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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Qubit
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# All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary
# Science Fair Project Encyclopedia
For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.
# Qubit
A qubit is not to be confused with a cubit, which is an ancient measure of length.
A qubit (quantum + bit; pronounced /kyoobit/ [1] ) is a unit of quantum information. That information is described by state in a 2-level quantum mechanical system, whose two basic states are conventionally labeled $|0 \rangle$ and $|1 \rangle$ (pronounced: ket 0 and ket 1). A pure qubit state is a linear quantum superposition of those two states. This is significantly different from the state of a classical bit, which can only take the value 0 or 1.
A qubit's most important distinction from a classical bit, however, is not the continuous nature of the state (which can be replicated by any analog quantity), but the fact that multiple qubits can exhibit quantum entanglement. Entanglement is a nonlocal property that allows a set of qubits to express superpositions of different binary strings (01010 and 11111, for example) simultaneously. Such "quantum parallelism" is one of the keys to the potential power of quantum computation.
A number of qubits taken together is a qubit register . Quantum computers perform calculations by manipulating qubits.
Similarly, a unit of quantum information in a 3-level quantum system is called a qutrit, by analogy with the unit of classical information trit. Other names have been suggested to designate a unit of information for higher-level systems, though there is not yet general agreement on terminology. For instance qudit has been suggested both as a term to denote both a digit of quantum information (that is in a 10-level quantum system) or as a term to denote a unit of quantum information in a d-level quantum system, although the latter usage now seems to be taking hold.
Benjamin Schumacher discovered a way of interpreting quantum states as information. He came up with a way of compressing the information in a state, and storing the information on a smaller number of states. This is now known as Schumacher compression. Schumacher is also credited with inventing the term qubit.
The state space of a single qubit register can be represented geometrically by the Bloch sphere. This is a space of dimension 2, which means essentially that the single qubit register space has two local degrees of freedom. An n-qubit register space has 2n+1 − 2 degrees of freedom which is much larger than what one would expect classically with no entanglement, that is 2n.
## External links
• [1] A good update on qubits in the Jan 2005 issue of Scientific American.
• [2] The organization cofounded by one of the pioneers in quantum computation, David Deutsch
03-10-2013 05:06:04
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http://mathhelpforum.com/number-theory/121271-basic-number-theory.html
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# Thread:
1. ## Basic number theory
I just started self learning basic number theory so please excuse me for any noob questions:
1. Prove that the fraction $%5Cfrac%7Bn%5E3+2n%7D%7Bn%5E4+3n%5E2+1%7D$ is in lowest terms for every positive integer $%5Cmbox%7Bn%7D$
2. Let $%5C%7Ba,b,c%5C%7D%20%5Cin%20%5Cmathbb%7BN%7D$, show that $%5Cfrac%7B[a,b,c]%5E2%7D%7B[a,b][b,c][c,a]%7D%20=%20%5Cfrac%7B%28a,b,c%29%5E2%7D%7B%28a,b%29%28b,c%29%28c,a%29%7D$
3. Prove that consecutive Fibonacci numbers are always relatively prime.
4. Show that $1%20+%20%5Cfrac%7B1%7D%7B2%7D%20+%20%5Cfrac%7B1%7D%7B3%7D%20+%20%5Ccdots%20+%20%5Cfrac%7B1%7D%7Bn%7D$ can never be an integer. (I'm thinking of showing $%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Csum_%7Bk=2%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bk%7D$ converges to a value... or something along those lines, probs wrong heh)
Can anyone show me how to do these questions? Thank you.
2. ## First question
I would try a little rearrangement with the first question.
I would factor out n in the numerator into $n(n^2 + 2)$ and add and subtract 1 in the denominator which would transform the denominator into $n^4 + 3n^2 + 2 - 1$ which factors into $(n^2 +1)(n^2 + 2) - 1$. Now notice the common factor $(n^2 + 2)$ in both numerator and denominator.
I'll let you take it the rest of the way.
3. ## With question #4
As you may know, it's a harmonic series which sums to infinity as n increases without bound. You need to find a representative term for the sum of the first n terms and show that this term can never be an integer no matter what n is.
4. For the first one, you want to find polynomials $p(n), q(n)$ having integer coefficients such that
$p(n)(n^3+2n)+q(n)(n^4+3n^2+1)=1$.
For the third one, use the identity $f_n^2-f_{n-1}f_{n+1}=(-1)^{n+1}$.
For number 4, do not use wonderboy's advice, as you will not find a closed form for the $n$th harmonic number. Here is a hint : consider the highest power of $2$ less than $n$, say $2^m$. Then there is no other integer between $1$ and $n$ which is divisible by $2^m$. So what would you get if you put $1+\frac{1}{2}+...+\frac{1}{n}$ in a fraction having lowest terms, say $\frac{p}{q}$? (show that $p$ is odd, while $q$ is even!)
5. for the second one, hint: [m,n](m,n)=mn
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http://unapologetic.wordpress.com/2007/08/23/internal-hom-functors/?like=1&source=post_flair&_wpnonce=eb8d9d939c
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# The Unapologetic Mathematician
## Internal Hom Functors
As Todd Trimble pointed out, things get really nice when a category is enriched over itself. That is, the morphisms from one object to another in $\mathcal{V}$ themselves have the structure of an object of $\mathcal{V}$. This trivially the case for $\mathbf{Set}$, because there’s a set of functions from one set to another. We also know that in $\mathbf{Ab}$ there’s an abelian group of homomorphisms from one abelian group to another. We say that the category has an “internal hom functor”, because the hom functor lands back inside the category itself, rather than in the category of sets.
For the moment, let’s consider a category $\mathcal{V}$ that is not only monoidal (which is needed to have an enriched category), but also symmetric and closed. Remember that “closed” means we have an adjunction $\underline{\hphantom{X}}\otimes B\dashv (\underline{\hphantom{X}})^B$ for each object $B$. In $\mathbf{Set}$ the set $A^B$ is the set of functions from $B$ to $A$, while in $\mathbf{Ab}$ it’s the abelian group of homomorphisms from $B$ to $A$. We see that these are already the internal hom functors we’re looking for in these situations.
So in general let’s take our symmetric, monoidal, closed category $\mathcal{V}$, with underlying ordinary category $\mathcal{V}_0$. The adjunction between the monoidal structure and the exponential has a counit — an arrow $A^B\otimes B\rightarrow A$ — which corresponds to “evaluation” in both of our sample cases. That is, it takes a function $f:B\rightarrow A$ and an element $b\in B$ and gives an element $f(b)\in A$. We can use this to build a category.
Start with the objects of $\mathcal{V}_0$, and define the hom-object from $B$ to $A$ as $A^B$ (using the exponential functor from the closed structure). We need to find arrows $A^B\otimes B^C\rightarrow A^C$ and $\mathbf{1}\rightarrow A^A$, and we’ll use the adjunction to do it. For composition, we have the arrow
$(A^B\otimes B^C)\otimes C\rightarrow A^B\otimes(B^C\otimes C)\rightarrow A^B\otimes B\rightarrow A$
where the first step is the associator and the other two are evaluations. This is an element of $\hom_{\mathcal{V}_0}((A^B\otimes B^C)\otimes C,A)$, so the adjunction sends it to an element of $\hom_{\mathcal{V}_0}(A^B\otimes B^C,A^C)$, as we require. For identities, we can just use the left-unit arrow $\mathbf{1}\otimes A\rightarrow A$ and pull the same trick. Now properties of adjoints give us the required relations to make this a category enriched over $\mathcal{V}$.
And finally we can check that $V(A^B)=\hom_{\mathcal{V}_0}(\mathbf{1},A^B)\cong\hom_{\mathcal{V}_0}(B,A)$, so the “underlying set” of $A^B$ is actually the set of morphisms from $B$ to $A$ in the underlying category $\mathcal{V}_0$. This justifies our suspicions that the $\mathcal{V}$-category we just built is in fact $\mathcal{V}$ itself, now as a category enriched over itself.
### Like this:
Posted by John Armstrong | Category theory
## 3 Comments »
1. Nice post, John. One thing I love about category theory is the way a few simple concepts continually reflect back on and through one another, creating these kaleidoscopic displays and, after a while, real depth of perception. What you can achieve with the concept of an adjunction (and closely intertwined concepts of representables, limits, monad, monoid, theory, …) is nothing short of astonishing.
A quick note: the archetypal example of a 2-category is Cat, as you’ve said. But what is a 2-category? A Cat-enriched category! So we are saying Cat is a Cat-enriched category. The deeper explanation, following the philosophy of today’s entry, is that Cat is a cartesian closed category! (Not a trivial result, that!)
What I really mean is, there’s this hierarchy: 0-categories (sets), 1-categories (categories), 2-categories, … where an (n+1)-category is an (n-Cat)-enriched category! And one can prove by induction that n-Cat is cartesian closed, hence enriched in itself, hence an (n+1)-category! [NB: I am speaking of *strict* n-categories, not the weak n-categories that are much the rage these days.] There are some lovely expositions of this sort of thing in old installments of John Baez’s This Week’s Finds.
Sorry, I get carried away. Please, carry on…
Comment by Todd Trimble | August 23, 2007 | Reply
2. Sorry, I get carried away.
No, it’s good to have a Greek chorus around sometimes. Yes, $\mathbf{Cat}$ is a $\mathbf{Cat}$-category (which is how I defined strict 2-categories) so it’s an example of this. There’s more to be said about these, so I’m not trying to get it all into one post.
I forget if I mentioned before, but when you first said “categories enriched in themselves” I somehow didn’t make the connection with “internal homs”, which is the language I’m more familiar with. Once I realized that, clearly I’m going to be talking about them. When I eventually (long way off) get to things like sheaves the concept of an internal hom is the best motivator for a lot of how they work.
Comment by | August 23, 2007 | Reply
3. [...] fact, the evaluation above is the counit of the adjunction between and the internal functor . This adjunction is a natural isomorphism of sets: . That is, left -modules are in [...]
Pingback by | October 24, 2008 | 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://www.impan.pl/cgi-bin/dict?indicate
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## indicate
[see also: demonstrate, show, point, express, convey, suggest, mark]
......, where the prime indicates that only terms with $p>0$ may appear.
We shall write “a.e. $[\omega]$” whenever clarity requires that the measure be indicated.
We now indicate some of the inherent difficulties.
The proof will only be indicated briefly.
We now show that $G$ is in the symbol class indicated.
for $k$ in the indicated range
Go to the list of words starting with: a b c d e f g h i j k l m n o p q r s t u v w y z
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http://mathhelpforum.com/algebra/118653-finding-polynomials.html
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# Thread:
1. ## Finding polynomials
Hello, I cant figure this problem out for the life of me. Could someone help me?
Find the polynomial P(x) of degree 4 with integer coefficients, and zeroes $3-2i$ and 3 with 3, a zero of multiplicity of 2.
Thanks in advance...
2. Originally Posted by Godzilla
Hello, I cant figure this problem out for the life of me. Could someone help me?
Find the polynomial P(x) of degree 4 with integer coefficients, and zeroes $3-2i$ and 3 with 3, a zero of multiplicity of 2.
Zeros are, therefore, 3-2i, 3+2i, 3, and 3.
$(x - 3 + 2i)(x - 3 - 2i)(x - 3)(x - 3)$
or
$x^4-12x^3+58x^2-132x+117$
3. can someone show me how to factor that? I am confused because of the (3-2i) and (3+2i)? Thanks!!
4. This polynomial:
Originally Posted by Haversine
$x^4-12x^3+58x^2-132x+117$
is actually a result of this factorization:
Originally Posted by Haversine
$(x - 3 + 2i)(x - 3 - 2i)(x - 3)(x - 3)$
That is, if I know the roots are A, B, C, and D, I just set up a series of polynomial factors (x - A)(x - B)(x - C)(x - D).
There's nothing special about the (x - 3 + 2i) factor, for instance. All it's saying is that I want a term that equals zero when "3 - 2i" is substituted for x.
The other imaginary term pops in there because the complex conjugate will always be a root.
5. The whole point is that $x^4-12x^3+58x^2-132x+117$ is from the product (x-3+2i)(x-3- 2i)(x-3)(x-3). For this problem, at least, it is not a matter of "factoring" but of "multiplying"!
As far as (x-(3- 2i))(x-(3+2i))= (x-3+2i)(x-3-2i) is concerned, think of it as a "product of sum and difference= difference of squares": $((x-3)- 2i)((x-3)+ 2i)= (x-3)^2- (2i)^2= x^3- 6x+ 9- 4= x^2- 6x+ 5$
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http://roy-t.nl/index.php/tag/computing-science/
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## Thesis: upper sets in partially ordered sets
Today I finally finished my thesis. It’s topic was to count the number of upper sets in partially ordered sets Which is quite a hard problem since it’s in the complexity class #P-Complete (that’s the class of counting the solutions to the decision problems in NP-complete). All and all I’m quite pleased with the result. Although the upper bound is still $O(2^{n})$, (can’t quite get under there without solving P=NP and winning a million dollars) I’ve manged to find a solution that has a best case of $O(n)$ both in time and memory complexity. With a particularly large data set the brute-force algorithm took over 2 hours to complete while my algorithm took 0.025 seconds. Now that’s what I’d call a speed gain (and yes it was a real life data set, no tricks here). You can see this for yourself in the graph at the bottom of this post the ‘naïeve algoritme’ is the brute force approach, the ‘Familiealgoritme zonder uptrie’ is the first version of my algorithm, the ‘Familiealgoritme met uptrie’ is the final version of my algorithm. It uses a trie like data structure to speed up searching and uses a lot less memory. Note that the graph has a logarithmic scale.
Unfortunately for most readers my thesis is in Dutch, but I’ve translated the abstract to English:
Counting the number of upper sets in partially ordered sets gives us a unique number that can be used to compare sets. This number is like the fingerprint of a set. Until now there isn’t, as to my knowledge, an efficient algorithm to calculate this number. This meant that the number had to be calculated either by hand or by using a brute force approach. Using a brute force approach leads quickly to problems, even for trivially small data sets since this means that you have to generate 2^n subsets and check each of these subsets on upwards closure. When calculating by hand you can use symmetry but this menial process can take a lot of time and is error prone. In this thesis I present an algorithm that can calculate exact, and usually fast, the number of upper sets in a partially ordered set.
You can download my thesis here: Upper sets in partially ordered sets (Bsc thesis Roy Triesscheijn) as I’ve said before the text is in Dutch, but the proofs and attached code should be readable enough. If you’ve got any questions feel free to ask below!
06
Jul 2012
CATEGORY
Blog, Personal
COMMENTS No Comments
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http://mathoverflow.net/questions/90464/uniqueness-of-equilibrium-from-infinite-strategies
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## Uniqueness of equilibrium from infinite strategies
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I took the following game from the Peter Winkler collection (chapter "Games"):
Two numbers are chosen independently at random from the uniform distribution on [0,1]. Player A then looks at the numbers. She must decide which one of them to show to player B, who upon seeing it, guesses whether it's the larger or smaller of the two. If he guesses right, B wins, otherwise A wins. Payoff to a player is his/her winning probability.
One easily identifies the following mixed strategy Nash equilibrium:
"Player A shows the larger number with prob 1/2 and player B guesses 'larger' with prob 1/2"
The book also suggests a smart pure strategy for A, which is in effect identical to her mixed strategy above (in the sense that locks her winning prob at 1/2 regardless of B's strategy):
"Player A shows the number which is closer to 1/2"
A little thought shows that B also has a pure strategy in like manner:
"Player B guesses larger iff the number he sees exceeds 1/2"
Together these two strategies form a pure strategy Nash equilibrium.
To be clear, let me define a pure strategy for B as a function `${f}_{B}:[0,1]\longmapsto \{larger, smaller\}$`, i.e., he assigns "larger" or "smaller" to every real in [0,1].
Similarly, A's pure strategy is a function `$f_A(\{x,y\})=x\ or\ y$`, i.e., she assigns x or y to every set `$\{x,y\}$`, where `$x,y\in [0,1]$`.
My question is: Is the above pure strategy Nash equilibrium unique? Given the above definition, there are infinite pure strategies for each player. Could other less obvious or highly artificial equilibria be constructed? How can we prove or disprove the uniqueness of equilibrium from those infinite strategies?
Edit: To avoid possible ambiguity in Steven's answer, I add the last sentence in the 2nd paragraph.
-
## 2 Answers
I think you're going to have to be careful about what you mean by equilibrium.
Definition 1: $(f_A,f_B)$ is an equilibrium if, taking $f_B$ as given, $f_A$ maximizes the expected value of $A's$ payoff to $(-,f_B)$ (and symmetrically).
Definition 2: $(f_A,f_B)$ is an equilibrium if, for every $(x,y)$, taking $f_B(x,y)$ as given, $f_A(x,y)$ maximizes $A's$ payoff to $(-,f_B(x,y))$ (and symmetrically).
In the first case, you are looking for the equilibrium in a single game. In the second case, you are looking for a family of equilibria in a family of games (parameterized by $(x,y)$).
If I take literally your request for equilibria in "this game" (singular), it would seem that Definition 1 applies. In that case, you can start with your Nash equilibrium, vary either player's strategy arbitrarily on any set of measure zero, and have another Nash equilibrium.
-
Thank you for reminding. Yes it's Definition 1 --- otherwise there's no need for the requirement "uniform distribution on [0,1]". But while varying ranges of equilibrium $f_A$ and $f_B$ on measure zero sets trivially leads you to a different equilibrium by the definition, is a more essential change possible? – unknown (google) Mar 7 2012 at 16:36
### 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.
Given that calculation of probability is involved. It is appropriate to restrict strategies to be measurable functions.
"Player B guesses larger iff the number he sees exceeds 1/2". Denote this strategy of B ${S}_{B}^{*}$.
The first part of the argument proves (I hope) that ${S}_{B}^{*}$ is the only kind of pure strategy (except by a difference of measure zero) that B can adopt in any pure strategy equilibrium. The second part proves A's corresponding pure strategies for ${S}_{B}^{*}$. Their combinations then form all possible pure strategy equilibria.
By definition, B's pure strategy is to choose a measurable set $B_L\subseteq[0,1]$ such that he report larger for $x\in B_L$ and smaller for $x\in [0,1]/B_L$. Now say if $m(B_L)=a\neq1/2$, A can adopt the following pure strategy:
"Show the smaller number if both $x,y\in B_L$, otherwise show the larger number".
which guarantees her a winning probability $\geq a^2+(1-a)^2>1/2$. To counter this strategy of A, B's better off reverting to ${S}_{B}^{*}$. Hence $m(B_L)\neq1/2$ can't be an equilibrium pure strategy for B.
Now let $B_L\subseteq[0,1]$ and $m(B_L)=1/2$. Define $B_S=[0,1]/B_L$. Consider the following incomplete specification of a strategy for A:
"Show the smaller number if both $x,y\in B_L$; show the larger if both $x,y\in B_S$"
which already guarantees winning probability $\geq1/2$ for A. What about the remaining situations, i.e., $x\in B_L$ and $y\in B_S$?
For any measurable $B\subseteq B_L$ with $m(B)>0$, we can define `$C=\{x\in B_S|x>y, \forall y\in B\} $`. Suppose there exists such a $B$ such that $m(C)>0$, then A can adopt the following pure strategy:
"Show the smaller number if both $x,y\in B_L$, otherwise show the larger number"
which will guarantee her winning probability $\geq 1/2+2m(C)m(B)>1/2$, which can't be an equilibrium pure strategy for the same reason as in the $m(A_L)\neq 1/2$ case. Because $B\subseteq B_L$ was arbitrary, we then know in a pure strategy equilibrium it is necessary that the set `$\{x\in B_S|x>y, \forall y\in B_L\} $` has measure zero. Hence, to conclude, necessary conditions for strategies of B in a pure equilibrium:
1. $m(B_L)=m(B_S)=1/2$
2. `$\{x\in B_S|x>y, \forall y\in B_L\} $` has measure zero.
However, we already know ${S}_{B}^{*}$ and its possible variations on measure zero sets are strategies that B can adopt in a pure strategy equilibrium. Since these are exactly those strategies that satisfy 1 and 2. We conclude the only pure equilibrium strategy for B is ${S}_{B}^{*}$ (except by a difference of a measure zero set).
Given ${S}_{B}^{*}$, as long as $x\in [0,1/2]$ and $y\in (1/2,1]$, choice of $f_A$ is irrelevant, and player B will guess correctly. Since this happens with probability 1/2, the best A can do is to salvage all remaining situations, which amounts to show the smaller one if both numbers fall into (1/2,1] and larger one if both into [0,1/2], and achieve a winning probability of 1/2.
And this should include all possible pure strategy equilibria.
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http://mathoverflow.net/questions/87649/curvatures-of-contours-of-solutions-of-3d-poissons-equation
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## Curvatures of contours of solutions of 3d Poisson’s equation
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $f(x,y,z)$ be a complex function in a 3d euclidian space that fulfill the Poisson's equation $$\frac{\partial^2}{\partial x^2} f + \frac{\partial^2}{\partial y^2} f + \frac{\partial^2}{\partial z^2} f = c,$$ where c is a real number.
Let $X$ be a surface being a contour surface of $|f|^2$, i.e. $|f(x,y,z)|^2=k$ for some $k\geq0$.
Can anything be said on the curvature (mean or gaussian) of $X$? (I.e. if it is bounded from above or below? If, within one surface, its changes are bounded?)
Motivation: I am interested in the penetration length of evanescent waves. Then the Poisson's equation can be interpreted as the wave equation for a monochromatic wave, where $f$ is one component of the electric field. Surface of the same $|f|^2$ are surfaces of the same intensity.
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## 1 Answer
There's nothing you can say locally, other than that the level surfaces of $|f|^2$ are real-analytic (when they are surfaces). Conversely, any embedded real-analytic surface $S$ in $\mathbb{R}^3$ has an open neighborhood $U$ on which there is a function $f:U\to\mathbb{C}$ that satisfies $\Delta f = c$ on $U$ and $|f|^2 = \lambda$ on $S$, where $\lambda>0$ is an arbitrarily specified constant. Moreover, you can arrange that the differential of $|f|^2$ along $S$ is nonzero, so that it is, effectively isolated.
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http://physics.stackexchange.com/questions/39456/the-z-torque-how-can-it-be-shown-intuitively-that-it-does-not-work/52398
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# The Z-Torque: how can it be shown intuitively that it does not work?
There is a new kickstarter project that claims to increase torque and power compared to a normal crank on a bicycle (Z-Torque on kickstarter). If this patented (US Patent Number 5899119) approach works it would be a revolution for personal transport.
The idea behind this is that a Z-shaped crank should allow somehow better leverage compared to a straight crank. There are quite a few comments that this should not work as intended but as in a lot of cases it sounds plausible at first but it is incompatible with physics. As the inventor and some of the supporters are convinced that this actually works I am looking for a clear and simple argument why this is not possible.
Using vectors and cross products ($\tau = \vec{r}\times\vec{F}$) this is easy to show but can it be shown with as few abstractions as possible?
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## 5 Answers
Since the crank is rigid and symmetrical, just draw a straight line connecting the footpads through the center. The radial distance of the footpad from the center is the length of the torque arm. If you were to replace the Z-crank with a standard crank that had the same torque arm length, the torque felt at the center of the crank would be the same.
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While this is absolutely correct, I have found that some (many?) people just don't feel that it is intuitive and it won't convince them. Fool. Money. The length of time they remain together. – dmckee♦ Oct 10 '12 at 4:24
I thought about a longer explanation. However, honestly one would need to do finite element analysis and have complete knowledge of the properties of the material they are using to do it properly. However, if one assumed identical material with identical properties, with identical per unit length densities etc, one could show the differences in energy losses between the expansion and contraction of the material in about an hour or two or so (which unfortunately I don't have at the moment :-( – Hal Swyers Oct 10 '12 at 9:02
@HalSwyers That is an awful lot of unnecessary work! The claim of reputable physics is that torque comes from the point of action. Regardless of how you reconstruct the mechanical connections, the torque is conserved. The beauty of something being conserved is that you don't have to do infinite work to show it in every case. The real problem is that physics has failed to convince someone of the merit of using the "point of action", which is a consequence of "opposite and equal" Newtonian mechanics. Conservation laws should seem like magic... to anyone who just learned them. – AlanSE Oct 15 '12 at 3:04
@AlanSE completely agree. The point I was getting at is that these sorts of claims often do not hinge on simple physical principles, but ultimately rely on some complex explanation. If the claim is efficiency, then it is possible that with the right selection of materials, the right distribution etc, one could show that thermodynamic losses due to molecular interaction in one design is better than the other. There is a derivable amount of heat lost in the design, and the difference could be quantified with proper analysis, and to be fair I just don't know which way the answer would go. – Hal Swyers Oct 15 '12 at 11:11
My way of saying this is: Pretend you're a foot. Something stiff is constraining you to move in a circle about a center. If that something is straight or crooked you can't even see, because you're a foot. – Mike Dunlavey Jan 29 at 12:38
For an observable demonstration, all you would have to do is have the rider stand on one pedal. With all of the rider's weight on one pedal oriented at the 6 o'clock position, it would be obvious that the force from the pedal is acting on the crank spindle in a straight line, and the "elbow" of the crank has no significance.
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For claims like these, I think it is important to get somewhat specific about what claims are actually made. The claims are argued with logic, and as such, the claimant will eventually run into contradictions - which are the same observations that led to development of real physics. I would argue that in most cases throughout human history, the claimant then alters the story to either:
1. temporarily avoid the contradiction, or "argue around it"
2. allude to credible science that actually describes something else
My favorite example of #2 is Thorium laser powered cars. If you start reading, you will find articles written by people with no relevant knowledge. However, if you follow the rabbit hole on the internet long enough you'll rind references to actual science that proposes accelerator driven subcritical Thorium reactors, where the "laser" is really charged particle beams, but the innovations that allows it to be done with lasers, without producing fission products (completely absurd), and small enough to put in a car - that's secret. This point is also suspiciously close to where you'll find the donate button. That was a bit of a tangent.
After watching the videos, I'm convinced this is a case of #1. I will argue this by using these two diagrams from his video.
This is the same thing shown in the question. I wanted to be sure that the interpretation of the claim was correct. It is. Here is the 2nd part of the argument.
I would argue this as somewhat of a "smoking gun". Let's say $\theta$ is the angle from the vertical where the foot lies. He is actually arguing that when $\theta=0$ the torque is non-zero when using the z-crank, although it is zero when using a normal crank.
If one's college level first-year physics class did its job, they should be able to disprove the claim with this information, as it is plenty sufficient. The above thinking represents a fundamental misunderstanding of the physics at work. We need not look at the rest of the claims, or his approach that you can use a:
greater portion of rotation for power generation
His point form the above two images is sufficient to refute the claims. The core misunderstanding is believing that the "torque" comes from the direction of the crank material locally to the shaft. No, torque is based on where the force is applied from. The foot applies the force. The pivot is still in the same place. The pivot is in the same place, the force is in the same place (establishing $\mathbf{r}_1=\mathbf{r}_2$), is the same magnitude, and is in the same direction (establishing $\mathbf{F}_1=\mathbf{F}_2$) - the torque of the two cases is the same ($\boldsymbol \tau = \mathbf{r}\times \mathbf{F}$). He says it's not. I'm finished, because the entire point of the Kickstarter project is built on the idea that they're not.
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Assume that the length of the crank in itself, not the length between the footpads and the center, is what matters in determining the force applied. Following this assumption, a two-meter bent-beyond-recognition crank would be even more effective. Showing that the "intuitive" belief that "longer crank = more power" does not hold.
Alternatively, following the initial, wrong, assumption, we could also construct a crank which was basically a square - four bends to the same direction, with a slight sideways curl to allow one side of the square to pass its cousin. This crank should also, according to the nutjob theory, be more efficient. Now, ask why a seriest of bolts connecting the two sides running right next to each other would make one lick of a difference...
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There's a very simple way to explain this to a skeptic.
Suppose the crank starts out as a simple rigid rod of some very strong material, like steel.
Then suppose some material is added to one side and removed from the other, changing its shape a little. Does that change anything?
Suppose that process continues, and a lot of material is removed from one side and added to the other. Does that change anything?
EDIT: Another way to look at it. Suppose you start with a diamond-shaped piece of steel. It acts like the original crank, right?
Then you remove material from the center, leaving a picture-frame shape. Does this act any different from the original crank?
Then cut away one side of it, leaving a V-shape. Now does it act any differently?
The behavior of the crank is the same, whether it is straight or misshapen, assuming the material is strong enough to be rigid. The distance between the two end pivot points remains the same, so from the viewpoint of the pedal and center pivots it is indistinguishable from the straight crank.
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I like the continuous evolution of the crank, this is hard to argue around. – Alexander Jan 28 at 16:30
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http://physics.stackexchange.com/questions/3861/how-do-we-resolve-operator-ordering-ambiguities-when-quantizing-generic-nonlinea?answertab=oldest
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# How do we resolve operator ordering ambiguities when quantizing generic nonlinear second-class constraints?
Dirac came up with a general theory of constraints, including second-class constraints. To quantize such systems, he first computed the Dirac bracket classically, and only then "promoted" the classical Dirac bracket into a commutator. However, this leads to operator ordering ambiguities over and above what already exists for the Poisson bracket.
Is there any more direct way of coming up with a quantum Dirac bracket and Hamiltonian operator which resolves such ambiguities? If the system is symmetrical, symmetry considerations can often single out the "correct" quantization, but what about asymmetrical systems?
Ideally, the correct quantum prescription ought to match what we get from using path integrals.
Similarly, if the second-class constraint is linear, the quantum commutator and Hamiltonian is also straightforward. However, even relatively simple models can have subtleties. Consider the toy model $$L=\frac{\dot{q}^2}{2} -f[q]\frac{F^2}{2}$$ where F is an auxiliary variable and $f$ is a function of $q$. The classical second-class constraints are given by $p_F \approx 0$ and $fF\approx 0$. Using path integrals as a guide, $$\int \mathcal{D}q\, \mathcal{D}F\, \exp\left( \frac{i\dot{q}^2}{2} - \frac{ifF^2}{2} \right) \propto \int f^{-1/2}\mathcal{D}q\, \exp\left( \frac{i\dot{q}^2}{2} \right),$$ which corresponds to the following operator ordering for the Hamiltonian: $$\widehat{H} = \frac{1}{2} \widehat{f}^{1/4}\widehat{p}\widehat{f}^{-1/2}\widehat{p}\widehat{f}^{1/4}.$$
But what about the generic nonlinear case?
On another note, is there a formalism where we can impose the Dirac bracket after quantization, rather than before?
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I think that there's no "God-given" ordering you should apply. A general quantum theory is not specified "purely" by its classical limit. In the context of quantum field theory, all the renormalization issues may be interpreted as "generalized ordering effects", so all the uncertainty about the total coupling and the right subtraction of the infinite parts may be interpreted as an ordering effect that doesn't have any unique canonical solution. – Luboš Motl Jan 25 '11 at 18:12
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A professor once said that going from the classical Poisson bracket to the quantum mechanical commutator, or that promoting classical observables to qm operators, always includes something akin to a leap of faith, because there is no rigorous way to get from classical mechanics to quantum mechanics. – Lagerbaer Jan 27 '11 at 17:55
## 2 Answers
If you think that there are simple and unambiguous ways to construct theories, then you were misled by someone.
Every achievement was reached by trying different possibilities. Only more-or-less successful constructions were left in the end (different in different particular cases).
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There's a nice little book that deals with this question of yours in a mathematically rigorous way: Mathematical Topics between Classical and Quantum Mechanics (Springer Monographs in Mathematics).
In the end of the day, it boils down to studying Symplectic geometry and defining more appropriately what is meant by 'quantization'.
You can look at Deformation (Weyl) quantization and Geometric Quantization, they attack your questions heads-on.
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http://math.stackexchange.com/questions/237292/are-all-4-regular-graphs-hamiltonian?answertab=active
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# Are all $4$-regular graphs Hamiltonian
It is easy to show that all connected $2$-regular graphs are Hamiltonian. The Petersen graph is a $3$-regular graph that is not Hamiltonian. Are there any $4$ regular graphs that are not Hamiltonian?
Thank you
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Thank you all. I up voted all answers. – Amr Nov 14 '12 at 20:41
## 2 Answers
Note that you didn't specify connected graph, which is certainly necessary for Hamiltonicity. More generally, a Hamilton cycle has the property that deleting any $k$ vertices breaks the cycle into at most $k$ connected components. In the case $k=1$ a Hamiltonian graph must be strongly connected.
Pick your favourite $4$-regular graph $G$, and delete one edge $e$ so it has two vertices of degree $3$. Take two copies of $G-e$, and create a new vertex that is adjacent to the four deficient vertices. The resulting graph is $4$-regular, connected, but not strongly connected.
Also, did you try Googling 4-regular non-Hamiltonian? The Meredith graph is very strongly connected, $4$-regular and still not Hamiltonian.
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The answer is no. The graph on the picture is the smallest counterexample
You can most likely generalize this construction for every $k > 4$ and deduce that you can always find connected $k$-regular graphs that are not Hamiltonian.
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http://math.stackexchange.com/questions/316206/two-bags-of-colored-marbles-different-colors-is-each-bag-how-many-combinations
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# Two bags of colored marbles, different colors is each bag, how many combinations of complete pairings?
I've reduced a different problem I have to something that can be explained as below, but I'm not sure how to solve this last part.
Assume there are two bags of colored marbles. Any color in the first bag will not be in the second bag. For example, the first bag has two blue marble and three green marbles, while the second bag has one red marble, one white marble, and three black marbles. Each marble in the first bag has to be paired up with a marble in the second bag. How many different complete pair sets (where every marble is paired with another) are there?
A general solution would be amazing, else a pointer in the right direction would be great.
Thanks for any help!
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What is a complete pair set? – muzzlator Feb 27 at 20:05
@muzzlator: Where every marble in the first bag is paired with a marble in the second bag. One such combination of pairs is what I am calling a "complete pair set". – Albeit Feb 27 at 20:21
Hm but the second bag in your example has only 4 marbles. What would it mean in that case? (Sorry about the delay, I'm on and off) – muzzlator Feb 27 at 21:08
@muzzlator: Sorry, typo, fixed now. – Albeit Feb 27 at 21:11
Please let me know if this is wrong. I believe I have a way to calculate the number of pairings but the order or the final pairs matters. $$pairs = \frac{n!^2}{\prod(a_j!)\prod(b_k!)}$$ where $a_j$ and $b_k$ are the components of each bag. This will give the number of possible pairs in a particular order. (aka aA bB is different from bB aA). This lapse is why this is a comment and not an answer. – kaine Feb 27 at 22:33
## 2 Answers
Hint: Start by imagining that the marbles are numbered, so are all distinguishable. If there are $n$ marbles in each bag, you can line the ones in the first bag up and there are $n!$ ways to match up the marbles from the second bag. Then if you have four blue marbles that are indistinguishable in the first bag, there are $24$ ways that get collapsed to $1$
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That's what I thought at first, too; but the problem is that there are sets of indistinguishable marbles in both bags; and some of the $n!$ permutations of the matching may coincide with permutations among those sets and others may not. – joriki Feb 27 at 21:24
@joriki: Good point. It will be more complicated. – Ross Millikan Feb 27 at 21:38
Yes, I have it that far - if I assume that all the marbles in one of the two bags is unique, I can find the number of permutations of the marbles in the second bag: N!/(n1!n2!n3!), N=total marbles, n1=marbles of color 1, etc... But as @joriki mentioned, there will be repeated permutations as the marbles in one bag are not unique. – Albeit Feb 27 at 21:41
This explanation will look at the cases of: a) one bag contains 2 colors and the other contains and arbitrary number of colors; b) one bag contains 3 colors and the other contains arbitrary number; and c) each bag contains arbitrary numbers.
For every part below there are two bags containing n marbles. $n_{i,j}$ refers to the number of marbles of the jth color in the ith bag. $c_i$ is the number of different colors in bag i.
If anyone is willing to help confirm accuracy of this, i would be very appreciative as I've derived all of this using the referenced question as the only guide and I don't use combinatorics much.
a) As long as $c_1=2$ (as with the question's example), this problem is equivalent to just picking $n_{1,1}$ balls from the second bag.
number of combinations without repetition with limited supply
The above question helps explain how to find the number of combinations of $n_{1,1}$ marbles from bag two. Essentially you analyze the following polynomial for the coefficients:
$$P = \prod_{j=1}^{c_2}{\sum_{k=0}^{n_{2,j}}x^k}$$
For the question example:
$$P=(1+x)(1+x)(1+x+x^2+x^3) = 1+3x+4x^2+4x^3+3x^4+x^5$$
and the answer is 4 because the coefficient of $x^2$ is 4. This makes sense as the potential cases can be described as the colors paired with blue (black/black, red/black, white/black, and red/white).
b) The problem becomes more complicated when each bag contains more than 2 types of marbles. For the case of 3 colors in bag 1, the simular case requires that one consider both an x and a y. Define a polynomial:
$$F = \prod_{j=1}^{c_2}{\sum_{k=0}^{n_{2,j}}(\sum_{l=0}^{k} x^ly^{k-l})}$$
Convert it into an expanded form and note that the coefficient of $x^{n_{1,1}}y^{n_{1,2}}$ is the number of relevant combinations.
For examples, if $n=5$, $c_1=3$, $c_2=3$, $n_{2,1}=2$, $n_{2,2}=2$, $n_{2,3}=1$ then:
$$F=(1+x+y+x^2+x*y+y^2)^2*(1+x+y)$$ $$F=x^5+3 x^4 y+3 x^4+5 x^3 y^2+8 x^3 y+5 x^3+5 x^2 y^3+11 x^2 y^2+11 x^2 y+5 x^2+3 x y^4+8 x y^3+11 x y^2+8 x y+3 x+y^5+3 y^4+5 y^3+5 y^2+3 y+1$$
If bag 2 contains 2 black, 2 red, and 1 green (for instance) there will be 11 combinations.
c) For the final case of an arbitrary number of colors in each bag, i am not sure how to write the notation. Each component in bag 1 must be assigned a letter (for instance x or y in the above examples). Define $x^ay^bz^c...$ as the state in which a are paired with color x and b are paired with color y and c are paired with color z etc. You can then define a state polynomial f such that it is the sum of all probablistic states that are possible for a particular color from bag 2 (aka f = xz+yz+zz+xy+y^2 means that the color in question can be paired with 1x and 1z, 1y and 1z, 2zs, 1x and 1y, or 2ys but 2xs are impossible. All states included in f must be mutually exclusive. Multipling all of these f state polynomials together will form an overall state polynomial F simular to those seen above. By this solution, a 1 in the f state polynomial would indicate that none of that color are chosen so would not appear in any problem where all marbles are paired up.
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http://math.stackexchange.com/questions/252632/the-importance-of-parallel-arrows-in-a-commutative-square/252993
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# The importance of parallel arrows in a commutative square
I noticed that whenever there is a commutative square, the relation it imposes on parallel morphisms is usually very important (e.g. natural transformations, pullbacks). In contrast, there's usually almost nothing to be said about the consecutive morphisms in these diagrams, even though the equation expressed by the square directly involves their composition. Is there a conceptual explanation for this in general?
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Commutativity says that it does not matter which of the two paths you follow. However, this makes no statement about where your intermediate stops are on the two paths. Saying that going from London via Oslo to Moscow fulfills the same travel purpose as going from London via Rome to Moscow (e.g. you can personally transmit greetings from Buckingham Palace to the Kreml) can hardly say a lot about the relation between Rome and Oslo. In fact, Naples or Stockholm might be just as nice. – Hagen von Eitzen Dec 6 '12 at 22:52
@HagenvonEitzen you don't say... – Alexei Averchenko Dec 6 '12 at 22:54
exact sequences are a special type of useful consecutive. – user51427 Dec 6 '12 at 23:07
@sunflower fixed. – Alexei Averchenko Dec 6 '12 at 23:51
@HagenvonEitzen Excellent metaphor! – magma Dec 7 '12 at 8:29
show 2 more comments
## 2 Answers
This is an interesting observation , but imho this is a case where there is less than meets the eye. Commutative squares by themselves do not really say much more about the constituent arrows - or pairs of arrows - than their commutative relation already says. In particular your examples - natural transformations and pulbacks/pushouts - derive their interest because of other/extra conditions imposed on the squares.
Take natural transformations: the real noteworthy fact is that you can build a commutative square for all pairs of objects in the domain category. In itself a single square is not that special. It is like saying: a single puzzle tile means nothing, it is not even a rectangle, but taken together in a special way, the whole collection of tiles has the remarkable property of forming a rectangular puzzle.
Same with pulbacks/pushouts: the real interesting thing is the the universal property.
Not convinced yet? Ok, let's dig deeper.
Take a very special commutative square: a commutative triangle.
It can be interpreted as either one of the following commutative squares with identity morphisms:
where $\text{Id}_X^{\mathbb{C}}$ stands for "the identity morphism of object X in category $\mathbf{C}$"
Now: in the first square, f is parallel with an identity - which is a very special morphism: iso, mono, epi... What insight can we get from this? None! f can be anything.
Same story with g in the other diagram.
Conclusion: the importance of parallel arrows in commutative squares? In general...none
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– magma Dec 7 '12 at 11:25
I think you have hit upon an important idea. We don't know the general story yet. But my feeling is that the commutative squares (for naturality, for instance) are drawn in a particular way for a good reason, which is unfortunately not captured by the theory. In particular, the arrows drawn horizontally play a different role from those drawn vertically. The horizontal arrows play the role of "functions", i.e., map inputs to outputs, whereas the vertical arrows play the role of "relationships", i.e., something upstairs is related to something downstairs. Naturality says that all "relationships" are preserved by "functions".
In Computer Science, we work with a concept called relational parametricity, which is more general than naturality, where "functions" and "relationships" are decoupled from each other. We don't talk about "composition" of functions with relationships, but use more general axioms of fibrations to relate the two. When "relationships" are specialized to "functions", these conditions do boil down to using composition of morphisms. These are very intriguing facts, and it would be great for more people to get involved in investigating them.
The foundational article on relational parametricity is Reynolds: Types, abstraction and parametric polymorphism. The categorical axiomatization I am talking about may be found in Dunphy and Reddy: Parametric Limits.
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http://mathoverflow.net/questions/83363/is-there-a-nice-application-of-category-theory-to-functional-complex-harmonic-ana
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## Is there a nice application of category theory to functional/complex/harmonic analysis?
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[Title changed, and wording of question tweaked, by YC, because the original title asked a question which seems different from the one people want to answer.]
I've read looked at the examples in most category theory books and it normally has little Analysis. Which, is strange as I've even seen lattice theory be used to motivate a whole book on category theory.
I was wondering is there a nice application of category theory to functional analysis?
It's weird as read that higher category theory is used in Quantum mechanics as it foundation, yet QM has heavy use of Hilbert spaces.
Sorry for cross posting to MSE. However, somebody there suggested I posted it here to get better response (well, more than two replies). I'm surprised that most books on category theory have very little mention of Analysis. Especially since Grothendieck originally studied functional analysis.
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Personally, I think this question is a bit vague - functional analysis is by now a very broad church - but in any case you should make this question "Community Wiki" by clicking the appropriate box. This is the preferred option for questions which seek a big list of rankable items, rather than a single "right" or definitive answer – Yemon Choi Dec 13 2011 at 19:42
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Also, please add a link to the cross-post (just so that people here don't unwittingly duplicate what has been said there). Thanks – Yemon Choi Dec 13 2011 at 19:44
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@Yemon Choi I was more looking for books that try to put a foundation on Mathematical Analysis via Category theory sort of like the Mathematics made difficult book where he tries to put elementary Maths in terms of Category theory. – Dan Dec 13 2011 at 19:52
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I think Dmitri Pavlov has a lot to say on this topic, see for eaxample <a href="mathoverflow.net/questions/20740/…; – Michael Greinecker Dec 13 2011 at 22:01
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@Dan: could you perhaps reword your question to make it clearer what kind of "uses" you have in mind. Your question suggests that you are looking for applications in analysis, but then your comment says you are looking for things which give analysis a category-theoretic foundation. If you make the question more precise then people may be able to give more focused/relevant answers – Yemon Choi Dec 14 2011 at 0:23
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## 15 Answers
In relatively mundane, but intensely useful and practical, ways, the naive-category-theory attitude to characterize things by their interactions with other things, rather than to construct (without letting on what the goal is until after a sequence of mysterious lemmas), is enormously useful to me.
E.g., it was a revelation, by now many years ago, to see that the topology on the space of test functions was a colimit (of Frechet spaces). Of course, L. Schwartz already worked in those terms, but, even nowadays, few "introductory functional analysis" books mention such a thing. I was baffled for some time by Rudin's "definition" of the topology on test functions, until it gradually dawned on me that he was constructing a thing which he would gradually prove was the colimit, but, sadly, without every quite admitting this. It is easy to imagine that it was his, and many others', opinion that "categorical notions" were the special purview of algebraic topologists or algebraic geometers, rather than being broadly helpful.
Similarly, in situations where a topological vector space is, in truth, a colimit of finite-dimensional ones, it is distressingly-often said that this colimit "has no topology", or "has the discrete topology", ... and thus that we'll ignore the topology. What is true is that it has a unique topology (since finite-dimensional vector spaces over complete non-discrete division rings such as $\mathbb R$ or $\mathbb C$ do, and the colimit is unique, at least if we stay in a category of locally convex tvs's). Also, every linear functional on it is continuous (!). But it certainly is not discrete, because then scalar multiplication wouldn't be continuous, for one thing. But, despite the prevalence of needlessly inaccurate comments on the topology, the fact that all linear maps from it to any other tvs are continuous mostly lets people "get by" regardless.
Spaces with topologies given by collections of semi-norms are (projective/filtered) limits of Banach spaces. Doctrinaire functional analysts seem not to say this, but it very nicely organizes several aspects of that situation. An important tangible example is smooth functions on an interval $[a,b]$, which is the limit of the Banach spaces $C^k[a,b]$. Sobolev imbedding shows that the (positively-indexed) $L^2$ Sobolev spaces $H^s[a,b]$ are {\it cofinal} with the $C^k$'s, so have the same limit: $H^\infty[a,b]\approx C^\infty[a,b]$, and such.
All very mundane, but clarifying.
[Edit:] Partly in response to @Yemon Choi's comments... perhaps nowadays "functional analysts" no longer neglect practical categorical notions, but certainly Rudin and Dunford-Schwartz's "classics" did so. I realize in hindsight that this might have been some "anti-Bourbachiste" reaction. Peter Lax's otherwise very useful relatively recent book does not use any categorical notions. Certainly Riesz-Nagy did not. Eli Stein and co-authors's various books on harmonic analysis didn't speak in any such terms. All this despite L. Schwartz and Grothendieck's publications using such language in the early 1950s. Yosida? Hormander?
I do have a copy of Helemskii's book, and it is striking, by comparison, in its use of categorical notions. Perhaps a little too formally-categorical for my taste, but this isn't a book review. :)
I've tried to incorporate a characterize-rather-than-construct attitude in my functional analysis notes, and modular forms notes, Lie theory notes, and in my algebra notes, too. Oddly, though, even in the latter case (with "category theory" somehow traditionally pigeon-holed as "algebra") describing an "indeterminate" $x$ in a polynomial ring $k[x]$ as being just a part of the description of a "free algebra in one generator" is typically viewed (by students) as a needless extravagance. This despite my attempt to debunk fuzzier notions of "indeterminate" or "variable". The purported partitioning-up of mathematics into "algebra" and "analysis" and "geometry" and "foundations" seems to have an unfortunate appeal to beginners, perhaps as balm to feelings of inadequacy, by offering an excuse for ignorance or limitations?
To be fair (!?!), we might suppose that some tastes genuinely prefer what "we" would perceive as clunky, irrelevant-detail-laden descriptions, and, reciprocally, might describe "our" viewpoint as having lost contact with concrete details (even though I'd disagree).
Maybe it's not all completely rational. :)
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Which functional analysts are these, then, who don't like to speak of Frechet spaces as (projective/inverse) limits of Banach spaces in an appropiate category of TVSes? – Yemon Choi Dec 14 2011 at 0:35
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Helemskii has a book "Lectures on Functional Analysis" (or some similar title) where he covers some of the results one would normally associate with Rudin Chapters 1 to 4, but with forgeful functors and the such-like. Unfortunately I haven't got access to a copy to check. – Yemon Choi Dec 14 2011 at 0:37
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Oh, and +1 (this is more or less my own use of category theory) – Yemon Choi Dec 14 2011 at 0:44
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I find it particularly helpful about the limits of vector spaces, that the dual of a limit is the colimit of the duals.... – Marc Palm Dec 14 2011 at 8:28
@pm Not in the category of Banach spaces and linear contractions it isn't... the dual of a coproduct is the product of the duals – Yemon Choi Dec 14 2011 at 9:33
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I've never completely understood what counts as "an application of category theory". With other areas of mathematics an "application" of area A to area B is generally a result which translates a problem in B into the language of A, solves the problem using the main theorems of A, and translates the solution back into the language of B. The problem is that I don't know what the main theorems of category theory are (or even if there are any "main theorems").
What I can say is that many interesting and nontrivial categories do arise in certain parts of functional analysis and it is useful to understand the structure of these categories. The specific part of functional analysis that I have in mind is the theory of operator algebras. For instance, in C*-algebra theory one considers a category whose objects are C*-algebras and whose morphisms are given by groups $KK(A,B)$ which simultaneously generalize K-theory and K-homology. Many of the deepest theorems in the subject are organized around the "Kasparov product" which is nothing more than the composition law $KK(A,B) \times KK(B,C) \to KK(A,C)$ in this category. KK-theory and its close cousin E-theory can be characterized according to homotopy invariance and various functorial properties.
On a related note, Connes' noncommutative geometry program (arguably part of functional analysis) relies heavily on the tools of category theory and homological algebra. Even at its inception the program was based on an analogy between de Rham cohomology and the periodic cyclic homology of a "smooth subalgebra" of a C*-algebra. In the process of investigating the relationship between cyclic homology and K-theory people have realized that it is useful to take seriously the category of projective modules over such smooth subalgebras rather than passing to K-theory. The work of Jonathan Block is particularly relevant; you might have a look here, for example.
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Concerning the first paragraph: There are some theorems in "pure" category theory which are nontrivial (General Adjoint Functor Theorem and the related Freyd's Representability Criterion, Beck's Monadicity Theorem, Recoqnition Theorems of locally presentable categories, Brown's Representability Theorem for triangulated categories,...). Unfortunately many people outside of category theory still believe that these general theorems are not useful at all, although the last decades have brought so many applications ... – Martin Brandenburg Dec 14 2011 at 13:04
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Thanks for the examples - based on this question I was actually contemplating posting a question of the form "What are the main theorems in category theory?" I confess that I don't know the statements of any of those results, and I have heard of only two of them. Thus I can make no intelligent statement about whether or not they have applications to analysis, except that it wouldn't surprise me if they come up in noncommutative geometry (given the flavor of some papers in that area). – Paul Siegel Dec 14 2011 at 16:01
Well, I decided to post that question after all: mathoverflow.net/questions/83437/…. I'm very curious now, and I hope you'll contribute! – Paul Siegel Dec 14 2011 at 16:53
To add just a little bit to this answer I would say that the whole issue of Morita equivalence in C*-algebras (and, if you wish, in its geometrical counterpart i.e. Poisson geometry) certainly springs out from categorical ideas. – Nicola Ciccoli Dec 15 2011 at 7:38
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For a simpler example of a "main theorem in category theory" there's the Yoneda lemma. – Noah Snyder Dec 15 2011 at 16:37
The chain rule for differentiation
$D(f \circ g ) = Df \circ Dg$
is the first example of functoriality one meets and counts as analyis I guess! Of course to properly interpret this it is best to think of f and g as maps between manifolds and Df and Dg as their tangent maps defined on the tangentbundles of the manifolds. The functor of differentiation is thus the functor of taking the tangent bundle (on objects) and tangent maps (on functions).
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The substantial book
Kriegl, A. and Michor, P.W., The convenient setting of global analysis, Mathematical Surveys and Monographs, Volume 53. American Mathematical Society, Providence, RI (1997).
does use categorical notions and methods; a pdf is available.
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Probably this is already known to many readers here, but I'll add it because we are in CW mode:
It is possible to construct and characterize $L^1[0,1]$ categorically, namely as the "smallest" pointed Banach space $X$ satisfying the fixed point $X \cong X \times X$. See this note by Tom Leinster and this MO discussion. The integral $\int_{0}^{1} : L^1[0,1] \to \mathbb{R}$ comes from the universal property.
Although I have do admit that this does not compute $\int_{0}^{1} \sin(\pi x) ~ dx$ or alike, I think this is a quite enlightening and strinkingly simple characterization of such a complicated (well at least it is not just a toy example) Banach space. All you have to know is that you can identify a function with two functions, the rest follows. So this example goes in the direction of Paul Garret's answer. I don't know if it simplifies computations or even enables us to do new ones, but it follows one of the goals of category theory: Unification.
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Nice example, but: "complicated" - really? I would say that Tom's result (spinning off from Freyd's) is a lovely characterization of a natural Banach space. Moreover, to me its force is that one gets the integral for free. – Yemon Choi Dec 14 2011 at 19:15
First, a disclaimer: I am not even close to being an analyst. Second, I don't know of any applications of category theory to the areas of analysis that you mention. I don't think we have got to that point yet, for the reason given below. But here is an answer to a more general question that I hope you'll find illuminating.
