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http://stats.stackexchange.com/questions/6731/using-singular-value-decomposition-to-compute-variance-covariance-matrix-from-li
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# Using Singular Value Decomposition to Compute Variance Covariance Matrix from linear regression model
I have a design matrix of p regressors, n observations, and I am trying to compute the sample variance-covariance matrix of the parameters. I am trying to directly calculate it using svd.
I am using R, when I take svd of the design matrix, I get three components: a matrix $U$ which is $n \times p$, a matrix $D$ which is $1\times 3$ (presumably eigenvalues), and a matrix $V$ which is $3\times 3$. I diagonalized $D$, making it a $3\times 3$ matrix with 0's in the off-diagonals.
Supposedly, the formula for covariance is: $V D^2 V'$, however, the matrix does not match, nor is it even close to R's built in function, `vcov`. Does anyone have any advice/references? I admit that I am a bit unskilled in this area.
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## 1 Answer
First, recall that under assumptions of multivariate normality of the linear-regression model, we have that $$\hat{\beta} \sim \mathcal{N}( \beta, \sigma^2 (X^T X)^{-1} ) .$$
Now, if $X = U D V^T$ where the right-hand side is the SVD of X, then we get that $X^T X = V D U^T U D V = V D^2 V^T$. Hence, $$(X^T X)^{-1} = V D^{-2} V^T .$$
We're still missing the estimate of the variance, which is $$\hat{\sigma}^2 = \frac{1}{n - p} (y^T y - \hat{\beta}^T X^T y) .$$
Though I haven't checked, hopefully vcov returns $\hat{\sigma}^2 V D^{-2} V^T$.
Note: You wrote $V D^2 V^T$, which is $X^T X$, but we need the inverse for the variance-covariance matrix. Also note that in $R$, to do this computation you need to do
````vcov.matrix <- var.est * (v %*% d^(-2) %*% t(v))
````
observing that for matrix multiplication we use `%*%` instead of just `*`. `var.est` above is the estimate of the variance of the noise.
(Also, I've made the assumptions that $X$ is full-rank and $n \geq p$ throughout. If this is not the case, you'll have to make minor modifications to the above.)
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Thank you! That did the trick. – Will Jan 31 '11 at 3:45
@Will, good. Glad it worked. You might consider accepting the answer then. Regards. – cardinal Jan 31 '11 at 3:58
+1 great answer. – mpiktas Jan 31 '11 at 5:04
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http://www.physicsforums.com/showthread.php?t=79376
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Physics Forums
## Group theory problems
Question 1 a
Find the order of
$$(\mathbb{Z}_4 \times \mathbb{Z}_2) \backslash \langle (2,1) \rangle$$.
Question 1 b
Find the order of
$$(\mathbb{Z}_2 \times \mathbb{Z}_4) \backslash \langle (1,1) \rangle$$.
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I was attempting these two questions from Section 14 of "A First Course in Abstract Algebra" by John Fraleigh (where I get all my questions from in fact). Although the two questions were slightly different, I was unable to distinguish them due to my lack of insight. For (a), I can write out $\mathbb{Z}_4\times\mathbb{Z}_2[/tex] as [tex]\mathbb{Z}_4\times\mathbb{Z}_2 = \{ (0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1) \}[/tex] Since I am asked to compute the factor group, I want to be able to picture a factor group in the same way as 'picking' bits out. In other words we have the 8 elements in [itex]\mathbb{Z}_4\times\mathbb{Z}_2[/tex] and I want to factor out [itex]\langle (2,1) \rangle$. Now $\langle (2,1) \rangle$ is the subgroup generated by $(2,1)$. So $$\langle (2,1) \rangle = \{ (2,1),(4,2),(6,3),(8,4),\dots\}$$ But $(4,2) \equiv (0,0)$, $(6,3) \equiv (2,0)$, $(8,4) \equiv (0,0)$, $(10,5) \equiv (2,1)$, etc. So $$\langle (2,1) \rangle = \{ (0,0),(2,1),(2,0) \}$$ What am I missing? Shouldn't this have 2 elements? EDIT: The set $\mathbb{Z}_4\times\mathbb{Z}_2$ is now ammended.
$$\mathbb{Z}_4\times\mathbb{Z}_2 = \{ (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3) \}$$
That looks more like Z_2 * Z_4 ;)
The computation of <(2, 1)> is indeed faulty. You've written that (6, 3) = (2, 0), which is wrong. It should be (6, 3) = (2, 1), so that <(2, 1)> consists of (0, 0) and (2, 1).
And btw, you don't need to explicitly calculate the factor groups.
## Group theory problems
The computation of <(2, 1)> is indeed faulty. You've written that (6, 3) = (2, 0), which is wrong. It should be (6, 3) = (2, 1), so that <(2, 1)> consists of (0, 0) and (2, 1).
AHA!! Geez I make some pretty stupid mistakes!
Ok so now I have
$$\langle (2,1) \rangle =\{ (0,0),(2,1) \}$$
So
$$|\langle (2,1) \rangle | = 2$$
Therefore the order of the factor group is
$$\frac{|\mathbb{Z}_4 \times \mathbb{Z}_2|}{|\langle (2,1) \rangle|} = \frac{8}{2} = 4$$
For part b we now have $$\mathbb{Z}_2 \times \mathbb{Z}_4 = \{ (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$$ Which is obviously slightly different to $\mathbb{Z}_4 \times \mathbb{Z}_2$. And now $$\langle (1,1) \rangle = \{ (1,1),(2,2),(3,3),\dots\}$$ $$= \{ (0,0),(1,1),(2,0),(3,1)\}$$ So the orde of the factor group is 2. The purpose of this exercise was to illustrate that factor groups of finite groups can be done by 'picking' out what you dont want.
And btw, you don't need to explicitly calculate the factor groups.
Ive shown you the way that I do it - it is rather long and exhausting (particularly with more complicated examples), could you show me exactly how you would do these problems? (just one will suffice).
Recognitions: Homework Help Science Advisor The questions you've been given have simply been to find the order of the factor groups. It should seem obvious to you that for a finite group G, |G/H| = |G|/|H|. So in your case, you just need to find the order |H|. You're given the generator of H in your problems, like (1,1). It's not hard to see that this has order 4, and that G has order 4 x 2 = 8, so you know |G/H| = |G|/|H| = 8/4 = 2.
The questions you've been given have simply been to find the order of the factor groups. It should seem obvious to you that for a finite group G, |G/H| = |G|/|H|. So in your case, you just need to find the order |H|. You're given the generator of H in your problems, like (1,1). It's not hard to see that this has order 4, and that G has order 4 x 2 = 8, so you know |G/H| = |G|/|H| = 8/4 = 2.
Yes I do see that G has order 8. But it is not easy for me, at this stage, to see that something like (1,1) has order 4, or (2,1) has order 2. THIS is what I lack. Some of my friends, and you, for instance, can just look at that and know its order...so how do you do it?
Recognitions:
Homework Help
Science Advisor
Yes I do see that G has order 8. But it is not easy for me, at this stage, to see that something like (1,1) has order 4, or (2,1) has order 2. THIS is what I lack. Some of my friends, and you, for instance, can just look at that and know its order...so how do you do it?
I don't know, maybe do a bunch of problems and it will start to come naturally two you. Well, I can give you the ideas but it's up to you to practice with them so they become natural. First, groups like $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \dots \times \mathbb{Z}_{n_k}$ are abelian. Secondly, any element $(a_1, \dots , a_k)$ is just $(a_1, 0, \dots , 0) + (0, a_2, 0, \dots , 0) + \dots + (0, \dots , 0, a_k )$. Finally, if you have an element $x_1x_2\dots x_k$, or, using additive notation, $x_1 + x_2 + \dots + x_k$, the order of this element is the least common multiple of the order of each of the $x_i$. So, for example:
(1,1)
= (1,0)(0,1) [OR if you like additive notation, (1,0) + (0,1)]
The order of (1,0) in $\mathbb{Z}_2 \times \mathbb{Z}_4$ is just the order of 1 in $\mathbb{Z}_2$, and the order of (0,1) in $\mathbb{Z}_2 \times \mathbb{Z}_4$ is just the order of 1 in $\mathbb{Z}_4$. Those numbers are, respectively, 2 and 4, whose least common multiple is 4. If we're looking at (1,2), then we have (1,2) = (1,0) + (0,2) and the order of (1,0) in $\mathbb{Z}_2 \times \mathbb{Z}_4$ is the order of 1 in $\mathbb{Z}_2$, which is 2, and the order of 2 in $\mathbb{Z}_4$ is also 2. The least common multiple of 2 and 2 is 2.
So it depends on how quickly you can find the order of a single number in a single cyclic group, and how quickly you can find the least common multiple a pair (or more) of numbers. It's pretty easy to do the first thing. If you want to find the order of x in $\mathbb{Z}_n$, just compute n/gcd(n,x). Most of the time, though, you can just look at it and tell. You can tell easily that 2 has order 2 in $\mathbb{Z}_4$. Finding the lcm of some numbers is also generally not too tough. If you need a precise method for finding it, you can ask, or look it up at mathworld, but in most cases, it's something you can tell by looking at it, again, like the lcm of 2 and 4.
The order of (1,0) in is just the order of 1 in , and the order of (0,1) in is just the order of 1 in . Those numbers are, respectively, 2 and 4, whose least common multiple is 4. If we're looking at (1,2), then we have (1,2) = (1,0) + (0,2) and the order of (1,0) in is the order of 1 in , which is 2, and the order of 2 in is also 2. The least common multiple of 2 and 2 is 2.
Perfect, that is exactly what I needed!!!
One more question concerning factor groups. Say we factor the group $H$ from the group $G$. So we have
[tex]G\backslash H[/itex]
Now, how come the order of $H$ always seems to be a divisor of $G$? I mean, if it wasn't, then the factor group would be impossible - because you cant have half an element.
This lead me to believe that $H$ must have a special property. That is, the order of $H$ must be a divisor of $G$. But how can we guarantee that in all these factoring problems, that our $|H|$ will divide $|G|$ with no remainder?
Would you care to explain this phenomenon? Do we only allow $H$ to be normal? As in the left and right cosets coincide?
Recognitions: Homework Help Science Advisor This is Lagrange's Theorem. A subgroup of a finite group G has order that divides |G|. The proof uses the idea of cosets. Suppose you have a subgroup H. Now take some g in G that isn't in H. gH must have the same size as H, obviously, and the two must be disjoint. Suppose they weren't disjoint, then for some h in H, and some h' in H, we'd have: gh = h' this being the element in the intersection of gH and H. But this gives us: $g = h'h^{-1}$ and since H is a subgroup, $h'h^{-1} \in H$, which means g is in H, contradicting our choice of g. Continuing this process, we must eventually fill out all of G with distinct disjoint cosets of H. It should be clear from here why any subgroup H has order that divides |G|.
Thankyou AKG, I will look into that. You're explanations are worth their weight in gold! (especilly since I have finals coming up). Question 2 Find all prime ideals of $\mathbb{Z}_6$ To find a prime ideal $N$ of $\mathbb{Z}_6$ I would begin by finding ideals such that by finding factor rings $\mathbb{Z}_6/N$ that are integral domains. In my opinion, all prime ideals will contain 0. Because by factoring out 0 from $\mathbb{Z}_6$ is a good start to assuring $\mathbb{Z}_6/N$ is an integral domain - because an integral domain is a commutative ring with unit $1\neq 0$ containing no divisors of 0. Now Im not sure how to continue.
A prime ideal of $\mathbb{Z}_6$ is $$N_1 = \{ 0,2,4 \}$$ Since (note that elements of $(\mathbb{Z}_6)$ commute under multiplication) $$0\times 2 = 0\times 4 = 0 \in N_1$$ $$2\times 2 = 4 \in N_1$$ $$2\times 4 = 8 \equiv 2 \in N_1$$ $$4\times 4 = 16 \equiv 4 \in N_1$$ Therefore $N_1$ is a prime ideal. The set $$N_2 = \{ 0,3 \}$$ is also a prime ideal for exactly the same reasons. My question now becomes: How do you find these prime ideals without trial-and-error? I mean, it took me a while to work out which elements belong to the prime ideal because I calculated every one of them - surely there is an easy way. And also, I found 2 prime ideals - is there a way of knowing that you found them all?
For a illustration as how this method become tedious. Consider $\mathbb{Z}_{12}$. A prime ideal is $$N_1 = \{ 0,2,4,8 \}$$ How do I know it is maximal? How do I know there is or isn't any more? If $N_1$ is a prime ideal, then isn't $$N_2 = \{ 0,2,4 \}$$ also a prime ideal? Surely there must be an easier way that to going through all different multiples of two numbers and checking to see if they belong in the prime ideal or not.
Recognitions: Gold Member Science Advisor Staff Emeritus N1 isn't even an ideal, let alone a prime ideal. Remember the definition of ideal -- adding any two things in an ideal is again in an ideal. Also, multiplying a number in the ideal by a number of the ring is again in the ideal!
Recognitions: Homework Help Science Advisor This is the first time I'm seeing prime ideals, but I'll try to answer your question. An ideal is a subset of your ring that forms an additive group, so we want to look at subgroups of $\mathbb{Z}_6$ where we treat $\mathbb{Z}_6$ as the additive group. These subgroups are just, <0>, <1>, <2>, <3>. Next, we want it so that whenever x is in $\mathbb{Z}_6$ and y is in our ideal, both xy and yx are in the ideal. But this is true for all those subgroups, so they're all ideals. For prime ideals, it must be that whenever xy is in the ideal, then either x or y are. <0> doesn't work. If x = 2, y = 3, then xy = 0, but neither of x nor y are in <0>. <1> works because it's all of $\mathbb{Z}_6$. Since 6 is even, if xy is in <2>, then x*y must be even, so one of x or y must be even, so one of them must be in <2>, so <2> is a prime ideal. Note I'm using x*y to denote normal multiplication, and xy to denote ring multiplication, which is multiplication modulo 6. Finally, if xy is in <3>, then x*y is a multiple of 3. If x*y = 0, then one of x or y is zero, so one of them is in <3> since zero is in <3>. Otherwise, x*y is a positive multiple of 3. Since 3 is prime, then one of x or y must be a multiple of 3, and since x and y are elements of {0,1,...,5}, then one of them must be 3 itself, and hence in <3>. Look here: http://mathworld.wolfram.com/PrimeIdeal.html The second sentence is: For example, in the integers, the ideal $\mathfrak{a} = \left \langle{p}\right\rangle$ is prime whenever p is a prime number. Here, we weren't dealing with $\mathbb{Z}$ but $\mathbb{Z}_6$, but I'd guess that a similar result holds.
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http://mathoverflow.net/questions/45422?sort=votes
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## maximal coordinate on a sphere
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
What is the easiest (preferably without calculations) way to see that the mean value of $\max(x_1,x_2,\dots,x_n)$ on the sphere $\mathbb{S}^{d-1}= \{ (x_1,\dots,x_n):\ x_1^2+\dots+x_n^2=1 \}$ behaves like $\sqrt{\log(n)/n}$, or at least that is is much more then $1/\sqrt{n}$ for large $n$? The same (and less or more a priori equivalent) question concerns the standard Gaussian measure and expectation of $\infty$-norm w.r.t. it.
The proofs I know (for example, the one which V. Milman attributes to Figiel) use too many integrals.
And by the way, how to put {,} in math here? \ { does not work for me
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1
\lbrace and \rbrace. – Gerry Myerson Nov 9 2010 at 12:01
1
Or \\{ and \\}. – Andrey Rekalo Nov 9 2010 at 12:02
Oh, thanks! $\ \$ – Fedor Petrov Nov 9 2010 at 12:50
doesn't the fact that on your hypersphere ($d$ should be $n$ in your question), $\|x\|_\infty \ge 1/\|x\|_1 \ge 1/\sqrt{n}$ help? – S. Sra Nov 10 2010 at 15:04
## 1 Answer
For some positive $c$ bounded away from zero, the probability that a standard gaussian variable is larger than $c\sqrt{\log n}$ is $1/n$. It follows that the probability that at least one variable out of $n$ independent standard gaussians is larger than $c\sqrt{\log n}$ is $1-(1-1/n)^n$ which tends to $1-1/e$. From that one gets that the expectation of the maximum of $n$ standard gaussians is at least (basically) $(1-1/e)c\sqrt{\log n}$. As you said the question about variables on the sphere is equivalent to that.
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Thanks, that's clear indeed. – Fedor Petrov Nov 9 2010 at 12:53
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http://mathoverflow.net/questions/121488?sort=votes
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## A sum involving mod(n) arithmetic
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi there, I have been studying the following set (in order to investigate the average of products of Ramanujan sums with some weights):
$$A=\lbrace (n,m) \in \mathbb{Z}/q\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z} \mid \text{ }nm\equiv a \mod{q},\text{ } n \equiv 1 \mod{d_{1}}, \text{ } m \equiv 1 \mod{d_{2}} \rbrace$$
where $q$ is any positive integer, $d_{1}$ and $d_{2}$ are proper divisors of $q$ and $(a,q)=1$.
For the case $a=1$, using a bit algebra I have deduced the elementary formula
$$\sum_{\substack{n_{1}n_{2}\equiv 1\mod{q}, \\ n_{1} \equiv 1 \mod{d_{1}}, \\ n_{2} \equiv 1 \mod{d_{2}}}}=\frac{\phi(q)}{\mathrm{lcm}(\phi(d_{1}),\phi(d_{2}))}.$$
Is it possible to derive a formula for $|A|$?
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4
The sum of WHAT, and what exactly is your question? – Seva Feb 11 at 16:17
Actually we want to count a set, which we abbreviate by summing 1s and putting constraints as indexes. – Egeselazuzyan Feb 11 at 17:49
There are no 1s in your question. And what is $k$ in the right hand side of the second formula? – Max Alekseyev Feb 11 at 17:53
2
Then the WHAT is 1, and you need to put it in the expression. Gerhard "Or Suffer The Community Wrath" Paseman, 2013.02.11 – Gerhard Paseman Feb 11 at 17:55
Question also posted to m.se, math.stackexchange.com/questions/299752/… – Gerry Myerson Feb 12 at 0:05
## 1 Answer
I believe the formula $|A|=\frac{\varphi(q)}{\mathrm{lcm}(\varphi(d_1),\varphi(d_2))}$ is not quite correct. In particular, for $q=15$, $d_1 = 3$, $d_2=5$, $a=1$, this formula gives $|A|=2$, while, in fact, $A = \{ (1,1) \}$ with $|A|=1$. Below I derive a correct formula.
First, it is clear that if $a\not\equiv1\pmod{\gcd(d_1,d_2)}$, then $|A|=0$. For the rest assume that $a\equiv1\pmod{\gcd(d_1,d_2)}$.
Let $p$ be a prime dividing $q$ and $t=\nu_p(q)>0$ (i.e., $t$ is the valuation of $q$ w.r.t. $p$) and $s_1 = \nu_p(d_1)$, $s_2 = \nu_p(d_2)$. It is easy to see that the number of elements modulo $p^t$ in $A$ is indeed $$\frac{\varphi(p^t)}{\mathrm{lcm}(\varphi(p^{s_1}),\varphi(p^{s_2}))} = \begin{cases} (p-1)p^{t-1},& \text{if}\ s_1=s_2=0\\ p^{t-\max\{s_1,s_2\}},&\text{otherwise}. \end{cases}$$
Now, for any $a$ such that $\gcd(a,q)=1$ and $a\equiv1\pmod{\gcd(d_1,d_2)}$, we have (thanks to CRT) $$|A| = \prod_{p|q} \frac{\varphi(p^{\nu_p(q)})}{\mathrm{lcm}(\varphi(p^{\nu_p(d_1)}),\varphi(p^{\nu_p(d_2)}))} = \frac{\varphi(q)}{\prod_{p|q} \mathrm{lcm}(\varphi(p^{\nu_p(d_1)}),\varphi(p^{\nu_p(d_2)}))}.$$ Notice that this product does not collapse into $\frac{\varphi(q)}{\mathrm{lcm}(\varphi(d_1),\varphi(d_2))}$.
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http://math.stackexchange.com/questions/49074/cauchy-schwarz-inequality-and-holders-inequality
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# Cauchy-Schwarz inequality and Hölder's inequality
It might sound silly, but I am always curious whether Hölder's inequality $$\sum_{k=1}^n |x_k\,y_k| \le \biggl( \sum_{k=1}^n |x_k|^p \biggr)^{\!1/p\;} \biggl( \sum_{k=1}^n |y_k|^q \biggr)^{\!1/q} \text{ for all }(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n.$$ can be derived from the Cauchy-Schwarz inequality.
Here $\frac{1}{p}+\frac{1}{q}=1$, $p>1$.
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3
How about the other way round? – Fabian Jul 2 '11 at 19:53
@Fabian: ... Cauchy-Schwartz is a special case of Hölder's Inequality ... $p=q=2$ – Mark Bennet Jul 2 '11 at 20:03
@Sunni: I deleted my answer "No, as far as I know (...)" because apparently it was wrong. – Américo Tavares Jul 2 '11 at 22:28
That is fine. You are very welcome to post your answers. – Sunni Jul 2 '11 at 22:43
## 2 Answers
Yes it can, assuming nothing more substantial than the fact that midpoint convexity implies convexity. Here are some indications of the proof in the wider context of the integration of functions.
Consider positive $p$ and $q$ such that $1/p+1/q=1$ and positive functions $f$ and $g$ sufficiently integrable with respect to a given measure for all the quantities used below to be finite. Introduce the function $F$ defined on $[0,1]$ by $$F(t)=\int f^{pt}g^{q(1-t)}.$$ One sees that $$F(0)=\int g^q=\|g\|_q^q,\quad F(1)=\int f^p=\|f\|_p^p,\quad F(1/p)=\int fg=\|fg\|_1.$$ Furthermore, for every $t$ and $s$ in $[0,1]$, $$F({\textstyle{\frac12}}(t+s))=\int h_th_s,\qquad h_t=f^{pt/2}g^{q(1-t)/2},\ h_s=f^{ps/2}g^{q(1-s)/2},$$ hence Cauchy-Schwarz inequality yields $$F({\textstyle{\frac12}}(t+s))^2\le\int h_t^2\cdot\int h_s^2=F(t)F(s).$$ Thus, the function $(\log F)$ is midpoint convex hence convex. In particular, $1/p=(1/p)1+(1/q)0$ with $1/p+1/q=1$ hence $$F(1/p)\le F(1)^{1/p}F(0)^{1/q},$$ which is Hölder's inequality $\|fg\|_1\le\|f\|_p\|g\|_p$.
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Wow, that is amazing. It is rarely seen in literatures mention Holder's inequality is implied by CS. Is this consideration known before? – Sunni Jul 2 '11 at 22:32
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That midpoint convex implies convex does require some regularity on a function such as measurability. There's very little chance it won't hold here of course but this is probably worth mentioning. – Zarrax Jul 2 '11 at 23:27
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@Zarrax: Right. In the present case $F$ is also convex on $[0,1]$ hence continuous on $(0,1)$. This is enough to prove that $\log F$ is convex (and continuous) on $(0,1)$ hence, rather than measurability issues which are not a problem here, the important missing step is that convexity of $\log F$ should be used on $1/p=(1/p)t+(1/q)(1-t)$, then the limits of $F(t)$ and $F(1-t)$ when $t\to0$ should be compared to $F(0)$ and $F(1)$. I deliberately omitted these technicalities in my post, which is the reason why the first paragraph announces some indications of the proof, rather than a proof. – Did Jul 2 '11 at 23:43
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@Sunni, I think this can interest you: YUAN-CHUAN LI AND SEN-YEN SHAW, A Proof of Hölder's Inequality using the C-S Inequality, Journal of Inequalities in Pure and Applied Mathematics. – leo Jul 3 '11 at 4:13
In the book: "THE CAUCHY–SCHWARZ MASTER CLASS" of J. MICHAEL STEELE, in the case of finite sums, as you post, is the exercise 9.3. – leo Jul 3 '11 at 4:23
show 3 more comments
I suggest the following consideration. We will prove the above inequality for rational $p,q\in(1,\infty)$ with $\frac1p+\frac1q= 1$, and the irrational cases follow by continuity.
If $p$ and $q$ are rational, let $p=\frac ab$ and $q=\frac ac$ with $b+c=a$ and $2^m\ge a$. Now by induction $$\sum |x^{(1)}\dots x^{(2^m)}| \le \left( \sum |x^{(1)}\cdots x^{(2^{m-1})}|^2 \right)^{\frac12}\cdot\left( \sum |x^{(2^{m-1}+1)}\cdots x^{(2^{m})}|^2 \right)^{\frac12}$$ $$\le \left( \sum |x^{(1)}|^{2^m} \right)^{\frac1{2^m}}\cdots \left( \sum |x^{(2^m)}|^{2^m} \right)^{\frac1{2^m}}$$ where $x^{(i)}$ are sequences of length $n$, whose indices are omitted. By plugging in $$x^{(1)}= \dots= x^{(b)}= x^{\frac a{b2^m}},\quad x^{(b+1)}= \dots= x^{(b+c)}= y^{\frac a{c2^m}},\quad x^{(b+c+1)}= \dots = x^{(2^m)}=(xy)^{\frac1{2^m}}$$ we get $$\sum |xy|= \sum |x^{\frac a{2^m}}y^{\frac a{2^m}}(xy)^{\frac {2^m-a}{2^m}}| \le \left( \sum |x|^{p} \right)^{\frac{b}{2^m}}\cdot \left( \sum |y|^{q} \right)^{\frac c{2^m}}\cdot \left( \sum |xy| \right)^{\frac {2^m-a}{2^m}}$$ implying the inequality we aimed for by dividing the third term of the RHS, and putting the inequality to the $\frac{2^m}{a}$-th power.
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Your $p, q$ should not be in $(0,1)$. – Sunni Jul 2 '11 at 20:41
yea they arent, i dont know why i put it there ... oops – Peter Patzt Jul 2 '11 at 20:54
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http://mathhelpforum.com/number-theory/198882-calculation-300-digit-sophie-germain-prime-number.html
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# Thread:
1. ## Calculation 300 digit Sophie Germain prime number
I have been set a challenge to produce a 300 digit Sophie Germain prime number using computer code.
Maths at this level is not something that I use a great deal and certain not with these size of numbers. So I apologise if this is all a bit "basic".
Working on the basis of A prime number p is a [COLOR=rgb(0.000000%, 69.400000%, 31.400000%)]Sophie Germain Prime, [/COLOR]when also 2p + 1 is a prime.
But, I need to generate this result in the most efficient method possible and would like any assistance or advice on how to produce such a number without making my starting point of 3 and working up to this size of number. Is there a more efficient method to achieve this?
Any assistance would be greatly appreciated.
2. ## Re: Calculation 300 digit Sophie Germain prime number
No! There's no reason to start with 3! Why not start with $10^{300}+1$? There's no magic formula that I know of to produce these things, but in the 300 digit range, about .14% of numbers are prime, so the starting odds of p and 2p+1 being prime are one in about 475000. Sure, that's a half a million primality tests, but it certainly beats starting at 3.
If it was me, I would start with a table of a million or so random 300 digit numbers, along with their "partners" giving you random pairs (p,2p+1) in the range you're looking for. Then sieve out the evens, multiples of 3, then 5, 7, 11, etc, until you get to about $10^6$. Whatever you have left over, use a high-powered primality test like Lucas-Lehmer to eliminate your pairs. You might be able to get away with only a few thousand tests after sieving.
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# Tagged Questions
1answer
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### Combining Proportions to get Newton's Law of Universal Gravitation
I've read a little on the history of Newton's Law of Gravitation and noticed that the formula can be separated into 3 distinct parts that lead to the end result of $F_g = G \frac{m_1 m_2}{r^2}$; the ...
1answer
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### Lorentz force law in Newtonian relativity
I know that in special relativity Electric and Magnetic fields mix together in different reference frames, but my question is about classical mechanics. It seems weird to me is that the Lorentz Force ...
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http://math.stackexchange.com/questions/4544/why-is-e-pi-sqrt163-almost-an-integer
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# Why is $e^{\pi \sqrt{163}}$ almost an integer?
The fact that Ramanujan's Constant $e^{\pi \sqrt{163}}$ is almost an integer ($262 537 412 640 768 743.99999999999925...$) doesn't seem to be a coincidence, but has to do with the $163$ appearing in it. Can you explain why it's almost-but-not-quite an integer in layman's terms (I'm not a mathematician)?
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mathoverflow.net/questions/30787 and mathoverflow.net/questions/4775 are relevant, but let's see somebody try to explain it without saying "modular". ;) – J. M. Sep 13 '10 at 16:01
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This is not really the kind of phenomenon with a layman's-terms explanation... – Qiaochu Yuan Sep 13 '10 at 16:01
@Qiaochu, indeed! – Mariano Suárez-Alvarez♦ Sep 13 '10 at 16:02
I observed yet another interesting (IMHO) "almost an integer" related to exp(Pi*sqrt(163)) result, which is: sum((1/(exp(Pi*sqrt(163))^k))*(120/(8*k+1)-60/(8*k+4)-30/(8*k+5)-30/(8*k+6)), k = 0 ... infinity) = 94.000000000000000014789449792044364408558923807659819... Regards, Alexander R. Povolotsky – Alex Jun 24 '12 at 15:12
## 3 Answers
I do not think "why" has a reasonable layman's-terms answer, but let me at least explain "how" with a simpler example of such a numerical coincidence. If one takes powers of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ it is not hard to see that they are close to integers. For example, $\phi^{20} = 15126.999934...$. One might ask an analogous question about why these numbers are close to integers. The answer is that
$$\phi^n + \varphi^n = L_n$$
where $L_n$ is the $n^{th}$ Lucas number (an integer), and where $\varphi = \frac{1 - \sqrt{5}}{2}$ has absolute value less than $1$. As $n$ gets larger, $\varphi^n$ gets smaller, so $\phi^n$ becomes a better and better approximation to the integer $L_n$.
A similar, but much more complicated, phenomenon is happening here. The reason $e^{\pi \sqrt{163}}$ is so close to an integer is that it, plus a small error term, gives a special formula for that integer. But "why" this formula exists is a rather long and complicated story (as Robin Chapman hints at) and I do not think there is any reasonable way to talk about it in layman's terms.
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– Qiaochu Yuan Sep 13 '10 at 18:02
accepted as answer for explaining how some numbers approach integers, albeit with a completely different example. I understand Rumanujan's Constant is way beyond my understanding of mathematics. – stevenvh Sep 14 '10 at 14:56
This is quite a challenge to express in "layman's terms", but the reason is that $$j\left(\frac{1+\sqrt{-163}}{2}\right)$$ is an integer where $j$ is the $j$-function. When you substitute $(1+\sqrt{-163})/2$ into the $q$-expansion (see the wikipedia page) of $j$, all terms save the first two are small, and the first two equal $$-\exp(\pi\sqrt{163})+744.$$
The reason that this $j$-value is an integer is due to the quadratic field $\mathbb{Q}(\sqrt{-163})$ having class number one, or equivalently that all positive-definite integer binary quadratic forms of discriminant $-163$ are equivalent.
Added I'll try to explain the connection with binary quadratic forms. Consider a quadratic form $$Q(x,y)=ax^2+bxy+cy^2$$ with $a$, $b$ and $c$ integers. I'll only consider forms $Q$ which are primitive, so that $a$, $b$ and $c$ have no common factor $> 1$, and positive-definite, that is $a > 0$ and the discriminant $D=b^2-4ac < 0$. There is a notion of equivalence of quadratic forms, and two primitive positive-definite forms $Q$ and $Q'(x,y)=a'x^2+b'xy+c'y^2$ (necessarily also of discriminant $D$) are equivalent if and only if $$j\left(\frac{b+\sqrt{-D}}{2a}\right) =j\left(\frac{b'+\sqrt{-D}}{2a'}\right).$$ For each possible discriminant there are only finitely many equivalence classes. Thus we get a finite set of $j$-values for each discriminant, and the big theorem is that they are the solutions of a monic algebraic equation with integer coefficients. When there is only one class the equation has the form $x-k=0$ where $x$ is an integer, and the $j$-value must be an integer.
My recommended reference for this is David Cox's book Primes of the form $x^2+ny^2$. But these results appear towards the end of this 350-page book.
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– J. M. Sep 13 '10 at 16:15
As QY pointed out this cannot be explained in layman terms: Here are the papers which you would like to see: www.isibang.ac.in/~sury/ramanujanday.pdf
http://www.isibang.ac.in/~sury/episq163.pdf
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http://mathoverflow.net/questions/30396?sort=newest
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## Derived Algebraic Geometry and Chow Rings/Chow Motives
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I recently heard a talk about Chow motives and also read Milne's exposition on motives. If I understand it correctly, the naive definition of the Chow ring would be that it simply consists of all algebraic cycles, but to define a multiplication one needs to impose a certain equivalence relation, either rational or numerical equivalence.
I wondered myself if an alternative definition using derived algebraic geometry would be possible. Regardless which framework of derived algebraic geometry you use, a feature should be that you can get always the correct intersection/fiber product. Therefore, one might try to define a derived Chow ring by considering 'derived algebraic cycles' (without any equivalence relation). One would probably get a space out of this instead of a set, but this wouldn't necessarily be a bad thing. Also, the associated category of 'derived Chow motives' would then be a simplicial category (or $(\infty,1)$-category).
What I would like to know is the following: Has somebody tried to build such a theory and if not, what are the problems of it or why is it perhaps a bad idea right from the start?
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I gather this is part of the motivation behind Jacob Lurie's paper on "Structured Spaces" (which you can find on his website) though I have not understood enough of it yet to see if he carries out this particular application. – Eric Finster Aug 22 2010 at 14:40
## 2 Answers
I can't give you a complete answer apart from saying that this is definitely a hard problem!
If you have a derived scheme $X$, you can always truncate to get an every-day scheme $t_0 (X)$. On the level of rings, this corresponds to truncating a simplicial ring $A$ to $\pi_0 (A)$. The canonical morphism $A \to \pi_0(A)$ is obviously 0-connected. A result of Waldhausen then tells you that $A$ and $\pi_0(A)$ have isomorphic $K_0$ and $K_1$.
Now the Chow-Ring is more or less the same as $K_0$ (after tensoring with Q and up to rational equivalence). So I guess what this means is that on the level of cycles it is impossible to tell the difference between a derived scheme and an ordinary scheme, because $CH(X) = CH(t_0 (X)).$
I think the fancy way of saying this is that the etale topoi of $X$ and $t_0(X)$ are the same, but that is slightly over my head. This result can be found both in Luries thesis and in HAG II by Toen-Vezzosi.
The situation immediately changes though if you work with Bloch´s higher Chow groups! They see the difference starting from $K_2$, since the higher Chow groups are tied to higher $K$-groups via a Riemann-Roch map. So on the level of these cycles you could tell a difference.
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I don't know much about this stuff, so instead of answering the question, I try to formulate more precise questions, in the hope someone else will take up these questions:
One of the main reason to look for cycles is that they give realizations (their fundamental class) in all cohomology theories, which happen to have special properties (e.g., are Hodge cycles or Tate cycles), and anytime you see a Hodge (or Tate) cycle in cohomology, you expect that it comes from an algebraic cycle (the Hodge or Tate conjecture) and hence similar phenomena should occur in all cohomology theories (i.e., there is a Hodge (or Tate) cycle in all realizations).
Now, if the following were true:
1) Any 'derived algebraic cycle' gives rise to virtual fundamental classes in all cohomology theories, which again are Hodge or Tate cycles.
2) It is not clear that the virtual fundamental classes of 'derived algebraic cycles' are already fundamental classes of real algebraic cycles,
then one might formulate a 'derived' Hodge or Tate conjecture, which would have the same consequences.
Your question has another aspect, which regards a possible framework for working with these motives; I leave this aside as I understand even less about how this should work.
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http://mathoverflow.net/questions/41784/roots-of-permutations/41866
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## roots of permutations
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Consider equation $x^2=x_0$ in symmetric group $S_n$, where $x_0\in S_n$ is fixed. Is it true that for each integer $n\geq 1$, the maximal number of solutions (square roots) has identity permutation? How far may it be generalized?
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I assume x_0 and x_o are the same? Also "identical permutation" means "identity permutation"? – JBL Oct 11 2010 at 13:52
of course, thank you – Fedor Petrov Oct 11 2010 at 14:19
The number of $k$-th roots of a permutation is the number of ways to collect its cycles into tuples according to certain rules. At least for prime $k$, I think it follows immediately that the maximum is achieved at the identity permutation. – JBL Oct 11 2010 at 14:34
## 2 Answers
The maximum of the function counting square roots is attained at $x_0=1$ and this statement generalises quite well.
Let $s(\chi)$ denote the Frobenius-Schur indicator of the irreducible character $\chi$. Thus, $s(\chi)=1$ if the representation of $\chi$ can be realised over $\mathbb{R}$, $s(\chi)=-1$ if $\chi$ is real-valued but the corresponding representation is not realisable over $\mathbb{R}$ and $s(\chi)=0$ if $\chi$ is not real-valued. Then, the number of square roots of an element $g$ in any group is equal to $$\sum_\chi s(\chi)\chi(g),$$ where the sum runs over all irreducible characters of the group. See below for a quick proof of this identity.
Of course, in $S_n$, all Frobenius-Schur indicators are 1, so the number of square roots of $x_0$ is just $\sum_\chi \chi(x_0)$. This proves that the maximal number of solutions is indeed attained by $x_0 = 1$, since each character value attains its maximum there. This generalises immediately to any group, for which any representation is either realisable over $\mathbb{R}$ or has non-real character, e.g. all abelian groups. Other cases would require more thought, but this is likely the right way to go about it.
Edit: One reference I have found for the identity expressing the number of square roots in terms of Frobenius-Schur indicators is Eugene Wigner, American Journal of Mathematics Vol. 63, No. 1 (Jan., 1941), pp. 57-63, "On representations of certain finite groups". Once you get used to the notation, you will recognise it in displayed formula (11). Since the notation is really heavy going, I will supply a quick proof here:
Claim: If $n(g)$ is the number of square roots of an element $g$ of a finite group $G$, then we have $$n(g) = \sum_\chi s(\chi)\chi(g),$$ where the sum runs over all characters of $G$ and the Frobenius-Schur indicator of $\chi$ is defined as $s(\chi)=\frac{1}{|G|}\sum_{h\in G}\chi(h^2)$.
Proof: It is clear that $n(g)$ is a class function. So let's just take inner products with all characters of $G$ to find their coefficients, noting that we can write $n(g) = \sum_h \delta_{g,h^2}$ (here $\delta$ is the usual Kronecker delta): $$\begin{align*} \left< n,\chi \right> &= \frac{1}{|G|}\sum_{g\in G}n(g)\chi(g) = \frac{1}{|G|}\sum_{g\in G}\sum_{h\in G}\delta_{g,h^2}\chi(g)=\\ &=\frac{1}{|G|}\sum_{h\in G}\sum_{g\in G}\delta_{g,h^2}\chi(g) = \frac{1}{|G|}\sum_{h\in G}\chi(h^2) \end{align*}$$ as required.
Edit 2: I got curious and ran a little experiment. The proof above applies to all finite groups that have no symplectic representations. So the natural question is: what happens for those that do? Among the groups of size $\leq 150$, there are 1911 groups that have a symplectic representation, and for 1675 of them, the square root counting function does not attain its maximum at the identity! There are several curious questions that impose themselves: is there a similar (representation-theoretic?) 2-line criterion that singles out those 300-odd groups that satisfy the conclusion but not the assumptions of the above proof? What happens for the others? Can we find an if and only if characterisation that provides more insight into the structure of the groups, whose square root counting functions is maximised by the identity? Following Pete's suggestion, I have started two follow-up questions on this business: one on square roots and one on $n$-th roots.
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Wow, neat: I didn't see that coming at all. Do you have a reference for your identity? – Pete L. Clark Oct 11 2010 at 14:55
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See also Exercise 7.69 of my book Enumerative Combinatorics, vol. 2. In particular, part (c) asserts that if $k$ is a positive integer and $r_k(w)$ is the number of $k$th roots of $w\in S_n$, then $r_k$ is a character of $S_n$. It follows that $r_k$ takes its maximum value at the identity permutation. – Richard Stanley Oct 11 2010 at 15:26
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@Alex: Your Edit 2 seems very interesting. I doubt it is getting optimal exposure here: please consider asking it as as a separate question. – Pete L. Clark Oct 13 2010 at 16:48
Pete, thank you for your encouragement! I might do that, but not now, since posting questions at 3 in the morning is asking for embarrassing situations. – Alex Bartel Oct 13 2010 at 17:48
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Two your last question - "how far may it be generalized" - Richard Stanley answered when you fix the equation ($X^2=c$) and vary the group. You may also wonder about other equations. The situation is interesting: There are equations and groups with the property that the identity is not the RHS yielding the most solutions. This is so even though the LHS has no constants, just variables.
One may rephrase the question as follows: given a word $w=w(X_1,X_2,\ldots,X_r)$ in the free group $F_r$ with variables $X_1,\ldots,X_r$, and given any finite group $G$, one may naturally consider $w$ as inducing a function $G^r \to G$ by plugging elements of $G$ as variables. This in turn defines a probability distribution on $G$: if you plug uniform random elements, what do you get? The most likely outcome is often, but not always, the identity.
In fact, the probability of getting the identity can be made arbitrarily small iff the group is non-solvable. I circulated this as a conjecture some years ago and it was proven by Miklos Abert (for the non-solvable case) and Nikolov and Segal (for the solvable one).
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In fact, Richard Stanley didn't fix the equation. He is talking about equations of the form $x^k = c$ in $S_n$. – Alex Bartel Oct 12 2010 at 8:39
True. My answer attempts to generalize further into equations in several variables. – Alon Amit Oct 12 2010 at 18:44
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http://crypto.stackexchange.com/questions/1646/is-it-possible-to-distinguish-a-securely-encrypted-ciphertext-from-random-noise/1648
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# Is it possible to distinguish a securely-encrypted ciphertext from random noise?
Say I have a bunch of data encrypted with a secure block cipher (such as AES). An attacker has unlimited access to this encrypted data. The attacker doesn't know whether the data is encrypted or if it's just purely random bits. Is it possible (even theoretically) for the attacker to distinguish the encrypted data from purely random bits?
There seems to be many questions asking whether or not it's possible to identify a particular encryption scheme from the ciphertext, but what I want to know is if it's even possible to determine that the data is encrypted in the first place (as opposed to being random bits).
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## 3 Answers
It... depends.
AES is a block cipher. It works over 128-bit blocks. For a given key, AES is a permutation of the $2^{128}$ possible values that 128-bit blocks may assume. As a purportedly secure block cipher, AES is supposed to be indistinguishable from a random permutation, i.e. a permutation selected randomly and uniformly among the $2^{128}!$ possible permutations of the space of 128-bit blocks.
If you consider AES used in counter mode (CTR): some piece of hardware encrypts the successive values of a counter with AES, and spews out the concatenation of the encrypted blocks. You challenge the attacker to distinguish between such a stream, and a purely random stream of equal length. Since AES is a permutation, the AES-CTR stream will never include twice the same block value (by encrypting two distinct counter values, you necessarily obtain two distinct block values). However, the purely random stream is also expected not to repeat the same 128-bit value, until you reach a length of about $2^{64}$ blocks, i.e. quite a lot.
In that sense, AES-CTR is supposed to be indistinguishable from random noise: if AES-CTR was distinguishable, then this would imply that AES (the block cipher) is not indistinguishable from a random permutation, and that would be viewed as a structural weakness in AES. With a $k$-bit key, the cost of distinguishing AES from a random permutation should be $2^{k-1}$, no less (that's the average cost of brute-forcing the key). No such structural weakness is known yet for AES (for AES-192 and AES-256 there are some related-key attacks, but they imply using several AES instances with specific keys which are linked to each other algebraically).
Now, although indistinguishability is academically important (a cipher is deemed weak if it cannot achieve it), it is rarely relevant to practical situations. Most protocols which use encryption very straightforwardly admit to using a specific encryption protocol. For instance, if you use SSL/TLS, the client and server announce in the initial handshake message what kind of encryption algorithm is used, and this is not a problem for practical security. If security crumbles when the algorithm is known, then this is also considered as a structural weakness of the algorithm.
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Yes and no.
The only cipher that provably has no such distinguisher is the one-time pad.
For practical symmetric ciphers (e.g., AES), we have no proof that no such distinguisher exists or does not exist. The best we can do is say "A bunch of really smart folks have been trying to find such a distinguisher in order to gain fame (and possibly fortune) for a long enough time and haven't found one." This leads us to believe that no such distinguisher will be found for decades to come, so we use the cipher. Could someone release a distinguisher like this for AES tomorrow? Sure.
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Assuming a randomly chosen key and a randomly chosen plaintext why is it impossible to prove that no distinguisher exists? Has this been proved? – Ethan Heilman Jan 12 '12 at 14:48
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– mikeazo♦ Jan 12 '12 at 15:05
In theory, there is a simple distinguisher for encrypted data: Try all the possible keys, decrypt the stream and look if the result is something which makes sense.
Of course, this will not work if you encrypted garbage (and one could say that encrypted random data is really indistinguishable from random data itself, ignoring block sizes).
And practically, the key space of all modern block ciphers is large enough that trying all (or even a large part of all) keys is impossible.
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Brilliant, it provably exists but it is as hard as cracking the key, because it is the key. Of course, that does not rule out better distinquishers, but I guess proving that would be about the same as cracking the cipher anyway - the one given by Thomas Pornin for CTR seems to be better than testing the entire key size anyway. – owlstead Jan 13 '12 at 0:57
I don't think it is a given that an AES message can be brute forced. OTP is a perfect cypher because it cannot be brute forced. As brute forcing it would generate all plausible decryptions. Would brute forcing a short AES encrypted message also produce many/most plausible decryptions? – deft_code Dec 28 '12 at 0:53
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@deft_code This depends on the length of the message. There are (for AES-128) $2^{128}$ possible keys, which means almost as many possible plaintexts for a given ciphertext. If the plaintext size is in a similar order of the key size, we'll get most of the possible plaintexts. But with increasing plaintext size, the probability of a false match shrinks. (And of course, this still depends on what is a "plausible" plaintext.) And in practice, there is no way to brute-force $2^{128}$ keys. – Paŭlo Ebermann♦ Jan 4 at 19:08
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http://www.physicsforums.com/showthread.php?t=110545
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Physics Forums
rayleigh jeans formula
I am to derive the incorrect Rayleigh-Jeans formula from the correct Planck formula to show why plank's constant does not appear in the Rayleigh-Jeans formula. I should also recall the Stefen-Boltmann Law
here's what I have but I'm stuck...
Rayleigh-Jeans formula: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}$$
Planks formula: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}$$
so I am thinking I am somehow supposed to get: $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = 1$$ but I dont know how to even begin. any ideas?
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Recognitions: Homework Help Science Advisor First, the Planck formula is $$u(\lambda) = \frac{8 \pi}{\lambda^4} \frac{ h c / \lambda}{e^{hc/\lambda k T} - 1}$$. You have an extra factor of $$k T$$ in your Planck formula which is the source of the confusion. It should now be a simple matter to obtain the Rayleigh-Jeans formula by taking the appropriate limit.
so that should lead me to prove that... $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ I am sorry but it is still not apparent how I should go about solving this... as T nears infinity, i get infinity on both sides, so does that prove that $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$? and how does the Stefan-Boltzmann Law come into play?
Mentor
rayleigh jeans formula
For Rayleigh-Jeans, you need to look at what happens the $\lambda$ becomes large.
For Stephan-Boltzmann, you need to look at the total contribution from all wavelengths, i.e., you need to look at
$$\int_{0}^{\infty} u \left( \lambda \right) d \lambda.$$
Make a change of integration variable so that $T$ does not appear explicitly in the integrand. Evaluating the resulting integral requires some specialized knowledge of special function. If you only need to show proportionality to $T^4$, then the integral need not be evaluated. If you need the proportionality constant, use software (e.g., Maple) or tables to evaluate the integral
Regards,
George
$$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ when lambda --> infinity, the left hand side becomes 0/0 so I applied hospital's rule to get: $$\frac{hc(-\frac{1}{ \lambda^2})}{\frac{hc}{kT} e^{hc/ \lambda K T}} = kT$$ $$\lambda \rightarrow \inf$$ $$\frac{hc}{\frac{hc}{kT}}=kT$$ $$kT=kT$$ I did not use the stefen-boltzmann law, did I do this correctly?
Mentor I guess I mistunderstood - I thought you wanted to derive Rayleigh-Jeans and Stefan-Boltzmann from Planck. To derive Rayleigh-Jeans, expand as a series the exponential in Physics Monkey's expression, and find what happens when $\lambda$ becomes large, but not infinite. Regards, George
the Rayleigh-Jeans formula is: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}$$ while Planck's formula consists of Rayleigh-Jeans but includes $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}$$ instead of $$kT$$ so what I did was set $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ and solved for when the wavelength was really big, which shows that indeed $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ plugging this into Planck's equation, I will get the Rayleigh-Jeans formula: $$u(\lambda)=\frac{8 \pi }{\lambda^{4}}\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1}$$ $$\frac{hc/ \lambda}{e^{hc/ \lambda K T} - 1} = kT$$ Rayleigh-Jeans formula: $$u(\lambda)=\frac{8 \pi k T }{\lambda^{4}}$$ did it do this right? was I supposed to include the stefen-boltzmann law somewhere in there?
Mentor I don't think so - it appears that you've just gone round in a circle. You need to start with $$u \left( \lambda \right) = \frac{8 \pi hc}{\lambda^{5} \left( e^{hc/ \lambda K T} - 1 \right)},$$ do what I suggested in my previous post, and arrive at $$u \left( \lambda \right) = \frac{8 \pi k T }{\lambda^{4}}.$$ Regards, George
Blog Entries: 9 Recognitions: Homework Help Science Advisor If T is large, then the exponent in $e^{hc/ \lambda kT}$ is small and you can use Bernoulli's formula $$e^{x}=1+x$$ valid for "x" very small. Daniel.
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http://math.stackexchange.com/questions/302465/half-sine-and-half-cosine-quaternions
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# half sine and half cosine quaternions
Something is a little bit unclear to me. In the image below you see that you need to divide the angle by a half. Acccording to wikipedia they say that this is so that I could rotate clockwise or counter clockwise. The sine function returns a value between 1 and -1. So I don't understand it. Could somebody explain why I need to divide my angle by a half?
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– Rahul Narain Feb 13 at 21:49
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To be honest, I don't find Wikipedia's explanation very illuminating at all. You can think about it this way instead: The actual rotation is defined by the map $\mathbf x\mapsto\mathbf q\mathbf x\mathbf q^*$. You get a $\theta/2$ from $\mathbf q$ on the left, and another $\theta/2$ from $\mathbf q^*$ on the right, which adds up to a $\theta$. Of course, this is not a rigorous explanation. – Rahul Narain Feb 13 at 21:52
Another reason: If it were $\cos\theta+\mathbf a\sin\theta$ instead of $\cos\theta/2+\mathbf a\sin\theta/2$, then rotation of $\pi$ about any axis would give you the same result. But you can readily verify that that's not true for rotations in 3D. – Rahul Narain Feb 14 at 2:41
## 1 Answer
In one sense, there is no "reason" as such. Sometimes you just need a constant factor. In this case, however, there is one way of thinking about it which helps: The quaternion represents a directed area, and you're rotating by that area.
Here's why it's the correct area: Consider a sector of a unit circle, where the angle of the sector is $\theta$. (Note that we're all grown-ups here, so we're working in radians.) Then the area of the sector is $\frac{\theta}{2}$. There's that division by two.
What's not obvious, though, is why a quaternion represents an area, or what that even means. That's a little bit too involved for this box, but I'll try to explain how things in 2D, then give an outline of how to fill in the gaps.
Consider a 2D plane with vectors. We'll add two basis vectors, called $\hat{e_1}$ and $\hat{e_2}$ (you may know them as $i$ and $j$ in some notation). We know that vectors have an inner product (a.k.a. dot product) which returns a scalar, commutes and so on. (I won't formally define a vector space or inner product space here; look it up.) The basis vectors are orthonormal:
$$e_1 \cdot e_1 = e_2 \cdot e_2 = 1$$ $$e_1 \cdot e_2 = 0$$
We also know that in 3D, there is another product, the outer product (a.k.a. cross product, Gibbs product), which obeys the laws of a Lie bracket. The outer product anticommutes (i.e. $x \times y = - y \times x$). We could, therefore, think of there being a more general product, where the inner product is the bit that commutes and the outer product is the bit that anticommutes.
So we will assume a geometric product $xy$, which in general neither commutes nor anticommutes. If $x$ and $y$ are vectors, we will define:
$$x \cdot y = \frac{1}{2}(xy + yx)$$ $$x \times y = \frac{1}{2}(xy - yx)$$
You can see that by construction, the inner product commutes and the outer product anticommutes. Moreover, if $x$ and $y$ are vectors:
$$xy = x \cdot y + x \times y$$
The first term is a scalar and the second term would, in 3D, be a vector. This is looking a bit like quaternions already, but let's not get ahead of ourselves. In 2D, there is no cross product, so we need to find out exactly what the outer product means.
Whatever it means, if $x$ is a vector, $xx$ must be a scalar, because the outer product part must be zero (after all, $x \times x = - x \times x$). In particular, ${\hat e_1}{\hat e_1} = {\hat e_2}{\hat e_2} = 1$.
