Says:
September 3, 2010 at 11:19 pm | Reply
[...] ICM2010 – Ngo laudatio (gowers.wordpress.com) [...]
13. Minhyong Kim Says:
September 12, 2010 at 6:17 am | Reply
In the course of this strange task I took on of explaining to a Korean journalist the significance of the fundamental lemma, I came upon this entry. Perhaps I can contribute one or two remarks on this rather complex topic.
As far as *direct applications* of the Langlands programme to Diophantine problems are concerned, they tend to be confined to the study of abelian varieties, that is, elliptic curves and their higher-dimensional generalizations. This derives from an approach to Diophantine equations that proceeds through the study of $L$-functions. That is, to an equation $X$, one can associate a function $L(X,s)$ in an complex variable $s$, that has the form of the Riemann zeta function in its Euler product form. It’s something of an article of faith that these functions encode deep arithmetic information about $X$. There are a number of situations where this faith seems reasonably well-justified, such as cubic equations in two variables (elliptic curves), and made precise in the conjectures of Birch and Swinnerton-Dyer. In any case, regardless of how devout you are, an embarrassing fact is that $L(X,s)$ is in general defined as a horrible infinite product over primes, and it’s not clear that it’s a reasonable function of any sort at all. Another article of faith, which comes in some sense before the previous one. is that this function is indeed nice: It should have a continuation to the whole plane, and satisfy a natural functional equation. One of the main points of the Langlands programme is to prove this statement through an identification
\$$L(X,s)=L(f,s),$\$
where $f$ is one of these things called an automorphic form. [Actually, it should in general be a more complicated object called an automorphic representation of a large group, \$GL_n\$ with entries in the adeles of an algebraic number field. A rather simple way to show sophistication in this subject, even without going into all the buzzwords mentioned at the end of the post, is to casually use the two terms `automorphic form' and `automorphic representation' interchangeably.] Because \$L(f,s)\$ is associated to a topological group, harmonic analysis has been used to show already all the nice analytic properties one could hope for.
Anyways, the point I am trying to make is that the Langlands programme does propose to contribute to a deep and systematic study of *all* equations. Unfortunately, the kind of information contained in these $L(X,s)$ tend to of the *abelianized* sort, something like the free-abelian group generated by the solutions modulo some geometric relations. The reason there are direct implications in the case of abelian varieties is because the naive solutions then form a group, so that the abelianized information and the `direct’ information coincide. It is a somewhat speculative but serious challenge for arithmeticians to come up with a non-abelian version of this whole picture, whereby one might come to a systematic (albeit conjectural) understanding of naive solution sets for rather general equations in a manner resembling the abelian situation.
Meanwhile, I might remark also that one can always try to use the understanding of elliptic curves or abelian varieties gained through $L$-functions to study other equations, for example, by getting solutions of an equation to *parametrize* elliptic curves or abelian varieties. This is what happened in the theorems of Wiles and of Faltings.
My feeling is this old ICM lecture of Langlands paints a good picture of the subject in very broad brushstrokes. Perhaps one sentence there worth retaining is
`However, all evidence indicates there are fewer $L$-functions than the definitions suggest, and that every $L$-function, motivic or automorphic, is equal to a standard $L$ function.
The mysterious term `motivic’ $L$-function refers (essentially) to one associated to a Diophantine equation, while a standard $L$- function is one coming from an automorphic represention of $GL_n$.
Finally, I believe the fundamental lemma itself has direct application to a version of the $p$-adic analogue of the BSD conjecture, as studied by Chris Skinner and Eric Urban. [I'm not entirely sure about this though, and I hope some real expert will confirm or deny this.]
14. Minhyong Kim Says:
September 12, 2010 at 6:26 am | Reply
One more remark: as implied in the Langlands quote, there are $L$-functions associated to automorphic representations of more general groups, such as unitary groups or symplectic groups. However, Langlands is saying that as far as $L$-functions are concerned, we should always be able to reduce to the $GL_n$ case, where everything is relatively simple. Ngo’s theorem implies some special but important cases of this last statement (and more).
15. Fundamental Lemma Proof – Ngo Bao Chau « INNOWORLD Says:
September 14, 2010 at 5:22 pm | Reply
[...] ICM2010 – Ngo laudatio (gowers.wordpress.com) [...]
16. Happenings – 2011 Oct 15 « Rip’s Applied Mathematics Blog Says:
October 15, 2011 at 7:51 pm | Reply
[...] certifiably one of the best mathematicians in the world. Let me grab a couple of sentences from him (here on Tim Gowers’ blog):. I don’t understand what an automorphic form is, but there are levels of non-understanding (I [...]
17. The Two Cultures of Mathematics: A Rebuttal | Persiflage Says:
December 12, 2012 at 5:50 am | Reply
[...] Gowers main point is that a significant part of the mathematical establishment looks down on combinatorics as not being “deep”, and that this attitude is both harmful and ignorant. On this point, I think that Gowers criticisms are fair, accurate, and valuable. It’s undeniably true that there are many graduate students who fall in love with formalism to the detriment of content, and milder forms of this predujice are pervasive throughout mathematics. To this end, I think Gowers’ essay is timely and relevant. However, I can’t help but sense a little that, perhaps after having spent a career defending combinatorics against ignorant snobs, Gowers suffers from the opposite prejudice, where “theory-builders” are a short distance away from empty formalists, sitting comfortably in their armchairs thinking deep thoughts, studying questions so self referential that they no longer have any application to the original questions which motivated them (this sense also comes from reading some of the remarks on the Langlands programme here). [...]
18. Research Life-Stories: Bobby Kleinberg | Windows On Theory Says:
March 21, 2013 at 3:22 pm | Reply
[...] for the mathematics community as a whole. My feelings on this subject were reinforced in 2010 when experts tried in vain to explain the work of Fields Medalist Ngô Bảu Châu to the mathematical [...]
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http://mathhelpforum.com/differential-equations/205992-laplace-s-equation-3-dimension-orthogonality-eigen-function.html
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1Thanks
• 1 Post By chiro
Thread:
1. Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)
Any help will be appreciated!
I am solving Laplace's Equation in 3-Dimensions using superposition.
The problem is that I don't know how to use orthogonality to solve for the constant when you have two Eigen functions.
$\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}=0$
$u(x,y,z)=v(x,y,z)+w(x,y,z)$
This is my attempt to get a solution for v(x,y,z).
$v(x,y,z)=A(x)B(y)C(z)$
The solution using the given homogeneous boundary conditions comes out to be:
$v(x,y,z)=\sum Ksinh(\lambda z)sin(n\pi x)sin(m\pi y)$
where $n:1\rightarrow \infty$ and $m:1\rightarrow \infty$
The remaining boundary condition is:
$v(x,y,1)=1$
Which gives:
$v(x,y,1)=\sum Ksinh(\lambda)sin(n\pi x)sin(m\pi y)=1$
How do I use orthogonality to find K when I have two eigen functions in the solution?
2. Re: Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)
Hey bilalsaeedkhan.
For the orthogonality condition, do you use the L^2 inner product and if so what are the limits of the integral?
3. Re: Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)
Originally Posted by chiro
Hey bilalsaeedkhan.
For the orthogonality condition, do you use the L^2 inner product and if so what are the limits of the integral?
I don't know what the L^2 inner product is so cant say anything about that. But I was able to solve it though. Thanks for taking a look at it.
4. Re: Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)
Just for your future reference, we can establish orthogonality between functions if we get <f,g> = 0 where <f,g> = Integral fg over the right interval.
Different inner products can have different intervals of integration and it depends on what your full space is (for example on space would be all functions on the interval [-1,1] while another might be all of R and subsequently every space has its own issues with how to construct a basis).
So with this if you have an inner product defined in terms of an L^2 integral inner product, you can check whether they are orthogonal and this is the kind of thing that happens in Fourier Analysis.
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http://www.physicsforums.com/showthread.php?p=4192868
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Physics Forums
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Mentor
## Does stock market create wealth?
Quote by OmCheeto You sound quite knowledgeable in this topic. I don't mean to change the subject, but I just lost the equivalent of 10% of my annual salary the other day in a stock. It's IPO was $2 billion. I was buying shares at$20 a pop, and had no fear, as they made an exceptional product, and planned on selling at $30/share. Over the last 4 years, the price went down to 5 cents a share. And now the company is bankrupt. Though shares are still being traded over the counter for 8 cents a share, yielding a market cap of around$11 million. The problem with the picture though, is that there is another company willing to buy my company for over \$500 million. Does anyone know what it means when there is a 50:1 difference like that?
My guess would be that this is a company who's value is heavily influenced by speculation. Typically, a company trying to buy another company must bid more than the current market cap because:
1. Via supply and demand, trying to buy a large quantity of stock is a new demand.
2. Knowing that someone wants to buy the company gives current and new prospective stockholders renewed confidence in the value.
Sorry, but I believe Russ was talking about me when he mentioned "stupid".
I wasn't referring to you and as I know very little about your finances, I have no idea if you are generally a smart or stupid investor, but if you send me a copy of all of your financial statements, I can provide a more informed judgement.
Mentor
Quote by Tosh5457 That argument was made before in the thread and I replied to it. If you exclude dividends, by owning shares of a company, and in a way that it doesn't allow you to control the company and therefore get the profits from the company by other mechanism than dividends (that doesn't happen in the stock market, that's a takeover) you don't own anything of the company. If you have 10% of the shares of a company, can you sell 10% of their capital? Can you even sell anything that the company owns to make a profit? You can't, because you don't own anything of it.
This is silly. Stock is by definition ownership of a company. The fact that you don't have the ability to make decisions about the company on your own based on your small share doesn't change that. Other forms of shared ownership can work the same way, such as some joint bank accounts or a married couple who owns a house.
If I define profit of an investor as money spent or received in the stock market, if I buy $1000 of a stock my profit will be -$1000 and the profit of the seller will be +\$1000. Using that definition it's obvious that it is going to be a zero-sum game in respect to the profits.
You've improperly defined "profit" based on a built-in assumption that the stock itself has no value. Your argument is circular.
If I define profit of an investor in a given trade as: In case of buying, profit = 0. In case of selling, profit = Price that the investor sold the stock - Price that the investor acquired the stock This is the usual and intuitive definition of profit. ... Sum of the variations of balances = \$0
This is better because at least you correctly define the company as having value, but it is still incomplete as it assumes the value to be fixed. Again, you are using circular reasoning to prove that it is zero-sum.
Conclusions: The sum of the money in the balances of the investors is equal to the money in the system. From that it follows that if the money in a given system doesn't vary, the sum of the variations of the balances of the investors is always \$0.
But we know that this is false, don't we? The amount of money in the system does vary. It grows. Therefore:
what one gains, came from another one's pockets, which was what I was attempting to argue, although with wrong terms.
And by implication, if one person gains another has to lose. It's still wrong. Doesn't matter how many times you say it and your argument is getting worse, since even as a Ponzi scheme the amount of cash in the system is not fixed.
Let me put a finer point on the not-zero-sum issue: Over the past 100+ years, the stock market has averaged a roughly 5% annual growth rate after inflation. That means that millions of people, over several generations, have put money into the stock market and then later in life have taken out roughly 4x as much as they put in.
I think it's a good idea to think of a stock purchase as an investment in a business. If the business does well, your investment increases in value. If the business fails, your investment loses value.
Mentor
A little more on this and the zero-sum assumption:
Quote by Tosh what one gains, came from another one's pockets
It is true that every individual transaction is zero-sum. This is necessary in a fair system: otherwise one party is cheating the other. But don't confuse this into meaning that the entire system has zero value. What you are missing is what happens in between those transactions. In between those transactions, the economy grows. And both sides of the transaction grow together. They are reflections of each other. The next buyer has more money because the economy grew and s/he earned it. The company s/he's buying is worth more because it sold more widgets and has earned more money.
Don't fret about growing stock value requiring a growing amount of cash available to buy stocks: there will be a growing amount of available cash as long as the economy is growing.
And just to mention the other side of the coin; buying a car and selling it later is a losing proposition. The value depreciates over time because the car wears out. The car isn't magically worth less the next time you sell it (there is no violation of the zero-sum game of a transaction), it really does have less value.
Mentor ...then consider fine art. Fine art has roughly zero inherrent value. Impressionist paintings were once considered amateurish, then people decided they like the style and vision. Now they have huge values. But what if peoples' sense of style changes again? At least impressionists display style and vision, if not skill. What about pop art? What if people decide a can of soup, painted with no particular skill is barely worthy of a high school art class? Boom. Worthless.
As for the definition of profit, it doesn't matter when you only look at balances.
But we know that this is false, don't we? The amount of money in the system does vary. It grows. Therefore:
It's ridiculous that you didn't quote the part where I assumed that the money varied... And it's also ridiculous that you're using the argument that because the money in the system varies, then it's not a zero-sum game. You're forgetting the fact that when money enters the system, there is 1 more player in the game, therefore my conclusion still holds. That argument of yours also works to "prove" poker is a non-zero sum game
What you're saying is that there is creation of money.
Let me put a finer point on the not-zero-sum issue: Over the past 100+ years, the stock market has averaged a roughly 5% annual growth rate after inflation. That means that millions of people, over several generations, have put money into the stock market and then later in life have taken out roughly 4x as much as they put in.
Dividends and more money coming into the market coupled with positive expectations due to the ever-growing economy (this last factor doesn't make it a non-zero sum game, while still allowing the investors that had stock to get profit).
What you are missing is what happens in between those transactions. In between those transactions, the economy grows. And both sides of the transaction grow together. They are reflections of each other. The next buyer has more money because the economy grew and s/he earned it. The company s/he's buying is worth more because it sold more widgets and has earned more money.
The money on the side of the buyers growing is the same as saying more players with money entered in the stock market game, and that alone doesn't mean it's not a zero-sum game, it still is. Now on the other side, the companies grow, that's true. But that richness they earn don't pass to the stockholders. You argue it does, and I argue the only mechanism that exists for that are dividends. You can't describe the mechanism by which that happens. Of course if you accept that stocks can grow in the long-term just because of offer and demand in the secondary market you don't have to assume that mechanism exists, but since you can't accept that you have to assume it does.
Quote by russ_watters ...then consider fine art. Fine art has roughly zero inherrent value. Impressionist paintings were once considered amateurish, then people decided they like the style and vision. Now they have huge values. But what if peoples' sense of style changes again? At least impressionists display style and vision, if not skill. What about pop art? What if people decide a can of soup, painted with no particular skill is barely worthy of a high school art class? Boom. Worthless.
Good that you bring up that example, because it shows how vague a term value is. It's so vague that I can say there is infinite or 0 value in the world and nobody can say if I'm right or wrong. You're trying to prove me wrong by using a vague term like that, you have to rethink what you're doing.
I propose a definition of value using money - the value of any product is the money that it can be sold for. Using that definition I arrive at the conclusions I said before. Nonetheless, I don't think that term even needs to be defined, because what I'm saying applies to traders' balances, which is in money. If you can give another definition for value and explain why it's a convenient definition we can continue the discussion, or else I can't discuss with you because you're using a ill-defined concept to prove me wrong.
I can't believe this is still going on. All the proof anyone needs is to accept that if a unit of stock has a right to a stream of cash flows earned in by a business in the real economy, which is not a zero sum game, then a market for these certificates is not a zero sum game. If owing privately held business is not a zero sum game then how can owning a partial interest in a public company on be one? Why don't you find some real economists who support your view and have peer-reviewed research supporting it?
Mentor
Quote by Tosh5457 As for the definition of profit, it doesn't matter when you only look at balances.
Yes, correct definitions and assumptions matter. If you assume things that are wrong, you will likely get wrong conclusions from the logic. That's a basic principle of logic.
It's ridiculous that you didn't quote the part where I assumed that the money varied...
I consider it ridiculous that you say you accept that the economy has a continuously increasing (over the long-term) total value, then build models that assume otherwise!
And it's also ridiculous that you're using the argument that because the money in the system varies, then it's not a zero-sum game.
It seems like you are losing focus on what your argument is here. The problem isn't the money in the market, the problem is with the value of the stocks. We both agree that investors are continuously pumping cash into the stock market. What we're disagreeing on is whether there is a corresponding change in real value of the stocks themselves based on the value of the companies.
You're forgetting the fact that when money enters the system, there is 1 more player in the game, therefore my conclusion still holds. That argument of yours also works to "prove" poker is a non-zero sum game
Er, no:
Standard tournament poker has a fixed initial value and a fixed starting number of players, with players dropping out as they run out of money. It is an inherently degenerative situation, which is why it doesn't find an equilibrium but rather results in a single person holding all the money.
Casino poker on the other hand has a continuously changing quantity and roster of players and continuously changing pool of money. If you were to remove the house "take", it would overall be zero-sum, with winners and losers exactly in balance. Many people erroneously believe the stock market follows this model. It doesn't because the betters in the casino are trading cards, not stocks. Cards have no inherent value, stocks have a value....that is increasing.
But you said a growing roster of players. A casino poker game does not have a growing roster of players over the long term, but the economy does. Perhaps that's where your problem lies:
The fact that the economy is growing partly due to population growth, which therefore leads to more investors adding more money to the system, may make it appear similar to a pyramid scheme. But a closer look shows that the economy and stock market grow faster than the population (indeed, many European countries are shrinking, not growing). Why? Added value. Time (labor), intellectual property creation, and cash dug out of the ground (crops and minerals) add value to the economy faster than consumables (food, cars, etc.) take money out.
The growing population is giving you a false impression that the market is a pyramid scheme. Even if the number of investors remained static, the value would still grow in a growing economy.
What you're saying is that there is creation of money.
There is creation of money and value! That's the entire issue we're arguing about!
I have to say, it seems like you're really shooting from the hip here and losing track of the argument -- like you're not trying to learn but are just arguing for the sake of arguing, regardless of where your argument leads you.
The money on the side of the buyers growing is the same as saying more players with money entered in the stock market game, and that alone doesn't mean it's not a zero-sum game, it still is.
That is true only if the amount of money each player brings in when he/she enters stays constant. It doesn't. If each new player brought in as much money as the last, it could be like a pyramid scheme because growth would require new investors. But again, the growth rate exceeds the rate of new investors entering because the economy is growing faster than the population.
Now on the other side, the companies grow, that's true. But that richness they earn don't pass to the stockholders. You argue it does, and I argue the only mechanism that exists for that are dividends. You can't describe the mechanism by which that happens. Of course if you accept that stocks can grow in the long-term just because of offer and demand in the secondary market you don't have to assume that mechanism exists, but since you can't accept that you have to assume it does.
You missed something here because I have never argued that the value passes or has to pass to the shareholders. Others have argued it, but I avoided it because it is an unnecessary complication: It doesn't have to for stocks to have or gain value. Dividends are really a separate source of revenue entirely. A company could grow at a rate equal to excess revenue, which results in it never turning a profit, but in growing it still gains value for the stockholders. Or it could stick the profits in a bank account, with the same result.
Again, you are making the mistake of thinking that a person has to be able to directly access the assets of the company in order for their stock to hold value. All that has to exist is the theoretical possibility that they could. I think you are letting the size of companies make you think there is something else going on in larger companies than happens in smaller companies. In smaller companies, it is easy to see:
Owners of companies are shareholders, regardless of the size of a company or if the owners do any of the work. But if you look at small, direct ownership, it becomes easy:
If two people each use $10,000 to buy equipment start a company together, each now owns half of a$20,000 company. If the company turns a $10,000 a year profit for 20 years, but the owners stick that money into a bank account instead of taking it out of the company in bonuses, the company now has$20,000 worth of equipment and a bank account with $200,000 in it, for a total value of$220,000.
Now one of the partners wants to retire. He has a piece of paper that says he owns half of the company. The partnership agreement and negotiation will determine exactly what he can do with it and how it works, but typically the options are:
1. Sell his half of the company to the other partner. Now the initial investment was $10,000 but since the sole owner could just sell the assets and pocket the money in the bank account, it doesn't make sense to sell$100,000 in cash and $10,000 in equipment for$10,000. The sale price has to be about \$120,000.
2. Sell half to a new shareholder. Same valuation.
3. Force the other shareholder to dissolve the company. Same valuation.
No matter how you slice it, when the shareholder wants out, he pockets $110,000 for a return on his investment of$100,000. Whether a new shareholder enters the game or not.*
Now here's the part where you seem to be slipping up: At any time during those 20 years, the shareholders could make the decisions above. The fact that they don't doesn't mean that the company has zero value (or a constant value of \$20,000) in the meantime. The company has a higher value because they could.
*This made me think of another important point. One of your issues here is that you think that money can only pass through the market from the company to the shareholders without dividends. Actually, it can: the company can buy back the stock. Sometimes they do that and the buyback price does not have to equal the original issuing price. I'm annoyed I didn't think of that example before.
Mentor
Quote by Tosh5457 Good that you bring up that example, because it shows how vague a term value is. It's so vague that I can say there is infinite or 0 value in the world and nobody can say if I'm right or wrong. You're trying to prove me wrong by using a vague term like that, you have to rethink what you're doing.
You completely missed the point of the example. The example was highlighting a difference not a similarity. I assumed it would be obvious, so I didn't explain the other side of the coin:
Fine art has zero inherrent value. Products with uses do have inherrent value -- even if that value is difficult to calculate. The value of a house can never be exactly zero because even if no one wants to buy it, you can still live in it. A gallon of oil can never have exactly zero value because even if the commodity price crashes, you can still heat your house with it. Etc. And a company has a bank account with money in it, physical assets and the potential for more in the future.
So yes, I can say you are wrong. And I can see that your main problem here is a misunderstanding of the concept of value. You've applied this misunderstanding to the stock market here, but as I suspected that's just entrance point to a larger problem.
I propose a definition of value using money - the value of any product is the money that it can be sold for. Using that definition I arrive at the conclusions I said before. Nonetheless, I don't think that term even needs to be defined, because what I'm saying applies to traders' balances, which is in money. If you can give another definition for value and explain why it's a convenient definition we can continue the discussion, or else I can't discuss with you because you're using a ill-defined concept to prove me wrong.
Er....if you do that, then you miss inflation and this becomes even easier. *POOF* more money. *POOF* more money. *POOF* more money. We literally print it and inject it into the economy. So no, I don't think you really meant that this is just about the number of dollars. The problem still is your understanding of the concept of value. That statement was just another symptom of it. You've backed the wrong horse: money (particularly paper money) has little inherrent value and in some cases actually has had enormous swings in useful value due to run-away inflation and collapsing countries. So money can have a high value (never infinite though) or essentially zero value. If you're looking for a stable measure of value you picked probably the worst way to measure it.
The reality is that money is just a carrier medium for value (or "wealth", the word you used in the title). If the world's economy collapsed and rendered all money valueless, the lack of cash would not change the actual value of needed products, just the way it is expressed (though changing priorities would change the actual value). Values might be compared directly: A house is worth four cars. A car is worth fourteen cows. A share of facebook stock is worth eleven chickens and half a squirrel, etc. All the elimination of money does is make stocks messier to trade.
If you meant to also include the assumption that inflation is assumed to be nonexistent for the purpose of the model then the value of money is fixed and the quantity of money in the market grows, reasonably accurately reflecting the growth in the actual value of the market.
Want to simplify matters even further? Assume P/E ratio is fixed as well, eliminating the effect of speculation. Then money has constant value and stocks still have increasing value, in direct and fixed proportion to the earnings of the company.
No, these assumptions do not help your model produce the outcome you are looking for, they are just causing you additional confusion.
Mentor
Quote by BWV Why don't you find some real economists who support your view and have peer-reviewed research supporting it?
Not a bad idea, even for our side of the argument. Peer reviewed may be a problem for such a basic issue, but there is no shortage of investor help sites and publications that discuss this. It is a very common question and googling "how stocks gain value" provides some good results that read as if they were written for this thread:
The first link discusses our entry-point into the problem, listing the two basic ways that stock values change (speculation that changes P/E ratio and increase in real value), a clearer description of what I was talking about in post #8:
Mathematically, we can divide all stock price changes into just two categories: 1. A stock's price can change because its multiple(s) change. This means that stock traders change their view of what a stock is worth without any underlying change in the stocks achieved revenues or earnings. For example the (trailing) P/E ratio or multiple changes, or the Price to Book value ratio changes. Generally this means that the outlook for future earnings has become more positive or more negative or the required rate of return on the stock has changed. 2. A stock's fundamentals change as a result of releasing updated financial data. For example the stock's book value, trailing 12 months revenue or trailing 12 month's earnings changes when it releases financial performance for the latest quarter. Category 1 (multiple changes) are responsible for almost all of the day-to-day, minute-to minute, movement in stock prices. Category 2 (fundamental growth) is responsible for most of the long term change in a stock's price over a period of years.
http://www.investorsfriend.com/price_increases.htm
But this doesn't explain the mechanism of how that value is responsible for the gain in stock price. But this one does and even better is about gaining value without considering dividends -- even asking why it isn't a pyramid scheme(!):
New investors often want to know: If a stock doesn't pay dividends, isn't buying it like participating in a Ponzi scheme because your return depends on what the next guy in line is willing to pay for your shares? That is a very good question and it's important you understand the answer.
This article is right on point and also includes an example that admittedly I based my above example on (my version is simpler). Bottom line remains the same:
On Wall Street, the same holds true for huge companies. Take Berkshire Hathaway. The stock has gone from $8 to more than$100,000 per share over 40+ years because Warren Buffett has reinvested the profits into other investments. When he took over, the company owned nothing but some unprofitable textile mills. Today, Berkshire owns 13.1% of American Express, 8.6% of Coca-Cola, 5.7% of ConocoPhillips, 1.1% of Johnson & Johnson, 8.9% of Kraft Foods, 3.1% of Procter & Gamble, 4.3% of U.S. Bancorp, 0.5% of Wal-Mart Stores, 18.4% of The Washington Post, 7.2% of Wells Fargo, and totally controls GEICO, Dairy Queen, MidAmerican Energy, Helzburg Diamonds, Nebraska Furniture Mart, Benjamin Moore Paints, NetJets, See's Candies, and much more. That doesn't even include the fact that the holding company just spent $44 billion to buy Burlington Northern Santa Fe. Is Berkshire worth$102,000+ per share? Absolutely. Even if it doesn't pay out those earnings now, it has hundreds of billions of dollars in assets that could be sold, and generates tens of billions of dollars in profit each year. That has value, even if the shareholders don't get the benefit in the form of cash, because the Board of Directors could literally turn on the spigot and start paying massive dividends tomorrow.
Berkshire Hathaway is a good example because it has never had a stock split and never paid a dividend. So its value is soley due to the value of the companies the fund owns.
Ok I'll just ignore your multiple fallacies and misunderstandings, I know you're absolutely convinced I'm wrong, and I'm absolutely convinced you're wrong. So let's just get to the fundamental point where we disagree, which you correctly pointed out:
What we're disagreeing on is whether there is a corresponding change in real value of the stocks themselves based on the value of the companies.
To be clear, we're talking about stock holders that don't use their voting rights and companies with no dividends, which applies to the great majority of the traders, and it's what I'm interested in. But first I need to know, what do you mean by real value? Is it the money it's worth on the secondary market, or other thing?
Mentor We don't need to argue about the concept of value. We are getting bogged down in details when the problem is really that you don't accept a fundamental fact: We both agree that if a company has a big, fat bank account, then by definition, the owners of the company - the stockholders - own the money in the bank account. You believe that since the stockholders can't access the money in the account, it can't or shouldn't affect the stock price. You are wrong on the premise and therefore wrong in the conclusion: they do, so it does. It may not be easy and few may choose to do it, but stockholders can access the wealth of the company. That's the fundamental fact that you refuse to accept. The complexity of large companies is probably what is tripping you up, so I encourage you to start by reading the examples given for small companies and accepting it for them.
Recognitions:
Science Advisor
Mathematically, we can divide all stock price changes into just two categories: 1. A stock's price can change because its multiple(s) change.... 2. A stock's fundamentals change as a result of releasing updated financial data...
No. The only reason a stock's price rises is because there are more buyers than sellers. Your quute is just two ways to rationalize WHY there are more buyers.
And the fundamental mistake of most economic theoriies is the assumption that decisions made by humans are always rational.