I think the thing to remember here is that category theory is 'structural mathematics'. That is, it seeks to understand mathematical objects and constructions purely in terms of abstract external structure, as opposed to internal details about how an object is put together. In areas like algebra and computer science, this sort of structure is already there and visible, so for example it's easy to define the notion of 'group object' or 'monoid action', and to discuss constructions like quotients and semidirect products and so on in purely structural terms.
My impression of analysis is that the structures involved are less unequivocal and less clearly visible, and consequently it's harder to give a broad unified structural picture of the kind that we're used to in algebra. So the category-theoretic or structural understanding of analysis is a good deal less well-developed. But there are some interesting facts about certain structures found in (very elementary) analysis and topology:
• Metric spaces are a particular kind of enriched category, and the notion of Cauchy completeness has a very neat definition in that context.
• Compact Hausdorff spaces are the algebras for a monad on Set (the ultrafilter monad).
• Topological spaces are lax algebras for the same monad on the bicategory Rel of relations.
• Normed vector spaces can be viewed as enriched categories with duals (Lawvere), or as compact-closed ordinary categories equipped with a certain kind of functor (see Geoff Cruttwell's thesis).
• C-star algebras are monadic over Set (although I gather not much more is known about this).
• Non-standard analysis has a nice interpretation in terms of the filter-quotient construction on toposes.
• There is a notion of Fourier or z-transform for Joyal's 'structure types'.
There are probably many more (and this answer is CW so passers-by are invited to add them).
Personally, I think that the structural and material viewpoints on mathematics complement each other very well, so I'd be delighted if someone could point out (or write!) a structural account of -- a sort of 'Mac Lane and Birkhoff' for -- even elementary analysis.
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Where can I read about C^* algebras being monadic? – David Carchedi Dec 13 2011 at 21:38
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@David: there is a paper of Pelletier and Rosicky which at least mentions this result, see my answer mathoverflow.net/questions/8550/… – Yemon Choi Dec 13 2011 at 22:02
@Yemon: yes, that's the one I was thinking of. – Finn Lawler Dec 13 2011 at 22:50
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Finn, I honestly don't know how you are using the word 'material' here, but I'm pretty most sure analysts don't care how the real numbers are defined, as long as you get a complete ordered field. That's structural, not material, in the sense I am familiar with (set theoretic constructions). Do you have a good example of what you mean by "material" in analysis? – Todd Trimble Dec 14 2011 at 2:55
@Todd: I didn't mean to suggest that analysis takes a material approach where, say, algebra takes a structural approach -- all I meant by the last paragraph was that, as Paul Garrett points out in his answer, texts on analysis typically eschew the structural point of view, and that it would be nice if there were more structurally-oriented accounts of analysis to complement the traditional ones. – Finn Lawler Dec 14 2011 at 19:27
At the suggestion of Yemon, I have moved my comment to an answer. The Gelfand representation gives an equivalence between the category of commutative, unital $C^*$-algebras and the opposite category of compact Hausdorff spaces.
Breifly, let $A$ be a $C^*$-algebra, and let $\Sigma$ be the collection of nonzero homomorphisms $A \rightarrow \mathbb{C}$. Then $\Sigma$ sits inside $A^*$, the dual of $A$. Thus we can endow it with the weak-$*$ topology. With this topology, if we consider $C(\Sigma)$, the algebra of continuous functions $\Sigma \rightarrow \mathbb{C}$, it turns out that we obtain a canonical isometric $*$-isomorphism $A \rightarrow C(\Sigma)$.
The functor $CommC^*Alg \rightarrow CptHdTop^{op}$ given by $A \mapsto \Sigma$ defined above is an equivalence of categories. A great reference for the details is Conway's book on functional analysis (but he doesn't mention categories or functors).
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As far as I understand, this is one of the jumping off points in certain non-commutative geometry/pseudo-geometry programs. When studying non-commutative C*-algebras, the thought is that they should correspond to some non-commutative space analogous to the commutative case. – Eric A. Bunch Dec 15 2011 at 2:13
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Without reference to the morphisms, of course, this theorem is in many if not most books on functional analysis. (The force of the categorical viewpoint is the relationship between various $C^\ast$-algebras, not just the fact that the Gelfand transform maps onto C(max ideal space))) – Yemon Choi Dec 15 2011 at 4:21
+1 because you mentioned it as an application to analysis. Knowing that a commutative C-algebra is the same as functions on a compact space is helpul, e.g. in spectral theory. The other direction is much less useful (knowing that a compact space is the same as the spectrum of a C-algebra does not tell you anything. – Johannes Ebert Dec 15 2011 at 12:21
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One can prove the universal property of the Stone-Čech compactification by provings its associated C^*-algebra has the appropriate property. To do this one should really look at bounded continuous functions and the spectrum as giving an adjunction which restricts to an equivalence on the usual suspects. – Benjamin Steinberg Dec 15 2011 at 20:19
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@Yemon: Definitely, reference to the morphisms is necessary to fully realize the categorical viewpoint. My category theorist friend would be horrified to see me make no mention of what the functor does on morphisms! I hope he can forgive me if he ever reads this. – Eric A. Bunch Dec 16 2011 at 0:16
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The following book treats parts of Banach space theory and harmonic analysis from the point of view of category theory: Johann Cigler, Viktor Losert, Peter W. Michor: Banach modules and functors on categories of Banach spaces. Lecture Notes in Pure and Applied Mathematics 46, Marcel Dekker Inc., New York, Basel, (1979), MR 80j:46112, Zbl 411.46044. Review in Bull. AMS 3,2 (1980) xv+282 pp., Scanned pdf here.
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I had been meaning to invest in a copy one day, so the generosity of making it available is appreciated by this semi-categorical analyst at least! – Yemon Choi Oct 2 at 9:39
The theory of interpolation spaces is one such example. The classical interpolation theorems of M. Riesz, Thorin and Marcinkiewicz and their generalizations by A. Zygmund, E. M. Stein, G. Weiss, and C. Fefferman are basic tools in harmonic analysis. The Riesz-Thorin theorem yields a quick proof of the Hausdorff-Young inequality on the $L^p \to L^{p'}$ boundedness of the Fourier transform, as well as the classic proof of the $L^p$-boundedness of the Hilbert transform by M. Riesz. The Marcinkiewicz interpolation theorem is effectively a generalization of the proof of the Hardy-Littlewood maximal inequality and is a crucial tool in proving the $L^p$-boundedness of a wide range of singular integrals---say, those of convolution type or the Calder\'{o}n-Zygmund operators.
The interpolation theorem of Riesz-Thorin and Marcinkiewicz have been generalized substantially by A. Calder\'{o}n and J. Peetre, respectively, and modern treatments of the subject---Bergh-L\"{o}fstr\"{o}m, Bennett-Sharpley, etc.---often utilize category theory to make the notion of interpolation precise. To this end, we need the notion of a Banach couple, which is a pair $(A_0,A_1)$ of Banach spaces such that there exists a Hausdorff topological vector space $X$ such that we have continuous linear embeddings $A_0 \hookrightarrow X$ and $A_1 \hookrightarrow X$. The collection of Banach couples forms a category, whose typical morphism $T:(A_0,A_1) \to (B_0,B_1)$ is a bounded linear operator $T:A_0 + A_1 \to B_0 + B_1$ such that the restrictions $T|_{A_0}$ and $T|_{A_1}$ are bounded linear operators mapping into $B_0$ and $B_1$, respectively. An interpolation space for a Banach couple $(A_0,A_1)$ is a Banach space $A$ such that
1. $A_0 \cap A_1 \hookrightarrow A \hookrightarrow A_0 + A_1$;
2. if $T:A_0 + A_1 \to A_0 + A_1$ is a bounded linear operator such that $T|_{A_0}:A_0 \to A_1$ and $T|_{A_1}: A_1 \to A_1$ are bounded linear operators, then the restriction of $T$ on $A$ is a bounded linear operator into $A$;
and an interpolation functor a functor from the category of Banach couples to the category of Banach spaces that sends Banach couples to corresponding interpolation spaces and transforms morphisms between Banach couples into the corresponding bounded linear operators between sums of Banach spaces. With this terminology, we can view an interpolation theorem as a construction of such a functor.
Since this definition is quite general, it is conceivable to have interpolation theorems that are not quite as nice as the classical ones. To characterize the nicer ones, we define exact interpolation spaces $A$ and $B$ to be interpolation spaces of Banach couples $(A_0,A_1)$ and $(B_0,B_1)$ such that
`$$ \|T\|_{A \to B} \leq \max \left( \|T\|_{A_0 \to B_0} , \|T\|_{A_1 \to B_1} \right)$$`
for all morphisms $T: (A_0,A_1) \to (B_0,B_1)$ and an exact interpolation functor an interpolation functor that produces exact interpolation spaces. Now, the theorem of Aronszajin and Gagliardo states that every interpolation space admits an exact interpolation functor that sends the corresponding Banach couple to the interpolation space. With the aid of category-theoretic methods, we thus have a guarantee that our efforts in finding a particular interpolation theorem will not go wasted.
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Nice example. On a related note, there was some work by Kaijser and Wick-Pelletier (like B-S-D, only two actual people there) on formulating interpolation couples in an even more overtly category-theoretic way. Unfortunately I've forgotten the details and don't have references to hand, but I've always thought it deserved to be better known. – Yemon Choi Dec 22 2011 at 8:13
Perhaps this example is too naive, but one can view the Riesz Representation Theorem categorically as saying that integration with respect to a measure is a natural equivalence of the functor which takes a compact Hausdorff space $X$ and produces the Banach space of finite signed measures on $X$, and the functor which takes and $X$ and produces the dual of $C(X)$. There is a lovely article on this by Hartig: http://www.jstor.org/pss/2975760
Another example that comes to mind is Pontryagin Duality as presented here: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102911979
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Let me expand on the second example of Lian. Category theory proves trivially that the category of profinite abelian groups is dual to the category of discrete torsion abelian groups. Indeed the former is the pro-completion of the category of finite abelian groups and the latter is the ind-completion. So one just needs the trivial fact that the category of finite abelian groups is self-dual.
Similarly, assuming the Peter-Weyl theorem (you can't get around this) one has that the category of compact abelan groups is the pro-completion of the category of compact abelian Lie groups. The category of discrete abelian groups is the ind-completion of the category of finitely generated abelian groups. So the duality between compact abelian groups and discrete abelian groups boils down to the structure theorem for finitely generated abelian groups, the finite case and that Z is dual to the circle.
So category theory organizes these proofs (the proof still boils down to the same nuts and bolts).
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I would suggest that the following three applications of category theory to functional analysis can be useful (they have points of contact with some of the earlier answers): they concern three topics (which are related) and at the least provide a unifying thread---in my opinion, they do more---to many themes in abstract analysis (spaces of measures, distributions, analytic functionals, the Riesz representation theorem and Gelfand-Naimark duality). These are: extensions of categories, free topological vector spaces and extended duality.
Extensions of categories. The basic example is the extension of the category of Banach spaces (with continuous linear mapings as morphisms) to the class of locally convex spaces (Wiener) and convex bornological spaces (Buchwalter and Hogbe-Nlend). The first has, of course, long occupied a place in the mainstream of functional analysis, the latter less so. They can be regarded as the categories obtained by "adding inductive, respectively, projective limits. This informal notion has been formalised in the first edition of the book "Saks spaces and applications to functional analysis". Exactly the same process can be applied to the category of metric spaces (where we obtain that of uniform spaces), that of Banach algebras (convex bornological algebras and locally multiplicatively convex algebras). The examples can be multiplied indefinitely. We mention two further extensions which we shall refer to below---compactological spaces (Buchwalter) (inductive limits of compact spaces) and Saks spaces (adding projective limits to the category of Banach spaces with linear contractions as morphisms---paradoxically, despite the fact that this category {\it has} limits). These example shows that this process often leads to us rediscovering America. However, it does have the advantage over Columbus that we are rediscovering a plethora of continents with a single expedition, often ones which are known but have hardly been investigated so that a rescrutinising may be worthwhile (after all, Columbus himself had been anticipated by native Americans).
Free locally convex spaces. The basic example here is the following. If we start with the unit interval $I$, then we can consider the free vector space over this set. If we then supply it with the finest locally convex topology which agrees with the original one on $I$ and complete it, we obtain a locally convex space which has the universal property that every continuous function from $I$ into a Banach spaces lifts to a unique continuous linear operator thereon (and is characterised by this property). Not surprisingly, this is just the space of Radon measures on $I$. Our point is that this construction can be carried out in an infinity of analogous situations and leads to a unified aproach to a large array of spaces which are of interest in analysis. We mention a small sample---compact spaces (universal property for continuous functions), compactologies and completely regular spaces (bounded, continuous functions), metric spaces (bounded Lipschitz-continuous functions), uniform spaces (bounded uniformly continuous functions), compact rectangles in $n$-space ($C^\infty$-functions), compact manifolds (again $C^\infty$-functions), open subsets of $n$-dimensional complex space (holomorphic functions). These lead to a long list of interesting spaces (of bounded Radon measures, of uniform measures, of distributions, of analytic functionals) and many of their basic properties can be deduced from general principles which arise from this method of construction (most important example, duality theorems). Again, many of these spaces are known but the historical path to their discovery was long and stony. It is of advantage to have a natural unified approach to their construction. Again, there aare further important cases which are indeed known but seem to have passed out of the mainstream despite an obvious demand for them. A particularly important and, in my opinion, unfortunate example (and a perennial favourite for queries in this forum) is the topic of extensions of the Riesz representation, which we shall now discuss.
Extensions of duality. As we have just mentioned, the classical example is the Riesz representation theorem for compacta. It was initially shown by Buck (for locally compact space and later, by other researchers, to the class of completely regular spaces) that this can be extended to the non-compact case by using the so-called strict topology which can be most succinctly described here as the finest locally convex topology on the space $C^b(S)$ of bounded, continuous fuctions on $S$ which agrees with that of compact convergence on the unit ball for the supremum norm. Again a number of queries on this forum suggest that this topic which barely entered into the mainstream despite the prominence of its proponents (Buck, Beurling and Herz---mainly motivated by questions in harmonic analysis) and, sadly, seems to have vanished without a trace. There is a simple and natural scheme at work here. If we have a duality between two of the central catogories of analysis, then we can extend it in an obvious way to one between say the category obtiained by adding inductive limits to the first one and projective limits to the second one. This leads almost automatically to the above extension of the Riesz representation to the case of completely regular spaces (or, better, compactological spaces). If we take as our starting point one of the central dualities of abstract analysis (that between a Banach space and its dual as a Banach space, or, if one wants a symmetric duality, as a Waelbroeck space, between a compact space and the Banach space of continuous functions thereon, resp. the same space regarded as a $C^*$-algebra, between a metric space and the space of bounded, Lipschitz functioons, then we can apply this process to obtain a large classes of extended dualities which are useful in abstract analysis.
Why is this useful? Lack of space prevents an elaborate justification of these three methods but I would like to mention the following version of Occam's rasor. One can speculate that one of the reasons for the fact that many of these extended dualities failed to enter into the mainstream of abstract analysis lies in the fact that to analysists accustomed to the Banach space settings, the structures employed here seemed unattractively elaborate and artifical, not to say baroque. (The strict topology is not only not normable, but is not a member of the accepted "nice" classes of locally convex spaces---Frechet, $DF$-, even barrelled or bornological. Of course, any such extension must necessarily be more elaborate than the original duality and the above considerations show that the ones presented here are the simplest that can succeed in the given situations and so are inevitable.
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Good stuff - much food for thought. Do you know if I could find further details in, say, the North-Holland book of Cooper? ;-) – Yemon Choi Nov 18 at 19:09
very much so. as noted in the text, the stuff on category theory is only in the first edition. – jbc Nov 19 at 8:56
Isn't the Banach-Mazur metric space on norms over ${\mathbb R}^n$ such an application?
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Of course the equivalence of categories mentioned by Eric A. Bunch is true only for commutative $C^\ast$-algebras. There however is a quite similar result for a wider category of non-commutative $C^\ast$-algebras: a continuous-trace $C^\ast$-algebra $A$ with Hausdorff spectrum $X$ is isomorphic to the $C^\ast$-algebra $\Gamma_0(X,\mathcal{A})$ of continuous sections vanishing at $\infty$ of some continuous $C^\ast$-bundle $\mathcal{A}\to X$; the latter is actually a Dixmier-Douady bundle, in the sense that it has typical fiber the algebra $\mathbb{K}(H)$ of compact operators on some separable Hilbert space. This construction yields an equivalence between the category of continuous-trace $C^\ast$-algebras with Hausdorff spectrum and pairs $(X,\mathcal{A})$.
When it comes to study $K$-theory of such $C^\ast$-algebras, the above equivalence may be very useful, for if $A$ corresponds to the pair $(X,\mathcal{A})$, then $K_\ast(A)$ is isomorphic to the twisted $K$-theory ${}^{\mathcal{A}}K^\ast(X)$, which in the finite-dimensional case may be interpreted in terms of geometric objects (twisted vector bundles). For instance, let $G$ be a compact Hausdorff group. Then the dual space of $G$ is homeomorphic to the the spectrum $X$ of the $C^\ast$-algebra $C^\ast(G)$ (which is actually continuous-trace). Now, for $k\in \mathbb{N}$, let $X_k$ be the (open) subspace of $X$ consisting of (equivalence classes of) irreducible representations of $G$ of rank $k$. Then, for each $k$, there is an Azumaya bundle $\mathcal{A}_k\to X_k$ (i.e. a Dixmier-Douady bundle of finite dimension), and there is an isomorphism $K_\ast(C^\ast(G))\cong \bigoplus_k {}^{\mathcal{A}_k} K^\ast(X_k)$ (cf. http://front.math.ucdavis.edu/0201.5207 for a generalization of this example).
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Here are four articles on category theory and analysis (see section IV of that collection). I liked this one: R. Boerger, Fubini's Theorem from a Categorical Viewpoint, manages to conclude Fubini's theorem from the fact that left adjoints preserve coproducts...
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I've only had a superficial look at the paper, but it seems to me that what people usually call Fubini's theorem is an analytical fact that under certain boundedness conditions one can exchange the order of two integrals. Is the paper cited just saying, at the end of the day, that one can exchange the order of two finite sums? Near the bottom of the first page the author says things "since we are not worried about convergence" (as he's in some finite world) -- but surely what people usually call Fubini's theorem is precisely an issue of convergence? – Kevin Buzzard Dec 13 2011 at 21:25
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@Kevin: shush, everyone knows that these kind of technicalities are of no concern to Proper Mathematicians... ;) – Yemon Choi Dec 13 2011 at 22:04
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Further to my previous comment: "unlike Modo, Sergeant Colon did understand the meaning of the word 'irony'. He thought it meant 'something like iron'." – Yemon Choi Dec 14 2011 at 0:25
"Is the paper cited just saying, at the end of the day, that one can exchange the order of two finite sums?" I don't think, the conclusion is just that. If you look at 6.2, they say the integral of the product measure is the double integral over each measure separately (for simple functions). So my take on the category theory, is that it is concluding the product measure acts functorially, as expected by the suspected diagram chase?! So I guess $\int_{A\times B}f(x,y)(\mu_A\times\mu_B)(x,y)=\int_A\int_B f(x,y)\mu_A(x)\mu_B(y)$ is the Fubini theorem. I agree there is no analysis. – Junkie Dec 14 2011 at 5:10
2
A "simple function" is a very restrictive class of function -- for example, it has finite image. This is why I'm wondering whether we're really doing sums, not integrals. – Kevin Buzzard Dec 14 2011 at 7:44
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http://unapologetic.wordpress.com/2008/11/18/the-category-of-representations-of-a-hopf-algebra/?like=1&source=post_flair&_wpnonce=8149a33672
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# The Unapologetic Mathematician
## The Category of Representations of a Hopf Algebra
It took us two posts, but we showed that the category of representations of a Hopf algebra $H$ has duals. This is on top of our earlier result that the category of representations of any bialgebra $B$ is monoidal. Let’s look at this a little more conceptually.
Earlier, we said that a bialgebra is a comonoid object in the category of algebras over $\mathbb{F}$. But let’s consider this category itself. We also said that an algebra is a category enriched over $\mathbb{F}$, but with only one object. So we should really be thinking about the category of algebras as a full sub-2-category of the 2-category of categories enriched over $\mathbb{F}$.
So what’s a comonoid object in this 2-category? When we defined comonoid objects we used a model category $\mathrm{Th}(\mathbf{CoMon})$. Now let’s augment it to a 2-category in the easiest way possible: just add an identity 2-morphism to every morphism!
But the 2-category language gives us a bit more flexibility. Instead of demanding that the morphism $\Delta:C\rightarrow C\otimes C$ satisfy the associative law on the nose, we can add a “coassociator” 2-morphism $\gamma:(\Delta\otimes1)\circ\Delta\rightarrow(1\otimes\Delta)\circ\Delta$ to our model 2-category. Similarly, we dispense with the left and right counit laws and add left and right counit 2-morphisms. Then we insist that these 2-morphisms satisfy pentagon and triangle identities dual to those we defined when we talked about monoidal categories.
What we’ve built up here is a model 2-category for weak comonoid objects in a 2-category. Then any weak comonoid object is given by a 2-functor from this 2-category to the appropriate target 2-category. Similarly we can define a weak monoid object as a 2-functor from the opposite model 2-category to an appropriate target 2-category.
So, getting a little closer to Earth, we have in hand a comonoid object in the 2-category of categories enriched over $\mathbb{F}$ — our algebra $B$. But remember that a 2-category is just a category enriched over categories. That is, between $H$ (considered as a category) and $\mathbf{Vect}(\mathbb{F})$ we have a hom-category $\hom(H,\mathbf{Vect}(\mathbb{F}))$. The entry in the first slot $H$ is described by a 2-functor from the model category of weak comonoid objects to the 2-category of categories enriched over $\mathbb{F}$. This hom-functor is contravariant in the first slot (like all hom-functors), and so the result is described by a 2-functor from the opposite of our model 2-category. That is, it’s a weak monoid object in the 2-category of all categories. And this is just a monoidal category!
This is yet another example of the way that hom objects inherit structure from their second variable, and inherit opposite structure from their first variable. I’ll leave it to you to verify that a monoidal category with duals is similarly a weak group object in the 2-category of categories, and that this is why a Hopf algebra — a (weak) cogroup object in the 2-category of categories enriched over $\mathbb{F}$ has dual representations.
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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://gilkalai.wordpress.com/2009/09/21/eran-nevo-the-g-conjecture-iii-algebraic-shifting/?like=1&source=post_flair&_wpnonce=756b7883e4
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Gil Kalai’s blog
## (Eran Nevo) The g-Conjecture III: Algebraic Shifting
Posted on September 21, 2009 by
This is the third in a series of posts by Eran Nevo on the g-conjecture. Eran’s first post was devoted to the combinatorics of the g-conjecture and was followed by a further post by me on the origin of the g-conjecture. Eran’s second post was about the commutative-algebra content of the conjecture. It described the Cohen-Macaulay property (which is largely understood and known to hold for simplicial spheres) and the Lefshetz property which is known for simplicial polytopes and is wide open for simplicial spheres.
## The g-conjecture and algebraic shifting
### Squeezed spheres
Back to the question from last time, Steinitz showed that
any simplicial 2-sphere is the boundary of a convex 3-polytope.
However, in higher dimension
there are many more simplicial spheres than simplicial polytopes,
on a fixed large number of vertices. We will need Kalai’s squeezed spheres (of dimension $\geq 4$) which demonstrate this.
It is not known whether the hard Lefschetz property holds for all simplicial spheres. Recently Satoshi Murai showed that that the hard Lefschetz property holds for squeezed spheres. This gives a refinement of Billera-Lee part of the $g$-theorem in terms of generic initial ideals. We will phrase it more combinatorially, via algebraic shifting.
### Symmetric algebraic shifting
Symmetric algebraic shifting is an operator on simplicial complexes, $K\rightarrow \Delta(K)$, defined by Kalai. $\Delta(K)$ carries the same information as the generic initial ideal of $I_K$.
$\Delta(K)$ has the same $f$-vector as $K$, and it is shifted (see our earlier post on shifting). What properties of $K$ can be read off $\Delta(K)$? Well, the hard Lefschetz property can be read!
Let $\Delta(d,n)$ be the maximal simplicial complex on the vertex set $[n]$ with all maximal faces of the same dimension $d-1$ (such complex is called pure such that it doesn’t contain any of the sets $T^d_d, T^d_{d-1}...,T^d_{\lceil d/2\rceil}$, where
$T^d_{d-k}=\{k+2,k+3,...,d-k,d-k+2,d-k+3,...,d+2\}$, $0\leq k\leq \lfloor d/2\rfloor.$
For example, $T^3_2=\{2,3,5\}$ and $T^3_1=\{4,5\}$ so the maximal faces in $\Delta(3,n)$ are the ones of the form $\{1,2,m\}$ or $\{1,3,m\}$ or $\{2,3,4\}$. In particular, $\Delta(d,n)$ is shifted, and actually it equals $\Delta(C(d,n))$, the symmetric shifting of the boundary of a cyclic polytope.
Now, our simplicial $(d-1)$-sphere on $n$ vertices $K$ has the hard Lefschetz property iff
$\Delta(K)\subseteq \Delta(d,n)$.
Murai showed the following: suppose that a simplicial complex $L$ is pure $(d-1)$-dimensional on $n$ vertices, with $h(L)$ symmetric and $L\subseteq \Delta(d,n)$. Then there exists a (squeezed) sphere $K$ such that $\Delta(K)=L$. $K$ is the squeezed sphere which Kalai constructed from the half-dimensional skeleton of $L$. Given an $M$-vector $g$, the Billera-Lee polytope corresponds to this construction, where the half-dimensional skeleton of $L$ is the compressed complex (w.r.t. the rev-lex order) with $f$-vector equals $g$.
### Van Kampen-Flores complexes
Kalai and Sarkaria (independently) conjectured that if a simplicial complex $L$ on $n$ vertices can be embedded in the $(d-1)$-sphere, then
$\Delta(L)\subseteq \Delta(d,n)$.
In particular, for $L$ a triangulation of the $(d-1)$-sphere, the $g$-conjecture would follow.
Note that $T^{2d+1}_d=\{d+3,d+4,...,2d+3\}$. $T^{2d+1}_d\in \Delta(L)$ iff the $d$-skeleton of the $(2d+2)$-simplex, $\sigma^{2d+2}_{\leq d}$, a.k.a the van Kampen-Flores complex, is contained in $\Delta(L)$, because $\Delta(L)$ is shifted. It is known that $\sigma^{2d+2}_{\leq d}$ does not embed in $S^{2d}$, and we would like to conclude that if $\sigma^{2d+2}_{\leq d}\subseteq \Delta(L)$ then $L$ does not embed in $S^{2d}$. Building on a result of Ed Swartz, if we could prove it, then it would follow that the $g$-vector of any piecewise linear sphere is an $M$-vector!
Say that a simplicial complex $H$ is a minor of a simplicial complex $L$ if you can obtain $H$ from $L$ by successive deletions and (admissible) contractions. Here deletion means taking a subcomplex, and contraction means identifying two vertices \$u,v\$ which satisfy the link condition, i.e. $lk(v)\cap lk(u) = lk(\{v,u\})$. If $H,L$ are one dimensional, this does recover the usual definition of minors for graphs.
We can show that if $\sigma^{2d+2}_{\leq d}$ is a minor of $L$ then $L$ does not embed in $S^{2d}$. Is it true that $\sigma^{2d+2}_{\leq d}\subseteq \Delta(L)$ implies that $\sigma^{2d+2}_{\leq d}$ is a minor of $L$? The answer is Yes for $d=0,1$, and we don’t know the situation for $d>1$. If it is true, then the $g$-conjecture for PL-spheres would follow.
We just mentioned PL-spheres. Can we solve the $g$-conjecture for special families of spheres? And what about other manifolds?? Next time…
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This entry was posted in Combinatorics, Convex polytopes, Guest blogger, Open problems and tagged g-conjecture, Shifting. Bookmark the permalink.
### 2 Responses to (Eran Nevo) The g-Conjecture III: Algebraic Shifting
1. Cialis says:
mlkMuV Excellent article, I will take note. Many thanks for the story!
2. Pingback: Satoshi Murai and Eran Nevo proved the Generalized Lower Bound Conjecture. | Combinatorics and more
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http://math.stackexchange.com/questions/264355/forecasting-lottery/264360
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# Forecasting Lottery
I do understand that all lottery has a negative mathematical expectation but I am wondering, if we have a set of historical data of the winning numbers, is it possible to increase the winning chance?
I also do understand that each round is a independent event but according to this theory where you roll a dice $n$ times, you each side should get $n/6$ times as $n \to \infty$.
So my questions is, does such a function exist to increase the winning chances of a lottery?
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Whats a negative mathematical expectation? – Ethan Dec 24 '12 at 1:38
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what do you mean?? (o)(o) – wtsang02 Dec 24 '12 at 1:40
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What he means by a "negative mathematical expectation" is a negative expected value of the winnings. – Joe Z. Dec 24 '12 at 1:44
## 4 Answers
While Joe Zeng's answer is certainly theoretically correct, as the practice shows sometimes it is quite possible to increase the chances to win a lottery. All of us (well, most) remember the story of Smoke Bellew by Jack London, that noticed that one roulette was dried and got some numbers with greater probability than others. Here is another story which shows that good knowledge of statistics might be very helpful. I will leave all the details to my link, the story is really worth reading.
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This is true; I was assuming that the reader was talking about a theoretically fair lottery, where every result was equally likely. – Joe Z. Dec 24 '12 at 2:18
No.
Because different events are (meant to be) independent, the probability of each lottery being a certain result is unaffected by previous results.
The theory you are referring to is the Law of Large Numbers, which is only a statistical trend (rule) and not a certainty.
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In many lotteries, one shares the grand prize with all the people who picked the winning set of numbers.
Lotteries in the main pick the winning numbers by a process that simulates randomness well. However, the number choices by lottery players are anything but random. If one stays away from the popular sorts of choices, one can increase one's probability of not having to share, in the highly unlikely case that one happens to pick the winning numbers.
A number of lotteries maintain and publish statistics on the numbers that bettors have chosen.
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can you explain "popular sorts of choices" – wtsang02 Dec 24 '12 at 8:10
I once saw a list of popular choices, you can probably hunt them down. – André Nicolas Dec 24 '12 at 8:17
About rolling the die, I will explain by example: Suppose that you have a six sided die, numered 1 - 6. The "expected value" of a roll is 3.5. This means that if you roll the die many times, say 1000, and write down the number that you get each time, the numbers will average out to 3.5. It does not mean that you can forecast that on the next roll, you will have the number 3.5 come up. If the person rolling the die is trying to make it land on a certain number, then it is equally likely to land on any one side when rolled.
The organisers of the lottery also calculate expected value. If many people play the lottery, then on average, they are making a profit.
I have read a story about a team calculating the physics of a roulette wheel and placing bets accordingly.
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http://math.stackexchange.com/questions/250687/quasi-linear-time-fully-homomorphic-encryption-using-p-adic-ring-homomorphism
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# Quasi-linear time fully homomorphic encryption using p-adic ring homomorphism
I recently encountered a breakthrough in FHE crypto, which claims to have a literally quasi-linear time FHE without any "lambda" factor in the keys and no noise in the cipher-text.
This fully homomorphic encryption works like for addition like $E(m_1)·E(m_2) = E(m_1+m_2)$ and for multiplication like $E(m_1)^{m_2} = E(m_1·m_2)$ ; they say they achieved it using p-adic exponential.
My Question is: What is exactly p-adic exponential?
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4
What you describe is not fully homomorphic. Notice that $m_2$ in the second equation is not encrypted. – mikeazo Dec 2 '12 at 23:00
Hasn't he defined it in P1.3? Were you looking for more than that? – Amzoti Dec 4 '12 at 13:42
2
A warning about the paper: it doesn't sound like the author knows much about crypto. There's the misrepresentation of FHE that Mike pointed out; in addition, he claims his system is Informationally Secure (which is impossible for any public key system); he claims that a message cannot be uniquely decrypted; if this were true, this would imply that someone with the private key cannot decrypt it. His claim of $O(n \log n)$ time is also bogus. I would approach this system only with extreme caution. – poncho Dec 4 '12 at 15:39
## 1 Answer
To answer the question (and ignoring the fluff), the $p$-adic exponential is a partial function on the $p$-adic numbers defined by the power series
$$\exp(x) = \sum_{i=0}^{\infty} \frac{x^n}{n!}$$
This power series converges if and only if $x$ is a $p$-adic integer divisible by $p$. It satisfies many of the properties you'd expect. For example, $\exp(x+y) = \exp(x) \exp(y)$, and it has an inverse, the $p$-adic logarithm.
If you're not familiar with the $p$-adic numbers, we can approximate them with modular arithmetic. If $x$ is an integer such that $x \equiv 0 \pmod p$, then the above power series also makes sense in the arithmetic of integers modulo $p^n$. For an example with the $3$-adic logarithm:
$$\begin{align} \exp(3) &\equiv 1 + 3 + \frac{9}{2} + \frac{27}{6} + \cdots& \pmod{27} \\ &\equiv 1 + 3 + \frac{9}{2} + \frac{9}{2} + 0& \pmod{27} \\ &\equiv 1 + 3 + 9 \cdot 14 + 9 \cdot 14 & \pmod{27} \\ &\equiv 13& \pmod{27} \end{align}$$
Note the remaining terms (indicated by $\cdots$) are all divisible by $27$, which is why I replaced withm with $0$ in the above calculation. And that $2$ and $14$ are inverses modulo $27$: i.e. $2 \cdot 14 \equiv 1 \pmod{27}$.
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http://nrich.maths.org/6507
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### Ball Bearings
If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.
### Air Routes
Find the distance of the shortest air route at an altitude of 6000 metres between London and Cape Town given the latitudes and longitudes. A simple application of scalar products of vectors.
### Epidemic Modelling
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.
# Crystal Symmetry
##### Stage: 5 Challenge Level:
Caesium Chloride assumes a 'body centred cubic structure' in which each caesium ion is surrounded by $8$ chlorine ions located at the vertices of a cube, and vice-versa.
Mathematically, we can choose cartesian coordinates such that the ions lie on the integer lattice comprising the points $(l, m, n)$ with $l, m,n$ integers. The caesium ions are located at points with $l+m+n$ even and the chlorine ions at points with $l-m-n$ odd.
To start, make sure that you understand why the lattice points represent correctly a body centred cubic structure.
I can transform the points ${\bf v}$ in a lattice by multiplying by a constant matrix $M$ or adding a constant vector ${\bf c}$ through
{\bf v} \rightarrow {\bf v} +{\bf c}\text{ or } {\bf v} \rightarrow M{\bf v}
Which of the following vectors and matrices preserve exactly the structure of the caesium chloride when they transform the lattice?
$$M =\pmatrix{1&0&0\cr 0&0&-1\cr 0&1&0}\,,\pmatrix{3&0&0\cr 0&1&0\cr 0&0&2}\,,\pmatrix{1&1&-1\cr -1&1&1\cr 1&-1&1}$$
$${\bf c} = \pmatrix{1\cr 0\cr 0}\,,\pmatrix{1\cr 1\cr 0}\,,\pmatrix{2\cr 2\cr 2}\,,\pmatrix{4\cr -2\cr -8}$$
In each case prove why the crystal structure is preserved, or explain what goes wrong.
Using your geometrical intuition as a guide, investigate the types of transformations which will leave the crystal invariant.
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://math.stackexchange.com/questions/182639/the-euler-characteristic-of-s1-a-paradox
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# The Euler characteristic of S1: a paradox
The Euler characteristic of $S^1$ should be equal to that of a triangle, the two being homeomorphic to each other. But it is $1$ for the triangle and zero for $S^1$; how does it make sense?
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9
how are you getting that a triangle has euler characteristic 1? – Eric O. Korman Aug 14 '12 at 23:21
It depends on your definition of a triangle. How many faces does your triangle have? – M Turgeon Aug 14 '12 at 23:21
## 1 Answer
If by “triangle” you mean “three points and the segments connecting each pair of them”, then the Euler characteristic of it is $0$ (it has three edges and three vertices).
If by “triangle” you mean the same as above, plus the part of plane enclosed by the segment, then it is certainly not homeomorphic to $S^1$, but to a two-dimensional disk $D^2$, and it has Euler characteristic $1$, just like the disk.
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http://mathoverflow.net/questions/62509/integral-identity-for-legendre-polynomials
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## Integral identity for Legendre polynomials
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
How does one prove the following integral identity, where $P_n(x)$ is the $n$th Legendre polynomial? ```$$
\int_0^1 P_n(2t^2-1) dt = \frac{(-1)^n}{2n+1}
$$```
### Notes & Background
• A variant of this appears in, for instance, Erdelyi et al "Higher transcendental functions" 10.10(49), but with nothing in the way of explanation.
• This comes up in harmonic analysis on $U(3)$, when comparing Gelfand-Tseltin bases associated to different choices of nested sequences $U(3) \supset U(2) \supset U(1)$.
• Eventually, I'll be looking for a $q$-analogue, related to harmonic analysis on $U_q(3)$, so a proof that will transport well would be my true desire.
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2
By means of power series, this transforms into something like $\int\limits_0^1\dfrac{1}{\sqrt{1-2\left(2x^2-1\right)t+t^2}}dx=\mathrm{arctan}t$ (note that I have changed your $t$ into an $x$). Maybe it's easier this way? – darij grinberg Apr 21 2011 at 9:30
2
$$\int_0^1 \frac 1 {\sqrt{1-2(2x^2-1)t^2+t^4}}dx = \frac {\arctan t} t$$ might work even better. – thei Apr 21 2011 at 10:30
## 1 Answer
Nice idea. As far as I'm concerned, the above comments are "answers", since they check out. I might as well record the details: ```$$
\int_0^1 \frac{dx}{\sqrt{1-2(2x^2-1)t + t^2}}
= \frac{1}{2\sqrt{t}} \int_0^1 \frac{dx}{\sqrt{\frac{(1+t)^2}{4t} -x^2}} = \frac{1}{2\sqrt{t}}\arcsin\left(\frac{\scriptstyle 2\sqrt{t}}{\scriptstyle1+t}\right).
$$``` The half-angle formula for $\sin$ reduces this to $$\frac{1}{\sqrt{t}} \arcsin \sqrt{\frac{t}{1+t}} = \frac{1}{\sqrt{t}} \arctan \sqrt{t},$$ which has the desired power series expansion.
Thanks.
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http://math.stackexchange.com/questions/45418/solving-left-frac-2x-64-right-le-5-for-x?answertab=oldest
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# Solving $\left| \frac{-2x-6}{4} \right| \le 5$ for $x$
Say I have a statement like:
$$\left| \frac{-2x-6}{4} \right| \le 5.$$
And I want to find the closed interval form of $x$. i.e. I want to know what the maximum and minimum $x$ can be. How do I do this?
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## 3 Answers
You can get rid of the - inside of the absolute value and have the inequality $${|2x + 6|\over 4} \le 5$$ or $$|x + 3| < 10.$$ Now you can use the fact that for real numbers $a$ and $b$, we have ${\rm distance}(a,b) = |a - b|$ and observe that $|x + 3| = {\rm distance}(x, -3)$.
You now have the inequality ${\rm distance}(x, -3) < 10$, which yields $-13 < x < 7$. This is a nice geometric way to think about the problem.
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You can translate the absolute value to two statements: $\frac{-2x-6}{4} \le 5$ and $-\frac{-2x-6}{4} \le 5$. Each one gives one end of the interval. Can you take it from here?
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I think so, am I trying to solve for 'x' in both of those? – InBetween Jun 15 '11 at 1:22
@InBetween, yes (what then?). – Gerry Myerson Jun 15 '11 at 1:35
Sweet! Got it, thanks for the help! – InBetween Jun 15 '11 at 3:02
$|z|\le a$ is the same thing as $-|a|\le z\le|a|$. If you take $z$ to be $(-2x-6)/4$, and $a$ to be $5$, that should get you started.
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http://mathoverflow.net/questions/tagged/characters
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## Tagged Questions
3answers
253 views
### Characters of p-groups
Berkovich mentioned the following result of Mann in his book on p-groups: The number of nonlinear irreducible characters of given degree in a p-group is divided by p-1. Do you kn …
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### Permutation character of the symmetric group on subsets of certain size
The symmetric group $S_n$ acts on $[n]:=\{1,\ldots,n\}$, thereby inducing an action on the set $$\wp_k(n)=\{\: A\subseteq[n] \::\: \#A=k \:\}$$ of subsets of cardinality $k$, simpl …
1answer
154 views
### Character table entries and sums of roots of unity
It is well-known that the entries of the character table of a finite group are sums of roots of unity. Question: Is the converse true? Explicitly, given \$z\in \mathbb{Z}[\mu_\inf …
1answer
356 views
### When does the modulus of a sum of an integer and an algebraic integer equal an integer?
Let say Z is a sum of n-roots of unity and thus an algebraic integer, and D is an rational integer. If |z+D| is an integer, what can we conclude regarding Z? can we say |Z| is in …
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### A problem about primitive Dirichlet characters
Show that for a primitive character $\chi\bmod q$, $q\gt1$ $\sum_{|n|\le q}(1-|n|/q)\chi(n)e(na)\ll\sqrt q$ uniformly in $a$, where $e(na)=\exp(2\pi ina)$ Thank you for any co …
0answers
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### Class function counting solutions of equation in finite group: when is it a virtual character?
Let $w=w(x_1,\dots,x_n)$ be a word in a free group of rank $n$. Let $G$ be a finite group. Then we may define a class function $f=f_w$ of $G$ by f_w(g) = |\{ (x_1,\dots, x_n)\in …
0answers
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### Interplay between two definitions of the transfer homomorphism.
The transfer homomorphism can be defined in a couple of ways. For the purposes of this question, assume that all groups are finite. Defintion 1. Let \$1\leqslant K'\leqslant L …
0answers
308 views
### Degrees of irreducible characters of groups of order 48 [closed]
Hello i wonder if there is a simple argument to show that no group of order 48 can have an irreducible character of degree larger than 4? Thanks, Karim
2answers
672 views
### Two Definitions of “Character” of topological groups
When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows: Let $G$ be a topological group. A characte …
1answer
146 views
### Induced character for non-injective homomorphisms
Any group homomorphism $\phi\colon H\to G$ gives rise to an induction/restriction adjunction between $G$-representations and $H$-representations: \hom_G(\phi_! M, N) \cong \hom_ …
2answers
456 views
### Character group of Frobenius kernels
Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic $p$ (e.g., $G=SL_n(k)$). Then $G$ is equal to its derived subgroup $[G,G]$. Consequ …
2answers
314 views
### Finite, abelian, yet “fugitive” orthogonal subgroups
A popular concept in quantum computation, used extensively to design algorithms for finite-abelian-groups, are the so-called orthogonal subgroups Let \$G=\mathbb{Z}_{d_1}\times\ …
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1k views
### Introduction to L-series and Dirichlet characters?
I'm looking for an introductory text on Dirichlet characters and the L-series of a field K, specifically for quartic extensions of $\mathbb{Q}$. I have Davenport's Multiplicative …
2answers
362 views
### When is a class function on a group G (finite abelian) into the rational numbers Q an element of the rational representation ring of G?
Given a class function $f: G \to \mathbb Q$, where $G$ is a finite abelian group, is there an easy way to decide whether $f$ is an element of the rational representation ring \$R_{\ …
2answers
483 views
### Is 2-sylow subgroup of a rational group also a rational group?
As we know, a finite group $G$ is a rational group if $\chi (g)\in\mathbb{Q}$, where $\chi$ is every irreducible charahter and $g\in G$. I have an interesting question that is "Is …
15 30 50 per page
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http://mathoverflow.net/questions/47474/function-that-sums-to-zero-over-cube-vertices
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## function that sums to zero over cube vertices
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Does anyone have an answer to the three-dimensional analogue of the 2009 Putnam Competition A1 problem, viz., if $f\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ satisfies $\sum_{i=1}^8 f(a_i) = 0$ whenever $a_1, \ldots, a_8$ are the vertices of a cube, must $f$ be identically zero?
A few thoughts: Since the cube is not self-dual above dimension $2$, the solution (well, at least, my solution) to the $2$-dimensional problem doesn’t generalize. To try to show that the answer to the $3$-dimensional problem is “no,” one might try letting $\Omega$ be the set of ordered pairs $(X, f_X)$ where $X \subseteq \mathbb{R}^3$ is a subset and $f_X\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ is a function that sums to zero over any eight points of $X$ that form the vertices of a cube. Ordering $\Omega$ in the usual way, we can invoke Zorn’s Lemma to obtain a maximal $(X, f_X) \in \Omega$. If $X \neq \mathbb{R}^3$, then every $x \in \mathbb{R}^3 \setminus X$ must, in two “incompatible” ways, be the eighth vertex of a cube whose remaining vertices lie in $X$. But I don't see how a contradiction arises from this (and of course one could make the same argument in two dimensions, where a contradiction does not arise).
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## 1 Answer
First prove that the sum of the values of your function at the vertices of any regular tetrahedron is zero. You can do this by considering a $2\times 2\times 2$ cube made out of $8$ smaller cubes that you can color in a checkerboard fashion. Sum all the vertices of the black cubes and subtract the sum of the vertices of all the white cubes.
Second you can show that the sum of the values at the vertices of a $x\times x\sqrt{2}$ rectangle is always zero. You need to consider two tetrahedra sharing an edge for this.
From here it is very similar to the Putnam problem you mention. (In fact some solutions apply in the same exact way.)
P.S. This shows that it is enough to have the information only for cubes whose sides are parallel to the axis.
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1
Nice! It seems that at step 2, we could alternatively consider two tetrahedra sharing a face to conclude that the function must be constant, hence zero. – Greg Marks Nov 27 2010 at 3:10
You can find solutions to Putnam problems at amc.maa.org/a-activities/a7-problems/…. – Richard Stanley Nov 27 2010 at 3:15
So, then, what about in dimension $n > 3$? – Greg Marks Nov 27 2010 at 3:52
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http://math.stackexchange.com/questions/220462/conformal-relation-for-2-dim-lorentz-space-times
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# Conformal relation for 2-dim Lorentz space-times
I have two 2-dimensional space-times ($\mathbb{S}^1\times\mathbb{R}$) with signature $(-,+)$. One of them is flat the other one has non-vanishing curvature (Riemann tensor), both have vanishing Ricci tensor. But they seem to have a similar global and causal structure. Of course, because of the 2-dimensional case they are local conformally flat.
I am looking for a global relation between them that could explain the similar causal and global structure and I think that a (global) conformal transformation would be a possible approach.
********************Edit in response to comments*******************
The two metrics in question are $$ds^{2}=Td\psi^{2}+2dTd\psi \quad\text{ defined on }\quad S^{1}\times\mathbb{R}$$ and $$ds^{2}=-(\frac{2m}{r}-1)d\nu^{2}+2d\nu dr \quad\text{ defined on }\quad S^{1}\times(0,\infty).$$
Note that $\psi$ and $\nu$ are the according periodic variables.
I already could calculate the local conformal relation. But what about the global relation?
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I don't really know any way to answer this, but what if you look at possible obstructions to construct a global map form a cover that gives you a local conformal equivalence to the Lorentz space at each chart? Was there any attempt to do this? – Yuri Vyatkin Oct 25 '12 at 5:29
1
A friend of mine gave to me two examples of metrics on $S^1 \times \Bbb R$, one is $d\theta^2-dx^2$, and the other is $-d\theta^2 + dx^2$. They are not globally conformally equivalent because the former does not admit closed timelike curves (it is causal), while the former does. Therefore, to show the existence you need to specify the metrics that you are using. – Yuri Vyatkin Nov 28 '12 at 2:48
## 1 Answer
The two metrics in question are $$ds^{2}=Td\psi^{2}+2dTd\psi \text{ defined on } S^{1}\times\mathbb{R}$$ and $$ds^{2}=-(\frac{2m}{r}-1)d\nu^{2}+2d\nu dr \text{ defined on } S^{1}\times(0,\infty).$$ Note that $\psi$ and $\nu$ are the according periodic variables.
I already could calculate the local conformal relation. But what about the global relation?
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This should have been an edit to the question, but apparently you have multiple unregistered accounts. – user53153 Feb 8 at 6:06
– Willie Wong♦ Feb 8 at 8:47
Now, while I copied the text verbatim into your question statement, there seems to be some typos on your part on the definition of your metrics: a metric is a nondegenerate quadratic form on the space of tangent vectors. Yet neither of your "metrics" are quadratic forms. The first one contains $T\psi^2$ which is a pure scalar term, and the latter one contains a linear term $(1-2m/r) d\nu$. – Willie Wong♦ Feb 8 at 8:53
Also, you should probably state which of the coordinates is the one in the $\mathbb{S}^1$ direction, and which is in the $\mathbb{R}_{(+)}$ direction. – Willie Wong♦ Feb 8 at 8:57
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http://en.wikipedia.org/wiki/Hydrostatic_pressure
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# Fluid statics
(Redirected from Hydrostatic pressure)
Table of Hydraulics and Hydrostatics, from the 1728 Cyclopaedia
Fluid statics or hydrostatics is the branch of fluid mechanics that studies fluids at rest. It embraces the study of the conditions under which fluids are at rest in stable equilibrium; and is contrasted with fluid dynamics, the study of fluids in motion.
Hydrostatics is fundamental to hydraulics, the engineering of equipment for storing, transporting and using fluids. It is also relevant to geophysics and astrophysics (for example, in understanding plate tectonics and the anomalies of the Earth's gravitational field), to meteorology, to medicine (in the context of blood pressure), and many other fields.
Hydrostatics offers physical explanations for many phenomena of everyday life, such as why atmospheric pressure changes with altitude, why wood and oil float on water, and why the surface of water is always flat and horizontal whatever the shape of its container.
## History
Some principles of hydrostatics have been known in an empirical and intuitive sense since antiquity, by the builders of boats, cisterns, aqueducts and fountains. Archimedes is credited with the discovery of the mathematical law that bears his name, that relates the buoyancy force to the volume and density of the displaced fluid. The Roman engineer Vitruvius warned readers about lead pipes bursting under hydrostatic pressure[1]
The concept of pressure and the way it is transmitted by fluids were formulated by the French mathematician and philosopher Blaise Pascal in 1647.
## Pressure in fluids at rest
Due to the fundamental nature of fluids, a fluid cannot remain at rest under the presence of a shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as an infinitesimally small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. If this were not the case, the fluid would move in the direction of the resulting force. Thus, the pressure on a fluid at rest is isotropic; i.e., it acts with equal magnitude in all directions. This characteristic allows fluids to transmit force through the length of pipes or tubes; i.e., a force applied to a fluid in a pipe is transmitted, via the fluid, to the other end of the pipe. This principle was first formulated, in a slightly extended form, by Blaise Pascal, and is now called Pascal's law.
### Hydrostatic pressure
See also: Vertical pressure variation
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity.[2] A fluid in this condition is known as a hydrostatic fluid. The hydrostatic pressure can be determined from a control volume analysis of an infinitesimally small cube of fluid. Since pressure is defined as the force exerted on a test area (p = F/A, with p: pressure, F: force normal to area A, A: area), and the only force acting on any such small cube of fluid is the weight of the fluid column above it, hydrostatic pressure can be calculated according to the following formula:
$p(z)=\frac{1}{A}\int_{z_0}^z dz' \iint\limits_A dx' dy'\, \rho (z') g(z') = \int_{z_0}^z dz'\, \rho (z') g(z')$,
where:
• p is the hydrostatic pressure (Pa),
• ρ is the fluid density (kg/m3),
• g is gravitational acceleration (m/s2),
• A is the test area (m2),
• z is the height (parallel to the direction of gravity) of the test area (m),
• z0 is the height of the zero reference point of the pressure (m).
For water and other liquids, this integral can be simplified significantly for many practical applications, based on the following two assumptions: Since many liquids can be considered incompressible, a reasonably good estimation can be made from assuming a constant density throughout the liquid. (The same assumption cannot be made within a gaseous environment.) Also, since the height h of the fluid column between z and z0 is often reasonably small compared to the radius of the Earth, one can neglect the variation of g. Under these circumstances, the integral boils down to the simple formula:
$\ p = \rho g h,$
where h is the height z − z0 of the liquid column between the test volume and the zero reference point of the pressure. Note that this reference point should lie at or below the surface of the liquid. Otherwise, one has to split the integral into two (or more) terms with the constant ρliquid and ρ(z')above. For example, the absolute pressure compared to vacuum is:
$\ p = \rho g H + p_\mathrm{atm},$
where H is the total height of the liquid column above the test area the surface, and patm is the atmospheric pressure, i.e., the pressure calculated from the remaining integral over the air column from the liquid surface to infinity.
Hydrostatic pressure has been used in the preservation of foods in a process called pascalization.[3]
In medicine, hydrostatic pressure in blood vessels is the pressure of the blood against the wall. It is the opposing force to oncotic pressure.
### Atmospheric pressure
Statistical mechanics shows that, for a gas of constant temperature, T, its pressure, p will vary with height, h, as:
$\ p (h)=p (0) e^{-Mgh/kT}$
where:
• g is the acceleration due to gravity
• T is the absolute temperature
• k is Boltzmann constant
• M is the mass of a single molecule of gas
• p is the pressure
• h is the height
This is known as the barometric formula, and may be derived from assuming the pressure is hydrostatic.
If there are multiple types of molecules in the gas, the partial pressure of each type will be given by this equation. Under most conditions, the distribution of each species of gas is independent of the other species.