So let's expand $x$ and $y$ in terms of the basis vectors and see what happens.
$$(x_1 \hat e_1 + x_2 \hat e_2)(y_1 \hat e_1 + y_2 \hat e_2) = (x_1 x_2 + y_1 y_2) + (x_1 y_2 + x_2 y_1) {\hat e_1}{\hat e_2}$$
The first part is the inner product; we already know that. The second part is the outer product of two vectors. We don't know what ${\hat e_1}{\hat e_2}$ is, but the thing it's multiplied by is the area of a parallelogram with adjacent sides $x$ and $y$. So it has something to do with area.
To save typing, we will call this quantity (known as a pseudoscalar) $I = {\hat e_1}{\hat e_2}$. Why $I$? Let's multiply it by itself using the geometric product and see what happens. Note that the expansion into inner and outer product is only defined for vectors, and this isn't a vector. But we do know what it does to basis vectors, and in particular that ${\hat e_2}{\hat e_1} = - {\hat e_1}{\hat e_2}$, and so:
$$I^2 = {\hat e_1}{\hat e_2}{\hat e_1}{\hat e_2} = -{\hat e_1}{\hat e_1}{\hat e_2}{\hat e_2} = -1$$
So the geometric product of two vectors can be thought of as a complex number. Again, this is looking a lot like quaternions.
To complete the connection, we need to work out what exponentiation does. We can use the well-known identity that if $I\theta = \theta I$ and $I^2=-1$ then $e^{I\theta} = \cos \theta + I \sin \theta$.
(Incidentally, if $J\theta = \theta J$ and $J^2 = 1$ then $e^{J\theta} = \cosh \theta + J \sinh \theta$. This turns out to be important when you extend this idea to 4D, which you need to do to analyse relativistic space-time.)
You can probably see immediately from this what $e^{I\theta}$ does to vectors in 2D, but I encourage you to work out the theory of it. Then you can impress your physicist friends by telling them you independently invented 2D spinors.
The final piece of the puzzle which connects this to quaternions is that if $v$ is a 3D unit vector represented as a quaternion with no real component, then $v^2 = -1$. The proof is left as an exercise.
You should be able to work the rest out for yourself if you want to. If you don't, I would suggest asking Google about "geometric algebra" or "Clifford algebra". There are some excellent tutorials around on the net.
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1
Interesting viewpoint on interpreting spinors as areas. How do you reconcile this idea with the construction of rotations by a pair of reflections? – Muphrid Feb 13 at 23:29
I don't want to go into the whole theory of it because this is already quite long. As a clue, think about what happens in higher dimensions. In Euclidean $n$-space, you reflect about a $n-1$-dimensional hyperplane, but you rotate about a $\frac{n(n-1)}{2}$-dimensional directed area. – Pseudonym Feb 14 at 1:41
Well yeah, that's the number of bivectors in any $n$ dimensional space. But if bivectors are directed areas, how can you geometrically say spinors are also directed areas? What do you do with the scalar part? – Muphrid Feb 14 at 4:24
– Pseudonym Feb 14 at 4:35
One more comment: If $\gamma$ is an even-graded element of $\hbox{Cl}^0_{2,0}$ and $u$ is a vector, then $\gamma u \gamma^* = \gamma^2 u$. This is another possible reason for the $\theta/2$. – Pseudonym Feb 14 at 4:37
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http://www.nag.com/numeric/CL/nagdoc_cl23/html/F11/f11grc.html
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NAG Library Function Documentnag_sparse_herm_basic_setup (f11grc)
1 Purpose
nag_sparse_herm_basic_setup (f11grc) is a setup function, the first in a suite of three functions for the iterative solution of a complex Hermitian system of simultaneous linear equations. nag_sparse_herm_basic_setup (f11grc) must be called before nag_sparse_herm_basic_solver (f11gsc), the iterative solver. The third function in the suite, nag_sparse_herm_basic_diagnostic (f11gtc), can be used to return additional information about the computation.
These three functions are suitable for the solution of large sparse complex Hermitian systems of equations.
2 Specification
#include <nag.h>
#include <nagf11.h>
void nag_sparse_herm_basic_setup (Nag_SparseSym_Method method, Nag_SparseSym_PrecType precon, Nag_SparseSym_Bisection sigcmp, Nag_NormType norm, Nag_SparseSym_Weight weight, Integer iterm, Integer n, double tol, Integer maxitn, double anorm, double sigmax, double sigtol, Integer maxits, Integer monit, Integer *lwreq, Complex work[], Integer lwork, NagError *fail)
3 Description
The suite consisting of the functions:
• nag_sparse_herm_basic_setup (f11grc),
• nag_sparse_herm_basic_solver (f11gsc),
• nag_sparse_herm_basic_diagnostic (f11gtc),
is designed to solve the complex Hermitian system of simultaneous linear equations $Ax=b$ of order $n$, where $n$ is large and the matrix of the coefficients $A$ is sparse.
nag_sparse_herm_basic_setup (f11grc) is a setup function which must be called before the iterative solver nag_sparse_herm_basic_solver (f11gsc). nag_sparse_herm_basic_diagnostic (f11gtc), the third function in the suite, can be used to return additional information about the computation. Either of two methods can be used:
1. Conjugate Gradient Method (CG)
For this method (see Hestenes and Stiefel (1952), Golub and Van Loan (1996), Barrett et al. (1994) and Dias da Cunha and Hopkins (1994)), the matrix $A$ should ideally be positive definite. The application of the Conjugate Gradient method to indefinite matrices may lead to failure or to lack of convergence.
2. Lanczos Method (SYMMLQ)
This method, based upon the algorithm SYMMLQ (see Paige and Saunders (1975) and Barrett et al. (1994)), is suitable for both positive definite and indefinite matrices. It is more robust than the Conjugate Gradient method but less efficient when $A$ is positive definite.
Both CG and SYMMLQ methods start from the residual ${r}_{0}=b-A{x}_{0}$, where ${x}_{0}$ is an initial estimate for the solution (often ${x}_{0}=0$), and generate an orthogonal basis for the Krylov subspace $\mathrm{span}\left\{{A}^{\mathit{k}}{r}_{0}\right\}$, for $\mathit{k}=0,1,\dots $, by means of three-term recurrence relations (see Golub and Van Loan (1996)). A sequence of real symmetric tridiagonal matrices $\left\{{T}_{k}\right\}$ is also generated. Here and in the following, the index $k$ denotes the iteration count. The resulting real symmetric tridiagonal systems of equations are usually more easily solved than the original problem. A sequence of solution iterates $\left\{{x}_{k}\right\}$ is thus generated such that the sequence of the norms of the residuals $\left\{‖{r}_{k}‖\right\}$ converges to a required tolerance. Note that, in general, the convergence is not monotonic.
In exact arithmetic, after $n$ iterations, this process is equivalent to an orthogonal reduction of $A$ to real symmetric tridiagonal form, ${T}_{n}={Q}^{\mathrm{H}}AQ$; the solution ${x}_{n}$ would thus achieve exact convergence. In finite-precision arithmetic, cancellation and round-off errors accumulate causing loss of orthogonality. These methods must therefore be viewed as genuinely iterative methods, able to converge to a solution within a prescribed tolerance.
The orthogonal basis is not formed explicitly in either method. The basic difference between the two methods lies in the method of solution of the resulting real symmetric tridiagonal systems of equations: the conjugate gradient method is equivalent to carrying out an $LD{L}^{\mathrm{H}}$ (Cholesky) factorization whereas the Lanczos method (SYMMLQ) uses an $LQ$ factorization.
Faster convergence can be achieved using a preconditioner (see Golub and Van Loan (1996) and Barrett et al. (1994)). A preconditioner maps the original system of equations onto a different system, say
$A-x-=b-,$ (1)
with, hopefully, better characteristics with respect to its speed of convergence: for example, the condition number of the matrix of the coefficients can be improved or eigenvalues in its spectrum can be made to coalesce. An orthogonal basis for the Krylov subspace $\mathrm{span}\left\{{\stackrel{-}{A}}^{\mathit{k}}{\stackrel{-}{r}}_{0}\right\}$, for $\mathit{k}=0,1,\dots $, is generated and the solution proceeds as outlined above. The algorithms used are such that the solution and residual iterates of the original system are produced, not their preconditioned counterparts. Note that an unsuitable preconditioner or no preconditioning at all may result in a very slow rate, or lack, of convergence. However, preconditioning involves a trade-off between the reduction in the number of iterations required for convergence and the additional computational costs per iteration. Also, setting up a preconditioner may involve non-negligible overheads.
A preconditioner must be Hermitian and positive definite, i.e., representable by $M=E{E}^{\mathrm{H}}$, where $M$ is nonsingular, and such that $\stackrel{-}{A}={E}^{-1}A{E}^{-\mathrm{H}}\sim {I}_{n}$ in (1), where ${I}_{n}$ is the identity matrix of order $n$. Also, we can define $\stackrel{-}{r}={E}^{-1}r$ and $\stackrel{-}{x}={E}^{\mathrm{H}}x$. These are formal definitions, used only in the design of the algorithms; in practice, only the means to compute the matrix-vector products $v=Au$ and to solve the preconditioning equations $Mv=u$ are required, that is, explicit information about $M$, $E$ or their inverses is not required at any stage.
The first termination criterion
$rkp ≤ τ bp + Ap × xkp$ (2)
is available for both conjugate gradient and Lanczos (SYMMLQ) methods. In (2), $p=1,\infty \text{ or }2$ and $\tau $ denotes a user-specified tolerance subject to $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,\sqrt{n}\right)\epsilon \le \tau <1$, where $\epsilon $ is the machine precision. Facilities are provided for the estimation of the norm of the matrix of the coefficients ${‖A‖}_{1}={‖A‖}_{\infty }$, when this is not known in advance, used in (2), by applying Higham's method (see Higham (1988)). Note that ${‖A‖}_{2}$ cannot be estimated internally. This criterion uses an error bound derived from backward error analysis to ensure that the computed solution is the exact solution of a problem as close to the original as the termination tolerance requires. Termination criteria employing bounds derived from forward error analysis could be used, but any such criteria would require information about the condition number $\kappa \left(A\right)$ which is not easily obtainable.
The second termination criterion
$r-k2 ≤ τ max1.0, b2 / r02 r-02 + σ1 A- × Δx-k2$ (3)
is available only for the Lanczos method (SYMMLQ). In (3), ${\sigma }_{1}\left(\stackrel{-}{A}\right)={‖\stackrel{-}{A}‖}_{2}$ is the largest singular value of the (preconditioned) iteration matrix $\stackrel{-}{A}$. This termination criterion monitors the progress of the solution of the preconditioned system of equations and is less expensive to apply than criterion (2). When ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is not supplied, facilities are provided for its estimation by ${\sigma }_{1}\left(\stackrel{-}{A}\right)\sim \underset{k}{\mathrm{max}}\phantom{\rule{0.25em}{0ex}}{\sigma }_{1}\left({T}_{k}\right)$. The interlacing property ${\sigma }_{1}\left({T}_{k-1}\right)\le {\sigma }_{1}\left({T}_{k}\right)$ and Gerschgorin's theorem provide lower and upper bounds from which ${\sigma }_{1}\left({T}_{k}\right)$ can be easily computed by bisection. Alternatively, the less expensive estimate ${\sigma }_{1}\left(\stackrel{-}{A}\right)\sim \underset{k}{\mathrm{max}}\phantom{\rule{0.25em}{0ex}}{‖{T}_{k}‖}_{1}$ can be used, where ${\sigma }_{1}\left(\stackrel{-}{A}\right)\le {‖{T}_{k}‖}_{1}$ by Gerschgorin's theorem. Note that only order of magnitude estimates are required by the termination criterion.
Termination criterion (2) is the recommended choice, despite its (small) additional costs per iteration when using the Lanczos method (SYMMLQ). Also, if the norm of the initial estimate is much larger than the norm of the solution, that is, if $‖{x}_{0}‖\gg ‖x‖$, a dramatic loss of significant digits could result in complete lack of convergence. The use of criterion (2) will enable the detection of such a situation, and the iteration will be restarted at a suitable point. No such restart facilities are provided for criterion (3).
Optionally, a vector $w$ of user-specified weights can be used in the computation of the vector norms in termination criterion (2), i.e., ${{‖v‖}_{p}}^{\left(w\right)}={‖{v}^{\left(w\right)}‖}_{p}$, where ${\left({v}^{\left(w\right)}\right)}_{\mathit{i}}={w}_{\mathit{i}}{v}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$. Note that the use of weights increases the computational costs.
The sequence of calls to the functions comprising the suite is enforced: first, the setup function nag_sparse_herm_basic_setup (f11grc) must be called, followed by the solver nag_sparse_herm_basic_solver (f11gsc). nag_sparse_herm_basic_diagnostic (f11gtc) can be called either when nag_sparse_herm_basic_solver (f11gsc) is carrying out a monitoring step or after nag_sparse_herm_basic_solver (f11gsc) has completed its tasks. Incorrect sequencing will raise an error condition.
4 References
Barrett R, Berry M, Chan T F, Demmel J, Donato J, Dongarra J, Eijkhout V, Pozo R, Romine C and Van der Vorst H (1994) Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods SIAM, Philadelphia
Dias da Cunha R and Hopkins T (1994) PIM 1.1 — the parallel iterative method package for systems of linear equations user's guide — Fortran 77 version Technical Report Computing Laboratory, University of Kent at Canterbury, Kent, UK
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
Hestenes M and Stiefel E (1952) Methods of conjugate gradients for solving linear systems J. Res. Nat. Bur. Stand. 49 409–436
Higham N J (1988) FORTRAN codes for estimating the one-norm of a real or complex matrix, with applications to condition estimation ACM Trans. Math. Software 14 381–396
Paige C C and Saunders M A (1975) Solution of sparse indefinite systems of linear equations SIAM J. Numer. Anal. 12 617–629
5 Arguments
1: method – Nag_SparseSym_MethodInput
On entry: the iterative method to be used.
${\mathbf{method}}=\mathrm{Nag_SparseSym_CG}$
Conjugate gradient method.
${\mathbf{method}}=\mathrm{Nag_SparseSym_SYMMLQ}$
Lanczos method (SYMMLQ).
Constraint: ${\mathbf{method}}=\mathrm{Nag_SparseSym_CG}$ or $\mathrm{Nag_SparseSym_SYMMLQ}$.
2: precon – Nag_SparseSym_PrecTypeInput
On entry: determines whether preconditioning is used.
${\mathbf{precon}}=\mathrm{Nag_SparseSym_NoPrec}$
No preconditioning.
${\mathbf{precon}}=\mathrm{Nag_SparseSym_Prec}$
Preconditioning.
Constraint: ${\mathbf{precon}}=\mathrm{Nag_SparseSym_NoPrec}$ or $\mathrm{Nag_SparseSym_Prec}$.
3: sigcmp – Nag_SparseSym_BisectionInput
On entry: determines whether an estimate of ${\sigma }_{1}\left(\stackrel{-}{A}\right)={‖{E}^{-1}A{E}^{-\mathrm{H}}‖}_{2}$, the largest singular value of the preconditioned matrix of the coefficients, is to be computed using the bisection method on the sequence of tridiagonal matrices $\left\{{T}_{k}\right\}$ generated during the iteration. Note that $\stackrel{-}{A}=A$ when a preconditioner is not used.
If ${\mathbf{sigmax}}>0.0$ (see below), i.e., when ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is supplied, the value of sigcmp is ignored.
${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$
${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is to be computed using the bisection method.
${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_NoBisect}$
The bisection method is not used.
If the termination criterion (3) is used, requiring ${\sigma }_{1}\left(\stackrel{-}{A}\right)$, an inexpensive estimate is computed and used (see Section 3).
Suggested value: ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_NoBisect}$.
Constraint: ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$ or $\mathrm{Nag_SparseSym_NoBisect}$.
4: norm – Nag_NormTypeInput
On entry: defines the matrix and vector norm to be used in the termination criteria.
${\mathbf{norm}}=\mathrm{Nag_OneNorm}$
Use the ${l}_{1}$ norm.
${\mathbf{norm}}=\mathrm{Nag_InfNorm}$
Use the ${l}_{\infty }$ norm.
${\mathbf{norm}}=\mathrm{Nag_TwoNorm}$
Use the ${l}_{2}$ norm.
Suggested values:
• if ${\mathbf{iterm}}=1$, ${\mathbf{norm}}=\mathrm{Nag_InfNorm}$;
• if ${\mathbf{iterm}}=2$, ${\mathbf{norm}}=\mathrm{Nag_TwoNorm}$.
Constraints:
• if ${\mathbf{iterm}}=1$, ${\mathbf{norm}}=\mathrm{Nag_OneNorm}$, $\mathrm{Nag_InfNorm}$ or $\mathrm{Nag_TwoNorm}$;
• if ${\mathbf{iterm}}=2$, ${\mathbf{norm}}=\mathrm{Nag_TwoNorm}$.
5: weight – Nag_SparseSym_WeightInput
On entry: specifies whether a vector $w$ of user-supplied weights is to be used in the vector norms used in the computation of termination criterion (2) (${\mathbf{iterm}}=1$): ${{‖v‖}_{p}}^{\left(w\right)}={‖{v}^{\left(w\right)}‖}_{p}$, where ${v}_{\mathit{i}}^{\left(w\right)}={w}_{\mathit{i}}{v}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$. The suffix $p=1,2,\infty $ denotes the vector norm used, as specified by the argument norm. Note that weights cannot be used when ${\mathbf{iterm}}=2$, i.e., when criterion (3) is used.
${\mathbf{weight}}=\mathrm{Nag_SparseSym_Weighted}$
User-supplied weights are to be used and must be supplied on initial entry to nag_sparse_herm_basic_solver (f11gsc).
${\mathbf{weight}}=\mathrm{Nag_SparseSym_UnWeighted}$
All weights are implicitly set equal to one. Weights do not need to be supplied on initial entry to nag_sparse_herm_basic_solver (f11gsc).
Suggested value: ${\mathbf{weight}}=\mathrm{Nag_SparseSym_UnWeighted}$.
Constraints:
• if ${\mathbf{iterm}}=1$, ${\mathbf{weight}}=\mathrm{Nag_SparseSym_Weighted}$ or $\mathrm{Nag_SparseSym_UnWeighted}$;
• if ${\mathbf{iterm}}=2$, ${\mathbf{weight}}=\mathrm{Nag_SparseSym_UnWeighted}$.
6: iterm – IntegerInput
On entry: defines the termination criterion to be used.
${\mathbf{iterm}}=1$
Use the termination criterion defined in (2) (both conjugate gradient and Lanczos (SYMMLQ) methods).
${\mathbf{iterm}}=2$
Use the termination criterion defined in (3) (Lanczos method (SYMMLQ) only).
Suggested value: ${\mathbf{iterm}}=1$.
Constraints:
• if ${\mathbf{method}}=\mathrm{Nag_SparseSym_CG}$, ${\mathbf{iterm}}=1$;
• if ${\mathbf{method}}=\mathrm{Nag_SparseSym_SYMMLQ}$, ${\mathbf{iterm}}=1$ or $2$.
7: n – IntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}>0$.
8: tol – doubleInput
On entry: the tolerance $\tau $ for the termination criterion.
If ${\mathbf{tol}}\le 0.0$, $\tau =\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(\sqrt{\epsilon },\sqrt{n}\epsilon \right)$ is used, where $\epsilon $ is the machine precision.
Otherwise $\tau =\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{tol}},10\epsilon ,\sqrt{n}\epsilon \right)$ is used.
Constraint: ${\mathbf{tol}}<1.0$.
9: maxitn – IntegerInput
On entry: the maximum number of iterations.
Constraint: ${\mathbf{maxitn}}>0$.
10: anorm – doubleInput
On entry: if ${\mathbf{anorm}}>0.0$, the value of ${‖A‖}_{p}$ to be used in the termination criterion (2) (${\mathbf{iterm}}=1$).
If ${\mathbf{anorm}}\le 0.0$, ${\mathbf{iterm}}=1$ and ${\mathbf{norm}}=\mathrm{Nag_OneNorm}$ or $\mathrm{Nag_InfNorm}$, then ${‖A‖}_{1}={‖A‖}_{\infty }$ is estimated internally by nag_sparse_herm_basic_solver (f11gsc).
If ${\mathbf{iterm}}=2$, then anorm is not referenced.
Constraint: if ${\mathbf{iterm}}=1$ and ${\mathbf{norm}}=\mathrm{Nag_TwoNorm}$, ${\mathbf{anorm}}>0.0$.
11: sigmax – doubleInput
On entry: if ${\mathbf{sigmax}}>0.0$, the value of ${\sigma }_{1}\left(\stackrel{-}{A}\right)={‖{E}^{-1}A{E}^{-\mathrm{H}}‖}_{2}$.
If ${\mathbf{sigmax}}\le 0.0$, ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is estimated by nag_sparse_herm_basic_solver (f11gsc) when either ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$ or termination criterion (3) (${\mathbf{iterm}}=2$) is employed, though it will be used only in the latter case.
Otherwise, sigmax is not referenced.
12: sigtol – doubleInput
On entry: the tolerance used in assessing the convergence of the estimate of ${\sigma }_{1}\left(\stackrel{-}{A}\right)={‖\stackrel{-}{A}‖}_{2}$ when the bisection method is used.
If ${\mathbf{sigtol}}\le 0.0$, the default value ${\mathbf{sigtol}}=0.01$ is used. The actual value used is $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{sigtol}},\epsilon \right)$.
If ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_NoBisect}$ or ${\mathbf{sigmax}}>0.0$, then sigtol is not referenced.
Suggested value: ${\mathbf{sigtol}}=0.01$ should be sufficient in most cases.
Constraint: if ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$ and ${\mathbf{sigmax}}\le 0.0$, ${\mathbf{sigtol}}<1.0$.
13: maxits – IntegerInput
On entry: the maximum iteration number $k={\mathbf{maxits}}$ for which ${\sigma }_{1}\left({T}_{k}\right)$ is computed by bisection (see also Section 3). If ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_NoBisect}$ or ${\mathbf{sigmax}}>0.0$, then maxits is not referenced.
Suggested value: ${\mathbf{maxits}}=\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(10,n\right)$ when sigtol is of the order of its default value $\left(0.01\right)$.
Constraint: if ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$ and ${\mathbf{sigmax}}\le 0.0$, $1\le {\mathbf{maxits}}\le {\mathbf{maxitn}}$.
14: monit – IntegerInput
On entry: if ${\mathbf{monit}}>0$, the frequency at which a monitoring step is executed by nag_sparse_herm_basic_solver (f11gsc): the current solution and residual iterates will be returned by nag_sparse_herm_basic_solver (f11gsc) and a call to nag_sparse_herm_basic_diagnostic (f11gtc) made possible every monit iterations, starting from iteration number monit. Otherwise, no monitoring takes place. There are some additional computational costs involved in monitoring the solution and residual vectors when the Lanczos method (SYMMLQ) is used.
Constraint: ${\mathbf{monit}}\le {\mathbf{maxitn}}$.
15: lwreq – Integer *Output
On exit: the minimum amount of workspace required by nag_sparse_herm_basic_solver (f11gsc). (See also Section 5 in nag_sparse_herm_basic_solver (f11gsc).)
16: work[lwork] – ComplexCommunication Array
On exit: the array work is initialized by nag_sparse_herm_basic_setup (f11grc). It must not be modified before calling the next function in the suite, namely nag_sparse_herm_basic_solver (f11gsc).
17: lwork – IntegerInput
On entry: the dimension of the array work.
Constraint: ${\mathbf{lwork}}\ge 120$.
Note: although the minimum value of lwork ensures the correct functioning of nag_sparse_herm_basic_setup (f11grc), a larger value is required by the other functions in the suite, namely nag_sparse_herm_basic_solver (f11gsc) and nag_sparse_herm_basic_diagnostic (f11gtc). The required value is as follows:
Method Requirements
CG ${\mathbf{lwork}}=120+5n+p$
SYMMLQ ${\mathbf{lwork}}=120+6n+p$
where
• $p=2×\left({\mathbf{maxits}}+1\right)$, when an estimate of ${\sigma }_{1}\left(A\right)$ (sigmax) is computed;
• $p=0$, otherwise.
18: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
6 Error Indicators and Warnings
NE_BAD_PARAM
On entry, argument number $〈\mathit{\text{value}}〉$ had an illegal value.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_CONSTRAINT
On entry, ${\mathbf{norm}}=〈\mathit{\text{value}}〉$, ${\mathbf{iterm}}=〈\mathit{\text{value}}〉$ and ${\mathbf{anorm}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{iterm}}=1$ and ${\mathbf{norm}}=\mathrm{Nag_TwoNorm}$, ${\mathbf{anorm}}>0.0$.
On entry, ${\mathbf{sigcmp}}=〈\mathit{\text{value}}〉$, ${\mathbf{maxitn}}=〈\mathit{\text{value}}〉$, ${\mathbf{sigmax}}=〈\mathit{\text{value}}〉$ and ${\mathbf{maxits}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$ and ${\mathbf{sigmax}}\le 0.0$, $1\le {\mathbf{maxits}}\le {\mathbf{maxitn}}$.
NE_ENUM_INT
On entry, ${\mathbf{method}}=〈\mathit{\text{value}}〉$ and ${\mathbf{iterm}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{method}}=\mathrm{Nag_SparseSym_CG}$, ${\mathbf{iterm}}=1$.
On entry, ${\mathbf{method}}=〈\mathit{\text{value}}〉$ and ${\mathbf{iterm}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{method}}=\mathrm{Nag_SparseSym_SYMMLQ}$, ${\mathbf{iterm}}=1$ or $2$.
NE_ENUM_REAL_2
On entry, ${\mathbf{sigcmp}}=〈\mathit{\text{value}}〉$, ${\mathbf{sigmax}}=〈\mathit{\text{value}}〉$ and ${\mathbf{sigtol}}=〈\mathit{\text{value}}〉$.
Constraint: if ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$ and ${\mathbf{sigmax}}\le 0.0$, ${\mathbf{sigtol}}<1.0$.
NE_INT
On entry, ${\mathbf{lwork}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{lwork}}\ge 120$.
On entry, ${\mathbf{maxitn}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{maxitn}}>0$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}>0$.
NE_INT_2
On entry, ${\mathbf{monit}}=〈\mathit{\text{value}}〉$ and ${\mathbf{maxitn}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{monit}}\le {\mathbf{maxitn}}$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_OUT_OF_SEQUENCE
nag_sparse_herm_basic_setup (f11grc) has been called out of sequence: either nag_sparse_herm_basic_setup (f11grc) has been called twice or nag_sparse_herm_basic_solver (f11gsc) has not terminated its current task.
NE_REAL
On entry, ${\mathbf{tol}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{tol}}<1.0$.
Not applicable.
8 Further Comments
When ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is not supplied (${\mathbf{sigmax}}\le 0.0$) but it is required, it is estimated by nag_sparse_herm_basic_solver (f11gsc) using either of the two methods described in Section 3, as specified by the argument sigcmp. In particular, if ${\mathbf{sigcmp}}=\mathrm{Nag_SparseSym_Bisect}$, then the computation of ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is deemed to have converged when the differences between three successive values of ${\sigma }_{1}\left({T}_{k}\right)$ differ, in a relative sense, by less than the tolerance sigtol, i.e., when
$max σ 1 k - σ 1 k - 1 σ 1 k , σ 1 k - σ 1 k - 2 σ 1 k ≤ sigtol .$
The computation of ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ is also terminated when the iteration count exceeds the maximum value allowed, i.e., $k\ge {\mathbf{maxits}}$.
Bisection is increasingly expensive with increasing iteration count. A reasonably large value of sigtol, of the order of the suggested value, is recommended and an excessive value of maxits should be avoided. Under these conditions, ${\sigma }_{1}\left(\stackrel{-}{A}\right)$ usually converges within very few iterations.
9 Example
This example solves a complex Hermitian system of simultaneous linear equations using the conjugate gradient method, where the matrix of the coefficients $A$, has a random sparsity pattern. An incomplete Cholesky preconditioner is used (nag_sparse_sym_chol_fac (f11jac) and nag_sparse_sym_precon_ichol_solve (f11jbc)).
9.1 Program Text
Program Text (f11grce.c)
9.2 Program Data
Program Data (f11grce.d)
9.3 Program Results
Program Results (f11grce.r)
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http://particlephd.wordpress.com/2009/02/07/the-spin-connection/
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# High Energy PhDs
A discussion of particle physics and strings
February 7, 2009
## The spin connection
Posted by Sparticles under Field Theory
[4] Comments
I’ve been spending some time thinking about spinors on curved spacetime. There exists a decent set of literature out there for this, but unfortunately it’s scattered across different `cultures’ like a mathematical Tower of Babel. Mathematicians, general relativists, string theorists, and particle physicists all have a different set of tools and language to deal with spinors.
Particle physicists — the community from which I hail — are the most recent to use curved-space spinors in mainstream work. It was only a decade ago that the Randall-Sundrum model for a warped extra dimension was first presented in which the Standard Model was confined to a (3+1)-brane in a 5D Anti-deSitter spacetime. Shortly after, flavor constraints led physicists to start placing fields in the bulk of the RS space. Grossman and Neubert were among the first to show how to place fermion fields in the bulk. The fancy new piece of machinery (by then an old hat for string theorists and a really old hat for relativists) was the spin connection which allows us to connect the flat-space formalism for spinors to curved spaces. [I should make an apology: supergravity has made use of this formalism for some time now, but I unabashedly classify supergravitists as effective string theorists for the sake of argument.]
One way of looking at the formalism is that spinors live in the tangent space of a manifold. By definition this space is flat, and we may work with spinors as in Minkowski space. The only problem is that one then wants to relate the tangent space at one spacetime point to neighboring points. For this one needs a new kind of covariant derivative (i.e. a new connection) that will translate tangent space spinor indices at one point of spacetime to another.
By the way, now is a fair place to state that mathematicians are likely to be nauseous at my “physicist” language… it’s rather likely that my statements will be mathematically ambiguous or even incorrect. Fair warning.
Mathematicians will use words like the “square root of a principle fiber bundle” or “repere mobile” (moving frame) to refer to this formalism in differential geometry. Relativists and string theorists may use words like “tetrad” or “vielbein,” the latter of which has been adopted by particle physicists.
A truly well-written “for physicists” exposition on spinors can be found in Green, Schwartz, and Witten, volume II section 12.1. It’s a short section that you can read independently of the rest of the book. I will summarize their treatment in what follows.
We would like to introduce the a basis of orthonormal vectors at each point in spacetime, $e^a_\mu(x)$, which we call the vielbein. This translates to `many legs’ in German. One will often also hear the term vierbein meaning `four legs,’ or `funfbein’ meaning `five legs’ depending on what dimensionality of spacetime one is working with. The index $\mu$ refers to indices on the spacetime manifold (which is curved in general), while the index $a$ labels the different basis vectors.
If this makes sense, go ahead and skip this paragraph. Otherwise, let me add a few words. Imagine the tangent space of a manifold. We’d like a set of basis vectors for this tangent space. Of course, whatever basis we’re using for the manifold induces a basis on the tangent space, but let’s be more general. Let us write down an arbitrary basis. Each basis vector has $n$ components, where $n$ is the dimensionality of the manifold. Thus each basis vector gets an undex from 1 to $n$, which we call $mu$. The choice of this label is intentional, the components of this basis map directly (say, by exponentiation) to the manifold itself, so these really are indices relative to the basis on the manifold. We can thus write a particular basis vector of the tangent space at $x$ as $e_\mu(x)$. How many basis vectors are there for the tangent space? There are $n$. We can thus label the different basis vectors with another letter, $a$. Hence we may write our vector as $e^a_\mu(x)$.
The point, now, is that these objects allow us to convert from manifold coordinates to tangent space coordinates. (Tautological sanity check: the $a$ are tangent space coordinates because they label a basis for the tangent space.) In particular, we can go from the curved-space indices of a warped spacetime to flat-space indices that spinors understand. The choice of an orthonormal basis of tangent vectors means that
$e^a_\mu (x) e_{a\nu}(x) = g_{\mu\nu}(x)$,
where the $a$ index is raised and lowered with the flat space (Minkowski) metric. In this sense the vielbeins can be thought of as `square roots’ of the metric that relate flat and curved coordinates. (Aside: this was the first thing I ever learned at a group meeting as a grad student.)
Now here’s the good stuff: there’s nothing `holy’ about a particular orientation of the vielbein at a particular point of spacetime. We could have arbitrarily defined the tangent space z-direction (i.e. $a = 3$, not \$\mu=3\$) pointing in one direction ($x_\mu=(0,0,0,1)$) or another ($x_\mu=(0,1,0,0)$) relative to the manifold’s basis so long as the two directions are related by a Lorentz transformation. Thus we have an $SO(3,1)$ symmetry (or whatever symmetry applies to the manifold). Further, we could have made this arbitrary choice independently for each point in spacetime. This means that the symmetry is local, i.e. it is a gauge symmetry. Indeed, think back to handy definitions of gauge symmetries in QFT: this is an overall redundancy in how we describe our system, it’s a `non-physical’ degree of freedom that needs to be `modded out’ when describing physical dynamics.
Like any other gauge symmetry, we are required to introduce a gauge field for the Lorentz group, which we shall call $\omega_{\mu\phantom{a}\nu}^{\phantom{mu}a}(x)$. From the point of view of Riemannian geometry this is just the connection, so we can alternately call this creature the spin connection. Note that this is all different from the (local) diffeomorphism symmetry of general relativity, for which we have the Christoffel connection.
What do we know about the spin connection? If we want to be consistent with general relativity while adding only minimal structure (which GSW notes is not always the case), we need to impose consistency when we take covariant derivatives. In particular, any vector field with manifold indices ($V^\mu(x)$) can now be recast as a vector field with tangent-space indices ($V^a = e^\mu_a(x)V^\mu(x)$). By requiring that both objects have the same covariant derivative, we get the constraint
$D_\mu e^a_\mu(x) = 0$.
Note that the covariant derivative is defined as usual for multi-index objects: a partial derivative followed by a connection term for each index. For the manifold index there’s a Christoffel connection, while for the tangent space index there’s a spin connection:
$D_\mu e^a_\mu(x) = \partial_\mu e^a_\nu - \Gamma^\lambda_{\mu\nu}e^a_\nu + \omega_{\mu\phantom{a}b}^{\phantom\mu a}e^b_\nu$.
This turns out to give just enough information to constrain the spin connection in terms of the vielbeins,
$\omega^{ab}_\mu = \frac 12 g^{\rho\nu}e^{[a}_{\phantom{a}\rho}\partial_{\nu}e^{b]}_{\phantom{b]}\nu}+ \frac 14 g^{\rho\nu}g^{\tau\sigma}e^{[a}_{\phantom{[a}\rho}e^{b]}_{\phantom{b]}\tau}\partial_{[\sigma}e^c_{\phantom{c}\nu]}e^d_\mu\eta_{cd}$,
this is precisely equation (11) of hep-ph/980547 (EFT for a 3-Brane Universe, by Sundrum) and equation ( 4.28 ) of hep-ph/0510275 (TASI Lectures on EWSB from XD, Csaki, Hubisz, Meade). I recommend both references for RS model-building, but note that neither of them actually explain where this equation comes from (well, the latter cites the former)… so I thought it’d be worth explaining this explicitly. GSW makes a further note that the spin connection can be using the torsion since they are the only terms that survive the antisymmetry of the torsion tensor.
Going back to our original goal of putting fermions on a curved spacetime, in order to define a Clifford algebra on such a spacetime it is now sufficient to consider objects $\Gamma_mu(x) = e^a_\mu(x)\gamma_a$, where the right-hand side contains a flat-space (constant) gamma matrix with its index converted to a spacetime index via the position-dependent vielbein, resulting in a spacetime gamma matrix that is also position dependent (left-hand side). One can see that indeed the spacetime gamma matrices satisfy the Clifford algebra with the curved space metric, $\{\Gamma_\mu(x),\Gamma_\nu(y)\} = 2g_{\mu\nu}(x)$.
There’s one last elegant thought I wanted to convey from GSW. In a previous post we mentioned the role of topology on the existence of the (quantum mechanical) spin representation of the Lorentz group. Now, once again, topology becomes relevant when dealing with the spin connection. When we wrote down our vielbeins we assumed that it was possible to form a basis of orthonormal vectors on our spacetime. A sensible question to ask is whether this is actually valid globally (rather than just locally). The answer, in general, is no. One simply has to consider the “hairy ball” theorem that states that one cannot have a continuous nowhere-vanishing vector field on the 2-sphere. Thus one cannot always have a nowhere-vanishing global vielbein.
Topologies that can be covered by a single vielbein are actually `comparatively scarce’ and are known as parallelizable manifolds. For non-parallelizable manifolds, the best we can do is to define vielbeins on local regions and patch them together via Lorentz transformations (`transition functions’) along their boundary. Consistency requires that in a region with three overlapping patches, the transition from patch 1 to 2, 2 to 3, and then from 3 to 1 is the identity. This is indeed the case.
Spinors must also be patched together along the manifold in a similar way, but we run into problems. The consistency condition on a triple-overlap region is no longer always true since the double-valuedness of the spinor transformation (i.e. the spinor transformation has a sign ambiguity relative to the vector transformation). If it is possible to choose signs on the spinor transformations such that the consistency condition always holds, then the manifold is known as a spin manifold and is said to admit a spin structure. In order to have a consistent theory with fermions, it is necessary to restrict to a spin manifold.
### 4 Responses to “The spin connection”
1. Sparticles Says:
February 8, 2009 at 5:07 am
For more information about the spin connection in the language of quantum field theory, Bertlemann’s “Anomalies in Quantum Field Theory” seems very comprehensive.
One can also refer to the usual mathematical physics references in the field: Nakahara and Eguchi-Gilkey-Hanson.
2. ncr Says:
February 9, 2009 at 1:28 am
Nice post
A couple of remarks:
- Pedantry: A few indices are not in their right places, e.g. vector field with tangent space indices, cov. derivative and its action on the vielbein.
- I think the connection needs to be antisymmetrised in nu and mu (lower indices in the first term).
One thing I don’t understand is “consistent with general relativity while adding only minimal structure”. Do you know if this means, for instance, that it does not change the geodesic equation?
3. Sparticles Says:
February 16, 2009 at 4:01 pm
Indeed, I think you’re right, NCR… but you’ll forgive me if I’m too lazy busy to fix the indices.
I also am unsure about ‘minimal structure’.
4. Jeremy Says:
May 5, 2010 at 2:58 pm
I say, that post is really a great job. Thank you very much… It has made some important things much more clear in my mind.
Comments are closed.
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http://mathoverflow.net/questions/10086/the-category-of-representations-of-a-group
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## The Category of Representations of a Group
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Do people study the category of representations of a compact finite group (not just irreducible ones)? I'm more interested in small cases like S_3 and SU(2) but I'd be curious about general cases like $S_n, SU(n)$. These must be tensor categories since - well... they admit tensor products and direct sums. Can these representations be considered a ring?
** What does this category look like in the case of S_3?
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Pick pretty much any textbook which deals with the representation of finite groups to find a description of the modules category of S_3... – Mariano Suárez-Alvarez Dec 30 2009 at 3:32
## 3 Answers
The category Rep(G) is a symmetric tensor category, and it is a theorem that this structure determines G (Tannaka-Krein duality, but I'm not familiar with it). Each object is dualizable because there is a dual representation, from which appropriate evaluation and coevaluation maps can be constructed. The unital object is the trivial representation.
(Incidentally, a symmetric tensor category is a kind of categorification of a commutative ring.)
In the finite case, It is a fusion category because there are finitely many simple objects, there is rigidity as stated above, and because $\mathbb{C}[G]$ (or more generally $k[G]$ for $k$ a field of characteristic prime to the order of $G$) is a semisimple algebra (Maschke's theorem). Semisimplicity is also true for continuous finite-dimensional representations of compact groups by the same "averaging" argument used in Maschke's theorem, though the group algebra is not necessarily semisimple. In the finite group case, the number of simple objects is equal to the number of conjugacy classes in G. In the infinite group case, for instance, the rotation group $SO(n)$ has infinitely many irreducible finite-dimensional representations obtained by the action on the spherical harmonics of various degrees (i.e. harmonic polynomials restricted to the sphere).
In the case of S_n, generators of this ring can be indexed by the Young diagrams of size n. The relations are given by the tensor product rules, and while the Pieri rule gives a special case of this, as far as I know, there is no a simple general way to express the tensor product of two representations associated to Young diagrams as a sum of Young diagrams.
However, there are apparently algorithms to do this.
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3
It's also worth mentioning that the dual and tensor structures "come from" the Hopf algebra structure of the group algebra, namely the antipode and comultiplication respectively. The same thing is true of the category of representations of any Hopf algebra. – Qiaochu Yuan Dec 30 2009 at 4:28
1
In the usual statements of TK duality, the "category of representations" data consists of a (symmetric, etc.) category along with a faithful functor to VECT. See for example the paper "An introduction to Tannaka duality and quantum groups" by Ross and Street. – Theo Johnson-Freyd Jan 4 2010 at 4:36
3
@Theo: Don't you mean Joyal and Street? (Ross is Street's first name.) You can find the paper here: maths.mq.edu.au/~street/CT90Como.pdf – José Figueroa-O'Farrill Feb 3 2010 at 23:50
### 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 what sense is $SU(n)$ a finite group? Perhaps your question is about compact (including finite) groups? In that case, the representation theory is very well understood. For compact groups, my favourite reference is the book by Bröcker and tom Dieck Represetations of compact Lie groups. Section II.7 describes the representation ring in general and introduces the Adams operations,... and then Section VI.5 contains the structure of the representation rings for the simply-connected classical Lie groups.
I should have added that the representations only form a semiring under direct sum and tensor product. The representation ring is obtained from this by the standard Grothendieck construction. You essentially have to add so-called virtual representations.
The book I mentioned does not treat the case of infinite-dimensional representations, though. Perhaps someone here can say more about this case.
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Do they not form a ring in the finite group case? – Grétar Amazeen Dec 30 2009 at 3:28
No, you don't have additive inverses. You need to allow an irreducible representation to occur with negative multiplicity. – Qiaochu Yuan Dec 30 2009 at 3:40
Of course, thanks. I was thinking of formal combinations, but somehow wasn't :) – Grétar Amazeen Dec 30 2009 at 3:48
In the finite case at least, the answer is yes. It is a tensor category, and one fundamental result is that the category of representations of G is equivalent, as tensor categories, to the category of $\mathbb{C}G$-modules, where $\mathbb{C}G$ is the group algebra. The same holds if you just look at the finite dimensional representations, and finite dimensional modules if I remember correctly.
As for considering it as a ring, the answer is also yes, if you allow formal combinations of them (so called virtual representations) (thanks for pointing that out Jose and Qiaochu!). It has a basis the irreducible representations, and the multiplication is the tensor product, with the trivial representation as 1.
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What's your definition of the group algebra for an infinite compact group? I've seen a few slightly different ones. – Qiaochu Yuan Dec 30 2009 at 3:09
I'm not sure, I'm only answering for the finite case. – Grétar Amazeen Dec 30 2009 at 3:11
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http://math.stackexchange.com/questions/8186/when-do-the-multiples-of-two-primes-span-all-large-enough-natural-numbers
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# When do the multiples of two primes span all large enough natural numbers?
It is well-known that given two primes $p$ and $q$, $pZ + qZ = Z$ where $Z$ stands for all integers. It seems to me that the set of natural number multiples, i.e. $pN + qN$ also span all natural numbers that are large enough. That is, there exists some $K>0$, such that $$pN + qN = [K,K+1,...).$$
My question is, given $p$ and $q$, can we get a upper bound on $K$?
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I assume you mean pN + qN contains [K, K+1, ...). It's not hard to see that below K there are gaps. – Qiaochu Yuan Oct 28 '10 at 15:01
5
This works the same whether p and q are prime or not, as long as their greatest common divisor is 1. – Ross Millikan Oct 28 '10 at 15:58
## 3 Answers
$K = pq + 1$ if $\mathbb{N} = \{ 1, 2, 3, ... \}$, and $K = pq - p - q + 1$ if $\mathbb{N} = \{ 0, 1, 2, 3, ... \}$. This is known as the coin problem, or Frobenius problem (and you only need $p, q$ relatively prime). It frequently appears on middle- and high-school math competitions.
Edit: I completely misremembered how hard the proof is. Here it is. If $n$ is at least $pq+1$, then the positive integers
$$n-p, n-2p, ... n-qp$$
have distinct residue classes $\bmod q$, so one of them must be divisible by $q$. On the other hand, it's not hard to see that $pq$ itself cannot be written in the desired way.
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Qiaochu: Typo? $K=pq-p-q+1$. – Byron Schmuland Oct 28 '10 at 15:16
1
@Byron: this would be true if the OP is allowing non-negative integer multiples, but the OP wants positive integer multiples. – Qiaochu Yuan Oct 28 '10 at 15:19
1
@Qiaochu: the OP said "nautral numbers". Please tell me you're not one of those misguided souls who doesn't think $0$ is a natural number! (More seriously, you should probably clarify this in your response. You might as well give the answer both ways.) – Pete L. Clark Oct 28 '10 at 16:22
@Pete: I'm agnostic. Generally I use Z_{\ge 0} for the non-negative integers. I'll add the clarification. – Qiaochu Yuan Oct 28 '10 at 16:29
1
Nice answer. However, to ensure that the coefficient of p is positive, I think you should be saying that n-p,n-2p,...,n-qp have distinct residue classes mod q. – George Lowther Nov 6 '10 at 20:59
show 1 more comment
HINT $\rm\ \ p\ (p^{-1}\: mod\ q) + q\ (q^{-1}\: mod\ p)\ =\ pq+1\$ since it's $1$ mod $\rm\:p,q\:$ and $\rm >2\:$ and $\rm < 2\:pq$
To represent largers numbers keep adding $\rm\ 1 = ap+bq\$ while, if need be, adding $\rm\:\pm (q ,-p)\:$ to the coefficients to keep them positive.
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This follows easily by Bézout's lemma (as Qiaochu notes, x and y only need to be coprime to generate the unit ideal (indeed, this is the proper generalization of the term "coprime")).
We can ever so-slightly strengthen this to show that $\mathbf{Z}$ is a PID (this is the true content of Bézout's lemma).
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http://physics.stackexchange.com/questions/8095/measuring-acceleration-of-earth-due-to-its-fall-around-the-sun?answertab=votes
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# Measuring acceleration of earth due to its fall around the sun
Every orbiting of a satellite around a mass is nothing else but a constant fall - and therefore acceleration - towards this mass. In a way it is a "falling around" that mass.
My question
Is it possible to measure this acceleration on earth due to its "falling around" the sun?
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– sum1stolemyname Apr 5 '11 at 14:24
We can measure the effects of differential accelleration, as tidal forces. – Omega Centauri Apr 5 '11 at 15:32
1
I can't wait to see someone answer something along the lines "because of the equivalence principle, one can measure only local accelerations relative to an inertial frame" and then someone mumble something back about Mach principle... – lurscher Apr 5 '11 at 17:19
## 2 Answers
The answer depends in part on what you mean by "measure". You can certainly calculate the acceleration using the laws of kinematics and dynamics, as Lawrence B. Crowell points out in his answer. Does that, in your mind, count as "measuring the acceleration"?
Here's a much more direct recipe that I think would uncontroversially count as measuring the acceleration. Pick a set of fixed stars, and measure the velocity of the Earth relative to them using the Doppler effect. Any one measurement only gives you one component of ${\bf v}$, but if you measure a few, you can get the full vector. Do this twice, at two different times, and calculate $\Delta{\bf v}/\Delta t$.
With current technology it would be very easy to do this to the required accuracy. In fact, astronomers are making precisely these required measurements all the time -- not with the purpose of measuring the acceleration, but for a wide variety of other reasons. In fact, astronomers often deliberately subtract out the time-varying Doppler shifts due to Earth's changing velocities, because what they're interested in are (often much smaller) changes due to the actual motions of the stars.
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Thank you. Could you also measure the acceleration itself on earth? Acceleration is always absolute and can be felt (like in an accelerating car), so it should be possible with very sensitive instruments, shouldn't it? – vonjd Apr 5 '11 at 18:19
3
No -- that's precisely what you can't do, because of the equivalence principle. The equivalence principle says, among other things, that objects in free fall "feel like" they're not accelerating. Since the Earth's motion around the Sun is a free-fall motion, experiments done locally on the Earth can't reveal this acceleration. – Ted Bunn Apr 5 '11 at 18:25
Thank you, yeah - that clarifies the matter! – vonjd Apr 5 '11 at 19:49
The acceleration can be found by a number of means. $F~=~ma$ with the Newtonian law of gravity and the centripetal acceleration $a~=~v^2/r$, $$m\frac{v^2}{r}~=~-\frac{GMm}{r^2}.$$ This gives $v~=~\sqrt{GM/r}$, which is $v~=~29.5km/s$ or $v~=~2.95\times 10^4m/s$, for the mass of the sun and $r~=~1.5\times 10^11m$. The acceleration is then $0.0058m/s^2$
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http://mathhelpforum.com/trigonometry/140871-tricky-sin-equation.html
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# Thread:
1. ## Tricky sin equation
Hello!
This is the equation Im stuck with:
$2sin (0.5x) = 3/3x+2$
Im not sure where to start, because I cant take out any of the x values,
should i let 0.5x be a completely new vale like 'y' and but then i will still be stuck
Thank you!
2. Originally Posted by appleseed
Hello!
This is the equation Im stuck with:
$2sin (0.5x) = 3/3x+2$
Im not sure where to start, because I cant take out any of the x values,
should i let 0.5x be a completely new vale like 'y' and but then i will still be stuck
Thank you!
What is required in the problem? Do you want to find the value of x? If yes, how much decimal places in x is required ?
In the form of infinite series sinθ can be written as
sinθ = θ - (θ)^3/3! + (θ)^5/5! and so on.
3. Originally Posted by sa-ri-ga-ma
What is required in the problem? Do you want to find the value of x? If yes, how much decimal places in x is required ?
In the form of infinite series sinθ can be written as
sinθ = θ - (θ)^3/3! + (θ)^5/5! and so on.
Hello!
Thank you so much for the reply!!!
Yes i am asked to find the value of x and the answer is required to 3 s.f.
It might be better if i give you the context of the equation.
''''''''''''There are two equations g(x) and f(x), There are two values of x for which the gradient of f is equal to the gradient
of g. Find both these values of x. '''''''''"
So i've already found the derivative of the two equations and i've checked that its correct, and then i equated the two derivatives, but i dont know how to find x.
I was thinking that maybe it could be done by trig identities, lol but i dont know how though
anyways Thank you! !!
4. Hello appleseed
There are infinitely many solutions to the equation
$2\sin(\tfrac12x)=\frac{3}{3x+2}$
which is what I assume you meant. (You should have written 3/(3x+2) if you don't know how to write it using LaTeX.)
Have a look at the diagram I've attached, where I've plotted the graph of each side separately.
You won't be able to solve an equation like this to get exact answers. You'll have to use a numerical method - e.g. Newton-Raphson. Do you know how?
Grandad
Attached Thumbnails
5. Originally Posted by Grandad
Hello appleseed
There are infinitely many solutions to the equation
$2\sin(\tfrac12x)=\frac{3}{3x+2}$
which is what I assume you meant. (You should have written 3/(3x+2) if you don't know how to write it using LaTeX.)
Have a look at the diagram I've attached, where I've plotted the graph of each side separately.
You won't be able to solve an equation like this to get exact answers. You'll have to use a numerical method - e.g. Newton-Raphson. Do you know how?
Grandad
Hi Grandad!!!
I think i've heard of Newton-Raphson before, but i cant remember exactly how to find the answer using it. Could you briefly go over the steps please?
Thanks!
6. ## Newton-Raphson method
Hello appleseed
Originally Posted by appleseed
Hi Grandad!!!
I think i've heard of Newton-Raphson before, but i cant remember exactly how to find the answer using it. Could you briefly go over the steps please?
Thanks!
If you've not had any practice with this method, this is not the easiest example to start with.
If you want to get a solution to the equation
$f(x) = 0$
and you have a first approximation, $x = a_1$, to the answer, then a second approximation, $a_2$, is given by
$a_2=a_1-\frac{f(a_1)}{f'(a_1)}$
In this case you'll have to let
$f(x) = 2\sin(\tfrac12x)-\frac{3}{3x+2}$
So
$f'(x) = \cos(\tfrac12x) +\frac{9}{(3x+2)^2}$
From the graph, there's a solution close to $x = 1$. So with $a_1 = 1$ we get
$a_2 = 1-\frac{f(1)}{f'(1)}$
$\approx 0.7$
... and so on.
The next approximation, $a_3$, uses $0.7$ as the starting value:
$a_3 = 0.7 -\frac{f(0.7)}{f'(0.7)}$
$\approx 0.73$ (to 2 d.p.)
Repeating the process, to 4 d.p., I make the answer $0.7314$.
The next positive solution is around $x = 6$. To 4 d.p., the solution converges very quickly to $6.1361$.
Grandad
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http://nrich.maths.org/7197/index
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nrich enriching mathematicsSkip over navigation
### Folium of Descartes
Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.
### Witch of Agnesi
Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.
### Quick Route
What is the quickest route across a ploughed field when your speed around the edge is greater?
# Patterns of Inflection
##### Stage: 5 Challenge Level:
This problem is a good place to put into action the analogies for calculus explored in Calculus Analogies, so you might wish to consider that problem first.
Draw sets of coordinate axes $x-y$ and sketch a few simple smooth curves with varying numbers of turning points.
Mark on each curve the approximate locations of the maxima $M$, the minima $m$ and the points of inflection $I$.
In each case list the order in which the maxima, minima and points of inflection occur along the curve.
What patterns do you notice in these orders? Make a conjecture.
Test out you conjecture on the cubic equations
3x^3-6x^2+9x+11\quad\mbox{ and } 2x^3-5x^2-4x
Prove your conjecture for any cubic equation.
