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http://math.stackexchange.com/questions/7266/why-does-this-process-when-iterated-tend-towards-a-certain-number-the-golden
# Why does this process, when iterated, tend towards a certain number? (the golden ratio?) 1. Take any number $x$ (edit: x should be positive, heh) 2. Add 1 to it $x+1$ 3. Find its reciprocal $1/(x+1)$ 4. Repeat from 2 So, taking $x = 1$ to start: • 1 • 2 (the + 1) • 0.5 (the reciprocal) • 1.5 (the + 1) • 0.666... (the reciprocal) • 1.666... (the + 1) • 0.6 (the reciprocal) • 1.6 (the + 1) • 0.625 • 1.625 • 0.61584... • 1.61584... • 0.619047... • 1.619047... • 0.617647058823.. etc. If we look at just the "step 3"'s (the reciprocals), we get: • 1 • 0.5 • 0.666... • 0.6 • 0.625 • 0.61584... • 0.619047... • 0.617647058823.. This appears to always converge to 0.61803399... no matter where you start from. I looked up this number and it is often called "The golden ratio" - 1, or $\frac{1+\sqrt{5}}{2}-1$. 1. Is there any "mathematical" way to represent the above procedure (or the terms of the second series, of "only reciprocals") as a limit or series? 2. Why does this converge to what it does for every starting point $x$? edit: darn, I just realized that the golden ratio is actually 1.618... and not 0.618...; I edited my answer to change what the result is apparently (golden ratio - 1). However, I think I could easily make it the golden ratio by taking the +1 "steps" of the original series, instead of the reciprocation steps of the original series: • 2 • 1.5 • 1.666... • 1.6 • 1.625 • 1.61584... • 1.619047... • 1.617647058823.. which does converge to $\frac{1+\sqrt{5}}{2}-1$ Explaining either of these series is adequate as I believe that explaining one also explains the other. - 1 – Hans Lundmark Oct 20 '10 at 8:07 ## 5 Answers We want to show that the function $f(x) = \frac{1}{1+x}$ has a unique fixed point to which it converges when iterated on positive $x$. If $x$ is positive, then $f(x)$ is between $1$ and $0$, so $f(f(x))$ is between $1$ and $\frac{1}{2}$. It is not hard to see that in fact $f$ fixes the interval $\left[ \frac{1}{2}, 1 \right]$. Now, for $x, y$ in this interval, $$\left| \frac{1}{x+1} - \frac{1}{y+1} \right| = \left| \frac{y-x}{(1+x)(1+y)} \right| \le \frac{4}{9} |x - y|$$ so on this interval $f$ satisfies the conditions of the Banach fixed point theorem. The unique fixed point to which everything always converges is the unique solution to $f(x) = x$, which you have already found. - 1 Note that this argument is extremely general; in the right intervals it applies to many other functions one might idly be tempted to iterate on a calculator... – Qiaochu Yuan Oct 20 '10 at 15:09 1 – J. M. Oct 20 '10 at 15:17 +1: For talking about Banach's fixed point theorem. – Aryabhata Oct 20 '10 at 16:32 1 The generality of this answer is pretty striking. It even showed why iterating $f(x) = \sqrt{x+1}$ also tends towards the golden ratio. and...basically every other function I've been trying out on my shiny new hp 35s rpn calculator. – Justin L. Oct 20 '10 at 16:49 1 Maybe it's also worth mentioning what happens for negative x. If x equals -F_{n+1}/F_n for any index n (where F_n is the Fibonacci sequence) then some iterate of x is equal to -1 and the sequence blows up. On the other hand, for x less than -2, f(x) is in (-1, 0] so f(f(x)) is positive. And for x greater than -1.5, f(x) is less than -2. The danger zone is [-2, -1.5] and there is some repelling behavior here away from -phi, in addition to the blowing up; note that on this interval f is expansive rather than contractive. – Qiaochu Yuan Oct 20 '10 at 21:42 You are iterating the operation $$x\mapsto\frac1{x+1}.$$ We can represent this in matrix terms. Set $$A=\begin{pmatrix} 0&1\\ 1&1\end{pmatrix}.$$ If $$A \begin{pmatrix} x\\ 1\end{pmatrix} =\begin{pmatrix} y_1\\ z_1\end{pmatrix}$$ then $1/(x+1)=y_1/z_1$. After $n$ iterations one gets $$A^n \begin{pmatrix} x\\ 1\end{pmatrix} =\begin{pmatrix} y_n\\ z_n\end{pmatrix}$$ and $x_n=y_n/z_n$ is got by applying the map $x\mapsto 1/(x+1)$ $n$ times to $x$. Naturally you won't be surprised to find that the eigenvalues of $A$ are $\tau=\frac12(1+\sqrt5)$ and $\tau=\frac12(1-\sqrt5)$. The eigenvectors are $v=(1\ \tau)^t$ and $w=(1\ \tau')^t$. We can write $$\begin{pmatrix} x\\ 1\end{pmatrix}=av+bw$$ and so $$\begin{pmatrix} y_n\\ z_n\end{pmatrix}=a\tau^n v+b\tau'^nw.$$ Now $\tau>1>|\tau'|$ so that for large $n$, $y_n$ and $z_n$ are very close to $a\tau^n$and $a\tau^{n+1}$ so that $x_n\to\infty$ if $a=0$. The only exception is when $a=0$ which only arises for $x=-\frac12(1+\sqrt5)$. (I bet you didn't test that one!) - 2 I also bet he didn't test $x=\frac(1-\sqrt{5})$. ;-) – Hans Lundmark Oct 20 '10 at 7:55 1 It's hard to test that $x = -(1+\sqrt{5})/2$ is a fixed point numerically, because it's a repelling fixed point. – Michael Lugo Oct 20 '10 at 16:31 1 (Just to clarify: My comment refers to an earlier version where there was a typo. It's been corrected now.) – Hans Lundmark Oct 20 '10 at 17:29 as I remark in the comments to my answer there are a few other exceptions, namely cases where z_n = 0 for some finite n. Of course one can continue the recursion beyond this if one is willing to work in P^1(R) instead of R... – Qiaochu Yuan Oct 21 '10 at 10:24 This was supposed to be an addendum to Hans's comment, but it got too long. The iteration Justin considered formally generates a continued fraction: $$\cfrac{1}{1+\cfrac{1}{1+\dots}}$$ and as I mentioned in this answer, the numerators and denominators of the nth convergent of a continued fraction can be computed recursively. If we apply the formula in that answer to this situation, we get $$\begin{bmatrix}C_n\\D_n\end{bmatrix}=\begin{bmatrix}C_{n-1}\\D_{n-1}\end{bmatrix}+\begin{bmatrix}C_{n-2}\\D_{n-2}\end{bmatrix}$$ with initial conditions $$\begin{bmatrix}C_{-1}\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix},\qquad \begin{bmatrix}C_{0}\\D_{0}\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$ From here, the setup is now similar to Robin's answer, since the two recursions in fact generate the Fibonacci numbers: $C_n=F_n$ and $D_n=F_{n+1}$ Your continued fraction then is the limit $$\lim_{n\to\infty}\frac{F_n}{F_{n+1}}$$ and the equivalence to Robin's answer is due to the Binet formula: $$F_n=\frac{\varphi^n-\left(-\varphi^{-1}\right)^n}{\sqrt{5}}$$ (which in fact can be derived from Robin's answer). Substituting that into the limit and evaluating gets you the answer. - Here is another way of looking at it. First consider the special case of starting with $1$. Consider what happens when $\displaystyle x = \frac{f_n}{f_{n+1}}$ where $f_n$ is the $n^{th}$ fibonacci number. You get $$\frac{1}{\frac{f_n}{f_{n+1}} + 1} = \frac{f_{n+1}}{f_n + f_{n+1}} = \frac{f_{n+1}}{f_{n+2}}$$ Since $\displaystyle 1 = \frac{f_1}{f_2}$ we see that after $n$ iterations, $\displaystyle x = \frac{f_{n+1}}{f_{n+2}}$ This can be generalized to any other starting value, by using a Fibonacci like sequence, which satisfies the recurrence $\displaystyle a_{n+2} = a_{n+1} + a_{n}$ and choosing appropriate $a_{2}$ and $a_{1}$ so that $\displaystyle \frac{a_1}{a_2}$ is the initial guess for $x$. The $n^{th}$ value for $x$ will be given by $\displaystyle \frac{a_n}{a_{n+1}}$ The general formula for such sequences is given by $a_{n} = A\alpha^n + B\beta^n$ where $\alpha,\beta$ are roots of $\displaystyle z^2 = z + 1$ and thus the limit of $\displaystyle \frac{a_n}{a_{n+1}}$ can be easily found, which will be one of $1/\alpha$ or $1/\beta$ (which you can also see, by assuming there is a limit $1/L$ and setting $\displaystyle 1/L = \frac{1}{1+1/L}$). - Define the function $f(x) = \frac1{1+x}$ so the process you defined is just going from x to f(x). let $x_0\in\mathbb{R}$ and $x_{i+1} = f(x_i)$ then your claim is that $\lim\; x_i = \frac {1+\sqrt{5}}{2}$. if this series converges then $x = \lim\; x_i = \lim\; f(x_{i-1}) = f( \lim\; x_{i-1} ) = f(x)$ so the limit is $x=\frac{1}{1+x}$ or in other words you have $x^2+x-1=0$. the solutions are $\frac{-1\pm \sqrt{5}}{2}$ this isn't exactly the golden ratio. if you take $\frac1{x-1}$ instead then you'll have it. you still need to find out for which $x_0$ this sequence converges - this could be done using numerical analysis (sorry, I'm a bit rusty on this, so I'll leave this part to someone else). -
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http://regularize.wordpress.com/tag/convex-sets/
# regularize Trying to keep track of what I stumble upon November 15, 2012 ## What is convexity? Posted by Dirk under Math, Optimization | Tags: convex functions, convex sets, convexity, non-convex optimization | 1 Comment There are several answers to the following question: 1. What is a convex set? For a convex set you probably know these definitions: Definition 1 A subset ${C}$ of a real vector space ${V}$ is convex if for any ${x,y\in C}$ and ${\lambda\in[0,1]}$ it holds that ${\lambda x + (1-\lambda)y\in C}$. In other words: If two points lie in the set, then every convex combination also lies in the set. While this is a “definition from the inside”, convex sets can also be characterized “from the outside”. We add closedness as an assumption and get: Definition 2 A closed subset ${C}$ of a real locally convex topological vector space ${V}$ is convex if it is the intersection of closed halfspaces (i.e. sets of the form ${\{x\in V\ :\ \langle a,x\rangle\geq c\}}$ for some ${a}$ in the dual space ${V^*}$ and ${c\in {\mathbb R}}$). Moreover, we could define convex sets via convex functions: Definition 3 A set ${C\subset V}$ is convex if there is a convex function ${f:V\rightarrow {\mathbb R}}$ such that ${C = \{x\ :\ f(x)\leq 0\}}$. Of course, this only makes sense once we have defined convex functions. Hence, we could also ask the question: 2. What is a convex function? We can define a convex function by means of convex sets as follows: Definition 4 A function ${f:V\rightarrow{\mathbb R}}$ from a real vector space into the real numbers is convex, if its epigraph ${\text{epi} f = \{(x,\mu)\ :\ f(x)\leq \mu\}\subset V\times {\mathbb R}}$ is convex (as a subset of the vector space ${V\times{\mathbb R}}$). The epigraph consists of the points ${(x,\mu)}$ which lie above the graph of the function and carries the same information as the function. (Let me note that one can replace the real numbers here and in the following with the extended real numbers ${\bar {\mathbb R} = {\mathbb R}\cup\{-\infty,\infty\}}$ if one uses the right extension of the arithmetic and the obvious ordering, but we do not consider this in this post.) Because epigraphs are not arbitrary convex sets but have a special form (if a point ${(x,\mu)}$ is in an epigraph, then every ${(x,\lambda)}$ with ${\lambda\geq \mu}$ is also in the epigraph), and because the underlying vector space ${V\times {\mathbb R}}$ comes with an order in the second component, some of the definitions for convex sets from above have a specialized form: From the definition “convex combinations stay in the set” we get: Definition 5 A function ${f:V\rightarrow {\mathbb R}}$ is convex, if for all ${x,y\in V}$ and ${\lambda\in [0,1]}$ it holds that $\displaystyle f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y).$ In other words: The secant of any two points on the graph lies above the graph. From the definition by “intersection of half spaces” we get another definition. Since we added closedness in this case we add this assumption also here. However, closedness of the epigraph is equivalent to lower-semicontinuity (lsc) of the function and since lsc functions are very convenient we use this notion: Definition 6 A function ${f:V\rightarrow{\mathbb R}}$ is convex and lsc if it is the pointwise supremum of affine functions, i.e., for some set ${S}$ of tuples ${(a,c)\in V^*\times {\mathbb R}}$ it holds that $\displaystyle f(x) = \sup_{(a,c) \in S} \langle a,x\rangle + c.$ A special consequence if this definition is that tangents to the graph of a convex function lie below the function. Another important consequence of this fact is that the local behavior of the function, i.e. its tangent plane at some point, carries some information about the global behavior. Especially, the property that the function lies above its tangent planes allows one to conclude that local minima of the function are also global. Probably the last properties are the ones which give convex functions a distinguished role, especially in the field of optimization. Some of the previous definitions allow for generalizations into several direction and the quest for the abstract core of the notion of convexity has lead to the field of abstract convexity. 3. Abstract convexity When searching for abstractions of the notion of convexity one may get confused by the various different approaches. For example there are generalized notions of convexity, e.g. for function of spaces of matrices (e.g. rank-one convexity, quasi-convexity or polyconvexity) and there are also further different notions like pseudo-convexity, invexity or another form ofquasi-convexity. Here we do not have generalization in mind but abstraction. Although both things are somehow related, the way of thinking is a bit different. Our aim is not to find a useful notion which is more general than the notion of convexity but to find a formulation which contains the notion of convexity and abstracts away some ingredients which probably not carry the essence of the notion. In the literature one also finds several approaches in this direction and I mention only my favorite one (and I restrict myself to abstractly convex functions and not write about abstractly convex sets). To me, the most appealing notion of an abstract convex function is an abstraction of the definition as “pointwise supremum of affine functions”. Let’s look again at the definition: A function ${f:V\rightarrow {\mathbb R}}$ is convex and lsc if there is a subset ${S}$ of ${V^*\times {\mathbb R}}$ such that $\displaystyle f(x) = \sup_{(a,c)\in S} \langle a,x\rangle + c.$ We abstract away the vector space structure and hence, also the duality, but keep the real-valuedness (together with its order) and define: Definition 7 Let ${X}$ be a set and let ${W}$ be a set of real valued function on ${X}$. Then a function ${f:X\rightarrow{\mathbb R}}$ is said to be ${W}$-convex if there is a subset ${S}$ of ${W\times {\mathbb R}}$ such that $\displaystyle f(x) = \sup_{(w,c)\in S} w(x) + c.$ What we did in this definition was simply to replace continuous affine functions ${x\mapsto \langle a,x\rangle}$ on a vector space by an arbitrary collection of real valued functions ${x\mapsto w(x)}$ on a set. On sees immediately that for every function ${w\in W}$ and any real number ${c}$ the function ${w+c}$ is ${W}$ convex (similarly to the fact that every affine linear function is convex). Another nice thing about this approach is, that it allows for some notion of duality/conjugation. For ${f:V\rightarrow{\mathbb R}}$ we define the ${W}$-conjugate by $\displaystyle f^{W*}(w) = \sup_{w\in W} \Big(w(x)- f(x) \Big)$ and we can even formulate a biconjugate $\displaystyle f^{W**}(x) = \sup_x \Big(w(x) - f^{W*}(w)\Big).$ We naturally have a Fenchel inequality $\displaystyle w(x) \leq f(x) + f^{W*}(w)$ and we may even define subgradients as usual. Note that a conventional subgradient is an element of the dual space which defines a tangent plane at the point where the subgradient is taken, that is, ${a\in V^*}$ is a subgradient of ${f}$ at ${x}$, if for all ${y\in V}$ it holds that ${f(y) \geq f(x) + \langle a,y-x\rangle}$ or $\displaystyle f(y) - \langle a,y\rangle\geq f(x) - \langle a,x\rangle.$ A ${W}$-subgradient is an element of ${W}$, namely we define: ${w\in\partial^{W}f(x)}$ if $\displaystyle \text{for all}\ y:\quad f(y) -w(y) \geq f(x) - w(x).$ Then we also have a Fenchel equality: $\displaystyle w\in\partial^W f(x)\iff w(x) = f(x) + f^{W*}(w).$ One may also take dualization as the starting point for an abstraction. 4. Abstract conjugation We could formulate the ${W}$-conjugate as follows: For ${\Phi(w,x) = w(x)}$ we have $\displaystyle f^{W*}(x) = \sup_w \Big(\Phi(w,x) - f(x)\Big).$ This opens the door to another abstraction: For some sets ${X,W}$ (without any additional structure) define a coupling function ${\Phi: X\times W \rightarrow {\mathbb R}}$ and define the ${\Phi}$-conjugate as $\displaystyle f^{\Phi*}(w) = \sup_x \Big(\Phi(w,x) - f(x)\Big)$ and the ${\Phi}$-biconjugate as $\displaystyle f^{\Phi**}(x) = \sup_x \Big(\Phi(w,x) - f^{W*}(w)\Big)$
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http://math.stackexchange.com/questions/138797/is-there-a-function-or-anything-else-that-gives-the-same-result-on-an-interval
# Is there a function or anything else that gives the same result on an interval? [duplicate] Possible Duplicate: Is there a function or anything else that gives the same result on a set? I want to know if there is a function, for example, that gives the same result on the interval $[10,15]$. I want to have these function values $$f(10)=25,\, f(11)=25,\, f(12)=25,\, f(13)=25,\, f(14)=25,\, f(15)=25$$ Thanks. - 1 How about $f: [10, 15]\to \mathbb R, x \mapsto 25$ – martini Apr 30 '12 at 9:00 2 Of course there is. For example the constant function $f(x) = 25$, for all $x \in \mathbb{R}$. – Martin Wanvik Apr 30 '12 at 9:03 Exuse me i just asked a more specific question thank you – Boubouh Apr 30 '12 at 9:28 1 @Boubouh: As I said, edit your original question. There is no need to ask another one. – Martin Wanvik Apr 30 '12 at 9:30 @Boubouh: You might also use $f(x)=25\cos (2\pi x)$ for $x\in\mathbb R$. – Dejan Govc Apr 30 '12 at 9:48 show 2 more comments ## marked as duplicate by Did, Willie Wong♦Apr 30 '12 at 10:55 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 1 Answer Actually, following Dejan Govc's lead, you could take any Fourier Series $$f(x) = a_0 + \sum_{n=1}^\infty a_n \cos\left(2\pi nx\right) + b_n \sin\left(2\pi nx\right)$$ whose cosine coefficients $a_n$ sum to $25$ (the $b_n$ are completely free). The constant function corresponds to $(a_0,a_1,\dots)=(25,0,0,\dots)$ and Dejan's to $(0,25,0,\dots)$, both with all $b_n=0$. The complete function space would look like the set of periodic functions on the unit interval with the boundary condition $f(0)=f(1)=f(k)=25$ for all $k\in\mathbb{Z}$. This is a pretty large space of functions! - +1. Nice observation! – Dejan Govc Apr 30 '12 at 10:17
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http://math.stackexchange.com/questions/201169/what-is-the-relevance-of-the-supremum-in-this-question/201182
# What is the relevance of the supremum in this question? Prove that $d$ is a metric on the set $X$. $d_u(f,g) = \sup\{|f(x) - g(x)|: x \in I \}, X = C(I)$ the set of all continuous functions from the closed bounded interval $I = [a,b]$ to $\mathbb{R}$ I don't understand what difference using the supremum of $|f(x) - g(x)|$ makes. $|f(x) - g(x)|$ will always be a real number so its seems taking the suprmemum of it is pointless. Am I missing something here? - ## 1 Answer The problem is that $|f(x) - g(x)|$ is a function of $x$; for each $x$, it returns the difference between $f$ and $g$ at that point. You want it to be a single number for all $x \in I$, that's why you take the supremum. You define the distance between two functions to be the farthest they are ever separated. - Cheers. Is it the curly brackets $sup\{...\}$ that indicate that I am taking the supremum of a set of values? – dukenukem Sep 23 '12 at 15:10 @dukenukem: Pretty much. Sometimes the supremum is written a little differently, like this: $\displaystyle \sup_{x\in I}|f(x)-g(x)|$. But it's the same thing. – Javier Badia Sep 23 '12 at 15:21 Would taking the max be the same as taking the supremum in this case> – dukenukem Sep 24 '12 at 6:45 – lazyhaze Sep 24 '12 at 10:23
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http://math.stackexchange.com/questions/218402/proper-orthogonal-matrix
# Proper Orthogonal Matrix Prove that, for any $\theta$, $\lambda$, $\mu$ $Q = \pmatrix{\cos{\lambda}\cos{\mu} - \cos{\theta}\sin{\lambda}\sin{\mu}&\sin{\lambda}\cos{\mu} + \cos{\theta}\cos{\lambda}\sin{\mu}&\sin{\theta}\sin{\mu}\\-\cos{\lambda}\sin{\mu} - \cos{\theta}\sin{\lambda}\cos{\mu}&-\sin{\lambda}\sin{\mu} + \cos{\theta}\cos{\lambda}\cos{\mu}&\sin{\theta}\cos{\mu}\\ \sin{\theta}\sin{\lambda}&-\sin{\theta}\cos{\lambda}&\cos{\theta}}$ is a proper orthogonal matrix and write down a formula for $Q^{-1}$ How will I be able to do this problem? I know that in order to be a proper orthogonal matrix it must have $\det(Q) = 1$ and it must form orthonormal columns, but how can I show that here? - – Ganesh Oct 22 '12 at 0:51 @Ganesh the book didn't mention anything about euler angles. – diimension Oct 22 '12 at 0:55 ## 1 Answer Just calculate $Q Q^T$ and simplify it to $I$, taking into account the trigonometric identities $\cos^2 \alpha + \sin^2 \alpha = 1$. It's a bit tedious, but has to work if your matrix is correct. - That was what I was afraid of. I was hoping there was an easier way. Toda Raba!! – diimension Oct 22 '12 at 1:02 Actually the $(1,3)$ element of your matrix is wrong: $\sin \lambda\; \sin \mu$ should be $\sin\theta\; \sin \mu$. – Robert Israel Oct 22 '12 at 1:05 Yup, you are right. Thanks for catching that. I edited it. – diimension Oct 22 '12 at 1:07
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http://unapologetic.wordpress.com/2010/07/26/the-measures-of-ordinate-sets/
# The Unapologetic Mathematician ## The Measures of Ordinate Sets If $(X,\mathcal{S})$ is a $\sigma$-algebra and $f:X\to\mathbb{R}$ is a Borel-measurable function we defined the upper and lower ordinate sets $V^*(f)$ and $V_*(f)$ to be measurable subsets of $X\times\mathbb{R}$. Now if we have a measure $\mu$ on $X$ and Lebesgue measure on the Borel sets, we can define the product measure $\lambda$ on $X\times\mathbb{R}$. Since we know $V^*(f)$ and $V_*(f)$ are both measurable, we can investigate their measures. I assert that $\displaystyle\lambda(V^*(f))=\int f\,d\mu=\lambda(V_*(f))$ It will be sufficient to establish this for simple functions, since for either the upper or the lower ordinate set we can approximate any measurable $f$ by a monotone sequence of simple $\{f_n\}$ so that $\lim\limits_{n\to\infty}V^*(f_n)=V^*(f)$ or $\lim\limits_{n\to\infty}V_*(f_n)=V_*(f)$. Then the limit will commute with $\lambda$ (since measures are continuous), and it will commute with the integral as well. So, we can assume that $f$ is simple, and write $\displaystyle f=\sum\limits_{i=1}^n\alpha_i\chi_{E_i}$ with the $\{E_i\}$ a pairwise-disjoint collection of measurable sets. But now if the equality holds for each of the summands then it holds for the whole function. That is, we can assume — without loss of generality — that $f=\alpha\chi_E$ for some real number $\alpha$ and some measurable subset $E\subseteq X$. And now the result should be obvious! Indeed, $V^*(f)$ is the measurable rectangle $E\times[0,\alpha]$, while $V_*(f)$ is the measurable rectangle $E\times[0,\alpha)$. Since the product measure on a measurable rectangle is the product of the measures of the two sides, these both have measure $\alpha\mu(E)$. On the other hand, we calculate the integral as $\displaystyle\int f\,d\mu=\int\alpha\chi_E\,d\mu=\alpha\mu(E)$ and so the equality holds for such rectangles, and thus for simple functions, and thus for all measurable functions. Of course, it should now be clear that the graph of $f$ has measure zero. Indeed, we find that $\displaystyle\lambda(V^*(f)\setminus V_*(f))=\lambda(V^*(f))-\lambda(V_*(f))=\int f\,d\mu-\int f\,d\mu=0$ These results put a precise definition to the concept of the integral as the “area under the graph”, which was the motivation behind our definition of the Riemann integral, way back when we introduced it. ### Like this: Posted by John Armstrong | Analysis, Measure Theory No comments yet. « Previous | Next » ## 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://mathhelpforum.com/discrete-math/6984-help-trees.html
# Thread: 1. ## Help with trees ok, so i'm supposed to draw all the trees with five vertices and their complements. now, i have all the trees with five vertices down. but i don't know how to draw their complements. can anyone help me with this and show me an example of one? i would be really grateful. 2. Originally Posted by clockingly ok, so i'm supposed to draw all the trees with five vertices and their complements. now, i have all the trees with five vertices down. but i don't know how to draw their complements. can anyone help me with this and show me an example of one? i would be really grateful. To find a complement you need to subtract from the complete graph on 5 vertices. Below is the tree on 5 vertices and its complemnt in red. Attached Thumbnails 3. thanks! sorry about my other post in the urgent forum. i didn't know it wasn't allowed. i'm still confused, though. the picture you uploaded is a graph. but i'm supposed to be drawing the complement of a tree (unless the one you uploaded is also a tree??). all the trees with five vertices are at the link below. how do i find the complements of those? reason i'm confused is because i didn't draw them in the pentagon shape. 4. Originally Posted by clockingly thanks! i'm still confused, though. the picture you uploaded is a graph. but i'm supposed to be drawing the complement of a tree (unless the one you uploaded is also a tree??). all the trees with five vertices are at the link below. how do i find the complements of those? [draw=100]Point (20,20); Point (40,20); Point (60,20); Point (80,20); Point (60,40); Segment (1,2); Segment (3,5); Segment (2,3); Segment (3,4);[/draw] That is the tree -----> [draw=100]Point (20,20); Point (40,20); Point (60,20); Point (80,20); Point (60,40); Segment (1,5); Segment (2,5); Segment (4,5);[/draw] That is the complement. ----> [EDIT] apparently not 5. thank you so much! that really helped! 6. Nice visuals Quick but I do not think that is the answer. First the number of edges need to be six. And the complement of a connected graph is a connected graph. 7. Originally Posted by ThePerfectHacker Nice visuals Quick but I do not think that is the answer. First the number of edges need to be six. And the complement of a connected graph is a connected graph. Oh, well then nevermind... You're going to have to explain the idea more clearly... 8. Originally Posted by ThePerfectHacker Nice visuals Quick but I do not think that is the answer. First the number of edges need to be six. And the complement of a connected graph is a connected graph. really? looking at wikipedia it says that all you do is connect the unconnected vertices... 9. hey Quick - so if your method is right and if i'm correct, then aren't both the last tree and its complement isomorphic? the last tree would look like a vertex in the middle (not connected to anything) with its four surrounding points connected in a diamond shape i think. 10. Originally Posted by clockingly hey Quick - so if your method is right and if i'm correct, then aren't both the last tree and its complement isomorphic? You're going to have to wait for a response from someone who's more "intune" with graphs, as all I know right now are circuits and paths 11. Originally Posted by clockingly hey Quick - so if your method is right and if i'm correct, then aren't both the last tree and its complement isomorphic? the last tree would look like a vertex in the middle (not connected to anything) with its four surrounding points connected in a diamond shape i think. First I do not think that is correct (though I am not expert on Graph theory). Because the complement is not connected while the original graph is connected. So how can they be isomorphic? 12. doesn't "isomorphic" just mean that the graphs have the same number of edges and vertices, though? 13. Originally Posted by clockingly doesn't "isomorphic" just mean that the graphs have the same number of edges and vertices, though? No, if two graphs are isomorphic then it means they have the same number of edges and vertices but not the other say around. Formally, given two graphs: $G=(V,E)$ und $G'=(V',E')$ They are isomorphic means that, There exists a bijection, $\phi:V\to V$ and $\phi(ab)=\phi(a)\phi(b)$ And, $\forall xy\in E' \exists a,b\in V$ such that $xy=\phi(a)\phi(b)$ 14. ok, so i get that the degrees of each vertice has to be the same in both graphs. i'm also confused because i'm supposed to find all the trees that are isomorphic to their complements. how do i approach this? aren't there infinitely many trees?
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http://physics.stackexchange.com/questions/55222/can-a-black-hole-bounce/55231
# Can a black hole bounce? Is there a limit to the amount of matter that a black hole can accrete per second and if so could a certain sized black hole bounce off a dense enough surface? - ## 2 Answers Let's answer the first part of your question first. In general, astrophysical objects are limited in the amount they can accrete by the Eddington limit. What happens is as matter is accreted onto the object (black hole in our case), it heats up due to conversion of gravitational potential energy into kinetic thermal energy. This hot matter emits photons whose total energy per time is called the luminosity of the object. If this luminosity is high enough (and you should get a higher luminosity for more accreted matter) then the outward pressure of the photons on the matter actually can overcome the gravitational inward pull. This point (rate of matter accretion) is called the Eddington limit. To partially answer the second part of your question, if the black hole had non-zero total net electric charge, then it could certainly bounce off another material of sufficient density. However if the charge on the black hole was identically zero, there would be no repulsive force to create a bounce. - – jeffdk Feb 26 at 23:32 Thankyou, the Edington limit was what I had in mind but could not recall. I'll have a go at calculating some masses and velocities now and see what I end up with. Would I get better bounces with smaller or larger BHs I wonder;-) – Jitter Feb 27 at 3:17 If you are taking the charged BH scenario, then the important quantity would be the ratio of charge, $q$ to BH mass, $M$. The larger $q/M$ the better the bounce. In fact you could probably calculate a minimum $q$ necessary to get a good bounce by requiring that the electrostatic repulsion is sufficiently great as to force the BH to turn around before it could get close enough to accrete any matter from what it is bouncing off of. – jeffdk Feb 27 at 10:49 Look, there's no limit to the accretion because the blackhole simply starts growing as it accretes more mass. I don't think it can "bounce off" any surface. Yes, if the gravitational forces are comparable, for example, if it encounters another black-hole, they can get in equilibrium, bounce off each other or simply merge. This happens in the universe. Every galaxy is found to have a supermassive black-hole at the center. We have evidence for galaxies merging along with their black-holes. Sometimes the galaxies just tear each other apart from tidal influence and the black-holes simply don't interact. Although, I have never heard of anything like bouncing or merging happening on garden-variety black-holes. The truth is, we haven't detected any significant sum of black-holes. All we have and are sure of are the black-holes at center of galaxies. - 2 LIGO's main purpose is to look for mergers of stellar-mass black holes. It's an expected event, it just hasn't been observed optically. – Jerry Schirmer Feb 26 at 23:02 @JerrySchirmer Exactly! – Cheeku Feb 26 at 23:03 but you can't say that stellar mass black holes are unobserverd. You have plenty of examples of those, such as things like Cygnus X1. – Jerry Schirmer Feb 26 at 23:07 @JerrySchirmer Yes, I didn't actually mean unobserved. I didn't use 'unobserved'. We have like millions of galaxies to run simulations and do data analysis, but not such a "significant sum(number)" of black-holes. – Cheeku Feb 26 at 23:11
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http://mathoverflow.net/questions/115458/smallest-lipschitz-constant-on-non-convex-domains/115459
## Smallest Lipschitz constant on non-convex domains ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is well known that if a function $f:U\to \mathbb C^n$, $U\subset \mathbb C^m$ satisfies $\sup_{x\in U}\|Df(x)\|_{\infty} = C < \infty$ uniformly on $U$ and $U$ is compact and convex, then $f$ is Lipschitz with smallest possible constant $C$. What if $U$ is non-convex, but still compact and connected? Is there any reasonable "measure of non-convexity" which can be used to bound the Lipschitz constant of such $f$? - ## 1 Answer You still have $\|f(x) -f(y)\| \le C L(x,y)$ where $L(x,y)$ is the length of the shortest path in $U$ from $x$ to $y$ (assuming such a path of finite length exists). The basic problem is that there can be points $x,y$ such that $\|x - y\|$ is small but $L(x,y)$ is large. An appropriate "modulus of non-convexity" would be the supremum of $L(x,y)/\|x-y\|$ for $x, y \in U$ with $x \ne y$, assuming that is finite. - Of course, should have figured this out myself. Thanks! – dima Dec 5 at 10:18
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http://mathhelpforum.com/differential-geometry/120270-sequence-lebesgue-space.html
# Thread: 1. ## Sequence, Lebesgue Space Show that if $|| f_n ||_1 \leq 2^{-n}$ for every $n \geq 1$ then $(f_n)_n$ converges to zero a.e. If $(f_n)_n$ is a sequence in $L^1[0, 1]$ is such that $\sum_{n=1}^{\infty} || f_n ||_1 < \infty$ then $\sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $s \in [0, 1]$. However, I don't see how to show this. I would appreciate some hints on how to proceed. 2. Originally Posted by eskimo343 Show that if $|| f_n ||_1 \leq 2^{-n}$ for every $n \geq 1$ then $(f_n)_n$ converges to zero a.e. If $(f_n)_n$ is a sequence in $L^1[0, 1]$ is such that $\sum_{n=1}^{\infty} || f_n ||_1 < \infty$ then $\sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $s \in [0, 1]$. However, I don't see how to show this. I would appreciate some hints on how to proceed. Not true, take $f_n(x)=1$ if $x=0$ and $f_n(x)=0$ otherwise. Then $\sum\parallel f_n\parallel_1<\infty$ but $\sum|f_n(0)|=\infty$ 3. Originally Posted by putnam120 Not true, take $f_n(x)=1$ if $x=0$ and $f_n(x)=0$ otherwise. Then $\sum\parallel f_n\parallel_1<\infty$ but $\sum|f_n(0)|=\infty$ That does not contradict the assertion that $\sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $s \in [0, 1]$. Originally Posted by eskimo343 Show that if $|| f_n ||_1 \leq 2^{-n}$ for every $n \geq 1$ then $(f_n)_n$ converges to zero a.e. Let $g_N = \sum_{n=N+1}^\infty|f_n|$. Then $\|g_N\|_1\leqslant\sum_{n=N+1}^\infty\|f_n\|_1\leq slant2^{-N}$ (sum of geometric series). Define $S_N = \{x:g_N(x)\geqslant2^{-N/2}\}$. Then $\mu(S_N)\leqslant2^{-N/2}$, where $\mu$ denotes the measure. If $x\notin S_N$ then $|f_n(x)|<2^{-N/2}$ for all n>N. Therefore if $x\notin\limsup_{N\to\infty}S_N$ then $f_n(x)\to0$ as $n\to\infty$. But $\limsup_{N\to\infty}S_N$ is a null set.