### Buoyancy
Main article: Buoyancy
Any body of arbitrary shape which is immersed, partly or fully, in a fluid will experience the action of a net force in the opposite direction of the local pressure gradient. If this pressure gradient arises from gravity, the net force is in the vertical direction opposite that of the gravitational force. This vertical force is termed buoyancy or buoyant force and is equal in magnitude, but opposite in direction, to the weight of the displaced fluid. Mathematically,
$F = \rho g V$
where ρ is the density of the fluid, g is the acceleration due to gravity, and V is the volume of fluid directly above the curved surface.[4] In the case of a ship, for instance, its weight is balanced by pressure forces from the surrounding water, allowing it to float. If more cargo is loaded onto the ship, it would sink more into the water – displacing more water and thus receive a higher buoyant force to balance the increased weight.
Discovery of the principle of buoyancy is attributed to Archimedes.
### Hydrostatic force on submerged surfaces
The horizontal and vertical components of the hydrostatic force acting on a submerged surface are given by the following:[4]
$F_H = p_cA$
$F_V = \rho g V$
where:
• pc is the pressure at the center of the vertical projection of the submerged surface
• A is the area of the same vertical projection of the surface
• ρ is the density of the fluid
• g is the acceleration due to gravity
• V is the volume of fluid directly above the curved surface
## Liquids (fluids with free surfaces)
Liquids can have free surfaces at which they interface with gases, or with a vacuum. In general, the lack of the ability to sustain a shear stress entails that free surfaces rapidly adjust towards an equilibrium. However, on small length scales, there is an important balancing force from surface tension.
### Capillary action
When liquids are constrained in vessels whose dimensions are small, compared to the relevant length scales, surface tension effects become important leading to the formation of a meniscus through capillary action. This capillary action has profound consequences for biological systems as it is part of one of the two driving mechanisms of the flow of water in plant xylem, the transpirational pull.
### Hanging drops
Without surface tension, drops would not be able to form. The dimensions and stability of drops are determined by surface tension.The drop's surface tension is directly proportional to the cohesion property of the fluid.
## See also
Continuum mechanics
Laws
Scientists
Look up hydrostatics in Wiktionary, the free dictionary.
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http://mathoverflow.net/questions/12606/universal-covering-space-of-wedge-products
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## Universal Covering Space of Wedge Products
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Today I was studying for a qualifying exam, and I came up with the following question;
Is there a simple description in terms of the subspaces universal covers for the universal cover of a wedge product?
This question came about after calculating universal covers of the wedge of spheres ($\mathbb{S}^1 \vee\mathbb{S}^1$ and $\mathbb{S}^1 \vee\mathbb{S}^n$) and the wedge of projective space with spheres. In these cases, the universal cover looks like the cross product of the sheets of the universal covers of each space in the wedge.
For the case of wedging two spheres, we can use the fact that $\pi_{n\geq2}\left(U\right)$ is isomorphic to $\pi_{n\geq2}\left(X\right)$ for $U$ covering $X$.
I googled around a bit to try and find something, but nothing appeared.
Thanks in advance!
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4
The universal covering of $S^1\vee S^1$ is quite different from the cross product of the universal covering spaces of the factors. – Mariano Suárez-Alvarez Jan 22 2010 at 4:59
1
(By the way, the wedge of two spaces is usually denoted \vee in LaTeX.) – Mariano Suárez-Alvarez Jan 22 2010 at 5:00
Mariano, I understand that it is the fractal "snowflake". I did not mean for that comment to be taken literally, however it looks somewhat like it in the sense that in every sheet of the helix, we connect another helix and at every sheet of it... etc. Perhaps I shouldn't of included that idea, it was mostly rough thinking. Thanks for pointing this out though. Also, thanks for the \vee, I was wondering about that :) – B. Bischof Jan 22 2010 at 5:07
1
Typically in the context of topology, $\vee$ is the wedge sum, and $A\wedge B := (A\times B)/(A\vee B)$ is called the smash product. Mariano, just for the record, the exterior product of modules is called the wedge product and uses $\wedge$. This notation has been around since the 1930s, long before the wedge sum notation. So I guess it's fair in this case to blame the topologists. – Harry Gindi Jan 22 2010 at 9:22
1
When our spaces are not pointed, the smash product becomes a specialized type of pushout, but I've only seen this used once in HTT, so I can't fill you in on the details, so to speak. – Harry Gindi Jan 22 2010 at 9:37
show 3 more comments
## 1 Answer
If $X$ and $Y$ are two reasonable spaces with universal covers $\tilde{X}$ and $\tilde{Y}$, there is a nice picture of the universal cover $\widetilde{X \vee Y}$ which has the combinatorial pattern of an infinite tree. The tree is bipartite with vertices labeled by the symbols $X$ and $Y$. The edges from an $X$ vertex are bijective with the fundamental group $\pi_1(X)$, and likewise for $Y$ vertices and $\pi_1(Y)$. To make $\widetilde{X \vee Y}$, replace each $X$ vertex by $\tilde{X}$ and each $Y$ vertex by $\tilde{Y}$. The base point of $X$ lifts to $|\pi_1(X)|$ points in $\tilde{X}$, and likewise for $Y$. In $\widetilde{X \vee Y}$, copies of $\tilde{X}$ are attached to copies of $\tilde{Y}$ at lifts of base points. For example, if $X = Y = \mathbb{R}P^2$, then the tree is an infinite chain and $\widetilde{X \vee Y}$ is an infinite chain of 2-spheres.
This tree picture nicely and dramatically generalizes to Bass-Serre theory.
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5
Your \wedge's should be \vee's. That's LaTeX's worst naming convention, to my knowledge :P – Mariano Suárez-Alvarez Jan 22 2010 at 7:21
Thanks! Another excellent answer. Also, thanks for the link, that looks very interesting. – B. Bischof Jan 22 2010 at 14:08
Thanks Mariano, I fixed it. – Greg Kuperberg Jan 22 2010 at 18:03
For the record, this picture is often credited to Scott and Wall in their paper: Topological methods in group theory. Homological group theory (Proc. Sympos., Durham, 1977), pp. 137--203, – HW Jan 22 2010 at 19:44
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http://mathoverflow.net/revisions/9076/list
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## Return to Answer
2 added 842 characters in body
Let's assume the characteristic is $0$.
Let $R$ be a normal $N$-graded ring of dimension 2 with homogenous maximal ideal $m$. Then $R$ has rational singularity if and only if the non-negative degrees part of the graded local cohomology module $H_m(R)$ vanish. That is $H_m(R)_{(i)}=0$ for $i\geq 0$. This is due to Watanabe, see Theorem 2.2 in: http://www.ams.org/tran/2003-355-03/S0002-9947-02-03186-0/home.html
It is safe to work in affine situation, so let $S=k[x,y]$ and $R=S^G$. Then $R$ is normal and generated by forms of positive degree. Since $H_{(x,y)}(S)_{(i)}=0$ for $i\geq 0$, it follows that $H_m(R)_{(i)}=0$ for $i\geq 0$ (one can compute local cohomology in $S$ by using a system of parameters which are elements in $R$).
In characteristic $p$ one probably has to use Frobenius. Note that Boutot's theorem fails in this case (I think it is still true for finite group though).
A truly easy proof is probably not easy to find unless one has a truly elementary definition of rationality.
EDIT: There are other ways to see this:
II) Again, assume $k$ is algebraically closed of characteristic $0$.Let $S=k[[x,y]]$ and $R=S^G$. Then the following 2 facts will suffice (using same notation as above):
1) There are only finitely many indecomposable reflexive modules over $R$. (Proof not hard, they have to be summands of $S$). In particular, the class group of $R$ is finite.
2) Since $R$ is complete, $R$ has rational singularity is equivalent to the class group of $R$ is finite. This is Theorem 17.4 in Lipman paper on rational singularity.
III) Finally, one can quote Prop 5.15 of Kollar-Mori book on birational geometry. It gave the exact statement, but the proof uses general machinery, and probably close to what you already knew.
1
Let's assume the characteristic is $0$.
Let $R$ be a normal $N$-graded ring of dimension 2 with homogenous maximal ideal $m$. Then $R$ has rational singularity if and only if the non-negative degrees part of the graded local cohomology module $H_m(R)$ vanish. That is $H_m(R)_{(i)}=0$ for $i\geq 0$. This is due to Watanabe, see Theorem 2.2 in: http://www.ams.org/tran/2003-355-03/S0002-9947-02-03186-0/home.html
It is safe to work in affine situation, so let $S=k[x,y]$ and $R=S^G$. Then $R$ is normal and generated by forms of positive degree. Since $H_{(x,y)}(S)_{(i)}=0$ for $i\geq 0$, it follows that $H_m(R)_{(i)}=0$ for $i\geq 0$ (one can compute local cohomology in $S$ by using a system of parameters which are elements in $R$).
In characteristic $p$ one probably has to use Frobenius. Note that Boutot's theorem fails in this case (I think it is still true for finite group though).
A truly easy proof is probably not easy to find unless one has a truly elementary definition of rationality.
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http://physics.stackexchange.com/questions/38297/the-radius-of-the-universe-and-time/38303
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# The radius of the universe and time
Ok - fair warning - Non-physicist asking dumb-assed questions here again. I've been reading a lot of Einstein, Feynmann, Ferris etc. I'm just loving this stuff.
But I suddenly found myself looking at the 4-D sphere concept of our expanding universe, as well as the distortion of spacetime caused by objects with mass and so on, and suddenly thought - this 4th spacial dimension (that we cannot comprehend) through which our universe is expanding - is this actually the dimension of time?
Is the time dilation experienced closer to bodies with mass not a literal curving of the skin of our 4-D sphere in toward its center? So at that point in spacetime the radius is increasing more slowly compared to the radius further from the mass?
I'm afraid my brain just keeps failing when I try to incorporate other concepts from GR into this view - such as time dilation with increasing speed. However the shortening of length of an object in the direction of travel kind of makes sense there as well but only if there was a similar unilateral distortion near bodies of mass.
And then I got to thinking about black holes and their extreme curvature - are they literally stopping universal expansion at their point of singularity such that deltaR=0 and therefore time stops? Or do they even stretch right back to R=0?
Ok, enough of my sillyness - can someone put this into perspective for me?
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1
You should probably condense this into a single question (I can count at least five). To clear up some points: - Our universe doesn't have four spatial dimensions; it has three spatial dimensions and one time dimension. - Time dilation and length contraction comes from Special relativity, not General (it has nothing to do with spacetime curvature). - Black holes do not stop universal expansion. – Dmitry Brant Sep 25 '12 at 14:02
## 3 Answers
The universe isn't expanding through a fourth spatial dimension. It's common to visualise a curved manifold by embedding it onto a space of one higher dimension, but this is a purely mathematical operation to make it easier for our 3D brains to grasp. There is no sense in which the universe around us curves through another spatial dimension.
To try and make this clearer, imagine a universe with only one spatial dimension and model this by a piece of elastic with an ant walking along it. You could stretch and/or compress the elastic while the ant is walking and the ant would see some bits of spacetime expanding and shrinking. However everything is just in one dimension.
In mathematical terms, if you feel an urge to Google coming on, bending through an extra dimension is extrinsic curvature while in our elastic band model the deformation is intrinsic curvature. In GR, and as far as we know in real life, the curvature of the universe is intrinsic.
This distinction kind of invalidates the other questions you asked. However if you would like to repost your question in the light of my answer I'd be happy to attempt an answer.
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In GR, Einstein had no understanding of the universe expanding at all. I understood the notion of our universe being the skin of a 4D sphere was a fairly well accepted concept (either that or a 4D torus). Is this not the case? Your 1D universe analogy has expansion/contraction within that dimension that you describe as intrinsic curvature. Of course if that length of elastic was curving out in a second dimension the ant would still see simple expansion even though the distance between the start and end points hadn't changed (in 2D). Is the radius of our universe the metric we see as time? – Gary Beilby Sep 26 '12 at 14:44
Einstein knew perfectly well that GR showed the universe had to be expanding or contracting. He had to add a cosomological constant to get a static solution, and he only did this because prevailing opinion was that the universe was static. – John Rennie Sep 26 '12 at 14:50
No general relativist I know considers the universe to be the skin of a 4D sphere. – John Rennie Sep 26 '12 at 14:51
The universe has no radius. The observable universe has a radius of 13.7 billin light years because it's 13.7 billion years (in our rest frame) since the Big Bang, but the universe is either infinite or curved on a length scale at least 100 times greater than 13.7 billion light years. We know this because experiment shows the universe is flat to within 1%. – John Rennie Sep 26 '12 at 14:53
It is best to only think of the expansion of the universe as happening over very large distance scales. Anything that is bound to something else, such as the way our finger is bound to our hand, or the way that Earth is bound to the Sun, is very decidedly not expanding, it is only when thing sare very far apart and the distance between them makes them essentially non-interacting that expansion starts to become a dominant effect.
Also, current observations say that the recollapsing spherical geometry is very unlikely to correspond with our physical universe.
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It's a little hard to follow the specific question, but since you are beginning with introductory texts we will make the assumption that when you are referring to 4D, you are actually reading about 4 dimensional spacetime, where there are 3 dimensions of space and 1 dimension in time. I will refer you to this answer for a discussion as to the interpretation of dimension of spacetime. What is important is that when discussion relativity, we define a metric for that spacetime. In special relativity the metric is the Minkowski metric which is viewed as a representation of flat spacetime.
General relativity is the theory that deals with curvature of spacetime, where the metric of spacetime is not constant and can change depending on one's location in spacetime. Most experimental evidence however indicates that spacetime is flat globally, meaning it is flat everywhere, and the behavior of the universe is still best described the the $\Lambda CDM$ model which assumes the existence of flat spacetime, cold dark matter, and a cosmological constant. Based on this model, and experimental observation, the observable universe is 13.7 billion years old, and because relativity allows us to equate time to distance, we say that the "radius" of the observable universe is 13.7 billion light-years.
Based on observation, we have determined that the cosmological constant $\Lambda$ is near zero but slightly positive in value, this means that the universe is expanding and accelerating. This means that even though there may be enough mass to reach a critical density to stop the universe if there were no cosmological constant, because there is some amount of vacuum energy and space is increasing in size, the vacuum energy will eventually overpower the effects of gravity and the universe will expand and ultimately suffer a heat death.
As far as time stopping, relativity tells us that there is a relationship between objects traveling at different relative speeds, where an object that appears to move at relative speeds approaching light also appears to have its clock slow down, from the perspective of a stationary observer. This is called time dilation. In essence, as a person accelerates to the speed of light relative to the universe, their local clocks appear to be moving slower than those of an observer that has remained stationary. If a massive object where able to reach the speed of light, then its local clock would in principle completely stop.
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Ok - I did get excited and throw in too many questions and concepts to begin with. I would like to re-clarify - Can you show me how time is not the radius of the universe? Can we be sure the dimension we see as the unidirectional passage of time is not simply the space through which the universe expands? – Gary Beilby Sep 26 '12 at 14:19
– Hal Swyers Sep 27 '12 at 10:21
cont. As far as expanding into a space, this is not the conception that astronomers and cosmologists like. First, expansion implies a time derivative, which if you equate our notion of time into a spatial dimension for expansion, you would still need another variable of time, which is unphysical. GR assume a 4-d spacetime manifold, and the expansion parameter is the cosmological constant $\Lambda$ in the equation, $R_{ij} - \dfrac{1}{2} Rg_{ij} + \Lambda g_{ij} = \kappa T_{ij}$ – Hal Swyers Sep 27 '12 at 10:33
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http://mathoverflow.net/questions/97058?sort=oldest
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## Do fixed point sets in equivariant crepant resolutions have the same cohomology? How about for the specific case of Nakajima quiver varieties?
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A crepant resolution $f:Y\to X$ is a resolution of singularities with $f^*(K_X)=K_Y$. Crepant resolutions do not always exist, and when they exist they may not be unique. However, different crepant resolutions $Y_1$ and $Y_2$ share many properties. In particular, Kontsevich introduced motivic integration to prove that the $Y_i$ have the same Betti numbers.
Suppose that $X$ has a $\mathbb{C}^*$ action, and that $f_i:Y_i\to X; i=1,2$ are equivariant crepant resolutions -- that is, the $Y_i$ have $\mathbb{C}^*$ actions and the $f_i$ are equivariant maps. Let $F_i\subset Y_i$ be the fixed point sets. Do the $F_i$ have the same Betti numbers?
## How about for Nakajima quiver varieties?
This may be too much to expect in general, so let me mention that the case of particular interest to me is when $Y$ and $X$ are Nakajima quiver varieties (see Ginzburg).
More specifically, $X$ is the space $(\mathbb{C}^2/G)^n/S_n$, where a generator $g\in G=\mathbb{Z}/r\mathbb{Z}$ acts on $\mathbb{C}^2$ by $g(x,y)=(\omega x, \omega^{-1} y)$ for $\omega$ a primitive $r$th root of unity, and the $Y_i$ are certain other quiver varieties. Nakajima considers the $\mathbb{C}^*$ action given by $t(x,y)=(tx,y)$. I care about a more general action $t(x,y)=(t^a x,t^b y), a,b>0$. This $X$ is a Nakajima quiver variety, and other Nakajima quiver varieties provide natural equivariant crepant resolutions.
Note that in this case it is known that the $Y_i$ are in fact diffeomorphic; however they are not equivariantly diffeomorphic for my torus action. Also note that using the action of the larger torus $(\mathbb{C}^*)^2$, we can see that $\chi(F_1)=\chi(Y_1)=\chi(F_2)$, so they at least have the same euler characteristic.
For these Nakajima quiver varieties you can compute the $H^*(F_i)$ on a case by case basis using the obvious $(\mathbb{C}^*)^2$ action, and the question seems to have an affirmative answer. However, actually proving it in general using this method boils down to a difficult combinatorial question about partitions that is what I was originally considering. I only recently came up with the current formulation of the question, which seems like quite a natural question, and I was hoping that I would find the answer to my question already in the literature (for instance, in Kaledin's work on the symplectic McKay correspondence or symplectic resolutions more generally), but have had no luck so far, and so I turn to MO.
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What do you mean by `$H^*(F_1) = H^*(F_2)$' ? There is no natural homomorphism between them. If you just compare the dimensions of RHS and LHS, they are the Euler numbers of $Y_1$ and $Y_2$, and hence are equal. – Hiraku Nakajima Jun 9 at 3:22
Hiraku, I could have been clearer. We certainly have $\chi(F_1)=\chi(X)=\chi(F_2)$ by localization. But it appears that more is true -- namely, the betti numbers of the $F_i$ are equal. I've edited the question to make this a bit clearer. – Paul Johnson Jun 15 at 11:11
## 2 Answers
For a Nakajima quiver variety satisfying Kirwan surjectivity on $H^2$ (which may well be all of them; I don't know a counterexample), all crepant resolutions are diffeomorphic, since they are quiver varieties themselves (you can check that you get every class in the Picard group from GIT reductions). If the $\mathbb{C}^*$ of interest to you comes from a linear action of the cotangent bundle you reduced to get the quiver variety, then you can make the diffeomorphism $S^1$-equivariant, and thus induce a diffeomorphism on fixed point sets (look at the proof of 3.4 in Proudfoot's thesis; that's for the abelian case, but the proof is the same.
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Thanks for the reference -- unfortunately, it appears that with my group action they aren't in fact equivariantly diffeomorphic. I'm really looking at just the ordinary hilbert scheme of points on C^2. Z_r acting antidiagonal on C^2 induces an action on the hilbert scheme, and the components of the fixed point set are the quiver varieties. The C^* action I have is the induced action from a C^\$ action on C^2. With the whole (C^*)^2, you know the fixed point sets agree -- they are finite sets. But for different resolutions the normal bundles of the fixed point sets are different. – Paul Johnson May 17 2012 at 17:46
### 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.
In the situation when Proposition 5.7 in my paper in Duke 1994 is applicable, the answer is YES: Apply the Poincare duality to $F_\alpha$ and use the formula for the Morse index $m_\alpha$. Then $\dim F_\alpha + m_\alpha = \frac12 \dim M$ is independent of $\alpha$, so the sum of $(\frac12 \dim M-i)$-th Betti numbers of $F_\alpha$ computes the $i$-th Betti number of $M$, which is independent of the choice of the crepant resolution.
For the $\mathbb C^*$-action $t(x,y) = (tx,y)$, the symplectic form is multiplied with weight $1$, but the condition that the orientation $\Omega$ contains no cycles is violated. So I must be more careful. (First of all, Poincare duality changes ordinary cohomology to cohomology with compact support.) I never considered this situation before, but may have a chance to say something using the same argument with care.
More general action, when the symplectic form is not necessarily multiplied with weight $1$, I do not know a clean formula for $m_\alpha$. So I do not have any idea why the Betti numbers are the same.
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It seems that arxiv.org/abs/1206.5640 gives a positive answer when $a+b=r$. – Hiraku Nakajima Jun 26 at 9:21
I understand the point of the argument. If $r$ is divisible by $a+b$, the action of $a+b$-th roots of $t$ is well-defined. So the symplectic form is of weight $1$, and the above argument works. – Hiraku Nakajima Jun 26 at 11:21
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http://mathhelpforum.com/calculus/130236-integrating-volume.html
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# Thread:
1. ## Integrating a volume
We just started today and a problem I have is...
Write a Riemann sum and then a definite integral representing the volume of the region, using the slice shown in the figure below. Evaluate the integral exactly. Use your work to answer the questions below.
What is the approximate volume of the slice with respect to y?
So I got l/14=(7-y)/7 and then I get 20y deltay is the volume of the slice but it is a circle, where would I put pi into this?
Thanks for the help,
Tyler
Attached Thumbnails
2. The area of the cut is a rectangle A=2x*10
Radius is 7 so x=7
$<br /> V = \int_{a}^{b}A(y)dy<br />$
In your case you have
$<br /> V = \int_{a}^{b} 2x * 10 dy = 20 \int_{a}^{b} x dy <br />$
So, you need to write the x in terms of y, since you have dy
The point (x,y) is on the circel
$x^{2} + y^{2} = r^{2}$
$x^{2} + y^{2} = 7^{2}$
So,
$x = \sqrt {49-y^{2}}$
$<br /> V = 20 \int_{0}^{7} \sqrt {49-y^{2}} dy <br />$
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http://mathoverflow.net/questions/10941?sort=votes
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## elementary classification of artinian rings
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
this may be too elementary for mathoverflow, but I'll give it a try.
rings are commutative here. it is well-known that every $0$-dimensional noetherian ring is artinian. the standard proof uses a filtration argument; then it's left to show that every finitely generated vector space is artinian (dimension!) and that extensions of artinian by artinian modules are artinian (tage the images and the preimages of the chain, finally both are stable). by a sheaf argument, it's easy to reduce to: every noetherian ring with exactly one prime ideal is artinian.
is there a proof which is somehow more direct? perhaps a clever manipulation of chains of ideals? I don't expect it, but it would be great for the students in my tutorial, which had to solve this as an exercise without knowing anything about artinian or noetherian rings going beyond the definitions.
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I don't think it's too elementary at all. I do think that you should use correct capitalization, but that's a separate issue :) – Theo Johnson-Freyd Jan 7 2010 at 2:51
## 2 Answers
My take on the standard proof can be found on pp. 62-63 of my notes on commutative rings:
http://math.uga.edu/~pete/integral.pdf
(The bit about ACC/DCC being preserved by extensions occurs on p. 57 and should probably be explicitly mentioned in the proof on p. 63.)
Altogether this takes about 1-1.5 pages. I have never seen anything substantially shorter or more direct.
By the way, that's a tough problem to ask a student to solve on his/her own!
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this is the standard proof I've mentioned above. so basically your answer is not concerned with my question. – Martin Brandenburg Jan 6 2010 at 18:31
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I acknowledged that it was the standard proof in my response. My point was that this is the best proof that I know of, so that someone who is curious about the standard proof can look it up there. I agree that it is little or no progress towards the answer to your question, but I think a downvote is a bit harsh. – Pete L. Clark Jan 6 2010 at 18:36
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What part of "I have never seen anything substantially shorter or more direct." is not an answer to "is there a proof which is somehow more direct?"? – Theo Johnson-Freyd Jan 7 2010 at 2:52
### 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.
Sorry if this is merely a reformulation of what has already been said (and doubtless it is a "standard proof"), but perhaps a suggestive hint for students would be to show that if an ideal $P$ of a commutative noetherian ring $R$ is maximal for the property that $R/P$ is non-artinian, then $P \subset R$ must be prime. A sort of philosophical underpinning for this hint is offered in a pretty paper by T. Y. Lam and M. L. Reyes, "A prime ideal principle in commutative algebra," J. Algebra 319 (2008), no. 7, 3006-3027.
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An afterthought: the Lam-Reyes Principle is better than just a "philosophical underpinning" in this case.  The set of ideals of a commutative noetherian ring with the property that the corresponding factor ring is artinian is a monoidal filter and hence both an Oka family and an Ako family (in Lam and Reyes's terminology); thus, by the main result of their paper, maximal members of the complementary set of ideals must be prime. – Greg Marks Jun 26 2010 at 0:55
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http://math.stackexchange.com/users/14480/jowen?tab=activity&sort=comments
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# JOwen
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| | 170 reputation | website | | member for | 1 year, 9 months |
| 8 badges | location | | seen | 2 hours ago | |
# 9 Comments
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| Apr21 | comment | Betting on the digits of a real numberI've updated the question in response to your comments. |
| Apr20 | comment | Betting on the digits of a real number@joriki: Yes, I see that I formulated the question slightly incorrectly. I think I meant "Is there a strategy such that there is no $x$ that 'defeats' it in the long run". My instinct initially was that there might be choices for $x$ that 'defeat' every strategy (but you have shown that to not be the case), which is why I made the error, I think. |
| Apr20 | comment | Betting on the digits of a real number@carlop: $0.5 - 2|x-0.5|$ is not always greater than zero for $x$ between 0 or 1, take $x=1$, for example. |
| Apr20 | comment | Betting on the digits of a real number@joriki: But there is not a unique $x$ given an initial string of digits, so what does "correct guess" even mean? I see the problem you are bringing up, but I guess when I'm talking about strategy, I mean some rule or algorithm to play this game no matter what $x$ is. |
| Apr20 | comment | Betting on the digits of a real numberI see. Does it clarify things if we define strategies to be functions mapping strings of digits (all the $nN$ revealed digits that I know at turn $n$) to guesses $p$? |
| Apr20 | comment | Betting on the digits of a real numberThanks for the comment. $x$ is kept secret, so the choice of $p$ must be made based only on what digits have been revealed in play. I have edited the question to clarify this. |
| Feb7 | comment | An approximate relationship between the totient function and sum of divisorsThanks. I find these kinds of facts remarkable. Do you know what are the prerequisites for Hardy and Wright? |
| Aug12 | comment | Recognizing and Using Chaitin's ConstantGreat answer, thanks. |
| Aug11 | comment | Recognizing and Using Chaitin's ConstantThank you all for your replies! |
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http://www.physicsforums.com/showthread.php?p=4171732
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## Fourth order PDE
Hi guys!
This is not a homework question, it was a question on my test a few days ago. I could not solve it.
Out of memory, the problem was a rod of length L with an end fixed in a wall and the other end free. Its motion satisfies the PDE ##a^4\frac{\partial ^4 u }{\partial x^4} + \frac{\partial ^2 u}{\partial t^2}=0## where ##a>0##.
The boundary conditions are ##u(0,t)=\frac{\partial u}{\partial x}(0,t)=0## and ##\frac{\partial ^2 u}{\partial x^2} (L,t)=\frac{\partial ^3 u}{\partial x^3 }(L,t)=0##.
I had to show that its eigenfrequencies satisfy the following relation: ##\cosh \left ( \frac{\sqrt \omega L }{a} \right ) = \sec \left ( \frac{\sqrt \omega L }{a} \right )##.
Attempt at solution: First off the problem struck me like a hammer since there was no similar problem in my assignments.
I tried to solve the PDE usng separation of variables, until the professor told us "there's no need to solve the PDE. Of course you can do it but it's extra work" which basically means to me that there's some trick I totally missed.
Anyway what I had reached after separation of variable ##u(x,t)=X(x)T(t)## is ##T(t)=A\cos (\lambda t )+B\sin (\lambda t)## where lambda is a constant of separation.
The remaining ODE I had to solve was ##a^4 \frac{X''''}{X}-\lambda ^2 =0##. I assumed the solution was of the form ##X(x)=Ae^{kx}+B^{-kx}## (now I realize it's wrong since it's of order 4 so there should be 4 linearly independent terms, not 2) and I reached after plugging it back into the ODE that ##k=\pm \frac{\lambda }{a}## which gave me ##X(x)##. I tried a few other things like trying to find out the constants of the general solution to the PDE (but I've got X(x) wrong since it's of order 4, not 2) but I reached nothing. Obviously I missed a trick.
So I don't really know how to proceed to solve the problem, even at home. Has anyone an idea?
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Quote by fluidistic since it's of order 4 so there should be 4 linearly independent terms, not 2
cos(kx) would also be a solution, yes?
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Whoops, I made a mistake in the equation and boundary condition (mixed ^3 for ^2 and ^2 for ^3, I edited the post)
Quote by haruspex cos(kx) would also be a solution, yes?
Yeah I'll have to check this out, thanks for the tip. I'll come back to this probably tomorrow.
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## Fourth order PDE
Quote by fluidistic The remaining ODE I had to solve was ##a^4 \frac{X''''}{X}-\lambda ^2 =0##. I assumed the solution was of the form ##X(x)=Ae^{kx}+B^{-kx}## (now I realize it's wrong since it's of order 4 so there should be 4 linearly independent terms, not 2)
I'll give you a slightly different angle than haruspex. What if you had just been given the ODE
$$a^4 X''- q X =0?$$
I've replaced ##\lambda^2## with ##q##. Say you don't know the sign of q. Do you remember how you first went about solving that equation to get the general solution?
If you can recall that, then suppose you were given the ODE you found,
$$a^4 X''''- q X =0.$$
Use the same trick to find the four terms that should be present in your general solution.
By the way, when you separated variables, how are you sure that the separation constant should be positive? Why can't it be negative?
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Ok thanks a lot guys!!!
Quote by Mute I'll give you a slightly different angle than haruspex. What if you had just been given the ODE $$a^4 X''- q X =0?$$ I've replaced ##\lambda^2## with ##q##. Say you don't know the sign of q. Do you remember how you first went about solving that equation to get the general solution?
Yes, absolutely. If q is negative you get oscillatory motion (Acos ##\sqrt q## x + B sin ##\sqrt q## x). If q=0 you get the equation of a straight line, namely Ax+b and if q is postive you get growing and decaying exponential whose arguments are plus and minus the square root of q times x.
If you can recall that, then suppose you were given the ODE you found, $$a^4 X''''- q X =0.$$ Use the same trick to find the four terms that should be present in your general solution. By the way, when you separated variables, how are you sure that the separation constant should be positive? Why can't it be negative?
Hmm ok I will try to see if it can help me for the fourth order ODE.
About the constant of separation, I reached ##a^4\frac{X''''}{X}+\frac{T''}{T}=0##. So that T''/T is worth a constant. In my mind it was clear that the rod would have a periodic motion in time because of the physical problem. So I assumed the constant of separation worth to be ##-\lambda ^2## so that I could express the solution ##T(t)## as oscillatory and without square root. If the sign of q, or lambda would have been different, then the motion of the rod would have been different.
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Quote by fluidistic Ok thanks a lot guys!!!Yes, absolutely. If q is negative you get oscillatory motion (Acos ##\sqrt q## x + B sin ##\sqrt q## x). If q=0 you get the equation of a straight line, namely Ax+b and if q is postive you get growing and decaying exponential whose arguments are plus and minus the square root of q times x.
Yes, that is correct, but do you remember how you derive those solutions for ##q \neq 0## in the first place? It's similar to what you tried to do to solve the fourth order ODE. Say that instead of plugging in a trial solution ##A\exp(kx) + B\exp(-kx)## you had only plugged in ##exp(kx)##... then what?
About the constant of separation, I reached ##a^4\frac{X''''}{X}+\frac{T''}{T}=0##. So that T''/T is worth a constant. In my mind it was clear that the rod would have a periodic motion in time because of the physical problem. So I assumed the constant of separation worth to be ##-\lambda ^2## so that I could express the solution ##T(t)## as oscillatory and without square root. If the sign of q, or lambda would have been different, then the motion of the rod would have been different.
Good. Why does the beam only oscillate in time, though? Why can't the oscillations decay? What kind of term in the PDE do you think would give you decaying oscillations? (Or possibly no oscillations at all, depending on parameter values)
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Quote by Mute Yes, that is correct, but do you remember how you derive those solutions for ##q \neq 0## in the first place? It's similar to what you tried to do to solve the fourth order ODE. Say that instead of plugging in a trial solution ##A\exp(kx) + B\exp(-kx)## you had only plugged in ##exp(kx)##... then what?
Ah yes, with the characteristic equation. So here if my equation is ##X''''\underbrace{-\frac{\lambda^2}{a^4}}_{=q}X=0## the char. eq. is ##r^4+qr=0## which yields ##r=0 \text{ or } r=(-q)^{3/2}##. Apparently I'll have to use variation of parameters to get the solution because of the triply repeated root and ##r=0##.
Good. Why does the beam only oscillate in time, though? Why can't the oscillations decay? What kind of term in the PDE do you think would give you decaying oscillations? (Or possibly no oscillations at all, depending on parameter values)
Hmm yeah of course the motion could be different from oscillating. If your q was positive then ##T(t)=Ae^{-\sqrt q t}## and the characteristic equation for ##X(x)## yields ##r=0## or ##r =\left ( \frac{q}{a^4} \right ) ^{3/2}## with again a triple root.
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Quote by fluidistic Ah yes, with the characteristic equation. So here if my equation is ##X''''\underbrace{-\frac{\lambda^2}{a^4}}_{=q}X=0## the char. eq. is ##r^4+qr=0## which yields ##r=0 \text{ or } r=(-q)^{3/2}##. Apparently I'll have to use variation of parameters to get the solution because of the triply repeated root and ##r=0##.
Careful! Only derivatives will bring down powers of r. The characteristic equation is actually just ##\alpha^4 r^4 +q = 0##. Also be careful when taking roots here. In your first attempt with the accidental extra factor of r, ##r(\alpha^4 r^3 + q) = 0##, the non-zero roots are ##r = (q/\alpha^4)^{1/3}##. This is not a triple root because the cube root actually produces three distinct roots in the complex plane, not a single triple root. Similarly, with the correct characteristic equation, you will have four distinct roots in the complex plane.
Hmm yeah of course the motion could be different from oscillating. If your q was positive then ##T(t)=Ae^{-\sqrt q t}## and the characteristic equation for ##X(x)## yields ##r=0## or ##r =\left ( \frac{q}{a^4} \right ) ^{3/2}## with again a triple root.
Ok, what I was actually getting at with that question was that there was no damping in the original PDE, so you wouldn't expect any decaying oscillatory behavior. I was just posing the question to give you some more practice thinking about the form of the PDE/ODEs in case a similar question ever came up on a future exam.
That said, while we're on the topic of what we expect the behavior of the solutions to be based on physical reasoning, are you sure the eigenfrequency relation you were asked to show was ##\cosh(\sqrt{\omega}L/\alpha) = \mbox{sec}(\sqrt{\omega}L/\alpha)##? That doesn't have any real solutions other than ##\omega = 0##.
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Quote by Mute Careful! Only derivatives will bring down powers of r. The characteristic equation is actually just ##\alpha^4 r^4 +q = 0##. Also be careful when taking roots here. In your first attempt with the accidental extra factor of r, ##r(\alpha^4 r^3 + q) = 0##, the non-zero roots are ##r = (q/\alpha^4)^{1/3}##. This is not a triple root because the cube root actually produces three distinct roots in the complex plane, not a single triple root. Similarly, with the correct characteristic equation, you will have four distinct roots in the complex plane.
Whoops, you are right. I made a stupid mistake here.
Ok, what I was actually getting at with that question was that there was no damping in the original PDE, so you wouldn't expect any decaying oscillatory behavior. I was just posing the question to give you some more practice thinking about the form of the PDE/ODEs in case a similar question ever came up on a future exam.
Ah nice, thank you. I'm preparing for an upcoming final exam worth 100% of my grade for the course. Here I sought oscillatory solutions for no particular reason. I did not notice that the PDE had no damping term so next time I'll be more careful.
That said, while we're on the topic of what we expect the behavior of the solutions to be based on physical reasoning, are you sure the eigenfrequency relation you were asked to show was ##\cosh(\sqrt{\omega}L/\alpha) = \mbox{sec}(\sqrt{\omega}L/\alpha)##? That doesn't have any real solutions other than ##\omega = 0##.
I'm 99% sure that yes. I can't access the test to confirm though but I'm sure that yes.
I just "solved" ##a^4X''''-\lambda ^2 X=0##. The characteristic equation gave me ##r= \pm \frac{\sqrt \lambda}{a}## or ##r=\pm \frac{\sqrt {-\lambda}}{a}##. Hmm so what does this tell me.
That ##X(x)=Ae^{\frac{\sqrt \lambda x}{a}}+Be^{\frac{-\sqrt \lambda x}{a}}+Ce^{\frac{i\sqrt \lambda x }{a}}+De^{\frac{-i\sqrt \lambda x}{a}}##. Now I guess I must toss out some coefficients.
Recognitions: Gold Member I just found out the problem in a book by chance! Riley's mathematical methods for physics and engineering. The problem is as I wrote but the relation is ##\cosh \left ( \frac{\sqrt \omega L }{a} \right ) = - \sec \left ( \frac{\sqrt \omega L }{a} \right )## (notice the - sign) where omega stands for the angular frequency of vibration. (So that there does not seem to have more than 1 omega).
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Quote by fluidistic I just found out the problem in a book by chance! Riley's mathematical methods for physics and engineering. The problem is as I wrote but the relation is ##\cosh \left ( \frac{\sqrt \omega L }{a} \right ) = - \sec \left ( \frac{\sqrt \omega L }{a} \right )## (notice the - sign) where omega stands for the angular frequency of vibration. (So that there does not seem to have more than 1 omega).
Ok, that's comforting, because before the only real solution was ##\omega = 0##, which would mean the displacement was either constant in time or linearly increasing in time! That didn't seem right, hence why I suspected there must be something missing. =)
Are you now able to derive that condition from applying the boundary conditions to your general solution?
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Quote by Mute Ok, that's comforting, because before the only real solution was ##\omega = 0##, which would mean the displacement was either constant in time or linearly increasing in time! That didn't seem right, hence why I suspected there must be something missing. =) Are you now able to derive that condition from applying the boundary conditions to your general solution?
I've changed my lambda for omega in order to get a temporal solution under the form ##T(t)=Ae^{i\omega t}+Be^{-i\omega t}##.
So that ##X(x)=Ce^{\frac{\sqrt \omega x}{a}}+De^{\frac{-\sqrt \omega x}{a}}+Ee^{\frac{i\sqrt \omega x }{a}}+Fe^{\frac{-i\sqrt \omega x}{a}}##.
The condition ##u(0,t)=0## gives ##C+D+E+F=0##. The other 3 boundary conditions gives slightly more complicated conditions over C, D, E and F. So I must solve a system of 4 homogeneous equations. I have basically ##\mathbf A \begin{bmatrix} C \\ D \\E \\F \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\0 \\0 \end{bmatrix}## where A is a 4x4 (not beautiful) matrix. Now if A is singular then the only solution is the trivial one (C=D=E=F=0) and I get no condition over omega. However if A is non singular, then I should obtain the desired condition over omega. The problem is that the A matrix is pretty "ugly" to me and it involves calculating its determinant and equate it to 0 and solve for omega. I've done a few terms and it's not like there's something that cancels out... sigh.
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Quote by fluidistic I've changed my lambda for omega in order to get a temporal solution under the form ##T(t)=Ae^{i\omega t}+Be^{-i\omega t}##. So that ##X(x)=Ce^{\frac{\sqrt \omega x}{a}}+De^{\frac{-\sqrt \omega x}{a}}+Ee^{\frac{i\sqrt \omega x }{a}}+Fe^{\frac{-i\sqrt \omega x}{a}}##. The condition ##u(0,t)=0## gives ##C+D+E+F=0##. The other 3 boundary conditions gives slightly more complicated conditions over C, D, E and F. So I must solve a system of 4 homogeneous equations. I have basically ##\mathbf A \begin{bmatrix} C \\ D \\E \\F \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\0 \\0 \end{bmatrix}## where A is a 4x4 (not beautiful) matrix. Now if A is singular then the only solution is the trivial one (C=D=E=F=0) and I get no condition over omega. However if A is non singular, then I should obtain the desired condition over omega. The problem is that the A matrix is pretty "ugly" to me and it involves calculating its determinant and equate it to 0 and solve for omega. I've done a few terms and it's not like there's something that cancels out... sigh.
What if you just try eliminating variables step by step from the four equations? I think I tried it a while back and it boiled down to C(something) = C(something else), though I wasn't being particularly careful.
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Quote by fluidistic Now if A is singular then the only solution is the trivial one (C=D=E=F=0) and I get no condition over omega. However if A is non singular, then I should obtain the desired condition over omega.
You meant that the other way about, right?
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Quote by haruspex You meant that the other way about, right?
Oops, yes. Change singular for non singular and non singular for singular. Or simply singular for inversible.
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http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_6&diff=31243&oldid=31242
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# User:Michiexile/MATH198/Lecture 6
### From HaskellWiki
(Difference between revisions)
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| | The natural transformation <math>\eta</math> is called the ''unit'' of the adjunction. | | The natural transformation <math>\eta</math> is called the ''unit'' of the adjunction. |
| | | | |
| | | + | This definition, however, has a significant amount of asymmetry: we can start with some <math>f:A\to U(B)</math> and generate a <math>g: F(A)\to B</math>, while there are no immediate guarantees for the other direction. However, there is a proposition we can prove leading us to a more symmetric statement: |
| | | | |
| | | + | '''Proposition''' For categories and functors |
| | | | |
| - | * Free and forgetful | + | [[Image:AdjointPair.png]] |
| - | * Curry and uncurry | + | |
| | | | |
| | | + | the following conditions are equivalent: |
| | | + | # <math>F</math> is left adjoint to <math>U</math>. |
| | | + | # For any <math>c\in C_0</math>, <math>d\in D_0</math>, there is an isomorphism <math>\phi: Hom_D(Fc, d) \to Hom_C(c,Ud)</math>, natural in both <math>c</math> and <math>d</math>. |
| | | + | moreover, the two conditions are related by the formulas |
| | | + | * <math>\phi(g) = U(g) \circ \eta_c</math> |
| | | + | * <math>\eta_c = \phi(1_{Fc}</math> |
| | | + | |
| | | + | '''Proof sketch''' |
| | | + | For (1 implies 2), the isomorphism is given by the end of the statement, and it is an isomorphism exactly because of the unit property - viz. that every <math>f:A\to U(B)</math> generates a unique <math>g: F(A)\to B</math>. |
| | | + | |
| | | + | Naturality follows by building the naturality diagrams |
| | | + | |
| | | + | [[Image:NaturalityDiagramsAdjointProposition.png]] |
| | | + | |
| | | + | and chasing through with a <math>f: Fc\to d</math>. |
| | | + | |
| | | + | For (2 implies 1), we start out with a natural isomorphism <math>\phi</math>. We find the necessary natural transformation <math>\eta_c</math> by considering <math>\phi: Hom(Fc,Fc) \to Hom(c, UFc)</math>. |
| | | + | |
| | | + | QED. |
| | | + | |
| | | + | By dualizing the proof, we get the following statement: |
| | | + | |
| | | + | '''Proposition''' For categories and functors |
| | | + | |
| | | + | [[Image:AdjointPair.png]] |
| | | + | |
| | | + | the following conditions are equivalent: |
| | | + | # For any <math>c\in C_0</math>, <math>d\in D_0</math>, there is an isomorphism <math>\phi: Hom_D(Fc, d) \to Hom_C(c,Ud)</math>, natural in both <math>c</math> an |
| | | + | # There is a natural transformation <math>\epsilon: FU \to 1_D</math> with the property that for any <math>g: F(c) \to d</math> there is a unique <math>f: c\to U(d)</math> such that <math>g = \epsilon_D\circ F(f)</math>, as in the diagram |
| | | + | |
| | | + | [[Image:NaturalityDiagramsDualAdjointProposition.png]] |
| | | + | |
| | | + | moreover, the two conditions are related by the formulas |
| | | + | * <math>\psi(f) = \epsilon_D\circ F(f)</math> |
| | | + | * <math>\epsilon_d = \psi(1_{Ud}</math> |
| | | + | |
| | | + | where <math>\psi = \phi^{-1}</math>. |
| | | + | |
| | | + | Hence, we have an equivalent definition with higher generality, more symmetry and more ''horsepower'', as it were: |
| | | + | |
| | | + | '''Definition''' An ''adjunction'' consists of functors |
| | | + | |
| | | + | [[Image:AdjointPair.png]] |
| | | + | |
| | | + | and a natural isomorphism |
| | | + | |
| | | + | [[Image:AdjointIso.png]] |
| | | + | |
| | | + | The ''unit'' <math>\eta</math> and the ''counit'' <math>\epsilon</math> of the adjunction are natural transformations given by: |
| | | + | * <math>\eta: 1_C\to UF: \eta_c = \phi(1_{Fc})</math> |
| | | + | * <math>\epsilon: FU\to 1_D: \epsilon_d = \psi(1_{Ud})</math>. |
| | | + | |
| | | + | ---- |
| | | + | |
| | | + | Some of the examples we have had difficulties fitting into the limits framework show up as adjunctions: |
| | | + | |
| | | + | The ''free'' and ''forgetful'' functors are adjoints; and indeed, a more natural definition of what it means to be free is that it is a left adjoint to some forgetful functor. |
| | | + | |
| | | + | Curry and uncurry, in the definition of an exponential object are an adjoint pair. The functor <math>-\times A: X\mapsto X\times A</math> has right adjoint <math>-^A: Y\mapsto Y^A</math>. |
| | | + | |
| | | + | ====Notational aid==== |
| | | + | |
| | | + | One way to write the adjoint is as a ''bidirectional rewrite rule'': |
| | | + | |
| | | + | <math>\frac{F(X) \to Y}{X\to G(Y)}</math>, |
| | | + | |
| | | + | where the statement is that the hom sets indicated by the upper and lower arrow, respectively, are transformed into each other by the unit and counit respectively. The left adjoint is the one that has the functor application on the left hand side of this diagram, and the right adjoint is the one with the functor application to the right. |
| | | | |
| | | | |
| Line 147: | | Line 215: | |
| | | | |
| | ===Homework=== | | ===Homework=== |
| | | + | |
| | | + | Complete homework is 6 out of 10 exercises. |
| | | | |
| | # Prove that an equalizer is a monomorphism. | | # Prove that an equalizer is a monomorphism. |
| Line 155: | | Line 225: | |
| | # What is the pullback in the category of posets? | | # What is the pullback in the category of posets? |
| | # What is the pushout in the category of posets? | | # What is the pushout in the category of posets? |
| | | + | # Prove that the exponential and the product functors above are adjoints. What are the unit and counit? |
| | | + | # (worth 4pt) Consider the unique functor <math>!:C\to 1</math> to the terminal category. |
| | | + | ## Does it have a left adjoint? What is it? |
| | | + | ## Does it have a right adjoint? What is it? |
| | | + | # Suppose [[Image:AdjointPair.png]] is an adjoint pair. Find a natural transformation <math>FUF\to F</math>. Conclude that there is a natural transformation <math>\mu: UFUF\to UF</math>. Prove that this is associative, in other words that the diagram |
| | | + | |
| | | + | :[[Image:AdjointMuAssociative.png]] |
| | | + | |
| | | + | :commutes. |
## Revision as of 02:34, 28 October 2009
IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ.
## Contents
### 1 Useful limits and colimits
With the tools of limits and colimits at hand, we can start using these to introduce more category theoretical constructions - and some of these turn out to correspond to things we've seen in other areas.
Possibly among the most important are the equalizers and coequalizers (with kernel (nullspace) and images as special cases), and the pullbacks and pushouts (with which we can make explicit the idea of inverse images of functions).
One useful theorem to know about is:
Theorem The following are equivalent for a category C:
• C has all finite limits.
• C has all finite products and all equalizers.
• C has all pullbacks and a terminal object. Also, the following dual statements are equivalent:
• C has all finite colimits.
• C has all finite coproducts and all coequalizers.
• C has all pushouts and an initial object.
For this theorem, we can replace finite with any other cardinality in every place it occurs, and we will still get a valid theorem.
====Equalizer, coequalizer==== Consider the equalizer diagram:
A limit over this diagram is an object C and arrows to all diagram objects. The commutativity conditions for the arrows defined force for us fpA = pB = gpA, and thus, keeping this enforced equation in mind, we can summarize the cone diagram as:
Now, the limit condition tells us that this is the least restrictive way we can map into A with some map p such that fp = gp, in that every other way we could map in that way will factor through this way.
As usual, it is helpful to consider the situation in Set to make sense of any categorical definition: and the situation there is helped by the generalized element viewpoint: the limit object C is one representative of a subobject of A that for the case of Set contains all $x\in A: f(x) = g(x)$.
Hence the word we use for this construction: the limit of the diagram above is the equalizer of f,g. It captures the idea of a maximal subset unable to distinguish two given functions, and it introduces a categorical way to define things by equations we require them to respect.
One important special case of the equalizer is the kernel: in a category with a null object, we have a distinguished, unique, member 0 of any homset given by the compositions of the unique arrows to and from the null object. We define the kernel Ker(f) of an arrow f to be the equalizer of f,0. Keeping in mind the arrow-centric view on categories, we tend to denot the arrow from Ker(f) to the source of f by ker(f).
In the category of vector spaces, and linear maps, the map 0 really is the constant map taking the value 0 everywhere. And the kernel of a linear map $f:U\to V$ is the equalizer of f,0. Thus it is some vector space W with a map $i:W\to U$ such that fi = 0i = 0, and any other map that fulfills this condition factors through W. Certainly the vector space $\{u\in U: f(u)=0\}$ fulfills the requisite condition, nothing larger will do, since then the map composition wouldn't be 0, and nothing smaller will do, since then the maps factoring this space through the smaller candidate would not be unique.
Hence, $Ker(f) = \{u\in U: f(u) = 0\}$ just like we might expect.
Dually, we get the coequalizer as the colimit of the equalizer diagram.
A coequalizer has to fulfill that iBf = iA = iBg. Thus, writing q = iB, we get an object with an arrow (actually, an epimorphism out of B) that identifies f and g. Hence, we can think of $i_B:B\to Q$ as catching the notion of inducing equivalence classes from the functions.
This becomes clear if we pick out one specific example: let $R\subseteq X\times X$ be an equivalence relation, and consider the diagram
where r1 and r2 are given by the projection of the inclusion of the relation into the product onto either factor. Then, the coequalizer of this setup is an object X / R such that whenever x˜Ry, then q(x) = q(y).
#### 1.1 Pullbacks
The preimage f − 1(T) of a subset $T\subseteq S$ along a function $f:U\to S$ is a maximal subset $V\subseteq U$ such that for every $v\in V: f(v)\in T$.
We recall that subsets are given by (equivalence classes of) monics, and thus we end up being able to frame this in purely categorical terms. Given a diagram like this:
where i is a monomorphism representing the subobject, we need to find an object V with a monomorphism injecting it into U such that the map $fi: U\to S$ factors through T. Thus we're looking for dotted maps making the diagram commute, in a universal manner.
The maximality of the subobject means that any other subobject of U that can be factored through T should factor through V.
Suppose U,V are subsets of some set W. Their intersection $U\cap V$ is a subset of U, a subset of V and a subset of W, maximal with this property.
Translating into categorical language, we can pick representatives for all subobjects in the definition, we get a diagram with all monomorphisms:
where we need the inclusion of $U\cap V$ into W over U is the same as the inclusion over V.
Definition A pullback of two maps $A \rightarrow^f C \leftarrow^g B$ is the limit of these two maps, thus:
By the definition of a limit, this means that the pullback is an object D with maps $\bar f: D\to B$, $\bar g: D\to A$ and $f\bar g = g\bar f : D\to C$, such that any other such object factors through this.
For the diagram $U\rightarrow^f S \leftarrow^i T$, with $i:T\to S$ one representative monomorphism for the subobject, we get precisely the definition above for the inverse image.
For the diagram $U\rightarrow W \leftarrow V$ with both map monomorphisms representing their subobjects, the pullback is the intersection.
#### 1.2 Pushouts
Often, especially in geometry and algebra, we construct new structures by gluing together old structures along substructures. Possibly the most popularly known example is the Möbius band: we take a strip of paper, twist it once and glue the ends together.
Similarily, in algebraic contexts, we can form amalgamated products that do roughly the same.
All these are instances of the dual to the pullback:
Definition A pushout of two maps $A\leftarrow^f C\rightarrow^g B$ is the co-limit of these two maps, thus:
Hence, the pushout is an object D such that C maps to the same place both ways, and so that, contingent on this, it behaves much like a coproduct.
### 2 Free and forgetful functors
Recall how we defined a free monoid as all strings of some alphabet, with concatenation of strings the monoidal operation. And recall how we defined the free category on a graph as the category of paths in the graph, with path concatenation as the operation.
The reason we chose the word free to denote both these cases is far from a coincidence: by this point nobody will be surprised to hear that we can unify the idea of generating the most general object of a particular algebraic structure into a single categorical idea.
The idea of the free constructions, classically, is to introduce as few additional relations as possible, while still generating a valid object of the appropriate type, given a set of generators we view as placeholders, as symbols. Having a minimal amount of relations allows us to introduce further relations later, by imposing new equalities by mapping with surjections to other structures.