Extension: Consider the same problem for polynomials of order 4 or greater.
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://mathhelpforum.com/differential-geometry/99114-induction.html
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# Thread:
1. ## induction
Suppose that $R > 0$, $x_{0}>0$, and
$x_{n+1}= \frac{1}{2}(\frac{R}{x_{n}}+x_{n})$, $n\geq0$
Prove: For $n\geq1$, $x_{n}>x_{n+1}>\sqrt{R}$ and
$x_{n}-\sqrt{R}\leq\frac{1}{2^n}\frac{(x_{o}-\sqrt{R})^2}{x_{o}}$
MY attempt at a solution:
I really don't know how to go about this... at first i tried to show that if $x_{n}>x_{n+1}>\sqrt{R}$ is true then $x_{n}-x_{n+1} > 0$.
So that means
$x_{0}- \frac{1}{2}(\frac{R}{x_{0}}+x_{0})>0$
$\frac{x_{0}}{2}-\frac{R}{2x_{0}}>0$
$x_{0}-\frac{R}{x_{0}}>0$
My problem is that i don't know how big R in comparison to $x_{0}$, so unless i restrict R to be less than $x_{0}$ i can't prove that $x_{n}>x_{n+1}>\sqrt{R}$
Basically, i need help with this question... can anyone help ?
I'm doing a self study by myself so i would appreciate a detailed solution if possible.. please and thank u.
2. I believe you are right, in fact if $x_0<\sqrt{2}$ (for
instance $x_0=1$ and $R=4$ then the sequence increases in the first terms $x_1>x_0$, but $x_1>x_2>2$ and then the hypothesis seems to hold (for
$n\geq 1$ as stated).
However, if $x_0=\sqrt{R}$ we have the constant sequence $(\sqrt{R})$
I conjecture without doing any more calculations that this is the unique problem with the hypothesis, we have to require $x_0\neq \sqrt{R}$.
I suggest
a) Try to prove the inequality if $x_0>\sqrt{R}$, it seems an easy induction at least for the first one.
b) Try to show using differential calculus that $f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right)$
maps $[0,\sqrt{R})$ into $(\sqrt{R},\infty)$. This would solve the case fos "small $x_0$'s"
because it would allow to apply a)
I haven't made the calculations but it seems to me the way of attacking it.
3. Sorry i'm not very sure about what you're asking me to do.
My point was that $x_{n}>x_{n+1}$ is not true for all n if R> $x_{0}$. So do i restrict R so that this case is satistfied ?
If i do this then it would seem that the sequence is decreasing at first but then it would blow up to infinity .
The question is puzzuling to me.
4. My idea is to divide the problem in the following cases:
a) If $x_0>\sqrt{R}$ then $x_0>x_1>\cdots x_n >\sqrt{R}$ and the sequence tends to $\sqrt{R}$. I believe it follows from the inequality you obtained in the first post, by an easy induction. Observe that the inequality you got is $x_0-x_1=\frac{x_0^2-R}{2x_0}$. If $x_0>\sqrt{R}$ then you have clearly $x_0>x_1>\sqrt{R}$ and you can apply induction. For showing $x_1>\sqrt{R}$ you have to use that $f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right)$ is increasing in $[\sqrt{R},\infty)$ and $f(\sqrt{R})=\sqrt{R}$.
b) If $x_0=\sqrt{R}$ then $x_n=\sqrt{R}$ for all $n$.
c) If $x_0<\sqrt{R}$ then $x_1>\sqrt{R}>x_0$ and then $x_1>x_2>\cdots x_n >\sqrt{R}$ by the same argument of a). The key for obtaining $x_1>\sqrt{R}$ is to observe that
$f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right)$ is decreasing in $(0,\sqrt{R}]$ and $f(\sqrt{R})=\sqrt{R}$.
I mean that $f(x):=\frac{1}{2}\left(\frac{R}{x}+x\right)$, $x\in \mathbb{R}^+$ has a unique attracting point , the minimum, and that any iteration goes to this point. The case b)
is an exception to your statement since it involves strict inequalities. c) is NOT an exception since you have to show $x_n>x_{n+1}>\sqrt{R}$ for $n\geq 1$,
If it is still not clear I will try to write any step detailedly, but I don't have now the necessary time.
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http://mathoverflow.net/questions/31663/distinct-numbers-in-multiplication-table
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## Distinct numbers in multiplication table
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider multiplication table for numbers $1,2,\cdots, n$. How many different numbers are there? That is how many different numbers of the form $ij$ with $1 \le i, j \le n$ are there?
I'm interested in a formulae or an algorithm to calculate this number in time less than $O(n^2)$.
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2
A nice paper on this with plenty of references: Multiples and Divisors, by Steven Finch, available online from CiteSeer. – SJR Jul 13 2010 at 6:04
## 1 Answer
This is the multiplication table problem of Erdos. According to Kevin Ford, Integers with a divisor in $(y,2y]$, Anatomy of integers, 65-80, CRM Proc. Lecture Notes, 46, Amer Math Soc 2008, MR 2009i:11113, the number of positive integers $n\le x$, which can be written as $n=m_1m_2$, with each $m_i\le\sqrt x$, is bounded above and below by a constant times $x(\log x)^{-\delta}(\log\log x)^{-3/2}$, where $\delta=1-(1+\log\log2)/\log2$.
Erdos' work on this problem can be found (in Russian) in An asymptotic inequality in the theory of numbers, Vestnik Leningrad Univ. Mat. Mekh. i Astr. 13 (1960) 41-49.
Another reference is http://www.research.att.com/~njas/sequences/A027424 where a PARI program is given.
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Thank you for references! The PARI program implements straightforward $O(n^2)$ algorithm. Is there any algorithm which works faster than $O(n^2)$? – falagar Jul 13 2010 at 8:43
Ford's Annals of Mathematics preprint is on the arXiv: arxiv.org/abs/math/0401223 – Charles Jul 22 2010 at 4:35
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http://mathhelpforum.com/advanced-algebra/124833-solved-show-r-b.html
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# Thread:
1. ## [SOLVED] Show that r|b
if r belongs integer and r is a non zerosolution of x^2+ax+b=0 (where a,b belongs to integers), prove that r|b.
2. Originally Posted by Deepu
if r belongs integer and r is a non zerosolution of x^2+ax+b=0 (where a,b belongs to integers), prove that r|b.
We are told that $r^2 + ar + b = 0$
$\Rightarrow r^2 + ar = -b$
$\Rightarrow r(r + a) = -b$
Now what can you say?
Spoiler:
Since $r,a \in \mathbb{Z}$, we have that $(r + a) \in \mathbb{Z}$, so that $r$ divides the left side of the equation. But this means it must also divide the right side. Hence, $r|b$
3. Originally Posted by Jhevon
We are told that $r^2 + ar + b = 0$
$\Rightarrow r^2 + ar = -b$
$\Rightarrow r(r + a) = -b$
Now what can you say?
Spoiler:
Since $r,a \in \mathbb{Z}$, we have that $(r + a) \in \mathbb{Z}$, so that $r$ divides the left side of the equation. But this means it must also divide the right side. Hence, $r|b$
Thank you..
4. Originally Posted by Deepu
Thank you..
the conclusion follows even more immediately if you recall that $m|n \Longleftrightarrow n = mk \text{ for some }k \in \mathbb{Z}$
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http://mathhelpforum.com/advanced-algebra/202870-separable-element-extension-field-also-field.html
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3Thanks
• 1 Post By Vlasev
• 1 Post By Deveno
• 1 Post By Deveno
# Thread:
1. ## A separable element in an extension field is also in the field
Last problem I had from my exam:
Let $F$ be an extension field of a field $K$ whose char is $p$. Let $c$ be an element of $F$ which is separable over $K$ and for which there exists a positive integer $n$ such that $c^{p^n} \in K$. Show that $c \in K$.
The first thing that bugged me was that it wasn't clear if $F$ or $K$ had char $p$, but then it seems that whichever has char $p$ means that the other would also have char $p$. I'm also tempted to assume that $p$ is prime, but that would be presumptuous of me
I'm not sure what to do. Any help would be much appreciated
2. ## Re: A separable element in an extension field is also in the field
The characteristic of a ring $R$ is usually defined as the smallest positive integer $n$ such that adding $1_R$ $n$ times to itself gives $0_R$. If no such number exists, we say the characteristic is 0. Another way to define it is as the smallest positive integer $n$ such that adding $r\in R$ to itself $n$ times produces $0_R$, for any $r\in R$. The two definitions are equivalent, can you prove it?
Now, onto your question! IF $F$ is a field of characteristic $n$, then either $n = 0$ or $n$ is prime. (You can prove this)
It's not hard at all to see that a field $F$ and any of its extensions $E$ must have the same characteristic. They share a multiplicative identity.
As a next step, you should take the definitions of the things you are using and play around with them to see if you can make any progress.
3. ## Re: A separable element in an extension field is also in the field
I'm familiar with the definition of the characteristic of a ring (although, just wanted to point out that your first definition assumes the ring has a multiplicative identity). The definitions are equivalent since $r = 1r$.
For the second part, I haven't formally proven it, but I see how it works. If the characteristic isn't prime, then you get zero divisors, which you can't have in fields.
Now, my real problem is trying to get that $c$ is indeed inside $K$.
The first case (trivial) is if $p=0$, then $c^{0^n}=c^1=c \in K$.
So, if $p \neq 0$ (i.e. $p$ is prime), then ... I'm not sure .
I've tried several things, and I think this is the closest:
$f(x)=x^{p^n}-c^{p^n} \in K[x]$. Clearly, $f(c)=0$, so one of the irreducible factors of $f(x)$ is the monic irreducible polynomial of $c$ (let's call it $g(x)$). Thus, $g(x)$ should divide $x^{p^n}-c^{p^n} = (x-c)^{p^n}$, so $g(x)=(x-c)^{m}$ for some $0 \leq m < p^n$. But $c$ is separable over $K$, so $m=0$, thus $g(x)=x-c \in K[x]$, which implies that $c \in K$.
The part I'm not sure about is if I can say that $g(x)$ should divide $(x-c)^{p^n}$.
Is what I've got okay?
4. ## Re: A separable element in an extension field is also in the field
i think you mean m = 1 (because (x - c)0 = 1, which has no roots).
the minimal polynomial for c, m(x) always divides any polynomial f(x) for which f(c) = 0.
to see this, write:
f(x) = m(x)q(x) = r(x), where either r = 0, or deg(r) < deg(m).
then 0 = f(c) = m(c)q(c) + r(c) = 0q(c) + r(c) = r(c).
since deg(r) < deg(m) contradicts m being the minimal polynomial of c, we must have r = 0.
5. ## Re: A separable element in an extension field is also in the field
oops! yeah! I meant 1, haha, it's 3 a.m. here
On that note, I think you meant m(x)q(x) + r(x)
I definitely need to sleep soon.
I'm confused. I thought g(x) was the min. poly. of c in K[x], if it is, then you may read the next paragraph onwards. Otherwise, how do we justify that g(x) is in K[x]?
Also, I'm okay with the fact that the minimal polynomial of c divides any polynomial for which c is a root. My "issue" was interpreting that polynomial in another form. Basically, I was fine with $g(x)$ divides $x^{p^n}-c^{p^n}$, but not too sure about $g(x)$ dividing $(x-c)^{p^n}$ because I wasn't sure how you would interpret that product in $K[x]$ (i.e. if you expand the $(x-c)^{p^n}$, in $F[x]$ the inner terms would go to zero, but how would you interpret the inner terms in $K[x]$ when we don't know if $c^m$ for $0 < m < p^n$ is in $K[x]$?) Like, how would you interpret the term $acx^{p^n-1}$ in $K[x]$ when we don't know that $c$ is in $K$, so we can't take the product $ac$ (since $p|a$, we can let $ac=0c$ but still, what does that mean in $K$ when we don't know that $c$ is in $K$)
That was basically my "issue" because the expansion of $(x-c)^{p^n}$ seems to imply that $c$ is in $K$. I think you could argue that since the inner terms goes to zero, and they are linearly independent, then each term must equal zero in $K$, which means that the product is "doable" in $K$, which means that the powers of $c$ should be in $K$.
6. ## Re: A separable element in an extension field is also in the field
the fact that the "inner terms disappear" is a function of the characteristic, not the field itself.
we know that g(x) is in K[x] (by definition). since g(x) = (x - c)m in F[x] (since f(x) splits in F), and since c is separable, g(x) must be linear. that is all you need.
we don't "expand" f(x) in K[x], we expand it in F[x]. since g(x) is in K[x], and F is an extension of K, g(x) is also in F[x].
it's like this: separability of an element c MEANS it's minimal polynomial has no repeated roots. yes, these roots may lie in a larger field, but if they do they are DISTINCT.
however, f(x) has only ONE root in F[x]: c. the polynomials of the form xpn - apn in a field of characteristic p are SPECIAL.
now if c were NOT separable, then we'd be stuck. it could be that some lesser power of c lies in K, but not c itself.
you should "play around" with say, Z3 and the field of order 9. differences of two cubes (or ninth powers) in such a field are "nice", in that they are very easy to factor.
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http://mathoverflow.net/questions/85497/on-the-existence-of-the-laplace-transform
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## On the existence of the Laplace Transform [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Most of the lectures on the Laplace transform :
• define the Laplace transform Lf as a complex-valued function ;
• give the same example of function that does not admit a Laplace transform : f(t) = exp(t²) ;
• give - with no more explanation - the following proof : the integral from zero to infinity of exp(t² - st) is divergent for every real s.
But of course, an integral parameter from zero to infinity of a function g(s,t) may be divergent for every real s and convergent for some complex s.
So, it seems to me that most of the lectures on the Laplace transform are misleading.
And my question is : Why there is no clear explanation in all these lectures ?
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1
I fear this is both not a real question AND subjective and argumentative, so I am voting to close. In addition, can you find a complex $s$ for which the integral converges? – Igor Rivin Jan 12 2012 at 15:22
I second Igor's comment. Furthermore, you may consider to post a reformulation your question at math.stackexchange.com. – Dirk Jan 12 2012 at 15:27
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http://physics.stackexchange.com/questions/33130/cyclic-co-ordinates-implying-the-constant-velocity-motion-of-center-of-mass-of-a
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# Cyclic co-ordinates implying the constant velocity motion of center of mass of a system of particles
I'm reading the section on Central Force in my textbook (Goldstein's Classical Mechanics has a similar argument in the chapter titled "The Central Force Problem", first section), where we have the following:
The Lagrangian for the system of two particles is found to be $$L=\frac{1}{2}M \dot R^2+\frac{1}{2}\mu \dot r^2-V(r)$$ where $R$ is the position vector of the center of mass of the particles.
The textbook says that since the three components of $R$ do not appear in the Lagrangian, they are cyclic.
(My first question is : Is it referring to the fact that $L$ is not a function of $(x,y,z)$? What about the $V(r)$ term. This introduces a position-dependence, doesn't it?)
We continue "..(therefore) the center of mass is either at rest or moving at a constant velocity, and we can drop the first term of the Lagrangian in our discussion. The effective Lagrangian is now given by $$L=\frac{1}{2}\mu \dot r^2-V(r)$$
"(end quote)
I don't quite see how we conclude that the center of mass is either at rest or moving at a constant velocity based on the fact that $L$ is not a function of ($x,y,z$).
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## 1 Answer
The three components of $R$ indeed do not appear in the Lagrangian; $V(r)$ is a function of only $r$ (i.e. the distance between the particles). If $V$ were a function of $R$ it would imply the presence of some external field and you wouldn't be dealing with the same two-body problem anymore.
That the center of mass is either at rest or moving at a constant velocity can easily be seen from the Euler-Lagrange equations for the original Lagrangian. For $R$ E-L equation reads: $$\ddot{R} = 0 .$$ In fact, in the absence of external forces, the center of mass of a system is always at rest or moving at a constant velocity.
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http://mathhelpforum.com/algebra/177553-help-project.html
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# Thread:
1. ## Help With Project
Very lost with a project in math class.
Need to find out and show work to see of this system is parallel, perpendicular or neither?
x-3y=-5
y=3x-9
Any Help on how to find this out?
2. You should know that an equation in the form $y = mx + c$ (or $y = ax + b$)
has the gradient as m and y-intercept as c.
If the gradient of the two lines are the same, then they are parallel.
If their product gives -1, it means they are perpendicular.
If they are neither the same, nor their product gives -1, then they are neither parallel nor perpendicular.
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http://en.wikipedia.org/wiki/Petersen_graph
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Petersen graph
From Wikipedia, the free encyclopedia
Jump to: navigation, search
Petersen graph
The Petersen graph is most commonly drawn as a pentagon with a pentagram inside, with five spokes.
Named after Julius Petersen
Vertices 10
Edges 15
Radius 2
Diameter 2
Girth 5
Automorphisms 120 (S5)
Chromatic number 3
Chromatic index 4
Fractional chromatic index 3
Properties Cubic
Strongly regular
Distance-transitive
Snark
In the mathematical field of graph theory, the Petersen graph is an undirected graph with 10 vertices and 15 edges. It is a small graph that serves as a useful example and counterexample for many problems in graph theory. The Petersen graph is named for Julius Petersen, who in 1898 constructed it to be the smallest bridgeless cubic graph with no three-edge-coloring.[1] Although the graph is generally credited to Petersen, it had in fact first appeared 12 years earlier, in a paper by A. B. Kempe (1886).
Donald Knuth states that the Petersen graph is "a remarkable configuration that serves as a counterexample to many optimistic predictions about what might be true for graphs in general."[2]
Constructions
Petersen graph as Kneser graph $KG_{5,2}$
The Petersen graph is the complement of the line graph of $K_5$. It is also the Kneser graph $KG_{5,2}$; this means that it has one vertex for each 2-element subset of a 5-element set, and two vertices are connected by an edge if and only if the corresponding 2-element subsets are disjoint from each other. As a Kneser graph of the form $KG_{2n-1,n-1}$ it is an example of an odd graph.
Geometrically, the Petersen graph is the graph formed by the vertices and edges of the hemi-dodecahedron, that is, a dodecahedron with opposite points, lines and faces identified together.
Embeddings
The Petersen graph is nonplanar. Any nonplanar graph has as minors either the complete graph $K_5$, or the complete bipartite graph $K_{3,3}$, but the Petersen graph has both as minors. The $K_5$ minor can be formed by contracting the edges of a perfect matching, for instance the five short edges in the first picture. The $K_{3,3}$ minor can be formed by deleting one vertex (for instance the central vertex of the 3-symmetric drawing) and contracting an edge incident to each neighbor of the deleted vertex.
The Petersen graph has crossing number 2.
The most common and symmetric plane drawing of the Petersen graph, as a pentagram within a pentagon, has five crossings. However, this is not the best drawing for minimizing crossings; there exists another drawing (shown in the figure) with only two crossings. Thus, the Petersen graph has crossing number 2. On a torus the Petersen graph can be drawn without edge crossings; it therefore has orientable genus 1.
The Petersen graph is a unit distance graph: it can be drawn in the plane with each edge having unit length.
The Petersen graph can also be drawn (with crossings) in the plane in such a way that all the edges have equal length. That is, it is a unit distance graph.
The simplest non-orientable surface on which the Petersen graph can be embedded without crossings is the projective plane. This is the embedding given by the hemi-dodecahedron construction of the Petersen graph. The projective plane embedding can also be formed from the standard pentagonal drawing of the Petersen graph by placing a cross-cap within the five-point star at the center of the drawing, and routing the star edges through this cross-cap; the resulting drawing has six pentagonal faces. This construction forms a regular map and shows that the Petersen graph has non-orientable genus 1.
Symmetries
The Petersen graph is strongly regular (with signature srg(10,3,0,1)). It is also symmetric, meaning that it is edge transitive and vertex transitive. More strongly, it is 3-arc-transitive: every directed three-edge path in the Petersen graph can be transformed into every other such path by a symmetry of the graph.[3] It is one of only 13 cubic distance-regular graphs.[4]
The automorphism group of the Petersen graph is the symmetric group $S_5$; the action of $S_5$ on the Petersen graph follows from its construction as a Kneser graph. Every homomorphism of the Petersen graph to itself that doesn't identify adjacent vertices is an automorphism. As shown in the figures, the drawings of the Petersen graph may exhibit five-way or three-way symmetry, but it is not possible to draw the Petersen graph in the plane in such a way that the drawing exhibits the full symmetry group of the graph.
Despite its high degree of symmetry, the Petersen graph is not a Cayley graph. It is the smallest vertex-transitive graph that is not a Cayley graph.[5]
Hamiltonian paths and cycles
The Petersen graph is hypo-Hamiltonian: by deleting any vertex, such as the center vertex in the drawing, the remaining graph is Hamiltonian. In addition, this drawing shows symmetry of order three, in contrast to the symmetry of order five visible in the first drawing above.
The Petersen graph has a Hamiltonian path but no Hamiltonian cycle. It is the smallest bridgeless cubic graph with no Hamiltonian cycle. It is hypohamiltonian, meaning that although it has no Hamiltonian cycle, deleting any vertex makes it Hamiltonian, and is the smallest hypohamiltonian graph.
As a finite connected vertex-transitive graph that does not have a Hamiltonian cycle, the Petersen graph is a counterexample to a variant of the Lovász conjecture, but the canonical formulation of the conjecture asks for a Hamiltonian path and is verified by the Petersen graph.
Only five vertex-transitive graphs with no Hamiltonian cycles are known: the complete graph K2, the Petersen graph, the Coxeter graph and two graphs derived from the Petersen and Coxeter graphs by replacing each vertex with a triangle.[6] If G is a 2-connected, r-regular graph with at most 3r + 1 vertices, then G is Hamiltonian or G is the Petersen graph.[7]
To see that the Petersen graph has no Hamiltonian cycle C, we describe the ten-vertex 3-regular graphs that do have a Hamiltonian cycle and show that none of them is the Petersen graph, by finding a cycle in each of them that is shorter than any cycle in the Petersen graph. Any ten-vertex Hamiltonian 3-regular graph consists of a ten-vertex cycle C plus five chords. If any chord connects two vertices at distance two or three along C from each other, the graph has a 3-cycle or 4-cycle, and therefore cannot be the Petersen graph. If two chords connect opposite vertices of C to vertices at distance four along C, there is again a 4-cycle. The only remaining case is a Möbius ladder formed by connecting each pair of opposite vertices by a chord, which again has a 4-cycle. Since the Petersen graph has girth five, it cannot be formed in this way and has no Hamiltonian cycle.
Coloring
A 4-coloring of the Petersen graph's edges
A 3-coloring of the Petersen graph's vertices
The Petersen graph has chromatic number 3, meaning that its vertices can be colored with three colors — but not with two — such that no edge connects vertices of the same color.
The Petersen graph has chromatic index 4; coloring the edges requires four colors. A proof of this requires checking four cases to demonstrate that no 3-edge-coloring exists. As a connected bridgeless cubic graph with chromatic index four, the Petersen graph is a snark. It is the smallest possible snark, and was the only known snark from 1898 until 1946. The snark theorem, a result conjectured by W. T. Tutte and announced in 2001 by Robertson, Sanders, Seymour, and Thomas,[8] states that every snark has the Petersen graph as a minor.
Additionally, the graph has fractional chromatic index 3, proving that the difference between the chromatic index and fractional chromatic index can be as large as 1. The long-standing Goldberg-Seymour Conjecture proposes that this is the largest gap possible.
The Thue number (a variant of the chromatic index) of the Petersen graph is 5.
Other properties
The Petersen graph:
• is 3-connected and hence 3-edge-connected and bridgeless. See the glossary.
• has independence number 4 and is 3-partite. See the glossary.
• is cubic, has domination number 3, and has a perfect matching and a 2-factor.
• has 6 distinct perfect matchings.
• is the smallest cubic graph of girth 5. (It is the unique $(3,5)$-cage. In fact, since it has only 10 vertices, it is the unique $(3,5)$-Moore graph.)[9]
• has radius 2 and diameter 2. It is the largest cubic graph with diameter 2.[10]
• has graph spectrum −2, −2, −2, −2, 1, 1, 1, 1, 1, 3.
• has 2000 spanning trees, the most of any 10-vertex cubic graph.[11]
• has chromatic polynomial $t(t-1)(t-2)\left(t^7-12t^6+67t^5-230t^4+529t^3-814t^2+775t-352\right).$[12]
• has characteristic polynomial $(t-1)^5(t+2)^4(t-3)$, making it an integral graph—a graph whose spectrum consists entirely of integers.
Petersen coloring conjecture
According to DeVos, Nesetril, and Raspaud, "A cycle of a graph G is a set C $\subseteq$ E(G) so that every vertex of the graph (V(G),C) has even degree. If G,H are graphs, we define a map φ: E(G) —> E(H) to be cycle-continuous if the pre-image of every cycle of H is a cycle of G. A fascinating conjecture of Jaeger asserts that every bridgeless graph has a cycle-continuous mapping to the Petersen graph. Jaeger showed that if this conjecture is true, then so is the 5-cycle-double-cover conjecture and the Berge-Fulkerson conjecture."[13]
Related graphs
The Petersen family.
The generalized Petersen graph G(n,k) is formed by connecting the vertices of a regular n-gon to the corresponding vertices of a star polygon with Schläfli symbol {n/k}.[14] For instance, in this notation, the Petersen graph is G(5,2): it can be formed by connecting corresponding vertices of a pentagon and five-point star, and the edges in the star connect every second vertex. The generalized Petersen graphs also include the n-prism G(n,1) the Dürer graph G(6,2), the Möbius-Kantor graph G(8,3), the dodecahedron G(10,2), the Desargues graph G(10,3) and the Nauru graph G(12,5).
The Petersen family consists of the seven graphs that can be formed from the Petersen graph by zero or more applications of Δ-Y or Y-Δ transforms. The complete graph K6 is also in the Petersen family. These graphs form the forbidden minors for linklessly embeddable graphs, graphs that can be embedded into three-dimensional space in such a way that no two cycles in the graph are linked.[15]
The Clebsch graph contains many copies of the Petersen graph as induced subgraphs: for each vertex v of the Clebsch graph, the ten non-neighbors of v induce a copy of the Petersen graph.
Notes
1. Knuth, Donald E.,
2. Babai, László (1995), "Automorphism groups, isomorphism, reconstruction", in Graham, Ronald L.; Grötschel, Martin; Lovász, László, Handbook of Combinatorics I, North-Holland, pp. 1447–1540, Corollary 1.8 .
3. Pegg, Ed, Jr. (2002), "Book Review: The Colossal Book of Mathematics", Notices of the American Mathematical Society 49 (9): 1084–1086, doi:10.1109/TED.2002.1003756
4. Hoffman, Alan J.; Singleton, Robert R. (1960), "Moore graphs with diameter 2 and 3", IBM Journal of Research and Development 5 (4): 497–504, doi:10.1147/rd.45.0497, MR0140437 .
5. Jakobson & Rivin (1999); Valdes (1991). The cubic graphs with 6 and 8 vertices maximizing the number of spanning trees are Möbius ladders.
6. Biggs, Norman (1993), Algebraic Graph Theory (2nd ed.), Cambridge: Cambridge University Press, ISBN 0-521-45897-8
7. Bailey, Rosemary A. (1997), Surveys in Combinatorics, Cambridge University Press, p. 187, ISBN 978-0-521-59840-8
References
Wikimedia Commons has media related to: Petersen graph
• Exoo, Geoffrey; Harary, Frank; Kabell, Jerald (1981), "The crossing numbers of some generalized Petersen graphs", 48: 184–188 .
• Coxeter, H. S. M. (1950), "Self-dual configurations and regular graphs", 56 (5): 413–455, doi:10.1090/S0002-9904-1950-09407-5 .
• Holton, D. A.; Sheehan, J. (1993), The Petersen Graph, Cambridge University Press, doi:10.2277/0521435943, ISBN 0-521-43594-3 .
• Jakobson, Dmitry; Rivin, Igor (1999). "On some extremal problems in graph theory". arXiv:math.CO/9907050.
• Kempe, A. B. (1886), "A memoir on the theory of mathematical form", Philosophical Transactions of the Royal Society of London 177: 1–70, doi:10.1098/rstl.1886.0002
• Lovász, László (1993), (2nd ed.), North-Holland, ISBN 0-444-81504-X .
• Petersen, Julius (1898), "Sur le théorème de Tait", L'Intermédiaire des Mathématiciens 5: 225–227
• Schwenk, A. J. (1989). "Enumeration of Hamiltonian cycles in certain generalized Petersen graphs". J. Combin. Theory B 47 (1): 53–59. doi:10.1016/0095-8956(89)90064-6.
• Valdes, L. (1991). "Extremal properties of spanning trees in cubic graphs". Congressus Numerantium 85: 143–160. .
• Watkins, Mark E. (1969). "A Theorem on Tait Colorings with an Application to the Generalized Petersen Graphs". 6 (2): 152–164. doi:10.1016/S0021-9800(69)80116-X.
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http://boolesrings.org/mpawliuk/2012/09/14/stevos-forcing-class-fall-2012-class-1/
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# Stevo’s Forcing Class Fall 2012 – Class 1
By | Published: September 14, 2012
(In the fall of 2012 I will be taking Stevo Todorcevic’s class in Forcing at the University of Toronto. I will try to publish my notes here, although that won’t always be possible.)
Show Summary
• Discuss examples of ccc posets whose product is not ccc.
• Prove a theorem of Baumgartner’s that relates the branches in a tree to its antichain structure.
• Display the differences in chain conditions.
This class is going to be held Wed 1-2 and Friday 3:30-5:30 in BA 6183.
The theme for today is “Things that are ccc, but whose products are not”. So first some examples:
Example 1. If $T$ is a Suslin tree, then $T^2$ is not a Suslin tree. (See Lemma III.2.18 in Newnen.)
Show Definitions
A Suslin Tree is a tree with uncountably many nodes, no uncountable chains and no uncountable antichains. The existence of these objects is independent of ZFC, but follows from the combinatorial principle ‘diamond’ or Godel’s axiom $V=L$. Easing the requirement from “no uncountable antichains” to “no uncountable levels” gives the definition of an Aronszajn tree, whose existence can be shown in ZFC.
Example 2. Let $\mathcal{A}(T) =$ finite antichains of $T$, where $T$ is a Suslin Tree. Then $\mathcal{A}(T)$ is powerfully ccc (its finite products are ccc), but $T \times \mathcal{A}(T)$ is not ccc.
“A Suslin tree is the skeleton of a Suslin continuum. If you take x-rays, that’s what you get.” – Stevo
This example was first studied by Baumgartner in his thesis:
Theorem (Baumgartner, 1969): TFAE for a tree $T$:
• $\mathcal{A}(T)$ is ccc;
• $T$ has no uncountable branches.
It is clear that NOT (2) imples NOT (1), as the singletons of a branch form an antichain.
“There are many proofs of this fact, but Baumgartner’s original proof is still the most useful.” – Stevo
proof of Theorem. Like most proofs that a poset is ccc, we start with an uncountable subset of the poset and refine, refine, refine.
Let $\{p_\xi : \xi \in \omega_1\} \sse \mathcal{A}(T)$, we want to find $\xi \neq \eta$ such that $p_\xi \cup p_\eta \in \mathcal{A}(T)$. We refine 3 times to an uncountable set $\Gamma \sse \omega_1$ so that:
1. (All conditions are the same length) $\vert p_\xi \vert = n$ for all $\xi \in \Gamma$,
2. ($\Gamma$ is a $\Delta$-sytem) There is an $r \in \mathcal{A}(T)$ such that for $\xi \neq \eta$ in $\Gamma$ we have $p_\xi \cap p_\eta = r$.
3. (Heights are increasing) The $ht(p_\xi) < ht(p_\eta)$ for $\xi < \eta$.
We want to assume that $r$ is actually empty, and we can do this by considering $q_\xi := p_\xi \setminus r \neq \emptyset$. Note that $q_\xi \not\perp q_\eta$ implies $p_\xi \not \perp p_\eta$.
WLOG, we get $r = \emptyset$.
We suppose that $p_\xi \perp p_\eta$ for all $\xi \neq \eta$ in $\Gamma$ (and look for a contradiction). This means that if $\xi < \eta$ then there is an $x \in p_\xi$ and a $y \in p_\eta$ such that $x <_T y$.
Now take a uniform ultrafilter $\mathcal{U}$ on $\omega_1$ such that $\Gamma \in \mathcal{U}$.
Show Ultrafilter Facts
An uniform ultrafilter is one where all the elements of $\mathcal{U}$ have the same cardinality. In this case they are all uncountable. Such ultrafilters exist in ZFC, and we can ensure that there is a uniform ultrafilter that contains any given uncountable subset of $\omega_1$.
Fix $\xi \in \Gamma$. Write $p_\xi = \{p_\xi (0), \dots , p_\xi (n-1) \}$.
There is an $i_\xi < n$ such that $\Gamma_\xi := \{\eta \in \Gamma : \exists y \in p_\eta, p_\xi (i_\xi) <_T y \} \in \mathcal{U}$.
Define $X := \{p_\xi (i_\xi) : \xi \in \Gamma\}$.
Question: Is there a large antichain in X?
Show Answer
No! There is no antichain of cardinality $n+1$, because $\Gamma_\xi = \bigcup_{j<n} \Gamma_\xi (j)$ which is in $\mathcal{U}$ by choice. Where $\Gamma_\xi (j) = \{\eta \in \Gamma_\xi : p_\xi (i_\xi) <_T p_\eta (j)\}$.
Now for each $\xi \in \Gamma$ there is a $j_\xi < n$ so that $\Gamma_\xi (j_\xi) \in \mathcal{U}$. Thus $\bigcap_{i \leq n+1} \Gamma_\xi (i_\xi) \in \mathcal{U}$ and we can choose an $\eta$ in this intersection so that $ht(p_\eta) > ht(p_{\xi_i})$ for all $i \leq n$.
Now if $j_\xi$ does not depend on $\xi$ then we are done. This follows from exercise 2. [QED]
Some Exercises:
Exercise 1: Suppose that $\mathbb{P}$ is a ccc poset and let $\{ p_\xi : \xi < \omega_1\} \sse \mathbb{P}$. Show that there is an infinite set $I \sse \omega_1$ such that $\{p_i : i \in I\}$ are pairwise compatible.
Exercise 2: Let $T$ be a tree with an uncountable subset that has no antichains of size $\geq N$. Show that $T$ has an uncountable chain.
Some chain conditions, listed from easiest to satisfy to hardest to satisfy:
• ccc
• powerfully ccc
• productively ccc
• $\sigma$-finite-cc
• $\sigma$-bounded-cc
• $\sigma$-2-linked
• $\sigma$-n-linked
• $\sigma$-n-linked ($\forall n$)
• $\sigma$-centred
• countable
Note that Borel posets can distinguish all of the blue and dark blue properties. By the following exercise, the dark blue posets are small.
Exercise 3. If $\mathbb{P}$ is an atomless $\sigma$-(2)-linked poset then $\vert \mathbb{P} \vert \leq \mathfrak{c}$.
hint. Use the Erdös-Rado theorem.
Question. Can a boolean algebra supporting a continuous submeasure algebra distinguish $\sigma$-finite-cc from $\sigma$-bounded-cc?
This entry was posted in Course Notes, Stevo's Forcing Course Fall 2012. Bookmark the permalink. Post a comment or leave a trackback: Trackback URL.
### 5 Comments
1. Posted October 18, 2012 at 8:58 am | Permalink
Thanks for sharing these notes Mike!
Small typo: In “[...] such that $\Gamma_\xi := \lbrace \eta \in \Gamma : \exists y \in P_\eta ,p_\xi(i_\xi ) <_T y \rbrace \in \mathcal{U}$" the first P should be lower case.
I also have a very hard time seeing the difference between "blue" and "dark blue" on my laptop. Could you use more contrasting colors?
• Micheal Pawliuk
Posted October 18, 2012 at 10:52 am | Permalink
Thanks for the feedback. These kind of comments are very helpful. I’ve made the light blue lighter, so hopefully there is more “pop”.
2. Anonymous
Posted November 27, 2012 at 4:06 pm | Permalink
Do we need to add some additional assumptions in exercise 3? (like having the poset be separative) As it stands we could just add a huge blob of mutually compatible guys at the top of some centered forcing, say Cohen forcing, and violate the conclusion of cardinality bounded by the continuum
3. Set Theorist
Posted January 25, 2013 at 12:15 am | Permalink
What is the ordering of the A(T) posets?
• Micheal Pawliuk
Posted January 30, 2013 at 8:14 pm | Permalink
It should just be inclusion. I.e. One antichain is compatible with another, if their union is an antichain.
### One Trackback
• By Stevo’s Forcing Class Fall 2012 – Class 2 on September 18, 2012 at 11:00 pm
[...] second lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the first lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your [...]
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http://mathhelpforum.com/advanced-algebra/189541-infinite-group-print.html
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# Infinite group
Printable View
• October 4th 2011, 05:38 AM
dwsmith
Infinite group
Prove that if $\Omega=\{1,2,....\}$, then $S_{\Omega}$ is an infinite group.
Let $\sigma:\Omega\to\Omega$. I am guessing I need to define a map but I don't know as what.
• October 4th 2011, 03:44 PM
NonCommAlg
Re: Infinite group
Quote:
Originally Posted by dwsmith
Prove that if $\Omega=\{1,2,....\}$, then $S_{\Omega}$ is an infinite group.
Let $\sigma:\Omega\to\Omega$. I am guessing I need to define a map but I don't know as what.
for every integer $n \geq 1$, there is an obvious injection $S_n \longrightarrow S_{\Omega}$. so $|S_{\Omega}| > |S_n| = n!$ and thus $|S_{\Omega}|= \infty$.
• October 4th 2011, 05:38 PM
Tinyboss
Re: Infinite group
Or you could note that it contains the transposition (n, n+1) for every natural number n.
All times are GMT -8. The time now is 10:00 AM.
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http://math.stackexchange.com/questions/314526/expressing-a-function-wrt-to-another-function
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# expressing a function wrt to another function
I am having trouble wrapping my head around this problem
given a function $f=h(g_1,g_2,g_3)$ if $h=x^2-yz$ and $f=h(x+y, y^2, x+z)$ then is it correct to apply the pointwise vector coordinate function and get $f=(x+y)^2-y^2(x+z)$
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Assuming $h(x,y,z)=x^2-yz$ then yes it is correct. – Maesumi Feb 26 at 3:39
Is there an explanation of what this means, expressing f in terms of h like this (physics or geometrically maybe) – knkumar Feb 26 at 17:53
## 1 Answer
All you are doing is "substitution". In this problem it is not clear what the "meaning" is, as no context is given.
In a standard substitution you typically change the item being measured. For example if you had $y=\sqrt{a^2-x^2}$ and you let $x=a\sin \theta$ then $y$ becomes $a\cos\theta$. Instead of measuring a length $x$ to get $y$ now you are measuring angle $\theta$ to get $y$.
In your problem $g_1$ and $g_2$ and $g_3$ specify your original input.
For $h$ you may want to use different lettering than $x,y,z$, e.g. $h(u,v,w)=u^2-vw$. Then say $u(x,y,z)=x+y,v(x,y,z)=y^2,w(x,y,z)=x+z$, and then $f(x,y,z)=h(u(x,y,z),v(x,y,z),w(x,y,z))$. Now it looks more convoluted but perhaps the role each item plays is less confusing.
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http://en.wikipedia.org/wiki/Autocorrelated
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# Autocorrelation
(Redirected from Autocorrelated)
A plot showing 100 random numbers with a "hidden" sine function, and an autocorrelation (correlogram) of the series on the bottom.
Visual comparison of convolution, cross-correlation and autocorrelation.
Autocorrelation is the cross-correlation of a signal with itself. Informally, it is the similarity between observations as a function of the time separation between them. It is a mathematical tool for finding repeating patterns, such as the presence of a periodic signal obscured by noise, or identifying the missing fundamental frequency in a signal implied by its harmonic frequencies. It is often used in signal processing for analyzing functions or series of values, such as time domain signals.
## Definitions
Different fields of study define autocorrelation differently, and not all of these definitions are equivalent. In some fields, the term is used interchangeably with autocovariance.
### Statistics
In statistics, the autocorrelation of a random process describes the correlation between values of the process at different times, as a function of the two times or of the time difference. Let X be some repeatable process, and i be some point in time after the start of that process. (i may be an integer for a discrete-time process or a real number for a continuous-time process.) Then Xi is the value (or realization) produced by a given run of the process at time i. Suppose that the process is further known to have defined values for mean μi and variance σi2 for all times i. Then the definition of the autocorrelation between times s and t is
$R(s,t) = \frac{\operatorname{E}[(X_t - \mu_t)(X_s - \mu_s)]}{\sigma_t\sigma_s}\, ,$
where "E" is the expected value operator. Note that this expression is not well-defined for all time series or processes, because the variance may be zero (for a constant process) or infinite. If the function R is well-defined, its value must lie in the range [−1, 1], with 1 indicating perfect correlation and −1 indicating perfect anti-correlation.
If Xt is a second-order stationary process then the mean μ and the variance σ2 are time-independent, and further the autocorrelation depends only on the difference between t and s: the correlation depends only on the time-distance between the pair of values but not on their position in time. This further implies that the autocorrelation can be expressed as a function of the time-lag, and that this would be an even function of the lag τ = s − t. This gives the more familiar form
$R(\tau) = \frac{\operatorname{E}[(X_t - \mu)(X_{t+\tau} - \mu)]}{\sigma^2}, \,$
and the fact that this is an even function can be stated as
$R(\tau) = R(-\tau).\,$
It is common practice in some disciplines, other than statistics and time series analysis, to drop the normalization by σ2 and use the term "autocorrelation" interchangeably with "autocovariance". However, the normalization is important both because the interpretation of the autocorrelation as a correlation provides a scale-free measure of the strength of statistical dependence, and because the normalization has an effect on the statistical properties of the estimated autocorrelations.
### Signal processing
In signal processing, the above definition is often used without the normalization, that is, without subtracting the mean and dividing by the variance. When the autocorrelation function is normalized by mean and variance, it is sometimes referred to as the autocorrelation coefficient.[1]
Given a signal $f(t)$, the continuous autocorrelation $R_{ff}(\tau)$ is most often defined as the continuous cross-correlation integral of $f(t)$ with itself, at lag $\tau$.
$R_{ff}(\tau) = (f(t) * \overline{f}(-t))(\tau) = \int_{-\infty}^\infty f(t+\tau)\overline{f}(t)\, {\rm d}t = \int_{-\infty}^\infty f(t)\overline{f}(t-\tau)\, {\rm d}t$
where $\overline{f}$ represents the complex conjugate and $*$ represents convolution. For a real function, $\overline{f} = f$.
The discrete autocorrelation $R$ at lag $j$ for a discrete signal $x_n$ is
$R_{xx}(j) = \sum_n x_n\,\overline{x}_{n-j}.$
The above definitions work for signals that are square integrable, or square summable, that is, of finite energy. Signals that "last forever" are treated instead as random processes, in which case different definitions are needed, based on expected values. For wide-sense-stationary random processes, the autocorrelations are defined as
$R_{ff}(\tau) = \operatorname{E}\left[f(t)\overline{f}(t-\tau)\right]$
$R_{xx}(j) = \operatorname{E}\left[x_n\,\overline{x}_{n-j}\right].$
For processes that are not stationary, these will also be functions of $t$, or $n$.
For processes that are also ergodic, the expectation can be replaced by the limit of a time average. The autocorrelation of an ergodic process is sometimes defined as or equated to[1]
$R_{ff}(\tau) = \lim_{T \rightarrow \infty} \frac{1}{T} \int_0^T f(t+\tau)\overline{f}(t)\, {\rm d}t$
$R_{xx}(j) = \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=0}^{N-1}x_n\,\overline{x}_{n-j}.$
These definitions have the advantage that they give sensible well-defined single-parameter results for periodic functions, even when those functions are not the output of stationary ergodic processes.
Alternatively, signals that last forever can be treated by a short-time autocorrelation function analysis, using finite time integrals. (See short-time Fourier transform for a related process.)
Multi-dimensional autocorrelation is defined similarly. For example, in three dimensions the autocorrelation of a square-summable discrete signal would be
$R(j,k,\ell) = \sum_{n,q,r} x_{n,q,r}\,x_{n-j,q-k,r-\ell}.$
When mean values are subtracted from signals before computing an autocorrelation function, the resulting function is usually called an auto-covariance function.
## Properties
In the following, we will describe properties of one-dimensional autocorrelations only, since most properties are easily transferred from the one-dimensional case to the multi-dimensional cases.
• A fundamental property of the autocorrelation is symmetry, $R(i) = R(-i)$, which is easy to prove from the definition. In the continuous case,
the autocorrelation is an even function
$R_f(-\tau) = R_f(\tau)\,$ when $f$ is a real function,
and the autocorrelation is a Hermitian function
$R_f(-\tau) = R_f^*(\tau)\,$ when $f$ is a complex function.
• The continuous autocorrelation function reaches its peak at the origin, where it takes a real value, i.e. for any delay $\tau$, $|R_f(\tau)| \leq R_f(0)$. This is a consequence of the Rearrangement inequality. The same result holds in the discrete case.
• The autocorrelation of a periodic function is, itself, periodic with the same period.
• The autocorrelation of the sum of two completely uncorrelated functions (the cross-correlation is zero for all $\tau$) is the sum of the autocorrelations of each function separately.
• Since autocorrelation is a specific type of cross-correlation, it maintains all the properties of cross-correlation.
• The autocorrelation of a continuous-time white noise signal will have a strong peak (represented by a Dirac delta function) at $\tau=0$ and will be absolutely 0 for all other $\tau$.
• The Wiener–Khinchin theorem relates the autocorrelation function to the power spectral density via the Fourier transform:
$R(\tau) = \int_{-\infty}^\infty S(f) e^{j 2 \pi f \tau} \, {\rm d}f$
$S(f) = \int_{-\infty}^\infty R(\tau) e^{- j 2 \pi f \tau} \, {\rm d}\tau.$
• For real-valued functions, the symmetric autocorrelation function has a real symmetric transform, so the Wiener–Khinchin theorem can be re-expressed in terms of real cosines only:
$R(\tau) = \int_{-\infty}^\infty S(f) \cos(2 \pi f \tau) \, {\rm d}f$
$S(f) = \int_{-\infty}^\infty R(\tau) \cos(2 \pi f \tau) \, {\rm d}\tau.$
## Efficient computation
For data expressed as a discrete sequence, it is frequently necessary to compute the autocorrelation with high computational efficiency. A brute force method based on the signal processing definition $R_{xx}(j) = \sum_n x_n\,\overline{x}_{n-j}$ can be used when the signal size is small. For example, to calculate the autocorrelation of the real signal sequence $x = (2,3,1)$ (i.e. x(0)=2, x(1)=3, x(2)=1, and x(i)=0 for all other values of i) by hand, we first recognize that the definition just given is nothing but the usual multiplication with right shifts, where each vertical addition gives the autocorrelation for particular lag values:
``` 2 3 1
× 2 3 1
-----------------
2 3 1
6 9 3
4 6 2
-----------------
2 9 14 9 2
```
Thus the required autocorrelation sequence is $R_{xx}=(2,9,14,9,2)$, where $R_{xx}(0)=14,$ $R_{xx}(-1)= R_{xx}(1)=9,$ and $R_{xx}(-2)= R_{xx}(2)=2,$ the autocorrelation for other lag values being zero. In this calculation we do not perform the carry-over operation during addition as is usual in normal multiplication. Note that we can halve the number of operations required by exploiting the inherent symmetry of the autocorrelation. If the signal happens to be periodic, i.e. $x=(...,2,3,1,2,3,1,...),$ then we get a circular autocorrelation (similar to circular convolution) where the left and right tails of the previous autocorrelation sequence will overlap and give $R_{xx}=(...,14,11,11,14,11,11,...)$ which has the same period as the signal sequence $x.$
While the brute force algorithm is order n2, several efficient algorithms exist which can compute the autocorrelation in order n log(n). For example, the Wiener–Khinchin theorem allows computing the autocorrelation from the raw data X(t) with two Fast Fourier transforms (FFT):[2]
FR(f) = FFT[X(t)]
S(f) = FR(f) FR*(f)
R(τ) = IFFT[S(f)]
where IFFT denotes the inverse Fast Fourier transform. The asterisk denotes complex conjugate.
Alternatively, a multiple τ correlation can be performed by using brute force calculation for low τ values, and then progressively binning the X(t) data with a logarithmic density to compute higher values, resulting in the same n log(n) efficiency, but with lower memory requirements.[citation needed]
## Estimation
For a discrete process with known mean and variance for which we observe $n$ observations $\{X_1,\,X_2,\,\ldots,\,X_n\}$, an estimate of the autocorrelation may be obtained as
$\hat{R}(k)=\frac{1}{(n-k) \sigma^2} \sum_{t=1}^{n-k} (X_t-\mu)(X_{t+k}-\mu)$
for any positive integer $k<n$. When the true mean $\mu$ and variance $\sigma^2$ are known, this estimate is unbiased. If the true mean and variance of the process are not known there are a several possibilities:
• If $\mu$ and $\sigma^2$ are replaced by the standard formulae for sample mean and sample variance, then this is a biased estimate.
• A periodogram-based estimate replaces $n-k$ in the above formula with $n$. This estimate is always biased; however, it usually has a smaller mean square error.[3][4]
• Other possibilities derive from treating the two portions of data $\{X_1,\,X_2,\,\ldots,\,X_{n-k}\}$ and $\{X_{k+1},\,X_{k+2},\,\ldots,\,X_n\}$ separately and calculating separate sample means and/or sample variances for use in defining the estimate.
The advantage of estimates of the last type is that the set of estimated autocorrelations, as a function of $k$, then form a function which is a valid autocorrelation in the sense that it is possible to define a theoretical process having exactly that autocorrelation. Other estimates can suffer from the problem that, if they are used to calculate the variance of a linear combination of the $X$'s, the variance calculated may turn out to be negative.
## Regression analysis
In regression analysis using time series data, autocorrelation of the errors is a problem. Autocorrelation of the errors, which themselves are unobserved, can generally be detected because it produces autocorrelation in the observable residuals. (Errors are also known as "error terms" in econometrics.)
Autocorrelation violates the ordinary least squares (OLS) assumption that the error terms are uncorrelated. While it does not bias the OLS coefficient estimates, the standard errors tend to be underestimated (and the t-scores overestimated) when the autocorrelations of the errors at low lags are positive.
The traditional test for the presence of first-order autocorrelation is the Durbin–Watson statistic or, if the explanatory variables include a lagged dependent variable, Durbin's h statistic. A more flexible test, covering autocorrelation of higher orders and applicable whether or not the regressors include lags of the dependent variable, is the Breusch–Godfrey test. This involves an auxiliary regression, wherein the residuals obtained from estimating the model of interest are regressed on (a) the original regressors and (b) k lags of the residuals, where k is the order of the test. The simplest version of the test statistic from this auxiliary regression is TR2, where T is the sample size and R2 is the coefficient of determination. Under the null hypothesis of no autocorrelation, this statistic is asymptotically distributed as $\chi^2$ with k degrees of freedom.
Responses to nonzero autocorrelation include generalized least squares and the Newey–West HAC estimator (Heteroskedasticity and Autocorrelation Consistent).[5]
## Applications
• One application of autocorrelation is the measurement of optical spectra and the measurement of very-short-duration light pulses produced by lasers, both using optical autocorrelators.
• Autocorrelation is used to analyze dynamic light scattering data, which notably enables to determine the particle size distributions of nanometer-sized particles or micelles suspended in a fluid. A laser shining into the mixture produces a speckle pattern that results from the motion of the particles. Autocorrelation of the signal can be analyzed in terms of the diffusion of the particles. From this, knowing the viscosity of the fluid, the sizes of the particles can be calculated.
• The small-angle X-ray scattering intensity of a nanostructured system is the Fourier transform of the spatial autocorrelation function of the electron density.
• In optics, normalized autocorrelations and cross-correlations give the degree of coherence of an electromagnetic field.
• In signal processing, autocorrelation can give information about repeating events like musical beats (for example, to determine tempo) or pulsar frequencies, though it cannot tell the position in time of the beat. It can also be used to estimate the pitch of a musical tone.
• In music recording, autocorrelation is used as a pitch detection algorithm prior to vocal processing, as a distortion effect or to eliminate undesired mistakes and inaccuracies.[6]
• Autocorrelation in space rather than time, via the Patterson function, is used by X-ray diffractionists to help recover the "Fourier phase information" on atom positions not available through diffraction alone.
• In statistics, spatial autocorrelation between sample locations also helps one estimate mean value uncertainties when sampling a heterogeneous population.
• The SEQUEST algorithm for analyzing mass spectra makes use of autocorrelation in conjunction with cross-correlation to score the similarity of an observed spectrum to an idealized spectrum representing a peptide.
• In Astrophysics, auto-correlation is used to study and characterize the spatial distribution of galaxies in the Universe and in multi-wavelength observations of Low Mass X-ray Binaries.
• In panel data, spatial autocorrelation refers to correlation of a variable with itself through space.
• In analysis of Markov chain Monte Carlo data, autocorrelation must be taken into account for correct error determination.
## References
1. ^ a b Patrick F. Dunn, Measurement and Data Analysis for Engineering and Science, New York: McGraw–Hill, 2005 ISBN 0-07-282538-3
2. Box, G. E. P., G. M. Jenkins, and G. C. Reinsel. Time Series Analysis: Forecasting and Control. 3rd ed. Upper Saddle River, NJ: Prentice–Hall, 1994.[]
3. Spectral analysis and time series, M.B. Priestley (London, New York : Academic Press, 1982)
4. Percival, Donald B.; Andrew T. Walden (1993). Spectral Analysis for Physical Applications: Multitaper and Conventional Univariate Techniques. Cambridge University Press. pp. 190–195. ISBN 0-521-43541-2.