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http://nrich.maths.org/5626
### Quaternions and Rotations Find out how the quaternion function G(v) = qvq^-1 gives a simple algebraic method for working with rotations in 3-space. ### Quaternions and Reflections See how 4 dimensional quaternions involve vectors in 3-space and how the quaternion function F(v) = nvn gives a simple algebraic method of working with reflections in planes in 3-space. # Two and Four Dimensional Numbers ##### Stage: 5 Challenge Level: To do this problem you only need to know how to add and multiply two by two matrices. The problem gives models for complex numbers and 4-dimensional numbers called quaternions. Although you don't need the information to do this problem you may like to read the NRICH article What are Complex Numbers? and the Plus article: Curious Quaternions . (1) Let $C^*$ be the set of $2 \times2$ matrices of the form $$\left( \begin{array}{cc} x& -y\\ y& x\end{array} \right)$$ where $x$ and $y$ are real numbers and addition and multiplication are defined according to the usual rules for adding and multiplying matrices. (a) Add and multiply the matrices $$\pmatrix {x & -y \cr y & x} {\rm \ and }\ \pmatrix {u & -v \cr v & u}.$$ (b) What are the identities and inverses for addition and multiplication? (c) Consider also the subset $R^*$ for which $y=0$. Investigate the arithmetic of $R^*$ (addition, subtraction, multiplication, division, identities, inverses, distributive law) and compare it with the arithmetic of real numbers. (d) Compare the arithemetic of $C^*$ with that of complex numbers. (e) The matrix $\pmatrix {-1 & 0 \cr 0 & -1}$ is equivalent to the real number -1. What is the matrix equivalent to $i = \sqrt {-1}$ in the set of complex numbers? (2) Complex numbers are two-dimensional numbers $x + iy$ where $x$ and $y$ are real numbers and $i = \sqrt {-1}$. Quaternions are four-dimensional numbers of the form $a+{\bf i}x+{\bf j}y+{\bf k}z$ where $a, x, y, z$ are real numbers and ${\bf i, j, k}$ are all different square roots of $-1$. Can such a number system exist? To answer this, let's assume that in part (1) you have established that if real numbers and 2 by 2 matrices exist then so does the complex number $i = \sqrt {-1}$. The model for the system of quaternions is the set of linear combinations of 2 by 2 matrices: $$aI +{\bf i}x + {\bf j}y + {\bf k}z\ = a\pmatrix {1 & 0\cr 0 & 1}+ x\pmatrix {i & 0\cr 0 & -i}+y\pmatrix {0 & 1\cr -1 & 0}+ z\pmatrix {0 & i\cr i & 0}.$$ where $a, x, y, z$ are real numbers. (a) Work out the matrix products: ${\bf i^2, \ j^2}$ and ${\bf k^2}$ showing that these matrices give models of three different square roots of -1. (b) Work out the matrix products: ${\bf i j}$, ${\bf j i}$, ${\bf j k}$, ${\bf k j}$, ${\bf k i}$ and ${\bf i k}$ showing that the three matrices ${\bf i,\ j,\ k}$ model unit vectors along the three axes in 3-dimensional space ${\bf R^3}$ with matrix products isomorphic to vector products in ${\bf R^3}$. You have now shown that this set of linear combinations of matrices models the quaternions. (3) Investigate the sequence: ${\bf i},\$ ${\bf i j},\$ ${\bf i j}$ ${\bf k},\$ ${\bf i j}$ ${\bf k i},\$ ${\bf i j}$ ${\bf k i}$ ${\bf j,\ ...}$ ... and the story continues in the Plus article: Ubiquitous Octonions The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://math.stackexchange.com/questions/29711/ra-leq-rbrc-given-a-leq-bc
# $r^a \leq r^b+r^c$ given $a \leq b+c$? Apologies for the (maybe obvious) question. It's come up as part of a proof of the triangle inequality for a metric function I'd like to work with. For real number $r \geq 1$ and positive integers a, b, c: Is it generally true that $r^a \leq r^b+r^c$? Given that $a \leq b + c$. Happy to accept a pointer in the right direction. Thanks - 3 have a look at $r=3$, $a=2$, $b=c=1$. – Alexander Thumm Mar 29 '11 at 12:01 well I feel somewhat stupid! Much obliged. – Tom Seaton Mar 29 '11 at 12:08 ## 1 Answer Since $x\mapsto r^x$ is nondecreasing when $r\geq 1$, you know that $r^a\leq r^{b+c}$, and nothing sharper than this can hold in general (because $x\mapsto r^x$ is strictly increasing when $r>1$, and $b+c$ might in fact equal $a$). Since $r^{b+c}=r^br^c$, the question amounts to whether $r^br^c\leq r^b+r^c$. Although you're restricting $b$ and $c$ to positive integers, you still should expect this to be false, because a product of positive numbers is not generally smaller than their sum. And in fact, it isn't hard to find counterexamples, one of which was given in a comment by Alexander Thumm. -
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http://www.arachnoid.com/relativity/index.html
Home | Science | * Evolution * Answers * Dark Energy * Gravitation Equations * Gravity * The Doubt Factory * What is Science? * Why Science Needs Theories Are You Overweight? Gravitation Physics / Conservation of Energy Physics / Relativity Reader Feedback for the article "Evolution" Satellite Finder Solar Computer Unicode Character Search Why is the Sky Dark at Night? Share This Page Physics / Relativity A concise relativity tutorial Introduction | Special Relativity | Spacetime Geometry Spacetime Causality | General Relativity | Conclusion References Most recent revision: 03.17.2013 (double-click any word to see its definition) Note: In this article, footnotes are marked with a light bulb over which one hovers. Introduction This article's goal is to give nonspecialist readers an intuitive grasp of some easily understood aspects of relativity theory, in particular the relationship between space and time. No specialized knowledge is assumed apart from some high-school mathematics. To help convey relativity's key ideas, to the extent possible the theory's ideas are modeled with graphics and animations. It's my hope that, rather than replacing mathematical ideas, the visual aids will make the mathematics easier to understand. It is often said that, when writing for a popular audience, each included equation cuts one's readership in half. It's my hope that the promise of some insight into relativity theory will suspend this rule. Special Relativity Let's start with a bit of history. In 1905 Albert Einstein published his first relativity paper (On the electrodynamics of moving bodies, A. Einstein, 1905), in which he described a theory later known as Special Relativity. In Part I of his paper Einstein included a number of equations meant to quantify the relationship between space and time "velocities" and dimensions. Here's a key equation from Einstein's paper: (1) $\beta = \frac{1}{\sqrt{1-v^2/c^2}}$ Where: • v = space velocity • c = speed of light • β = a dimensionless term that quantifies changes in space and time at velocity v At the time Einstein published his paper, he was able to use such equations to predict changes in time "velocity" and spatial dimensions that would result from provided space velocities. But when Einstein's math teacher Hermann Minkowski read his former student's paper, he noticed something Einstein did not — Minkowski realized the equations united space and time in a four-dimensional entity Minkowski called spacetime. About this realization, Minkowski later said, "Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality." To see what Minkowski saw, let's write a simpler version of equation (1) that includes terms for space (v) and time (t) velocities: (2) $c^2 = t^2 + v^2$ Where: • c = speed of light • t = time "velocity" • v = space velocity Those familiar with geometry will recognize that equation (2) can be used to describe the relationship between a right triangle's hypotenuse (c) and its other two sides (t and v), based on the Pythagorean Theorem. Using this equation, the relationship between space (v) and time (t) velocities is easily derived: Figure 1: Interactive spacetime diagram (3) $v^2 = c^2 - t^2$ (4) $t^2 = c^2 - v^2$ Based on equations (3) and (4), if c is held constant, we see that any increase in v must cause a decrease in t — they're bound together in such a way that their combined values must produce this constant result: (5) $c = \sqrt{t^2 + v^2}$ Examine Figure 1 and notice about variables v (space) and t (time) that they're at right angles to each other — they lie in different dimensions. If your browser supports the "canvas" feature, you may drag your mouse cursor horizontally across Figure 1 and notice that: • The c value, the speed of light (and the hypotenuse length), remains constant (a requirement of relativity theory). • When space velocity v increases, time velocity t decreases. Figure 1 shows the relationship between space and time velocities in special relativity — when v equals 0, when there is no space velocity, the "velocity" of time equals c, the speed of light. Conversely, when space velocity equals c, there is no time velocity — time has stopped. This leads to these points: • The special relativity equations imply that space and time represent related dimensions in spacetime (three dimensions for space, one for time). • In relativity theory, the speed of light defines the relationship between the space and time dimensions. • Relativity theory requires that c (the speed of light) remain constant in all frames of reference, therefore (as shown in Figure 1) any increase in space velocity must produce a decrease in time velocity. • Figure 1 shows that, when space velocity equals zero, time's "velocity" is the speed of light. • At all nonzero space velocities, time's velocity declines proportionally. • If we somehow could move at the speed of light, time would stop. • For reasons provided in the next section, objects with mass can't approach the speed of light. • Photons, the carrier particles of the electromagnetic field, have no mass and do travel at the speed of light, as a result of which they don't experience time. Spacetime Geometry Let's look at some consequences of the space and time changes described above. Time changes Imagine that astronauts on board a spacecraft (call it spacecraft b) are traveling at a speed of $\frac{1}{2} c$. According to the equations above, the rate at which time passes should be: (6) $t' = t \sqrt{1 - v^2/c^2} = \sqrt{1 - (1/2)^2} = 0.866 t$ Where: • t = time rate at velocity 0 • t' = time rate at velocity v Figure 2: Spacecraft in relative motion Equation (6) means that, for astronauts traveling at $\frac{1}{2} c$, because of relativistic time dilation the passage of an hour should require 69 minutes and 17 seconds. Can the astronauts in spacecraft b detect this time change? Well, no, they can't — their clocks will slow down, their heartbeats will slow down, even supremely accurate atomic clocks will slow down, all in step. This means any tests the astronauts conduct within spacecraft b will show perfect agreement between various sources of information about the passage of time. However, if astronauts on a relatively stationary spacecraft (call it spacecraft a) were to observe the clocks on board spacecraft b, by comparing their own clocks with those on board the moving spacecraft, they would notice spacecraft b's time dilation — indeed, they could use the time difference to determine spacecraft b's velocity relative to their own. Mass changes The above equations rule out travel at the speed of light, using this reasoning: • In space, in the absence of gravity, an object's mass is measured by its resistance to changes in speed (acceleration) — more massive objects require more force (or more time) to change their speed. • Imagine an experiment aboard spacecraft b that pushes a mass with a spring that exerts a known force, as in Figure 2. The experiment should be able to measure the object's mass, based on this expression of Newton's second law of motion: (7) $m = \frac{f}{a}$ Where: • m = mass, kilograms • f = force, Newtons • a = acceleration, meters per second • According to equation (7), if a mass is pushed by a one-Newton force for one second, and accelerates to a speed of one meter per second, the object has a mass of one kilogram. • Let's say the astronauts aboard spacecraft b conduct this experiment and determine that an object has a mass of one kilogram. • Let's also say the astronauts aboard spacecraft a observe the experiment aboard spacecraft b, but a's crew comes to a different conclusion. In the frame of reference of spacecraft a and because of spacecraft b's time dilation, the mass aboard spacecraft b only acquired a speed of 0.866 meters per second, therefore it weighs: (8) $m' = \frac{m}{\sqrt{1 - v^2/c^2}} = \frac{m}{\sqrt{1 - (1/2)^2}} = 1.154 m$ Where: • m = mass at velocity 0 • m' mass at velocity v This result means that, at a velocity of $\frac{1}{2}c$, masses on board spacecraft b weigh 15% more. • Does this relativistic mass increase have any practical consequences? Yes, it does — the experimental mass weighs more, but so does the entire spacecraft. In order to get to its destination, the spacecraft's engines have to deliver more power to accelerate the heavier spacecraft. • Let's imagine a more extreme example. Let's say we want to move at 99% of the speed of light. The time dilation for this speed is: (9) $t' = t \sqrt{1 - v^2/c^2} = \sqrt{1 - 0.99^2} = 0.141 t$ This means an hour aboard spacecraft b requires seven hours and five minutes aboard spacecraft a (but the astronauts on board spacecraft b subjectively experience one hour of time). • At 0.99c, the mass change is: (10) $m' = \frac{m}{\sqrt{1 - v^2/c^2}} = \frac{m}{\sqrt{1 - 0.99^2}} = 7.08 m$ This means, for a given acceleration, the spacecraft's engines must deliver seven times more power. And as the spacecraft moves closer to the speed of light, its mass increases without bound and the power required to increase speed also increases without bound. • At the speed of light, the above equations break down — they indicate a time rate of zero and an infinite mass. • This is why massive objects cannot travel at the speed of light. Readers may wonder whether the mass increase is real — isn't it a coincidental side effect of time dilation, with no independent reality? The answer is that both the time and mass effects are real and interchangeable — one can derive time dilation from mass increase or vice versa. Both interpretations are equally valid. Spacetime Causality Spacetime interval Hermann Minkowski's contribution to relativity theory — his spacetime interpretation — changed how we picture the relationship between causes and effects. We now know there is a clear demarcation between effects and their possible causes, and the speed of light is the gatekeeper. Expressed simply, if a cause at spacetime location a can propagate to spacetime location b at less than or equal to c (the speed of light), then cause a may produce effect b, otherwise not. To understand this, we need to introduce the idea of a spacetime interval: (11) $s^2 = \Delta r^2 - c^2 \Delta t^2$ (12) $\Delta r = r_a - r_b$ (space difference) (13) $\Delta t = t_a - t_b$ (time difference) Where: • s = four-dimensional spacetime interval • c = speed of light • $r_a$ = spatial location of event a • $r_b$ = spatial location of event b • $t_a$ = temporal location of event a • $t_b$ = temporal location of event b Because a spacetime interval takes both space and time dimensions into account and with respect to causes and effects, its meaning is unambiguous: Value of s2 Basis Comment $s^2 \lt 0$ $\Delta r^2 \lt c^2 \Delta t^2$ Time-like interval: Cause a can produce effect b $s^2 = 0$ $\Delta r^2 = c^2 \Delta t^2$ Light-like interval: Cause a can produce effect b $s^2 \gt 0$ $\Delta r^2 \gt c^2 \Delta t^2$ Space-like interval: Cause a cannot produce effect b Light cone diagram Because visualizing four-dimensional spacetime intervals may be difficult at first, readers may want to play with Figure 3, an interactive light cone diagram — change the viewing angle and size with your mouse: Anaglyphic () mode:    Inverted (black background): Figure 3: Interactive light cone diagram Some notes on Figure 3: • This resource is much better viewed in 3D, using anaglyphic  glasses. • The green plane at the center represents the space dimensions and the present time. • The axis perpendicular to the green plane represents the time dimension. • The central point o, where the cones intersect, is the "origin", the present location and time — the x,y,z (space dimensions) and t (time dimension) all equal zero. • It's helpful to picture the past and future cones as expanding into the space dimensions at the speed of light as they extend through time: • Spacetime coordinate points a - e in Figure 3 are located at the same space position, but different time positions. • Spacetime coordinate c is located in the present (it's on the surface of the green plane), but is separated from the origin o by a spatial distance, therefore it cannot influence events at the origin o. • Coordinate b has the same space position as c, but a different time position. Because coordinate b is outside the past light cone, like coordinate c it cannot influence events in the present. • Coordinate a has the same space position as c, but it's located farther in the past than b. Because coordinate a is inside the past light cone, it can influence events in the present. • The same relationship applies to coordinates d and e — coordinate d, located outside the future light cone, cannot be influenced by events in the present, but coordinate e can. Causality Causality, the relationship between causes and effects, is a central issue in physics, and the plausibility of a new theory can be measured in part by whether it allows effects to precede their causes. For example, some readings of General Relativity allow violations of causality, and those issues are a matter of much debate. There is no experimental confirmation of these effects, and such confirmation is unlikely. There are quantum effects such as entanglement that, at first glance, allow for instantaneous communication at superluminal speeds, but as it turns out, this isn't so — entanglement really does cause two particles to interact at any distance, but this fact cannot be used to circumvent limitations posed by the speed of light. General Relativity Special vs. general relativity Special relativity deals with a subset of physical effects — those that don't involve accelerations. General relativity is a much broader theory, and because of its scope and experimental confirmations, it is regarded by some as the crowning achievement of twentieth century physics. In order for special relativity to explain effects that prior theories could not, it had to contradict certain assumptions about everyday reality like the idea that time is the same for everyone. General relativity follows this pattern, but because it explains more, it contradicts more common-sense assumptions. In the prior special relativity section we saw that quantities like velocity, time and mass depend on one's reference frame, but in the midst of these perceptions, space retained its overall shape. In general relativity, the shape of space itself is a term in the equations and depends on one's frame of reference. One might say that in general relativity, there's more relativity. In special relativity, a force called gravity pulls the moon in a curved orbit around the earth. In general relativity, gravity isn't a force, and the moon moves in a straight line through curved spacetime. Figure 4: Gravitational starlight deflection In special relativity, only objects possessing mass feel the force of gravity, and massless particles like photons travel in straight lines through a space that has no role in gravitation. In general relativity, as physicist John Wheeler famously said, "Matter tells space how to curve. Space tells matter how to move." Spacetime curvature: starlight deflection When Einstein published general relativity in 1916, because it had no experimental confirmation it was met with healthy skepticism. In an effort to make his theory less theoretical and more empirical, Einstein suggested that a field of stars be photographed during a solar eclipse and compared to the same field without the effect of the sun. The idea is that spacetime curvature near the massive sun should change the apparent positions of those stars nearest the sun (Figure 4). Figure 5: Optical lens geometry After a number of failed attempts, in 1919 the effort succeeded and (in spite of observational difficulties and marginal results) the outcome supported Einstein's theory and confirmed the idea that space curved around masses. Spacetime curvature: Einstein ring Since that early experiment and because of further theoretical and observational work, another prediction has been confirmed — under special circumstances in which a light source, a large mass and the earth are in precise alignment, spacetime curvature can produce something called an Einstein ring. Figure 6: Gravitational lens geometry Einstein rings and similar optical phenomena arise from what is called gravitational lensing, consisting of light following lines of spacetime curvature, but Figure 6 shows that the resulting ray traces have little in common with an optical lens (Figure 5) — indeed, the geometry of a gravitational lens is in some ways the opposite of an optical lens. It's important to understand about Figure 6 that it doesn't show photon paths curving through a flat background spacetime — it shows spacetime being curved by mass. The photon paths are in essence straight lines through curved spacetime. The reason for the word "ring" in Einstein ring is that, unlike the optical lens shown in Figure 5, there is only one pathway that allows light to travel from the source to the viewer, shown in Figure 6 as blue dotted lines. Remember that Figure 6 is a flat representation of a three-dimensional system, and the blue lines represent a cylinder in three dimensions — a cylinder that the viewer sees as a ring of light. Because of how gravitational lensing works, the requirements for a visible, complete Einstein ring are rather severe — three bodies must be in precise alignment, and the mass and location of the deflecting body must also meet strict requirements. In most cases, an astronomer will see one or more short arcs representing a light source behind a massive body, or sometimes an elongated arc resembling a horseshoe. Spacetime curvature: simulator Black hole mass: Figure 7: Interactive gravitational lensing diagram For readers who want to explore the effects predicted by general relativity, Figure 7 is a gravitational curvature simulator that models a supermassive black hole deflecting the light from a background galaxy. To use the model, change the location of the black hole (red dot) by dragging your mouse across the figure, and change the black hole's mass with your mouse wheel. The physical model in Figure 7 is correct, but it represents an unlikely set of circumstances. To get this effect in reality, a very large black hole, such as is thought to live at the center of most galaxies, would have to be moving freely between galaxies, and would have to align itself between the viewed galaxy and Earth — not very likely. Most real Einstein rings and arcs result from one galaxy being aligned by chance with another, farther away, so that we see some of the distant galaxy's light deflected around the foreground galaxy. But so far, we haven't observed any perfect Einstein rings. To produce an Einstein ring with Figure 7, drag the black hole (the red dot) to the position of a bright spot, two of which appear near the top center of the galaxy image. With some care, you can align the black hole over a bright spot and produce a perfect ring. Figure 8: Black hole spacetime curvature Event Horizon Close examination of the geometry modeled by Figure 7 shows that, as one approaches a sufficiently dense, massive object, spacetime curvature increases without bound. The area nearest the black hole in Figure 7 (red dot) is empty because the spacetime curvature is greater than 90° (shown as red lines in Figure 8), so those paths don't intersect any visible objects. Even closer to the black hole, near the event horizon's radius, spacetime curvature is such that a photon will perpetually orbit the black hole. Remember about relativity that all perceptions are local. This means that, to an observer distant from the curvature shown in Figure 8, it might be possible to observe it more or less as shown, but to an observer approaching a black hole's event horizon, instead of seeing spacetime curvature increase, the observer would instead see the event horizon's curvature decrease (assuming the horizon were visible at all). At a radius of $\frac{3}{2}r$ (r = event horizon radius), the black hole's surface would become a flat plane, infinite in extent, with eyelines extending around the horizon — indeed, in all directions at a distance of $3 \pi r$, the observer would see the back of his own head. Microlensing Gravitational lensing plays a part in the search for planets around other stars as well as free-roaming masses. There are a number of ways to detect a planet orbiting a star that's too distant to image directly: • Detect a small reduction in starlight caused by a planet passing in front of the star (a method used by the Kepler spacecraft). • Detect a shift in the spectrum of a star caused by its motion toward, and away from, the earth, as it orbits the common center of mass of itself and an orbiting planet. • Detect a small brightening of starlight caused by a planet or other mass moving between a star and earth. The third method in the above list is called gravitational microlensing. It's only rarely used to detect planets near a star — it's more suitable for detecting dark masses distant from stars or even galaxies. As it turns out, when a massive object passes in front of a distant star, even when the range is too great to resolve an Einstein ring, because of microlensing the amount of light received can be momentarily greater than without the intervening object. This method can be used to detect objects that emit no light of their own. Conclusion At this point it should be clear that relativity's key ideas are completely accessible and easy to understand, and it's a shame they're not better understood by nonspecialists. I say this in particular because of how often one hears the claim that superluminal space travel is just around the corner — all we need to do is find a wormhole, or a "tear in the fabric of spacetime," as any number of movie script writers have put it. These fantasies underestimate the importance of causality in physical theory. If superluminal speeds were possible, time would lose its conventional meaning, effects could precede their causes and energy would not be conserved — apart from the fact that it opens the door to any number of temporal paradoxes. Rather than read about how the task of jumping directly and instantaneously across the universe is only a matter of building the right spacecraft from parts available at your local hardware store, I would prefer it if people understand how this is supremely unlikely, and better, that people understand why. It's my hope that this article will serve as a brief guide to relativity's unfamiliar territory, and perhaps reduce the number of science fiction ideas that masquerade as science. References 1. On the electrodynamics of moving bodies, A. Einstein, 1905 — Einstein's first relativity paper. 2. Special Relativity — an encyclopedia summary of the restricted version of relativity theory. 3. Hermann Minkowski — Einstein's math teacher. 4. Spacetime — the unification of three space dimensions, and one time dimension, into a four-dimensional entity with common properties. 5. Pythagorean Theorem — a defining principle of geometry 6. Photon — the carrier particle of the electromagnetic field. 7. Newton's second law of motion — the net force on an object is proportional to the rate of change of its linear momentum. 8. Spacetime interval — a four-dimensional spacetime distance used in causality computations. 9. light cone — a visualization aid for picturing spacetime intervals. 10. Causality — the relationship between causes and effects. 11. Quantum entanglement — a bizarre effect of quantum theory in which two or more particles become interdependent, regardless of their spatial separation. 12. General relativity — special relativity's successor theory that includes accelerations and gravitation. 13. Tests of General Relativity: Deflection of light by the Sun — an early confirmation of general relativity. 14. Black hole — an object so massive and dense that surface escape velocity exceeds the speed of light. 15. Einstein ring — a somewhat dramatic optical manifestation of spacetime curvature. 16. Gravitational lens — an effect of general relativity in which spacetime curvature produces distinct images. 17. Kepler (spacecraft) — an ambitious and successful program to detect planets orbiting distant stars. 18. Gravitational microlensing — an observational method that relies on spacetime curvature, moving masses, and starlight. 19. Event horizon — the radius of a black hole at which escape velocity is equal to the speed of light. 20. Temporal paradox — a class of logical problems associated with time travel. Home | Science | * Evolution * Answers * Dark Energy * Gravitation Equations * Gravity * The Doubt Factory * What is Science? * Why Science Needs Theories Are You Overweight? Gravitation Physics / Conservation of Energy Physics / Relativity Reader Feedback for the article "Evolution" Satellite Finder Solar Computer Unicode Character Search Why is the Sky Dark at Night? Share This Page
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http://mathoverflow.net/questions/3776?sort=newest
## When are two proofs of the same theorem really different proofs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Many well-known theorems have lots of "different" proofs. Often new proofs of a theorem arise surprisingly from other branches of mathematics than the theorem itself. When are two proofs really the same proof? What I mean is this. Suppose two different proofs of theorem are first presented formally and then expanded out so that the formal proofs are presented starting from first principles, that is, starting from the axioms. Then in some sense two proofs are the same if there are trivial operations on the sequence of steps of the first formal proof to transform that proof into the second formal proof. (I'm not sure what I mean by "trivial") - ## 14 Answers The other posters have well pointed out that the proof identity problem can be approached from various directions. If you're interested in proof theory and are willing to delve into natural deduction and category theory, you might be interested in two proposals for addressing the proof identity problem: the Normalization Conjecture and the Generality Conjecture. See Dozen's "Identity of proofs based on normalization and generality" for a nice introduction to these two ways of viewing the proof identity problem. - ### 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. Some additional recent references on "equivalence" or "homotopy" between proofs include 1) S. Awodey, Type theory and homotopy, also on the arXiv 2) Various notes by V. Voevodsky Not that I understand much of these - certainly not much enough to see a down-to-earth example of two explicit proofs of some elementary statement that are not homotopic, as detected by some invariant. Is anyone able to provide such an example? 3) J. Conant and O. Thistlethwaite, Boolean formulae, hypergraphs and combinatorial topology Then, there are two very basic things to note. First, the question itself dates back to Hilbert's (would-be) 24th Problem, which R. Thiele discovered in Hilbert's notebooks a century later (translation and remarks by Thiele, boldface mine): The 24th problem in my Paris lecture was to be: Criteria of simplicity, or proof of the greatest simplicity of certain proofs. Develop a theory of the method of proof in mathematics in general. Under a given set of conditions there can be but one simplest proof. Quite generally, if there are two proofs for a theorem, you must keep going until you have derived each from the other, or until it becomes quite evident what variant conditions (and aids) have been used in the two proofs. Given two routes, it is not right to take either of these two or to look for a third; it is necessary to investigate the area lying between the two routes. Attempts at judging the simplicity of a proof are in my examination of syzygies and syzygies between syzygies. The use or the knowledge of a syzygy simplifies in an essential way a proof that a certain identity is true. Because any process of addition [is] an application of the commutative law of addition etc. [and because] this always corresponds to geometric theorems or logical conclusions, one can count these [processes], and, for instance, in proving certain theorems of elementary geometry (the Pythagoras theorem, [theorems] on remarkable points of triangles), one can very well decide which of the proofs is the simplest. [Part of the last sentence is not only barely legible in Hilbert’s notebook but also grammatically incorrect. Corrections and insertions that Hilbert made in this entry show that he wrote down the problem in haste.] Second, there is a good reason why the question has been traditionally treated in an intuitionistic setup, from Seely to Awodey. (Note that intuitionistic proofs are perhaps less scary if thought of as computer programs, via the Curry-Howard correspondence.) The reason is that in classical logic, with a standard formalization of the notion of "proof", every two proofs of the same statement must be equivalent with every reasonable notion of equivalence. The idea is in Kevin Buzzard's answer. For a rigorous explanation see Yves Lafont's Appendix B in Girard's Proofs and Types (The standard sequent calculus notation used in that appendix is introduced in the very beginning of the book.) It looks like Alessio Guglielmi has some way of overcoming this difficulty by using a non-standard proof-theoretic setup which I wish I understood better. - Two proofs are different if one of them relies on CFSG and the other doesn't. This is not an exhaustive list of criteria. - 1 CFSG = ? $\mbox{}$ – Joseph O'Rourke Dec 4 2010 at 14:59 What does "relies on" mean? – JBL Dec 4 2010 at 15:01 1 Classification of finite simple groups. – JBL Dec 4 2010 at 15:01 Sorry, CFSG is "Classification of Finite Simple Groups". "Relies on" means "uses or uses a thereom whose known proof uses it, etc." – Jonathan Kiehlmann Dec 4 2010 at 18:17 2 How can you tell if a proof really relies on CFSG in an essential way? – JBL Dec 5 2010 at 5:40 show 3 more comments This might be of interest: http://arxiv.org/pdf/cs.LO/0610123 • Straßburger, Lutz (20 October 2006), "Proof Nets and the Identity of Proofs", Technical Report 6013, INRIA - You can express any Proof as a typed Lambda-Term, looking at the theorem as a Type. This term can be normalized. I would say, if two of these Proof-Terms have the same normal form, then they name the same proof. - Maybe this might be of interest: Andreas Blass, Nachum Dershowitz, Yuri Gurevich: WHEN ARE TWO ALGORITHMS THE SAME? - 2 Thanks for referring to this. Since people who look only at the title and abstract wouldn't see anything about proofs, it may be worthwhile to refer specifically to Section 7.2 (starting at the bottom of page 20 of the version you linked to), which discusses equality of proofs. – Andreas Blass Dec 4 2010 at 19:43 This of course is a deep question in the philosophy of mathematics. The program mentioned by Tom Leinster is certainly a very interesting contribution to this, but if it proceeds at a purely mathematical level then at most it can define an equivalence relation on the class of proofs. There's still a further question whether this equivalence relation really is "the right one" to capture the notion of "same" or "different" proofs. Also, note that there's an open question as to whether mathematical proofs really are the sort of thing studied by proof theorists. Certainly the sort of thing that is published in a math journal is not the sort of thing that is studied by proof theorists. To cite the most obvious differences, the former have words of English in them (or French or Japanese or Russian or some other language) while the latter don't. But for more significant differences, note that the former also cite well-known results from the literature, and skip steps that are sufficiently obvious to the reader, while the latter don't. You can avoid this problem by assuming that published proofs are converted into formal proofs by means of spelling out all the steps in the proof of the well-known theorem, or the obvious fact. But this might not preserve the notion of "same proof". For instance, consider a theorem that in some sense only has one proof, which happens to rely essentially on quadratic reciprocity. Do we really want to say that this theorem actually has just as many distinct proofs as quadratic reciprocity does? There are lots of interesting questions here about the relation of proof theory to actual proofs, and what light it can shed on this intuitive notion of sameness of proof. And of course, there is probably also light to be shed in the other direction too, as our technical mathematical results in proof theory and category theory absorb results from the intuitive ideas we have about proof sameness. - 7 Right. For example, one intuition I have about proofs is that they have "tangent vectors" associated to them: different proofs of quadratic reciprocity, for example, point in different directions for further generalization. But it's unclear how one would go about formalizing this notion. – Qiaochu Yuan Nov 6 2009 at 15:57 This doesn't affect your main point, but it's not true of HoTT approach that ‘at a purely mathematical level then at most it can define an equivalence relation on the class of proofs’. It makes the class of proofs into a space with a homotopy type, and each equivalence class (as a connected space) may still have interesting structure. But you are correct that it remains to be seen if ‘this equivalence relation really is "the right one"’. – Toby Bartels Jan 12 at 0:08 The HoTT approach should also be able to handle this: ‘For instance, consider a theorem that in some sense only has one proof, which happens to rely essentially on quadratic reciprocity. Do we really want to say that this theorem actually has just as many distinct proofs as quadratic reciprocity does?’. It is simply relative homotopy. But this may be an example of ‘our technical mathematical results in proof theory and category theory absorb results from the intuitive ideas we have about proof sameness’ if it might not otherwise occur to HoTT people to consider that! – Toby Bartels Jan 12 at 0:11 Some other answers have alluded to this, but just to spell it out explicitly: The Curry-Howard isomorphism, in one of its simpler forms, says that objects of the free cartesian closed category CCC[S] on a set S of objects correspond to statements of the multiplicative fragment of intuitionistic logic (things we can build from /\ and ⇒) with free variables from S, and there is at least one morphism P → Q in CCC[S] iff P ⇒ Q is a theorem. Thus we can regard a morphism P → Q as a "proof" of P ⇒ Q. There may be several morphisms from P to Q; for instance if A ∈ S and P = A × A, Q = A, then there are exactly two morphisms from P to Q (projection to the first or second factor), which we can regard as two different proofs of the theorem (A /\ A) ⇒ A. Probably the easiest way to see what the different proofs are in this system is to use the third part of the Curry-Howard isomorphism: morphisms P → Q in CCC[S] correspond to functions in the simply typed lambda calculus of type P ⇒ Q, where × in CCC[S] is interpreted as the product of types and the internal Hom as a function type. For instance there are two functions of type (A * A) → A, namely λ(a, b). a and λ(a, b). b. A more interesting example: the theorem (A ⇒ A) ⇒ (A ⇒ A) has one proof for every natural number, corresponding to λ f. λ x. f (f (... (f x)...)). See This Weeks' Finds week 240 for more along these lines. - If we have two proofs of the same theorem such that each proof has a different normal forms, can we modify the set of axioms so that there is only one normal form proof of the theorem, yet the universe of theorems remains unchanged from the original set of axioms? More generally, can we select a set of axioms that minimizes the number of normal form proofs for each and every theorem in the original axiom set? Taking this train of thought to the limit, for any axiom system, does there exist another axiom system with the same universe of theorems but which admits only one normal form proof of each theorem? Such a system of axioms could be called a "tight" set of axioms for a given universe of theorems. - It is indeed an open task of proof theory to give a good formal definition of when two proofs should be considered equivalent. A usual thing is to consider a category with formulas as objects and equivalence classes of proofs as morphisms, where two proofs are considered equivalent if they have the same normal form (in many logics every proof can be brought into a unique normal form, i.e. a chain of deductions of which the first half are e.g. elimination rules and the second half introduction rules). Moreover this transformation of a proof into normal form can often be done algorithmically and is then described by a rewriting system. This provides the link of syntactic proof theory to homotopy theory, mentioned by Tom Leinster, it can be made very plausible via rewriting systems, see e.g Y. Lafont's homepage or the corresponding sections of P.-A. Mellies' homepage. Also check out the "Categorical Semantics of Linear Logic" paper on Mellies' page - there he considers invariants of proofs, each of which should yield a notion of equivalence! However all of these are syntactical notions of equivalence and, as Terry Tao mentions in his comment at Gowers' blog (see the link in Justin's answer), there is also a semantic notion of equivalence saying that two proofs are equivalent if they have the same degree of generalizability. And while the syntactic notions of equivalence capture quite well the formal operations by which one can relate different proofs, the real challenge is (imho) to give a formal definition of semantic equivalence and recognize it syntactically! The earliest published attempt I know of are two articles by Lambek, this and J. Lambek, Deductive systems and categories II, in: Lecture Notes in Mathematics 86 (Springer, Berlin, 1969), especially the second where, if I remember well, he does in fact try to give a syntactic characterization of semantic equivalence. - In combinatorics it is often useful to find bijections between two combinatorial structures one is studying. An example is a bijection between 321-avoiding permutations and 132-avoiding permutations. A lot of different bijections have been shown to exist and the paper Classification of bijections between 321- and 132-avoiding permutations by Claesson, Kitaev shows that some of these are related by "trivial" bijections. Maybe this is a very special case of what Tom Leinster mentions in his answer, about one proof (bijection in this case) is deformed into another by a sequence of trivial steps (trival bijections in this case). - You've hit on an area of research that's picking up some momentum at the moment. It involves connections between proof theory, homotopy theory and higher categories. The idea is that a proof or deduction is something like a path (from the premiss to the conclusion), and when you "deform" one proof into another by a sequence of trivial steps, that's something like a homotopy between paths. Or, in the language of higher-dimensional categories, a deduction is a 1-morphism, and a deformation of deductions is a 2-morphism. You can keep going to higher deductions. There are close connections with type theory too. If you have the right kind of background, the following papers might be helpful: Awodey and Warren, Homotopy theoretic models of identity types, http://arxiv.org/abs/0709.0248 Van den Berg and Garner, Types are weak omega-groupoids, http://arxiv.org/abs/0812.0298 - 8 Has anyone proposed a specific proof theory (even a toy model) where the "proof spaces" can be seen not to be homotopy discrete? – Reid Barton Nov 3 2009 at 4:36 4 Presumably if you take intuitionistic propositional calculus then you get such a model. The proof terms correspond to the simply-typed $\lambda$-calculus. A $\beta$-reduction from one proof to another is a path between proofs, or a 2-cell in a suitable 2-categorical sense. I think it was Robert Seely who wrote about this originally. – Andrej Bauer Dec 4 2010 at 12:11 My opinion, and it's only an opinion, is that it would be very difficult to formalise what it means for two proofs to be different. Here's an intuitive reason why. If I give you two proofs of theorem X, and both proofs are exactly the same, except that one proof had a couple of extra lines in the middle which proved an intermediate result which was of no relevance, then surely these two proofs would be "the same". So surely any sort of "sameness" equivalence relation that one is trying to formally set up on the set of proofs of a statement would have to allow for deleting or adding lines to a proof that aren't used. But now there's perhaps a problem, because proof A and proof B of theorem X might both be "the same" as proof C, where proof C is the disjoint union of proofs A and B. On the other hand it's manifestly clear that sometimes two proofs of a fact are "different" on an intuitive level. For example I remember doing the exercise as an undergraduate that the map SL(2,Z)-->SL(2,Z/nZ) was surjective, but I used the fact that there was infinitely many primes in an AP. A couple of days later I found a proof that didn't use this and was entirely elementary. Clearly the proofs were "different". All I'm saying is that although this is in some sense obvious, what I'm saying is that it might be tough to formalise. - Perhaps this issue could be explained using homotopy theory (as Tom mentioned). Adding an irrelevant couple of lines certainly doesn't change the homotopy type of a path, and it's possible that the disjoint union of two sufficient proofs could be interpreted as the concatenation of their paths. (I'll readily admit that I wasn't able to make much headway in the papers Tom linked, so I have no idea if this is actually how things are done...) – Aaron Mazel-Gee Nov 3 2009 at 2:49 3 I follow you up to the point where you transform "(1) argument A; (2) argument B; (3) conclude X from 1" to "(1) argument A; (2) argument B; (3) conclude X from 2". I can imagine that not being allowed as a "proof homotopy". But certainly you'd have to be careful to ensure that that was the case. – Reid Barton Nov 3 2009 at 4:33 My understanding is that the difficulty is well-known, and the standard solution has been to switch from classical logic to intuitionistic. The references are in my answer in this thread, mathoverflow.net/questions/3776/… – Sergey Melikhov Dec 5 2010 at 2:53 Maybe this blog entry by Gowers will be of interest. - 1 Thank you for the interesting blog. I was thinking "sameness" in a more formal, proof theoretic manner. – Martyguy Nov 2 2009 at 10:41 There is some discussion in the comments of that blog post about the formal version of this question. Kenny's comment (near the beginning) is especially illuminating, concerning technical difficulties. – S. Carnahan♦ Nov 3 2009 at 20:37
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http://mathhelpforum.com/calculus/2699-e-question.html
# Thread: 1. ## e question How are $\lim_{x \to \infty}(1+\frac{1}{x})^x$ and $1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} ...$ similar or related to each other, other than their results being $e$? 2. It should be $\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right )^x$ and I don't really know how to answer your question. Is this out of a book, looking for something specific or just you pondering? 3. Originally Posted by cinder How are $\lim_{x \to \infty}(1+\frac{1}{x})^x$ and $1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} ...$ similar or related to each other, other than their results being $e$? Consider instead: $<br /> e_1(x)=\lim_{k \to \infty}e_1(x,k)<br />$ where: $<br /> e_1(x,k)=\left(1+\frac{x}{k}\right)^k<br />$ and: $<br /> e_2(x)=1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} ...<br />$ expand $e_1(x,k)$ and look at the behaviour of the coefficients of powers of $x$ that occur in both, and consider what happens when $k\to \infty$. Maybe? RonL 4. Originally Posted by cinder How are $\lim_{x \to \infty}(1+\frac{1}{x})^x$ and $1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} ...$ similar or related to each other, other than their results being $e$? Well, we definie $\ln x=\int^x_1 \frac{dx}{x}$ for $x>0$. And define, $e$ such as $\ln e=1$ (it does exists and well-defined.) Then we can prove that, $e^x=1+x+\frac{x^2}{2!}+...$ Where $e^x$ is the inverse function of $\ln x$ because this function is bijective. Thus if, $x=1$ then, $e=1+1+\frac{1}{2!}+\frac{1}{3!}+...$ 5. You don't have to define exp(x) as the inverse of log(x) which you defined through an integral. That's certainly possible (and is often done) but is not of any use here since we can just as well start by defining exp(x) (and after that define log(x) as its inverse, though that's not necessary here). What the topic starter was asking, I think, is to prove the equivalence between these two possible definitions. Of course, this follows trivially from the equivalence of the analogous definition for exp(x), but proving that equivalence would be a bit 'overkill' to prove this one. 6. Originally Posted by TD! You don't have to define exp(x) as the inverse of log(x) which you defined through an integral. That's certainly possible (and is often done) but is not of any use here since we can just as well start by defining exp(x) (and after that define log(x) as its inverse, though that's not necessary here). What the topic starter was asking, I think, is to prove the equivalence between these two possible definitions. Of course, this follows trivially from the equivalence of the analogous definition for exp(x), but proving that equivalence would be a bit 'overkill' to prove this one. I never met two mathematicians from Analysis that define some function the same way. Every person has his own way, everything else is biconditional. I just happen to like the the integral defintion, because it demonstrates the importance of the fact that countinous functions are Riemann integrable. 7. That's a bit strange, there aren't too many ways on how log(x) and exp(x) are conventionally defined. I know at least 3 books (and thus authors, i.c. professors) who start by defining log(x) as that integral; then defining exp(x) as its inverse (which is why I remarked that it is often done ). One of them being "my professor" (the one I had for analysis); on the other hand you have for example Rudin (from 'Principles of Mathematical Analysis') who defines exp(x) first, as the series. There are indeed different possibilities, but fundamentally there aren't too many different definitions in use (log through the integral, the series, the limit, the initial value probem y' = y with y(0) = 1). 8. No I was implying that about functions not log and exp. For example, you can define arctan as an integral, or inverse of tan function. Also you can define it intuitively as a trigometric interprestation. This happens with a lot of functions. 9. I see, that's true of course although I still think you'll find that many functions are often defined in a similar way - within the rather limited number of fundamentally different ways of defining them of course. Apart from that, I just read the original question carefully. What are you exactly looking for? As said before, both expressions are completely equivalent which means, in this case, that they're the same - being e as you said already. You can prove this, but is that what you're looking for? 10. I couldn't tell you exactly what I'm looking for. My calculus teacher mentioned it might be part of a bonus question on a test, but wanted us to do figure it out. After looking but not really finding anything, I figured I'd ask here.
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http://mathhelpforum.com/calculus/87631-help-finding-rate-increase-using-product-quotient-rule.html
# Thread: 1. ## Help with finding rate of increase using product/quotient rule I've been having trouble with a particular problem from my Calculus book in the Product and Quotient rules chapter. I was hoping someone on this forum could help me out. Thanks in advance! Question: Suppose the price of an object is \$14 and 12,000 units are sold. The company wants to increase the quantity sold by 1200 units per year, while increasing the revenue by \$20,000 per year. At what rate would the price have to be increased to reach these goals? 2. Originally Posted by elong I've been having trouble with a particular problem from my Calculus book in the Product and Quotient rules chapter. I was hoping someone on this forum could help me out. Thanks in advance! Question: Suppose the price of an object is \$14 and 12,000 units are sold. The company wants to increase the quantity sold by 1200 units per year, while increasing the revenue by \$20,000 per year. At what rate would the price have to be increased to reach these goals? If x is the price of each object, than the price, P, times x is the revenue. So $R=Px$. For the first part, P=\$14 and x=12,000 so $R=14 \times 12,000=168,000$. Now if the amount sold has to been increased by 1,200 units per year this means that the new total objects sold is 12,000+1,200=13,200. Follow? So now we know the units sold and the desired revenue, but not the price. Let's set up the equation again: $R=Px$ so $20,000=P(13,200)$. Now solve for P and solve for how much this new price has increased from the old. edit: Your problem might be a little different than this. I don't know if the first line means those products were sold once or that's the yearly amount. 3. Thank you. I do follow what you are saying and it does seem logical, but what is throwing me off is the fact that I think we are suppose to either use the product or quotient rule to solve this problem. The following is a similar question that I solved correctly. Ques: Suppose the price of an object is \$20 and 20,000 units are sold. If the price increases at a rate of \$1.25 per year and the quantity sold increases at a rate of 2000 per year, at what rate will revenue increase? Answer: R' = Q'P + QP' (Product Rule) At a certain moment of time (call it t0) we are given P(t0) = 20. Q(t0) = 20,000 (items) P'(t0) = 1.25 Q'(t0) = 2,000 (items/year) --> R'(t0) = 2,000(20) + (20,000)1.25) = 65,000 \$ per year
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http://physics.stackexchange.com/questions/48528/evaporation-of-water-content-from-a-solid-material-by-applying-low-pressure?answertab=oldest
# Evaporation of water content from a solid material by applying low pressure I have a raw material which melts at $96\ ^\circ C$. My aim is to make water content evaporate at temperature below this temperature. I can apply vaccume oven for this. I want to know at what pressure the water evaporates by keeping low temperature so that the material doesn't melt. - – dmckee♦ Jan 7 at 14:02 1 I think water evaporates even at room temperature but slowly. The boiling point of water can go below $96\ ^\circ C$ if its vapor pressure drops below 0.86 atm. – Tariq Jan 7 at 17:01 ## 1 Answer Water will evaporate as long as the relative humidity of the surrounding air is below 100%, although the process may be (very) slow. You seem to be confusing evaporation with boiling, but you don't need to boil the water in the material to dry it. It depends a lot on your configuration, but you are probably better off blowing hot dry air over your material than creating a vacuum, because then water vapor molecules will be moving away from the surface of your material by diffusion, which is normally a much slower process than convection. Still, if you want to boil the water content of your material, your guide should be the vapor pressure of water at your temperature of choice, which you can check here. As an example, the vapor pressure at $90\ ^\circ\mathrm{C}$ is $70\ \mathrm{kPa}$, so if you heat your sample to that temperature and have a vacuum of less than $0.7\ \mathrm{atm}$, there will be boiling. -
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http://mathoverflow.net/questions/43286?sort=newest
## How and how much do the notations and diagrams influence our understanding of mathematical concepts? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) How and how much do the notations and diagrams influence our understanding of mathematical concepts? This question was stimulated by the MathOverflow questions Thinking and Explaining and Suggestions for good notation. - 3 Maybe community wiki? – Francesco Polizzi Oct 23 2010 at 11:24 3 This reminds me of a quote from the AMS Notices of Oct 2010. In the article about Gronthendieck's school, a point was made about writing $X$ over $S$, with vertical arrows, instead of $X$ to the left of $S$, with horizontal arrows, and how this brings in a different mode of thinking. (I don't have my copy handy right now, so can't say what context this was given in.) – Willie Wong Oct 23 2010 at 11:55 @Willie Wong, does it really bring in a different mode of thinking, or does it merely indicate that the vertical arrows are viewed as being different from the horizontal arrows, and that this difference is viewed as representing a different mode of action? In other words, unfamiliar objects or symbols are viewed as being novel, exciting certain parts of the brain more than familiar objects would. Familiar objects are already "wired up", whereas novel objects are not pre-conditioned to elicit a particular point of view, thus allowing a different approach to be considered. – sleepless in beantown Oct 23 2010 at 13:32 3 A lot, but this is a better blog post than a question. – Theo Johnson-Freyd Oct 23 2010 at 19:30 An interesting post related to this subject appeared today, by Dick Lipton: rjlipton.wordpress.com/2010/11/30/… – Cristi Stoica Nov 30 2010 at 13:32 ## 4 Answers I think a good answer to this comes from category theory, linear logic, diagrams and the geometry of tensor calculus (Joyal-street). We often talk about category theory through the use of diagrams which are planar graphs (objects as nodes morphisms as arrows). Written down, these diagrams can be seen as fishing nets, kind of embedded in a plane, so we don't care about how one line crosses over another. These diagrams can be rewritten in different ways that respect the topology of the network. These deformations are exactly what Joyal and Street were talking about in the geometry of tensor calculus. We know from further work that, (and excuse my poor explanation ) the geometry of tensor calculus is a model of linear logic. This would mean that the axioms of a symmetric monoidal category can support the axioms of linear logic...(please excuse the poor understanding, I have come to these thoughts with little help). The long and short is that, if we talk about category theory in terms of diagrams, then we are most likely thinking in terms of linear logic. This would be in contrast to a model of the theory of categories in Set. In that case, we have a set of objects and set of morphisms, the axioms of (some kind of ) Set theory and further the axioms of the category we want to talk about. This would be thinking in a different kind of logic. - ### 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. To support the last remark of Donu Arapura, the following anecdote might be helpful: The late Beno Eckmann, one of the key players of the early developments in algebraic topology in the 40ies and 50ies, was asked to explain, why the revolution in algebraic topology happened in the 50ies. You can find his answer in his "Mathematical Miniatures". In short, he explains that the idea to represent a function by an arrow, and a composition of functions by a diagram was completely unknown until the late 1940ies (!!!), when Leray introduced this notation. There seems to be no doubt that even the formulation of modern algebraic topology would have been impossible without the idea of an arrow and/or a diagram! - This example illustrates a point of view that I learnt from Dennis Sullivan. In general, when speaking of free (or projective) resolutions of $R$-modules one writes $\cdots\to F_n\to\cdots\to F_1\to F_0\stackrel{\varepsilon}{\rightarrow} M\to 0$ where $F_n$'s are free (or projective) $R$-modules. For calculating $Ext$ or $Tor$ we use one such resolution of $M$ and then show that the resulting answer doesn't depend on the resolution chosen. The resolution above can be rewritten as a map $\varphi$ between two chain complexes of $R$-modules : $\cdots\to F_n\to\cdots\to F_1\to F_0\to 0$ $\cdots\to 0\hspace{0.2cm} \to \hspace{0.2cm} \cdots\hspace{0.2cm}\to 0\to M\to 0$ where the only non-trivial vertical map is $\varepsilon:F_0\to M$. Notice that $\varphi$ is a quasi-isomorphism and saying that the derived functors are independent of the resolution chosen is akin to saying that any two chain complex of $R$-modules representing $M$ are quasi-isomorphic. To an algebraic topologist (and possibly for others too) this is so much more natural. This point of view emphasizes that the non-triviality of the module $M$ gets coded into the differentials between $F_n$'s. The $F_n$'s themselves contain almost no information since the rank is the only possible invariant for $F_n$ and even that be made to change by adding a copy of $R$ to $F_{n+1}$ and $F_n$. - 1 (Your parenthetic remark applies to other too.) One could argue more broadly that homological algebra (and many other things) would be impossible without the shift from thinking about equations to thinking in terms of diagrams (exact sequences, complexes and commutative diagrams). – Donu Arapura Oct 23 2010 at 21:04 1 I would add that diagrammatic reasoning = category theory has changed the shape of algebra for good in the last sixty years. – Leo Alonso Nov 12 2010 at 12:23 "To an algebraic topologist this is so much more natural" --- more natural than what? I don't know any other way of viewing the situation... – Kevin Lin Nov 12 2010 at 22:02 One example, slightly outside of mathematics, is the Feynman diagram which represent the interaction between particles in a field over a period of time. They're slightly different from space-time diagrams, but make it easier to see the contributions of particles and anti-particles to an interaction. Another example, in chemistry and mathematics, is that a series of chemical reactions can be described by a series of equations • $A + B \to C + D$, • $D + E \to F + G$, • $G + H \to J + A$ which does not clearly describe the "ins and outs" of this chemical cycle as well as a graph (directed graph) diagram does: ````: A : J-<-- / \ --<--B : \/ \/ : H-->/ \---> C : / \ : / \ : G-----------D : / \ : F <-- -<--E ```` which clearly illustrates that the cyclic nature of this series of actions or the catalytic nature of some of these moities. Sometimes, rewriting the steps of a proof simplifies understanding of it at an earlier level of our education, while it seems wholly un-necessary at later parts of our education. For example, $(p+q)^3$ is easily expanded in our heads to 1, 3, 3, 1, and $p^3, p^2q, pq^2, q^3$ and recombined into $p^3+3p^2p+3pq^2+q^3$, even though the initial and final $1$ of the binomial coefficients are effectively silent. But for beginning algebra students in high-school, putting the ones back into the expansion makes it easier to comprehend. In the general case about encountering a new notational technique the first time may simply be about how the mind deals with novelty. In the example of Willie Wong about the Grothendieck's school's vertical arrows, does it really bring in a different mode of thinking, or does it merely indicate that the vertical arrows are viewed as being different from the horizontal arrows, and that this difference is viewed as representing a different mode of action? In other words, unfamiliar objects or symbols are viewed as being novel, exciting certain parts of the brain more than familiar objects would. Familiar objects are already "wired up", whereas novel objects are not pre-conditioned to elicit a particular point of view, thus allowing a different approach to be considered. - Indeed, Feynman diagrams are very descriptive. Understanding them as terms in a perturbation expansion leads naturally to the idea of sum over histories. It seems to me that they guided Feynman in conceptualizing point particles as fundamental among other possible bases in the Hilbert space, although the expansion can be done as well in terms of other propagators, such as the momentum space ones. Does anybody know if Feynman-like diagrams are used in other applications of perturbation theory? – Cristi Stoica Oct 23 2010 at 17:53 @Cristi Stoica: The Mayer cluster expansion is a similar diagrammatic expansion en.wikipedia.org/wiki/Cluster_expansion that comes up in statistical mechanics. Though the wikipedia article doesn't show the pictures you usually draw, see this paper for an intro combinatorics.org/Volume_11/PDF/v11i1r32.pdf – jc Nov 12 2010 at 22:13
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http://cogsci.stackexchange.com/questions/3187/how-to-adjust-sse-or-rmse-for-the-number-of-free-parameters-in-the-model/3191
# How to adjust SSE or RMSE for the number of free parameters in the model? How do I adjust SSE (sum of squared errors) or RMSE (root-mean-square errors) for the number of free parameters in the model? Is there an "adjusted" RMSD metric similar to the adjusted r-squared metric? - please don't assume that we know what RMSD stands for. The more effort you invest in carefully phrasing your question, the more motivated people will feel to answer it. – Artem Kaznatcheev Feb 8 at 5:04 4 – Jeromy Anglim♦ Feb 8 at 5:13 Not sure to understand what you mean by "adjusted" here ? Do you mean adjusted as for "goodness of fit" or as correction in regards to some parameters ? – Cheatboy2 Feb 8 at 14:30 @Cheatboy2 I believe he is referring to a correction in the number of parameters, as in adjusted $r^2$ – Jeff Feb 8 at 23:17 ## 2 Answers To my knowledge, there is no adjusted RMSD. RMSD, unlike $R^2$, isn't typically used to compare models across the literature. $R^2$ represents the proportion of variance explained by the model, a construct which translates well across different experimental designs. Adjusted $R^2$ distorts this by accounting for the number of parameters in your model, but is a better estimate of the proportion of variance in the population explained by your model. RMSD is in whatever units you happen to be using. It should be interpreted against some benchmark for what you think a good model should be, or whatever has been deemed an "acceptable error rate". It's not clear to me that adjusting for the number of parameters in RMSD would provide a number that is inherently meaningful, but I'm not sure if this is true. If you would like to select among several model candidates, you may want to compute AIC (Akaike Information Criterion) or BIC (Bayesian Information Criterion), both of which are attempts to maximize the likelihood of your model while minimizing the number of parameters. Both metrics are commonly used in cognitive modeling, and I think this may be what you are looking for. While I think this question is on-topic here, you will likely get better answers on stats.stackexchange.com (as Jeromy points out) because your question is not really specific to cognitive modeling. - There are at least 3 ways to discount SSE (or RMSE) by the number of free params: $$\text{adjusted RMSE} = \sqrt{\frac{SSE}{n - k}}$$ $$AIC = n \times ln\left(\frac{SSE}{n}\right) - k \times ln(n)$$ $$BIC = n \times ln\left(\frac{SSE}{n}\right) - 2 \times k$$ or in computer code style: ````k = number of free params n = number of DV's SSE = sum of squared errors adjusted RMSE = sqrt ( SSE / n - k ) AIC = n * ln(SSE/n) - k * ln(n) BIC = n * ln(SSE/n) - 2 * k ```` -
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http://physics.stackexchange.com/questions/53374/solving-the-time-independent-schrodinger-equation-does-a-complex-solution-make?answertab=oldest
# Solving the time independent Schrodinger equation: Does a complex solution make sense? In my notes, I have the Time Independent Schrodinger equation for a free particle $$\frac{\partial^2 \psi}{\partial x^2}+\frac{p^2}{\hbar^2}\psi=0\tag1$$ The solution to this is given, in my notes, as $$\Large \psi(x)=C e^\left(\frac{ipx}{\hbar}\right)\tag2$$ Now, since (1) is a second order homogeneous equation with constant coefficients, given the coefficients we have, we get a pair of complex roots:$$r_{1,2}=\pm \frac{ip}{\hbar}\tag3$$ Thus, the most general solution looks something like:$$\psi(x)=c_1 \cos \left(\frac{px}{\hbar}\right)+c_2 \sin \left(\frac{px}{\hbar}\right)\tag4$$ However, instead of writing the solution as a cosine plus a sin, the professor seems to have taken a special case of the general solution (with $c_1=1$ and $c_2=i$) and converted the resulting $$\psi(x)=\cos \left(\frac{px}{\hbar}\right)+ i\sin \left(\frac{px}{\hbar}\right)\tag5$$ into exponential form, using $$e^{i\theta}=\cos \theta + i\sin \theta \tag6$$ to get (2). The main question I have concerning this is: shouldn't we be going after real solutions, and ignoring the complex ones for this particular situation? According to my understanding $\Psi(x,t)$ is complex but $\psi(x)$ should be real. Thanks in advance. - The wavefunction needn't and shouldn't be real. – Chris Feb 8 at 10:38 There are cases where you can get away with a real wavefunction, but the complex case is more general and fundamental. The free particle Hamiltonian $\hat{H}$ commutes with reflection $x\rightarrow -x$,$p\rightarrow -p$, so states with momenta $\pm p$ are both solutions. In equation (2) they have chosen the solution which is an eigenvalue of the momentum operator $\hat{p}$ with a plus sign $+$. The other sign is also a solution, representing a wave going in the opposite direction. Your real solution contains both left moving and right moving waves. – Michael Brown Feb 8 at 11:05 If you look at the particle current $\vec{j}\propto \psi^\star \nabla \psi - \psi \nabla \psi^\star$ you'll see that real wavefunctions correspond to states where there is no net current, so you can only really expect them to turn up when you have bound states. If there is nothing to reflect a particle back the way it came then it is free to move off to infinity and the current can't vanish, so the wavefunction can't be real. – Michael Brown Feb 8 at 11:10 Related: The book of Griffiths, Intro to QM, Problem 2.1b, p.24; and this Phys.SE post. – Qmechanic♦ Feb 8 at 15:54 ## 1 Answer There is no need for the solution $\psi(x)$ to be real. What must be real is the probability density that is "carried" by $\psi(x)$. In some loose and imprecise intuitive way, you may think about a TV image carried by electromagnetic waves. The signal that travels is not itself the image, but it carries it, and you can recover the image by decoding the signal properly. Somewhat similarly, the complex wave function that is found by solving Schrödinger equation carries the information of "where the particle is likely to be", but in an indirect manner. The information on the probability density $P(x)$ of finding the particle is recovered from $\psi(x)$ simply by multiplying it times its complex conjugate: $$\psi(x)^*\psi(x) = P(x)$$ that gives a real function as a result. Note that it is a density: what you compute eventually is the probability of finding the particle between $x=a$ and $x=b$ as $\int_{a}^{b} P(x) dx$ As you know, when you multiply a complex number(/function) times its complex conjugate, the information on the phase is lost: $$\rho e^{i \theta}\rho e^{-i \theta}=\rho^{2}$$ For that reason, in some places one can (not quite correctly) read that the phase has no physical meaning (see footnote), and then you may wonder "if I eventually get real numbers, why did not they invent a theory that directly handles real functions?". The answer is that, among other reasons, complex wave functions make life interesting because, since the Schrödinger equation is linear, the superposition principle holds for its solutions. Wave functions add, and it is in that addition where the relative phases play the most important role. The archetypical case happens in the double slit experiment. If $\psi_{1}$ and $\psi_{2}$ are the wave functions that represent the particle coming from the hole number $1$ and $2$ respectively, the final wave function is $$\psi_{1}+\psi_{2}$$ and thus the probability density of finding the particle after it has crossed the screen with two holes is found from $$P_{1+2}= (\psi_{1}+\psi_{2})^{*}(\psi_{1}+\psi_{2})$$ That is, you shall first add the wave functions representing the individual holes to have the combined complex wave function, and then compute the probability density. In that addition, the phase informations carried by $\psi_{1}$ and $\psi_{2}$ play the most important role, since they give rise to interference patterns. Comment: Feynman is quoted to have said "One of the miseries of life is that everybody names things a little bit wrong, and so it makes everything a little harder to understand in the world than it would be if it were named differently." It is quite similar here. Every book says that the phase of the wave function has no physical meaning. That is not 100% correct, as you see. -
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http://mathoverflow.net/questions/21453/for-which-lie-groups-is-the-convolution-of-any-two-nonzero-integrable-compactly-s
## For which Lie groups is the convolution of any two nonzero integrable compactly supported functions nonzero? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Titchmarsh convolution theorem implies that the convolution of two nonzero functions $f,g\in L^1(\mathbb R)$ with compact support is nonzero. There is a generalization of this theorem to the case of $f,g\in L^1(\mathbb R^n)$ with compact support. Now, consider a Lie group $G$ endowed with its Haar measure, and $p\geq1$. We may formulate a property (T1) If $f,g\in L^p(G)$ are nonzero and compactly supported, the convolution $f*g$ is nonzero. This is true for $G=\mathbb R^n$, but fails for abelian groups of the form $G=\mathbb R^n \times \mathbb T^m$, where $\mathbb T^m$ is the $m$-dimensional torus, and $m>0$. [Counterexample: let $f$ be the characteristic function of $K\times \mathbb T^m$, where $K\subseteq \mathbb R^n$ is compact, and $g(x,y)=f(x,y)h(y)$, where $h\in L^p(\mathbb T^m)$ and $\int_{\mathbb T^m}h =0$]. However, we may strengthen the assumptions in (T1), obtaining (T2) There exists an open set $U\subseteq G$ with compact closure, such that if $f,g\in L^p(G)$ are nonzero and supported in left translates of $U$, the convolution $f*g$ is nonzero. This is actually true for any abelian Lie group $G$, since in this case $G$ is locally isomorphic to $\mathbb R^k$. Of course, if (T1) or (T2) hold for some $p$, then they also hold for all $q>p$. The most interesting cases are $p=1$ and $p=2$. So, my question is: Which Lie groups are known to satisfy property (T1) or (T2) with $p=1$ or $p=2$? - This nice question is of course related to the zero-divisor conjecture for discrete groups (considered as zero-dimensional Lie groups if you want). Asking this for connected topological groups (or Lie groups), a first more specific question would be: Does the existence of torsion in $G$ lead to zero-divisors in $L^1(G)$? – Andreas Thom Aug 13 2010 at 12:18
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http://en.wikibooks.org/wiki/Engineering_Analysis/MISO_Transformations
# Engineering Analysis/MISO Transformations Consider now a system with two inputs, both of which are random (or pseudorandom, in the case of non-deterministic data). For instance, let's consider a system with the following inputs and outputs: • X: non-deterministic data input • Y: disruptive noise • Z: System output Our system satisfies the following mathematical relationship: $Z = g(X, Y)$ Where g is the mathematical relationship between the system input, the disruptive noise, and the system output. By knowing information about the distributions of X and Y, we can determine the distribution of Z.