One of the first observations in each of the cases we can do is that such a map ends up being completely determined by where the generators go - the symbols we use to generate. And since the free structure is made to fulfill the axioms of whatever structure we're working with, these generators combine, even after mapping to some other structure, in a way compatible with all structure.
To make solid categorical sense of this, however, we need to couple the construction of a free algebraic structure from a set (or a graph, or...) with another construction: we can define the forgetful functor from monoids to sets by just picking out the elements of the monoid as a set; and from categories to graph by just picking the underlying graph, and forgetting about the compositions of arrows.
Now we have what we need to pinpoint just what kind of a functor the free widget generated by-construction does. It's a functor $F: C\to D$, coupled with a forgetful functor $U: D\to C$ such that any map $S\to U(N)$ in C induces one unique mapping $F(S)\to N$ in D.
For the case of monoids and sets, this means that if we take our generating set, and map it into the set of elements of another monoid, this generates a unique mapping of the corresponding monoids.
This is all captured by a similar kind of diagrams and uniquely existing maps argument as the previous object or morphism properties were defined with. We'll show the definition for the example of monoids.
Definition A free monoid on a generating set X is a monoid F(X) such that there is an inclusion $i_X: X\to UF(X)$ and for every function $X\to U(M)$ for some other monoid M, there is a unique homomorphism $g:F(X)\to M$ such that f = U(g)iX, or in other words such that this diagram commutes:
We can construct a map $\phi:Hom_{Mon}(F(X),M) \to Hom_{Set}(X,U(M))$ by $\phi: g\mapsto U(g)\circ i_X$. The above definition says that this map is an isomorphism.
### 3 Adjunctions
Modeling on the way we construct free and forgetful functors, we can form a powerful categorical concept, which ends up generalizing much of what we've already seen - and also leads us on towards monads.
We draw on the definition above of free monoids to give a preliminary definition. This will be replaced later by an equivalent definition that gives more insight.
Definition A pair of functors,
is called an adjoint pair or an adjunction, with F called the left adjoint and U called the right adjoint if there is natural transformation $\eta: 1\to UF$, and for every $f:A\to U(B)$, there is a unique $g: F(A)\to B$ such that the diagram below commutes.
The natural transformation η is called the unit of the adjunction.
This definition, however, has a significant amount of asymmetry: we can start with some $f:A\to U(B)$ and generate a $g: F(A)\to B$, while there are no immediate guarantees for the other direction. However, there is a proposition we can prove leading us to a more symmetric statement:
Proposition For categories and functors
the following conditions are equivalent:
1. F is left adjoint to U.
2. For any $c\in C_0$, $d\in D_0$, there is an isomorphism $\phi: Hom_D(Fc, d) \to Hom_C(c,Ud)$, natural in both c and d.
moreover, the two conditions are related by the formulas
• $\phi(g) = U(g) \circ \eta_c$
• ηc = φ(1Fc
Proof sketch For (1 implies 2), the isomorphism is given by the end of the statement, and it is an isomorphism exactly because of the unit property - viz. that every $f:A\to U(B)$ generates a unique $g: F(A)\to B$.
Naturality follows by building the naturality diagrams
Image:NaturalityDiagramsAdjointProposition.png
and chasing through with a $f: Fc\to d$.
For (2 implies 1), we start out with a natural isomorphism φ. We find the necessary natural transformation ηc by considering $\phi: Hom(Fc,Fc) \to Hom(c, UFc)$.
QED.
By dualizing the proof, we get the following statement:
Proposition For categories and functors
the following conditions are equivalent:
1. For any $c\in C_0$, $d\in D_0$, there is an isomorphism $\phi: Hom_D(Fc, d) \to Hom_C(c,Ud)$, natural in both c an
2. There is a natural transformation $\epsilon: FU \to 1_D$ with the property that for any $g: F(c) \to d$ there is a unique $f: c\to U(d)$ such that $g = \epsilon_D\circ F(f)$, as in the diagram
Image:NaturalityDiagramsDualAdjointProposition.png
moreover, the two conditions are related by the formulas
• $\psi(f) = \epsilon_D\circ F(f)$
• εd = ψ(1Ud
where ψ = φ − 1.
Hence, we have an equivalent definition with higher generality, more symmetry and more horsepower, as it were:
Definition An adjunction consists of functors
and a natural isomorphism
The unit η and the counit ε of the adjunction are natural transformations given by:
• $\eta: 1_C\to UF: \eta_c = \phi(1_{Fc})$
• $\epsilon: FU\to 1_D: \epsilon_d = \psi(1_{Ud})$.
Some of the examples we have had difficulties fitting into the limits framework show up as adjunctions:
The free and forgetful functors are adjoints; and indeed, a more natural definition of what it means to be free is that it is a left adjoint to some forgetful functor.
Curry and uncurry, in the definition of an exponential object are an adjoint pair. The functor $-\times A: X\mapsto X\times A$ has right adjoint $-^A: Y\mapsto Y^A$.
#### 3.1 Notational aid
One way to write the adjoint is as a bidirectional rewrite rule:
$\frac{F(X) \to Y}{X\to G(Y)}$,
where the statement is that the hom sets indicated by the upper and lower arrow, respectively, are transformed into each other by the unit and counit respectively. The left adjoint is the one that has the functor application on the left hand side of this diagram, and the right adjoint is the one with the functor application to the right.
### 4 Homework
Complete homework is 6 out of 10 exercises.
1. Prove that an equalizer is a monomorphism.
2. Prove that a coequalizer is an epimorphism.
3. Prove that given any relation $R\subseteq X\times X$, its completion to an equivalence relation is the kernel of the coequalizer of the component maps of the relation
4. Prove that if the right arrow in a pullback square is a mono, then so is the left arrow. Thus the intersection as a pullback really is a subobject.
5. Prove that if both the arrows in the pullback 'corner' are mono, then the arrows of the pullback cone are all mono.
6. What is the pullback in the category of posets?
7. What is the pushout in the category of posets?
8. Prove that the exponential and the product functors above are adjoints. What are the unit and counit?
9. (worth 4pt) Consider the unique functor $!:C\to 1$ to the terminal category.
1. Does it have a left adjoint? What is it?
2. Does it have a right adjoint? What is it?
10. Suppose is an adjoint pair. Find a natural transformation $FUF\to F$. Conclude that there is a natural transformation $\mu: UFUF\to UF$. Prove that this is associative, in other words that the diagram
commutes.
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http://cs.stackexchange.com/questions/1957/master-theorem-not-applicable
|
# Master theorem not applicable?
Given the following recursive equation
$$T(n) = 2T\left(\frac{n}{2}\right)+n\log n$$ we want to apply the Master theorem and note that
$$n^{\log_2(2)} = n.$$
Now we check the first two cases for $\varepsilon > 0$, that is whether
• $n\log n \in O(n^{1-\varepsilon})$ or
• $n\log n \in \Theta(n)$.
The two cases are not satisfied. So we have to check the third case, that is whether
• $n\log n \in \Omega(n^{1+\varepsilon})$ .
I think the third condition is not satisfied either. But why? And what would be a good explanation for why the Master theorem cannot be applied in this case?
-
7
– Saeed Amiri May 20 '12 at 21:06
4
Case three is not satisfied because $\log n$ is not $\Omega(n^\epsilon)$ for any $\epsilon > 0$. Use l'Hôpital's rule on the limit $\frac{\log n}{n^{\epsilon}}$ – sdcvvc May 20 '12 at 21:26
1
Once you show that neither case applies, that is proof that you can not apply the master theorem as stated. – Raphael♦ May 20 '12 at 21:53
Who needs the Master Theorem? Use recursion trees. – JeffE Jun 17 '12 at 11:27
– Raphael♦ Jun 18 '12 at 18:27
## 2 Answers
The three cases of the Master Theorem that you refer to are proved in the Introduction to Algorithms by Thomas H. Cormen,Charles E. Leiserson,Ronald L. Rivest and Clifford Stein (2nd Edition, 2001).
It is correctly observed that the recurrence in question falls between Case 2 and Case 3. That is $f(n) = n \log n$ grows faster than $n$ but slower than $n^{1+\varepsilon}$ for any $\varepsilon > 0$.
However the theorem can be generalized to cover this recurrence. Consider
Case 2A: Consider $f(n) = \Theta (n^{\log_b a }\log_b^k n)$ for some $k\geq 0$.
This case reduces to Case 2 when $k = 0$. It is intuitively clear that along each branch of the recurrence tree $f(x)$ is being added $\Theta (\log_b n)$ times. A sketch of a more formal proof can be found below. The final result is that
$$T(n) = \Theta (n^{\log_b a }\log_b^{k+1} n)$$.
In the Introduction to algorithms this statement is left as an exercise.
Applying this statement to the recurrence in question we finally get $$T(n) = \Theta(n \cdot \log^2 n).$$
More details on the Master Theorem can be found in the excellent (imho) Wikipedia page.
As @sdcvvc points in the comments to prove that Case 3 does not apply here one can invoke L'Hospital's rule that says that $$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f^\prime(x)}{g^\prime(x)}$$ for any functions $f(x)$ and $g(x)$ differentiable in the vicinity of $c$. Applying this to $f(n) = n \log n$ and $g(n) = n^{1+\varepsilon}$ one can show that $\log n \not\in \Theta (n^{1+\varepsilon}).$
Sketch of the Proof of the Master Theorem for Case 2A.
This is a reproduction of parts of the proof from Introduction to Algorithms with the necessary modifications.
First we prove the following Lemma.
Lemma A:
Consider a function $$g(n) = \displaystyle \sum_{j=0}^{\log_b {n-1}} a^j h(n/b^j)$$ where $h(n) = n^{\log_b a} \log_b^k n.$ Then $g(n) = n^{\log_b a} \log_b^{k +1} n.$
Proof: Substituting the $h(n)$ into the expression for $g(n)$ one can get $$g(n) = \displaystyle n^{\log_b a} \log_b^k n \; \sum_{j=0}^{\log_b{n-1}} \left(\frac{a}{b^{\log_b a}}\right)^j = n^{\log_b a} \log_b^{k +1} n.$$
QED
If $n$ is an exact power of $b$ given a recurrence $$T(n) = a T(n/b) + f(n),\quad T(1) = \Theta(1)$$ one can rewrite it as $$T(n) = \Theta(n^{\log_b a}) + \displaystyle \sum_{j=0}^{\log_b n - 1} a^j T(n/b^j).$$ Substituting $f(n)$ with $\Theta(n^{\log_b a} \log_b^k n)$, moving $\Theta$ outside and applying Lemma A we get
$$T(n) = \Theta (n^{\log_b a }\log_b^{k+1} n).$$
Generalizing this to an arbitrary integer $n$ that is not a power of $b$ is beyond the scope of this post.
-
The Akra-Bazzi theorem is a strict generalization of the master theorem. As a bonus its proof is a blizzard of integrals that will make your head spin ;-)
In any case, Sedgewick in his "Introduction to the analysis of algorithms" argues convincingly that one should strive to prove $T(n) \sim g(n)$ type asymptotics.
-
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http://mathoverflow.net/questions/101536?sort=oldest
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## is there a solution to system of linear Diophantine equations?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a matrix A \in Z^{n \by m}, where m > n and a vector b \in Z^n. Then, under what conditions does an integer solution exist to the equation
Ax = b.
Is there a way to bound the norm of the solution vector x in terms of norms of A and b?
Essentially I want something like Siegel's lemma but for the inhomogeneous case.
I am not an expert on this and will appreciate any help.
Thanks!
-
2
$A$ gives a map from $\mathbb Z^n$ to $\mathbb Z^m$. To check existence of a solution, you first want to know the cokernel of this map. You can compute the size of this cokernel by taking the greatest common divisor of the $m\times m$ minors. To check if there is a solution, you just need to check modulo all the primes dividing this. Since that is over a field you can use standard linear algebra. I don't know how to find out how far away the closest root is from the origin. – Will Sawin Jul 6 at 23:07
## 3 Answers
I don't know the answer to the second question (bounding the norm), but for the first, just compute the Smith normal form of A (and transform $b$ appropriately).
-
Igor: I don't think this works since the inverse of an integer matrix need not be an integer matrix. Can you elaborate on your solution? – Vidit Nanda Jul 7 at 5:11
2
Igor, you probably meant Hermite n.f., not Smith n.f. Vel, see my comment for a reference to details on this. – Dima Pasechnik Jul 7 at 5:28
1
@Vel: I dont understand you comment. The "conjugating" matrices in SNF are invertible over $\mathbb{Z}$. So, if $M D N = A,$ then $A x = b$ can be written as $D (N x) = M^{-1} b,$ from which the existence or lack thereof to the original system is clear. – Igor Rivin Jul 7 at 13:29
### 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.
This kind of questions arise very often in integer linear optimization. It is well-known that the bitsize of a solution will be polynomially bounded by the sizes of of $A$ and $b$, i.e. by the maximum bitsize of entries of $A$ and $b$, and by $\max(n,m)$. See e.g. Corollary 5.2a in the A.Schrijver's book.
There are many sufficient conditions known for the existence of an integer solution, e.g. $A$ being totally unimodular (i.e. each square submatrix has determinant 0,1,or -1) is one.
Your problem is in fact easier (in optimization one often assumes $x\geq 0$), and Will gives a good suggestion in his comment above. The book in Sect. 5.3 also gives more details on the algorithm Igor described in his answer.
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Thanks! This is very helpful. --SAK
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Dear SAK, I've merged your accounts. If you register, you won't have to worry about unintentionally creating new accounts. – S. Carnahan♦ Jul 11 at 4:51
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http://physics.stackexchange.com/questions/24534/noise-level-units-confusion?answertab=oldest
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# noise level (units confusion)
i had a question in one of my classes regarding SNR in underwater acoustic channels. There are a couple of terms with the unit `dB re uPa`. I know it stands for dB with reference to uPa but I am not exactly sure what it means. Can I convert it to dB. If yes, how?
Thanks in advance!!
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## 1 Answer
dB$\mu$Pa would mean "dB's with respect to 1 $\mu$Pa".
If I have a pressure of X Pascals, then to express it in dB$\mu$Pa, I would compute
$$20log_{10}(\frac{X}{1\mu Pa})$$
i.e
$$20log_{10}(10^6X)$$
since there are $10^6$ $\mu$Pa per Pa.
So the rule is: compute how many micropascals you have and take $20log_{10}$ of that number.
For the second part of your question, "converting it to dB", you would have to specify exactly which dB quantity you were interested in. Wikipedia lists a rather large number of options.
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2
The general rule about finding dB is a little more complicated than this suggests. dBmW is "dB's with respect to 1 milliwatt." If we have a power M mW, in dB we have: $10 log_{10}(M/1 mW)$. We use a factor of 20 for units we have decided are Amplitude and a factor of 10 for units we have decided are Intensity. The general idea is 10 X as much energy is 10 dB higher, energy $\propto$ Ampitude$^2$. – mwengler Apr 29 '12 at 17:12
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http://particlephd.wordpress.com/2009/02/15/a-brief-introduction-to-instantons/
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# High Energy PhDs
A discussion of particle physics and strings
February 15, 2009
## A Brief Introduction to Instantons
Posted by lanzr under Field Theory
1 Comment
This brief instroduction is based on David Tong’s TASI Lectures on Solitons Lecture 1:Instantons.
We’ll first talk about the instantons arise in SU(N) Yang-Mills theory and then explain the connection between them and supersymmetry. By the end we’ll try to explain how string theory jumps in this whole business.
Instantons are nothing but a special kind of solution for the pure SU(N) Yang-Mills theory with action $S=\frac{1}{2 e^2}\int d^4x Tr F_{\mu\nu}F^{\mu\nu}$. Motivated by semi-classical evaluation of path integral, we search for finite action solutions to the Euclidean equations of motion, $\mathcal{D}_{\mu} F^{\mu\nu}=0$. In order to have a finite action, we need field potential $A_{\mu}$ to be pure gauge at the boundary $\partial R^4=S^3$, i.e. $A_{\mu}=ig^{-1}\partial_{\mu}g$.
Then the action will be given roughly by a surface integral which depends on the third fundamental group of SU(N), given by $\Pi_3(SU(N))=k$, k is usually called the charge of the instanton. For the original action to be captured by this number k, we need to have self-dual or anti self-dual field strength.
A specific solution of $k=1$ for SU(2) group can be given by $A_{mu}=\frac{\rho^2(x-X)_{\nu}}{(x-X)^2((x-X)^2+\rho^2)}\bar{\eta_{\mu\nu}}^i(g\sigma^i g^{-1})$ where $Xs$ are coordinate parameter $\rho$ is scale parameter and together with the three generator of the group, we have 8 parameters called collective coordinates. $\eta$ is just some matrix to interwine the group index $i$ with the space index $\mu$.
For a given instanton charge $k$ and a given group SU(2), an interesting question is how many independent solutions we have. The number is usually counted by given a solution $A_{\mu}$and we try to find how many infinitisiaml perturbation of this solution $\delta_\alpha A_\mu$, known as zero modes, $\alpha$ is the index for this solution space, usually called moduli space.
When we consider a Yang-Mills theory with an instanton background instead of a pure Yang-Mills theory. We’d like to know if we still have non-trivial solutions, and especially if these solutions will give rise to even more collective coordinates. This is where fermion zero modes and supersymmetry comes in. For $\mathcal{N}=2$ or 4 supersymmetry in $D=4$, it’s better to promote the instanton to be a string in 6 dimentiona or a 5 brane in 10 dimensions respectively. The details of how to solve the equation will be beyond the scope of this introduction, and We’ll refer the reader to the original lecture notes by David Tong.
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http://nanoexplanations.wordpress.com/2011/12/02/polygon-rectangulation-part-1-minimum-number-of-rectangles/
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the blog of Aaron Sterling
# Polygon rectangulation, part 1: Minimum number of rectangles
Posted on December 2, 2011
Over the next few posts, I will consider problems of polygon rectangulation: given as input $P$ an orthogonal polygon (all interior angles are 90 or 270 degrees), decompose $P$ into adjacent, nonoverlapping rectangles that fully cover $P$. Different problems impose different conditions on what constitutes a “good” rectangulation. Today we will discuss how to find a rectangulation with the least number of rectangles.
Polygon decomposition is a method often used in computer graphics and other fields, in order to break a (perhaps very complex) shape into lots of small manageable shapes. Polygon triangulation may be the best-studied decomposition problem. (When triangulating, we don’t require that the input polygon be orthogonal, and our objective is to cut the polygon into triangles according to some notion of “best” decomposition.) There is an extensive literature on polygon rectangulation as well, because of its connection to VLSI design. Suppose, for example, that our input $P$ represents a circuit board, and we want to subdivide the board by placing as little additional “ink” on the board as possible, in order to save money on each unit. However, because of mechanical limitations, we can only place ink horizontally or vertically — i.e., only create rectangulations of $P$. Many questions in VLSI design are closely related to finding a rectangulation of minimum total length, which I will discuss in a future post. The algorithm for minimum-length rectangulation is more complicated than the one I will present today for minimum-number-of-rectangles rectangulation, so today’s post can be considered a warm-up.
The attendees of the recent Midwest Theory Day know that Ming-Yang Kao and I found an application of rectangulation to DNA self-assembly. I will blog about that in the new year. The only other application of rectangulation to self-assembly that I know about is A Rectangular Partition Algorithm for Self-Assembly, by Li and Zhang, which appeared in a robotics conference. (Readers interested in the latest Midwest Theory Day are invited to check out a “workshop report” I wrote on the CSTheory Community Blog.)
These slides (pdf format) by Derrick Stolee contain many lovely pictures about polygon rectangulation. I think they may be a bit hard to follow all the way through as there is no ”attached narrative,” but I recommend them anyway.
We can divide orthogonal polygons into two classes: those that contain holes, and those that are hole-free. Hole-free polygons are also called simple polygons, and we will limit ourselves to rectangulation of simple orthogonal polygons for the remainder of this post.
By convention, we assume the input size to our rectangulation problem is $n$, the number of vertices of polygon $P$. Note that this means that a square with a huge interior is “smaller” than an L-shape with a small interior: complexity of the boundary is more important than a shape’s total area. The justification for this is that a rectangulation algorithm is likely to receive a set of vertices as input, and we assume that each pair of vertex coordinates can be stored in some constant amount of memory, so the complexity of executing a rectangulation is identical for every polygon that is similar (in the Euclidean sense) to $P$.
In an orthogonal polygon, there are two types of vertices: those at the point of a 90 degree interior angle (convex vertices), and those at the point of a 270 degree interior angle (concave vertices). In many rectangulation algorithms, the number of concave vertices is a fundamental parameter; indeed, at the end of this post, we will change the input size $n$ to the number of concave vertices of $P$, not the total number of vertices of $P$. At first, this may seem pointless, since the asymptotic complexity of the algorithm remains the same, regardless of the definition of $n$, as there are exactly 4 fewer concave vertices than convex vertices in any orthogonal polygon. To gain an intuition for why authors focus on concave vertices, we will build a rectangulation algorithm for any polygon from a rectangulation algorithm that assumes the concave vertices of $P$ have a niceness property.
Let $v$ and $w$ be concave vertices of $P$ that do not share the same edge of $P$. Call $v$ and $w$ cogrid if they lie on the same horizontal or vertical line. A chord is a line segment fully contained in $P$ that connects two cogrid (concave) vertices. The follolwing is a simple algorithm that finds a rectangulation for any $P$ that contains no chords. I believe it appeared first in “Minimal Rectangular Partition of Digitized Blobs,” by Ferrari et al. — a paper whose title I love, but which I can only find behind paywalls, unfortunately.
Algorithm 1
1. Input: $P$ a simple orthogonal polygon with no chords.
2. For each concave vertex, select one of its incident edges. (Two edges are incident to each concave vertex.)
3. Extend this edge until it hits another such extended edge, or a boundary edge of $P$.
4. Return the extensions of edges as the rectangulation.
So, if $P$ has no chords, the number of segments we need to draw is exactly the number of concave vertices of $P$. We will now use this algorithm as a subroutine in a larger algorithm to rectangulate any simple $P$.
Rectangulation of simple polygons with chords
Now let’s assume $P$ has 1 or more chords. Let $b$ be the size of the largest set of nonintersecting chords, and let’s redefine $n$ to be the number of concave vertices (no longer just the number of vertices). Ferrari et al. showed that the number of rectangles in a minimum-rectangle partition of $P$ is $n-b+1$. Here is their algorithm to achieve that.
Algorithm 2
1. Input $P$ a simple orthogonal polygon.
2. Find the chords of $P$.
3. Construct a bipartite graph with edges between vertices in the sets $V$ and $H$, where each vertex in $V$ corresponds to a vertical chord, and each vertex in $H$ corresponds to a horizontal chord. Draw an edge between vertices $v \in V$ and $h \in H$ iff the chords corresponding to $v$ and $h$ intersect.
4. Find a maximum matching $M$ of the bipartite graph.
5. Use $M$ to find a maximum independent set $S$ of vertices of the bipartite graph. (This set corresponds to a maximum set of nonintersecting chords of $P$.)
6. Draw the chords corresponding to $S$ in $P$. This subdivides $P$ into $|S|+1$ smaller polygons, none of which contains a chord.
7. Using Algorithm 1, rectangulate each of the chordless polygons.
8. Output the union of the rectangulations of the previous step.
This algorithm is discussed in Graph-Theoretic Solutions to Computational Geometry Problems by David Eppstein, where he talks about work that was done to optimize the algorithm through a better maximum matching algorithm, and using a specialized geometric data structure. Ferrari et al. used the Hopcroft/Karp maximum matching algorithm to obtain a rectangulation algorithm that runs in time $O(n^{2.5})$. Rather than go into more detail, I will now outline a method that yields a strictly better bound.
Rectangulation in time $O(n \log \log n)$
Through a different method, Liou et al. demonstrated an algorithm with the property stated in the title of their article: Minimum Partitioning Simple Rectilinear Polygons in $O(n \log \log n)$ Time. The algorithm is technical and lengthy; I will focus on a few points.
Liou et al. sidestep the (relatively) expensive step of building the bipartite graph in Algorithm 2. The outline of their algorithm is:
Algorithm 3 (outline)
1. Input $P$ a simple orthogonal polygon.
2. Find a maximum matching of the chords of $P$ without constructing the bipartite graph, by taking advantage of the specialized geometry of those chords. (This requires $O(n \log \log n)$ time.)
3. Given the maximum matching, find a set of maximum nonintersecting chords of $P$. (This step is essentially the same as before, and requires $O(n)$ time.)
4. Use Algorithm 1 to partition each chord-free subpolygon obtained from the nonintersecting chords found in Step 3, (Requires $O(n)$ time.)
So Algorithm 3, overall, runs in $O(n \log \log n)$ time.
W.T. Liou, J.J.M. Tan, & R.C.T. Lee (1989). Minimum Partitioning Simple Rectilinear Polygons in O(n log log n) Time Symposium on Computational Geometry, 344-353 : 10.1145/73833.73871
FERRARI, L., SANKAR, P., & SKLANSKY, J. (1984). Minimal rectangular partitions of digitized blobs Computer Vision, Graphics, and Image Processing, 28 (1), 58-71 DOI: 10.1016/0734-189X(84)90139-7
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### 3 Responses to Polygon rectangulation, part 1: Minimum number of rectangles
1. Derrick Stolee
Thanks for using my slides! I put those together because the multi-dimensional problem (shown briefly at the end) arose during work for my undergrad thesis. As far as I know, there is no polynomial-time algorithm OR NP-completeness proof for three or more dimensions.
I would be very interested to see an algorithm or completeness proof to the multi-dimensional case.
2. Pingback: Polygon rectangulation, part 2: Minimum number of fat rectangles | Nanoexplanations
3. Pingback: Polygon rectangulation wrap up | Nanoexplanations
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http://math.stackexchange.com/questions/116042/finding-the-component-of-a-vector-tangent-to-a-circle
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# Finding the component of a vector tangent to a circle
## Problem
Given a vector and a circle in a plane, I'm trying to find the component of the vector that is tangent to the circle. The location of the tangent vector is unimportant; I only need to know its magnitude and whether it is directed clockwise or counter clockwise about the circle.
## Background (just in case you're interested)
I'm writing an iPhone app that displays a dial. The user is allowed to spin the dial by dragging it around with his finger. If the user then flicks his finger in some direction (and lets go), the dial should continue to spin before coming to a stop. For added realism, the amount of spin should be directly proportional to the velocity of the flick. In other words, I'm simulating the inertia and momentum of the dial.
The OS already provides a vector describing the velocity and direction of the flick. Note that if it happened to be tangential to the dial, I could simply find the magnitude of the vector and use it directly as the amount of spin. But the flick could be in any direction -- even outside the dial. A flick along the radius of the dial, for example, should result in no motion. Therefore, I need to find:
1. The component of the flick vector that is tangential to the dial
2. Whether this tangent vector is clockwise or counter clockwise around the dial
With this information, I can calculate how much spin should be put on the dial by finding the magnitude of the tangent vector.
## Illustration
That might not be clear, so here's a diagram to illustrate:
• $V$: flick vector
• $P$: start point of the vector
• $D$: midpoint of the vector
• $T$: tangent vector I'm trying to find
• $E$: point where the tangent vector touches the circle
• $R$: radius of the circle
• $C$: center of the circle
• $T'$: another way of looking at the tangent vector (for an alternate approach described later)
• $V_{2}$, $V_{3}$: other possible flick vectors
## My approach
My first approach was to derive an equation for the tangent line ($T$).
The midpoint $D$ would be given by:
$D = ( P_{x} + \frac{V_{x}}{2}, P_{y} + \frac{V_{y}}{2})$
The slope $m$ of line $\overline{C D}$ would be:
$m = \frac{D_{y} - C_{y}}{D_{x} - C_{x}} = \frac{P_{y}+\frac{V_{y}}{2} - C_{y}}{P_{x} + \frac{V_{x}}{2} - C_{x}}$
And then the equation for line $\overline{C D}$ would be:
$y - C_{y} = m(x - C_{x})$
$E$ would be the intersection of that line and the circle:
$(x - C_{x})^{2} + (y - C_{y})^{2} = R^{2}$
By solving for $x$ and $y$ in the $\overline{C D}$ equation and substituting into the circle equation, I get:
$x = C_{x} ± \frac{R}{\sqrt{1 + m^{2}}}$
$y = C_{y} ± \frac{R}{\sqrt{1 + m^{2}}}$
These $x, y$ values are the coordinates of point $E$.
Now I finally have an equation for the tangent line $T$. I simply use the perpendicular slope of line $\overline{C D}$ and the coordinates of point $E$:
$y - E_{y} = - \frac{1}{m}(x - E_{x})$
I've verified that my work is correct by plugging in some numbers, but I'm not sure what to do next. I still don't have the magnitude of the tangent vector.
## Alternate approach
As an alternate approach, I thought of ignoring $T$ altogether and considering $T'$ instead, since I only need to know the magnitude of the tangent vector. But in the right triangle I only know the length of the hypotenuse ($|V|$). That's not enough information to determine the length of $T'$.
Or, if I could somehow determine the $x, y$ components of $T'$, then I believe the dot product ($V \cdot T'$) would give me the magnitude of the tangential component of the vector V. (Please correct me if I'm wrong.) But I'm not sure how to get those values in the first place.
How can I proceed? Thanks for any suggestions.
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## 2 Answers
The vector $T$ is the component of $V$ perpendicular to $CD$. Assuming the center of the circle $C$ is the origin, this is just $$T = V - \frac{D(D\cdot V)}{\lVert D\rVert^2}.$$
If you want the signed magnitude of the vector, with positive being anticlockwise and negative being clockwise, you should take the cross product of $D/\lVert D\rVert$ and $V$ instead. In two dimensions, the cross product of two vectors is a scalar, $$u \times v = \begin{vmatrix}u_x & v_x \\ u_y & v_y\end{vmatrix} = u_xv_y - u_yv_x.$$
If the circle is not centered at the origin, just replace $D$ with $D - C$ in all of the above expressions.
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Thanks for your answer! When you say $D$, I assume you mean the vector from C to D. I applied your formula for $T$ and I seem to be getting the correct values back. However, I'm not understanding your technique for getting the circular direction. I thought the cross product is always a vector, not a scalar. In any case, I applied your function and got a negative value for V (clockwise -- correct) but also a negative value for $V_{2}$ (clockwise -- incorrect). Any thoughts? Thanks. – vocaro Mar 5 '12 at 6:19
Strictly speaking, the cross product is a function from pairs of 3D vectors to 3D vectors, but if you have two 2D vectors $\vec u$ and $\vec v$, the cross product of $(u_x, u_y, 0)$ and $(v_x, v_y, 0)$ is the vector $(0, 0, w)$ where $w = u_xv_y - u_yv_x$, so one can think of $w$ as "the cross product in 2D" of $\vec u$ and $\vec v$. I don't know if it has a more specific name, but it is a fairly natural thing to do. From the picture, I don't think the value for $\vec V_2$ should be negative. Can you tell me the coordinates of $\overrightarrow{CD}$ and $V$ in that case? – Rahul Narain Mar 5 '12 at 17:04
Sorry, my calculations were wrong. $V_{2}$ is indeed positive as you predicted. I have accepted your answer; thank you for your help! – vocaro Mar 6 '12 at 5:25
I think that the simplest way would be like this:
Firstly, find the angle between (the line from the center of the circle to the midpoint of the Vector) and (the Vector itself). (Do you know how to do this? If not, comment and we can expand). I'm capitalizing the V in the Vector we start with to distinguish it.
Once you have this angle $\theta$, then the length of the tangent you're looking for is given by $L\sin^{} \theta$, where $L$ is the length of the Vector. And whether it's clockwise or counterclockwise should come from your calculation of the angle.
Or we could do everything with complex numbers. If the angle that the midpoint-line makes with the positive real axis is $\theta$, then multiply the complex representatives of the endpoints of the Vector by $e^{-i\theta}$, and calculate the imaginary part of the difference $d$ between the endpoints. Then positive and negative correspond to positive and negative imaginary part, respectively. And for that matter, the tangent vector is given by $e^{i \theta} \text{Im}(d)$, where Im stands for the imaginary part.
If you know some complex, then you'll see we rotated the vector so that our tangent point on the circle is always $(1,0)$, so we took the $y$ component and rotated it back.
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Thanks for your answer! Unfortunately I'm not sure which angle you're calling θ. Do you mean CDP or CPD? I can compute either one using the dot product formula, but after that I'm not sure what to do. For example, I plugged in numbers and got 1.3 radians (75 degrees) for angle CDP, which looks right. But I can't plug this value into your $Lsin^{-1}θ$ formula because the inverse sine only allows values from -1 to 1. By the way, what formula is that? How was it derived? Thanks. – vocaro Mar 4 '12 at 4:09
@Vocaro: Whoops! I didn't mean to include the inverse part of the inverse sine. That was a terrible mistake, and of course you shouldn't get any real value for that part. I just meant to refer to the triangle, that's all. – mixedmath♦ Mar 4 '12 at 6:05
Thanks for the correction. When I apply the dot product formula to find the angle CDP, it looks correct, and when I apply $Lsinθ$, the tangent length looks correct also. However, you said "whether it's clockwise or counterclockwise should come from your calculation of the angle." The dot product formula I'm using is $arccos(\frac{A_{x}B_{x}+A_{y}B_{y}}{|A||B|})$, which will always give a positive angle. Am I using the right formula here? – vocaro Mar 4 '12 at 20:03
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http://physics.stackexchange.com/questions/20704/what-is-the-most-efficient-way-to-use-hand-dryer
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# What is the most efficient way to use hand dryer?
What's the most efficient way to place your hands under the hand dryer?
Let's assume that dryer creates simple donward flow of hot air.
Here are some examples:
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The leftmost one. In the other two your hands are overlapped so that area isn't getting any hot air. :) – Pratik Deoghare Feb 8 '12 at 13:21
Of course, if you are talking about energy efficient , then its a moot point as most bathroom hand dryers stay on for a while after you finish (or turn off too early, in which canse you have to turn them on again). – Manishearth♦ Feb 8 '12 at 13:54
@Manishearth I'm talking about speed. The goal is to minimize time you need to dry your hands – Poma Feb 8 '12 at 14:28
@PratikDeoghare You can hold your hands this way without overlapping. And of course those variants are just an example, feel free to provide your own solutions. – Poma Feb 8 '12 at 14:30
@Poma I was kidding. – Pratik Deoghare Feb 8 '12 at 15:36
## 3 Answers
If you want to dry your hands completely, you need to turn them over and over anyways.
But the best approach would be if you have your palms facing up/down (can't tell which of these diagrams it is). You need to maximise the surface area of the exposed part, as heat you receive will be $\propto (surface\space area)\times(time)\times\cos(angle\space of\space tilt)$.
So the best way would be to keep your palms separately, and rotate them slowly. Surprisingly, the heat needs to fall on the sides of your hand for the same amount of time as it needs to fall on your palms. This is because the sides of your hands absorb less heat (less surface area), but they have less water that needs to be evaporated. So just turning your hands in circles is the best idea. No need to touch them together and try to squeeze the water out (You're drying your hands, not washing them with soap).
EDIT: If you want the most efficient way to dry your hands, then yes, rubbing them together helps (not squeezing, though). The most efficient way would be that you dry your palms, on both sides. Then, you rub your palms against the sides of your hands. Now repeat. Spreading the water around speeds up drying, as the water film reduces its thickness uniformly. Shaking your hands before drying also speeds it up.
In the comments, you mentioned that "air flow" causes evaporation, and not heat. This is only partially true. When water evaporates, it "steals" heat from the surroundings to do so (even if it doesn't heat up to 100 degrees C, it still has to take latent heat). If the air is stagnant, it will slowly lose heat and cool down, reducing the rate at which it can give heat, thus reducing evaporation. Aside from that, it will become more humid, again reducing the rate. When you have a wind, the old air (cold and humid) is constantly replaced by the new air (warm and dry), so you get a good constant rate of evaporation. Yes, a faster airflow will increase the rate of evaporation (mainly due to the surface humidity that it manages to carry away), but temperature is also a big factor.
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When I place them perpendicular to floor I feel air flow on both sides but it has less pressure – Poma Feb 8 '12 at 14:32
First thing, its heat you want, not pressure. Anyways, I agree that its not effective in drying your palms, but it is required if you want to dry the sides of your hands. Read the third paragraph carefully. – Manishearth♦ Feb 8 '12 at 14:42
The important thing is that any part of your hand should spend the same time underneath the dryer, regardless of the size. Of course, when rotating your hands, the shape of your hands does come into the picture (due to the $\cos\theta$ term) and makes the optimization complicated. – Manishearth♦ Feb 8 '12 at 14:45
When palms become dry I can rub my hands to redistribute water – Poma Feb 8 '12 at 14:55
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And also I think the main thing that causes water to evaporate is actually air flow and not heat – Poma Feb 8 '12 at 14:56
show 4 more comments
I believe that both answers so far are partially completed. So, I will try to complete then.
Two determinant factors must be considered. On one hand, the water evaporation. On the other hand, the blowing of the water to the floor (by air flux). A typical drier uses them both.
As @Alexander pointed out, AirBlade technology is the current most efficient way of drying hands, and it basically uses air blowing. The pertinent point here is that the hands are parallel to the gravity force vector. That way, the work needed to make the water to leave is minimized.
For the case of a 'old' dryer, this idea stands: hands must be the closer to parallel the better: it minimizes the total needed work to pick all the elements of water and send them to the floor.
With this said, the correct answer will be the left one: the hands are parallel to the air flux, thus minimizing the work necessary to transport an element of water to the ground. All the other cases will just increase the necessary work to be done by the air flux.
However, I guess there will be a point where blowing does not compensate, mainly due to water viscosity. On that point, evaporation should be the main drying factor. On that case, @Manishearth point is the rule of tomb: perpendicular to the air flux, maximizing evaporation.
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1
In my experience they don't blow water droplets off your hand. And if they do, its more efficient to shake the bulk water off first manually. – Manishearth♦ Feb 26 '12 at 16:49
(Also @Manishearth) I'm not entirely convinced that putting your hands perpendicular to the flow is best for evaporation. For a cube at least, I've seen the correlation that $\text{Nu}=0.71\,\text{Re}^{0.52}\$ for the front and $\text{Nu}=0.12\,\text{Re}^{0.70}\$ for the sides. So for sufficiently turbulent flow, it'd make sense that the parallel arrangement offers better convection of heat and humidity. – rdhs Feb 26 '12 at 18:19
Neither of these examples are efficient by any standards. Instead of heating up the air which is consuming a lot of energy it is much better to increase the pressure and blow of the water instead of evaporating it. The currently most efficient hand dryer is the Dyson Airblade.
There are similar dryers available but the idea is always the same.
You can try this in any workshop that has compressed air. With a relatively small amount of air with a pressure of 8 bar you can dry your hands within seconds.
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The OP was asking for the most efficient way to dry your hands under a dryer. I doubt we are allowed to install a new dryer in the process.. – Manishearth♦ Feb 8 '12 at 14:48
Yes, the goal is to minimize time you need to dry your hands using given bathroom dryer – Poma Feb 8 '12 at 14:58
The airblade is awesome! – Lagerbaer Feb 23 '12 at 6:34
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http://math.stackexchange.com/questions/205482/what-is-the-fastest-algorithm-to-check-if-a-number-has-only-3-divisors
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# What is the fastest algorithm to check if a number has only 3 divisors?
Which is the fastest way to check if a number has only 3 unique factors ? Any help will be highly appreciated?
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check it's a square of another prime? – Patrick Li Oct 1 '12 at 16:12
@PatrickLi: another? – Marc van Leeuwen Oct 1 '12 at 16:41
@MarcvanLeeuwen check it's the square of a prime number. :) – Patrick Li Oct 1 '12 at 16:47
## 2 Answers
The fastest method will depend on the types of numbers you intend to check and how you are running the tests (pencil and paper, x86_64, etc.). But basically, check if the number has any small prime factors (say, the first 25: 2, 3, 5, ..., 97). If so, the number has exactly 3 divisors if and only if the number is the square of that prime. Otherwise you have a number with no tiny prime factors; now check if the number is a square. You can first test mod some promising moduli, like 63, 64, and 65. If it passes that test, take the integer square root $s$ and test if that, squared, gives the number. If not, the number has more than 3 divisors. If so, test if the remaining number is prime. The first step is using a probable-prime test, maybe checking if the number is a b-strong pseudoprime for some randomly-chosen $1 < b < s-1.$ If that fails the number is composite and so $s^2$ has more than three divisors. Otherwise apply a primality-proving test to $s$.
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Supposing that by "only 3 divisors" you mean "exactly 3 divisors", the only such numbers are numbers of the form $p^2$, where $p$ is prime; they are divisible by $1, p,$ and $p^2$. So the first thing to try is to test if the number $n$ is a perfect square, which you can do efficiently by calculating the integer square root $i = \lfloor \sqrt n\rfloor$ and then checking to see if $i^2=n$. If not, $n$ fails. If so, $n$ has the desired property if and only if $i$ is prime, which you can check using any of the usual primality tests.
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http://math.stackexchange.com/questions/19291/f1-fx-fx/19294
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# $f(1-f(x))=f(x)$
Find all continuous $f:[0,1] \rightarrow [0,1]$ such that $f(1-f(x))=f(x)$.
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3
Why? What have you tried? – Matthew Conroy Jan 27 '11 at 23:20
$f(x) = 1-x$, $\forall x \in [0,1]$ or $f(x) = 0$, $\forall x \in [0,1]$ – user17762 Jan 27 '11 at 23:26
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@bobokinks, it really looks like you are posting random functional equations! :) If you explained why you want to know the answer, what you have tried, why you expect that there is a sensible/interesting answer, &c, it'd be nice. – Mariano Suárez-Alvarez♦ Jan 27 '11 at 23:36
2
They're fun problems. As long as it doesn't become excessive I don't see why there has to be an explanation. – Zarrax Jan 27 '11 at 23:41
## 1 Answer
Let $m,M$ be the minimum and maximum $f$ achieves on $[0,1]$ (there are such since f is continuous). From the intermediate value theorem, for each $m\leq y \leq M$ there is an $x\in [0,1]$ such that $f(x)=y$, so $f(1-y)=f(1-f(x))=f(x)=y$. This shows that if $m \leq y\leq M$ then $f(1-y)=y$.
for the $[0,1] - [1-M, 1-m]$ you can extend $f$ any way you want as long as its range is in $[m,M]$.
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http://mathhelpforum.com/advanced-statistics/174729-unbiased-estimators.html
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# Thread:
1. ## unbiased estimators
suppose that a random variable X has a geometric distribution (pq^x) for which the parameter p is unknown (0<p<1). Find a statistic that will be an unbiased estimator of 1/p
Any help on this would be greatly appreciated, thank you so much
2. Do you have a sample or just one observation?
You do realize E(X)=1/p depending on how you define your distribution
http://en.wikipedia.org/wiki/Geometric_distribution
So, in that setting X+1 is the answer
3. Is there a way to prove X+1 is an unbiased estimator of 1/p? We usually write out proofs which I have not gotten the hang of yet
4. IF you are using $P(X=x)=pq^x$ for x=0,1,2,3...
Then E(X)=(1/p)-1, do you need a proof of that?
THEN E(X+1)=E(X)+1=(1/p)-1+1=1/p.
5. OR for $P(X=x)=pq^{x-1}$ for x=1,2,3...
then E(X)=1/p. IN that case X is the answer
But do you have a sample X_1, X_2,....? or just one X?
6. i think it is a sample
7. Then use $\bar X+1$
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http://electronics.stackexchange.com/questions/14708/maximise-q-factor-in-low-pass-sallen-and-key-filter?answertab=oldest
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# Maximise Q factor in low pass Sallen and Key Filter
How do I maximise the Q-factor of the low pass sallen and key design.
I have this equation: $Q = \sqrt{ \frac {(R1 \times R2 \times C1 \times C2)} {R1 \times C1 \times (1-A) + (R1 \times C2) + (R2 \times C2)}}$
but I'm not sure of the best way to maximise it!
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## 1 Answer
I'm not sure why you would want to do this - and presumably you have constraints such as the cutoff frequency and LF gain, but with an LF gain of 1 you can obtain an arbitrarily large Q (theoretically) by making C1 much larger than C2 whilst maintaining the product C1C2 to preserve the cutoff frequency. There will of course be a practical limit to both capacitor values.
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http://math.stackexchange.com/questions/81531/limits-of-integration-for-random-variable
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# Limits of integration for random variable
Suppose you have two random variables $X$ and $Y$. If $X \sim N(0,1)$, $Y \sim N(0,1)$ and you want to find k s.t. $\mathbb P(X+Y >k)=0.01$, how would you do this? I am having a hard time finding the limits of integration. How would you generalize $\mathbb P(X+Y+Z+\cdots > k) =0.01$? I always get confused when problems involve multiple integrals.
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## 1 Answer
Hint: Are the random variables independent?
If so, you can avoid integration by using the facts
• the sum of independent normally distributed random variables has a normal distribution
• the mean of the sum of random variables is equal to the sum of the means
• the variance of the sum of independent random variables is the sum of the variances
• for a standard normal distribution $N(0,1)$: $\Phi^{-1}(0.99)\approx 2.326$
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And if they are not independent, then knowing the distribution of each of them is not enough anyway. – Henning Makholm Nov 13 '11 at 5:37
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http://math.stackexchange.com/questions/9491/function-asymptotic-where-fx-fraca-o-frac1-sqrtxb-o-frac1
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# function asymptotic where $f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$
If $a$ and $b$ are positive real numbers, and if $f(x)$ has the following asymptotic property
$f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$
then is the following true?
$f(x) = \frac{a}{b} + O(\frac{1}{\sqrt{x}})$
This might look like homework but it isn't.
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I suppose you are interested in the asymptotics as $x \to \infty$? – Aryabhata Nov 8 '10 at 22:48
## 2 Answers
Yes. One way to see this is to actually do the long division (like the kind you learned in elementary school)! Unfortunately, typesetting that in full on this forum will overtax my LaTeX powers.
Anyway, dividing $b + O\left(\frac{1}{\sqrt{x}}\right)$ into $a + O\left(\frac{1}{\sqrt{x}}\right)$ yields $\frac{a}{b}$ with a remainder of $O\left(\frac{1}{\sqrt{x}}\right)$. So we have $$\frac{a + O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + \frac{O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + O\left(\frac{1}{\sqrt{x}}\right),$$ since $b + O\left(\frac{1}{\sqrt{x}}\right) = O(1).$
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It is true. In the spirit of epsilon/delta, you are challenged to prove that $|f(x)-\frac{a}{b}| \lt \frac{M}{\sqrt{x}}$ for $x\gt x_0$ where your challenger gives M and you have to find an $x_0$ that works. But you get to challenge back saying the numerator should be within $\frac{N}{\sqrt{x}}$ of $a$ and similarly the denominator should be within $\frac{P}{\sqrt{x}}$ of $b$. So take $N=\frac{M}{2b}$ and $P=\frac{aM}{2b^2}$ and take the larger of the $x_0$'s that come back.
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1
Upvoted for a good explanation and an interesting narrative. Also, use of the epsilon delta proof here is probably more rigorous. – AnonymousCoward Nov 9 '10 at 0:24
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@GottfriedLeibniz: Ross's answer is a good answer, but I would disagree that it's more rigorous. Big-O expressions have perfectly good (i.e., rigorously justified) algebraic rules that they follow. (See, for example, Chapter 9 of Concrete Mathematics.) My answer is based on those. – Mike Spivey Nov 9 '10 at 0:37
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@Mike fair enough. As someone less familiar with asymptotics his also made more sense to me because it was more 'first principles' than your answer. – AnonymousCoward Nov 9 '10 at 4:24
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@GottfriedLeibniz: That's what I suspected, which is partly why I answered your comment as I did. Big-O notation can seem quite mysterious at first, but if you work with it enough the mystery starts to disappear. That said, I still make mistakes. In fact, I even once made a Big-O mistake in a research paper that was subtle enough that none of the referees or editors caught it, either. Three years after the paper was published someone else wrote to the editor pointing out the mistake! – Mike Spivey Nov 9 '10 at 4:56
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http://mathoverflow.net/revisions/18779/list
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## Return to Answer
4 deleted 202 characters in body
[EDIT: As Jim pointed out, I interpreted "semi-simple group" as "semi-simple Lie group." "Semi-simple group" actually has several different interpretations depending on what category you like.]
The short answer is: no. In theory, there are explicit formulae for branchings based on the Weyl character formula, but no reasonable person would call these simple.
There are interesting results which give you some information about branching multiplicities, but all involve real work. For instance, if G and H are compact, you can obtain asymptotic information about H-invariants in the representations $V_{n\lambda}$ as a function of $n$: it's a polynomial, whose leading order is the dimension of $\mathcal{O}_{\lambda}/\!\!/H$, the symplectic reduction of the coadjoint orbit through $\lambda$ by $H$, and whose leading coefficient is the symplectic volume of this manifold.
If you prefer algebraic geometry, this polynomial is the Hilbert polynomial of the corresponding GIT quotient.
If $H$ is a root subalgebra, then life is a bit easier, and you can use combinatorial methods like crystals, but this is still not easy.
3 added 206 characters in body
[EDIT: As Jim pointed out, I interpreted "semi-simple group" as "semi-simple Lie group." "Semi-simple group" actually has several different interpretations depending on what category you like.]
The short answer is: no. In theory, there are explicit formulae for branchings based on the Weyl character formula, but no reasonable person would call these simple.
There are interesting results which give you some information about branching multiplicities, but all involve real work. For instance, if G and H are compact, you can obtain asymptotic information about H-invariants in the representations $V_{n\lambda}$ as a function of $n$: it's a polynomial, whose leading order is the dimension of $\mathcal{O}_{\lambda}/\!\!/H$, the symplectic reduction of the coadjoint orbit through $\lambda$ by $H$, and whose leading coefficient is the symplectic volume of this manifold.
If you prefer algebraic geometry, this polynomial is the Hilbert polynomial of the corresponding GIT quotient.
If $H$ is a root subalgebra, then life is a bit easier, and you can use combinatorial methods like crystals, but this is still not easy.
2 added 398 characters in body; added 251 characters in body; added 3 characters in body
The short answer is: no. In theory, there are explicit formulae for branchings based on the Weyl character formula, but no reasonable person would call these simple.
There are interesting results which give you some information about branching multiplicities, but all involve real work. For instance, if G and H are compact, you can obtain asymptotic information about H-invariants in the representations $V_{n\lambda}$ as a function of $n$: it's a polynomial, whose leading order is the dimension of $\mathcal{O}_{\lambda}/\!\!/H$, the symplectic reduction of the coadjoint orbit through $\lambda$ by $H$, and whose leading coefficient is the symplectic volume of this manifold.
If you prefer algebraic geometry, this polynomial is the Hilbert polynomial of the corresponding GIT quotient.
If $H$ is a root subalgebra, then life is a bit easier, and you can use combinatorial methods like crystals, but this is still not easy.
1
The short answer is: no. In theory, there are explicit formulae for branchings based on the Weyl character formula, but no reasonable person would call these simple.
There are interesting results which give you some information about branching multiplicities, but all involve real work.
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http://mathhelpforum.com/geometry/172440-triangle.html
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# Thread:
1. ## Triangle
Where does the square root 3 come from. I get the height to s^2 = H^2 + (s/2)^2 ==>
H = s/2 ==> A = B * H (B = s ) so (S*S)/(2 * 2) ==> A = s^2 / 4 ,,,where do i
get the square root 3.
2. Being able to see the triangle would be nice...
3. All sides are equal,,they are all s
4. Then use Heron's formula
Heron's formula - Wikipedia, the free encyclopedia
5. I just want to know where the square root 3 comes from in the formula,, the formula is for calculating the Area of a trinagle where all sides are the same lenght.
6. Originally Posted by Makron
I just want to know where the square root 3 comes from in the formula,, the formula is for calculating the Area of a trinagle where all sides are the same lenght.
remember the relationships betwen the sides of a 30-60-90 triangle?
7. Originally Posted by Makron
Where does the square root 3 come from. I get the height to s^2 = H^2 + (s/2)^2 ==>
H = s/2 ==> A = B * H (B = s ) so (S*S)/(2 * 2) ==> A = s^2 / 4 ,,,where do i
get the square root 3.
Of course, if you'd bothered to do as suggested, and make use of Heron's Formula, the appearance of $\displaystyle \sqrt{3}$ is obvious...
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http://mathoverflow.net/questions/3559/colloquial-catchy-statements-encoding-serious-mathematics/104057
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## Colloquial catchy statements encoding serious mathematics
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
As the title says, please share colloquial statements that encode (in a non-rigorous way, of course) some nontrivial mathematical fact (or heuristic). Instead of giving examples here I added them as answers so they may be voted up and down with the rest.
This is a community-wiki question: One colloquial statement and its mathematical meaning per answer please!
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Recent answers suggest this question is getting a bit long in the tooth. – S. Carnahan♦ Jul 6 at 5:38
## 53 Answers
Another one: not so much a catchphrase, but a nifty interpretation of a theorem:
"Suppose a human is walking a dog on the leash and they encounter a lamp post. Then, if the leash is kept short enough, the human and the dog wind around the post the same number of times."
I learned this interpretation of Rouche's Theorem from the textbook in complex analysis by Saff and Snider. They include pictures, too.
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By “the leash is kept short enough”, do you mean it can't be flung over the lamp? – Zsbán Ambrus Jul 5 at 21:31
### 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.
Truth is undefinable,
which is a statement of Tarski's theorem. More precisely,
Truth in a context where one can do arithmetic is undefinable in that context.
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A straight line is the shortest distance between two points.