5. Christopher F. Baum (2006). An Introduction to Modern Econometrics Using Stata. Stata Press. ISBN 1-59718-013-0.
6. Tyrangiel, Josh (2009-02-05). "Auto-Tune: Why Pop Music Sounds Perfect". Time Magazine.
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http://www.onemathematicalcat.org/algebra_book/online_problems/practice_points.htm
|
PRACTICE WITH POINTS
• Jump right to the exercises!
• Want more basic practice with the coordinate plane and points first?
Locating Points in Quadrants and on Axes
As discussed in Locating Points in Quadrants and on Axes,
an ordered pair $\,(x,y)\,$ is a pair of numbers, separated by a comma, and enclosed in parentheses.
The number that is listed first is called the first coordinate or the $\,{x}$-value.
The number that is listed second is called the second coordinate or the $\,y$-value.
The coordinate plane (also called the $\,xy$-plane) is a device to ‘picture’ ordered pairs.
Each ordered pair corresponds to a point in the coordinate plane,
and each point in the coordinate plane corresponds to an ordered pair.
For this reason, ordered pairs are often called points.
Recall that the origin is the point $\,(0,0)\,$.
This section gives you practice with the connection between ordered pairs and movement in the coordinate plane.
Moving Up/Down/Left/Right
Left/right movement is controlled by the $\,x$-value (the first coordinate).
Moving to the right increases the $\,x$-value; moving to the left decreases it.
For example, if a point $\,(a,b)\,$ is moved two units to the right, then the new coordinates are $\,(a+2,b)\,$.
If a point $\,(a,b)\,$ is moved three units to the left, then the new coordinates are $\,(a-3,b)\,$.
Up/down movement is controlled by the $\,y$-value (the second coordinate).
Moving up increases the $\,y$-value; moving down decreases it.
For example, if a point $\,(a,b)\,$ is moved four units up, then the new coordinates are $\,(a,b+4)\,$.
If a point $\,(a,b)\,$ is moved five units down, then the new coordinates are $\,(a,b-5)\,$.
Recall that the origin is the point $\,(0,0)\,$.
Reflecting a Point about a Line
Suppose you have a line drawn on a piece of paper.
On this same piece of paper, you have a point.
For the moment, suppose that the point does not lie on the line.
To ‘reflect the point about the line’ means, roughly,
that you want the point that is the same distance from the line,
but on the other side of the line.
This idea is illustrated below:
| | | |
|------------------------------------------------------------------------------------------------------------------------------------------------------------------------|------------------------------------------------------------------------------------------------------|------------------------------------------------------------------------------------------------------|
| | | |
| reflecting about an arbitrary line: the distance from P to Q is the same as the distance from P' to Q. (the coordinates of the reflected pointare not so easy to find) | reflecting about the $\,x$-axis (the $x$-value stays the same; take the opposite of the $\,y$-value) | reflecting about the $\,y$-axis (the $y$-value stays the same; take the opposite of the $\,x$-value) |
Notice that if you fold the piece of paper along the line of reflection,
then the original point and its reflection will land right on top of each other.
It's kind of like reflecting in a mirror—except instead of the mirror reflecting ‘back at you’,
it instead projects to the other side.
If a point actually lies on the line that you're reflecting about,
then the reflection of the point is itself.
EXAMPLES:
Question: Start at the origin.
Move to the right $\,2\,$, and down $\,4\,$.
At what point are you?
Solution: $\,(2,-4)\,$
Why?
$(0,0) \ \ \overset{\text{right 2}}{\rightarrow}\ \ (0+2,0) = (2,0) \ \ \overset{\text{down 4}}{\rightarrow}\ \ (2,0-4) = (2,-4)$
Question: Start at the point $\,(1,-3)\,$.
Move to the left $\,4\,$, and up $\,2\,$.
At what point are you?
Solution: $\,(-3,-1)\,$
Why?
$(1,-3) \ \ \overset{\text{left 4}}{\rightarrow}\ \ (1-4,-3) = (-3,-3) \ \ \overset{\text{up 2}}{\rightarrow}\ \ (-3,-3+2) = (-3,-1)$
Question: Start at the point $\,(-2,5)\,$.
Stay in the same quadrant,
but double your distance from the $\,x$-axis,
and triple your distance from the $\,y$-axis.
At what point are you?
Solution: $\,(-6,10)\,$
Why?
The up/down info (the $\,y$-value) gives distance from the $\,x$-axis.
The left/right info (the $\,x$-value) gives distance from the $\,y$-axis.
Thus:
$(-2,5)$ original point:currently $\,5\,$ units from the $\,x$-axis (above)and $\,2\,$ units from the $\,y$-axis (to the left)
$(-2,5\cdot 2) = (-2,10)$ double distance from $\,x$-axis; stay above
$(-2\cdot 3,10) = (-6,10)$ triple distance from $\,y$-axis, stay to the left
Question: Start at the point $\,(-3,7)\,$.
Reflect about the $\,x$-axis.
At what point are you?
Solution: $\,(-3,-7)\,$
To reflect about the $\,x$-axis,
the $\,x$-value stays the same,
and you take the opposite of the $\,y$-value.
Question: Start at the point $\,(-3,7)\,$.
Reflect about the $\,y$-axis.
At what point are you?
Solution: $\,(3,7)\,$
To reflect about the $\,y$-axis,
the $\,y$-value stays the same,
and you take the opposite of the $\,x$-value.
Question: Start at the point $\,(-3,7)\,$.
Reflect about the $\,x$-axis, and then reflect about the $\,y$-axis.
At what point are you?
Solution: $\,(3,-7)\,$
Why?
$(-3,7) \ \ \overset{\text{reflect about x-axis}}{\rightarrow}\ \ (-3,-7) \ \ \overset{\text{reflect about y-axis}}{\rightarrow}\ \ (3,-7)$
Question: Start at the point $\,(-3,1)\,$.
Reflect about the vertical line that passes through the $\,x$-axis at $\,2\,$.
At what point are you?
Solution: $\,(7,1)\,$
Why?
On this problem, you need to stop and think.
It might be helpful to make a sketch (see below).
• Since you're reflecting about a vertical line, the $\,y$-value won't change.
It was $\,1\,$, and it will remain $\,1\,$.
• Figure out the distance from the point to the line: $2 - (-3) = 5$
You need to go this far on the other side of the vertical line: $2 + 5 = 7$
So, the new $\,x$-value is $\,7\,$.
Question: Start at the point $\,(-3,1)\,$.
Reflect about the horizontal line that passes through the $\,y$-axis at $\,4\,$.
At what point are you?
Solution: $\,(-3,7)\,$
Why?
Again, you need to stop and think.
Again, a sketch may be helpful (see below).
• Since you're reflecting about a horizontal line, the $\,x$-value won't change.
It was $\,-3\,$, and it will remain $\,-3\,$.
• Figure out the distance from the point to the line: $4 - 1 = 3$
You need to go this far on the other side of the horizontal line: $4 + 3 = 7$
So, the new $\,y$-value is $\,7\,$.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
the Pythagorean Theorem
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
(MAX is 12; there are 12 different problem types)
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http://mathoverflow.net/questions/762/roots-of-analytic-functions/2274
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## roots of analytic functions
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $z$ be a complex variable and $f(z)$ be a formal power series with rational coefficients (an element in $\mathbb Q[[z]]$), with a finite radius of convergence, and assume $f(z)$ has a meromorphic continuation to the whole complex plane (so it has at most countably many poles). What do we know about the number-theoretic property of the roots and poles? Are they algebraic numbers? If they are, are they stable under the action of the Galois group of the rationals?
More generally, if the coefficients of $f(z)$ are algebraic numbers, and let $\sigma$ be an automorphism of the algebraic closure of the rationals, then what is the relation between roots of $f(z)$ and roots of $f^{\sigma}(z),$ where $f^{\sigma}(z)$ is the power series obtained by applying sigma to the coefficients of $f(z)?$
Without assuming the finiteness of radius of convergence, $\sin(z)$ gives a counter-example.
Edit: Let me give a second try, by imposing more requirements on $f(z).$ I'm thinking about the case where $f(z)$ is the zeta function of an algebraic stack over a finite field, so let's assume $f(z)$ has an infinite product expansion over $\ell$-adic numbers, like $\prod P_{odd}(z)/\prod P_{even}(z),$ where each $P_i(z)$ is a polynomial over $\mathbb Q_{\ell}$ with constant term 1. Assume they have distinct weights, e.g. reciprocal roots of $P_i(z)$ have weights $i.$ Then can we conclude that all coefficients of $P_i(z)$ are rational numbers? Thanks.
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## 2 Answers
Allowing the coefficients to be rational (with unbounded denominators) allows for too many possibilities, since for any power series with real coefficients, we may approximate the terms by terms with rational coefficients and with errors that tend as fast to zero as we wish in any bounded region of the plane (use the Dirichlet theorem on Diophantine approximation). Thus you can get a power series with rational coefficients from any power series with real coefficients by adding a suitable entire function. The class of power series with real coefficients is so wide that your requirement of meromorphic continuation does not meaningfully constrain. Unbounded denominators is the problem here.
If the denominators are bounded, your question ties in with a very nice theorem of Fritz Carlson from the early 1920s: A power series with integer coefficients and radius of convergence 1 either sums to a rational function or else has the unit circle as a natural boundary (analytic continuation across the unit circle is not possible anywhere on the circle). Of course, a natural boundary is the generic case in Carlson´s theorem.
-
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The power series for (1/(z+1))*sin(z) has rational coefficients and a finite radius of convergence, but its zeros are not algebraic. You can construct more interesting examples too : for instance, consider sin(z)/cos(z+2i)
-
Thanks. Let me give a second try, by imposing more requirements on f(z). I'm thinking about the case where f(z) is the zeta function of an algebraic stack over a finite field, so let's assume f(z) has an infinite product expansion over \ell-adic numbers, like \prod P_{odd}(z)/\prod P_{even}(z), where each P_i(z) is a polynomial over Q_{\ell} with constant term 1. Assume they have distinct weights, e.g. reciprocal roots of P_i(z) have weights i. Then can we conclude that all coefficients of P_i(z) are rational numbers? Thanks. – shenghao Oct 16 2009 at 17:30
OK, that was something that should have gone in an edit of the question, not a comment. – Ben Webster♦ Oct 16 2009 at 18:22
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http://mathoverflow.net/revisions/32672/list
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## Return to Answer
3 added 43 characters in body
Write $X_{B,b} = \{\alpha \in \mathbb{Z}^B : \sum_j \alpha_j = b\}$. Now using the convention $0^0 \equiv 1$, define the matrix $W_{\alpha, \alpha'} := \alpha^{\alpha'}$. For arbitrary $f:X_{B,b} \rightarrow \mathbb{K}$ we can write $f_\alpha \equiv \sum_{\alpha'} c_{\alpha'}$ from which it follows that $c = W^{-1}f$. The facts that this procedure is well-defined, and that $W$ possesses an inverse, follow from a result in multivariate interpolation assuring us that the Lagrange interpolation problem on $X_{B,b}$ is "poised".
Actually, although generic discrete point sets admit a specific multivariate Lagrange interpolation protocol that satisfies many desirable properties, only $X_{B,b}$ does it so beautifully. As a result, we obtain a Lagrange interpolation: $f_{\mathfrak{I}}(x) := \sum_\alpha (W^{-1}f)_\alpha x^\alpha$ which satisfies $f_{\mathfrak{I}}(\alpha) = f(\alpha)$.
You can use this to define differencing schemes on the a triangular (or hexagonal by suitable dual hand-waving) grid by considering $B = 3$. An example of the interpolation is shown.
Define $d_{\mathfrak{I}} f := d(f_{\mathfrak{I}})|_{X_{B,b}}$. Note (e.g.) that
$\partial_j f_{\mathfrak{I}} = \sum_\alpha (W^{-1} f)_\alpha \partial_j x^\alpha = \sum_\alpha (W^{-1} f)_\alpha \frac{\alpha_j}{x_j} \partial_j x^\alpha$
(suitably interpreted) is easy to compute in silico. Explicitly, set
$\left(W_{(\partial_j)}\right)_{\alpha, \alpha'} := \frac{\alpha'_j}{\alpha_j} \alpha^{\alpha'}, \quad \left(\mathcal{W}_{(\partial_j)}\right)_{x, \alpha'} := \frac{\alpha'_j}{x_j} x^{\alpha'}.$
Then
$\partial_j f_{\mathfrak{I}} = \mathcal{W}_{(\partial_j)} W^{-1} f, \quad \partial_j f \equiv W_{(\partial_j)} W^{-1} f.$
2 added 659 characters in body
Write $X_{B,b} = \{\alpha \in \mathbb{Z}^B : \sum_j \alpha_j = b\}$. Now using the convention $0^0 \equiv 1$, define the matrix $W_{\alpha, \alpha'} := \alpha^{\alpha'}$. For arbitrary $f:X_{B,b} \rightarrow \mathbb{K}$ we can write $f_\alpha \equiv \sum_{\alpha'} c_{\alpha'}$ from which it follows that $c = W^{-1}f$. The facts that this procedure is well-defined, and that $W$ possesses an inverse, follow from a result in multivariate interpolation assuring us that the Lagrange interpolation problem on $X_{B,b}$ is "poised".
Actually, although generic discrete point sets admit a specific multivariate Lagrange interpolation protocol that satisfies many desirable properties, only $X_{B,b}$ does it so beautifully. As a result, we obtain a Lagrange interpolation: $f_{\mathfrak{I}}(x) := \sum_\alpha (W^{-1}f)_\alpha x^\alpha$ which satisfies $f_{\mathfrak{I}}(\alpha) = f(\alpha)$.
You can use this to define differencing schemes on the hexagonal grid by considering $B = 3$. An example of the interpolation is shown.
Define $d_{\mathfrak{I}} f := d(f_{\mathfrak{I}})|_{X_{B,b}}$. Note (e.g.) that
$\partial_j f_{\mathfrak{I}} = \sum_\alpha (W^{-1} f)_\alpha \partial_j x^\alpha = \sum_\alpha (W^{-1} f)_\alpha \frac{\alpha_j}{x_j} \partial_j x^\alpha$
(suitably interpreted) is easy to compute in silico. Explicitly, set
$\left(W_{(\partial_j)}\right)_{\alpha, \alpha'} := \frac{\alpha'_j}{\alpha_j} \alpha^{\alpha'}, \quad \left(\mathcal{W}_{(\partial_j)}\right)_{x, \alpha'} := \frac{\alpha'_j}{x_j} x^{\alpha'}.$
Then
$\partial_j f_{\mathfrak{I}} = \mathcal{W}_{(\partial_j)} W^{-1} f, \quad \partial_j f \equiv W_{(\partial_j)} W^{-1} f.$
1
Write $X_{B,b} = \{\alpha \in \mathbb{Z}^B : \sum_j \alpha_j = b\}$. Now using the convention $0^0 \equiv 1$, define the matrix $W_{\alpha, \alpha'} := \alpha^{\alpha'}$. For arbitrary $f:X_{B,b} \rightarrow \mathbb{K}$ we can write $f_\alpha \equiv \sum_{\alpha'} c_{\alpha'}$ from which it follows that $c = W^{-1}f$. The facts that this procedure is well-defined, and that $W$ possesses an inverse, follow from a result in multivariate interpolation assuring us that the Lagrange interpolation problem on $X_{B,b}$ is "poised".
Actually, although generic discrete point sets admit a specific multivariate Lagrange interpolation protocol that satisfies many desirable properties, only $X_{B,b}$ does it so beautifully. As a result, we obtain a Lagrange interpolation: $f_{\mathfrak{I}}(x) := \sum_\alpha (W^{-1}f)_\alpha x^\alpha$ which satisfies $f_{\mathfrak{I}}(\alpha) = f(\alpha)$.
You can use this to define differencing schemes on the hexagonal grid by considering $B = 3$. An example of the interpolation is shown.
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http://mathhelpforum.com/differential-geometry/172275-question-about-norms-lp-spaces-chebyshev-norm-print.html
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# Question About Norms in Lp spaces and Chebyshev Norm
Printable View
• February 22nd 2011, 05:22 PM
DmitroMoroz
Question About Norms in Lp spaces and Chebyshev Norm
Here is my question; I appreciate any and all help and hints.
We assume that $A$ is a finite dimensional linear subspace of the normed linear space $C[0,1]$ of continuous functions on the interval $[0,1]$. We let $f$ be in $C[0,1]$ and define $a(p)$ to be the best approximation from $A$ to $f$ as defined by the $p$ norm for functions. We define $a*$ to be the best approximation from $A$ to $f$ as defined by the Chebyshev norm. Does the sequence ${a(1),a(2),...}$ of best approximations to $f$ in $Lp$ converge to $a*$ as [LaTeX ERROR: Convert failed] approaches infinity?
Best Regards, Dmitro.
All times are GMT -8. The time now is 12:59 PM.
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http://cogsci.stackexchange.com/questions/1826/should-always-selecting-the-same-response-on-the-iowa-gambling-test-result-in-a
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# Should always selecting the same response on the IOWA Gambling Test result in a good value?
I was taking the Iowa Gambling Task. I examined what happens when the same option is selected on every run. To my surprise I got a pretty good value when choosing options C and D (\$4500) in every run and a pretty bad one ($-500) with options A and B.
Is this intended by the test or is it a bug / feature of my downloaded program?
(I got \$4700 when mixing some A's among plenty of C's)
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## 1 Answer
In the standard version of the task there are "good" decks (overall positive payoff) and "bad" decks (overall negative payoff), so the optimal strategy is to figure out which ones are the good decks and stick with those. In principle, which deck is which should be randomized, but I don't know if your version implements that. There are also many variations that manipulate amount of gain/loss (e.g., rare big gains vs. consistent small gains) or change the payoffs in the middle, so the optimal strategy can vary.
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But what if I would know the good deck, call it C. Is there a reason for the \$4500? – draks ... Nov 6 '12 at 11:22
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I'm not sure I understand your question. The key aspect of the task is the relative values of the payoffs, not the absolute values. So (typically) the optimal strategy is to always choose the good deck, whether that gets you $4 or$40 or \$4000 is an essentially arbitrary design decision. Depending on how you set up your study, you may want to use big numbers (make it more exciting?) or small numbers (actually pay participants their winnings?), the key piece is still the relative value of the "good" vs. "bad" decks. – Dan M. Nov 6 '12 at 15:13
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http://math.stackexchange.com/questions/206941/basis-for-the-solution-space
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# basis for the solution space
Find the basis for the solution space of the system and describe all solutions:
$3x_1 - x_2 + x_4 = 0$
$x_1 + x_2 + x_3 + x_4 = 0$
I row reduce the matrix:
$\begin{pmatrix} 1&1&1&1\\ 0&-4&-3&-2 \end{pmatrix}$
And from here I do not know what to do.
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## 1 Answer
The form of the reduced matrix tells you that everything can be expressed in terms of the free parameters $x_3$ and $x_4$. It may be helpful to take your reduction one more step and get to $$\pmatrix{4&0&1&2\cr0&4&3&2\cr}$$ Now writing $x_3=s$ and $x_4=t$ the first row says $x_1=(1/4)(-s-2t)$ and the second row says $x_2=(1/4)(-3s-2t)$. If we don't like fractions, we can instead write $x_3=-4u$, $x_4=-4v$, whence $x_1=u+2v$, $x_2=3u+2v$. So we have $$(x_1,x_2,x_3,x_4)=(u+2v,3u+2v,-4u,-4v)=u(1,3,-4,0)+v(2,2,0,-4)$$ Now you can read off the basis, $\{{\,(1,3,-4,0),(2,2,0,-4)\,\}}$.
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What happened to the Tex? Some of your letters overlap and I cannot read it that well. – CodeKingPlusPlus Oct 4 '12 at 3:53
Nvm. It is rendering fine now. But, where does the $u$ and $v$ come from exactly? – CodeKingPlusPlus Oct 4 '12 at 3:59
Could you provide a short explanation? – CodeKingPlusPlus Oct 4 '12 at 4:05
If you don't like $u$ and $v$, you can use $s$ and $t$, instead. You'll get an answer with fractions in it. There's nothing wrong with that, but I prefer to avoid fractions, so I multiplied everything by 4. If $a,b$ is a basis for a vector space, so is $4a,4b$. – Gerry Myerson Oct 4 '12 at 4:06
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http://mathoverflow.net/questions/70162?sort=newest
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Why Lawvere theories have finite products? and more
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According to Wikipedia, a Lawvere theory consists of a small category $L$ with (strictly associative) finite products and a strict identity-on-objects functor $I:\aleph_0^\text{op}\rightarrow L$ preserving finite products.
Why a Lawvere theory have n-products for any n finite? For example, why isn't a Lawvere theory for monoids a category T with four elements: $1$, $T$, $T^2$ and $T^3$, and morphisms $e:1 \to T$ and $*:T^2 \to T$ making the appropiate diagrams commute and such that they are products of each other as expected? ($T^3$ is needed in order to state these diagrams)
Also, why do they usually use the 'free' maps between the desired objects to model and not just the operators that the desired family of algebras has?
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1
If you only had $1, T, T^2, T^3$, how would you describe a product of four things? – Qiaochu Yuan Jul 12 2011 at 19:05
I think this is the point where I get confused. Why would I need a product of four things? If model for $T$ would be a functor $F$ from $T$ to $Set$, then the product would be $F * : F T^2 \to F T$. For example if $F T = \mathbb{Z}_n$, then $F * : \mathbb{Z}_n^2 \to \mathbb{Z}_n$ would be the product (+) of my monoid (Z_n, +, 0) in Set, wouldn't it? – tryingotounderstand Jul 12 2011 at 19:46
If you want your theory to describe monoids, that is, sets equipped with a multiplication $m(a, b)$ satisfying some axioms, and that theory cannot even describe $m(m(m(a, b), c), d)$, then it is not a very good theory. – Qiaochu Yuan Jul 12 2011 at 20:12
But isn't the monoid described just by the functor that tells what's the set ($F T$), the operation ($F *$) and the neuter ($F e$)? – tryingotounderstand Jul 12 2011 at 20:25
I don't think so. If you want to consider models in categories other than $\text{Set}$, then I think it is possible to find a functor (model) $F$ such that $F(T)^4$ doesn't exist in the target category, and so one can't talk about the map $m(m(m(a, b), c), d)$ in the target category, which would just be silly. – Qiaochu Yuan Jul 12 2011 at 21:56
2 Answers
I think you must have misunderstood how Lawvere theories are supposed to work. Let me try to motivate them from an algebraist point of view, and answer your question in passing.
An algebraic structure such as a group or a ring is usually described in terms of operations and equations they satisfy. But why do we choose certain operations and certain equations? There are many ways of axiomatizing any given algebraic structure. For example, in a ring we could take the unary operation negation $-x$ as basic, or the binary operation subtraction $x - y$ as basic. And why not both? Why is multiplication taken as basic in the theory of groups, rather than division?
Can we describe an algebraic structure in a canonical way, so that no preference is made about which operations and equations count as "basic"? If we are not allowed to prefer any particular choice of operations and axioms, then we must favor them all equally! So, a "canonical" description should include all operations and all equations.
Let us consider the theory of groups to see how this works. We customarily start with three basic operations: a nullary operation (constant) unit $1$, a unary operation inverse ${}^{-1}$, and a binary operation multiplication $\cdot$. There are five axioms:
• $x \cdot (y \cdot z) = (x \cdot y) \cdot z$
• $x \cdot 1 = x$
• $1 \cdot x = x$
• $x \cdot x^{-1} = 1$
• $x^{-1} \cdot x = 1$
To get a canonical theory of groups which does not prefer the above operations and axioms, we should generate all possible operations and equations out of the basic ones, and then take them all as basic. Some generated operations will have names, e.g., division $x/y = x \cdot y^{-1}$, squaring $x^2 = x \cdot x$, etc., but others will not. An example of such an operation might be a ternary operation $p(x_1,x_2,x_3) = ((x_1 \cdot 1) \cdot x_2^{-1}) \cdot (x_3 \cdot x_2)$. In fact, for each $n$ there will be infinitely many $n$-ary operations, represented by expressions built from basic operations and variables $x_1, \ldots, x_n$. As new "axioms" we will simply take all equations that follow logically from the above five axioms. If we now "forget" that we started with the three basic operations and five axioms, we will get a somewhat unusual theory with infinitely many operations and infinitely many axioms, which nevertheless still describes what a group is.
This all sounds very messy. Are we supposed to invent notation for infinitely many operations? And what use is there in having so many axioms that they already are closed under logical deduction? If we truly are algebraists at heart, we must free ourselves of the shackles of syntax.
Let us make a category $\mathcal{T}$ out of the messy description of groups we generated above. Remember that we are not trying to construct the category of groups, but rather a category which nicely organizes the infinitely many operations and equations. The idea is simple enough. The morphisms of $\mathcal{T}$ should correspond to the operations of the theory. The equations should correspond to the fact that certain morphisms are equal.
An $n$-ary operation can be thought of as a map $G^n \to G$ where $G$ is the carrier set of a group and $G^n$ is the $n$-fold product of $G$'s. Thus, for each $n$ there should be an object $\mathtt{G}^n$ in $\mathcal{T}$ and the morphisms $\mathtt{G}^n \to \mathtt{G}^1$ should correspond to the $n$-ary operations. We therefore posit that the objects of $\mathcal{T}$ are $$\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$$ where you must not think of $\mathtt{G}^n$ as any kind of set, or an $n$-fold product of anything. We formally write $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2, \ldots$, but we could have as well declared that the objects of $\mathcal{T}$ are the natural numbers and write them simply but confusingly as $0, 1, 2, \ldots$.
The morphisms $\mathtt{G}^n \to \mathtt{G}^1$ are expressions built from $n$ variables $x_1, \ldots, x_n$ and the operations $1$, $\cdot$, and ${}^{-1}$. Two such expressions $p(x_1, \ldots, x_n)$ and $q(x_1, \ldots, x_n)$ are considered to be the same morphism if the theory of groups proves $p(x_1, \ldots, x_n) = q(x_1, \ldots, x_n)$. What about the morphisms $\mathtt{G}^n \to \mathtt{G}^m$? Since we expect that $\mathtt{G}^m$ will in fact end up being the $m$-fold product of $\mathtt{G}^1$'s, a morphism $\mathtt{G}^n \to \mathtt{G}^m$ corresponds uniquely to an $m$-tuple of morphisms $\mathtt{G}^n \to \mathtt{G}^1$. In other words, the morphisms $\mathtt{G}^n \to \mathtt{G}^m$ are $m$-tuples of expressions in variables $x_1, \ldots, x_n$, where again two such $m$-tuples represent the same morphism if the theory of groups proves them equal.
Composition in $\mathcal{T}$ is performed by substitution. For example, the composition of $(x_1^{-1}, x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^2$ and $(x_1 \cdot x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$ is $(x_1^{-1} \cdot x_1^{-1} \cdot (x_2 \cdot x_1) : \mathtt{G}^2 \to \mathtt{G}^1$. The identity morphism from $\mathtt{G}^n \to \mathtt{G}^n$ is the $n$-tuple $(x_1, x_2, \ldots, x_n)$.
You might think that a better $\mathcal{T}$ would contain just three objects $\mathtt{G}^0, \mathtt{G}^1, \mathtt{G}^2$ and morphisms corresponding to the basic operations $1$, ${}^{-1}$ and $\cdot$. But that would not even be a category, and if somehow you managed to make one, you would still have to make sure that $\mathtt{G}^2 = \mathtt{G}^1 \times \mathtt{G}^1$ and that $\mathtt{G}^0$ is the terminal object. It would not really be any prettier.
We can now actually show that $\mathtt{G}^m$ is the $m$-fold product of $\mathtt{G}^{1}$'s. The $k$-th projection $\pi_k : \mathtt{G}^m \to \mathtt{G}^1$ is (represented by) the expression $x_k$. I leave the rest as an exercise. Another exercise is to show that $\mathcal{T}$ has finite products computed as $\mathtt{G}^n \times \mathtt{G}^m = \mathtt{G}^{n + m}$.
We are still doing a lot of syntax disguised as category theory, but that is a necessary step that allows us to see what sort of category $\mathcal{T}$ is. We are ready to define when a category in general is the description of an algebraic theory. I am phrasing the following definition a bit imprecisely without the technical distraction of requiring a "strict identity on objects functor from $\aleph_0^\mathrm{op}$ ...":
Definition [Lawvere]: An algebraic theory is a category with distinct objects $\mathtt{A}^0, \mathtt{A}^1, \mathtt{A}^2, \ldots$ such that $\mathtt{A}^n$ is the $n$-fold product of $\mathtt{A}^1$'s.
It turns out that the models of such a theory/category $\mathcal{T}$ are precisely those functors $F : \mathcal{T} \to \mathsf{Set}$ which preserve finite products. Every such functor is already determined by how it maps $\mathtt{A}^1$ and morphisms $\mathtt{A}^n \to \mathtt{A}^1$, which of course correspond to the $n$-ary operations. In a particular case $F$ might be determined by even fewer pieces of information. For example, if $\mathcal{T}$ is the category which describes the theory of groups, $F$ will be determined already by how it maps the morphisms $1 : \mathtt{G}^0 \to \mathtt{G}^1$, $(x_1 \cdot x_2) : \mathtt{G}^2 \to \mathtt{G}^1$, and $(x_1^{-1}) : \mathtt{G}^1 \to \mathtt{G}^1$, because these generate all other morphisms (except projections and pairings, but $F$ preserves products).
The whole point of the exercise was to arrive at a non-syntactic notion of "algebraic theory". The next step is to look for examples which really are non-syntactic in nature. Here is one: the category whose objects are Euclidean spaces $\mathbb{R}^n$ and the morphisms are smooth maps. This theory describes what is known as smooth algebras.
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Thank you, very clear. I was misunderstanding it all, not thinking in syntax at all, just thinking about the models. – tryingotounderstand Jul 13 2011 at 6:38
Great answer! I knew nothing about Lewver theories and now I have a very decent grasp at their heuristic motivation! – Qfwfq Jul 13 2011 at 11:19
Andrej, I quote from nLab article on Lawvere theories: "Remark. For T a Lawvere theory, we are to think of the hom-set T(n,1) as the set of n-ary operations defined by the theory. For instance for T the theory of abelian groups, we have T(2,1)={+,−} and T(0,1)={0}." Why is T(2,1) just {+,-},? Wouldn't it include other 'unnamed' operations as well? – tryingotounderstand Jul 13 2011 at 16:31
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It's a mistake on nLab, if you ask me. – Andrej Bauer Jul 14 2011 at 4:53
I just saw these comments. Ugh. I won't say who pulled that boner, but the nLab article has now been corrected. – Todd Trimble Feb 5 2012 at 17:40
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 began as a comment but got too long. I agree with Andrej's answer, but let me add a few remarks. (1) The requirement, in the quoted definition, that the objects `$A^n$` be distinct is, I think, a red herring. Usually they'll be distinct because they won't even be isomorphic, but in those unusual algebraic theories where two of them can be isomorphic, it would do no harm to allow them to be equal. (2) Although Lawvere's definition said that the `$A^n$` should be the only objects in an algebraic theory, this amounts to requiring algebras to be single-sorted. It does no harm to allow multi-sorted theories. So one can take (and I'd be inclined to take) algebraic theories to be just categories with finite products. (Maybe I should say small categories here.) (3) I think one can give a technical foundation to the question about using only a few of the powers of $A$. The fact that, for example, monoids can be defined by means of operations taking at most 2 arguments and equations involving at most 3 variables manifests itself in the category-theoretic approach as follows. Consider the algebraic theory $M$ of monoids and the full subcategory $N$ whose objects are `$A^0,A^1,A^2,A^3$`. Then, if I'm not overlooking something, $M$ is the universal example of an algebraic theory equipped with a functor from $N$ preserving all the finite products that exist in $N$. So the subcategory $N$ suggested in the question suffices to generate, in a fairly natural way, the algebraic theory $M$.
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I think you're right about $N$ and $M$. But tryingtounderstand should be warned that the "official" definition has all $A^n$'s, not just the generating ones, and this is why a Lawvere theory has products. – Andrej Bauer Jul 13 2011 at 5:00
Thanks, after reading Andrej's post, now (3) makes sense. – tryingotounderstand Jul 13 2011 at 6:41
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Well, that's a minor niggle. Even without the distinctness requirement, given a single-sorted Lawvere theory, there is no ambiguity about what the n-ary operations are, for any n (they're the morphisms from $A^n$ into $A^1$). And if a morphism should happen to encode simultaneously operations of different arities, what of it? We may then think of "operations (with a definite arity)" as not merely identified with morphisms into $A^1$, but rather, as pairs of an arity n and a morphism from $A^n$ into $A^1$, and all is well. – Sridhar Ramesh Jul 13 2011 at 7:58
1
Suppose we have $A^2 = A^3$ and $f : A^2 \to A^1$. Then in the theory we would have a binary operation $(f,2)$ and a ternary operation $(f,3)$. We should also have an equation saying that $(f,2)$ and $(f,3)$ are equal. But such an equation cannot be written down. (In what context, with two or with three variables?) Can we really get away without the equation? I think not, because then the resulting theory will have different models than the category we started with. – Andrej Bauer Jul 13 2011 at 8:09
1
I suppose the real concern is that we do not want functors/models to have to preserve the equality between $A^2$ and $A^3$, only the isomorphism. So, you're right; if what we want to do is faithfully model algebraic theories as given by operations of finite arities and universal equations between them, then we basically want the objects to be (in correspondence with) the finite arities. [That, or, just as well (I think), the objects can be whatever, we'll use anafunctors for models, and no one will ever speak about equality...] – Sridhar Ramesh Jul 13 2011 at 9:59
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http://www.territorioscuola.com/wikipedia/en.wikipedia.php?title=Linear_system
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Linear system - Wikipedia, the free encyclopedia
# Linear system
A linear system is a mathematical model of a system based on the use of a linear operator. Linear systems typically exhibit features and properties that are much simpler than the general, nonlinear case. As a mathematical abstraction or idealization, linear systems find important applications in automatic control theory, signal processing, and telecommunications. For example, the propagation medium for wireless communication systems can often be modeled by linear systems.
A general deterministic system can be described by operator, $H$, that maps an input, $x(t)$, as a function of $t$ to an output, $y(t)$, a type of black box description. Linear systems satisfy the properties of superposition and scaling or homogeneity. Given two valid inputs
$x_1(t) \,$
$x_2(t) \,$
as well as their respective outputs
$y_1(t) = H \left \{ x_1(t) \right \}$
$y_2(t) = H \left \{ x_2(t) \right \}$
then a linear system must satisfy
$\alpha y_1(t) + \beta y_2(t) = H \left \{ \alpha x_1(t) + \beta x_2(t) \right \}$
for any scalar values $\alpha \,$ and $\beta \,$.
The system is then defined by the equation H(x(t)) = y(t), where y(t) is some arbitrary function of time, and x(t) is the system state. Given y(t) and H, x(t) can be solved for. For example, a simple harmonic oscillator obeys the differential equation:
$m \frac{d^2(x)}{dt^2} = -kx$
If $H(x(t)) = m \frac{d^2(x(t))}{dt^2} + kx(t)$, then H is a linear operator. Letting y(t) = 0, we can rewrite the differential equation as H(x(t)) = y(t), which shows that a simple harmonic oscillator is a linear system.
The behavior of the resulting system subjected to a complex input can be described as a sum of responses to simpler inputs. In nonlinear systems, there is no such relation. This mathematical property makes the solution of modelling equations simpler than many nonlinear systems. For time-invariant systems this is the basis of the impulse response or the frequency response methods (see LTI system theory), which describe a general input function $x(t)$ in terms of unit impulses or frequency components.
Typical differential equations of linear time-invariant systems are well adapted to analysis using the Laplace transform in the continuous case, and the Z-transform in the discrete case (especially in computer implementations).
Another perspective is that solutions to linear systems comprise a system of functions which act like vectors in the geometric sense.
A common use of linear models is to describe a nonlinear system by linearization. This is usually done for mathematical convenience.
## Time-varying impulse response
The time-varying impulse response h(t2,t1) of a linear system is defined as the response of the system at time t = t2 to a single impulse applied at time t = t1. In other words, if the input x(t) to a linear system is
$x(t) = \delta(t-t_1) \,$
where δ(t) represents the Dirac delta function, and the corresponding response y(t) of the system is
$y(t) |_{t=t_2} = h(t_2,t_1) \,$
then the function h(t2,t1) is the time-varying impulse response of the system.
## Time-varying convolution integral
### Continuous time
The output of any continuous time linear system is related to the input by the time-varying convolution integral:
$y(t) = \int_{-\infty}^{\infty} h(t,s) x(s) ds$
or, equivalently,
$y(t) = \int_{-\infty}^{\infty} h(t,t-\tau) x(t-\tau) d \tau$
### Discrete time
The output of any discrete time linear system is related to the input by the time-varying convolution sum:
$y[n] = \sum_{k=-\infty}^{\infty} { h[n,k] x[k] }$
or equivalently,
$y[n] = \sum_{m=-\infty}^{\infty} { h[n,n-m] x[n-m] }$
where
$k = n-m \,$
represents the lag time between the stimulus at time m and the response at time n.
## Causality
A linear system is causal if and only if the system's time varying impulse response is identically zero whenever the time t of the response is earlier than the time s of the stimulus. In other words, for a causal system, the following condition must hold:
$h(t,s) = 0\text{ for }t < s \,$
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http://math.stackexchange.com/questions/197909/any-interesting-number-theoretic-results-properties-concerning-particularly-valu?answertab=votes
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# Any interesting number-theoretic results/properties concerning particularly values of $n$ in $2n+1$? [closed]
So this question has strange origins: I was looking at the Leibniz series for $\pi$, and I started to wonder about the relationship between the partial sum and the parity of the value $n$ in the denominator: $2n+1$. (Since these partial sums are related to trig-functions, which ultimately help understand the proof of Euler's formula, I thought this question might be worth thinking about.)
In any case, is there anything particularly interesting about about the value of $n$ in odd-integers? That is, if for some odd number $2n+1$, are the interesting results if $n$ is even or $n$ is odd? (Mersennne primes sort of come to mind, but those are of the form $2^{p}-1$.)
Perhaps as a general direction, can we say something about whether a number is prime, is not prime, is near primes, given we know something about this $n$ (perhaps we could consider cases where $n$ is prime)?
Just wondering. Any thoughts, papers, or further reading would appreciated!
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A.G., I can't follow your question. What do you mean by the value of $n$? Is the result "$2n$ is not prime unless $n=1$" something that would interest you? – Yuval Filmus Sep 17 '12 at 7:28
Looks like you're asking if there are any interesting differences between odd numbers of the form $4m+1$ and those of the form $4m+3$. Is that it? If so the answer is yes, plenty. – Marc van Leeuwen Sep 17 '12 at 11:55
## closed as not a real question by William, Yuval Filmus, Noah Snyder, Norbert, J. M.Oct 10 '12 at 17:04
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
## 1 Answer
If $n$ is even, then $2n+1=4m+1$ for some $m$; if $n$ is odd, then $2n+1=4m-1$ for some $m$.
This does have some interesting consequences, e.g., an odd prime is a sum of two squares if and only if it is $4m+1$. But it doesn't have the kind of consequence you have suggested, concerning primality; asymptotically, the number of $4m+1$ primes is the same as the number of $4m-1$ primes.
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http://mathoverflow.net/questions/50275/a-puzzling-remark-of-manin-icm-1978
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## A puzzling remark of Manin (ICM 1978)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Manin ends his 1978 ICM talk with this remark:
I would also like to mention I. M. Gel'fand's suggestion that the $\zeta$-functions of certain special differential operators should have an arithmetic meaning. The first class to consider is that of the Schrödinger operators $-d^2/dx^2 + u(x)$ with algebro-geometric potentials $u(x)$ arising as solutions of the Korteweg-de Vries equation, for example: $u(x) = 2\wp(x)$ where $\wp$ is the Weierstrass function of an elliptic curve over $Q$. In fact, it seems that the values of this zeta-function at negative integers, which can be calculated explicitly, admit a $p$-adic interpolation.
Simply put, I'd be grateful to anyone who can explain to me what he's talking about. In particular, has "Gel'fand's suggestion" found explicit form, either as theorem or conjecture, anywhere in the literature?
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3
Let me formulate a different but related question. Assume $u$ is $P$-periodic. Floquet's theory tells us that the spectrum of $-d^2/dx^2+u(x)$ over $L^2(\mathbb R)$ is the union of intervals $[a_{2j},b_{2j}]$ and $[b_{2j+1},a_{2j+1}]$ with $a_k$ and $b_k$ non-decreasing. The $a_k$'s (resp. $b_k$'s) are e.v. associated with periodic (resp. anti-periodic) e.f. If $u=2\wp$, these eigenvalues have double multiplicity, except for $a_0$. This is the only case without gaps in the spectrum. Is there any relation with the fact that $\zeta$ has good properties ? – Denis Serre Dec 24 2010 at 8:47
@Denis I would certainly benefit from more details or a reference. For example, I can't decipher "e.v." or $P$-periodic (I'll guess that P is a lattice). – David Feldman Dec 25 2010 at 18:26
"eigenvalues associated with periodic (respectively anti-periodic) eigenfunctions", I'd guess. – David Loeffler Dec 26 2010 at 12:18
Right, thanks David Loeffler. – David Feldman Dec 26 2010 at 19:07
## 1 Answer
From Introduction to “Scattering Theory for Automorphic Functions” by Lax and Phillips (1976) and some other sources I could suppose, that proper reference would be the Gelfand presentation on 1962 ICM, but I do not have access to it, and so not quite certain.
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3
Gelfand's article is here mathunion.org/ICM/ICM1962.1/Main/… . See mathoverflow.net/questions/20071/… on how to quickly find ICM articles. – Gil Kalai Apr 27 2011 at 18:02
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http://nrich.maths.org/2373/note
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The graph of $y=\sin x$ lies above the line $y=2x/\pi$ joining the origin to the point $(\pi/2, 1)$. The graph of $\sin x$ lies below the line $y=x$. Why? What does this tell you about $\sin x$?
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http://math.stackexchange.com/questions/28390/is-any-transposition-a-product-of-simple-transpositions/28396
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# Is any transposition a product of simple transpositions?
Is any transposition a product of simple transpositions? If yes, how can you prove this?
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A simple transposition is a transposition in the form $(i,i+1)$ so (7,8) is a simple transposition but (7,10) is not. – Vafa Khalighi Mar 21 '11 at 23:47
## 3 Answers
Every element of $S_n$ can be written as a product of simple transpositions. This fact can be proven by induction on $n$.
As a first step, prove that $(1,m)$ is always a product of simple transpositions by induction of $m$. The case $m = 2$ is trivial, so we assume that $(1,m-1)$ is a product of simple transpositions. But $(1,m) = (m-1,m)(1,m-1)(m-1,m)$, so $(1,m)$ is a product of simple transpositions.
Now we show that all transpositions in $S_n$ are products of transpositions of the form $(1,m)$. The case $n = 2$ is trivial, so we assume that any transposition of $S_{n-1}$ is such a product. Given the transposition $t$, we must have some $m$, possibly $1$ which exchanges position with 1. If we perform the transposition $(1,m)$, we only have $n-1$ positions left to permute, which can be done using simple transpositions by the inductive hypothesis. This completes the proof.
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There is an algorithm that takes a permutation and writes it as a product of simple transpositions. It's called bubble sort.
What you do is : if $f(i) > f(i+1)$, write $f$ = $f' \circ (i,i+1)$. Repeat on $f'$ until you get the identity and are left with a product of simple transposition.
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Yes. One way to see this is to notice what happens when you conjugate transpositions by transpositions. For example, $(5,6)(3,6)(5,6)=(3,5)$ shows that you can get one step closer to a simple transposition, from $(3,6)$ to $(3,5)$, by conjugating by $(5,6)$. Then $(4,5)(3,5)(4,5)=(3,4)$, so we've gotten to a simple transposition after $2$ conjugations. To write $(3,6)$ as a product of simple transpositions, you can just unwind this: $(3,6)=(5,6)(4,5)(3,4)(4,5)(5,6)$.
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In other words assume without loss $i+1<j$ and write $(i\ j)^{(i\ i+1)} = (i+1\ j)$. Repeat until satisfied. – Myself Mar 21 '11 at 23:59
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http://electronics.stackexchange.com/questions/37431/help-finding-transfer-function-for-root-locus-using-matlab
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# Help finding transfer function for root locus using Matlab
I'd like to pick $k_d$ using root locus method, but have problems deriving the necessary transfer function of the system presented below. Assume $k_p$ is fixed. The question originates from Randal Beard's paper: "Quadrotor dynamics and control", p.42. The answer is actually given there, but for a slightly different block diagram and with no derivation. So it's the derivation that matters to me most.
As far as I understand the method in question, I need obtain an equation:
$1 + k_d P(s) = 0$,
but don't know how to derive $P(s)$.
If you're able and willing to help, please don't only provide the solution - I need to know how the solution was obtained to be able to help myself in the future. Any hints appreciated.
EDIT: What I already tried out is simplifying the above block diagram into the form below:
Then we have:
$\Large L(s) = \frac{G(s)H(s)}{1 + k_d G(s)H(s)}$ %Inner loop's transfer function
$\Large R(s) = \frac{L(s)/s}{1 + k_p L(s) / s} = \frac{L(s)}{s + k_p L(s)}$, substituting for $L(s)$ we have:
$\Large R(s) = \huge \frac{\frac{G(s)H(s)}{1 + k_d G(s)H(s)}}{s + k_p \frac{G(s)H(s)}{1 + k_d G(s)H(s)}} = \Large \frac{G(s)H(s)}{s(1+k_d G(s)H(s)) + k_p G(s)H(s)}$.
So the point is, how to convert $R(s)$ denominator into: $1 + k_d P(s) = 0$.
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## 3 Answers
OK, I got it. To obtain the necessary transfer function in Evan's form, one has to assume $\phi_c = 0$. Then, the block diagram in question can be converted into:
So $P(s) = \Large \frac{p}{\alpha} = \frac{G(s)H(s)}{1 + \frac{k_p}{s} G(s)H(s)}$.
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Lets take a look at the transfer function of the inner loop $$I(s) = \frac{p}{\Phi_e}$$ To understand don't use formulas, do it (at least the until you understand) by hand. So lets build the transfer function (just follow your diagram): $$p = H(s) G(s) (k_p \Phi_e - k_d p)$$ Rearrange it: $$p \left(1+k_d H(s) G(s)\right) = k_p H(s) G(s) \Phi_e$$ So our tranfer function is: $$I(s) = \frac{p}{\Phi_e} = \frac{k_p H(s) G(s)}{1+k_d H(s) G(s)}$$ To find the roots of the characteristic equation you need to find the solution to: $$1+k_d H(s) G(s) = 0$$ So in your case, when you compare the formula with your formula $$1+k_d P(s) = 0$$ it is easy to see what's P(s): $$P(s) = H(s) G(s)$$
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Thanks PetPaulsen, I edited my question again - mostly adding what I already figured out and mentioning Beard's paper the question originates from. I think the outer loop's influence is missing in your answer. I just figured out that maybe the block diagram should be transformed assuming $\phi_c = 0$ to obtain the answer I'm looking for. I'll try that out and let know. – mmm Aug 9 '12 at 7:24
@mmm - Yes, now I see what you mean. I am used to lay out the inner loop first. Then the inner loop is fixed (just another transfer function I add to the plant) and I lay out the outer loop. So I am afraid I cannot help you here. Anyway if I have addiditonal information I will let you know. – PetPaulsen Aug 9 '12 at 7:30
Break this into two pieces. The inner loop (G(s), H(s) and the feedback term kd) and the outer loop, Kp onto the inner loop.
Inner loop first. The transfer function is the forward path divided by 1 + the loop gain.
I(s) = transfer function of the inner loop = $\dfrac{G(s) H(s) \dfrac {1}{s}} {(1+G(s)H(s)k_d)}$
Simplifying this; $I(s)= \dfrac {s} {s G(s) H(s) + sk_d}$ Not 100% sure on the math here, did it real quick.
Now the entire loop. $T(s)= \dfrac {k_p I(s)} {1 + k_p I(s)}$
If i was doing this, I would maybe simplifiy this down, but you could just plug T(s) into Matlab and call the RLTool.
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Can you please calrify what your inner loop is? At first glance I did expect $I(s) = p/\Phi_e$, but you got no $k_p$ in your formula, and the term (1/s) confuses me, too. That doesn't look right. – PetPaulsen Aug 9 '12 at 6:40
Thanks for your time Michael. I think, there is a mistake in your answer, as PetPaulsen pointed out, though. The inner loop shouldn't contain the $1/s$ term - it should be added to the $T(s)$ transfer function instead. But the main question is how to transfer this into Evan's form for root locus based $k_d$ design, and it's still unanswered. – mmm Aug 9 '12 at 6:49
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http://mathoverflow.net/questions/33958/generalization-of-schurs-lemma-update
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## Generalization of Schur’s lemma (Update)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am not a mathematician nor physicist. I just know the basics of the representation theory. In my research, I realized that there is an orthogonality relation between the unitary group matrix elements as follows:
$$I_1 = \int \mathrm{D}\mathbf{U} \; U_{i j}^{(\mathbf{r})} U_{k l }^{*(\mathbf{r}^{\prime})} = \frac{1}{ d_{ \mathbf{r} } } \delta_{\mathbf{r} \mathbf{r}^{\prime} } \delta_{i k} \delta_{j l}$$
where $\mathbf{U} \in \mathcal{U}(N)$, $\mathrm{D}\mathbf{U}$ is the standard Haar measure, $U_{ij}^{(\mathbf{r})}$ denotes the $(i,j)$-th element of the representation matrix of $\mathbf{U}$, and $d_{ \mathbf{r} }$ is the dimension of the irreducible representation $\mathbf{r}$.
Now, I need to know the answer for this integral:
$$I_2 = \int \mathrm{D} \mathbf{U} \; U_{i_1 j_1}^{(\mathbf{r})} U_{ k_1 l_1 }^{ * ( \mathbf{r} ) } U_{ i_2 j_2 }^{(\mathbf{r}^{\prime})} U_{ k_2 l_2 }^{* ( \mathbf{r}^{ \prime \prime } ) }$$
I appreciate any help.
p.s. Here is my conjecture for the answer:
```$$
I_2 = \delta_{ \mathbf{r}^{\prime} \mathbf{r}^{\prime \prime} } \times \left\{
\eqalign{
\frac{1}{ d_{ \mathbf{r} } d_{ \mathbf{r}^{ \prime } } -1 } \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 } ( 1- \delta_{ \mathbf{r} \mathbf{r}^{\prime} } ) \\
+ \delta_{ \mathbf{r} \mathbf{r}^{\prime} } \left[
\eqalign{
\frac{ 1 }{ d_{ \mathbf{r} }^2 -1 }
( \delta_{ i_1 k_1 } \delta_{ j_1 l_1 } \delta_{ i_2 k_2 } \delta_{ j_2 l_2 }
+ \delta_{ i_1 k_2 } \delta_{ j_1 l_2 } \delta_{ i_2 k_1 } \delta_{ j_2 l_1 } ) \\
- \frac{ 1 }{ d_{ \mathbf{r} } ( d_{ \mathbf{r} }^2 -1 ) } ( \delta_{ i_1 k_1 } \delta_{ j_1 l_2 } \delta_{ i_2 k_2 } \delta_{ j_2 l_1 } + \delta_{ i_1 k_2 } \delta_{ j_1 l_1 } \delta_{ i_2 k_1 } \delta_{ j_2 l_2 } ) }
\right]
} \right\} $$```
UPDATE:
I have been advised that it might be helpful if I can find the tensor product of two irreducible representations, $\mathbf{s} = \mathbf{r} \otimes \mathbf{r}^{\prime}$, which most likely leads to a reducible representation, and then I need to decompose $\mathbf{s}$ into its irreducible components (by using the Clebsch–Gordan coefficients, according to wikipedia), to be able to use the Schur's lemma to get the answer!!!
However, it is hard for me to do this, and needs awful background.
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Why do call this a generalisation of the Schur's Lemma? It seems to me like a generalisation of the Peter-Weyl theorem instead. – José Figueroa-O'Farrill Jul 30 2010 at 23:30
I don't know the answer to your question, but this seems to me like the sort of calculation that lattice gauge theorists might know how to do. – José Figueroa-O'Farrill Jul 30 2010 at 23:32
Thanks for the hint. p.s. The mathematicians I have talked to so far call I_1 the Schur's lemma. – Alireza Jul 31 2010 at 21:01
Do you really want two ${\bf r}$'s, one ${\bf r'}$ and one ${\bf r''}$ in your expression? Or is there a typo? – Peter Shor Aug 1 2010 at 15:46
1
I don't think your conjecture can be right. Look at $SU(2)$. Suppose ${\bf r}=3$, ${\bf r}'=5$ and ${\bf r''=7}$. Then the tensor products decompose into irreducible representations as: $R_3\otimes R_5=R_3\oplus R_5 \oplus R_7$ and $R_3\otimes R_7=R_5\oplus R_7\oplus R_9$, where $R_n$ is the irreducible representation of dimension $n$. When you sum over your indices $i_{1,2},\, j_{1,2},\, k_{1,2},\, l_{1,2}$, you should get $12=5+7$, not $0$, since these overlap in $R_5$ and $R_7$. How this sum of $12$ is distributed over $i_{1,2},\,j_{1,2},\,k_{1,2},\,l_{1,2}$ is a mystery to me. – Peter Shor Aug 2 2010 at 20:29
show 6 more comments
## 1 Answer
I'm afraid I do not know the answer to your problem, but here's a counterexample to your conjecture that the result is zero if $r'\ne r''$. take $r=(2,1)$ (i.e., the Young diagram with two boxes one the first row and one on the second). then in the decomposition into irreducible representations of $r=(2,1)$ with its dual $r^\star=(...,-1,-2)$ one finds $(2,...,-1,-1)$ and $(1,1,...,-2)$. now these occur naturally as the tensor product of $r'=(2)$ and the dual of $r''=(1,1)$, or vice versa. ergo, $r\otimes r^\star\otimes r'\otimes r''^\star$ contains the trivial representation so that the integral of some of its matrix elements will be non-zero.