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http://physics.stackexchange.com/questions/54949/gaussian-wave-packet/54969
# Gaussian wave packet At our QM intro our professor said that we derive uncertainty principle using the integral of plane waves $\psi = \psi_0(k) e^{i(kx - \omega t)}$ over wave numbers $k$. We do it at $t=0$ hence $\psi = \psi_0(k) e^{ikx}$ $$\psi = \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k$$ where $\psi_0(k)$ is a $k$-dependent normalisation factor (please correct me if I am wrong). This dependency was said to be a Gaussian function $$\psi_0(k)= \psi_0 e^{i(k-k_0)^2/4\sigma k^2}$$ where $\psi_0$ is an ordinary normalisation factor (please correct me if I am wrong). QUESTION 1: Why do we choose $\psi_0(k)$ as a Gaussian function? Why is this function so appropriate in this case? QUESTION 2: I don't know how did our professor get a Gaussian function with an imagnary number $i$ in it. His Gaussian is nothing like the one on Wikipedia which is $$f(x) = a e^{-(x-b)^2/2c^2}$$ QUESTION 3: We used the first integral i wrote down to calculate the Heisenberg's uncertainty principle like shown below, but it seems to me that most of the steps are missing and this is the reason i don't understand this. Could anyone explain to me step by step how to do this. $$\begin{split} \psi &= \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k\\ \psi &= \int\limits_{-\infty}^{+\infty} \psi_0 e^{i(k-k_0)^2/4\sigma k^2} \cdot e^{ikx} \, \textrm{d} k\\ \psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2} \end{split}$$ I think this is connected to a Gaussian integral, but it doesn't look quite like it to me. Well in the end our professor just says that out of the above it follows that $$\boxed{\delta x \delta k = \frac{1}{2}}$$ I don't understand this neither. It was way too fast or me. - For question 1): first of all, your teacher was probably thinking about wave packets, which intuitively are more or less Gaussian - it's not a stretch to think that a wavepacket should have some mean wavenumber $k_0$ and some variance $\sigma$. There's also a mathematical reason: the bound you're deriving holds for all possible $L^2$ functions, and the Gaussian is the only one that saturates it. So apart from doing the full proof, it's the most interesting test case you can work with. – Vibert Feb 24 at 11:51 1 – Emilio Pisanty Feb 24 at 17:17 ## 1 Answer It is hard to know what exact the professor has done, but from what I have understood made an effort to help the situation. Q1. A Gaussian function is chosen to represent a freely expanding wave function for several reasons: (i) The Gaussian function represents a normal probability distribution function. Since $|\psi(x)|^2$ represents a probability distribution function for a huge class of particles moving in a similar way, and having momentum within a certain range, the large numbers theorem points towards a Gaussian function, to represent the wave function of particles within a region of space determined by the width, $\sigma$, of the Gaussian function. The $\sigma$ also happens to be the standard deviation, i.e. the uncertainty in the position of the particle. (ii) The expectation value of the position of the particle turns out to be the value of $b$ in your Gaussian wave packet. (iii) The Gaussian wave packet also contains the wavy bit shown by the phase factor $e^{ikx}$. This is what makes Gaussian wave packets an excellent representation of particles, which are known to be located within some region of width $w$, but nevertheless they are all moving as plane waves while the packet travels along in space. (iv) The wave packet is made of an infinitely large number of momentum values in a momentum width which relates to $\sigma$ by a Fourier transformation. Q2. So to put some order in all these, let us consider the general Gaussian function $\psi(x)=\psi_0e^{-A(x-x_0)^2+Bx+C}$ which has a Fourier transformation $\psi(k)=\psi_0\sqrt{\frac{\pi}{A}}e^{\frac{B^2}{4A}+Bx_0+C}$. The normalisation constant $\psi_0$ is given by an integration in the range [-$\infty, +\infty$] and has the value $\sqrt{\frac{A}{\pi}}$ but leave it as $\psi_0$. Q3. To make contact with your professor, try to apply these results using the following: $A=\frac{1}{2\sigma^2}$ $B=ik$ $x_0=b$ $C=0$ - – 71GA Feb 25 at 19:19 @71GA Thank you for your notes. I don't think the square rooting should make any difference to the form of the uncertainty principle, since it is only itroducing a scale factor of 1/2. Not sure why should one prefer to go through the square root though. – JKL Feb 25 at 21:36
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http://math.stackexchange.com/questions/77348/how-to-prove-that-lim-limits-h-to-0-fracah-1h-ln-a/77368
# How to prove that $\lim\limits_{h \to 0} \frac{a^h - 1}{h} = \ln a$ In order to find the derivative of a exponential function, on its general form $a^x$ by the definition, I used limits. $\begin{align*} \frac{d}{dx} a^x & = \lim_{h \to 0} \left [ \frac{a^{x+h}-a^x}{h} \right ]\\ \\ & =\lim_{h \to 0} \left [ \frac{a^x \cdot a^h-a^x}{h} \right ] \\ \\ &=\lim_{h \to 0} \left [ \frac{a^x \cdot (a^h-1)}{h} \right ] \\ \\ &=a^x \cdot \lim_{h \to 0} \left [\frac {a^h-1}{h} \right ] \end{align*}$ I know that this last limit is equal to $\ln(a)$ but how can I prove it by using basic Algebra and Exponential and Logarithms properties? Thanks - $a^h=\exp(\ln a \cdot h)$. Are you able to Taylor expand the exponential function? – anon Oct 30 '11 at 23:33 Well, can we use the definition but be a bit witty? Perhaps we could use the definition (along with the chain rule) on $\log y = \log (a^x)$? – mixedmath♦ Oct 30 '11 at 23:43 I think it is worth mentioning that some authors (such as E. Landau) define the logarithm that way. – Peter Tamaroff May 2 '12 at 21:23 I consider this to be a rather tricky issue. Let me explain. Let $a$ be positive real number. Then to \textit{define} $a^x$ algebraically is easy on the rationals, just ask that the standard exponent laws hold. But if you want to define $a^x$ for irrational values, you need to do some analysis. In particular you need to prove that $a^x$ is continuous on the rationals. – Baby Dragon Jun 3 '12 at 8:04 ## 2 Answers First, we prove it for $a=e$. Let $\frac{1}{t} = e^h - 1$. Then $e^h = 1+\frac{1}{t}$, so $h = \ln(1 + \frac{1}{t})$. As $h\to 0$, we have $\frac{1}{t}\to 0$, so $t\to \infty$ (if $h\to 0^+$) or $t\to-\infty$ (if $h\to 0^-$). We have: $$\begin{align*} \lim_{h\to 0^+}\frac{e^h-1}{h} &= \lim_{t\to\infty}\left(\frac{1/t}{\ln(1+\frac{1}{t})}\right)\\ &= \lim_{t\to\infty}\frac{1}{t\ln(1+\frac{1}{t})}\\ &= \lim_{t\to\infty}\frac{1}{\ln\left( (1+\frac{1}{t})^t\right)}\\ &= \frac{1}{\ln\left(\lim_{t\to\infty}(1 + \frac{1}{t})^t\right)}\\ &= \frac{1}{\ln(e)}\\ &= 1. \end{align*}$$ Since we also have $$\begin{align*} \lim_{t\to-\infty}\left(1 + \frac{1}{t}\right)^t &= \lim_{n\to\infty}\left(1 - \frac{1}{n}\right)^{-n}\\ &= \lim_{n\to\infty}\frac{1}{(1 + \frac{-1}{n})^n}\\ &= \frac{1}{e^{-1}}\\ &= e,\end{align*}$$ a similar calculation shows that $$\lim_{h\to 0^-}\frac{e^h-1}{h} = 1.$$ Hence, $$\lim_{h\to 0}\frac{e^h-1}{h} = 1.$$ Now for arbitrary $a\gt 0$, $a\neq 1$, we rewrite $a^h = e^{h\ln(a)}$, and make the substitution $t = h\ln(a)$. Then the result follows from the case $a=h$ as Gerry Myerson shows. - It depends a bit on what you're prepared to accept as "basic algebra and exponential and logarithms properties". Look first at the case where $a$ is $e$. You need to know that $\lim_{h\to0}(e^h-1)/h)=1$. Are you willing to accept that as a "basic property"? If so, then $a^h=e^{h\log a}$ so $$(a^h-1)/h=(e^{h\log a}-1)/h={e^{h\log a}-1\over h\log a}\log a$$ so $$\lim_{h\to0}(a^h-1)/h=(\log a)\lim_{h\to0}{e^{h\log a}-1\over h\log a}=\log a$$ -
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http://math.stackexchange.com/questions/46316/angle-brackets-for-tuples?answertab=active
# Angle brackets for tuples I've recently noticed that use of angle brackets for writing tuples, e.g. $\langle x, y \rangle$ instead of the usual round brackets in a few books I've been reading — Lawvere's Sets for Mathematics, Mac Lane's Categories for the Working Mathematician, Forster's Logic, induction and sets, for example. I've also seen occasional use of it in Hartshorne's Algebraic Geometry, but there round brackets seem predominant. Is there some subtle distinction between the two notations I've missed, and what might the reasons for not using round brackets be? Is this practice peculiar to a particular tradition in mathematics (say, foundations)? - I'm also curious, since I've encountered abstract algebra texts using angled brackets, e.g., to denote groups <set, operation>: e.g. $\langle \mathbb Z, + \rangle$. – amWhy Jun 19 '11 at 17:01 In Strooker's Introduction to categories, homological algebra, and sheaf cohomology, he uses $\textopencorner x,y\textcorner$ for tuples, but I haven't seen that anywhere else (edit: the corners doesn't seem to be working, they are on page 9 here). – Zev Chonoles♦ Jun 19 '11 at 18:09 8 It may just be a matter of parenthesis being overloaded. – Arturo Magidin Jun 19 '11 at 19:18 1 @amWhy: Formally speaking, a structure is a tuple... – Asaf Karagila Nov 10 '11 at 12:10 1 – Martin Sleziak Nov 10 '11 at 14:51 show 1 more comment ## 4 Answers I don't think this is quite what you are asking about, but at the level of ordered tuples of elements, I would use parentheses to emphasize that I only care about the tuple as a point object, and angles to emphasize that I care about the tuple as a direction, or vector. For example, I might have a line parametrized by $t$ with equation$$(x,y,z)=(x_0,y_0,z_0)+t\langle a,b,c\rangle$$ - 1 Is this convention in use by anyone other than you? I have never encountered it, but then I haven't read a lot of mathematical texts. – Rahul Narain Jul 24 '12 at 6:50 @Rahul I don't know. I picked it up somewhere. Stewart uses angle brackets for vectors. But in the vector chapters, points often get angle brackets too. – alex.jordan Jul 25 '12 at 7:10 Some analysts (in a wide sense) write $\langle x,y\rangle$ (angle brackets), for $x$ and $y$ elements of a same set $X$, to denote the ordered pair element of $X\times X$. A more classical (to me) notation is $(x,y)$ (parenthesis), but, if for example $X=\mathbb R$ and $x\leqslant y$, the notation $(x,y)$ may refer to the open interval $\{z\in\mathbb R\mid x<z<y\}$. Hence the bracket notation might have been designed as a way to avoid the confusion. Bourbaki use the notation $]x,y[$ for open intervals and $[x,y]$ for segments. This notation, of frequent use in the mathematical literature written in French (and in others), removes the risk of confusion mentioned above. I do not know how useful brackets are for objects like $\langle\mathbb Z,+\rangle$, since the objects inside the brackets are of a different nature. - Kind of @Arturo's point in a comment. – Did Nov 10 '11 at 12:17 What do those analysts (in a wide sense) do in order to avoid confusion of ordered pairs of vectors from a space with a scalar product? Oh, well... There are only finitely many symbols. Further trivium: were taught the Bourbaki-style intervals in elementary school in Switzerland. – t.b. Nov 10 '11 at 12:28 They write scalar products as $x\cdot y$. // Replaced French by written in French. // Thanks for the remarks. – Did Nov 10 '11 at 12:35 – t.b. Nov 10 '11 at 12:49 @t.b. I see... I was not aware of that (and I first thought you were referring to the French speaking part of the country). Thanks. – Did Nov 10 '11 at 13:15 Herbert Enderton --and other logicians-- use angle brackets to indicate a structure, that is, a set with underlying functions, relations and constants. For example, we denote the usual natural numbers with addition, multiplication, an identity 0 for addition and an identity 1 for multiplication (in that order) as $\left\langle\mathbb{N},+^{\mathbb{N}},\cdot^{\mathbb{N}},0^{\mathbb{N}},1^{\mathbb{N}}\right\rangle$. In general, for a structure $\mathfrak{A}$ with relations $r_1,\dots,r_i$ (each of a given arity), functions $f_1,\dots,f_j$ (each of a given arity) and constants $c_1,\dots,c_k$, an interpretation of this structure with domain $A$ is denoted $\left\langle A,r_1^{\mathfrak{A}},\dots,r_i^{\mathfrak{A}},f_1^{\mathfrak{A}},\dots,f_j^{\mathfrak{A}},c_1^{\mathfrak{A}},\dots,c_k^{\mathfrak{A}}\right\rangle$. I don't know wheter this notation came from algebra first, but, as commented above, it appears that some authors use angle brackes for algebraic structures in general (e.g. $\langle\mathbb{Z},+\rangle$, $\langle\mathbb{R},+,\cdot\rangle$). Another use of angle brackets is to indicate an inner product in a vector space, to distinguish the inner product with other well-known operations between the elements of the space, for example, composition of automorphisms of a given set and multiplication of numbers in a ring, which are usually denoted with common syntactic concatenation ($fg$ for the composition of $f$ and $g$, $xy$ for the (ring-)product of $x$ and $y$). - I'm not an expert, but I'm quite sure the difference between angle and round brackets depends on the individual notation conventions of each author/paper/work. Sometimes I see angle brackets instead of round ones without any particular meaning, while (in example) in the computability and complexity class I attended only round brackets were used for classic tuples: the angle brackets were used as a shortcut to mean the encoding for a turing machine of that tuple. Unfortunately I have no clue about the origins of the angle brackets, it would be interesting. -
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http://math.stackexchange.com/questions/87601/techniques-for-visualising-n-dimensional-space/87749
# Techniques for visualising $n$-dimensional space Can you guys point me to the types of things I should read about (I'm a maths lay person really) if I want to learn about the current thinking how $n$ dimensional spaces can be visualised and thus navigated? Sorry this question seems a bit vague, but I'm trying think of how I navigate through data in many dimensions and I feel that mathematicians would have solved these problems many many years ago. Edit: Thank you for the informed debate: Let me clarify what I'm after. I'm looking at model for navigating data using a computer screen (so 2 dimensions). Add a third dimension (think zoom in zoom out). Ok so what about $n$ dimensions, what techniques could help me navigate them while confining my self o the 3d world of the computer model. Reading the link below to mathsoverflow was interesting and I now have some interesting ideas coming out. - 1 When you say "maths would have solved these problems many many years ago" what do you mean? What's the problem exactly? I think there's a meta-problem, in that you haven't specified a problem. For example, how do you visualize in $\mathbb R^3$? – Ryan Budney Dec 2 '11 at 3:18 6 The old joke is, you first visualize an infinite-dimensional space, then cut down to $n$ dimensions. – Gerry Myerson Dec 2 '11 at 3:19 I'm assuming you mean an n-manifold. Mathematician have various techniques that make it easy to handle larger dimensions. It doesn't mean you can visualize it. 2-manifolds are impossible to visualize i.e. projective space $RP_1$ – simplicity Dec 2 '11 at 3:23 Is this a dupe of this or this? – J. M. Dec 2 '11 at 3:53 – Tony Dec 2 '11 at 3:57 show 2 more comments ## 7 Answers When somebody says "high-dimensional space is hard to visualize", they are thinking of visualizing with the eyes. But mathematicians visualize with the brain! I highly recommend the AMS article The World of Blind Mathematicians. Who could be better at visualizing things they can't see? - Thank you I will look into this. – Preet Sangha Dec 2 '11 at 8:49 I'm looking at model for navigating data using a computer screen (so 2 dimensions). Add a third dimension (think zoom in zoom out). Ok so what about n dimensions, what techniques could help me navigate them while confining my self o the 3d world of the computer model. There is a very large body of published work (both theoretical and applied) on multidimensional data visualization (also called multivariate or $n$-dimensional). Try a web search with these terms! Some example titles: • Visualization of Multi-variate Scientific Data (Bürger & Hauser) • A Taxonomy of Glyph Placement Strategies for Multidimensional Data Visualization (Ward) • Handbook of Data Visualization (Chen, Härdle & Unwin) and so on. Ward's paper (p. 10) offers some advice when viewing N-dimensional data with a 2-dimensional display, re selecting from the N(N − 1)/2 possible orthogonal views in the N-dimensional space (not including rotational variations). - I know there is. However as a maths layperson a lot of these are ou t of my depth. Thank you though - I'll follow these up. – Preet Sangha Dec 2 '11 at 21:23 I'm not so experienced with problems in N dimensions but i can suggest one thing: do not confuse the geometry with an analitical process. Also do not confuse when you are talking about mathematical analysis visualizing it with the help of the geometry and viceversa. The geometry is a really old branch of the math, sometimes someone describe it as a science apart, the thing is that this discipline was born when the human knowledge had to dial with nature very closely and directly, like was during the ancient egipt or in greece. The geometry was born from the observation of the reality, nothing more and nothing less, and the way to express geometry consist in the use of the mathematical language because it is universal and unambiguous. The analitic process is born to approach and to try to solve other kinds of problems and has to dial with concepts that are not properly available in nature like the concepts of approximation, N dimensions, the laws that we think are the correct ones to describe the natures, the concept of infinite, the infinitely small and the infinitely big, and so on. The analitic process is really focused on the human needs and only this, if you should describe the lenght of a piece of wood to make a bridge with your hands, you probably do not need to have a really precise measurement, you probably do not need it at all, but if you have to abstract that bridge into a project you probably have to deal with an approximation of that measure, so you need a value, simply because you have to put a quantity on a piece of paper, and thanks to this you can skip from geometry to an analitical process . You simply can't imagine more than 3 dimension because N>3 does not belong to the world you are in, geometrically speaking. - N>3 string theorist would debate that. Personally I think you can simulate a being that lives in 4 dimensions on a computer, assuming AI progresses enough to that stage. Then, that being would be able to see in 4 dimensions. With time I think humans could see in higher dimensions. It would be pointless through. – simplicity Dec 2 '11 at 4:08 you say it right "theorist", Preet Sangha is looking for a pratical solution not theories. The human knowledge has switched from geometry to analitical analysis much time ago, the analisys is good when we have to dial with complex problems but is pratically impossible to visualize over N>3, this is the price to pay for us. Also do not confuse the terms, usually the physicist use the verb "describe" and not "visualize" when they talk about N dimensions. – Micro Dec 2 '11 at 4:16 Your best thinking in $\mathbb{R}^2$ and 3 (never $\mathbb{R}$) for counterexamples, but that is it really. I mean everything's 'the same', especially if you consider (I know this is lin. algebra but the point is emphasized more strongly with open balls) open balls in $\mathbb{R}^n$. Literally everything you need to 'know' from an open ball will be 'seen' from a picture of a disc or sphere [i.e $\mathbb{R}^2$ = union of open balls centre zero and radius n; similar results for $\mathbb{R}^n$]. You can also 'move' open balls as $B(r,a) = a + rB(1,0)$. I don't want to go into too much detail and these examples are terrible but basically $\mathbb{R}^2$ and $\mathbb{R}^3$ will usually be fine for counterexamples, there is an obvious decomposition (though not 'best') of $\mathbb R^n$ into direct sums of the subspaces $\mathbb{R}^2$ and $\mathbb{R}^3$ (and $\mathbb{R}$), etc. It is obviously best not to work in $\mathbb{R}^n$ often though, $(\mathbb{Z}_p)^n$; vectors with entries modulo $p$ for prime $p$ is an important example. There are two other obvious important vector spaces, but both of these are easy to visualize for counterexamples. - I know this was a poor answer (due to the poor 'problem' posted - a meta problem), but downvotes with critism are more useful than downvotes for everyone. – Adam Dec 2 '11 at 3:29 This is very confusing. Firstly, open balls in $R^4$ are wildly different than open balls in $R^3$. Even, then there are objects in $R^4$ that are impossible in $R^3$ like Klein bottle. – simplicity Dec 2 '11 at 3:30 ?? It is trivial to come up with specific examples to emphasize how things are different but they are not wildly different (whatever that means). Consider proving that a compact set is bounded in R^n. Well note from the example of R^2, R^n is the union of open balls centered at 0 with integer radius so it is clear we can find a big enough ball to contain the set hence bounded. Say we want to prove the union of subspaces is not a vector space - take two different lines in R^2, easily extended (by embedding R^2 into R^n) ... I could go on, I think you are missing the point however. – Adam Dec 2 '11 at 3:35 However, I think you missed what OP wanted. Also, that doesn't work all the time so it's misleading. There is probably crazy objects in $R^4$ that don't look like anything in $R^3$. Well, I can give several if you like. – simplicity Dec 2 '11 at 3:52 Check out http://en.wikipedia.org/wiki/Parallel_coordinates if your data is discrete, it maybe very useful. If it is not, there are extensions but they are not always insightful. - I sometimes like to pervert the standard "perspective" 2-D rendering of the $x,y$ & $z$ axes in $\mathbb{R}^3$ by drawing $n$ axes through a common origin all with highly acute angles between them, which I imagine are all pairwise perpendicular (for each of the $\binom{n}{2}=\frac{n(n-1)}{2}$ pairs of axes) and sometimes even indicate this with the little right angle indicators (parallel to each alternate axis in a pair) -- as if I am looking from some yet other unspecified, distant dimension. But it requires of course some imagination, and complements rather than replaces other, non-visual ways of conceptualizing $\mathbb{R}^n$. - I would like to say that you can't visualize higher dimension in Mathematics. Even for n=4 it is impossible. Several barriers that stop you from doing it. If you take the 2-manifolds you have really three objects i.e. the sphere, the torus and finally projective space. The problem is even for closed surfaces you would think that it's easy to visualize, however the projective space can't be realized properly in $R^3$ i.e. can't be embedded into it(this mean it crosses itself like the Klein Bottle does). You would need to go into the fourth dimension. This is a popular image of the Klein bottle. But, in reality this shouldn't intersect itself. However, it's impossible to visualize it because you would need to think in 4 dimensions. Mathematics has got around this problems by developing topology and algebraic methods. The naive way to visualize the Klein Bottle is to find the points in which it intersects itself in $R^3$ and then give them another colour and say they are in the 4th dimension. 3-manifolds are even more impossible to visualize. 3-spheres would look crazy There is no true way to visualize it. Everything is really just algebra mainly group theory mixed with topology. If you really care to learn all this it's best to pick up any decent book on Topology and read it. Armstrong Basic Topology is the best place to start. Would like to add that the 4th dimension is the hardest dimension possible you can get. http://en.wikipedia.org/wiki/Exotic_R4 - thank you. I will look up topology – Preet Sangha Dec 2 '11 at 8:36
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http://physics.stackexchange.com/questions/12599/will-a-hole-cut-into-a-metal-disk-expand-or-shrink-when-the-disc-is-heated/22436
# Will a hole cut into a metal disk expand or shrink when the disc is heated? Suppose you take a metal disc and cut a small, circular hole in the center. When you heat the whole thing, will the hole's diameter increase or decrease? and why? - 4 Sometimes answers become clear when you think about various extremes. Instead of a small hole, make it big. Almost as big as the disk, coming very close to the rim, stopping short by just a paper-thin margin. You'll have a hoop of metal. How will that change when heated? – DarenW Mar 16 '12 at 3:15 i have a doubt.Consider the particles on the outer circumference of the disc expand outwards..and the inner circumference expand inwards....so the volume is increasing with the mass remaining constant...so with the explanation in one of the above answers it is possible that the hole's diameter decreases as the density is decreasing(in one of the answers it was mentioned that as density increases it is not possible)..but it does not happen.why? – user8673 Apr 14 '12 at 9:02 1 Here is a way to think about it that makes the answer obvious. Replace 'cut a hole into with 'draw a circle on' and consider what happens to the drawn circle when the disc is heated. – Benjamin Franz Apr 14 '12 at 13:51 @BenjaminFranz That does not make the answer obvious (at least to me). It presumes that the "cut a hole into" and "draw a circle on" scenarios are equivalent, but that's begging the question. Why are they equivalent? – jamesdlin May 2 at 2:39 ## 8 Answers Good question! Assuming the disc is uniform and isotropic (the same in different directions), the hole will expand in the same ratio as the metal. You can see this because the thermal expansion equation $$\mathrm{d} L = L\alpha\mathrm{d}T$$ applies to all lengths associated with the metal, including the circumference of the hole, since the edge of the hole is made out of metal. And if the circumference of the hole expands, so does the diameter. If you have a disc with different regions that are made of different types of metal, or if the metal that makes up your disc has an anisotropic crystal structure (so that it expands by different factors in different directions), then the analysis is more complicated. But in both cases, I think the hole would still get larger since the overall change in size is still an expansion. In order to get the hole to shrink, you would need to use a material with a negative thermal expansion coefficient $\alpha < 0$, which means it gets smaller as the temperature gets higher. In that case the entire disc would shrink as it heats up. Wikipedia has an entry on these kinds of materials (h/t Kevin Reid). - 1 – Kevin Reid Jul 22 '11 at 13:51 Consider the disc made of isotropic material is heated uniformly across its area... the metal near the outer circumference, tries to expand outward because of resistance from its disk-side neighboring molecules. Similarly, the part that is near the inner circumference of the hole (cut-out part) tend to expand to its free side, and thereby covering up the hole – Mallik Mar 15 '12 at 17:50 @Mallik: But it would also try to expand along the ring, and if it moves inward, it has even less space. Therefore it has to move outwards in order to expand (with the outer parts moving outwards more). – celtschk Apr 15 '12 at 21:54 David Zaslavski's answer is correct and complete. But I want to propose a different way to look at the problem. Think of the disc that was cut out, and imagine that you heat it too, exactly as you heat the plate. After heating, the disc will fit in exactly to the hole, just as if it was first heated and then cut out. Therefore, the hole will expand. - 1 I feel like this answer is begging the question. Why must the whole disc expand in the same way whether or not a smaller disc was cut out of its center? How do you know that the cut-out disc will fit exactly into the hole? For example, I could argue that without the expanding center piece, there is nothing providing an outward force to make the hole bigger. – jamesdlin May 2 at 2:32 @jamesdlin I agree that it's heuristic and as such one could argue differently. The real solution is sketched by David Z. If you want more justification, you can say that the thermo-elastic equations with $T=const$ and stress-free BC will result in a stress free configuration (easily verified). Therefore, having the disc in place or cut out has no effect on the surrounding - the "interaction" is the stress, and therefore stress free boundary conditions are equivalent to not having a disc at all. Again, I stress that this can be solved analytically and then there's no ambiguity. – yohBS May 2 at 6:50 Instead of a circular hole, let's think of a square hole. You can get a square hole two ways, you can cut it out of a complete sheet, or you can get one by cutting a sheet into 9 little squares and throwing away the center one. Since the 8 outer squares all get bigger when heat it, the inner square (the hole) also has to get bigger: Same thing happens with a round hole. This is confusing to people because the primary experience they have with stuff getting larger when heated is by cooking. If you leave a hole in the middle of a cookie and cook it, yes, the cookie gets bigger and the hole gets smaller. But the reason for this is that the cookie isn't so solid. It's more like a liquid, it's deforming. And as Ilmari Karonen points out, the cookie sheet isn't expanding much so there are frictional forces at work. - 1 Awesome way to explain it! – Manishearth♦ Mar 16 '12 at 1:36 2 The reason why cookies don't expand uniformly when baked has probably more to do with the fact that they're in semi-adhesive contact with the baking tray, which is both much stiffer than cookie dough and expands much less during the process, and which thus provides an external force counteracting the dough's expansion and deforming it. I suspect that, if you were to bake a cookie with a hole in free-fall, suspended in air without a tray, the hole would expand. – Ilmari Karonen Jul 24 '12 at 19:58 (In fact, it occurs to me that, in the absence of a convenient free-fall cookie oven, one might be able to conduct an equivalent experiment by (slowly) deep-frying the annular cookie instead. Hmm... it might be time for some science, here.) – Ilmari Karonen Jul 24 '12 at 20:03 2 Nice comment, however, I've seen doughnuts where the hole closes up due to frying. – Carl Brannen Jul 25 '12 at 18:08 If you worked in an auto shop, you'd know the answer already. When an axle gets stuck in a ball bearing, one way to pull it out is to heat up the bearing with a welding torch. The whole bearing, including the hole in the middle, expands and allows you to pull the axle free. - 1 You're exactly right, but I thought the spelling was "axle". – Mike Dunlavey Dec 17 '11 at 16:22 6 ... and you have to be careful to heat the bearing more than you heat the axle. – Mike Dunlavey Dec 17 '11 at 16:27 Thank you, fixed. – Florin Andrei Dec 18 '11 at 5:27 1 Or even if you were a medieval blacksmith fitting a metal rim on a wheel. – Martin Beckett Sep 23 '12 at 4:35 Im a machinist. We commonly heat holes to expand them in vatious applications. For example to install bearings that demand press. We use liquid nitrogen to also freeze the bearings. When both objects return to ambient temps the results are the hole shrinks. We can even control within some tolerance how much. Consider on an atomic level what is taking place. End of day? Heat a hole it expands. When it cools it will shrink - I think that there is an important assumption at work here. The hole will expand as long as the material is sufficiently rigid; since most things that we want to expand are rigid (jar lids and axle bearings, for example), and since a disk is likely to be made from a comparatively rigid alloy such as steel, it is generally fair to say that the hole would expand. But I think that you could also create a disk in which the hole would shrink; I would expect a hole in a disk made from a malleable material with a high thermal expansion coefficient (such as gold or lead) to shrink. - This is analogous to the reason that the holes in cookies get smaller, not larger, when baked. – dotancohen Oct 2 '12 at 12:04 The answer might lie in the ratio of ring thickness to circumference. In the case of an infinitely large disk compared to hole size the hole may get smaller. Remember the extremes of the theory must be answered. - 2 Be careful with how you visualize infinity. You seem to imagine that the disk can't grow outward, but it can. There is no "wall" at infinity preventing every atom from moving away from the center of your coordinates. In fact, a finite hole in an infinite plane will expand when everything is heated. – Chris White Jan 14 at 7:26 I know it has been answered already. Just a different perspective. When a solid body is heated, it expands as if we look though a magnifying glass - everything looks bigger, including the hole in the disc. Hence the hole in the expands when it is heated up. - ## protected by Qmechanic♦Jan 22 at 16:19 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://unapologetic.wordpress.com/2008/06/03/the-category-of-matrices-ii/?like=1&source=post_flair&_wpnonce=671450efa2
# The Unapologetic Mathematician ## The Category of Matrices II As we consider the category $\mathbf{Mat}(\mathbb{F})$ of matrices over the field $\mathbb{F}$, we find a monoidal structure. We define the monoidal product $\boxtimes$ on objects by multiplication — $m\boxtimes n=mn$ — and on morphisms by using the Kronecker product. That is, if we have an $m_1\times n_1$ matrix $\left(s_{i_1}^{j_1}\right)\in\hom(n_1,m_1)$ and an $m_2\times n_2$ matrix $\left(t_{i_2}^{j_2}\right)\in\hom(n_2,m_2)$, then we get the Kronecker product $\left(s_{i_1}^{j_1}\right)\boxtimes\left(t_{i_2}^{j_2}\right)=\left(s_{i_1}^{j_1}t_{i_2}^{j_2}\right)$ Here we have to be careful about what we’re saying. In accordance with our convention, the pair of indices $(i_1,i_2)$ (with $1\leq i_1\leq m_1$ and $1\leq i_2\leq m_2$) should be considered as the single index $(i_1-1)m_2+i_2$. It’s clear that this quantity then runs between ${1}$ and $m_1m_2$. A similar interpretation goes for the index pairs $(j_1,j_2)$. Of course, we need some relations for this to be a monoidal structure. Strict associativity is straightforward: $\left(\left(r_{i_1}^{j_1}\right)\boxtimes\left(s_{i_2}^{j_2}\right)\right)\boxtimes\left(t_{i_3}^{j_3}\right)=\left((r_{i_1}^{j_1}s_{i_2}^{j_2})t_{i_3}^{j_3}\right)=\left(r_{i_1}^{j_1}(s_{i_2}^{j_2}t_{i_3}^{j_3})\right)=\left(r_{i_1}^{j_1}\right)\boxtimes\left(\left(s_{i_2}^{j_2}\right)\boxtimes\left(t_{i_3}^{j_3}\right)\right)$ For our identity object, we naturally use ${1}$, with its identity morphism $\left(1\right)$. Note that the first of these is the object the natural number ${1}$, while the second is the $1\times1$ matrix whose single entry is the field element ${1}$. Then we can calculate the Kronecker product to find $\left(t_i^j\right)\boxtimes\left(1\right)=\left(t_i^j\right)=\left(1\right)\boxtimes\left(t_i^j\right)$ and so strict associativity holds as well. The category of matrices also has duals. In fact, each object is self-dual! That is, we set $n^*=n$. We then need our arrows $\eta_n:1\rightarrow n\boxtimes n$ and $\epsilon_n:n\boxtimes n\rightarrow1$. The morphism $\eta_n$ will be a $1\times n^2$ matrix. Specifically, we’ll use $\eta_n=\left(\delta^{i,j}\right)$, with $i$ and $j$ both running between ${1}$ and $n$. Again, we interpret an index pair as described above. The symbol $\delta^{i,j}$ is another form of the Kronecker delta, which takes the value ${1}$ when its indices agree and ${0}$ when they don’t. Similarly, $\epsilon_n$ will be an $n^2\times1$ matrix: $\epsilon_n=\left(\delta_{i,j}\right)$, using yet another form of the Kronecker delta. Now we have compatibility relations. Since the monoidal structure is strict, these are simpler than usual: $(\epsilon_n\otimes1_n)\circ(1_n\otimes\eta_n)=1_n$ $(1_{n^*}\otimes\epsilon_n)\circ(\eta_n\otimes1_{n^*})=1_{n^*}$ But now all the basic matrices in sight are various Kronecker deltas! The first equation reads $\left(\delta_a^b\delta^{c,d}\right)\left(\delta_{b,c}\delta_d^e\right)=\delta_a^e$ which is true. You should be able to verify the second one similarly. The upshot is that we’ve got the structure of a monoidal category with duals on $\mathbf{Mat}(\mathbb{F})$. ## 1 Comment » 1. [...] The Category of Matrices III At long last, let’s get back to linear algebra. We’d laid out the category of matrices , and we showed that it’s a monoidal category with duals. [...] Pingback by | June 23, 2008 | Reply « Previous | Next » ## 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://mathhelpforum.com/differential-geometry/183728-absolute-value-convergence-vs-convergence-they-same-print.html
# Absolute value convergence vs. convergence (Are they the same?) Printable View • June 27th 2011, 11:42 PM CountingPenguins Absolute value convergence vs. convergence (Are they the same?) The question is if the absolute value of a sequence converges then does the sequence itself converge. My gut reaction to this is no because you could have something with opposite signs (+ - + -) that would converge in absolute value but diverge otherwise, or even converge to different values. I know that lim n-> infinity |n/-n|=1 and n/-n =-1 but they both converges. No luck on the convergent, divergent scenario. (Keep in mind this is just some random idea about this problem from my head). What do you think? • June 27th 2011, 11:46 PM girdav Re: Absolute value convergence vs. convergence (Are they the same?) That's the idea indeed. Maybe you could take an explicit example $u_n:=(-1)^n$. • June 28th 2011, 12:01 AM CountingPenguins Re: Absolute value convergence vs. convergence (Are they the same?) Thanks. I know that putting |1/(-1)^n| converges to 1 and without the absolute value it doesn't. Thanks again. • June 28th 2011, 02:03 AM Prove It Re: Absolute value convergence vs. convergence (Are they the same?) Quote: Originally Posted by CountingPenguins The question is if the absolute value of a sequence converges then does the sequence itself converge. My gut reaction to this is no because you could have something with opposite signs (+ - + -) that would converge in absolute value but diverge otherwise, or even converge to different values. I know that lim n-> infinity |n/-n|=1 and n/-n =-1 but they both converges. No luck on the convergent, divergent scenario. (Keep in mind this is just some random idea about this problem from my head). What do you think? A series is said to be convergent if you can show that it satisfies a convergence test. Testing for absolute convergence is merely one of those tests. In other words, if a function is absolutely convergent, it is convergent, however, if a function is convergent it may or may not be absolutely convergent. A perfect example is the alternating harmonic series. It is convergent and has a value of $\displaystyle \ln{2}$, but it is not absolutely convergent, because the series of absolute values is the harmonic series, which is divergent. All times are GMT -8. The time now is 10:00 AM.
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http://mathoverflow.net/revisions/61079/list
## Return to Answer UPDATE: There is an error at the end of this answer, addressed here and in comments. I am leaving the original answer here because it is discussed elsewhere. Ugh, this turned out to be a nasty bit of brute forcing and I found myself grappling with notation before I got it (I think) right. Write $J := (j_1,\dots,j_N)$ and $p^\otimes := p^{(1)} \otimes \dots \otimes p^{(N)}$ for the tensor product measure. Besides the obvious variations of the above notation, write `$f^\otimes_{J^\land_m} := \prod_{\ell \ne m} f^{(\ell)}_{j_\ell}$`. Finally, write $\partial_\ell \equiv \partial_{f_\ell}$. Now for $\mbox{Var}_p \ne 0$ we have `$\partial_\ell \left( \mathcal{D}/\mbox{Var}_p \right) = 0 \iff -\partial_\ell \mathcal{D} \cdot \mbox{Var}_p + \mathcal{D} \cdot \partial_\ell \mbox{Var}_p = 0$` and `$\partial_\ell \mathcal{D} = -\sum_j \left( p_j Q_{j\ell} + p_\ell Q_{\ell j} \right) f_j,$` while if we assume w/l/o/g that $\sum_j p_j f_j = 0$ throughout then `$\partial_\ell \mbox{Var}_p = 2p_\ell f_\ell$`. It follows that the equalities in the first equation above are equivalent to `$\sum_j (p_j Q_{j\ell} + p_\ell Q_{\ell j})f_j \cdot \mbox{Var}_p(f) + 2\mathcal{D}(f) \cdot p_\ell f_\ell = 0$`. For $F: [\dim Q^\otimes] \rightarrow \mathbb{R}$ (not necessarily of the form $f^\otimes \equiv f^{\otimes N}$), set `$(F^\land_{L^\land_m})_j := F_{\ell_1,\dots,\ell_{m-1},j,\ell_{m+1},\dots,\ell_N}$`. Now if $Q^{(m)} = c_m Q$ then $\mathcal{D}^\otimes(F) = \frac{1}{2} \sum_{JK} p^\otimes_J Q^\otimes_{JK} (F_J - F_K)^2 = \frac{1}{2} \sum_{m,J^\land_m,j_m,k} p^\otimes_{J^\land_m} p_{j_m} c_m Q_{j_m k} \left ( (F^\land_{J^\land_m})_{j_m} - (F^\land_{J^\land_m})_{k} \right )^2$ $= \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} \mathcal{D}(F^\land_{J^\land_m})$ and `$\partial_L \mathcal{D}^\otimes(F) = -\sum_J (p^\otimes_J Q^\otimes_{JL} + p^\otimes_L Q^\otimes_{LJ}) F_J = -\sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})(F^\land_{L^\land_m})_j$`. Since $([f^\otimes]^\land_{L^\land_m})_j = f_j \cdot f^\otimes_{L^\land_m}$ and $\mathcal{D}([f^\otimes]^\land_{J^\land_m}) = (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f)$ it follows from the above that `$\left ( -\partial_L \mathcal{D}^\otimes \cdot \mbox{Var}_{p^\otimes} + \mathcal{D}^\otimes \cdot \partial_L \mbox{Var}_{p^\otimes} \right ) |_{f^\otimes}$` `$= \sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot f^\otimes_{L^\land_m} \cdot \mbox{Var}_p^N(f) + \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f) \cdot 2 p^\otimes_L f^\otimes_L$` ```$= \mbox{Var}_{p}^{N-1}(f) \cdot \sum_m c_m p^\otimes_{L^\land_m} f^\otimes_{L^\land_m} \cdot \left [ \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot \mbox{Var}_{p}(f) + 2\mathcal{D}(f) \cdot p_{\ell_m} f_{\ell_m} \right ]$```. Comparison with above shows that the term in brackets is zero iff `$\partial_{\ell_m} \left( \mathcal{D}/\mbox{Var}_{p} \right) = 0$`. Inspection of the above equations also shows that `$\mathcal{D}^\otimes/\mbox{Var}_{p^\otimes}$` has a local minimum (resp. maximum) at $f^\otimes$ iff $\mathcal{D}/\mbox{Var}_p$ has a local minimum (resp. maximum) at $f$. Finally, the uniqueness of extrema follows from viewing $\mathcal{D}/\mbox{Var}_p$ as the (well-behaved) quotient of two homogeneous quadratic polynomials in the entries of $f$. 1 [made Community Wiki] Ugh, this turned out to be a nasty bit of brute forcing and I found myself grappling with notation before I got it (I think) right. Write $J := (j_1,\dots,j_N)$ and $p^\otimes := p^{(1)} \otimes \dots \otimes p^{(N)}$ for the tensor product measure. Besides the obvious variations of the above notation, write `$f^\otimes_{J^\land_m} := \prod_{\ell \ne m} f^{(\ell)}_{j_\ell}$`. Finally, write $\partial_\ell \equiv \partial_{f_\ell}$. Now for $\mbox{Var}_p \ne 0$ we have `$\partial_\ell \left( \mathcal{D}/\mbox{Var}_p \right) = 0 \iff -\partial_\ell \mathcal{D} \cdot \mbox{Var}_p + \mathcal{D} \cdot \partial_\ell \mbox{Var}_p = 0$` and `$\partial_\ell \mathcal{D} = -\sum_j \left( p_j Q_{j\ell} + p_\ell Q_{\ell j} \right) f_j,$` while if we assume w/l/o/g that $\sum_j p_j f_j = 0$ throughout then `$\partial_\ell \mbox{Var}_p = 2p_\ell f_\ell$`. It follows that the equalities in the first equation above are equivalent to `$\sum_j (p_j Q_{j\ell} + p_\ell Q_{\ell j})f_j \cdot \mbox{Var}_p(f) + 2\mathcal{D}(f) \cdot p_\ell f_\ell = 0$`. For $F: [\dim Q^\otimes] \rightarrow \mathbb{R}$ (not necessarily of the form $f^\otimes \equiv f^{\otimes N}$), set `$(F^\land_{L^\land_m})_j := F_{\ell_1,\dots,\ell_{m-1},j,\ell_{m+1},\dots,\ell_N}$`. Now if $Q^{(m)} = c_m Q$ then $\mathcal{D}^\otimes(F) = \frac{1}{2} \sum_{JK} p^\otimes_J Q^\otimes_{JK} (F_J - F_K)^2 = \frac{1}{2} \sum_{m,J^\land_m,j_m,k} p^\otimes_{J^\land_m} p_{j_m} c_m Q_{j_m k} \left ( (F^\land_{J^\land_m})_{j_m} - (F^\land_{J^\land_m})_{k} \right )^2$ $= \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} \mathcal{D}(F^\land_{J^\land_m})$ and `$\partial_L \mathcal{D}^\otimes(F) = -\sum_J (p^\otimes_J Q^\otimes_{JL} + p^\otimes_L Q^\otimes_{LJ}) F_J = -\sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})(F^\land_{L^\land_m})_j$`. Since $([f^\otimes]^\land_{L^\land_m})_j = f_j \cdot f^\otimes_{L^\land_m}$ and $\mathcal{D}([f^\otimes]^\land_{J^\land_m}) = (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f)$ it follows from the above that `$\left ( -\partial_L \mathcal{D}^\otimes \cdot \mbox{Var}_{p^\otimes} + \mathcal{D}^\otimes \cdot \partial_L \mbox{Var}_{p^\otimes} \right ) |_{f^\otimes}$` `$= \sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot f^\otimes_{L^\land_m} \cdot \mbox{Var}_p^N(f) + \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f) \cdot 2 p^\otimes_L f^\otimes_L$` ```$= \mbox{Var}_{p}^{N-1}(f) \cdot \sum_m c_m p^\otimes_{L^\land_m} f^\otimes_{L^\land_m} \cdot \left [ \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot \mbox{Var}_{p}(f) + 2\mathcal{D}(f) \cdot p_{\ell_m} f_{\ell_m} \right ]$```. Comparison with above shows that the term in brackets is zero iff `$\partial_{\ell_m} \left( \mathcal{D}/\mbox{Var}_{p} \right) = 0$`. Inspection of the above equations also shows that `$\mathcal{D}^\otimes/\mbox{Var}_{p^\otimes}$` has a local minimum (resp. maximum) at $f^\otimes$ iff $\mathcal{D}/\mbox{Var}_p$ has a local minimum (resp. maximum) at $f$. Finally, the uniqueness of extrema follows from viewing $\mathcal{D}/\mbox{Var}_p$ as the (well-behaved) quotient of two homogeneous quadratic polynomials in the entries of $f$.