Recognitions:
Gold Member
Quote by russ_watters Owners of companies are shareholders, regardless of the size of a company or if the owners do any of the work. But if you look at small, direct ownership, it becomes easy: If two people each use $10,000 to buy equipment start a company together, each now owns half of a$20,000 company. If the company turns a $10,000 a year profit for 20 years, but the owners stick that money into a bank account instead of taking it out of the company in bonuses, the company now has$20,000 worth of equipment and a bank account with $200,000 in it, for a total value of$220,000.
If the company instead had a loss of 10k a year for 20 years do the investors now owe that money?
Is the disconnect only one way? >debt not mine as an investor...but profits are?
There is a very very clear segregation between what a stock holder is entitled to, and it absolutely isn't (not that it couldn't be written) the cash in a bank account.
At what point is the there a direct connection from a companies equity to the stock price?
Even an IPO is valuation. There is no calculation assets - liabilities = equity / number of shares = share price.
Of course it is up the buyers to determine what the value is.
There is no way to account for all variables.
For you simplified example there could be a host of issue. Was the retiring partner an expert who was the real value of the company?
Is their product/service now obsolete? Will there be higher then ever demand for their product/service.
Mentor
Quote by nitsuj If the company instead had a loss of 10k a year for 20 years do the investors now owe that money?
Investors paid the initial start-up money to the company and that's their primary risk. It's basically a loan. If a bank also loaned money to the company, it took a similar risk. Or if the company didn't pay its vendors, it would owe them. A company can't just run a continuous loss: the money has to be loaned to the company to cause the loss. At some point, people stop loaning the company money and it goes bankrupt.
Who gets what back is complicated and based on bankruptcy laws. Typically, shareholders are among the last ones to get money back. They have the most to gain if the company does well and the most to lose if it does poorly.
There is a very very clear segregation between what a stock holder is entitled to, and it absolutely isn't (not that it couldn't be written) the cash in a bank account.
Do you have a reference to that assertion? It contradicts the source I linked above and logic: That's why investors complain if a company has a large cash reserve, but isn't paying a dividend.
Even an IPO is valuation. There is no calculation assets - liabilities = equity / number of shares = share price. Of course it is up the buyers to determine what the value is.
Er, no: that calculation is done (much more complicated, of course) and the company sets the IPO price. You have that exactly backwards.
Didn't you follow the Facebook IPO debacle? The controversy was over the fact that the investment company doing the paperwork to make the IPO happen was also the one who calculated the selling price, which gives them a conflict of interest. When they jacked-up the price right before the IPO, then the stock plunged right after, the initial investors felt they got cheated.
Mentor
Quote by AlephZero No. The only reason a stock's price rises is because there are more buyers than sellers. Your quute is just two ways to rationalize WHY there are more buyers.
Yes, that's an explanation of why. There is nothing to be arguing about there. Even though you said "no", you didn't actually disagree.
And the fundamental mistake of most economic theoriies is the assumption that decisions made by humans are always rational.
I doubt any economist ever makes such a mistake. It certainly doesn't appear here: #1 in that quote is all about irrationality. So I don't know why you'd bring that up. Doesn't seem to have any relevance.
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## Why do we teach calculus students the derivative as a limit?
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I'm not teaching calculus right now, but I talk to someone who does, and the question that came up is why emphasize the $h \to 0$ definition of a derivative to calculus students?
Something a teacher might do is ask students to calculate the derivative of a function like $3x^2$ using this definition on an exam, but it makes me wonder what the point of doing something like that is. Once one sees the definition and learns the basic rules, you can basically calculate the derivative of a lot of reasonable functions quickly. I tried to turn that around and ask myself if there are good examples of a function (that calculus students would understand) where there isn't already a well-established rule for taking the derivative. The best I could come up with is a piecewise defined function, but that's no good at all.
More practically, this question came up because when trying to get students to do this, they seemed rather impatient (and maybe angry?) at why they couldn't use the "shortcut" (that they learned from friends or whatever).
So here's an actual question:
What benefit is there in emphasizing (or even introducing) to calculus students the $h \to 0$ definition of a derivative (presuming there is a better way to do this?) and secondly, does anyone out there actually use this definition to calculate a derivative that couldn't be obtained by a known symbolic rule? I'd prefer a function whose definition could be understood by a student studying first-year calculus.
I'm not trying to say that this is bad (or good), I just couldn't come up with any good reasons one way or the other myself.
EDIT: I appreciate all of the responses, but I think my question as posed is too vague. I was worried about being too specific, so let me just tell you the context and apologize for misleading the discussion. This is about teaching first-semester calculus to students straight out of high school in the US, most of whom have already taken a calculus course in high school (and didn't do well or retake it for whatever reason). These are mostly students who have no interest in mathematics (the cause for this is a different discussion I guess) and usually are only taking calculus to fulfill some university requirement. So their view of the instructor trying to get them to learn how to calculate derivatives from the definition on an assignment or on an exam is that they are just making them learn some long, arbitrary way of something that they already have better tools for.
I apologize but I don't really accept the answer of "we teach the limit definition because we need a definition and that's how we do mathematics". I know I am being unfair in my paraphrasing, and I am NOT trying to say that we should not teach definitions. I was trying to understand how one answers the students' common question: "Why can't we just do this the easy way?" (and this was an overwhelming response on a recent mini-evaluation given to them). I like the answer of $\exp(-1/x^2)$ for the purpose of this question though.
It's hard to get students to take you seriously when they think that you're only interested in making them jump through hoops. As a more extreme example, I recall that as an undergraduate, some of my friends who took first year calculus (depending on the instructor) were given an oral exam at the end of the semester in which they would have to give a proof of one of 10 preselected theorems from the class. This seemed completely pointless to me and would only further isolate students from being interested in math, so why are things like this done?
Anyway, sorry for wasting a lot of your time with my poorly-phrased question. I know MathOverflow is not a place for discussions, and I don't want this to degenerate into one, so sorry again and I'll accept an answer (though there were many good ones addressing different points).
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What competing definition do you have in mind? – Pete L. Clark Sep 27 2010 at 5:34
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Maybe I misunderstand your question. But what would be the point of teaching students the symbolic rules as axioms without explaining to them how they are derived? Would you advocate teaching maths undergraduates the combinatorial properties satisfied by character tables of finite groups, so that they can work out the tables in most cases, without proving any of the properties or maybe even without explaining what a character is? – Alex Bartel Sep 27 2010 at 5:41
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I think your only alternative is to present the "magic" differentiation rules with no justification. It is already common for students to have a black-box view of mathematics; I don't think you want to encourage it. Perhaps you want to begin with the definition via limits and then derive the rules from there. Emphasize to your students that "Why didn't we just use the rule from the start?" is not a valid question. The rule is a consequence of the definition, not a self-evident truth. – Austin Mohr Sep 27 2010 at 5:42
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there is an article by Solomon Friedberg entitled "Teaching mathematic graduate students how to teach" in the Notices of the AMS (52) 2005, where the question you ask and its didactical implications is part of a "case study". – Holger Partsch Sep 27 2010 at 11:35
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If calculus class were devoted to the project of getting students to learn to appreciate mathematics by a process that resembles mathematics (which they aren't, perhaps with good reason), then one could do this by simply holding off on the introduction of the power, product, chain and quotient rules. The geometric problem of computing tangent lines is natural and easy to motivate; the limit definition is reasonably easy to motivate from the geometric problem; and then students could spend reasonable amount of time flailing around trying to compute derivatives of different functions. (Cot'd) – JBL Sep 27 2010 at 14:04
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## 29 Answers
This is a good question, given the way calculus is currently taught, which for me says more about the sad state of math education, rather than the material itself. All calculus textbooks and teachers claim that they are trying to teach what calculus is and how to use it. However, in the end most exams test mostly for the students' ability to turn a word problem into a formula and find the symbolic derivative for that formula. So it is not surprising that virtually all students and not a few teachers believe that calculus means symbolic differentiation and integration.
My view is almost exactly the opposite. I would like to see symbolic manipulation banished from, say, the first semester of calculus. Instead, I would like to see the first semester focused purely on what the derivative and definite integral (not the indefinite integral) are and what they are useful for. If you're not sure how this is possible without all the rules of differentiation and antidifferentiation, I suggest you take a look at the infamous "Harvard Calculus" textbook by Hughes-Hallett et al. This for me and despite all the furor it created is by far the best modern calculus textbook out there, because it actually tries to teach students calculus as a useful tool rather than a set of mysterious rules that miraculously solve a canned set of problems.
I also dislike introducing the definition of a derivative using standard mathematical terminology such as "limit" and notation such as $h\rightarrow 0$. Another achievement of the Harvard Calculus book was to write a math textbook in plain English. Of course, this led to severe criticism that it was too "warm and fuzzy", but I totally disagree.
Perhaps the most important insight that the Harvard Calculus team had was that the key reason students don't understand calculus is because they don't really know what a function is. Most students believe a function is a formula and nothing more. I now tell my students to forget everything they were ever told about functions and tell them just to remember that a function is a box, where if you feed it an input (in calculus it will be a single number), it will spit out an output (in calculus it will be a single number).
Finally, (I could write on this topic for a long time. If for some reason you want to read me, just google my name with "calculus") I dislike the word "derivative", which provides no hint of what a derivative is. My suggested replacement name is "sensitivity". The derivative measures the sensitivity of a function. In particular, it measures how sensitive the output is to small changes in the input. It is given by the ratio, where the denominator is the change in the input and the numerator is the induced change in the output. With this definition, it is not hard to show students why knowing the derivative can be very useful in many different contexts.
Defining the definite integral is even easier. With these definitions, explaining what the Fundamental Theorem of Calculus is and why you need it is also easy.
Only after I have made sure that students really understand what functions, derivatives, and definite integrals are would I broach the subject of symbolic computation. What everybody should try to remember is that symbolic computation is only one and not necessarily the most important tool in the discipline of calculus, which itself is also merely a useful mathematical tool.
ADDED: What I think most mathematicians overlook is how large a conceptual leap it is to start studying functions (which is really a process) as mathematical objects, rather than just numbers. Until you give this its due respect and take the time to guide your students carefully through this conceptual leap, your students will never really appreciate how powerful calculus really is.
ADDED: I see that the function $\theta\mapsto \sin\theta$ is being mentioned. I would like to point out a simple question that very few calculus students and even teachers can answer correctly: Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians. In my department we audition all candidates for teaching calculus and often ask this question. So many people, including some with Ph.D.'s from good schools, couldn't answer this properly that I even tried it on a few really famous mathematicians. Again, the difficulty we all have with this question is for me a sign of how badly we ourselves learn calculus. Note, however, that if you use the definitions of function and derivative I give above, the answer is rather easy.
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I emphatically agree that students don't know what a function is. But then again, it is a deceptively deep concept. As for modern treatments that emphasize other things than the standard, did you ever look at "Calculus in Context", the five colleges calculus? Available at math.smith.edu/Local/cicintro/cicintro.html My problem with this approach though is that even if you can convince me easily that it's the right thing to do mathematically, how will it mesh with the courses in other disciplines that students take, which will expect much more traditional material. – Thierry Zell Sep 27 2010 at 12:55
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Any changes to math courses should of course be done only in close collaboration with other departments who rely on the math courses. But I think you'll find that, with the computational tools available, many of them will quite sympathetic to a "concept first, hand computation second" approach. Besides, I don't argue against teaching symbolic computation, just delaying it. And I do also like the Calculus in Context book but have not had experience using it. I suspect it works best with students with a stronger background than the ones I teach. – Deane Yang Sep 27 2010 at 13:39
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Harry, that is exactly how any pure mathematician, including me, would do it. But that's the hard way. For an engineer or physicists, who thinks in units and dimensional analysis and views the derivative as a "sensitivity" as I've described above, the answer is dead obvious. – Deane Yang Sep 27 2010 at 17:51
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Alexander and Pietro, unfortunately I said "I would like to see...", which means I don't really get to banish symbolic methods for a whole semester. In fact, I advise being pragmatic and teaching in a fashion that will not alienate you from your department or school administration. That said, if you want to slip in more understanding (which I claim actually helps students learn the symbolic methods better), I recommend taking problems from the Harvard calculus textbook, as well as their precalculus text ("Functions Modeling Change"). Especially those where no formula is given for the function. – Deane Yang Sep 28 2010 at 2:07
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Your description of a function as a box seems to miss the most important part: that whenever you put a given number in, you always get the same output. That is, the box behavior should be single-valued. (Otherwise, we might imagine a black-box that accepts a given input and outputs a random number, perhaps different every time, and although this accords with your description, it is not a function.) – Joel David Hamkins Jan 24 2011 at 11:36
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I'm teaching Calc 1 this semester, and I've stumbled onto something that I like very much.
First of all, I start (always) by having my students draw bunches of tangent lines to graphs, compute slopes and draw the "slope graphs" (they also do "area graphs", but that's not relevant to this answer). They build up a bit of intuition about slope and slope graphs.
Then (after a few days of this) I ask them to give me unambiguous instructions about how to draw a tangent line. They find, of course, that they are stumped.
In the past, I went from this to saying "we can't get a tangent line, but maybe we can get an approximately tangent line" and develop the limit formula.
This semester, I said, "we have an intuitive notion of tangency; suppose someone offered a definition of tangency -- what properties would it satisfy?" We had a discussion with the following result: tangency at point $x = a$ should satisfy:
1. tangency (of one function with another) should be an equivalence relation
2. if two linear functions are tangent at $x= a$, they are equal.
3. a quadratic has a horizontal tangent line at its vertex.
4. if $f$ and $g$ are tangent at $x = a$, then $f(a) = g(a)$.
5. if $f_1$ is tangent to $f_2$ at $x = a$ and $g_1$ is tangent to $g_2$ at $x = a$ then $f_1 + g_1$ is tangent to $f_2 + g_2$ at $x = a$ and similarly for the products.
6. the evident rule for composition.
Using these rules, we showed that if $f$ has a tangent line at $x = a$, it has only one. So we can define $f'(a)$ to be the slope of the tangent line at $x = a$, if it exists!
The axioms are enough to prove the product rule, the sum rule and the chain rule. So we get derivatives of all polynomials, etc., assuming only that tangency can be defined.
Then (limits having presented themselves in the computation of area) I defined $f$ to be tangent to $g$ if $\lim_{x\to a} {f(x) - g(x) \over x-a} = 0$. We derive the limit formula for the derivative, and check the axioms.
EDIT: Here's some more detail, in case you're wondering about implementing this yourself. I had the initial discussion about tangency in class, writing on the board. A day or so later, I handed out group projects in which the axioms were clearly stated and numbered, and the basic properties (as outlined above) given as problems.
The students' initial impulse is to argue from common sense, but I insisted on argument directly from the axioms. There was one day that was kind of uncomfortable, because that is very unfamiliar thinking. I had them work in class several days, and eventually they really took to it.
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This is very nice. – Deane Yang Sep 27 2010 at 14:20
I second Deane's comment. – Mark Meckes Sep 27 2010 at 17:12
Is there really a unique equivalence relation satisfying these rules? I do not see how these rules could ever access a function which is not a polynomial. If not, saying that you can define f'(a) to be the slope of the tangent line at x=a presupposes that you have chosen one of the many equivalence relations which satisfy these properties. – Steven Gubkin Nov 3 2010 at 15:41
So see that I can get more than polynomials: By using the chain rule and the product rule I can actually get any algebraic function. But I still do not see how to get any transcendental function. – Steven Gubkin Nov 4 2010 at 3:01
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See this MO question: mathoverflow.net/questions/44774/… – Steven Gubkin Nov 4 2010 at 14:23
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I'm going to answer this part:
does anyone out there actually use this definition to calculate a derivative that couldn't be obtained by a known symbolic rule?
Yes. $sin(x)$.
My point is that of course we can just learn the derivative of this function, but then we could just learn the derivative of any function. So looking for a "complicated function" that needs the limit definition is pointless: we could just extend our list of examples to include this function. It's a bit like the complaint that there's no closed form for a generic elliptic integral: all we really mean is that we haven't given it a name yet.
In fact, one could do $x^2$ like this, or even $x$, but I think that $sin(x)$ has a good pedagogical value. If you can get them first to ponder the question, "What is $sin(x)$?" then it might work. I'm teaching a course at the moment where I'm trying to get my students out of the "black box" mentality and start thinking about how one builds those black boxes in the first place. One of my starting points was "What is $sin(x)$?". Or more precisely, "What is $sin(1)$?". If you take that question, it can lead you to all sorts of interesting places: polynomial approximation of continuous functions, for example, and thence to Weierstrass' approximation theorem.
Many students will just want the rules. But if the students refuse to learn, that's their problem. My job is to provide them with an environment in which they can learn. Of course, I should ensure that what they are trying to learn is within their grasp, but they have to choose to grasp it. So I'm not going to give them a full exposition on the deep issues involving the ZF axioms if all I want is for them to have a vague idea of a "set" and a "function", but I am going to ensure that what I say is true (or at the least is clearly flagged as a convenient lie).
Here's a quote from Picasso (of all people) on teaching:
So how do you go about teaching them something new? By mixing what they know with what they don't know. Then, when they see vaguely in their fog something they recognise, they think, "Ah, I know that." And it's just one more step to, "Ah, I know the whole thing.". And their mind thrusts forward into the unknown and they begin to recognise what they didn't know before and they increase their powers of understanding.
We all remember professors who forgot to mix the new in with the old and presented the new as completely new. We must also avoid the other extreme: that of not mixing in any new things and simply presenting the old with a new gloss of paint.
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+1 for "if the students refuse to learn, that's their problem. My job is to provide them with an environment in which they can learn." – Mark Meckes Sep 27 2010 at 14:44
+1 for the same – Sean Rostami Sep 28 2010 at 4:42
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@Mark and Sean, I have to admit that I'm a little put off by the cheerleading for this particular phrase -- stripped from the context Andrew provided it, it comes across rather as, "I don't do a bad job of teaching, my students do a bad job of learning." I think this is an attitude those of us who teach should be careful to avoid, in general. (Of course, everyone who has ever taught has come across specific cases where it might be applied.) – JBL Sep 28 2010 at 13:12
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@JBL: point taken, although I disagree with your restatement. "My job is to provide them with an environment in which they can learn" is the sentiment of someone who takes doing that job well seriously. I really liked the line because of several recent conversations about students who don't take notes, skip class, rush through homework, and don't ask questions. Such students are the exception rather than the rule, but they can get under one's skin. For times like that, I thought Andrew's line would be a good substitute for the glib old saw about a horse and water. – Mark Meckes Sep 28 2010 at 13:57
@Mark: Yes, I didn't think that either you or Sean agreed with my rephrasing, just that it seems to me that this statement (in isolation) has a little of that ring to it. Students who behave as you describe are extremely irritating, but I think teachers would do well to avoid sounding like we think this is the norm :) – JBL Sep 28 2010 at 15:06
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I agree with the above comments.
The point of my comment-question "What competing definition do you have in mind?" was to emphasize something that seems to be under-emphasized in the question itself: the reason we speak of derivatives as limits is because that's the definition of the derivative, and we want to give a definition of the concept that is going to be discussed for much of the semester.
[It is possible to give other definitions of a derivative, but they are all variations on the same theme and, in particular, all use either the concept of limit or the (equivalent!) concept of continuity. For instance, Caratheodory has a nice definition of the derivative in terms of functions vanishing to first order, but this is not going to be any more palatable to the freshman calculus student.]
[Added: I admit that I forgot about nonstandard analysis when I wrote the above paragraph. That indeed has a somewhat different feel from the usual limits and continuity. One the one hand, although I have never taught calculus this way, I rather doubt that doing so would suddenly make the difficult concepts of continuity and differentiability go over easily. On the other hand, I certainly couldn't decide to teach a nonstandard approach to calculus because it would be...nonstandard. The curriculum among different sections, different classes and different departments has to have a certain minimal level of coherence, and at the moment the majority of the grad students and faculty in every math department I have ever seen are not familiar enough with nonstandard analysis to field questions from students who have learned calculus by this approach.]
If we don't give a definition of the most important concept in the course, then we lose all pretense of developing things in a logical sequence. In particular, it's hard to see how to discuss the derivations of any of the basic rules the students will actually be using to compute derivatives, and thus we would be forced to reduce calculus to a (long!) list of algorithms based on certain unexplained rules.
Nevertheless I take your question seriously, since I have taught a fair amount of freshman calculus in recent years. It is absolutely correct that a lot of students get impatient, angry and/or confused at the limit definition of the derivative (or really, at anything having to do with limits and/or continuity). I do derivations of things like the product rule and the power rule rather quickly in class, because I know that something like half the class isn't following and doesn't care to follow. And yet I do them anyway (not all of them, but more than half) because, to me, not to do them makes the course something I could not bring myself to teach (and, by the way, would put it well below the level of the AP calculus class I had in high school: I feel honorbound to give to my calculus students not too much less than was given to me). Thus there is a real disconnect between the calculus class that I want to teach and the calculus class that something like half of the students want to take. It's discouraging.
I would be happy to hear that I am making a false dichotomy between giving the limit definition of the derivative and just giving algorithms to solve problems. I definitely experiment with different kinds of explanation beyond (and instead of!) just a formal proof. Here are some things I have tried:
1) Take the definition of continuity as primary, and define the limit of a function at a point as the value at which one can (re)define the function to make it continuous. I think this should be helpful, since I think most people have an intuitive idea of a "continuous, unbroken curve" and much less of the limit of a function at a point.
2) Emphasize physical reasoning. The last time I taught freshman calculus, I spent the entire first day talking about velocities: first average velocity, then instantaneous velocity. If a differentiation rule has a plausible physical interpretation -- e.g. the chain rule says that rates of change should multiply -- then I often give it.
3) Emphasize "chemical reasoning", i.e., dimensional analysis:. I often give the independent variable and the dependent variable units and emphasize that the units of the derivative are different from the units of the original function. In this way one can see that the conjectured product rule $(fg)' = f'g'$ is dimensionally wrong and thus nonsense. (And again, the chain rule is "obvious" from a unit conversion perspective.) Similarly dimensional analysis should stop you from saying that the volume of a cylinder is $\pi rh$.
Unfortunately none of these things have worked with the portion of the class that doesn't want to hear anything but how to solve the problems.
Added: To more directly address your specific question: yes, there are problems one can ask of freshman calculus students which require them to use the limit definition of the derivative rather than (just) the differentiation rules, but I do not recommend asking many of these questions, since the students find them very difficult. A personal example: when I was teaching Math 1A (first semester calculus) as a graduate student at Harvard, we had communal exams but the course head (who was a tenured professor of mathematics, hence a very brilliant person) had the final say. On the first exam, we decided that one of the questions was too hard, so at the last minute the course head replaced it with the following one (which he did not show to us):
Consider the function $f(x)$ defined as $x^a \sin(\frac{1}{x^2})$ for $x \neq 0$ and $f(0) = 0$. What is the smallest integer value of $a$ such that $f$ is (i) continuous, (ii) differentiable, (iii) twice differentiable?
I had the good fortune to grade this problem. Out of $200$ or so exams, the median score was $0.5$ out of $12$. About three students wrote down the right numerical answer for part (iii), but this was not supported by any work or reasoning whatsoever.
Added: by the way, it's not as though the above question is "bad" in the sense that it's not testing mathematical competence and depth of understanding of calculus. I think it absolutely is, just at a level way above that which one should be testing in a freshman class for non math majors. For the next few years, when the story came up in a social setting involving mathematical hotshots, after telling it I would press them for an answer to part c) on the spot. Most people I asked did not get it. (Note that I would not of course give them pen and paper and a quiet spot to think about the problem for some period of time. I generally required an answer after a minute or so. Let's hold PhD mathematicians to higher standards than freshman non-majors after all!) For instance, I watched a cloud pass over one Fields Medalist's face as he got very confused. After a while though I stopped using this as a pop quiz in addition to a story: I can't explicitly remember why, but I'd like to think it dawned me how obnoxious it was to put people on the spot like that...
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+1. Differentiation can be done without limits too, en.wikipedia.org/wiki/Formal_derivative I interpreted the question as distinguishing between derivatives in analysis and "generating functions". Even most trigonometric functions have combinatorial meaning and so their derivatives can be computed formally. But as you say that misses the point of calculus (continuity, physical reasoning etc.). – Gjergji Zaimi Sep 27 2010 at 8:37
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The freshman-level example is especially biased because students tend to believe that any problem with two letters in it is very hard. But I've had a reasonable degree of success with a problem of this type, using a value for a (disclaimer: at a good school, though no Harvard). But it's also because I'd spent some time on this in class; you can't spring this on students out of the blue like that professor did and expect they'll do well. – Thierry Zell Sep 27 2010 at 11:41
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Teaching undergraduate calculus using nonstandard analysis is not out of the question. I haven't done it but I know others who have, using for example Henle and Kleinberg's Infinitesimal Calculus. – Timothy Chow Sep 27 2010 at 14:50
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@Mariano: no, I was dead serious. The point was that this person was far too bright to realize that this was a ridiculously hard question for freshman calculus. – Pete L. Clark Sep 27 2010 at 19:33
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@Thierry: I like your quote of "any problem with two letters in it is too hard." – drvitek Sep 27 2010 at 20:36
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While I think that ideally, even in a freshman course of calculus, students should receive some historical notions about the development of the ideas of infinitesimal calculus, I think that, even in a freshman course of calculus, the true definition of derivative of a function should be given, that is, via the first order approximation. A function $f:(a,b)\to\mathbb{R}$ is differentiable at $x$ if there exists $m$ such that $$f(x+h)=f(x)+mh+o(h)\quad \mathrm{as}\ \ h\to0.$$
The fact that the coefficient $m$ (the derivative) can be characterized, and sometimes efficiently computed, as a limit of a quotient, has certainly to be observed, and should be applied immediately to treat some elementary functions like $x^2$, $1/x$ or $e^x$, as usual. But I would never give it as a definition.
I think there is a philosophical issue here. It may seem simpler to define something as the result of a procedure for getting it, compared with defining it via a characteristic property. But the latter way is superior, and on a long distance, simpler. And in the case of students who will stop there their mathematical education, then, I prefer they at least see the true idea behind, rather that being able to compute the derivative of $\cos(e^x)$ : when will that be of use for them?
The definition via first order expansion is very natural, and more understandable to the freshman students. It has a more direct geometrical meaning. It reflects the physical idea of linearity of small increments (like in Hooke's law of elasticity, etc). It is much closer to the practical use of derivatives in approximations. It makes easier all the elementary theorems of calculus (consider how needlessly complicated becomes the proof of the theorem for the derivative of a composition by introducing a useless quotient). Finally, it is closer to the generalization to Fréchet differential, which is a good thing for those students that will continue their study in maths.
A funny remark, from my experience. Ask students that received the definition of derivative as limit of incremental quotient, to compute $\lim_{x\to 0 }\sin(x)/x$. Will anybody say, it's the derivative of $\sin(x)$ at $0$, that is $\cos(0)=1$? No, they will try and use the "rule of de L'Hopital"!
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From my prof back in the days: "Some of you might have heard of a thing called L'Hopital's rule. It has a lot of hypotheses that no-one ever checks, and students always apply it when the quotient is in the wrong form, so I won't teach it and you'd better not use it." And now I do the same... (I didn't mind when he said that because I was one of the ones who'd never heard of it.) – Thierry Zell Sep 27 2010 at 12:42
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Yes! I think that introducing the differential of a function of several variables would be much easier for students if they had this point of vue on derivative. – Benoît Kloeckner Sep 27 2010 at 15:57
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@Andrew: and if I was the chairman in another maths department, I'd immediately engage him with double salary. De'LHopital himself would be embarassed to know somebody's still wasting time with such an awkward theorem like that thing that brings his name. Theorems, like cakes, don't always come out well; that thing came out very badly, and left a mess in the oven. Today, it may be at most of some historical interest. Teach the Landau notation instead! (Btw, as you probably know, Edmund Landau was fired from Göttingen in 1933, with the pretext of his way of teaching a calculus course.) – Pietro Majer Sep 27 2010 at 19:13
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@Pietro: what you give is essentially Caratheodory's definition, as alluded to in my answer. It's so close to the usual definition that I don't really believe that students have a significantly easier time with it. However, I believe that when you teach calculus, this definition inspires you and you do a very good job teaching it, more so than you would with the standard definition. I suspect that most "the students find it easier when..." statements are like this, but that's fine -- finding the version that you can get behind enthusiastically and explain well is part of good teaching. – Pete L. Clark Sep 27 2010 at 19:49
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About that funny remark: even saying $\lim_{x\to 0} \sin(x)/x$ is the derivative of $\sin(x)$ at 0 may be viewed as cheating, since the typical textbook approach is to use a geometric argument to prove $\lim_{x\to 0} \sin(x)/x = 1$ and then use that limit to prove that $\frac{d}{dx} \sin(x) = \cos(x)$. – Mark Meckes Sep 28 2010 at 14:27
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I wanted to add one further point to the many good answers already given here: "black box" symbolic computation, in the absence of understanding the formal definitions, can work when everything goes right, but is very unstable with respect to student errors (which are sadly all too common). Knowledge of definitions provides a crucial extra layer of defence against such errors. (Of course, it is not the only such layer; for instance, good mathematical or physical intuition and conceptual understanding are also very important layers of defence, as is knowledge of key examples. But it is a key layer in situations which are too foreign, complicated, or subtle for intuition or experience to be a good guide.)