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http://math.stackexchange.com/questions/112980/pretty-simple-question-about-running-time/112982
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# Pretty simple question about running time
What is the smallest value of $n$ such that an algorithm whose running time is $100n^2$ runs faster than an algorithm whose running time is $2^n$ on the same machine?
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1
– Arturo Magidin Feb 24 '12 at 19:41
Just doing some arithmetic, it's $n=15$. – David Mitra Feb 24 '12 at 19:42
Thanks @ArturoMagidin didn't know about Lambert's W function. Actually that's the solution. – Randolf R-F Feb 24 '12 at 19:46
## 2 Answers
On the one hand, you could just plug in some numbers and see what pops up. Here, you know you only care about 4 digit values of $2^n$, so perhaps start at $2^{12}$ or something. It turns out that $n=15$ is the smallest integer.
However, we should also know that runtimes are usually expressed up to a constant of multiplication. So perhaps it's actually $16$ or $14$ (though both are unlikely here, because of the large differences).
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I find that the first integer such that $100n^2 < 2^n$ is $n=15$. Is this what you wanted?
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http://wiki.math.toronto.edu/TorontoMathWiki/index.php/2009-2010_FAWG_Seminar
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# 2009-2010 FAWG Seminar
### From TorontoMathWiki
The Fields Analysis Working Group Seminar takes place (usually) at the Fields Institute Room 210 on Thursdays at 12h10. The seminar is co-organized by J. Colliander, Colin Decker, R.J. McCann, and L. Guth.
The seminar also has a page at the Fields Institute.
Of related interest in Toronto (and sometimes cross-listed):
Archive: 2008-2009_FAWG_Seminar, 2007-2008_FAWG_Seminar, 2006-2007_FAWG_Seminar. 2008-2009 FAWG Seminar
## July 23, Denzler, Friday, 15h10, Fields Institute Room 210 *Please Note the Unusual Time*
• Jochen Denzler (University of Tenesse at Knoxville)
Title: Asymptotics of fast diffusion via dynamical systems.
Abstract: This gives an outline of how to implement rigorously a formalism that gives the convergence rate of fast diffusion to the (self-similar) Barenblatt solution along with ideas coming from linearized stability of dynamical systems. Joint work with Robert McCann and Herbert Koch.
## April 29, Dotterrer, 12:00-1:00 @Fields Institute Room 210
| | | | | | | |
|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|-------------------------------|----------|----------|------------|---------------------------|------------|
| Dominic Dotterrer homepage (U. Toronto) | Fields Analysis Working Group | Thursday | April 29 | 12:00-1:00 | Fields Institute Room 210 | |
| Title: The probabilistic method in geometry: Bourgain's Theorem | | | | | | |
| Abstract: For quite some time now, the probabilistic method has been yielding geometric fruit. By studying "typical" geometric structures, we can sometimes understand the "extremal" ones. This is exemplified in the proof of Bourgain's embedding theorem, which states that any finite metric space on n points can be embedded in Euclidean space with less than O(log n) metric distortion. I will give a careful, annotated proof of this theorem and show that the estimate is sharp. | | | | | | |
| [<arxiv> arXiv] | 2010_04_29_Dotterrer_Notes | | | | | 2010_04_29 |
## April 15, Rochon, Thursday 1:10-2:00, Fields Institute Room 210 *Please Note the Unusual Time*
• Frederic Rochon (University of Toronto)
Title: Ricci flow and the determinant of the Laplacian on non-compact manifolds, Part 2.
Abstract: After introducing the notion of determinant of the Laplacian on a non-compact surface with ends asymptotically isometric to a cusp or a funnel, we will show that in a given conformal class (with 'renormalized area' fixed), this determinant is maximal for the metric of constant scalar curvature, generalizing a well-known result of Osgood, Phillips and Sarnak in the compact case. This will be achieved by combining a corresponding Polyakov formula with some long time existence result for the Ricci flow for such metrics. This is a joint work with P. Albin and C.L. Aldana.
2010_04_08_FAWG_Rochon_Notes, arXiv
## April 8, Rochon, Thursday 12:10-1:00, Fields Institute Room 210
• Frederic Rochon (University of Toronto)
Title: Ricci flow and the determinant of the Laplacian on non-compact manifolds
Abstract: After introducing the notion of determinant of the Laplacian on a non-compact surface with ends asymptotically isometric to a cusp or a funnel, we will show that in a given conformal class (with 'renormalized area' fixed), this determinant is maximal for the metric of constant scalar curvature, generalizing a well-known result of Osgood, Phillips and Sarnak in the compact case. This will be achieved by combining a corresponding Polyakov formula with some long time existence result for the Ricci flow for such metrics. This is a joint work with P. Albin and C.L. Aldana.
2010_04_08_FAWG_Rochon_Notes, arXiv
## April 1, Chugunova, Thursday 12:10-1:00, Fields Institute Room 230
• Marina Chugunova (University of Toronto)
Title: On the speed of the propagation of the thin film interface
Abstract: The equation $u_t+[u^n(u_{xxx}+\alpha^2 u_x - sin(x))]_x=0$ with periodic boundary conditions is a model of the evolution of a thin liquid film on the outer surface of a horizontal cylinder in the presence of gravity field. We use energy-entropy methods to study different properties of generalized weak solutions of this equation. For example: finite speed of the compact support propagation for n(13) is proved by application of local energy-entropy estimates. Joint work with A. Burchard, M. Pugh, B. Stephens, and R. Taranets
## March 11, Richards, Thursday 12:10-1:00, Fields Institute Room 210
• Geordie Richards (University of Toronto)
Title: Almost Sure Local Well-posedness for the Stochastic KdV-Burgers Equation.
Abstract: We consider the stochastic KdV-Burgers equation on the 1-d torus as a toy model for a stochastic Burgers equation. The stochastic Burgers equation we model is obtained by differentiating the well-known Kardar-Parisi-Zhang (KPZ) equation in space. We present almost sure local well-posedness in $H^{-1/2-}(\mathbb{T})$ for the stochastic KdV-Burgers equation. Time permitting, we will discuss the issue of global existence in time. This is a joint work with Tadahiro Oh and Jeremy Quastel.
## March 4, Pass, Thursday 12:10-1:00, Fields Institute Room 210
• Brendan Pass (University of Toronto)
Title: Rectifiability of optimal transportation plans.
Abstract: I will prove the following result (which represents joint work with Robert McCann and Micah Warren): any solution to a Kantorovich optimal transportation problem on two smooth n-dimensional manifolds $X$ and $Y$ is supported on an n-dimensional Lipschitz submanifold of the product $X \times Y$, provided the cost is $C^2$ and nondegenerate. If time permits, I will discuss how this generalizes to the multi-marginal problem.
## February 25, Pass, Thursday 12:10-1:00, Fields Institute Room 230
• Brendan Pass (University of Toronto)
Title: The multi-marginal optimal transportation problem.
Abstract: I consider an optimal transportation problem with more than two marginals. I will discuss how the signature of a certain pseudo-Riemannian form provides an upper bound for the dimension of the support of the optimal measure. Time permitting, I will also discuss conditions on the cost function that ensure existence and uniqueness of an optimal map.
## January 28, Li, Thursday 12:10-1:00, Fields Institute Room 230
• Jiayong Li (University of Toronto)
Title: New examples on spaces of negative sectional curvature satisfying Ma-Trudinger-Wang conditions.
Abstract: When the domain of the optimal transportation problem is a Riemannian manifold, an interesting problem is to analyze the regularity of the optimal map, with the transport cost related to the Riemannian distance. There have been extensive studies about the Riemannian distance squared on the sphere and the quotient of the sphere. In this talk, we discuss the regularity of the optimal map on a manifold with constant sectional curvature, with the transport cost given by a real-valued function composed with the Riemannian distance. We will show the relation between the Jacobi vector field and the Ma-Trudinger-Wang tensor, which is an important quantity for the regularity of the optimal map. As a consequence of this relation, we give new examples of cost functions satisfying the Ma-Trudinger-Wang conditions, and a perturbative result of the distance squared on the Euclidean space. This is joint work with P. Lee.
## January 21, Zwiers, Thursday 12:10-1:00, Fields Institute Room 230
• Ian Zwiers (University of Toronto)
Title: Minimal Navier-Stokes Singularities
Abstract: Suppose that in three dimensions there exists a solution to Navier-Stokes that forms a singularity in finite time (for data in $\dot{H}^\frac{1}{2}$). Then there exists such data of minimal norm. This is a recent result of Rusin and Šverák, building on Lemarié-Rieusset's development of local Leray solutions.
References:
W. Rusin and V. Šverák, Minimal initial data for potential Navier-Stokes singularities.
http://arxiv.org/abs/0911.0500
P.G. Lemarié-Rieusset, Recent developments in the Navier-Stokes problem, Chapman & Hall / CRC, 2002
(Math Library)
## January 14, Burchard, Thursday 12:10-1:00. Fields Institute Room 210
• Almut Burchard (University of Toronto)
Title: Competing Symmetries and convergence of sequences of random symmetrizations
Abstract: Rearrangements change the shape of a function while preserving its size. The symmetric decreasing rearrangement, which is used for finding extremals of functionals that involve gradients or convolutions, replaces a given function f with a radially decreasing function f * .
The symmetric decreasing rearrangement can be approximated by sequences of simpler rearrangements, such as Steiner symmetrizations or polarization. In this talk, I will discuss the convergence of random Steiner symmetrizations to the symmetric decreasing rearrangement. The Competing Symmetries technique of Carlen and Loss will be explained in detail.
References:
A. Volcic, Random Steiner symmetrization of measurable sets.
http://arxiv.org/abs/0902.0462
A. Burchard, Short course on rearrangements (Section 3.2).
http://www.math.utoronto.ca/almut/rearrange.pdf
A. Burchard, Steiner Symmetrization is continous in W^{1,p}
(Theorem 3 and Sections 6-7). GAFA 7 (1997), 823-860.
http://www.math.utoronto.ca/almut/preprints/steiner.ps
## January 7, Pocovnicu, Thursday 12:10-1:00, Fields Institute Room 230
• Oana Pocovnicu (Orsay)
Title: Traveling waves for the cubic Szegö equation on the real line
Abstract: We consider the cubic Szegö equation on the real line. This equation was introduced by Gérard and Grellier on the circle as a toy model for non-dispersive evolution equations in studying the nonlinear Schrödinger equation on a sub-Riemannian manifold. It turns out that this equation is completely integrable. i.e., it has a Lax pair and there is an infinite sequence of conserved quantities. In this talk, after discussing its well-posedness in the Hardy space on the upper half-plane, we show that the only traveling waves are of the form C / (x − p), with Im p < 0. Moreover, they are shown to be stable, in contrast to the situation on the circle where some traveling waves were shown to be unstable.
(Some background information is available in the paper of Gérard-Grellier.)
## December 3, Erdös, Thursday 12:10-1:00. Fields Institute Room 210
• Laszlo Erdös (Munich)
Title: Dynamical formation of correlations in a Bose-Einstein condensate
Abstract: We consider the evolution of N bosons interacting with a repulsive short range pair potential in three dimensions. The potential is scaled according to the Gross-Pitaevskii scaling, i.e. it is given by N2V(N(xi − xj)). We monitor the behavior of the solution to the N-particle Schrödinger equation in a spatial window where two particles are close to each other. We prove that within this window a short scale interparticle structure emerges dynamically. The local correlation between the particles is given by the two-body zero energy scattering mode. This is the characteristic structure that was expected to form within a very short initial time layer and to persist for all later times, on the basis of the validity of the Gross-Pitaevskii equation for the evolution of the Bose-Einstein condensate. The zero energy scattering mode emerges after an initial time layer where all higher energy modes disperse out of the spatial window.
This is a joint work with A. Michelangeli and B. Schlein.
The paper of the same title is linked here.
Slides from the talk.
Numerical simulation of correlations.
## November 26, Jerrard, Thursday 12:10-1:00, Fields Institute Room 210
• Robert Jerrard (University of Toronto)
Title: Partial regularity for hypersurfaces minimizing elliptic parametric integrands Part II
Abstract: I will give one or two expository talks on an old paper of Schoen, Simon, and Almgren in which they prove that hypersurfaces in $R^{n+1}$ that solve certain geometric variational problems are smooth away from a closed set of $n-2$ dimensional Hausdorff measure 0. The geometric variational problems in question -- the "parametric elliptic integrands" mentioned in the title -- should be thought of as generalizations of the minimal surface problem.
1977_Schoen_Simon_Almgren_Acta
mathscinet
## November 19, Jerrard, Thursday 12:10-1:00, Fields Institute Room 210
• Robert Jerrard (University of Toronto)
Title: Partial regularity for hypersurfaces minimizing elliptic parametric integrands
Abstract: I will give one or two expository talks on an old paper of Schoen, Simon, and Almgren in which they prove that hypersurfaces in $R^{n+1}$ that solve certain geometric variational problems are smooth away from a closed set of $n-2$ dimensional Hausdorff measure 0. The geometric variational problems in question -- the "parametric elliptic integrands" mentioned in the title -- should be thought of as generalizations of the minimal surface problem.
1977_Schoen_Simon_Almgren_Acta, :mathscinet
## November 5, Lieb, Thursday 12:10-1:00. Fields Institute Room 210
• Elliott Lieb (Princeton University)
Title: Mathematics of the Bose Gas: A truly quantum-mechanical many-body problem
Abstract:The peculiar quantum-mechanical properties of the lowest energy states of Bose gases that were predicted in the early days of quantum-mechanics have finally been verified experimentally recently. The mathematical derivation of these properties from Schroedinger's equation has also been difficult, but much progress has been made in the last few years and some of this will be reviewed in this talk. For the low density gas with finite range interactions these properties include the leading order terms for the lowest state energy, the validity of the Gross-Pitaevskii equation in traps (including rapidly rotating traps), Bose-Einstein condensation and superfluidity, and the transition from 3-dimensional behavior to 1-dimensional behavior as the cross-section of the trap decreases. The phenomena described are highly quantum-mechanical, without a classical physics explanation, and it is very satisfying that reality and these mathematical predictions agree.
## October 29, Colliander, Thursday 12:10-1:00. Fields Institute Room 210
• J. Colliander (University of Toronto)
Title: Towards partial regularity for nonlinear Schrödinger?
Abstract: What happens at the end of life of an exploding solution of a nonlinear Schrödinger equation? This talk will describe ideas toward a description of the set of points where the solution becomes singular. In particular, a heuristic argument suggesting Hausdorff dimension upper bounds on the singular set will be presented. These upper bounds are saturated by recent examples of blowup solutions with thick singular sets. Comparisons with corresponding results for Navier-Stokes and other equations will also be discussed.
## October 22, Lee, Thursday 12:10-1:00. Fields Institute Room 210
• Paul Lee (University of California Berkeley)
Title: The Ma-Trudinger-Wang conditions for natural mechanical actions.
Abstract: The Ma-Trudinger-Wang conditions are important necessary conditions for the regularity theory of optimal transportation problems. In this talk, we will discuss new costs arising from natural mechanical actions which satisfy this condition. This is a joint work with R. McCann.
## October 15, McCann, Thursday 12:10-1:00. Fields Institute Room 210
• R.J. McCann (University of Toronto)
Title: An optimal multidimensional price strategy facing informational asymmetry, Part III
Abstract: The monopolist's problem of deciding what types of products to manufacture and how much to charge for each of them, knowing only statistical information about the preferences of an anonymous field of potential buyers, is one of the basic problems analyzed in economic theory. The solution to this problem when space of products and of buyers can each be parameterized by a single variable (say quality X, and income Y) garnered Mirrlees (1971) and Spence (1974) their Nobel prizes in 1996 and 2001, respectively. The multidimensional version of this question is a largely open problem in the calculus of variations described in Basov's book "Multidimensional Screening". I plan to give a couple of lectures explain recent progress with A Figalli and Y-H Kim, which identifies structural conditions on the value b(X,Y) of product X to buyer Y, which reduce this problem to a convex program in a Banach space--- leading to uniqueness and stability results for its solution, confirming robustness of certain economic phenomena observed by Armstrong (1996) such as the desirability for the monopolist to raise prices enough to drive a positive fraction of buyers out of the market, and yielding conjectures about the robustness of other phenomena observed Rochet and Chone (1998),such as the clumping together of products marketed into subsets of various dimension. Ideas from differential geometry / general relativity and optimal transportation are relevant to passage to several dimensions.
## October 8, Decker, Thursday 12:10-1:00. Fields Institute Room 230
• Colin Decker (University of Toronto)
Title: Uniqueness of matching in the marriage market.
Abstract: Economists are interested in studying marriage behaviour because it provides insight into a basic economic unit, the household, and because changes in marital behaviour offer insight into other social and economic variables of interest. Given agents described by multi dimensional discrete types, and their preferences, a competitive model of the marriage market describes how individuals will arrange themselves in marriage. Whether this described arrangement is unique is a key question and one that recurs in the study of matching markets.
Choo and Siow (2006) introduced a competitive model of the marriage market that incorporates several important features from economic theory. It is not known whether the Choo-Siow model predicts a unique marital arrangement given the preferences of agents. I will identify sufficient conditions on the preferences of agents to guarantee the existence of a unique marital arrangement. To achieve this result, Robert McCann, Ben Stephens and I adapted the continuity method, commonly used in the study of elliptic PDE, to the setting of isolating the positive roots of a system of polynomial equations.
## October 1, McCann, Thursday 12:10-1:00. Fields Institute Room 230
• R.J. McCann (University of Toronto)
Title: An optimal multidimensional price strategy facing informational asymmetry, Part II.
Abstract: See below.
## September 24, McCann, Thursday 12:10-1:00. Fields Institute Room 210
• R.J. McCann (University of Toronto)
Title: An optimal multidimensional price strategy facing informational asymmetry.
Abstract: The monopolist's problem of deciding what types of products to manufacture and how much to charge for each of them, knowing only statistical information about the preferences of an anonymous field of potential buyers, is one of the basic problems analyzed in economic theory. The solution to this problem when space of products and of buyers can each be parameterized by a single variable (say quality X, and income Y) garnered Mirrlees (1971) and Spence (1974) their Nobel prizes in 1996 and 2001, respectively. The multidimensional version of this question is a largely open problem in the calculus of variations described in Basov's book "Multidimensional Screening". I plan to give a couple of lectures explain recent progress with A Figalli and Y-H Kim, which identifies structural conditions on the value b(X,Y) of product X to buyer Y, which reduce this problem to a convex program in a Banach space--- leading to uniqueness and stability results for its solution, confirming robustness of certain economic phenomena observed by Armstrong (1996) such as the desirability for the monopolist to raise prices enough to drive a positive fraction of buyers out of the market, and yielding conjectures about the robustness of other phenomena observed Rochet and Chone (1998),such as the clumping together of products marketed into subsets of various dimension. Ideas from differential geometry / general relativity and optimal transportation are relevant to passage to several dimensions.
Organizational discussion and lunch to follow.
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http://mathhelpforum.com/differential-geometry/115591-continuity-1-x.html
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# Thread:
1. ## Continuity of 1/x
Show that 1/x is continuous on at any c =/= 0
hint: chose delta to stay away from 0
i have a general idea of how it works but I'm not really sure how to
"use delta = min{[c]/2,c^2(e/2)} "(answer from book)
2. Originally Posted by mtlchris
Show that 1/x is continuous on at any c =/= 0
hint: chose delta to stay away from 0
i have a general idea of how it works but I'm not really sure how to
"use delta = min{[c]/2,c^2(e/2)} "(answer from book)
Did you try writing out the definition?
$\left|\frac{1}{x}-\frac{1}{c}\right|=\left|\frac{c-x}{xc}\right|$....how look that the $\delta$. It's $\min$ because obviously we want to restrict our attention so that we may apply some estimation (i.e. on "whatever" neighborhood of c it is always true that "whatever" is less than or equal to "whatever else"). Also, use that in conjunction with the fact that you really only care about bounding $\frac{1}{|xc|}$ for one can make $|c-x|$ as arbitrarily small as one wants.
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http://crypto.stackexchange.com/questions/2970/can-i-pre-define-the-points-in-shamirs-secret-sharing-algorithm?answertab=active
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# Can I pre-define the points in Shamir's Secret Sharing algorithm
With Shamir's secret sharing is it possible to pre-define the points returned by the algorithm?
For simplicity if I have (k, n) where k=2, and n=4, and I have the points X,Y,Z, and Q. Can I create the coefficients so that X,Y,Z, and Q are the points returned from the algorithm.
If yes will it still be secure (or secure enough)?
Edit:
Change my example to k=2 instead of k=1 to keep the comments from degenerating into a discussion on the merits of only needing a single piece of knowledge to access the secret.
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k=1 doesn't require a sharing scheme? – emboss Jun 14 '12 at 17:21
If I have a secret S, and I want 4 people to have access to it, then I can make k=1. Each person then has their own 'password' to the secret. Maybe not a likely scenario but a valid one. – grieve Jun 14 '12 at 18:32
But to my understanding k=1 means having one share alone allows you to reconstruct the secret directly, without needing another share. In SSS, this results in handing S to the 4 people directly, because a_0 = S and there are no random coefficients to be chosen. – emboss Jun 15 '12 at 13:38
I just set K=1 as an example, my question still applies where k > 1. I think focusing on that is not productive, so I am editing the question. – grieve Jun 18 '12 at 20:13
Is there a way to move it there? – grieve Jun 18 '12 at 21:02
show 2 more comments
## 3 Answers
As poncho and Maeher have noted, this isn't possible with straightforward Shamir's secret sharing.
In fact, it's pretty obvious, once you think about it, that there's no way to choose more than $k$ shares independently in advance and get a consistent secret out of them with any unconditionally secure $k$-out-of-$n$ threshold secret sharing scheme, even if you allow the secret to be random.
In particular, imagine you have $k+1$ independently chosen shares $s_1, s_2, \dotsc, s_{k+1}$. By definition, either of the $k$-tuples $(s_1, s_2, \dotsc, s_k)$ and $(s_2, s_3, \dotsc, s_{k+1})$ are enough to fully reconstruct the secret, whereas the $k-1$-tuple $(s_2, s_3, \dotsc, s_k)$ alone provides no information about the share. This means that, for every pair of secrets $S$ and $S'$ and any given $k-1$ shares $s_2, s_3, \dotsc, s_k$, there must exist shares $s_1$ and $s_{k+1}$ such that $(s_1, s_2, \dotsc, s_k)$ yields the secret $S$ while $(s_2, s_3, \dotsc, s_{k+1})$ yields $S'$.
All that said, however, there is a simple way to modify Shamir's secret sharing scheme so that the shares can be chosen independently in advance.
Namely, let $s_1, s_2, \dotsc, s_n$ denote the shares chosen in advance, and let $S$ be the secret you wish to share. Now compute $n$ "auxiliary" shares $a_1, a_2, \dotsc, a_n$ using ordinary Shamir's secret sharing on $S$, and add them to the corresponding pre-chosen shares to get $$p_i := a_i \oplus s_i$$ for all $i \in \{1, 2, \dotsc, n\}$, where $\oplus$ denotes addition in the finite field the shares belong to.
Finally, publish the results $p_1, p_2, \dotsc, p_n$.
Now, to reconstruct the secret $S$, any $k$ participants may each subtract their share $s_i$ from the public $p_i$ (or just reveal their $s_i$ and let others combine it with $p_i$) to obtain the corresponding auxiliary shares $a_i$ and reconstruct $S$ from them.
Meanwhile, assuming that the shares $s_1, s_2, \dotsc, s_n$ are randomly chosen from the finite field used, merely knowing $p_i$ but not $s_i$ for any $i$ provides no information about the corresponding auxiliary share $a_i$. Thus, knowing less than $k$ of the shares will not reveal any information about the secret.
(Of course, if the pre-chosen shares are not uniformly chosen random values, then the unconditional security property will not hold. One may still obtain at least computational security, up to the limits set by brute force guessing attacks, by feeding the pre-chosen shares through a secure cryptographic hash function before combining them with the auxiliary shares.)
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With Shamir's secret sharing for the case of $k$ out of $n$, you construct a random polynomial of degree $k-1$, because $l$ points uniquely identify a polynomial of degree $k-1$.
The usual way to construct said polynomial is to uniformly choose coefficients $a_1$ through $a_{k-1}$ and setting $a_0=s$.
There is an alternative to that, where you choose $k-1$ points uniformly at random. The $k$th point is then $(0,s)$, you can compute the polynomial defined by those points and then compute the remaining $n-(k-1)$ points. There is absolutely no difference between the two approaches.
So in summary: No that is not possible. You can freely choose at most $k-1$ shares. And also for the security to hold, those must be chosen uniformly at random.
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Well, no, in general, you won't be able to select arbitrary points $X, Y, Z$ and $Q$ that would work in a Shamir secret sharing scheme (unless $k\ge4$).
Here's why; Shamir's scheme assumes that the points are on some polynomial of degree $k-1$ or less; for $k=2$, this means that the points all must be solutions of a linear equation $y = a_1 x + a_0$ for some constants $a_1, a_0$. There will usually not be any such constants that will simultaneously work for all four points, and so Shamir's scheme will not work.
BTW: what problem are you trying to solve? You mention "security"; what do you mean by that? If you want to prevent someone from doing something, what is that something?
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I have a client/server architecture, where each client can have multiple username/passwords pairs. I wanted to encrypt data on the server side such that any client user logging in would be able to decrypt it. The server would randomly generate the encryption key, and then use a combination of the passwords of the users and Samir's Secret Sharing to allow the users to decrypt the encryption key, and then the data. I actually found a completely different way to do this, after posting this question, but was still interested to see if it was possible. – grieve Jun 20 '12 at 19:16
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http://math.stackexchange.com/questions/221841/poisson-or-binomial/221845
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# Poisson or Binomial?
In a fabric company, the fixed probability of a machine producing a bad fabric is $p$ independent on the fabric previously produced. At the output of the machine, $n$ fabrics are taken at random. $X$ is the random variable equal to the number of bad fabrics in a selection of $n$ fabrics at the mouth of the machine.
• What probability distribution law does $X$ follow and why?
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## 3 Answers
This fits perfectly the definition of a binomial distribution. There are $n$ independent trials, and in each trial the probability of "success," if you can call it success, is $p$. It is exactly like tossing a funny coin $n$ times, with the probability of a head equal to $p$, and our random variable the number of heads.
To put it another way, let $X_i=1$ if the $i$-th fabric is defective, and $0$ if it is not. Then $$X=\sum_{i=1}^n X_i,$$ and the sum of $n$ independent identically distributed Bernoulli random variables has binomial distribution.
Remark: There is some connection with the Poisson. If $n$ is large and $p$ is small, with $np$ of modest size, then the binomial distribution of this problem is well-approximated by the Poisson with parameter $\lambda=np$. But it is still a binomial.
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I understand this, but in my question, one only needs the number of defective fabrics whatever the order of arrangement. Unless what was considered was the order at which the fabrics come out of the machine? – F'Ola Yinka Oct 27 '12 at 1:29
The Binomial distribution measures the number of successes. The probability that $X=k$ is $\binom{n}{k}p^k(1-p)^{n-k}$. It is true that when we derive this formula, we consider all the possible orders in which the successes and/or failures could occur. – André Nicolas Oct 27 '12 at 1:34
so where does this "order" that was considered falls in my question. Is it at the selection of the fabrics from the mouth of the machine or is it the order at which the fabrics came out of the machine? – F'Ola Yinka Oct 27 '12 at 1:36
It is an arbitrary order. After we have chosen our $n$ fabrics, we line them up in a row without testing them for quality. But remember, this is how we derive the formula. Once we have derived the formula, we need not consider order at all. – André Nicolas Oct 27 '12 at 1:43
I guess I just have to accept it that way. Even if the order is arbitrary, it is at least acceptable. Thank you for your answer and time. – F'Ola Yinka Oct 27 '12 at 1:46
Defective vs. not defective usually implies binomial. If $n$ is sufficiently large and $p$ (probability of success) is very small, the Poisson distribution can be used to approximate the binomial distribution.
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Use binomial when each observation is one of two possible outcomes, e.g. a random sample of size $n$ is drawn and the number of defects found in the sample is recorded. Use Poisson when dealing with rates of occurrence e.g. if the number of defects per length of fabric exceeds $L$ then etc. – glebovg Oct 27 '12 at 1:30
Binomial would be more appropriate here. See the machine has a Bernoulli trial that has succes probability $p$. Because of independance, each of the $n$ fabrics you took have a $p$ chance of being bad. The number of bad fabrics is then given by the binomial distribution.
Poisson would give you an approximation of this, but $n$ would have to be large enough, and $np$ small enough.
Poisson distribution are more related to problems of the type : A machine produces fabrics at random. In $5$ minutes, it produces on average $3$ fabrics. What is the probability that you have $5$ bad fabrics after $10$ minutes?
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but doesn't binomial always consider the arrangement of the fabric? – F'Ola Yinka Oct 27 '12 at 1:22
What do you mean by arrangement? – Jean-Sébastien Oct 27 '12 at 1:24
sorry. what i meant was that the order of selection is put into consideration when using binomial. – F'Ola Yinka Oct 27 '12 at 1:25
Unless you know any kind of information about the fabrics before taking your sample, you can imagine your sample as soon as it come out of the machine, as i the experience had just been realized. If however you wait the end of the day, you know that the machine produced say $k$ bad fabrics and you take a sample of $n$,then $X$ will be Hypergeometric – Jean-Sébastien Oct 27 '12 at 1:28
I understand you but you didn't answer my question. When binomial is used, the order of selection is always considered which I don't see in this case. – F'Ola Yinka Oct 27 '12 at 1:33
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http://mathhelpforum.com/statistics/186700-desired-exam-question-paper-probability-question.html
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# Thread:
1. ## The desired exam question paper (probability question)
There is an exam. There are 25 exam question papers for 25 students who are lined up outside of the classroom. Each time, a students gets into the classroom, pick one paper (RANDOMLY) and go, sits on its desk and start his exam.
You are the 5th person in the line, who has to go to the class and pick a paper. Among those 25 papers, there is only 1 paper which questions are familiar to you.
Now the problem is: What is the probability that you will go into the classroom, and pick that one paper you want. (In other word, the probability that NO other student before you, will pick that exam paper which questions are familiar to you)
Thank you in advance.
2. ## Re: The desired exam question paper (probability question)
Hello, Narek!
The problem seems to be very simple.
Perhaps I'm misinterpreting it?
There is an exam. There are 25 different exam papers
for 25 students who are lined up outside of the classroom.
Each student enters the room, randomly picks a paper and starts his exam.
You are the 5th person in the line, who has to enter the room and pick a paper.
Among those 25 papers, there is only ine paper whose questions are familiar to you.
What is the probability that you will enter the classroom and pick that one paper?
That is, the probability that NO other student before you will pick that exam paper.
Let that special exam paper be called the "Exam".
For you to get the Exam, it must be the fifth one selected.
The probability that the Exam is the fifth one chosen is: . $\frac{1}{25}$
We can check this answer.
For you to get the Exam, the first four students must NOT get the Exam,
. . then you must get the Exam.
This probability is: . $\frac{24}{25}\cdot \frac{23}{24}\cdot\frac{22}{23}\cdot \frac{21}{22}\cdot \frac{1}{21} \;=\;\frac{1}{25}$
3. ## Re: The desired exam question paper (probability question)
that was an elegant way to solve the problem SOROBAN....
4. ## Re: The desired exam question paper (probability question)
Originally Posted by Narek
There is an exam. There are 25 exam question papers for 25 students who are lined up outside of the classroom. Each time, a students gets into the classroom, pick one paper (RANDOMLY) and go, sits on its desk and start his exam.
You are the 5th person in the line, who has to go to the class and pick a paper. Among those 25 papers, there is only 1 paper which questions are familiar to you.
Now the problem is: What is the probability that you will go into the classroom, and pick that one paper you want. (In other word, the probability that NO other student before you, will pick that exam paper which questions are familiar to you)
Thank you in advance.
It does not matter than some students have already gone in and taken some of the papers. The probability of you getting the one you want is still 1/25 however many have already gone.
CB
5. ## Re: The desired exam question paper (probability question)
Thank you for the answers. It is completely clear.
6. ## Re: The desired exam question paper (probability question)
Exactly, imagine, you are the last person entering the room, the odds of you getting that table will be 1/25. It is like playing poker, you presume that no one has the cards you are looking for
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http://mathoverflow.net/questions/107796/integrating-a-product
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## Integrating a product
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
By trying to find a marginal distribution I came accross integration of the product series. For the sake of generality, lets assume the integral is of following form: $$\int \prod_{k=1}^{n}\left ( x+a_{k} \right )^{b_{k}}dx.$$ $a_{k}$ is a real coefficient and $b_{k}$ is positive integers. Is there any method, that could be used to integrate this analyticaly? It is of no problem to calculate it by some numerical method. But the first thing I would like to try is to find its analytical expression.
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2
And $b_k$ are what? Positive integers? If so, answer is a polynomial. Integers not necessarily positive? Use partial fractions. In most other cases, closed-form expressions are not likely. – Robert Israel Sep 21 at 19:35
1
What do you mean in the question by 'a series of products'? You may already know that each term $(x+a_k)^{b_k}$ may be expanded as a binomial series which has a positive but finite radius of convergence if $b_k$ is not a positive integer. These series may be multiplied, and then you can integrate term by term to get a series expansion of your antiderivative. – Stopple Sep 21 at 19:54
What are the $b_k$ and what are the limits of integration? – Alexandre Eremenko Sep 22 at 1:06
## 2 Answers
Since the derivative of such a product has a similar form $$\left(\sum_{k=1}^n\frac{b_k}{x+a_k} \right)\prod_{k=1}^{n}\left ( x+a_{k} \right )^{b_{k}},$$ One might hope that something similar is true of some antiderivative. (I am implicitly assuming that the $b_k$ are positive integers, although the formula is valid more generally. However there are cases that will involve some $\ln |x+a_k|$ . I will continue to neglect these.)
If we imagine that we have some one antiderivative $$F(x)=\int \prod_{k=1}^{n}\left ( x+a_{k} \right )^{b_{k}}dx$$ and let $F_k=F(x)-F(-a_k),$ then $F_k(x)$ is also an antiderivative and is divisible by $(x+a_k)^{b_k+1}$
That is not particularly deep.
We can use integration by parts to obtain an answer without any explicit multiplying although marching through to the end will not be very pretty. It is more satisfying to know that we can do it than to actually do so.
1. As we know, $\int(x+a_1)^{b_1}dx=\frac{1}{b_{1}+1}(x+a_1)^{b_1+1}$
2. We can get to the above form by $b_2$ iterations of $\int(x+a_1)^{b_1}(x+a_2)^{b_2}dx=\frac{1}{b_{1}+1}\left((x+a_1)^{b_1+1}(x+a_2)^{b_2}-b_2\int(x+a_1)^{b_1+1}(x+a_2)^{b_2-1}dx\right)$
3. Similarly, we can eventually get from three factors to two etc.
It would be only slightly more complicated (but also only slightly more satisfying) to have products of factors $(m_kx+a_k)^{b_k}$.
A large table of integrals might have formulas such as this one.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The magic words are "Lauricella Hypergeometric Function".
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http://en.wikipedia.org/wiki/Gibbs'_phase_rule
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# Gibbs' phase rule
Gibbs' phase rule[1] [2] was proposed by Josiah Willard Gibbs in the 1870s as the equality
$F\;=\;C\;-\;P\;+\;2$
where C is the number of components and P (alternatively π or Φ) is the number of phases in thermodynamic equilibrium with each other. Typical phases are solids, liquids and gases. A system involving one pure chemical is an example of a one-component system. Two-component systems, such as mixtures of water and ethanol, have two chemically independent components. F is the number of degrees of freedom, which means the number of intensive properties such as temperature or pressure, which are independent of other intensive variables. There are rare cases when it does not work, for instance when two substances have the same triple point, in which case, the equation predicts that there are -2 degrees of freedom.[citation needed]
## Foundations
• A phase is a form of matter that is homogeneous in chemical composition and physical state. Typical phases are solid, liquid and gas. Two immiscible liquids (or liquid mixtures with different compositions) separated by a distinct boundary are counted as two different phases, as are two immiscible solids.
• The number of components (C) is the number of chemically independent constituents of the system, i.e. the minimum number of independent species necessary to define the composition of all phases of the system.[2] For examples see component (thermodynamics).
• The number of degrees of freedom (F) in this context is the number of intensive variables which are independent of each other.
The basis for the rule (Atkins and de Paula,[2] justification 6.1) is that equilibrium between phases places a constraint on the intensive variables. More rigorously, since the phases are in thermodynamic equilibrium with each other, the chemical potentials of the phases must be equal. The number of equality relationships determines the number of degrees of freedom. For example, if the chemical potentials of a liquid and of its vapour depend on temperature (T) and pressure (p), the equality of chemical potentials will mean that each of those variables will be dependent on the other. Mathematically, the equation μliq(T, p) = μvap(T, p), where μ = chemical potential, defines temperature as a function of pressure or vice versa. (Caution: do not confuse p = pressure with P = number of phases.)
To be more specific, the composition of each phase is determined by C – 1 intensive variables (such as mole fractions) in each phase. The total number of variables is (C–1)P + 2, where the extra two are temperature T and pressure p. The number of constraints are C(P–1), since the chemical potential of each component must be equal in all phases. Subtract the number of constraints from the number of variables to obtain the number of degrees of freedom as F = (C–1)P + 2 – C(P–1) = C – P + 2.
The rule is valid provided the equilibrium between phases is not influenced by gravitational, electrical or magnetic forces, or by surface area, and only by temperature, pressure, and concentration.
## Consequences and examples
### Pure substances (one component)
For pure substances C = 1 so that F = 3 – P. In a single phase (P = 1) condition of a pure component system, two variables (F = 2), such as temperature and pressure, can be chosen independently to be any pair of values consistent with the phase. However, if the temperature and pressure combination ranges to a point where the pure component undergoes a separation into two phases (P = 2), F decreases from 2 to 1. When the system enters the two-phase region, it becomes no longer possible to independently control temperature and pressure.
Carbon dioxide pressure-temperature phase diagram showing the triple point and critical point of carbon dioxide
In the phase diagram to the right, the boundary curve between the liquid and gas regions maps the constraint between temperature and pressure when the single-component system has separated into liquid and gas phases at equilibrium. If the pressure is increased by compression, some of the gas condenses and the temperature goes up. If the temperature is decreased by cooling, some of the gas condenses, decreasing the pressure. Throughout both processes, the temperature and pressure stay in the relationship shown by this boundary curve unless one phase is entirely consumed by evaporation or condensation, or unless the critical point is reached. As long as there are two phases, there is only one degree of freedom, which corresponds to position along the phase line.
The critical point is the black dot at the end of the liquid-gas boundary. As this point is approached, the liquid and gas phases become progressively more similar until, at the critical point, there is no longer a separation into two phases. Above the critical point and away from the phase boundary curve, F = 2 and the temperature and pressure can be controlled independently. Hence there is only one phase, and it has the physical properties of a dense gas, but is also referred to as a supercritical fluid.
Of the other two-boundary curves, one is the solid–liquid boundary or melting point curve which indicates the conditions for equilibrium between these two phases, and the other at lower temperature and pressure is the solid–gas boundary.
Even for a pure substance, it is possible that three phases, such as solid, liquid and vapour, can exist together in equilibrium (P = 3). If there is only one component, there are no degrees of freedom (F = 0) when there are three phases. Therefore, in a single-component system, this three-phase mixture can only exist at a single temperature and pressure, which is known as a triple point. Here there are two equations μsol(T, p) = μliq(T, p) = μvap(T, p), which are sufficient to determine the two variables T and p. In the diagram for CO2 the triple point is the point at which the solid, liquid and gas phases come together, at 5.2 bar and 217 K. It is also possible for other sets of phases to form a triple point, for example in the water system there is a triple point where ice I, ice III and liquid can coexist.
If four phases of a pure substance were in equilibrium (P = 4), the phase rule would give F = −1, which is meaningless, since there cannot be −1 independent variables. This explains the fact that four phases of a pure substance (such as ice I, ice III, liquid water and water vapour) are not found in equilibrium at any temperature and pressure. In terms of chemical potentials there are now three equations, which cannot in general be satisfied by any values of the two variables T and p, although in principle they might be solved in a special case where one equation is mathematically dependent on the other two. In practice, however, the coexistence of more phases than allowed by the phase rule normally means that the phases are not all in true equilibrium.
### Two-component systems
For binary mixtures of two chemically independent components, C = 2 so that F = 4 – P. In addition to temperature and pressure, the other degree of freedom is the composition of each phase, often expressed as mole fraction or mass fraction of one component.
Boiling Point Diagram
As an example, consider the system of two completely miscible liquids such as toluene and benzene, in equilibrium with their vapours. This system may be described by a boiling-point diagram which shows the composition (mole fraction) of the two phases in equilibrium as functions of temperature (at a fixed pressure).
Four thermodynamic variables which may describe the system include temperature (T), pressure (p), mole fraction of component 1 (toluene) in the liquid phase (x1L), and mole fraction of component 1 in the vapour phase (x1V). However since two phases are in equilibrium, only two of these variables can be independent (F = 2). This is because the four variables are constrained by two relations: the equality of the chemical potentials of liquid toluene and toluene vapour, and the corresponding equality for benzene.
For given T and p, there will be two phases at equilibrium when the overall composition of the system (system point) lies in between the two curves. A horizontal line (isotherm or tie line) can be drawn through any such system point, and intersects the curve for each phase at its equilibrium composition. The quantity of each phase is given by the lever rule (expressed in the variable corresponding to the x-axis, here mole fraction).
For the analysis of fractional distillation, the two independent variables are instead considered to be liquid-phase composition (x1L) and pressure. In that case the phase rule implies that the equilibrium temperature (boiling point) and vapour-phase composition are determined.
Liquid-vapour phase diagrams for other systems may have azeotropes (maxima or minima) in the composition curves, but the application of the phase rule is unchanged. The only difference is that the compositions of the two phases are equal exactly at the azeotropic composition.
## Phase rule at constant pressure
For applications in materials science dealing with phase changes between different solid structures, pressure is often imagined to be constant (for example at one atmosphere), and is ignored as a degree of freedom, so the rule becomes
F = C − P + 1 .
This is sometimes misleadingly called the "condensed phase rule", but it is not applicable to condensed systems which are subject to high pressures (for example in geology), since the effects of these pressures can be important.
## References
1. Gibbs, J.W., Scientific Papers (Dover, New York, 1961)
2. ^ a b c Atkins, P.W.; de Paula, J. (2006). Physical chemistry (8th ed.). Oxford University Press. ISBN 0-19-870072-5. Chapter 6
## Further reading
• Mogk, David: Teaching Phase Equilibria. Gibbs' Phase Rule: Where it all Begins (The phase rule in geology)
• Predel, Bruno; Hoch, Michael J. R; Pool, Monte. Phase Diagrams and Heterogeneous Equilibria : A Practical Introduction. Springer. ISBN 3-540-14011-5.
• White, Mary Anne. Properties of Materials. Oxford University Press (1999). ISBN 0-19-511331-4. Chapter 9. Thermodynamics Aspects of Stability
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http://physics.stackexchange.com/questions/54861/in-klein-gordon-why-should-infinite-downwards-photon-cascades-be-possible
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# In Klein-Gordon, why should infinite downwards photon cascades be possible?
Here is a simple point about the standard interpretation of the Klein-Gordon equation that for the life of me I've never been able to understand:
Why would the existence of true negative energy states result in an infinite downward cascade of photon emissions?
I believe it was Dirac who originated this concept, but no matter how I turn it around in my head, I don't understand the logic of it.
A suite of true negative energy particle solutions to Klein-Gordon (not antimatter, which experimentally shows positive energy) would by exact symmetry behave identically to their positive energy equivalents. Thus they would radiate negative energy photons until they reach their own ground states.
So, if viewed from that true negative energy perspective, a downward cascade of positive energy photons would look like an ordinary electron raising its energy indefinitely by emitting negative energy photons. But we don't see that from within out positive particle perspective, so why should we expect the downward version to exist, either?
Is there some critical conceptual and/or mathematical rationale that I'm missing for why infinitely-downward photon cascades should exist?
I guess by implication that I may also be asking this: Why is Klein-Gordon not simply interpreted as having four solutions, rather than just two? They would include the known positive energy "pro" and "anti" matter solutions that share positive energy photons, but also a set of true negative energy versions of matter, antimatter, and photons. Surely this idea has at least been explored and then dismissed for some specific reason? I've looked some, but have not had any luck finding references.
To balance out the total energy of the universe, most current theories seem to assign the negative energy to space itself, again for reasons I truly do not understand. Such an approach leads unavoidably to complex balancing acts of trying to get the very differently structured positive and negative energies always to match in magnitude. Wouldn't it be both easier and simpler to have a true, exactly symmetric negative-energy universe scurrying away from us in the $t^-$ direction, with space simply as the zero-energy balancing point between the two?
Again, there must be some specific analysis that led to the current reliance on the more complicated "no true negative energy particles allowed" approach to balancing the energy of the universe (multiverse really), but I've had no luck uncovering it. What am I missing? Who made the decision to go to negative-energy-space balancing, and why?
2013-02-26 - Some relevant references
To my surprise, Paul Dirac's 1933 Nobel Lecture does a clear and remarkably succinct job of explaining his logic not just about negative energy states, but how he wound up at antimatter by contemplating them. At one point I immediately thought of tachyons. After checking a bit, sure enough, those ideas later became at least an aspect of tachyon theory. Odd how that feature of his negative energy sea is almost never mentioned.
As you can likely guess from the tachyon connection, Dirac interpreted the kinetics of negative energy states in an intriguing way that I was not expecting at all:
If one looks at Eq. (1), $[\frac{W^2}{c^2} - p_r^2 - m^2c^2]\psi = 0$, one sees that it allows the kinetic energy $W$ to be either a positive quantity greater than $mc^2$ or a negative quantity less than $-mc^2$. This result is preserved when one passes over to the quantum equation (2), $[\frac{W}{c} - {\alpha_r}p_r - {\alpha_0}mc]\psi = 0$, or (3) {$\alpha_\mu=I$; $\alpha_{\mu}\alpha_{\nu} + \alpha_{\nu}\alpha_{\mu} = 0$; $\alpha_{\mu}\neq\alpha_{\nu}$; $\mu,\nu = 0, 1, 2, 3$}. These quantum equations are such that, when interpreted according to the general scheme of quantum dynamics, they allow as the possible results of a measurement of $W$ either something greater than $mc^2$ or something less than $-mc^2$.
Now in practice the kinetic energy of a particle is always positive. We thus see that our equations allow of two kinds of motion for an electron, only one of which corresponds to what we are familiar with. The other corresponds to electrons with a very peculiar motion such that the faster they move, the less energy they have, and one must put energy into them to bring them to rest.
(I must remark that it is delightful to observe the little mutations in notation, with $W$ in place of $E$ and $\alpha$ in place of $\gamma$. Einstein's famous $E=mc^2$ is another example; it looked quite different in the original paper.)
Seeing the original version also reminds me of the remarkable number of choices that Dirac had to make to get to his matrices. The reasons for some of his choices were obvious, but for others they were not. In any case, his choice of an inverted-velocities interpretation of the kinematics provides some opportunity for understanding his original logic more meaningfully.
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## 1 Answer
The issue here is with this sentence:
"Thus they would radiate negative energy photons until they reach their own ground states."
Negative energy means that there is no ground state for these particles, in the sense that they would eventually stop cascading at some energy, say $-E_0<0$. In actuality we generally only care that the spectrum be bounded below, since we can add a constant to all energies to set that bound equal to zero anyway. So a negative energy state is more or less defined as a state that is unstable against this cascading process towards lower and lower energy.
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alexarvanitakis, thanks. I am pondering what you said... – Terry Bollinger Feb 23 at 20:16
... and still thinking. My image is a bit like like mirror image number lines overlaid on top of each other, sharing zero, but distinguished e.g. by the chiral structure of the two particle sets. This is not a typical image, I realize. :) More later, must go now. – Terry Bollinger Feb 23 at 20:35
Αρβανιτάκης Αλέξανδρος, thanks again for addressing this; I see you are well versed in topics such as Banach spaces. I am indeed proposing (without having intended to) a full symmetry that cannot be expressed with a single number line. I do not know a good terminology for that; sorry. Yet the idea remains very simple conceptually: Each side of zero perceives the other side as negative, and the sum of two particles is not energy, but a null state. It would be akin to a virtual electron/positron pair. The two universes moving apart along $t+$ and $t-$ would quite literally form a virtual pair. – Terry Bollinger Feb 24 at 2:37
And again, my intent was not to propose new physics. Surely this has been discussed and analyzed... somewhere? The generalization of virtual pairs to null-energy universe pairs just seems way too obvious not to have been addressed.. somewhere, surely? I feel like I must be looking in all the wrong places, nothing more. – Terry Bollinger Feb 24 at 2:39
That's another guy, no relation. Anyway this new proposal of yours doesn't make any sense to me. \shrug – alexarvanitakis Feb 24 at 3:40
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http://mathoverflow.net/questions/8396?sort=newest
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Does every ODE comes from something in physics?
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Not sure if this is appropriate to Math Overflow, but I think there's some way to make this precise, even if I'm not sure how to do it myself.
Say I have a nasty ODE, nonlinear, maybe extremely singular. It showed up naturally mathematically (I'm actually thinking of Painleve VI, which comes from isomonodromy representations) but I've got a bit of a physicist inside me, so here's the question. Can I construct, in every case, a physical system modeled by this equation? Maybe even just some weird system of coupled harmonic oscillators, something. There are a few physical systems whose models are well understood, and I'm basically asking if there's a construction that takes an ODE and constructs some combination of these systems that it controls the dynamics of.
Any input would be helpful, even if it's just "No." though in that case, a reason would be nice.
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I'm wondering where your interest in Painleve VI comes from. Frobenius manifolds, perhaps? Is this question related to anything else, or is it just something you're wondering about? – Kevin Lin Dec 10 2009 at 0:37
Well, the question of physical systems was just idle curiosity. My interest in Painleve VI is coming from a couple of papers by Boalch (one of which I linked to below) that talk about it, irregular connections, etc, and some relations of that stuff to Higgs bundles and the Hitchin System – Charles Siegel Dec 10 2009 at 0:44
You know,this is a good question and I never thought of it before. I suppose in principle,sure-but it's kind of like building an amusement park on the ocean floor that people couldn't actually use.In principle,you certainly could.But why would you? – Andrew L Aug 13 2010 at 22:04
7 Answers
It is possible to solve a large class of ODEs by means of analog computers. Each of the pieces of the differential equation corresponds to an electronic component and if you wire them up the right way you get a circuit described by the ODE. Wikipedia has lots of information on the subject and a link like this one gives explicit examples of circuits. It's not hard to build circuits for things like the Lorenz equation and see a nice Lorenz attractors on an oscilloscope display.
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1
From the setup, it seems like it can only handle certain types of nonlinearity, and it seems with constant coefficients. The specific equation I'm looking at is on page three of this paper: arxiv.org/pdf/math/0308221v3 and it seems nastier than any of the examples by a rather large amount. It might still work, but I don't see it. – Charles Siegel Dec 10 2009 at 0:11
I just pointed you to the simplest example of how to derive a circuit from an ODE, but sophisticated analog computers can add, subtract, multiply, integrate, differentiate, and even divide (using logarithms), so your differential equation is in the closure of this set of possible operations, apart from the singularities. If you expect to pass through the singularities then you'll probably have to transform coordinates and it might get messy. – Dan Piponi Dec 10 2009 at 0:19
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No. Let us take an ODE with a blowing up solution (x'=x^2,x(0)=1) o let us take an ODE without uniqueness of solutions (x'=x^(1/3),x(0)=0). There is no possible physical interpretation for such kinds of equations. Obviously, the addition of new terms in such equations can yield a physically implementable problem, but in such a case the resulting ODE is not the same as the original one.
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1
There are physical situations where uniqueness of solutions does not hold because of symmetry breaking, for example, the buckling of a beam under the force directed along its axis. – Victor Protsak Aug 14 2010 at 4:44
For the general question I vote no. As you formulated I see it, potentially, as a question of differential algebra. Although there is no definition of physical proccess I imagine it can be formalised through constraints on the field extensions allowed to solve your equation. The prototypical result I have in mind is Liouville's Theorem.
If instead you specialize the general question to Painleve's equation then I would bet that the answer is yes. Painlevé equation is one of these ubiquitous objects in Mathematics and I would not be surprised if it models a physical phenomena. I already crossed with a Springer Lecture Notes relating it to the geometry of surfaces.
As a side remark let me notice that Painlevé's equations were originally found not as equations governing isomonodromic deformations but instead as non-linear second order equations having the so called Painlevé property(absence of movable singular points).