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http://mathoverflow.net/questions/55085/new-proofs-to-major-theorems-leading-to-new-insights-and-results/55093
## New proofs to major theorems leading to new insights and results? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am wondering, historically, when has a new proof of an old theorem been particularly fruitful. A few examples I have in mind (all number theoretic) are: First example is classical... which is Euler's proof of Euclid's theorem which asserts that there exist infinitely many primes. Here is when the factorization $\displaystyle \prod_p (1-p^{-s})^{-1} = \sum_{n=1}^\infty \frac{1}{n^s}$ was first introduced, leading of course to what is now known as the Riemann Hypothesis. Second example is when Hardy and Littlewood gave an alternative proof of Waring's problem, which was done by Hilbert earlier. Their proof introduced what is now known as the Hardy-Littlewood Circle Method and gave an exact asymptotic for the Waring bases, which is stronger than Hilbert's result which only asserted that every sufficiently large positive integer can be written as the sum of a bounded number of $k$th powers. Later on the Hardy-Littlewood method proved very fruitful in other results, namely Vinogradov's Theorem asserting that every sufficiently large odd positive integer can be written as the sum of three primes. Third example is Tim Gowers' alternate proof to Szemeredi's Theorem asserting that every subset of the positive integers with positive upper density contains arbitrarily long arithmetic progressions. This advance, namely the introduction of Gowers uniformity norms, led eventually to the Green-Tao Theorem proving the existence of arbitrarily long arithmetic progressions in the primes. So I am wondering if there exist other incidences (number theory related or not) where a new proof really gave legitimate new insights, perhaps even a proof of a (major) new result. Edit: I am primarily interested in examples where a new proof sparked off a new direction in research. This is best supported by having a major new theorem proved using techniques inspired by the new proof. An example of something that I am not interested in is something like Donald Newman's proof of the prime number theorem, which while elegant and 'natural' as he puts it, has saw limited generalization to other areas and one is hard pressed to apply the same technique to other problems. - 9 This question seems a little broad to me. – Qiaochu Yuan Feb 10 2011 at 23:31 6 Vojta's proof of Faltings's theorem. – David Hansen Feb 10 2011 at 23:37 9 Community wiki? – J.C. Ottem Feb 10 2011 at 23:53 4 I would guess that any new proof of major results require new insights. – Hailong Dao Feb 11 2011 at 2:09 2 Please remember to mark posts asking for lists of answers as community wiki! I've hit this one with the wiki hammer. – Scott Morrison♦ Feb 11 2011 at 4:38 ## 9 Answers Here are a few examples from the 19th century. 1. Unsolvability of the quintic equation. Abel (1826) proved this by algebraic ingenuity, but without clarifying the concepts involved. Galois (1830) gave a proof that introduced the concepts of group, normal subgroup, and solvability (of groups), thus laying the foundations of group theory and Galois theory. 2. Double periodicity of elliptic functions. Abel and Jacobi established this (1820s) mainly by computation. Riemann (1850s) put elliptic functions on a clear conceptual basis by showing that the underlying elliptic curve is a torus, and that the periods correspond to independent loops on the torus. 3. Riemann-Roch theorem. Riemann (1857) discovered this theorem using Riemann surfaces, but applying physical intuition (the "Dirichlet principle"). This principle was not made rigorous until 1901. In the meantime, Dedekind and Weber (1882) gave the first rigorous and complete proof of Riemann-Roch, by reconstructing the theory of Riemann surfaces algebraically. In the process they paved the way for modern algebraic geometry. - I'd like to offer a slightly different related perspective on your second example: Legendre spent a great deal of his life mucking about with elliptic integrals, but Abel and Jacobi realized that it was much cleaner to work with their inverses, the elliptic functions! – David Hansen Feb 12 2011 at 5:45 David, you make a valid point, but it is harder to view the step beyond Legendre taken by Abel and Jacobi as a new proof of a known theorem. – John Stillwell Feb 12 2011 at 6:45 ### 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. A very nice example in my eyes is Serre's proof of Riemann-Roch: Sometimes, you are just not satisfied with existing proofs, and you look for better ones, which can be applied in different situations. A typical example for me was when I worked on the Riemann-Roch theorem (circa 1953), which I viewed as an "Euler-Poincare" formula (I did not know then that Kodaira-Spencer had had the same idea.) My first objective was to prove it for algebraic curves - a case which was known for about a century! But I wanted a proof in a special style; and when I managed to find it, I remember it did not take me more than a minute or two to go from there to the 2-dimensional case (which had just been done by Kodaira). He is speaking, of course, of the sheaf-theoretic proofs, which are usually presented today. This was the period where he was working on FAC, GAGA and his duality theorem, which revolutionized algebraic geometry. - Witten's supersymmetric proof of the Atiyah-Singer index theorem. - 2 Could you give a reference. – MBN Feb 11 2011 at 15:35 This is not the original reference: MR0836727 (87h:58207) Getzler, Ezra . A short proof of the local Atiyah-Singer index theorem. Topology 25 (1986), no. 1, 111--117 – Bruce Westbury Feb 16 2011 at 12:57 The Manin-Mumford Conjecture, first proved by Raynaud in 1983, states that the points on a curve $X$ of genus 2 or more that are torsion when embedded into its Jacobian are finite in number. The bound is also independent of how you embed the curve (that is, it is independent of the "base point"). Ken Ribet reproved this using the notion of an "almost rational torsion point". This is cool because it can lead to explicit versions of Manin-Mumford. The idea is that the set we're interested in is contained in the set of almost rational torsion points, which Ribet proves is finite, together with hyperelliptic branch points if the curve happens to be hyperelliptic. So finding the almost rational torsion on jacobians of curves might help make things explicit. Doing this for modular jacobians, Ribet proved that `$X_0(N) \cap J_0(N)_{tors} = \{0,\infty\}$`, unless $X_0(N)$ is hyperelliptic, in which case you simply add the branch points. To be honest I should say that this theorem was first proved by Matthew Baker (who actually gives two proofs), and independently around the same time by A. Tamagawa. Ribet's approach is similar to Bakers' second approach. For a survey, google "Torsion points on modular curves and Galois theory" to summon a pdf by Ribet and Minhyong Kim. - The classic example from mathematical physics is Richard Feynman's Space-Time approach to nonrelativistic quantum mechanics (1948), which (in essence) proved that the Green function of the Schroedinger equation was equal to a path integral. The article begins: It is a curious historical fact that modern quantum mechanics began with two quite different mathematical formulations: the differential equation of Schroedinger, and the matrix algebra of Heisenbert. [...] This paper will describe what is essentially a third formulation of non-relativistic quantum theory. As for the value of seeking multiple derivations, we have Feynman's Nobel Address The Development of the Space-Time View of Quantum Electrodynamics (1965): There is always another way to say the same thing that doesn't look at all like the way you said it before. I don't know what the reason for this is. I think it is somehow a representation of the simplicity of nature. [...] Perhaps a thing is simple if you can describe it fully in several different ways without immediately knowing that you are describing the same thing. In a classical context, we have Saunders Mac Lane in Hamiltonian mechanics and geometry (1970) presenting new geometric analyses of old dynamical problems: Mathematical ideas do not live fully till they are presented clearly, and we never quite achieve that ultimate clarity. Just as each generation of historians must analyse the past again, so in the exact sciences we must in each period take up the renewed struggle to present as clearly as we can the underlying ideas of mathematics. In the mid-1970s these various derivations came together as Fadeev and Popov's (1974) Covariant quantization of the gravitational field, which provided the foundations for todays' gold-standard method of BRST quantization, for which van Holten's Aspects of BRST quantization (2002) is a good review: Quite often the preferred dynamical equations of a physical system are not formulated directly in terms of observable degrees of freedom, but in terms of more primitive quantities [...] Out of these roots has grown an elegant and powerful framework for dealing with quite general classes of constrained systems using ideas borrowed from algebraic geometry. By this 90-year process of successive rederivations, we nowadays have arrived at a more nearly global appreciation—encompassing both classical and quantum dynamics—of the ideas that Terry Tao's essay What is a Gauge? discusses. Cutting-edge research in classical, quantum, and (increasingly common) hybrid dynamical systems uses all of these mathematical approaches, each formally equivalent to all the others ... but with very different ideas behind them. The resulting naturality has lent new passion to the longstanding romance between mathematics and physics. - 3 Heisenbert!! I love that guy. – David Hansen Feb 11 2011 at 5:54 3 I personally prefer Heisenernie, but to each his own. – Igor Rivin Feb 11 2011 at 6:25 2 Lol ... I was gonna fix it ... but once I discovered that apparently no one named "Heisenbert" has ever written any scientific article (at least, none indexed on MathSciNet, Inspec, or Pubmed) ... well, I couldn't bring myself to terminate the poor guy ... so his name now is a link to a picture. Like Edward Abbey's character George Hayduke, like WWII's Kilroy, and like the immortal Bourbaki ... "Heisenbert lives!" :) – John Sidles Feb 11 2011 at 6:59 Hrushovski's Model-Theoretic proof for the Mordell-Lang conjecture over function fields is an example. - Hilbert basis theorem, a nonconstructive proof replacing and generalizing a monstrous explicit calculation ("this is not mathematics, this is theology"). - Perhaps, this can be expanded a bit. In classical invariant theory, the generators of the rings of invariants were found by explicit calculation. Whereas Hilbert showed that finite sets of generators must exist on a priori grounds. This was indeed a paradigm shift. – Donu Arapura Feb 11 2011 at 15:58 I guess that this may not be an answer to the question however. – Donu Arapura Feb 11 2011 at 16:05 If you're interested in something that's expected to do this (here, with a related paper here), but is a very current project, there's Lazic's proof of the finite generation of (log) canonical rings. I don't know of any great insights gained from the new proof, other than the surprising fact that it's POSSIBLE to prove it this way, and that the method, rather than requiring the Mori program to prove the theorem, allows a proof of many important theorems in the Mori program from it. This is, though, quite a work in progress. - Serre's construction of the $p$-adic zeta function, using the fact that the values of the Riemann zeta function at negative integers are constant terms of Eisenstein series. This paved the way for a lot a mathematics including the proof of Iwasawa main conjecture by Mazur and Wiles. -
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http://mathhelpforum.com/advanced-applied-math/18815-temperature-room.html
# Thread: 1. ## Temperature in a Room Say I have a rectangular room (so its volume is easy to find) and an air conditioner that cools this room. I know the initial state of the room without the air conditioner and cool this room eventually to the same temperature as the air conditioner. I turn off the air conditioner. How much time does it take for the room to heat up to its initial state? My idea: Maybe we can use Newton's Law of Cooling (or Heating) here somehow? But I think the volume has something to do with the solution to this problem. 2. Originally Posted by ThePerfectHacker Say I have a rectangular room (so its volume is easy to find) and an air conditioner that cools this room. I know the initial state of the room without the air conditioner and cool this room eventually to the same temperature as the air conditioner. I turn off the air conditioner. How much time does it take for the room to heat up to its initial state? My idea: Maybe we can use Newton's Law of Cooling (or Heating) here somehow? But I think the volume has something to do with the solution to this problem. You would normaly assume that the rate of change of temprature with the air conditioner turned off is: $<br /> \frac{dT_{room}}{dt}=k(T_{ambient}-T_{room})<br />$ and (oversimplifying) that when the airconditioner is on that: $<br /> \frac{dT_{room}}{dt}=k(T_{ambient}-T_{room})-K<br />$ The essential assumption is that the air conditioner removes heat at a constant rate, while heat flows into the room at a rate proportional to the temprature difference. The problem is that the thermal capacity and heat loss constant for a room are difficult things to calculate. There are tools to do so but I doubt you will want to look for them. RonL 3. Originally Posted by ThePerfectHacker Say I have a rectangular room (so its volume is easy to find) and an air conditioner that cools this room. I know the initial state of the room without the air conditioner and cool this room eventually to the same temperature as the air conditioner. I turn off the air conditioner. How much time does it take for the room to heat up to its initial state? My idea: Maybe we can use Newton's Law of Cooling (or Heating) here somehow? But I think the volume has something to do with the solution to this problem. And a spherical room would be much simpler to work with. -Dan
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http://math.stackexchange.com/questions/141579/associativity-of-norms-in-inseparable-extensions
# Associativity of norms in inseparable extensions Let $K$ be a field. Let $L/K$ and $E/L$ be finite extensions. Let $α$ be an element of E. Let $N_{E/K}(α)$ be the norm of $α$, i.e. the determinant of the regular representaion matrix of $α$. It is well known that $N_{E/K}(α)$ $=$ $N_{L/K}(N_{E/L}(α))$ if $E/K$ is separable. I tried to find a proof of this formula in inseparable extensions, but failed. Where can I find it? It'd be also nice if someone provides a sketch of the proof, here. - ## 1 Answer The proof can be found here. - Thanks. That's a nice proof. – Makoto Kato May 8 '12 at 2:06 By the way, I think the proof of Lemma 1.1 in the above link is incorrect. – Makoto Kato May 8 '12 at 2:38 – Makoto Kato May 14 '12 at 0:58
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http://math.stackexchange.com/questions/7511/functions-that-are-their-own-nth-derivatives-for-real-n/7517
# Functions that are their Own nth Derivatives for Real n Consider (non-trivial) functions that are their own nth derivatives. For instance $\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x$ $\frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-x} = e^{-x}$ $\frac{\mathrm{d}^3}{\mathrm{d}x^3} e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2}) = e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2})$ $\frac{\mathrm{d}^4}{\mathrm{d}x^4} \sin x = \sin x$ $\cdots$ Let $f_n(x)$ be the function that is it's own nth derivative. I believe (but I'm not sure) for nonnegative integer n, this function can be written as the following infinite polynomial: $f_n(x) = 1 + \cos(\frac{2\pi}{n})x + \cos(\frac{4\pi}{n})\frac{x^2}{2!} + \cos(\frac{6\pi}{n})\frac{x^3}{3!} + \cdots + \cos(\frac{2t\pi}{n})\frac{x^t}{t!} + \cdots$ Is there some sense in which this function can be extended to real n using fractional derivatives? Would it then be possible to graph $z(n, x) = f_n(x)$, and would this function be smooth and continuous on both n and x axes? Or would it have many discontinuities? - You need to specify, by the way, what particular variant of 'fractional derivatives' you have in mind... – Mariano Suárez-Alvarez♦ Oct 22 '10 at 16:55 – mellamokb Oct 22 '10 at 17:38 ## 5 Answers The set of functions $f$ such that $f^{(n)}-f=0$ is a vector space of dimension $n$, spanned by the functions $e^{\lambda t}$ with $\lambda$ an $n$th root of unity. In particular, there are many such functions, not just one: the general such function is of the form $$f(t)=\sum_{k=0}^{n-1}a_ke^{\frac{2\pi i k}{n}t}.$$ This is explained in every text on ordinary diffential equations; I remember fondly, for example, Theory of Ordinary Differential Equations by Earl A. Coddington and Norman Levinson, but I am sure you can find more modern expositions in every library. - Wow, thanks! I might have to pick up a copy of a textbook like that. I'm going off my basic calculus background into an area I know almost nothing about. – mellamokb Oct 22 '10 at 17:15 2 Since $e^{\frac{2 \pi i k}{n}}$ are the $n$th roots of unity for $0 \le k < n$, shouldn't this be $f(t) = \sum_{k=0}^{n-1}{ a_k e^{e^{\frac{2 \pi i k }{n}} t}}$ ? – I. J. Kennedy Oct 23 '10 at 12:51 The solutions to a homogeneous linear equation with constant coefficients $$a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0$$ correspond to roots of the "auxiliary polynomial" $a_nt^n + \cdots + a_0$. If the polynomial has roots $r_1,\ldots,r_k$ (in complex numbers), with multiplicities $a_1,\ldots,a_k$, then a basis for the solutions is given by $${ e^{r_1x}, xe^{r_1x},\ldots, x^{a_1-1}e^{r_1x},e^{r_2x},\ldots,x^{a_k-1}e^{r_kx}}.$$ Here, the complex exponential is used, so that if $a$ and $b$ are real numbers and $i$ is the square root of $-1$, we have $$e^{a+bi} = e^a(\cos(b) + i \sin(b)).$$ In your case, you are looking at the polynomial $t^n - 1 = 0$, whose roots are the $n$th roots of unity. They are all distinct; so you can either take the complex exponentials, or the real and complex parts. So if $\lambda$ is a primitive $n$th root of $1$, then a basis for the space of solutions is $$e^x, e^{\lambda x}, e^{\lambda^2 x},\ldots,e^{\lambda^{n-1}x}.$$ The general solution is a linear combination of these with complex coefficients: $$f(x) = a_0e^x + a_1e^{\lambda x} + a_2e^{\lambda^2x}+\cdots+a_{n-1}e^{\lambda^{n-1}x},\qquad a_0,\ldots,a_{n-1}\in\mathbb{C}.$$ If you don't want complex values, you can take a general form as above, and take the real and complex parts separately. - How would you go about generalizing this to fractional derivatives, and further finding a distinct solution for each value of n? – mellamokb Oct 22 '10 at 17:40 @mellamokb: first question: following what you write in response to Mariano, you are still dealing with the nullspace of an operator (but instead of the operator being a polynomial in $D$, the differential operator, you are dealing with an operator $T$ whose $n$th power is $D$ for some $n$), so the solution set is still a vector space described by a basis. You would need to determine the dimension and appropriate basis, presumably, but I've never done fractional derivatives so I cannot tell you. – Arturo Magidin Oct 22 '10 at 18:25 @mellamokb: Second question: you have the basis: if you pick a primitive $n$-th root of unity $\lambda$, then $e^{\lambda t}$ will be different for each $n$, which gives you a distinct solution for each $n$. It is still not the only solution, though. – Arturo Magidin Oct 22 '10 at 18:26 What you are interested here are what Spanier and Oldham term "cyclodifferential functions", functions that are regenerated after being differintegrated to the appropriate order (or to put it another way, functions that are eigenfunctions of the differintegration operator). For the cyclodifferential equation $${}_0 D_x^{\alpha}y=y$$ (or in more familiar notation, $$\frac{\mathrm{d}^{\alpha}y}{\mathrm{d}x^{\alpha}}=y$$, but the problem with this notation in the setting of differintegrals is that it neglects to take the lower limit that is present in both the Caputo and Riemann-Liouville definitions into account). as you might have seen, $\exp(x)$ is a cyclodifferential function for ${}_0 D_x^1 y$ (i.e. $\exp(x)$ is its own derivative), $\cosh(x)$ and $\sinh(x)$ are cyclodifferentials for ${}_0 D_x^2 y$ (differentiating those two functions twice gives you the originals); $$\frac1{\sqrt{\pi x}}+\exp(x)\mathrm{erfc}(-\sqrt{x})$$ ($\mathrm{erfc}(x)$ is the complementary error function) is a cyclodifferential for ${}_0 D_x^{\frac12} y$ (it is its own semiderivative), and in general $$x^{\alpha-1}\sum_{j=0}^\infty \frac{C^j x^{\alpha j}}{\Gamma(\alpha(j+1))}$$ for $\alpha > 0$ and $C$ an appropriate eigenvalue, is a cyclodifferential for ${}_0 D_x^{\alpha}$. - After you read Spanier/Oldham, the books of Miller/Ross and Podlubny might be of interest. – J. M. Oct 22 '10 at 23:44 Sorry to give yet another answer that does not address the issue of fractional $n$ [it seems that fractional derivatives are not such a familiar topic to many research mathematicians; certainly they're not to me], but: There is a little issue here which has not been addressed. By the context of the OP's question, I gather s/he is looking for real-valued functions which are equal to their $n$th derivative (and not their $k$th derivative for $k < n$). Several answerers have mentioned that the set of solutions to $f^{n} = 0$ forms an $n$-dimensional vector space. But over what field? It is easier to identify the space of such complex-valued functions, i.e., $f: \mathbb{R} \rightarrow \mathbb{C}$: namely, a $\mathbb{C}$-basis is given by $f(x) = e^{2 \pi i k/n}$ for $0 \leq k < n$. But what does this tell us about the $\mathbb{R}$-vector space of real-valued solutions to this differential equation? The answer is that it is $n$-dimensional as a $\mathbb{R}$-vector space, though it does not have such an immediately obvious and nice basis. Let $W$ be the $\mathbb{R}$-vector space of real-valued functions $f$ with $f^{(n)} = 0$ and $V$ the $\mathbb{C}$-vector space of $\mathbb{C}$-valued functions $f$ with $f^{(n)} = 0$. There is a natural inclusion map $W \mapsto V$. Phrased algebraically, the question is whether the induced map $L: W \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V$ is an isomorphism of $\mathbb{C}$-vector spaces. In other words, this means that any given $\mathbb{R}$-basis of $W$ is also a $\mathbb{C}$-basis of $V$. This is certainly not automatic. For instance, viewing the Euclidean plane as first $\mathbb{R}^2$ and second as $\mathbb{C}$ gives a map $\mathbb{R}^2 \rightarrow \mathbb{C}$ which certainly does not induce an isomorphism upon tensoring with $\mathbb{C}$, since the first space has (real) dimension $2$ but the second space has (complex) dimension $1$. For more on this, see Theorem 1.6 of http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf It turns out that this is actually a problem in Galois descent: according to Theorem 2.14 of the notes of Keith Conrad already cited above, the map $L$ is an isomorphism iff there exists a conjugate-linear involution $r: V \rightarrow V$, i.e., i.e., a map which is self-inverse and satisfies, for any $z \in \mathbb{C}$ and $v \in V$, $r(zc) = \overline{z} r(c)$. But indeed we have such a thing: an element of $V$ is just a complex-valued function $f$, so we put $r(f) = \overline{f}$. Note that this stabilizes $V$ since the differential equation $f^{(n)}) = 0$ "is defined over $\mathbb{R}$": or more simply, the complex conjugate of the $n$th derivative is the $n$th derivative of the complex conjugate. Thus we have "descent data" (or, in Keith Conrad's terminology, a G-structure) and the real solution space has the same dimension as the complex solution space. It is a nice exercise to use these ideas to construct an explicit real basis of $W$. - 1 "There is a little issue here which has not been addressed." Actually, I see it was addressed in Arturo's answer, but clearly I was looking for an excuse to talk about Galois descent... – Pete L. Clark Oct 24 '10 at 0:08 Heh :) ${}$ – Mariano Suárez-Alvarez♦ Mar 15 '11 at 4:34 When you say non-trivial it looks like you mean such that the $m^{th}$ derivative of $f$ is not equal to $f$ for every $m < n$. Unfortunately, this is still not enough to define $f_n$ uniquely; for any positive integer $1 \le k \le n$ such that $\gcd(k, n) = 1$, any nontrivial linear combination of the $\phi(n)$ functions $e^{ \frac{2\pi i k}{n} x}$ works. One way to make a consistent choice across all values of $n$ is to take $k = 1$; in that case, you might want to choose $$f_n(x) = e^{ \frac{2\pi i}{n} x } = \cos \frac{2\pi x}{n} + i \sin \frac{2\pi x}{n}.$$ This is well-defined for any complex number $n \neq 0$ and all complex numbers $x$, and it is probably true that the $n^{th}$ fractional derivative of $f_n$ is equal to $f_n$. There is an essential singularity at $n = 0$ for fixed nonzero $x$. - Thanks for the clear explanation. That was indeed my intent for using the word 'non-trivial'. I haven't studied to any depth into theory of differential equations, so this is all novel to me. – mellamokb Oct 22 '10 at 17:32
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http://mathhelpforum.com/algebra/185809-speed-earth-around-sun.html
# Thread: 1. ## Speed of the Earth around the Sun. "The Earth revolves around the sun in an elliptical orbit, however for this problem you can assume that the orbit is circular with a radius of 9.3 * 10^7 miles. In what speed in feet per second does Earth travel around the sun?" I know this problem has something to do with angular velocity, but I'm confused as to how to start solving the problem. 2. ## Re: Astronomy problem Maybe you can use the formula: $v=\frac{2\pi R}{T}$ with $R$ is the radius of the circle and $T$ the period. 3. ## Re: Astronomy problem Originally Posted by explodingtoenails "The Earth revolves around the sun in an elliptical orbit, however for this problem you can assume that the orbit is circular with a radius of 9.3 * 10^7 miles. In what speed in feet per second does Earth travel around the sun?" I know this problem has something to do with angular velocity, but I'm confused as to how to start solving the problem. 1. You have to determine the circumference of a circle with a radius of approximately 93.000.000 miles. Convert the result into feet. (I've got 3085295313240 ft but of course the pretended accuracy of this number is ridiculous) 2. You are asked to find the duration of one year. There are a lot of different "years" in astronomy. I would take the tropical year. According to Wikipedia (have alook here: Year - Wikipedia, the free encyclopedia) "The mean tropical year is approximately 365 days, 5 hours, 48 minutes, 45 seconds" Convert this time into seconds (I've got 31,556,925 s) 3. Calculate the speed (I've got $v \approx \dfrac{3,085,300,000,000}{31,556,925}\approx 97770\ \frac{ft}s$ )
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http://mathhelpforum.com/differential-geometry/97767-differential-geometry-straight-lines-shortest-connections.html
# Thread: 1. ## Differential Geometry: Straight lines as shortest connections I started this problem, but got stuck shortly after beginning Basically, I started with the middle peice of math from the in/equation above. Since v is a constant vector, I took it out the front of the integral and used the fundamental theorem of calculus such that the integral of the derivative collapses to γ(t) with the terminals b and a. Subbing in the bounds, leads to γ(b) - γ(a), and given the equations in the equation, this simplies to v[Q-P], notice how this is different to the left hand side of the in/equality (PQ)v. Is my method on the right track, or have I gone completely off the rails? 2. Use the Cauchy-Schwartz inequality: $\frac{d\gamma}{dt}\cdot v\leq \left\vert\frac{d\gamma}{dt} \right\vert \vert v\vert =\left\vert\frac{d\gamma}{dt} \right\vert$ and integrate. Ps. To show that straight lines are distance minimizers, choose $v=\frac{1}{|PQ|}PQ$ and conclude that the minimum of curve distances is attained if $\gamma$ is a straight line. 3. Hi, thanks for the reply. Just curious as to know if there are any alternative methods, as I do not beleive Cauchy-Schwartz inequality has been covered in this unit (although in past years I have seen glimpses of it) 4. The Cauchy-Schwartz inequality is pretty basic. I don't think you need to prove it in every new course.
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http://mathforum.org/mathimages/index.php/Talk:Differentiability
# Talk:Differentiability ## Anna 6/26 Can you say that $y=x^2$ is an everywhere differentiable function? I feel like that a subtlety missing in this page. Also I'd retitle the "basic description" section to be something else, for the same reasons in Alan's comment below. ## David 6/25 I think you have a typo in the final term for the tangent line section you have an h were it should be an a. ## Alan 6/18 I think your basic information needs to be revised. If a hypothetical reader has no calculus backgrounds; the basic information would be over his or her head. The limit definition is very mathematical and could be off putting to someone not familiar with it. Saying something like "A point is differentiable if you can draw a tangent at this point...." and then have some tangent line pictures would probably be a mathematically gentler introduction. Your basic information should perhaps go into the "a more mathematical description" part of the page. Also I think it would better to have the page in the helper page format (with no main image). The parabola seems to be something ubiquitous in middle and high school curriculums, so I don't think people will be attracted to a picture of it. Everything else looks good though. ## Anna (6/9) In the first example, instead of saying that the function "skips" that point, have you considered using the term undefined? If you're trying to avoid really mathematical language, maybe say "behaves differently". Also, I don't feel like it's clear that you are taking two separate one sided limited to come up with that $\infty$ and $- \infty$. Have you considered writing out the two limits separately? Are you planning on adding a couple more sentences with examples of functions that are continuous but not differentiable, or at least pointing back to your corner function?
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http://mathoverflow.net/questions/64843?sort=oldest
## Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've been reading Hatcher's survey "A short exposition of the Madsen-Weiss theorem". In it, he outlines a nice proof of the "generalized Mumford conjecture", which asserts that the stable cohomology of the mapping class group is the same as the cohomology of $\Omega^{\infty} AG_{\infty,2}^{+}$. Here $AG_{n,m}$ is the space of affine $m$-planes in $\mathbb{R}^n$ and the "plus" sign indicates the 1-point compactification. Now, the Mumford conjecture I know and love asserts that the (rational) stable cohomology ring of the mapping class group is $\mathbb{Q}[e_1,e_2,\ldots]$, where the $e_i$ are the MMM classes. Can anyone explain to me why the rational cohomology ring of $\Omega^{\infty} AG_{\infty,2}^{+}$ is a polynomial ring with 1 generator in each even dimension? This appears to be explained in Madsen-Tillmann's paper introducing the generalized Mumford conjecture, but that paper is rather formidable and I have been unable to extract an answer to the above question from it (indeed, it doesn't really appear to be talking about the affine Grassmannian at all!). - In addition to Neil's answer: I explained this in my preprint: math.uni-bonn.de/people/ebert/papers/… (page 5-10) – Johannes Ebert May 13 2011 at 8:35 Thanks Jonannes! – Alan H May 13 2011 at 17:27 ## 1 Answer Most of this is not special to the case of $AG_{\infty,2}^+$. For any spectrum $X$, we have a Hurewicz map $h:\pi_{\ast}(X)\to H_{\ast}(X)$, which induces a map $h':\mathbb{Q}\otimes\pi_{\ast}(X)\to\mathbb{Q}\otimes H_{\ast}(X)$. Standard calculations show that $h'$ this is an isomorphism when $X=S^n$ for some $n$, and it follows by induction up the skeleta that $h'$ is an isomorphism for all $X$. Next, the homotopy groups $\pi_{\ast}(X)$ are (essentially by definition) the same as the homotopy groups of the space $\Omega^\infty(X)$, so we have an unstable Hurewicz map $h'':\pi_{\ast}(X)=\pi_{\ast}(\Omega^\infty(X))\to H_{\ast}(\Omega^\infty(X))$. By combinining these we get a map $\mathbb{Q}\otimes H_\ast(X)\to \mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$. Next, every infinite loop space is a homotopy-commutative H-space, and this makes $H_\ast(\Omega^\infty(X))$ into a graded-commutative (and graded-cocommutative) Hopf algebra. Now let $A_*$ be the free graded-commutative ring generated by $\mathbb{Q}\otimes H_\ast(X)$, with the Hopf algebra structure for which the generators are primitive. More explicitly, $A_\ast$ is a tensor product of polynomial algebras (one for each even-dimensional generator in $\mathbb{Q}\otimes H_\ast(X)$) and exterior algebras (one for each odd-dimensional generator). It is now quite formal to construct a canonical map $A_\ast\to\mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$ of bicommutative Hopf algebras. One can then show that this map is always an isomorphism. Indeed, it is possible to reduce to the case $X=S^n$ again, and the groups $\mathbb{Q}\otimes H_\ast(\Omega^k S^{n+k})$ can be calculated by repeated use of Serre spectral sequences, and one can then let $k$ tend to infinity. As $\mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$ is free on primitive generators, it is a standard fact that the dual Hopf algebra $\mathbb{Q}\otimes H^\ast(\Omega^\infty(X))$ is also free on primitive generators in the same degrees. For the Madsen-Weiss case, you now just need to know the groups $\mathbb{Q}\otimes H_\ast(AG_{\infty,2}^+)$. This is not hard, because $AG^+_{\infty,2}$ is the Thom space of a virtual vector bundle over $BSO(2)=\mathbb{C}P^\infty$, so we can use the Thom isomorphism theorem. -
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http://mathoverflow.net/questions/119377?sort=newest
Automorphism of finite groups and Hurwitz spaces Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If $G$ is a finite group, embedded as a transitive subgroup of $S_n$ for some $n$, will every automorphism of $G$ extend to an inner automorphism of $S_n$? I'm trying to connect the language that's used in many different sources on Hurwitz spaces and Nielsen classes. On Michael Fried's website, he notes that "absolute equivalence" on a Nielsen class $(g_1,\ldots,g_r)$ is an equivalence given by componentwise conjugation by an element $h\in N_{S_n}(G)$. On the other hand, in some other sources, eg Volklein's chapter on "Moduli Spaces of Covers of the Riemann Sphere" in his book Groups as Galois groups, and in Fried's paper "The Inverse Galois Problem and Rational Points on Moduli Spaces", their notion of absolute equivalence on Nielsen classes seems to allow for componentwise application of any automorphism of $G$. Also, does anyone have a pdf of Fried's 1977 paper "Fields of Definition of Function Fields and Hurwitz Families - Groups as Galois Groups". It's in the 1977 Volume of Communications in Algebra, but my university doesn't seem to have access to issues from that journal prior to 2000. thanks • will - Its definitely true that if you choose the regular representation of $G$ then every automorphism extends to an inner automorphism in $S_{|G|}$ (so $G$ has at least one transitive representation with the desired property), although in your case it seems you need the result to hold for a far wider class of transitive representations. – ARupinski Jan 20 at 2:01 3 Answers Such automorphisms need not extend to inner automorphisms of $S_n$. Take the dihedral group $D_8$ sitting transitively inside $S_4$, say as the following set of permutations: $\{(1)(2)(3)(4), (13)(2)(4), (1)(3)(24), (12)(34), (14)(23), (1234), (13)(24), (1432)\}$. Now the sets of reflections $\{(13)(2)(4), (1)(3)(24)\}$ and $\{(12)(34),(14)(23)\}$ are equivalent to one another via an outer automorphism of $D_8$, but clearly this automorphism is not inner in $S_4$. - 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. While the answers by Eric and ARupinski give negative examples for your question, here is the precise characterization for when the answer is yes: Let $\alpha$ be an automorphism of the transitive subgroup $G\le S_n$, and $G_1$ be the stabilizer of $1$ in $G$. Then $\alpha$ extends to an automorphism of $S_n$ if and only if $G_1$ and $\alpha(G_1)$ are conjugate in $S_n$. The necessity of the condition is clear, and the sufficiency is a nice exercise. I believe the result is also in the permutation groups book by Dixon-Mortimer. - No, this isn't even true if $G$ is $S_n$ itself: the symmetric group $S_6$ has an outer automorphism. -
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http://mathhelpforum.com/calculus/176425-find-limits-integral.html
Thread: 1. Find limits for integral why does the unit circle with center in (1,0) have period $-\frac{\pi}{2} , \frac{\pi}{2}$ and not $0, 2\pi$ ? 2. The problem statement here makes no sense. Can you please either give the context of the problem, or better yet, simply state the original problem? Thanks! 3. As Ackbeet said, your question makes no sense. Functions have periods, not geometric objects. 4. Ok, sorry. The integral is: $\int \int_R (x^2+y^2)^{(3/2)} dxdy$ R is the circle $(x-1)^2+y^2 \leq 1$ 5. Are you asking why the limits in the integral are $-\pi/2$ to $\pi/2?$ I'm still not completely understanding the question. 6. im supposed to find the limits and the "blueprint" says (-pi/2 to pi/2) for theta and (0 to 2cos(theta) for r. I cant see why -pi/2 to pi/2 and not 0 to 2pi.. it is a circle 7. My guess is that the limits are using some sort of symmetry condition in the integral, whereby you're only integrating over half the area, or something along those lines. What substitutions do you want to use? 8. im just suppose to solve it using polar cordinates. thats all. And im suppose to integrate over the circle $(x-1)^2+y^2\leq 1$ so i guess they mean the whole circle. the only thing is that it has its center at (1,0) and not (0,0). $(r\cos(\theta)-1)^2 + r^2\sin^2(\theta) \leq 1 \Rightarrow r \leq 2\cos(\theta)$, so thats for r, but I cant see how they get to $-\pi/2 \leq \theta \leq \pi/2$ 9. Ignore my comment in post # 7. The reason the angle limits are what they are is that the entire circle is in the right half-plane, which is described by those limits, and not 0 to 2pi. 10. ahh.. ok . that made sense. so if the circle had been in (0,1) (the upper-pane) the limits would have been 0 to pi? 11. Assuming a radius of 1, you are correct. 12. ok, thank you for your time! 13. You're welcome! Have a good one.