For instance, without knowing the formal definition of the derivative, a student could very easily start with a true formula such as
$$(x^2)' = 2x$$
and do something like "substitute x=3" to obtain the false formula
$$(9)' = 6.$$
(An example that I have actually seen: someone attempted to prove Fermat's last theorem by starting with the equation
$$a^n + b^n = c^n$$
and then differentiating with respect to $n$. Ironically, a variant of this type of trick actually works when solving FLT over polynomial rings, but that's another story...)
Now, without bringing in the definition of a derivative (and of a function), how could you explain to the student what went wrong here in a way that the student will actually remember? Saying that one can use the law of substitution or the trick of differentiating both sides in some situations, but not in others, is likely to be recalled inaccurately, if at all (and may have the side effect that the student may view such basic moves as substitution as somehow being "suspect", thus avoiding it in the future).
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People say I'm mean for asking for the derivative of $\pi^2$, but I think it's a memorable example for the students. Another place the blind symbol manipulation goes wrong is on $\sin^{-1} x = \arcsin x$. Many students are willing to assume that if $f(x)=g(x)$ then $f'(x)=g'(x)$, which is only a consequence for some meanings of the first equation. – Douglas Zare Nov 6 2010 at 16:43
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@Terry: I agree that the black-box use is an issue, but some might argue that it's possible to fix without necessarily going all the way to a limit definition, because the root problem is conceptual understanding of functions, and more precisely of the derivative as a function. A simple aphorism would suffice: "chug then plug, don't plug and chug!" More seriously, one could discuss the difference between the derivative as a function and the derived number at a point (slope) purely graphically, with no references to limits. – Thierry Zell Nov 7 2010 at 0:01
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@Douglas: I didn't think the derivative of $\pi^2$ was a mean thing to ask! Who thinks so? students? colleagues? Never mind, I'll be sure to borrow it for next time. (Even meaner would be to ask for the derivative of $e^2$, btw.) As for the blind symbol manipulation, I cannot believe that it took me all these years before seeing for the first time (in an exam) that the derivative of $\arctan x$ was $\mathrm{arcsec}^2 x$. In retrospect, I should have been expecting this for a long time! – Thierry Zell Nov 7 2010 at 0:09
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I think this type of error is avoidable if you introduce $x$ not as a variable, but as a special symbol for identity function. Even more, you can use bold $\mathbf{x}$ for the identity function and normal $x$ for a value. – Anixx Jan 2 2011 at 18:35
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When my father was a judge at a high school math fair, a student gave a presentation on calculus. During the question period after the presentation, he asked the student "If f(x) = 3^2, what is the derivative of f(x)?" The student said "6". My father then asked "If f(x) = 9, what is the derivative of f(x)?" The student said "0". He asked the first question again and the student still said "6". Of course this student did not go on to the next round of the math fair. – KConrad Jan 23 2011 at 20:31
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The derivative of $x|x|$ is best computed via the "limit" definition. A more general example is $xf(x)$ where $f$ is any continuous function, and we are computing the derivative at $x=0$.
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This is a good example -- Steven, take note :). (Though I would tweak it to get something with nonzero derivative, if possible.) – JBL Sep 27 2010 at 16:08
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If you want example with nonzero derivative consider $x|x|+x$. – Igor Belegradek Sep 27 2010 at 17:12
For the more general example, one should also ask that $f$ is not differentiable at $0$. (By varying such $f$, we can get any derivative at all, or none. Taking $f(x) = x \sin(1/x)$, extended continuously, gives a very interesting example that's come up elsewhere on this page.) – Toby Bartels Apr 3 2011 at 23:08
This example can be done straightforwardly without limits, using infinitesimals (either informally or in NSA). – Ben Crowell Oct 6 at 2:10
I am surprised that no answer has explicitly mentioned the fundamental theorem calculus yet: that is a classic, and important, instance of calculating the derivative using the limit definition. So, for example, the integral sine function
$$\int_0^x \frac{\sin t}{t} dt$$
has important applications in signal processing and the cumulative distribution function of the normal distribution $N(a,\sigma^2)$
$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^x e^{-\frac{(t-a)^2}{2\sigma^2}}dt$$
is the bread and butter of probability and statistics. Both functions are not elementary and their derivatives, while significant, would be impossible to calculate by other means.
I also disagree with the comment that piecewise defined functions "are not good at all" for illustrating the definition of the derivative based on limits. In fact, piecewise polynomial functions, in the form of splines, are used in mechanical engineering (e.g. to design the shape of the car body), and provide a neat opportunity to relate conceptual and computational aspects of derivatives.
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+1, the guys who deal in splines always make a big deal out of left and right continuity for making the approximant $c^n$ (for whatever value of $n$ is needed by the application). I too feel that the error function and the sine integral are too important not to at least be given a passing mention in the context of the Fundamental Theorem. – J. M. Sep 28 2010 at 1:55
I agree with both points here. – Deane Yang Sep 28 2010 at 2:08
The definition of derivatives is useful in exercises about functional equations. Ever solve $f(x+y)=f(x)f(y)$ ? A more elaborate one is $[f(x):f(y):f(z):f(t)]=[x:y:z:t]$, functions preserving the cross-ratio (= anharmonic ratio).
However, we should not neglect the interest of the black-box side of mathematics. We should remember that it is this aspect which has made mathematics so much unavoidable in Science. Somehow, it participates to the unreasonable effectiveness of Mathematics in the Natural Sciences'' (E. Wigner's famous statement). After all, the definition of derivatives has the same status as the constructions of ${\mathbb Z},{\mathbb Q},{\mathbb R},{\mathbb C}$. One can spend a year without thinking about them, while using these fundamental objects every hour, by applying rules. Do you remember the construction of the polynomial algebra $k[X]$ ? How would you define $\pi$ ? In a more advanced situation, chemists have efficient rules to deal with characters of representations of finite groups, and they do not need to read a justification, or to remember it, even though the first Chapter of J.-P. Serre's book was intended to be read by his chemist wife. Mathematics is the tool box of Science. It is even a tool box for itself, in the sense that new topics use the older ones. To go further, we must accept older truths. Of course, it is way better to accept them for good reasons, that is, because we have completely understood the definitions. But if the half of a classroom, who does not intend to do mathematical research, neglects the definition and prefer focussing on the rules, there is no problem at all, provided they apply the rules correctly. There are many ways to learn rules, one of them being solving a lot of exercises.
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+1. But, I think a point some other people have been making is that students should at least see the formal definition. But I agree that you shouldn't beat it over the head of students who aren't happy... – Matthew Daws Sep 27 2010 at 12:07
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If scientists want their students to learn a sequence of formal rules, I think they should be the ones to teach them. Personally I want to teach mathematics, and mindless manipulation of symbols is not math. – Steven Gubkin Nov 11 2010 at 15:31
The way that Calculus is traditionally taught gives a false impression that every function worth looking at can be differentiated using the rules of differentiation. This comes from a misconception that any function worth looking at can be described by an algebraic formula, or using trigonometric or logarithmic functions.
That's just not the case: the most common everyday functions don't have any formulas. Some examples:
1. Price of a company stock over several decades.
2. Volume of water in a water tower over the course of a week
3. Median price of a house in your area (adjusted for inflation), over the course of 100 years.
4. US National Debt over the last two hundred years.
5. US Deficit
For such functions, rate of change has a very real meaning. I find that students who had Calculus in high-school are stumped if I give them an example like that and ask them to graph the rate at which, say, the US national debt has changed throughout US history, and how that relates to the deficit.
Understanding the derivative as both rate of change and the slope of the tangent line helps, and the only good way to tie those concepts is with using limits.
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@Anna: I agree heartily. To take it a step further: I have often thought that the traditional calculus sequence lacks an applied component which severely limits its usefulness to those who are not going on to physics and math. As you say, when given a real world function of interest, you are generally not given an algebraic expression for it. Rather, in order to apply the methods of calculus in a quantitative way, there needs to be a step where you create a mathematical model of the function. I was never taught how to do this step myself, and it seems not to be at all trivial... – Pete L. Clark Jan 29 2011 at 22:38
Once or twice I tried to remedy this by asking exam questions like: "Write down an explicit function which has local minima at $\pm 2$ and approaches infinity as $x$ approaches $\pm infinity$." I was hoping that a student would see that a simple function satisfying these conditions is a fourth degree polynomial with double roots at $\pm 2$, thus $f(x) = (x-2)^2 (x+2)^2$. But they had a lot of trouble with this, and the answer to my question "Does teaching the standard curriculum give them the tools to answer this question?" was "No." But wouldn't it be great if students could do this? – Pete L. Clark Jan 29 2011 at 22:44
@Pete: I have been thinking a lot about how the calculus we teach is too neat and not applied enough these days, and I've been thinking about it not because of my calculus classes, but because of my higher-level courses where I've had to use "messy" or "exotic" stuff like Taylor expansions pretty often to motivate some avenues of investigation. I wish my students were more comfortable with this, and expected less of the neat closed-form formula results. Though of course, we should teach them to appreciate when closed-form stuff happens too! – Thierry Zell Mar 12 2011 at 1:25
I appreciate your point, but your examples are all discontinuous functions, so in fact the notion of a limit doesn't quite work. For example, the US deficit can only change in steps of one cent, the amount of water only in steps of one molecule. These could all be used as examples to show that the Cauchy-Weierstrass limit does not actually connect to reality. Perhaps a better point to make with these examples would be that students could benefit by understanding discrete/numerical calculus as well as the classical calculus of continuous real functions. – Ben Crowell Oct 6 at 2:47
An example I like is $\exp(-\frac{1}{x^2})$ and the "bump functions" one can construct with it.
First of all, this example is important in differential geometry (e.g. Whitney's embedding theorem) and complex analysis (as an example of a real $C^\infty$ function which isn't holomorphic).
In second place, even in first year calculus it's an important illustration of the concept of derivative and of Taylor's theorem. It's important in my opinion to understand why all derivatives at zero are zero (i.e. because it goes to zero faster then any polinomial) but even so the function is changing values.
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The example is well-chosen, but your parenthesis sounds misleading to me :there exist functions that go to zero faster than any polynomial at zero, while they are not even twice derivable. – Benoît Kloeckner Sep 27 2010 at 15:53
You're right! Thank you. To be honest, I didn't consider this possibility at the time of writing. I'll leave the parenthesis as is, since it still has some content and you're comment is right below. – Pablo Lessa Sep 27 2010 at 16:57
@ Benoi: Are there examples of such functions which are continuous in a neighborhood of 0? I ask because the only examples I can find are along the lines of the characteristic function of the rationals time exp(-1/x^2). – Steven Gubkin Nov 2 2010 at 22:59
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Take $f(x) = \exp(-1/x^2)$ and $g$ a continuous nowhere differentiable function. The function $h(x) = f(x)g(x)$ is continuous, goes to zero faster then any polynomial when $x \to 0$ but isn't differentiable at any point other than $0$. Hence $h$ isn't twice differentiable. – Pablo Lessa Nov 9 2010 at 12:10
This is a question that I also struggle with sometimes. On the one hand, I understand the value of sweeping things under the carpet when students are not ready for them yet. When I learned Calculus in High School, we talked about -- but never properly defined -- limits (I'm can't recall if we did the limit derivatives). Yet, we managed to go pretty far into the material, e.g. establishing recurrence relations for integrals of the type $\int_a^b e^{-x}\sin(n x)$. This lack of definition was a very frustrating point for me, and when I finally learned about $(\epsilon,\delta)$ two years later, a wave of relief washed over me. Yet, I'm pretty sure that my cohorts did not feel the same way, hence my sympathy for teachers who want to keep things simple by hiding the definition.
At the same time, I don't want my Calc course to be a series of magic tricks, so I always insist on the logical construction of the course: we want to investigate slopes of tangents. We want to work exactly, not approximately. This is why we'll get into limits in the first place (not very historical, but a logical development). So what do I do?
• I briefly cover $(\epsilon,\delta)$ without really applying it. Just to show the difference between a "wordy" definition and a mathematical one.
• I insist on the fact that the limit laws are derived fro this rigorous definition. (You can sketch the proof for the sum of limits for instance).
• This sets up for students how the mathematical edifice is built: abstract definitions to formalize intuition, big gun theorems proved rigorously from these definition (limit laws, derivative laws...). Add a few examples to the mix and then you're set up for practical, mechanical computations (the stuff that computers do).
• I am upfront about the fact that I don't expect my students to use the $(\epsilon,\delta)$ definition, though I like them to memorize it. The payoff will be later.
• I also stress that, historically, calculus was done without this definition for a long time: so it can be done, they will be able to do it, but it also has its limitations when dealing with more abstract material.
In a course that is set up in this way, it is quite natural to cover the limit definition of derivatives. There are a lot of good reasons why one should do that anyway, some of which have already been addressed. Functions which are defined piecewise do require this, and that includes important examples like $\exp(-1/x^2)$ and fun ones like $x^2\sin(1/x)$. The rigorous derivation of the derivative of $\sin x$ is another good example.
There are also wrong ways of doing this. In the comments, Holger pointed to the case study in the Notices article Teaching mathematics graduate students how to teach. Here, the problem asked to use the definition of derivative to compute the slope of a certain cubic at a point. By the time the exam rolls around, you have easier ways of doing this, so of course the students would feel that this is an arbitrary and confusing question.
[Actually, I took so long to write this that I've been ninja'd by Pietro on this example.] One example that I have yet to see though is Taylor series: defining the derivative in this way makes it obvious that $$f(a+h) \approx f(a)+f'(a)h+o(h)$$ and sets you up for the higher order ones. Yes, you can see that from the graph too, but at that level most of my students have a terrible time reasoning from an abstract graph.
Given how fundamental these ideas are, especially in Physics, I can never stress enough these kinds of relationship in my course.
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So far no one's mentioned (or did I miss it?) that if you make students compute $$\lim_{w\to 5} \frac{w^6 - 5^6}{w - 5},$$ then some of them will use L'Hopital's rule to do that, if you don't tell them not to.
Here's an example of something I have students do with the limit definition of the derivative: http://wnk.hamline.edu/~mjhardy/1170/notes/quiz.10.19.pdf
They find all sorts of creative ways of getting things wrong when doing this. Here's another: http://wnk.hamline.edu/~mjhardy/1170/homework/13th.pdf
I think after they've done several like this, they actually do learn what this is for, and that it's not being used as a way to avoid quick and efficient ways of computing derivatives.
But I have them thinking about instantaneous rates of change without using limits on the first day of the course: http://wnk.hamline.edu/~mjhardy/1170/handouts/September.8.pdf
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Fantastic handouts! I think you would get a lot more upvotes on this answer if people clicked on the links. – Steven Gubkin Jan 28 2011 at 15:20
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Sorry about the broken links. Those were for the duration of the semester. I've reinstated the first one. To be continued....... – Michael Hardy Mar 12 2011 at 21:36
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OK, for now the one that ends with "13th.pdf" can be considered superseded by this: wnk.hamline.edu/~mjhardy/1170/homework/5th.pdf – Michael Hardy Mar 12 2011 at 21:38
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THe "handout" referred to there is here: wnk.hamline.edu/~mjhardy/1170/handouts/… – Michael Hardy Mar 12 2011 at 21:38
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All links give 404 errors. – Julien Puydt May 19 2012 at 13:06
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I hope my answer is read as a response to the question asked, rather than as either a defense of or disagreement with the choices the pedagogists (is that a word?) make.
I think one of the main reasons to teach derivatives in terms of the $h\to 0$ limit is that it captures the dual notions of "instantaneous velocity" and "slope", which are respectively physical and geometric.
(Ok, now I will mention some personal opinions about teaching calculus. I love physics, and sometimes pretend to be a physicist, so for me the geometric/physical meanings of calculus are very important. So I would love if they were emphasized more. Unfortunately, we do not do enough in introductory calculus classes in that direction, and it is very hard to present functions and ask students to find the slopes of their graphs without essentially teaching them these black-box techniques. So I don't know whether it's worth it: maybe we should just do the algebraic part of calculus --- it's the only thing we tend to test anyway. I also don't really think that MO is the best place to get into that discussion, though, and I don't think that OP intended as such.)
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Interesting answer. I must stress the connections with Physics than I realized, since a student asked me just last week if I taught Physics also. You mention that we tend to only test the algebraic part of calculus; clearly, we should put our money where our mouth is. Because I emphasize the theory, I make sure there are at least some non-algebraic questions on my tests (you can make up some easy ones). If a prof thinks that the audience can only handle the algebraic stuff, I'm ok with it, but then you should be teaching only that. – Thierry Zell Sep 27 2010 at 12:00
I agree that there should be at least somewhat more physics in calculus courses than their currently is. To do it otherwise seems like another example of the decortication of calculus: the subject takes place in a certain intellectual context and as a response to certain scientific problems. A lot of students get really worried when you talk physics though, since they think you're expecting them to have some outside knowledge that they in fact don't have... – Pete L. Clark Jan 29 2011 at 22:18
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...The irony of this for me personally is that I haven't taken a physics class since high school, so all the physics I now know is that which comes up when explaining closely related topics in mathematics. For instance, if I am supposed to talk about first order differential equations, I feel compelled to talk at least a little about second order differential equations, especially those of the form F(x) = c x'', because this is the best motivation I know as to what solving differential equations means and why it's important:... – Pete L. Clark Jan 29 2011 at 22:21
...the equation gives a law governing the behavior of some object over time, and the solution to the equation tells you what the consequences of that law are for the behavior of the object. In particular, my favorite example is m x'' = - kx. To solve this equation without writing down Hooke's Law, a picture of a spring, carefully explaining the physical significance of the minus sign...what a disservice that would be! By making it physical, you give the student a chance to use her intuition: "Well, if I were forced to satisfy this differential equation, what would I do?" – Pete L. Clark Jan 29 2011 at 22:27
Well, the definition of derivative is probably one of the best application of the notion of limit, from a didactical point of view. If you define the derivative as a limit process then students who understand it will not miss the geometric flavour: the slope of the tangent line is the limit as $h\rightarrow 0$ of the slope of the line through $(x, f(x))$ and $(x+h, f(x+h))$. I think this is beautiful and relatively simple, once you get the students to think about it for a minute. Plus, it answers the question "When do we agree that the graph of $f$ admit the existence of a tangent line at $(x, f(x))$"?
Of course one has to keep in mind that for most students the useful thing to learn is how to compute practically a derivative without using the definition but rather applying a collection of rules. Nevertheless I think it is important to give them an idea of where all these rules come from. Think about those students who want to get a a math major? No?
In Italy in the so called "scientific high school", the schools that provide you with the widest and most basic education (you learn a bit of everything) with a focus in math, physics, chemistry perhaps, ecc.. we are taught the limit using the $\epsilon-\delta$ definition, and the derivative from its definition. This is to say that I think it is possible to have students learn this theoretical aspects of calculus, if high school kids do.
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As you probably know, the Italian government is now planning to gradually change the high school teaching programs into three main topics: "Religion", "Use of guns", "Commercials". So the content of the maths programs of the Italian high school is becoming soon a historical topic. If this is the trend, I guess (I hope) Italy itself will become soon a historical topic too :-( – Pietro Majer Sep 27 2010 at 12:22
I also learnt calculus in high school using the $\epsilon$-$\delta$ definition. This was an ordinary public (state-run) high school in the United States. – Toby Bartels Apr 3 2011 at 21:11
The standard definitions of limit, continuity and derivative are things of beauty mathematically - flexible and well-honed like fine woodworking tools. But to get calculus students to care, and appreciate their meaning and significance, takes some motivation.
A pretty good way to motivate $\epsilon$-$\delta$ is that it has to do with determining what control on input error ($\delta$) is needed to guarantee meeting a given tolerance for output error ($\epsilon$). How accurately do you have to aim a spacecraft to ensure it enters Martian orbit without burning up the way Beagle 2 did, costing hundreds of millions? Students can appreciate this is a serious question, and that it is fair to insist they be able to handle simple examples like $f(x)=-100x+50$, $\epsilon=10^{-2}$. (In large lectures for freshmen, I wouldn't do much more than Lipschitz examples or something carefully designed so $\delta$ is easy to find without cases. Many calculus students are adults but, ahem, need practice with inequalities.)
One can tell engineering students who just want the formulas that they'll be surprised to find that in a couple of years they'll be estimating sensitivity coefficients'' numerically from black-box software or experiment. Gee, sensitivity coefficients are just derivatives, and they'll be estimated from the definition, not symbol-pushing.
Speaking of which, it's nice to express the error in the definition explicitly: $$\frac{f(x)-f(a)}{x-a} = f'(a)+E_a(x),$$ and do the algebra that occurs to few to write $$f(x)=f(a)+ f'(a)(x-a)+ E_a(x)(x-a)$$ This makes the nature of linear approximation a bit more apparent.
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It's funny that actually many students believe that the symbiosis is always the other way around, i.e. derivatives are used to compute limits (l'Hopital etc.). My favorite example of an elegant calculation of a derivative using the limit definition comes from basic physics. Ask the students why the acceleration of an object performing uniform circular motion is always perpendicular to the velocity. One could come up with a non-elegant solution by writing the equations of motion and using a derivatives table, or one could observe the nice geometric proof of considering an infinitesimal isosceles triangle formed by the two velocity vectors that are a few seconds apart and notice that $\Delta \vec{v}$ is the base of this triangle and points toward the center of the circle.
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If you use the derivative table for a scalar product (applied to the square product), from v.v=cte, you get 2v.a=0 -- that doesn't look that non-elegant. In fact, I use this very example to show my students (which are more into physics) that this rule gets results they're already well-acquainted with. That convinces them more than the proof... – Julien Puydt May 19 2012 at 13:16
I believe it gives a good conceptual or practical reasoning to why we would want to study calculus in the first place. As it was first introduced to me, we can always talk about the average speed a car has traveled over a certain distance. Even very small distances, but apart from a speedometer, how can we say 'I'm travelling XX km/h right now'. There enters the limit definition, where we want the instantaneous rate of change!
If we only presented the formal rules for differentiation, we run in to the same problem as high school students who dislike math present "But my calculator can just do it! Why do I need to learn this?!". If the fundamentals are not taught, one day they will be forgotten.
There are certainly other rigorous approaches to the derivative out there. The delta-epsilon method, which most students in their first year simply struggle to grasp as easily as the $h \rightarrow 0$ definition. This approach is typically reserved for the math majors who go on to take a course in analysis, not the general first calculus course for all science majors.
While I do not use this definition in practice, I am primarily not calculating derivatives, so take that for what it's worth I suppose.
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A problem I like to give students to solve shortly after introducing the derivative is to evaluate $f'(2)$ for $f(x) = x^x$. Of course, this function can be rewritten as $f(x) = e^{x\ln x}$ but in my experience students don't think of this. In fact, students who have seen Calculus before almost universally reach the solution $f'(2) = 4$ which they get from the mistaken idea that $f'(x) = x\cdot x^{x-1} = x^x$. The only students that usually get this problem correct are those that haven't yet learned any of the computational methods and only know the definition.
I teach the limit definition and emphasize the physical and geometric interpretations, and then move from that to the concept of the tangent line and linear approximation. I think these concepts encapsulate most of what is significant (intuitively) about the definition. I dislike exam questions that require students to compute derivatives using the limit definition when they know a "better" way to do it. It isn't too hard to write a problem where no formula for the function is given and ask students questions about the sign or approximate magnitude of the derivative or whether or not the function should even have a derivative. For students who to do not intend to pursue mathematics, this seems appropriate to me. Even those who become mathematicians will almost surely see these ideas again in complete detail in an elementary analysis course.
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How do you differentiate x^x using only the limit definition? This seems like a tall order to me. – Steven Gubkin Nov 11 2010 at 15:35
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I only meant that they would numerically estimate it at x = 2 using the limit definition. It is easy to estimate using the definition, but if they try to differentiate and plug in 2 they will probably get the wrong answer. – Jeremy West Nov 11 2010 at 18:13
It is worth noting that there is a lot of historical precedent for teaching it as a limit, which occurs already in Euclid. I.e. Euclid characterizes the tangent to a circle as the unique line such that between it and any other line through the same point, one can interpose a secant (Prop. 16, Book III). (Strictly, he says equivalently that one cannot interpose another line between the tangent and the circle itself, i.e. every other line through the point is a secant.) Thus the tangent is the limit of those secants. Thus I believe one can easily say that the limiting point of view is the original one of Euclid. From this point of view, the idea of limit is the one used so fruitfully by the Greeks, and the contribution of the mathematicians of later times is to make that notion more precise.
On the other hand, if you want to avoid the conceptual difficulty students have with limits, you can follow Descartes instead, at least for derivatives of polynomials, and characterize the tangent line as the unique line such that subtracting its equation from the original function gives a polynomial with a double root at the given point. This leads to motivating the Zariski cotangent space, as M/M^2.
Both points of view also have a nice dynamic interpretation as realizing the tangent as the unique line intersecting the curve doubly at the point, understood as the limit of the two secant intersections,and measured by the presence of a double root.
But if you want a defense of limits, I suggest Euclid Prop. 16, Book III as ample precedent.
If you want a defense of making students practice using the limit definition, I propose that as noted above, this is the only way to get them to appreciate the fundamental theorem of calculus. That theorem cannot be appreciated by memorizing rules for derivatives, One must understand the definition and apply it to an abstractly defined area function. I suggest that one reason many students do not understand why the fundamental theorem of calculus is true, is that (again as noted above) they have not grasped either what an abstractly defined function is, nor what a derivative truly means.
So if you want them to understand the relation between the derivative and the integral, then I agree with others that they need to know what a function is and derivative is. The reasoning here is that once someone understands something, he can use it in more settings than could possibly be covered by any set of rules.
Another practical benefit of testing the use of the h-->0 definition to obtain derivatives of simple functions, is that it forces practice in algebra, trig identities, and exponentials, skills which most of my students are almost completely lacking.
However, I recommend you teach it any way that makes sense to you. after all you understand it, so whatever you say based on that understanding will be useful. Make up your mind what seems important to you, and go for it!
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The answer I give my students is that mathematicians want to know what a word (in this case 'derivative') means in all cases, and the definition of the derivative is a communal agreement about what to say in strange cases such as the absolute value function. (Well, since I banish symbolic stuff from the first two weeks, I say 'function whose graph has a sharp corner like this one (draws on board)'.)
If students press further, I point out that in a literature class they are expected to learn the communal agreement on the difference between a 'simile' and a 'metaphor'. It helps that I am at a liberal arts institution and not a technical one.
Let me also use this opportunity to share a pedagogical trick:
I find it helpful (third time I've tried it) to break up the definition of $f^\prime(2)$ into two parts:
1) Define a new function $E_2$ by the formula $$E_2(x)=\frac{f(x)-f(2)}{x-2}.$$ 2) Take the limit of $E_2$ at 2.
To pull this off, you do need to take the function $E_2$ somewhat seriously; graph it, write formulas for it, et c.
Rationale:
1) It always helps to break up complicated definitions into smaller pieces.
2) It emphasizes that you take limits of functions (in the sense of machines that accept a single number as input and gives a single number of output) rather than of symbolic expressions.
3) Students get to really understand why a discontinuous function or something like the absolute value function is not differentiable (at the relevant point).