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This is very far from my expertise, but I believe that the point of Universal Differential Equations is that any behavior exhibited by any differential equation can be found in a UDE for certain choices of parameter. (Think by analogy to a universal Turing machine, which can mimic any Turing machine.) So, if you are willing to idealize a lot, it would be enough to find a physical model for a UDE.
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If you're going to do that then we could just use zeta function universality, and the fact that it is known how to build an analogue computer to compute the zeta function, to claim we can solve any differential equation. :-) – Dan Piponi Dec 10 2009 at 2:27
I am not sure it would be enough. Indeed I suspect that the generic differential equation (in any reasonable sense) is an universal differential equation. There are results which ensures that any germ of holomorphic map f:(0)(0) can be approximated by a holonomy map of a rational differential equations of the form dydx=P(xy)Q(xy) . See "The generic rational differential equation dw/dz=P(z,w)/Q(z,w) on $\mathbb{C}\mathbb{P}^2$ carries no interesting transverse structure" by Belliart, Liousse, and Loray. Erg. Theory and Dyn. Syst., 21 (2001), p.1599-1607. – jvp Dec 10 2009 at 2:55
I vote no. Let n be a large enough positive integer that larger numbers can't be reasonably written using primitive recursion. Take a generic nonlinear ODE involving about n terms with derivatives of order around n. I'd claim that this doesn't model anything physical, and it can't be integrated by any device that fits in the universe.
If you demand that the ODE be reasonably small (e.g., mathematically interesting), then it tautologically models the behavior of a device set up to integrate it. I don't really have an answer to the broader question of why certain differential equations show up in areas of mathematics close to physics where you don't really expect them.
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1
I'd argue that the notion of "physical modeling" is independent of whether a physical model can be realized in the real world. After all, physicists talk about infinite system sizes all the time, or idealized behavior of particles and matter in regimes which can never be tested. From Charles's question, it seems he's willing to accept at least some degree of artificiality in the models. – jc Dec 10 2009 at 1:39
In that case, I suppose one can always construct a device in an infinite-volume universe in which information propagates at infinite speed (and there is no gravity). – S. Carnahan♦ Dec 10 2009 at 1:47
A fairly silly answer is that the answer is obviously "Yes", since one can build a computer to integrate numerical solutions to your ODE. (now that I think about it, Sigfpe's answer is essentially the same as mine.)
Going along these lines I guess one can find more "physical" models of my suggestion (in the sense that doing physics is often finding toy models that contain the essence of the phenomena etc) by proposing various lattice models or cellular automata which are known to have universal constructors. Or by designing circuits made out of balls and springs.
I spent a little bit of time trying to put down the right words which would make your question more precise and more in line with your intent, but I think ultimately it boils down to what kind of physical models you'd be satisfied with.
As much of physics can be described in terms of ODEs, any sufficiently powerful type of model is going to contain the sort of answer I described above. I think the right question is what's the "simplest" (or perhaps "weakest") known physical model for Painlevé VI.
One kind of answer to that question would be finding a physical system for which some solution of Painlevé VI gives some physically measurable function - along these lines, I know that Painlevé functions are highly useful in various integrable models / lattice models, e.g. famously, the spin-spin correlation function in the 2D Ising model in the scaling limit is a solution to Painlevé III - thus, my guess is that Painlevé VI shows up in one of these contexts, but the literature is pretty vast.
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For particular parameters, Painlevé VI describes the spin-spin correlations of the square lattice Ising model with diagonal separation between spins. (No scaling limit has been taken.) See "Studies on holonomic quantum fields, XVII". Michio Jimbo and Tetsuji Miwa, Proc. Jpn. Acad., Ser. A, Vol. 56, 405-410 (1980). A crucial erratum appears as an appendix to "The $\tau$ function of the Kadomtsev-Petviashvili equation transformation groups for soliton equations, I". Masaki Kashiwara and Tetsuji Miwa, Proc. Jpn. Acad., Ser. A, Vol. 57, 342-347 (1981). – Will Orrick Dec 10 2009 at 1:24
Nice find, though I don't have access. Amazing that the series of papers goes all the way from I to XVII. – jc Dec 10 2009 at 1:46
This may not be the answer that you are looking for, but I believe that you should be able to write the Painlevé VI equation as a hamiltonian system, in which case it would govern the dynamics of some "physical" system. The reason for the double quotes is that this is perhaps not a system arises in nature. Most likely -- although I don't know for sure -- it will not be just coupled harmonic oscillators.
Less directly, Painlevé equations arise in the study of integrable hierarchies, some of which (e.g., KdV, nonlinear Schrödinger,...) are used to model natural phenomena.
Edit
An explicit form for the Hamiltonian of Painlevé VI can be found here, right after Theorem 2.1. Although it is polynomial, it does depend explicit on 'time'. Hence as a hamiltonian system it is certainly not very natural. For one thing, energy is not conserved. This is to be expected, since Painlevé VI is itself not integrable, which it would have to be if you could find a conserved quantity as it is a one-dimensional system.
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I'm having trouble putting into words quite what I'm looking for, but I think it's roughly an algorithm to generate a "physical" system (in physics language, not just a Hamiltonian system) from the ODE. – Charles Siegel Dec 10 2009 at 0:12
I think I understand what you mean, even though I agree it is difficult to formalise. My guess is that not every ODE arises out of a "physical" (in your sense) system. – José Figueroa-O'Farrill Dec 10 2009 at 0:18
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http://mathoverflow.net/revisions/4663/list
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## Return to Answer
2 admit defeat
Update: in response to comment below, I'm not sure anymore the probability in question was 0.
Let's try the measure that gives an equal weight for any true statement of fixed length $N$ (written in mathematical English).
Then we have statements of the form "S or 1+2=3" which form more then $1/10^{20}$th of all true statements of a given length. On the other hand, the statements "S and Z" (where `Z` is an undecidable problem) form some positive (bounded below) measure subset in the statements of given length as well.
So, yes, for the measure described above the measure of provable statements is within some positive bounds $[a, b]$ for any $N$ greater than some $N_0$.
But that measure is proportional, for a fixed $N$, to the characteristic function of the set of all true statements, so, no, my answer doesn't give a "natural" measure.
I think the probability of "a random true statement is provable" is 0 in any good formalization of your problem.
Here's the reasoning: consider a true undecidable statement `S`. Now I would say that in any reasonable definition of probability the long statement will contain `S` with probabilitiy that goes to 1 as its length increases. One can't prove it until there's no definition of this probability, but the related fact for strings is straightforward:
Consider any string `S`. Then the probability $P(n) := \{$the random string of length $n$ contains `S`$\}$ goes to 1 as $n\to \infty$.
The proof can be done by considering the random strings of the form (chunk 1)(chunk 2)...(chunk `N`) where all chunks are of the same length as `S`.
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I think the probability of "a random true statement is provable" is 0 in any good formalization of your problem.
Here's the reasoning: consider a true undecidable statement `S`. Now I would say that in any reasonable definition of probability the long statement will contain `S` with probabilitiy that goes to 1 as its length increases. One can't prove it until there's no definition of this probability, but the related fact for strings is straightforward:
Consider any string `S`. Then the probability $P(n) := \{$the random string of length $n$ contains `S`$\}$ goes to 1 as $n\to \infty$.
The proof can be done by considering the random strings of the form (chunk 1)(chunk 2)...(chunk `N`) where all chunks are of the same length as `S`.
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http://mathhelpforum.com/pre-calculus/99140-finding-formula-function.html
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# Thread:
1. ## Finding formula for a function
I need to find the formula for this function, and i'm not sure what to do...
Code:
```x f(x)
-2 -25.22
0 3.50
2 32.22
4 60.94
6 89.66```
I've looked all over the internet and can't find anything to help me out
2. Originally Posted by JAMESveeder
I need to find the formula for this function, and i'm not sure what to do...
Code:
```x f(x)
-2 -25.22
0 3.50
2 32.22
4 60.94
6 89.66```
I've looked all over the internet and can't find anything to help me out
We can model the function using a linear equation.
Note that when x increases by 2, f(x) increases by 28.72
Since $f(0)=3.5=b$ and the slope of the line is $m=\frac{28.72}{2}=14.36$, it follows that the equation of the line is $f(x)=mx+b=14.36x+3.50$.
3. Originally Posted by JAMESveeder
I need to find the formula for this function, and i'm not sure what to do...
Code:
```x f(x)
-2 -25.22
0 3.50
2 32.22
4 60.94
6 89.66```
I've looked all over the internet and can't find anything to help me out
Try plotting it!
It's a straight line passing through (0,3.5) and with slope 14.36
CB
4. Great explanation, thank you!
You can't use the same method for an exponential function though, can you?
5. Originally Posted by JAMESveeder
Great explanation, thank you!
You can't use the same method for an exponential function though, can you?
No. For an exponential function, you would need to find a common RATIO. In other words, you need to find what each number is MULTIPLIED by to get the next.
6. So with this:
Code:
```x g(x)
.5 -1
1 0
2 1
4 1
8 3```
Each number is multiplied by 2
So how would I apply that to the explanation that was said before?
7. Originally Posted by JAMESveeder
So with this:
Code:
```x g(x)
.5 -1
1 0
2 1
4 2
8 3```
Each number is multiplied by 2
So how would I apply that to the explanation that was said before?
you would find the line through the data: $x,\ \log_2(g(x))$ .. or on second thoughts $\log_2(x),\ g(x)$
CB
8. Ah, thanks CaptainBlack
This helps a lot, thanks very much guys.
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http://physics.stackexchange.com/questions/5643/how-is-it-possible-for-orbits-to-maintain-stability
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# How is it possible for orbits to maintain stability?
According to $a = v^2/R$, the circular velocity and radial distance between two attracting objects (such as planets), must remain in perfect proportion in order for orbital motion to take place. How is it possible for objects in nature to achieve this proportion perfectly?
Not only that, to maintain an orbit seems to be impossible. For example, assuming that the moon is orbiting the earth 'perfectly'. Let us say the moon is then hit by a series of meteorites. This would shift the balance slightly, and cause the moon's orbit to decay? Apparently not... How is it possible for the moon to remain in orbit for so long?
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## 3 Answers
If the velocity of a satellite differs from the right velocity of a circular orbit, Newton's equations imply that the object will simply move along a non-circular orbit, an ellipse. This fact as well as the detailed parameters of this ellipse were already known to Johannes Kepler.
All planets and moons in the real world orbit around their stars or planets along ellipses and there is no fine-tuning here whatever. The deviation from a circular orbit is known as "eccentricity" of the ellipse and it is nonzero for all real celestial objects: none of them has a fine-tuned velocity. For any initial position or velocity, one finds an ellipse (which may be a circle if someone, e.g. NASA, fine-tunes the parameters) or a hyperbola or a parabola (if the speed exceeds the escape speed or is equal to it) and the object will move along it, in agreement with Newton's laws of motion.
All the elliptical trajectories of the 2-body system are stable (and the elliptical ones are periodic): a small perturbation of the initial state only leads to equally small perturbations of the final state. This proposition has to be modified for 3 bodies and larger numbers (chaotic behavior) and for nearby orbits around very heavy objects in general relativity that may be unstable. But in Newton's theory for 2 bodies, everything is easy.
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"If the velocity of a satellite differs from the right velocity of a circular orbit, Newton's equations imply that the object will simply move along a non-circular orbit, an ellipse" aren't other conic sections like parabola or hyperbola permitted? – user1355 Feb 22 '11 at 16:32
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@sb1: not without seriously modifying the energy of the orbit --- parabola and hyperbola solutions require you to be able to escape to infinity. – genneth Feb 22 '11 at 17:02
Thanks, @genneth, that's what I meant, indeed. The other conic sections are also discussed in my answer. – Luboš Motl Feb 23 '11 at 6:51
Let's treat the case of one single particle going around a 1/r potential (i.e. Moon around Earth). In the rotating frame of the Moon's orbit, there is an effective potential given by: $$V(r) = -\frac{GMm}{r} + \frac{J^2}{2mr^2},$$ where $J$ is the angular momentum of the orbit (and is conserved). The problem then reduces to that of a single particle moving in a one-dimensional potential, which has a well-defined minimum:
Thus any small displacement is stable, and simply results in an oscillation of the orbital radius.
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Of course, this completely ignores the effect of other objects. Some of the trajectories which one assumes to be independent in the Solar system could get close (e.g. in early Solar system, or for comets) and that would change the picture drastically. – Marek Feb 22 '11 at 12:04
@Marek: I read the OP as asking for the 2 body case --- it is a mis-understanding of how basic orbits work as opposed to the much more sophisticated question of how many-body orbits are stable (do we even have a complete answer to that? My knowledge stops at KAM theorems...) – genneth Feb 22 '11 at 14:13
I guess you're right, I misinterpreted the question. And yeah, my knowledge also stops at KAM (or rather, even before; basically I just know that the theory exists and that's it). But I'd like to learn more one day. – Marek Feb 22 '11 at 16:40
That is the conditions for circular orbits.
Orbits have no trouble existing in non-circular (elliptical) varieties. In this case the velocity and radius vary in such a way as to keep the angular momentum constant.
You can find the angular momentum at a particular point in time as $L = m * v_t$ where $v_t$ is the velocity transverse to the line connecting the two bodies. From the considerations you stated for circular orbits you should be able to deduce when the orbit is heading outward and when inward.
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– Marek Feb 22 '11 at 10:30
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http://www.maths.manchester.ac.uk/~jm/wiki/Vortices/Ring?action=browse
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# Ring
Other configurations
## A ring of n point vortices
### Region of Stability
for $$\theta\in[0,\pi]$$ the co-latitude, the ring is stable if $$\begin{array}{r:l}n=3 & \forall \theta\cr n=4 & \cos^2\theta > 1/3 \cr n=5 & \cos^2\theta > 1/2 \cr n=6 & \cos^2\theta > 4/5 \end{array}$$
For $$n\geq 7$$ the ring is always unstable (there are real eigenvalues)
### Bifurcations
###### Where do bifurcations occur?
The eigenvalues of the $$\theta$$-part of the reduced Hessian are (for $$\mu\neq0$$ when $$\ell=1$$ as reduced space is smaller) $$\lambda^{(\ell)} = \frac{n}{2\sin^2\theta_0}[-(n-\ell-1)(\ell-1) + (n-1)\cos^2\theta_0].$$ The $$\phi$$-part is positive definite. Therefore a bifurcation occurs when $$\cos^2\theta_0 = \frac{1}{n-1}(n-\ell-1)(\ell-1).$$ (Recall $$\mu=n\cos\theta_0$$.)
On the $$\ell=1$$ mode the Hessian is positive definite.
Since $$2\leq\ell\leq[n/2]$$, and $$\cos^2\theta_0<1$$, we have that bifurcations with an eigenvalues passing through 0 only occur for
• $$\ell=2:\; n\geq 4$$
• $$\ell=3:\; n=6$$
• $$\ell\geq4$$ never
Note: For $$\ell\geq4$$ (so $$n\geq8$$), the $$\theta$$-part of the Hessian is always negative definite and so the eigenvalues of the linear system are all real (and non-zero).
Also, for $$n=7,\,\ell=3$$ the expression simplifies to $$\lambda^(3) = -21$$ (independent of $$\theta$$), and again the eigenvalues of the linear system are all real.
###### Bifurcation types:
• For $$\ell=n/2$$ and n even, a 2-dimensional mode, the action has kernel $$\mathbb{Z}_{n/2}$$ so is effectively a rep of $$\mathbb{Z}_2\times \mathbb{Z}_2$$ ($$=D_n/\mathbb{Z}_{n/2}$$). The quadratic invariants are therefore $$(\delta\phi)^2+\lambda (\delta\theta)^2$$ which has imaginary or real eigenvalues (it's 2-dimensional so there are no other possibilities anyhow). This gives rise (generically) to a $$\mathbb{Z}_2$$-pitchfork bifurcation in the $$\theta$$-direction. (The genericity condition is a non-zero coefficient in $$(\delta\theta)^4$$.)
• When the mode is 4-dimensional (in particular $$\ell=2$$), the $$\phi$$-space and the $$\theta$$-space are invariant and Lagrangian (ie the symplectic rep is of the from $$V\oplus V$$, with V abs. irred.). The quadratic invariant is therefore $$|p|^2+\lambda|q|^2$$, which is stable when $$\lambda\gt0$$ and has all eigenvalues real when $$\lambda\lt0$$. The bifurcation is a $$D_{n'}$$-pitchfork, where $$n'=n$$ if n is odd, and $$=n/2$$ if n is even. What is the genericity condition? (And which type of pitchfork occurs when $$n'=4$$?) See here for pictures
###### Summary:
$$n=3$$: no bifurcations (only have $$\ell=1$$ mode, and eigenvalues are always imaginary).
$$n=4,\; \ell=2$$: The ring loses stability at $$\cos^2\theta=1/3$$ through this mode, which is 2-dimensional. On this mode, the D4 action factors through one of $$\mathbb{Z_2\times Z_2}$$ (as rotation by $$\pi$$ acts trivially on this mode), so it's a reducible representation, and one pair of the eigenvalues passes through zero, becoming real. This suggests a $$\mathbb{Z}_2$$-pitchfork bifurcation. The broken symmetry is the rotation by $$\pi/2$$.
$$n=5,\; \ell=2$$: a 4 dimensional mode, with positive definite Hessian for $$\cos^2\theta_0 \gt 1/2$$. When $$\cos^2\theta_0$$ passes through 1/2, the system undergoes a $$D_5$$-pitchfork bifurcation.
$$n=6$$:
• $$\ell=3$$: a 2 dimensional mode, with positive definite Hessian for $$\cos^2\theta_0>4/5$$ and the linearization has real eigenvalues if $$\cos^2\theta<4/5$$. is it sub- or super-critical?
• $$\ell=2$$: a 4 dimensional mode, with positive definite Hessian for $$\cos^2\theta_0>3/5$$. When $$\cos^2\theta_0$$ passes through 3/5, the system undergoes a $$D_3$$-pitchfork bifurcation (such bifurcations are generically transcritical).
$$n=7,\;\ell=2$$: a 4 dimensional mode, positive definite if $$\cos^2\theta_0>2/3$$ When $$\cos^2\theta_0$$ passes through 2/3, the system undergoes a $$D_7$$-pitchfork bifurcation. What type?
$$n=8,\;\ell=2$$: a 4 dimensional mode, positive definite if $$\cos^2\theta_0>5/7$$. When $$\cos^2\theta_0$$ passes through 5/7, the system undergoes a $$D_4$$-pitchfork bifurcation. Which type? Transcritical or sub-/super-critical?
$$n=9,\; \ell=2$$: a 4 dimensional mode, positive definite if $$\cos^2\theta_0>3/4$$ When $$\cos^2\theta_0$$ passes through 3/4, the system undergoes a $$D_9$$-pitchfork bifurcation. What type?
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http://mathoverflow.net/revisions/43036/list
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Return to Answer
Post Made Community Wiki by Ben Webster♦
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Using $(a, b, ... )$ is handy to denote a column vector, which is the transpose of the row vector $[a, b, ... ]$, especially in linear text. Correspondingly, all displayed matrices should be written with brackets, not parentheses. This notation agrees with the usual identification of coordinates with column vectors.
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http://mathoverflow.net/questions/79420?sort=votes
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## Generalising Gelfand’s spectral theory
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This is primarily a request for references and advices.
Question (edited on 10/29/2011). What's known about comprehensive generalisations of Gelfand's spectral theory for unital [associative] normed algebras [over the real or complex field] (*)?
Here, a generalisation should be meant as a framework, say, with the following distinctive features (among the others):
1. It should be founded on somehow different bases than the classical theory - especially to the extent that the notion itself of spectrum isn't any longer defined in terms of, and cannot be reduced to, the existence of any inverse in some unital algebra.
2. It should recover (at least basic) notions and results from the classical theory for unital Banach algebras in some appropriate incarnation (more details on this point are given below), for which the "generalised spectrum" does reproduce the classical one.
3. It should be unsensitive to completeness [under suitable mild hypotheses] in any setting where completeness is a well-defined notion (**), so yielding as a particular outcome that an element in a unital normed algebra, $\mathfrak{A}$, shares the same spectrum as its image in the Banach completion of $\mathfrak{A}$.
(*) If useful to know, my absolute reference here is the (let me say) wonderful book by Charles E. Rickart: General Theory of Banach Algebras (Academic Press, 1970).
(**) At least in principle, the kind of generalisation that I've in mind is tailored on the properties of topological vector spaces, though I've worked it out only in the restricted case of normed spaces.
Historical background.
As acknowledged by Jean Dieudonné in his History of Functional Analysis, the notion of spectrum (along with the foundation of modern spectral theory) was first introduced by David Hilbert in a series of articles inspired by Fredholm's celebrated work on integral equations (the word spectrum seems to have been lent by Hilbert from an 1897 article by Wilhelm Wirtinger) in the effort of lifting properties and notions from matrix theory to the broader (and more abstract) framework of linear operators. Especially, this led Hilbert to the discovery of complete inner product spaces (what we call, today, Hilbert spaces just in his honour). In 1906, Hilbert himself extended his previous analysis and discovered the continuous spectrum (already present but not fully recognised in earlier work by George William Hill in connection with his own study of periodic Sturm-Liouville equations).
A few years later, Frigyes Riesz introduced the concept of an algebra of operators in a series of articles culminating in a 1913 book, where Riesz studied, among the other things, the algebra of bounded operators on the separable Hilbert space. In 1916 Riesz himself created the theory of what we call nowadays compact operators. Riesz's spectral theorem was the basis for the definitive discovery of the spectral theorem of self-adjoint (and more generally normal) operators, which was simultaneously accomplished by Marshall Stone and John von Neumann in 1929-1932.
The year 1932 is another important date in this story, as it saw the publication of the very first monography on operator theory, by Stefan Banach. The systematic work of Banach gave new impulse to the development of the field and almost surely influenced the later work of von Neumann on the theory of operator algebras (developed, partly with Francis Joseph Murray, in a series of articles starting from 1935). Then came the seminal work of Israil Gelfand (partly in collaboration with Georgi E. Shilov and Mark Naimark), who introduced Banach algebras (under the naming of normed rings) and elaborated the corresponding notion of spectrum starting with a 1941 article in Matematicheskii Sbornik.
Now, it is undoubtable that Gelfand's work has deeply influenced the subsequent developments of spectral theory (and, accordingly, functional analysis). Yet, as far as I can understand in my own small way, something is still missing. I mean, something which may still be done, on the one hand, to clean up some inherent "defects" (or better fragilities) of the classical theory and, on the other, to make it more abstract and, then, portable to different contexts.
Naïve stuff.
As I learned from an anonymous user on MO (here), the term spectrum, in operator theory as well as in the context of normed algebras, is seemingly derived from the Latin verb spècere ("to see"), from which the root spec- of the Latin word spectrum ("something that appears, that manifests itself, vision"). Furthermore, the suffix -trum in spec-trum may come from the Latin verb instruo (like in the English word "instrument", which follows in turn the Latin noun instrumentum). So, the classical (or, herein, Gelfand) spectrum may be really considered, even from an etymological perspective, as a tool to inspect (or get improved knowledge of) some properties. I like to think of it as a sort of magnifying glass; we can move it through the algebra, zoom in and out on its elements, and get local information about them and/or global information about the whole structure.
Now, taking in mind (some parts of) another thread on this board about "wrong" definitions in mathematics, we are likely to agree that the worth of a notion is also measured by its sharpness (let me be vague on this point for the moment). And the classical notion of spectrum is, in fact, so successful because it is sharp in an appropriate sense, to the extent that it reveals deep underlying connections, say, between the algebraic and topological structures of a complicated object such as a Banach algebra (which is definitely magic, at least in my view). On another hand, what struck my curiosity is the consideration that the same conclusion doesn't hold (not at least with the same consistency) if Banach algebras are replaced by arbitrary (i.e. possibly incomplete) normed algebras, where the spectrum of a given element, $\mathfrak{a}$, can be scattered through the whole complex plane (in the complete case, as it is well-known, the spectrum is bounded by the norm of $a$, and indeed compact). So the question is: Why does this happen? And my answer is: essentially because the classical notion of spectrum is too algebraic, though completeness can actually conceal its true nature and make us even forgetful of it, or at least convinced that it must not be really so algebraic (despite of its own definition!) if it can dialogue so well with the topological structure. Yes, any normed algebra can be isometrically embedded (as a dense subalgebra) into a Banach one, but I don't think this makes a difference in what I'm trying to say, and it does not seriously explain anything. Clearly enough, the problem stems from the general failure in the convergence of the Neumann series $\sum_{n=0}^\infty (k^{-1}\mathfrak{a})^n$ for $k$ an arbitrary scalar with modulus greater than the norm of $\mathfrak{a}$. And why this? Because the convergence of such a Neumann series follows from the cauchyness of its partial sums, which is not a sufficient condition to convergence as far as the algebra is incomplete. According to my humble opinion, this is something like a "bug" in the classical vision, but above all an opportunity for getting a better understanding of some facts.
Motivations.
In the end, my motivation for this long post is that I've seemingly developed (the basics of) something resembling a spectral theory for linear (possibly unbounded) operators between different normed spaces. To me, this stuff looks like a sharpening of the classical theory in that it removes some of its "defects" (including the one addressed above); and also as an abstraction since, on the one hand, it puts standard notions from the operator setting (such as the ones of eigenvalue, continuous spectrum, and approximate spectrum) on a somehow different ground (so possibly foreshadowing further generalisations) and, on the other, it recovers familiar results (such as the closeness, the boundedness, and the compactness of the spectrum as well as the fact that all the points in the boundary are approximate eigenvalues) as a special case (while revealing some (unexpected?) dependencies).
Then, I'd really like to know what has been already done in these respects before putting everything in an appropriate form, submitting the results to any reasonable journal, and being answered, possibly after several months, that I've just reinvented the wheel. It would be really frustrating... Yes, of course, I've already asked here around (in Paris), but I've got nothing more concrete than contrasting (i.e. negative and positive) feelings. Also, I was suggested to contact a few people, and I've done it with one of them some weeks ago (sending him something like a ten page summary after checking his availability by an earlier email), but I've got no reply so far and indeed he seems to have disappeared... Then, I resolved to come here and consult the "oracle of MO" (as I enjoy calling this astounding place). :-)
Thank you in advance for any help.
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Small point: I don't think there's anything unnatural about the idea that if A is a dense subalgebra of B (neither necessarily complete) then the $A$-spectrum of an element $a\in A$ can differ from the $B$-spectrum of $a$ in $B$. That's just a "fact of life" – Yemon Choi Oct 28 2011 at 22:04
Could you perhaps edit your "post" to highlight what your specific question or request is? There seems to be a lot of background justification, and not much in the way of an actual "question" – Yemon Choi Oct 28 2011 at 22:06
I'm also somewhat unconvinced by this passage: "So the question is: Why does this happen? And my answer is: essentially because the classical notion of spectrum is too algebraic, though completeness can actually conceal its true nature and make us even forgetful of it, or at least convinced that it must not be really so algebraic (despite of its own definition!) if it can dialogue so well with the topological structure" – Yemon Choi Oct 28 2011 at 22:07
@Yemon Choi. Thanks for your comments. I see your point, and I've nothing to object. Yet, my view is different, or - better - complementary. I'm mainly interested in approximation theory, where the (still naïve) basic idea is, say, that you're not really interested in solving your equations in an "exact way", but you content to do it within the bounds of an arbitrary $\varepsilon > 0$. So, in this setting (as well as in others), you may want to rule out any unnecessary complications arising from the fact that favorable properties are simply lacking (as a "fact of life"), and hence make (...) – Salvo Tringali Oct 29 2011 at 10:28
(...) your framework somehow more flexible (starting with your definitions). That's that. Also, I should have probably remarked that my reflection about completeness is not the true point here, though it served as a trigger (and it is indeed a feature of the kind of generalisation that I've in mind). One thing for sure, I'm going to edit the OP to second your suggestion, highlight my requests, and (try to) clean up some passages. As for the rest, could you elaborate your last comment above? Thanks again. – Salvo Tringali Oct 29 2011 at 10:40
## 1 Answer
There exists an approach by Berezansky and Brasche to define a kind of generalized selfadjointness and develop some eigenfunction expansions for operators between different Hilbert spaces. This is done within the theory of rigged Hilbert spaces. Their motivation was apparently different from yours - they wanted to cover the case of Schroedinger operators with distribution potentials. Nevertheless there can be some connection. See
Yu.M. Berezansky and J. Brasche, Generalized selfadjoint operators and their singular perturbations. Methods Funct. Anal. Topol. 8, No. 4, 1-14 (2002);
Yu.M. Berezansky, J. Brasche, and L. P. Nizhnik, On generalized selfadjoint operators on scales of Hilbert spaces, Methods Funct. Anal. Topol. 17, No. 3, 193-196 (2011).
The second paper is available online: http://www.imath.kiev.ua/~mfat/
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@Anatoly Kochubei. Thanks for your answer and the references. Indeed, the 2nd paper is interesting to me, though it's still, say, too "localised" with respect to the kind of stuff that I've in mind. In any case, I will read it with more attention. Also, let me remark once more that I've proved so far only basic results (essentially the ones listed in the OP and a few minor others). For one thing, I did not even have a way to show that, under reasonable assumptions, this "generalized spectrum" is non-empty. – Salvo Tringali Oct 29 2011 at 15:27
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http://math.stackexchange.com/questions/72976/how-will-this-equation-imply-pnt
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How will this equation imply PNT
So we have $$\sum_{n \leq x} \frac{\Lambda (n)}{n}=\log{x}+C+o(1)$$ where $C$ is a constant, its partial summation is $$\sum_{n \leq x} \frac{\Lambda (n)}{n}=\frac{\psi(x)}{x}+\int_1^x \frac{\psi (t)}{t^2} dt$$ How should I go from here to prove that $\psi(x) \sim x$, which is a equivalent form of PNT.
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I don't know. What makes you think it is possible to go from those two equations to $\psi(x)\sim x$? – Gerry Myerson Oct 16 '11 at 11:57
1 Answer
Do you mean $O(1)$ or $o(1)$ in your first equation? If the former, then (a) it doesn't make sense to include the $C$ term, since it can be absorbed into the error and (b) I don't think that estimate is strong enough to prove PNT.
If you mean $o(1)$, then see my blog post, especially part 3.
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FYI part 4. there includes $1/\log1$ summands. – anon Oct 16 '11 at 21:37
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http://mathoverflow.net/questions/9885/the-convergence-of-eisenstein-series-of-weight-zero
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## The convergence of Eisenstein series of weight zero
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider Eisenstein series of weight zero, i.e.
$E_{\mathfrak{a}}(z,\ s,\ \chi) = \sum_{ \gamma \in \Gamma_{\mathfrak{a}} \backslash \Gamma } \bar{\chi}(\gamma) (Im\sigma_{\mathfrak{a}}^{-1} \gamma z)^s$,
where $\chi$ is a multiplier system of weight zero ( $\chi\ :\ \Gamma \rightarrow \mathbb{C}^*$ is a group homomorphism) singular at cusp $\mathfrak{a}$. Then my first question is that why this series converges absolutely in $Re(s)>1$?
My second question is how to calculate the following summation:
$\sum_{d\ (mod c)}\ \epsilon_d(\frac{c}{d})$, where $\gamma =$ $[ \begin{pmatrix} a & b\ c & d \end{pmatrix} ]$ $\in \Gamma_0(4)$, $(\frac{c}{d})$ is the extended quadratic residue symbol and $c = b^2.$
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Any chance you could flesh out that question a bit? Specifically, can you define more explicitly all the notation that you use (and not just some of it) - it might be obvious to you since you're familiar with it, but to someone else learning from a different book, perhaps not. (e.g. what does $\epsilon$ mean, what is $\sigma_{a}$, etc). – Vinoth Dec 28 2009 at 1:36
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I voted to close. If you provide more background and explain why you're interested in this question, it might not seem like you're just trying to get someone else to do your work for you. – Qiaochu Yuan Dec 28 2009 at 8:10
Well, I think that my problem is not fitful or complete. Actually, my problem comes from Automorphic Form, and you guys can refer to this place: mathoverflow.net/questions/2515/… – Alex Dec 31 2009 at 14:07
## 1 Answer
It seems you have started reading something from the middle.
Hint for the first one: Do it for $SL_2(\mathbb{Z})$ first. Note that
$Im \frac{az+b}{cz+d} = y/|cz+d|^2$.
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I got it! Thanks – Alex Dec 31 2009 at 14:17
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http://www.aimath.org/textbooks/beezer/MISLEsection.html
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Matrix Inverses and Systems of Linear Equations
We begin with a familiar example, performed in a novel way.
Example SABMI: Solutions to Archetype B with a matrix inverse.
The matrix $B$ of the previous example is called the inverse of $A$. When $A$ and $B$ are combined via matrix multiplication, the result is the identity matrix, which can be inserted "in front" of $\vect{x}$ as the first step in finding the solution. This is entirely analogous to how we might solve a single linear equation like $3x=12$. \begin{equation*} x=1x=\left(\frac{1}{3}\left(3\right)\right)x=\frac{1}{3}\left(3x\right)=\frac{1}{3}\left(12\right)=4 \end{equation*} Here we have obtained a solution by employing the "multiplicative inverse" of $3$, $3^{-1}=\frac{1}{3}$. This works fine for any scalar multiple of $x$, except for zero, since zero does not have a multiplicative inverse. Consider seperately the two linear equations,
\begin{align*} 0x&=12 & 0x&=0 \end{align*}
The first has no solutions, while the second has infinitely many solutions. For matrices, it is all just a little more complicated. Some matrices have inverses, some do not. And when a matrix does have an inverse, just how would we compute it? In other words, just where did that matrix $B$ in the last example come from? Are there other matrices that might have worked just as well?
## Inverse of a Matrix
Definition MI (Matrix Inverse) Suppose $A$ and $B$ are square matrices of size $n$ such that $AB=I_n$ and $BA=I_n$. Then $A$ is invertible and $B$ is the inverse of $A$. In this situation, we write $B=\inverse{A}$.
Notice that if $B$ is the inverse of $A$, then we can just as easily say $A$ is the inverse of $B$, or $A$ and $B$ are inverses of each other.
Not every square matrix has an inverse. In Example SABMI the matrix $B$ is the inverse the coefficient matrix of Archetype B. To see this it only remains to check that $AB=I_3$. What about Archetype A? It is an example of a square matrix without an inverse.
Example MWIAA: A matrix without an inverse, Archetype A.
Let's look at one more matrix inverse before we embark on a more systematic study.
Example MI: Matrix inverse.
We will now concern ourselves less with whether or not an inverse of a matrix exists, but instead with how you can find one when it does exist. In Section MINM:Matrix Inverses and Nonsingular Matrices we will have some theorems that allow us to more quickly and easily determine just when a matrix is invertible.
## Computing the Inverse of a Matrix
We've seen that the matrices from Archetype B and Archetype K both have inverses, but these inverse matrices have just dropped from the sky. How would we compute an inverse? And just when is a matrix invertible, and when is it not? Writing a putative inverse with $n^2$ unknowns and solving the resultant $n^2$ equations is one approach. Applying this approach to $2\times 2$ matrices can get us somewhere, so just for fun, let's do it.
Theorem TTMI (Two-by-Two Matrix Inverse) Suppose \begin{equation*} A= \begin{bmatrix} a&b\\ c&d \end{bmatrix} \end{equation*} Then $A$ is invertible if and only if $ad-bc\neq 0$. When $A$ is invertible, then \begin{equation*} \inverse{A}=\frac{1}{ad-bc} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix} \end{equation*}
Proof.
There are several ways one could try to prove this theorem, but there is a continual temptation to divide by one of the eight entries involved ($a$ through $f$), but we can never be sure if these numbers are zero or not. This could lead to an analysis by cases, which is messy, messy, messy. Note how the above proof never divides, but always multiplies, and how zero/nonzero considerations are handled. Pay attention to the expression $ad-bc$, as we will see it again in a while (Chapter D:Determinants).
This theorem is cute, and it is nice to have a formula for the inverse, and a condition that tells us when we can use it. However, this approach becomes impractical for larger matrices, even though it is possible to demonstrate that, in theory, there is a general formula. (Think for a minute about extending this result to just $3\times 3$ matrices. For starters, we need 18 letters!) Instead, we will work column-by-column. Let's first work an example that will motivate the main theorem and remove some of the previous mystery.
Example CMI: Computing a matrix inverse.
Notice how the five systems of equations in the preceding example were all solved by exactly the same sequence of row operations. Wouldn't it be nice to avoid this obvious duplication of effort? Our main theorem for this section follows, and it mimics this previous example, while also avoiding all the overhead.
Theorem CINM (Computing the Inverse of a Nonsingular Matrix) Suppose $A$ is a nonsingular square matrix of size $n$. Create the $n\times 2n$ matrix $M$ by placing the $n\times n$ identity matrix $I_n$ to the right of the matrix $A$. Let $N$ be a matrix that is row-equivalent to $M$ and in reduced row-echelon form. Finally, let $J$ be the matrix formed from the final $n$ columns of $N$. Then $AJ=I_n$.
Proof.
We have to be just a bit careful here about both what this theorem says and what it doesn't say. If $A$ is a nonsingular matrix, then we are guaranteed a matrix $B$ such that $AB=I_n$, and the proof gives us a process for constructing $B$. However, the definition of the inverse of a matrix (Definition MI) requires that $BA=I_n$ also. So at this juncture we must compute the matrix product in the "opposite" order before we claim $B$ as the inverse of $A$. However, we'll soon see that this is always the case, in Theorem OSIS, so the title of this theorem is not inaccurate.
What if $A$ is singular? At this point we only know that Theorem CINM cannot be applied. The question of $A$'s inverse is still open. (But see Theorem NI in the next section.) We'll finish by computing the inverse for the coefficient matrix of Archetype B, the one we just pulled from a hat in Example SABMI. There are more examples in the Archetypes (appendix A) to practice with, though notice that it is silly to ask for the inverse of a rectangular matrix (the sizes aren't right) and not every square matrix has an inverse (remember Example MWIAA?).
Example CMIAB: Computing a matrix inverse, Archetype B.
## Properties of Matrix Inverses
The inverse of a matrix enjoys some nice properties. We collect a few here. First, a matrix can have but one inverse.
Theorem MIU (Matrix Inverse is Unique) Suppose the square matrix $A$ has an inverse. Then $\inverse{A}$ is unique.
Proof.
When most of us dress in the morning, we put on our socks first, followed by our shoes. In the evening we must then first remove our shoes, followed by our socks. Try to connect the conclusion of the following theorem with this everyday example.
Theorem SS (Socks and Shoes) Suppose $A$ and $B$ are invertible matrices of size $n$. Then $AB$ is an invertible matrix and $\inverse{(AB)}=\inverse{B}\inverse{A}$.
Proof.
Theorem MIMI (Matrix Inverse of a Matrix Inverse) Suppose $A$ is an invertible matrix. Then $\inverse{A}$ is invertible and $\inverse{(\inverse{A})}=A$.
Proof.
Theorem MIT (Matrix Inverse of a Transpose) Suppose $A$ is an invertible matrix. Then $\transpose{A}$ is invertible and $\inverse{(\transpose{A})}=\transpose{(\inverse{A})}$.
Proof.
Theorem MISM (Matrix Inverse of a Scalar Multiple) Suppose $A$ is an invertible matrix and $\alpha$ is a nonzero scalar. Then $\inverse{\left(\alpha A\right)}=\frac{1}{\alpha}\inverse{A}$ and $\alpha A$ is invertible.
Proof.
Notice that there are some likely theorems that are missing here. For example, it would be tempting to think that $\inverse{(A+B)}=\inverse{A}+\inverse{B}$, but this is false. Can you find a counterexample? (See exercise MISLE.T10.)
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http://physics.stackexchange.com/questions/463/list-of-the-most-fundamental-assumptions-of-quantum-mechanics/471
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# List of the most fundamental assumptions of quantum mechanics?
Can anybody give a concise list of the most fundamental assumptions of quantum mechanics in plain english ?
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what do you mean with assumptions ? – Stefano Borini Nov 10 '10 at 0:39
Just like we have a dimensionless point particle, the constant linear flow of time, and an immovable space fabric having straight lines in the classical Newtonian mechanics ? – mumtaz Nov 10 '10 at 0:45
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Sadly, quantum mechanics is not written in plain English. – Mark Eichenlaub Nov 10 '10 at 1:57
@MarkE And there does not exist a continuous mapping of quantum mechanics into everyday language. – Mark C Nov 10 '10 at 6:04
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as a quantum chemist would say, yes, it exists, but its projection in the english space has lower dimension, thus has a larger error due to the variational theorem. – Stefano Borini Nov 10 '10 at 8:39
show 2 more comments
## 5 Answers
Uncertain Principles: 7 essential elements of QM
Sorry but I can't be made plainer than that in English.
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Especially well told in point #5 is that the nonlocal effect is seen only when the measurements are brought together in one place. That detail is too often overlooked by writers trying to convey quantum mechanics to non-physicists. – DarenW Aug 1 '11 at 7:25
I cannot agree with Orzel, mainly because he is adding interpretation to the principles, which muddies the waters too much to be an acceptable answer. – joseph f. johnson Jan 14 '12 at 16:59
In a rather concise manner, Shankar describes four postulates of nonrelativistic quantum mechanics.
I. The state of the particle is represented by a vector $|\Psi(t)\rangle$ in a Hilbert space.
II. The independent variables $x$ and $p$ of classical mechanics are represented by Hermitian operators $X$ and $P$ with the following matrix elements in the eigenbasis of $X$
$$\langle x|X|x'\rangle = x \delta(x-x')$$
$$\langle x|P|x' \rangle = -i\hbar \delta^{'}(x-x')$$
The operators corresponding to dependent variable $\omega(x,p)$ are given Hermitian operators
$$\Omega(X,P)=\omega(x\rightarrow X,p \rightarrow P)$$
III. If the particle is in a state $|\Psi\rangle$, measurement of the variable (corresponding to) $\Omega$ will yield one of the eigenvalues $\omega$ with probability $P(\omega)\propto |\langle \omega|\Psi \rangle|^{2}$. The state of the system will change from $|\Psi \rangle$ to $|\omega \rangle$ as a result of the measurement.
IV. The state vector $|\Psi(t) \rangle$ obeys the Schrödinger equation
$$i\hbar \frac{d}{dt}|\Psi(t)\rangle=H|\Psi(t)\rangle$$
where $H(X,P)=\mathscr{H}(x\rightarrow X, p\rightarrow P)$ is the quantum Hamiltonian operator and $\mathscr{H}$ is the Hamiltonian for the corresponding classical problem.
After that, Shankar discusses the postulates and the differences between quantum mechanics and classical mechanics, hopefully in plain english. Probably you want to take a look at that book: Principles of Quantum Mechanics
There are, of course, other sets of equivalent postulates.
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+1, I was thinking of the same thing, I just couldn't remember the list of assumptions in detail. – David Zaslavsky♦ Nov 10 '10 at 2:26
Ah, the first mention of Shankar I've seen! – Mark C Nov 10 '10 at 6:04
Thanks for the conciseness and reference to Shankar's book. I will definitely have a go at it. – mumtaz Nov 10 '10 at 8:27
This is hardly in plain english ;) – Stefano Borini Nov 10 '10 at 8:28
@Stefano ...yeah its not , thats why i up-voted it but could not mark it as the answer ;) – mumtaz Nov 10 '10 at 8:43
show 1 more comment
Here are some ways of looking at it for an absolute beginner. It is only a start. It is not the way you will finally look at it, but there is nothing actually incorrect here.
First, plain English just won't do, you are going to need some math. Do you know anything about complex numbers? If not, learn a little more before you begin.
Second, in quantum mechanics, we don't work with probabilities, you know, numbers between 0 and 1 that represent the likelihood of something happening. Instead we work with the square root of probabilities, called "probability amplitudes", which are complex numbers. When we are done, we convert these back to probabilities. This is a fundamental change; in a way the complex number amplitudes are just as important as the probabilities. The logic is therefore different from what you are used to.
Third, in quantum mechanics, we don't imagine that we know all about objects in between measurements, rather we know that we don't. Therefore we only focus on what we can actually measure, and possible results of a particular measurement. This is because some of the assumed imagined properties of objects in between measurements contradict one another, but we can get a theory where we can always work with the measurements themselves in a consistent way. Any measurement (meter reading, etc.) or property in between measurements has only a probability amplitude of yielding a given result, and the property only co-arises with the measurement itself.
Fourth, if we have several mutually exclusive possibilities, the probability amplitude of either one or the other happening is just the sum of the probability amplitudes of each happening. Note that this does not mean that the probabilities themselves add, the way we are used to classically. This is new and different.
If we add up the probabilities (not the amplitudes this time but their "squares") of all possible mutually exclusive measurements, gotten by "squaring" the probability amplitudes of each mutually exclusive possibilities, we always get 1 as usual. This is called unitarity.
This is not the whole picture, but it should help somewhat. As a reference, try looking at three videos of some lectures that Feynman once gave at Cornell. That should help more.
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thanks for the link to vids – mumtaz Nov 10 '10 at 8:52
The probability amplitudes are not quite « square roots » of the probabilities, they differ from the square root by a phase factor which is important. Perhaps a parenthetical remark should be put into the third paragraph to this effect. – joseph f. johnson Jan 14 '12 at 17:07
@josephf.johnson: One can declare them to be signed square roots if one implements "operator i" and doubles the size of the Hilbert space. This is a clearer point of view for those who do not like complex numbers sneaking into physics without a physical argument. – Ron Maimon Apr 27 '12 at 18:22
The suggestion you sketch here will indeed work, but doesn't change the issue I brought up, of the phase factors. Relative phase factors are physical, whether you use the formalism of complex numbers or use an alternative real space with double the dimension (this procedure is called base change). – joseph f. johnson Apr 28 '12 at 15:36
I don't really understand exactly what you mean with assumptions, but I guess you ask about the physical nature and behavior of the quantistic world. I am going to be very inaccurate in the following writing, but I need some space to get things understood. What you are going to read is a harsh simplification.
The first assumption is the fact that particles at the quantistic level don't have a position, like real-life objects, they don't even have a speed. Well, they sort of have it, but in reality they are "fuzzy". Fuzzy in the sense that you cannot really assign a position to them, or a speed.
To be fair, you perfectly know a real-world case scenario where you have to give up the concepts of speed and position for an entity. Take a guitar, and pluck a string. Where is the "vibration" ? Can you point at one specific point in the string where the "vibration" is, or how fast this vibration is traveling. I guess you would say that the vibration is "on the string" or "in the space between one end of the string, and the other hand". However, there are parts of the string that are vibrating more, and parts that are not vibrating at all, so you could rightfully say that a larger part of the "vibration" is in the center of the string, a smaller part is on the lateral parts close to the ends, and no vibration is at the very ends. In some sense, you will experience higher "presence" of the vibration in the center.
With electrons, it's exactly the same. Think "electron" as "vibration" in the previous example. Soon, you will realize that you cannot really claim anything about the position of an electron, for the fact that has wave nature. You are forced to see the electron not as a charged ball rolling here and there, but as a diffuse entity totally similar to the concept of "vibration". As a result, you cannot say anything about the position of an electron, but you can claim that there are zones in the space where the electron has "higher presence" and zones that have "smaller presence". These zones are probabilistic in nature, and this probability is directly obtained by a mathematical description called the wavefunction.
The wavefunction mainly depends on the external potential, meaning that the presence of charges, such as protons and other electrons, will affect the perceived potential and will have an effect on the probability distribution in space of an electron.
A second assumption is the mapping between physical quantities (such as energy, or momentum) and quantistic operators. Take for example the momentum of an everyday object: it is given by its mass times its speed. In the quantistic world, you don't really have the concept of position, hence you don't have the concept of speed, hence you have to reinterpret the whole thing in terms of probability (rememeber the string ?), which means that what you have about a quantistic object is its wavefunction, and you have to extract the information about the momentum from this wavefunction. How do you do it ?
Well, there's a dogma in quantum mechanics, that if you apply a magic "operator" to the wavefunction, it gives you the quantity you want. There are (simple) rules to generate these operators, and you can apply them to a wavefunction to query the momentum, or the total energy, the kinetic energy and so on.
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so the first thing we notice down there is a clear shift from a deterministic to a probabilistic notion of existence ? – mumtaz Nov 10 '10 at 0:57
I would not say "probabilistic existence". It's not like if you say "the electron has, say, 40% probability of being here" does not mean that it spends its time 40% here and 60% somewhere else. I think the plucked string is the most appropriate real-world example. It's not like in the middle of the string there's all the vibration 40 % of the time, and 60% is somewhere else. The vibration is just 40 % there and 60 % on the rest of the string. – Stefano Borini Nov 10 '10 at 1:05
so we dont have real dimensionless point particles then. These are extended in space , right ? ...but wait are we talking about the same space that we have an intuitive notion of from our empirical experience or is the space down there also funky ? – mumtaz Nov 10 '10 at 1:21
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@mumtaz:The electron is technically handled as a dimensionless point, but it's irrelevant. What you work on is the coordinate of an electron, but what you have is a coordinate of space where an electron's wavefunction has a given value. when you say wavefunction $\psi(x)$, your $x$ is a dimensionless point, but it's not really like you have a infinitesimal particle stuck in point $x$. – Stefano Borini Nov 10 '10 at 8:36
Dirac seems to have thought that the most fundamental assumption of Quantum Mechanics is the Principle of Superposition I read somewhere, perhaps an obituary, that he began every year his lecture course on QM by taking a piece of chalk, placing it on the table, and saying, now I paraphrase:
One possible state of the piece of chalk is that it is here. Another possible state is that it is over there (pointing to another table). Now according to Quantum Mechanics, there is another possible state of the chalk in which it is partly here, and partly there.
In his book he explains more about what this « partly » means: it means that the properties of a piece of chalk that is partly in state 1 (here) and partly in state 2 (there), are in between the properties of state 1 and state 2.
EDIT: I find I have compiled the basic explanations given by Dirac in the first edition of his book. Unfortunately, it is relativistic, but I am not going to re-type it all.
« we must consider the photons as being controlled by waves, in some way which cannot be understood from the point of view of ordinary mechanics. This intimate connexion between waves and particles is of very great generality in the new quantum mechanics. ...
« The waves and particles should be regarded as two abstractions which are useful for describing the same physical reality. One must not picture this reality as containing both the waves and particles together and try to construct a mechanism, acting according to classical laws, which shall correctly describe their connexion and account for the motion of the particles.
« Corresponding to the case of the photon, which we say is in a given state of polarizationn when it has been passed through suitable polarizing apparatus, we say that any atomic system is in a given state when it has been prepared in a given way, which may be repeated arbitrarily at will. The method of preparation may then be taken as the specification of the state. The state of a system in the general case then includes any information that may be known about its position in space from the way in which it was prepared, as well as any information about its internal condition.
« We must now imagine the states of any system to be related in such a way that whenever the system is definitely in one state, we can equally well consider it as being partly in each of two or more other states. The original state must be regarded as the result of a kind of superposition of the two or more new states, in a way that cannnot be conceived on classical ideas.
...
« When a state is formed by the superposition of two other states, it will have properties that are in a certain way intermediate between those of the two original states and that approach more or less closely to those of either of them according to the greater or less `weight' attached to this state in the superposition process.
« We must regard the state of a system as referring to its condition throughout an indefinite period of time and not to its condition at a particular time, ... A system, when once prepared in a given state, remains in that state so long as it remains undisturbed.....It is sometimes purely a matter of convenience whether we are to regard a system as being disturbed by a certain outside influence, so that its state gets changed, or whether we are to regard the outside influence as forming a part of and coming in the definition of the system, so that with the inclusion of the effects of this influence it is still merely running through its course in one particular state. There are, however, two cases when we are in general obliged to consider the disturbance as causing a change in state of the system, namely, when the disturbance is an observation and when it consists in preparing the system so as to be in a given state.
« With the new space-time meaning of a state we need a corresponding space-time meaning of an observation. [, Dirac has not actually mentioned anything about `observation' up to this point...]
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http://mathhelpforum.com/advanced-algebra/125873-does-mean-matrix-singular.html
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# Thread:
1. ## does this mean a matrix is singular?
If a matrix A $\neq$ 0 and $A^k=0$ where k is a positive integer, does that mean A is singular?
2. A must be singular. Suppose not, then there exists $A^{-1}$ satisfying $AA^{-1}=A^{-1}A = I$. Then $A^k (A^{-1})^k = I = 0$, giving a contradiction.
3. so the =0 means a matrix full of zeros right?
4. Originally Posted by superdude
so the =0 means a matrix full of zeros right?
Yes, that is the 0 matrix.
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http://math.stackexchange.com/questions/69199/how-can-i-verify-the-pythagorean-trig-identity-using-approximations-for-sin-x/69203
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# How can I verify the Pythagorean Trig Identity using approximations for $\sin x$ and $\cos x$ derived from their infinite series representations?
How can I verify the Pythagorean Trig Identity using approximations for $\sin x$ and $\cos x$ derived from their infinite series representations?
I can see that the infinite series of $\sin x$ and $\cos x$ should mean that for small angles, $\sin x$ is approximately $x$, and $\cos x$ is approximately $1-.5x^2$. I'd like to use these approximations to verify that $\sin^2x + \cos^2x$ is approximately $1$.
I seem to be off on the wrong track with my approach to this task. First, I simply substituted the approximations derived from the first (or first and second) terms of the series into the Pythagorean identity. Perhaps my algebra is off, but two attempts yielded the result $x^4 = 0$ which fails to confirm either the Pythagorean identity or the approximations. Of course, we know intuitively the results should verify truly.
Can you please suggest an appropriate way to approach this?