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How can a line be a distance? – Rasmus May 10 2011 at 8:11
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Seems to me you're trying to make it fit into the conventions used by mathematicians, not shared by others. – Michael Hardy May 10 2011 at 21:41
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That's possible. The google result shocked me. ;) – Rasmus May 11 2011 at 12:14
show 3 more comments
I'm surprised that nobody has mentioned the famous,
"Four colors suffice."
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There is no free lunch. Refers to risk/reward in financial investment and the fact that an efficient market moves to the point where you can only make money by taking risk.
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Kronecker was wrong: God did not make the integers. He only made the empty set. Then He made mathematicians so they could make the integers from the empty set.
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Assume there was no empty set. Consider the set of all empty sets, ... [seen years ago in Martin Gardner; not sure of the original source] – Noam D. Elkies Jul 6 at 4:37
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The consequences of this little harmless looking statement are deep enough that I worry that the majority will think that it is provably wrong.
Simple statement:
• "You can pick a real number at random between 0 and 1, so that any number is as likely as any other."
more colloquially, (Freiling):
• "You can throw a dart at the unit square."
more computer-scienc-y:
• "The process of coin flipping to determine the binary digits of a real number converges to a unique well defined real number answer in the limit of infinitely many throws".
The interpretation of this statement is not that the probability distribution of the result is a well defined function, nor that statements about whether this number is in this or that Borel set can be assigned a probability--- both these assertions are true and boring. The assertion above is that a real number "x" produced in this way actually exists as an element of the mathematical universe, and every question you can ask of it, including "does x belong to this arbitrary subset S of [0,1]" gets a well defined yes or no answer in the limit. If you believe this assertion is self-evidently true, as I do, beware the implications!
• The continuum hypothesis is false. (Sierpinsky,Freiling)
For contradiction, well order [0,1] with order type aleph-1, then choose two numbers x,y at random in [0,1]. What is the probability that $y\le x$ in the well-ordering? Since the set ${ z|z\le y}$ is countable for any y, the answer is 0. The same thing works whenever sets of cardinality less than the continuum always have zero Lebesgue measure.
• Every subset of [0,1] has well-defined Lebesgue measure. (Solovay, more or less)
Make a countable list of independent numbers $x_i$, and ask for each one whether the number is in the set S or not. The fraction of random picks which land in S will define the Lebesgue measure of S. In more detail, if you write down a "1" every time $x_i$ is in S, and write down a zero when $x_i$ is not in S, then the number of ones divided by the number of throws converges to a unique real number, which defines the Lebesgue measure of S.
In this forum, somewhere or other, someone had the idea that this process will not converge for sets S which are not measurable, alternating between long strings of "0"s and long strings of "1"s in such a way that it will not have an average frequency of 1's. This is impossible, because the picks are independent. That means that any permutation of the 0's and 1's is as likely as any other. If you have a long string of N zeros and ones, the only permutation invariant of these bits is the number of ones. Any segregation of zeros or ones that has oscillating mean has less than epsilon probability whenever the mean number of ones after M throws, deviates by more than a few times $\sqrt{\ln \epsilon}/\sqrt{N}$ from the mean established by the first N throws.
It is astonishing to me that someone here simultaneously holds in their head the two ideas: "there exists a non measurable subset of [0,1]" and "you can choose a real number at random between [0,1]". The negation of the first statement is the precise statement of the second.
(Solovay defined this stuff precisely, but did not accept the resulting model as true. Others take the axiom of determinacy, thereby establishing that all subsets of R are measurable and that choice fails for the continuum, but determinacy is a stronger statement than "you can pick in [0,1]".)
• The axiom of choice fails, already for sets of size the continuum.
Since the axiom of choice easily gives a non-measurable set.
• The continuum has no well order.
This is because you could then do choice on the reals. So the first bullet on this list should really be rephrased as "the continuum hypothesis is just a stupid question".
• Sorry, you can NOT cut up a grape and rearrange the pieces so that it is bigger than the sun.
Simply because if you put a grape next to the sun, and pick a random point in a big box that surrounds both, the probability that the random point lands in the grape is less than the probability that it lands in the sun. The Lebesgue measure of the pieces is well defined, and invariant under translations and rotations, so it never amounts to more than the measure of the grape.
• The reals which are in "L", the Godel constructable universe, have measure zero.
When the axiom of Choice holds for all elements of the powerset of Z (i.e. R), then the pea can be split up and rearranged to make the sun. The axiom of choice holds in L, so that the Godel constructible L-points in the pea can be cut up and rotated and translated to fit over the L-points of the sun. This means that these points make a measure zero set, both in the pea and in the sun, when considered as a sub-collection of the real numbers which admit random picks.
To understand the Godel constructible universe, and choice, I will pretend that the phrase "Godel constructible" simply means "computable." This is a bald-faced lie. the Godel constructible universe contains many non-computable numbers, but they all resemble computable numbers, in that they are defined by a process which takes an ordinal number of steps and at each step uses only text sentences of ZF acting on previously defined objects. If you replace ordinal by "omega" and "text sentences acting on previously defined elements" by "arithmetical operations defined on previously defined memory", you get computable as opposed to Godel-constructible. To well order the Godel universe, you just order the objects constructed at each ordinal step by alphabetical order and ordinal birthday. To well order the computable reals, you just order their shortest program alphabetically (like the well-ordering of the Godel-universe, this ordering is explicitly definable, but not computable).
• That stupid hat trick doesn't work in the random-pick real numbers
There is a recently popularized puzzle: A demon puts a hat, either red or green, on the head of a countably infinite number of people. Each person sees everyone else's hat, and is told to simultaneously guess the color on their heads. If infinitely many get this wrong, everybody loses. If only finitely many people get the answer wrong, everybody wins.
When the demon picks the hat color randomly on each person's head, they lose. Each person has 50% chance of getting their hat right. End of story. Nothing more to say. Really. This is why set theory has nothing to do with weather prediction.
• The stupid hat trick does work over the computable reals, but is intuitive.
If the demon is forced to place hats according to a fixed definite computer program, there are only countably many different programs, the demon must pick a program, and stick with it. Then it is reasonable that each person can figure out the program used from the infinite answers at their disposal, up to a finite number of errors.
Supposing the people are provided just with some halting oracles and a prearranged agreement regarding computer programs. They do not need a choice function on the continuum. The people see the other hats, and they test the computer programs one by one, in lexicographic order, until they find the shortest program consistent with what they see. They then go through all the programs again, until they find the shortest program on integers which will give be only different from what they see in finitely many places (this requires a stronger oracle, but it still doesn't require a choice function). Then they answer with the value of this program at their own position.
(more precisely, to see everyone else's hat means that the demon provides a program which will give the value of everyone elses hat. You use the halting oracle to test whether each program successively will answer correctly on everyone else's hat, until you find the shortest program that does so.)
This version also has application to weather prediction: by studying the weather long enough, you can guess that it is obeying the Navier Stokes equations. Then you can simulate these equations to predict the weather. Come to think of it, this is exactly what we humans did.
• The stupid hat trick is also intuitive in L, so long as you always think inside a countable model of ZF(C).
The demon again is constrained to definable reals below omega one, which is now secretly a countable ordinal (but ZF doesn't know it). So there is very little difference between the conceptual method to guess the definable real, except that now it isn't so easy to interpret things in terms of oracles.
• There is no problem with "$R_L$, the L version of R, coexisting inside $R_R$, the actual version of R, in your mental model of the universe.
The axiom of choice is true in L, which includes a particular model for the real numbers (and all powersets). This model is fine for interpreting all the counterintuitive statements of ZFC, since they are just plain true in L. When you read a choicy theorem, you just imagine little "L" subscripts on the theorem, and then it is true (this is called relativizing to L in logic). But you always keep in mind that L is measure zero. Then that's it. There are no more intuitive paradoxes.
Your mental axiom system for no-intuitive-paradox mathematics than can be this
ZF (but you interpret powerset as "L-powerset")
V=L (and therefore Choice for all sets in your universe, and the continuum hypothesis for the constructible reals, and for the constructible powersets)
For each set S in the universe L, (which is well-ordered by V=L), there is a non-well-orderable proper class of subsets of S, the true power-class. Every subclass of a powerclass has a real valued Lebesgue measure, and every subset (meaning well-orderable sub-SET, not a non-well-orderable subCLASS) of a powerclass has measure zero. All powerclasses are the same size, since they are not powerclasses of previous powerclasses, just powerclasses of dinky little sets. The measure of the proper-class completion of the dinky little measure-zero L-Borel sets is the same as the measure assigned to these Borel sets in L.
This system does nothing but shuffle the intuition around. There is no new real mathematics here (no new arithmetic theorems). But with this in your head, you banish all the choice paradoxes to the dustbin of history. No more puzzles, no more paradoxes, no more nothing. This has been a public service announcement.
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This sentence is false.
The famous liar paradox. As the wiki article explains:
If "this sentence is false" is true, then the sentence is false, which would in turn mean that it is actually true, but this would mean that it is false, and so on ad infinitum.
Similarly, if "this sentence is false" is false, then the sentence is true, which would in turn mean that it is actually false, but this would mean that it is true, and so on ad infinitum.
Alternately,
All Cretans are liars.
or as pointed out in the comment below the Barber paradox:
The barber shaves only those men in town who do not shave themselves.Who shaves the barber?
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Don't forget the (male) barber who shaves only those who do not shave themselves: en.wikipedia.org/wiki/Barber_paradox This encodes Russell's antinomy. – Margaret Friedland Jul 4 at 20:48
Surprised this one hasn't appeared yet:
The flap of a butterfly's wings in Brazil can set a tornado in Texas.
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Being compact is the next best thing to being finite.
As seen for example by the fact that the (uniform) limit of a sequence of continuous real-valued functions on a compact space is continuous. Or the even more down to earth example: The extreme value theorem.
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I've heard a minor variation: "Compact is the new finite." – S. Carnahan♦ Aug 6 at 0:47
You Can’t Unscramble Scrambled Eggs
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I understand the others, but how does this relate to P vs. NP? – Michael Lugo Nov 3 2009 at 15:56
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Ah, but theorem: Consider a compact pan with some unscrambled eggs in a closed (no "external" i.e. time-varying physics) classical Newtonian (energy is kinetic, which is positive-def quadratic in velocity, plus potential, which depends only on position) universe. The eggs may be in the process of scrambling. Then at some time in the future (indeed, after some precisely integer number of years, where how long you have to wait can be given an explicit absolute bound in terms of epsilon), the eggs will be within epsilon of unscrambled. – Theo Johnson-Freyd Dec 24 2009 at 20:43
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Non-mathematically, this reminds me of the cryptic crossword clue: gegs (9,4). – Andrew Lobb Feb 22 2010 at 21:40
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It's better to be lucky than good.
P != NP. A nondeterministic polynomial is one which is always "lucky".
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Given that `P!=NP` is unproven, perhaps I hope it's better to be lucky than good. – Ilya Nikokoshev Nov 1 2009 at 20:48
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Or "better lucky than smart" – vonjd Feb 23 2010 at 14:24
There is a way to cut up a pea and rearrange the pieces to get the sun.
As a corollary of the Banach-Tarski paradox, we have that if A and B are bounded subsets of R^n (n > 2) with nonempty interior, there exists a partition of A into k pieces {A1, ..., Ak} and isometries of R^n {f1, ... , fk} so {f1A1, ... , fkAk} partitions B.
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This is not an immediate consequence, and the pea-to-sun statement is a highly misleading translation of the statement of the theorem. Both the pea and sun are roughly spherical, but the theorem shows that you can geometrically decompose a sphere into two spheres of the same radius, not into a sphere of a different radius. If you disagree, please tell me how many pieces you will use to go from a ball of radius 1 into a ball of radius 2. I would use 9 pieces to decompose a sphere into two of the same radius, although that's not minimal. – Douglas Zare Feb 19 2010 at 21:15
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Maybe it shoud be formulated as: Give me one pea and I'll feed the world. :-) – Yiftach Barnea Apr 19 2011 at 8:41
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@Douglas: There is a generalization of the Banach-Tarski-theorem that applies to almost arbitrary subsets of IR^n. (I think the precise condition is that they have non-empty interior) – Johannes Hahn Apr 19 2011 at 10:14
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@Johannes: I think that, in addition to having non-empty interior, the sets need to be bounded. – Andreas Blass Jul 5 at 1:01
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Visualize world peas? – Noam D. Elkies Jul 6 at 4:47
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There's no such thing as a free lunch.
This refers to the No Free Lunch theorem. The theorem states that it's impossible to develop a search optimization algorithm that works well for all possible problems. Rather, for every class of problems which a given algorithm performs well at, there is a complementary class for which it does not. Thus, you may think you're getting a free lunch, but you're really just paying for it somewhere else.
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I like the phrase counting in two different ways'' -- which I learned in a math camp. If you're not sure what it means, I suggest the exercise of trying to prove that $2^n = \sum_{i=0}^n {n\choose i}$ by looking for a proof which could be aptly described by this phrase.
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Even quite irregularly shaped objects, such as tables and chairs, become approximately spherical if you wrap them in enough newspaper.
(I think this is by J. H. Conway but I heard it through Bill Thurston, who we recently lost. RIP.)
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What is the serious mathematics that this encodes? – Michael Lugo Aug 25 at 15:02
Poisson arrivals see time averages.
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Non-random is a special case of random.
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Nothing contains everything.
or
There is no universe.
This is how Halmos (pp. 6-7) summarizes the answer of axiomatic set theory to Russell's paradox.
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GAGA the acronym for Serre's famous Geometrie algebrique geometrie analytique.
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Numbers are mutually friendly if they share their abundancy
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An ancient Greek question: What is the difference between a mathematician and a policeman?
A mathematician tries to square the circle. A policeman tries to circle the square.
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It's a win win situation...the probability of a desirable event is 1
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These phrases have very different meanings for me. – Douglas Zare Jul 10 at 3:41
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http://mathoverflow.net/questions/62571/vague-question-on-pic0/62575
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## Vague question on $Pic^0$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
For a smooth variety $X$ when $Pic^0(X)$ is trivial, we get an isomorphism between $N^1(X)$ and Picard group and life become easier.
My question is whether in general there are theorems, criteria ... which results in triviality of $Pic^0$ in some useful examples ?? Like for hyeprsurfaces in projective space, or in toric varieties, or for complete intersections, or in Fano varieties,...
Please tell, if you know any theorem or criteria like this, or if you know any method which helps one to answer this question in a specific example.
Don't get mad, I know it is vague question.
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Lefschetz hyperplane theorem tells us that $H^1(X)=0$ for a hypersurface $X\subset P^n$ when $n>2$ which implies $Pic^0(X)$ is trivial. – Jim Bryan Apr 21 2011 at 21:03
## 2 Answers
Working over $\mathbb{C}$:
We have the short exact sequence of sheaves `$0 \to \mathbb{Z} \to \mathcal{O} \to \mathcal{O}^* \to 0$` and we thus have `$$H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}) \to \mathrm{Pic}(X) \to H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}).$$` Note that the kernel of the last map is discrete, while the cokernel of the first map is connected. So `$\mathrm{Pic}^0(X) = \mathrm{CoKer}(H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}))$` and `$NS(X) = \mathrm{Ker}(H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}))$`.
Moreover, $H^1(X, \mathcal{O})$ is a $\mathbb{C}$-vector space, and $H^1(X, \mathbb{Z})$ is a finitely generated abelian group. So `$\mathrm{Pic}^0(X)$` is trivial if and only if $H^1(X, \mathcal{O})=0$.
I know two main situations where you can conclude that $H^1(X, \mathcal{O})$ is $0$. If $X$ is Stein (for example, affine), then $H^1(X, \mathcal{O})$ vanishes because cohomology of any coherent sheaf on $X$ vanishes. Also, if $X$ is compact Kahler (for example, projective) then $H^1(X, \mathcal{O}) \cong H^{1,0}(X)$ and $\dim H^{1,0}(X) = (1/2) \dim H^1(X, \mathbb{C})$. So $H^1(X, \mathcal{O})$ vanishes if $X$ has no first cohomology. Most (I think all) of the examples you mention are of the second type.
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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.
By definition, for $X$ a compact Kähler manifold, $$Pic^0(X)=H^1(X,\mathcal O_X)/H^1(X,\mathbb Z)$$ and $H^1(X,\mathbb Z)$ is a lattice of maximal rank. Thus the vanishing of $Pic^0(X)$ is equivalent to the vanishing of $H^1(X,\mathcal O_X)$, which is, by Hodge theory, purely topological: $H^1(X,\mathcal O_X)\simeq H^{0,1}(X;\mathbb C)$ and $h^{1,0}(X)=h^{0,1}(X)=b_1(X)/2$.
Hence you are asking for Kähler manifolds whose first Betti number vanishes.
Now, $H^1(X,\mathbb C)$ is the complexification of the abelianization of $\pi_1(X)$. So a compact Kähler manifold has zero Jacobian variety if and only if the abelianization of its fundamental group vanishes after complexification, i.e. it is completely torsion.
A first rich class of examples is given, as Jim said, by the Lefschetz hyperplane theorem. Take an $n$-dimensional smooth projective variety $X$ with zero $H^1(X,\mathcal O_X)$ (for instance a simply connected smooth projective manifold, e.g. $\mathbb P^n$), and $D\subset X$ any smooth ample divisor. Then one has isomorphisms ```$$
H^k(X,\mathbb Z)\to H^k(D,\mathbb Z),\quad k<n-1
$$``` and it is injective when $k=n-1$. In particular $H^1(D,\mathbb Z)=0$ and all such hypersurfaces will have zero Jacobian variety, provided $n>2$. You can figure out a similar construction in the smooth complete intersection case.
Fano varieties are known to be rationally connected and hence simply connected, therefore provide other examples (to see it more elementary if $-K_X$ is ample then $H^q(X,O_X)=0$ if $q>0$ by Kodaira's vanishing. Thus $H^1(X,\mathcal O_X)=0$).
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The abelianization of $\pi_1(X)$ is $H_1(X, \mathbb{Z})$, not $H^1(X, \mathbb{Z})$. For instance, if $X$ is an Enriques surface, one has $H^1(X, \mathbb{Z})=0$ but $H_1(X, \mathbb{Z})=\pi_1(X)=\mathbb{Z}_2$. – Francesco Polizzi Apr 21 2011 at 23:19
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You are right. But since here I was looking just at the dimensions of the spaces this does not modify very much my answer. I'll correct slightly my answer, then. Thank you anyway! – diverietti Apr 22 2011 at 8:37
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http://mathoverflow.net/revisions/91869/list
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2 edited tags
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# closed set and z-ultrafilter on normal space
Let $X$ be a completely regular, Hausdorff topological space and let $\cal F$ be a $z$-ultrafilter on $X$. Then for each zero set $W$ in $X$, either $W\in \cal F$ or there exists $Z\in \cal F$ such that $Z$ does not meet $W$ (this is the $z$-ultrafilter property). Now suppose that $X$ is additionally normal. Then is it true that for every closed set $W$ in $X$ either $W$ contains an element $Z$ of $\cal F$ or there exists $Z\in \cal F$ such that $Z$ does not meet $W$?
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http://stats.stackexchange.com/questions/29477/how-to-write-a-linear-model-formula-with-100-variables-in-r/29483
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# How to write a linear model formula with 100 variables in R
Is there an easy way in R to create a linear regression over a model with 100 parameters in R? Let's say we have a vector Y with 10 values and a dataframe X with 10 columns and 100 rows In mathematical notation I would write `Y = X[[1]] + X[[2]] + ... + X[[100]]`. How do I write something similar in R syntax?
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are there 100 or 1000? Also, you would normally have the columns be the variables and the rows be observations (it appears that is reversed here) – Macro May 30 '12 at 13:09
100 the extra 0 was a typo – Christian May 30 '12 at 13:11
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Really? Are you sure you want to do this? I'd be concerned about overfitting and correlation between linear combinations of the predictors. Not only that, with 100 predictors but only 10 observations, you have $p>n$ and linear regression isn't going to work at all. – Aaron May 31 '12 at 1:47
## 4 Answers
Try this
````df<-data.frame(y=rnorm(10),x1=rnorm(10),x2=rnorm(10))
lm(y~.,df)
````
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Great answers!
I would add that by default, calling `formula` on a `data.frame` creates an additive formula to regress the first column onto the others.
So in the case of the answer of @danas.zuokas you can even do
````lm(df)
````
which is interpreted correctly.
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+1 I never knew this. Interesting – Macro May 30 '12 at 13:34
Still, this answer does not work if you want to mix in interaction terms. Yours does (+1). – gui11aume May 30 '12 at 13:36
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I'm continually amazed at how overloaded most of `R`'s operators are :) – Macro May 30 '12 at 13:38
If each row is an observation and each column is a predictor so that $Y$ is an $n$-length vector and $X$ is an $n \times p$ matrix ($p=100$ in this case), then you can do this with
````Z = as.data.frame(cbind(Y,X))
lm(Y ~ .,data=Z)
````
If there are other columns you did not want to include as predictors, you would have to remove them from `X` before using this trick, or using `-` in the model formula to exclude them. For example, if you wanted to exclude the 67th predictor (that has the corresponding name `x67`), then you could write
````lm(Y ~ .-x67,data=Z)
````
Also, if you want to include interactions, etc.. you will need to add them manually as (for example)
````lm(Y ~ .+X[,1]*X[,2],data=Z)
````
or make sure they are entered as columns of `X`.
-
You can also use a combination of the `formula` and `paste` functions.
Setup data: Let's imagine we have a data.frame that contains the predictor variables `x1` to `x100` and our dependent variable `y`, but that there is also a nuisance variable `asdfasdf`. Also the predictor variables are arranged in an order such that they are not all contiguous in the data.frame.
````Data <- data.frame(matrix(rnorm(102 * 200), ncol=102))
names(Data) <- c(paste("x", 1:50, sep=""),
"asdfasdf", "y", paste("x", 51:100, sep=""))
````
Imagine also that you have a string containing the names of the predictor variables. In this case, this can easily be created using the `paste` function, but in other situations, `grep` or some other approach might be used to get this string.
````PredictorVariables <- paste("x", 1:100, sep="")
````
Apply approach: We can then construct a formula as follows:
````Formula <- formula(paste("y ~ ",
paste(PredictorVariables, collapse=" + ")))
lm(Formula, Data)
````
• the `collapse` argument inserts `+` between the predictor variables
• `formula` converts the string into an object of class formula suitable for the `lm` function.
-
1
+1. I use this technique all the time. Occasionally, however, having the formula stored in a variable causes issues. See stackoverflow.com/a/7668846/210673 for use of `do.call` evaluate the formula before calling `lm`. – Aaron May 31 '12 at 1:44
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|
http://mathoverflow.net/questions/111066/is-there-any-techniques-for-solving-a-differential-equation-including-iterated-fu/111078
|
## Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am trying to solve this differential equation but I have no idea how.
$f ' (x) = f( f( x ) )$
Although I don't think this differential equation is solvable, I'd like to know if there is any interesting approach on solving a differential equation of this kind, or at least a non-trivial solution of the equation.
P.S. I don't think chain rule is useful for this
-
I think this is more suitable for math.stackexchange.com – Beni Bogosel Oct 30 at 10:49
4
Beni, why do you say that? – Vidit Nanda Oct 30 at 11:57
@Vel Nias I think math analysis questions are not welcome here. – Anixx Nov 1 at 11:03
6
@Anixx. For the best of my knowledge, this claim is plain wrong. However, I agree that the remarks like "this is homework" or "ask that on MSE" that are not supported by any evidence that the person making them can solve the problem himself can be somewhat irritating... As to Beni's recommendation itself, MSE is not a bad site per se but it is just DROWNED in "homeworks" nowadays. MO and AoPS are much better choices for something nontrivial IMHO. – fedja Nov 1 at 12:33
Is that a delay differential equation? – Zsbán Ambrus Nov 1 at 14:03
show 2 more comments
## 5 Answers
Nothing is new under the Moon...
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=321705
-
wow! Thank you very much! Did you solve it by yourself? You're amazing! – frigen Nov 1 at 3:17
I see no solution following the link. – Anixx Nov 1 at 3:27
1
Meaning you haven't scrolled down or you failed to understand what is written there? In the latter case you are welcome to ask questions. – fedja Nov 1 at 3:32
@fedja I only see the supposed proofs of existence. Regarding the solution, you yourself wrote "I have no hope for an explicit elementary formula for it." – Anixx Nov 1 at 3:34
2
Existence and uniqueness of a one-parameter family of solutions, to be exact ;). Do you believe one can come with a formula? We can, probably, give it a shot and try to prove that the functions in question are not elementary but that is quite another story (and, most likely, quite a non-trivial one given that there exist formal elementary pseudo-solutions like the ones you mentioned). If you are interested in such a project, I can think of what might be the right approach here (but a bit later :)). – fedja Nov 1 at 3:46
show 3 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.
There are two closed form solutions:
$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$
The solution technique can be found in this paper.
For a general case, solution of the equation
$$f'(z)=f^{[m]}(z)$$
has the form
$$f(z)=\beta z^\gamma$$
where $\beta$ and $\gamma$ should be obtained from the system
$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$
In your case $m=2$.
-
See also my answer regarding real solutions below. – Anixx Nov 2 at 10:36
I don't know, but one answer is $f(x)=ax^c$ where $a=\frac12(\sqrt {3}+i){ e^{\frac16\pi\sqrt {3}}}$ and $c=\frac12+\frac12i\sqrt{3}$. Another is obtained by taking the complex conjugate of both $a$ and $b$.
-
3
The only unclear thing is where this function is defined. Note that complex powers of real numbers are complex and complex powers of complex numbers are branching like crazy... – fedja Nov 1 at 2:41
And regarding real solutions to the question, Alex Gavrilov is completely correct. A Taylor expansion at fixed point $p$ gives us the real solution. Existence of this solution is proven in the paper which I already referenced from my another answer.
$$f(z)=\sum_{n=0}^\infty \frac{d_n (z-p)^n}{n!}$$
where $d_n$ is defined as follows:
$$d_0=p$$ $$d_{n+1}=\sum _{k=0}^n d_k \operatorname{B}_{n,k}(d_1,...,d_{n-k+1})$$
where $B_{n,k}$ are the Bell polynomials
This gives the following starting coefficients:
$$d_1=p^2$$ $$d_2=p^3+p^4$$ $$d_3=p^4 + 4 p^5 + p^6 + p^7$$ $$d_4=p^5 + 11 p^6 + 11 p^7 + 8 p^8 + 4 p^9 + p^{10} + p^{11}$$
etc.
The fixed point $p$ here serves as a parameter, which determines the family of solutions. According the linked theorem, the expansion should converge in the neighborhood of $p$ for $0 < |p| < 1$ or $p$ being a Siegel number.
-
1
Anixx, this becomes boring. Yeah, if $p$ is small enough, this has a chance to work (though I wonder how you prove that there are no other solutions). However, for large $p$, you have a polynomial with the leading term $p^N$ with $N\approx k^2$ for the $k$'th coefficient. The miraculous cancellations can shave only something like $8^{N}$ off it for a typical $p$ (Remez). So, you are left with $(p/8)^{ck^2}$ which eats up the factorial and the geometrical progression for breakfast and happily flies to infinity by the lunchtime if $p$ is like $-20$. – fedja Nov 2 at 2:50
@fedja yes, the proof in the linked paper requires p<1 or a Siegel number. I wiil add this to the answer. – Anixx Nov 2 at 10:25
For what I know, the standard method is the Taylor series expansion at a fixed point, i.e. at a point $x=a$ such that $f(a)=a$.
-
Yes. +1 And I will give the expansion in another answer. – Anixx Nov 1 at 11:04
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|
http://cs.stackexchange.com/questions/tagged/integers
|
# Tagged Questions
Questions about properties of, working with and algorithms on integers.
1answer
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### Algorithm to determine if a number is perfect on a Turing Machine
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### Solve Integer Factoring in randomized polynomial time with an oracle for square root modulo $n$
I'm trying to solve exercise 6.5 on page 309 from Richard Crandall's "Prime numbers - A computational perspective". It basically asks for an algorithm to factor integers in randomized polynomial time ...
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### How to guess the value of $j$ at the end of the loop?
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### Shifting subset sum solution by constant positive integer
While reading the Wikipedia article about the subset sum problem I came across this example: "is there a non-empty subset whose sum is zero? For example, given the set $\{ −7, −3, −2, 5, 8 \}$, the ...
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### Comparing rational numbers
Given $a,b,c,d \in \mathbb N$ and $b,d \notin \{0\}$, $$\begin{eqnarray*} \frac a b < \frac c d &\iff& ad < cb \end{eqnarray*}$$ My questions are: Given $a,b,c,d$ Assuming we can ...
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### Calculate storage requirements for a data set
I have a simple problem. I can't seem to even find the right search terms to get me pointed in the direction I need to be heading. I'm writing a bunch of integers to disk. Lot's of them. Starting ...
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### Fastest square root method with exact integer result?
I am dealing with the problem of computing $s = \lfloor sqrt(x)\rfloor$ with $x \in [0,30000^2]$. The common sqrtf(x) on C language is too slow for this case, ...
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### Get specified bit in addition of two large binary numbers
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### Converting function to bitwise only?
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### How to compute linear recurrence using matrix with fraction coefficients?
What I'm trying to do is generate Motzkin numbers mod a large number $10^{14} + 7$ (not prime), and it needs to compute the $n$th Motzkin number as fast as possible. From Wikipedia, the formula for ...
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### Run time of product of polynomially bounded numbers
Let $M$ denote a set of $n$ positive integers, each less than $n^c$. What is the runtime of computing $\prod_{m \in M} m$ with a deterministic Turing machine?
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What is the complexity of computing $n^{n^2},\;n \in \mathbb{N}$?
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I'm having a very hard time trying to figure out how to solve this problem efficiently. Let me describe how it goes: "A hard working mom bought several fruits with different nutritional values for ...
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### Most efficient algorithm to print 1-100 using a given random number generator
We are given a random number generator RandNum50 which generates a random integer uniformly in the range 1–50. We may use only this random number generator to ...
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Given a sequence of natural numbers, you can add any natural number to any number in the sequence such that their xor becomes zero. My goal is to minimize the sum of added numbers. Consider the ...
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### What is the bitwise xor of an interval?
Let $\oplus$ be bitwise xor. Let $k,a,b$ be non-negative integers. $[a..b]=\{x\mid a\leq x, x\leq b\}$, it is called a integer interval. What is a fast algorithm to find \$\{ k\oplus x\mid x\in ...
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### Representing Negative and Complex Numbers Using Lambda Calculus
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### Detecting overflow in summation
Suppose I am given an array of $n$ fixed width integers (i.e. they fit in a register of width $w$), $a_1, a_2, \dots a_n$. I want to compute the sum $S = a_1 + \ldots + a_n$ on a machine with 2's ...
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### Overflow safe summation
Suppose I am given $n$ fixed width integers (i.e. they fit in a register of width $w$), $a_1, a_2, \dots a_n$ such that their sum $a_1 + a_2 + \dots + a_n = S$ also fits in a register of width $w$. ...
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### Language of the graph of an affine function
Write $\bar n$ for the decimal expansion of $n$ (with no leading 0). Let : be a symbol distinct from any digit. Let $a$ and $b$ ...
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### Language of the values of an affine function
Write $\bar n$ for the decimal expansion of $n$ (with no leading 0). Let $a$ and $b$ be integers, with $a > 0$. Consider the language of multiples of $a$ plus a ...
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|
http://math.stackexchange.com/questions/239207/hessian-matrix-of-a-quadratic-form
|
# Hessian matrix of a quadratic form
I need a help with one example. I have to proove that hessian matrix of a quadratic form $f(x)=x^TAx$ is $f^{\prime\prime}(x) = A + A^T$. I am not even sure how the Jaxobian looks like (I never did one for $x \in R^n$). Thanks for any advice.
-
## 2 Answers
So let's compute the first derivative, by definition we need to find $f'(x)\colon\mathbb R^n \to \mathbb R^n$ such that $$f(x+h) = f(x) + f'(x)h + o(h), \qquad h \to 0$$ We have \begin{align*} f(x+h) &= (x+h)^tA(x+h)\\ &= x^tAx + h^tAx + x^tAh + h^tAh\\ &= f(x) + x^t(A + A^t)h + h^tAh \end{align*} As $|h^tAh|\le \|A\||h|^2 = o(h)$, we have $f'(x) = x^t(A + A^t)$ for each $x \in \mathbb R^n$. Now compute $f''$, we have \begin{align*} f'(x+h) &= x^t(A + A^t) + h^t(A + A^t)\\ &= f(x) + h^t(A + A^t) \end{align*} So $f''(x) = A + A^t$.
-
Intuitively, the gradient and Hessian of $f$ satisfy \begin{equation} f(x + \Delta x) \approx f(x) + \nabla f(x)^T \Delta x + \frac12 \Delta x^T Hf(x) \Delta x \end{equation} and the Hessian is symmetric.
In this problem, \begin{align*} f(x + \Delta x) &= (x + \Delta x)^T A (x + \Delta x) \\ &= x^T A x + \Delta x^T A x + x^T A \Delta x + \Delta x^T A \Delta x \\ &= x^T A x + \Delta x^T(A + A^T)x + \frac12 \Delta x(A + A^T) \Delta x. \end{align*}
Comparing this with the approximate equality above, we see that $\nabla f(x) = (A + A^T) x$ and $Hf(x) = A + A^T$.
-
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http://math.stackexchange.com/questions/184770/what-effect-does-multiplication-by-a-constant-have-on-a-functions-plot-in-gene
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# What effect does multiplication by a constant have on a function's plot, in general?
I am looking at the equation of a line and its plot. I can intuitively see what these operations do to the plot's form: adding a constant, multiplying by a constant, adding a parametrised value, etc. For some simple conic functions, I am able to do the same, almost as intuitively, although not always. When I consider higher order polynomials, the effects vary to the point of not being able to recognize the original plot anymore. Is there a general definition for what effect multiplying for a constant has on a function's plot? I am using gnuplot and/or octave to play around with this (should I be using something else?) and still, for some function, I can see the effects and understand them (e.g. This "amplifies the oscillation range", this "contracts" this part of the curve, etc. but is there a more general way of making estimations about what will happen to the function's plot?)
-
When you say "multiplication by a constant", do you mean $y=k\times f(x)$, $y= f(k \times x)$, or something else? – Henry Aug 20 '12 at 20:48
y = f(x) k, sorry. Is k not called a multiplicative constant in this case? – Robottinosino Aug 20 '12 at 20:50
If $k > 1$ (resp. $0 < k < 1$) then it stretches (resp. shrinks) the plot in the $y$-axis direction. If $k$ is negative then it's the same $+$ mirroring about $x$-axis. – user2468 Aug 20 '12 at 21:02
## 1 Answer
For a positive value $k$, the graph of $y = kf(x)$ is obtained from the graph of $y=f(x)$ by scaling in the vertical direction. In coordinates, if $(x_0, y_0)$ is on the graph of $y=f(x)$, then that point moves to $(x_0, ky_0)$ on the graph of $y=kf(x)$. Any point on the $x$-axis remains fixed (as then $y=0$). When $k > 1$, the graph is stretched vertically (away from the $x$-axis), while if $0 < k < 1$, the graph is compressed vertically (toward the $x$-axis).
Another way to think about this: The graph of $y=kf(x)$ is exactly the same as the graph of $y=f(x)$ if the $y$-axis is scaled so that $1$ is relabeled $k$, $2 \mapsto 2k$, $-1 \mapsto -k$, etc.
Hope this helps!
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http://en.wikipedia.org/wiki/Medial
|
Medial
Medial magmas
In abstract algebra, a medial magma (or medial groupoid) is a set with a binary operation which satisfies the identity
$(x \cdot y) \cdot (u \cdot v) = (x \cdot u) \cdot (y \cdot v)$, or more simply, $xy\cdot uv = xu\cdot yv$
using the convention that juxtaposition denotes the same operation but has higher precedence. This identity has been variously called medial, abelian, alternation, transposition, interchange, bi-commutative, bisymmetric, surcommutative, entropic etc.[1]
Any commutative semigroup is a medial magma, and a medial magma has an identity element if and only if it is a commutative monoid. Another class of semigroups forming medial magmas are the normal bands.[2] Medial magmas need not be associative: for any nontrivial abelian group and integers m ≠ n, replacing the group operation $x+y$ with the binary operation $x \cdot y = mx+ny$ yields a medial magma which in general is neither associative nor commutative.
Using the categorial definition of the product, one may define the Cartesian square magma M × M with the operation
(x, y)∙(u, v) = (x∙u, y∙v) .
The binary operation ∙ of M, considered as a function on M × M, maps (x, y) to x∙y, (u, v) to u∙v, and (x∙u, y∙v) to (x∙u)∙(y∙v) . Hence, a magma M is medial if and only if its binary operation is a magma homomorphism from M × M to M. This can easily be expressed in terms of a commutative diagram, and thus leads to the notion of a medial magma object in a category with a Cartesian product. (See the discussion in auto magma object.)
If f and g are endomorphisms of a medial magma, then the mapping f∙g defined by pointwise multiplication
$(f\cdot g)(x) = f(x)\cdot g(x)$
is itself an endomorphism.
Bruck–Toyoda theorem
The Bruck–Toyoda theorem provides the following characterization of medial quasigroups. Given an abelian group A and two commuting automorphisms φ and ψ of A, define an operation ∗ on A by
x ∗ y = φ(x) + ψ(y) + c
where c some fixed element of A. It is not hard to prove that A forms a medial quasigroup under this operation. The Bruck-Toyoda theorem states that every medial quasigroup is of this form, i.e. is isomorphic to a quasigroup defined from an abelian group in this way.[3] In particular, every medial quasigroup is isotopic to an abelian group.
Generalizations
The term medial or (more commonly) entropic is also used for a generalization to multiple operations. An algebraic structure is an entropic algebra[4] if every two operations satisfy a generalization of the medial identity. Let f and g be operations of arity m and n, respectively. Then f and g are required to satisfy
$f( g(x_{11}, \ldots, x_{1n}), \ldots, g(x_{m1}, \ldots, x_{mn}) ) = g( f(x_{11}, \ldots, x_{m1}), \ldots, f(x_{1n}, \ldots, x_{mn}) ).$
References
1. Yamada, Miyuki (1971), "Note on exclusive semigroups", Semigroup Forum 3 (1): 160–167, doi:10.1007/BF02572956 .
2. Kuzʹmin, E. N. and Shestakov, I. P. (1995). "Non-associative structures". Algebra VI. Encyclopaedia of Mathematical Sciences 6. Berlin, New York: Springer-Verlag. pp. 197–280. ISBN 978-3-540-54699-3.
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http://mathoverflow.net/questions/38187/completeness-of-an-infinite-mixture-of-gaussians-representation
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## Completeness of an “infinite mixture of gaussians” representation
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Is there a complete "infinite mixture of gaussians representation" for densities? What I mean is, is there, for any reasonably big class of densities $\phi(x)$ I can come up with a function $c(\mu, \tau)$ such that:
$\displaystyle\phi(x) = \int d\mu d\tau\; c(\mu, \tau) \exp\left(-\frac{\tau (x-\mu)^2}{2}\right)$??
($\tau = \frac{1}{\sigma}$ is the inverse of the variance and the normalization of the gaussian was absorbed in $c(\mu,\tau)$)
I'm using the word "complete" in the sense theoretical physicists talk about the completeness of a basis for a vector space. I'm not sure if it's an adequate use of the term for mathematicians (probably not).
If yes, does it generalize for multivariate distributions? Also, is there a known "inverse integral transform" for this representation? Something like:
$\tilde{\phi}(\Sigma, \mu) = \int dx \; \tilde{\mathcal{N}}(\Sigma, \mu | x) \phi(x)$
$\phi(x) = \int d\Sigma \mathcal{N}(x | \Sigma, \mu) \tilde{\phi}(\Sigma,\mu)$
where $\Sigma$ and $\mu$ are the covariance matrix and the vector of means.
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## 1 Answer
How about this: the set of finite mixtures of (non-degenerate) Gaussians is weakly dense in the space of probability measures on $\mathbb{R}$. Proof: the set of finite mixtures of constants is certainly weakly dense. But a constant can be approximated as a Gaussian with small variance.
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So, the gaussians have even more "degrees of freedom" than necessary to represent all densities? I guess this kills my hope for an inversion formula... – Rorsa Sep 9 2010 at 23:38
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http://math.stackexchange.com/questions/25739/maximal-multiplicative-set-implies-complement-is-a-prime-ideal
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Maximal multiplicative set implies complement is a prime ideal
Let $R$ be a commutative ring with identity. I've been trying to prove the following:
If $M \subset R$, i.e. $M \neq R$, is a maximal multiplicative set, then $R \setminus M$ is a prime ideal of $R$.
I have spent some time in it, but nothing is coming to my mind. Any idea/solution will be appreciated.
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I think you should be clear here that you want all your multiplicative subsets to be disjoint from $\{0\}$ (otherwise in my argument $S^{-1} R$ is the zero ring, and in general things don't behave very well). – Pete L. Clark Mar 8 '11 at 8:30
I understand that a set is multiplicative if it is subset of $R$ that is closed under multiplication and contains the identity. Does this implies that it does not contains 0? (I don't see it) Or is it something that you have to add to the definition for things to work? – 4math Mar 8 '11 at 9:15
no, a multiplicative subset can contain zero, and as far as the general definition goes you do want to allow this: in particular for any element $f \in R \setminus \{0\}$ you want to be able to localize at the multiplicative subset generated by $f$, and if $f$ is nilpotent, this contains zero. But in practice the case of $0 \in S$ is often excluded. To see that your assertion is not true unless zero is excluded, consider $\mathbb{Z} \setminus \{-1\}$: this is a maximal proper multiplicative subset of $\mathbb{Z}$ and it is not of the form you want. – Pete L. Clark Mar 8 '11 at 13:55
3 Answers
I assume that all multiplicative subsets discussed here are disjoint from $\{0\}$.
Let me write $S$ for the maximal multiplicative subset of $R$.
Step 1: I claim that $S^{-1} R$ is a local ring.
Indeed, if not, there would be at least two maximal ideals in the localization, i.e., two prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, each maximal with respect to being disjoint from $S$. Thus $S$ is contained in both $R \setminus \mathfrak{p}_1$ and $R \setminus \mathfrak{p}_2$, but these are two different sets so at least one of the containments must be proper, contradicting the maximality of $S$.
Step 2: Therefore there is a unique maximal ideal in the localization, which corresponds to a prime ideal $\mathfrak{p}$ of $R$ which is disjoint from $S$. Arguing as above, maximality of $S$ implies $S = R \setminus \mathfrak{p}$.
Remark 1: When $R$ is a domain, the unique maximal multiplicative subset is obviously $R \setminus \{0\}$. In this case it is tempting to construe "maximal multiplicative subset" to mean "multiplicative subset maximal with respect to being properly contained in $R \setminus \{0\}$."
Remark 2: Conversely, the primes $\mathfrak{p}$ such that $R \setminus \mathfrak{p}$ is maximal are precisely the primes which are minimal with respect to containing $\{0\}$. (When $R$ is a domain and the question is reconstrued as above, we want $\mathfrak{p}$ to be minimal with respect to properly containing $\{0\}$.)
Mariano's answer makes this clear as well.
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Very instructive answer. I had come up to something similar, but I hadn't realized your step 1. Thank you! – 4math Mar 8 '11 at 8:19
@PeteL.Clark +1 Very nice answer. I was stuck on a similar problem in Atiyah - Macdonald. – BenjaLim Apr 25 '12 at 23:30
Another way to proceed is to show
• First, that every multiplicative subset $S$ of a ring is contained in a saturated multiplicative subset $S'$, that is, one such that $$ab\in S'\implies a\in S'\wedge b\in S'.$$ In fact, there is a smallest saturated multiplicative subset containing $S$, called the saturation.
• Second, that a saturated multiplicative subset $S$ of a ring (which does not contain $0$) is the intersection of the complements of the prime ideals which are disjoint from it. (The hard part here is to show that such primes actually exist: you can construct them as the maximal elements in the family of ideals disjoint from $S$)
Once you have these two facts, your claim follows easily.
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+1: I like this argument better than mine. – Pete L. Clark Mar 8 '11 at 8:20
+1 @MarianoSuárez-Alvarez Very nice answer. – BenjaLim Apr 26 '12 at 0:00
We must assume $\rm 0\not\in M$ since $\rm\: 0\in M\: \Rightarrow\: 0 \not\in M' := R\setminus M\:,\:$ so $\rm\:M'\:$ is not an ideal. A simple Zorn lemma argument (see below) shows that, since $\rm\:M\:$ is multiplicatively closed, the ideal $\rm\{0\}\subset M'\:$ can be enlarged to an ideal $\rm\:P\:$ maximal w.r.t. to exclusion of $\rm\:M\:,\:$ and such an ideal must be prime. Hence $\rm\: P = M'\:,\:$ else we could enlarge $\rm\:M\:$ to the monoid $\rm\:P'\:,\:$ contra maximality of $\rm\:M\:.\quad$ QED
Note $\:$ The prime $\rm\:P\:$ above may be alternatively constructed as the contraction of a maximal ideal $\rm\:Q\:$ of the localization $\rm\: R_M = M^{-1} R\:.\:$ $\rm\:Q\:$ exists since $\rm\ 0\not\in M\ \Rightarrow\ R_M \ne \{0\}\:.\:$ Generally, there is a bijective order-preserving correspondence between all prime-ideals in $\rm\:R_M\:$ and all prime ideals in $\rm R$ disjoint from $\rm\:M\:,\:$ see Theorem 34 in Kaplansky's Commutative Rings. His Theorem 1, pp. 1-2, is the above-invoked form of this result employing no localization theory. I've appended it below.
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http://math.stackexchange.com/questions/222025/notation-for-the-set-of-all-polynomials-inconsistent-with-the-notation-of-a-subr?answertab=oldest
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# Notation for the set of all polynomials inconsistent with the notation of a subring generated by a set
Let $R$ be a ring and $S\subseteq R$. Then usually $R[S]$ denoted the subring generated by $S$ (especially in the context of field extensions I have seen this notation to be used).
My question is: How is this notation consistent with the notation $R[X]$ for the set of all polynomials with coefficients in $R$ ? Since if $R$ is the set of all polynomials, then for the element $X\in R$, the set $R[X]$ (in the sense of the definition above) is, since it is the smallest subring containing the element $X$, the set $\{ \sum_i X^i \mid i\in \mathbb{N}\}$.
So obviously $R\neq R[X]$ and therefore the notation $R[X]$ for the set of all polynomials doesn't seem to me to be consistent with the notation $R[S]$ for the smallest subring containing $S$.
Thus I wonder, why one doesn't use a different notation for the set of all polynomials. After all, when (rigorously) defining the set of all polynomials with coefficients in $R$, one usually defines $R[X]$ to be the set of all function $\mathbb{N} \rightarrow R$, that are are nonzero only for finitely many values (here $X$ is just a letter with not any mathematical meaning yet) and then one shows that if one defines $X$ as $X=(0,1,0,0,\ldots )$, that one can write every such function $f$ as $\sum_n f(n)X^n$, which yields, when making yet another definition $f(n)=a_n$, a mathematical object that comprises what we usually understand a polynomial to be.
But, as I showed, even if $X=(0,1,0,0,\ldots )$, one can't say that $R[X]$ can also be understood as the subring generated by $X$, so $R[X]$ has to remain just a string of symbol, if one wants to remain contradiction free (where this contradiction is on a syntactical/notational level and not a mathematical level).
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I interpret $R[S]$ as "the smallest $R$-subalgebra of some context-specified $R$-algebra generated by $S$". Under this reading $R[X]$ makes perfect sense, since $X$ generates $R[X]$ as an $R$-subalgebra of $R[X]$. – Zhen Lin Oct 27 '12 at 13:56
@ZhenLin And is it also custom to denote with "$R[S]$" the smallest $R$-algebra generated by $S$ ? – temo Oct 27 '12 at 14:35
No further comments/answer from anybody :( ? – temo Oct 27 '12 at 16:22
Ok, seems I have to prepare myself for a bounty...(bump!!) – temo Oct 28 '12 at 17:14
For $S\subseteq R$, you would have $R[S] = R$. Most likely you have seen this when you have two rings $k$ and $R$ with $k\subseteq S\subseteq R$. Then $k[S]$ is used for the smallest subring of $R$ containing $S$. If $k$ lies in the center of $R$ then both $R$ and $k[S]$ are $k$-algebras. – Marc Olschok Oct 29 '12 at 19:50
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http://physics.stackexchange.com/questions/18739/question-based-on-galileos-law-of-falling-bodies/18741
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# Question Based On Galileo's Law Of Falling Bodies
Galileo discovered that the distance fallen is proportional to the square of the time it has been falling.Why is it proportional to the square of the time and not just time? i.e $d \propto t^2$ why not just $d \propto t$
I know this question involves some common sense which i'am not able to use for some reason!
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I'm not sure what you mean by common sense here? If it was that obvious, we wouldn't be praising Galileo for discovering the law. It's also something contingent to a certain extent, i.e. it does not follow from some reasoning starting from "common sense" principles. – Raskolnikov Dec 25 '11 at 7:36
I tend to get shy when asking questions that "looks elementary" to professional physicists! – alok Dec 25 '11 at 7:48
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Don't be shy to ask. I'm just saying that someone had to do the experiment. You can't just reason out what nature is like. – Raskolnikov Dec 25 '11 at 9:55
@Raskolnikov: Indeed, the reason Galileo's law of falling bodies wasn't discovered two thousand years earlier was because no one bothered to do a simple experiment to test "common sense." There's a lesson to be learned here. – David H Nov 26 '12 at 1:20
## 2 Answers
It might be worth taking a look to the original text, Galileo's Discourses on Two New Sciences. The reasoning you're looking for is on the Third Day, a translation of which may be found online. The relevant parts are labelled Theorem I and Theorem II in the above-linked translation.
To derive that distance in a uniformly accelerated motion (e.g. free fall) goes as time squared, Galileo first argues that
The time in which any space is traversed by a body starting from rest and uniformly accelerated is equal to the time in which that same space would be traversed by the same body moving at a uniform speed whose value is the mean of the highest speed and the speed just before acceleration began.
This is argued on a graphical basis (see above link).
However, even though the pictures may look pretty similar to modern functional representations (e.g. of velocity vs. time) and arguments involve finding equal areas in different situations, the arguments never involve an actual calculation of an "area" with mixed units, which wasn't yet conceivable at the time (e.g. $m/s \cdot s = m$). In fact, the whole Third Day seems very convoluted precisely because the notion of velocity wasn't clearly numerical yet, since only commensurable (same unit) quantities could conceivably be operated with (added, divided,...). Proportions of non-commensurable quantities, however, could be compared (today we'd say they are dimensionless), as in
If a moving object traverses two distances in equal intervals of time, these distances will bear to each other the same ratio as the speeds
(earlier in the Third Day)
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Because the velocity increases with time, and the total time increases with time, so that the velocity times time increase as time-squared. It is dimensional analysis--- if you gain an increment of speed every time unit, you must multiply by the time twice to get a distance.
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Ah.Dimensional analysis.Right.But still having little bit of trouble understanding it. – alok Dec 25 '11 at 7:33
Maye it will help to know that the distance travelled is the area under the curve of the graph of the velocity vs. time, and this graph is a triangle for constant velocity? This is Galileo's argument from way back when. – Ron Maimon Dec 25 '11 at 14:11
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http://mathoverflow.net/questions/993/is-every-finite-group-a-group-of-symmetries/19541
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## Is every finite group a group of “symmetries”?
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I was trying to explain finite groups to a non-mathematician, and was falling back on the "they're like symmetries of polyhedra" line. Which made me realize that I didn't know if this was actually true:
Does there exist, for every finite group G, a positive integer n and a convex subset S of R^n such that G is isomorphic to the group of isometries of R^n preserving S?
If the answer is yes (or for those groups for which the answer is yes), is there a simple construction for S?
I feel like this should have an obvious answer, that my sketchy knowledge of representations is not allowing me to see.
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There's a lot of variants of your question. Every finite group can also be realized as the group of isometries of some compact hyperbolic surface. Similarly, it can be realized as the group of symmetries of a hyperbolic 3-manifold. – Ryan Budney Feb 16 2010 at 8:10
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Every group is the automorphism group of some graph. – Kevin O'Bryant Feb 16 2010 at 19:05
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Every group is the automorphism group of some topology. – Kevin O'Bryant Feb 16 2010 at 19:05
@RB and KO'B: I was aware of the variants you mention, and there are certainly more one can cook up. What I wanted to say to the non-mathematician is the most literal generalization of symmetry in the colloquial sense: that every finite group is the isometry group of some polytope in R^n. – Andrew McIntyre Feb 16 2010 at 20:04
@RB: Do you have a reference for every finite group being the automorphism group of a Riemann surface? I seem to recall hearing this as folklore, but it occurs to me that I have no idea how it is proved. That suggests another question: is there an "equivariant" embedding theorem for manifolds with symmetries? I.e. if a hyperbolic surface X has nontrivial automorphisms, is there an embedding of X in R^n for some n, such that every automorphism of X is induced by an isometry of R^n preserving the image of X? – Andrew McIntyre Feb 16 2010 at 20:12
show 5 more comments
## 8 Answers
The permutohedron may have additional symmetries. For example, the order 3 permutohedron {(1,2,3),(1,3,2),(2,1,3),(3,1,2),(3,2,1)} is a regular hexagon contained in the plane x+y+z=6, which has more than 6 symmetries.
I think we can solve it as follows:
Let G be a group with finite order n thought via Cayley's representation as a subgroup of S_n.