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This semester I am teaching Calculus I following Rogawski's textbook. The chapter on limits spends some serious effort on getting comfortable with average rates of change, going as far as creating a table of values (easy to do with a graphing calculator) and numerically estimating the limit, before going on to derivatives in the next chapter devoted to derivatives. By the way, I don't see how this approach could accomplish a worthy goal stated in Rationale 3. – Victor Protsak Sep 29 2010 at 7:12
If it is just a question of definition but not a question of computation, I have heard when I was a student the following definition:
Let $f$ be a real continuous function, class $C^0$. Let ```$$
\Delta f(x,x') = \frac{f(x') - f(x)}{x'-x} \quad\mbox{defined on}\quad {\bf R}^2 -\{x=x'\}
$$``` If $\Delta f$ admits a continuous extension on the diagonal `$\{x=x'\}$` then it is unique, and $f$ is said to be of class $C^1$. The function $f'x) = \Delta f(x,x)$ is then called the derivative of $f$
Of course this is the standard definition, nothing new under the sun, but the $\epsilon-\delta$ calculus if hidden, and of course, not for long :-)
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It's an interesting alternative to the usual, but ultimately I'm afraid it's even more challenging than the usual one. Of course, here I'm not talking about using the definition to perform computations, but even to grasp the intuitive meaning, resorting to functions of two variables strikes me as more than a student can comfortably handle. – Thierry Zell Jan 27 2011 at 16:14
This also has the technical problem of not covering the case when a function is differentiable but not continuously so. (Of course, this might appear in a context where one is uninterested in such functions.) We can fix this by making it slightly more complicated: ask whether $\Delta{f}(a,-)$, the restriction of $\Delta{f}$ to a given vertical line $\{x = a\}$, admits a continuous extension; if so, then this is unique, and we define $f'(a)$ to be the new value $\Delta{f}(a,a)$. – Toby Bartels Apr 3 2011 at 21:22
$\frac{\sin x}{x}$ at $x = 0$ should be a good example.
P.S.: Talking about esoteric definitions, if you can introduce stationary point without derivatives, you can then introduce derivatives using sheaves, like you would introduce vectors on a smooth manifold. It would broaden the consciousness of your freshmen, he-he ^_^
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One way to avoid limits without losing too much is to teach the calculus of finite differences. Conceptually, the move from numbers to lists-of-numbers as first-class mathematical objects is easier than the move from numbers to real-valued-functions-of-a-real-variable, and the easier move also forms a good stepping stone to the harder one. One can develop the calculus of finite differences mutatis mutandis and thereby make the transition to infinitesimal calculus essentially painless. (So, for example, one should work not with polynomials per se, but with linear combinations involving rising or falling powers).
All the black box rules have their analogues, and all are reasonably easy to see and/or prove. Passing the limit, when it happens, comes as a welcome simplification.
Aside from the conceptual challenge of functions themselves, students find limits difficult because of their quantifier complexity. I have never understood why standard algebra pedagogy suppresses quantifiers, thus, for example, leaving many students unable to distinguish between unknowns (literals bound by existential quantifiers), variables (literals bound by universal quantifiers) and constants (literals that belong to the language itself). Students who miscalculate the derivative of $\pi^2$, mentioned elsewhere, don't get this distinction. People who become mathematicians usually "got it" without anyone spelling all this out, and then they learned about quantifiers studying logic in college, so they regard quantifiers as sophisticated and advanced. But most students don't "get it," and I think this accounts for the huge attitude downturn when they get to algebra.
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There was a recent article in the American Math Monthly, Analysis with Ultrasmall Numbers, that might be of interest. For a summary of its implementation in a high school classroom, see http://maths.york.ac.uk/www/sites/default/files/odonovan-slides.pdf.
A quick skim of its implementation seems to suggest that it provides a groundwork for some of the informal manipulations used in calculus-based physics classes.
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Another alternative way of teaching calculus is via infinitesimals (for example the book Elementary Calculus, An Infintesimal Approach by Keisler). The way of thinking about calculus via infinitesimals is obviously very natural, and mathematicians (e.g. the pioneers of calculus, Euler etc...) have used arguments using infinitesimals long before they should have been really allowed to do so. Keisler's book (and in general the area of `Non-standard analysis') makes rigorous our intuition regarding infinitesimals, and is a set of rules that teach us how to formally reason with them. In my opinion this system is intuitive, but the student can never really have a proper understanding of what they are doing "from the ground up" with out some basic knowledge of model theory. The limit approach is less intuitive, but at least a student doesn't have to just accept some rules without truly understanding what's behind them. Possibly this infinitesimal approach is a half way house between teaching it properly with limits and just teaching rules of differentiation to people who aren't interested.
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I am a STUDENT in 11th grade who has just finished BC Calculus.
I don't have a PhD or even a high school diploma, obviously.
But I think that beginning with h->0 is essential. Otherwise, we don't have any definition of a derivative.
But more to the point, I don't think it's strictly necessary to learn the concepts completely before you do symbolic calculations.
How would the teacher make sure that the students eventually learn them before it's too late, then: Leibniz's lovely notation.
Newton's notation is essentially a meaningless shorthand. Prime is arbitrarily chosen to mean a derivative. I don't like that. (I like it as a shorthand, but there is no real meaning behind it).
But Leibniz's has actual meaning: dy/dx is analogous to delta-y/delta-x.
If we are always used to writing dy/dx=... or df(x)/dx=..., then it is no great stretch to write things like df(x)=...dx. And this leads us nicely into differential equations by separation of variables, and concepts such as substituting variables when you integrate.
In my humble opinion, using Newton's notation should be avoided as much as possible, because it doesn't make it clear what you are doing and turns people into robots, mindlessly following the rules of differentiation.
I don't think that Newton's method id all bad. I think it may be good when you are taking derivatives of higher orders, because once you have the concept of derivatives down, it's more important to see that you are taking the derivative of another derivative. (The "exponents" in Leibniz's notation make it a bit confusing).
If I were a Calculus teacher (and I very well may become one someday), I would all but scrap Newton's notation.
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What puzzled me as a calculus student decades ago was the meaning of the dx and the dy in dy/dx. I thought I'd solved the problem by understanding these components not to have independent meaning outside the "quotient" dy/dx . But then I met differentials in the context of multi-variable calculus and felt despair. Even worse, I saw differentials presented very formally - as "expressions" that "varied" in particular ways under coordinate changes. Only years later, in graduate school, did I finally understand dy and dx as linear variables living in cotangent spaces. – David Feldman May 19 2012 at 6:49
Which is all to say Leibniz's notation has its pedagogical dangers too! – David Feldman May 19 2012 at 6:49
What's wrong with viewing a dx as "a small change in x" and a dy as "a small change in y (resulting from the small change in x)"? – Steven Gubkin May 19 2012 at 7:22
I'm interested in the differentials-based approach advocated by Dray and Manogue at Bridging the Vector Calculus Gap. This is for multivariable calculus, but they do discuss the one-variable version (pdf). As they mention, people have reviewed calculus (especially for science courses) in these terms, but has anybody lately taught it this way?
(Also, the theory behind this approach is a little unclear beyond the first derivative, which is what led me to this question.)
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I suppose that I should mention now that I have taught it this way, twice, in a terminal mathematics course intended for non-STEM majors (especially business): tobybartels.name/MATH-1400. – Toby Bartels Dec 24 2011 at 6:44
Dear friends (and foes;-), limits are not needed to understand and do calculus. You can read my article at http://www.mathfoolery.com/Article/simpcalc-v1.pdf and my recent translation of a 1981 talk by V.A. Rokhlin at http://mathfoolery.wordpress.com/2011/01/01/a-lecture-about-teaching-mathematics-to-non-mathematicians/ I hope it will get you thinking.
The following is my response to fedja, it may be of interest to those who haven't read my article. First, let me indicate briefly my suggestions on how to approach calculus. The most important principle (of V.I. Arnold, explicitly stated by him in his recent Lectures on Partial Differential Equations) is to concentrate on examples, calculations, and applications, staying away from the generalities before they become necessary and the ideas behind them are well understood in concrete situations.
I will mostly discuss differentiation, see my article and my talk slides at http://www.mathfoolery.com/talk-2010.pdf for more details. In high school they teach kids how to factor algebraic expressions, long and synthetic division of polynomials, the fact that if $f(a)=0$ then $x-a$ divides $f(x)$ (for a polynomial $f$). Why not use it and develop differentiation of polynomials? Indeed, if you ask a high school student to make sense of $\frac{x^2-a^2}{x-a}$ for $x=a$, (s)he is very likely just to factor the numerator, cancel $x-a$ and stick in $x=a$ to get $2a$ (or $2x$). This stuff is purely algebraic and all the differentiation rules are immediate.
Parenthetically, once they are established for polynomials, they are forced upon us for any reasonable understanding of differentiation, at least for uniformly differentiable because of the Weierstrass approximation theorem, for example. Also differentiation as an aspect of factoring becomes apparent.
We also can similarly develop differentiation of rational functions and roots, and use implicit differentiation to do other algebraic functions. This gives us already a lot to play with and to apply. The sine function can be treated geometrically, as an aspect of kinematics of the uniform rotation. This broadens the range of potential applications.
To get to the geometric and intuitive meaning of differentiation, we can notice that $x^n-a^n-na^{n-1}(x-a)$ has a double root at $x=a$, or we can look at the expression $f(x+h)$, $f$ being a polynomial, as a polynomial in $h$ with coefficients depending on $x$. It has a constant term $f(x)$, the linear term $f'(x)h$ and all the higher order terms, so $f(x+h)-f(x)-f'(x)h=h^2r(x,h)$ where $r$ is a polynomial. This way, if we restrict $x$ and $h$ to some finite interval, we arrive at our basic estimate, uniform in $x$ and $h$:
$$|f(x+h)-f(x)-f'(x)h| \le Kh^2$$
that indicates how close is a polynomial to its affine approximation using its differential.
This inequality allows us to explain why polynomials with positive derivatives are increasing. We simply notice first that if $f' \gt C$ and $0 \lt h \lt C/K$, then $f(x) \lt f(x+h)$, and therefore $f(A) \lt f(B)$ when $A \lt B$. Then, by applying this fact to $f(x)+Cx$, we see that $f(B)-f(A) \gt C(A-B)$ for any $C \gt 0$ when $f' \ge 0$, and therefore $f(A) \le f(B)$. This is called the monotonicity principle, and it is the most complicated theorem in this approach to calculus. Everything else follows from it.
Now, to broaden our scope, we promote our basic estimate to the definition status and call the functions that satisfy this definition (uniformly) Lipschitz differentiable (LD). Derivatives of polynomials are polynomials, and differentiation of polynomials is related to their factoring. Likewise, derivatives of LD functions are Lipschitz. Indeed, we can rewrite our basic estimate as $|\frac{f(x)-f(a)}{x-a}-f'(a)|\le K|x-a|$, then notice that $\frac{f(x)-f(a)}{x-a} = \frac{f(a)-f(x)}{a-x}$ and conclude that $|f'(x)-f'(a)| \le 2K|x-a|$. Moreover, $f$ is LD if and only if $f(x)-f(a)$ factors through $x-a$ in the class of Lipschitz functions of 2 variables, $x$ and $a$. Differentiation rules are straight forward.
I suggest to develop integration in parallel with differentiation (since they work and are understood better together) starting with simple examples of Newton-Leibniz theorem, and working our way to approximating definite integrals by approximating the integrands by the functions that are easy to integrate, say, piecewise-linear functions showing integrability of, say, Lipschitz function, positivity of integral and proving Newton-Leibniz.
Now I can get to fedja's objections. The Lipschitz theory takes care of all the piecewise-analytic functions, and that's almost everything that we deal with in elementary calculus. When we run into functions that don't fit into the Lipschitz theory ($x^{3/2}$, for example), we broaden our definitions by replacing $h^2$ in our basic estimate (with $|h|^{3/2}$ for our example). Then we observe that the theory still holds for the weaker estimates. The Holder estimates i.e. the ones we get by replacing $h^2$ with $|h|^{1+\gamma}$, $0 \lt \gamma \lt 1$ gives us much more room to play. All moduli of continuity are not needed for any problem involving only a finite number of functions, and it covers the vast majority of problems we encounter in calculus.
Now, in the classical treatment the extreme value theorem is used to prove the Lagrange theorem that is used to prove that a function with positive derivative is increasing. But in our approach we have a direct proof of this fact, so we don't need it. And we don't need the intermediate value theorem to prove the Newton-Leibniz since it can be proven directly using positivity of the integral. By the way, both of these theorems are non-constructive.
One may ask about minima and maxima within this approach. The monotonicity principle takes care of this topic, since it assures us that the point where the derivative changes its sign form plus to minus will be a local maximum; the similar obvious result is true for a local minimum.
Fedja also said that the inverse function theorem fails miserably within Lipschitz functions. My guess is that he was talking about the theorem that says that the inverse of a monotonic continuous function on a closed interval is continuous. This is true, of course, but it may be a good thing, since it raises our awareness of the fact that the inverses of very nice functions can be computationally horrendous. It also can be a motivation to consider some other moduli of continuity. As for the inverse function theorem about local invertibility of the differentiable functions, its treatment within the Lipschitz class is not much different from the standard, and is somewhat simpler.
In any case, fedja and I should probably take our dialogue elsewhere. Thanks for all the comments.
I also want to mention that similar approaches to calculus and introductory analysis have been tried with a good measure of success by Hermann Karcher at Bonn University and Mark Bridger of Notheastern University. See Karcher's lecture notes with an English summary at http://www.math.uni-bonn.de/people/karcher/MatheI_WS/ShellSkript.pdf and 2007 book "Real Analysis: a Constructive Approach" by Mark Bridger, where he defines differentiation via factoring of $f(x)-f(a)$ into $x-a$ in the class of continuous functions. Karcher said (in a recent e-mail to Dick Palais): "I taught my last Calculus course by first using only Lipschitz continuity. At the end of the first semester I reached uniform convergence of functions and continuity. From the second semester on the procedure was the standard one. The students liked it a lot, I still meet one or the other and they still smile." There is also a nice book by Peter Lax "Calculus with Applications," where he deals with uniform instead of the pointwise notions.
Added on 1/28/2011. I have just got an e-mail from Hermann Karcher. It says: "it was nice to hear from you again. I read Rokhlin's talk. Maybe one has to go even farther: Sometimes I think, everybody needs his own explanation, and a successful teacher is good in guessing what each individual child needs.I also read your paper. You won't be surprised that I am familiar with your arguments. I am now retired for 7 years. My last three semester beginners course was my most successful one. Today I attended the PhD colloquium of the student of a younger colleague. That student (and some of the younger people in the audience) had been in that last course of mine. They were happy to see me again and say how much fun those three semesters had been. For various reasons I was then in a situation where I could completely ignore, how the standard course in analysis proceeds. During the first semester I did my own stuff,ending up at continuous functions and their uniform convergence at the END of the first term. The next two semesters proceeded as usual - but all the fun we had came from building the foundations differently and the fun stayed with us. I wish you similar experiences. Don't try to convert too many grown ups, just enjoy teaching. Best regards --Hermann Karcher."
By the way, an English translation of Karcher's lecture notes is in progress. I also heard today via facebook form Ursula Weiss who is a math professor in Germany (we were both graduate students at Brandeis in the late 1980s) that she has just finished translating the first chapter.
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"As small as you want" is exactly the expression you tried to eliminate. Once you start saying "there exists $a$ such that $f'(a)$ is as small as you want", I see no difference with "for every $\varepsilon>0$..." and you defeat the whole purpose of the alternative exposition. Moreover, in place of one standard "for every ..., there exists...", you introduce several nonstandard ones. The main aim of the whole exercise is to kill this construction, not to multiply it. – fedja Jan 26 2011 at 6:21
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@Misha: If you interpret "What do you mean when you write X?" as picking on you, we cannot have a conversation. It's also distressing that you don't find the fact that a professor of mathematics finds your writings on calculus hard to follow as being anything other than criticism. As I said, I came to this answer with some interest in your point of view. After receiving your reaction, this is no longer the case. – Pete L. Clark Jan 26 2011 at 8:02
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@Misha: please don't say that I don't want to understand the ideas. I have said multiple times that I do want to do so, so suggesting otherwise is essentially accusing me of intellectual dishonesty and/or bad faith. That's pretty insulting given the demonstrable amount of time I have put into asking and answering questions on this website. If you want to claim I am not competent to follow your arguments: that's fine; go ahead. Perhaps you could clarify what your audience is, though, if you do not intend your calculus writings to be easily understandable by a PhD mathematician. – Pete L. Clark Jan 26 2011 at 11:34
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As to your second question, I believe that yes, this is being read, and, moreover, you, probably, have about as much attention now from wide mathematical audience to your teaching ideas as you are ever going to get. You certainly turned some people away (Pete, say) by the moment and what happens next depends on what you say, but I find public discussions more useful than private e-mails (full public record of speeches and independent arbitrage are big pluses. Besides, you can really win some people you don't know to your side). Of course, if you do not care, we can stop here. – fedja Jan 27 2011 at 19:21
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«I probably understood him a bit better than he cared for...» Well, that's surely going to attract good will! – Mariano Suárez-Alvarez Jan 27 2011 at 23:57
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### Delta Dirac Charge Density question
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What is the difference among linear charge density, surface charge density and volume charge density.
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http://en.citizendium.org/wiki/Integer
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Welcome to Citizendium - the Citizens' Compendium!
# Integer
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The integers (Latin word integer means "untouched") are the natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and zero.
More formally, the integers are the only integral domain whose positive elements are well-ordered, and this order is preserved by addition. Like the natural numbers, the integers form a countably infinite set. The set of all integers is usually denoted by a boldface Z (or blackboard bold $\mathbb{Z}$, or Unicode ℤ), which stands for Zahlen (German for "numbers").[1]
The term rational integer is used in algebraic number theory to distinguish these "ordinary" integers, embedded in the field of rational numbers, from other "integers" such as the algebraic integers.
## Algebraic properties
Like the natural numbers, Z is closed below the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. Also, unlike the natural numbers, it is closed under subtraction. Z is not closed under the operation of division, since the quotient of two integers (e.g., 1 divided by 2), need not be an integer.
The following lists some of the basic properties of addition and multiplication for any integers a, b and c.
| | | |
|-----------------------------------|-----------------------------------|-------------------------------------------------|
| | addition | multiplication |
| closure: | a + b is an integer | a × b is an integer |
| associativity: | a + (b + c) = (a + b) + c | a × (b × c) = (a × b) × c |
| commutativity: | a + b = b + a | a × b = b × a |
| existence of an identity element: | a + 0 = a | a × 1 = a |
| existence of inverse elements: | a + (−a) = 0 | |
| distributivity: | a × (b + c) = (a × b) + (a × c) | |
| No zero divisors: | | if ab = 0, then either a = 0 or b = 0 (or both) |
In the language of abstract algebra, the first five properties listed above for addition say that Z under addition is an abelian group. As a group under addition, Z is a cyclic group, since every nonzero integer can be written as a finite sum 1 + 1 + ... 1 or (−1) + (−1) + ... + (−1). In fact, Z under addition is the only infinite cyclic group, in the sense that any infinite cyclic group is isomorphic to Z.
The first four properties listed above for multiplication say that Z under multiplication is a commutative monoid. However, note that not every integer has a multiplicative inverse; e.g. there is no integer x such that 2x = 1, because the left hand side is even, while the right hand side is odd. This means that Z under multiplication is not a group.
All the properties from the above table, except for the last, taken together say that Z together with addition and multiplication is a commutative ring with unity. Adding the last property says that Z is an integral domain. In fact, Z provides the motivation for defining such a structure.
The lack of multiplicative inverses, which is equivalent to the fact that Z is not closed under division, means that Z is not a field. The smallest field containing the integers is the field of rational numbers. This process can be mimicked to form the field of fractions of any integral domain.
Although ordinary division is not defined on Z, it does possess an important property: given two integers a and b with b ≠ 0, there exist unique integers q and r such that a = q × b + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b. The integer q is called the quotient and r is called the remainder, resulting from division of a by b. This is the basis for the Euclidean algorithm for computing greatest common divisors.
Again, in the language of abstract algebra, the above says that Z is a Euclidean domain. This implies that Z is a principal ideal domain and any positive integer can be written as the products of primes in an essentially unique way. This is the fundamental theorem of arithmetic.
## Types of integers
• Positive integers is a set ℤ+ ∈ ℤ which contains only positive numbers (same as natural numbers);
• Negative integers is a set ℤ- ∈ ℤ which contains only negative numbers;
• Non-positive integers is a subset of ℤ which contains negative integers and zero (ℤ- ∪ {0});
• Non-negative integers is a subset of ℤ which contains positive integers and zero (ℤ+ ∪ {0});
• Almost integer is a number that is very close to an integer;
• Radical integers is a set of numbers, that is closed under addition, multiplication, subtraction and root extraction.
## Order-theoretic properties
Z is a totally ordered set without upper or lower bound. The ordering of Z is given by
... < −2 < −1 < 0 < 1 < 2 < ...
An integer is positive if it is greater than zero and negative if it is less than zero. Zero is defined as neither negative nor positive.
The ordering of integers is compatible with the algebraic operations in the following way:
1. if a < b and c < d, then a + c < b + d
2. if a < b and 0 < c, then ac < bc. (From this fact, one can show that if c < 0, then ac > bc.)
It follows that Z together with the above ordering is an ordered ring.
## Conversion of real numbers to integers
.
It is possible to convert a real number to an integer. One way is called rounding, when a real number is rounded down if its fractional part less than 5, and up if more than or equal to 5. Also there are two functions called floor and ceiling functions. A floor function of x (denoted as ⌊x⌋) has a value of the biggest number not more than x:
$\lfloor x \rfloor=\max\, \{n\in\mathbb{Z}\mid n\le x\}.$
A ceiling function of x (denoted as ⌈x⌉) has a value of the smallest number not less than x:
$\lceil x \rceil=\min\,\{n\in\mathbb{Z}\mid n\ge x\}.$
Unicode entities for floor function symbols are ⌊ and ⌋, for ceiling function are ⌈ and ⌉.
## Construction
The integers can be constructed from the set of natural numbers by defining them to be the set of equivalence classes of pairs of natural numbers $\mathbb{N} \times \mathbb{N}$ under the equivalence relation
$(a,b) \sim (c,d)$ if $a+d = b+c.\,$
Taking 0 to be a natural number, the natural numbers may be considered to be integers by the embedding that maps $n\,$ to $[(n, 0)],\,$ where $[(a,b)]\,$ denotes the equivalence class having $(a, b)\,$ as a member.
Addition and multiplication of integers are defined as follows:
$[(a,b)]+[(c,d)] := [(a+c,b+d)].\,$
$[(a,b)]\cdot[(c,d)] := [(ac+bd,ad+bc)].\,$
It is easily verified that the result is independent of the choice of representatives of the equivalence classes.
Typically, $[(a,b)]\,$ is denoted by
$\begin{cases} n, & \mbox{if } a \ge b \\ -n, & \mbox{if } a < b, \end{cases}$
where
$n = |a-b|.\,$
If the natural numbers are identified with the corresponding integers (using the embedding mentioned above), this convention creates no ambiguity.
This notation recovers the familiar representation of the integers as $\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}.$
Some examples are:
$0 = [(0,0)] = [(1,1)] = \dots = [(k,k)] = \dots$
$1 = [(1,0)] = [(2,1)] = \dots = [(k,k-1)] = \dots$
$-\!$ $1 = [(0,1)] = [(1,2)] = \dots = [(k,k+1)] = \dots$
$2 = [(2,0)] = [(3,1)] = \dots = [(k,k-2)] = \dots$
$-\!$ $2 = [(0,2)] = [(1,3)] = \dots = [(k,k+2)] = \dots$
## Integers in computing
An integer (sometimes known as an "int", from the name of a datatype in the C programming language) is often a primitive datatype in computer languages. However, integer datatypes can only represent a subset of all integers, since practical computers are of finite capacity. Also, in the common two's complement representation, the inherent definition of sign (mathematics) distinguishes between "negative" and "non-negative" rather than "negative, positive, and 0". (It is, however, certainly possible for a computer to determine whether an integer value is truly positive.)
Variable-length representations of integers, such as bignums, can store any integer that fits in the computer's memory. Other integer datatypes are implemented with a fixed size, usually a number of bits which is a power of 2 (4, 8, 16, etc.) or a memorable number of decimal digits (e.g., 9 or 10).
In contrast, theoretical models of digital computers, such as Turing machines, typically do not have infinite (but only unbounded finite) capacity.
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http://mathoverflow.net/questions/75372/automorphism-of-enrique-surface/75488
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## automorphism of Enrique surface
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
What is the fixed point set of an order two automorhism group of an Enriques surface.
-
1
Are there two questions here? 'Why?' is perhaps a bit too broad for MO. Would you accept "it follows from the axioms of ZFC"? :P – David Roberts Sep 14 2011 at 3:24
Ill edit it, if that bothers you. I was wondering if someone could analyze this if he/she doesnt have an answer. – Ru Sep 14 2011 at 5:05
What do you want to know about the fixed point set? – J.C. Ottem Sep 14 2011 at 9:32
That what it is? Curves?points? how many? – Ru Sep 15 2011 at 1:40
## 1 Answer
In general, the fixed locus of an involution $\iota$ on a smooth complex surface $S$ is the union of a smooth curve $D$ and of $k$ isolated points. This follows by Cartan's Lemma that says that in suitable holomorphic coordinates near a fixed point the action is linear.
There are trace formulae that relate these and the action of $\iota$ on the cohomology of $S$. (Holomorphic Fixed Point Formula): $$\sum_{i=0}^2(-1)^i\text{Trace}(\iota|H^i(S,{\mathcal O}_S)) = \frac{k-D\cdot K_S}{4}$$
(Topological Fixed Point Formula): $$\sum_{i=0}^4(-1)^i\text{Trace}(\iota|H^i(S,\mathbb C)) = k+e(D)$$ where $e(D) = -D^2-D\cdot K_S$ is the topological Euler characteristic of $D$.
In the case of Enriques surface, $h^i({\mathcal O}_S)=0$ for $i=1,2$ and $h^1(S,{\mathbb Z})=0$, $h^2(S,{\mathbb Z})=10$. So the formulae above give you $k=4$ and the relation ${Trace}(\iota|H^2(S,\mathbb C)) = 2-D^2$.
References: (1) the holomorphic fixed point formula can be found at pg.566 of M.F.Atiyah, I.M.Singer, The index of elliptic; (2) for the topological fixed point formula, see (30.9) of M. Greenberg, Algebraic Topology: A First Course, W. A. Benjamin Publ., Reading, Mass. 1981. operators: III, Ann. of Math. 87 (1968), 546-604.
-
Dear Rita,Thanks. I will be pleased if you could tell me where I can find the proof of the above formulas and that the fixed point locus are isolated curves and points? – Ru Sep 21 2011 at 18:53
I've edited the answer and inserted the references. – rita Sep 21 2011 at 20:21
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http://mathhelpforum.com/statistics/72719-basic-probability-questions.html
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# Thread:
1. ## Basic Probability Questions.
1.) In the first half of the year, you are asked to read 10 of the short stories. Your teacher provides you with 25 short stories. How many different ways are there to choose your 10 stories?
2.) In the second half of the year, your teacher has 33 short stories to choose from. You are asked to read 12 stories this time. How many different ways are there to choose the 12 stories to read?
3.)The probability that you will make the hockey team is 2/3. The probability that you will make the swimming team is 3/4. If the probability that you make both teams is 1/2, what is the probability that you at least make one of teams? that you make neither of the team?
4.) In order to choose a mascot for a new school, 2755 students were surveyed: 896 chose a falcon, 937 chose a ran, and 842 chose a panther. The remaining students did not vote. A student is chosen at random.
- What is the probability that the student's choice was a panther?
- Was not a ram?
- Was either a falcon or a ram?
5.)
35 students in an Algebra 2 class took a test: 9 received A's, 18 Received B's, and 8 Received C's. Find the prob of a given even.
- IF a student from the class is chosen at random, what is the probbaility that the student did not receive a C?
- If the teacher randomly chooses 3 test papers, what is the probability that the teacher chose tests with grades A, B, and C in that order?
for 5, im guessing... add the prob of A and B?
for 4, 842/2755?
1.) 10(25) ---?????
2.) 12(33) -----?
2. For Q 1 & 2 it depends on whether the order of reading is important or not. IE is reading A then B different from reading B then A. From the wording it sounds like order is not important and therefore you are interested in the combination. In that case
(Q1)
$C = \frac{25!}{10!(25-10)!} = 3,268,760$
(Q2)
$C = \frac{33!}{12!(33-12)!} = 354,817,320$
I guess the lesson is a small change in the initial parameters can lead to a big difference in the number of combinations.