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http://math.stackexchange.com/questions/223032/smooth-monotone-mathbbr3-curve-with-constant-nontrivial-curvature?answertab=votes
# Smooth Monotone $\mathbb{R}^3$ curve with constant (nontrivial) curvature So I was trying to construct a closed curve in $\mathbb{R}^3$ with constant positive curvature and non-trivial torsion. To do this I tried to glue two helices together in a smooth way with a curve that is: Smooth, Monotone, and has the same curvature as a helix $(\cos(t),\sin(t),t)$. Anyway this type of curve should exist but I cannot construct it.. Alternativly, I was thinking could we reconstruct the curve from its torion and curvature functions; since they determine a unique curve (up to rigid motion) in Euclidean space. If so, the curve would have to satisfy $k(s)=1/\sqrt 2$ and $t(s)=1-2s$. Many thanks in advance! :) - ## 1 Answer Any torus knot works. That is, any torus knot embedded with constant slope. -
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http://math.stackexchange.com/tags/classifying-spaces/hot
Tag Info Hot answers tagged classifying-spaces 13 What functor does $K(G, 1)$ represent for nonabelian $G$? K(G,1) aka BG classifies G-bundles — i.e. G-coverings, if G is discrete. (Details can be found e.g. in May's Concise Course in Algebraic Topology.) Usual definition of Cech cohomology works for $H^1(X;G)$ even in non-abelian case (but it's just the usual cocycle definition of G-bundle). As for universal coefficient theorem, even if $H_1(X;\mathbb Z)$ is ... 9 Why is the cohomology of a $K(G,1)$ group cohomology? Akhil, you're thinking of this the opposite of how I think group cohomology was discovered. The concept of group cohomology originally centered around the questions about the (co)homology of $K(\pi,1)$-spaces, by people like Hopf (he called them aspherical rather than $K(\pi,1)$ spaces, and Hopf preferred homology to cohomology at that point). I think the ... 5 Finite dimensional Eilenberg-Maclane spaces There is never a $K(G, n)$ for $n \geq 2$ which is a finite complex (and $G$ nontrivial). In fact, a finite complex has finitely generated homotopy groups in all dimensions (by Serre's mod $\mathcal{C}$ theory applied to the universal cover). So one reduces to seeing that a $K(\mathbb{Z}, n)$ or a $K(\mathbb{Z}/p^k, n)$ cannot be a finite complex. In both ... 5 Group structure on Eilenberg-MacLane spaces The standard classifying space functor $B$ from topological groups to topological spaces is product preserving, so it takes abelian topological groups to abelian topological groups. Start with an abelian group $G$ as a discrete topological group, so a $K(G,0)$. Apply the functor $B$ iteratively $n$ times to reach $B^nG$, which is an abelian topological ... 3 Group structure on Eilenberg-MacLane spaces Well, an interval is a $K(\{e\},1)$, where $\{e\}$ is the trivial group, and an interval has no group structure that can make it a topological group (every continuous map has a fixed point). So there certainly are some $K(G,n)$'s out there that cannot be made into topological groups. Maybe if you refine the question there could be some sort of answer. 3 Why is the cohomology of a $K(G,1)$ group cohomology? Well, you'll probably want a more conceptual proof, but one thing you can do is check they are computed by the same chain complex: for $K(G,1)$ take the simplicial construction of the classifying space $BG$ and compute its cohomology in the usual way for simplicial sets (using the dual to the complex of formal linear combinations of simplices); for the group ... 3 Why is the cohomology of a $K(G,1)$ group cohomology? This is really a comment on ryan'sanswer: I have to disagree with Ryan. Group cohomology was in its early stages before Eilenberg and Maclane came along. There are awful and ugly formulations of just $H^1$ and $H^2$ that lead me to believe that they must have been formulated before E&M did their work. I am thinking of factor sets and cocycle conditions ... 2 Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference? The homotopy theoretic proof is as follows: Let $E \longrightarrow \Sigma$ be a principal $G$-bundle over a surface $\Sigma$. Such a bundle is determined by a homotopy class $[f_E] \in [\Sigma, BG]$ by classifying space theory. Since $G$ is simply connected (and presumably connected), the classifying space $BG$ is $2$-connected (i.e. connected, simply ... 2 Classifying space of the reals If we can construct a space, which we will call $EG$ such that our group acts transitively and freely, then we may set $BG=EG/G$. But if we have $\mathbb{R}$ act on $\mathbb{ER}=\mathbb{R}^2$ by $r∗(x,y)=(x+r,y)$, this will do the trick. Thus we get that $B\mathbb{R}=\mathbb{R}$. Their are other models as well. 2 Is there a classifying space for covering maps? The answer to both your questions is yes, and Qiaochu gave the basic idea. The base space is $BS_n$ and the fiber is $ES_n$. You can make this concrete (very analogous to Grassmannians) by using the model $BS_n \equiv C_n(\mathbb R^\infty) / S_n$ and $ES_n = C_n(\mathbb R^\infty)$ where $C_n$ indicates the configuration space of $n$ labelled points in ... 1 Why is the group of covering transformations relative to the quotient map isomorphic to a subgroup of the Fundamental Group? By the end of step 2, you know that $(\tilde{X},q)$ is the universal covering of $Y$, so there is a map $\Psi : \Pi_1(Y,y_0) \to CoveringMaps(\tilde{X},q)$ defined just like $\Phi$ : For any loop $\gamma : [0;1] \to Y$, $\Psi(\tilde{\gamma})(x_0)$ is the endpoint of the lift of $\gamma$ starting from $x_0$. Furthermore, $\Psi$ is a group isomorphism between ... 1 Homotopy-Fibre Sequence of Classifying Spaces Assume you have the exact sequence $$1\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 1$$Then it induces a fibration $$BH\rightarrow BG\rightarrow B(G/H)$$ as we imagine some large enough total space $EG$ whose quotient by $G$ is $BG$, by $H$ is $BH$, etc. Now assume we know $B(G/H)$ and $B(H)$ but do not know $BG$, then we need certain invariants to ... 1 How to correct a wrong proof about the Birman exact sequence? See Andy Putman's answer to this math overflow question. This at least gives you good references. To answer your question about torsion, almost all values of $s,g$, and $r$ will have torsion in the mapping class group, and moduli space will only be a rational classifying space. Just build a symmetric looking surface with those values and isometries of that ... 1 Cohomology of $\Bbb CP^{\infty}=BU_1, BU_2,\dots$ : A reference request Which class are you taking? This material is not easy. You should ask your professor to ask for a proof or some hints. There is a "simple" proof not using spectral sequences at here and is quite readable. Notice there is an obvious mistake in the proof. I hope David Speyer or someone else can give an answer on the spectral sequence part(which I do not ... Only top voted, non community-wiki answers of a minimum length are eligible
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http://mathoverflow.net/questions/57870/count-the-number-of-homogeneous-polynomials/57874
## Count the number of homogeneous polynomials ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a general way of counting the number of homogeneous polynomials of certain degree in a complex projective space or a weighted complex projective space, mod the ideal generated by some homogeneous polynomials with smaller degrees? As an example, consider the degree 18 homogeneous polynomials in $W\mathbb{P}_{[2,2,2,4]}^3$, mod the ideal generated by two degree 8 homogeneous polynomials $P_1=x^4_1$ and $P_2=x_2^4$, where $x_1$ and $x_2$ are the first and second coordinates of $W\mathbb{P}_{[2,2,2,4]}^3$. I can count the number of equivalent classes by directly listing all of such homogeneous polynomials; I would like to know if there is a more general and efficient way of doing this. Edit: to avoid possible confusion, I have replaced "polynomial" by "homogeneous polynomial". - ## 1 Answer I guess what you are after is the Hilbert function of ideal. More precisely, if you have a multigraded polynomial ring $k[x_1,\ldots,x_n]$ and the ideal is given by $I=(f_1,\ldots,f_r)$, the Hilbert function is simply the number $$h(\alpha)=\dim_k (k[x_1,\ldots,x_n]/I)_\alpha.$$Here $\alpha$ may take values in the multigrading. When $|\alpha|$ is large, the Hilbert-Serre theorem says that $h(\alpha)$ is actually a polynomial function in $\alpha$ and so the generating function is actually a rational function. There are many algorithms to compute the Hilbert function of such rings based on the theory of Gröbner basis and you could try them out in Macaulay2. In certain special cases there are other alternatives though. As in your case, you can often turn this problem into a counting problem. Note that monomials of degree $n$ in $\mathbb{P}_{2,2,2,4}$ correspond bijectively to non-negative solutions of the equation $$2a+2b+2c+4d=n$$Let $s(n)$ denote this number. Then since the ideal $I=(P_1,P_2)$ is a complete intersection of two degree 8 polynomials, we get that the dimension of polynomials of degree $n$ modulo $I$ is exactly $$s(n)-2s(n-8)+s(n-16)$$If $n=18$, we get dimension $125-2\cdot 35+3=60$. In particular, if the ideal is c.i., this argument shows that it suffices to know the number of degree $k$ polynomials in the polynomial ring. In the general case however, you might not be so lucky that your ideal is a c.i. and then perhaps Hilbert-functions are better suited. - Don't you need to add s(n-16), or am I confused? – Alexander Woo Mar 9 2011 at 1:55 I used Macaulay2 as suggested, and I got 60. I suppose we need to add s(n-16). Thank you guys. – Moduli Mar 9 2011 at 5:00 Ah, yes that's right! – J.C. Ottem Mar 9 2011 at 7:15
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http://en.wikipedia.org/wiki/Quaternions
# Quaternion (Redirected from Quaternions) Quaternion multiplication × 1 i j k 1 1 i j k i i −1 k −j j j −k −1 i k k j −i −1 In mathematics, the quaternions are a number system that extends the complex numbers. They were first described by Irish mathematician William Rowan Hamilton in 1843[1][2] and applied to mechanics in three-dimensional space. A feature of quaternions is that the product of two quaternions is noncommutative. Hamilton defined a quaternion as the quotient of two directed lines in a three-dimensional space[3] or equivalently as the quotient of two vectors.[4] Quaternions can also be represented as the sum of a scalar and a vector. Quaternions find uses in both theoretical and applied mathematics, in particular for calculations involving three-dimensional rotations such as in three-dimensional computer graphics and computer vision. They can be used alongside other methods, such as Euler angles and rotation matrices, or as an alternative to them depending on the application. In modern mathematical language, quaternions form a four-dimensional associative normed division algebra over the real numbers, and thus also form a domain. In fact, the quaternions were the first noncommutative division algebra to be discovered.[5] The algebra of quaternions is often denoted by H (for Hamilton), or in blackboard bold by $\mathbb H$ (Unicode U+210D, ℍ). It can also be given by the Clifford algebra classifications Cℓ0,2(R) ≅ Cℓ03,0(R). The algebra H holds a special place in analysis since, according to the Frobenius theorem, it is one of only two finite-dimensional division rings containing the real numbers as a proper subring, the other being the complex numbers. The unit quaternions can therefore be thought of as a choice of a group structure on the 3-sphere S3 that gives the group Spin(3), which is isomorphic to SU(2) and also to the universal cover of SO(3). Graphical representation of quaternion units product as 90°-rotation in 4D-space, ij = k, ji = −k, ij = −ji ## History Main article: History of quaternions Quaternion plaque on Brougham (Broom) Bridge, Dublin, which says: Here as he walked by on the 16th of October 1843 Sir William Rowan Hamilton in a flash of genius discovered the fundamental formula for quaternion multiplication i2 = j2 = k2 = ijk = −1 & cut it on a stone of this bridge Quaternion algebra was introduced by Hamilton in 1843.[6] Important precursors to this work included Euler's four-square identity (1748) and Olinde Rodrigues' parameterization of general rotations by four parameters (1840), but neither of these writers treated the four-parameter rotations as an algebra.[7][8] Carl Friedrich Gauss had also discovered quaternions in 1819, but this work was not published until 1900.[9][10] Hamilton knew that the complex numbers could be interpreted as points in a plane, and he was looking for a way to do the same for points in three-dimensional space. Points in space can be represented by their coordinates, which are triples of numbers, and for many years Hamilton had known how to add and subtract triples of numbers. However, Hamilton had been stuck on the problem of multiplication and division for a long time. He could not figure out how to calculate the quotient of the coordinates of two points in space. The great breakthrough in quaternions finally came on Monday 16 October 1843 in Dublin, when Hamilton was on his way to the Royal Irish Academy where he was going to preside at a council meeting. While walking along the towpath of the Royal Canal with his wife, the concepts behind quaternions were taking shape in his mind. When the answer dawned on him, Hamilton could not resist the urge to carve the formula for the quaternions, i2 = j2 = k2 = ijk = −1, into the stone of Brougham Bridge as he paused on it. On the following day, Hamilton wrote a letter to his friend and fellow mathematician, John T. Graves, describing the train of thought that led to his discovery. This letter was later published in the London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, vol. xxv (1844), pp 489–95. On the letter, Hamilton states, And here there dawned on me the notion that we must admit, in some sense, a fourth dimension of space for the purpose of calculating with triples ... An electric circuit seemed to close, and a spark flashed forth. Hamilton called a quadruple with these rules of multiplication a quaternion, and he devoted most of the remainder of his life to studying and teaching them. Hamilton's treatment is more geometric than the modern approach, which emphasizes quaternions' algebraic properties. He founded a school of "quaternionists", and he tried to popularize quaternions in several books. The last and longest of his books, Elements of Quaternions, was 800 pages long; it was published shortly after his death. After Hamilton's death, his student Peter Tait continued promoting quaternions. At this time, quaternions were a mandatory examination topic in Dublin. Topics in physics and geometry that would now be described using vectors, such as kinematics in space and Maxwell's equations, were described entirely in terms of quaternions. There was even a professional research association, the Quaternion Society, devoted to the study of quaternions and other hypercomplex number systems. From the mid-1880s, quaternions began to be displaced by vector analysis, which had been developed by Josiah Willard Gibbs, Oliver Heaviside, and Hermann von Helmholtz. Vector analysis described the same phenomena as quaternions, so it borrowed some ideas and terminology liberally from the literature of quaternions. However, vector analysis was conceptually simpler and notationally cleaner, and eventually quaternions were relegated to a minor role in mathematics and physics. A side-effect of this transition is that Hamilton's work is difficult to comprehend for many modern readers. Hamilton's original definitions are unfamiliar and his writing style was wordy and difficult to understand. However, quaternions have had a revival since the late 20th Century, primarily due to their utility in describing spatial rotations. The representations of rotations by quaternions are more compact and quicker to compute than the representations by matrices. In addition, unlike Euler angles they are not susceptible to gimbal lock. For this reason, quaternions are used in computer graphics,[11] computer vision, robotics, control theory, signal processing, attitude control, physics, bioinformatics, molecular dynamics, computer simulations, and orbital mechanics. For example, it is common for the attitude-control systems of spacecraft to be commanded in terms of quaternions. Quaternions have received another boost from number theory because of their relationships with the quadratic forms. Since 1989, the Department of Mathematics of the National University of Ireland, Maynooth has organized a pilgrimage, where scientists (including the physicists Murray Gell-Mann in 2002, Steven Weinberg in 2005, and the mathematician Andrew Wiles in 2003) take a walk from Dunsink Observatory to the Royal Canal bridge. Hamilton's carving is no longer visible. ## Definition As a set, the quaternions H are equal to R4, a four-dimensional vector space over the real numbers. H has three operations: addition, scalar multiplication, and quaternion multiplication. The sum of two elements of H is defined to be their sum as elements of R4. Similarly the product of an element of H by a real number is defined to be the same as the product in R4. To define the product of two elements in H requires a choice of basis for R4. The elements of this basis are customarily denoted as 1, i, j, and k. Every element of H can be uniquely written as a linear combination of these basis elements, that is, as a1 + bi + cj + dk, where a, b, c, and d are real numbers. The basis element 1 will be the identity element of H, meaning that multiplication by 1 does nothing, and for this reason, elements of H are usually written a + bi + cj + dk, suppressing the basis element 1. Given this basis, associative quaternion multiplication is defined by first defining the products of basis elements and then defining all other products using the distributive law. ### Multiplication of basis elements The equations $i^2=j^2=k^2=ijk=-1$, where i, j, and k are basis elements of H, determine all the possible products of i, j, and k. For example right-multiplying both sides of −1 = ijk by k gives $\begin{align} -k & = i j k k = i j (k^2) = i j (-1), \\ k & = i j. \end{align}$ All the other possible products can be determined by similar methods, resulting in $\begin{alignat}{2} ij & = k, & \qquad ji & = -k, \\ jk & = i, & kj & = -i, \\ ki & = j, & ik & = -j, \end{alignat}$ which can be expressed as a table whose rows represent the left factor of the product and whose columns represent the right factor, as shown at the top of this article. ### Hamilton product For two elements a1 + b1i + c1j + d1k and a2 + b2i + c2j + d2k, their Hamilton product (a1 + b1i + c1j + d1k)(a2 + b2i + c2j + d2k) is determined by the products of the basis elements and the distributive law. The distributive law makes it possible to expand the product so that it is a sum of products of basis elements. This gives the following expression: $a_1a_2 + a_1b_2i + a_1c_2j + a_1d_2k$ ${}+ b_1a_2i + b_1b_2i^2 + b_1c_2ij + b_1d_2ik$ ${}+ c_1a_2j + c_1b_2ji + c_1c_2j^2 + c_1d_2jk$ ${}+ d_1a_2k + d_1b_2ki + d_1c_2kj + d_1d_2k^2.$ Now the basis elements can be multiplied using the rules given above to get:[6] $a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2$ ${}+ (a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2)i$ ${}+ (a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2)j$ ${}+ (a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2)k.$ ### Ordered list form Using the basis 1, i, j, k of H makes it possible to write H as a set of quadruples: $\mathbf{H} = \{(a, b, c, d) \mid a, b, c, d \in \mathbf{R}\}.$ Then the basis elements are: $\begin{align} 1 & = (1, 0, 0, 0), \\ i & = (0, 1, 0, 0), \\ j & = (0, 0, 1, 0), \\ k & = (0, 0, 0, 1), \end{align}$ and the formulas for addition and multiplication are: $(a_1,\ b_1,\ c_1,\ d_1) + (a_2,\ b_2,\ c_2,\ d_2) = (a_1 + a_2,\ b_1 + b_2,\ c_1 + c_2,\ d_1 + d_2).$ $\begin{align} (a_1,\ b_1,\ c_1,\ d_1)&(a_2,\ b_2,\ c_2,\ d_2) = \\ & = (a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2, \\ & {} \qquad a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2, \\ & {} \qquad a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2, \\ & {} \qquad a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2). \end{align}$ ### Scalar and vector parts A number of the form a + 0i + 0j + 0k, where a is a real number, is called real, and a number of the form 0 + bi + cj + dk, where b, c, and d are real numbers, and at least one of b, c or d is nonzero, is called pure imaginary. If a + bi + cj + dk is any quaternion, then a is called its scalar part and bi + cj + dk is called its vector part. The scalar part of a quaternion is always real, and the vector part is always pure imaginary. Even though every quaternion is a vector in a four-dimensional vector space, it is common to define a vector to mean a pure imaginary quaternion. With this convention, a vector is the same as an element of the vector space R3. Hamilton called pure imaginary quaternions right quaternions[12][13] and real numbers (considered as quaternions with zero vector part) scalar quaternions. If a quaternion is divided up into a scalar part and a vector part, i.e. $q = (r,\ \vec{v}),\ q\in\mathbf{H},\ r\in\mathbf{R},\ \vec{v}\in\mathbf{R}^3$ then the formulas for addition and multiplication are: $(r_1,\ \vec{v}_1) + (r_2,\ \vec{v}_2) = (r_1 + r_2,\ \vec{v}_1+\vec{v}_2)$ $(r_1,\ \vec{v}_1) (r_2,\ \vec{v}_2) = (r_1 r_2 - \vec{v}_1\cdot\vec{v}_2, r_1\vec{v}_2+r_2\vec{v}_1 + \vec{v}_1\times\vec{v}_2)$ where "·" is the dot product and "×" is the cross product. ### Remarks #### Noncommutativity of multiplication Noncommutativity of quaternion multiplication. × 1 i j k 1 1 i j k i i −1 k −j j j −k −1 i k k j −i −1 Unlike multiplication of real or complex numbers, multiplication of quaternions is not commutative. For example, ij = k, while ji = −k. The noncommutativity of multiplication has some unexpected consequences, among them that polynomial equations over the quaternions can have more distinct solutions than the degree of the polynomial. The equation z2 + 1 = 0, for instance, has infinitely many quaternion solutions z = bi + cj + dk with b2 + c2 + d2= 1, so that these solutions lie on the two-dimensional surface of a sphere centered on zero in the three-dimensional subspace of quaternions with zero real part. This sphere intersects the complex plane at two points i and −i. The fact that quaternion multiplication is not commutative makes the quaternions an often-cited example of a strictly skew field. #### Historical impact on physics P.R. Girard’s essay The quaternion group and modern physics[14] discusses some roles of quaternions in physics. It "shows how various physical covariance groups: SO(3), the Lorentz group, the general relativity group, the Clifford algebra SU(2), and the conformal group can be readily related to the quaternion group" in modern algebra. Girard began by discussing group representations and by representing some space groups of crystallography. He proceeded to kinematics of rigid body motion. Next he used complex quaternions (biquaternions) to represent the Lorentz group of special relativity, including the Thomas precession. He cited five authors, beginning with Ludwik Silberstein who use a potential function of one quaternion variable to express Maxwell's equations in a single differential equation. Concerning general relativity, he expressed the Runge–Lenz vector. He mentioned the Clifford biquaternions (split-biquaternions) as an instance of Clifford algebra. Finally, invoking the reciprocal of a biquaternion, Girard described conformal maps on spacetime. Among the fifty references, Girard included Alexander Macfarlane and his Bulletin of the Quaternion Society. In 1999 he showed how Einstein's equations of general relativity could be formulated within a Clifford algebra that is directly linked to quaternions.[15] A more personal view of quaternions was written by Joachim Lambek in 1995. He wrote in his essay If Hamilton had prevailed: quaternions in physics: "My own interest as a graduate student was raised by the inspiring book by Silberstein". He concluded by stating "I firmly believe that quaternions can supply a shortcut for pure mathematicians who wish to familiarize themselves with certain aspects of theoretical physics."[16] #### Sums of four squares Main article: Lagrange's four-square theorem Quaternions are also used in one of the proofs of Lagrange's four-square theorem in number theory, which states that every nonnegative integer is the sum of four integer squares. As well as being an elegant theorem in its own right, Lagrange's four square theorem has useful applications in areas of mathematics outside number theory, such as combinatorial design theory. The quaternion-based proof uses Hurwitz quaternions, a subring of the ring of all quaternions for which there is an analog of the Euclidean algorithm. ## Conjugation, the norm, and reciprocal Conjugation of quaternions is analogous to conjugation of complex numbers and to transposition (also known as reversal) of elements of Clifford algebras. To define it, let q = a + bi + cj + dk be a quaternion. The conjugate of q is the quaternion a − bi − cj − dk. It is denoted by q*, q,[6] qt, or $\tilde q$. Conjugation is an involution, meaning that it is its own inverse, so conjugating an element twice returns the original element. The conjugate of a product of two quaternions is the product of the conjugates in the reverse order. That is, if p and q are quaternions, then (pq)* = q*p*, not p*q*. Unlike the situation in the complex plane, the conjugation of a quaternion can be expressed entirely with multiplication and addition: $q^* = - \frac 1 2 (q + iqi + jqj + kqk).$ Conjugation can be used to extract the scalar and vector parts of a quaternion. The scalar part of p is (p + p*)/2, and the vector part of p is (p − p*)/2. The square root of the product of a quaternion with its conjugate is called its norm and is denoted ||q||. (Hamilton called this quantity the tensor of q, but this conflicts with modern usage. See tensor.) It has the formula $\lVert q \rVert = \sqrt{qq^*} = \sqrt{q^*q} = \sqrt{a^2 + b^2 + c^2 + d^2}.$ This is always a non-negative real number, and it is the same as the Euclidean norm on H considered as the vector space R4. Multiplying a quaternion by a real number scales its norm by the absolute value of the number. That is, if α is real, then $\lVert\alpha q\rVert = |\alpha|\lVert q\rVert.$ This is a special case of the fact that the norm is multiplicative, meaning that $\lVert pq \rVert = \lVert p \rVert\lVert q \rVert.$ for any two quaternions p and q. Multiplicativity is a consequence of the formula for the conjugate of a product. Alternatively it follows from the identity $\det \Bigl(\begin{array}{cc} a+ib & id+c \\ id-c & a-ib \end{array}\Bigr) = a^2 + b^2 + c^2 + d^2,$ (where i denotes the usual imaginary unit) and hence from the multiplicative property of determinants of square matrices. This norm makes it possible to define the distance d(p, q) between p and q as the norm of their difference: $d(p, q) = \lVert p - q \rVert.$ This makes H into a metric space. Addition and multiplication are continuous in the metric topology.[clarification needed] A unit quaternion is a quaternion of norm one. Dividing a non-zero quaternion q by its norm produces a unit quaternion Uq called the versor of q: $\mathbf{U}q = \frac{q}{\lVert q\rVert}.$ Every quaternion has a polar decomposition q = ||q|| Uq. Using conjugation and the norm makes it possible to define the reciprocal of a quaternion. The product of a quaternion with its reciprocal should equal 1, and the considerations above imply that the product of q and q*/||q||2 (in either order) is 1. So the reciprocal of q is defined to be $q^{-1} = \frac{q^*}{\lVert q\rVert^2}.$ This makes it possible to divide two quaternions p and q in two different ways. That is, their quotient can be either p q −1 or q −1 p. The notation p/q is ambiguous because it does not specify whether q divides on the left or the right. ## Algebraic properties Cayley graph of Q8. The red arrows represent multiplication on the right by i, and the green arrows represent multiplication on the right by j. The set H of all quaternions is a vector space over the real numbers with dimension 4. (In comparison, the real numbers have dimension 1, the complex numbers have dimension 2, and the octonions have dimension 8.) The quaternions have a multiplication that is associative and that distributes over vector addition, but which is not commutative. Therefore the quaternions H are a non-commutative associative algebra over the real numbers. Even though H contains copies of the complex numbers, it is not an associative algebra over the complex numbers. Because it is possible to divide quaternions, they form a division algebra. This is a structure similar to a field except for the commutativity of multiplication. Finite-dimensional associative division algebras over the real numbers are very rare. The Frobenius theorem states that there are exactly three: R, C, and H. The norm makes the quaternions into a normed algebra, and normed division algebras over the reals are also very rare: Hurwitz's theorem says that there are only four: R, C, H, and O (the octonions). The quaternions are also an example of a composition algebra and of a unital Banach algebra. Because the product of any two basis vectors is plus or minus another basis vector, the set {±1, ±i, ±j, ±k} forms a group under multiplication. This group is called the quaternion group and is denoted Q8.[17] The real group ring of Q8 is a ring R[Q8] which is also an eight-dimensional vector space over R. It has one basis vector for each element of Q8. The quaternions are the quotient ring of RQ8 by the ideal generated by the elements 1 + (−1), i + (−i), j + (−j), and k + (−k). Here the first term in each of the differences is one of the basis elements 1, i, j, and k, and the second term is one of basis elements −1, −i, −j, and −k, not the additive inverses of 1, i, j, and k. ## Quaternions and the geometry of R3 Because the vector part of a quaternion is a vector in R3, the geometry of R3 is reflected in the algebraic structure of the quaternions. Many operations on vectors can be defined in terms of quaternions, and this makes it possible to apply quaternion techniques wherever spatial vectors arise. For instance, this is true in electrodynamics and 3D computer graphics. For the remainder of this section, i, j, and k will denote both imaginary[18] basis vectors of H and a basis for R3. Notice that replacing i by −i, j by −j, and k by −k sends a vector to its additive inverse, so the additive inverse of a vector is the same as its conjugate as a quaternion. For this reason, conjugation is sometimes called the spatial inverse. Choose two imaginary quaternions p = b1i + c1j + d1k and q = b2i + c2j + d2k. Their dot product is $p \cdot q = b_1b_2 + c_1c_2 + d_1d_2.$ This is equal to the scalar parts of p*q, qp*, pq*, and q*p. (Note that the vector parts of these four products are different.) It also has the formulas $p \cdot q = \textstyle\frac{1}{2}(p^*q + q^*p) = \textstyle\frac{1}{2}(pq^* + qp^*).$ The cross product of p and q relative to the orientation determined by the ordered basis i, j, and k is $p \times q = (c_1d_2 - d_1c_2)i + (d_1b_2 - b_1d_2)j + (b_1c_2 - c_1b_2)k.$ (Recall that the orientation is necessary to determine the sign.) This is equal to the vector part of the product pq (as quaternions), as well as the vector part of −q*p*. It also has the formula $p \times q = \textstyle\frac{1}{2}(pq - q^*p^*).$ In general, let p and q be quaternions (possibly non-imaginary), and write $p = p_s + \vec{p}_v,$ $q = q_s + \vec{q}_v,$ where ps and qs are the scalar parts, and $\vec{p}_v$ and $\vec{q}_v$ are the vector parts of p and q. Then we have the formula $pq = (pq)_s + (\vec{pq})_v = (p_sq_s - \vec{p}_v\cdot\vec{q}_v) + (p_s\vec{q}_v + \vec{p}_vq_s + \vec{p}_v \times \vec{q}_v).$ This shows that the noncommutativity of quaternion multiplication comes from the multiplication of pure imaginary quaternions. It also shows that two quaternions commute if and only if their vector parts are collinear. For further elaboration on modeling three-dimensional vectors using quaternions, see quaternions and spatial rotation. ## Matrix representations Just as complex numbers can be represented as matrices, so can quaternions. There are at least two ways of representing quaternions as matrices in such a way that quaternion addition and multiplication correspond to matrix addition and matrix multiplication. One is to use 2×2 complex matrices, and the other is to use 4×4 real matrices. In each case, the representation given is one of a family of linearly related representations. In the terminology of abstract algebra, these are injective homomorphisms from H to the matrix rings M(2, C) and M(4, R), respectively. Using 2×2 complex matrices, the quaternion a + bi + cj + dk can be represented as $\begin{bmatrix}a+bi & c+di \\ -c+di & a-bi \end{bmatrix}.$ This representation has the following properties: • Complex numbers (c = d = 0) correspond to diagonal matrices. • The norm of a quaternion (the square root of a product with its conjugate, as with complex numbers) is the square root of the determinant of the corresponding matrix.[19] • The conjugate of a quaternion corresponds to the conjugate transpose of the matrix. • Restricted to unit quaternions, this representation provides an isomorphism between S3 and SU(2). The latter group is important for describing spin in quantum mechanics; see Pauli matrices. Using 4×4 real matrices, that same quaternion can be written as $\begin{bmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix}= a \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}.$ In this representation, the conjugate of a quaternion corresponds to the transpose of the matrix. The fourth power of the norm of a quaternion is the determinant of the corresponding matrix. Complex numbers are block diagonal matrices with two 2×2 blocks. ## Quaternions as pairs of complex numbers Main article: Cayley–Dickson construction Quaternions can be represented as pairs of complex numbers. From this perspective, quaternions are the result of applying the Cayley–Dickson construction to the complex numbers. This is a generalization of the construction of the complex numbers as pairs of real numbers. Let C2 be a two-dimensional vector space over the complex numbers. Choose a basis consisting of two elements 1 and j. A vector in C2 can be written in terms of the basis elements 1 and j as $(a + bi)1 + (c + di)j.\$ If we define j2 = −1 and ij = −ji, then we can multiply two vectors using the distributive law. Writing k in place of the product ij leads to the same rules for multiplication as the usual quaternions. Therefore the above vector of complex numbers corresponds to the quaternion a + bi + cj + dk. If we write the elements of C2 as ordered pairs and quaternions as quadruples, then the correspondence is $(a + bi,\ c + di) \leftrightarrow (a, b, c, d).$ ## Square roots of −1 In the complex numbers, there are just two numbers, i and −i, whose square is −1 . In H there are infinitely many square roots of minus one: the quaternion solution for the square root of −1 is the surface of the unit sphere in 3-space. To see this, let q = a + bi + cj + dk be a quaternion, and assume that its square is −1. In terms of a, b, c, and d, this means $a^2 - b^2 - c^2 - d^2 = -1,$ $2ab = 0,$ $2ac = 0,$ $2ad = 0.$ To satisfy the last three equations, either a = 0 or b, c, and d are all 0. The latter is impossible because a is a real number and the first equation would imply that a2 = −1. Therefore a = 0 and b2 + c2 + d2 = 1. In other words, a quaternion squares to −1 if and only if it is a vector (that is, pure imaginary) with norm 1. By definition, the set of all such vectors forms the unit sphere. Only negative real quaternions have an infinite number of square roots. All others have just two (or one in the case of 0). The identification of the square roots of minus one in H was given by Hamilton[20] but was frequently omitted in other texts. By 1971 the sphere was included by Sam Perlis in his three page exposition included in Historical Topics in Algebra (page 39) published by the National Council of Teachers of Mathematics. More recently, the sphere of square roots of minus one is described in Ian R. Porteous's book Clifford Algebras and the Classical Groups (Cambridge, 1995) in proposition 8.13 on page 60. Also in Conway (2003) On Quaternions and Octonions we read on page 40: "any imaginary unit may be called i, and perpendicular one j, and their product k", another statement of the sphere. ### H as a union of complex planes Each pair of square roots of −1 creates a distinct copy of the complex numbers inside the quaternions. If q2 = −1, then the copy is determined by the function $a + b\sqrt{-1} \mapsto a + bq.$ In the language of abstract algebra, each is an injective ring homomorphism from C to H. The images of the embeddings corresponding to q and −q are identical. Every non-real quaternion lies in a subspace of H isomorphic to C. Write q as the sum of its scalar part and its vector part: $q = q_s + \vec{q}_v.$ Decompose the vector part further as the product of its norm and its versor: $q = q_s + \lVert\vec{q}_v\rVert\cdot\mathbf{U}\vec{q}_v.$ (Note that this is not the same as $q_s + \lVert q\rVert\cdot\mathbf{U}q$.) The versor of the vector part of q, $\mathbf{U}\vec{q}_v$, is a pure imaginary unit quaternion, so its square is −1. Therefore it determines a copy of the complex numbers by the function $a + b\sqrt{-1} \mapsto a + b\mathbf{U}\vec{q}_v.$ Under this function, q is the image of the complex number $q_s + \lVert\vec{q}_v\rVert i$. Thus H is the union of complex planes intersecting in a common real line, where the union is taken over the sphere of square roots of minus one, bearing in mind that the same plane is associated with the antipodal points of the sphere. ### Commutative subrings The relationship of quaternions to each other within the complex subplanes of H can also be identified and expressed in terms of commutative subrings. Specifically, since two quaternions p and q commute (pq = qp) only if they lie in the same complex subplane of H, the profile of H as a union of complex planes arises when one seeks to find all commutative subrings of the quaternion ring. This method of commutative subrings is also used to profile the coquaternions and 2 × 2 real matrices. ## Functions of a quaternion variable Main article: quaternion variable Like functions of a complex variable, functions of a quaternion variable suggest useful physical models. For example, the original electric and magnetic fields described by Maxwell were functions of a quaternion variable. ### Exponential, logarithm, and power Given a quaternion, q = a + bi + cj + dk = a + v, the exponential is computed as $\exp(q) = \sum_{n=0}^\infty \frac{q^n}{n!}=e^{a} \left(\cos \|\mathbf{v}\| + \frac{\mathbf{v}}{\|\mathbf{v}\|} \sin \|\mathbf{v}\|\right)$ and $\ln(q) = \ln \|q\| + \frac{\mathbf{v}}{\|\mathbf{v}\|} \arccos \frac{a}{\|q\|}$.[21] It follows that the polar decomposition of a quaternion may be written $q=\|q\|e^{\hat{n}\theta},$ where the angle θ and the unit vector $\hat{n}$ are defined by: $a=\|q\|\cos(\theta)$ and $\mathbf{v}=\hat{n} \|\mathbf{v}\|=\hat{n}\|q\|\sin(\theta).$ Any unit quaternion may be expressed in polar form as $e^{\hat{n}\theta}$. The power of a quaternion raised to an arbitrary (real) exponent is given by: $q^\alpha=\|q\|^\alpha e^{\hat{n}\alpha\theta}.$ ## Three-dimensional and four-dimensional rotation groups Main articles: Quaternions and spatial rotation and Rotation operator (vector space) The term "conjugation", besides the meaning given above, can also mean taking an element a to rar−1 where r is some non-zero element (quaternion). All elements that are conjugate to a given element (in this sense of the word conjugate) have the same real part and the same norm of the vector part. (Thus the conjugate in the other sense is one of the conjugates in this sense.) Thus the multiplicative group of non-zero quaternions acts by conjugation on the copy of R3 consisting of quaternions with real part equal to zero. Conjugation by a unit quaternion (a quaternion of absolute value 1) with real part cos(θ) is a rotation by an angle 2θ, the axis of the rotation being the direction of the imaginary part. The advantages of quaternions are: 1. Nonsingular representation (compared with Euler angles for example). 2. More compact (and faster) than matrices. 3. Pairs of unit quaternions represent a rotation in 4D space (see Rotations in 4-dimensional Euclidean space: Algebra of 4D rotations). The set of all unit quaternions (versors) forms a 3-dimensional sphere S3 and a group (a Lie group) under multiplication, double covering the group SO(3, R) of real orthogonal 3×3 matrices of determinant 1 since two unit quaternions correspond to every rotation under the above correspondence. For more details on this topic, see Point groups in three dimensions. The image of a subgroup of versors is a point group, and conversely, the preimage of a point group is a subgroup of versors. The preimage of a finite point group is called by the same name, with the prefix binary. For instance, the preimage of the icosahedral group is the binary icosahedral group. The versors' group is isomorphic to SU(2), the group of complex unitary 2×2 matrices of determinant 1. Let A be the set of quaternions of the form a + bi + cj + dk where a, b, c, and d are either all integers or all rational numbers with odd numerator and denominator 2. The set A is a ring (in fact a domain) and a lattice and is called the ring of Hurwitz quaternions. There are 24 unit quaternions in this ring, and they are the vertices of a 24-cell regular polytope with Schläfli symbol {3,4,3}. ## Generalizations Main article: quaternion algebra If F is any field with characteristic different from 2, and a and b are elements of F, one may define a four-dimensional unitary associative algebra over F with basis 1, i, j, and ij, where i2 = a, j2 = b and ij = −ji (so (ij)2 = −ab). These algebras are called quaternion algebras and are isomorphic to the algebra of 2×2 matrices over F or form division algebras over F, depending on the choice of a and b. ## Quaternions as the even part of Cℓ3,0(R) The usefulness of quaternions for geometrical computations can be generalised to other dimensions, by identifying the quaternions as the even part Cℓ+3,0(R) of the Clifford algebra Cℓ3,0(R). This is an associative multivector algebra built up from fundamental basis elements σ1, σ2, σ3 using the product rules $\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 1,$ $\sigma_i \sigma_j = - \sigma_j \sigma_i \qquad (j \neq i).$ If these fundamental basis elements are taken to represent vectors in 3D space, then it turns out that the reflection of a vector r in a plane perpendicular to a unit vector w can be written: $r^{\prime} = - w\, r\, w.$ Two reflections make a rotation by an angle twice the angle between the two reflection planes, so $r^{\prime\prime} = \sigma_2 \sigma_1 \, r \, \sigma_1 \sigma_2$ corresponds to a rotation of 180° in the plane containing σ1 and σ2. This is very similar to the corresponding quaternion formula, $r^{\prime\prime} = -\mathbf{k}\, r\, \mathbf{k}.$ In fact, the two are identical, if we make the identification $\mathbf{k} = \sigma_2 \sigma_1, \mathbf{i} = \sigma_3 \sigma_2, \mathbf{j} = \sigma_1 \sigma_3$ and it is straightforward to confirm that this preserves the Hamilton relations $\mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{i} \mathbf{j} \mathbf{k} = -1.$ In this picture, quaternions correspond not to vectors but to bivectors, quantities with magnitude and orientations associated with particular 2D planes rather than 1D directions. The relation to complex numbers becomes clearer, too: in 2D, with two vector directions σ1 and σ2, there is only one bivector basis element σ1σ2, so only one imaginary. But in 3D, with three vector directions, there are three bivector basis elements σ1σ2, σ2σ3, σ3σ1, so three imaginaries. This reasoning extends further. In the Clifford algebra Cℓ4,0(R), there are six bivector basis elements, since with four different basic vector directions, six different pairs and therefore six different linearly independent planes can be defined. Rotations in such spaces using these generalisations of quaternions, called rotors, can be very useful for applications involving homogeneous coordinates. But it is only in 3D that the number of basis bivectors equals the number of basis vectors, and each bivector can be identified as a pseudovector. Dorst et al. identify the following advantages for placing quaternions in this wider setting:[22] • Rotors are natural and non-mysterious in geometric algebra and easily understood as the encoding of a double reflection. • In geometric algebra, a rotor and the objects it acts on live in the same space. This eliminates the need to change representations and to encode new data structures and methods (which is required when augmenting linear algebra with quaternions). • A rotor is universally applicable to any element of the algebra, not just vectors and other quaternions, but also lines, planes, circles, spheres, rays, and so on. • In the conformal model of Euclidean geometry, rotors allow the encoding of rotation, translation and scaling in a single element of the algebra, universally acting on any element. In particular, this means that rotors can represent rotations around an arbitrary axis, whereas quaternions are limited to an axis through the origin. • Rotor-encoded transformations make interpolation particularly straightforward. For further detail about the geometrical uses of Clifford algebras, see Geometric algebra. ## Brauer group Further information: Brauer group The quaternions are "essentially" the only (non-trivial) central simple algebra (CSA) over the real numbers, in the sense that every CSA over the reals is Brauer equivalent to either the reals or the quaternions. Explicitly, the Brauer group of the reals consists of two classes, represented by the reals and the quaternions, where the Brauer group is the set of all CSAs, up to equivalence relation of one CSA being a matrix ring over another. By the Artin–Wedderburn theorem (specifically, Wedderburn's part), CSAs are all matrix algebras over a division algebra, and thus the quaternions are the only non-trivial division algebra over the reals. CSAs – rings over a field, which are simple algebras (have no non-trivial 2-sided ideals, just as with fields) whose center is exactly the field – are a noncommutative analog of extension fields, and are more restrictive than general ring extensions. The fact that the quaternions are the only non-trivial CSA over the reals (up to equivalence) may be compared with the fact that the complex numbers are the only non-trivial field extension of the reals. ## Quotes This section . Please help improve the article by editing it to take facts from excessively quoted material and rewrite them as sourced original prose. Consider transferring direct quotations to Wikiquote. (January 2012) • "I regard it as an inelegance, or imperfection, in quaternions, or rather in the state to which it has been hitherto unfolded, whenever it becomes or seems to become necessary to have recourse to x, y, z, etc." — William Rowan Hamilton (ed. Quoted in a letter from Tait to Cayley). • "Time is said to have only one dimension, and space to have three dimensions. […] The mathematical quaternion partakes of both these elements; in technical language it may be said to be "time plus space", or "space plus time": and in this sense it has, or at least involves a reference to, four dimensions. And how the One of Time, of Space the Three, Might in the Chain of Symbols girdled be." — William Rowan Hamilton (Quoted in R.P. Graves, "Life of Sir William Rowan Hamilton"). • "Quaternions came from Hamilton after his really good work had been done; and, though beautifully ingenious, have been an unmixed evil to those who have touched them in any way, including Clerk Maxwell." — Lord Kelvin, 1892. • "I came later to see that, as far as the vector analysis I required was concerned, the quaternion was not only not required, but was a positive evil of no inconsiderable magnitude; and that by its avoidance the establishment of vector analysis was made quite simple and its working also simplified, and that it could be conveniently harmonised with ordinary Cartesian work." Oliver Heaviside, Electromagnetic Theory, Volume I, pp. 134–135 (The Electrician Printing and Publishing Company, London, 1893). • "Neither matrices nor quaternions and ordinary vectors were banished from these ten [additional] chapters. For, in spite of the uncontested power of the modern Tensor Calculus, those older mathematical languages continue, in my opinion, to offer conspicuous advantages in the restricted field of special relativity. Moreover, in science as well as in every-day life, the mastery of more than one language is also precious, as it broadens our views, is conducive to criticism with regard to, and guards against hypostasy [weak-foundation] of, the matter expressed by words or mathematical symbols." — Ludwik Silberstein, preparing the second edition of his Theory of Relativity in 1924. • "… quaternions appear to exude an air of nineteenth century decay, as a rather unsuccessful species in the struggle-for-life of mathematical ideas. Mathematicians, admittedly, still keep a warm place in their hearts for the remarkable algebraic properties of quaternions but, alas, such enthusiasm means little to the harder-headed physical scientist." — Simon L. Altmann, 1986. ## Notes 1. "On Quaternions; or on a new System of Imaginaries in Algebra (letter to John T. Graves, dated October 17, 1843)". 1843. 2. 3. 4. 5. Journal of Theoretics http://www.journaloftheoretics.com/articles/3-6/qm-pub.pdf `|url=` missing title (help). 6. ^ a b c See Hazewinkel et. al. (2004), p. 12. 7. Conway, John Horton; Smith, Derek Alan (2003). On quaternions and octonions: their geometry, arithmetic, and symmetry. p. 9. ISBN 1-56881-134-9. 8. Robert E. Bradley, Charles Edward Sandifer (2007). Leonhard Euler: life, work and legacy. p. 193. ISBN 0-444-52728-1. . They mention Wilhelm Blaschke's claim in 1959 that "the quaternions were first identified by L. Euler in a letter to Goldbach written on May 4, 1748," and they comment that "it makes no sense whatsoever to say that Euler "identified" the quaternions in this letter... this claim is absurd." 9. Simon L. Altmann (December 1989). "Hamilton, Rodrigues, and the Quaternion Scandal". Mathematics Magazine 62 (5): 306. 10. C. F. Gauss, "Mutationen des Raumes" [Transformations of space] (c. 1819) [edited by Prof. Stäckel of Kiel, Germany] in: Martin Brendel, ed., Carl Friedrich Gauss Werke [The works of Carl Friedrich Gauss] (Göttingen, Germany: Königlichen Gesellschaft der Wissenschaften [Royal Society of Sciences], 1900), vol. 8, pages 357-361. 11. Ken Shoemake (1985). "Animating Rotation with Quaternion Curves". Computer Graphics 19 (3): 245–254. doi:10.1145/325165.325242.  Presented at SIGGRAPH '85. (1996) is often cited as the first mass-market computer game to have used quaternions to achieve smooth three-dimensional rotations. See, for example, Nick Bobick's, "Rotating Objects Using Quaternions", Game Developer magazine, July 1998 12. 13. 14. Girard, P. R. The quaternion group and modern physics (1984) Eur. J. Phys. vol 5, p. 25–32. doi:10.1088/0143-0807/5/1/007 15. Lambek, J. If Hamilton had prevailed: quaternions in physics (1995) Math. Intelligencer, vol. 17, #4, p. 7—15. doi:10.1007/BF03024783 16. "quaternion group". Wolframalpha.com. 17. 18. Hamilton (1899). Elements of Quaternions (2nd ed.). p. 244. ISBN 1-108-00171-8. 19. Quaternions and Geometric Algebra. Accessed 2008-09-12. See also: Leo Dorst, Daniel Fontijne, Stephen Mann, (2007), , Morgan Kaufmann. ISBN 0-12-369465-5 ## External articles and resources Look up quaternion in Wiktionary, the free dictionary. ### Books and publications • Hamilton, William Rowan. On quaternions, or on a new system of imaginaries in algebra. Philosophical Magazine. Vol. 25, n 3. p. 489–495. 1844. • Hamilton, William Rowan (1853), "Lectures on Quaternions". Royal Irish Academy. • Hamilton (1866) Elements of Quaternions University of Dublin Press. Edited by William Edwin Hamilton, son of the deceased author. • Hamilton (1899) Elements of Quaternions volume I, (1901) volume II. Edited by Charles Jasper Joly; published by Longmans, Green & Co.. • Tait, Peter Guthrie (1873), "An elementary treatise on quaternions". 2d ed., Cambridge, [Eng.] : The University Press. • Michiel Hazewinkel, Nadiya Gubareni, Nadezhda Mikhaĭlovna Gubareni, Vladimir V. Kirichenko. Algebras, rings and modules. Volume 1. 2004. Springer, 2004. ISBN 1-4020-2690-0 • Maxwell, James Clerk (1873), "A Treatise on Electricity and Magnetism". Clarendon Press, Oxford. • Tait, Peter Guthrie (1886), "Quaternion". M.A. Sec. R.S.E. Encyclopaedia Britannica, Ninth Edition, 1886, Vol. XX, pp. 160–164. (bzipped PostScript file) • Joly, Charles Jasper (1905), "A manual of quaternions". London, Macmillan and co., limited; New York, The Macmillan company. LCCN 05036137 //r84 • Macfarlane, Alexander (1906), "Vector analysis and quaternions", 4th ed. 1st thousand. New York, J. Wiley & Sons; [etc., etc.]. LCCN es 16000048 • 1911 encyclopedia: "Quaternions". • Finkelstein, David, Josef M. Jauch, Samuel Schiminovich, and David Speiser (1962), "Foundations of quaternion quantum mechanics". J. Mathematical Phys. 3, pp. 207–220, MathSciNet. • Du Val, Patrick (1964), "Homographies, quaternions, and rotations". Oxford, Clarendon Press (Oxford mathematical monographs). LCCN 64056979 //r81 • Crowe, Michael J. (1967), A History of Vector Analysis: The Evolution of the Idea of a Vectorial System, University of Notre Dame Press. Surveys the major and minor vector systems of the 19th century (Hamilton, Möbius, Bellavitis, Clifford, Grassmann, Tait, Peirce, Maxwell, Macfarlane, MacAuley, Gibbs, Heaviside). • Altmann, Simon L. (1986), "Rotations, quaternions, and double groups". Oxford [Oxfordshire] : Clarendon Press ; New York : Oxford University Press. LCCN 85013615 ISBN 0-19-855372-2 • Altmann, Simon L. (1989), "Hamilton, Rodrigues, and the Quaternion Scandal". Mathematics Magazine. Vol. 62, No. 5. p. 291–308, December 1989. • Adler, Stephen L. (1995), "Quaternionic quantum mechanics and quantum fields". New York : Oxford University Press. International series of monographs on physics (Oxford, England) 88. LCCN 94006306 ISBN 0-19-506643-X • Trifonov, Vladimir (1995), "A Linear Solution of the Four-Dimensionality Problem", Europhysics Letters, 32 (8) 621–626, doi:10.1209/0295-5075/32/8/001 • Ward, J. P. (1997), "Quaternions and Cayley Numbers: Algebra and Applications", Kluwer Academic Publishers. ISBN 0-7923-4513-4 • Kantor, I. L. and Solodnikov, A. S. (1989), "Hypercomplex numbers, an elementary introduction to algebras", Springer-Verlag, New York, ISBN 0-387-96980-2 • Gürlebeck, Klaus and Sprössig, Wolfgang (1997), "Quaternionic and Clifford calculus for physicists and engineers". Chichester ; New York : Wiley (Mathematical methods in practice; v. 1). LCCN 98169958 ISBN 0-471-96200-7 • Kuipers, Jack (2002), "Quaternions and Rotation Sequences: A Primer With Applications to Orbits, Aerospace, and Virtual Reality" (reprint edition), Princeton University Press. ISBN 0-691-10298-8 • Conway, John Horton, and Smith, Derek A. (2003), "On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry", A. K. Peters, Ltd. ISBN 1-56881-134-9 (review). • Kravchenko, Vladislav (2003), "Applied Quaternionic Analysis", Heldermann Verlag ISBN 3-88538-228-8. • Hanson, Andrew J. (2006), "Visualizing Quaternions", Elsevier: Morgan Kaufmann; San Francisco. ISBN 0-12-088400-3 • Trifonov, Vladimir (2007), "Natural Geometry of Nonzero Quaternions", International Journal of Theoretical Physics, 46 (2) 251–257, doi:10.1007/s10773-006-9234-9 • Ernst Binz & Sonja Pods (2008) Geometry of Heisenberg Groups American Mathematical Society, Chapter 1: "The Skew Field of Quaternions" (23 pages) ISBN 978-0-8218-4495-3. • Vince, John A. (2008), Geometric Algebra for Computer Graphics, Springer, ISBN 978-1-84628-996-5. • For molecules that can be regarded as classical rigid bodies molecular dynamics computer simulation employs quaternions. They were first introduced for this purpose by D.J. Evans, (1977), "On the Representation of Orientation Space", Mol. Phys., vol 34, p 317. • Zhang, Fuzhen (1997), "Quaternions and Matrices of Quaternions", Linear Algebra and its Applications, Vol. 251, pp. 21–57. ### Links and monographs • Hazewinkel, Michiel, ed. (2001), "Quaternion", , Springer, ISBN 978-1-55608-010-4 • Matrix and Quaternion FAQ v1.21 Frequently Asked Questions • "Geometric Tools documentation" (frame; body) includes several papers focusing on computer graphics applications of quaternions. Covers useful techniques such as spherical linear interpolation. • Patrick-Gilles Maillot Provides free Fortran and C source code for manipulating quaternions and rotations / position in space. Also includes mathematical background on quaternions. • "Geometric Tools source code" (frame; body) includes free C++ source code for a complete quaternion class suitable for computer graphics work, under a very liberal license. • Doug Sweetser, Doing Physics with Quaternions • Quaternions for Computer Graphics and Mechanics (Gernot Hoffman) • The Physical Heritage of Sir W. R. Hamilton (PDF) • D. R. Wilkins, Hamilton’s Research on Quaternions • Quaternion Julia Fractals 3D Raytraced Quaternion Julia Fractals by David J. Grossman • Quaternion Math and Conversions Great page explaining basic math with links to straight forward rotation conversion formulae. • John H. Mathews, Bibliography for Quaternions. • Quaternion powers on GameDev.net • Representing Attitude with Euler Angles and Quaternions: A Reference, Technical report and Matlab toolbox summarizing all common attitude representations, with detailed equations and discussion on features of various methods.[] • Charles F. F. Karney, Quaternions in molecular modeling, J. Mol. Graph. Mod. 25(5), 595–604 (January 2007); doi:10.1016/j.jmgm.2006.04.002; E-print arxiv:0506177. • Johan E. Mebius, A matrix-based proof of the quaternion representation theorem for four-dimensional rotations., arXiv General Mathematics 2005. • Johan E. Mebius, Derivation of the Euler-Rodrigues formula for three-dimensional rotations from the general formula for four-dimensional rotations., arXiv General Mathematics 2007. • NUI Maynooth Department of Mathematics, Hamilton Walk. • OpenGL:Tutorials:Using Quaternions to represent rotation • David Erickson, Defence Research and Development Canada (DRDC), Complete derivation of rotation matrix from unitary quaternion representation in DRDC TR 2005-228 paper. Drdc-rddc.gc.ca • Alberto Martinez, University of Texas Department of History, "Negative Math, How Mathematical Rules Can Be Positively Bent",Utexas.edu • D. Stahlke, Quaternions in Classical Mechanics Stahlke.org (PDF) • Morier-Genoud, Sophie, and Valentin Ovsienko. "Well, Papa, can you multiply triplets?", arxiv.org describes how the quaternions can be made into a skew-commutative algebra graded by Z/2 × Z/2 × Z/2. • Curious Quaternions by Helen Joyce hosted by John Baez. • Luis Ibanez "Tutorial on Quaternions" Part I Part II (PDF) • R. Ghiloni, V. Moretti, A. Perotti (2013) "Continuous slice functional calculus in quaternionic Hilbert spaces," Rev.Math.Phys. 25 1350006. An expository paper about continuous functional calculus in quanternionic Hilbert spaces useful in rigorous quaternionic quantum mechanics.