Let S={A_1,...,A_n} be the set of vertices of a regular simplex centered at the origin in a (n-1)-dimensional real inner product space V. Let r be the distance between the origin and A_1. The set of vertices S is an affine basis for V.
First unproven claim: If a closed ball that has radius r contains S, then it is centered at the origin. Let B be this ball.
The group of isometries that fix S hence contains only isometries that fix the origin and permute the vertices, which can be identified with S_n in the obvious way. The same is true if we replace S by its convex hull.
Now G can be thought of as a group containing some of the symmetries of S.
Let C=k(A_1+2A_2+3A_3+...+nA_n)/(1+2+...+n), with k a positive real that makes the distance between C and the origin a number r' slightly smaller than r.
Let GC={g(C) / g in G}. It has n distinct points, as a consequence of being S an affine basis of V.
Let P be the convex hull of the points of S union GC.
Remark: A closed ball of radius r contains P iff it is B. The intersection of the border of B and P is S.
Second unproven claim: The extremal points of P are the elements of S union GC.
Claim: G is the group of symmetries of P.
If g is in G, g is a symmetry of GB and of S, and it is therefore a symmetry of P. If T is a symmetry of P, then T(P)=P, and in particular, T(P) is contained in B, and hence T(0)=0 (i.e. T is also a symmetry of B). T must also fix the intersection of P and the border of B, so T permutes the points of S, and it can be thought of as an element s of S_n sending A_i to A_s(i). And since T fixes the set of extremal points of P, T also permutes GC. Let's see that s is in G.
Since T(C) must be an element of g(C) of GC, we have T(C)=g(C). But since T is linear, T(C/k)=g(C/k). Expanding,
(A_s(1)+2A_s(2)+...+nA_s(n)/(1+...+n)=(A_g(1)+2A_g(2)+...+nA_g(n))/(1+...+n).
For each i in {1,...,n} the coefficient that multiplyes A_i is s⁻1(i)/(1+...+n) in the left hand side and g⁻1(i) in the right hand side. It follows that s=g.
I think that, taking n into account, the ratio r'/r can be set to substantiate the second unproven claim. The first unproven claim may be a consequence of Jung's inequality.
EDIT: With the previous argument, we can represent a finite group of order n as the group of linear isometries of a certain polytope in an n-1 dimensional real inner product space.
Now, if a finite group G of linear isometries of an (n-1)-dimensional inner product space V is given, can we define a polytope that has G as its group of symmetries? Yes. I'll give a somehow informal proof.
Let G={g_1,...,g_m}. Let A={a_1,...,a_n} be the set of vertices of a regular n-simplex centered at the origin of V. Let S be the sphere centered at the origin that contains A, and let C be the closed ball. Notice that C is the only minimun closed ball contaiing A.
(Remark: The set A need not be a regular simplex. It may be any finite subset of S that intersects all the possible hemispheres of S. C will then still be only minimum closed ball containing it.)
Remark: An isometry of V is linear iff it fixes the origin.
Before proceeding, we need to be sure that the m copies of A obtained by making G act on it are disjoint. If that is not the case, our set A is useless but we can find a linear isometry T such that TA does de job. We consider the set M of all linear isometries with the usual operator metric, and look into it for an isometry T such that for all (g,a) and (h,b) distinct elements of GxA the equation g(Ta)=h(Tb) does not hold. Because each of the n*m(n*m-1) equations spoils a closed subset of M with empty interior(*), most of the choices of T will do.
Let K={ga/g in G, a in A}. We know that it has n*m points, which are contained in the sphere S. Now let e be a distance that is smaller than a quarter of any of the distances between different points of K. Now, around each vertex a=a_i of A make a drawing D_i. The drawing consists of a finite set of points of the sphere S, located near a (at a distance smaller than e). One of the points must be a itself, and the others (if any) should be apart from a and very near each other, so that a can be easily distinguished. Furthermore, for i=1 the drawing D_i must have no symmetries, i.e, there must be no linear isometries fixing D_1 other than the identity. For other values of i, we set D_i={a_i}. The union F of all the drawings contains A, but has no symmetries. Notice that each drawing has diameter less than 2*e,
Now let G act on F and let Q be the union of the m copies obtained. Q is a union of n*m drawings. Points of different drawings are separated by a distance larger than 2*e. Hence the drawings can be identified as the maximal subsets of Q having diameter less than 2*e. Also, the ball C can be identified as the only sphere with radius r containing Q. S can be identified as the border of C.
Let's prove that the set of symmetries of Q is G. It is obvious that each element of G is a symmetry. Let T be an isometry that fixes Q. It must fix S, so it must be linear. Also, it must permute the drawings. It must therefore send D_1 to some gD_i with g in G and 1<=i<=n. But i must be 1, because for other values of i, gD_i is a singleton. So we have TD_1=gD_1. Since D_1 has no nontrivial symmetries, T=g.
We have constructed a finite set Q with group of symmetries G. Q is not a polytope, but its convex hull is a polytope, and Q is the set of its extremal points.
(*) To show that for any (g,a) and (h,b) distinct elements of GxA the set of isometries T satisfying equation g(Ta)=h(Tb) has empty interior, we notice that if an isometry T satisfies the equation, any isometry T' with T'a=Ta and T'b=/=Tb must do (since h is injective). Such T' may be found very near T, provided dimV>2. The proof doesn't work for n=1 or 2, but these are just the easy cases.
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This is a really nice argument. I like that it is effective: given an explicit description of a group, one could easily find the vertices of the polyhedron P. It's even almost visualizable. – Andrew McIntyre Feb 22 2010 at 15:39
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This construction shows that a finite group G of order n can be realized as exactly the isometries of a convex polyhedron in R^n. Now, suppose m is the least positive integer such that G has a faithful representation in GL(m,R). Typically m is much smaller than n. I am curious whether it is possible to use your strategy to produce a polyhedron in R^m, whose symmetries are exactly the elements of G, considered as elements in GL(m,R). If that could be done, it would be a solution to the original problem with minimal dimension. – Andrew McIntyre Feb 22 2010 at 15:50
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Here's a sketch of an ugly argument. First construct an undirected graph whose automorphism group is G. You can do this by starting with a vertex vg for each element g of G and gluing in a path of length n(g-1h) from vg to vh with an extra leaf attached to the internal vertex of this path nearest to vg. Here n is an injective function from G to {3, 4, 5, ...}. (This construction might need |G| >= 4 or so, but clearly we can handle the smaller cases by hand.)
Now make the vertices of this graph into a metric space where the distance between two nonadjacent vertices is 1 and the distance between two adjacent vertices is 1-ε. Now we need to embed this metric space into some R^N. You can either appeal to some results about embedding finite metric spaces into R^N, or convince yourself that the map (x1, ..., xn) in (R^N)^n |-> (d(xi, xj)^2)_{1 <= i < j <= n} is a submersion near the regular simplex of edge length 1.
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Thanks, this is ugly but also neat! I am convinced now. I am still wondering whether there is a "nice" construction. I don't know how to define "nice", but I am getting the feeling that, for any reasonable candidate for "nice", one could construct counterexamples (I suspect that any construction that is too "simple" should end up with accidental symmetries, as noted by A.G. below). – Andrew McIntyre Oct 18 2009 at 16:05
I think I have a solution that will work, but I'm not 100% certain.
Let V be a faithful representation of G (so the map G→GL(V) has no kernel). Pick a "sufficiently generic" point of V and consider the convex hull of the orbit of that point. G includes into the group of symmetries of this polytope, but it could have additional symmetries. For example, Z/4 acts on the plane by rotation. If you take any point, its orbit is a square, which has additional symmetries (reflections).
You can fix this by modifying the above construction as follows. Replace the point by a set of points S which is totally asymmetric (has no symmetries at all in GL(V)). Think of this set S as being a small cluster of points very far from the origin, so all the images of S under the action of G are easily distinguishable. Since S was totally asymmetric, the only symmetries of the union of these images should be elements of G.
If you're careful with how you chose your S, you shouldn't "lose too much asymmetry" when you take the convex hull of the union of the images of S.
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Thanks! I was thinking basically along those lines, but I got stuck on trying to prove the "shouldn't lose too much asymmetry" step. I also couldn't convince myself that every rep in GL(V) would reduce to the orthogonal group (this is probably easy, but I couldn't see it). – Andrew McIntyre Oct 18 2009 at 15:58
That argument's a standard trick: if you start with an inner product (u, v), define a new one by (u, v)' = 1/|G| sum (gu, gv). This inner product is G-invariant so the representation is orthogonal with respect to it. – Qiaochu Yuan Oct 18 2009 at 21:30
You know, I use that trick at least once a week in the context of Poincaré series. I don't know why I didn't see it here. Thanks! – Andrew McIntyre Oct 19 2009 at 2:45
Here is an explicit (and short) version of Anton Geraschenko's answer.
Let $G$ act on a regular $(n-1)$-dimensional simplex $\Delta$ permuting its vertices as in Cayley's theorem ($n$ in the order of $G$). Let $p_0,\dots,p_{n-1}$ be the vertices of this simplex.
Cut off a small simplex $\Delta'$ (located near $p_0$) constructed as follows: for $k=1,\dots,n-1$, let `$p'_k$` be the point on the edge `$[p_0p_k]$` at the distance $k\varepsilon$ from $p_0$, where $\varepsilon=1/100n$. The simplex $\Delta'$ is the convex hull of `$p_0,p'_1,\dots,p'_{n-1}$`, you cut is off from $\Delta$ by the hyperplane through points `$p'_1,\dots,p'_{n-1}$`. Also, cut off all images of $\Delta'$ under the action of $G$.
The resulting polytope does not have any extra symmetries, so $G$ is its group of self-isometries.
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Suppose one has a convex 4-gon in the plane. What symmetry groups can it have?
The graph of this convex 4-gon is a 4-cycle so as a graph its automorphism group is the dihedral group which I will denote D(4) which has 8 elements. Now there is a convex 2-dimensional polygon which has this as its group, namely a square. However, The group D(4) has a cyclic subgroup of order 4, yet there is NO convex 4-gon which has the cyclic group of order 4 as its set of isometries. There is a rectangle with unequal sides which has a group of order 4 as its symmetry group but this is the Klein group, not the cyclic group, of order 4.
For 3-dimensions, a similar thing can happen. It is known that the vertex-edge graph of any 3-dimensional convex polytope is a planar and 3-connected graph and the converse holds. This is Steinitz's Theorem. Suppose H is such a graph (e.g. planar and 3-connected) and the (full) automorphism group of H is G. There is a beautiful theorem of Peter Mani's which states H can be realized in 3-space by a metric polytope P which has group G as its group of isometries. However, it does not follow that for any subgroup I of G that there is a 3-dimensional convex polyhedron whose group of isometries is I. In fact, for the group with 48 elements which is the isometry group of the graph of the 3-cube there is a subgroup of order 24, the rotation group, but there is no 3-dimensional convex polyhedron which is combinatorially a 3-cube which has 24 isometries.
Here is the reference for Mani's paper:
P. Mani, Automorphismen von polyedrischen Graphen, Math. Ann. 192 (1971) 279–303.
This is a generalization of this theorem to complexes, as mentioned in this paper:
http://arxiv.org/abs/math/0310165
If you restrict your attention to graphs rather than polytopes there is a nice theorem of Roberto Frucht.
For any finite group H, there is a graph G(H) such that the automorphism group of G(H) is H.
There are extensions of Frucht's Theorem including to 3-valent graphs. If there was an extension of Frucht's theorem to planar 3-connected graphs than via Steinitz's Theorem the original question would be answered. I am not sure if this has been done or not.
A survey paper (about graphs with specified automorphism groups and related matters) of Babai's is available:
www.cs.uchicago.edu/files/tr_authentic/TR-94-10.ps
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Neat! Do you know if there are analogous results in higher dimensions? A refinement of the original question, (assuming that previous commenters and I are all correct, and the answer is yes), would be to find lower and upper bounds for the integer n, as a function of the order of the group (or some other data from the group). This would be sort of analogous to embedding theorems for manifolds. If the permutohedron argument suggested by Mariano S-A works, we would have the order of the group as an upper bound. – Andrew McIntyre Feb 19 2010 at 15:03
I have edited my post above to include some additional information which address some of what your raise above. – Joseph Malkevitch Feb 19 2010 at 20:39
This has troubled my sleep for a long time, and I keep wanting to offer a variant of the following answer: Consider the regular representation $\mathbb R G$ of $G$. For any 'natural' choice of inner product on $\mathbb R G$ (such as one making $G$ an orthonormal basis), the action is by isometries, and preserves some obvious convex bodies (such as the simplex spanned by $G$); but, for most such choices, $G$ is obviously far smaller than the full group of symmetries.
Inner products on $\mathbb R G$ are the same as $G$-square matrices satisfying certain conditions; the one that I suggested corresponds to the identity matrix. I tried to come up with an argument that showed that a small but suitably 'deranged' perturbation of that matrix would give us a pairing such that $G$ still acted by isometries, but any other symmetries of the basic simplex were killed off. Well --no luck so far, and it's not much different from Anton's suggestion; but, as a representation theorist, I just couldn't resist the appeal of using an 'obvious' vector space (albeit with a non-obvious Euclidean structure).
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Hee hee, I'm glad it troubled your sleep! I have an intuitive feeling that the inner product has too few degrees of freedom to kill symmetries as flexibly as needed. But I can't back up that feeling with an actual proof. It would be lovely if one could find a clean, representation theoretical answer to this question! – Andrew McIntyre Feb 19 2010 at 15:08
I consulted Laszlo Babai (University of Chicago) who has written extensively on the subject of automorphism groups, on the following issue:
Given any group H is there always a planar 3-connected graph G (hence a 3-polytopal graph by Steinitz's Theorem) such that the automorphism group of G is H? (By a theorem of Peter Mani's if such a graph G existed for H then there would be a realization of the graph G as the vertex-edge graph of a 3-polytope P such that the isometries of P would be H.)
Professor Babai informed me that there is no universal theorem here. He offered the 8 element quaternion group as an example of a group H which does not arise as the automorphism group of a planar 3-connected graph.
Here are references to some of his papers that treat aspects of this and related issues:
L. Babai, and W. Imrich: On groups of polyhedral graphs, Discrete Math. 5 (1973), 101-103.
L. Babai: Automorphism groups of planar graphs I, Discrete Math. 2 (1972), 285-307.
L. Babai: Automorphism groups of planar graphs II, in, Infinite and finite sets (Proc. Conf. Keszthely, Hungary, 1973, A. Hajnal et al eds.) Bolyai - North-Holland (1975), 29-84.
L. Babai: Groups of graphs on given surfaces, Acta Math. Acad. Sci. Hung. 24 (1973), 215-221.
L. Babai: Automorphism groups of graphs and edge-contraction, Discrete Math. 8 (1974), 13-20.
L. Babai: Vertex-transitive graphs and vertex-transitive maps, J. Graph Theory 15 (1991), 587--627.
It also turns out there is a paper by: Jurgen Bokowski, G. Ewald, and P. Kleinschmidt, On the combinatorial and affine automorphisms of polytopes, Israel J. of Math., 47 (1984) 123-130 with the following abstract, which may be of interest to those thinking about this circle of ideas:
Abstract We disprove the longstanding conjecture that every combinatorial automorphism of the boundary complex of a convex polytope in euclidean space E d can be realized by an affine transformation of Ed.
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From Wikipedia: "In group theory, Cayley's theorem, named in honor of Arthur Cayley, states that every group G is isomorphic to a subgroup of the symmetric group on G. This can be understood as an example of the group action of G on the elements of G."
So if I well understand, every finite group G is a subgroup of symmetric group G, so if You have symmetric group, and can visualise it, probably by some kind of graph, then there the same picture is useful for visualising subgroup. There is no matter which geometric object is under this picture... So the my answer ( of course partial ) is: try to visualise symmetric group only.
Some remarks: 1: Your question should state that You are interested in symmetries which are exactly from group G, and not any more. If You allow convex bodies for which given group G is only a part of its group of symmetries, the answer would be YES, and sphere gives You model for all finite groups which is trivial.
2:You should do not allow to "degenerate symmetries" that is for example to count some axis symmetries twice ( for example triangle may have 3 axis of symmetry - which may represent for some group elements of order 2, and You do not want them to represent 6 symmetries of order 2 paired together)
Then: I have read the whole question, and I think that the answer is - in general case - NO, asuming You consider 1,2 as requirements, and YES if You from 1,2 above from the list of requirements. Full symmetric group is the case where is not possible to point such convex subset, because if something has such big symmetry it has also have more symmetries inside, but I cannot show any proof for that statement.
However it would be very interesting to see what numbers n of needed R^n are in the series when You order finite groups by their size....
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You did not read the whole question. As you mention, any finite group is a subgroup of a symmetric group; answering the question for the symmetric group would answer it for all finite groups. However, I was not looking to visualize the group as symmetries of "some kind of graph" (which is straightforward), but as isometries of a convex set in R^n. Note that the previous commenters have given good answers. – Andrew McIntyre Feb 15 2010 at 15:48
Is the group of Euclidean symmetries of a permutohedron (built as the convex hull of the point $(1,2,\dots,n)$ and all its permutations in $\mathbb R^n$ larger that $S_n$? – Mariano Suárez-Alvarez Feb 15 2010 at 20:59
ok, I was wrong. Thank You. – kakaz Feb 16 2010 at 9:19
@Mariano S-A: Thanks, I didn't know about the permutohedron! I'll have to check details, but I think your suggestion answers the second part of my original question, giving a simple construction for the required convex set. Just give the group G as a subset of S_n, and then truncate the permutohedron appropriately to restrict its symmetries to the subgroup. – Andrew McIntyre Feb 16 2010 at 19:46
@kakaz: the conditions 1 and 2 you say should be added to the original question are, if I understand them correctly, contained in the word "isomorphic". Reid B and Anton G's answers above show that the answer is yes, with 1 and 2 as requirements. – Andrew McIntyre Feb 16 2010 at 19:49
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http://physics.stackexchange.com/questions/27695/colour-decomposition-of-n-gluon-tree-amplitude
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# Colour decomposition of $n-$gluon tree amplitude
I have here a $SU(N_c)$ Yang-Mill's theory and let the index $i$, label the $n$-gluons, and $\{k_i, \lambda_i, a_i\}$ be its momenta, helicity and colour index and $\cal{A}_n^{tree/1-loop}(\{k_i, \lambda_i, a_i\})$ be the tree/1-loop-level amplitude for their scattering. Then apparently the following two equations hold,
• ${\cal A}_n^{tree}(\{k_i, \lambda_i, a_i\}) = g^{n-2}\sum_{\sigma \in S_n/\mathbb{Z}_n} Tr[T^{a_\sigma (1)}\ldots T^{a_\sigma (n)}] A_{n}^{tree}(\sigma(1^{\lambda_1})\ldots\sigma(n^{\lambda_n}))$
• ${\cal A}_n^{1-loop}(\{k_i, \lambda_i, a_i\}) = g^n [ \sum_{\sigma \in S_n/\mathbb{Z}_n} N_c Tr[T^{a_\sigma(1)}\ldots T^{a_\sigma(n)}] A_{n;1}^{tree}(\sigma(1^{\lambda_1})\ldots\sigma(n^{\lambda_n})) +$ $\sum _ {c=2} ^{[\frac{n}{2}] +1} Tr[T^{a_\sigma (1)}\ldots T^{a_\sigma(c-1)}] Tr[T^{a_\sigma (c)}\ldots T^{a_\sigma(n)}] A_{n;c}^{tree}(\sigma(1^{\lambda_1})\ldots\sigma(n^{\lambda_n}))]$
I want to know the proof for the above two equations.
It seems that this lecture note tries to sketch some argument for the first of the above two expressions but then its not very clear.
• Thought I haven't seen this clearly written anywhere but I guess that the factors of $A_n^{tree}$ and $A_{n;1}$ and $A_{n;c}$ that occur on the RHS of the above two equations are what are called "colour ordered amplitudes". It would be great if someone can say something about this idea too. (..i do plan to put up another separate question later focusing on that aspect..)
{..my LaTeX seems all garbled! It would be great if someone can edit that and put in a line as to what has gone wrong..}
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## 1 Answer
You can show that the amplitude has that form by thinking about the Feynman rules. I'll only discuss particles in the adjoint representation (this is suitable for supersymmetric theories), but this can be done more generally. Note that if you have particles transforming in the fundamental representation of $SU(N)$ then the amplitude does not have the form in your question.
Think about the three-gluon vertex. It contains a $f^{a b c}$ factor which, up to a constant, can be written as $\operatorname{tr}(T^a [T^b, T^c]) = \operatorname{tr}(T^a T^b T^c) - \operatorname{tr}(T^a T^c T^b)$. So this can be written as a combination of traces of products of Lie algebra generators. Now, think about joining together two such triple vertices. We have to compute quantities like $\operatorname{tr}(T^a T^b T^c) \operatorname{tr}(T^c T^d T^e)$, where the index $c$ is summed over.
Now, for $SU(N)$ we have that $$(T^a)_i^j (T^a)_k^l = \delta^j_k \delta^l_i - \frac 1 N \delta^i_j \delta^k_l,$$ where we sum over the index $a$. For $U(N)$ the last term in the right hand side is absent. Using this, you'll see that the color factor for joining two triple vertices can also be written as a single trace. The same can done for the quartic vertex.
For a tree level scattering, your can do this recursively. You start with some triple vertex and you keep joining other vertices and by using the two identities above you can always rewrite the answer as a single trace.
At loop level, this doesn't work because you can get things like $\operatorname{tr}(T^a \cdots T^b T^c) \operatorname{tr}(T^c T^d \cdots T^b)$, with sum over $b$ and $c$. Using the $SU(N)$ identity you get a contribution $\operatorname{tr}(T^a \cdots T^b T^d \cdots T^b)$. Now using the identity again you get $\operatorname{tr}(T^a \cdots) \operatorname{tr}(T^d \cdots)$. In general, at $\ell$ loops you can have $\ell+1$ traces if you have a large enough number of particles.
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Can you kindly expand a bit more on how the constraint on the representation of \$ – user6818 Feb 20 '12 at 1:29
[Ignore the garbled type above!] Can you kindly expand a bit more on how the constraint on the representation of $SU(N)$ comes in? Like if you could write some explicit equations for what you call the fundamental representation - in standard Yang-Mill's theory is there a constraint on which representation should the fermions be in? (..like the gauge fields are constrained to be in the adjoint representation..) The last equation on the first page of my linked lecture note already gives a plausibility argument for this decomposition. – user6818 Feb 20 '12 at 1:36
Can you kindly explain as to how does the third equation on the second page of my linked notes match against the usual Feynman rules? I did not get the part of your argument at the 1-loop level. Can you kindly give a reference for what you are saying? – user6818 Feb 20 '12 at 1:38
@Anirbit, the fermions can be in an arbitrary representation. In my answer I considered fermions in the adjoint for simplicity. If they are in the fundamental of $SU(N)$, then you get a color factor of $(T^a)_i^j$ for an interaction with gauge fields, while you had an $f^{abc}$ in the case of the adjoint representation. As a consequence, you can now also get strings of gauge algebra generators, $(T^a T^b \cdots)_i^j$. For references you can try this review article or these lecture notes. – Sidious Lord Feb 21 '12 at 7:56
If i understand you right then you are saying that the third equation on the second page of my linked lecture notes reduces to the usual Feynman rules when written for the adjoint representation? Can you make it a little more explicit as to what you have in mind as the "adjoint" and the "fundamental" representation? Its getting confusing. The usual normalization stated on the top of the first page of my linked notes when used on the trace factors of the third equation of the second page of my linked notes then it seems to produce an extraneous factor of "-2". – user6818 Feb 23 '12 at 0:22
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http://mathforum.org/mathimages/index.php?title=Polar_Coordinates&oldid=20647
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# Polar Coordinates
### From Math Images
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Polar Coordinates are coordinates which locate points by their distance from the origin and their rotation from the positive x-axis. Traditionally, $r$ represents distance from the origin and $\theta$ represents rotation in radians from the positive x-axis.
For example, the point $(x,y) = \left (\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}} \right )$ is 1 unit from the origin (as can be seen by the Pythagorean Theorem) and is a rotation of $\tfrac{\pi}{4}$ from the positive x-axis, so can be represented in polar form by: $(r,\theta)= \left (1,\tfrac{\pi}{4} \right )$.
Locating a point using its distance from the origin and its rotation for the + x-axis
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http://physics.stackexchange.com/questions/tagged/scattering+conservation-laws
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# Tagged Questions
1answer
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### Scattering problem: Finding the speed of the scatterer after collision
A particle of mass $M$ moving in a straight line with speed $v$ collides with a stationary particle of the same mass. In the center of mass coordinate system, the first particle is deflected by 90 ...
1answer
414 views
### Scattering problem: Converting the two-body lab frame problem into a one-body center-of-mass frame problem
I'm reading the section on scattering in Goldstein's Classical Mechanics, and I have a rather basic question about this. It says that scattering in the laboratory is a two-body problem because of ...
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http://mathhelpforum.com/calculus/89652-instantaneous-velocity-print.html
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# Instantaneous Velocity
Printable View
• May 19th 2009, 07:44 AM
bearhug
Instantaneous Velocity
A flare is shot up from the deck of a ship, the initial upward velocity of the flare is 30 m/s. The height of the flare is given by $h=-5t^{2}+30t+10$
The Question is:
Determine the instantaneous velocity of the flare at t=3 seconds by using the slopes of the secants.
In other examples I found the instantaneous velocity by creating a table and using numbers that get closer and closer to 3 seconds until I can reach a conclusion.....but how do I use the slopes of the secants to find it?
Thank you
• May 19th 2009, 09:05 AM
Isomorphism
Quote:
Originally Posted by bearhug
A flare is shot up from the deck of a ship, the initial upward velocity of the flare is 30 m/s. The height of the flare is given by $h=-5t^{2}+30t+10$
The Question is:
Determine the instantaneous velocity of the flare at t=3 seconds by using the slopes of the secants.
In other examples I found the instantaneous velocity by creating a table and using numbers that get closer and closer to 3 seconds until I can reach a conclusion.....but how do I use the slopes of the secants to find it?
Thank you
Ideally one would differentiate h and find the instantaneous velocity. Apparently "slope of secants" is the approximate idea of differentiation. So my guess is, "I found the instantaneous velocity by creating a table and using numbers that get closer and closer to 3 seconds until I can reach a conclusion" is called "slope of secants" method.
Lets call the height h(t), since its a function of time, then instantaneous velocity at t=3 is $\frac{h(3 + x) - h(3)}{x}$, where x can be any small number(like 0.01,0.001 etc)
• May 19th 2009, 10:34 AM
HallsofIvy
Notice that $\frac{h(3+x)- h(3)}{x}$ is the slope of the secant line: the line between (3, h(3)) and (3+x, h(3+x)). Do that calculation and see what happens for very small values of x.
All times are GMT -8. The time now is 05:10 AM.
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http://mathhelpforum.com/differential-geometry/129138-solved-application-cauchy-s-integral-formula.html
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Thread:
1. [SOLVED] Application of Cauchy's Integral Formula
Here's what I got (since $\frac{1}{2 - z} = \frac{-1}{z-2}$).
The book says the answer is $2\pi i$, not $-2\pi i$. Who is right?
2. The book is right. Check you have got the right singularity.
With "right" i mean the one who is inside your curve and apply the theorem as you just did.
3. Originally Posted by mabruka
The book is right. Check you have got the right singularity.
With "right" i mean the one who is inside your curve and apply the theorem as you just did.
Oh, because the curve is |z+1|= 2, a = -2 is inside the curve, and a = 2 is not. Therefore, I get $\int \! \frac{z^2 (\frac{1}{2 - z}) \, dz}{z - (-2)}$
$= f(-2) = 2\pi i(-2)^2(\frac{1}{2 - (-2)}) = 2\pi i$
Cool, thanks.
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http://mathoverflow.net/questions/54071/dimensions-of-orbit-spaces/54131
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## Dimensions of orbit spaces
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be a compact Lie group acting effectively on a compact, Hausdorff topological space $X$. I am looking for results of the type
If $X$ is a ... and the action is ... then $\dim(X/G)\leq \dim(X)-\dim(G)$.
Here $\dim$ denotes the covering dimension.
For example, $X$ is a smooth manifold and $G$ acts smoothly and almost-freely (all isotropy groups finite).
Can anyone point me to a more general theorem of this type?
-
1
Something to do with positive-dimensional stabilisers, I'm sure would help here. – David Roberts Feb 2 2011 at 7:22
## 2 Answers
I'm loathe to answer my own question, but...
Theorem IV.3.8 of Bredon's book Introduction to compact transformation groups (which sadly is on the page google books decided not to include!) seems satisfying.
Let $G$ be a compact Lie group acting locally smoothly on the manifold $M$, such that $M/G$ is connected. If $P$ is a principal orbit (an orbit of maximum dimension) then
$$\dim(M/G)=\dim(M)-\dim(P).$$
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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.
There is a finer approach to the dimension of the quotient when the action of $G$ on $M$ doesn't define a fibration, but has singular orbits. You'll find it here. Of course for the principal orbit you have your formula, but on singular orbits it gives something else. I didn't give a general formula (I should do it however) but I treated the examples $\Delta_n = {\bf R}^n/{\rm SO}(n)$. Topologically, for any $n$ these quotients are homeomorphic to the half line $[0,\infty[$, but not diffeologically, and if we have well (according to your formula) ${\rm dim}_x(\Delta_n) = 1$ for $x \neq 0$, we have however ${\rm dim}_0(\Delta_n) = n$; which proves by the way, that for 2 different quotients, if they are homeomorphic, they are not diffeomorphic. I'm not sure it will help you very much, because it seems that you are interested only in the principal orbits, but your question was too tempting to not underline this construction.
-
Pattrick, thanks for your answer. Diffeological spaces are new to me! Are they related to the notion of stratifold (as described in Matthias Kreck's book on Geometric Algebraic Topology)? Actually, I am really only interested in the topological (covering) dimension of the quotient, but its always good to learn about new notions of dimension. (Also, its not that I'm only interested in principal orbits, just that I can't think of examples with $\dim(P)\neq\dim(G)$!?) – Mark Grant Feb 3 2011 at 6:51
Or rather I can't think of effective actions. – Mark Grant Feb 3 2011 at 6:52
Hi Mark, diffeology is more general than stratifolds. Because also, it is not aimed to answer the same questions. A diffeological space $X$ is defined by claiming what parametrizations (maps defined on open subset of numerical vector spaces) in $X$ are "smooth". These smooth parametrizations (called plots) must satisfy 3 axioms (covering, smooth compatibility, locality). For $\dim(P)\neq\dim(G)$ what about the example os $\Delta_n$, you want something more singular? – Patrick I-Z Feb 3 2011 at 8:01
Ah yes of course, in your example the principal orbits are spheres, which checks out with Bredon's theorem :) I have been too busy thinking about circle actions I forgot about the classical symmetry groups. Thanks. – Mark Grant Feb 3 2011 at 8:13
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http://rjlipton.wordpress.com/2009/07/28/oblivious-turing-machines-and-a-crock/
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## a personal view of the theory of computation
by
The Fischer-Pippenger Theorem, oblivious Turing Machines and a strange FOCS talk
Michael Fischer has made a huge impact on at least two quite separate areas of complexity theory. In the first part of his career he made important and seminal contributions to the basic understanding of Turing Machines (TM). This work is quite technically challenging and still stands as some of the deepest work done on the behavior of TM’s. While today we are more interested, often, in the behavior of general algorithms, the care and feeding of TM’s is important. In my opinion, there remain many nuggets to be mined regarding their behavior.
In the second part of his career, he switched to the area of distributed computing. He has made many important contributions to this area: both upper and lower bounds. Much of the work—especially in the beginning–was joint work with Nancy Lynch. It is interesting to note that Nancy started her career in recursion theory, before also switching to help create the theory of distributed systems. She was awarded the prestigious Donald Knuth Prize in 2007 for her work on distributed computing. I plan on talking more about that in the future.
Today I plan to talk about a wonderful theorem that is due to Mike Fischer and Nicholas Pippenger. Their result showed a tight relationship between time complexity of a TM and boolean complexity. The bound they proved three decades ago is still the best known. Amazing.
At the FOCS 1975, which was held that year at Berkeley, I had two great memories. One concerns discovering on checking into my room that I was sharing with Rich DeMillo, that there was another occupant sharing the room. There was a mouse racing around our room. A mouse. The meeting was being held in the old Claremont Hotel, and when we told the desk that we had cleverly trapped the mouse under a wastebasket, their response was, “yes, and what would you like us to do?” We, Rich and I, changed rooms—I believe the mouse stayed.
The other memory, from that FOCS, is one about Mike and the word “crock”. See this for the meaning if you are unsure what “crock” means, but I prefer not to directly explain it. That way we can keep our G rating for this blog.
Here is what happened. I was walking the hall when a talk entitled “Economy of Descriptions by Parsers, DPDA’s, and PDA’s” was about to start. I had no intention of going to this talk; it was great work no doubt, but it was not something I knew about, nor cared to know about. As I was trying to find a quiet place to think I ran into Professor Dr. Fischer. I was junior and he was senior. Mike told me that the next talk was on an important topic—one that he had worked on—and I should come hear it. So off we went and sat down in the middle of the hotel ballroom.
The talk was given by Matt Geller and his co-authors were: Harry Hunt, Thomas Szymanski, and Jeffrey Ullman. As soon as the talk started Mike proceeded to fall asleep. I figured that I could not leave in the middle of the talk, and since I had a blank pad and a pen, I started to try to solve some problem I was working on at the time. Talks can be a quite relaxing place to think, all the more true if you are not really listening to the speaker. I knew Matt well, but as I said the talk’s topic was not one I followed.
The talk ended and Matt proceeded to get the usual technical questions. He answered them in turn, and all seemed fine to the naïve observer—me. Somehow Matt then said something that hit a raw nerve with his co-authors. Ullman, recall he was one of the co-authors, raised his hand, and was called on by the session chair. Jeff said, “Matt did you just say X?” Matt immediately answered yes he had. With that Jeff said,
“That is a crock, and I want to completely disassociate myself from that.”
As soon as “crock” was uttered, Mike Fischer’s head popped up and he was fully awake. Is there some subconscious mechanism we have for certain words? Anyway, Mike quickly began to whisper to me what was going on? I started to try to explain the context the best I could while the session chair quickly pointed out that were on a coffee break.
The next thing that happened was unique. Matt was soon surrounded by a circle of experts and all his co-authors. They had a heated argument about was X true. The more Matt said the further, I noticed, that his co-authors moved away—the circle around him keep growing in size. I never did find out what the issue was.
By the way, Geller moved out of theory a few years later and announced that he was moving to Hollywood to become a screenwriter. He actually had some success with TV shows, and there on the screen from time to time was, screenplay by Matt Geller.
Let’s turn now to a wonderful theorem on TM’s.
Oblivious Turing Machines
A TM operates on an input ${x}$ of length ${n}$. The TM can move its heads based on the values of the input. However, we call a TM oblivious (OTM) if the motions of the heads are independent of the input ${x}$. Thus, the head motion can only depend on the length ${n}$ of the input, but not on the individual bits of the input.
Oblivious TM’s are important for several reasons. First, there is a strong connection between the time an OTM takes and the size of the boolean circuit that the machine computes. Second, there have been applications to cryptography over the years of OTM’s, the fact that the heads move the same way independent of the input bits is useful in certain information hiding schemes.
Suppose a TM accepts some language ${L \subseteq \{0,1\}^{*}}$. In a natural way there is a boolean function ${f_{n}}$, for each ${n}$, so that for all ${x \in \{0,1\}^{n}}$,
$\displaystyle f_{n}(x) = 1 \text{ if and only if } x \in L.$
By the boolean complexity of ${f_{n}(x)}$ we mean the size of the smallest general circuit that computes ${f_{n}(x)}$.
Theorem: Suppose that ${M}$ is a deterministic oblivious TM that runs in time ${T(n)}$. Then, the language it accepts has boolean complexity at most ${O(T(n))}$.
The famous Fischer-Pippenger Theorem is:
Theorem: Suppose that ${M}$ is a deterministic TM that runs in time ${T(n)}$. Then, there is a deterministic TM that accepts the same language as ${M}$ and is oblivious. More, it runs in time ${O(T(n)\log T(n))}$.
See this for a modern description of these results. Earlier in 1972 John Savage proved essentially the first theorem and the second with the weaker bound of ${O(T(n)^{2})}$ in his paper “Computational work and time on finite machines.” John’s paper is a very important early paper that had many cool ideas in it. I think it does not get as much attention as it deserves because John tried to make connections between “real” complexity measures and more theoretical ones. Yet it is an early important piece of work.
The Proof
I will not give a detailed proof of their theorem. The reference does a great job, but I do want to say something about how it works in general.
The central idea of simulating a TM and making the tapes oblivious is really a neat kind of amortized complexity argument. This was before the concept of amortized complexity was invented. Mike and Nick had to cleverly move the Turing heads in a fixed manner, but yet do an arbitrary computation.
I think the following may help explain what they needed to do, without getting into the details of Turing tapes. Consider the problem of sorting ${n}$ numbers
$\displaystyle x_{1}, \dots, x_{n}.$
There are many algorithms for sorting ${n}$ numbers in ${O(n \log n)}$ comparisons. However, the data motion is quite adaptive, that is what is stored where and when depends on the value of the data.
At a cost of only an extra ${O(\log n)}$ one can make the data motion independent of the data values. The trick is to use a sorting network, such as: Batcher odd-even mergesort, bitonic sort, or Shell sort. In a sorting network the basic primitive is swap: compare two data items and move the larger to the “bottom” position and the smaller to the “top” position. The order of which items to perform the swap operation on is fixed independent of the values of the items. Here is an example of a sorting network and a trace of how it works on a given input.
See this for details on sorting networks.
In a natural sense a sorting network is oblivious, since which data items are compared at each step are determined independent of the data values. Of course, it is possible to do even better with a more complex sorting network, but I will leave that for another day.
Open Problems
Improve the Fischer-Pippenger Theorem, or prove that it is optimal. A separate question is to improve the circuit simulation result. There is no reason that a TM that runs in time ${n^{100}}$ cannot always be computed by a linear size circuit. I have commented earlier that the famous Andrey Kolmogorov may have believed this. I have tried over the years to improve both these theorems with no success. I hope you are more clever and can see a new approach.
### Like this:
from → History, People, Proofs
9 Comments leave one →
1. anonymous
July 28, 2009 6:11 pm
If we drop the oblivious requirement from the Fischer-Pippenger Theorem, do we know if we can improve upon the log T(n) term?
• rjlipton *
July 28, 2009 9:34 pm
Not sure what you mean. They prove that time T goes to TlogT oblivious time. Then, TlogT oblivious time goes to circuit size O(TlogT).
If you mean can we avoid the step of going to oblivious and go directly to circuit size, the answer is unknown. But I think if this is what you mean it a great idea. If we could somehow avoid making the TM oblivious then perhaps could get a better bound.
2. July 28, 2009 11:39 pm
The proof given in the Vollmer book referenced is not actually based on the original Pippenger-Fischer paper (“Relations Among Complexity Measures”), but is rather based on the paper “The Network Complexity and the Turing Machine Complexity of Finite Functions”, by C.P. Schnorr. From what I can parse, Schnorr based his results off of lectures given by Fischer. Schnorr’s proof seems much simpler to me — the Pippenger-Fischer proof uses a recursive method that I can’t seem to unwind (while Schnorr’s paper gives a straightforward procedure).
Also, I think the above anonymous is referring to how efficient we can simulate TM’s if we drop the oblivious requirement. The T\log T simulation (without obliviousness) was first achieved by Hennie and Stearns in “Two-Tape Simulation of Multitape Turing Machines” (in 1966), and this implied a better deterministic time hierarchy theorem: that for f with g\log g=o(f), TIME(g) is strictly contained in TIME(f). This hasn’t been improved to my knowledge, so I think the answer is that the T\log T simulation (even without obliviousness) is the best we know.
It might also be worth mentioning that the oblivious T\log T simulation comes up in other areas, such as showing time-space lower bounds for SAT.
3. August 6, 2009 12:55 pm
When I introduce this theorem, I diagram the oblivious walk first:
Jag(1)Jag(3)Jag(1)Jag(7)Jag(1)Jag(3)Jag(1)Jag(15)Jag(1)… where
Jag(m) = 0->1->…->m->(m-1)->…->0->(-1)->…->(-m)->(-m+1)->…->0.
I think the one-sided version of this was employed by Axel Thue—I have to re-hunt the reference—so I call it the above the “two-way Thue walk.” Then I claim that n steps of the TM M can be done in n “jags”, and prove that n jags take O(nlogn) steps, with a small constant on the O. This top-down visual approach lets me shortcut the proof of the claim, especially useful when the class has 50-minute lectures and I’m anxious to segue from the bit-level underpinnings to the higher-level NP theory. For Cook’s Theorem I then use Schnorr’s O(nlogn) reduction directly to 3SAT from the same paper, though translating z = x NAND y to clauses (x v z) && (y v z) && (-x v -y v -z).
Savitch’s original O(n^2)-sized circuits have one important virtue: they can be easily stamped out on a 2-dim square mesh, which entails that the distance-d neighborhood of any cell has O(d^2) neighbors . The O(n log n) circuits—whose essence one can grasp by drawing the Thue walk and adding vertical lines connecting successive visits to cell j (for all j)—have exp(d)-sized neighborhoods. The question that most interests me is, suppose we allow a cubic mesh, or more generally allow O(d^m)-sized neighborhoods for general m > 2. Can we achieve circuit size better than O(n^2), such as O(n^{m/(m-1)), perhaps ignoring log factors? This says O(n^2) for m=2, O(n^1.5) for m=3, O(n^1.333…) for m=4, …, and yes O-tilde(n) for “m=infinity”.
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# Tagged Questions
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http://unapologetic.wordpress.com/2010/04/30/measurable-extended-real-valued-functions/?like=1&source=post_flair&_wpnonce=a832ba55c5
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# The Unapologetic Mathematician
## Measurable (Extended) Real-Valued Functions
For a while, we’ll mostly be interested in real-valued functions with Lebesgue measure on the real line, and ultimately in using measure to give us a new and more general version of integration. When we couple this with our slightly weakened definition of a measurable space, this necessitates a slight tweak to our definition of a measurable function.
Given a measurable space $(X,\mathcal{S})$ and a function $f:X\to\mathbb{R}$, we define the set $N(f)$ as the set of points $x\in X$ such that $f(x)\neq0$. We will say that the real-valued function $f$ is measurable if $N(f)\cap f^{-1}(M)$ is a measurable subset of $X$ for every Borel set $M\in\mathcal{B}$ of the real line. We have to treat ${0}$ specially because when we deal with integration, ${0}$ is special — it’s the additive identity of the real numbers.
The entire real line $\mathbb{R}$ is a Borel set, and $f^{-1}(\mathbb{R})=X$. Thus we find that $N(f)$ must be a measurable subset of $X$. If $E$ is another measurable subset of $X$, then we observe
$\displaystyle E\cap f^{-1}(M)=((E\cap N(f))\cap f^{-1}(M))\cup((E\setminus N(f))\cap f^{-1}(M))$
The second term on the right is either empty or is equal to $E\setminus N(f)$. And so it’s clear that $E\cap f^{-1}(M)$ is measurable. We say that the function $f$ is “measurable on $E$” if $E\cap f^{-1}(M)$ is measurable for every Borel set $M$, and so we have shown that a measurable function is measurable on every measurable set.
In particular, if $X$ is itself measurable (as it often is), then a real-valued function is measurable if and only if $f^{-1}(M)$ is measurable for every Borel set $M\in\mathcal{B}$. And so in this (common) case, we get back our original definition of a measurable function $f:(X,\mathcal{S})\to(\mathbb{R},\mathcal{B})$.
The concept of measurability depends on the $\sigma$-ring $\mathcal{S}$, and we sometimes have more than one $\sigma$-ring floating around. In such a case, we say that a function is measurable with respect to $\mathcal{S}$. In particular, we will often be interested in the case $X=\mathbb{R}$, equipped with either the $\sigma$-algebra of Borel sets $\mathcal{B}$ or that of Lebesgue measurable sets $\overline{\mathcal{B}}$. A measurable function $f:(\mathbb{R},\mathcal{B})\to(\mathbb{R},\mathcal{B})$ will be called “Borel measurable”, while a measurable function $f:(\mathbb{R},\overline{\mathcal{B}})\to(\mathbb{R},\mathcal{B})$ will be called “Lebesgue measurable”.
On the other hand, we should again emphasize that the definition of measurability does not depend on any particular measure $\mu$.
We will also sometimes want to talk about measurable functions taking value in the extended reals. We take the convention that the one-point sets $\{\infty\}$ and $\{-\infty\}$ are Borel sets; we add the requirement that a real-valued function also have $f^{-1}(\{\infty\})$ and $f^{-1}(\{-\infty\})$ both be measurable to the condition for $f$ to be measurable. However, for this extended concept of Borel sets, we can no longer generate the class of Borel sets by semiclosed intervals.
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1. [...] Measurable Real-Valued Functions We want a few more convenient definitions of a measurable real-valued function. To begin with: a real-valued function on a measurable space is measurable if and only if for [...]
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3. [...] sets and Lebesgue measure, and to tweak the definition of a measurable function to this space like we did before to treat the additive identity specially. Then we could set up products (which we will eventually [...]
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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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# Curvature of spacetime in only required to explain tidal forces?
I'm a bit confused about the equivalence principle in GR.
I'm quoting from Wikipedia:
An observer in an accelerated reference frame must introduce what physicists call fictitious forces to account for the acceleration experienced by himself and objects around him. One example, the force pressing the driver of an accelerating car into his or her seat, has already been mentioned; another is the force you can feel pulling your arms up and out if you attempt to spin around like a top. Einstein's master insight was that the constant, familiar pull of the Earth's gravitational field is fundamentally the same as these fictitious forces
Later it is written:
The equivalence between gravitational and inertial effects does not constitute a complete theory of gravity. When it comes to explaining gravity near our own location on the Earth's surface, noting that our reference frame is not in free fall, so that fictitious forces are to be expected, provides a suitable explanation. But a freely falling reference frame on one side of the Earth cannot explain why the people on the opposite side of the Earth experience a gravitational pull in the opposite direction
Here are some things I hope I understand correctly:
• A particle in free fall is in an inertial frame of reference
• Curvature of spacetime in only required in order to explain tidal forces, as long as you ignore tidal forces, you can explain gravity without curvature.
• Gravity is a fictious force experienced in a non-inertial reference frame
My Questions (2 very related questions)
• 1) The statement that curvature of spacetime in only required to explain tidal forces seems weird to me. In the case that there is no curvature of spacetime, what explains gravity? I mean, if gravity is a "fictitious-force", what is the "real cause" of it? (Again this question stems from the statement that curvature is only needed to explain tidal forces, and not all of gravity).
Last example from Wikipedia:
For gravitational fields, the absence or presence of tidal forces determines whether or not the influence of gravity can be eliminated by choosing a freely falling reference frame
• 2) If I'm in outer space and I'm freely falling towards earth, let's say I'm very small and I don't experience tidal forces, both me and earth are freely falling and thus in inertial reference frames, and yet I see the earth accelerating towards me, in my frame is it said that "gravity is eliminated"? just because I feel no tidal forces?
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## 1 Answer
I've only skimmed the Wikipedia article you link to. From a quick look I'd say the paragraphs you quote are making points about what a theory of gravity needs to look like. For example you say "Curvature of spacetime in only required in order to explain tidal forces", but what that really means is that it's impossible to have a theory of gravity without curvature. That's because any theory of gravity inevitably has to describe tidal forces. You go on to say "as long as you ignore tidal forces, you can explain gravity without curvature", but you can't ignore tidal forces so you can't explain gravity without curvature.
To take your two specific questions:
Question 1. Gravity i.e. General Relativity isn't a theory of forces: it's a theory of curvature. By focussing on the "fictitious forces" you're getting the wrong idea of how GR works. When you solve the Einstein equation you get the geometry (curvature) of space. This predicts the path a freely falling object will take. We call this a geodesic and it's effectively a straight line in a curved spacetime. If you want the object to deviate away from a geodesic then you must apply a force - and there's nothing fictitious about it.
For example, GR predicts that spacetime is curved at the surface of the Earth, and if you and I were to follow geodesics we'd plummet to the core. That we don't do so is evidence that a force is pushing us away from the geodesic, and obviously that's the force between us and the Earth. But, and it's important to be clear about this, the force is not the force of gravity, it's the force between the atoms in us and the atoms in the Earth resisting the free falling motion along a geodesic.
Question 2. Again this is really just terminology. When you're free falling "gravity" is not eliminated. Remember that "gravity" is curvature, and in fact the curvature is the same for all observers regardless of their motion. That's because the curvature tensor is the same in all co-ordinate frames. The existance of tidal forces is proof that gravity/curvature is present.
When you're free falling you are moving along a geodesic. It is true to say that there are no forces acting, but this is always the case when you are moving along a geodesic. Remember a geodesic is a straight line and objects move in a straight line when no forces are acting. There would only be a force if you deviated from the geodesic e.g. by firing a rocket motor.
Response to fiftyeight's comment: this got a bit long to put in a comment so I thought I'd append it to my original answer.
I'm guessing your thinking that if you accelerate a spaceship it changes speed, so when you stop something has happened, but when the Earth accelerates you nothing seems to happen. The Earth can apply a force to your for as long as you want, and you never seem to go anywhere or change speed. Is that a fair interpretation of your comment?
If so, it's because of how you're looking at the situation. Suppose you and I start on the surface of the Earth, but you happen to be above a very deep mine shaft (and in a vacuum so there's no air resistance - hey, it's only a thought experiment :-). You feel no force because you're freely falling along a geodesic (into the Earth), while I feel a force between me and the Earth. From your point of view the force between me and the Earth is indeed accelerating me (at 9.81ms$^{-2}$). If you measure the distance between us you'll find I am accelerating away from you, which is exactly what you'd expect to see when a force is acting. If the force stopped, maybe because I stepping into mineshaft as well, then the acceleration between us would stop, though we'd now be moving at different velocities. This is exactly what you see when you stop accelerating the spaceship.
It's true that a third person standing alongside me doesn't think I'm accelerating anywhere, but that's because they are accelerating at the same rate. It's as though, to use my example of a spaceship, you attach a camera to the spaceship, then decide the rocket motor isn't doing anything because the spaceship doesn't accelerate away from the camera.
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great answer, thank you, one clarification if I may: In the case of Me and You standing on Earth, once the forces of the floor acting on my feet have given me some acceleration and deviated me from my geodesic, am I not on a new geodesic now through spacetime in which I am just floating above the floor. What I mean is I don't understand why the forces of the floor need to keep accelerating me all the time, because usually once a force have acted for some time you can stop the force and the object will keep moving in an inertial way. – fiftyeight May 7 '12 at 15:07
Hi fiftyeight, I've appended a reply to the original question because it got a bit long to put in a comment. – John Rennie May 7 '12 at 18:00
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http://mathoverflow.net/revisions/71206/list
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2 edited tags
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# Discrete-analytic functions
I do not know if such concept already exists but lets consider functions which are equal to its Newton series.
We know that functions which are equal to their Taylor series are called analytic, so lets call functions that are equal to their Newton series "discrete analytic".
The formula is alalogious to Taylor series but uses finite differences instead of dirivatives, so for any discrete-analytic function:
$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$
It is known that for a functional equation $\Delta f=F\,$there are infinitely many solutions which differ by any 1-periodic function. But it appears that there is only one (up to a constant) discrete-analytic solution, i.e. all discrete-analytic solutions differ only by a constant term.
Thus I have the following questions:
• Do discrete-analytic functions express special properties on the complex plane?
• Is there a method to extend the notion of discrete analiticity to a range of functions for which Newton series does not converge (so to make it possible to choose the distinguished solution to the abovementioned equation)?
For the second part of the question, as I know there is at least one one similar attempt, the Mueller's formula:
If $$\lim_{x\to{+\infty}}\Delta f(x)=0$$ then $$f(x)=\sum_{n=0}^\infty\left(\Delta f(n)-\Delta f(n+x)\right)$$
although it seems not to be universal and I do not now whether it is always useful.
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http://gilkalai.wordpress.com/2008/06/10/hellys-theorem-hypertrees-and-strange-enumeration-i/?like=1&source=post_flair&_wpnonce=81d32a40f7
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Gil Kalai’s blog
## Helly’s Theorem, “Hypertrees”, and Strange Enumeration I
Posted on June 10, 2008 by
### 1. Helly’s theorem and Cayley’s formula
Helly’s theorem asserts: For a family of n convex sets in $R^d$, n > d, if every d+1 sets in the family have a point in common then all members in the family have a point in common.
Cayley’s formula asserts: The number of trees on n labelled vertices is $n ^{n-2}$.
In this post (in two parts) we will see how an extension of Helly’s theorem has led to high dimensional analogs of Cayley’s theorem.
left: Helly’s theorem demonstrated in the Stanford Encyclopedia of Philosophy (!), right: a tree
### 2. Background
This post is based on my lecture at Marburg. The conference there was a celebration of new doctoral theses in discrete mathematics, so I decided to devote my lecture to my own graduate years. I studied at the Hebrew University of Jerusalem and my supervisor was Micha. A Perles. I did my master’s thesis and some research as an undergraduate with Micha’s supervision, and it is correct to say that Perles educated me as a mathematician.