(Q3)
These two questions sum to one. So the chance of make at least one team = 1 - the chance of making none.
Chance of making none is
$p_0 = (1-2/3)(1-3/4) = 1/12<br />$
Therfore the chance of making at least one is
$p_1 = 11/12$
Another way of seeing this is to draw a Venn diagram.
(Q4)
This is just a straightforward division of number by population so
$p_{panther} = 842/2755$
$p_{not ram} = 1 - p_{ram} = 1 - 937/2755$
$p_{falcon|ram} = p_{falcon} + p_{ram} = (896 + 937)/2755$
(Q5)
This is similar to Q4.
$P_{not C} = 1 - p_C = 1 - 8/35$
In part two you are after an A, B and C in a specific order, so in contrast to Q1 this is a permutation and not a combination.
$p = 9/35 \times 18/34 \times 8/33 = 216/6545$
3. Originally Posted by NYCKid09
1.) In the first half of the year, you are asked to read 10 of the short stories. Your teacher provides you with 25 short stories. How many different ways are there to choose your 10 stories?
Mr F says: ${\color{red} ^{25}C_{10}}$
2.) In the second half of the year, your teacher has 33 short stories to choose from. You are asked to read 12 stories this time. How many different ways are there to choose the 12 stories to read?
Mr F says: ${\color{red} ^{33}C_{12}}$
3.)The probability that you will make the hockey team is 2/3. The probability that you will make the swimming team is 3/4. If the probability that you make both teams is 1/2, what is the probability that you at least make one of teams? that you make neither of the team?
Mr F says:
${\color{red} \frac{2}{3} + \frac{3}{4} - \frac{1}{2}}$.
1 - (above answer).
4.) In order to choose a mascot for a new school, 2755 students were surveyed: 896 chose a falcon, 937 chose a ran, and 842 chose a panther. The remaining students did not vote. A student is chosen at random.
- What is the probability that the student's choice was a panther?
- Was not a ram?
- Was either a falcon or a ram?
Mr F says: Draw a Venn diagram.
5.)
35 students in an Algebra 2 class took a test: 9 received A's, 18 Received B's, and 8 Received C's. Find the prob of a given even.
- IF a student from the class is chosen at random, what is the probbaility that the student did not receive a C?
Mr F says: ${\color{red} 1 - \frac{8}{35}}$
- If the teacher randomly chooses 3 test papers, what is the probability that the teacher chose tests with grades A, B, and C in that order?
Mr F says: (9/35)(18/34)(8/33) = ....
for 5, im guessing... add the prob of A and B?
for 4, 842/2755?
1.) 10(25) ---?????
2.) 12(33) -----?
..
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http://mathhelpforum.com/algebra/26467-confused.html
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# Thread:
1. ## confused :(
Ok, I am graphing P=(1,2) with slope m=3. This is how I worked out
$y-2=3(x-1)$
= $y-2=3x-3$
= $y=3x-3+2$
= $y=3x-1$
If x=0 then;
$y=3(0)-1$
= $y=-1$
and so (0,-1)
this would be my second point no, or is this my run? In checking with the book's answer to see if I had done this right it showed my second point to be (2,5) with a run of 1 and a rise of 3......Did I do something wrong in the calculations!?
Thanks!!
2. You are right. When graphing a line by getting 2 points and drawing a line through them you can use any 2 points on a line. The book simply chose its points differently.
3. Originally Posted by badgerigar
You are right. When graphing a line by getting 2 points and drawing a line through them you can use any 2 points on a line. The book simply chose its points differently.
.....oh, ok. So in this case, having my second points (0,-1) I can then find the run of the slope no?!
Thanks !!!
4. So in this case, having my second points (0,-1) I can then find the run of the slope no?!
The run is the x-value of the 2nd point minus the x-value of the first point. The rise is the y-value of the second point minus the y-value of the first point.
They are pretty much only useful to find the slope using $m = \frac {rise}{run}$
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http://nrich.maths.org/6876/index?nomenu=1
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## 'Matrix Meaning' printed from http://nrich.maths.org/
### Show menu
This problem involves the algebra of matrices and various geometric concepts associated with vectors and matrices. As you consider each point, make use of geometric or algebraic arguments as appropriate. If there is no definitive answer to a given part, try to give examples of when the question posed is or is not true.
In the five questions below: $R, S$ are rotation matrices; $P, Q$ are reflection matrices; $M,N$ are neither rotations nor reflections. Consider each part in 2D and 3D.
1. Is it always the case that $M+N = N + M$?
2. It it always the case that $RS= SR$?
3. It it always the case that $RP= PR$?
4. It it always the case that $PQ= QP$?
5. Is it ever the case that $MN = NM$?
How do the values of the determinants of the various matrices affect the results of these questions?
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http://understandinguncertainty.org/comment/681
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# Can we say whether a drug would have enabled someone to live longer? Sadly not.
Submitted by david on Tue, 04/20/2010 - 17:43
In the first televised election debate last Thursday, David Cameron stated that “I have a man in my constituency … who had kidney cancer who came to see me with seven others. Tragically, two of them have died because they couldn't get the drug Sutent that they wanted..”. How reasonable was it to claim that two would not have died had they had access to Sutent? Some statistical analysis can give us an insight.
After a process of price negotiation with manufacturers Pfizer, Sutent (the market name for sunitinib) was licensed in February 2009 by NICE (National Institute for Health and Clinical Excellence) for some but not all patients with kidney cancer.
NICE did not recommend the drug for all cases due to lack of evidence on its benefits:
No data were presented to the Committee on the clinical or cost effectiveness of sunitinib compared with best supportive care as a first-line treatment for people with a poor prognosis who were unsuitable for immunotherapy. In the absence of robust data, the Committee concluded that sunitinib could not be considered a clinically effective first-line treatment for people with a poor prognosis who are unsuitable for immunotherapy (paragraph 4.3.16).
I do not know the details of the patients that came to see Cameron, and in any case it is never possible to say what exactly would have happened had something else been done - what is known as a 'counterfactual'. However, presumably the patients were in the group that was not recommended for Sutent. They may well have felt they too could have benefited, but even if Sutent was as effective for them as it is for the those suitable for treatment, this unfortunately still does not guarantee they would have lived longer with the new drug.
As can be seen from the technical details in the NICE appraisal, survival benefits of drugs are often expressed in terms of the hazard ratio of, say, 0.80, which would mean that, each month, someone on the new drug has 80% of the chance of dying in the following month as they would were they not taking the drug. Rather remarkably, by a neat bit of mathematics, we can show that knowing the hazard ratio means we can work out the chance that an individual getting the treatment will outlive someone who does not.
Specifically, if the hazard ratio is [math]h[/math], then someone taking the drug has a probability [math]1/(1+h) [/math] of outliving someone who is not taking the drug. So if [math]h [/math]= 0.5, that is the monthly risk of dying is halved by the drug, then the probability of living longer by taking the drug is 1/1.5 = .67, or 67%. This extraordinarily simple result does not seem to be widely known, although discussed in Henderson and Keiding’s excellent paper on applying survival models to individuals. The result is proved at the bottom of this post for those with an interest in survival analysis.
So what is the hazard ratio for Sutent? The NICE Committee assumed median overall survival on the current standard treatment to be 27 months, compared to 37 months with Sutent (paragraph 4.3.9). They assumed Weibull survival models which means that the hazard ratio is not constant across time, but if we make an assumption of exponential distributions then the ratio of their median survival is also their hazard ratio, i.e. 0.73, which seems reasonable given the hazard ratios quoted elsewhere in the report.
So even if Sutent was as effective for those currently not recommended by NICE, it is sad but true that there would still have been only 1/1.73 = 58% chance that someone with access to the drug would have survived longer than a similar person denied the drug. Drugs may be beneficial on average, but this does not mean that we can be confident that patients given the drug will live longer than those denied it.. [NB the preceding bold text replaces the following italic text which is inappropriate -see helpful comment below would have survived longer had they had access to the drug. Drugs may be beneficial on average, but this does not mean that we can be confident that a patient will live longer if they take the drug.].
#### Proof of the result (for consenting statisticians only).
For an unknown survival time $X_0$ without the drug, denote the survival function at time $t$ by $S_0(t) = P(X_0>t)$, the survival density by $f_0(t)$, and the hazard function by $h_0(t)$, where $h_0(t) = f_0(t)/S_0(t)$, essentially the risk of dying having survived until time $t$.
Then the integrated hazard $H_0(t) = - \log S_0(t)$, and so $S_0(t) = e^{-H_0(t)}$. If $\lambda$ is the hazard ratio associated with a drug, assumed independent of $t$ (a proportional hazards assumption), then someone taking drug has integrated hazard $H_1(t) = \lambda H_0(t)$ and so $S_1(t) =S_0^\lambda(t).$
So the probability that the survival time $X_1$ for a random individual given the drug is greater than the survival time $X_0$ for someone not given the drug is
$$P(X_1 > X_0) = \int P(X_1 > t) f_0(t) dt = \int S_1(t) f_0(t) dt = \int S_0^\lambda (t) dS_0(t) = - \left[\frac{S_0^{\lambda+1}}{\lambda+1}\right]_1^0 = \frac{1}{\lambda+1}.$$
Levels:
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## Comments
Anonymous (not verified)
Sat, 04/24/2010 - 15:57
### Hazard Ratios
I suspect the result is so simple to state and to prove because, if you are actually in the position of knowing the hazard ratio accurately, most of the difficult maths has already been done. Even so, it's a useful result, and useful to know a little more about how NICE makes its decisions.
carcinoma cancer
Wed, 07/25/2012 - 07:51
### carcinoma cancer
A fantastic blog always comes-up with new and helpful information and while reading I have feel that this blog is really have all those quality that qualify a blog to be a good one. Cancer is until today the most feared disease of all.Today everybody want to prevent himself dangerous disease like carcinoma cancer .So people should be aware from its symptoms. To learn about acs
Anonymous (not verified)
Mon, 04/26/2010 - 08:00
### Not quite right
>>> we can show that knowing the hazard ratio means we can work out the chance that an individual getting the treatment will outlive someone who does not. that's correct. >>> the probability of living longer by taking the drug is 1/1.5 = .67 that's not correct. The problem is that we cannot tell from the hazard rate alone whether the drug slightly extends the life of all patients taking it, or extends the life of some and shortens the life of others (with a gain on average). If we know the distribution of lifespan of patients on the drug, and the distribution of lifespan of patients not on the drug, we can calculate the probability that patient A on the drug will outlive patient B not on the drug. The uniform hazard ratio is a simply case of this. But this is not the same as the probability that patient A on the drug will outlive the same patient A not on the drug. To estimate that, we'd need to know the distribution of effects on individual patients.
david
Mon, 05/10/2010 - 08:43
### Absolutely right
yes you are right, and it shows how easy it is to use the wrong language of 'counterfactuals' (what would have happened if someone had done something different), which can never be determined and so should not be given probabilities. The probability should only be applied to a statement that is, in principle, confirmable: eg that patient X will live longer than patient Y.
I have edited the blog.
mrc7
Wed, 06/02/2010 - 17:49
### Lifespan distributions
This exchange is very interesting and seems to me to be very relevant to a number of new (and very expensive) biologicals that are in clinical trials, or recently introduced to the market. Quite a few of these drugs seem to exhibit bi-modal distributions in that a proportion of patients show little or no benefit and another proportion can receive a great benefit. The problem is of course that patients who are selected to receive the new drug normally then do not receive the previous standard therapy. If that previous therapy had shown a wider acting but more modest activity, including some benefit for patients who fail on the new drug, then a proportion of individual patients end up being much worse off than before the new drug was introduced. The problem is that patient pressure groups tend to highlight the successes and not the downside for some patients. I guess this is where personalised medicine comes in to play! Mike Clark
Anonymous (not verified)
Mon, 02/28/2011 - 01:46
### My experience
Incredible Article. As a middle school student I am unable to understand some aspects of the article, but my family recently had this issue with my grandma as she was in the hospital.
grg
Mon, 05/07/2012 - 11:05
### relative hazards
Nice proof of the relative hazard result. Here is another using only words (but also calculations offline). It makes no difference if we assume that X_0 has the exponential distribution with parameter 1. Then X_1 is exponential with parameter lambda, and the result follows. The full proof, when written out, is longer than yours, but this one connects to things taught in elementary probability courses.
seoearning
Sat, 08/04/2012 - 07:39
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http://math.stackexchange.com/questions/4841/inverse-of-a-triangular-matrix-both-upper-lower-is-triangular
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# Inverse of a triangular matrix( both upper & lower ) is triangular
How can we prove that inverse of upper(lower) triangular matrix is upper(lower) triangular...
Can anybody answer to this question.......
Thanks in advance.........
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1
Why not try a constructive proof? Better yet, look at the 2-by-2 case first, and figure out how you can generalize your observations from it. – J. M. Sep 17 '10 at 8:11
1
A further hint: look up "forward elimination" and "backsubstitution", and figure out how to use these to find the inverse of a triangular matrix. – J. M. Sep 17 '10 at 8:12
## 4 Answers
Personally, I prefer arguments which are more geometric to arguments rooted in matrix algebra. With that in mind, here is a proof.
First, two observations on the geometric meaning of an upper triangular invertible linear map.
1. Define $S_k = {\rm span} (e_1, \ldots, e_k)$, where $e_i$ the standard basis vectors. Clearly, the linear map $T$ is upper triangular if and only if $T S_k \subset S_k$.
2. If $T$ is in addition invertible, we must have the stronger relation $T S_k = S_k$.
Indeed, if $T S_k$ was a strict subset of $S_k$, then $Te_1, \ldots, Te_k$ are $k$ vectors in a space of dimension strictly less than $k$, so they must be dependent: $\sum_i \alpha_i Te_i=0$ for some $\alpha_i$ not all zero. This implies that $T$ sends the nonzero vector $\sum_i \alpha_i e_i$ to zero, so $T$ is not invertible.
With these two observations in place, the proof proceeds as follows. Take any $s \in S_k$. Since $TS_k=S_k$ there exists some $s' \in S_k$ with $Ts'=s$ or $T^{-1}s = s'$. In other words, $T^{-1} s$ lies in $S_k$, so $T^{-1}$ is upper triangular.
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Your last $T$ should be $T^{-1}$, shouldn't it? – Agustí Roig Sep 18 '10 at 6:53
Yes - corrected now that I reworded the last few sentences. – alex Sep 18 '10 at 17:08
This is my preferred proof also. It explicitly exhibits the group of invertible upper triangular matrices as the group of symmetries of something, which (to my mind) is always the most natural way to define a group. – Qiaochu Yuan Sep 18 '10 at 20:23
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Another method is as follows. An invertible upper triangular matrix has the form $A=D(I+N)$ where $D$ is diagonal (with the same diagonal entries as $A$) and $N$ is upper triangular with zero diagonal. Then $N^n=0$ where $A$ is $n$ by $n$. Both $D$ and $I+N$ have upper triangular inverses: $D^{-1}$ is diagonal, and $(I+N)^{-1}=I-N+N^2-\cdots +(-1)^{n-1}N^{n-1}$. So $A^{-1}=(I+N)^{-1}D^{-1}$ is upper triangular.
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1
Just a tiny terminology note: $N$ in your answer would be termed a "strictly upper triangular matrix"; the definition of "strictly lower triangular matrix" is similar. – J. M. Sep 18 '10 at 10:40
@Robin: How can you say that $I+N$ has upper triangular inverses? – ramanujan_dirac Feb 23 at 13:38
I'll add nothing to alext87 answer, or J.M. comments. Just "display" them. :-)
Remeber that you can compute the inverse of a matrix by reducing it to row echelon form and solving the simultaneous systems of linear equations $(A \vert I)$, where $A$ is the matrix you want to invert and $I$ the unit matrix. When you have finished the process, you'll get a matrix like $(I\vert A^{-1})$ and the matrix on the right, yes!, is the inverse of $A$. (Why?)
In your case, half of the work is already done:
\begin{pmatrix} a^1_1 & a^1_2 & \cdots & a^1_{n-1} & a^1_n & 1 & 0 & \cdots & 0 & 0 \\ & a^2_2 & \cdots & a^2_{n-1} & a^2_n & & 1 & \cdots & 0 & 0 \\ & & \ddots & \vdots & \vdots & & & \ddots & \vdots & \vdots \\ & & & a^{n-1}_{n-1} & a^{n-1}_n & & & & 1 & 0 \\ & & & & a^n_n & & & & & 1
\end{pmatrix}
Now, what happens when you do back substitution starting with $a^n_n$ and then continuing with $a^{n-1}_{n-1}$...?
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Suppose that $U$ is upper. The $i$th column $x_i$ of the inverse is given by $Ux_i=e_i$ where $e_i$ is the $i$th unit vector. By backward subsitution you can see that $(x_i)_j=0$ for $i+1\leq j\leq n$. I.e all the entries in the $i$th column of the inverse below the diagonal are zero. This is true for all $i$ and hence the inverse $U^{-1}=[x_1|\ldots|x_n]$ is upper triangular.
The same thing works for lower triangular using forward subsitution.
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http://johncarlosbaez.wordpress.com/2011/01/21/information-geometry-part-6/
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# Azimuth
## Information Geometry (Part 6)
So far, my thread on information geometry hasn’t said much about information. It’s time to remedy that.
I’ve been telling you about the Fisher information metric. In statistics this is nice a way to define a ‘distance’ between two probability distributions. But it also has a quantum version.
So far I’ve showed you how to define the Fisher information metric in three equivalent ways. I also showed that in the quantum case, the Fisher information metric is the real part of a complex-valued thing. The imaginary part is related to the uncertainty principle.
You can see it all here:
• Part 1 • Part 2 • Part 3 • Part 4 • Part 5
But there’s yet another way to define the Fisher information metric, which really involves information.
To explain this, I need to start with the idea of ‘information gain’, or ‘relative entropy’. And it looks like I should do a whole post on this.
So:
Suppose that $\Omega$ is a measure space — that is, a space you can do integrals over. By a probability distribution on $\Omega$, I’ll mean a nonnegative function
$p : \Omega \to \mathbb{R}$
whose integral is 1. Here $d \omega$ is my name for the measure on $\Omega$. Physicists might call $\Omega$ the ‘phase space’ of some classical system, but probability theorists might call it a space of ‘events’. Today I’ll use the probability theorist’s language. The idea here is that
$\int_A \; p(\omega) \; d \omega$
gives the probability that when an event happens, it’ll be one in the subset $A \subseteq \Omega$. That’s why we want
$p \ge 0$
Probabilities are supposed to be nonnegative. And that’s also why we want
$\int_\Omega \; p(\omega) \; d \omega = 1$
This says that the probability of some event happening is 1.
Now, suppose we have two probability distributions on $\Omega$, say $p$ and $q$. The information gain as we go from $q$ to $p$ is
$S(p,q) = \int_\Omega \; p(\omega) \log(\frac{p(\omega)}{q(\omega)}) \; d \omega$
We also call this the entropy of $p$ relative to $q$. It says how much information you learn if you discover that the probability distribution of an event is $p$, if before you had thought it was $q$.
I like relative entropy because it’s related to the Bayesian interpretation of probability. The idea here is that you can’t really ‘observe’ probabilities as frequencies of events, except in some unattainable limit where you repeat an experiment over and over infinitely many times. Instead, you start with some hypothesis about how likely things are: a probability distribution called the prior. Then you update this using Bayes’ rule when you gain new information. The updated probability distribution — your new improved hypothesis — is called the posterior.
And if you don’t do the updating right, you need a swift kick in the posterior!
So, we can think of $q$ as the prior probability distribution, and $p$ as the posterior. Then $S(p,q)$ measures the amount of information that caused you to change your views.
For example, suppose you’re flipping a coin, so your set of events is just
$\Omega = \{ \mathrm{heads}, \mathrm{tails} \}$
In this case all the integrals are just sums with two terms. Suppose your prior assumption is that the coin is fair. Then
$q(\mathrm{heads}) = 1/2, \; q(\mathrm{tails}) = 1/2$
But then suppose someone you trust comes up and says “Sorry, that’s a trick coin: it always comes up heads!” So you update our probability distribution and get this posterior:
$p(\mathrm{heads}) = 1, \; p(\mathrm{tails}) = 0$
How much information have you gained? Or in other words, what’s the relative entropy? It’s this:
$S(p,q) = \int_\Omega \; p(\omega) \log(\frac{p(\omega)}{q(\omega)}) \; d \omega = 1 \cdot \log(\frac{1}{1/2}) + 0 \cdot \log(\frac{0}{1/2}) = 1$
Here I’m doing the logarithm in base 2, and you’re supposed to know that in this game $0 \log 0 = 0$.
So: you’ve learned one bit of information!
That’s supposed to make perfect sense. On the other hand, the reverse scenario takes a bit more thought.
You start out feeling sure that the coin always lands heads up. Then someone you trust says “No, that’s a perfectly fair coin.” If you work out the amount of information you learned this time, you’ll see it’s infinite.
Why is that?
The reason is that something that you thought was impossible — the coin landing tails up — turned out to be possible. In this game, it counts as infinitely shocking to learn something like that, so the information gain is infinite. If you hadn’t been so darn sure of yourself — if you had just believed that the coin almost always landed heads up — your information gain would be large but finite.
The Bayesian philosophy is built into the concept of information gain, because information gain depends on two things: the prior and the posterior. And that’s just as it should be: you can only say how much you learned if you know what you believed beforehand!
You might say that information gain depends on three things: $p$, $q$ and the measure $d \omega$. And you’d be right! Unfortunately, the notation $S(p,q)$ is a bit misleading. Information gain really does depend on just two things, but these things are not $p$ and $q$: they’re $p(\omega) d\omega$ and $q(\omega) d\omega$. These are called probability measures, and they’re ultimately more important than the probability distributions $p$ and $q$.
To see this, take our information gain:
$\int_\Omega \; p(\omega) \log(\frac{p(\omega)}{q(\omega)}) \; d \omega$
and juggle it ever so slightly to get this:
$\int_\Omega \; \log(\frac{p(\omega) d\omega}{q(\omega)d \omega}) \; p(\omega) d \omega$
Clearly this depends only on $p(\omega) d\omega$ and $q(\omega) d\omega$. Indeed, it’s good to work directly with these probability measures and give them short names, like
$d\mu = p(\omega) d \omega$
$d\nu = q(\omega) d \omega$
Then the formula for information gain looks more slick:
$\int_\Omega \; \log(\frac{d\mu}{d\nu}) \; d\mu$
And by the way, in case you’re wondering, the $d$ here doesn’t actually mean much: we’re just so brainwashed into wanting a $d x$ in our integrals that people often use $d \mu$ for a measure even though the simpler notation $\mu$ might be more logical. So, the function
$\frac{d\mu}{d\nu}$
is really just a ratio of probability measures, but people call it a Radon-Nikodym derivative, because it looks like a derivative (and in some important examples it actually is). So, if I were talking to myself, I could have shortened this blog entry immensely by working with directly probability measures, leaving out the $d$‘s, and saying:
Suppose $\mu$ and $\nu$ are probability measures; then the entropy of $\mu$ relative to $\nu$, or information gain, is
$S(\mu, \nu) = \int_\Omega \; \log(\frac{\mu}{\nu}) \; \mu$
But I’m under the impression that people are actually reading this stuff, and that most of you are happier with functions than measures. So, I decided to start with
$S(p,q) = \int_\Omega \; p(\omega) \log(\frac{p(\omega)}{q(\omega)}) \; d \omega$
and then gradually work my way up to the more sophisticated way to think about relative entropy! But having gotten that off my chest, now I’ll revert to the original naive way.
As a warmup for next time, let me pose a question. How much is this quantity
$S(p,q) = \int_\Omega \; p(\omega) \log(\frac{p(\omega)}{q(\omega)}) \; d \omega$
like a distance between probability distributions? A distance function, or metric, is supposed to satisfy some axioms. Alas, relative entropy satisfies some of these, but not the most interesting one!
• If you’ve got a metric, the distance between points should always be nonnegative. Indeed, this holds:
$S(p,q) \ge 0$
So, we never learn a negative amount when we update our prior, at least according to this definition. It’s a fun exercise to prove this inequality, at least if you know some tricks involving inequalities and convex functions — otherwise it might be hard.
• If you’ve got a metric, the distance between two points should only be zero if they’re really the same point. In fact,
$S(p,q) = 0$
if and only if
$p d\omega = q d \omega$
It’s possible to have $p d\omega = q d \omega$ even if $p \ne q$, because $d \omega$ can be zero somewhere. But this is just more evidence that we should really be talking about the probability measure $p d \omega$ instead of the probability distribution $p$. If we do that, we’re okay so far!
• If you’ve got a metric, the distance from your first point to your second point is the same as the distance from the second to the first. Alas,
$S(p,q) \ne S(q,p)$
in general. We already saw this in our example of the flipped coin. This is a slight bummer, but I could live with it, since Lawvere has already shown that it’s wise to generalize the concept of metric by dropping this axiom.
• If you’ve got a metric, it obeys the triangle inequality. This is the really interesting axiom, and alas, this too fails. Later we’ll see why.
So, relative entropy does a fairly miserable job of acting like a distance function. People call it a divergence. In fact, they often call it the Kullback-Leibler divergence. I don’t like that, because ‘the Kullback-Leibler divergence’ doesn’t really explain the idea: it sounds more like the title of a bad spy novel. ‘Relative entropy’, on the other hand, makes a lot of sense if you understand entropy. And ‘information gain’ makes sense if you understand information.
Anyway: how can we save this miserable attempt to get a distance function on the space of probability distributions? Simple: take its matrix of second derivatives and use that to define a Riemannian metric $g_{ij}$. This Riemannian metric in turn defines a metric of the more elementary sort we’ve been discussing today.
And this Riemannian metric is the Fisher information metric I’ve been talking about all along!
More details later, I hope.
This entry was posted on Friday, January 21st, 2011 at 8:02 am and is filed under information and entropy. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### 21 Responses to Information Geometry (Part 6)
1. John Sidles says:
John, the two themes of your previous (excellent) essay #5 (namely dissipative physics and partition functions) can be combined with the theme of this essay #6 (namely probability measures) as follows.
Let us consider a spin-j particle in a magnetic field (so that it has equally-spaced energy levels) that is in dynamical contact with a thermal reservoir, so that its relaxation is described by Bloch parameters (so the harmonic oscillator is simply the infinite-j limit). Then unravel the dynamics, by the methods of Carmichael and Caves, in a Lindblad gauge that has no quantum jumps.
Now ask three natural questions: (1) What is the geometric submanifold of Hilbert space onto which the quantum trajectories are dynamically compressed? (2) What is the pulled-back symplectic structure of that submanifold? (3) What is the induced P-representation of the thermal density matrix?
The answers are simple and geometrically natural: (1) The manifold is the familiar Riemann sphere of the Bloch equations. (2) The symplectic structure associated to Bloch flow is canonical symplectic structure of the Riemann sphere. (2) The positive P-representation is given in terms of the thermal Q representation by
$P_{j}(x) = 1/Q_{j+1}(-x).$
These geometric relations are reflections of the general principle that Wojciech Zurek calls einselection … can be verified algebraically, yet the motivation for their derivation arises wholly in geometric ideas associated to Kählerian structure and Hamiltonian/Lindbladian flow.
It is perfectly feasible to systematically translate the whole of Nielsen and Chang’s Quantum Computation and Quantum Information from its original algebraic language into the geometric language of Kählerian/Hamiltonian/Lindbladian flow.
The resulting algebraic and geometric representations of quantum dynamics are rigorously equivalent … and yet they “sit in our brains in different ways” (in Bill Thurston’s phrase). In practical quantum systems engineering both the algebraic and geometric dynamical pictures come into play … it’s fun! :)
• John Baez says:
Your comment sounds interesting but there are a lot of things I don’t understand about it, so it’s quite mysterious. Here are a few questions. I could try to look up answers on the all-knowing Google, but it’s probably better for other people if I just ask.
I should probably start with just one. If you answer that, nice and simply, without saying a whole lot of stuff that raises new questions in my mind, I can move on to the next.
You write:
Let us consider a spin-j particle in a magnetic field (so that it has equally-spaced energy levels) that is in dynamical contact with a thermal reservoir, so that its relaxation is described by Bloch parameters (so the harmonic oscillator is simply the infinite-j limit).