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http://sciencehouse.wordpress.com/2009/10/
# Scientific Clearing House Carson C. Chow ## Archive for October, 2009 ### Hurdles for mathematical thinking October 30, 2009 From my years as both a math professor and observer of people, I’ve come up with a list of  hurdles for mathematical thinking.  These are what I believe to be the essential set of skills  a  person must have if they want to understand and do mathematics.  They don’t need to have all these skills to use mathematics but would need most of them if they want to progress far in mathematics.  Identifying what sorts of conceptual barriers people may have could help in improving mathematics education. I’ll first give the list and then explain what I mean by them. 1. Context dependent rules 2. Equivalence classes 3. Limits and infinitesimals 4. Formal  logic 5. Abstraction (more…) Posted in Mathematics, Philosophy, Physics | 4 Comments » ### Darts and Diophantine equations October 24, 2009 Darts is a popular spectator sport in the UK. I had access to cable television recently so I was able to watch a few games.  What I find interesting about professional darts is that the players must solve a Diophantine equation to win.  For those who know nothing of the game, it involves throwing a small pointed projectile at an enumerated target board that looks like this: A dart that lands on a given sector on the board obtains that score.  The center circle of the board called the bulls eye is worth 50 points.  The ring around the bulls eye is worth 25 points.  The wedges are worth the score ascribed by the number on the perimeter.  However, if you land in the inner ring then you get triple the score of the wedge and if you land in the outer ring you get double the score.  Hence, the maximum number of points for one dart is the triple twenty worth 60 points. (more…) ### The NP economy October 21, 2009 I used to believe that one day all human labour would be automated (e.g. see here).  Upon further reflection, I realize that I am wrong.  The question of whether or not machines will someday replace all humans depends crucially on whether or not P is equal to NP.   Jobs that will eventually be automated will be the ones that can be solved easily with an algorithm.  In computer science parlance, these are problems in the computational complexity class P (solvable in polynomial time).   For example, traditional travel agents have disappeared faster than bluefin tuna because their task is pretty simple to automate.  However, not all travel agents will disappear.  The ones that survive will be more like concierges that put together complex travel arrangements or require negotiating with many parties. Eventually, the jobs that humans will hold (barring a collapse of civilization as we know it) will involve solving problems in the complexity class NP (or harder).  That is not to say that machines won’t be doing some of these jobs, only that the advantage of machines over humans will not be as clear cut.  While it is true that if we could fully reproduce a human and make it faster and bigger then it could do everything that a human could do better but as I blogged about before, I think it will be difficult to exactly reproduce humans.  Additionally, for some very hard problems that don’t even have any good approximation schemes, blind luck will play an important role in coming up with solutions.  Balancing different human centric priorities will also be important and that may be best left for humans to do.   Even if it turns out that P=NP there could still be some jobs that humans can do like working on undecidable problems. So what are some jobs that will be around in the NP economy?  Well, I think mathematicians will still be employed. Theorems can be verified in polynomial time but there are no known algorithms in P to generate them.   That is not to say that there won’t be robot mathematicians and mathematicians will certainly use automated theorem proving programs to help them (e.g. see here). However, I think the human touch will always have some use.  Artists and comedians will also have jobs in the future.  These are professions that require intimate knowledge of  what it is like to be human .  Again, there will be machine comics and artists but they won’t fully replace humans.  I also think that craftsmen like carpenters, stone masons, basket weavers and so forth could also make a comeback.  They will have to exhibit some exceptional artistry to survive but the demand for them could increase since some people will always long for the human touch in their furniture and houses. The question then is whether or not there will be enough NP jobs to go around and whether or not everyone is able and willing to hold one.  To some, an NP economy will be almost Utopian – everyone will have interesting jobs.    However, there may be some people who simply don’t want or can’t do an NP job.   What will happen to them?  I think that will be a big (probably undecidable) problem that will face society in the not too distant future, provided we make it that far. Posted in Computer Science, Economics, Optimization | 5 Comments » ### Arts and Crafts October 16, 2009 There is an opinion piece by  Denis Dutton in the New York Times today on Conceptual Art, which presents some views that I am very sympathetic to.   All creative endeavours involve some inspiration and perspiration – There is the idea and then there is the execution of that idea.  Conceptual art essentially removes the execution aspect of art and makes it a pure exercise in cleverness.  In some sense it does crystallize the essence of art but I’ve always found it lacking.   I just can’t get that inspired by a medicine cabinet.  I’ve always found that the craft of a work of art to be as compelling (if not more) as the idea itself.  In many cases the two are inseparable.  Dutton argues that the craft aspect of art will never disappear because people intrinsically enjoy witnessing virtuosity.  I’m inclined to agree.  So while Vermeer or Caravaggio will remain timeless Damien Hirst may just fade away in time. Corrected the spelling of Damien Hirst’s name on May 15,2012 Posted in Art, Philosophy | 4 Comments » ### Retire the Nobel Prize October 12, 2009 I’ve felt for sometime now that perhaps we should retire the Nobel Prize.  The money could be used to fund grants, set up an institute for peace and science, or even have a Nobel conference like TED.  The prize puts too much emphasis on individual achievement and in many instances misplaced emphasis.  The old view of science involving the lone explorer seeking truth in the wilderness needs to be updated to a new metaphor of the sandpile, as used to described self-organized criticality by Per Bak, Chao Tang, and Kurt Wiesenfeld.  In the sandpile model, individual grains of sand are dropped on the pile and every once in awhile there are “avalanches” where a bunch of grains cascade down.  The distribution of avalanche sizes is a power law.  Hence, there is no scale to avalanches and there is no grain that is more special than any other. This is just like science.  The contributions of scientists and nonscientists are like grains of sand dropping on the sandpile of knowledge and every once in awhile a big scientific avalanche is triggered.  The answer to the question of who triggered the avalanche is that everyone contributed to it.  The Nobel Prize rewards a few of the grains of sand that happened to be proximally located to some specific avalanche (and sometimes not) but the rewarded work always depended on something else. (more…) Posted in Opinion, Science | 10 Comments » ### Talk at MBI October 6, 2009 I’m currently at the Mathematical Biosciences Institute for a workshop on Computational challenges in integrative biological modeling.  The slides for my talk on using Bayesian methods for parameter estimation and model comparison are here. Title: Bayesian approaches for parameter estimation and model evaluation of dynamical systems Abstract: Differential equation models are often used to model biological systems. An important and difficult problem is how to estimate parameters and decide which model among possible models is the best. I will argue that Bayesian inference provides a self-consistent framework to do both tasks. In particular, Bayesian parameter estimation provides a natural measure of parameter sensitivity and Bayesian model comparison automatically evaluates models by rewarding fit to the data while penalizing the number of parameters. I will give examples of employing these approaches on ODE and PDE models. Posted in Bayes, Bioinformatics, Biology, Pedagogy, Probablity, Talks | 1 Comment » ### Human scale October 2, 2009 I’ve always been intrigued by how long we live compared to the age of the universe.  At 14 billion years, the universe is only a factor of $10^8$ older than a long-lived human.  In contrast, it is immensely bigger than us.  The nearest star is 4 light years away, which is a factor of $10^{16}$ larger than a human, and the observable universe is about 25 billion times bigger than that.    The size scale of the universe is partly dictated by the speed of light which at $3 \times 10^8$ m/s is coincidentally (or not) the same order of magnitude faster than we can move as the universe is older than we live. Although we are small compared to the universe, we are also exceedingly big compared to our constituents. We are comprised of about $10^{13}$ cells, each of which are about $10^{-5}$ m in diameter.  If we assume that the density of the cell is about that of water ($1 {\rm g/ cm}^3$) then that roughly amounts to $10^{14}$ molecules.  So a human is comprised of something like $10^{27}$ molecules, most of it being water which has an atomic weight of 18.  Given that proteins and organic molecules can be much larger than that a lower bound on the number of atoms in the body is $10^{28}$. The speed at which we can move is governed by the reaction rates of metabolism.  Neurons fire at an average of approximately 10 Hz, so that is why awareness operates on a time scale of a few hundred milliseconds.  You could think of a human moment as being one tenth of a second.  There are 86,400 seconds in a day so we have close to a million moments in a day although we are a sleep for about a third of them.  That leads to about 20 billion moments in a lifetime. Neural activity also sets the scale for how fast we can move our muscles, which is a few metres per second.  If we consider a movement every second then that implies about a billion twitches per lifetime.  Our hearts beat about once a second so that is also the number of heart beats in a lifetime. The average thermal energy at body temperature is about $10^{-19}$ Joules, which is not too far below the binding energies of protein-DNA and protein-protein interactions required for life.   Each of our  cells can translate about 5 amino acids per second, which is a lot of proteins in our lifetime.  I find it completely amazing that  a bag of $10^{28}$ or more things, incessantly buffeted by noise, can stay coherent for a hundred years.  There is no question that evolution is the world’s greatest engineer.  However, for those that are interested in artificial life this huge expanse of scale does pose a question -  What is the minimal computational requirement to simulate life and in particular something as complex as a mammal?  Even if you could do a simulation with say $10^{32}$ or more objects,  how would you even know that there was something living in it? The numbers came from Wolfram Alpha and Bionumbers. Posted in Computer Science, Philosophy, Physics | 2 Comments »
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http://www.physicsforums.com/showthread.php?t=592229
Physics Forums Mentor ## Simple question about simple functions I feel like this should be really easy, but for some reason I don't see how to finish it. I'm probably missing something obvious. The integral of an integrable simple function ##f=\sum_{k=1}^n a_k\chi_{E_k}## is $$\int f\,\mathrm{d}\mu=\sum_{k=1}^n a_k\mu(E_k),$$ where the right-hand side is interpreted using the convention ##0\cdot\infty=0##. If we want to use something like this as the definition of the integral, we need to make sure that there's no ambiguity in it. (The problem is that there are many ways to write a given integrable simple function as a linear combination of characteristic functions of measurable sets). One way to do that is to specify the a_k and E_k. We define the a_k by writing the range of f as ##f(X)=\{a_1,\dots,a_n\}## and the E_k by ##E_k=f^{-1}(a_k)##. Now there's no ambiguity in the formula above, so it's safe to use it as a definition. This raises the question of whether ##f=\sum_{k=1}^n a_k\chi_{E_k}=\sum_{i=1}^m b_i\chi_{F_i}## implies that \sum_{k=1}^n a_k\mu(E_k)=\sum_{i=1}^m b_i\mu(F_i). If the E_k are defined as above, it's easy to see that ##\{E_k\}_{k=1}^n## is a partition of X (the underlying set of the measure space). If the ##\{F_i\}_{i=1}^m## is another partition of X, then I find it easy to prove the equality above. However, it holds even when ##\{F_i\}_{i=1}^m## is not a partition of X, right? I first thought that it doesn't, but I failed to find a counterexample, so now I think it does. My problem is that I don't see how to do the proof in the general case. If ##\{E_k\}_{k=1}^n## and ##\{F_i\}_{i=1}^m## are both partitions of X, then we just write \begin{align} & \sum_{k=1}^n a_k\mu(E_k) =\sum_{k=1}^n a_k\mu\bigg(\bigcup_{i=1}^m E_k\cap F_i\bigg) =\sum_{k=1}^n \sum_{i=1}^m a_k \mu\big(E_k\cap F_i\big),\\ & \sum_{i=1}^m b_i\mu(F_i) =\sum_{i=1}^m b_i\mu\bigg(\bigcup_{k=1}^n E_k\cap F_i\bigg) =\sum_{k=1}^n \sum_{i=1}^m b_i \mu\big(E_k\cap F_i\big), \end{align} and then we can easily prove that the left-hand sides are equal by showing that the right-hand sides are equal term for term. Let k,i be arbitrary. If ##E_k\cap F_i=\emptyset##, then ##a_k\mu(E_k\cap F_i)=0=b_i\mu(E_k\cap F_i)##. If ##E_k\cap F_i\neq\emptyset##, then let ##x\in E_k\cap F_i## be arbitrary. We have ##a_k=f(x)=b_i##, and this obviously implies ##a_k\mu(E_k\cap F_i)=b_i\mu(E_k\cap F_i)##. So...anyone see how to prove or disprove the general case? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus You can always force the $F_i$ to be a partition: If $F_i\cap F_j\neq \emptyset$, then you can write $$b_i\chi_{F_i}+b_j\chi_{F_j}=b_i\chi_{F_i\setminus F_j}+b_j\chi_{F_j\setminus F_i}+(b_i+b_j)\chi_{F_i\cap F_j}$$ And if $F=\bigcup F_j$, then you just need to add a term $0\chi_{X\setminus F}$. Quote by Fredrik So...anyone see how to prove or disprove the general case? One way of proving this is to split the $\{F_i\}_{i = 1}^m$ into disjoint subsets. This gives you some refinement $\{F_i'\}_{i = 1}^{k}$ of the $\{F_i\}_{i=1}^m$. Now write out your simple function in terms of the original $a_1,\dots,a_m$ and the $\{F_i'\}_{i=1}^{k}$. It is fairly simple to check that these have the same integral and that completes the proof. Mentor ## Simple question about simple functions Thanks guys. Forgive me for being perhaps slower than usual today, but I still don't see how to turn these ideas into a calculation that looks like this: $$\sum_{i=1}^m b_i\mu(F_i)=\cdots=\sum_{k=1}^n a_k\mu(E_k).$$ I don't even see an easy way to define the refinement in jgens's post. I've been thinking things like this: Define I={1,...,m}. For each x in X, define ##I_x=\{i\in I|x\in F_i\}## and ##G_x=\bigcap_{i\in I_x}F_i##. Now ##\{G_x|x\in X\}\cup \{F^c\}## (where F was defined in micromass' post) should be finite, and be a partition. But I still need to prove that (probably not too hard), and then find a way to use it in the calculation. Mentor I think I figured this out. I thought this was really hard actually. Makes me wonder if I'm making it more complicated than it needs to be. These are the essentials of my proof: Define I={1,...,m}, and ##F=\bigcup_{i\in I} F_i##. For each x in F, define ##I_x=\{i\in I|x\in F_i\}## and ##G_x=\bigcap_{i\in I_x}F_i##. For all x in F, ##x\in G_x##, so ##\{G_x|x\in F\}## covers F. ##z\in G_x## implies ##G_z\subset G_x##. A corollary of this is that for all x,y in F, either ##G_x\cap G_y=\emptyset## or ##G_x=G_y##. So the collection ##\{G_x|x\in F\}## is mutually disjoint, and it's also finite, because $$\big|\{G_x\}\big|\leq\big|\{I_x\}\big| \leq\big|\mathcal P(I)\big|=2^m.$$ Since it's finite, there's a bijection ##H:\{1,\dots,p\}\to\{G_x|x\in F\}##. I'll write ##H_j## instead of H(j). Define ##H_0=F^c##. Now ##\{H_j\}_{j=0}^p## is a partition of X, and we have $$\begin{align} \sum_{k=1}^n a_k\mu(E_k) &=\sum_{j=0}^p\sum_{k=1}^n a_k \mu(E_k\cap H_j)\\ \sum_{i=1}^m b_i\mu(F_i) &=\sum_{j=0}^p\sum_{i=1}^m b_i \mu(F_i\cap H_j). \end{align}$$ To prove that the left-hand sides are equal, we prove that the sums over j on the right-hand sides are equal term for term. Let j be arbitrary. Let x be a member of F such that ##H_j=G_x##. Let ##k_0## be the positive integer such that ##G_j\subset E_{k_0}##. Note that ##H_j\subset F_i## when ##i\in I_x## and ##F_i\cap H_j=\emptyset## when ##i\notin I_x##. These things imply that $$\sum_{k=1}^n a_k\mu(E_k\cap H_j) =a_{k_0}\mu(H_j)=\sum_{i\in I_x} b_i\mu(H_j)= \sum_{i=1}^m b_i\mu(F_i\cap H_j).$$ Hm, now that I'm looking at this, I'm thinking that there probably isn't an easier way. Thread Tools | | | | |-------------------------------------------------------------|----------------------------------|---------| | Similar Threads for: Simple question about simple functions | | | | Thread | Forum | Replies | | | Calculus & Beyond Homework | 2 | | | Precalculus Mathematics Homework | 3 | | | Calculus | 3 | | | Calculus & Beyond Homework | 6 | | | Calculus & Beyond Homework | 2 |
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http://mathhelpforum.com/math-challenge-problems/95364-quadratic-residues.html
Thread: 1. Quadratic Residues Let $p$ be a prime such that $p>(n^2+n+k)^2+k$. ( $<br /> n,k \in \mathbb{Z}^ + <br />$ ) Consider the sequence: $<br /> n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k<br />$. Prove that there's a pair $(m,m+k)$ of integers from our sequence, such that $<br /> \left( {\tfrac{m}<br /> {p}} \right)_L = \left( {\tfrac{{m + k}}<br /> {p}} \right)_L = 1<br />$ This is a lovely problem, have fun!. PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too) 2. Originally Posted by PaulRS Let $p$ be a prime such that $p>(n^2+n+k)^2+k$. ( $<br /> n,k \in \mathbb{Z}^ + <br />$ ) Consider the sequence: $<br /> n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k<br />$. Prove that there's a pair $(m,m+k)$ of integers from our sequence, such that $<br /> \left( {\tfrac{m}<br /> {p}} \right)_L = \left( {\tfrac{{m + k}}<br /> {p}} \right)_L = 1<br />$ This is a lovely problem, have fun!. PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too) if $\left(\frac{n^2+k}{p} \right)=1,$ choose $m=n^2$ and if $\left(\frac{(n+1)^2+k}{p} \right)=1,$ then choose $m=(n+1)^2.$ so we may assume that $\left(\frac{n^2+k}{p} \right)=\left(\frac{(n+1)^2+k}{p} \right)=-1.$ now we have this cute identity: $(n^2+k)((n+1)^2+k)=(n^2+n+k)^2+k.$ therefore: $\left(\frac{(n^2+n+k)^2+k}{p} \right)=1$ and we choose $m=(n^2+n+k)^2. \ \Box$ Remark: to put the above result in simple words: for any positive integers $n,k$ and any prime number $p > (n^2+n+k)^2+k,$ at least one of the numbers $n^2+k, \ (n+1)^2+k, \ (n^2+n+k)^2+k$ is a quadratic residue modulo $p.$
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http://mathoverflow.net/questions/84824?sort=newest
## an operation on binary strings ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Recently, as part of some joint research, Tom Roby was led to a curious operation on strings of L's and R's which he calls "bounce-reading": We start by reading the string at the left. When the symbol we have just read is an L, the next symbol we read is the leftmost unread symbol; but when the symbol we have just read is an R, the next symbol we read is the rightmost unread symbol. For instance, if our string of L's and R's is LLRRLRLR, we read the positions in the order 1,2,3,8,7,4,6,5, obtaining the bounce-reading LLRRLRRL. It may be more natural to think of this as an action on circular words; when we read a letter, we delete it, close up the circle, and move either to the left or the right of the deletion-point. Has this operation been studied before? - Ultimately it does not matter, but I think it would be nicer to start by reading the symbol on the right (which immediately becomes the leftmost symbol of the next string.) Otherwise we have some initial substring ($L^kR$ for some $k \ge 0$) which is invariant and then the rest of the word which bounces according to the rule I suggest. The description with circular words makes some sense but it has to be (oriented) circular words with a distinguished start letter. After all, $RRLL$ and $RLLR$ behave differently. – Aaron Meyerowitz Jan 10 2012 at 21:52 ## 3 Answers Well, it has now, since I just sank my morning into studying it. I sure am a sucker for a naive combinatorics problem. Here's what I know, or can conjecture: • The map you describe is a bijection on words of length $n$, because it's easy to write down its inverse. I've included python code below. • Let $B_L$ be the bounce-reading algorithm. Let $B_R$ be the bounce-reading algorithm with the following change: replace the phrase "we start by reading the string at the left" with "we start by reading the string at the right". Then $B_LB_R^{-1}(w)$ seems to shift $w$ cyclically by one letter. Sometimes the shift is the left, and sometimes to the right; which of these things happens depends on $w$ in a manner which I don't understand. I noticed this because the maximal orbit sizes of $B_LB_R^{-1}(w)$ for each $n$ seem to match the OEIS sequence https://oeis.org/A027375. • Let $S_n$ be the set of words $w$ of length $n$ for which $B_L(w) = B_R(w)$. Then the sequence ${|S_{n+1}| - |S_n|}$ seems to be the Fibonacci numbers, at least for $n\geq 2$. This probably means $S_n$ has some nice structure. None of the other obvious statistics that describe this map are in the OEIS yet. Here are naive python implementations of $B_L, B_R, B_L^{-1}, B_R^{-1}$ if you want to check these assertions. ````def bounce_left(w): if len(w) == 1: return w leftchar = w[0] x = w[1:] if(leftchar == "L"): return leftchar + bounce_left(x) else: return leftchar + bounce_right(x) def bounce_right(w): if len(w) == 1: return w last_index = len(w) - 1 rightchar = w[last_index] x = w[:last_index] if(rightchar == "L"): return rightchar + bounce_left(x) else: return rightchar + bounce_right(x) def unbounce(w): if w == "": return "" output = "" n = len(w) - 1 for index in reversed(range(n)): if w[index] == "L": output = w[index+1] + output else: output = output + w[index+1] return output def unbounce_left(w): return unbounce("L" + w) def unbounce_right(w): return unbounce("R" + w) ```` - 1 Do I understand it correctly from the code that your $B_R$ reads things from the right but still writes the result from the left? (It is not quite clear from the word description of $B_R$, and was not my naive guess about it before reading the code.) – Vladimir Dotsenko Jan 4 2012 at 13:37 1 Yes, you understand it correctly. I'll edit the description so that it is less ambiguous. Thanks. – Benjamin Young Jan 4 2012 at 13:55 2 I mean, it's sort of odd that a procedure for reading a string at both ends should always start at the left end. That's the only reason I even thought about $B_R$ at all. It also might be interesting to think about reversing the string you write down. – Benjamin Young Jan 4 2012 at 14:03 Great, thanks a lot! – Vladimir Dotsenko Jan 4 2012 at 15:24 Hi Ben! This is a really silly remark but you're missing out on perhaps the best piece of Python syntax, negative indexing: pyfaq.infogami.com/what-s-a-negative-index – Dan Petersen Jan 11 2012 at 7:17 show 1 more comment ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I suspect that it has not been studied, but it should be! I base this on the fact that various statistics one might expect to be associated with it do not appear in the OEIS (even with an appeal to superseeker@oeis.org). For example $$2, 3, 6, 7, 19, 27, 36, 79, 130, 384, 473, 710, 2903, 4197$$ gives the maximum size orbit for strings of lengths up to $16.$ It turns our that $RRRRRLRLLLLLL$ has an orbit of size $473$ and that is the longest orbit for any string of length 13. - It may be convenient to consider an extension of the map to words in a double alphabet $A=\{L,R\}\times \{L,R\}$ or, equivalently, pairs of words in $\{L,R\}$ of the same length. (Your situation is obtained when you consider a pair: a word and the reverse word.) The advantage of the extension is that then the map is given by a Mealy automaton. Although I do not know whether exactly this transformation was studied, groups generated by such transformations have been of course studied as "automaton groups". See papers and books by Grigorchuk, Nekrashevych and others. -
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http://mathhelpforum.com/geometry/170578-perimeter-equilateral-triangle.html
# Thread: 1. ## Perimeter of an equilateral triangle The perimeter of an equilateral triangle exceeds the perimeter of a square by 1989cm. The length of each side of the triangle exceeds the length of each side of the square by d cm. The square has perimter greater than 0. How many positive integers are not possible values for d? t for side of triangle s for side of square 3t = 4s + 1989 t - s = d t= s +d 3s + 3d = 4s + 1989 3d = s + 1989 d = s/3 + 663 it doesn't seem to make sense. Pls help. 2. Originally Posted by Veronica1999 The perimeter of an equilateral triangle exceeds the perimeter of a square by 1989cm. The length of each side of the triangle exceeds the length of each side of the square by d cm. The square has perimter greater than 0. How many positive integers are not possible values for d? t for side of triangle s for side of square 3t = 4s + 1989 t - s = d t= s +d 3s + 3d = 4s + 1989 3d = s + 1989 d = s/3 + 663 it doesn't seem to make sense. Pls help. $\displaystyle d=\frac{s}{3}+663$ $\displaystyle s>0 \implies d>663$ 663 positive integers are not possible values for d. They are: 1, 2, 3, ... , 663.
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http://mathforum.org/mathimages/index.php?title=Parametric_Equations&diff=6423&oldid=6415
# Parametric Equations ### From Math Images (Difference between revisions) | | | | | |----------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|----------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| | | | | | | Line 20: | | Line 20: | | | | | | | | | Click to see a circle drawn parametrically:{{hide|{{#eqt: Circledraw.swf|450}}}} | | Click to see a circle drawn parametrically:{{hide|{{#eqt: Circledraw.swf|450}}}} | | | | + | | | | | + | The butterfly curve in this page's main image uses a similar method, but with more complicated parametric equations. | | | |Pre-K=No | | |Pre-K=No | | | |Elementary=No | | |Elementary=No | | Line 25: | | Line 27: | | | | |HighSchool=Yes | | |HighSchool=Yes | | | |ImageDesc= | | |ImageDesc= | | - | Sometimes curves which would be very difficult or even impossible to graph in terms of elementary functions of x and y can be graphed using a parameter. One example is the butterfly curve. | + | Sometimes curves which would be very difficult or even impossible to graph in terms of elementary functions of x and y can be graphed using a parameter. One example is the butterfly curve, as shown in this page's main image. | | | | | | | | This curve uses the following parametrization: | | This curve uses the following parametrization: | ## Revision as of 14:15, 29 June 2009 Butterfly Curve The Butterfly Curve is one of many beautiful images generated using parametric equations. Butterfly Curve Field: Calculus Created By: Direct Imaging # Basic Description We often graph functions by letting one coordinate be dependent on another. For example, graphing the function $f(x) = y = x^2$ has y values that we trace out depend upon x values. However, it is very useful to consider functions where each coordinate is equal to an equation of an independent variable, known as a parameter. Changing the value of the parameter can change value of any coordinate being used. We choose a range of values for the parameter, and the values that our function takes on as the parameter varies traces out a curve, known as a parametrized curve. Parametrization is the process of finding a parametrized version of a function. ### Parametrized Circle One curve that can be easily parametrized is a circle of radius one: We use the variable t as our parameter, and x and y as our normal Cartesian coordinates. We now let $x = cos(t)$ and $y = sin(t)$, and let t take on all values from $0$ to $2\pi$. When $t=0$, the coordinate $(1,0)$ is hit. As t increases, a circle is traced out as x initially decreases, since it is equal to the cosine of t, and y initially increases, since it is equal to the sine of t. The circle continues to be traced until t reaches $2\pi$, which gives the coordinate $(1,0)$ once again. It is also useful to write parametrized curves in vector notation, using a coordinate vector: $\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} cos(t) \\ sin(t)\\ \end{bmatrix}$ The butterfly curve in this page's main image uses a similar method, but with more complicated parametric equations. # A More Mathematical Explanation Note: understanding of this explanation requires: *Linear Algebra [Click to view A More Mathematical Explanation] Sometimes curves which would be very difficult or even impossible to graph in terms of elementary fun [...] [Click to hide A More Mathematical Explanation] Sometimes curves which would be very difficult or even impossible to graph in terms of elementary functions of x and y can be graphed using a parameter. One example is the butterfly curve, as shown in this page's main image. This curve uses the following parametrization: $\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} \sin(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right) \\ \cos(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right)\\ \end{bmatrix}$ Parametric construction of the butterfly curve ## Parametrized Surfaces and Manifolds In the above cases only one independent variable was used, creating a parametrized curve. We can use more than one independent variable to create other graphs, including graphs of surfaces. For example, using parameters s and t, the surface of a sphere can be parametrized as follows: $\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} sin(t)cos(s) \\ sin(t)sin(s) \\cos(t) \end{bmatrix}$ While two parameters are sufficient to parametrize a surface, objects of more than two dimensions will require more than two parameters. These objects, generally called manifolds, may live in higher than three dimensions and can have more than two parameters, so many times cannot be visualized. Nevertheless they can be analyzed using the methods of vector calculus and differential geometry. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://www.physicsforums.com/showthread.php?t=79794
Physics Forums ## Common source amplifier Just out of interest guys have I got this right? Function of certain components of a common source amplifier: Capacitors at input and output: They are coupling capacitors and simply block the DC voltage, but allow the AC voltage through. Gate Resistance: Is the purpose of this to limit the amount of current reaching the gate of the JFET?? Source resistance: This reduces the gain of the circuit, which thereby reduces distortion. Drain resistance: Limits the maximum voltage gain? I think that the source and capacitor explanations are right, but I'm not that sure about the Gate and Drain resistors. PhysOrg.com engineering news on PhysOrg.com >> Researchers use light projector and single-pixel detectors to create 3-D images>> GPS solution provides 3-minute tsunami alerts>> Single-pixel power: Scientists make 3-D images without a camera Quote by Soilwork Just out of interest guys have I got this right? Function of certain components of a common source amplifier: Capacitors at input and output: They are coupling capacitors and simply block the DC voltage, but allow the AC voltage through. Gate Resistance: Is the purpose of this to limit the amount of current reaching the gate of the JFET?? Source resistance: This reduces the gain of the circuit, which thereby reduces distortion. Drain resistance: Limits the maximum voltage gain? I think that the source and capacitor explanations are right, but I'm not that sure about the Gate and Drain resistors. The Gate resistor provides a ground reference to the Gate. The source resistor provides a positive voltage to the source. The combination of those factors makes VGS < 0. This is a requirement for the functioning of the FET. The source resistor really reduces the gain, but if this is undesirable you can bypass it with a large capacitor. The drain resistor really is responsible for the gain. If your source resistor is not bypassed the gain is approximately $$A_V = \frac {R_D}{R_S}$$. If it is bypassed the gain is approximately $$A_V = g_m R_D$$ Cheers for that :) Another question that I have been struggling finding an answer for is what effect a finite input impedance has on the response of an amplifier? I mean I've been reading up on how to use op-amps and the differences between ideal and real, and can't find the reason for this one thing. I did find somewhere that a finite input impedance puts an upper bound on the resistances in the feedback circuit...but how this effects the response of the amplifier I'm not sure. Sorry for asking stupid questions, but I'm new at this :( Mentor ## Common source amplifier They're not stupid questions by any means. I'd suggest that you take a look at the following book -- it's a really great intro electronics book that will answer a lot of your questions. It's also the most practical intro electronics book that I've ever seen. "The Art of Electronics" by Horowitz and Hill http://www.amazon.com/exec/obidos/AS...051601-9854513 Quote by Soilwork Cheers for that :) Another question that I have been struggling finding an answer for is what effect a finite input impedance has on the response of an amplifier? I mean I've been reading up on how to use op-amps and the differences between ideal and real, and can't find the reason for this one thing. I did find somewhere that a finite input impedance puts an upper bound on the resistances in the feedback circuit...but how this effects the response of the amplifier I'm not sure. Sorry for asking stupid questions, but I'm new at this :( A finite input impedance will be part of the feedback circuit. ahh k so it will change the amount of feedback in the circuit. Cheers :) Thanks for the book suggestion. I will probably order it next week. Also, the input impedance should be much higher than the source impedance (internal impedance of the source). This not only guarantees that the voltage signal to be amplified is efficiently transferred in a useable form to the amplifier inputs, but also is critically important for lowering distortion of the waveform. It works by limiting the current draw from the source which can distort the voltage waveform and affect frequency response as well. If the current varies significantly in the input loop including the source of the signal, you begin to affect the signal you are trying to measure. When discussing 'impedance matching' in amplifier stages (especially audio), what is meant is keeping the input impedance of the next stage high enough not to affect the circuit behaviour and signal of the preceding stage. For high frequency applications, stray capacitance from physical layout and stage coupling becomes a dominant issue. Quote by Rogue Physicist Also, the input impedance should be much higher than the source impedance (internal impedance of the source). This not only guarantees that the voltage signal to be amplified is efficiently transferred in a useable form to the amplifier inputs, but also is critically important for lowering distortion of the waveform. It works by limiting the current draw from the source which can distort the voltage waveform and affect frequency response as well. If the current varies significantly in the input loop including the source of the signal, you begin to affect the signal you are trying to measure. When discussing 'impedance matching' in amplifier stages (especially audio), what is meant is keeping the input impedance of the next stage high enough not to affect the circuit behaviour and signal of the preceding stage. For high frequency applications, stray capacitance from physical layout and stage coupling becomes a dominant issue. I think you didn't understand clearly the post. Source here refers not to power source or signal source, but to the terminal of the FET called source. Hi, I am new to this website. I am having troubles in designing a common source amplifier with a voltage gain, Av=26 dB (no bypass capacitor) and a gain of 40 DB (with a bypass capacitor) and a frequency of 100-1000HZ. Can anyone please guide me to calculate the value of the bypass capacitor? I searched everywhere for the methods of finding the capacitors values but still in vain. thank you very much for all your help. thanks, peace Quote by berkeman They're not stupid questions by any means. I'd suggest that you take a look at the following book -- it's a really great intro electronics book that will answer a lot of your questions. It's also the most practical intro electronics book that I've ever seen. "The Art of Electronics" by Horowitz and Hill http://www.amazon.com/exec/obidos/AS...051601-9854513 Hey I have that book! allthough i have not read it yet... Quote by peace_lp Hi, I am new to this website. I am having troubles in designing a common source amplifier with a voltage gain, Av=26 dB (no bypass capacitor) and a gain of 40 DB (with a bypass capacitor) and a frequency of 100-1000HZ. Can anyone please guide me to calculate the value of the bypass capacitor? I searched everywhere for the methods of finding the capacitors values but still in vain. thank you very much for all your help. thanks, peace Basically, the impedance of the bypass capacitor at the lowest operation frequency (100 Hz) should be much smaller than the resistance of the source resistor (generally 10 times smaller). If you have the source resistor, you can calculate the capacitance. 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http://en.wikibooks.org/wiki/Cellular_Automata/Global_Dynamics
Cellular Automata/Global Dynamics Attractor Basin When discussing global dynamics of cellular automata the word state is used to describe configurations. Graphic conventions The following diagrams/captions are based on the original figure State-space and basins of attraction on the DDLab website. Sub-trees, basins of attraction, and the entire basin of attraction field (for cellular automata, random Boolean networks, and discrete dynamic networks in general) can be computed and drawn with the DDLab software. The state space is made of all available states of a cellular automaton. It's size is $S = v^{n}$ where $n$ is the the size of the (finite) lattice, and $v$ is the value-range or alphabet, $v=2$ for binary systems. $B$ is one of the states. The trajectory $A \rightarrow B \rightarrow C$ is defined by the global transition function. State $A$ is a predecessor and state $C$ is the successor of state $B$. If the global transtition function is not injective, than states may have more than one predecessor (preimage). The in-degree of the state $B$ is the number of its predecessors. A state may have no predecessors, than it is called garden of Eden. A trajectory may encounter a state occurred previously (it must if the lattice is finite), thus it has entered an attraction cycle. A point attraction cycle has the length of a single cell. By recursively searching predecessors from a root state on the attraction cycle (the predecessor on the cycle is excluded) till all garden of Eden states are reached a transient tree is constructed. A subtree can be constructed from an arbitrary state on the tree, the root of the subtree. By adding transient trees to each state on the attraction cycle, a basin of attraction is constructed. Some attraction cycles may have no transient trees, this is especially true for reversible cellular automata. By grouping all the states in the state space into respective basins of attraction, a basin of attraction field is constructed. Injectivity, surjectivity and bijectivity The CA rule is reversible, if the global transition function $F$ is bijective (both injective and surjective). Injectivity of the global transition function $F$ is injective, there is at most one preimage for any CA configuration. $F: X \mapsto Y \qquad \forall x_1,x_2 \in \Omega \; [x_1 \neq x_2 \Rightarrow F(x_1) \neq F(x_2)]$ So there are no configurations with more than one predecessor. Surjectivity of the global transition function $F$ is injective, there is at least one preimage for any CA configuration. $F: X \mapsto Y \qquad \forall y \in \Omega \; \exists x [ y = F(x)]$ So there are no configurations without a predecessor (so called Garden of Eden). It is easy to determine the existence of garden of Eden states for one dimensional CA using De Bruijn Diagrams. Bijectivity of the global transition function $F$ is bijective if it is both injective and surjective. For any configuration there is exactly one predecessor.
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http://math.stackexchange.com/questions/tagged/absolute-value
# Tagged Questions The absolute-value tag has no wiki summary. 1answer 28 views ### Simplifying $\left|\left|\sqrt{-x^2}-1\right|-2\right|$ How do we simplify the expression $\left|\left|\sqrt{-x^2}-1\right|-2\right|$? This is very confusing. Do they cancel out and become just simply $\sqrt{-x^2}-1-2$? 0answers 34 views ### Calculation of the sub gradient of the first norm of a matrix Lets say I have a matrix X and its first norm $||X||_1$. How do I calculate the subgradient of this norm with respect to matrix X itself. 1answer 19 views ### How to graph an absolute value equation? How would you graph: $|x+y|=1$ ? I can do the normal $y=|x+1|$ and all that. But how would you do a question with two of these unknowns in the absolute value? Any help would be greatly appreciated, ... 1answer 36 views ### Is any norm on $\mathbb R^n$ invariant with respect to componentwise absolute value? Given $\mathbf{x}=(x_1,...,x_n) \in \mathbb{R}^n$ , define $\mathbf{x}'=(|x_1|,...,|x_n|)$ . Then, is it $||\mathbf{x}'|| = ||\mathbf{x}||$ for every norm on $\mathbb{R}^n$ ? NB: The answer ... 1answer 19 views ### How to linearize the following LP I want to minimize $|d_1-d_2|+e1+e2+e3$ where $d_1,d_2,e_1,e_2,e_3>=0$ and $|.|$ denotes the absolute value, for some linear constraints. Is there any way I can linearize the objective function? 1answer 59 views ### Absolute values in $\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$ in my math class we were given a list of indefinite integrals, and one of them was: $$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$$ My working: \int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}=\int ... 1answer 20 views ### how to find absolute value for complex fraction I have a Fourier transfer equation $H(jw) = \frac{jwL}{(jw)^2LC+jw\frac{L}{R}+1}$, and I need to find frequency to make $|H(jw)|$ is max. I know I should take the derivative of $|H(jw)|$ then find ... 2answers 50 views ### Prove That $|a +b| = |a| +|b|$ if $a$ and $b$ Have Same Signs, And $|a +b| < |a| + |b|$ if $a$ and $b$ Have Opposite Signs (Proved Differently) [duplicate] My Proof: This problem has mainly four cases, they are as follows: 1) $a, b > 0$ 2) $a, b < 0$ 3) $a > 0 > b$ 4) $a < 0 < b$ Let suppose that the sum of the real numbers \$a ... 1answer 41 views ### What is the modulus of a number? What is the exact definition of the modulus of a number? As far as I know, it is the distance between the origin and the point associated with this number. So if \$z=a+bi \in \Bbb ... 5answers 76 views ### Prove That $|a +b| = |a| +|b|$ if $a$ and $b$ Have Same Signs, And $|a +b| < |a| + |b|$ if $a$ and $b$ Have Opposite Signs My Proof: $|a +b| = |a| +|b|$ ..... $(i)$ $|a +b| < |a| + |b|$ ..... $(i)$ If $'a'$ and $'b'$ have same signs: Let $a$ and $b$ be equal to $-x$. Replacing $a$ and $b$ with $-x$ in the equation ... 2answers 72 views ### Absolute Value of $|-3 -2|$ $|-3 -2|$ is the distance between the points $-3$ and $-2$. If we solve it further then, in one way I get $|-5| = 5$. But $5$ is the distance between $0$ and $-5$ in this case. In other way, \$2 ... 2answers 33 views ### Absolute Value Problem, Solution and Method Please check my method and also if I have solved the following problem correctly: Problem: $f(x) = |x - \frac12| + |x + \frac12|$ If $x = -1$, then: $f(-1) = |-1 - \frac12| + |-1 + \frac12|$ From ... 3answers 36 views ### Question about absolute value in inequalities My book presents the following: $$7 \le x \le 9$$ so $$-1 \le x - 8 \le 1$$ and $$|x-8| \le 1$$ I usually get confused with the way that taking the absolute value of an expression works. Could ... 3answers 41 views ### finding values for absolute convergence Find all values of real number p or which the series converges: $$\sum \limits_{k=2}^{\infty} \frac{1}{\sqrt{k} (k^{p} - 1)}$$ I tried using the root test and the ratio test, but I got stuck on ... 1answer 43 views ### Absolute of a trig function Consider the function $$f(x) = 1\dfrac{1}{2} - 3\sin \left(\dfrac{1}{2}x \right).$$ I need to find the absolute of this function, which to my eye would just be f(x) = 1\dfrac{1}{2} + 3\sin ... 2answers 43 views ### Absolute Convergence of a Series Find all values of real number p for which the series converges absolutely $$\sum_{k=2}^{\infty} \frac{1}{k\, (\log{k})^p}$$ 2answers 74 views ### Calculating the integral $\int_{0}^{5} { \frac{|x-1|}{|x-2| + |x-4|} } dx$ How do we calculate the following integral: $$\int_{0}^{5} { \frac{|x-1|}{|x-2| + |x-4|} } dx$$ 2answers 86 views ### How to evaluate the inequality $|x+1|<-1$? Okay perhaps the title isn't specific enough, I didn't know how to word it exactly. I'm finding the interval of convergence for a power series and i know the answer to be (-2,0] I end up with the ... 2answers 43 views ### Can summations distribute across absolute values? Can I distribute a summation as follows? $$k\sum_{x \in X} \left| x - b \right| = \left| \left(k\sum_{x \in X}x \right) - \left( k\sum_{x \in X}b \right) \right|$$ 0answers 76 views ### Continuous, differentiable, continuously differentiable I came across the following problem: Let $\alpha \in \mathbb R$. Where is the function continuous, differentiable, continuously differentiable? f(x) = \begin{cases} x|x|^\alpha & ... 2answers 43 views ### Finding absolute max and min values of function Function given as $f(x,y) = 3x^2 + 2xy^2$. If $(x,y)$ lies in the region inside including edges of the triangle in the first quadrant given by $x\ge0, y\ge0, y\le2-x$. Reduce $f$ to a single variable ... 1answer 82 views ### For what $\alpha \in \mathbb R$ is $|x|^\alpha$ differentiable in $x=0$? I came across the following question: For what $\alpha \in \mathbb R$ is $|x|^\alpha$ differentiable in $x=0$? What I have tried: Since for $\alpha = 1$ is clearly non-differentiable in ... 1answer 26 views ### Is this (or when) does this equality hold? Let $a,b,c,d \in \mathbb{R}$ and $x,y$ are variables which are also real numbers $$|ax + by|^2 + |cx + dy|^2 + 2|ax + by||cx + dy| = (ax + by)^2 + (cx + dy)^2 + 2(ax + by)(cx + dy)$$ Is this always ... 2answers 59 views ### proving $|x - 1| < {1\over4} \Rightarrow |2x - 1| \geq {1\over 2}$ I tried solving the above, consider that: ($x \in R)$, I know it's not a complicated problem to solve though I struggle getting on with this question, What I've done far is: \$|x-1|<{1\over4} ... 3answers 55 views ### Find $x$ for absolute value inequalities I'm trying to figure out this inequality: $|x+1| + |x| \leq x^2$ I thought about trying it with two cases: $(x = -x)$ and $(x = +x)$ but I don't seem to find out how to go through from here, ... 1answer 54 views ### Finding The Contour Maps Of A Function Of Two Variables I am given the function $f(x,y) = \ln|y-x^2|$, and am suppose to find the contour maps. Let $z = c = f(x,y)$. $c = \ln|y-x^2| \rightarrow e^c = e^{\ln|y-x^2|} \rightarrow e^c = |y-x^2|$ I know I ... 3answers 67 views ### Solving $2|x+1|>|x+4|$ I'm trying to solve the following equations and inequalities for $x\in\mathbb R$: $$2|x+1|>|x+4|$$ I know I'm supposed to consider the intervals $(-\infty,-4), [-4,-1]$ and $(-1,\infty)$ but ... 3answers 103 views ### Determine all solutions to $|x+12|+|x-5|=15$ Determine all solutions to the following. $$\lvert x+12\rvert +\lvert x-5\rvert =15.$$ 1answer 146 views ### Finding the points of the curve where the tangent line is horizontal The curve given is $\displaystyle y = \ln|x-2| + x + \frac{12}{x-2}$. Find the points of the curve where the tangent line is horizontal. My first stumbling block is the absolute value function. I ... 3answers 105 views ### How does one calculate the integral of the sum of two absolute values? I know how to find the integral of just one absolute value, but this problem presents the integral of the sum of two absolute values. Help! I want to evaluate: $$\int_a^b{(|x-1| + |x+1|) dx}$$ 2answers 55 views ### Prove $||a| - |b|| \leq |a - b|$ [duplicate] I'm trying to prove that $||a| - |b|| \leq |a - b|$. So far, by using the triangle inequality, I've got: $$|a| = |\left(a - b\right) + b| \leq |a - b| + |b|$$ Subtracting $|b|$ from both sides yields, ... 2answers 86 views ### Question regarding usage of absolute value within natural log in solution of differential equation The problem from the book. $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 -y$ I understand the solution till this part. $\ln \vert 6 - y \vert = x + C$ The solution in the book is $6 - Ce^{-x}$ ... 3answers 108 views ### Exposition On An Integral Of An Absolute Value Function At the moment, I am trying to work on a simple integral, involving an absolute value function. However, I am not just trying to merely solve it; I am undertaking to write, in detail, of everything I ... 4answers 170 views ### Truth set of $-|x| \lt 2$? An exercise in my Algebra I book (Pearson and Allen, 1970, p. 261) asks for the graph of the truth set for $-\left|x\right| \lt 2; x \in \mathbb{R}$. I've re-stated the inequality in the equivalent ... 3answers 179 views ### Proof of triangle inequality I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :) 2answers 174 views ### How to find critical points of an absolute values function I am asked to find How many critical points does the function $g(x) = |x^2 − 4|$ have? I know that the result is $3$ but I can only find $2$. What I do, is to equal the equation to $0$, so $x^2-4=0$ ... 1answer 38 views ### Integral of a mode squared [closed] Hi could someone show me how to calculate $\psi_0$ out of the equation below? $$\int \limits^{}_{V} \big|\psi_0 \sin (\omega t - kx) \big|^2 \, \textrm{d} V = 1\\$$ 2answers 43 views ### Absolute ratios I'm curious about the following idea: suppose we have two values $P$ and $Q$, and the magnitude of the ratio $\frac{P}{Q}$ is between $0$ and $\infty$. If $P$ is smaller, then it's between $0$ and ... 3answers 106 views ### Adding equations in Triangle Inequality Proof Inequality to prove: $|a+b|\leq |a| + |b|$ Proof: $-|a| \leq a \leq |a|$ $-|b| \leq b \leq |b|$ Add 1 and 2 together to get: $-(|a|+|b|)\leq a+b\leq|a|+|b|$ $|a+b|\leq|a|+|b|$ What is the ... 1answer 57 views ### Integration Involving the Absolute Function How do I integrate the double integral of the form $|x^2-y|$ with the boundaries $-1\leq x\leq 1$ and $-1\leq y\leq 1$? 2answers 85 views ### Solving $| ax + b | \gt c$ Does $| a x + b | > c$ always result in two solutions, $x \gt \dfrac{c - b}{a}$, and $x \lt\dfrac{-c - b}{a}$? If I understand correctly, the first solution, $x > \dfrac{c - b}{a}$, is only ... 1answer 36 views ### minimum of absolute value If we consider the following problem $$\mathbb{E}[(Y-y)^2 | X=x]$$ I can easily show that the minimum with respect to $y$ occurs at $$y=\mathbb{E}[Y |X=x]$$ How can I find the minimum of ... 3answers 85 views ### Graphing Absolute Value Functions Given: $y = -|2x + 1|-3$ I came up with the graph of... $1, -7$ $0, -5$ $-1, -3$ $-2, -1$ $-3, 1$ If you were to graph this, it would turn out to be an entirely straight line. This is an ... 2answers 68 views ### Is there a lower-bound version of the triangle inequality for more than two terms? The triangle inequality $|x+y|\leq|x|+|y|$ can be generalized by induction to $$|x_1+\ldots+ x_n|\leq|x_1|+\ldots+|x_n|.$$ Can we generalize the version $|x+y|\geq||x|-|y||$ to $n$ terms too? I need ... 2answers 80 views ### Log laws and modulus If you have the log of a modulus, (like after integration), how do the log laws work? So if you have $a\ln\left|2x-3\right|$ does it become: $\ln\left|(2x-3)^a\right|$ or $\ln(\left|2x-3\right|)^a$, ... 1answer 74 views ### Absolute function continuous implies function piecewise continuous? I have a simple true/false question that I am not sure on how to prove it. If $|f(x)|$ is continuous in $]a,b[$ then $f(x)$ is piecewise continuous in $]a,b[$ Anyone that can point me in the ... 2answers 68 views ### True/false question: limit of absolute function I have this true/false question that I think is true because I can not really find a counterexample but I find it hard to really prove it. I tried with the regular epsilon/delta definition of a limit ... 1answer 83 views ### double integral of an absolute function I'm just a little unsure of how to tackle this one. I understand that typically you would separate the integral into two for where x is positive or negative, I'm just unsure of how to separate it for ... 1answer 59 views ### Separable first-order linear equation and absolute value removal We can use the integral of $\frac{1}{x}$ in order to solve a separable first-order linear equation like this: $\frac{dy}{dt} + f(t) y = 0$ $ln |y| = \left(-\int f(t)\,dt\right) + C$ and then: ... 2answers 177 views ### Why is the derivative of $\frac{|x|}{x}$ equal to $\emptyset$ at $x=0$? I got a bit of a confusion here. If $\varphi(x)=\frac{|x|}{x}$, then \varphi(x) = \left.\Bigg\{ \begin{array}{cc} 1 &if \ x>0\\ \emptyset & if \ x=0\\ -1 & if \ x <0 \end{array} ...