Added: I see from the answers below, that I should not have set the substitution in the Pythagorean Trig Identity equal to 1, because that very value of 1 is what I am verifying. Instead, the Pythagorean Trig Identity should be calculated using the approximation substitutions, then compared to the value of 1. If they are approximately close, the verification is done, and the Pythagorean Trig Identity is indeed approximately equal to 1.
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## 4 Answers
If $\sin x$ is approximately $x$, and $\cos x$ is approximately $1−.5x^2$, then $(\sin x)^2 + (\cos x)^2$ is approximately $$x^2 + (1-.5 x^2)^2 = x^2 + 1 - x^2 + .25 x^4 = 1 + .25 x^4.$$
Note that the $x^2$ term cancels out, so the result is $1$ with a term of order $x^4$.
As you use more terms of the power series for sin and cos, the result will be closer to 1.
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Marty, I see what you're saying. I set mine = 1, in the style of the Pythagorean Trigonometric Identity, which leads to cancellation of the 1, and x^4 = 0. How do I justify not placing the 1 as I have, and as you did not? – FreeTrader Oct 2 '11 at 5:51
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I see from Brian M. Scott's added comment that I should NOT set equal to 1 per the Pythagorean Trig Identity, but rather, should see what answer I get from the substitution, then compare my result to the Pythagorean Trig Identity's result of 1. Marty, you have told me my answer is 1 plus a term of order x^4 which will improve in accuracy the further along the infinite series I am willing to go in calculating my approximation. I understand this now. Thank you very much. – FreeTrader Oct 2 '11 at 6:02
Your algebra seems to be off:
$$\sin^2x + \cos^2 x \approx x^2 + \left(1-\dfrac{x^2}2\right)^2 = x^2 + 1 - x^2 + \dfrac{x^4}4 = 1 + \dfrac{x^4}4,$$ which is approximately $1$ for small values of $x$.
Added: Note that you can’t expect to get exactly $1$, since you’re using an inexact approximation. If, as anon suggests, you were setting this equal to $1$ and solving for $x$, you weren’t vitiating the result: you were just showing that the approximation is perfect only at $x=0$.
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I imagine OP's algebra might have been to set that equal to 1 and then derive $x^4=0$. – anon Oct 2 '11 at 5:41
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@anon: Now that you mention it, that does sound likely. Failure of interpretation, not failure of algebra. – Brian M. Scott Oct 2 '11 at 5:44
I did the algebra as you did, up until the right side of your final =. anon is correct, I did set original = 1, which derives x^4 = 0. How do you get the positive 1 + x^4/4 on the right hand side without addressing the 1 in the Pyth Trig ID, Brian? – FreeTrader Oct 2 '11 at 5:53
Brian, I see from your added comment that I should NOT set equal to 1 per the Pythagorean Trig Identity, but rather, should see what answer I get from the substitution, then compare my result to the Pythagorean Trig Identity's result of 1. Thank you! – FreeTrader Oct 2 '11 at 6:00
I would like to accept both answers here, as they combine to produce a solid understanding of the right way to approach this type of problem. Thank you very much. – FreeTrader Oct 2 '11 at 6:03
Why not look at the formal series for sine and cosine, $s(x)$ and $c(x)$, exactly the series you’ve written, and notice that $s'=c$, $c'=-s$, and differentiate the formal expression $s^2+c^2$ to get zero. Reflect on that, and realize that you’ve just shown that the formal series $s^2+c^2$ is constant (not for calculus reasons, but because a nonconstant formal series has nonzero derivative). And the constant in question is clearly $1$.
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$x^4=0$ is correct if interpreted to mean "the identity holds up to terms of degree 4 or higher". This is because terms of degree $\geq 4$ were dropped from the power series for $\sin$ and $\cos$ at the start of the calculation. If terms of degree $\geq n$ are dropped from the series then this order $n$ approximation to $\sin^2 + \cos^2$ will be equal to $1 + ax^n + bx^{n+1} + \dots$ for some numerical coefficients $a,b,\dots$ (that depend on the value of $n$ used).
Near $x=0$ the higher degree terms are smaller and an identity being correct up to higher order means greater accuracy.
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Thanks for the helpful comment. I appreciate it. – FreeTrader Oct 2 '11 at 6:11
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http://www.math.uah.edu/stat/applets/VarianceEstimateExperiment.html
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### The Variance Estimation Experiment
Confidence Interval
Sampling Distribution
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|-----------------------|--------------------|----------------------|
| Sampling Distribution | Pivot Distribution | Success Distribution |
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#### Description
The experiment is to select a random sample of size $$n$$ from a specified distribution, and then to construct an approximate confidence interval for the standard deviation $$\sigma$$ at a specified confidence level.
The distribution can be chosen with a list box; the options are
In each case, the appropriate parameters and the sample size can be varied with scroll bars. The probability density function of the selected distribution is shown in blue in the first graph, as well as a bar, centered at the distribution mean and extending $$\sigma$$ to the left and right.
The confidence level can be selected from a list box, as can the type of interval--two sided, upper bound, or lower bound. The pivot variable $$V$$ has the chi-square distribution with $$n - 1$$ degrees of freedom. The probability density function and the critical values of $$V$$ are shown in blue in the second graph.
Random variables $$L$$ and $$R$$ denote the left and right endpoints of the confidence interval for $$\sigma$$, and $$I$$ indicates the event that the confidence interval contains the $$\sigma$$. The theoretical probability density function of $$I$$ is shown in blue in the third graph.
On each run, the sample density and the confidence interval are shown in red in the first graph, and the value of $$V$$ is shown in red in the second graph. Note that the confidence interval contains $$\sigma$$ in the first graph if and only if $$V$$ falls between the critical values in the second graph. The third graph shows the proportion of successes and failures in red. The first table records the sample standard deviation $$S$$, the lower and upper confidence bounds $$L$$ and $$R$$, the pivot variable $$V$$, and $$I$$. Finally, the second table gives the theoretical and empirical probability density functions of $$I$$.
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http://quant.stackexchange.com/questions/843/risk-neutral-probability-in-binomial-lattice-option-coming-greater-than-1-what?answertab=oldest
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# Risk neutral probability in binomial lattice option coming greater than 1…what's wrong?
I am substituting reasonable values in the below fomula (like r=0.12, T=20, nColumn=16, sigma=0.004)...why is probability coming out to be greater than 1? Any help? Thanks!
````del_T=T./nColumn; % where n is the number of columns in binomial lattice
u=exp(sigma.*sqrt(del_T));
d=1./u;
p=(exp(r.*del_T)-d)./(u-d); % risk neutral probability
````
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could you please five a reference to this formula in the literature? It seems that this is an approximation of probability in the sense, that it can be $<1$ only under some conditions, not always. – Ilya Mar 30 '11 at 6:29
– S_H Mar 30 '11 at 20:14
Getting a prob > 1 (or < 0) means that you've chosen your lattice spacing in a way that makes the algorithm numerically unstable. – Foster Boondoggle Dec 20 '11 at 3:54
## 2 Answers
Yeah, I've found this formula. So you just need to put $$\Delta t < \frac{\log{u}}{r}.$$
Edited: To avoid arbitrage one should have $0<d<1+r<u$ - (Shreve, Stochastic Calculus for Finance I), or in you case $0<d(\Delta t)<\mathrm{e}^{r\Delta t}<u(\Delta t)$. Only under this condition your formula $$p = \frac{\mathrm{e}^{r\Delta t}-d}{u-d}$$ is valid and the probability will be less than $1$ and greater than $0$ - in fact I told you the same from the beginning. Using the formula for $u(\Delta t)$ we have that for a time step $$\Delta t < \frac{\sigma^2}{r^2}.$$
It's strange that this conditions are not presented in wikipedia. Moreover they abuse notation for $u(\Delta t)$ and $d(\Delta t)$ using there $t$ rather than $\Delta t$.
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Where did you find this formula? Citation needed. – chrisaycock♦ Mar 31 '11 at 15:57
It comes from the definition of the risk-neutrality. In fact in CRR model for the riskless income $1+r$ is used rather then $e^{r\Delta t}$. – Ilya Mar 31 '11 at 16:37
The first condition doesn't do any good either. Usually `u~[1.001,1.5]`, and as I said in comment to `quant_dev` that by putting `nColumn=1000`, which gives `del_T=0.02`, then you can calculate that your given condition is satisfied, however the `p` is still greater than 1. – S_H Apr 1 '11 at 18:08
@H_S: Using your parameters you will get that if $nColumn = 1000$ then $\frac{\log u}{r} = 0.0047$ - so $del_T = 0.02$ does not satisfy the first condition. Please do you calculations properly before writing that something does not work. First condition is right but not convenient since by changing $del_T$ you also change $u$, therefore I also provide the second condition. If you apply it you will see that $nColumn\geq 20 000$. – Ilya Apr 4 '11 at 6:16
Simple: decrease the time step. The binomial tree is just an approximation, and you can't really call $p$ a genuine probability.
Your parameters are also rather far from "typical". I would choose:
$r = 0.05$ (still higher than the current risk-free rate)
$\sigma = 0.2$ (more typical volatility value)
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It doesn't. Even if I make `nColumn=1000` and `T=20`, still `p=4.1291`. – S_H Mar 29 '11 at 21:21
For small enough time steps it will. The binomial tree is just an approximation, and you can't really call $p$ a genuine probability. – quant_dev Mar 29 '11 at 21:37
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http://mathoverflow.net/questions/97777/non-characteristic-maps-ala-d-modules
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## Non-characteristic maps (ala D-modules)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am trying to understand a ``well known' fact (see Kashiwara's`Introduction to microlocal analysis page 63, remark 4.8) about non-characteristic morphisms. Here is the setup:
All varieties are over the complex numbers. Given a smooth variety $X$, write $T^* X$ for its cotangent bundle, and $T^*_XX \subseteq T^*X$ for the zero section.
Let $f: X \to Y$ be a morphism of smooth varieties. Write $f_{\pi}: T^*Y \times_Y X \to T^*Y$ for the projection map, and let $f_d: T^*Y \times_Y X \to T^* X$ be the map dual to the derivative. Let $\Lambda \subseteq T^* Y$ be a closed $\mathbb{C}^*$ stable subvariety ($\mathbb{C}^*$ acting on $T^*Y$ in the evident way). Then $f$ is called non-characteristic for $\Lambda$ if
$f_{\pi}^{-1}(\Lambda) \cap f_d^{-1}(T^*_XX) \subseteq T^*_YY \times_Y X$
The `well known' fact: if $f$ is non-characteristic for $\Lambda$, then $f_d$ restricted to $f_{\pi}^{-1}(\Lambda)$ is finite.
I would be grateful if someone could explain the truth of this to me.
Some remarks:
a) The statement is actually an if and only if, but the converse is straightforward, since the fibres of $f_d$ are $\mathbb{C}^*$ stable.
b) I believe I understand how to show that $f_d$ restricted to $f_{\pi}^{-1}(\Lambda)$ is quasi-finite (using the the $\mathbb{C}^*$ stability and the upper semi-continuity of fibre dimension). But that's as far as I have got.
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Difference_operator
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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.
# Difference operator
In mathematics, a difference operator maps a function f(x) to another function f(x + a) − f(x + b).
The forward difference operator
Δf(x) = f(x + 1) - f(x)
occurs frequently in the calculus of finite differences, where it plays a role formally similar to that of the derivative, but used in discrete circumstances. Difference equations can often be solved with techniques very similar to those for solving differential equations. Analogously we can have the backward difference operator
$\nabla f(x)=f(x)-f(x-1).\,$
When restricted to polynomial functions f, the forward difference operator is a delta operator, i.e., a shift-equivariant linear operator on polynomials that reduces degree by 1. For any polynomial function f we have
$f(x)=\sum_{k=0}^\infty\frac{(\Delta^k f)(0)}{k!}(x)_k$
where
$(x)_k=x(x-1)(x-2)\cdots(x-k+1)$
is the "falling factorial" or "lower factorial" and the empty product (x)0 defined to be 1. (Warning: In the theory of special functions, the notation (x)k is often used for rising factorials; the former notation, however, is universal among combinatorialists.) In analysis with p-adic numbers, the assumption that f is a polynomial function can be weakened all the way to the assumption that f is merely continuous. That is Mahler's theorem.
Note that only finitely many terms in the above sum are non-zero: Δk f = 0 if k is greater than the degree of f. Note also the formal similarity of this result and Taylor's theorem.
03-10-2013 05:06:04
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http://mathhelpforum.com/calculus/212843-evaluating-limit-area-under-graph-continuous-function.html
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3Thanks
• 1 Post By hollywood
• 1 Post By hollywood
• 1 Post By hollywood
Thread:
1. Evaluating limit of an area under the graph of a continuous function.
The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles:
(a) Use this definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit
A= lim n-->infinity ( f(x1)1/n +f(x2)1/n + ...+f(xn)1/n )
x1= x1+deltax = 1*deltax
x2= 2deltax
x3= 3deltax
xi=ideltax= i(1/n)
My answer =
(b) Use the following formula for the sum of the cubes of the first integers to evaluate the limit in part (a).
I am not sure at all how to solve this part.
Any help greatly appreciated!!
2. Re: Evaluating limit of an area under the graph of a continuous function.
Starting from your expression $\lim_{n\to\infty} \sum_{i=1}^n \left(\frac{i}{n}\right)^3 \frac{1}{n}$, you can factor $\frac{1}{n^4}$ out of the sum, since it doesn't depend on i:
$\lim_{n\to\infty}\frac{1}{n^4}\sum_{i=1}^ni^3$
and now the sum is exactly $1^3+2^3+\dots+n^3$, so you can use the formula you were given.
- Hollywood
3. Re: Evaluating limit of an area under the graph of a continuous function.
Thanks Hollywood! Since the numbers are infinite, how do I know where to stop adding? It's asking for an exact numerical answer some I'm unsure of the exact amount.
4. Re: Evaluating limit of an area under the graph of a continuous function.
You first evaluate the *finite* sum to get a function of n, then take the limit of this function as n goes to infinity.
- Hollywood
5. Re: Evaluating limit of an area under the graph of a continuous function.
Ok, after plugging in 1^3 and 2^3 into ( n(n+1)/2 )^2
I got 1+1296 which equals 1297.
Now I use the 1/n^4 summation i^3 as n-->infinity ?
Would the answer be 0 because 1/large = approaches zero?
6. Re: Evaluating limit of an area under the graph of a continuous function.
0 wasn't right; scratch that thought
7. Re: Evaluating limit of an area under the graph of a continuous function.
$\lim_{n\to\infty}\frac{1}{n^4}\sum_{i=1}^ni^3 =$
$\lim_{n\to\infty}\frac{1}{n^4}\left(\frac{n(n+1)}{ 2}\right)^2 =$
$\lim_{n\to\infty}\frac{1}{n^4}\frac{n^4+2n^3+n^2}{ 4}$
Can you take it from there?
- Hollywood
8. Re: Evaluating limit of an area under the graph of a continuous function.
Thanks for your help hollywood; this calculus is new to me so I'm slow at learning it. thanks for the patience. Do we have to plug in 1^3 and 2^3 into that formula?
Or if we are solving for n here is my work:
1/n^4 is zero since a small # over an infinitely large number = 0. However, the n^4+2n^3 +n^2/4 part would be infinity since a very large # over a small number goes to infinity. 0*infinity would be zero though? Am I overlooking something? Sorry for my naivety.
9. Re: Evaluating limit of an area under the graph of a continuous function.
Yes, both the numerator and denominator go to infinity as n goes to infinity. But you can simplify the expression to something that you can take the limit of:
$\lim_{n\to\infty}\frac{1}{n^4}\frac{n^4+2n^3+n^2}{ 4}=$
$\lim_{n\to\infty}\frac{n^4+2n^3+n^2}{ 4n^4}=$
$\lim_{n\to\infty}\frac{1}{4}+\frac{1}{2n}+\frac{1} {4n^2}$
and now you can see that $\frac{1}{2n}$ and $\frac{1}{4n^2}$ go to zero as n goes to infinity, and of course $\frac{1}{4}$ stays $\frac{1}{4}$, so the answer is $\frac{1}{4}$.
- Hollywood
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http://mathhelpforum.com/differential-equations/68633-direction-field-diiferental-equation.html
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Thread:
1. direction field-diiferental equation
Hi, i know how to Sketch the direction field but don't unbdestand the part
"Using the direction field decide whether a solution y(x) tends to a finite limit when x goes to infinity. How does this limit depend on the value of the initial condition y(0) ?" (Please look at the attachment)
2. Originally Posted by nerdo
Hi, i know how to Sketch the direction field but don't unbdestand the part
"Using the direction field decide whether a solution y(x) tends to a finite limit when x goes to infinity. How does this limit depend on the value of the initial condition y(0) ?" (Please look at the attachment)
If you sketched the direction field, the answers should be obvious: sketch a few solutions with various initial conditions, look at what is happening... The direction field gives you the slopes of the solution, so you know when they increase of decrease, this is the main point. Look in particular at what happens when $y(0)=1$ or $y(0)=2$. These are the "critical" cases.
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http://physics.stackexchange.com/questions/4147/covariant-description-of-light-scattering-at-a-fastly-rotating-cylinder?answertab=votes
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# Covariant Description of Light Scattering at a fastly rotating Cylinder
Let us consider the following Gedankenexperiment:
A cylinder rotates symmetric around the $z$ axis with angular velocity $\Omega$ and a plane wave with $\mathbf{E}\text{, }\mathbf{B} \propto e^{\mathrm{i}\left(kx - \omega t \right)}$ gets scattered by it.
We assume to know the isotropic permittivity $\epsilon(\omega)$ and permeability $\mu(\omega)$ of the cylinder's material at rest. Furthermore, the cylinder is infinitely long in $z$-direction.
The static problem ($\Omega = 0$) can be treated in terms of Mie Theory - here, however, one will need a covariant description of the system for very fast rotations (which are assumed to be possible) causing nontrivial transformations of $\epsilon$ and $\mu$.
Hence my question:
### What is the scattering response to a plane wave on a fastly rotating cylinder?
Thank you in advance
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What is "infinite" on that cylinder? – Georg Jan 29 '11 at 13:35
Thank you @Georg for pointing out to the misleading formulation. I mean infinitely in $z$-direction. I will change it in a second :) Greets – Robert Filter Jan 29 '11 at 13:38
4
@Carl: You might consider that there are still some things in classical electrodynamics which are somehow basic but not standard homework problems. To my mind, the covariant description of electrodynamics in media belongs to this class. Greets – Robert Filter Jan 30 '11 at 12:24
1
Robert, it might be useful to begin with the case for light impinging on a moving half-infinite media (i.e. infinite plane dividing space into two different materials). That case solves trivially (just boost the case for non moving material), and can be summed up (I think) to give a limiting case for the rotating cylinder (in the limit of small wave length). But it's been 30 years since I took E&M and it was never my "best" subject. – Carl Brannen Jan 30 '11 at 23:51
1
@Carl: Thank you for the hint. The difference to the reflection problem at a half-space is the rotational character of the system. One attempt to solve the problem is to go into a co-rotating coordinate system and transform the plane wave accordingly - In this case I am not sure if such a framework is physically correct. The other way would be to just covariantly transform the medium - this is much more general since we would learn about the special relativistic relation of $\epsilon$ and $\mu$. Greets – Robert Filter Jan 31 '11 at 9:13
show 10 more comments
## 2 Answers
First of all, I don't quite understand the following phrase: "The static problem (Ω=0) can be treated in terms of Mie Theory". The Mie theory is for diffraction on a homogeneous sphere, not a cylinder. The complete solution of the problem of diffraction of electromagnetic waves on an infinite homogeneous cylinder was obtained in J. R. Wait, Can. Journ. of Phys. 33, 189 (1955) (or you may find the outline of the Wait's solution for a cylindrical wave in http://arxiv.org/abs/physics/0405091 , Section III). This solution is rather complex, so I suspect your problem can only be solved numerically, as it seems significantly more complex. The Wait's problem is a special case of your problem, so the solution of the latter problem cannot be simpler than the Wait's solution. In particular, it seems advisable to expand your plane wave into cylindrical waves, following Wait. It seems that the material equations for the rotating cylinder can be obtained following http://arxiv.org/abs/1104.0574 (Am. J. Phys. 78, 1181 (2010)). However, the cylinder will not be homogeneous (the material properties will depend on the distance from the axis and may be anisotropic). I suspect the problem can be solved using numerical solution of an ordinary differential equation for the parameters of the cylindrical waves.
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Can you at least solve the problem analytically for some special cases where still $\Omega\neq 0$? – Alexey Bobrick Mar 18 '12 at 15:10
Probably. For example, for a perfectly conducting cylinder, the radiation will not penetrate significantly into the cylinder, so the problem would be pretty much equivalent to that for a homogeneous cylinder. This case may look relatively trivial though. Anyway, I am afraid I don't have much time or motivation to solve this problem. For example, I am not enthusiastic about studying the AM. J. Phys. article trying to determine the electric properties of the rotating cylinder. With all due respect, the author of the question may be in a better position to do that. – akhmeteli Mar 18 '12 at 17:19
Thank you @akmeteli for your input. I however think that the determination of the properties of the rotating cylinder is at the core of this problem - how do $\epsilon$ and $\mu$ transform? Greets – Robert Filter Apr 22 '12 at 10:06
@Robert Filter: I agree. However, this issue is discussed, e.g., in the Am. J. Phys. I cited. I am not sure though that it would be possible to find an exact solution of the diffraction problem for the inhomogeneous cylinder (or the exact solution can be too complex to be useful). – akhmeteli Apr 22 '12 at 16:07
Look here for some details Some remarks on scattering by a rotating dielectric cylinder Also articles that cite them.
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We're looking for long answers that provide some explanation and context. Don't just give a one-line answer: please explain why you're recommending it as a solution. Answers that don't explain anything will be deleted. See Good Subjective, Bad Subjective for more information.
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http://math.stackexchange.com/questions/70110/limit-of-sums-of-iid-random-variables-which-are-not-square-integrable
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# Limit of sums of iid random variables which are not square-integrable
The Central Limit Theorem tells us that for an iid sequence of random variables $(X_n)_{n\geq 0}$ of finite variance $\sigma^2$ and zero mean
$$\lim_{n\to\infty}\frac{S_n}{\sqrt{n}}=^d N(0,\sigma^2)$$
where $S_n=X_1+\cdots+X_n$.
Suppose we have a similar sequence, except now we suppose that $X_n$ has infinite variance. Then is it possible for the sequence $\frac{S_n}{\sqrt{n}}$ to converge in distribution? Is there always some $\alpha$ such that $n^\alpha S_n$ converges to a non-constant distribution?
(It seems to me that the answer to the first question should be no, but I'm having trouble showing this.)
Thank you.
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There is the Cauchy distribution for instance. But there are many other distributions with infinite variance. You'll need to specify more to have a definite limiting distribution. – Raskolnikov Oct 5 '11 at 17:37
1
– ShawnD Oct 5 '11 at 18:02
## 1 Answer
The generalized central limit theorem states (see this for a summary), that if $X_i$ are i.i.d. such that its density function has left tail power-law asymptotic $\mathbb{P}(X < -x) \sim d x^{-\mu}$ and right tail asymptotic $\mathbb{P}(X > x) \sim 1- c x^{-\mu}$ as $x \to +\infty$, then there exist sequences $a_n$ and $b_n$ such that the random variate $Z_n = ((\sum_{i=1}^n X_i) - a_n )/b_n$ converges in probability to a stable distribution with stability index $\alpha = \min(\mu, 2)$ and asymmetry parameter $\beta = \frac{c-d}{c+d}$.
Details on the constructive choice of sequences $a_n$ and $b_n$ are given in the table found at the link above. Also see page 62 of Zolotarev and Uchaikin on Google books.
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http://unapologetic.wordpress.com/2009/04/23/the-polarization-identities/?like=1&source=post_flair&_wpnonce=c1c9c39881
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# The Unapologetic Mathematician
## The Polarization Identities
If we have an inner product on a real or complex vector space, we get a notion of length called a “norm”. It turns out that the norm completely determines the inner product.
Let’s take the sum of two vectors $v$ and $w$. We can calculate its norm-squared as usual:
$\displaystyle\begin{aligned}\lVert v+w\rVert^2&=\langle v+w,v+w\rangle\\&=\langle v,v\rangle+\langle v,w\rangle+\langle w,v\rangle+\langle w,w\rangle\\&=\lVert v\rVert^2+\lVert w\rVert^2+\langle v,w\rangle+\overline{\langle v,w\rangle}\\&=\lVert v\rVert^2+\lVert w\rVert^2+2\Re\left(\langle v,w\rangle\right)\end{aligned}$
where $\Re(z)$ denotes the real part of the complex number $z$. If $z$ is already a real number, it does nothing.
So we can rewrite this equation as
$\displaystyle\Re\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v+w\rVert^2-\lVert v\rVert^2-\lVert w\rVert^2\right)$
If we’re working over a real vector space, this is the inner product itself. Over a complex vector space, this only gives us the real part of the inner product. But all is not lost! We can also work out
$\displaystyle\begin{aligned}\lVert v+iw\rVert^2&=\langle v+iw,v+iw\rangle\\&=\langle v,v\rangle+\langle v,iw\rangle+\langle iw,v\rangle+\langle iw,iw\rangle\\&=\lVert v\rVert^2+\lVert iw\rVert^2+\langle v,iw\rangle+\overline{\langle v,iw\rangle}\\&=\lVert v\rVert^2+\lVert w\rVert^2+2\Re\left(i\langle v,w\rangle\right)\\&=\lVert v\rVert^2+\lVert w\rVert^2-2\Im\left(\langle v,w\rangle\right)\end{aligned}$
where $\Im(z)$ denotes the imaginary part of the complex number $z$. The last equality holds because
$\displaystyle\Re\left(i(a+bi)\right)=\Re(ai-b)=-b=-\Im(a+bi)$
so we can write
$\displaystyle\Im\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v\rVert^2+\lVert w\rVert^2-\lVert v+iw\rVert^2\right)$
We can also write these identities out in a couple other ways. If we started with $v-w$, we could find the identities
$\displaystyle\Re\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v\rVert^2+\lVert w\rVert^2-\lVert v-w\rVert^2\right)$
$\displaystyle\Im\left(\langle v,w\rangle\right)=\frac{1}{2}\left(\lVert v-iw\rVert^2-\lVert v\rVert^2-\lVert w\rVert^2\right)$
Or we could combine both forms above to write
$\displaystyle\Re\left(\langle v,w\rangle\right)=\frac{1}{4}\left(\lVert v+w\rVert^2-\lVert v-w\rVert^2\right)$
$\displaystyle\Im\left(\langle v,w\rangle\right)=\frac{1}{4}\left(\lVert v-iw\rVert^2-\lVert v+iw\rVert^2\right)$
In all these ways we see that not only does an inner product on a real or complex vector space give us a norm, but the resulting norm completely determines the inner product. Different inner products necessarily give rise to different norms.
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 5 Comments »
1. [...] the other hand, what if we have a norm that satisfies this parallelogram law? Then we can use the polarization identities to define a unique inner [...]
Pingback by | April 24, 2009 | Reply
2. [...] now we have the polarization identities to work with! The real and imaginary parts of are completely determined in terms of expressions [...]
Pingback by | July 13, 2009 | Reply
3. [...] if for all vectors , then we can use the polarization identities to conclude that [...]
Pingback by | August 5, 2009 | Reply
4. [...] at a given point, while the line element is a quadratic function of a single vector. However, the polarization identities will allow you to recover the bilinear function from the quadratic [...]
Pingback by | October 21, 2009 | Reply
5. [...] verify our assertion for the product , we turn and recall the polarization identities from when we worked with inner products. Remember, they told us that if we know how to calculate [...]
Pingback by | May 7, 2010 | 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://math.stackexchange.com/questions/179950/what-are-the-points-of-discontinuity-of-tan-x/179985
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# What are the points of discontinuity of $\tan x$?
$f(x) = \tan x$ is defined from $\mathbb R - \{\frac{\pi}{2} (2n+1) \mid n \in \mathbb Z\}$ to $\mathbb R$. For every $x$ in its domain, $$f(x) = \frac{\sin x}{\cos x}$$ where $\cos x$ is never 0. Thus, (in short) $\tan x$ is defined for all points in its domain. Now the question remains, is $\tan x$ discontinuous at $x = \pi/2$ (which is outside its domain)?
The question arises because the test for continuity in a textbook mentions that $f(x)$ is continuous at $x = c$ when:
1. $f(c)$ exists.
2. $\lim_{x \to c} f(x)$ exists.
3. $f(c) = \lim_{x \to c} f(x)$.
And my teacher says failure of any of the above results in $x = c$ being a point of discontinuity. Yet, according to me, first test above merely tests the point for its domain and should be the criteria for any point of discontinuity too.
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## 5 Answers
There are no “right” or “wrong” definitions, but there are “standard” and “non-standard” definitions. In my opinion, the definition of continuity from your textbook is non-standard when applied to functions defined on sets with isolated points. For example, the function $f:{\mathbb Z} \to {\mathbb R}$ defined by $f(x) = x$ is a continuous function according to the standard definition but it is discontinuous according to the definition from your textbook.
Here is a standard definition.
Let $D \subset {\mathbb R}$ and $f:D \to {\mathbb R}$. We say that $f$ is continuous at point $c$ if one of the following condition holds:
1. $c$ is a limit point of $D$ and $\lim_{x\to c} f(x) = c$, or
2. $c$ is an isolated point of $D$.
Some textbooks define essential discontinuities even for points in $\bar D$ as follows.
Let $E \subset {\bar D}$. We say that $c\in E$ is an essential discontinuity of $f$ on $E$ if there is no function $\hat f: E \to {\mathbb R}$ such that
1. ${\hat f}(x) = f(x)$ for $x\in D\cap E$,
2. $\hat f(x)$ is continuous at point $c$.
According to this definition, $\pi/2$ is an essential discontinuity of $\tan x$ on $\mathbb R$. Of course, it's important what the set $E$ is, in this definition. For example, consider the the function $g(x):{\mathbb R} \setminus {\mathbb Z}\to \mathbb R$ defined by $g(x) = x - \lfloor x \rfloor$. Then 0 is not an essential discontinuity of $g(x)$ on $[0, 1)$, nor on $(-1,0]$. But 0 is an essential discontinuity of $g(x)$ on $(-1,1)$.
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So you are saying any discreet function is continuous? That is logical if one comes to think about it. Plus, I am sorry I couldn't get the part of essential discontinuity. What does the notation D-dash mean? – Anurag Kalia Aug 8 '12 at 3:12
1
(a) Yes, absolutely. Every function defined on a discrete set is continuous. (b) ${\bar D}$ is the closure of $D$, that is, the set that contains $D$ and all limit points of $D$ (in other words, $\bar D$ is the smallest closed set that contains $D$). – Yury Aug 8 '12 at 3:44
This is a real nightmare, for university teachers. Every student of mine comes to my class from high school and is sure that $x \mapsto 1/x$ is discontinuous at $0$. The reason why calculus textbooks are so ambiguous is that their authors do not like to leave something undiscussed. For some reason, the answer to any question should be "yes" or "no"; hence they tend to formally state that functions are discontinuous outside their domain of definition.
In my opinion, this is a very bad approach: it is a matter of fact that discontinuous should not be read as the negative of continuous. The domain of definition makes a difference, and the most useful idea is that of continuous extension.
Almost any mathematician would say that the tangent function is continuous inside its own domain of definition.
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I am afraid that the definition of the continuity of a function $f$ in given point $x$ requires that function to be defined in $x$. If $x$ is outside of the domain of $f$, then you can say that neither f is continuous in $x$ nor that $f$ is discontinuous in $x$. Definition just do not cover such a situation.
You may also want to consult the Definition of continuity or Classification of discontinuities.
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Actually, the definition in any textbook I referred to give the definition of points of continuity only. They are dead silent when talking about discontinuity, whether f(c) needs to be defined or not adding to the confusion. – Anurag Kalia Aug 8 '12 at 3:07
$\tan$ is continuous. As others have mentioned, considering continuity outside domain doesn't make sense.
The continuity can be made more apparent if you consider $\tan$ as a restriction of its natural extension to a function from $\mathbf R$ to the one-point compactification of reals $\mathbf R\cup \{\infty\}$. The extension makes full "circles" in a continuous way, and a restriction of a continuous function is again continuous.
In general, quotients of continuous functions are continuous where defined, because the function $\varphi:(x,y)\mapsto x/y$ is continuous for nonzero $y$, so for continuous $f,g$ we can consider $f/g$ as a composition $\varphi\circ (f,g)$, and composition of continuous functions is continuous.
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I read about the extended real line in calculus as $\mathbb R \bigcup \infty$. That is, they treat $\infty$ and $-\infty$ as the same. How can they be same? There is at least a minus sign as a point of difference between the two. Wikipedia on compactification asks to treat real line as a circle where its open ends meet at $\infty$. Yet real line is infinite in length! Do we treat the radius of this circle to be infinite too? Isn't it just more practical to add two points, $+\infty$ and $-\infty$ to real line, and more intuitive too? – Anurag Kalia Aug 8 '12 at 3:04
1
@AnuragKalia: that depends on what you want to obtain. The two-point compactification $\mathbf R\cup\lbrace +\infty,-\infty\}$ is sometimes considered, too, although, from what I know, not quite as often, as it is not always more natural, for example, consider a line with slope $\alpha$ on a plane. For $\alpha$ real, different $\alpha$ yield different lines. Yet clearly for $\alpha$ infinite, the sign does not matter, so it does not make sense to differentiate the infinities. This is the basic idea behind $\mathbf R\cup \{\infty\}$ as the projective line $\mathbf{RP^1}$. – tomasz Aug 8 '12 at 12:02
One way to interpret point 1. is to say that if $x=c$ is not in the domain of $f$, then $f$ is not continuous at $x=c$.
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1
However what can be said is that $\tan x$ has no continuous extension $\mathbb R\to \mathbb R$ because every such extension has to be discontinuous in $\pi/2 + n\pi$. Note that this no longer holds if you extend the codomain, see tomasz's answer. – celtschk Aug 7 '12 at 16:30
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http://mathematica.stackexchange.com/questions/tagged/fourier-analysis
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# Tagged Questions
Questions on the discrete and continuous Fourier analysis functions of Mathematica, as well as the FourierSeries` package.
1answer
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### Fourier transform of sampled data
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### Discrete FFT of non-periodic signal excited by short pulse
Say I have a signal F(t) which represents a excitation by a pulse (so it is not periodic and declining to zero amplitude). According to this answer, for instance, ...
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I have been trying to obtain the Fourier transform for time domain root-raised-cosine pulse using Mathematica. The answer I'm getting from Mathematica is not the correct answer. When I input the ...
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### How to get Contours/maxima for an image in a useful form [closed]
I have some images in which I would like to find the locations of various maxima, plus the contour lines at a level e.g., 50% of that maximum intensity. As a first step, I did a FourierDCT on the ...
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104 views
### Finding Peaks in an Image using Fourier — Problems with Derivatives
Rest of OP is below, but I am editing to add that the real problem seems to be in the way my functions are being defined. Sorry, but the one reply below is not clear enough to help here. My function ...
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### Fourier transformation of solution of differential equation
I'm trying to plot a Fourier transform of solution of differential equation. I have tried with: ...
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### Plotting the frequency spectrum of a data series using Fourier
testData = Table[N@Sin[500 x], {x, 0, 100}]; ListLinePlot[Abs[Fourier[testData]], PlotRange -> Full] Gives me Which I do not expect because the Fourier ...
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270 views
### How can I obtain a zero-filled Fourier transform, as produced by MATLAB's fft?
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2answers
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### Obtaining the Fourier transform of an operator
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### Speeding up numerical Fourier Transform
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### Fourier Coefficients in Mathematica [duplicate]
I'm calculating Fourier Coefficients by hand and trying to verify them in Mathematica. However, in Mathematica I get them wrong by a factor of 2. Is there some part of the Mathematica functions I'm ...
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### How to calculate FDCT
I'm new to Mathemtica and I'm trying to calculate Discrete Cosine Transformation FDCT. I found the FourierDCT built-in function, but not DCT, so I need to implement it. I have tried couple of ideas ...
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### Proving (self) similarity with Mathematica - Reccurrence Plots, Similarity Plots etc
I posted this question in math.se but given the sheer tumultous number of questions that keep appearing on math.se and also given that I am trying to accomplish this in Mathematica, I thought I'd post ...
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### Mathematica's definition of Fourier Transform
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1k views
### Fourier Transform of a Step Function
I'm trying to obtain the form of a sinc function that I know I'm suppose to get in Mathematica. I'm doing this because I intend to do a lot with Fourier Transforms and I'd like to know I'm not missing ...
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403 views
### discrete Fourier transform
As an answer to my code I will get a periodic discrete time dependent function called "data" which I want to get a Discrete Fourier transform of it using just one period of it,but I think some thing ...
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763 views
### How to make Fourier behave like FourierTransform?
I'm not very experienced with Fourier Transforms, so there may be something inherently wrong with attempting to do this, but how can I make the discrete Fourier behave like the continuous ...
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### Computing the minimum number of terms required in a Fourier series to achieve a particular upper bound on the error
In a Fourier series, the maximum error bound is the difference of the function and the partial sum of its Fourier series. Within an interval, as we increase the number of terms of partial sums, the ...
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### Speed up Fourier for Booleans
I'm calculating a bit of the power spectrum of a two color cellular automaton using: ...
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### Coulomb potential as a Fourier transform
It is well known from theory that the Coulomb potential can be obtained as a Fourier transform in the following way: \int \frac{\mathrm{d}^3p}{\left( 2 \pi \right)^3} \frac{e^{\mathrm{i} ...
3answers
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### Can one find the beat of a tune with Fourier analysis?
I'm trying to find out if it's possible to find the beat of a tune by Fourier analysis with Mathematica. I'm taking a 44.1 kHz sample sound and hoping that I might get a nice peak for a frequency ...
2answers
232 views
### Parallelization of distinct array write access from subkernels
I'm working on an implementation of a multivariate FFT, which is (or at least should be) highly parallelizable due to the row-column-algorithm. However, i can't figure out how to implement that. The ...
3answers
2k views
### Plotting Partial Sums of Fourier Series
I'd like to plot some partial sums for a Fourier Series problem, but I am not sure if the output I am getting is correct. I want to be able to plot the partial sums and the function on the same graph. ...
4answers
1k views
### Numerical Fourier transform of a complicated function
Say I have a function $f(x)$ that is given explicitly in its functional form, and I want to find its Fourier transform[1]. If $f$ is too complicated to have an analytic expression for $\hat f(k)$, how ...
2answers
2k views
### Fast Fourier Transform
Is there a Fast Fourier Transform in Mathematica? Although looking in the help I could not find one. I am looking to implement the equivalent of fft in MATLAB.
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http://mathhelpforum.com/algebra/17031-stumbled-upon-answer.html
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# Thread:
1. ## Stumbled upon an answer....
Here's something I found interesting. "If $f(x) = 3x + 7$ and $g(x) = -2x + 5$, find the abscissa of a point which is on the graphs of both $f$ and $g$."
I really didn't know how to do it; the book certainly didn't spell it out anywhere. I started by graphing both equations, getting a rough idea of where the equations crossed, and plugged in a couple of points until I found the point $(-\frac{2}{5}, 5\frac{4}{5})$. Then I just happened to notice that earlier I had tried simplifying $3x + 7 = -2x + 5$, which simplifies to $5x = -2$, and that those numbers corresponded to the abscissa of the point in question. (Never mind that they also correspond to the numbers of the function $g(x)$).
I tried a few other simple linear equations and got the same results, so I feel like I've stumbled upon a consistent solution to the problem. Obviously I haven't discovered anything new, but I wonder, is this solution valid to all sets of two linear equations? If so, is there a named theorem for it?
2. Originally Posted by earachefl
Here's something I found interesting. "If $f(x) = 3x + 7$ and $g(x) = -2x + 5$, find the abscissa of a point which is on the graphs of both $f$ and $g$."
I really didn't know how to do it; the book certainly didn't spell it out anywhere. I started by graphing both equations, getting a rough idea of where the equations crossed, and plugged in a couple of points until I found the point $(-\frac{2}{5}, 5\frac{4}{5})$. Then I just happened to notice that earlier I had tried simplifying $3x + 7 = -2x + 5$, which simplifies to $5x = -2$, and that those numbers corresponded to the abscissa of the point in question. (Never mind that they also correspond to the numbers of the function $g(x)$).
I tried a few other simple linear equations and got the same results, so I feel like I've stumbled upon a consistent solution to the problem. Obviously I haven't discovered anything new, but I wonder, is this solution valid to all sets of two linear equations? If so, is there a named theorem for it?
It isn't so much a theorem as a process.
Consider the graphs of f(x) and g(x). Assume they are both linear functions and the slopes are different, so they cross.
Call the coordinates of the point where they cross (x, y). Then this same point (x, y) satisfies both:
y = f(x)
y = g(x)
So we have that
y = f(x) = g(x)
This is the step that you mentioned.
-Dan
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http://mathhelpforum.com/differential-equations/182376-laplace-equation.html
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# Thread:
1. ## laplace equation
$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0, u=-K \quad\mbox{at} \quad x=\pm b, y=\pm h$
How to solve this?
2. If you let $u = v - K$ then $v$ satisfies
$\dfrac{\partial^2 v}{\partial x^2} + \dfrac{\partial^2 v}{\partial y^2} = 0$
where $v = 0$ on the boundary of the rectangle. The solution of this second problem is $v \equiv 0$. Thus, the solution of your problem us $u = -K$ on the entire region.
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http://mathhelpforum.com/differential-equations/64389-general-solution-differential-equations.html
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# Thread:
1. ## general solution for differential equations
Hi I wasn't sure whether to post this in the linear algebra forum or the calculus forum, but it deals with differential equations. I'm having a hard time understanding this one question and how it differs from the normal case. The question is:
Find the general real-valued solution for the system of equations:
$y'_1 = 3y_1 + y_2$
$y'_2 = y_1 + 3y_2$
Now, I know how to solve equations of the form
$y' = 3y + f$
where f is just some function or constant. The confusing part for this problem is that $y_1$ & $y_2$ are in both equations, so you can't just solve each equation outright like you typically would when there is just y and not the y1 and y2.
Once the general equation is known, I know its pretty easy to solve the initial value problem I need to solve after, but I just need help on how to approach the problem.
Am I supposed to somehow substitute to get one equation with all y1 and the other with all y2 or how would I go about doing it? It just seems like a weird problem to me that I've never encountered before.
2. We could write it as:
$x'=2x+y$
$y'=x+3y$
Now, we can find the eigenvalues:
$\begin{bmatrix}3-{\lambda}&1\\1&3-{\lambda}\end{bmatrix}$
${\lambda}^{2}-6{\lambda}+8=0$
${\lambda}=2, \;\ {\lambda}=4$
Now, can you proceed?.
Start by plugging the eigenvalues we just found back in for lambda in the matrix above.
3. Yes, thanks that makes A LOT more sense lol, thank you very much.
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http://math.stackexchange.com/questions/67863/a-criterion-for-a-group-to-be-abelian
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# A criterion for a group to be abelian
I noted a discussion on groups being abelian under a certain restriction on powers of elements, e.g. http://tiny.cc/chs45. Maybe this result (probably not too well-known) concludes it all.
Let and $m$ and $n$ be coprime natural numbers. Assume that $G$ is a group such that $m$-th powers commute and $n$-th powers commute (that is for all $g, h$ $\in$ $G$: $g^mh^m=h^mg^m$ and $g^nh^n=h^ng^n$). Then $G$ is abelian.
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So what is your question? – Derek Holt Sep 27 '11 at 9:00
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@Derek: Let $G$ be a finite group. Assume that there are two coprime integers $m$ and $n$ such that for all $g, h \in G$ holds (1) $g^mh^m = h^mg^m$ and (2) $g^nh^n = h^ng^n$. How do you prove that $G$ is abelian? – Someone Sep 27 '11 at 14:06
## 5 Answers
This is false. In any nonabelian group of exponent $m$, $m$th powers and $n$th powers commute for any $n$, in particular any $n$ coprime to $m$.
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Maybe I was not clear enough: I did not mean the $m$-powers commuting with the $n$-powers, but the $m$-powers among each other and the $n$-powers among each other. – Nicky Hekster Sep 27 '11 at 7:59
$(m,n)=1\implies pn+qm=1$.
$(g^nh^m)^{np}=g^n(((h^mg^n)^p)^n(h^mg^n)^{-1})h^m= (h^mg^n)^{pn}g^n(h^mg^n)^{-1}h^m=(h^mg^n)^{pn}$.
$(g^nh^m)^{mq}=g^n((h^mg^n)^{mq} (h^mg^n)^{-1})h^m=g^n((h^mg^n)^{-1} (h^mg^n)^{mq})h^m=(h^mg^n)^{qm}$.
$(g^nh^m)^{np}(g^nh^m)^{mq}=(h^mg^n)^{np}(h^mg^n)^{qm}\implies g^nh^m=h^mg^n$.
$gh=(gh)^{pn+mq}=(hg)^{pn+mq}=hg$.
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I am also confused with the question ! Isn`t the term " Abelian " synonymous of " "commutative" ? Meaning that a group G is " Abelian " if $ab=ba$ for all $a,b \in G$
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The hypothesis is not "$ab=ba$ for all $a,b\in G$." Only special cases of elements commuting are assumed. – anon Sep 27 '11 at 11:31
Assuming that $G$ is finite (looking at the tags), you can prove that for any fixed prime $p$ all $p$-elements commute ($p$ does not divide $m$ or $n$).
Then conclude that the $p$-Sylow sugroups are normal and abelian for all $p$.
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G does not have to be finite. Let $M \subset G$ be the subgroup generated by all m-th powers and let $N \subset G$ be the subgroup generated by all n-th powers. These subgroups are clearly abelian normal subgroups. Since m and n are coprime $G = MN$ and $M \cap N$ is contained in the center $Z(G)$ of $G$. To prove that G is abelian it suffices to show that M and N commute, that is $[M,N]=1$. Note that $[M,N] \subset (M \cap N)$. Let $a \in M$ and $b \in N$. Then $[a, b] = a^{−1}b^{−1}ab \in M \cap N$. Hence $[a, b] = z$ with $z \in Z(G)$. Hence $b^{−1}ab = za$, whence $b^{−1}a^nb=z^na^n$. Since $a^n \in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.
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Why do you tag the question finite-groups, if you don't assume your groups to be finite? – jug Sep 27 '11 at 19:40
What question? As far as I can see, no question was asked. – Derek Holt Sep 27 '11 at 22:00
I'm confused - were you looking for help, or just posting a result for all to see? The latter isn't really the intended use of this site. – Alon Amit Sep 27 '11 at 22:37
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http://mathoverflow.net/questions/75522?sort=oldest
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## How can I write down a point in the Berezinian of a super vector space?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A vector space $V$ of dimension $n$ has an associated determinant line $Det(V)$.
An element of $Det(V)$ is represented as a (formal limear combination) of expresstions of the form $v_1 \wedge v_2 \wedge \ldots \wedge v_n$, subject to the usual multilinearity and antisymmetry relations.
I'm wondering what is analog of the above fact/construction in the world of super vector spaces.
Let $V$ be a supervector space of dimension $n|m$. Then there is a line $Ber(V)$ called the Berezinian of $V$, that behaves like a super-determinant.
Here's a naive description of the Berezinian: for $V=V_0\oplus V_1$, it is given by $$Ber(V)=Det(V_0)\otimes Det(V_1)^*.$$ That's clearly not a good description of $Ber(V)$, as it relies on the decomposition of $V$ into even and odd parts, which is not a $GL(V)$-invariant thing to do.
I want to make sure that I don't get non-invariant answers. To ensure that, I'll do things in families (and thus make the question more complicated $-$ sorry for that):
Let $\Lambda=\Lambda(\theta_1,\ldots,\theta_n)$ be a Grassmann algebra (=exterior algebra) on $n$ variables, and let $V$ be a $Spec(\Lambda)$-parametrized family of super vector spaces, i.e., a super vector bundle $V\to Spec(\Lambda)$. How can I describe concretely a section of the associated line bundle $Ber(V)\to Spec(\Lambda)$?
For those who don't like the above language, I can translate into algebra. Let $\Lambda=\Lambda(\theta_1,\ldots,\theta_n)$, and let $V$ be a free $\Lambda$-module on $n$ even generators and $m$ odd generators. How can I describe concretely an even element of the rank one $\Lambda$-module $Ber(V)$?
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Did you mean tensor product instead of direct sum in the formula for Ber(V)? Also it seems like V_2 should be replaced by V_1. – Dmitri Pavlov Sep 15 2011 at 16:03
Are you aware of the description of Ber(V) as Ext*_{Sym(V*)}(R,Sym(V*))? (Here R is the ring of scalars and Sym(V*) acts on R by augmentation.) If you combine this description with any explicit description of Ext, would this count as an explicit description of the Berezinian? – Dmitri Pavlov Sep 15 2011 at 16:14
@Dmitri: I was unaware of this description, and I like it! You should leave it (maybe with more details or references, please) as an answer, because I would like to vote it up. – Theo Johnson-Freyd Sep 15 2011 at 16:58
Thank you Dmitri for the small typos. Yes, I was aware (since today) of that description. But it's so much more complicated than the definition of Det(V)... I suspect that there is some simple answer involving Schur functors. Namely, that Ber(V) is just some appropriate Schur functor applied to V. But I don't know whether that's true or not, nor do I know which Schur functor to apply. – André Henriques Sep 15 2011 at 18:14
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It took me a while to realize that $\mathbb R\left(\theta_1,...,\theta_n\right)$ does not mean the field of rational functions over $\mathbb R$ in the $\theta_i$ here. As a notation for an $n$-dimensional Grassmann algebra, it is unknown to me. (Grassmann=exterior? There are many things that can be called Grassmann algebras...) – darij grinberg Sep 15 2011 at 21:00
show 5 more comments
## 2 Answers
From page 61 of Deligne and Morgan's article Notes on supersymmetry (following Joseph Bernstein):
"A basis $\{e_1,\ldots,e_p,e_{p+1},\ldots,e_{p+q}\}$ of $L$ defines a one-element basis $[e_1,\ldots,e_p,e_{p+1},\ldots,e_{p+q}]$ of $Ber(L)$."
That's an answer to the question.
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Well, a complete answer should probably include a description of how such elements transform under a change of basis. (In the purely even case such a description amounts to saying that Ber(L)=Λ^top(L).) – Dmitri Pavlov Nov 5 2011 at 10:56
I known that I'm being pedantic... but I can't help it: "how such elements transform" is physics lingo. In math, I would talk about the "linear relation that two such elements satisfy". – André Henriques Nov 5 2011 at 16:00
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From my point of view, bases are a part of physics lingo :-), so if you talk about bases it makes sense to use other parts of the physics lingo too. More seriously, didn't you want to get an expression of Ber(L) in terms of some Schur functor? Your current answer simply amounts to saying that Ber is functorial with respect to isomorphisms of vector spaces and dim Ber(R^{m|n}) = 1. – Dmitri Pavlov Nov 5 2011 at 19:34
You're completely right. What I really wanted was some connection between the Berezinian and Schur functors... Even worse, my answer provides zero information on the Berezinian actually is (well... it gives $\varepsilon>0$ information: in your comment, you describe very precisely how much information it gives). However, strictly speaking, that quote from Deligne-Morgan does answer my original question as it provides a notation for elements of $Ber(L)$... – André Henriques Nov 5 2011 at 21:08
Hi Andre. Did you figure this out yet? I didn't realize you had asked this and have just asked a similar question: mathoverflow.net/questions/81191/… – David Carchedi Nov 17 2011 at 19:48
### 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 Koszul complex introduced by Yu. Manin in his book "Gauge field theory and complex geometry". The cohomology of this complex yields the Berezinian.
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Welcome to MO!! – Alexander Chervov Jan 27 at 11:50
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http://math.stackexchange.com/questions/245158/isomorphic-abelian-groups-which-are-not-isomorphic-r-modules
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# Isomorphic abelian groups which are not isomorphic R-modules
I'm looking for an example of of two isomorphic abelian groups, which are not isomorphic $R$ modules for some ring $R$.
I suppose we can just the same abelian group $M$ twice, and use a different operation $R\times M \rightarrow M$ so the $R$ modules aren't isomorphic. I can't think of such a group $M$ and ring $R$ to make this possible, though. Any ideas? Thanks.
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## 2 Answers
All finite dimensional real vector spaces are isomorphic as abelian groups. You can even add the countably infinite dimensional ones.
Later. $\mathbb R$ is an infinite dimensional $\mathbb Q$-vector space, so a direct sum of finitely many copies (or countable many) of $\mathbb R$ is isomorphic to $\mathbb R$ as a $\mathbb Q$-vector space, and therefore as an abelian group. This depends on vector spaces having bases (so on the axiom of choice, more or less) and on the fact that if $A$ is an infinite set then there is a bijection between $A\times\{0,1\}$ and $A$, which probably also depends on having choice at hand. Since we do have choice, there is no problem :-)
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The dual numbers over a field come to mind.
Let $k[\epsilon]=k[t]/(t^2)$ be considered as a $k[t]$-module. As a $k$-module (i.e. vector space), $k[\epsilon]\cong k^2.$ This implies that $k[\epsilon]\cong k^2$ as abelian groups. However, we can give $k^2$ the trivial $k[t]$-module structure whereby $t\cdot(a,b)=(0,0)$ for every $a,b\in k.$ This structure is different from the previous, since in $k[\epsilon]$ we have $t\cdot (a,0)=(0,a\epsilon)\neq (0,0)$ for $a\in k^\times.$
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http://quant.stackexchange.com/questions/4834/threshold-calculation-for-buying-a-mean-reverting-asset/4849
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# Threshold calculation for buying a mean-reverting asset
I am trying to figure-out an optimal policy for buying a unit when its price follows a mean-reverting price process (Ornstein–Uhlenbeck), when I have a finite time deadline for buying the unit.