I know why a spin-j particle in a magnetic field has (2j+1) equally spaced energy levels. But I don’t know what its “Bloch parameters” are. What are they?
I know about the Bloch sphere for a spin-1/2 particle. Each mixed state has expectation values of angular momentum operators $J_x, J_y, J_z$ lying in some ball. That would also be true for higher spins, though the ball would have a bigger radius. Is that what you’re talking about?
2. John Sidles says:
Yep! The Bloch parameters are commonly denoted $T_1$ and $T_2$ … in the magnetic resonance literature these are often called the longitudinal and transverse relaxation times. In the infinite-j limit $T_1 = \pi Q/f$, where $f$ is the harmonic oscillator frequency and $Q$ is the quality. The details are worked out in Spin microscopy’s heritage, achievements, and prospects and references therein.
It is instructive to tell this same story as it might have unfolded in a universe in which quantum systems engineers embraced category theory early and fervently … you can regard the following as the (real-time!) first draft of a lecture that I’ll give at the Asilomar Experimental Nuclear Magnetic Resonance Conference (ENC) this April … comments and corrections are welcome!
We’ll begin by noticing that “Mac Lane has said that categories were originally invented, not to study functors, but to study natural transformations.” (Baez and Dolan, Categorification, 1998). In seeking an original text for this quote, the best that I have found (so far) is Mac Lane’s “But I emphasize that the notions category and functor were not formulated or put in print until the idea of a natural transformation was also at hand.” (Mac Lane, The development and prospects for category theory, 1996). John (B), is there a better text than this one?
Another terrific article by Mac Lane is his Chauvenet Lecture Hamiltonian mechanics and geometry (1970), which closes with the idea that “More use of categorical ideas may be indicated [in dynamics] … . Here, as throughout mathematics, conceptual methods should penetrate deeper to give clearer understanding.”
For engineers, one major obstruction to wider categoric-theoretic applications is that commutative diagrams express equivalences that are exact. Exactitude is wonderful when it reflects exact conservation laws or exact gauge-type invariances. But how can category theoretic methods help us in the case (overwhelmingly more common for engineers) of calculations that are not exact?
One thing we can do is go all the way back to the very first commutative diagram (or is it?) that appeared in the mathematical literature, namely the one in Section 5 of Mac Lane and Eilenberg’s Natural isomorphisms in group theory (1942). We alter it to express not a commutative relation, but rather a projective relation (so that the arrow-heads are cyclic rather commutative). The resulting category-theoretic notion of projective naturality is equally rigorous to the traditional notion of commutative naturality, but less stringent in its implications and thus broader in its applications … this is exactly what engineers need.
The details are worked through in a draft preprint Elements of naturality in dynamical simulation frameworks for Hamiltonian, thermostatic, and Lindbladian flows on classical and quantum state-spaces (arXiv:1007.1958) … Tables I and II recast the commutative diagrams of Eilenberg and Mac Lane into the projective form that is more suited to practical engineering, and the remainder of the preprint systematically translates Chapters 2 and 8 of Nielsen and Chuang’s Quantum Computation and Quantum Information—including the Bloch equations—into this geometric/category-theoretic framework.
We’re now reworking this arXiv:1007.1958 preprint (prior to its presentation at Asilomar) to pay tribute to its category-theoretic origins on the one had, and yet to make clear its rigorous formal equivalence to traditional algebraic formulations of quantum dynamics on the other hand. Hence our desire to reference original texts whenever possible … suggestions from category-theorists in this regard are very welcome.
We have found that many wonderful articles on geometric dynamics are readily appreciated in terms of these category-theoretic ideas. In particular, Ashtekar and Schilling’s Geometrical formulation of quantum mechanics (1999) and J.W. vanHolten’s Aspects of BRST quantization (2005) both are founded upon geometric ideas that are well-expressed in a category-theoretic notation.
One comes away from this reading with the idea that, for practical purposes at least, the effective state-space of quantum mechanics is not a flat, non-dynamic Hilbert space of exponentially large dimensionality, but rather is a curved, dynamical state-space of polynomial dimensionality … very much like the state-space of classical mechanics. This is good news for 21st century mathematicians, scientists, and engineers: it means that there is plenty of both practical and fundamental work to do.
Does nature herself exploit the mathematics of projective naturality to render herself easy to simulate? Is nature’s state-space of quantum dynamics a Hilbert space only in a small-dimension local approximation? That is a fun question—and a great virtue of category-theoretic method is that they allow us to pose this question cleanly.
From an “Azimuthal” point-of-view, the main objective is perhaps more engineering-oriented: to develop 21st century spin metrology methods that “see every atom in its place,” precisely as von Neumann and Feynman once envisioned. What their 20th century envisioned as an utopian capability, our 21st century has reasonable prospects of achieving in practice … and that is exciting.
• John Baez says:
Hmm. You said how the Bloch parameters are denoted, and what they’re called, but not what they are. I guessed that they were the expectation values of $J_x, J_y$ and $J_z$ for a spin-j particle, and you said:
Yep!
but then you just mentioned two things, denoted $T_1$ and $T_2$… and you said they’re called relaxation times. So I don’t think I understand, yet… and my guess doesn’t sound right.
So: what are the Bloch parameters, exactly? I can’t get into all the fancier stuff you’re discussing before I understand the basics.
We’ll begin by noticing that “Mac Lane has said that categories were originally invented, not to study functors, but to study natural transformations.” (Baez and Dolan, Categorification, 1998). In seeking an original text for this quote, the best that I have found (so far) is Mac Lane’s “But I emphasize that the notions category and functor were not formulated or put in print until the idea of a natural transformation was also at hand.” (Mac Lane, The development and prospects for category theory, 1996). John (B), is there a better text than this one?
That could easily be the best. A quick Google scan reveals that everyone is running around saying a version of the quote sort of like mine, but without giving a source. Perhaps they’re all quoting me!
But I know I got it from somewhere reputable, and maybe it was this… or maybe just verbally, from someone who knew Mac Lane. (I met him a few times, but he didn’t say this to me!)
By the way, you can do LaTeX here if you put ‘latex’ right after the first dollar sign, and don’t do double dollar signs or anything too fancy. Thus,
\$latex \int e^x dx = e^x + C\$
gives
$\int e^x dx = e^x + C$
I’ve been LaTeXifying some of your comments thus far, just for fun. But this is how one does it oneself.
3. John Sidles says:
Thank you, John, for the LaTeXing hint!
With regard to the Bloch parameters, the Wikipedia pages on (1) the Bloch equations, (2) the Bloch-Torrey equations, and (3) the Landau-Lifshitz Gilbert (LLG) equations give all the definitions and dynamical equations … but unfortunately the Wikipedia version of the dynamical equations are not expressed in the geometrically natural language of symplectic and metric flow (per your post Information Geometry #5) … this translation one has to do for one’s self.
Ok, so let’s do it. If the Hamiltonian potential associated to the Bloch equation is the usual quantum-state expectation of the Hamiltonian operator on the submanifold of coherent states, then what might be the symplectic form on the Bloch sphere of states? Well, what else could it be … but the canonical $S^2$ volume form?
Guided by this intuition, it is easy to work out that the symplectic/Hamiltonian flow on the Bloch sphere is identical to Wikipedia’s Bloch equations, which with the addition of a metric flow component is identical to the LLG equations, etc. (and we have a student writing up these equivalences).
In other words, the Bloch equations, BLoch-Torrey equations, LLG equations, etc. all are geometrically natural from a category-theoretic point-of-view. Any other result would be wholly unexpected … yet it is remarkable how few textbooks (none AFAICT) discuss the symplectic and metric structures that are naturally associated to Bloch/LLG dynamics.
• John Baez says:
So you’re not going to tell me what the Bloch parameters are?
That’s too bad, because it’s just the first of many questions I had, but I can’t really proceed without knowing the answer to this one.
I can of course look things up, but I’m too busy, so I’d been hoping to have a conversation of the old-fashioned sort, where mathematician A says “What are the Bloch parameters?” and mathematician B says “The Bloch parameters are…”
• Justin says:
This is not my expertise per se, but here is the general picture:
T1 is spin-lattice relaxation time. It is a characteristic time of re-equilibration of spin states responding to thermal fluctuations, leading to thermalization.
T2 is spin-spin relaxation time. It is a characteristic time of re-equilibration of spin states responding to quantum fluctuations, leading to decoherence.
4. John Sidles says:
John, I’m really trying … it’s tough to balance rigorous but too-long, against brief but non-rigorous. Here’s another try.
In any linear model of density matrix relaxation, the relaxation is described by a set of rate constants … these rate constants are the Bloch parameters broadly defined.
A particularly common case is a spin-1/2 particle in a magnetic field, whose relaxation is transversely isotropic with respect to the field direction. Then the Bloch relaxation is described by precisely two Bloch parameters … traditionally these parameters are called $T_1$ and $T_2$, and the canonical reference is Slichter’s Principles of Magnetic Resonance.
Very much more can be said … the discussion of the “Lindblad form” in Nielsen and Chuang Ch. 8.4 is a pretty good start … one has to be ready for the practical reality that different authors—partly by history and partly by practical necessity—embrace very different terminology and notation in describing these ideas.
5. John Sidles says:
As a followup, once one has a reasonable working grasp on the relaxation of density matrices via Lindblad processes, then a logical next step is to abandon density matrix formalisms … in favor of quantum trajectory formalisms … see in particular the unravelling formalism introduced by Carmichael in his An Open Systems Approach to Quantum Optics (1993) … as far as I know, this is the work that introduced the now-popular concept of “unravelling” to the quantum literature (see Carmichael’s Section 7.4, page 122).
Here the guiding mathematical intuition is that density matrices are not defined on non-Hilbert quantum state-spaces (for example, on tensor network state-spaces) … but unravelled ensembles of trajectories are well-defined … thus the pullback (of dynamical structures) and the pushforward (of trajectories and data-streams) is seen to be geometrically natural … in which respect Carmichael-type quantum unravelling formalisms are more readily generalizable & geometrically natural than density matrix formalisms.
The preceding strategy for making mathematical and physical sense of a vast corpus of literature is driven mainly by considerations of geometric and category-theoretic naturality … no doubt many alternative approaches are feasible … this diversity is what’s fun about quantum dynamics. I hope this helps! :)
6. John Sidles says:
Continuing the above discussion, on Dick Lipton’s weblog Gödel’s Lost Letter and P=NP, the intimate relation of the above ideas to modern-day (classical) [quantum] {informatic} simulation codes is discussed.
7. David Corfield says:
In this article, this will be referred to as the divergence from P to Q, although some authors call it the divergence “from Q to P” and others call it the divergence “between P and Q” (though note it is not symmetric as this latter terminology implies).
Still it seems odd where you write
The information gain as we go from $p$ to $q$ is …,
I think I’d prefer it the other way around.
Later you write
So, relative entropy does a fairly miserable job of acting like a distance function.
There’s an argument that relative entropy is the square of a distance, agreed a few comments below. Look for Pythagorean theorem in Elements of Information theory here.
• John Baez says:
David wrote:
Still it seems odd where you write
The information gain as we go from $p$ to $q$ is …
I think I’d prefer it the other way around.
Oh, definitely! That was just a typo — I’ve fixed it now, thanks. I should mind my $p$‘s and $q$‘s.
There’s an argument that relative entropy is the square of a distance, agreed a few comments below.
Only after you symmetrize it, of course. The relative entropy is not symmetric:
$S(p,q) \ne S(p,q)$
so it can’t give a metric (of the traditional symmetric sort) when you take its square root. But Suresh Venkat is claiming that the symmetrized gadget, which he calls the ‘Jensen-Shannon distance’:
$\frac{1}{2} S(p,q) + \frac{1}{2} S(q,p)$
does give a metric when you take its square root. I’ll have to check this, or read about it somewhere if I give up. You just need to check the triangle inequality… the rest is obvious.
But as your n-Café comment notes, there’s also another nice metric floating around. If we define a version of relative entropy based on Rényi entropy instead of the usual Shannon kind, we get something called the ‘Rényi divergence’ which is symmetric only for the special case $\alpha = 1/2$… and while it still doesn’t obey the triangle inequality in that case, it’s a function of something that does. Not-so-coincidentally, I was just reading about this today. This thesis has a nice chapter on Rényi entropy and the corresponding version of relative entropy:
• T. A. L. van Even, When Data Compression and Statistics Disagree: Two Frequentist Challenges For the Minimum Description Length Principle, Chap. 6: Rényi Divergence, Ph.D. thesis, Leiden University, 2010.
Yeah, it’s on page 179:
Only for $\alpha = 1/2$ is Rényi divergence symmetric in its arguments. Although not itself a metric, it is a function of the square of the Hellinger distance
$\mathrm{Hel}^2(p,q) = \sum_{i = 1}^n (\sqrt{p_i} - \sqrt{q_i})^2$
[Gibbs and Su, 2002].
Anyway, my goal in this post was pretty limited. I just wanted to explain the concept of relative entropy a little bit, so people can following what I’m saying when I explain how it’s related to the Fisher information metric.
8. Now I want to describe how the Fisher information metric is related to relative entropy [...]
9. Remember what I told you about relative entropy [...]
10. For more on relative entropy, read Part 6 of this series [...]
11. Luisberis Velazquez says:
To my knowledge, the notion of relative entropy:
$S(p,q)= -\int \left[d\mu(x)/d\nu(x)\right]\log\left[d\mu(x)/d\nu(x)\right]d\nu(x)$
was early proposed by Jaynes to extend the notion of information entropy to the framework of continuous distributions. However, I think that the relative entropy is not a fully satisfactory generalization of this concept. A natural question here is how to introduce the measure:
$d\nu(x)=q(x)dx$
when no other information is available, except the continuous distribution:
$d\mu(x)=p(x)dx.$
Recently, I proposed a way to overcome this difficulty in the framework of Riemannian geometry of fluctuation theory:
http://iopscience.iop.org/1751-8121/45/17/175002/article
This geometry approach introduces a distance notion:
$ds^{2}=g_{ij}(x)dx^{i}dx^{j}$
between two infinitely close points $x$ and $x+dx$, where the metric tensor is obtained from the probability density $p(x)$ as follows:
$g_{ij}=-\frac{\partial^{2}\log p}{\partial x^{i}\partial x^{j}}+ \Gamma^{k}_{ij}\frac{\partial\log p}{\partial x^{k}} +\frac{\partial\Gamma^{k}_{jk}}{\partial x^{i}}-\Gamma^{k}_{ij}\Gamma^{l}_{kl}$.
Here, the symbol $\Gamma^{k}_{ij}$ denote the Levi-Civita affine connections. Formally, this is a set of covariant partial differential equations of second order in terms of the metric tensor. The measure $d\nu(x)$ can be defined as follows:
$d\nu(x)=\sqrt{\left|g_{ij}(x)/2\pi \right| }dx.$
Apparently, the relative entropy that follows from this ansatz is a global measure of the curvature of the manifold $\mathcal{M}$ where the continuous variables $x$ are defined. Some preliminary analysis suggest that curvature $R(x)$ is closely related to the existence of irreducible statistical correlations among the random variables $x$, that is, statistical correlations that survive any coordinate transformations $y=\phi(x)$.
• John Baez says:
Hi! How are you constructing your Levi-Civita connection starting from $p?$ I know how to construct a Levi-Civita connection starting from a metric, but you’ve defined your metric starting from a Levi-Civita connection. The fact that you speak of ‘the’ Levi-Civita connection makes me a little nervous, since there are many.
12. Luisberis Velazquez says:
By the way, this is a great blog!
• John Baez says:
Thanks!
13. Luisberis Velazquez says:
Levi-Civita affine connections or the metric connections only depend on the metric tensor:
$\Gamma _{ij}^{k}\left( x|\theta \right) =g^{km}(x|\theta)\frac{1}{2}\left[\frac{\partial g_{im}(x|\theta)}{\partial x^{j}}+\frac{\partial g_{jm}(x|\theta)}{\partial x^{i}}-\frac{\partial g_{ij}(x|\theta)}{\partial x^{m}}\right]$,
and they are the only affine connections with a tonsion-less covariant derivative $D_{k}$ that obeys the condition of Levi-Civita parallelism:
$D_{k}g_{ij}(x)=0$.
The relation between the metric tensor and the probability density represents a set of covariant partial equations of second-order in terms of the metric tensor. This equation is non-linear and self-consistent. As expected, it is difficult to solve in most of practical situations. However, this ansatz allows to introduce a measure $d\nu(x)$ for the relative entropy only considering the probability density of interest.
Surprinsingly, the value of this relative entropy can be expressed in an exactly way as follows:
$S(p|q)=\log Z-n/2$,
where $n$ is the dimension of the manifold $\mathcal{M}$ and $Z$ a certain normalization constant. Precisely, a direct consequence of this set of covariant differential equations is the possibility to rewrite the original distribution function:
$d\mu(x)=p(x)dx$
as follows:
$d\mu(x)=\frac{1}{Z}\exp\left[-\frac{1}{2}\ell^{2}(x,\bar{x})\right]d\nu(x)$.
Here, $d\nu(x)$ is the invariant measure:
$d\nu(x)=\sqrt{\left| g_{ij}(x)/2\pi \right|}dx$;
$\bar{x}$ denotes the most likely point of the distribution, while $\ell^{2}(x,\bar{x})$ denotes the arc-length of geodesics that connects the points $x$ and $\bar{x}$. Formally, this is a Gaussian distribution defined on a Riemannian manifold $\mathcal{M}$. The normalization constant $Z$ is reduced to the unity if the manifold $\mathcal{M}$ is diffeomorphic to the n-dimensional Euclidean real space $\mathbb{R}^{n}$, and it takes different values when the manifold exhibits a curved geometry. Consequently, this type of relative entropy is a global measure of the curvature of the manifold $\mathcal{M}$ where the random variables $x$ are defined. To my present understanding, curvature accounts for the existence of irreducible statistical correlations, something analogous to the irreducible character of gravity in different references frames due to its connection with curvature.
14. Luisberis Velazquez says:
Sorry with type-errors. The symbol $\theta$ in the Levi-Civita connections comes from my original work, with deals with parametric family of continuous distributions with a number $m$ of shape parameters $\theta=(\theta^{1},\theta^{2},\ldots \theta^{m})$. Certainly, most of differential equations of Riemannian geometry are difficult to solve. I spend a lot of time (years) to decide to explore this type of statistical geometry. Recently, I realize that some mathematical derivations are not difficult to perform despite the exact mathematical form of the metric tensor $g_{ij}(x)$ is unknown for a concrete distribution $d\mu(x)=p(x)dx$.
My fundamental interest on this geometry are its applications on classical statistical mechanics. This type of geometry was inspired on Ruppeiner’s geometry of thermodynamics. The metric tensor of this last formulation was modified replacing the usual derivative $\partial_{i}$ by the Levi-Civita covariant derivative:
$g_{ij}(x)=-D_{i}D_{j}\mathcal{S}(x)$.
Here, $\mathcal{S}(x)$ is the thermodynamic entropy. Since the entropy is a scalar function, the above relation guarantees the covariance of the metric tensor. Moreover, the thermodynamic entropy $\mathcal{S}(x)$ enters in mathematical expression of Einstein postulate of classical fluctuation theory, which should be extended as follows:
$d\mu(x)=\exp\left(\mathcal{S}(x)\right)d\nu(x)$
to guarantee the scalar character of the entropy. The gaussian distribution that I commented above is an exact improvement of gaussian approximation of classical fluctuation theory of statistical mechanics.
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http://math.stackexchange.com/questions/86015/find-a-cubed-function-knowing-the-maximum-and-the-point-of-inflection
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# Find a cubed function knowing the maximum and the point of inflection?
Is it possible to find the function of a cubed line if we know its maximum and its point of inflection?
if yes, can some one explain me?
Thank you very much!
-
What's a "cubed line"? – J. M. Nov 27 '11 at 11:38
And if you mean "cubing function", what do you mean by "maximum"? – David Mitra Nov 27 '11 at 11:50
You have had two requests for clarification, and one answer, and we've not heard anything back from you - are you still there? – Gerry Myerson Dec 1 '11 at 6:35
graphtheory92: remember to upvote, and/or accept answers you find to be helpful. You can accept one answer per question, but you can upvote any/all answers that are correct, that help, etc. ;-) – amWhy Feb 14 at 14:36
## 1 Answer
I'm going to assume that there's an equation $y=ax^3+bx^2+cx+d$ where $a,b,c,d$ are unknowns to be found - if that's not what you have in mind, please clarify.
I also assume you know there is a local maximum at $(r,s)$, and a point of inflection at $(u,v)$.
So what you know is $y(r)=s$, $y'(r)=0$, $y(u)=v$, and $y''(u)=0$. Well, that's four linear equations in four unknowns, I'm sure you can handle that.
EDIT: As J.M. notes in the comments, this is an example of Hermite interpolation.
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This is in fact a Hermite interpolation problem... – J. M. Nov 27 '11 at 12:08
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http://mathoverflow.net/questions/27861?sort=newest
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## Periodic Automorphism Towers
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In Scott's classic textbook on Group Theory, he asks:
Suppose that $G$ is a finite group. Is the sequence of isomorphism types of the groups $Aut^{(n)}(G)$ for $n \in \mathbb{N}$ eventually periodic?
Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence $|Aut^{(n)}(G)|$ always bounded above?
It apparently remains opens whether the sequence of automorphism types of $Aut^{(n)}(G)$ is in fact always eventually constant. (A wonderful theorem of Wielandt says that if $G$ is a finite centerless group, then the sequence is eventually constant.) So I would like to ask:
Does there exists a finite group such that $Aut(G) \not \cong G$ but $Aut^{(n)}(G) \cong G$ for some $n \geq 2$?
Edit: Joel has pointed out that my question is perhaps even open for infinite groups. This sounds like an interesting question which doesn't seem amenable to the standard tricks.
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2
You might want to look at the answers to mathoverflow.net/questions/5635/… – David Speyer Jun 11 2010 at 20:10
@David: I hadn't noticed that this question had already been asked in mathoverflow.net/questions/5635/…. But it's a good question so it's worth asking again! – Simon Thomas Jun 11 2010 at 20:30
How are the answers for 5635 not adequate? – Kevin O'Bryant Jun 11 2010 at 21:21
6
For everyone's information, some of the best answers to the other question amount to providing links to Simon Thomas' articles and book on the subject. – Joel David Hamkins Jun 11 2010 at 21:26
1
Mmmm ... maybe the infinite case is worth thinking about! – Simon Thomas Jun 12 2010 at 0:12
show 8 more comments
## 1 Answer
I think this anwers the question for infinite groups:
MR0470091 (57 #9858) Collins, Donald J. The automorphism towers of some one-relator groups. Proc. London Math. Soc. (3) 36 (1978), no. 3, 480--493. 20F55
Theorem (ii) states that if $G=\langle a,b \mid a^{-1}b^ra=b^s \rangle$ is a Baumslag-Solitar group with $r-s$ even, then $Aut(Aut(G))$ is isomorphic to $G$ and $G$ has an outer automorphism.
Moreover, when $r=1$, $G$ is the semidirect product $\mathbf Z \ltimes \mathbf Z[\frac 1 s]$, where $\mathbf Z$ acts via multiplication by $\frac 1 s$. Then $G$ is torsionfree, but $Aut(G)$ has an element of order 2 (see his lemma 3). If $G$ is represented as a matrix group, ```$(a,b) \mapsto \begin{pmatrix} s^a & b \\
0 & 1 \end{pmatrix}$```, then this outer automorphism is explicitely given by conjugation by $diag(i,-i)$, where $i$ is a square root of -1.
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This looks like an example but there is one part of your argument that doesn't seem correct. By Lemma 1 of Collins, when $r=1$, the group $G$ is centreless. This means that $Aut(G)$ is also centreless and hence embeds in $Aut(Aut(G)) \cong G$. Thus $Aut(G)$ cannot have an element of finite order. Of course, given the statement of Collins' Theorem, it seems almost certain that one of his groups does provide an answer to the infinite case of my question. – Simon Thomas Jun 12 2010 at 12:26
Indeed, after reading the paper more closely I think Theorem (ii) should read that Aut(Aut(G)) is isomorphic to Aut(G), so this paper doesn't seem to provide an example. Sorry about that. – Guntram Jun 12 2010 at 13:15
Looking more carefully at Collins' paper, I believe there is a typo in the statement of Theorem 1. Combining Proposition A and Proposition B, we see that $Aut(Aut(G)) \cong Aut(G)$ instead of $Aut(Aut(G)) \cong G$. And my previous comment shows that $Aut(Aut(G)) \not\cong G$. So I no longer believe that Collins' paper answers the infinite case of my question. – Simon Thomas Jun 12 2010 at 13:23
I see that we spotted the typo simultaneously! – Simon Thomas Jun 12 2010 at 13:24
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http://mathoverflow.net/questions/59067/frobenius-splitting-and-derived-cartier-isomorphism/66514
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## Frobenius splitting and derived Cartier isomorphism
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Let $X$ be a smooth algebraic variety over an algebraically closed field $k$ of characteristic $p>\dim X$. The motivation for my question comes from the following results.
1. If $X$ is Frobenius split (the $p$-th power map `$\mathcal{O}_X \to F_* \mathcal{O}_X$` admits n $\mathcal{O}_X$-linear splitting) then the Kodaira vanishing theorem holds for $X$.
The proof uses nothing but Serre vanishing and the projection formula.
2. If the complex `$F_* \Omega^\bullet_X$` is quasi-isomorphic to a complex with zero differentials, then the Kodaira-Akizuki-Nakano vanishing theorem holds for $X$.
The proof uses Cartier isomorphism, hypercohomology spectral sequences, Serre vanishing and the projection formula and is similar to that of 1.
3 (Deligne-Illusie 1987). If $X$ lifts to $W_2(k)$, then the complex `$F_* \Omega^\bullet_X$` is quasi-isomorphic to a complex with zero differentials.
4 (Buch-Thomsen-Lauritzen-Mehta 1995). If $X$ is strongly Frobenius split (that is, `$0\to B_1\to Z_1\to \Omega^1_X\to 0$` splits, where $Z_i$ and $B_i$ are cocycles/coboundaries in `$F_* \Omega^\bullet_X$`), then $X$ and $F$ lift to $W_2(k)$ and the Bott vanishing theorem holds for $X$.
My (maybe incorrect) feeling is that strong Frobenius splitting and lifting of the Frobenius to $W_2(k)$ are quite uncommon, Frobenius splitting is a common behavior "on the Fano side" and that lifting of $X$ to usually $W_2(k)$ exists.
Question. Are there examples of Frobenius split varieties for which `$F_* \Omega^\bullet_X$` is not quasi-isomorphic to a complex with zero differentials (for example, because the Hodge spectral sequence does not degenerate, see also this question on the Hodge spectral sequence)? If yes (that's my intuition here), does Frobenius splitting imply some weaker property of $F_* \Omega^\bullet_X$ which implies Kodaira vanishing?
Edit. Note that Frobenius splitting just states that the complex `$F_* \Omega^\bullet_X$` is quasi-isomorphic to a complex whose first differential $C^0 \to C^1$ is zero.
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I believe that even if Frobenius doesn't lift you still get that a Frobenius split variety (with nice assumptions) will lift to $W_2(k)$. This is because even though the obstruction to lifting Frobenius $\eta\in Ext^1(\Omega^1, B^1)$ is non-zero, the obstruction to merely lifting $X$ is the image of $\eta$ under the connecting homomorphism $Ext^1(\Omega^1, B^1)\to Ext^2(\Omega^1, \mathcal{O}_X)$. By the splitting assumption this map is $0$ and hence there is no obstruction to lifting $X$. See V. Srinivas Decomposition of the de Rham Complex. – Matt May 29 2011 at 19:10
Great! Why don't you post this as an answer? – Piotr Achinger May 30 2011 at 11:23
## 2 Answers
Keeping the notation of the question $X$ is a smooth variety over an algebraically closed field $k$ of characteristic $p>\dim X$. Just following along from Decomposition of the de Rham Complex by V Srinivas, we see that the obstruction to lifting the pair $(X,F)$ to a pair $(X^{(2)}, F^{(2)})$ where $X^{(2)}$ is a lift of the variety to $W_2(k)$ and $F^{(2)}$ is a lift of Frobenius consistent with all diagrams is exactly the class $\zeta\in \mathrm{Ext}^1(\Omega_{X/k}^1, B_X^1)$ that corresponds to the sequence $0\to B_X^1\to Z_X^1\to \Omega_{X/k}^1\to 0$.