Among my older “academic brothers” are Meir Katchalski, Michael Kallay (in Hebrew we have the same last name), and Nati Linial who graduated a couple of years before me. Noga Alon was an academic twin, and, among academic brothers whose graduate years overlapped with mine, were Ido Shemer and Yaakov Kupitz. Later, Perles and I had three joint graduate student,, the first of whom was Ron Adin. We had (and still do) a lovely “combinatorics and convexity seminar” on Monday mornings, and many other mathematical activities.
### 3. The Katchalski-Perles conjecture
Suppose you have a family of n convex sets in $R^d$ such that no more than m of them have a point in common (m>d). What is the maximum number of intersecting (d+1)-tuples of sets from the family?
The proposed extreme example was this: Take m-d copies of $R^d$ itself and n-m+d hyperplanes in general position. Among the hyperplanes, at most d have non empty intersection. Therefore, altogether you will not have more than m sets in the family with non-empty intersection. So the proposed answer is: ${{n} \choose {d+1}} - {{n-m+d} \choose {d+1}}$.
This conjecture was proposed by Katchalski and Perles around 1980. Note the following two extreme cases of the conjecture: If m=n-1 you go back to Helly’s theorem. The other extreme case is m=d+1, and in the rest of this post we will discuss this case.
### 4. Nerves and face numbers
The nerve N of a family of sets $A_1, A_2, \dots, A_n$ in an abstract simplicial complex whose vertices are labelled with {1,2,…,n} which record the intersection pattern of the sets. S belongs to the nerve if $\cap_{i \in S} A_i \ne \emptyset$. The nerve is a simplicial complex, namely a family of sets closed under taking subsets. Another piece of useful notation is this: for a simplicial complex K, denote by $f_i(K)$ the number of i-faces of K. An i-face is simply a set $S \in K$ with $|S| =i+1.$ For the nerve N of a family of sets, $f_i(N)$ is the number of (i+1)-tuples of sets in the family with non empty intersection.
### 5. Collapse of nerves
Suppose we have a family of n convex sets in $R^d$ and no d+2 of them have a point in common. A theorem of Wegner asserts that the nerve N of the family can be collapsed to its (d-1)-dimensional skeleton. In other words, we can delete all the d-dimensional faces (sets of size d+1) from the complex by repeating the following operation, which is called an elementary collapse.
Find a (d-1)-face that belongs to a unique d-face and delete both these faces from the simplicial complex.
In particular, this implies that the d-th homology of the nerve (say with coefficients modulo 2) vanishes. This vanishing of homology follows also (even in the greater generality of nerves of good covers of a subset of $R^d$ ) from the nerve lemma.
### 6. Perles’ observation
Now consider the incidence matrix A between (d-1)-faces and d-faces of the nerve N. Think about this matrix over the field of two elements Z/2. Namely, A is a matrix whose rows correspond to (d-1)-faces of N, and whose columns correspond to d-faces of N, and the entry is 1 if the d-face that corresponds to the row contains the (d-1)-face corresponding to the column, and 0 otherwise. The d-th homology with Z/2 coefficients corresponds to linear dependencies among the columns. (Because we assumed that there are no faces of dimension >d.)
Our nerve has n vertices. (Because the family of convex sets has n members.) If you take the incidence matrix for all possible (d-1)-faces (d-subsets) and all possibled-faces (=(d+1)-subsets) the rank of this matrix is $n-1 \choose d$. To see that the rank is not larger, note that the rows corresponding to sets containing ‘n’, linearly depend on the other rows. To see that the rank is not smaller note that rows corresponding to all d-sets not containing n together with columns corresponsing to all (d+1)-sets containing ‘n’ give us an identity square matrix of size $n-1 \choose d$.
Now, for the incidence matrix A coming from the nerve N of our family of convex sets, we know that the columns are linearly independent. Therefore, an upper bound on the rank gives us the same upper bound on the number of columns. This proves that a family of n convex sets in $R^d$ whose nerve N satisfies $\dim N= d$ (i.e., no d+2 sets in the family have a point in common), satisfies $f_d(N) \le{{n-1} \choose {d}}$. This is Perles’ proof of the case m=d+1 of his conjecture with Katchalski.
What are the cases of equality?, we can ask.
### 7. What are hypertrees and a conjecture about their number
For the very special case of d=1, m=3, the topological condition of being collapsible (or of vanishing first homology group) gives us the family of forests. If we insist on having $f_1(N)=n-1$ we get the family of trees. Those are nice objects that we like! (and like to count.) (Nerves N of families of intervals on the line, no three of them intersect, with $f_1(N)=n-1$, are special types of trees called caterpillars.) Also, for larger dimensions, if we have equality $f_{d-1}(N)=$ $n-1 \choose d$ it follows (using the Euler-Poincaré formula) that the (reduced) Betti numbers of N vanish.
So we can speculate that collapsible complexes, or perhaps acyclic simplicial complexes K of dimension d which satisfy $f_{d-1}(N)=$ $n-1 \choose d$ are some sort of “hypertrees” and that we should try to enumerate them á la Cayley. Some inspections suggest that their number should be: $n^{{n-2} \choose {d}}$.
Cayley’s theorem, the case of at most 4 vertices. This picture (from Wikipedia) describes the labelled trees on 2,3 and 4 vertices. On 4 labelled vertices there are four 2-dimensional “hypertrees”, each corresponding to taking 3 out of the 4 triangles in the boundary of a tetrahedron. This agrees with the conjectural formula.
### 8. Why the conjecture cannot work
The first place to try this conjecture is when the dimension is 2. The conjecture is true when n is at most 5. With six vertices there are some difficulties. The number of collapsible two dimensional simplicial complexes with complete first dimensional skeletons on 6 verticesis 46608. This comes 48 short of the conjecture which is $6^6$ = 46656. There are also 12 triangulations of the real 2-dimensional projective plane. Their status is negotiable. We may consider them as two dimensional hypertrees since their second homology groups vanish. But even with them we are short. All other simplicialcomplexes with 6 vertices, 10 triangles and complete one-dimensional skeletons have non zero second homology and are disqualified as hypertrees. Ethan Bolkercame to the same conjecture several years earlier while studying generalization of the transportation polytopes. He had also made similar conjectures for multicolored “hypertrees”. But the case n=6 convinced Bolker that these conjectures are false.
The 6-vertex triangulation of $RP^2$ is obtained by identifying opposite verices in the Icosahedron. The group of automorphism of this triangulation has 60 elements and therefore there are 12 such triangulations on 6 labeled vertices.
There is something even worse. The ratio between $n^{{n-2} \choose {d}}$ the number of “hypertrees” according to our conjecture and the total number of simplicial complexes with precisely ${n-1} \choose {d}$ d-faces (and complete (d-1)-dimensional skeletons), behaves like $(((d+1)/e)^{{n} \choose {d}}$. For d>1 this is a huge number. If any sort of enumeration of hypertrees gives the formula we conjectured, this ratio should better have been below 1.
We have strong evidence against a $n^{{n-2} \choose {d}}$ conjecture.
### What can we do??? (…to be continued)
(few typos corrected, June 18)
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### 5 Responses to Helly’s Theorem, “Hypertrees”, and Strange Enumeration I
1. Leonard Schulman says:
Hi Gil. We’re waiting in suspense. In the meantime: I suppose the line “If m=n you go back to Helly’s theorem” in sec. 3 par. 2 should read “If m=n-1 you go back to Helly’s theorem”. Best – Leonard
2. Gil Kalai says:
Dear Leonard, thanks! typo fixed. greetings from Jerusalem –Gil
3. Pingback: Plans and Updates « Combinatorics and more
4. Pingback: A Beautiful Garden of Hypertrees « Combinatorics and more
5. Pingback: Nerves of Convex Sets – A Recent Result by Martin Tancer « Combinatorics and more
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http://mathoverflow.net/questions/34369/on-the-behaviour-of-sinn-pi-x-when-x-is-irrational/34372
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## On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi, I'm interested in the behaviour of the sequence $(\sin(n!\pi x))$, when $x$ is irrational, as $n$ tends to infinity.
1) Is the sequence dense in $(-1,1)$?
or
2) Is it possible that for some irrational $x$, $\sin(n!\pi x)$ tends to $0$ as $n$ tends to infinity?
Any reference would be appreciated,
Thank you
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1
The case $x=e$ is a popular riddle. A variation is $n\sin(2\pi en!)\to2\pi$ (whence $e\notin\mathbb{Q}$). Apart from these elementary cases, it sounds like a question immediately going into wild open problems... – Pietro Majer Aug 3 2010 at 12:42
However, I guess it should not be difficult to buid an x for which 1) holds. This immediately produces a dense set $x+\mathbb{Q}$; and in fact I imagine that one can also prove that 1) holds generically. – Pietro Majer Aug 3 2010 at 13:19
1
Homework question. Seriously! When I was an undergraduate at Cambridge this was on one of the first problem sheets in the first analysis course. I can't help but think it was intended to scare people away. It generated much discussion and I think the consensus was that there are no nice answers to this question. – Dan Piponi Aug 3 2010 at 21:47
@sigfpe That wasn't a Hungarian-drafted problem sheet was it? – Yemon Choi Aug 3 2010 at 23:26
## 2 Answers
For $x=e=\sum 1/{{i}!}$, the sequence $sin(n!\pi x)$ tends to zero since the sequence of fractional parts {$n!x$} tends to zero. Hence generally answer on your question is negative.
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2
also note that the sequence $\sin(\pi xn!)$ only changes by finitely many terms if a rational is added to $x$; so in particular 2) holds in a dense set. – Pietro Majer Aug 3 2010 at 13:24
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Moreover, it easily generalizable to $\sum_{i\in I}1/i!$, where $I$ is any infinite subset of $\mathbb N$. That gives us a continuum points in $\mathbb R / \mathbb Q$. – Petya Aug 3 2010 at 15:21
Petya, I hope you don't mind my minor edit. I stared at your post for a full five minutes before I understood that you didn't mean to define $e$ as $\sum 1/i! \sin(n!\pi x)$... – Willie Wong Aug 3 2010 at 20:36
Thank you! I also stared on your comment for a few minutes! But now I understand. – Petya Aug 4 2010 at 22:01
### 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.
Denote by G the set of all x for which $sin(n! pi x)$ approaches 0 as n approaches infinity. Every real number $0 < x < 1$ can be uniquely written as $\sum_{n \ge 2} \frac{x(n)}{n!}$ where $0< x(n) < n$. From this it can be immediately seen that $x \in G$ iff $\frac{x(n)}{n}$ approaches 0. This immediately shows that there are continuum many points of G in every interval.
It can also be shown that G is an F-sigma-delta (countable union of countable intersections of open sets) additive subgroup of reals but it's neither G-delta nor F-sigma. One may iterate this construction to obtain additive groups of reals at arbitrarily high finite levels of Borel hierarchy in the following way: Let $G_0 = G$. Let $G_{n+1}$ be the set of all x such that the fractional part of $(n!x)$ converges to a point in $G_n$. Then $G_n$'s form an increasing chain of additive subgroups of reals and their union is a Borel additive subgroup of reals which is not at any finite level of Borel hierarchy.
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http://physics.stackexchange.com/questions/25794/what-prevents-stars-in-globular-clusters-from-merging-over-time-to-form-a-black
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# What prevents stars in globular clusters from merging over time to form a black hole?
Globular clusters are apparently very very old, and the density of these clusters appears to increase as one approaches the center of a cluster. Orbits are bound to be chaotic, since there is no particular orbital plane, unlike a spiral galaxy. From tidal effects alone it seems that over time many of the stars in the middle ought to have merged, forming new stars of greater and greater size. Eventually one should have seen supernovae occuring inside these clusters, or at least so it would seem, and there ought to be black holes in some of them. But it appears that this does not happen, and the stars in these clusters do not merge. What is keeping this from happening?
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1
Angular momentum – Andrew Apr 6 '12 at 11:01
## 1 Answer
There are three things I'll try to explain in my answer. First, globular clusters actually take a long time to evolve, on the order of $10^8$ to $10^9$ years. Second, many encounters between a binary and a wandering star can eject the wanderer with only a small tightening of the binary, thereby depleting the core. Third, it's currently contested that some globular clusters actually do have black holes at their centres.
The rough timescale for the evolution of a globular cluster is the crossing time. That is, the time an average star takes to traverse the whole cluster. The Wikipedia entry for globular clusters says "the mean value is on the order of $10^9$ years". So it would take some time to get a really big black hole to build up.
It's true that there are many interactions between stars in the dense core. However, a lot of these lead to the formation of binary systems rather than collisions or mergers. When a third star interacts with a binary, a common outcome is that the stars in the binary move closer together and energy is conserved by ejecting the third star at high speed. I recently saw a seminar where the speaker did a detailed calculation of how much "radiation" this produces and it turns out quite a lot. Something like, if a binary forms, it can eject 20 or so stars before the binary merges into a single star.
Finally, don't forget that people have already claimed to have evidence for black holes of a few thousand solar masses at the centres of M15 and $\omega$ Cen. They've also suggested such black holes at the centres of globular clusters around other galaxies. Mayall II is one candidate, orbiting Andromeda. The recent intermediate-mass black hole candidate HLX-1 seems to have a population of stars around it, so it might be a globular cluster (but it might also be the stripped core of a galaxy that recently interacted with ESO 243-49).
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http://physics.stackexchange.com/questions/tagged/uncertainty-principle?sort=active&pagesize=30
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This tag is for Heisenberg quantum mechanical uncertainty principle.
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### Observation of violation of the uncertainty principle?
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### Uncertainty Principle and Energy range for an electron in an atom
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Dilemma The uncertainty principle of energy and the 2nd law of thermodynamics don't add up : the uncertainty principle of energy says that \$\Delta \tau \cdot \Delta E \ge \frac{h}{4\pi} = ...
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### Velocity of measurement
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### Experimental perspective in understanding the Heisenberg Uncertainty Principle
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### In QM, does random data “come from anywhere”? Also, what are the properties of the data?
I have only taken a basic quantum mechanics course (this book, so you know where I'm coming from), but I've been wondering about something. If we set up a quantum system in a known state and take a ...
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### Heisenberg uncertainty principle derivation - unexplained factor of $4 \sigma_k^2$ in Gaussian
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### Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time?
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### Gaussian wave packet
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### Proof of Canonical Commutation Relation (CCR)
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### What is $\Delta t$ in the time-energy uncertainty principle?
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### reference for wavepackets and uncertainty relation
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### Theoretical penetration limit for evanescent waves
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http://www.physicsforums.com/showthread.php?p=3761409
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Physics Forums
## Help needed, converting weight to a falling force.
Hey all,
Just a quick overview of myself, I have recently started rock climbing (6 month) and I'm finding myself becoming more curious about the technical equipment involved.
I'm finding it hard to word the questions I'm trying to find the answers too. So I'm going to try and I'm hoping someone will understand.
I weigh 100kg's, How much of an impact force will I create If i was to free fall over different distances with the top distance being no more than 100m.
Can you convert the weight of a moving person into Kn ?(This being the impact force a rope can take, Most ropes can take an average of 9Kn.)
Below is a few things about the Climbing rope i'm currently using. hope this helps
Impact Force (kN): 8.90
Static Elongation (%): 9.10
Dynamic Elongation (%): 33.00
I will be honest although physics interests me I'v never really understood most of it and only done what was taught in school. So any direct answers would be appreciated, But answers with a description as to why would be even better.
Thanks all
Craig
P.s I'm hoping I'v worded this correctly, I doubt I have so any feed back or questions are more than welcome.
Thanks again
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Homework Help As you fall, you gravitational potential energy is converted to kinetic energy - so you do $mgh=E_K$ neglecting air resistance. In general you can't so this will give you an over-estimate. You fall the length of your rope - the rope then stretches to take the strain - storing the kinetic energy ... then you bounce around a bit. It is very common to model the rope+climber as an undamped harmonic oscillator. For more details on how to use the rope impact http://en.wikipedia.org/wiki/Fall_factor There are usually lots of rules of thumb for this. The fastest way to find out if you'll break the rope is to tie a weight to the end of it and drop it off a bridge or something.
When you fall you gain speed at roughly 10m/s^2 so after 1 sec your speed is 10m/s, after 2 sec your speed is 20m/s and so on. This means that you gain MOMENTUM ( mass x velocity) so after 1 sec of falling your momentum is 100 x 10 = 1000 Ns after 2 sec of falling your momentum is 2000 Ns and so on When you are brought to a halt (by rope, air bag, seat belt.... whatever) the force on you x the time for which it acts brings the momentum to zero (you stop!!) If the rope stops you in 1 sec then after falling for 1 sec the force to stop you is 1000N.... the same as your weight. You could call this '1g' After 2 sec of falling the force to stop you in 1 sec is 2000N.... 2g and so on You can see that it is important to extend the time to come to a stop for as long as possible to reduce the force you need to experience. This is the basic physics behind elastic ropes, seat belts, air bags, crumple zones etc..... to extend the time.
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Science Advisor
## Help needed, converting weight to a falling force.
Does "Dynamic elongation" tell you the percentage stretch when 8.9kN is applied (at the end of your fall)?
Interesting. Maybe I can use the above maths to work out my question on my thread ??? Thats if no one has answerd it yet. Wayne
Recognitions:
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Quote by waynexk8 Interesting. Maybe I can use the above maths to work out my question on my thread ??? Thats if no one has answerd it yet. Wayne
Unfortunately, "falling force" doesn't really mean anything, as a term in Physics and I was trying to ascertain exactly what the terms in the OP really mean (hence my question).
In any case, no muscles are involved in this question. I have a feeling that we won't be dealing with acres and acres of non-Physics reasoning, once we get to the bottom of the question.
Recognitions: Gold Member Science Advisor OK I've looked up the terms and done a rough calculation based on the gravitational potential energy lost on the journey to the bottom of your fall being equal to the elastic energy stored on the stretched rope. For your rope, if you (100kg) fall 100m to the end of the rope, it would appear to stretch by about 31m (!!!! boinnnng!!!) and the force on you, at the very bottom, will be about 8.4kN. This means you will be experiencing a pull of about 8g. It will come on gradually, of course, as the rope tightens. If your harness doesn't fit properly (or even if it does) that's a pretty horrific force it will be exerting on you. I will check this tomorrow and try to give you a way of working out the forces for any drop height.
Recognitions: Gold Member Science Advisor Ha!!! I see it now. All other things being equal, it makes no difference how far you fall to the max force on your harness. That's why the figure they publish is so useful. Hard to believe? Read on. . . . . I went to the trouble of calculating for a 10m rope length and got the same sort of answer as for a 100m rope and then started thinking and calculating. Here's the arm-waving argument that may explain it. The Kinetic Energy gained as you fall is proportional to the height fallen. This energy gets stored in the stretched rope. (Thankfully, there is significant internal friction so you don't just bounce up and down for ever!). A longer rope will stretch by the same percentage of its length as a similar short rope (the short rope will be 'stiffer', with a higher 'spring constant') so you will take a proportionally longer distance to come to a halt. The force at the end of the ride will be the same in all cases. But this isn't the whole story. As far as the experience goes, you will suffer a lot less from a short fall on a short rope, because the braking forces will be acting on you for a much shorter time so a long fall is not to be recommended. Of course, a short fall on an already long rope, will be much more gentle - which is what we would expect and is why your rope is always pulled in as much as poss.
Tags
gravity, impact force, weight
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http://mathoverflow.net/revisions/91685/list
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## Return to Question
3 added 11 characters in body
The odds of two random elements of a group commuting is the number of conjugacy classes of the group
$$\frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2} = \frac{c(G)}{|G|}$$
If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound).
Is there a character-theoretic proof of this fact? What is a generalization of this result... maybe it's a result about algebras semisimple-algebras rather than groups?
2 added 72 characters in body
The odds of two random elements of a group commuting is the number of conjugacy classes of the group
$$\frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2} = c(G)$$\frac{c(G)}{|G|}
If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound).
Is there a character-theoretic proof of this fact? What is a generalization of this result... maybe it's a result about algebras rather than groups?
1
# 5/8 bound in group theory
The odds of two random elements of a group commuting is the number of conjugacy classes of the group
$$\frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2} = c(G)$$
If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound).
Is there a character-theoretic proof of this fact? What is a generalization of this result... maybe it's a result about algebras rather than groups?
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http://mathoverflow.net/questions/97819?sort=votes
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## Box-dual of a partition - what is it called?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Fix natural numbers $n,m\in\mathbb{N}$. Given a partition $\lambda\vdash d$ with at most $n$ rows (and at most $m$ columns), we can define a partition $\lambda^\ast=(\lambda^\ast_1,\ldots,\lambda^\ast_n)$ by setting $\lambda^\ast_i:=m-\lambda_{n+1-i}$. Graphically, it is obtained by taking the complement of the Young diagram of $\lambda$ inside the $n\times m$ $-$ box $(m^n)$. Now, I'd be curious to learn about any combinatorial results involving this construction - but I don't know what it is called. If someone could tell me, that would be great - this way I'd not have such a hard time searching.
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A partition of $mn$ with at most $n$ rows and the most $m$ columns IS the $n\times m$ box, no? – Vladimir Dotsenko May 24 at 7:27
How silly. That was supposed to be some $d$, not $nm$. I fixed it. – Jesko Hüttenhain May 24 at 7:31
3
I've heard both "reverse partition" and "complementary partition" used to describe this. – Gjergji Zaimi May 24 at 8:10
That looks good, thanks a bunch. – Jesko Hüttenhain May 24 at 11:49
## 1 Answer
The "right" answer was given by Gjergji Zaimi in his comment above: Both "reverse partition" and "complementary partition" are terms that seem to be in use.
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http://physics.stackexchange.com/questions/46424/what-is-the-bump-near-m-mu-mu-approx-30-text-gev/46445
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What is the 'bump' near $M_{\mu\mu}\approx 30\text{ GeV}$
In this (attached) Summer 2011 plot from CMS (twiki page), they have a plot of the dimuon invariant mass spectrum across 3 orders of magnitude in energy. There seems to be a 'bump' near $M_{\mu\mu}\approx 30\text{ GeV}$ which I have indicated in the plot. Does anyone know where the gentle but distinct rise is coming from?
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Hmmm...nothing in my rather outdated home copy of the Particle Physics Booklet. Be aware that not every bump is a particle, but I haven't a clue. – dmckee♦ Dec 10 '12 at 5:38
Yeah, I know its not a particle. I just wanted to know what's going on in that region in terms of the dynamics. Or maybe is it because they are counting EVENTS and not absolute cross section? So there's a little bit of spectral luminosity dependence? – QuantumDot Dec 10 '12 at 6:29
1
One needs a Monte Carlo to be sure, but I think it is dimuons made up by a muon from a Z or W decay and the second one a random from some other process, even another Z or W . W and Z open at 80 and 90, GeV 30 GeV seems a reasonable sort of threshold for misidentified pairs to appear in bulk. – anna v Dec 10 '12 at 7:13
2 Answers
I would say that this is not physics but rather a detector/instrumentation effect. More specifically this could be due to the fact that the grey distribution is one that is collected with a high-$p_\mathrm{T}$ muon trigger. These triggers kick in at around 15-20 GeV. Depending on the topology of the event (angles) the invariant mass would of this range as well.
It is instructive to have a look at plots using different triggers.
1) in fact in the plot you reference there a a number of different trigger configurations (setup specifically for the resonances plus two general low/high pt trigger paths) superimposed over each other.
2) here's a similar plot with muons trigger that do not have a pt threshold. notice the lack of a bump in the region you specified.
3) here a similar plot by ATLAS (from 2010) where only events passing the high pt muon threshold are plotted. Notice how the bump seems to reappear.
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If you go to the preprint of CMS at this link the uncorrected dimuon plot is shown in fig1 on the left. The bump is there. The ratio of data to Monte Carlo is correctly around 1. From the monte carlo it is seen that the 30GeV bump comes from collective QCD related effects.
Figure 1: The observed dimuon (left) and dielectron (right) invariant mass spectra. No corrections are applied to the distributions. The points with error bars represent the data, while the various contributions from simulated events are shown as stacked histograms. By "EWK" we denote W to lepton neutrino and diboson production. The "QCD" contribution results from processes associated with QCD and could be genuine or misidentied leptons. The lower panels show the ratios between the measured and the simulated distributions including the statistical uncertainties from both.
The dielectron distribution does not show the bump, a good reason to accept that it is misidentification and kinematic balance effect.
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regarding the arxiv paper: given that this is a log scale, the QCD contribution in the muon plot seems not too large, maybe 10% at most. – luksen Dec 10 '12 at 16:51
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http://polymathprojects.org/2012/09/10/polymath7-research-threads-4-the-hot-spots-conjecture/?like=1&source=post_flair&_wpnonce=a75300a21d
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# The polymath blog
## September 10, 2012
### Polymath7 research threads 4: the Hot Spots Conjecture
Filed under: hot spots,research — Terence Tao @ 7:28 pm
It’s time for another rollover of the Polymath7 “Hot Spots” conjecture, as the previous research thread has again become full.
Activity has now focused on a numerical strategy to solve the hot spots conjecture for all acute angle triangles ABC. In broad terms, the strategy (also outlined in this document) is as follows. (I’ll focus here on the problem of estimating the eigenfunction; one also needs to simultaneously obtain control on the eigenvalue, but this seems to be to be a somewhat more tractable problem.)
1. First, observe that as the conjecture is scale invariant, the only relevant parameters for the triangle ABC are the angles $\alpha,\beta,\gamma$, which of course lie between 0 and $\pi/2$ and add up to $\pi$. We can also order $\alpha \leq \beta \leq \gamma$, giving a parameter space which is a triangle between the values $(\alpha,\beta,\gamma) = (0,\pi/2,\pi/2), (\pi/4,\pi/4,\pi/2), (\pi/3,\pi/3,\pi/3)$.
2. The triangles that are too close to the degenerate isosceles triangle ${}(0,\pi/2,\pi/2)$ or the equilateral triangle ${}(\pi/3,\pi/3,\pi/3)$ need to be handled by analytic arguments. (Preliminary versions of these arguments can be found here and Section 6 of these notes respectively, but the constants need to be made explicit (and as strong as possible)).
3. For the remaining parameter space, we will use a sufficiently fine discrete mesh of angles $(\alpha,\beta,\gamma)$; the optimal spacing of this mesh is yet to be determined.
4. For each triplet of angles in this mesh, we partition the triangle ABC (possibly after rescaling it to a reference triangle $\hat \Omega$, such as the unit right-angled triangle) into smaller subtriangles, and approximate the second eigenfunction $w$ (or the rescaled triangle $\hat w$) by the eigenfunction $w_h$ (or $\hat w_h$) for a finite element restriction of the eigenvalue problem, in which the function is continuous and piecewise polynomial of low degree (probably linear or quadratic) in each subtriangle; see Section 2.2 of these notes. With respect to a suitable basis, $w_h$ can be represented by a finite vector $u_h$.
5. Using numerical linear algebra methods (such as Lanczos iteration) with interval arithmetic, obtain an approximation $\tilde u$ to $u_h$, with rigorous bounds on the error between the two. This gives an approximation to $w_h$ or $\hat w_h$ with rigorous error bounds (initially of L^2 type, but presumably upgradable).
6. After (somehow) obtaining a rigorous error bound between $w$ and $w_h$ (or $\hat w$ and $\hat w_h$), conclude that $w$ stays far from its extremum when one is sufficiently far away from the vertices A,B,C of the triangle.
7. Using $L^\infty$ stability theory of eigenfunctions (see Section 5 of these notes), conclude that $w$ stays far from its extremum even when $(\alpha,\beta,\gamma)$ is not at a mesh point. Thus, the hot spots conjecture is not violated away from the vertices. (This argument should also handle the vertex that is neither the maximum nor minimum value for the eigenfunction, leaving only the neighbourhoods of the two extremising vertices to deal with.)
8. Finally, use an analytic argument (perhaps based on these calculations) to show that the hot spots conjecture is also not violated near an extremising vertex.
This all looks like it should work in principle, but it is a substantial amount of effort; there is probably still some scope to try to simplify the scheme before we really push for implementing it.
## 11 Comments »
1. It seems that one of the weaker links in the above process, at least at our current level of understanding, is Step 6. If we had L^2 bounds on the residual $\Delta w_h - \lambda w_h$, then one could get H^2 (and thence L^infty) bounds on the error $w-w_h$, but with piecewise polynomial approximations that are only continuous at the edges, $\Delta w_h$ is going to have some singular components on the edges of the triangle and so L^2 bounds for the residual are not available.
On the other hand, if we get really good bounds for Step 6, then maybe these bounds would also work for perturbations of the triangle and we could skip Step 7.
Comment by — September 10, 2012 @ 7:36 pm
• There are two pieces to Step 6.
part a. Denote $w_h$ as the discrete approximant of $w$ from the finite-dimensional subspace $V_h \subset H^1(PQR)$, where $w_h$ is computed by solving the discrete eigenvalue problem DVP2 exactly.
part b. Denote $v_h$ as the approximation of $w_h$, computed using an iterative process for computing eigenfunctions of discrete systems.
What we need is a rigorous $L^2$ estimate on the residual $\|\Delta v_h -\lambda v_h\|$, since what we have in hand is $v_h$.
I think one can use inverse and duality estimates to compute the finite element residual $\| \Delta w_h -\lambda w_h\|$, since $w_h$ is approximating a $C^2$ function, and can try to post some notes/references.
If we denote the quantity $v_h-w_h = z_h$, then we want $\Delta z_h - \lambda z_h$ to be well-controlled. We only have in hand $\lambda_h$ and $v_h$, so the error estimate we want will have to boot-strap from these.
Apart from using interval arithmetic, I’m not sure how to proceed on this. Insight into part b. is key.
Comment by — September 12, 2012 @ 5:11 pm
• One possibility is to not work with residuals, but instead with the Rayleigh quotient of $w_h$ (which is finite, because it only requires square integrability of the first derivative of w_h, not square integrability of the second derivative). If this Rayleigh quotient is known to be close to the true second eigenvalue $\lambda_2$, then $w_h$ is necessarily close in H^1 norm to a true second eigenfunction. On the plus side, this lets us skip Steps 6 and 7 (since the Rayleigh quotient is quite stable wrt perturbation of the triangle) and would be relatively simple to implement.
On the minus side, it only gives H^1 control on the true eigenfunction, which isn’t quite enough by itself to give L^infty control. But perhaps we can combine it with some elliptic regularity or something. For instance, in any disk in the triangle ABC, one can write the value of the eigenfunction at the centre of the the disk in terms of an integral on the disk involving a Bessel function. So if one has H^1 control on the eigenfunction in that disk, one gets pointwise control at the centre of the disk. If one is near an edge of the triangle, but not near a corner, one can do something similar after first applying a reflection. Things get worse near the corners, but we were planning to use a different argument for that case anyway (Step 8).
Another minus is that I suspect the dependence of the error on the mesh size is poor (something like the square root of the mesh size, perhaps). And it requires quite precise control on the second eigenvalue. But it may still be the lesser of two evils.
Comment by — September 14, 2012 @ 5:18 am
• Let’s see. For a symmetric positive-definite $A$, the Lanzcos iteration for the smallest eigenvalue of $Ax = \mu x$ proceeds until a certain error criteria is met. This error criteria is based on the discrete Rayleigh quotient $v^TAv$ (\$v\$ is assumed to be normalized).
In our setting, the quantity $w_h^T A w_h$ is the quantity $(\nabla I_h w_h, \nabla I_h w_h) + (I_h w_h, I_h w_h)$, where I am using the notation $I_hw_h$ to be the element of $H^1$ constructed by using the coefficients $w_h$ in some basis.
In other words, the algorithm already minimizes the discrete Rayleigh quotient. The nature of the matrix $A$ tells us that the true discrete eigenvalue converges to the true one (for the continuous one) quadratically in the mesh size. The asymptotic constant depends on the true eigenvalue and the spectral gap. Eigenvectors converge quadratically with tesselation size in the H1 norm, eigenvalues converge quartically.
Backward error analysis tells us that the computed $w_h$ is close to the true $w_h$ for a given matrix size. The asymptotic constant again depends on the spectral gap. http://web.eecs.utk.edu/~dongarra/etemplates/node151.html
So I think we already have $H^1$ control, and I should attempt to explicitly string together the constants.
I suspect you may be right: since the asymptotic constants above for any FIXED triangle PQR depend on the spectral gap, and since this spectral gap varies as we change the angles of PQR, we may have poor control in terms of the angle parameters. But staring at the spectral gap plot
http://people.math.sfu.ca/~nigam/polymath-figures/SpectralGap-07.jpg
I am optimistic that away from the equilateral triangle one could actually make the dependence of asymptotic constants precise.
What’s harder to get a hand on is the residual of the computed (ie, not the finite element, but actually computed) eigenvector in terms of the direct application of the Laplacian.
Comment by — September 14, 2012 @ 6:45 pm
• Here are some brief computations illustrating how an L^2 bound on the error in an approximate eigenfunction can be turned into L^infty bounds, at least if one is away from corners, thus avoiding the need to have to understand the residual.
More precisely, suppose that we have a true eigenfunction $-\Delta u = \lambda u$ and some discretised approximation $\tilde u$ to this eigenfunction, and that we have somehow obtained a bound $\| u - \tilde u \|_{L^2(ABC)} \leq \varepsilon$ on the L^2 error (e.g. we should be able to get this if the Rayleigh quotient of $\tilde u$ is close to the minimum value of $\lambda$).
Now suppose that the triangle ABC contains the disk $B(0,R)$ for some radius R. A bit of messing around with polar coordinates shows that the function
$\overline{u}(r) := \frac{1}{2\pi} \int_0^{2\pi} u(r,\theta)\ d\theta$
is smooth on $[0,R]$, equal to $u(0)$ at the origin, and obeys the Bessel differential equation $\partial_{rr} \overline{u}(r) + \frac{1}{r} \partial_r \overline{u}(r) + \lambda \overline{u}(r) = 0$
which can be solved using Bessel functions as
$\overline{u}(r) = J_0(\sqrt{\lambda} r) u(0)$
thus
$\int_0^{2\pi} u(r,\theta)\ d\theta = 2\pi J_0(\sqrt{\lambda} r) u(0)$
for all $0 \leq r \leq R$. Integrating this against a test function $f(r)\ r dr$ we get
$\int_0^{2\pi} \int_0^{R} u(r,\theta) f(r)\ r dr d\theta = 2\pi u(0) \int_0^R J_0(\sqrt{\lambda} r) f(r)\ r dr.$
Splitting $u$ into $\tilde u$ and $u-\tilde u$ and using Cauchy-Schwarz to bound the latter term, we obtain
$|\int_0^{2\pi} \int_0^{R} \tilde u(r,\theta) f(r)\ r dr d\theta - 2\pi u(0) \int_0^R J_0(\sqrt{\lambda} r) f(r)\ d dr| \leq \varepsilon (2\pi \int_0^R |f(r)|^2\ r dr)^{1/2}$.
This gives a computable upper and lower bound for $u(0)$. From the Cauchy-Schwarz inequality, we see that the optimal value of $f$ here is $f(r) = J_0(\sqrt{\lambda} r)$, leading to the bound
$|u(0) - A^{-1} \int_0^{2\pi} \int_0^{R} \tilde u(r,\theta) J_0(\sqrt{\lambda} r)\ r dr d\theta| \leq A^{-1/2} \varepsilon$ (1)
where
$A := 2\pi \int_0^R |J_0(\sqrt{\lambda} r)|^2\ r dr.$ (2)
Of course, this estimate works best when R is large, and we can only fit a big ball B(0,R) inside the triangle ABC if the origin 0 is deep in the interior of ABC. But if instead 0 is near an edge of ABC, but far from the vertices, one can still get a big ball inside the domain after reflecting the triangle across the edge 0 is near (but this multiplies the L^2 norm of the residual by a factor of sqrt(2)). In principle, this gives L^infty control on u everywhere except near the corners. (Of course, one still has to numerically integrate the expressions in (1) and (2), but this looks doable, albeit messy.)
Comment by — September 20, 2012 @ 10:42 pm
• This is really useful!
Here’s how I was trying to proceed:
Suppose u is the exact eigenfunction, and $R_hu$ its exact projection onto the finite dimensional subspace $V_h$. I was trying to compute the precise asymptotic constant $c$ in the sup-norm estimate $\|u-R_h u\|_\infty \leq c h^{\mu}$ (this is the kind of FEM estimate one gets, eg. page 3 of http://people.math.sfu.ca/~nigam/polymath-figures/Validation+Numerics.pdf), but I think your argument above may help with this.
One then has to estimate $\|R_h u- \tilde{u}\|_\infty$, where both now live in $V_h$. I think this can be controlled using the inverse and backward error estimates for eigenvalue problems – the trick is getting the asymptotic constants decently. One then puts these estimates together
I’m also (still!) trying to debug the code to locate the extrema of the piecewise quadratics on the smaller triangles. This would circumvent the interpolation-to-linears step. The calculations are so very messy (not hard), so I’m going very slowly.
Comment by — September 21, 2012 @ 6:13 pm
2. So unfortunately, nothing we discussed panned out. But the following were some ideas considered:
1) Perhaps it is the case that the nodal line is the image of a line or circle (hyberbolic geodesic?) under the Schwarz–Christoffel mapping. Or if not the image then close to it. Bartlomiej made some computer simulations and in many cases the fit was very close but for the equilateral triangle the fit was a fair bit off.
2) Terry’s re-proof of a coupling-argument using the vector maximum principle begs the question: Is it possible to reprove the Hot-Spots conjecture for obtuse triangles (the paper of Banuelos and Burdzy) using such a non-probabilistic argument? If so, perhaps the proof could be extended to the non-obtuse case. Of course we can see by computer simulation that the gradient field of certain acute triangles is much nastier… but I think it is still the case that the gradient vectors all lie in a half space? In fact I think the hot-spots conjecture would hold if we could show that the gradient vectors all lie in a (non-convex) cone of angle less than 360 degrees. In summary: Maybe something can be said about the gradient field.
3) There suggestion of Mihai Pascu involving the nodal line dividing the triangle into star-like domains with respect to hyperbolic geodesics. But I didn’t quite follow/remember… maybe Barlomiej can better explain.
4) There was the suggestion of Baneulos that we might examine the function $\phi_2(x)/\phi_1(x)$ where $\phi_i(x)$ is the $i$-th Dirichlet eigenfunction of the triangle. Apparently this ratio has properties similar to the second Neumann eigenfunction.
That’s all I can remember… hopefully I didn’t miss any.
Edit: I remembered another interesting comment:
5) Chris Burdzy pointed out that any argument used has to be one that wouldn’t work on a manifold. He gave the following counter-example: Consider the surface of a long baguette. The hot spots would then be at the
tips of the baguette”. This is a manifold without boundary but then you can cut a small hole in the baguette to make the surface into a manifold with boundary. If the hole is small enough, the hot spots wouldn’t move much and would therefore not be on the boundary.
Comment by — September 16, 2012 @ 10:56 pm
3. The observation (1) is interesting! let’s ignore the equilateral triangle case for the time being: is the nodal line always a line/circle under the Schwarz-Christoffel map? Is there some simple way to see this?
Comment by — September 17, 2012 @ 10:02 pm
• Presumably one can do explicit calculations for the 45-45-90 triangle. If the vertices are (0,0), (1,0), (1,1) then the second eigenfunction is $\cos(\pi x) + \cos(\pi y)$ and the nodal line is the bisecting altitude connecting (1/2,1/2) with (1,0). I guess this transforms to a horocycle or something under Schwarz-Christoffel?
The other explicit example is the 30-60-90 triangle, but the nodal line there is a bit more complicated. If we use the triangle with vertices (0,0,0), (1,-1,0), (1,-1/2,-1/2), the second eigenfunction is $\cos(\pi(x-y)/3) + \cos(\pi(y-z)/ 3) + \cos(\pi(z-x)/3)$, and I can’t work out the nodal line for that in my head…
Comment by — September 17, 2012 @ 10:44 pm
• Bartlomiej’s numerical simulations suggested conjecture (1) was false for equilateral triangles but true (or perhaps just coincidentally close) for the 30-60-90 triangle. As the 30-60-90 triangle is half the equilateral triangle perhaps there is a nice analytic way to compare the images of lines/circles under the SC-map for the two regions.
More specifically, does it hold that the image-of-lines/circles for the equilateral triangle are the reflections of those for the 30-60-90 triangle?
As a separate question: Suppose we could show that the nodal line lies in some specific “epsilon band”. That begs the question of how to prove a claim such as the following:
Claim: Consider an “almost-arc” with two straight Neumann-sides meeting at an acute angle at the corner $A$ and one wiggly Dirichlet-side such that the distance from any point of the Dirichlet-side to the corner $A$ lies in the interval $(R-\epsilon,R + \epsilon)$. Then for suitably large $R$ and small $\epsilon$, the maximum of the first eigenfunction should occur at the corner $A$.
(Instead of taking an “epsilon band” around a curved arc we might also take one around a straight line and consider an “almost-triangle”)
Our proofs using coupled-BM/vector-valued maximum principle won’t work if the Dirichlet-side is wiggly and yet intuitively the claim still ought to hold. Of course if $\epsilon\to 0$ it should hold by continuity, but it ought to hold for $\epsilon$ not too small…
Comment by — September 18, 2012 @ 4:39 pm
4. I’ve been playing around a bit trying to estimate the $X$ in Terry’s notes, in particular the $\frac{d^2}{dt^2} A(t)$ term, where $A(t)$ denotes the area of the triangle. I scanned in some notes at
http://www.math.missouri.edu/~evanslc/Polymath/Oct26th.pdf
Looking at the graph of the function in Wolfram Alpha it seems that $f(x)=\Gamma(\frac{x}{\pi})^2\sin(x)$ and its first and second derivatives will blow up when $x$ is close to $0$… but that is old news I suppose and we can presumably handle the case where one of the angles is close to $0$ by other means.
Despite its nice structure I didn’t see an easy way to read off the eigenvalues/vectors of the Hessian of $F$… but such information would yield a bound. In fact, using that our perturbation is restricted to a plane might give us a better bound.
Comment by — October 26, 2012 @ 8:59 pm
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http://math.stackexchange.com/questions/297482/integration-relation-between-2-integrals
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Integration : relation between 2 integrals
Let me define $J=\int_{0}^{1} \frac{\ln(t)\ln^{2}(1-t)}{t}dt$ and the function $g$ defined on $[0,1]$ : $g(x) = \int_{0}^{x} \frac{\ln(1-t)}{t}dt$ where $\ln^{2}(1-t)$ means $\ln(1-t) \times \ln(1-t)$. In the exercice I was trying to do, we first prove that for all $x \in ]0,1]$, $g(x) = \sum_{n=1}^{+\infty} \frac{x^{n}}{n^{2}}$. Then we prove that $\sum_{1 \leq m < n} \frac{1}{m^{2}n^{2}}$ can be expressed using $\zeta(2)$ and $\zeta(4)$. I found that $\sum_{1 \leq m < n} \frac{1}{m^{2}n^{2}} = \frac{\zeta(2)^{2}-\zeta(4)}{2}$.
In the last question of the exercice, I have to compute $J$. (To do so, I assumed there was some relation between $J$ and $g$... but I couldn't find it). So, my question is : how can I use $g$ to compute $J$ ?
Thanks !
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3
What's $f$ in this problem? – Ron Gordon Feb 7 at 20:30
Sorry, I made a mistake. I want to find a relation between $J$ and $g$. There is no $f$. – user61409 Feb 7 at 20:48
@RustynYazdanpour Of course, thanks. I skipped this exponent... – julien Feb 7 at 22:56
What does $\ln^2 (1-t)$ suppose to be ? $(\ln (1-t))(\ln (1-t))$ or $\ln (\ln (1-t))$ ? – Arjang Feb 8 at 7:53
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http://mathoverflow.net/questions/114333?sort=newest
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## Maximal soluble subgroups in a parabolic subgroup of finite classical simple group
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Let $G$ be a classical simple group over a finite field $GF(q)$ and $P$ a parabolic subgroup of $G$ stabilizing an isotropic subspace. Is the Borel subgroup of $G$ maximal soluble in $P$ and is there any maximal soluble subgroup in $P$ besides the Borel subgroup? Here by maximal soluble I mean a maximal one among soluble subgroups of $P$.
Any feedback is appreciated!
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1
Solvable subgroups not containing all points of a Borel are out of control (so best to focus of $B$ being maximal solvable in $G$ for "large" $p$, without mentioning $P$). To see this, for any $p$ and any finite solvable $\Gamma$ you can put $\Gamma$ into ${\rm{GL}}_n(\mathbf{F}_p)$ for large $n$ (e.g. permutations) and view ${\rm{GL}}_n$ as a proper parabolic in ${\rm{SL}}_{n+1}$ in an evident way. If $\Gamma$ doesn't have normal $p$-Sylows or does yet quotients by them are not abelian then $\Gamma$ has "nothing" to do with Borel subgroups (e.g., not contained in one) when $G$ is split. – xuhan Nov 24 at 19:04
@Binzhou: Geoff has answered your basic question by pointing out that the rational points of a larger parabolic than `$B$` can itself be solvable. This occurs only for some `$P$`, of course. (The surrounding discussions add other interesting points, but of course the determination of all maximal solvable subgroups in finite groups of Lie type is a much more open-ended question.) – Jim Humphreys Nov 25 at 14:37
@Jim: Many thanks to you and Geoff! I'm new to algebraic groups, so maybe I should read some text book or papers for fundation about this subject. By the way, if I mean a parabolic subgroup by the full stabilizer of an isotropic subspace, should I use the term maximal parabolic subgroup? – Binzhou Xia Nov 26 at 3:11
@Binzhou: I'm not sure about the current terminology used in finite classical groups, but it needs to be consistent with the general language of algebraic groups to avoid confusion. (In either case, by the way, "parabolic" is a term which originates far away from the application to subgroups here.) – Jim Humphreys Nov 26 at 22:22
## 2 Answers
I think there are examples when the Borel subgroup is not maximal solvable in a parabolic. One can occur when $q \leq 3,$ so for example when $G = {\rm GL}(3,2)$ both the maximal parabolics are themselves solvable (isomorphic to $S_{4}$). Similarly when $p= 3$, the group ${\rm GL}(3,3)$ has a solvable maximal parabolic $P$ with unipotent radical $U$ such that $P/U \cong {\rm GL}(2,3).$ Such behavior perpetuates itself in higher ranks to give solvable parabolic subgroups strictly containing the Borel in characteristic $2$ or $3$. Later comments: Really, from an inductive point of view, it is probably best not to assume that $G$ is simple as an abstract group, and then one can argue inductively on the rank of the associated $BN$-pair.This seems to reduce us to the rank $1$ case and then it would appear that the only case to worry about is when the whole group is solvable (otherwise the Borel intersects the unique component in a maximal subgroup).
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@Geoff: Is this phenomenon limited to $q = 2, 3$ (and not more general $q$ with $p \in \{2, 3\}$, let alone larger $p$)? More specifically, consider a split connected semisimple $G$ over $k = \mathbf{F}_q$ so that $G$ is simply connected and $k$-simple, with either: $q > 3$, $q = 3$ and $G \ne {\rm{SL}}_2$, or $q = 3$ and $G \ne {\rm{SL}}_2, {\rm{Sp}}_4$. By BN-pair stuff, $G(k)$ has center $\mu(k)$ where $\mu$ is the center of $G$, and $G(k)/\mu(k)$ is simple as an abstract group. For a Borel $k$-subgroup $B$ of $G$, is $B(k)$ maximal as a solvable subgroup of $G(k)$? – xuhan Nov 24 at 19:25
I meant "$q = 2$ and $G \ne {\rm{SL}}_2, {\rm{Sp}}_4$" at the end of the 2nd sentence in the preceding comment. – xuhan Nov 24 at 19:27
Yes, it is probably limited to $q = 2,3.$ – Geoff Robinson Nov 24 at 20:34
I am talking here about maximal solvable subgroup containing a Borel subgroup – Geoff Robinson Nov 24 at 22:48
@xuhan: you can also have parabolics $P$ such that $P/U$ is isomorphic to a product of several copies of ${\rm GL}(2,3)$ when $p = 3$, for example. – Geoff Robinson Nov 25 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.
EDIT: This responds to Geoff's comment (and my carelessness), but also adds a couple of other remarks.
Geoff's answer and the comments might be clarified a little as follows. You are looking at rational points of a parabolic subgroup over a finite field (where it's not important whether the algebraic group is of classical type or not). In the split (Chevalley group) case, it's easy to check orders of the various finite groups involved. Since a Levi factor of a proper parabolic group is just the product of a central algebraic torus and a lower rank semisimple group, the rational points sometimes yield a solvable group even if the original parabolic is not itself a solvable algebraic group (a Borel subgroup). This also involves the (nilpotent) group of rational points of the unipotent radical, so it gets complicated. To assemble a complete list of possibilities, look at the simple algebraic groups having solvable groups of rational points over a field of small characteristic. The torus part is already commutative, so the product and therefore the Levi subgroup of a parabolic can be solvable.
Naturally it gets more complicated to treat all types of twisted groups as well as Suzuki or Ree groups, but in principle it looks reasonable to assemble a complete list of solvable finite parabolics. (If there is enough motivation to do so.)
Concerning the determination of all maximal solvable subgroups of a finite simple group of Lie type, that's asking for a lot. The BN-structure is a valuable organizing tool, but the groups have considerably richer subgroup structure than can be identified from Lie theory alone. Even in the study of linear algebraic groups, older work of V.P. Platonov shows that it is tricky to pin down all their not necessarily connected maximal closed solvable subgroups.
[Terminology: for me language like "the Borel subgroup" is a bit jarring, since there are many conjugate choices starting with the algebraic group.]
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@Jim: Do you really mean " a proper parabolic is just the product of a central algebraic torus and a lower rank semisimple group", or are you talking about the Levi factor after the unipotent radical is factored out? – Geoff Robinson Nov 24 at 22:36
@Geoff: I was multi-tasking at the time. See my revised version. By the way, thanks for adapting to the weird American spelling "solvable". For us, most powders are soluble while some problems are solvable. – Jim Humphreys Nov 24 at 23:09
@Jim: Yes, I have always preferred "solvable" myself, partly a rebellious reaction of a postgraduate student against the Oxford tendency to use "soluble"- though Burnside and P. Hall also used "soluble" and they were in Cambridge long ago. I remember once David Sibley gave some lectures in Oxford about the Character Theory of the Odd Order Paper, which he started with " A finite group is soluble if and only if it can be dissolved in water"! I envy the ability to multitask, and I note your (correct) comment about THE Borel subgroup. – Geoff Robinson Nov 25 at 1:12
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http://nrich.maths.org/2735
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### Rotating Triangle
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Pericut
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Tri-split
A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture?
# Triangles Within Squares
##### Stage: 4 Challenge Level:
The diagram above shows that: $$8 \times T_2 + 1 = 25 = 5^2$$
Use a similar method to help you verify that: $$8 \times T_3 + 1 = 49 = 7^2$$ Can you generalise this result?
Can you find a rule in terms of $T_n$ and a related square number?
Can you find a similar rule involving square numbers for $T_{n}, T_{n+2}$ and several copies of $T_{n+1}$?
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://math.stackexchange.com/questions/109032/singing-bird-problem/109048
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# Singing Bird Problem
For my algorithms class, the professor has this question as extra credit:
````The Phicitlius Bauber bird is believed to never sing the same song twice.
Its songs are always 10 seconds in length and consist of a series of
notes that are either high or low pitched and are either 1 or 2 seconds long
How many different songs can the Bauber bird sing?
````
This strikes me as a combinatorial problem, but I'm having trouble figuring out the exact formula to use to fit the included variables of 10 seconds length, 2 notes and 2 note periods.
At first, I thought that $\binom{10}{2}*\binom{10}{2}$ can be a solution, since it fits finding all combinations of both and multiplies the total.
My issue is this results in a total of 2025 possible songs, and this just seems a little on the low side.
Any assistance is appreciated.
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## 3 Answers
I'm going to work under the assumption that two successive 1-second notes of the same pitch is distinct from a single 2-second note of that pitch.
Let $f(n)$ be the number of distinct $n$-second songs the bird can sing. The bird has four options for starting the song: 1-second low, 1-second high, 2-second low, and 2-second high. This gives the recursion $$f(n) = 2f(n-1) + 2f(n-2),$$ with $f(0) = 1$ and $f(1) = 2$.
The associated characteristic equation is $x^2 - 2x - 2 = 0$, which has roots $x = 1 \pm \sqrt{3}$. Thus, we get the general solution $$f(n) = A(1 + \sqrt{3})^n + B(1 - \sqrt{3})^n.$$
Using the initial conditions, we have the system $$\begin{align*} 1 &= A + B\\ 2 &= A(1 + \sqrt{3}) + B(1 - \sqrt{3}), \end{align*}$$ which has the unique solution $A = \frac{3 + \sqrt{3}}{6}$ and $B = \frac{3 - \sqrt{3}}{6}$.
Finally, we have $$f(n) = \frac{3 + \sqrt{3}}{6}(1 + \sqrt{3})^n + \frac{3 - \sqrt{3}}{6}(1 - \sqrt{3})^n.$$
For the problem at hand, we want $$f(10) = 18,272.$$
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1
That's strange, f(2)=7. But as Mark Eichenlaub says, it should be 6. I think you made a mistake somewhere. – Lopsy Feb 13 '12 at 21:52
2
The roots should be $1\pm\sqrt3$, and $f(10)$ is considerably smaller; a hasty calculation made it $18272$. @Lopsy means that your formula for $f(n)$ gives $f(2)=7$. – Brian M. Scott Feb 13 '12 at 21:55
Right, but your final equation with the radicals gives 7. I think I figured out what happened: the roots of that quadratic are 1+sqrt(3) and 1-sqrt(3), not 2+sqrt(3) and 2-sqrt(3) as you claim. – Lopsy Feb 13 '12 at 21:56
Thanks for everyone's keen eyes. I believe I've caught all the errors now. – Austin Mohr Feb 13 '12 at 22:16
2
Now it’s good. I’d add just one comment: $10$ is a small enough number that in this case it’s easier just to use the recurrence than it is to find the general solution. – Brian M. Scott Feb 13 '12 at 22:18
Hint: use recursion.