I tried to search the literature for it, but couldn't find anything. I would very much appreciate any help.
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An interesting paper could be: Yingdong Lv & Bernhard K. Meister: Application of the Kelly-Criterion to Ornstein-Uhlenbeck Processes. But they don't have the a finite time deadline. However, you could use their results to estimate the optimal trading strategy and do some Monte Carlo to incorporate the finite time horizon. However, I have to say that I didnt read the paper in detail. – philippe Feb 27 at 11:07
## 4 Answers
you find theoretical results for the Ornstein-Uhlenbeck process if you search for "pairs trading". In pairs trading it is assumed that the ratio of the pair is mean reverting. Then one often models this ratio as Ornstein–Uhlenbeck process.
You find something on page 11 here
Further theoretical results that might be of interest can be found here.
All these results are theoretical and you can play with them. I don't know how much they help you in practice.
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mean-reversion may be the underlying assumption of those that employ "pairs-trading" approaches, however, other than that there is no relation at all to the topic under discussion. A Pairs trading strategy exhibits completely different dynamics than a mean-reverting process, even though each individual asset may follow a mean-reverting process. Its like someone asks about Black Scholes and I answer with using a completely unrelated stochastic model just because both are driven by Brownian motions. – Freddy Jan 28 at 0:20
@Freddy I disagree, Alon looks for rules for the Ornstein-Uhlenbeck process and this is what I deliver. The headline "pairs trading" does not change this. – Richard Jan 28 at 8:05
I think it does matter, the dynamics of the OU-process driving a single asset are entirely different than the dynamics that describe a combined short and long position of a pairs trading approach even when each underlying asset is driven by an OU-process. I am sure your referenced papers describe the properties of OU-processes, however that is something the wiki page describes as well. You yourself pointed out OP is asking a specific question in regards to an optimal trading strategy, and I agree with you. But it does not concern pairs trading and goes beyond the OU-process itself. – Freddy Jan 28 at 8:45
The OU-process has $3$ parameters - what can be different if I speak of theoretical results concerning stopping times and theresuch? OU process is OU process no matter the context - at least if I speak of stopping times - of course the parameters will be different and all "caveats" too. – Richard Jan 28 at 9:53
But most importantly - let us here what @Alon needs. Are you with us, Alon? – Richard Jan 28 at 9:54
show 2 more comments
I think a good way to think about your problem is the example of finding an optimal VWAP trading strategy. You basically have a finite point in time by which you must have performed your transaction and you trade a similar asset than the one you are considering, one with the same underlying assumptions of mean-reversion (I make such assumption in the same way than you make the assumption of mean-reversion).
With this assumption in mind and given you must at some point in time transact you are now faced with the following optimization problem: By how much does the asset have to traverse away from whatever you define as mean point in order to induce you to transact and in what size?
Also, contrary to a pairs trading strategy you do not want to transact at the point where an asset moved away from its mean but in the same direction as your order direction. You believe in mean-reversion and assume you can transact the asset more optimally at a later point.
I cannot provide an optimization function (because its very closely related to something I have been working on in the past and do not want to make it public) but here couple points I would consider:
• Does the asset really mean-revert more often than it trends in order to overcome each and every cost of execution, including commission, market impact,...? Does it pay to consider implementing a hybrid strategy in which you measure regime changes and only employ a mean-reversion approach when price dynamics favor such approach?
• Get a firm grasp at how volatile the asset is. By how many standard deviations does the asset trade away from its mean?
• Are you willing to take on more proprietary risk in that you are willing to potentially transact the full size of your order at once given the asset diverted sufficiently much away from your defined mean? Or do you want to split the order into many child orders and trade smaller sizes at smaller diversions?
I would first try to answer and consider those points before proceeding. Please note I am sharing my own experience here and do not present an academic approach. I implemented a VWAP strategy with systematic proprietary overlay that performed at a close to zero tracking error over a longer period of time in several Asian equity markets, including names that were generally not considered to be executed through standard DMA engines either for lack of liquidity, or other anomalies.
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I think the question is worth a better answer. Your answer is at most a comment. He asks for rules for the Ornstein-Uhlenbeck process. This is a clear question. – Richard Jan 27 at 20:24
ok, let me elaborate then, fair point made, I tried to improve on the answer. – Freddy Jan 28 at 0:22
thanks for improving the answer. – Richard Jan 28 at 8:18
Following references from the answer provided by @Richard, we see that the optimality condition for a continuous process in general (and therefore an OU process in particular) is covered in Section 2 concluding on page 6 of Thompson 2002, where he also represents the solution in terms of the Hamilton-Jacobi-Bellman equations.
If you change the limits of the integral on the top of that page (and its antecedents) to $\min( T,H_S \wedge H_B )$ and then solve (which I don't think is necessarily possible in closed form) then you will have your optimum for the finite time horizon $T$.
If you actually try to trade this, pay close attention to the practical issues raised by @Freddy.
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I'm seriously trying to figure out the exact same thing for my dissertation. I can easily solve for reservation (threshold) prices when offered prices are independent, but I haven't yet solved for the case of mean reversion.
There's an example in Bertsekas (1987) page 83 with an autocorrelated asset sale model, but it's too brief for me to follow all the way.
Here are my first steps. The asset must be sold before period $T$. We know the final reservation price is zero: $RP_{T} = 0$. In the next to last period, the agent compares the payoff with selling in period $T-1$ or waiting until period $T$. The value function is
$J(T-1) = \max[P_{T-1},\beta E[P_T|P_{T-1}]$,
where $\beta$ is a discount factor. The threshold price at time $T-1$ is the value that makes the asset holder indifferent to selling in either of the two periods. Substituting the expected value of the OU process,
$P_{T-1}= \beta(\mu+e^{-\eta}\left(P_{T-1}-\mu\right))$,
Where $\eta$ is the level of mean reversion. Solving for $P_{T-1}$ yields the reservation price:
$RP_{T-1}=\frac{\beta \mu (1-e^{-\eta})}{1-\beta e^{-\eta}}$.
(check the algebra, but I think it's correct). Then, I derived the remainder of the reservation prices using the equation
$J(t) = \max[P_t,\beta E[J(t+1)|P_t]]$
where
$E[J(t+1)|P_t] = \mbox{Pr}\left(P_{t+1}\geq RP_{t+1}\right)\times\left(E\left[P_{t+1}|P_{t+1}\geq RP_{t+1}\right]\right) + \mbox{Pr}\left(P_{t+1}<RP_{t+1}\right)\times\left(RP_{t+1}\right)$.
For the OU process,
$P_{t+s}|P_{t}\sim N\left(\mu+e^{-\eta s}\left(P_{t}-\mu\right),\frac{\sigma^{2}}{2\eta}\left(1-\exp\left(-2\eta s\right)\right)\right).$
I used `R`'s `etruncnorm` function to calculate the probabilities in the value equation.
I have more details in my dissertation, pages 35-41: http://people.clemson.edu/~campbwa/dissertation/WAC_dissertation_3-15-2013.pdf
I have derived a full set of reservation prices, but they're too high. If I shift them down in the simulation model, profits increase!
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http://mathoverflow.net/questions/72412/why-are-the-following-two-constructions-of-zeta-functions-equal
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## (why) Are the following two constructions of zeta functions equal?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $X$ be a variety defined over $\mathbb{Q}$. One has the usual Hasse-Weil zeta function.
Now, let's do a different construction. Base change $X$ to $\mathbb{C}$: $X_{\mathbb{C}}$. Now look at its pure numerical motive $hX_{\mathbb{C}}$ (living in $\mathcal{M}_{num}(\mathbb{C})$). I am given to believe that there is a way to define zeta functions on pure numerical motives (although I can't say I understand the construction).
Are these two the same? Could this be? It seems like one would lose a lot of (arithmetic) information when base changing to $\mathbb{C}$. For example, doesn't this imply that one would get the same function if one were to take the Hasse-Weil zeta function for any other $\mathbb{Q}$-model of $X_{\mathbb{C}}$?
Am I misunderstanding, or is this truly an uncanny phenomena? Is the construction conjectural? If not -- then it must mean that $X_{\mathbb{C}}$, a geometric object, contains all of the arithmetic information of a $\mathbb{Q}$-model of it. Why should that be true?
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Half of your instincts are correct: you can't recover the zeta function from what happens at $\mathbb{C}$. You can see this already by just considering Dedekind zeta functions: certainly they carry more information than the degree of the extension. The zeta function is rather something of an amalgamation of what happens at every prime, or a kind of generating function for the number of solutions to some equation or what have you. – Moosbrugger Aug 9 2011 at 2:01
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I think Dipendra Prasad and C.S. Rajan conjecture something close to what you write, that is, that the zeta function for smooth projective curves should almost be determined by $X_{\mathbb{C}}$. Maybe something like: If $X$ and $Y$ over a number field $F$ are isospectral for every complex embedding of $F$, then there is a finite extenstion $K$ of $F$ such that the zeta functions of $X_K$ and $Y_K$ are the same. Here, the spectrum refers to that of the Laplacian, when computed using a constant curvature metric. Obviously, such a statement, if true, is meant to be difficult and deep. – Minhyong Kim Aug 9 2011 at 2:41
I'm beginning to think that I did misunderstand. I guess the two things that should be equal are the Hasse-Weil zeta function of $X$ and the zeta function of the motive $hX$ living in $\mathcal{M}_{num}(\mathbb{Q})$, rather than $\mathcal{M}_{num}(\mathbb{C})$... – James D. Taylor Aug 9 2011 at 2:58
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http://physics.stackexchange.com/questions/tagged/lienard-wiechert+electromagnetic-radiation
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# Tagged Questions
0answers
40 views
### Mathematical equivalence between Liénard-Wiechert potential and 4-potential in Rindler coordinates
I'm studying the problem of the radiation of an uniformly accelerated point charge: $$x^{\mu}(\lambda)\to(g^{-1}\sinh g\lambda,0,0,g^{-1}\cosh g\lambda)$$ I found that when a point charge is moving ...
2answers
240 views
### What is the physical meaning of retarded time?
Consider this figure Now, when I measure a field produced by the charge $e$ at the point $\mathbf r$, at the time $t=t_1$, it means that the charge sent the signal field at the time $t=t_r$, where ...
1answer
230 views
### why is advanced radiation absent?
the Lienard-Wiechert green functions have future and past null cones of radiation. Maxwell equations allow for a continuous range of mixtures between the retarded and advanced components, but we have ...
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http://mathoverflow.net/questions/88563/mathematical-research-inspired-in-fundamental-part-by-mathoverflow
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## Mathematical research inspired in fundamental part by mathoverflow
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Mathoverflow has led in several instances to new mathematical work that arose directly from mathematical ideas, questions or answers posted here. Articles containing such work, inspired in fundamental part by mathoverflow material, might be characterized as having been born on mathoverflow.
Let us collect together here references to such work. In each case, please include a link to the relevant mathoverflow post or posts, a link to the mathematical work, for example an article at the math arxiv or the relevant journal, and a very brief summary abstract. (More detailed abstracts would presumably be available for those following the links to the article.)
In order that this question will not become burdened with excessive posts, let's agree to the following principles:
1. Please makes posts here only about essentially completed articles, rather than works-in-progress.
2. Please make posts here only about articles whose main existence was inspired by mathoverflow, such as a case where the topic of a question or the essence of an answer became the main substance of an article.
3. Please do not make posts here concerning articles simply because they cite mathoverflow, or simply because mathoverflow was involved at a critical step of the article, since such citations will eventually become so numerous as to be unremarkable. Rather, make a post here only in connection with an article that could be characterized as essentially about the topic or ideas expressed on mathoverflow and the authors were directly inspired by that mathoverflow material.
Please note the related meta-thread for more general discussion about references to mathoverflow, and Gower's question on breakthroughs that seeks examples of situations where mathoverflow helped a researcher make a critical advance.
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I don't see why we cannot keep this on meta. – Thierry Zell Feb 15 2012 at 21:21
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MO has a great impact on the mathematical community as a whole - highly disputable, and I speak as someone who likes the site. – Yemon Choi Feb 15 2012 at 21:31
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It seems fine to me to have this list here on the main site, perhaps with the (modified) understanding that only completed papers, with links to journals or to the math arxiv, would be appropriate for posting. I would prefer it to the meta site discussion, in part because it will be nice to see it periodically pop up to the top, when someone completes a new MO-inspired paper. But with the modified understanding I proposed, this won't be so often as to become annoying. So I voted to re-open. – Joel David Hamkins Feb 16 2012 at 0:56
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In particular, the meta-site forum for this information is rather discursive, and doesn't organize the information as well as MO question/answer format would. – Joel David Hamkins Feb 16 2012 at 0:59
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Rather than having this comment thread get out of control, I recommend we move the discussion to meta. I've started a thread: meta.mathoverflow.net/discussion/1312/… Please upvote this comment so it appears "above the fold" – David White Feb 16 2012 at 20:10
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## 2 Answers
The paper "The mate-in-n problem of infinite chess is decidable" by D. Brumleve, J. D. Hamkins, and me was inspired by a question asked on Mathoverflow. The paper is available on the arxiv. Please see J. D. Hamkins' blog for the abstract or if you would like to post a comment, here's a short version of the abstract:
"Infinite chess is chess played on an infinite edgeless chessboard. The familiar chess pieces move about according to their usual chess rules, and each player strives to place the opposing king into checkmate. The mate-in-n problem of infinite chess is the problem of determining whether a designated player can force a win from a given finite position in at most n moves. The main theorem of this article, confirming a conjecture of the second author and C. D. A. Evans, establishes that the mate-in-n problem of infinite chess is computably decidable, uniformly in the position and in n. The proof proceeds by showing that the mate-in-n problem is expressible in what we call the first-order structure of chess, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable."
I'm looking forward to hearing about other papers inspired by Mathoverflow.
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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.
Ben Green's paper on (not) computing the Möbius function arose from this question on MathOverflow.
Abstract. Any function `$F : \{1,\dots,N\} \rightarrow \{-1,1\}$` such that $F(x)$ can be computed from the binary digits of $x$ using a bounded depth circuit is orthogonal to the Möbius function $\mu$ in the sense that $\frac{1}{N} \sum_{x \leq N} \mu(x)F(x) = o_{N \rightarrow \infty}(1)$. The proof combines a result of Linial, Mansour and Nisan with techniques of Kátai and Harman-Kátai, used in their work on finding primes with specified digits.
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Thanks, Joel: I updated my post accordingly. – GH Mar 4 2012 at 20:58
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http://crypto.stackexchange.com/questions/2538/is-my-hmac-secure-if-i-have-a-complete-series-of-hmacd-prefix-strings/2607
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# Is my HMAC secure if I have a complete series of HMAC'd prefix strings
Let's say I have a long sentence, like "The quick brown fox jumped over the lazy dog." Let's further say that I need to keep this string encrypted, so I use an HMAC. Let's further further say I want to be able to do prefix searches for this string, so I also store all possible prefixes of this string, like HMAC("T"), HMAC("Th"), HMAC("The"), HMAC("The "), etc.
I recognize there are some weaknesses between rows here involving overlap, but what I'm interested in is whether having this one row's set of HMAC values is enough to make things insecure. Could an adversary work out the key, or any part of the plaintext, given a series of HMAC values for all of a given input's prefix values?
Feel free to use any flavor of HMAC if it'll help you argue that there's a weakness.
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Please note that MACs are not designed for encryption - they provide message authentication only, and may naturally leak some bits of information about the message. – yarek May 4 '12 at 20:40
@yarek: You're absolutely correct. However, for HMAC specifically, one might be able to prove message confidentiality based on the security properties of the underlying hash function. – Ilmari Karonen May 4 '12 at 21:59
How do you expect to decrypt your "ciphertext"? – yarek May 4 '12 at 22:11
HMAC (instantiated with a particular hash function, e.g. HMAC-MD5) is a function with two arguments: the key, which is supposed secret; and the message to authenticate. Therefore HMAC("T") is ill defined. – fgrieu May 5 '12 at 8:58
@fgrieu: I'd assume that the OP is using $\operatorname{HMAC}(m)$ as a shorthand for $\operatorname{HMAC-H}_K(m)$ for some fixed hash $\operatorname{H}$ and key $K$. It would at least seem to make sense in context. – Ilmari Karonen May 6 '12 at 22:25
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## 2 Answers
I'll start by assuming that your instance of HMAC is a secure MAC function, which comes down to making various assumptions about the hash function it is instantiated with.
If this assumption holds, then an adversary will not be practically able to recover the key, even if they have access to a large number of MACs and the corresponding plaintexts, or even if they get to choose their own plaintexts — if it was possible for them to obtain the key, they'd be able to compute their own MACs and thus commit existential forgery, contradicting the assumed security property.
As for plaintext recovery, in general there's nothing that prevents a MAC function which is secure against existential forgery from leaking some information about the input plaintext, and indeed it's not hard to delibrately construct examples of such MACs (as Ricky Demer's answer shows).
However, the security proof of HMAC given by Bellare actually proves a stronger property, namely that HMAC is a PRF as long as the compression function of the hash it is instantiated with is also a PRF (or, more generally, than HMAC is a privacy preserving MAC (PP-MAC) as defined by Bellare, as long as the compression function is also a PP-MAC and the hash function itself is computationally almost universal (cAU); see remark on Lemma 4.2). This is a rather strong security property, and it indeed shows that, as long as the hash function satisfies the assumptions of the proof, HMAC will not leak information about the plaintext to an adversary who does not possess the key.
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Have F be any HMAC. $\:$ Have G be given by [the first bit of G's output is [if the plaintext is not the empty string then the first bit of the plaintext else 0] and the rest of G's output is the same as F's output]. $\:$ G's output is exactly one bit longer than F's output, and G is also a MAC (since someone with black box access to F could easily simulate having black box access to G and then have their guess for F be the result of dropping the first bit from their simulated guess for G). $\:$ One can trivially work out the first bit of a non-empty plaintext from G's output on that plaintext, as the first bit of G's output on that plaintext.
It is not required that HMAC's provide privacy, it is only required that HMAC's provide authenticity.
What would work is a [pseudorandom function family
whose instances have full domain] instead of just an HMAC.
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Please note that HMAC is a quite specific algorithm, not the name of a type of function - I think you mean MAC here where you wrote HMAC. – Paŭlo Ebermann♦ May 4 '12 at 11:46
I thought HMAC was a specific few algorithms (i.e., HMAC-MD5,HMAC-SHA1,HMAC-SHA256, perhaps others). – Ricky Demer May 4 '12 at 13:09
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You can build a HMAC from every (cryptographic) hash function (and the HMAC specification says how - this is what I meant with "specific algorithm"). But your construction of G quite certainly does not yield such a HMAC function, while it might create a valid (and secure) MAC. – Paŭlo Ebermann♦ May 4 '12 at 18:46
fixed ${}{}{}\:$ – Ricky Demer May 4 '12 at 22:32
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http://math.stackexchange.com/questions/155694/when-does-this-sequence-start-repeating-itself/155720
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When does this sequence start repeating itself?
Given the sequence $a_j = b^j \mod q$, where $1 < b, q < 2^n$, how can I prove that the sequence starts repeating itself at some term $a_k$ where $k \leq n$?
I have been looking at this problem for hours and am completely stuck on how to do it :/ Any help would be appreciated!
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I am unclear about the meaning of a term. Look for example at the sequence $1,2,3,4,5,1,2,3,4,5,1,2,\dots$. When should this sequence be considered to start repeating itself? – André Nicolas Jun 8 '12 at 17:26
@AndréNicolas: that sequence starts repeating itself on the first term. Actually in the problem they ask us to find the digit before it starts repeating itself, so I will change the statement of the question to "$k \leq n$." – badatmath Jun 8 '12 at 17:30
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You can find a formula for k in terms of the prime factorization of gcd(b,q) (namely it is just the max of the exponents appearing). Try a few examples with q = 100 and b ≤ 5 to see what will happen. – Jack Schmidt Jun 8 '12 at 17:34
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@Yuki: I think he is looking for the "burn in" not the "period". A sequence like this has an erratic initial segment, followed by a periodic segment. He is looking for the length of the initial segment. For instance, 2 mod 100 has initial segment [2] and periodic segment [4, 8, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52]. He wants an answer like "length of [2] is 1", I think at least. – Jack Schmidt Jun 8 '12 at 17:36
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@JackPenny: ow... get it! Thanks for clarifying! =p and sorry for that! – Yuki Jun 8 '12 at 17:41
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3 Answers
Note that if $\gcd(b,q)=1$, then (by Euler's Theorem) there exists $n_0$ such that $b^{n_0}\equiv1\mod q$, so, in this case, $a_1=a_{n_0+1}$, and $k=1$.
Otherwise, let $m=\gcd(b,q)$. By condition $q<2^n$, we have that $q$ has at most $n-1$ products of $m$, (since $m^n\ge2^n>q$), so if $q$ and $b$ has exactly the same primes in their prime factorization, then $q=lb$ (we can assume $b<q$ since otherwise, we take $b_0=b\mod q$), with $l<2^n$, and then, $b^{n-1}>l$ and then $b^{n-1}\equiv 0\mod q$, so, the sequence starts repeating for $k=n-1<n$. Otherwise, for $p=(n-1)+l$, for any $l\ge0$, we have $$a_p=b^{n-1}b^l\mod q$$ Define $m'=\gcd(b^{n-1},q), q'=q/m'>1,b'=b^{n-1}/m'$. In this case, $\gcd(b,q')=1$, so there exists $n_1$ such that $b^{n_1}\equiv 1\mod q'$, that is $$b'b^{n_1}\equiv b'\mod q'$$ So (by properties of $\textit mod$): $$b^{n-1}b^{n_1}\equiv b^{n-1}\mod q$$ and so, $a_{n-1}=a_{(n-1)+n_1}$, so if we set $k=n-1<n$ the sequence starts repeating, as required.
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Thank you for the answer! – badatmath Jun 8 '12 at 18:29
You're welcome. =p But I've missed a detail (Steven Stadnicki example): suppose $q$ not multiple of $b$ (so, you guarantee that $q'>1$)! – Yuki Jun 8 '12 at 18:36
Do it by the Pigeonhole principle. The sequence $b^j \bmod q$ can get at most $q$ values, namely $\{0,1,...,q-1\}$. So there is $j_0$ such that $b^{j_0} = b^1$ and it then clear that the sequence starts repeat itself (since e.g. $b^{j_0 + 1} \bmod q = b^{j_0} b^1 = b^1 b^1 = b^2 \bmod q$ and so on). I, of course, assumed that $b > 1$ is an integer.
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Thanks for your answer! Why must there be a $j_0$ such that $b^{j_0} = b^1$? Why can't the repetition be in the later parts of the sequence? – badatmath Jun 8 '12 at 17:56
but you have to also guarantee that $j_0\le n$ (one of the conditions of the question) – Yuki Jun 8 '12 at 17:59
By the Pigeonhole principle (see link). There are at most $q$ values that $b^j$ can have, so for $j_0 > q$ it must have a value appearing twice. – Zachi Evenor Jun 8 '12 at 18:02
Zachi: what about the sequence $2^j\bmod 8$? Note the original question doesn't assume that $(b,q)=1$. Your broad argument guarantees a loop, but it doesn't guarantee a loop starting 'at the start'. – Steven Stadnicki Jun 8 '12 at 18:28
This guarantees there is a burn-in (so "k" exists) but it does not give a very tight bound on k (it proves k ≤ q for any deterministic "markov" sequence, but in fact for this sequence k ≤ log2(q)). – Jack Schmidt Jun 8 '12 at 18:33
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If $\gcd(b,q)=1$, there is repetition from the beginning. Suppose that $\gcd(b,q)=d>1$. Let $q=q'r$, where $\gcd(q',b)=1$. Then any prime that divides $r$ must divide $b$.
Let $e$ be the smallest exponent such that $r$ divides $b^e$. Since $q \lt 2^n$, we have $e \lt n$. The powers of $b$ are periodic modulo $q'$ from the beginning. But they are congruent to $0$ modulo $r$ from $b^e$ on, so they are periodic modulo $q$ from $b^e$ on.
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http://unapologetic.wordpress.com/2010/01/07/change-of-variables-in-multiple-integrals-iii/
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# The Unapologetic Mathematician
## Change of Variables in Multiple Integrals III
Today we finish up the proof of the change of variables formula for multiple integrals:
$\displaystyle\int\limits_Xf(x^1,\dots,x^n)\,d(x^1,\dots,x^n)=\int\limits_{g^{-1}(X)}f(g(u^1,\dots,u^n))\left\lvert\frac{\partial(x^1,\dots,x^n)}{\partial(u^1,\dots,u^n)}\right\rvert\,d(u^1,\dots,u^n)$
So far, we’ve shown that we can chop $X$ up into a collection of nonoverlapping regions $T_{(k)}$ and $g^{-1}(X)$ into their preimages $A_{(k)}=g^{-1}(T_{(k)})$. Further, within each $A_{(k)}$ we can factor $g(u)=\theta_{(k)}(\phi_{(k)}(u))$, where $\phi_{(k)}$ fixes each component of $u$ except the last, and $\theta_{(k)}$ fixes that one. If we can show the formula holds for each such region, then it will hold for arbitrary (compact, Jordan measurable) $X$.
From here we’ll just drop the subscripts to simplify our notation, since we’re just concerned with one of these regions at a time. We’ve got $T$ and its preimage $A=g^{-1}(T)$. We’ll also define $B=\phi(A)$, so that $T=\theta(B)$. For each real $\xi$ we define
$\displaystyle\begin{aligned}T(\xi)&=\{(x^1,\dots,x^{n-1}\vert(x^1,\dots,x^{n-1},\xi)\in T\}\\B(\xi)&=\{(t^1,\dots,t^{n-1})\vert(t^1,\dots,t^{n-1},\xi)\in B\}\end{aligned}$
Then $(T(\xi),\xi)=\theta(B(\xi),\xi)$, since $\theta$ preserves the last component of the vector. We also define
$\displaystyle\begin{aligned}c&=\inf\{\phi^n(u)\vert u\in A\}\\d&=\sup\{\phi^n(u)\vert u\in A\}\end{aligned}$
The lowest and highest points in $T$ along the $n$th coordinate direction. Now we can again define $F(x)=f(x)\chi_T(x)$ and set up the iterated integral
$\displaystyle\int\limits_TF(x)dx=\int\limits_c^d\int\limits_{T(x^n)}F(x^1,\dots,x^{n-1},x^n)\,d(x^1,\dots,x^{n-1})\,dx^n$
We can apply the inductive hypothesis to the inner integral using $x=\theta(t)$, which only involves the first $n-1$ coordinates anyway. If we also rename $x^n$ to $t^n$, this gives
$\displaystyle\int\limits_TF(x)dx=\int\limits_c^d\int\limits_{B(t^n)}F(\theta(t^1,\dots,t^{n-1},t^n))\lvert J_\theta(t)\rvert\,d(t^1,\dots,t^{n-1})\,dt^n$
Which effectively integrates as $t$ runs over $B=\phi(A)$. But now we see that $B(t^n)$ lies within the projection $A_n$ of $A$, as we defined when we first discussed iterated integrals. We want to swap the order of integration here, so we have to rewrite the limits. To this end, we write $A=[a^1,b^1]\times\dots\times[a^n,b^n]$, $A_n=[a^1,b^1]\times\dots\times[a^{n-1},b^{n-1}]$, and define
$B^*(u^1,\dots,u^{n-1})=\{\phi^n(u^1,\dots,u^{n-1},u^n)\vert a^n\leq u^n\leq b^n\}$
which runs over the part of $B$ above some fixed point in $A_n$. Then we can reverse the order of integration to write
$\displaystyle\int\limits_TF(x)dx=\int\limits_{A_n}\int\limits_{B^*(u^1,\dots,u^{n-1})}F(\theta(t^1,\dots,t^{n-1},t^n))\lvert J_\theta(t)\rvert\,dt^n\,d(t^1,\dots,t^{n-1})$
Now we can perform the one-dimensional change of variables on the inner integral and swap out the variables $u^1=t^1$ through $u^{n-1}=t^{n-1}$ to write
$\displaystyle\int\limits_TF(x)dx=\int\limits_{A_n}\int\limits_{a^n}^{b^n}F(\theta(\phi(u)))\lvert J_\theta(\phi(u))\rvert\lvert J_\phi(u)\rvert\,du^n\,d(u^1,\dots,u^{n-1})$
But now we recognize the product of the two Jacobian determinants as the Jacobian of the composition:
$\displaystyle J_\theta(\phi(u))J_\phi(u)=J_{\theta\circ\phi}(u)$
and so we can recombine the iterated integral into the $n$-dimensional integral
$\displaystyle\int\limits_TF(x)dx=\int\limits_AF([\theta\circ\phi](u))\lvert J_{\theta\circ\phi}(u)\rvert\,du$
Finally, since $g=\theta\circ\phi$ we can replace the composition. We can also replace $F$ by $f$ since there’s no chance anymore for any evaluation of $f$ to go outside $T$. We’re left with
$\displaystyle\int\limits_Tf(x)dx=\int\limits_{g^{-1}(T)}f(g(u))\lvert J_g(u)\rvert\,du$
Essentially, what we’ve shown is that we can always arrange to first change $n-1$ variables (which we handle by induction) and then by the last variable. The overall scaling factor is the $n-1$-dimensional scaling factor from the first transformation times the one-dimensional scaling factor from the second, and this works because of how the Jacobian works with compositions.
This establishes the change of variables formula for regions within which we can write $g$ as the composition of two functions, one of which fixes all but the last coordinate, and the other of which fixes that one. Since we established that we can always cut up a compact, Jordan measurable set $X$ into a finite number of such pieces, this establishes the change of variables formula in general.
### Like this:
Posted by John Armstrong | Analysis, Calculus
## 4 Comments »
1. John,
What are the assumptions on f? I missed them if you stated them.
Can the assumptions on g be relaxed?
–rich
Comment by rich | January 2, 2011 | Reply
2. $f$ should only need to be integrable, while $g$ needs to be differentiable (in the higher-dimensional sense) to even define the Jacobian determinant. I don’t think I had to strengthen that assumption, but it certainly can’t be weakened.
Comment by | January 2, 2011 | Reply
3. I think g differentiable is a sufficient condition. I wonder whether g can be differentiable in a some sort weak sense since all takes place under and integral.
Comment by Rich Lehoucq | January 4, 2011 | Reply
4. It wouldn’t entirely surprise me if there were some give to the results, but whatever improvement is possible is likely not general. That is, we can weaken one condition by strengthening some other condition. Analytic results often have complicated and fussy “boundaries”, especially as compared with algebraic or categorical ones.
Comment by | January 4, 2011 | 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://mathoverflow.net/questions/61984?sort=votes
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rational points on algebraic curves over $Q^{ab}$
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Motivation:
Let $Q_{\infty,p}$ be the field obtained by adjoining to $Q$ all $p$-power roots of unity for a prime $p$. The union of these fields for all primes is the maximal cyclotomic extension $Q^{cycl}$ of $Q$. By Kronecker-Weber, $Q^{cycl}$ is also the maximal abelian extension $Q^{ab}$ of $Q$.
A well known conjecture due to Mazur (with known examples) asserts, for an elliptic curve $E$ with certain conditions, that $E(Q_{\infty,p})$ is finitely generated. This is the group of rational points of $E$ over $Q_{\infty,p}$ (not a number field!).
A theorem due to Ribet asserts the finiteness of the torsion subgroup $E(Q^{ab})$ for certain elliptic curves.
Questions:
(a) Can one expect to find elliptic curves (or abelian varieties) $A$ with $A(Q^{ab})$ finitely generated?
(c) Can one expect to find curves $C$ of genus $g >1$ with $C(Q^{ab})$ finite?
Thanks!
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Re question (c), Pete Clark has shown there are curves of every genus $g\geq 4$ with no points over ${\mathbb Q}^{ab}$: math.uga.edu/~pete/plclarkarxiv8v2.pdf. Fingers crossed and he'll appear here soon... – dke Apr 17 2011 at 3:14
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Well, here I am, but you've already said most of what I would have. Nevertheless I left an answer, the main point being that there is something to say in the genus one case as well... – Pete L. Clark Apr 17 2011 at 7:58
Thanks for all these interesting comments! – SGP Apr 17 2011 at 12:01
To answer (c), $\mathbb{Q}^{ab}$ is a large field (in the sense of Florian Pop) so any variety that has a $\mathbb{Q}^{ab}$-rational point, has infinitely many. So the answer to (c) is that either, as people pointed out, there are no points -- or there are nec. infinitely many. – Makhalan Duff Apr 17 2011 at 16:44
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$\mathbb{Q}^{ab}$ is only conjectured to be large. Otherwise Shafarevich's conjecture would be known (see math.upenn.edu/~harbater/patch35.pdf , pages 55-56.) – H. Hasson Apr 17 2011 at 18:20
show 2 more comments
3 Answers
Actually Ken Ribet proved that if $K$ is a number field and $K(\mu_{\infty})$ is its infinite cyclotomic extension generated by all roots of unity then for every abelian variety $A$ over $K$ the torsion subgroup of $A(K(\mu_{\infty}))$ is finite: http://math.berkeley.edu/~ribet/Articles/kl.pdf .
On the other hand, Alosha Parshin conjectured that if $K_{p}$ is the extension of $K$ generated by all $p$-power roots of unity (for a given prime $p$) then the set $C(K_{p})$ is finite for every $K$-curve $C$ of genus $>1$: http://arxiv.org/abs/0912.4325 (see also http://arxiv.org/abs/1001.3424 ).
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Thanks a lot for these very interesting papers! And thanks for pointing out the correct version of Ribet's result (I had the mistaken impression that it was only for elliptic curves). – SGP Apr 17 2011 at 12:37
You are welcome. – Yuri Zarhin Apr 17 2011 at 14:54
I think I am misremembering things and this is right and what I wrote in my answer (which I am going to delete) is wrong. – Felipe Voloch Apr 17 2011 at 16:08
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
As dke mentioned, I have a paper in which I construct various kinds of varieties $X_{/\mathbb{Q}}$ without abelian points (i.e., with $X(\mathbb{Q}^{\operatorname{ab}}) = \varnothing$). Here is a brief summary:
If $X$ admits a $2:1$ map to a variety $Y$ with infinitely many $\mathbb{Q}$-rational points, then $X$ itself has infinitely many quadratic points -- i.e., points defined over the union of all quadratic extensions of $\mathbb{Q}$. This certainly lives inside $\mathbb{Q}^{\operatorname{ab}}$, so gives infinitely many abelian points.
Now I call a curve $X$ hyperelliptic if it admits a $2:1$ map down to $\mathbb{P}^1$. (I say "I call" because I am not making any genus restrictions and requiring the map to be defined over $\mathbb{Q}$. Standard terminology is taking a little while to catch up to me here...) Now any curve of genus $0$ or $2$ is hyperelliptic, as is any curve of genus one with a rational point. So they all have infinitely many abelian points.
If $E$ is an elliptic curve over $\mathbb{Q}$, then what I'm saying is that if it is given as $y^2 = x^3 + Ax + b$, then take $x$ to be any rational number and extract the square root: that will give you an abelian point. One can see that only finitely many of these quadratic points are torsion points, so we are certainly getting positive rank this way. Do these quadratic points already give infinite rank? I'm not sure (but I feel like I am forgetting something here). [Added: I think I was forgetting what is in Dror Speiser's nice comment below!] Note that here I am -- anemically -- addressing your question a).
A genus one curve without rational points need not be hyperelliptic, and in my paper I construct lots of genus one curves over $\mathbb{Q}$ without elliptic points. This is the key part, actually, because using this I construct curves of every genus $g \geq 4$ over $\mathbb{Q}$ without abelian points.
This leaves genus $3$, which I was frustratingly unable to deal with in the paper and still can't. (In an appendix, I show that there are genus $3$ curves over some field without points over the maximal abelian extension of that field, unlike in the hyperelliptic cases. So my guess is that this should be possible over $\mathbb{Q}$ as well.)
I didn't think at all about the problem of infinitely versus finitely many abelian points, probably because it cannot be attacked using the methods of my paper. But of course it is interesting too.
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"Do these quadratic points already give infinite rank?" - yes. Say we have $n$ linearly independent points over the union of $\mathbb{Q}(\sqrt{d_i})$. Take $p$ a large prime that doesn't divide any one of the $d_i$, and that $x^3+Ax+b=0\pmod{p}$ has a solution $x_0$ (Chebotarev). Then $(x_0,\sqrt{f(x_0)})$ is a new point, and these $n+1$ points are again linearly independent: if a relation exists, just add to it its galois conjugate relation ($\sqrt{f(x_0)}\mapsto -\sqrt{f(x_0)}$), getting a relation on the $n$ points. The answer to a) (for jacobians of hyperelliptic curves) is "no". – Dror Speiser Apr 17 2011 at 9:14
Thank you! Your paper is a real eye-opener! – SGP Apr 17 2011 at 11:59
@Speiser: thanks for the explanation! – SGP Apr 17 2011 at 12:07
I admit that I haven't read it carefully, but in this paper E. Kobayashi conjectures that $E(\mathbb Q^{\rm ab})$ has infinite rank for all elliptic curves $E$ defined over $\mathbb Q^{\rm ab}$. In particular, assuming the "weak" Birch and Swinnerton-Dyer conjecture for $E$ and certain properties of twisted Hasse-Weil $L$-functions of $E$, she shows that if $E$ is defined over a number field $K$ of odd degree then $E(K\cdot\mathbb Q^{\rm ab})$ has infinite rank (Theorem 2 in Kobayashi's article).
This result for elliptic curves seems to suggest that the answer to your question (a) might be "no".
As for a simple abelian variety $A$ over a number field $K$, it is perhaps worth pointing out that Zarhin proved here that the torsion subgroup of $A(K^{\rm ab})$ is finite if and only if $A$ is not of CM-type over $K$ (when $K=\mathbb Q$ this reduces to an earlier theorem of Ribet).
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Thanks! I did not know any of these results! – SGP Apr 17 2011 at 12:00
You're welcome! – Stefano V. Apr 17 2011 at 12:35
As for a simple abelian variety $A$ over a number field $K$, the torsion subgroup of $A(K^{ab})$ is finite if and only if $A$ is not of CM-type over" $K$. (in other words, $\bar{K}$ on the last line of your comment should be replaced by $K$.) – Yuri Zarhin Apr 17 2011 at 14:54
@Yuri Zarhin: Actually, there is no $\bar{K}$ in the last line of my comment: it's just the underline to the word "here" in the second to last line. Or perhaps I'm misunderstanding what you mean, sorry... – Stefano V. Apr 17 2011 at 20:10
OOPS! Sorry. I've mistook the underline as bar over K. – Yuri Zarhin Apr 17 2011 at 20:27
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http://math.stackexchange.com/questions/152209/related-measures
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# Related Measures
($\Omega, F,P)$ is a measure space, $\ Q \ll P$ ($Q$ is related to $P$) i.e. $\ Q(A)= \int 1_ADdP$ where $\ D=dQ/dP$. Then $X$ is integrable wrt $Q$ if and only if $XD$ is integrable wrt $P$ and $\int XdQ = \int XDdP$. How do I prove this? I can only do it for $X$ a simple Borel function, not a general one.
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## 1 Answer
• Let $X$ integrable wrt $Q$. Write $X=X^+-X^-$, where $X^+$ and $X^-$ are the positive and negative parts of $X$. Let $X_n$ a sequence of step functions which converge pointwise to $X^+$ and $X_n\leq X^+$. We can show by Fatou's lemma that $$\int X^+DdP=\int \liminf_n X_nDdP\leq \liminf_n\int X_nDdP=\liminf_n\int X_ndQ\leq \int X_n^+dQ,$$ hence $DX^+$ is $P$ integrable. Do the same for $X^-$ to get the result. Use the same idea to get the converse, using the fact that $D$ is non-negative.
• You got $\int XdQ=\int XDdP$ for each $X$ simple. By a monotone convergence argument, you have to show that this equality holds if $X\geq 0$ and is integrable. Then you will deduce the general case.
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Ok, so for the second one I can say that for any non negative borel function X there exits a monotone sequence Xi where Xi are simple for all i such that $\ X=Sup_i(X_i)$ then $\int X dQ = \ int sup_iX_i dQ =sup_i \ intX_idQ$ via monotone convergence which is then $\sup_i \ intX_iDdP$ But now what can I do with this? – Rosie May 31 '12 at 21:53
You can write a Borel-measurable function as a difference of two measurable non-negative functions. – Davide Giraudo Jun 2 '12 at 13:02
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http://cms.math.ca/10.4153/CJM-1999-046-7
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Canadian Mathematical Society
www.cms.math.ca
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location: Publications → journals → CJM
Abstract view
# The Homology of Abelian Covers of Knotted Graphs
Read article
[PDF: 356KB]
http://dx.doi.org/10.4153/CJM-1999-046-7
Canad. J. Math. 51(1999), 1035-1072
Published:1999-10-01
Printed: Oct 1999
• R. A. Litherland
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Format: HTML LaTeX MathJax PDF PostScript
## Abstract
Let $\tilde M$ be a regular branched cover of a homology 3-sphere $M$ with deck group $G\cong \zt^d$ and branch set a trivalent graph $\Gamma$; such a cover is determined by a coloring of the edges of $\Gamma$ with elements of $G$. For each index-2 subgroup $H$ of $G$, $M_H = \tilde M/H$ is a double branched cover of $M$. Sakuma has proved that $H_1(\tilde M)$ is isomorphic, modulo 2-torsion, to $\bigoplus_H H_1(M_H)$, and has shown that $H_1(\tilde M)$ is determined up to isomorphism by $\bigoplus_H H_1(M_H)$ in certain cases; specifically, when $d=2$ and the coloring is such that the branch set of each cover $M_H\to M$ is connected, and when $d=3$ and $\Gamma$ is the complete graph $K_4$. We prove this for a larger class of coverings: when $d=2$, for any coloring of a connected graph; when $d=3$ or $4$, for an infinite class of colored graphs; and when $d=5$, for a single coloring of the Petersen graph.
MSC Classifications: 57M12 - Special coverings, e.g. branched 57M25 - Knots and links in $S^3$ {For higher dimensions, see 57Q45} 57M15 - Relations with graph theory [See also 05Cxx]
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http://skullsinthestars.com/2012/07/16/optics-basics-refraction/
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The intersection of physics, optics, history and pulp fiction
## Optics basics: refraction
Posted on July 16, 2012 by
In all of my discussions of basic principles of optics, I’ve so far neglected to talk about one of the most fundamental and important: refraction! In short, refraction is the bending of a ray of light when it passes from one medium to another. Unlike optical phenomena like diffraction and interference, which usually require careful experimental preparation to be observable, refraction is readily seen all around us. If you’ve ever seen a seemingly bent straw in a glass of water, you’ve seen the effect of refraction.
A straw which seemingly has a kink in it, thanks to refraction. The red line highlights the kink.
So we’ve all seen refraction, and can recognize it, but how does it work, exactly? And why does it happen? And what can we do with it? All of these questions are surprisingly non-trivial, and we’ll tackle them in this post.
Let’s start by getting the technical aspects of refraction out of the way. The figure below illustrates the basic idea and introduces the notation we need to describe refraction.
We have a ray of light traveling from the bottom of the picture in medium 1, making an angle $\theta_1$ with the normal to the surface. On crossing the interface into medium 2, the direction of the ray of light changes, and it now is traveling at an angle $\theta_2$ with respect to the normal to the surface.
One can show that the relationship between the “incident angle” $\theta_1$ and the “transmission angle” $\theta_2$ is dictated by the mathematical equation now known as Snell’s law,
$n_1\sin\theta_1 = n_2\sin\theta_2$.
In this equation, “sin” represents the trigonometric sine function. The quantities $n_1$ and $n_2$ represent the refractive indices of the two media.
The refractive index requires a bit of explanation. It represents (in a very specific sense) the fraction by which the speed of light is decreased in that particular medium. In other words, the speed of light in vacuum is $c = 3 \times 10^8$ meters/second, and the speed of light in the medium is $v = c/n$. The refractive index is a property of the medium in question: $n=1.33$ for water, and $n = 1.5$ for common glass. A medium with a higher refractive index is a “denser” medium, while the medium with a lower refractive index is a “rarer” medium.
For those who are not that comfortable with trigonometry, we can summarize the properties of Snell’s law in words. When a ray of light is travelling from a rarer to a denser medium, its direction gets bent towards the optical axis (the dashed perpendicular line). When a ray of light is travelling from a denser to a rarer medium, its direction gets bent away from the optical axis. We compare the two cases below.
You may notice the only difference between the two pictures is the direction of the arrows! This is an important observation: refraction is a reversible process! If a ray coming from a rare medium at angle $\theta_1$ transmits at angle $\theta_2$, then a ray coming from the dense medium at angle $\theta_2$ transmits at angle $\theta_1$. This observation will turn out to have an important implication momentarily.
Snell’s law is named after the Dutch astronomer Willebrord Snellius (1580-1626), though it is now known that it was discovered much earlier: the first correct description seems to have been achieved by the Baghdad scientist Ibn Sahl (c. 940-1000), in his book On Burning Mirrors and Lenses, published in 984.
People probably had an intuitive understanding of refraction much, much earlier than this, however. Because light is refracted when it exits a body of water, things underwater appear to be in a different position than they actually are. Spear fishermen figured out long ago that one must strike at a spot closer than the fish appears in order to spear it.
So we now have a description, both mathematical and observational, of what refraction is, and the next natural question to ask is: why does it happen? We have mentioned that refraction occurs because the light changes speed as it passes the interface between two media; this is clear from Snell’s law itself, which depends on the refractive indices of the two media. It is not clear, however, how a change in speed can also result in a change of direction. After all, when I step on the brakes of my car, my car (hopefully) doesn’t go swerving off to the side!
The answer to this conundrum lies in the wave nature of light. Though in most ordinary circumstances one can visualize light as traveling along definite lines (rays), as we have shown in the pictures above, in reality light propagates as a wave which is spread out through space. A snapshot of what refraction looks like for a wave is shown below, with the ray direction drawn on top.
This picture assumes a plane wave coming from a rare medium from below and traveling into a denser medium like glass. The bright colors represent the peaks of the wave (to be referred to as the wavefronts), which the dark colors represent the troughs. It should be noted that the direction of propagation of the wave (indicated by the ray direction) is perpendicular to the wavefronts.
It should be noted that the wavelength (distance between wavefronts) gets smaller when the wave enters the dense medium. This is the natural result of the wave speed decreasing: just as the separation between cars gets smaller when the speed limit drops, the separation between wavefronts gets smaller when the wave speed drops.
The wave, however, must be continuous across the interface between the two media: the wavefronts must match up on either side of the interface, as they do in the picture above. Because the wavefront spacing is different on either side, the only way they can match up is if the wavefronts are tilted in the denser medium — the wave must change its direction of propagation! An animation of refraction in action is shown below.
This physical explanation may not seem entirely satisfactory — is there a simpler way to explain why the wave would change direction? To some degree we have to accept that refraction “is what it is”: not every physical phenomenon, even one as mathematically simple as refraction, can be explained in everyday terms. If we’re willing to be a little silly, however, we can come up with a crude analogy.
Let’s imagine a piece of wood sliding along a sheet of ice. This piece of wood might be a hockey stick dropped by a player who was checked into the boards particularly hard. As it travels, this piece of wood ends up crossing a “sticky” line, which might be a very poorly painted blue line at an ice rink. The starting situation is as follows.
So what happens when the stick starts to cross the blue line? The right side, which crosses first, slows down, but the left side keeps moving. The net result? The stick rotates as it crosses the line, and its direction of travel naturally must change along with it!
By the time the entire stick crosses the blue line, every part of it has been slowed equally. It is now moving along a straight path again, but in a different direction than it was to begin with, just like a light wave changes direction on crossing the boundary between two media.
The stick in this case is analogous to the wavefronts of light, and the blue line represents the interface between two optical media. Though this example is crude, it does highlight an important aspect of refraction: it only happens for “wide” objects, such as long sticks or spread-out wavefronts. A tiny marble rolling across the same blue line would slow down but its direction of travel would be unaffected. With this in mind, we can say that refraction is a natural consequence of the “spread-out” wave nature of light.
So now that we understand what refraction is, what can we do with it? Well, pretty much all of conventional lens optics — eye glasses, contact lenses, microscopes, cameras, telescopes, magnifying glasses — is based on refraction. Light passing through a simple convex lens, for instance, will be refracted at both surfaces. Parallel light rays coming through the lens will be bent to converge to a focal point; this is illustrated below.
We can also draw a picture to illustrate the behavior of the wavefronts of light on passing through a lens of this sort.
You might notice that I haven’t drawn the wavefronts all the way to the point where they would, presumably, converge at the focus. The wave behavior of light in the region of focus is actually quite complicated, in large part due to interference and diffraction effects that arise in that region.
There is one curious aspect of the theory of refraction that we should address before concluding. In going from a rare medium to a dense medium, the biggest angle $\theta_2$ we can achieve comes when the incident ray is nearly parallel to the interface, or $\theta_1=90^\circ$.
The areas shaded in red represent all possible areas that the incident and transmitted rays can propagate. But now we have a problem: we have said that refraction is a reversible process, and of course we can send in a ray from the denser medium from any direction. What will happen to the ray pictured below?
The light coming in at this angle falls outside of Snell’s law, at least as we have described it so far. It turns out that the ray will be completely reflected at the interface, and no light will be transferred into the rarer medium! The angle $\theta_2$ described above is known as the critical angle, and any light propagating at this angle or greater from the denser medium will be perfectly reflected. This is a phenomenon known as total internal reflection, and it forms the basis of fiber optics, in which light is transmitted over long distances through transparent glass fibers. Optical fibers are now used to convey most of the information that gets transmitted over the internet.
Though the proper description of the optics of fibers requires some sophisticated mathematics, we can understand it roughly as shown in the following picture.
Light enters the glass fiber from the left at such an angle that it suffers from total internal reflection every time it bounces off of the boundary of the fiber. Essentially no light can escape from the fiber due to refraction, though a significant amount will be absorbed by the glass itself.
There’s so much more we can say about refraction! For instance, certain materials exhibit what is known as double refraction (or birefringence), as exhibited by this piece of optical calcite (source Wikipedia):
For optical calcite, there are two refracted rays that are transmitted by the material, instead of one, producing a double image. One of the refracted rays is independent of the orientation of the crystal, while the other rotates its position with the crystal, as shown in the short video below.
Double refraction arises because the unusual atomic crystalline structure of calcite causes different polarizations of light to be refracted differently.
Another unusual and surprising refractive effect is the recently-proposed idea of negative refraction. In principle, it is possible to fabricate materials which have a negative refractive index; a light ray entering or exiting such a material would be refracted in a sense opposite to a normal material.
I’ve gone on far too long about refraction, but anyone interested in more discussion of negative refraction can refer to this earlier post of mine.
Refraction, though one of the oldest optical phenomena to be observe and quantified, possesses a surprising amount of complexity and subtlety to it! The same thing can be said for reflection… but I’ll defer that discussion for another post.
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### 4 Responses to Optics basics: refraction
1. Frank Lee MeiDere says:
When I was in grade sic (I believe), the teacher told us to think of it as a column of soldiers marching along concrete toward a sandy beach. When they reach the sand, they cannot march as fast, and so each soldier slows down. If they approach the beach straight on, then they will continue to move in the same direction, but somewhat more slowly. If they approach at an angle, however, then those soldiers who reach the sand first will slow down first, and the trajectory of the column of soldiers will be changed.
I always found it a good way to think of refraction. The one thing I didn’t understand — and still don’t — is that if light slows down when entering the medium, does time also slow down?
• skullsinthestars says:
Thanks for the comment! A collection of soldiers marching is a good way to think about it, though it actually is a little subtle — one has to assume that the later soldiers adjust their direction to stay in line with the slower soldiers, otherwise they end up traveling in their original direction but at a different angle.
The one thing I didn’t understand — and still don’t — is that if light slows down when entering the medium, does time also slow down?
Time is completely unaffected when light passes into a medium — in fact, refraction is derived from the wave theory of light by using the assumption that the frequency of wave oscillation does not change on passing into the medium. The idea that time is affected comes, I suspect, from thinking about the special theory of relativity. In special relativity, the speed of light is said to be the same for all observers… but this is in a vacuum, not in a medium! In matter, light can slow down pretty much the same way a car or any ordinary object can.
2. Pingback: Waves,Properties and Their Applications. « PhysicsRUs
3. Clifford Andersen says:
If this is true then why do you not see light build up like water coming up to a dam?
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http://math.stackexchange.com/questions/258228/probability-that-the-10th-class-this-semester-is-the-3rd-class-cancelled-when-ch/258231
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# Probability that the 10th class this semester is the 3rd class cancelled when chances are independent with 0.05 probability each day?
Every day, a lecture may be cancelled due to inclement weather with probability 0.05. Class cancellations on different days are independent. Compute the probability that the tenth class this semester is the 3rd class cancelled?
This is a practice exam so the exact answer isn't near as important as the proper solution. My assumption is P(3rd cancelled is 10th class) = P(2 cancellations in 9 classes)*P(cancelled)
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## 1 Answer
The procedure you propose is correct.
Call a cancelled class a success. So we want the probability of exactly $2$ successes in $9$ trials, followed on the $10$-th trial by a success.
The probability of success is $p=0.05$. So the probability of exactly $2$ successes in $9$ trials is $\dbinom{9}{2}p^2(1-p)^7$.
Multiply by $p$ to get the desired probability.
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