If we look at the sequence `$0\to \mathcal{O}_X\to F_* \mathcal{O}_X\to B_X^1\to 0$` and take $\mathrm{Hom}(\Omega^1, -)$ we get a connecting homomorphism in the long exact sequence $\mathrm{Ext}^1(\Omega^1, B^1)\stackrel{\delta}{\to} \mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)$. It is well known that the obstruction to lifting lies in $\mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)\simeq H^2(X, \mathcal{T}_X)$, but what is not well-known is that the obstruction class in this case is exactly the image of $\zeta$ under $\delta$. So $\delta$ acts as sort of a forgetful map for obstruction to lifting the pair to obstruction for lifting the variety without lifting Frobenius.
While it is possible that a Frobenius split variety has non-zero obstruction class $\zeta$ (no example comes to mind right now) and hence the pair doesn't lift, this splitting assumption actually gives us lots of information when coupled with the above information.
We see that `$0\to \mathcal{O}_X\to F_* \mathcal{O}_X\to B_X^1\to 0$` splitting gives us $\mathrm{Ext}^1(\Omega^1, F_*\mathcal{O}_X)\twoheadrightarrow \mathrm{Ext}^1(\Omega^1, B^1)\stackrel{\delta}{\to} \mathrm{Ext}^2(\Omega^1, \mathcal{O}_X)$, so in fact $\delta=0$. This means that it doesn't matter whether or not we can lift the pair, all we had to know was that the obstruction to lifting $X$ was the image of $\zeta$ under $\delta$ which is $0$.
Thus any smooth Frobenius split variety lifts to $W_2(k)$ and since we assumed $p>\dim X$ we also get that the Hodge-de Rham spectral sequence degenerates at $E_1$ by work of Deligne and Illusie.
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It is indeed strange that in the Brion-Kumar book, the main reference on the subject, they prove Kodaira vanishing but don't even remark that the full Kodaira-Akizuki-Nakano vanishing holds. – Piotr Achinger May 31 2011 at 7:15
Do you know any examples of Frobenius split varieties which do not lift all the way to characteristic zero? – mdeland May 31 2011 at 13:18
I would love to know such an example. This is something I've been thinking a lot about lately. If you get a lift of the pair to $W_2$ then you should be able to repeat a similar argument to get a lift all the way up, but you'd still also need algebraizability. My guess is that there must be some fairly easy examples. – Matt May 31 2011 at 16:03
I had wondered that also but I could never prove it. Maybe lifting the pair to W_2 only guarantees a lifting of X to W_3? – mdeland May 31 2011 at 17:42
That is quite possible (although my gut says if you can do the next lift, then there is some sort of common argument that should keep going through). The main candidate for a Frobenius split that does not lift to characteristic $0$ is something that lifts to $W_2(k)$ but doesn't lift to characteristic $0$, but these are rather hard to come by. – Matt May 31 2011 at 21:15
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I just found a nice reference for this question: K. Joshi "Exotic Torsion, Frobenius Splitting and the Slope Spectral Sequence" (Canad. Math. Bull. Vol. 50 (4), 2007), section 9.
Theorem 9.1 (unpublished work of V. B. Mehta). Let X be a smooth, projective, F-split variety over an algebraically closed field $k$ of characteristic $p>0$. Then for all $i+j < p$, the Hodge to de Rham spectral sequence degenerates at $E^{i,j}_1$. In particular, for $i+j = 1$ we have the following exact sequence $$0\to H^0(X, \Omega^1_X)\to H^1_{DR}(X/k)\to H^1(X, \mathcal{O}_X)\to 0.$$ Moreover, any F-split variety with \$dim(X)
Corollary 9.2. Let $X/k$ be a smooth, projective, Frobenius split variety. Then $X$ admits a flat lifting to $W_2$.
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http://www.scholarpedia.org/article/Linear_multistep_method
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Linear multistep method
From Scholarpedia
Ernst Hairer and Gerhard Wanner (2010), Scholarpedia, 5(4):4591.
Curator and Contributors
1.00 - Gerhard Wanner
0.50 -
Eugene M. Izhikevich
0.17 -
Barbara Zubik-Kowal
0.17 -
James Meiss
0.17 -
Ernst Hairer
Dr. Ernst Hairer accepted the invitation on 3 October 2008 (self-imposed deadline: 3 April 2009).
Linear multistep methods constitute an important class of numerical integrators for ordinary differential equations, and particular methods are well suited for solving non-stiff and stiff equations as well as Hamiltonian systems over long time intervals. In contrast to one-step methods, multistep methods increase efficiency by using the information from several previous solution approximations.
Methods for nonstiff problems
Consider an initial value problem $\tag{1} \dot y=f(t,y), \quad y(t_0)=y_0 ,$
where the solution $$y(t)$$ can be scalar- or vector-valued. Let $$t_0,t_1,t_2,\ldots$$ be a sequence of grid points, which for simplicity is supposed to be equidistant with step size $$h=t_{n+1}-t_n\ .$$ A numerical method is an algorithm that yields approximations $$y_n$$ to the solution $$y(t_n)$$ at the grid points.
Figure 1: Explicit Adams methods
Explicit Adams methods
These methods are introduced by J.C. Adams (1883) for solving practical problems of capillary action. They are based on the following idea: suppose that $$k$$ values $$y_{n-k+1},\,y_{n-k+2},\ldots, y_{n}$$ approximating $$y(t_{n-k+1}),\, y(t_{n-k+2}),\ldots,y(t_{n})$$ are known. Compute the derivatives $\tag{2} f_{n+j}=f(t_{n+j},y_{n+j}), \quad j=-k+1,\ldots ,0$
and replace in the integrated form of (1) $\tag{3} y(t_{n+1})=y(t_{n})+\int _{t_{n}}^{t_{n+1}}f(t,y(t))\,dt$
the integrand $$f(t,y(t))$$ by the polynomial $$p(t)$$ interpolating the values (2). Then evaluate the integral analytically and obtain the next approximation to the solution, $$y_{n+1}\ .$$ After advancing the scheme by one step, this procedure can be repeated to obtain $$y_{n+2},\, y_{n+3}\ ,$$ and so on. See Figure 1 for an illustration using the logistic growth equation $$\dot y=a y (1-y)\ .$$
Newton's formula for polynomial interpolation can be written as $$\tag{4} p(t) =f_n + \frac{1}{h}(t-t_n)\nabla f_n + \frac{1}{2 h^2}(t-t_n)(t-t_{n-1})\nabla^2f_n + \ldots$$
where $$\nabla$$ is the backward difference operator $$\nabla f_n = f_n - f_{n-1}\ ,$$ $$\nabla^2 f_n = \nabla f_n - \nabla f_{n-1} = f_n -2f_{n-1}+f_{n-2}\ ,$$ etc. This leads to the following equations (for more details see Hairer, Nørsett & Wanner (1993), p. 357-358): $\tag{5} \begin{array}{ll} k=1:\quad & y_{n+1}=y_n+hf_n\\ k=2: \quad & y_{n+1}=y_n+h\bigl( {3\over 2}f_n- {1\over 2}f_{n-1}\bigr)\\[1mm] k=3: \quad & y_{n+1}=y_n+h\bigl( {23\over 12}f_n-{16\over 12}f_{n-1}+{5\over 12}f_{n-2} \bigr)\\[1mm] k=4: \quad & y_{n+1}=y_n+h\bigl( {55\over 24}f_n-{59\over 24}f_{n-1}+{37\over 24}f_{n-2} -{9\over 24}f_{n-3}\bigr) . \end{array}$
Implicit Adams methods
If $$f_{n+1}=f(t_{n+1},y_{n+1})$$ were known, the interpolation polynomial could be stretched over the entire interval $$t_{n-k+1},\ldots , t_{n+1}$$ and thereby an improved precision would be obtained. The corresponding formulas have the form $\tag{6} \begin{array}{ll} k=0:\quad & y_{n+1}=y_n+hf_{n+1} \\[1mm] k=1: \quad & y_{n+1}=y_n+h\bigl( {1\over 2}f_{n+1}+ {1\over 2}f_n\bigr)\\[1mm] k=2: \quad & y_{n+1}=y_n+h\bigl( {5\over 12}f_{n+1}+{8\over 12}f_n-{1\over 12}f_{n-1} \bigr)\\[1mm] k=3: \quad & y_{n+1}=y_n+h\bigl( {9\over 24}f_{n+1}+{19\over 24}f_n -{5\over 24}f_{n-1} +{1\over 24}f_{n-2}\bigr) . \end{array}$
and represent an implicit equation for computing $$y_{n+1}\ .$$ The special cases $$k=0$$ and $$k=1$$ are the implicit Euler method and the trapezoidal rule, respectively. They are actually one-step methods.
Figure 2: Predictor-corrector scheme
Predictor-Corrector schemes
An elegant way of solving the implicit equations in (6) is to predict a value $$\widehat y_{n+1}$$ by the explicit method (5) with the same value of $$k\ ,$$ and replace in the corrector (6) the missing value $$f_{n+1}$$ by $$\widehat f_{n+1}=f(t_{n+1},\widehat y_{n+1})\ .$$ This is the commonly used implementation of the implicit Adams methods for non-stiff problems. The complete algorithm created by this idea is illustrated in Figure 2.
General formulation and consistency
For theoretical investigations it is convenient to use a general framework that comprises all explicit and implicit Adams methods as well as the BDF schemes considered below. Following the seminal work of Dahlquist (1956), a linear multistep method (linear $$k$$-step method) is written in the form $\tag{7} \sum_{j=0}^k \alpha_j \,y_{n+j} =h \sum_{j=0}^k \beta _j \, f_{n+j}$
(notice that the indices are shifted when compared to the formulas (5) and (6)). The generating polynomials for its coefficients, $\tag{8} \rho (\zeta )=\sum_{j=0}^k \alpha_j \,\zeta^j , \qquad \sigma (\zeta ) = \sum_{j=0}^k \beta _j \,\zeta^j$
play an important role in the theory for these methods. Adams methods are characterized by the fact that $$\rho (\zeta ) = (\zeta -1)\,\zeta^{k-1}\ .$$ A linear multistep method is called consistent if it reproduces the exact solution for the differential equation $$\dot y =1\ ,$$ when exact starting approximations are used. This is equivalent to $$\rho (1)=0$$ and $$\rho ' (1)=\sigma (1)\ .$$ If it is exact for the differential equations $$\dot y =m x^{m-1}$$ with $$m=1,2,\ldots ,r\ ,$$ the method is said to have order $$r\ .$$ In terms of the generating polynomials, this means that $\rho (e^h ) - h \sigma (e^h ) = C_{r+1} h^{r+1} + O (h^{r+2}) \qquad \mbox{for}\quad h\to 0 .$ The accuracy of methods having the same order can be distinguished by their error constant $C = \frac{C_{r+1}}{\sigma (1)}.$
The basic theory for linear multistep methods has been developed by Dahlquist (1959) and is presented in the classical book of Henrici (1962). The main difference to one-step methods is that consistency alone does not guarantee convergence. A linear multistep method is called stable (or "zero-stable"), if for the differential equation $$\dot y =0$$ all solutions of the difference equation (7) remain bounded. This is equivalent to the property that all zeros of $$\rho (\zeta )$$ satisfy $$|\zeta |\le 1$$ with those lying on the unit circle being simple. The highest attainable order of linear $$k$$-step methods is $$2k\ .$$ However, for $$k\ge 3$$ these methods are of no practical use because of the following result.
First Dahlquist barrier. The order $$r$$ of a stable linear $$k$$-step method satisfies
• $$\quad r\le k+2\quad$$$$\quad$$ if $$k$$ is even,
• $$\quad r\le k+1$$$$\quad$$ if $$k$$ is odd,
• $$\quad r\le k$$$$\qquad$$$$\qquad$$$$\qquad$$ if $$\beta_k/\alpha_k \le 0\ ,$$ in particular if the method is explicit.
Convergence theorem. If a linear multistep method (7) is stable and of order $$r\ ,$$ then it is convergent of order $$r\ .$$
Convergence of order $$r$$ means that for sufficiently accurate starting approximations $$y_0,\ldots ,y_{k-1}$$ the global error satisfies $$y_{n}-y(t_n) = O (h^r)$$ on compact intervals $$[0,nh]\ ,$$ where $$nh \le T\ .$$ The constant symbolized by the $$O (h^r)$$ notation is bounded by $$\exp (TL)\ ,$$ where $$L$$ is a Lipschitz constant of the vector field $$f(t,y)\ .$$
Methods for stiff problems
Figure 3: Explicit Adams method for stiff problem
Continuing the computation of the above logistic growth equation $$\dot y=a y (1-y)\ ,$$ using $$ha= 5/7\ ,$$ with the explicit Adams process of order 3 leads to a bad surprise. Instead of approaching the steady-state solution $$y\approx 1\ ,$$ the numerical solution exhibits a violent instability phenomenon (see Figure 3).
Stiff differential equations are characterized by the fact that the use of explicit methods requires very small step sizes, and certain implicit methods (like the implicit Euler method) perform much better. Stiff differential equations arise frequently as singularly perturbed problems in chemical reaction systems and in electrical circuitry, and as space discretizations of parabolic partial differential equations.
Stiff stability analysis
The foregoing stability analysis for nonstiff problems, which is uniquely based on the polynomial $$\rho(\zeta)\ ,$$ is not sufficient for stiff differential equations. There are many attempts for studying the stability of methods for stiff problems. The most popular approach, initiated by Dahlquist (1963), is motivated by linearization of $$\dot y = f(y)$$ in the neighborhood of a stationary solution, translation and, in the case of systems, diagonalization. It is based on $\tag{9} \dot y=\lambda y ,$
where $$\lambda$$ represents an eigenvalue of the Jacobian of $$f(y)$$ and can be a complex number. In the above case, we have $$\lambda =-a=-5/7\ .$$ Method (7) applied to this equation gives $$\sum_{j=0}^k (\alpha_j-h\lambda\beta _j) y_{n+j} =0\ ,$$ whose stability depends on the roots of
Figure 4: Root-locus curve for explicit Adams method
$\tag{10} \rho(\zeta) -\mu\sigma(\zeta)=0 ,\qquad \mu=h\lambda .$
The stability domain $$S$$ is the set of complex numbers $$\mu\ ,$$ for which all roots of (10) satisfy $$|\zeta |\le 1$$ with those lying on the unit circle being simple. For the explicit Adams method with $$k=3\ ,$$ equation (10) becomes $\tag{11} { \zeta ^3-\zeta ^2-\mu\Bigl( \frac{23}{12}\zeta ^2-\frac{16}{12}\zeta+\frac{5}{12}\Bigr) =0},$
or $\tag{12} \mu=\frac{\zeta ^3-\zeta ^2}{\frac{23}{12}\zeta ^2-\frac{16}{12}\zeta +\frac{5}{12}} .$
Figure 5: Root-locus curve for predictor-corrector
An elegant way of avoiding the high degree equation in (11) is to insert the limiting values $$\zeta=e^{i\phi}\ ,$$ for $$0\le\phi\le 2\pi$$ real, into (12). This leads to the root locus curve for $$\mu\ ,$$ parts of which must form the boundary of the stability domain $$S\ .$$ These domains are drawn for four explicit Adams methods in Figure 4 and shrink rapidly with increasing $$k\ .$$ In the example from the beginning of this section, the value of $$\mu$$ was $$h\lambda=-5/7\ ,$$ far beyond the stability bound $$-6/11\ .$$
Implicit Adams methods for $$k\ge 2$$ have finite stability domains as well,ƒ but they are considerably larger than those for explicit Adams methods. However, if they are solved by a predictor-corrector approach, a good deal of the stability is lost, as seen in Figure 5 for the case $$k=3\ .$$
This class of methods was discovered, together with the phenomenon of stiffness, by Curtiss & Hirschfelder (1952) in calculations for chemistry, and their extreme importance for stiff problems has been recognized since the work of Gear (1971). In contrast to the Adams methods, where polynomial interpolations were used for the slopes $$f_i\ ,$$ we here interpolate the $$y$$-values $$y_{n-k+1},\ldots ,y_n$$ and require for the interpolating polynomial $$q(t)$$ the collocation condition $$\dot q (t_{n+1})= f(t_{n+1},q(t_{n+1}))\ .$$ The differentiation of Newton's interpolation formula (4) then leads to the methods $\tag{13} \sum_{j=1} ^k {1 \over j}\nabla^j y_{n+1} = hf_{n+1} ,$
($$\nabla$$ denotes again the backward difference operator) or, when expanded, $\tag{14} \begin{array}{ll} k =1:\quad & y_{n+1} - y_n = hf_{n+1} \\[1mm] k =2:\quad & {3 \over 2}y_{n+1} - 2y_n + {1 \over 2}y_{n-1} = hf_{n+1} \\[1mm] k =3:\quad & {11 \over 6}y_{n+1} - 3y_n + {3 \over 2}y_{n-1} - {1 \over 3}y_{n-2} = hf_{n+1} \\[1mm] k =4:\quad & {25 \over 12}y_{n+1} - 4y_n + 3y_{n-1} - {4 \over 3}y_{n-2} + {1 \over 4}y_{n-3} = hf_{n+1} \\[1mm] k =5:\quad & {137 \over 60}y_{n+1} - 5y_n + 5y_{n-1} - {10 \over 3}y_{n-2} + {5 \over 4}y_{n-3} - {1 \over 5}y_{n-4} = hf_{n+1} \\[1mm] k =6:\quad & {147 \over 60}y_{n+1} - 6y_n + {15 \over 2}y_{n-1} - {20 \over 3}y_{n-2} + {15 \over 4}y_{n-3} - {6 \over 5}y_{n-4} + {1 \over 6}y_{n-5} = hf_{n+1} . \end{array}$
Figure 6: BDF method for stiff problem
These are implicit methods and work, for $$k\le 6\ ,$$ significantly better for stiff problems, as illustrated in the movie of Figure 6. For $$k\ge7$$ they become unstable. Their stability domains, always the outside of the root locus curve, extend to $$-\infty$$ and can be seen in Figure 7. For stiff problems, their implementation requires some Newton-like procedures for the implicit equations, because a predictor-corrector approach would destroy this good stability property. For more details we refer to the entry on BDF methods. These methods are furthermore a very important tool for the treatment of differential-algebraic equations.
A-stability
The analytical solution of Dahlquist's test equation (9) is stable if and only if $$\hbox{Re}\lambda\le 0\ .$$ A very desirable property of a numerical method is that, under this condition, the numerical solution remains stable as well, i.e., that $\tag{15} S\supset {\Bbb C}^- .$
Figure 7: Root-locus curve for BDF
This property is called A-stability (Dahlquist 1963). Inspecting the above stability domains, we observe that the implicit Adams methods for $$k=1,2$$ (which are the implicit Euler and trapezoidal method) as well as the BDF methods for $$k=1,2$$ (the first one is implicit Euler again) have this property. A close inspection of the stability domains for the higher order BDF methods shows that some parts close to the imaginary axis, which correspond to oscillatory problems, do not produce stable solutions. This observation is part of the famous second Dahlquist barrier (Dahlquist 1963):
Second Dahlquist barrier. An A-stable multistep method must be of order $$r\le 2\ .$$ Among all multistep methods of order 2, the trapezoidal rule has the smallest error constant.
This severe order bound was the origin of numerous efforts for breaking it by generalizing multistep methods in various directions: Enright's second derivative methods, second derivative BDF methods, blended multistep methods (Skeel & Kong), super future points (Cash 1980), multistep collocation methods, multistep methods of Radau type, which all fall into the class of general linear methods. For more details, consult Section V.3 of Hairer & Wanner (1996). Also these methods, if A-stable, have their order barrier, which was for many years an open conjecture (Daniel-Moore 1968): An A-stable generalized multistep method, with $$s$$ implicit stages per step, must be of order $$r\le 2s\ .$$ The smallest error constant for such methods is that of the corresponding Gauss-Runge-Kutta method.
Figure 8: Order star for BDF3
Order stars. An elegant idea for proving the second Dahlquist barrier and the Daniel-Moore conjecture, together with many other stability results, is the consideration of the corresponding order star $$A\ .$$ Instead of the stability domain $$S\ ,$$ where the complex roots $$\zeta (\mu)$$ of (10) were compared, in absolute values, to 1, we now compare them to $$|e^{\mu}|\ ,$$ the absolute value of the exact solution, which they tend to approximate, i.e., $$A=\{ \mu\, ;\, |\zeta (\mu)| >|e^{\mu}| \} \ .$$ Since $$\zeta (\mu)$$ is multi-valued, this set has to be considered on the associated Riemann surface. The order star gives much insight into the structure of the function $$\zeta(\mu)\ ,$$ because it transforms analytic properties into geometric properties as follows: the star-shaped structure with $$r+1$$ sectors at the origin indicates that the method is of order $$r\ ,$$ and A-stability is characterized by the fact that poles are in the right half-plane and the intersection of $$A$$ with the imaginary axis is empty. The order star for the BDF3 method is drawn in Figure 8, and explains that such a method of order 3 cannot be A-stable. Details can be found in Hairer & Wanner (1996). The use of order arrows (Butcher 2008) is another way for observing that BDF3 is not A-stable.
A$$(\alpha )$$-stability. This is a concept, weaker than A-stability, but nevertheless very useful. It requires that the stability domain contains a sector with angle $$2 \alpha$$ in the negative half plane; more precisely, a linear multistep method (7) is called A$$(\alpha )$$-stable if $$S\supset \{ z\, ; \, |\arg (-z)|<\alpha, \, z\ne 0\}\ .$$ BDF methods are A$$(\alpha )$$-stable with angle given by $\begin{array}{c|cccccc} k~&1&2&3&4&5&6\\[3pt] \alpha~&~90^\circ~&~~90^\circ&86.03^\circ&73.35^\circ&51.84^\circ&17.84^\circ \end{array}$
Convergence
The interest of applying BDF methods (or other A-stable or A$$(\alpha )$$-stable multistep methods) relies in the fact that they can be applied to stiff differential equations with step sizes that are much larger than the inverse of a Lipschitz constant of $$f(y)\ .$$ In this situation, classical convergence theory (for non-stiff problems) does not provide any information. It is nevertheless possible to prove estimates $$y_n-y(t_n) =O(h^r)$$ of the global error (for sufficiently accurate starting approximations and for A$$(\alpha )$$-stable methods of order $$r$$) in the following situations:
• singularly perturbed problems $$\dot y =f(y,z),\, \varepsilon \dot z = g(y,z)\ :$$ if the eigenvalues of $$g_z(y,z)$$ are in the negative half plane, then the constant in the error estimation $$O(h^r)$$ can be chosen independently of $$\varepsilon >0\ .$$
• discretized parabolic equations and stiff differential equations satisfying a one-sided Lipschitz condition.
Precise statements and proofs can be found, for example, in Sections V.7, V.8, and VI.2 of the monograph Hairer & Wanner (1996).
Methods for Hamiltonian systems
For long-time integration of conservative differential equations special integrators are usually more appropriate. The study of such integrators is the main topic of geometric numerical integration (Hairer, Lubich & Wanner 2006), and there are interesting geometric integrators among multistep methods.
Second order Hamiltonian systems
An important class of conservative systems are second order differential equations $\tag{16} \ddot q = f(q), \qquad f(q)=-M^{-1}\nabla U(q)$
with a constant, symmetric, positive definite mass matrix $$M\ ,$$ and a smooth potential $$U(q)\ .$$ This equation is Hamiltonian with energy $$\textstyle H(p,q)={1\over 2}\, p^{\mathsf T} M^{-1} p +U(q)\ ,$$ where $$p=M\dot q$$ is the momentum. The exact flow is known to be symplectic, and the energy is preserved along solutions. Typical examples are $$N$$-body systems, e.g., planetary motion in astronomy or simulations in molecular dynamics.
Symmetric linear multistep methods
It is possible to write (16) as a first order system and to apply any multistep method considered above. Eliminating the velocity leads to a formula $\tag{17} \sum_{j=0}^k \alpha_{j}\,q_{n+j} = h^2 \sum_{j=0}^k \beta_{j}\, f(q_{n+j}) ,$
with coefficients $$\alpha_{j},\beta_{j}$$ that are different from those above. We again introduce generating polynomials $\rho (\zeta ) = \sum_{j=0}^k \alpha_{j}\, \zeta^j , \qquad \sigma (\zeta ) = \sum_{j=0}^k \beta_{j}\, \zeta^j .$ A linear multistep method (17) has order $$r$$ for differential equations (16) if $\rho (e^h ) - h^2 \sigma (e^h ) = O (h^{r+2}) \qquad \hbox{for}\quad h\to 0 ,$ and it is stable if all zeros of $$\rho (\zeta )$$ satisfy $$|\zeta |\le 1$$ with those lying on the unit circle being at most double. Order $$r$$ and stability guarantee convergence, and the global error satisfies $$q_{n}-q(t_{0}+nh) = O (h^r)$$ on compact intervals $$nh \le T\ .$$
For the integration of Hamiltonian systems, we are mainly interested in long-time integration. For example, the solution of the harmonic oscillator $$\ddot q = -\omega^2 q$$ satisfies $$\textstyle\frac 12 (\dot q(t) ^2 + \omega^2 q(t)^2)= const$$ and remains bounded for all times. Can we have a similar behavior for the numerical solution, when $$h$$ is fixed and $$n\to\infty\ ?$$ When applied to the harmonic oscillator, method (17) becomes a linear difference relation with characteristic equation $$\rho (\zeta ) - (h\omega)^2 \sigma (\zeta ) =0\ .$$ It turns out that near energy conservation for the harmonic oscillator can be achieved only if all roots have modulus one. This in turn implies that the method has to be symmetric, i.e., $\tag{18} \alpha_{j} = \alpha_{k-j}, \qquad \beta_{j}=\beta_{k-j}.$
Lambert & Watson (1976) initiated the study of the long-time behavior of symmetric multistep methods. Quinlan & Tremaine (1990) constructed high order methods (17) and applied them successfully to long-time integrations of the outer solar system. One of their methods of order $$8$$ with $$k=8$$ is given by $\tag{19} \begin{array}{rcl} \rho (\zeta ) &\!\! = \!\! & (\zeta -1)^2 (\zeta^6 + 2\zeta^5 +3\zeta^4 + 3.5 \zeta^3 + 3\zeta^2 +2\zeta +1) ,\\[2mm] 120960\,\sigma (\zeta ) &\!\! = \!\! & 192481 (\zeta^7 + \zeta ) + 6582 (\zeta^6 +\zeta^2) +816783 (\zeta^5 + \zeta^3) - 156812 \zeta^4 . \end{array}$
Notice that the method is explicit (i.e., $$\beta_{k}=0$$) and therefore its implementation with constant steps is straight-forward. Due to possible numerical resonances, symmetric multistep methods are recommended only for computations with high accuracy.
Long-time behavior
For the study of qualitative and quantitative features of the numerical solution on intervals that are far beyond the applicability of classical convergence theory, backward error analysis is the appropriate tool. It turns out that a symmetric method has to be $$s$$-stable (i.e., apart from the double root at $$1\ ,$$ all zeros of $$\rho (\zeta )$$ are simple and of modulus one). It has been shown in Hairer & Lubich (2004) that symmetric, $$s$$-stable multistep methods (17) of order $$r$$ have a long-time behavior that it very similar to that of symplectic one-step methods. In particular, their numerical solutions satisfy:
• for a Hamiltonian system (16) the total energy $$\textstyle H(p,q)=\frac 12 p^{\mathsf T} M^{-1} p +U(q)$$ is conserved up to $$O (h^r)$$ on intervals of length $$O (h^{-r-2})\ ;$$
• quadratic first integrals of the form $$\dot q^{\mathsf T} A q$$ (such as the angular momentum in $$N$$-body problems) are conserved up to $$O (h^r)$$ on intervals of length $$O (h^{-r-2})$$ (here, numerical approximations $$\dot q_{n}$$ for the derivative are computed by symmetric finite differences);
• for completely integrable Hamiltonian systems, action variables are in general conserved up to $$O (h^r)$$ and angle variables are bounded by $$O (t_{n} h^r)$$ on intervals of length $$O (h^{-r})\ .$$
Implementation and software
The efficient implementation of linear multistep methods is a nontrivial task. A code has to be able to automatically select the step size (which can be done either by formulas with grid-dependent coefficients or by using a Nordsieck vector formulation) and to choose an optimal order. Among the available software (most of it can be downloaded from http://www.netlib.org/ode/) let us mention the following:
• ODE/STEP is a widely used code for the solution of non-stiff differential equations. It is based on Adams methods and documented in the monograph of Shampine & Gordon (1975).