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http://mathhelpforum.com/differential-geometry/180307-singularity.html
# Thread: 1. ## singularity $f(z)=\frac{1}{sinh(z)}$ It's ok that z=0 is a singularity, but why $2ik\pi$, where k is a integer, are singularities too? sinh(z)=0 only when z=0 Regards 2. $\sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$ 3. Originally Posted by FernandoRevilla $\sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$ $sinh(ki\pi)=0$ so, why the argument of the sinh must be an even multiple of $i\pi$ ? 4. Originally Posted by hurz $sinh(ki\pi)=0$ so, why the argument of the sinh must be an even multiple of $i\pi$ ? You asked if $2k\pi i$ were singularities of $f(z)$ and the answer is yes. If you want to find all of them: $\sinh z=\dfrac{e^z-e^{-z}}{2}=0\Leftrightarrow\ldots\Leftrightarrow e^{2z}=1\Leftrightarrow\ldots\Leftrightarrow z=k\pi i\quad (k\in\mathbb{Z})$
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http://en.m.wikibooks.org/wiki/Space_Transport_and_Engineering_Methods/Orbital_Mechanics
# Section 1.2 - Orbital Mechanics ## Introduction Astrodynamics or Orbital Mechanics is mainly concerned with motions under gravity, either purely as a single force, or in combination with forces like thrust, drag, lift, light pressure, and others. As a topic it has a long history, with the motions of the planets, Moon, and Sun studied since ancient times, and with a scientific base starting about 500 years ago. With the advent of human-built spacecraft it has shifted from a merely observing the motions of natural bodies, to planning and executing missions. The relevance to Space Systems engineering is, of course, the need to deliver hardware to a desired destination or orbit. We will present some of the key ideas here. A more detailed and advanced introduction can be found in the Wikibook Astrodynamics and a set of Wikipedia articles at Astrodynamics - A Compendium. A good introductory printed textbook is Fundamentals of Astrodynamics, and there is an MIT Astrodynamics open course with downloadable materials. ↑Jump back a section ## Orbits An Orbit is the path that an object will follow when only affected by gravity. Orbits are Conic Sections, which are shaped generated by slicing a cone. In order of eccentricity these are circle, ellipse, parabola, and hyperbola. Circular and elliptical orbits are bound to the body being orbited and will repeat. Parabolic and hyperbolic orbits are not bound to the body, although influenced by its gravity. They will not repeat. Simple orbit calculations only consider the nearest massive body. More detailed and accurate calculations have to consider all bodies affecting the object enough to influence the accuracy of the result. Since gravity varies as the inverse square of distance, it never falls to zero, and every object in the Universe attracts every other object. But for the purpose of a given calculation only sufficiently near and massive objects will make enough of a difference to affect the result, and the rest of the Universe can be ignored. First let's consider the ideal case of a single massive object being orbited. Circular orbits have a constant velocity and distance from the center of mass of the body. This also means they have a constant Orbital Period, the time to complete one revolution around the body and return to the starting point. The circular orbit velocity, vo, for any body can be found from: $v_o = \sqrt(GM/r)$ Where G is the Gravitational constant (6.67 x 10-11 Nm2/s<supl>2), M is the Mass of the body orbited (in kg), and r is the radius to the center of the body orbited (in meters). G is a universal constant, and the mass of the Earth is essentially constant (neglecting falling meteors and things we launch away from Earth), so often the product GM = K = 3.986 x 10^14 m3/s2 is used. The orbital period, or time to complete one orbit, $T\,$ of a small body orbiting a central body in a circular or elliptic orbit is $T = 2\pi\sqrt{a^3/K}$ Escape velocity, the velocity required to escape from a body's gravity to infinity, or ve is found by $v_e = \sqrt(2GM/r)$ Since this formula is the same as that for circular orbit, except by a factor of 2 in the square root term, escape velocity is the square root of 2 (1.414+) times circular orbit velocity. Elliptical orbits will have a velocity at the nearest point to the body, or perigee, in between that of circular and escape. ### Orbit Parameters Several parameters are required to describe the location and orientation of an orbit, the shape of the orbit, and the position of an orbiting body at a given time. Axes - Periodic orbits are generally ellipses. An ellipse has a major and minor axis, which are the longest and shortest distances across the center of the ellipse. These axes are perpendicular to each other. Half of these axes, or the distances from center to edge of the ellipse, are called the Semi-major and Semi-minor axes respectively, with symbols a and b. The Semi-major axis is the value usually used to describe the overall size of an orbit. Eccentricity - The foci of an ellipse are the points along the Semi-major axis such that the sum of the distances from the foci to any point on the ellipse is constant. An orbit of a small body around a more massive one will have the massive one located at one focus of the elliptical orbit. The Focal length, f, is the distance from the focus to the center of the ellipse, and the shape of the orbit is measured by Eccentricity, e, which is defined as: $e = f/a$ The higher the eccentricity, the narrower is the ellipse relative to the semi-major axis, and the greater the difference between the nearest and farthest points of the body from the one it is orbiting. Perigee and Apogee - The prefixes peri- and ap- refer to the nearest and farthest points of an orbit from the center of the body being orbited. Different suffixes are used to indicate what body is being orbited, but perigee and apogee both means lowest and highest points of an Earth orbit, and generically that of any orbit. The general symbols are q for perigee and Q for apogee, and can be found from the formulas: $q = a-f = a(1-e)$ $Q = a+f = a(1+e)$ $f = ae$ ### Lagrange Points Given two large bodies, such as the Sun and Jupiter, and a third small body, such as an asteroid, there are five points relative to the large bodies where the net forces keep the small body in the same position relative to the two larger ones. Three of these, named L1, L2, and L3, are unstable. If you move slightly away from the exact point, you will tend to move further away. The other two, L4 and L5, are stable. Slight movements around these points will not cause the small body to drift away. L1, L2, and L3 are located between, behind, and opposite the second of the large bodies, respectively. L4 and L5 are located in the same orbit as the second large body, 60 degrees ahead and behind it. As the largest planet, Jupiter has the largest collection of asteroids in it's Lagrange points. Such natural objects are called "Trojans", since the first few at Jupiter were named after characters in the Trojan War. ### Rotation Nearly every natural body in orbit also rotates, so that a point on the surface of the body has a velocity greater than zero. This has several effects, discussed below: Rotation Period - This is the time it takes the body to complete one rotation with respect to the stars, the Sun, or the planet if it is a satellite of one. The most obvious effect of the rotation period is the day-night cycle on Earth. Some objects become locked into a rotational resonance with the parent body they orbit. This means the rotation period is a simple fraction of the orbit period. When the resonance is 1:1, it is called Tidally Locked, and the Moon is the most obvious example of that. The result is one side always faces the Earth, with a small wobble. Axial Tilt - Rotation defines an axis of rotation. The places where the axis meets the surface of the body are called poles, and midpoint of the surface between the poles is called the Equator. On smaller bodies with irregular shape, the Equator may not be well defined. On larger bodies which are more or less round, the Equator has the largest distance from the rotation axis. Axial Tilt is the angle the body axis makes with the axis of the body's orbit. The rotational inertia of large bodies causes their rotation axis to remain relatively fixed relative to the stars. As it orbits, first one pole, then the other, points towards the Sun, causing seasonal changes. Rotational Velocity - When you are on the surface of a rotating body, the circular motion about the axis produces an acceleration which can reduce gravity. The velocity and acceleration depend on the distance from the axis and the rotation period. For example, at the Earth's equator the rotation velocity is 465 m/s, which generates an acceleration of 0.0338 m/s2, or about 3% of gravity. Thus the apparent weight is less at the equator than the poles. Large bodies, more than about 1000 km in diameter, have internal forces greater than the strength of the internal materials. Since rotation lowers gravity in some parts relative to others, the body flows into an ellipsoidal, or flattened, shape. This is called hydrostatic equilibrium. The rotation of any body lowers the difference between orbit velocity and surface velocity when they are in the same direction. In the case of Earth it is 5.9% less, but in the case of some asteroids, like 4 Vesta, it can be 36% or higher. Very small objects which do not have structural flaws can even rotate faster than orbital velocity around them, producing regions where you cannot remain on the surface without mechanical aid. ### Perturbations Gravity forces extend to infinity. Therefore nearby bodies, such as the Moon and Sun for the Earth, also add an acceleration component to the gravity of the Earth. This varies over time as their direction and distance changes. The difference in gravity between the near and far sides from the nearby bodies are called Tidal Forces because they are the source of ocean tides on Earth. They also distort the shape of the solid part of the body. The gravity forces of other bodies also affect the orbit of the body as a whole. The forces besides the strongest one are called Perturbations, because they perturb the orbit caused by the strongest gravity force. On long time scales perturbations can drastically affect an orbit. This is most obvious in the case of Jupiter and comets. Comets are often near escape velocity, so small changes can easily change the orbit, and Jupiter has the most mass to cause such changes. ↑Jump back a section ## Velocity Map In space, physical distance does not matter as much as velocity, since space is mostly frictionless, and what costs you fuel is changing velocity. This graph shows the minimum ideal velocity relative to escape for the Sun's gravity well on the horizontal axis, and for planetary wells and some satellites and asteroids out to Jupiter on the vertical axis. There is no absolute reference frame against which to measure velocity. We choose escape as the zero point since it has the physical meaning of "to leave this gravity well, you must add this much velocity". Since you must add velocity to leave, the values are negative. If you have more than enough velocity to leave a gravity well, that is called excess velocity, and is measured infinitely far away. Determining Total Velocity Total mission velocity is the sum of vertical and horizontal velocity changes on the graph, both in km/s. To travel from Earth to Mars, for example, you first have to add velocity to climb out of the Earth's gravity well, add velocity to change orbit within the Sun's gravity well, then subtract velocity to go down Mars' gravity well. On the graph, that means taking the vertical axis velocity change from Earth surface to the top line, which is Solar System orbits (11.18), plus the horizontal segment to go from the Earth's orbit to Mars' orbit (2.3), plus the vertical change to go to the Martian surface (5.03). That gives a total mission velocity of 18.5 km/s, which has to be accounted for by various propulsion systems. To return to Earth, you then reverse the process. The graph shows theoretical values (single impulse to escape). Real changes in velocity (delta-V) will be higher because (1) maneuvers are not perfectly efficient, (2) orbits are elliptical and inclined, and (3) propulsion systems are not perfectly efficient in performing a given maneuver. Various losses are measured by the difference between ideal velocity, the velocity you would reach in a vacuum with no gravity well present, and the actual velocity you reach in a given circumstance. Therefore this chart is not an exact method for mission planning. It is intended to give a rough estimate as a starting point from which more detailed planning can start. Velocity Bands There are two velocity regions on the vertical axis for each planet or satellite. The lower sub-orbital region is where you have enough velocity to get off the object, but not be in a stable orbit. Those orbits will intersect the body again. So they can be used to travel from point to point on the body's surface, but not to stay in motion for multiple orbits. The higher orbital band, shown with a thicker line, indicates enough velocity for a repeating orbit. The shape of the orbit matters too, but for circular orbits the lowest point in this band is an orbit just above the surface of the body, and the highest point is an orbit just fast enough to escape from it's gravity well. Since gravity varies as the inverse square of distance, relatively large velocity changes are needed for small altitude changes near the surface of a body. Conversely, near escape velocity, relatively small velocity changes can produce large changes in altitude, and at escape it produces a theoretically infinite change. In reality there are multiple gravity wells that overlap, so escape from Earth merely places you in the larger Solar gravity well, and escape from the Sun places you in the larger gravity well of the Galaxy. Solar Orbits The top blue line represents orbits around the Sun away from local gravity wells. The surface of the two largest asteroids, 4 Vesta (-0.35) and 1 Ceres (-0.51) are marked, but the orbital bands for these objects, and the entire gravity well for most smaller asteroids, are too small to show. Instead, the range of Solar velocities are shown for Near Earth Objects and the Main Belt between Mars and Jupiter. In reality the velocities of small objects in the Solar System are spread across the entire chart. The two marked ranges are just of particular interest. The surface of Jupiter and the Sun, and their sub-orbital ranges are off the scale of this chart because of their very deep gravity wells. ↑Jump back a section ## Powered Flight ### Ascent Trajectories Circular orbit velocity at the earth's surface is 7910 meter/sec. At the equator, the Earth rotates eastward at 465 meters/sec, so in theory a transportation system has to provide the difference, or 7445 meters/sec. The Earth's atmosphere causes losses that add to the theoretical velocity increment for many space transportation methods. In the case of chemical rockets, they normally fly straight up initially, so as to spend the least amount of time incurring aerodynamic drag. The vertical velocity thus achieved does not contribute to the circular orbit velocity (since they are perpendicular), so an optimized ascent trajectory rather quickly pitches down from vertical towards the horizontal. Just enough climb is used to clear the atmosphere and minimize aerodynamic drag. The rocket consumes fuel to climb vertically and to overcome drag, so it would achieve a higher final velocity in a drag and gravity free environment. The velocity it would achieve under these conditions is called the 'ideal velocity'. It is this value that the propulsion system is designed to meet. The 'real velocity' is what the rocket actually has left after the drag and gravity effects. These are called drag losses and gee losses respectively. A real rocket has to provide about 9000 meters/sec to reach orbit, so the losses are about 1500 meters/sec, or a 20% penalty. Boost From a Non-rotating Body To go from a non-rotating body's surface to orbit requires that a rocket change its velocity from a rest velocity (zero) to a velocity that will keep the payload in orbit. If our rocket maintains a constant thrust during its ascent, then the total velocity change is $\int_0^{t_{orbit}} a\,dt = \int_0^{t_{orbit}} {T \over m}-{D \over m} - g\,dt$ where $a$ is the acceleration, $D$ is the drag, and $g$ is the planet's gravitational pull. Boost From Rotating Body ### Mass Ratio: Tsiolkovsky Rocket Equation For any rocket which expels part of its mass at high velocity to provide acceleration, the total change in velocity delta-v can be found from the exhaust velocity v(e) and the initial and final masses m(0) and m(1) by $\Delta v = v_\text{e} \ln \frac {m_0} {m_1}$ The difference between the initial mass m(0) and the final mass m(1) represents the propellant or reaction mass used. The ratio of the initial and final masses is called the Mass Ratio. The final mass consists of the vehicle hardware plus cargo mass. If the cargo mass is set to zero, then a maximum delta-v is reached for the particular technology, and missions that require more than this are impossible. ### Staging A certain fraction of a vehicle's loaded initial mass will be the vehicle's own hardware. Therefore from the above rocket equation there is a maximum velocity it can reach even with zero payload. When the required mission velocity is near or above this point, dropping some of the empty vehicle hardware allows continued flight with a new mass ratio range based on the smaller hardware mass. This is known as Staging, and the components of the vehicle are numbered in order of last use as first stage, second stage, etc. Last use is mentioned because stages can operate in parallel, so the one to be dropped first gets the lower stage number. The velocity to reach Earth orbit is approximately twice the exhaust velocity of the best liquid fuel mixes in use, thus the rocket equation yields a mass ratio of e^2 or 7.39, and a final mass of 13.5%. This percentage is close to the hardware mass of typical designs, so staging has commonly been used with rockets going to Earth Orbit. We desire a rocket with a number of stages that optimizes the economic efficiency (cost per payload unit mass). The economic efficiency depends on a number of factors, the mass efficiency being only one factor. Let us assume that we desire to launch a payload of weight P. The weight of each stage in the stack is $W_i = Pw_i$ where $w_i$ is a normalized weight for the stage. The total stack weight is thus $W = P \left (1+ \sum_{n=1}^N w_i \right )$ The change of velocity per unit mass for each stage is $\Delta v_i = I_{sp_i}\ln \mu_i$ where $\mu_i$ is the ratio of the weight before the burn of the ith stage to the weight after the burn of that stage. Thus, $\mu_i$ will always have a value greater than 1. The total change in velocity per unit mass for all the stages $\Delta v = \sum_{n=1}^N I_{sp_i}\ln \mu_i$ ↑Jump back a section ## References ↑Jump back a section
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http://mathhelpforum.com/advanced-statistics/170256-uniformly-integrable.html
# Thread: 1. ## uniformly integrable Hallo! Is the following sequence of integrable random variables $(X_h)_{h \in [0,1]}$ also uniformly integrable? $X_h=\prod\limits_{n=1}^{\frac{T}{h}}(1+\alpha_{(n-1)h}(\exp\{\mu_{(n-1)h}h+ \sigma (W_{nh}-W_{(n-1)h})\} -1))^\gamma$ whereas $W$ is a standard Brownian motion, $\sigma$ a constant $\in \matbb{R}_+,$ $\alpha_t$ is a random variable with values in [0,1], $\mu_t$ is a standard-normal distributed random variable and $\mu_t$ is continous in $t$ [0,T] is the time interval and it is required that $N:=\frac{T}{h} \in \matbb{N}$ and $\gamma$ is a constant $\in (0,1)$ I also know that $X_h$ converges in probability for $h \to 0$ to a integrable random variable $X$. I've no idea how to show it. Can anybody help me? I found out that $\mathbb{E}[X_h]\leq \mathbb{E}[X]<\infty \; \forall h \in [0,1] \quad$ and $X_h$ converges in probability to this random variable $X.$ Can I now conclude that $(X_h)_{h \in [0,1]}$ is uniformly integrable?
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http://math.stackexchange.com/questions/tagged/continued-fractions?page=3&sort=newest&pagesize=15
# Tagged Questions For questions on continued fractions. 1answer 221 views ### Is the Iterated Continued fraction from Convergent​s for Pi/2 exactly 3/2? Iterated continued fraction from convergents are described at https://oeis.org/wiki/Convergents_constant and https://oeis.org/wiki/Table_of_convergents_constants. Do you think there is any error in ... 3answers 1k views ### Extract a Pattern of Iterated continued fractions from convergents I have been working on an article at https://oeis.org/wiki/Table_of_convergents_constants where I posted a table of "convergents constants" (defined at https://oeis.org/wiki/Convergents_constant) ... 1answer 213 views ### What causes the convergence of Iterated continued fractions from convergents? Here is a small discovery I stumbled across a few weeks ago. I hope at least one person will find it interesting enough to help me. The iterated continued fractions from convergents (or convergents ... 2answers 708 views ### Motivation behind this eccentric Ramanujan Identity I just visited the MathJaX page due to the Math.SE website showing some problems while loading the page. I saw some demo math equations samples at this page, when this identity actually caught my ... 0answers 198 views ### How to find the number of continued fraction from a periodic representation? Problem Find the number that represented by $[2,2,2 \ldots]$ I know it wasn't difficult, but I was absent the last two classes. So I just want to make sure that I got it right. My attempt was, ... 1answer 128 views ### How to find continued fraction of the form $a\sqrt{b}$? For the form $\sqrt{b}$, I could just apply the recursive quadratic formula: $$P_{k+1} = a_kQ_k - P_k$$ $$Q_{k+1} = \dfrac{d - P^2_{k+1}}{Q_k}$$ $$\alpha_k = \dfrac{P_k + \sqrt{d}}{Q_k}$$ ... 1answer 322 views ### Deriving a trivial continued fraction for the exponential Lately, I learned about the following continued fraction for the exponential function: $$\exp(x)=1+\cfrac{x}{1-\cfrac{x/2}{1+x/2-\cfrac{x/3}{1+x/3-\cfrac{x/4}{1+x/4-\dots}}}}$$ I thought it was ... 2answers 188 views ### A question on continued fraction Let $a$ be a positive irrational number. Let $p_k/q_k, p_{k+1}/q_{k+1}$ be two consecutive convergents of its simple continued fraction, where $k\ge 1$. Is it possible that both ... 3answers 473 views ### Proving the continued fraction representation of $\sqrt{2}$ There's a question in Spivak's Calculus (I don't happen to have the question number in front of me in the 2nd Edition, it's Chapter 21, Problem 7) that develops the concept of continued fraction, ... 1answer 230 views ### Solving equations using continued fractions? We solve the pell equation using the continued fraction for square root of 2. What equations can we solve using the continued fraction of cube roots (and other numbers too)? 0answers 387 views ### What can Euler's identity teach us about (generalised) continued fractions? We know that $$e^{i \pi} = -1 .$$ We can transform all of the components of this identity into (generalized) continued fractions. When we start of with $\pi$, we see that \Big(3+ ... 1answer 213 views ### What is the length of a continued fraction expansion of a rational number? I was reviewing quantum factorization and am slightly unclear on a classical detail of order-finding. Given a (suitably nice) periodic function $f$ with unknown period $r$ and a power of two \$N > ... 1answer 381 views ### Continued Fraction expansion of tan(1) Prove that continued fraction of tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,...]. I tried using the same sort of trick used for finding continued fractions of quadratic irrationals and trying to find a ... 2answers 251 views ### Adding integers to an infinite continued fraction expansion doesn't change the value? I'm learning about continued fractions, and I've enjoyed them so far, but I'm unsure if I've done the following correctly. I have no real experience with analysis, so I'm not sure if my reasoning is ... 1answer 348 views ### Continued Fraction of an Infinite Sum What is the continued fraction for $\displaystyle\sum_{i=1}^n\frac{1}{2^{2^i}}$ It seems to be "almost" periodic, but I can't figure out the exact way to express it. 3answers 571 views ### Why are some mathematical constants irrational by their continued fraction while others aren't? Catalan's Constant and quite a few other mathematical constants are known to have an infinite continued fraction (see the bottom of that webpage). On wikipedia (I'm sorry, I can't post anymore ... 2answers 829 views ### How do I prove the partial denominators formula of the Bauer-Muir transformation of a generalized continued fraction? Notation: $b_{0}+\underset{n=1}{\overset{\infty }{\mathbb{K}}}\left( a_{n}/b_{n}\right)$ is the Gauss Notation for generalized continued fractions. Description of the Bauer-Muir transformation ... 2answers 666 views ### Continued fraction for $\frac{1}{e-2}$ A couple of years ago I found the following continued fraction for $\frac1{e-2}$: $$\frac{1}{e-2} = 1+\cfrac1{2 + \cfrac2{3 + \cfrac3{4 + \cfrac4{5 + \cfrac5{6 + \cfrac6{7 + \cfrac7{\cdots}}}}}}}$$ ... 1answer 573 views ### A nicer proof of Lagrange's 'best approximations' law? Let $p_N/q_N$ be the $N^\text{th}$ convergent of the continued fraction for some irrational number $\alpha$. It turns out that for any other approximation $p/q$ (with $q \le q_N$) which isn't a ...
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http://math.stackexchange.com/questions/39735/induced-map-on-eilenberg-maclane-space
# Induced map on Eilenberg-MacLane space Let $X$ be an $n-1$ connected space. Why does a map $X\rightarrow K(\pi_n(X),n)$ that induces an isomorphism on $\pi_n$ exist and what is this map? - ## 3 Answers If $X$ is a CW-complex, then you can build a $K(\pi_n(X),n)$ by attaching cells (of dimension $n+2$ and above) to $X$ to kill off the higher homotopy groups. The map you are thinking of is then an inclusion of $X$ into $K(\pi_n(X),n)$ as a sub-complex. - As stated, the statement is false. For example, let $X$ be the Hawaiian earring. Then $X$ is 0-connected. But $X$ can't have a map to an Eilenberg-Maclane space that's an isomorphism on $\pi_1$, because as a compact space it would have to have compact image, which would necessarily have finitely-generated $\pi_1$ in a CW complex like an Eilenberg-Maclane space, and $X$ has infinitely generated fundamental group. The statement is true for if $X$ is a CW-complex, as wckronholm explains. - It's a bad habit that a lot of us get into--thinking that all spaces are (almost) CW-complexes. – Josh May 17 '11 at 22:28 1 Indeed - whenever I see a homotopy theoretic claim about "all spaces", I immediately think, "Can I break this with the topologist's sine curve or the Hawaiian earring?" – MartianInvader May 18 '11 at 13:42 Let me assume $n>1$ and —so that your claim is true— that $X$ is a CW complex. The Hurewicz theorem, in view of the hypothesis, gives us an isomorphism $$\phi:H_n(X,\mathbb Z)\to\pi_n(X).$$ On the other hand, in view of the hypothesis made on $X$, the universal coefficient theorem for cohomology gives us an isomorphism $$\alpha:H^n(X,\pi_n(X))\to \hom(H_n(X,\mathbb Z), \pi_n(X))$$ Moreover, there is a canonical bijection $$\beta:[X,K(\pi_n(X), n)]\to H^n(X,\pi_n(X))$$ with $[X,K(\pi_n(X), n)]$ the set of homotopy classes of maps $X\to K(\pi_n(X),n)$. The map you are looking for is any $f:X\to K(\pi_n(X),n)]$ whose homotopy class $[f]$ is such that $\alpha(\beta([f]))=\phi$. (This works for any $K(\pi_n(X),n)$ that you pick) -
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http://quant.stackexchange.com/questions/3789/are-there-any-standard-techniques-for-adding-realistic-synthetic-microstructure?answertab=oldest
# Are there any standard techniques for adding realistic synthetic microstructure noise to a price series? This may seem like a strange question, but for my particular application we need to actually add synthetic microstructure noise to real time charts. The signal should still be representative of the aggregate market direction. I expect that a good technique would be something related to signal processing in electronics or sound engineering. They have white noise generators that can be restricted to a band. I would rather something far less complicated though. Is it perhaps good enough to take a random percentage of the actual change from the last difference? - 1 Can you expand on the application, the motivation or, if that's sensitive information, on general scenarios? In particular, (1) the relation (formal or heuristic) to band restricted WN generators and (2) what do you have in mind as a benchmark for "good enough" (again, formally or heuristically). – Ryogi Jul 11 '12 at 18:16 It's basically a gaming like situation where I need to present charts that respect the pattern and have real time feed updates but the prices must never actually be the real prices. I don't actually have a defined statistical envelope for the distance to the real price though. – barrymac Jul 12 '12 at 9:26 ## 3 Answers Why not use a Fair value model to predict prices and then randomly sample from the difference of the model and actual prices. The errors would be as good as your model - 2 Fair value is hardly a concept applied to frequencies wherein microstructure noise is relevant. – Ryogi Jul 11 '12 at 18:17 I'm not familiar with fair value models but from my first scan around the net about it doesn't look like a dynamic enough technique. I did come across something relating to VIX futures though so I am happy to be corrected. Perhaps you could add a bit more information? – barrymac Jul 12 '12 at 9:30 I assume, this is not for real-time display, so you can use the price from future. If this is not the case, this answer is irrelevant. I don't know about a standard technique, but this is my suggestion: $p_{noise} = p_{current} + \nu * (p_{future} - p_{current})$ where $p_{future}$ is future price for some horizon, and $\nu$ is a zero-mean Gaussian noise. - BTW, this can also be applied on correlated portfolio. You need to generate the noise $\nu$ to reflect the portfolio's covariance – Serg Jul 12 '12 at 18:07 it is for real time display as it happens but in any case I think you can still do something like this based on past prices and would be fine, as long as V is in the right range. – barrymac Jul 13 '12 at 9:50 @Serg Is this (using the difference between future and current) any better than using the difference between previous and current? I.e. your answer implies that if the market is about to get more volatile (bigger jumps between prices) then there is more noise in current. Is that an observable phenomena? – Darren Cook Jul 17 '12 at 22:45 Maybe the accepted answer to this earlier thread and the more detailed description on my blog might be of use to you. Within the FFT you could just manipulate the higher frequency components to create your synthetic microstructure noise. -
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http://math.stackexchange.com/questions/tagged/computational-mathematics+computer-algebra-systems
# Tagged Questions 2answers 53 views ### What free software can I use to solve a system of linear equations containing an unknown? Question: What free software can I use to solve a system of linear equations $M\mathbf{x}=\mathbf{y}$ where the entries of $\mathbf{y}$ vary with an unknown quantity $n$? Presumably I could do ... 1answer 60 views ### What is the fastest computational graph theory package? What is the fastest computational graph theory package with respect to executing algorithms and computing graph theoretic data? I am aware of this related question, which requests graph theory ... 1answer 47 views ### How do I determine if two of my software's representation of algebraic numbers are equal? I have software which stores information about algebraic numbers with absolute precision. If you build it up by creating instances of a Python representation of an integer, float, Decimal, or string, ... 1answer 73 views ### Best graphing program for Mac or PC? I just bought the highest end iMac, with a student discount, of course, and was wondering what is the best graphing program out there. A program that can graph any equation that I throw at it AND one ... 0answers 107 views ### A (contour?) integration (even if by using Mathematica!) I need to be able to calculate integrals of this type where the sum over $R$ is the sum over representations of a Lie group $G$ on whom $dU$ is the Haar measure and $\chi _ R ()$ is the character of ... 1answer 101 views ### What's the best way to detect an algebraic number? Suppose you calculate the first few (dozen, hundred) digits of a number which you believe to be a rational number. You can calculate the continued fraction for the number and truncate after a large ... 1answer 676 views ### Plot Y-Range on Mathematica I have a plot that I would like to slightly manipulate in Mathematica. Here is the code I am entering: Plot[{x, 2^x, log_2(x)}, {x, -1, 3}] As you can see $x$, $2^x$, and $log_2(x)$ are all ... 1answer 308 views ### Computing with ideals: over $K$ or over $\mathbb{Q}\subseteq K$? does it matter? I'm beginning to learn to use SINGULAR, the computer algebra system (CAS) for commutative algebra. NOTATION: If $K$ is a field of characteristic $0$, then $\mathbb{Q}\subseteq K$; otherwise ... 0answers 56 views ### Computing relations on the columns of a matrix Given an $m\times n$ (with $n>m)$ matrix $M$ over a polynomial ring $R=k[x_1,...,x_n]$, suppose that every column of $M$ is an $R$-linear combination of $m$ specified columns. I would like to ... 1answer 97 views ### Find $k^{th}$ root of $M \in GL(n,F_2)$ Given $M \in GL(n,F_2)$ which is known to have a $k^{th}$ root. How can I find a root algorithmically? Can I find all roots? Other than being invertible and having a $k^{th}$ root I know nothing of ...
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http://mathoverflow.net/questions/70657/hyperbolic-coxeter-polytopes-and-del-pezzo-surfaces/71606
## Hyperbolic Coxeter polytopes and Del-Pezzo surfaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Added. In the following link there is a proof of the observation made in this question: http://dl.dropbox.com/u/5546138/DelpezzoCoxeter.pdf I would like to find a reference for a beautiful construction that associates to Del-Pezzo surfaces hyperbolic Coxeter polytopes of finite volume and ask some related questions. Recall that a hyperbolic Coxeter polytope is a domain in $\mathbb H^n$ bounded by a collection of geodesic hyperplanes, such that each intersecting couple of hyperplanes intersect under angle $\frac{\pi}{n}$ ($n=2,3,...,+\infty$). Del Pezzo surface is a projective surface obtained from $\mathbb CP^2$ by blowing up (generically) at most $8$ points. Now, the construction Del Pezzo $\to$ Coxeter polytope goes as follows. Consider $H_2(X,\mathbb R)$, this is a space endowed with quadratic form of index $(1,n)$ (the intersection form), and there is a finite collection of vectors $v_i$ corresponding complex lines on $X$ with self-intersection $-1$. It is well known, for example that on a cubic surface in $\mathbb CP^3$ there are $27$ lines and this collection of lines has $E_6$ symmetry (if you consider it as a subset in $H_2(X,\mathbb Z)$). Now we just take the nef cone of $X$, or in simple terms the cone in $H_2(X,\mathbb R)$ of vectors that pair non-negatively with all vectors $v_i$. This cone cuts a polytope from the hyperbolic space corresponding to $H_2(X,\mathbb R)$, and it is easy to check that this polytope is Coxeter, with angles $\frac{\pi}{2}$ and $0$ (some points of this polytope are at infinity, but its volume is finite). Indeed, angles are $\frac{\pi}{2}$ and $0$ since $v_i^2=-1$, $v_i\cdot v_j=0 \;\mathrm{or}\; 1$. Example. If we blow up $\mathbb CP^2$ in two points this construction produces a hyperbolic triangle with one angle $\frac{\pi}{2}$ and two angles $0$. The connection between algebraic surfaces an hyperbolic geometry is very well-known, and exploited all the time but for some reason I was not able to find the reference to this undoubtedly classical fact (after some amount of googling). So, Question 1. Is there a (nice) reference for this classical fact? This question is motivated in particular by the following article http://maths.york.ac.uk/www/sites/default/files/Preprint_No2_10_0.pdf where the polytope corresponding to the cubic surface is used. The authors mention the relation of the polytope to 27 lines on the cubic, but don't say that the relation is in fact almost canonical. Question 2. The group of symplectomorhpisms (diffeos) of each Del-Pezzo surface $X$ is acting on $H_2(X,\mathbb R)$, let us denote by $\Gamma$ its image in the isometries of corresponding hyperbolic space. What is the relation between $\Gamma$ and the group generated by reflections in the faces of the corresponding Coxeter polytope? PS It one considers rational surfaces with semi-ample anti-canonical bundles, i.e. surfaces that can have only rational curves with self-intersection $-1$ and $-2$ one gets more examples of Coxeter polytopes; the faces of such polytopes intersect under angles $(\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{4}, 0)$. Here is a reference on "Algebraic surfaces and hyperbolic geometry" (but I don't think that the answer to question one is contained there) : http://www.dpmms.cam.ac.uk/~bt219/algebraic.pdf - ## 3 Answers Perhaps the nice article "Reflection groups in Algebraic Geometry" of Dolgachev describes this: link http://www.math.lsa.umich.edu/~idolga/reflections.pdf Other places to look: papers of Vinberg and Nikulin, and Manin's book on "Cubic forms". - 1 Dear Abhinav, this is a nice article indeed. I went through it (very) quickly before posting my question. Unfortunately I have not found in it yet the following statement: Nef cone of a rational surface with semi-ample anti-canonical bundle is a Coxeter polytope of finite volume. I feel like this statement should be included in every book mentioning Fano surfaces. But I have not found it yet (I have not found it in Manin's book either). So I would really appreciate a precise reference (at least for the case of Del -Pezzo surface). – Dmitri Jul 19 2011 at 15:17 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I tried to investigate the question for some time and sent emails to several experts in the field. For the moment the conclusion seems to be the following: The statement is well known to experts, was used many times, but was never written as a lemma, propostition, or a theorem in any article, or book. The only relatively unambiguous reference that I managed to find are lines 14-0 from the bottom of page 4 in the article of Viacheslav V. Nikulin "On algebraic varieties with finite polyhedral Mori cone" that you can find here http://arxiv.org/abs/math/0305040 It might be, that looking better one would be able to find an actual reference, and I would beg anyone who knows such a reference to give the answer to the question... Meanwhile I would like to formulate a beautiful (and apparently strongest) known result in this direction, that is omnipresent in the book of Alexeev and Nikulin (I would like to thank both authors for pointing out this to me), but again never stated explicitly.The name of the book is Classification of log del Pezzo surfaces of index $\le 2$, you can find it here: http://arxiv.org/abs/math/0406536 Theorem, Alexeev, Nikulin. Let $S$ be a Log del Pezzo surface of index 2. Then $S$ admits a particular (possibly non-minimal) resolution of singularities so that the set of unit vectors in the Nef cone of the resolution is a hyperbolic Coxeter polytope of finite volume. Here is an explanation of the terminology. log-del Pezzo surface is a (possibly singular) rational surface with ample anti-canonical bundle. The index of such surface is the minimal number $n$ such that $nK$ is a Cartier divisor. Log del Pezzo surface of index two have only quotient singularities, isomorphic to $\mathbb C^2/\Gamma$ where $\Gamma$ is a finite group with elements with determinants $\pm 1$. The proof of the theorem is a combination of the Cone theorem (this gives the finiteness of the volume) and the following simple Observation. Suppose we have vectors $v_i$ in $\mathbb R^{1,n}$ such that $v_i\cdot v_j=0,1$, $v_i\cdot v_i=-1,-2,-4$, and $v_i \cdot v_j=0$ if $v_i\cdot v_i=-2$, $v_j\cdot v_j=-4$. Then the angles between hyperplanes orthogonal to vectors $v_i$ can be only $\pi/2$, $\pi/3$, $\pi/4$, or $0$. Finally one just needs to find an appropriate resolution of singularities of $S$. - For $Q2$, in my paper joint with T.-J. Li http://arxiv.org/abs/1012.4146 there's a description by reflections of $\Gamma$, and Shevchishin proved the same conclusion in http://arxiv.org/abs/0904.0283 but his language is more Coxeter. We actually dealt with all rational surfaces. -
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http://www.scholarpedia.org/article/Aloha_random_access
# Aloha random access From Scholarpedia Norman Abramson (2009), Scholarpedia, 4(10):7020. Jump to: navigation, search Curator and Contributors 1.00 - Norman Abramson Figure 1: Multiple Transmitters Sharing an ALOHA channel. Aloha random access is a widely used technique for coordinating the access of large numbers of intermittent transmitters in a single shared communication channel. In an ALOHA channel each transmitter sharing the channel transmits data packets at random times. In most ALOHA channels the transmitters then rely on some protocol (such as repetition) to handle the case of packets lost due to interference by other packets. An ALOHA channel may also just provide a best effort delivery mechanism and leave it to the receiver to deal with lost packets. ## History ALOHA channels were originally analyzed and implemented in the AlohaNet at the University of Hawaii in 1970. The AlohaNet utilized UHF radio channels to connect computer resources on the islands of Oahu, Maui and Hawaii in the state of Hawaii (Abramson, 1970). In 1973 ALOHA was demonstrated in PacNet, a Pacific Ocean experimental satellite network involving NASA, the University of Hawaii and the University of Alaska, Tohoku University and the University of Electro-Communications in Japan, and the University of Sydney in Australia (Abramson, 1985). An ALOHA random access channel was used by Dr. Robert Metcalfe in 1973 as the basis of the Xerox cable-based Alto ALOHA Network later renamed and developed as Ethernet by 3COM (Metcalfe and Boggs, 1976). Since the 1980’s ALOHA has been the primary random access mechanism utilized by mobile telephone networks, satellite data networks, DOCSIS based cable data networks, Ethernet, WiFi and WiMAX. ## Operation of an ALOHA Channel An ALOHA channel provides access to a common communication channel from multiple independent packet transmitters by the simplest of all mechanisms. When each transmitter is ready to transmit its packet, it simply transmits the packet burst without any coordination with other transmitters using the shared channel. If each user of the ALOHA channel is required to have a low duty cycle, the probability of a packet from one user overlapping and thus interfering with a packet from another user is small as long as the total number of users on the shared ALOHA channel is not too large. As the number of users on the shared ALOHA channel increases the number of packet overlaps increase and the probability that a packet will be lost due to an overlap with another packet on the same channel also increases. Figure 2: Packets in an ALOHA Random Access Channel. The key question of how many such users can share an ALOHA random access channel is dealt with in Section 4. ## Throughput of an ALOHA Channel The start times of the packets in an ALOHA channel may be modeled as a Poisson point process with parameter $$\lambda$$ packets/second. If each packet in the channel lasts $$\tau$$ seconds, the normalized channel traffic can be defined as $\tag{1} G = \lambda \tau$ If only those packets which do not overlap with any other packet are received correctly, there is packet rate $$\lambda'< \lambda$$ defining the rate of occurrence of packets received correctly. Then the normalized channel throughput of the ALOHA channel can be defined as $\tag{2} S = \lambda' \tau$ and the normalized throughput of an ALOHA random access channel is given by (Abramson, 1970) $\tag{3} S = G e^{-2G}$ The maximum value of the normalized throughput of an ALOHA channel is equal to $$\frac{1}{2e}=0.184$$ and occurs when the traffic G is equal to 0.5. Figure 3: ALOHA Channel Throughput vs. Channel Traffic. It is possible to modify the completely unsynchronized operation of the transmitters on an ALOHA random access channel in order to increase the maximum throughput of the channel. If a synchronized time base is established in the ALOHA channel to define a sequence of slots of the same duration as a packet transmission and each transmitter in the ALOHA random access channel is required to start any packet transmission at the start of a slot, the resulting channel is referred to as a slotted ALOHA channel (Roberts, 1975; Abramson, 1977). In a slotted ALOHA channel any overlap of two or more packets is a complete overlap and the elimination of partial packet overlaps results in an increase of channel throughput in a slotted ALOHA channel. The maximum throughput of a slotted ALOHA channel occurs when the channel traffic is equal to 1.0 and the maximum throughput is equal to 1/e = 0.368 or exactly twice the value for the unslotted ALOHA channel. In practice the use of the slotted ALOHA channel can result in less improvement than this result might indicate or even in a decrease in the channel throughput. If the transmitter packets are not all of the same duration then the loss of throughput due to wasted portions of fixed length slots can be greater than the factor of two improvement promised by slotting (Abramson, 1977). A wide variety of reservation techniques have been proposed and implemented which can increase the maximum throughput of an ALOHA channel by reserving packet transmission times when the transmitter has a long sequence of packets to transmit (Crowther, 1973). Slotted ALOHA and reservation ALOHA are sometimes referred to as S-Aloha and R-ALOHA respectively. Beginning in 1990 the connection between spread spectrum code division multiple access (CDMA) and ALOHA random access has been of interest. The combination of these two access technologies for both satellite and terrestrial wireless channels is referred to as Spread ALOHA (Abramson, 1990). ## Applications The first commercial application of ALOHA channels was launched in 1976 by Comsat General in the Marisat maritime satellite communications system. At about the same time Metcalfe working with a group from DEC, Intel and Xerox (the DIX Group) formulated an open Ethernet standard based on the Alto ALOHA network. Since 1983 ALOHA channels have been adopted for use in all major mobile telephone standards (1G, 2G and 3G) as the control channel and then for a variety of packet data channels integrated into these voice networks (e.g. GPRS and UMTS). ALOHA has also been adopted for use in a variety of protocols used in wired networks, CSMA/CD and CSMA/CA in single channel local area networks and DOCSIS for commercial cable networks. ## References [1] Abramson, N. (1970) The ALOHA System – Another Alternative for Computer Communications, AFIPS Conference Proceedings, Vol. 37, pp.281-285, November, 1970. [2] Abramson, N. (1977) The Throughput of Packet Broadcasting Channels, IEEE Transactions on Communications, Vol. COM-25, No. 1, pp 117-128, January 1977. [3] Abramson, N. (1985) Development of the AlohaNet, IEEE Transactions on Information Theory, Vol. IT-31, No. 2, pp. 119-123, March 1985. [4] Abramson, N. (1990), VSAT Data Networks, Proceedings of the IEEE, vol.78, no. 7, pp 1267-1274, July, 1990. [5] Abramson, N. (2009), The AlohaNet -- Surfing for Wireless Data, IEEE Communications Magazine, vol.47, no. 12, pp 21-25, December, 2009. [6] Binder, R. et al. (1975), ALOHA Packet Broadcasting – A Retrospect, Proceedings of the National Computer Conference, Vol. 44, pp. 203-215, AFIPS Press. [7] Crowther, W. et al. (1973), A System for Broadcast Communication: Reservation ALOHA, Proceedings of the 6th HICSS, University of Hawaii, Honolulu, January, 1973. [8] Metcalfe, Robert M. and Boggs, David R. (1976), Ethernet: Distributed Packet Switching for Local Computer Networks, Communications of the ACM, Vol. 19, No. 7, July 1976. [9] Roberts, Lawrence G., (1975), ALOHA Packet System with and without Slots and Capture, Computer Communications Review, vol. 5, No. 2, pp. 28-42, April, 1975. Internal references • Lajos Hanzo (2008) Global system for mobile communications (GSM). Scholarpedia, 3(8):4115. • Arkady Pikovsky and Michael Rosenblum (2007) Synchronization. Scholarpedia, 2(12):1459. ## See also • GPRS (General Packet Radio Service) • EDGE (Enhanced Data Rates for GSM Evolution) • UMTS (Universal Mobile Telecommunications System) • WiFi (Wireless Fidelity) • WiMAX (Worldwide Interoperability for Microwave Access) • Zigbee • DOCSIS (Data Over Cable Service Interface Specification)
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http://math.stackexchange.com/questions/271527/region-of-convergence-of-z-transform-connected-area
# Region of convergence of Z-Transform connected area? Shouldn't the Region of Convergence of the Z transform be a connected area ? In Oppenheim solution manual, I've found this answer of a question that asks to determine the different forms of the impulse response of a certain system in the (d) option it shows that the ROC might be $\frac{1}{2} < |z| or |z| > 2$ Why is that ? - ## 1 Answer It looks definitely wrong to me (besides, the last term should be $u[-n-1]$ instead of $u[n-1]$). In any case, the $\frac{2}{3}2^n u[n]$ part only is valid for $|z|>2$, it's makes no sense to state it as valid solution for $|z|>2$ or $|z|<1/2$. Is that from the "official" solution book? -
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http://www.physicsforums.com/showthread.php?p=4158805