You know that the first note in the Bauber bird's song is either $1$ or $2$ seconds long. In the first case, the remainder of the song is $9$ seconds long; in the second case, the remainder of the song is $8$ seconds long.
Therefore, the number of $10$-second songs equals twice (remember, the last note can be low or high) the number of $9$-second songs plus the number of $8$-second songs. Can you see how to solve the problem from here?
Also, note that it's generally a bad idea to encourage "looking for things to do with the numbers in the problem" like you said you did with your first try, $\binom{10}{2}*\binom{10}{2}$. Although formula-hunting does have its uses (esp. when doing dimensional analysis, where it's essential), if you're going to try to find a formula first, it's a good idea to check the answer for small cases (like $1$ second, $2$ seconds, etc) and then extrapolate off of that than trying to guess formulae blind.
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Here is a non-recursive solution.
Suppose there are $a$ 1s songs and $b$ 2s songs, so $a + 2b = 10$. The total number of songs is $10 - b$.
We can count the number of such songs by first ignoring pitch and counting the ways the 1s and 2s songs can be ordered, then multiplying by $2^{a+b}$.
There are $a+1$ gaps between the $a$ 1s songs, so we want to know how many ways $b$ 2s songs can fit into $a+1$ gaps. This is $\binom{a+b}{b} = \binom{10-b}{b}$, as shown by Akash Kumar in this answer.
So we want
$$\sum_{b=0}^{5} 2^{10 - b}\dbinom{10 - b}{b} = 18,272$$
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2
But you can’t break the ten-second interval into $2$-second chunks, because a $2$-second note can overlap chunks. – Brian M. Scott Feb 13 '12 at 21:59
@BrianM.Scott Yes, you're right, thanks. Tried a new answer. – Mark Eichenlaub Feb 13 '12 at 23:04
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http://math.stackexchange.com/questions/187056/spotting-maximum-minimum-maximal-and-minimal-elements-in-a-poset/187062
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# Spotting maximum, minimum, maximal and minimal elements in a poset
Let $(\mathbb N^* \times \mathbb N^*, \varphi)$ be a poset defined as follows:
$$\begin{aligned} (a,b)\varphi(c,d)\Leftrightarrow ab<cd \text{ or } (a,b) = (c,d)\end{aligned}$$
Check if $\varphi$ is a total order and determine maximum, minimum, maximal and minimal elements.
It's easy to prove that $\varphi$ is not total because as we consider $(a,b),(c,d) \in \mathbb N^* \times \mathbb N^* : a = d \text{ and } b = c$ then
$$\begin{aligned} ab \nless cd \text{, } cd \nless ab \text{ and } (a,b) \neq (c,d)\end{aligned}$$
What I am having a very hard time with is spotting maximum, minimum, maximal and minimal elements. In my opinion the poset doesn't have any maximum or maximal elements as $\mathbb N^* \times \mathbb N^*$ is infinite, but it does have a minimal element which is $(1,1)$. Given that $\nexists (\varepsilon, \varepsilon') : 1 = \varepsilon' \text{ and } 1 = \varepsilon$ and $(1,1) \neq (\varepsilon, \varepsilon')$ then $(1,1)$ is comparable to all elements in $\mathbb N^* \times \mathbb N^*$, does this make it also the minimum?
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Pardon my ignorance, but what is $\mathbb{N}^*$ again? – Daniel Pietrobon Aug 26 '12 at 11:12
@DanielPietrobon $\mathbb N^* = \mathbb N \backslash \{0\}$. – haunted85 Aug 26 '12 at 11:15
## 1 Answer
I think your reasoning is sound enough.
If there is a maximal element, $(c,d)$ say, then $(a,b) \varphi (c,d)$ for all other elements $(a,b)$ in $\mathbb{N}^* \times \mathbb{N}^*$ which implies that: \begin{equation} ab < cd \text{ or } (a,b)=(c,d) \end{equation} But then $(c,d) \varphi (c+1,d)$ since $cd < (c+1)d$ contradicting the maximality of $(c,d)$.
Finally, $(1,1)$ is the minimal element as $1 \cdot 1=1$ is the smallest a product of two elements in $\mathbb{N}^*$ can be. That is, any other element $(e,f)$ has $e>1$ or $f>1$ In the first case, $(1,1) \varphi (e,f)$ since $1\cdot 1 < e \leq ef$. The second case is similar.
Also I like your profile: "Just a guy who hopes to live through his algebra exam." +1 for having realistic goals, haha.
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1
Also, the existence of a maximum element implies the existence of a maximal element. Your proof shows there is no maximal element, so by the contraposition, there is no maximum element. – ladaghini Aug 26 '12 at 11:46
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http://en.wikipedia.org/wiki/Species_diversity
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# Species diversity
Species diversity is the effective number of different species that are represented in a collection of individuals (a dataset). The effective number of species refers to the number of equally-abundant species needed to obtain the same mean proportional species abundance as that observed in the dataset of interest (where all species may not be equally abundant). Species diversity consists of two components, species richness and species evenness. Species richness is a simple count of species, whereas species evenness quantifies how equal the abundances of the species are.[1][2][3]
## Calculation of diversity
Species diversity in a dataset can be calculated by first taking the weighted average of species proportional abundances in the dataset, and then taking the inverse of this. The equation is:[1][2][3]
${}^q\!D={1 \over \sqrt[q-1]{{\sum_{i=1}^S p_i p_i^{q-1}}}}$
The denominator equals mean proportional species abundance in the dataset as calculated with the weighted generalized mean with exponent q - 1. In the equation, S is the total number of species (species richness) in the dataset, and the proportional abundance of the ith species is $p_{i}$. The proportional abundances themselves are used as weights. The equation is often written in the equivalent form:
${}^q\!D=\left ( {\sum_{i=1}^S p_i^q} \right )^{1/(1-q)}$
The value of q defines which kind of mean is used. q = 0 corresponds to the harmonic mean, and gives values equal to species richness because the $p_{i}$ values cancel out. q = 1 corresponds to the geometric mean and q = 2 to the arithmetic mean. As q approaches infinity, the generalized mean approaches the maximum $p_{i}$ value. In practice, q modifies species weighting, such that increasing q increases the weight given to the most abundant species, and fewer equally-abundant species are hence needed to reach mean proportional abundance. Consequently, large values of q lead to smaller species diversity than small values of q for the same dataset. If all species are equally abundant in the dataset, changing the value of q has no effect, but species diversity at any value of q equals species richness.
Negative values of q are not used, because then the effective number of species (diversity) would exceed the actual number of species (richness). As q approaches negative infinity, the generalized mean approaches the minimum $p_{i}$ value. In many real datasets, the least abundant species is represented by a single individual, and then the effective number of species would equal the number of individuals in the dataset.[2][3]
The same equation can be used to calculate the diversity in relation to any classification, not only species. If the individuals are classified into genera or functional types, $p_{i}$ represents the proportional abundance of the ith genus or functional type, and qD equals genus diversity or functional type diversity, respectively.
## Diversity indices
Often researchers have used the values given by one or more diversity indices to quantify species diversity. Such indices include species richness, the Shannon index, the Simpson index and the complement of the Simpson index (also known as the Gini-Simpson index).[4][5][6]
When interpreted in ecological terms, each one of these indices corresponds to a different thing, and their values are therefore not directly comparable. Species richness quantifies the actual rather than effective number of species. The Shannon index equals log(qD), and in practice quantifies the uncertainty in the species identity of an individual that is taken at random from the dataset. The Simpson index equals 1/qD and quantifies the probability that two individuals taken at random from the dataset (with replacement of the first individual before taking the second) represent the same species. The Gini-Simpson index equals 1 - 1/qD and quantifies the probability that the two randomly taken individuals represent different species.[1][2][3][6][7]
## Sampling considerations
Depending on the purposes of quantifying species diversity, the dataset used for the calculations can be obtained in different ways. Although species diversity can be calculated for any dataset where individuals have been identified to species, meaningful ecological interpretations require that the dataset is appropriate for the questions at hand. In practice, the interest is usually in the species diversity of areas so large that not all individuals in them can be observed and identified to species, but a sample of the relevant individuals has to be obtained. Extrapolation from the sample to the underlying population of interest is not straightforward, because the species diversity of the available sample generally gives an underestimation of the species diversity in the entire population. Applying different sampling methods will lead to different sets of individuals being observed for the same area of interest, and the species diversity of each set may be different. When a new individual is added to a dataset, it may introduce a species that was not yet represented. How much this increases species diversity depends on the value of q: when q = 0, each new actual species causes species diversity to increase by one effective species, but when q is large, adding a rare species to a dataset has little effect on its species diversity.[8]
In general, sets with many individuals can be expected to have higher species diversity than sets with fewer individuals. When species diversity values are compared among sets, sampling efforts need to be standardised in an appropriate way for the comparisons to yield ecologically meaningful results. Resampling methods can be used to bring samples of different sizes to a common footing.[9] Species accumulation curves and the number of species only represented by one or a few individuals can be used to help in estimating how representative the available sample is of the population from which it was drawn.[10][11]
## Trends in species diversity
The observed species diversity is affected not only by the number of individuals but also by the heterogeneity of the sample. If individuals are drawn from different environmental conditions (or different habitats), the species diversity of the resulting set can be expected to be higher than if all individuals are drawn from a similar environment. Increasing the area sampled increases observed species diversity both because more individuals get included in the sample and because large areas are environmentally more heterogeneous than small areas.
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http://en.wikisource.org/wiki/Page:Newton's_Principia_(1846).djvu/450
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# Page:Newton's Principia (1846).djvu/450
From Wikisource
444 [Book III.
the mathematical principles
the sine of the mean inclination, to four times the radius; that is, seeing the mean inclination is about 5° 8½, as its sine 896 to 40000, the quadruple of the radius, or as 224 to 10000. But the whole variation corresponding to BD, the difference of the sines, is to this horary variation as the diameter BD to the arc Gg, that is, conjunctly as the diameter BD to the semi-circumference BGD, and as the time of 20797⁄10 hours, in which the node proceeds from the quadratures to the syzygies, to one hour, that is as 7 to 11, and 20797⁄10 to 1. Wherefore, compounding all these proportions, we shall have the whole variation BD to 33" 10'" 33iv. as 224 $\scriptstyle \times$ 7 $\scriptstyle \times$ 20797⁄10 to 110000, that is, as 29645 to 1000; and from thence that variation BD will come out 16' 23½".
And this is the greatest variation of the inclination, abstracting from the situation of the moon in its orbit; for if the nodes are in the syzygies, the inclination suffers no change from the various positions of the moon. But if the nodes are in the quadratures, the inclination is less when the moon is in the syzygies than when it is in the quadratures by a difference of 2' 43", as we shewed in Cor. 4 of the preceding Prop.; and the whole mean variation BD, diminished by 1' 21½", the half of this excess, becomes 15' 2", when the moon is in the quadratures; and increased by the same, becomes 17' 45" when the moon is in the syzygies. If, therefore, the moon be in the syzygies, the whole variation in the passage of the nodes from the quadratures to the syzygies will be 17' 45"; and, therefore, if the inclination be 5° 17' 20", when the nodes are in the syzygies, it will be 4° 59' 35" when the nodes are in the quadratures and the moon in the syzygies. The truth of all which is confirmed by observations.
Now if the inclination of the orbit should be required when the moon is in the syzygies, and the nodes any where between them and the quadratures, let AB be to AD as the sine of 4° 59' 35" to the sine of 5° 17' 20", and take the angle AEG equal to double the distance of the nodes from the quadratures; and AH will be the sine of the inclination desired. To this inclination of the orbit the inclination of the same is equal, when the moon is 90° distant from the nodes. In other situations of the moon, this menstrual inequality, to which the variation of the inclination is obnoxious in the calculus of the moon's latitude, is balanced, and in a manner took off, by the menstrual inequality of the motion of the nodes (as we said before), and therefore may be neglected in the computation of the said latitude.
SCHOLIUM.
By these computations of the lunar motions I was willing to shew that by the theory of gravity the motions of the moon could be calculated from their physical causes. By the same theory I moreover found that the annual equation of the mean motion of the moon arises from the various
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http://mathhelpforum.com/differential-geometry/207832-uniform-continuity.html
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# Thread:
|x|
2. ## Re: Uniform Continuity
Originally Posted by lovesmath
Prove that the function is uniformly continuous on the set of real numbers.
f(x)=|x|
I think I have to use the reverse triangle inequality, but I am not sure how to do it.
There you have it.
$\left| {\left| x \right| - \left| {x_0 } \right|} \right| \leqslant \left| {x - x_0 } \right| < \delta = \varepsilon$
3. ## Re: Uniform Continuity
Here is an additional question.
Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set.
4. ## Re: Uniform Continuity
Originally Posted by lovesmath
Here is an additional question.
Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set.
What have you tried so far?
-Dan
5. ## Re: Uniform Continuity
I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.
Not sure what to do next.
6. ## Re: Uniform Continuity
Originally Posted by lovesmath
I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.
The key to this is the fact that $D$ is bounded.
That means that $\exists M>0$ such that $D\subset [-M,M]$.
In your text material you should have a proof that a continuous function is bounded on a closed interval of real numbers. This new proof follows that proof very closely.
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http://physics.stackexchange.com/questions/6530/rigor-in-quantum-field-theory
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# Rigor in quantum field theory
Quantum field theory is a broad subject and has the reputation of using methods which are mathematically desiring. For example working with and subtracting infinities or the use of path integrals, which in general have no mathematical meaning (at least not yet) ect. My question is a little vague, but i am interested in hearing what is the status of rigor in QFT. What is known to be mathematically rigorous and consistent, what is known to be not rigorous? Any examples and references are welcome.
Added: Just to clarify by rigorous I meant anything that a mathematician would find satisfactory. Also my question wasn't for books with rigorous (in some sense) approach, although that was welcomed. It was about specific examples of what is considered mathematically satisfactory and what not. For example the quantization of free fields satisfying the Klein-Gordon equation can be done rigorously. There is no mathematical definition in general of the Feynman path integral and so on.
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Given that path integrals are a system for manipulating symbols that follow well defined rules, it would seem that "mathematical meaning" means "thought up by a mathematician"? Or am I just being catty? – dmckee♦ Mar 8 '11 at 21:11
3
Strangely enough, rigor is one of the least well-defined and most vague words one can use. I predict you'd find a wide range of opinions based on personal tastes, training, and other psychological factors... – user566 Mar 8 '11 at 22:10
9
@Moshe: one definition would be that you can (in principle) let a computer program prove your theorem. Which can indeed be done for rigorous mathematical statements and it is believed that most of the proofs can be converted into that form. On the contrary, what physicists do is often just hand-waving, far from mathematical standards. In my opinion, rigor is pretty well-defined and also quite ignored in physics :) – Marek Mar 8 '11 at 22:18
5
@Moshe: what does incompleteness have to do with any of this? I did not say that everything can be proved. I just said that everything that mathematicians care about and believe that has been proved in some papers can indeed be proved from the lowest level (axioms + logical inference). Anyhow, no mathematician would agree with you. Rigor means that at least in principle the statement is absolutely 100% correct (and can be proved so). If there is even a slightest chance of loophole then it's not rigorous by any means. In physics it's just these 99% likelyhood statements most of the time. – Marek Mar 8 '11 at 23:00
3
– user566 Mar 9 '11 at 1:47
show 8 more comments
## 9 Answers
working with and subtracting infinities ... which in general have no mathematical meaning
is not really correct, and seems to have a common misunderstanding in it. The technical difficulties from QFT do not come from infinities. In fact, ideas basically equivalent to renormalization and regularization have been used since the beginning of math--see, e.g., many papers by Cauchy, Euler, Riemann, etc. In fact, G.H. Hardy has a book published on the topic of divergent series:
http://www.amazon.com/Divergent-AMS-Chelsea-Publishing-Hardy/dp/0821826492
There is even a whole branch of math called "integration theory" (of which things like Lebesgue integration is a subset) that generalizes these types of issues. So having infinities show up is not an issue at all, in a sense, they show up out of convenience.
So the idea that infinities have anything to do with making QFT axiomatic is not correct.
The real issue, from a more formal point of view, is that you "want" to construct QFTs via some kind of path integral. But the path integral, formally (i.e., to mathematicians) is an integral (in the general sense that appears in topics like "integration theory") over a pretty pathological looking infinite dimensional LCSC function space.
Trying to define a reasonable measure on an infinite dimensional function space is problematic (and the general properties of these spaces doesn't seem to be particularly well understood). You run into problems like having all reasonable sets being "too small" to have a measure, worrying about measures of pathological sets, and worrying about what properties your measure should have, worrying if the "$\mathcal{D}\phi$" term is even a measure at all, etc...
At best, trying to fix this problem, you'd run into an issue like you have in the Lebesgue integral's definition, where it defines the integral and you construct some mathematically interesting properties, but most of its utility is in letting you abuse the Riemann integral in the way you wanted to. Actually calculating integrals from the definition of the Lebesgue integral is not generally easy. This isn't really enough to attract the attention of too many physicists, since we already have a definition that works, and knowing all of its formal properties would be nice, and would certainly tell us some surprising things, but it's not clear that it would be all that useful generally.
From an algebraic point of view, I believe you run into trouble with trying to define divergent products of operators that depend on renormalization scheme, so you need to have some family of $C^*$-algebras that respects renormalization group flow in the right way, but it doesn't seem like people have tried to do this in a reasonable way.
From a physics point of view, we don't care about any of this, because we can talk about renormalization, and demand that our answers have "physically reasonable" properties. You can do this mathematically, too, but the mathematicians are not interested in getting a reasonable answer; what they want is a set of "reasonable axioms" that the reasonable answers follow from, so they're doomed to run into technical difficulties like I mentioned above.
Formally, though, one can define non-interacting QFTs, and quantum mechanical path integrals. It's probably the case that formally defining a QFT is within the reach of what we could do if we really wanted, but it's just not a compelling topic to the people who understand how renormalization fixes the solutions to physically reasonable ones (physicists), and the formal aspects aren't well-understood enough that it's something one could get the formalism for "for free."
So my impression is that neither physicists or mathematicians generally care enough to work together to solve this problem, and it won't be solved until it can be done "for free" as a consequence of understanding other stuff.
Edit:
I should also add briefly that CFTs and SCFTs are mathematically much more carefully defined, and so a reasonable alternative to the classic ideas I mentioned above might be to start with a SCFT, and define a general field theory as some kind of "small" modification of it, done in such a way to keep just the right things well-defined.
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I've got Hardy's book and quote it against what you've said. (I just don't have it with me). Hardy was a good mathematician and knew that how you choose to "regularize" a divergent series drastically effects the resulting sum. The reason QFT gets away with it is that there's an underlying assumption that the functions involved are complex and analytic. – Carl Brannen Mar 9 '11 at 1:16
Yes, that's part of what I meant by saying we want to fix our answers against "physically reasonable" solutions. Although complex analytic is actually too strong an analyticity property in general for us, and you do need some extra technical assumptions to make sure things are "physically reasonable." But worrying about properties in terms of analyticity is problematic from the infinite dimensional POV (think about the topological and measure theoretic properties of analytic subsets of these infinite dimensional LCSC spaces). – Mr X Mar 9 '11 at 1:21
Also, the space of paths you integrate over are the Brownian motion like ones, which aren't differentiable anywhere. But you still run into problems because other spaces than the obvious one are pathological ;). I believe you can approach ODEs and PDEs from this point of view (I don't know if much has been done with this because it's a pretty perverse thing to do), but thinking about them brings up a whole host of problems that are only worse in this case from an analytic POV. – Mr X Mar 9 '11 at 1:24
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very good summary; although i have to say that i find extremely sad and discouraging when i hear bright physicists say stuff like "From a physics point of view, we don't care about any of this, because we can talk about renormalization, and demand that our answers have physically reasonable properties...but the mathematicians are not interested in getting a reasonable answer". This might be right from a numerical (maybe numerological?) perspective, but its the completely wrong mindset to begin with. Mathematical consistency (or a clear pathway to it) is never a luxury. Avoiding it is – lurscher Jul 14 '11 at 15:17
Although, speaking as a mathematician, I feel I must correct you: the proper phrase is measure theory not "integration theory". – Alex Nelson Jun 28 '12 at 17:39
First: There is no rigorous construction of the standard model, rigorous in the sense of mathematics (and no, there is not much ambivalence about the meaning of rigor in mathematics).
That's a lot of references that Daniel cited, I'll try to classify them a little bit :-)
Axiomatic (synonymous: local or algebraic) QFT tries to formulate axioms for the Heisenberg viewpoint (states are static, observables are dynamic). There are three sets of axioms known:
Roughly, the Wightman axioms describe how fields relate to observables, the Osterwalder-Schrader axioms are the Wightman axioms for Euclidean field theory, and the Haag-Kastler axioms dodge fields entirely and describe the observables per se. All three sets of axioms are roughly equivalent, meaning that the equivalence has been proven, sometimes with additional assumptions that physicists deem to be irrelevant.
"PCT, Spin and Statistics, and All That" was the first introduction to the Wightman axioms.
"Local Quantum Physics: Fields, Particles, Algebras" is an introduction to the Haag-Kastler axioms, as is "Mathematical Theory of Quantum Fields".
"Perturbative Quantum Electrodynamics and Axiomatic Field Theory" is a description of QED from the viewpoint of the Haag-Kastler axioms.
"Introduction to Algebraic and Constructive Quantum Field Theory" is about the quantization of given classical equations in the spirit of Haag-Kastler.
"Quantum Physics: A Functional Integral Point of View" uses the Osterwalder-Schrader axioms.
2D conformal field theory can be axiomatized using the Osterwalder-Schrader axioms, for example.
Functorial quantum field theory axiomatizes the Schrödinger viewpoint, see e.g. hnLab on FQFT.
This includes for example topological quantum field theories, these describe essentially theories with finite degrees of freedom. This branch has had a lot of impact in mathematics, especially with regard to differential geometry, and here to the theory of 3D and 4D smooth manifolds. I'd put
Daniel S. Freed (Author), Karen K. Uhlenbeck: "Geometry and Quantum Field Theory"
in this category.
"Geometry and Quantum Field Theory"
Quantization of classical field theories: Note that the axiomatic approaches don't depend on classical field theories that need to be quantized, they open the doors for a direct construction of quantum systems without classical mirror. The Lagrangian approach to QFT is an example of an ansatz that starts with a classical field theory that needs to be quantized, for which different means can be used.
Ticciati: "Quantum Field Theory for Mathematicians" is actually a quite canonical introduction to Lagrangian QFT, without much ado.
There is a lot of material about the geometry of classical field theories and variants to quantize them, like "geometric quantization".
The book Welington de Melo, Edson de Faria: "Mathematical Aspects of Quantum Field Theory" is an example of this.
Much more advanced is "Quantum Fields and Strings: A Course for Mathematicians (2 vols)"
For the path integral there are two points of view:
• The path integral - along with the Feynman rules - is a book keeping device for a game called renormalization, that lets you calculate numbers according to arcane rules,
• the path integral is a mathematical construct like a "measure" - but not a measure in the sense of measure theory known today - that needs to be discovered and defined appropriately.
AFAIK there has not been much progress with the second viewpoint, but there are people working on it, for example the authors of the book "Mathematical Theory of Feynman Path Integrals: An Introduction". You can find a lot more material about the mathematical theory of path integrals on the nLab here.
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I thought the Osterwalder-Schrader axioms were describing the Euclidean path integral approach...not the Heisenberg picture. Also, there are some ambiguities with quantizing a classical field (even in quantum mechanics, there is ambiguities in the quantization procedure; see, e.g., the Groenewald-van Hove "no-go" theorem). – Alex Nelson Jun 28 '12 at 17:42
Here is my answer from a condensed matter physics point of view:
Quantum field theory is a theory that describes the critical point and the neighbor of the critical point of a lattice model. (Lattice models do have a rigorous definition).
So to rigorously define/classify quantum field theories is to classify all the possible critical points of lattice models, which is a very important and very hard project.
(One may replace "lattice model" in the above by "non-perturbatively regulated model")
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Thanks, can you point out a general exposition/overview article about lattice models and QFT. Or any sourse that can give me an idea. – MBN May 31 '12 at 9:09
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– mbq♦ Jun 1 '12 at 15:08
There are several books that approach QFT (and/or Gauge Theory) from different levels of 'mathematical rigor' (for some definition of "mathematical rigor" — that Moshe would approve ;-).
So, let me give you some sort of 'preliminary list'… it's by no means complete, and it's in no particular order either, but i think it may pave the way to further work.
In any case… there's much more out there, not only in terms of topics (renormalization, etc) but also in terms of articles, books and so on.
So, there's plenty of "mathematical rigor" in QFT (and String Theory, for that matter), including different 'levels' of it, that should please and suit varied tastes.
PS: There are other topics here that deal with this topic in a form or another, e.g., Haag's Theorem and Practical QFT Computations. So, don't be shy and take a look around. :-)
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There is not "plenty" of mathematical rigor, as the rigorous work is utter crap, barely capable of repeating the stuff that was current in the 1950s. One central problem is that mathematicians are stupid with regard to defining measures, so that the field of "measure theory" is wrong. They need to reaxiomatize the field to make every set of reals measurable before they can do path integrals, and they won't do that, so tough luck. – Ron Maimon Jun 28 '12 at 19:14
I'll give a reference that I didn't (yet) managed to finish myself. But it looks really rigorous:
N. N. Bogoliubov, A. A. Logunov, A. I. Oksak, I. T. Todorov (1990): General Principles of Quantum Field Theory. (ISBN 0-7923-0540-X. ISBN 978-0-7923-0540-8.)
It's what I call "Bogoliubov approach" to QFT.
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I'll take a look. I've heard about it, and it is the one approach I have never looked at. Not that I have read much about the others, I have a very superficial knowledge of the subject. – MBN Mar 8 '11 at 21:18
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This is not rigorous at all. I read it. Nothing by physicists is rigorous except for Glimm and Jaffe. – Ron Maimon Jun 28 '12 at 19:15
I think everything is sufficiently rigorous when you do it according to the math rules.
Cheating starts when they say: "The integral of the delta-function squared, although looks as infinity, must be determined from the experimental data". It is just funny.
Once I encountered a similar infinity in a simpler but exactly solvable problem. First, I wanted to do renormalizations (determining the integral value from experimental data) but fortunately I managed to choose a better initial approximation and decrease the perturbative corrections. So the problem is in the initial approximation. If it is good, then the perturbative corrections are small. Otherwise they are large.
I also found an explanation why subtractions (discarding corrections) work sometimes. From my current point of view, the QFT needs reformulating since it is badly constructed. Reformulated QFT does not need repairing its solutions on the go.
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I would like to point out that there are several different problems coming from different points of view on the subject. It would be very complicated to comment on all of them, so let me restrict to a particular one.
As a first remark, I have to state that nobody that works in mathematics may have a doubt about what "rigorous" means. I will not comment on this since it seems that it was already explained in a clear manner.
Concerning your question, I would like to state that QFT is not a "unique" theory, but a bunch of several different ones which are more less related to each other due to some intrinsic descriptions. For instance, the "behavior" and construction of the (real or complex) scalar field theory and of the gauge theory is rather different. This is a kind of natural consequence of the fact that Classical Field Theory (ClFT) (which is completely rigorous up to some extent, even though it still contains several nontrivial problems) is also a collection of several different theories, which share a general geometrical description, but which have their own particular difficulties: as a particular setting of ClFT we may obtain classical mechanics, electromagnetism or even nonabelian gauge theory, etc. Let me also add that the general philosophy underlying ClFT appears, in some sense, as the only manner to construct relativistic extensions of the free situation, as a major difference with classical mechanics, in which you may add any constraint to a free particle without breaking any fundamental principle of the theory. I'm only rephrasing what P. Deligne and D. Freed state in the first volume of "QFT and Strings for Mathematicians", which was already mentioned.
Concerning now the problem of the quantization of each of the particular settings you may consider in ClFT, there are several problems to deal with. Let me consider two different aspects of the problem: perturbative and nonperturbative QFT. We may say that the former is (morally) a shadow of the latter. Moreover, the perturbative QFT (pQFT) can be developed in a mathematically rigorous manner in lots of situations. You may see the article by R. Borcherds in the arXiv "Renormalization and quantum field theory" (even though some of the ideas were already present in other texts in the literature, and, in my opinion, they are lurking behind some of the constructions and proofs by the author, see for instance the articles by O. Steinmann, which were also considered by R. Brunetti, K. Fredenhagen, etc). In this situation he defines in a rigorous manner an object which behaves like the Feynmann measure ("via the Riesz theorem"), and he gives a very complete account of how the pQFT should be described in several situations. The problem stays however in giving a correct formulation of nonperturbative QFT. This is a major problem, and only a few rigorous constructions up to dimension 2 (also dimension 3, but really few as far as I know. It would nice to hear the experts in this point) were performed. You may see the book by J. Glimm and A. Jaffe "Quantum physics – a functional integral point of view". In fact the major problem comes when trying to quantize the gauge theory, as a sub collection of situations of QFT. The lack of such a general picture means in fact that we actually do not know what a Quantum Field Gauge Theory really looks like (or just is, if you want). In particular (I state this because some people argue that the following is a consequence of having only a perturbative description), two major claims of physicists about the standard model (which are in some sense related), the mass gap and the quark confinement, are not proved (the former in fact constitutes one of the Millennium Prize Problems). Needless is to say that none of the physical heuristic arguments are clearly sufficient.
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Mathematicians are very silly when it comes to "rigorous" regarding measure theory and this is why they are stuck. The problem starts when you have to axiomatize measure theory to define random picks. There should be no hard work involved in defining a constructive measure (a picking you can do on a computer, or a limit thereof), but there is. – Ron Maimon Jun 28 '12 at 19:17
The use of disqualifying adjectives is completely unnecessary and misleading, since the people involved here is in some sense irrelevant, meaning that what is important is the subject of discussion (i.e. the rigor in QFT). On the other hand, the axioms of measure theory are completely clear and well-known even to a 2nd/3rd year undergraduate student of mathematics. This is not the problem we are taking about. What is being mentioned here is the (apparent) incapability of (some of) the present mathematical tools to provide a complete and correct formulation of nonperturbative QFT in general. – Estanislao Jun 29 '12 at 14:04
The language is necessary to shame people to motivate change. The "measure theory axioms" are not the problem, the problem is that measure theory involved needs axioms at all! You need a sigma algebra on the space and there is no simple sigma algebra on the unknown space of field distributions a-priori. This means that people define the measure in a stupid roundabout way, while there is a simple logic result (Solovay's theorem) that guarantees that this is no problem at all. Other problems remain, but the issue becomes one of probability analysis, the measure theory is trivial. – Ron Maimon Jun 30 '12 at 2:15
What I mean by that is the following: "A free quantum field theory: consider picking every fourier transform value f(k) of a random function to be a Gaussian with a (specific) variance $\sigma(k)$. This is the (imaginary time) quantum field." Did I just define free quantum fields? Not for mathematicians, because a random picking algorithm, no matter how convergent, does not define a measure. You need a sigma algebra to define a measure. You can't say "the measure of a set is the probability that this random function lands in the set" because this only makes sense in a Solovay universe. – Ron Maimon Jun 30 '12 at 2:18
I'm not sure I understand what you try to say, because in my limited experience the Solovay thm (further extended by Krivine, Shelah, etc) is just a manner of stating that the construction of nonmeasurable Lebesgue sets depends on the axiom of choice. All these results are far from being simple in my opinion. In any case, this discussion seems to me to be somehow misleading because which measure-like objects are needed is not completely hidden: measures are in some sense too restrictive, and prodistributions seem much more well-adapted objects, as studied by P. Cartier and C. De Witt-Morette. – Estanislao Jul 1 '12 at 17:37
show 10 more comments
QFT's reputation for using methods which are mathematically unsound isn't really deserved these days. Certainly, not everything is under perfect analytic control, but the situation isn't really much worse than it is in fluid dynamics.
In particular, the 'subtraction of infinities' thing isn't really considered to be an issue anymore. The mathematicians who've looked at it recently (like Borcherds & Costello) have basically come to the conclusion that Wilsonian effective field theory resolves these difficulty. You can make all computations solely in terms of long-distance 'effective' quantities, which are the things left behind when physicists subtract infinities. Short distance infinities therefore don't present a problem for defining correlation functions; there's nothing inconsistent about the basic path integral formalism.
This is really the same conclusion the constructive field theorists came to, studying lower dimensional examples in the 70s & 80s.
The challenge in rigorous QFT is dealing with infrared divergences. If your spacetime has infinite volume, then your field system can have degrees of freedom of arbitrarily large size. Coupling to these degrees of freedom can give you infinities. There are real mathematical problems here, but they're more like describing the solutions of an equation than describing the equation itself. (Really non-trivial things can happen. In QCD, for example, there is confinement: many of the observables you'd naively expect to be integrable with respect to the path integral measure -- like the observable representing a free quark or a free gluon -- aren't. Instead, the integrable observables are complicated mixtures of quarks and gluon, like protons, neutrons, and glueballs.) Most of the heavy lifting in Glimm & Jaffe, for example, comes not from constructing the 2d $\phi^4$ path integral measure, but from proving that its $n$-point correlation functions actually exist.
Naturally, this means that most computations of observable expectation values -- like in lattice gauge theory -- are not under tight analytic control. Convergence in simulation is mostly a matter of good judgement, for now.
Saying anything rigorously about this stuff almost certainly will require mathematicians to get a better grip on renormalization in non-perturbative settings (i.e., on the lattice). There are a good number of mathematicians actively working on this stuff. Geometers and topologists are getting more sophisticated about topological field theory, while the analysts have taken up statistical field theory.
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Absolute lack of rigor in this aspect :
Integration over an infinite volume or infinite time ?
It could be correct math but it is physically incorrect.
Einstein, relativity, light cone... should bring some light to this issue.
For instance, the computation of the lagrangian of a single particle, immobile, is something along this: Kinetic energy = 0, Potential Energy = time growing value because the field can not be setted in ALL the space at once (since the begining/borning of the particle the field/energy, able to produce work, is spreading away at 'c' speed).
We make a blind approximation when we integrate over infinite because in this way the lagrangian appears constant, not being so, and this is wrong because one should always respect the limits of integration.
The make the total energy a constant value we have to put the internal energy of the particle into the lagrangian and this energy must decrease thru time as much as the potential energy grows thru time. By this reasoning one must conclude that the matter contents of the particles are always decreasing. We are not able to measure this effect in the lab because... you guess, the lab is shrinking also.
The belief that the space is expanding (without probable cause) is widespread for ages. The alternative, matter shrinking, offers a probable cause and an explanation for the measured 'space expansion', and the 'Dark Energy' becames 'nothing' at all, idem to the beloved 'cosmological constant', inflation period,...
The cosmological redshift is caused by bigger atoms of the past and are not rushing away.
May be Relativity is closer to QM than we suspected.
A self-similar model of the Universe unveils the nature of dark energy formally explains this shift of paradigm (with my modest contribution).
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http://physics.stackexchange.com/questions/9252/has-quantum-entanglement-been-demonstrated-to-be-able-to-take-place-over-infinit/9253
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Has quantum entanglement been demonstrated to be able to take place over infinite distances?
In my poor understanding of quantum physics, quantum entanglement means that certain properties of one of two 'entangled' quantum particles can lead to change over infinitely large distances when the other particles' properties are changed.
Disregarding this already mind-boggling event taking place over say 10 meters distance; how have physicists been able to demonstrate, beyond reasonable doubt, that this can take place over infinitely large distances?
For instance: have they done some of these tests between ISS and Earth perhaps?
How can they be so sure?
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Just as soon as we put two detector infinitely far apart. – dmckee♦ Apr 28 '11 at 21:30
Since the ISS is only around 350 km above Earth, entanglement has been shown at a reasonable fraction of this distance. Of course, this is nowhere near infinity. – Peter Shor Apr 28 '11 at 21:31
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Infinitely large distances should be replaced by arbitrarily large finite distances. – Qmechanic♦ Apr 28 '11 at 21:37
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The answers below are good, but noise in the apparatus ensures that it gets increasingly difficult to construct an experiment that verifies the violation of Bell inequalities as the distance increases. The ESA experiment is not desktop technology. Beating the effects of noise makes longer range experiments progressively more expensive and requires progressively more ingenuity. We have to introduce not only more effective shielding from noise but also more sophisticated ways to allow for the effects of the noise that can't be shielded. – Peter Morgan Apr 28 '11 at 22:17
2 Answers
First of all, it is very important to note that quantum entanglement is not a spooky action-at-a-distance as Einstein once called it! It is a strong correlation of measurements that is stronger than any classical correlation could ever be. This has been experimentally verified by the so-called violation of Bell's inequalities.
Second, quantum entanglement has been successfully demonstrated for distance of some 100 km, as the Wikipedia article on quantum teleportation states (references in Journales such as Nature are given there)
Zeilinger’s experiments on the distribution of entanglement over large distances began with both free-space and fiber-based quantum communication and teleportation between laboratories located on the different sides of the river Danube. This was then extended to larger distances across the city of Vienna and most recently over 144 km between two Canary Islands, resulting in a successful demonstration that quantum communication with satellites is feasible. His dream was to bounce entangled light off of satellites in orbit.[1] This was achieved during an experiment at the Italian Matera Laser Ranging Observatory.[4]
Admittedly, that is not a demonstration at infinite distance, but then, nothing has ever been demonstrated at that length scale and we would never be able to do and to demand such a demonstration is unreasonable. But please not the different scales involved here! Usually, we think of quantum mechanics to govern the microscopic world, involving length scales of under a micrometer, i.e. $10^{-6} m$. Achieving quantum effects on a length scale of $100 km = 10^5 m$ means you span 11 orders of magnitude. You can jokingly call this "infinite for all practical purposes".
The misunderstanding is that nothing changes in the moment of your measurement. There is a classical analogue that is wrong on the details but emphasizes the main idea: Suppose you take two playing cards, one is a king and the other a queen. You shuffle them and let your friend pick one at random while you keep the other one. Now, you have the king with probability $1/2$ and the queen also with probability 1/2, but you don't know which one you have without looking. You also don't know which one your friend has. But as soon as you look at your card, you immediately know which one your friend has: If you got the king, he must have got the queen. There is no spooky action at a distance that somehow changes what the card of your friend looks like!
In quantum mechanics, it is admittedly a bit more complex, because not only would you not know what card you have without looking, it isn't even determined until you look. It is, however, possible to go through the math and the mechanisms and come up with a satisfactory explanation that does not involve faster-than-light communication and processes.
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Good answer! One additional point. As you say, you can never test at infinite distances, so the best you can do is test at very large distances. When someone says "large", you should always ask "large relative to what"? In the case of these experiments, all of the natural length scales associated with the particles (wavelengths, etc.) are microscopic, so doing an experiment over 100 km is a truly enormous distance -- probably billions of times bigger than any other relevant length scale. So while this isn't infinite distance, it's very nearly as good! – Ted Bunn Apr 28 '11 at 21:17
TY. Added that. – Lagerbaer Apr 28 '11 at 21:49
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My intro mechanics professor used to say "for our purposes, infinity is 5, maybe 10" ;-) – David Zaslavsky♦ Apr 29 '11 at 2:49
The fundamental point is that QM is not based on probability, but on probability waves (or fields) which, differently from normal statistics, exhibit interference. – Sklivvz♦ Apr 30 '11 at 16:44
@Sklivvz There's a lecture and article by Scott Aaronson where he argues that all the axioms of QM are exactly what you expect when you define a notion of probability that uses the 2-norm instead of the 1-norm – Lagerbaer Sep 20 '11 at 16:46
Since my Alma Mater was involved I can point to this:
A team of European scientists has proved within an ESA study that the weird quantum effect called 'entanglement' remains intact over a distance of 144 kilometres.
(Source)
In September 2005, the European team aimed ESA's one-metre telescope on the Canary Island of Tenerife toward the Roque de los Muchachos Observatory on the neighbouring island of La Palma, 144 kilometres away. On La Palma, a specially built quantum optical terminal generated entangled photon pairs, using the SPDC process, and then sent one photon towards Tenerife, whilst keeping the other for comparison.
There is an effort for Quantum Entanglement in Space Experiments.
Experimentally is has not been proven to work over an infinite distance, since that would be impossible.
To better understand Quantum Entanglement (and why distance is not a problem) I suggest starting with the EPR Paradox.
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http://mathhelpforum.com/pre-calculus/132297-find-largest-real-solution.html
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# Thread:
1. ## Find the largest real solution
The largest real solution for x^2/3 - 5x^1/3 + 4 = 0 would be x =
For this one I only got two solutions, 64 and 1. I thought 64 was the correct answer but apparently I got something wrong. The real answer is 8, does anyone know what I did wrong here?
2. Originally Posted by nandokommando
The largest real solution for x^2/3 - 5x^1/3 + 4 = 0 would be x =
For this one I only got two solutions, 64 and 1. I thought 64 was the correct answer but apparently I got something wrong. The real answer is 8, does anyone know what I did wrong here?
Let $u=x^{1/3}$, then we have
$u^2-5u+4=0$
$(u-4)(u-1)=0\Rightarrow{u}=1,4\Rightarrow{x}=1,64$.
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http://stats.stackexchange.com/questions/29697/testing-survival-against-frequency-of-some-event
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# Testing survival against frequency of some event
I have the following data for 200 cases/subjects:
• time to death over a 15 year period, $t$. A datum with a value of 180 months means that the subject did not die over the 15 year period.
• the frequency of three particular types of event associated with each subject $a_{i}$, $b_{i}$ and $c_{i}$,
Apart from the obvious survival statistics and KM curves, how can I test the association between $t$ and each of $a_{i}$, $b_{i}$ and $c_{i}$. I am hypothesizing that a profile of $t$ will be distinct for each of $a_{i}$, $b_{i}$ and $c_{i}$.
One way I was thinking of testing this hypothesis was to simply do Pearson's correlation between $t$ and each of $a_{i}$, $b_{i}$ and $c_{i}$. I could then visualize this with a scatter plot of $t$ vs each of $a_{i}$, $b_{i}$ and $c_{i}$. Any suggestions on a more sophisticated analysis including variations of KM analysis? Code examples with R appreciated.
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Do the events of type $a$,$b$,$c$ occur before the start of the follow-up period, or during it? In the latter case I think you would need their exact timing. – Aniko Jun 3 '12 at 16:07
@Aniko all events have been counted before the follow-up period. – user1202664 Jun 3 '12 at 16:22
## 2 Answers
Here is a good way to start in R.
````library(survival)
# Assume your survival data is in the vector time
surv_time <- Surv(time, time < 180)
````
With that you can do the popular Cox Proportional Hazard model
````# a, b, and c are the vectors with frequency data.
summary(coxph(surv_time ~ a+b+c))
````
You will get a complete test of the effect of your 3 variables on the survival. The Cox PH model is
$$\lambda(t|X) = \lambda(t)\exp(X\beta)$$
where $\lambda(t|X)$ is the hazard function, interpreting as the chance of dying at time $t$, and $X$ is a set of measures on the individual. What the model says is that this hazard function can be anything (not necessary exponentially decreasing), but that increase/decrease of the variables in $X$ will increase/decrase the hazard in a proportional way. Variables significant in the test above will tend to multiply or divide the hazard a lot.
For a very didactic introduction to this (and many other) topic, I recommend Frank Harrell's Regression Modeling Strategies.
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Thanks but I suppose I'm looking for something that is alternative to or extends Cox PH – user1202664 Jun 3 '12 at 16:30
Why so? Anything wrong with Cox PH, or you are looking for complementary alternatives? – gui11aume Jun 3 '12 at 16:38
Looking for complementary alternatives. – user1202664 Jun 3 '12 at 19:26
Why not treat this as a competing risks model? The three types of event could be looked at as different outcomes. There is literature on this going back to the 1970s. A lot of recent work has been done by Jason Fine of UNC and Robert Gray. You can look for the Fine-Gray model. The cumulative incidence function is the generalization of Kaplan-Meier to competing risks.
Here is a link for a presentation that give background and other information. http://www.stata.com/meeting/australia09/aunz09_gutierrez.pdf
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http://quant.stackexchange.com/questions/tagged/itos-lemma
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# Tagged Questions
The itos-lemma tag has no wiki summary.
1answer
131 views
### Ito's Lemma - Integrand depends on upper limit of integration
A problem I came across while practicing using Ito's Lemma had a process with an integral whose integrand depends on the upper limit of integration (the goal is to find $dZ_{t}$): ...
1answer
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### Derivation of Ito's Lemma
My question is rather intuitive than formal and circles around the derivation of Ito's Lemma. I have seen in a variety of textbooks that by applying Ito's Lemma, one can derive the exact solution of a ...
1answer
170 views
### How to measure a non-normal stochastic process?
If I understand right, Itô's lemma tells us that for any process $X$ that can be adapted to an underlying standard normal Wiener measure $\mathrm dB_t$, and any twice continuously differentiable ...
1answer
241 views
### Demonstration of Ito's correction term/lemma in binomial tree
I am preparing an undergraduate QuantFinance lecture. I want to demonstrate the ideas of Ito's correction term and Ito's lemma in the most accessible manner. My idea is to take the "working horse" of ...
3answers
435 views
### How to use Itô's formula to deduce that a stochastic process is a martingale?
I'm working through different books about financial mathematics and solving some problems I get stuck. Suppose you define an arbitrary stochastic process, for example $X_t := W_t^8-8t$ where \$ W_t ...
1answer
354 views
### How to perform basic integrations with the Ito integral?
From the text book Quantitative Finance for Physicists: An Introduction (Academic Press Advanced Finance) I have this excercise: Prove that ...
3answers
3k views
### What is Ito's lemma used for in quantitative finance?
Further to my question asked here: prior post and which left some points unanswered, I have reformulated the question as follows: What is Ito's lemma used for in quantitative finance? and when is it ...
5answers
684 views
### Monte carlo methods for vanilla european options and Ito's lemma.
I understand that by applying Ito's lemma to the following SDE $$dX=\mu\,X\,dt+\sigma\,X\,dW$$ one obtains a solution to the above SDE which is as follows: {X}\left( t\right) =\mathrm{X}\left( ...
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http://nrich.maths.org/6166
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### Epidemic Modelling
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.
### Very Old Man
Is the age of this very old man statistically believable?
### bioNRICH
bioNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of the biological sciences, designed to help develop the mathematics required to get the most from your study of biology at A-level and university.
# Blood Buffers
##### Stage: 5 Challenge Level:
In this question [$A$] means the concentration of the chemical $A$ at equilibrium.
For a balanced chemical equation, where $A, B, C$ and $D$ are chemicals in aqueous solution and $a, b, c, d$ are whole numbers,
$$aA + bB \rightleftharpoons cC + dD+ eH_2O$$
the law of mass action tells us that for a fixed temperature, there is a constant $K$ (called the equilibrium constant) such that
\frac{[C]^c[D]^d}{[A]^a[B]^b} = K
(note the absence of the solvent concentration $[H_2O]$ ).
In the blood, the carbonic-acid-bicarbonate buffer prevents large changes in the pH of the blood. Chemically, it consists of two reactions which are simultaneously in equilibrium
$$H^+ +HCO^-_3+H_2O\rightleftharpoons^{K_1} H_2CO_3+H_2O \rightleftharpoons^{K_2} 2H_2O+CO_2$$
Show that
pH =pK -\log \left(\frac{[CO_2]}{[HCO^-_3]}\right)\quad \mbox{where }K=\frac{1}{K_1K_2}
Think about this equation. It shows that the pH of the blood is dependent on the ratio of the concentrations of $CO_2$ and $HCO^-_3$. These are large in the blood, so small changes in the relative concentrations leads to very small changes in the pH of the blood. They act as a 'buffer' against pH change.
Now, make a new variable $x$ to be the fraction of the buffer in the form of $HCO^-_3$. Thus,
x = \frac{[HCO^-_3]}{[HCO^-_3]+[H_2CO_3]+[CO_2]}
Show that
pH = pK-\log\left(\frac{1}{x}-1 -K_1[H^+]\right)
By taking the value $pK=6.1$ and treating $K_1[H^+]$ as very small, reproduce the titration curve
Extension:
1. Why is it numerically valid to ignore the $[H^+]$ term in the equation giving rise to the graph?
2. With the assumption that $K_1[H^+]=0$, use calculus to show that the second derivative of the pH is zero when $x=0.5$. Graphically, what does this correspond to?
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://physics.stackexchange.com/questions/8072/combining-chemical-elements
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# Combining chemical elements
Prof Brian Cox mentioned on Wonders of the Universe when discussing chemical elements that Carbon 12 consists of 6 protons and 6 neutrons, he also mentioned that Helium consists of 2 protons and 2 neutrons. He then went on to say that by combining certain elements we can create others - can anyone explain what he means by this with a practical example?
Can I get myself some Carbon and some Helium and somehow combine them? :)
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## 1 Answer
Brian Cox was talking about stellar nucleosynthesis. In theory you could combine the nucleus of a Helium atom (also called an alpha particle) and the nucleus of a Carbon atom to produce an Oxygen nucleus and some energy in the form of gamma radiation
$$^4_2He + ^{12}_6C \rightarrow ^{16}_8O +\gamma$$ This is an alpha process, and does happen inside certain stars.
In practice, you are unlikely to be able to do this personally, as you probably cannot get the nuclei moving fast enough or accurately enough to enable fusion.
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brilliant, that last sentence clarified exactly my query! – 78lro Apr 5 '11 at 11:22
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http://math.stackexchange.com/questions/15749/stochastic-integral-and-stieltjes-integral
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# Stochastic integral and Stieltjes integral
My question is on the convergence of the Riemann sum, when the value spaces are square-integrable random variables. The convergence does depend on the evaluation point we choose, why is the case. Here is some background to make this clearer. Suppose $f\colon \Re \mapsto \Re$ is some continuous function on $[a,b]$, the Stieltjes integral of $f$ with respect to itself $f$ is $\int^{b}_{a} f(t)df(t)$ if we take a partition $\Delta_n = \{t_0, t_1, \cdots, t_n \}$ of $[a,b]$ the Riemmans sums is $$L_{n} = \sum^{n}_{i=1} f(t_{i-1})(f(t_{i})-f(t_{i-1}))$$
Now if the limit exists say $\lim \limits_{n\to\infty} L_{n}= A$, then if we choose the evaluation point $t_{i}$ then the sum
$$R_{n} = \sum^{n}_{i=1} f(t_{i})(f(t_{i})-f(t_{i-1}))$$
will also converge to $A$ so $$\lim_{n\to\infty}L_{n} = \lim_{n\to\infty}R_{n} .$$ Now we apply same idea for a stochastic integral. Here $W(t)$ is a wiener process and we wish to find $$\int^{b}_{a}W(t)dW(t)$$ $$L_{n} = \sum^{n}_{i=1} W(t_{i-1})(W(t_{i})-W(t_{i-1}))$$
$$R_{n} = \sum^{n}_{i=1} W(t_{i})(W(t_{i})-W(t_{i-1}))$$ in $L^2$ norm the limits of $L_{n}$ and $R_{n}$ exist but are different $$\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = b-a$$ can someone explain why the limits are different ? If the limit exists which in this case it does. I would have expected $\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = 0$ in $L^2$ norm.
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Maybe it has to do with the $L^2$ norm. After all, $||\textbf{x}|| = \sqrt{x_{1}^2+ \cdots + x_{n}^2}$. – PEV Dec 28 '10 at 16:59
– George Lowther Dec 28 '10 at 21:52
@George Lowther thank your answer is indeed very helpful – almost_sure Dec 28 '10 at 23:36
## 3 Answers
Limits of $R_n$ and $L_n$ coincide when Stieltjes integral exists. Existence of the Stieltjes integral does not follow from the existence of these limits. In general existence and definition of Stieltjes integral can be messy business as the Figure 2.1 in page 6 (page 10 of the ps file) of this document can attest.
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thank you. I made the naive assumption if the limits $R_n$ and $L_n$ coincide then the stieltjes integral exists. thanks of the document – almost_sure Dec 28 '10 at 20:41
@almost_sure, you are welcome. I did not know that Stieltjes integrals can be complicated until I took a course where the professor was one of the authors of that document. – mpiktas Dec 28 '10 at 20:47
First write $$R_n - L_n = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 },$$ then consider Quadratic variation of Brownian motion.
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Thanks i understand why $R_n - L_n \to {\rm E}(Z_1^2)$ my question is, why does the evaluation point make such a difference when the sums converge in $L^2$ norm? but not in the real valued function case. like other posters said is it just have to do with the Norm. If the limit existed to me it seemed intuitive for $R_n - L_n \to 0$ in $L^2$ I am still not sure why this is not the case, the hint below have got me thinking it just the norms are different – almost_sure Dec 28 '10 at 20:20
Hint 1:
The basis of the explanation is in the different behaviour of the increments $f(t_i)-f(t_{i-1})$ and $W(t_i)-W(t_{i-1})$. The increments of the first are $O(t_i-t_{i-1})$ those of the second are $O(\sqrt{t_i-t_{i-1}})$.
Hint 2: $$R_n - L_n =\sum_{i=1}^n \left[ f(t_i)-f(t_{i-1}) \right]^2$$ and likewise for $W(t)$.
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