• DIVA is an Adams integrator by Fred Krogh. It can directly integrate second order differential equations.
• LSODE is the Livermore Solver of A.~Hindmarsh. It solves stiff or nonstiff differential equations of first order.
• DASSL is a differential-algebraic equation solver, based on BDF schemes (see Brenan, Campbell & Petzold 1989). The most recent version and various modifications can be downloaded from the web page of L.~Petzold at http://www.cs.ucsb.edu/~cse/software.html
• MEBDF is an implementation of modified extended BDF methods (Cash & Considine 1992). Various modifications can be obtained from http://www2.imperial.ac.uk/~jcash/
References
• Adams J.C. (1883) Appendix in: F. Bashforth (1883) An attempt to test the theories of capillary action by comparing the theoretical and measured forms of drops of fluid. With an explanation of the method of integration employed in constructing the tables which give the theoretical form of such drops". Cambridge Univ. Press.
• Brenan K.E., Campbell S.L. and Petzold L.R. (1989 and 1996) Numerical solution of initial-value problems in differential-algebraic equations. North-Holland, New York, and SIAM, Philadelphia.
• Butcher J.C. (2008) Numerical Methods for Ordinary Differential Equations. John Wiley & Sons Ltd., Chichester, England.
• Cash J.R. and Considine S. (1992) An MEBDF code for stiff initial value problems. ACM Trans. Math. Software 18:142-158.
• Curtiss C.F. and Hirschfelder J.O. (1952) Integration of stiff equations. Proc. Nat. Acad. Sci. 38:235-243.
• Dahlquist G. (1956) Convergence and stability in the numerical integration of ordinary differential equations. Math. Scand. 4:33-53.
• Dahlquist G. (1959) Stability and error bounds in the numerical integration of ordinary differential equations. Trans. of the Royal Inst. of Techn., Nr. 130, Stockholm, Sweden.
• Dahlquist G. (1963) A special stability problem for linear multistep methods. BIT 3:27-43.
• Gear C.W. (1971) Numerical initial value problems in ordinary differential equations. Prentice Hall.
• Hairer E and Lubich C (2004) Symmetric multistep methods over long times. Numer. Math. 97:699-723.
• Hairer E, Lubich C. and Wanner, G. (2006) Geometric Numerical Integration. Structure-Preserving Algorithms for Ordinary Differential Equations. Second edition. Springer-Verlag, Berlin. [1]
• Hairer E, Nørsett S.P. and Wanner, G. (1993) Solving Ordinary Differential Equations I: Nonstiff Problems. Second edition. Springer-Verlag, Berlin. [2]
• Hairer E. and Wanner, G. (1996) Solving Ordinary Differential Equations II: Stiff and Differential-Algebraic Problems. Second edition. Springer-Verlag, Berlin. [3]
• Henrici P. (1962) Discrete Variable Methods in Ordinary Differential Equations. Volume 1, John Wiley & Sons, New York.
• Lambert J.D. and Watson I.A. (1976) Symmetric multistep methods for periodic initial value problems. J. Inst. Maths. Applics. 18:189-202.
• Quinlan G.D. and Tremaine S. (1990) Symmetric multistep methods for the numerical integration of planetary orbits. Astron. J. 100:1694-1700.
• Shampine L. and Gordon M. (1975) "Computer Solution of Ordinary Differential Equations: The Initial Value Problem." Freeman.
Internal references
• Bill Gear (2007) Backward differentiation formulas. Scholarpedia, 2(8):3162.
• Stephen L. Campbell, Vu Hoang Linh, Linda R. Petzold (2008) Differential-algebraic equations. Scholarpedia, 3(8):2849.
• James Meiss (2007) Dynamical systems. Scholarpedia, 2(2):1629.
• Zdzislaw Jackiewicz (2007) General linear methods. Scholarpedia, 2(4):2852.
• James Meiss (2007) Hamiltonian systems. Scholarpedia, 2(8):1943.
• Lawrence F. Shampine and Skip Thompson (2007) Initial value problems. Scholarpedia, 2(3):2861.
• Andrei D. Polyanin, William E. Schiesser, Alexei I. Zhurov (2008) Partial differential equation. Scholarpedia, 3(10):4605.
• Jeff Moehlis, Kresimir Josic, Eric T. Shea-Brown (2006) Periodic orbit. Scholarpedia, 1(7):1358.
• Philip Holmes and Eric T. Shea-Brown (2006) Stability. Scholarpedia, 1(10):1838.
• Lawrence F. Shampine and Skip Thompson (2007) Stiff systems. Scholarpedia, 2(3):2855.
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http://mathhelpforum.com/number-theory/212663-alternating-series-test.html
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1Thanks
• 1 Post By johng
# Thread:
1. ## Alternating Series Test
* I hope its alright that I reposted this. I realized that Calculus might not be the best category for this. *
Show by example that the hypothesis b1>=b2>=...bn>=0 cannot be replaced by bk>=0 and limit k-->infinity =0
hint: use |ab|< 1/2(a2+b2)
I've found an example: bk=(1/k2 + 1/k) which satisfies the limit going to zer and all terms being positive, which diverges. I'm getting stuck with a rigorous proof of this, I thought about using the Dirichlet Test to prove this, but I'm getting hung up I think. Any help?
2. ## Re: Alternating Series Test
I'm afraid your counterexample is actually convergent. That is $\sum^\infty_{k=1}(-1)^{k+1}(1/k^2+1/k)$ is the sum of two convergent alternating series. I really don't see what example the hint is implying. However, here's an example and a few more comments.
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http://physics.stackexchange.com/questions/12975/observed-angular-velocity
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# “Observed” angular velocity
I have a homework question based on the following diagram:
I need to find the angular velocity of the object as seen by an observer at the origin of the frame. The question says that the observed angular velocity is given by $\omega_0 = \Delta \varphi_0 / \Delta t$, where $\Delta t = t_2 - t_1$ and $\Delta \varphi_0 = \varphi_0(t_1^*) - \varphi_0(t_2^*)$ is the change of the angle $\varphi$, the polar angle which the object had at times $t_1^*$ and $t_2^*$. These are the times when the light detected by the observer (at $t_1$ and $t_2$) was emitted. The object is located at $(x_1, y_1)$ at time $t_1^*$ and it is located at $(x_2, y_2)$ at time $t_2^*$.
I started to answer this question by just finding an expression for $\Delta \varphi_0$ in terms of the coordinates ($\Delta \varphi_0 = \arctan{(y_1/x_1)} - \arctan{(y_2/x_2)}$) and dividing this by an expression for $\Delta t$. However, I was told that this was not correct. Apparently, I have to get an expression for the angular momentum in terms of the angle $\theta$ in the diagram. My lecturer told me that this was could be done with some basic geometry. I can't quite see what to do though. Also, I was told that the angle $\Delta \varphi_0$ was assumed to be very small. I have a feeling that what the lecturer wants us to derive is the expression $\omega = \frac{|\vec{v}|\sin{(\theta)}}{|\vec{r}|}$, but I can't see how to get this. Can anyone help?
Edit: This question is from a special relativity course, but I believe this can be answered without using any knowledge of special relativity. As I said, the lecturer told me that it was basically just geometry.
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## 2 Answers
Here's something that might help you out: this problem is motivated by astronomical observations of distant stars or galaxies. If you think about it, when you look through a telescope at an astronomical object, the only thing you can measure (without a spectroscope) is its angular position $\varphi_0$. So the way you obtain information about the object's motion is to record its angular position at one time $t_1$, and then later at another time $t_2$. You can then calculate the ratio $\Delta \varphi_0/\Delta t$, and since for an astronomical object $\Delta \varphi_0$ is very small (just think about how little the stars appear to move through the sky over any human-length timescale), this ratio is a pretty good approximation to the derivative $\frac{\mathrm{d}\varphi_0}{\mathrm{d}t} = \omega$, the observed angular velocity of the object.
However, what you see through a telescope is only part of the story. $\Delta \varphi_0/\Delta t$ is not necessarily the actual angular velocity of the object, because even though you observed two light rays a time $\Delta t$ apart, it doesn't mean that they were emitted $\Delta t$ apart. The only way the time between their emission would be the same as the time between their observation is if both light rays had traveled the same distance before reaching you. But for many astronomical objects, that's not the case. Since they have motion parallel to your line of sight as well as perpendicular to it, it's very likely that one of the light rays was emitted closer to you than the other, which means it spent less time in transit. You need to take that difference of transit time into account, which is why the answer you should be getting is a little more complicated than $\omega = \frac{v\sin\theta}{r}$.
As far as actually solving the problem, I'd suggest that you start by finding expressions for the the travel time of each light ray - that is, find an expression for $t_1 - t_1^*$, and similarly for $t_2 - t_2^*$. Also, because the angle $\Delta\varphi_0$ is so small, you can do calculations parallel to and perpendicular to the line of sight separately. I would definitely suggest that you ignore the $x$ and $y$ coordinates entirely - think of it the way someone would be observing this from Earth, where they wouldn't know anything about those coordinates.
By the way, here's the reason you're doing this in a special relativity course: it's a common calculation in astrophysics to work this problem backwards, i.e. to determine the actual velocity of the distant object given observations of $\Delta\varphi_0$ and $\Delta t$, along with spectroscopic data that gives you the parallel component of the velocity. In certain cases, the velocity works out to be faster than light if you use the naive formula $\frac{v\sin\theta}{r}$. You need to account for the difference in light travel time to show that these objects actually do obey special relativity.
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Thank you very much for this informative answer. I assume the travel time of each light ray would just be the distance from the object to the origin divided by $c$ (i.e. $d_1/c$ and $d_2/c$) but I'm not sure. Also, I was wondering if you could clarify the following comment: "because the angle $\Delta \varphi_0$ is so small, you can do calculations parallel to and perpendicular to the line of sight separately". What kind of calculations would these be? – saurs Jul 30 '11 at 10:20
You get to figure out what kind of calculations they are ;-) Although in retrospect, perhaps that particular hint caused more confusion than it's worth. All I'm saying is, there isn't anything particularly tricky about this problem. – David Zaslavsky♦ Jul 30 '11 at 19:51
1
David Zaslavsky has laid out the main ideas very well. One additional hint about relating $\Delta\varphi_0$ to $\theta$: There are two right triangles in the diagram that share a common side. Express the length of that common side in terms of $\theta$ and given quantities, and also in terms of $\Delta\varphi_0$ and given quantities. (This is a slightly more explicit variation on David's suggestion about treating parallel and perpendicular components separately.) – Ted Bunn Jul 30 '11 at 22:26
@Ted Bunn: Thank you for that hint. Now that I think about it, however, I'm quite confused by the diagram. What does $\Delta r$ represent? If $\vec{r}$ is the position vector of the particle, then shouldn't $\Delta \vec{r}$ be pointing in the direction of $\vec{v}$? So I suppose the common side is given by $\Delta r / \cos{(\theta)}$. I'm not sure how to get an expression for that side in terms of $\Delta \varphi_0$ - would I use the sin rule or something like that? Sorry for all the questions, I know this should be an easy problem. – saurs Jul 31 '11 at 2:25
2
$\Delta r$ is the change in the magnitude of the vector $\vec r$. It's not the magnitude of the vector difference -- that is, $\Delta r\ne |\Delta\vec r|$. If you decompose $\vec r$ into radial and tangential components, then $\Delta r$ is the difference in the radial components. For small $\Delta\varphi_0$, that's the length of the line segment shown in the diagram. – Ted Bunn Jul 31 '11 at 2:46
show 4 more comments
If you split the velocity into two components (as show above), one being radial (I call it $v_r$) and the other in the hoop direction (I call it $v_{\varphi}$), then by inspection (trig properties) you can see the hoop velocity is $$v_{\varphi} = v\,\sin\theta$$ now by nature of angular velocity the hoop velocity is also $$v_{\varphi} = \omega\;r$$
Combine the two to get your answer.
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http://mathoverflow.net/questions/41912/computation-of-inverses-modulo-p-followup/41925
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Computation of inverses modulo p followup
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In responding to
http://mathoverflow.net/questions/40997/fast-computation-of-multiplicative-inverse-modulo-q/41631#41631
I mentioned an algorithm for computing the inverse of $a \mod p$ different from the extended Euclidean algorithm, hoping that someone could tell me how its speed stacks up against other algorithms. Since no one did, I'm asking directly if someone can tell me.
To compute the inverse of a modulo p, you can run the Euclidean algorithm starting with $p^2$ and $ap+1$, comparing the size of each remainder with $p$. The first remainder less than $p$ that appears will be an inverse for $a \mod p$.
This will always take either the same number of steps to reach the inverse as it takes to reach $\gcd(a,p)=1$ using the Euclidean algorithm with $a$ and $p$, or else one additional step, depending on whether the least positive residue of an inverse of $a$ is greater than $p/2$ or less than $p/2$. Thus, it requires approximately the same number of computations as the first half of the extended Euclidean algorithm (albeit with bigger numbers initially), excepting an extra comparison with $p$ at each step.
Question: How does the speed of this algorithm compare to others?
Aside: Pedagogically, this is nice since the second half of the extended Euclidean algorithm is the one my students tend to mess up. However, assuming our ultimate goal is for students to understand why they're doing what they're doing, perhaps the extended Euclidean algorithm is preferable.
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Quadratic complexity. Compared to standard quadratic algorithm: half the iterations, on twice-the-size numbers. Since the operations are more than linear in size, I would guess that the constant is larger than the constant of the standard algorithm. – Dror Speiser Oct 12 2010 at 16:10
2
If you examine AVS's answer to the question you linked, you will see that your algorithm is substantially slower than Fast Euclidean. – S. Carnahan♦ Oct 12 2010 at 16:23
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I didn't mention this in my answer to the linked question, but the half-gcd algorithm can be used to "jump" directly to any particular step of the extended Euclidean computation (not necessarily the last one) using a logarithmic number of recursive calls. One could perform a binary search to find the greatest remainder less than p using half-gcds, yielding a quasi-linear running time that is slower than the fast Euclidean algorithm by only a log factor, and even this might be avoided (at least on average) by jumping to the middle step and searching linearly from there. – AVS Oct 12 2010 at 18:39
1
The algorithm sketched in my comment above was only meant to show that your approach could be implemented in quasi-linear time. But this comes at the cost of making the algorithm much more complicated (which may be a pedagogical flaw or feature, depending on one's point of view), and it will still be slower than just computing the extended gcd. – AVS Oct 12 2010 at 18:45
Ah, thanks AVS! I wasn't expecting that performing the algorithm exactly as I wrote it would give a speed-up, but I was thinking that one of the other speed-ups might improve the algorithm I wrote to be comparable to others. One knows that the first remainder less than p will occur almost exactly half-way through the Euclidean algorithm with $p^2$ and $ap+1$, so jumping to the middle step will put you very close. – Barry Oct 12 2010 at 18:45
show 1 more comment
2 Answers
Here is a tidy way to solve $ax+by=gcd(a,b)$. Start with the matrix ```$$\left(\begin{array}{ccc}
1&0&a\\
0&1&b\\
\end{array}\right)
$$``` Suppose `$a\ge b$`. Then replace row 1 by row 1 minus $t$ times row 2, where $t=\lfloor a/b\rfloor$. Repeat this operation until the last entry in one of the two rows is zero. If the other row is $x,y,d$ then $ax+by=d$ and $d=gcd(a,b)$. This is very simple to program, and avoids the back-substitutions that students find confusing.
Most of the speedups of the Euclidean Algorithm described in Knuth's Art of Computer Programming v2 work here also.
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I use this way to teach it as well. One has the invariant that a typical row is $r s ar+bs$. In case of an error one can zero in on the line where the error occurred by checking back and forth to see where the invariant failed. When working in $\mathbb{Z}[\sqrt{2}]$ or $\mathbb{Z}[i]$ that can be helpful. – Aaron Meyerowitz Oct 13 2010 at 7:58
@Aaron: Strangely, I've never seen it in print. – SJR Oct 13 2010 at 11:31
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
As you mention pedagogy: I solve $a x + b y = \gcd(a,b)$ by simple continued fractions, both by hand and in C++ code. For consecutive convergents $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}},$ the standard relation is $p_n q_{n+1} - p_{n+1} q_n = (-1)^{n+1}$ where the starting point (from Khinchin's little book, Dover reprint) is $$\frac{p_{-2}}{q_{-2}} = \frac{0}{1}$$ and the fake but necessary $$\frac{p_{-1}}{q_{-1}} = \frac{1}{0}$$ For a rational number $\frac{a}{p}$ the final little 2 by 2 determinant may give the wrong sign (and does about half the time), in which case I negate everything.
So, this is the Euclidean algorithm but it all moves forward on the page, no back-substitutions or whatever is in the "second half" of extended gcd. And continued fractions are good for other things.
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http://stats.stackexchange.com/questions/12490/lognormal-distribution-using-binned-or-grouped-data
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# Lognormal distribution using binned or grouped data
I understand the Max likelihood estimators for mu and sigma for the lognormal distribution when data are actual values. However I need to understand how these formulas are modified when data are already grouped or binned (and actual values are not available). Specifically, for mu, the mle estimator is the sum of the logs of each X (divided by n which is the number of points). For sigma squared, the mle estimator is the sum of (each log X minus the mu, squared); all divided by n. (Order of operations is taking each log X minus the mu; square that; sum that over all X's; then divide by n). Now suppose data in bins b1, b2, b3, and so on where b1 to b2 is the first bin; b2 to b3 second bin and so on. What are the modified mu and sigma squared? thank you.
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## 1 Answer
Let $\Phi$ be the cumulative standard normal distribution function. The probability that a value $Y$ drawn from a lognormal distribution with log mean $\mu$ and log SD $\sigma$ lies in the interval $(b_i, b_{i+1}]$ therefore is
$$\Pr(b_i \lt Y \le b_{i+1}) = \Phi \left( \frac{\log(b_{i+1}) - \mu}{\sigma} \right) - \Phi \left( \frac{\log(b_{i}) - \mu}{\sigma} \right).$$
Call this value $f_i(\mu, \sigma)$.
When the data consist of independent draws $Y_1,Y_2, \ldots, Y_N$, with $Y_i$ falling in bin $j(i)$ and the bin cutpoints are established independently of the $Y_i$, the probabilities multiply, whence the log likelihood is the sum of the logs of these values:
$$\log(\Lambda(\mu, \sigma)) = \sum_{i=1}^{N} \log(f_{j(i)}(\mu, \sigma)).$$
It suffices to count the number of $Y_i$ falling within each bin $j$; let this count be $k(j)$. By collecting the $k(j)$ terms associated with bin $j$ for each bin, the sum condenses to
$$\log(\Lambda(\mu, \sigma)) = \sum_{j} k(j) \log(f_{j}(\mu, \sigma)).$$
The MLEs are the values $\hat{\mu}$ and $\hat{\sigma}$ that together maximize $\log(\Lambda(\mu, \sigma))$. There is no closed formula for them in general: numerical solutions are needed.
### Example
Consider data values known only to lie within the even intervals $[0,2]$, $[2,4]$, etc. I randomly generated 100 of them according to a Lognormal(0,1) distribution. In Mathematica this can be done via
````With[{width = 2},
data = width {Floor[#/width], Floor[#/width] + 1} & /@
RandomReal[LogNormalDistribution[0, 1], 100]
];
````
Here are their tallies:
````Interval Count
[0, 2] 77
[2, 4] 16
[4, 6] 5
[6, 8] 1
[16,18] 1
````
Finding the MLE for data like this requires two procedures. First, one to compute the contribution of a list of all 100 intervals to the log likelihood:
````logLikelihood[data_, m_, s_] :=
With[{f = CDF[LogNormalDistribution[m, s], #] &},
Sum[Log[f[b[[2]]] - f[b[[1]]]], {b, data}]
];
````
Second, one to numerically maximize the log likelihood:
````mle = NMaximize[{logLikelihood[data, m, s], s > 0}, {m, s}]
````
The solution reported by Mathematica is
````{-77.0669, {m -> -0.014176, s -> 0.952739}}
````
The first value in the list is the log likelihood and the second (evidently) gives the MLEs of $\mu$ and $\sigma$, respectively. They are comfortably close to their true values.
Other software systems will vary in their syntax, but typically they will work in the same way: one procedure to compute the probabilities and another to maximize the log likelihood determined by those probabilities.
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ok this is a great answer. What are the "numerical solutions" refrred to? Is there a text that shows a worked out example? Or can you offer a simple example with just a few data points on this site? I would greatly appreciate any help or reference. Ajay – user5243 Jul 1 '11 at 0:06
@Ajay I edited the solution to give a simple worked example. – whuber♦ Jul 1 '11 at 14:24
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http://www.cfd-online.com/W/index.php?title=Introduction_to_turbulence/Statistical_analysis/Multivariate_random_variables&oldid=5866
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# Introduction to turbulence/Statistical analysis/Multivariate random variables
### From CFD-Wiki
Revision as of 11:02, 3 June 2006 by Michail (Talk | contribs)
### Joint pdfs and joint moments
Often it is importamt to consider more than one random variable at a time. For example, in turbulence the three components of the velocity vector are interralated and must be considered together. In addition to the marginal (or single variable) statistical moments already considered, it is necessary to consider the joint statistical moments.
For example if $u$ and $v$ are two random variables, there are three second-order moments which can be defined $\left\langle u^{2} \right\rangle$ , $\left\langle v^{2} \right\rangle$ , and $\left\langle uv \right\rangle$. The product moment $\left\langle uv \right\rangle$ is called the cross-correlation or cross-covariance. The moments $\left\langle u^{2} \right\rangle$ and $\left\langle v^{2} \right\rangle$ are referred to as the covariances, or just simply the variances. Sometimes $\left\langle uv \right\rangle$ is also referred to as the correlation.
In a manner similar to that used to build-up the probabilility density function from its measurable counterpart, the histogram, a joint probability density function (or jpdf),$B_{uv}$ , can be built-up from the joint histogram. Figure 2.5 illustrates several examples of jpdf's which have different cross correlations. For convenience the fluctuating variables $u'$ and $v'$ can be defined as
$u' = u - U$ (2)
$v' = v - V$ (2)
where as before capital letters are usd to represent the mean values. Clearly the fluctuating quantities $u'$ and $v'$ are random variables with zero mean.
A positive value of $\left\langle u'v' \right\rangle$ indicates that $u'$ and $v'$ tend to vary together. A negative value indicates value indicates that when one variable is increasing the other tends to be decreasing. A zero value of $\left\langle u'v' \right\rangle$ indicates that there is no correlation between $u'$ and $v'$. As will be seen below, it does not mean that they are statistically independent.
It is sometimes more convinient to deal with values of the cross-variances which have ben normalized by the appropriate variances. Thus the correlation coefficient is defined as:
$\rho_{uv}\equiv \frac{ \left\langle u'v' \right\rangle}{ \left[ \left\langle u'^{2} \right\rangle \left\langle v'^{2} \right\rangle \right]^{1/2}}$ (2)
The correlation coefficient is bounded by plus or minus one, the former representing perfect correlation and the latter perfect anti-correlation.
As with the single-variable pdf, there are certain conditions the joint probability density function must satisfy. If $B_{uv}\left( c_{1}c_{2} \right)$ indicates the jpdf of the random variables $u$ and $v$, then:
• Property 1
$B_{uv}\left( c_{1}c_{2} \right) > 0$ (2)
always
• Property 2
$Prob \left\{ c_{1} < u < c_{1} + dc_{1} , c_{2} < v < c_{2} + dc_{2} \right\} = B_{uv}\left( c_{1}c_{2} \right) dc_{1}, dc_{2}$ (2)
• Property 3
$\int^{\infty}_{ - \infty} \int^{\infty}_{ - \infty} B_{uv}\left( c_{1}c_{2} \right) dc_{1} dc_{2} = 1$ (2)
• Property 4
$\int^{\infty}_{ - \infty} B_{uv}\left( c_{1}c_{2} \right) dc_{2} = B_{u}\left( c_{1} \right)$ (2)
where $B_{u}$ is a function of $c_{1}$ only
• Property 5
$\int^{\infty}_{ - \infty} B_{uv}\left( c_{1}c_{2} \right) dc_{1} = B_{v}\left( c_{2} \right)$ (2)
where $B_{v}$ is a function of $c_{2}$ only
The functions $B_{u}$ and $B_{v}$ are called the marginal probability density functions and they are simply the single variable pdf's defined earlier. The subscript is used to indicate which variable is left after the others are integrated out. Note that $B_{u}\left( c_{1} \right)$ is not the same as $B_{uv}\left( c_{1},0 \right)$. The latter is only a slice through the $c_{2}$ - axis, whale the marginal distribution is weighted by the integral of the distribution of the other variable. Figure 2.6. illustrates these differences.
If the joint probability density function is known, the joint moments of all orders can be determined. Thus the $m,n$ -th joint moment is
$\left\langle \left( u- U \right)^{m} \left( v - V \right)^n \right\rangle = \int^{\infty}_{-\infty} \int^{\infty}_{-\infty} \left( c_{1} - U \right)^{m} \left( c_{2} - V \right)^{n} B_{uv}\left( c_{1} , c_{2} \right) dc_{1} dc_{2}$ (2)
In the preceding discussions, only two random variables have been considered. The definitions, however, can easily be geberalized to accomodate any number of random variables. In addition, the joint statistics of a single random at different times or at different points in space could be considered. This will be done later when stationary and homogeneous random processes are considered.
dssd
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http://math.stackexchange.com/questions/155055/is-there-any-sense-in-zero-padding-a-matrix-to-make-it-n-times-n-and-find-its/155073
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# Is there any sense in zero-padding a matrix to make it $n\times n$ and find its eigenvalues?
I am debuging my Kalman filter and the Jacobian matrix of partial derivatives of h(measurement function) with respect to x(state) is not n×n, it is 13×16.
$\displaystyle \quad\ \bf H_{[i,j]}$ = $\bf \frac{\partial h_{[i]}}{\partial x_{[j]}}(\tilde x_k,0)$
I would like to extract as much information as possible and one the things I thought of was eigenvalues, but it is not square. I wonder if I can zeropadd and make it square.
Is there any sense?
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## 1 Answer
That depends on what you are doing (I have no idea about your application). But a priori filling with zeros does imo make sense. A $n\times m$ matrix can be understood as a linear map $\mathbb R^m\to \mathbb R^n$. Here you have $n=13<m=16$. You have two options now which eventually should give the same result.
First you realise that you can perform row and column operations to reduce your matrix to a non-trivial $13\times 13$ matrix and a $13\times 3$ block of zeros. This means that you split the bigger $\mathbb R^m$ into two subspaces, one of dimension $3$ which is mapped to $0$ and one of dimension $13$ which still maps surjectively on the image of the original linear map. All information is then concentrated in the $13\times 13$ block and this should give you some information.
Or you fill the matrix with zeros to get a $16\times 16$ matrix. This corresponds to changing your target space where the additional dimensions are not hit by your map. The image will then still live in a $13$-dimensional subspace of $\mathbb R^{16}$. But diagonalising will give you a nontrivial $13\times 13$ block in the $16\times 16$ matrix again. This will look like the $13\times 13$ block as above.
Edit: Just a comment on a possible application (I still don't get it, and won't without spending a lot more time on it, whcih I won't;P ). All row and column operations perform base changes on the source or the target. This somehow assumes that all dimensions have equal rights. So if all dimensions correspond to positions for example, then changing the base will just correspond to a coordinate change. If some dimensions correspond to positions and some, say, to velocities you will end up with mixed expressions after changing your base. This might or might not make sense.
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Thanks. I will have give a try to the first option. And respect to my application that matrix is a Jacobian that maps the state of my system, wich has 13 components (position, orientation and velocities) with 8 measured points (x and y). The realation is given by writing each state component as a fucntion of the points projected by the homography matrix. – Jav_Rock Jun 7 '12 at 11:03
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