Physics Forums Page 3 of 4 < 1 2 3 4 > The problem with infinity. Quote by bahamagreen How about addressing my previous post #28 first? Quote by bahamagreen I'm suggesting that is a flaw because you are defining some persons as both a non-customer and a customer - because the naturals and reals share some members in common (all naturals are members of the reals, some reals are members of the naturals). Yes, that is why I specified multiplying by an irrational, which is a subset of the real numbers but not a subset of the natural numbers. Hence the symmetry is complete. Quote by my_wan I can also relabel all natural numbers as real numbers simply by multiplying their name tags with an irrational number and assigning them that number. Recap: Set of all people (customers and potential customer) = real numbers Set of all actual customers = natural numbers Subset of non-customers (potential customers) = irrational numbers Given that the set of all potential customers is larger than the set of all actual customers, whether irrationals are contained in the set of reals or not, there remains more potential customers than actual customer. I specified an irrational for the explicit purpose of of avoid a clash with your naturals (set of actual customers), in spite of the fact that there are more reals than naturals, which technically mooted the rebuttal anyway. bahamagreen, I can sympathize with your difficulty on the hotel paradox. I have tried thinking through possible ways of getting around it. All of which involve refining definitions more than the paradox makes explicit. I'll try to construct a version of your argument that is harder to deconstruct. Though I will not offer any proof either. Neither does it reject infinities. For instance, if you compare the statements: (1) Hilbert's hotel contains an infinite number of rooms in which each room contains an occupant. (2) Hilbert's hotel contains an infinite number of rooms and occupied by an infinite number of guest, for which the cardinal numbers are equal. The hotel paradox essentially assumes these statements are equivalent. I suspect that this is not fully justified. There is only one countably infinite cardinal, $\aleph_0$, but there are uncountably many countably infinite ordinals ω. By definition in statement (1) we have assigned a one to one correspondence between the number of occupants and the number of rooms. Thus the one to one correspondence is in reference to ordinals rather than cardinals. Now the equivalence of the above 2 statements is predicated on the fact that ω + 1 = ω, i.e., addition and multiplication are not commutative. Seems straightforward enough, just as 0*1=0 and 1 + 0 = 1. However, if we look to calculus, 0 may not equal 0, but rather an infinitesimal ΔL, the inverse of an infinity. In calculus we must make use of these limits specifically to avoid these self same ordinal properties we associate with 0, and inversely infinity. If calculus requires us to avoid this property with respect to zero, why is 1/ΔL special? ΔL simply has the equivalence class of 0, wrt a finite interval. This wouldn't change much mathematically in operational terms, but would allow us to make a distinction between statements (1) and (2). In both ΔL and 1/ΔL the only thing that changes is the ordinal, not the cardinal. This would dictate that if the ordinal by definition has a one to one correspondence then there simply is no room to add another guest, though the cardinal remains the same up to $\aleph_1$. We can also still accommodate more guest, when ω_1 = ω_2, under the condition that switching rooms requires some finite time interval, or a time interval with a cardinal number less than the cardinal number of guest. Can anybody destruct that argument? It would be interesting to try and prove also. Quote by my wan How do you propose the reinstate consistency if you reimpose a finiteness condition on the physical world? I am not trying to impose a finiteness condition on the physical world. In fact I would reason that the cosmos (= everything that exists) must be infinite, otherwise we would not be here. The Universe may be finite; I believe there are very good arguments that it is; but this is simply the result of our limited perception. Here I am not talking about the limits of our equipment, but the fact that we observe everything within the restrictions of our 3+1 dimensions, which must not match the dimensionality of the infinite cosmos. Think of a spider walking through Flatland. i agreed that universe is expanding regularly.. thats why we cant get the actual shape of it dude Something happens before something and something happens after something. It means there is always beginning for the beginning and there is always beginning after ending according to Thermodynamics law. Universe, Universes, Multiverses, Infiniverses and etc are incomprehensibly Infinite ∞ It's been a little while, but I think I know what my problem is with Hilbert's Hotel. The problem posed in the story is that the infinity of rooms are each already occupied. So the question is how to assign a room to the new guest... I'm seeing an equivalence between the new guest and his new room. In looking at how both the guest and the room might be potential members of their infinite sets, the additional guest is allowed to exist and show up, but the additional room is not allowed to exist and thereby causes the assignment problem for the hotel manager... Why is it that in spite of an infinite number of guests already assigned to rooms, another guest is allowed to exist, yet of the infinite number of rooms, another one is not allowed to exist and be found? Another way to look a this is to break the thing into two independent questions: 1] Given an infinite hotel of rooms, are there any additional ones out there? Hilbert says, "No"... 2] Given an infinite world of guests, are there any additional ones out there? Hilbert says, "Yes". From that difference he presents the paradox of solving the match up of guests to rooms... but why the two answers to the same kind of question? That is the premise flaw I see here... Unless I'm still missing something. Recognitions: Homework Help Science Advisor Why is it that in spite of an infinite number of guests already assigned to rooms, another guest is allowed to exist, yet of the infinite number of rooms, another one is not allowed to exist and be found? You can look at a different setup where an additional guest shows up and an additional room is built, but that is trivial to solve. Hilbert does not say "there is". The question is "imagine that, ... , how can we solve it?". By the way: To make room for the additional guest, an infinite number of guests have to move. This is certainly annoying for them. And "a bit annoying" for an infinite amount of guests is worse than "very annoying" (sleeping in the corridor) for one guest ;). "The question is "imagine that, ... , how can we solve it?"." If that is the case, then it is solved by noticing that the premise is based on a clear logical inconsistency. If you imagine a hotel with infinite rooms, you have to apply the same logic to imagining an infinite population of guests... they are equivalent and need to be treated identically when considering the existence of an additional element. The point of the premise is that two different conditions are being applied to two logically identical objects, one condition allows no new elements and the other does allow new elements. Basically, the premise includes a guest without a room showing up, then more guests, then bus loads of guests, etc. If there are infinite guests already in the rooms of the hotel, and more guests are allowed to appear, then the same applies to the rooms; there are an infinite number of rooms, but more can be found. To say no more rooms can be found is the same as saying no additional guests can appear. "Infinite" may or may not entail "all", but either way needs to be applied to both the rooms and the guests... If infinite means "all", then the infinity of guests in rooms already means no additional guest can show up. If infinite does not mean "all", then there is at least one additional unoccupied room in the hotel. Recognitions: Homework Help Science Advisor @bahamagreen: That does not make sense. Imagine I have 3 bananas and give them to 3 monkeys. Each monkey gets a banana. Imagine I have 4 bananas and give them to 3 monkeys. You can imagine that, right? Even after I added a banana. It is an imaginary situation, I can use any numbers I like. I can distribute infinite bananas on 3 monkeys - at least one monkey has to get an infinite amount of bananas, so what? I can distribute infinite bananas on an infinitely many monkeys. And then I can take another banana and give it to monkeys. Quote by bahamagreen Basically, the premise includes a guest without a room showing up, then more guests, then bus loads of guests, etc. If there are infinite guests already in the rooms of the hotel, and more guests are allowed to appear, then the same applies to the rooms; there are an infinite number of rooms, but more can be found. To say no more rooms can be found is the same as saying no additional guests can appear. It doesn't matter whether new rooms can be found or not, because you can comfortably fit the new guests in the rooms that are already occupied. You're not getting the point. Forget about the methods of putting extra guests into the rooms. Look at the facts of the problem: Hotel has Infinity of guests Infinity of rooms If you allow that an additional guest (without a room) can exist, you must also allow that an additional room (without a guest) can exist. These two things are the logically identical, therefore the premise that the new guest has no room is false. If a hotel with infinity of rooms does not have an empty room available, then likewise, with an infinity of guests with rooms, there will not be possible a new guest without a room. Look at it this way; what if the original paradox had been reversed? The hotel has an infinity of rooms and guests... Then one day a new empty room is discovered, but the manager needs to report full occupancy to be paid his bonus. So he shuffles the guests through the rooms (including the new one). The simple solution is for anther guest to appear (like in the original). See, the discovery of a new guest is just like the discovery of a new room. If the universe were infinite and existed for an infinite amount of time and the universe weren't expanding, and the universe had the same kind of distribution of galaxies everywhere, then night would be bright as day because no matter what direction you look there would be light coming from that direction. But, due to red-shift, far away galaxies may no longer appear in the visible spectrum (but you could measure it). And if the Universe expanded fast enough outer parts could become "disconnected". So it's a little harder to figure out. Bahamageen, it’s refreshing to find someone who thinks along the same lines as I do about infinity. I believe the main problem is that since Cantor “tamed” infinity there have been two distinct perspectives; the mathematical infinity and the real infinity. Much confusion arises because there is such a profound difference between the two, yet we can discuss infinity with one person using one form and the other person using the other form. Mathematical infinities are simply mathematical concepts that are of value in some calculations. There exists an infinite number of mathematical infinities which can be of different sizes. Your arguments apply to real infinity. There can be only one, and it must include everything. If you try to introduce numbers to such an infinity your calculations lead to nonsensical answers. . Take for example the (UK) national Lottery. In an infinite universe, an infinite lottery becomes possible, and therefore inevitable, not only that, it must occur an infinite number of times. So, what would this infinite lottery be like? There would be an infinite number of people taking part, the staked money would be infinite, therefore, the jackpot (being a percentage of the stake) would also be infinite, the jackpot winners (being a percentage of the infinite number of people taking part) would be infinite, as would the number of losers. We can see from this that an infinite number of people would win an infinite share of an infinite amount of money, but, paradoxically, the same infinite number of people would not be winners at all. Mathematicians can find their way, logically, through Hilbert’s Hotel, followed by infinite guests grumpily changing rooms; but in the real world it calls to mind the well known debate about angels and pins. Quote by bahamagreen You're not getting the point. Forget about the methods of putting extra guests into the rooms. Look at the facts of the problem: Hotel has Infinity of guests Infinity of rooms If you allow that an additional guest (without a room) can exist, you must also allow that an additional room (without a guest) can exist. These two things are the logically identical, therefore the premise that the new guest has no room is false. If a hotel with infinity of rooms does not have an empty room available, then likewise, with an infinity of guests with rooms, there will not be possible a new guest without a room. . You're not getting the point that it's not important whether there is a room for the guest or not: you don't need it even if there is. You can fit the new guest into one of the old rooms and not use any new room. The premise that there is no room for the guest is not a premise of the mathematics of the paradox but only of the "story" behind it. The simple mathematical idea is wrapped in a textual anecdote and that anecdote requires the "no room" premise for the problem of fitting the new guest to exist. You can just as well state that the hotel manager is a freak who loves to move guests around. Nothing changes the fact that you can find a bijective function from integers to naturals (for example) which is all this "problem" is about. I understand it's just a story... but the "story" has a flaw, an inconsistent logical treatment of the rooms vs the guests. It makes no sense to allow for a "new" guest to be found and them maintain that a "new" room can't be found. I understand that IF you take the first statement as a given, THEN yes, the story acts fine as a puzzle for how to place the new guest in a room, in spite of all the infinite rooms already being occupied by the infinite guests. The problem with this story is that anyone who thinks about the origin of the new guest must logically conclude that the first statement is inconsistent... that statement being that no new rooms can exist. I'm pointing out a separation between two things: The story as presented (which may be fine for setting up how to solve a problem with infinities) and Looking at the problem on its face and realizing that it is a false problem. The reason I think this is important is because if the question (story) is allowed to contain illogical inconsistencies, what rules then prevent the answer from also using logical inconsistencies? If the set up for the story is "wrong", then how is any answer "right"? Or how can any answer that just magically and illogically "solves" the problem be legitimately denied? I expect the rules that apply to evaluating the answer to be the same rules that must apply to the construction of the question. Hilbert's Hotel does not meet that expectation; the problem needs to be logically repaired before being asked, and after the repair it is no longer a problem, no paradox. For learning purposes, maybe the problem should be altered so as not to raise the question about where the new guest came from by having the hotel originally occupied by an infinity of women, and the new guest is a man (a lucky man!), then do the subsequent movements of occupants to insure one person per room is enforced... (not so lucky man). Perhaps an analogy you will understand: Consider the number 10/9. It is 1.1111111... with an infinite number of 1's after the decimal point. Now you may argue that 1/9 is the 10th of 10/9, so the decimal digits get shifted by one to the right and therefore there must be one additional 1 after the decimal point. So, you could argue that 10/9-1/9 would need to be slightly below 1. But it's not. It's exactly 1 because 10/9-1/9 = 9/9 = 1. (This also follows from the definition of decimal using limits) Quote by bahamagreen You're not getting the point. Forget about the methods of putting extra guests into the rooms. Look at the facts of the problem: Hotel has Infinity of guests Infinity of rooms If you allow that an additional guest (without a room) can exist, you must also allow that an additional room (without a guest) can exist. Nah. The story tells you that all of the rooms are full. There is no room without a guest. Page 3 of 4 < 1 2 3 4 > Thread Tools | | | | |-------------------------------------------------|----------------------------|---------| | Similar Threads for: The problem with infinity. | | | | Thread | Forum | Replies | | | Calculus & Beyond Homework | 6 | | | Calculus & Beyond Homework | 8 | | | Cosmology | 28 | | | General Math | 13 |
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http://math.stackexchange.com/questions/37142/integral-int-0a-frac-exp-2-pi-iwxx-idx
Integral $\int_{0}^{A}\frac{\exp(-2\pi iwx)}{x-i}dx$ Here $A>0$, $w$-real,$\mathtt{i}$-complex. Mathematica gives the answer: $$\frac{1}{2}e^{2\pi w}(\mathtt{i}\pi+2\Gamma(0,2\pi w)-2\Gamma(0,2(1+\mathtt{i}A)\pi w)+2\ln(-\mathtt{i}+A)+2\ln(w)-2\ln(w+iA))$$ My questions: 1. How to obtain this results without mathematica? 2. Why does this integral grow exponentially as a function of $w$? Mostly I don't understand why this integral explodes.The real part is:$$\int_{0}^{A}\frac{x\cos(2\pi wx)+\sin(2\pi wx)}{1+x^{2}}dx$$ and geometrically I don't understand why this integral grows so fast - What do you mean by Gamma[0,2πw]? – AD. May 5 '11 at 8:26 incomplete gamma function (upper) – Katja May 5 '11 at 8:37 @AD and others: For clarification the Mathematica Manual say $\Gamma(a,z_0,z_1)$ is the generalized incomplete gamma function $\Gamma(a,z_0)-\Gamma(a,z_1)$. – Listing May 5 '11 at 8:37 1 For now, I'll only note that your results can be recast in terms of sine and cosine integrals; I'm peering at your problem now, and will see what comes up. – J. M. May 5 '11 at 10:50 1 In particular, have a look at this and this... – J. M. May 5 '11 at 11:13 show 2 more comments 1 Answer You have a typo in your post. The first exponential in the answer given by Mathematica should read $e^{-2\pi w}$. In total the solution does NOT grow exponentially. The exponential growth of the $\Gamma$-function is "cured" by the factor $e^{-2\pi w}$. To show that the integral does not grow exponential, it is possible to obtain an asymptotic expression for $w\to\infty$ without resorting to the explicit expression given in your post. To this end, we use integration by parts $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx = \frac{ie^{-2\pi iwx}}{2\pi w(x-i)} \Biggr|_{x=0}^A + \frac{i}{2\pi w} \int_0^A \frac{e^{-2\pi iwx}}{(x-i)^2} dx.$$ Repetitive integration by parts yields an asymptotic expansion for $w\to\infty$. The boundary term is the leading term, thus we have $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right).$$ For fun, I also give the next order term (if one is only interested in large $w$ this asymptotic expansion may prove more useful than the exact expression in terms of not so elementary functions). The next integration by parts yields $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx=\frac{ie^{-2\pi iwx}}{2\pi w(x-i)} \Biggr|_{x=0}^A - \frac{e^{-2\pi iwx}}{[2\pi w(x-i)]^2} \Biggr|_{x=0}^A- \frac{1}{(2\pi w)^2} \int_0^A \frac{e^{-2\pi iwx}}{(x-i)^3} dx,$$ which gives the next term in the asymptotic expansion $$\int_{0}^{A}\frac{e^{-2\pi iwx}}{x-i}dx \sim \frac{1}{2\pi w} \left( 1- \frac{e^{-2\pi i w A}}{1+i A}\right) - \frac{1}{(2\pi w)^2} \left( 1+ \frac{e^{-2\pi i w A}}{ (A-i)^2}\right).$$ - hm, now I am confused. I have checked,Mathematica gives me $e^{2\pi w}$. I would prefer $e^{-2\pi w}$ too – Katja May 5 '11 at 11:08 @Katja: $e^{2\pi w}$ is wrong. My post shows how the integral should behave for large $w$. – Fabian May 5 '11 at 11:09 @Katja: the two terms which I have given in my post approximate the integral reasonable well for almost all values of $w\gtrsim 1$ and $A$. – Fabian May 5 '11 at 11:26 Tsk, I already upvoted this (and thus cannot upvote it again), but I agree; if $w$ is big, this is a better choice for the numerics. – J. M. May 5 '11 at 23:47
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http://mathoverflow.net/questions/51008?sort=oldest
## Limit of a sequence of polygons. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Begin with a polygon $P_0$. Place two points on every edge of the polygon such that they divide each side equally into three parts. Create a new polygon $P_1$ by connecting all new points with lines. If we begin with a square and iterate this process, what is the limit as the number of iterations approaches infinity? It is clearly not a circle, but what the correct answer is, I do not know. - So $P_1$ is an octagon with alternate sides 1/3 and sqrt(2)/3 and all angles 3pi/4, right? – Ross Millikan Jan 3 2011 at 12:02 I don't understand what you mean by "connecting all new points with lines." I don't get a (convex) polygon if I interpret this naively. – Qiaochu Yuan Jan 3 2011 at 12:02 1 I think the convex hull of the new points is the desired polygon at each step. Each corner will be cut off, so the number of sides will double each step. The midpoints of the sides of the original square will always be part of the polygon, so the area is bounded below by 1/2 – Ross Millikan Jan 3 2011 at 12:13 1 On each step you add 4 midpoints: the first 4 come from the square, the next 4 are the newer midpoints of the octagon $P_1$, then 4 come from $P_2$, etc. All these points have to lie on the limit, but it's really hard to control the corresponding equations. – Wadim Zudilin Jan 3 2011 at 12:32 1 @William Zudilin: I believe you add (after the second step) twice as many midpoints each generation. And they do stay permanently – Ross Millikan Jan 3 2011 at 12:49 show 3 more comments ## 4 Answers Corrected per Thorny's comment: The limit curve does not appear smooth, as is visible in this picture of the first 12 polygons obtained by the process, which become visually indistiguishable at the end: Despite appearances, and despite my initial assertion, the limit curve actually is smooth. Here is a sequence of closeups of one of the apparent corners, enlarging by a factor of 2 in successive images: As Thorny pointed out, for the midpoint of any edge, there is an affine transformation fixing the point and taking the limit curve to itself, shrinking the edge by a factor of 3, and shrinking the plane mod the tangent to the edge by a factor of 3, and with only one eigenvector, which is tangent to the edge. Therefore, despite my initial impression, the limit curve is actually tangent to the $k$th stage polygon at the midpont of each of its edges. It looks sharp because the slopes converge slowly. To understand this better, think of a new process scaled up by a factor of 3. If we describe the polygon by a sequence of vectors ${v_1, \dots, v_k}$ representing the sides, at each stage we interpolate the sum of each pair of successive vectors in circular order. The slopes of the vectors are rational, and the interpolation is by Fary addition (adding numerators and denominators). For instance, if we start with the vectors $(1,0)$ and $(0,1)$, then after 4 subdivisions we get this sequence of slopes: {0, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 1, 4/5, 3/4, 5/7, 2/3, 5/8, 3/5, 4/7, 1/2, 3/7, 2/5, 3/8, 1/3, 2/7, 1/4, 1/5} All rational slopes are obtained in this way. Since the curves are convex and the slopes are dense in the limit, the limiting curve is $C^1$, but it is not twice differentiable everywhere. You can't expect this kind of curve to have a name (in most cases). - 3 Just a little error: there are no angles in the limit curve. It is composed of self-similar arcs under some affine transformations (with a linear part with matrix entries 1/3, 1/3, 0, 1/3, say), which are not similarities, so no nontrivial angles can be present. – Thorny Jan 5 2011 at 14:22 @Thorny: Thanks for your comment, that was an embarassing oversight on my part. I've fixed the discussion. – Bill Thurston Jan 6 2011 at 17:50 ### 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. The left figure shows the first three iterations, the right after 10 iterations, when the polygon has 4096 vertices. - You've described de Rham's trisection method (Un peu de mathématiques à propos d'une courbe plane. Elemente der Math. 2, (1947). 73–76, 89–97.) It's the first example of a corner-cutting method. For an analysis of convergence, see Carl de Boor's Cutting corners always works (Comput. Aided Geom. Design 4 (1987), no. 1-2, 125–131). See also Chaikin's corner-cutting method and subdivision surfaces. - Interesting. According to the link, Chalkin's method uses the 1/4, 3/4 points for subdivision, so this one is a variation rather than a special case. According to the link, Chalkin's rule converges to a quadratic B-spline curve, so for these ratios the limit is $C^1$ smooth. – Bill Thurston Jan 3 2011 at 16:26 @Bill, yes, thanks for the correction. I've edited my answer. – lhf Jan 3 2011 at 16:47 As noted, goes back to de Rham. Found also in: Georges de Rham, "Sur quelques courbes définies par des equations functionnelles". Univ. e Politec. Torino. Rend. Sem. Mat. 16 1956/1957 101–113. English tranlation in by book CLASSICS ON FRACTALS. -
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http://unapologetic.wordpress.com/2009/09/08/the-special-linear-group-and-others/?like=1&source=post_flair&_wpnonce=c1c9c39881
# The Unapologetic Mathematician ## The Special Linear Group (and others) We’ve got down the notion of the general linear group $\mathrm{GL}(V)$ of a vector space $V$, including the particular case of the matrix group $\mathrm{GL}(n,\mathbb{F})$ of the space $\mathbb{F}^n$. We also have defined the orthogonal group $\mathbb{O}(n,\mathbb{F})$ of $n\times n$ matrices over $\mathbb{F}$ whose transpose and inverse are the same, which is related to the orthogonal group $\mathrm{O}(V,B)$ of orthogonal transformations of the real vector space $V$ preserving a specified bilinear form $B$. Lastly, we’ve defined the group $\mathrm{U}(n)$ of unitary transformations on $\mathbb{C}^n$ — $n\times n$ complex matrices whose conjugate transpose and inverse are the same. For all of these matrix groups — which are all subgroups of some appropriate $\mathrm{GL}(n,\mathbb{F})$ — we have a homomorphism to the multiplicative group of $\mathbb{F}$ given by the determinant. We originally defined the determinant on $\mathrm{GL}(n\mathbb{F})$ itself, but we can easily restrict it to any subgroup. We actually know that for unitary and orthogonal transformations the image of this homomorphism must lie in a particular subgroup of $\mathbb{F}^\times$. But in any case, the homomorphism must have a kernel, and this kernel turns out to be important. In the case of the general linear group $\mathrm{GL}(V)$, the kernel of the determinant homomorphism consists of the automorphisms of $V$ with determinant ${1}$. We call this subgroup of $\mathrm{GL}(V)$ the “special linear group” $\mathrm{SL}(V)$, and transformations in this subgroup are sometimes called “special linear transformations”. Of course, we also have the particular special linear group $\mathrm{SL}(n,\mathbb{F})\subseteq\mathrm{GL}(n,\mathbb{F})$. When we take the kernel of any of the other groups, we prepend the adjective “special” and an $\mathrm{S}$ to the notation. Thus we have the special orthogonal groups $\mathrm{SO}(V,B)$ and $\mathrm{SO}(n,\mathbb{F})$ and the special unitary group $\mathrm{SU}(n)$. In a sense, all the interesting part of the general linear group is contained in the special linear subgroup. Outside of that, what remains is “just” a scaling. It’s a little more complicated than it seems on the surface, but not much. ### Like this: Posted by John Armstrong | Algebra, Linear Algebra ## 6 Comments » 1. [...] Shears Generate the Special Linear Group We established that if we restrict to upper shears we can generate all upper-unipotent matrices. On the other hand if we use all shears and scalings we can generate any invertible matrix we want (since swaps can be built from shears and scalings). We clearly can’t build any matrix whatsoever from shears alone, since every shear has determinant and so must any product of shears. But it turns out that we can use shears to generate any matrix of determinant — those in the special linear group. [...] Pingback by | September 9, 2009 | Reply 2. [...] shears alone generate the special linear group. Can we strip them down any further? And, with this in mind, how many generators does it take to [...] Pingback by | September 11, 2009 | Reply 3. [...] this gives a geometric meaning to the special orthogonal group . Orthogonal transformations send orthonormal bases to other orthonormal bases, which will send [...] Pingback by | November 3, 2009 | Reply 4. [...] to turn our heads and translate the laws of physics to compensate exactly. These rotations form the special orthogonal group of orientation- and inner product-preserving transformations, but we can also throw in reflections [...] Pingback by | November 10, 2009 | Reply 5. Sorry for commenting OT … what WP theme do you use? It’s looking great!! Comment by | December 7, 2009 | Reply 6. /me checks the bottom of the page “Theme: Andreas04 by Andreas Viklund.” Comment by | December 7, 2009 | Reply « Previous | Next » ## 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://mathhelpforum.com/pre-calculus/178851-need-help-changing-irrational-number-into-rational-number-involving-polar-coord.html
# Thread: 1. ## Need help changing an irrational number into a rational number, involving polar coord I am working with problems of changing from Polar to Cartesian coordinates. the problem is (-5, pi/3) I know x=-5cos pi/3 = -5/2 and y = -5sin pi/3 \approx 4.33 or = 5sqrt3/2 but I only know that it can be written as 5sqrt3/2 because after I got 4.33 I check the answer. I need to know how to convert numbers like 4.330127019 in 5sqrt3/2 thanks! 2. or another exmaple pi + tan^-1 (-1) approx 2.35619 or 3pi/4 (done in radians) 3. Originally Posted by colorado or another exmaple pi + tan^-1 (-1) approx 2.35619 or 3pi/4 (done in radians) Apart from recognizing a few common ones, ie 1.570796... = pi/2, I think you will just have to rely on the trigonometry. Also, please note that expressions like $\frac{5 \sqrt{3}}{2}$ are not rational numbers. A better description of what you are trying to say is probably "closed form" or "in the form of a fraction." -Dan 4. okay, hopefully approx values will work. Do you know where I might find a list of common ones? If not I'll try to find one. Thanks for the reply!!! 5. Originally Posted by colorado okay, hopefully approx values will work. Do you know where I might find a list of common ones? If not I'll try to find one. Thanks for the reply!!! Frankly I don't even try to remember the approximate forms. If I have a decimal number that I need to, say, find the inverse sine of I let my calculator deal with it. -Dan
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http://physics.stackexchange.com/questions/54534/is-it-possible-for-delta-x-sigma-x-of-any-free-particle-wave-packet-to-b/54537
# Is it possible for $\Delta x$ ($\sigma_x$) of any free particle wave packet to be decreasing at any time? Consider any wave packet describing a free particle (so no potential or other forces acting on it). Then it can be shown that $\Delta p$ does not change in time. However, my question is what happens with $\Delta x$ as we go forward in time? Does it have to increase at all times? Or is there a counter-example where the uncertainty in position is decreasing, if even for a short time period? My initial guess is $\Delta x$ must always increase, because $p \neq 0$, so that $\Delta p \neq0$ and hence $\Delta v = \frac{\Delta p}{m} \neq 0$. But if there's a spread in velocities, then the wave packet must also spread. Is this logic correct? Or could we have a wave packet where the back of it would move forward faster than the front, and for a certain period until that back end catches up with the front one, it would actually be narrower than at the outset, i.e. reducing $\Delta x$? If yes, how would one describe such a free particle (wave packet)? So it does seem to me that every wave packet describing a free particle will eventually spread, but the question is whether there can be a time period in its evolution when it is actually becoming narrower. edit: In particular, if it does not have to increase at all times, can this be shown without appealing to time reversal? - 2 – David Zaslavsky♦ Feb 20 at 18:16 @DavidZaslavsky Yeah, I saw that already, and something similar to my question was discussed in the comments there, as well. However, it was never stated whether $\Delta x$ must increase (or in the ideal case, stay the same) at all times, not just eventually. – Ryker Feb 20 at 18:21 ## 3 Answers If $\Psi(x,t)$ solves the Schrodinger equation, so does $\Psi^*(x,-t)$ , so no, there is nothing at all that must increase. - 2 Can you elaborate on that? Are you saying that if the uncertainty is increasing on, say, $t \in (t_{1}, t_{2})$, then there is a time period where the uncertainty in position is decreasing, namely $t \in (-t_{2}, -t_{1})$? – Ryker Feb 20 at 19:02 2 I am saying that if $\Psi(x,t)$ has increasing uncertainty from t1 to t2, then define $\bar{\Psi}(x,t) = \Psi(x,-t)$, and $\bar{\Psi}$ will have decreasing uncertainty from t1 to t2. – Mark Eichenlaub Feb 20 at 19:09 1 I find it extremely obvious, to be honest. I'm just saying you can time-reverse anything. If something goes up when time goes forward, it goes down when time goes backward, but the Schrodinger equation can't tell the difference between "time goes forward" and "time goes backward", so anything that can increase for one solution can decrease for another solution. – Mark Eichenlaub Feb 20 at 19:11 2 Or, if you want an example easy to visualize, take two Gaussians with a very wide separation but moving towards each other as an example. Then the uncertainty in x is basically just the separation, and the spread of the Gaussians is ignored. Evolve that forward in time some. The Gaussians get closer together, reducing the uncertainty in x, although they individually spread a bit. – Mark Eichenlaub Feb 20 at 21:32 3 Yes, the two Gaussians describe a free particle. If one Gaussian does, then two Gaussians can by superposition. It is easy to adjust the width of the Gaussians and the separations so the statement is correct. – Mark Eichenlaub Feb 20 at 21:42 show 17 more comments I don't know the answer to this yet, but here is a calculation that others might find useful in determining the answer. Let's compute the time derivative of $\sigma_X$. First note that $$\frac{d}{dt}\sigma_X^2 = 2\sigma_X\dot\sigma_X, \qquad \sigma_X = \langle X^2\rangle - \langle X\rangle^2$$ so \begin{align} \dot\sigma_X &= \frac{1}{2\sigma_X}\frac{d}{dt}\left(\langle X^2\rangle - \langle X\rangle^2\right) \\ &= \frac{1}{2\sigma_X}\left(\frac{d}{dt}\langle X^2\rangle - 2\langle X\rangle\frac{d}{dt}\langle X\rangle \right)\\ \end{align} Now use the general relation $$\frac{d}{dt}\langle O\rangle = \frac{i}{\hbar}\langle[H,O]\rangle$$ for an operator with no explicit time-dependence. It follows that \begin{align} \frac{d}{dt}\langle X\rangle = \frac{i}{\hbar}\langle[P^2/2m, X]\rangle =\frac{i}{2m\hbar}\langle -2i\hbar P\rangle = \frac{\langle P\rangle}{m} \end{align} so that also \begin{align} \frac{d}{dt}\langle X^2\rangle &= \frac{i}{\hbar} \left\langle\left[\frac{P^2}{2m}, X^2\right]\right\rangle = \frac{i}{2m\hbar}(-2i\hbar)\langle\{P,X\}\rangle = \frac{\langle \{P,X\}\rangle}{m} \end{align} where $\{P, X\}=PX+XP$, and therefore $$\dot\sigma_X = \frac{1}{2m\sigma_X}\Big(\langle \{P,X\}\rangle - 2\langle P \rangle \langle X \rangle\Big)$$ We want to know if there is a state for which at some time, $\dot\sigma_X<0$. Since $\sigma_X>0$, the expression just derived leads us to an equivalent question; is there a state for which the following inequality holds at some time? $$\langle\{P,X\}\rangle < 2\langle P\rangle \langle X \rangle\qquad ?$$ Let me know if you find any errors in the math as I did it quickly. - 1 Looks fine to me. Apply Mark Eichenlaub's approach to this. Suppose a state exists such that $\dot{\sigma}_x>0$. Then by your calculation $2\left\langle P\right\rangle \left\langle X\right\rangle > \left\langle \{P,X\}\right\rangle$. Now, under time reverse $P$ changes sign but not $X$, so under T: $2\left\langle -P\right\rangle \left\langle X\right\rangle > \left\langle \{-P,X\}\right\rangle$ Now multiply through by a minus and you get the desired result. – Michael Brown Feb 20 at 21:29 @MichaelBrown Cool thanks. Somehow I would still be more satisfied with a calculation for a particular state in which the bound were explicitly verified. It would be nice to see how it works out. – joshphysics Feb 20 at 21:56 3 You could take an example state of two Gaussians that are coming towards each other, with x=0 in between. Then <p> = <x> = 0, but the quadratic operator doesn't get killed by the mirror-image symmetry. – Mark Eichenlaub Feb 21 at 0:20 @MarkEichenlaub That's a nice example thanks. – joshphysics Feb 21 at 0:28 If my brain hasn't broken (not enough coffee today) then $|0>-|2>$ (in harmonic oscillator eigenfunctions) should be another example. Not too confident it works atm, but if it doesn't it shouldn't be hard to tweak into an example. :) – Michael Brown Feb 21 at 1:54 show 2 more comments 1) We reformulate OP's title question (v1) as follows: Show that for all possible wave packets $\psi(p,t)$ of a free particle, the position variance $$\tag{1} {\rm Var}(\hat{x}) ~=~\langle \hat{x}^2\rangle-\langle \hat{x}\rangle^2$$ decreases in at least some interval $[t_1,t_2]$ of time. As Mark Eichenlaub correctly observes in his answer and comments, a solution with decreasing ${\rm Var}(\hat{x})$ in $[t_1,t_2]$ can be mapped by time reversal symmetry to a a solution with increasing ${\rm Var}(\hat{x})$ in $[-t_2,-t_1]$. Here it is used that the Hamiltonian $\hat{H}=\frac{\hat{p}^2}{2m}$ for a free particle, and the position operator $\hat{x}$ both commute with the time reversal operator $\hat{T}$. See e.g. this Phys.SE answer for further details. However, time reversal symmetry does logically not rule out alone the possibility that a solution has monotonically increasing ${\rm Var}(\hat{x})$ for all time. (It only implies that if this is the case, then there will also be a solution with monotonically decreasing ${\rm Var}(\hat{x})$ for all time.) 2) Explicit calculation. Consider an arbitrary wave packet $$\tag{2} \psi(p,t)~=~A(p)e^{-i\theta(p,t)}$$ that represents the general solution to the time-dependent Schrödinger equation in the momentum representation. Here the angle $$\tag{3} \theta(p,t)~=~\theta_0(p) + \frac{p^2 }{2m}\frac{t}{\hbar},\qquad A(p),\theta_0(p)~\in~\mathbb{R},$$ is affine in $t$. We assume that the wave packet is normalized $$\tag{4} 1~=~||\psi(t)||^2~=~ \langle \psi(t)| \psi(t)\rangle~=~\int_{\mathbb{R}} \!dp~A^2 .$$ The position operator in the momentum representation reads $$\tag{5} \hat{x}~=~i\hbar\frac{\partial }{\partial p}.$$ Therefore the position average $$\tag{6}\langle \hat{x}\rangle~=~ \langle \psi(t)| \hat{x}|\psi(t)\rangle ~=~\hbar\int_{\mathbb{R}} \!dp~A^2\theta^{\prime}$$ is affine in $t$, and the average of the squared position $$\tag{7} \langle \hat{x}^2\rangle~=~ \langle \psi(t)| \hat{x}^2|\psi(t)\rangle ~=~\hbar^2 \int_{\mathbb{R}} \!dp~(A^{\prime 2} +A^2\theta^{\prime 2} )$$ is quadratic in $t$. Here primes denote differentiation wrt. $p$. Thus we immediately know that the position variance $$\tag{8} {\rm Var}(\hat{x})~=~at^2+bt+c$$ is quadratic in $t$ as well, where $a$, $b$, and $c$ are constant coefficients. It is straightforward to see from the Cauchy-Schwarz inequality $$\tag{9}\left(\int_{\mathbb{R}} \!dp~A^2\frac{p}{m} \right)^2 ~\leq~ \int_{\mathbb{R}} \!dp~\left(A\frac{p}{m}\right)^2 \cdot \int_{\mathbb{R}} \!dp~A^2$$ that the second-order coefficient $$\tag{10} a~=~\int_{\mathbb{R}} \!dp~\left(A\frac{p}{m}\right)^2-\left(\int_{\mathbb{R}} \!dp~A^2\frac{p}{m} \right)^2~\geq ~0,$$ is non-negative, cf. the normalization condition (4). In fact one may show that $a>0$ is strictly positive, and hence the ${\rm Var}(\hat{x})$ is a decreasing parabola for $t$ sufficiently negative, as we wanted to show. Sketched indirect proof of $a\neq 0$: The second-order coefficient $a$ becomes zero $~\Leftrightarrow~$ the Cauchy-Schwarz inequality (9) becomes an equality $~\Leftrightarrow~$ $A(p) p$ is proportional to $A(p)$ $~\Leftrightarrow~$ $A(p)$ is proportional to a delta function $\delta(p-p_0)$. But this does not correspond to a normalizable wave packet. 3) Example. Two mutually approaching wave trains are an easy intuitive example where the position variance ${\rm Var}(\hat{x})$ diminishes in some time-interval $[t_1,t_2]$. But this is, in some sense, a lazy example, which sort of betrays just how universal and omnipresent this behavior is for quantum mechanics. For instance, as we know from the general analysis in Section 2, already the simplest possible wave packet, i.e. a single Gaussian wave packet, displays this behavior. However, that is much less intuitive, and thus that much more fascinating/mind boggling to try to understand. Let us for simplicity set $\hbar=1=m$. A single Gaussian wave packet at $t=0$ is of the form $$\tag{11} \psi(p,t=0)~=~N \exp\left(-ipc - \frac{p^2}{2}\tau\right) ,$$ where $\tau>0$ and $c=a+ib\in\mathbb{C}$ are constants. Here $N>0$ is a normalization constant. The normalization condition (4) implies that $$\tag{12} N~=~\left(\frac{\tau}{\pi}\right)^{\frac{1}{4}}\exp\left(-\frac{b^2}{2\tau}\right).$$ Thus for arbitrary time $t$, the Gaussian wave packet reads $$\tag{13} \psi(p,t)~=~\psi(p,0)\exp\left(-i\frac{p^2}{2}t\right) ~=~N \exp\left(-ipc - \frac{p^2}{2}(\tau+it)\right) .$$ In the position representation, the Gaussian wave packet becomes $$\tag{14} \psi(x,t)~=~\int_{\mathbb{R}} \frac{dp}{\sqrt{2\pi}} \exp\left(ipx\right) \psi(p,t)~=~\frac{N}{\sqrt{\tau+it}} \exp\left(-\frac{(x-c)^2}{2(\tau+it)}\right).$$ The position probability distributions becomes $$|\psi(x,t)|^2~=~\frac{N^2}{|\tau+it|} \exp\left(-{\rm Re}\frac{(x-c)^2}{\tau+it}\right)$$ $$~=~\frac{N^2}{\sqrt{\tau^2+t^2}} \exp\left(-\frac{[(x-a)^2-b^2]\tau-2(x-a)bt}{\tau^2+t^2}\right)$$ $$\tag{15}~=~\sqrt{\frac{\tau}{\pi(\tau^2+t^2)}} \exp\left(-\frac{(x-a-\frac{bt}{\tau})^2\tau }{\tau^2+t^2} \right).$$ Therefore the position average $$\tag{16} \langle \hat{x}\rangle~=~\int_{\mathbb{R}} \!dx~x|\psi(x,t)|^2 ~=~a+\frac{bt}{\tau}$$ is an affine function of $t$, while the position variance $$\tag{17} {\rm Var}(\hat{x}) ~=~\int_{\mathbb{R}} \!dx~(x-\langle \hat{x}\rangle)^2|\psi(x,t)|^2 ~=~\frac{\tau^2+t^2}{2\tau}$$ is a quadratic function of $t$. The time symmetric profile (17) of the position variance ${\rm Var}(\hat{x})$ of a single Gaussian wave packet is probably somewhat surprising to anyone who borrows his intuition from classical physics. -
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http://mathoverflow.net/questions/9733?sort=oldest
## Matrix factorizations and physics ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have heard during some seminar talks that there are applications of the theory of matrix factorizations in string theory. A quick search shows mostly papers written by physicists. Are there any survey type papers aimed at an algebraic audience on this topic, especially with current state/open questions motivated by physics? Background: A matrix factorization of an element $x$ in a ring $R$ is a pair of square matrices $A,B$ of size $n$ such that $AB=BA=xId_n$. For more see, for example, Section 3 of this paper . - 1 Great question! I hope you get good answers. – Graham Leuschke Dec 25 2009 at 13:44 ## 3 Answers Indeed matrix factorizations come up in string theory. I don't know if there are good survey articles on this stuff, but here is what I can say about it. There might be an outline in the big Mirror Symmetry book by Hori-Katz-Klemm-etc., but I am not sure. When we are considering the B-model of a manifold, for example a compact Calabi-Yau, the D-branes (boundary states of open strings) are given by coherent sheaves on the manifold (or to be more precise, objects of the derived category of coherent sheaves). Matrix factorizations come up in a different situation, namely, they are the D-branes in the B-model of a Landau-Ginzburg model. Mathematically, a Landau-Ginzburg model is just a manifold (or variety) $X$, typically non-compact, plus the data of a holomorphic function $W: X \to \mathbb{C}$ called the superpotential. In this general situation, a matrix factorization is defined to be a pair of coherent sheaves $P_0, P_1$ with maps $d : P_0 \to P_1$, $d : P_1 \to P_0$ such that $d^2 = W$. I guess you could call this a "twisted (or maybe it's 'curved'? I forget the terminology) 2-periodic complex of coherent sheaves". When $X = \text{Spec}R$ is affine, and when the coherent sheaves are free $R$-modules, this is the same as the definition that you gave. The relationship between matrix factorization categories and derived categories of coherent sheaves was worked out by Orlov: http://arxiv.org/abs/math/0503630 http://arxiv.org/abs/math/0503632 http://arxiv.org/abs/math/0302304 I believe that the suggestion to look at matrix factorizations was first proposed by Kontsevich. I think the first paper that explained Kontsevich's proposal was this paper by Kapustin-Li: http://arXiv.org/abs/hep-th/0210296v2 There are some interesting recent papers regarding the relationship between the open-string B-model of a Landau Ginzburg model (which is, again, mathematically given by the matrix factorizations category) and the closed-string B-model, which I haven't described, but an important ingredient is the Hochschild (co)homology of the matrix factorizations category. Take a look at Katzarkov-Kontsevich-Pantev http://arxiv.org/abs/0806.0107 section 3.2. There is a paper of Tobias Dyckerhoff http://arxiv.org/abs/0904.4713 and a paper of Ed Segal http://arxiv.org/abs/0904.1339 which work out in particular the Hochschild (co)homology of some matrix factorization categories. The answer is it's the Jacobian ring of the superpotential. This is the correct answer in terms of physics: the Jacobian ring is the closed state space of the theory. Katzarkov-Kontsevich-Pantev also has some interesting stuff about viewing matrix factorization categories as "non-commutative spaces" or "non-commutative schemes". Edit 1: I forgot to mention: Kontsevich's original homological mirror symmetry conjecture stated that the Fukaya category of a Calabi-Yau is equivalent to the derived category of coherent sheaves of the mirror Calabi-Yau. Homological mirror symmetry has since been generalized to non-Calabi-Yaus. The rough expectation is that given any compact symplectic manifold, there is a mirror Landau-Ginzburg model such that, among other things, the Fukaya category of the symplectic manifold should be equivalent to the matrix factorizations category of the Landau-Ginzburg model. For example, if your symplectic manifold is $\mathbb{CP}^n$, the mirror Landau-Ginzburg model is given by the function $x_1+\cdots+x_n + \frac{1}{x_1\cdots x_n}$ on $(\mathbb{C}^\ast)^n$. This is sometimes referred to as the Hori-Vafa mirror http://arxiv.org/abs/hep-th/0002222 I think that various experts probably know how to prove this form of homological mirror symmetry, at least when the symplectic manifold is, for example, a toric manifold or toric Fano manifold, but it seems that very little of this has been published. There may be some hints in this direction in Fukaya-Ohta-Oh-Ono http://arxiv.org/abs/0802.1703 http://arxiv.org/abs/0810.5654, but I'm not sure. There is an exposition of the case of $\mathbb{CP}^1$ in this paper of Matthew Ballard http://arxiv.org/abs/0801.2014 -- this case is already non-trivial and very interesting, and the answer is very nice: the categories in this case are equivalent to the derived category of modules over a Clifford algebra. I quite like Ballard's paper; you might be interested in taking a look at it anyway. Edit 2: Seidel also has a proof of this form of homological mirror symmetry for the case of the genus two curve. Here is the paper http://arxiv.org/abs/0812.1171 and here is a video http://www.maths.ed.ac.uk/~aar/atiyah80.htm of a talk he gave on this stuff at the Atiyah 80 conference. - 2 I forgot to mention that matrix factorizations are also related to knot theory! e.g. arxiv.org/abs/math/0401268 – Kevin Lin Dec 25 2009 at 16:19 Very interesting answer Kevin, thanks! – Hailong Dao Dec 25 2009 at 17:22 ### 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. "Derived categories of coherent sheaves and triangulated categories of singularities" by D. Orlov. "Landau-Ginzburg to Calabi-Yau dictionary for D-branes" by Aspinwall."Triangulated categories of matrix factorizations" by Kajiura Saito and Takahashi. Also, I personally have many open questions in this area. Perhaps we should talk! - 1 I would also be very interested in hearing about the open questions you have :-) – Kevin Lin Dec 25 2009 at 15:50 Here's a video of a talk by Kentaro Hori at MSRI: Matrix Factorizations and Complexes of Vector Bundle ---- An Approach from 2d QFT with Boundary. It's not strictly what you're asking for, but does seem to be mostly expository, and does explain a little bit why matrix factorizations come into the equation; they're standins for a certain derived category, of course. -
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http://mathoverflow.net/questions/78627?sort=newest
## Is it consistent relative to ZF that $\frak c = \aleph_\omega$? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In ZFC we know that the continuum cannot have cofinality $\omega$. However, in the Feferman-Levy model we have that $\frak c=\aleph_1$, and that $\operatorname{cf}(\omega_1)=\omega$. In fact in the Feferman-Levy model, $\aleph_\omega^L=\aleph_1^V$. Is it consistent with ZF that $\frak c=\aleph_\omega$? Does that mean that the only restriction in ZF on the cardinality of the continuum is $\aleph_0<\frak c$? - Asaf, I think you are mistaken. Here is a proof in ZF that if ${\mathbb R}$ is a countable union of countable sets (as in the Feferman-Levy model) then every well-orderable subset of ${\mathbb R}$ is countable: If $\omega_1$ injects into ${\mathbb R}$, then so does the set ${\omega_1}^\omega$, because $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. But then, by Schröder-Bernstein, ${\omega_1}^\omega$ has the same size as ${\mathbb R}$. (Cont.) – Andres Caicedo Oct 20 2011 at 1:53 (II). Now, here is the nice bit, essentially following the argument for König's lemma: If $X$ is a countable union of countable sets, then no map from $X$ to $\omega_1^\omega$ is onto. For if $f$ is a map and $X=\bigcup_n X_n$ with each $X_n$ countable, then each $f[X_n]$ is countable, so $T_n=\omega_1\setminus\{f(x)(n)\mid x\in X_n\}$ is non-empty. Let $\Phi:\omega\to\omega_1$ be the map that at $n$ picks the minimum of $T_n$. Then $\Phi$ is not in the range of $f$. – Andres Caicedo Oct 20 2011 at 1:58 The fact that $\omega_1$ doesn't embed into $\mathbb{R}$ in the Feferman-Levy model is apparently due to Cohen - math.wisc.edu/~miller/res/two-pt.pdf – François G. Dorais♦ Oct 20 2011 at 2:55 @Andres: Thanks a lot, as I wrote to Joel in a comment, it seems that this was a dream that convinced me that in the Feferman-Levy $2^\omega=\omega_1$, as I can't find that reference anywhere. @Francois: Thanks for the reference, it seems interesting and I'll give it a read. – Asaf Karagila Oct 20 2011 at 8:36 ## 1 Answer The answer is no. The continuum cannot be $\aleph_\omega$, and this can be proved in ZF, that is, without using the axiom of choice. To see this, suppose towards contradiction that $P(\omega)$ is equinumerous with $\aleph_\omega$. Since $P(\omega)$ is equinumerous with $P(\omega)^\omega$, and this does not require AC, it follows that there is a bijection $f:\aleph_\omega\cong (\aleph_\omega)^\omega$. Let $g(n)$ be the first ordinal $\alpha\lt\aleph_\omega$ that is not among $f(\beta)(n)$ for any $\beta\lt\aleph_n$. Since there are fewer than $\aleph_\omega$ many such $\beta$, it follows that there are fewer than $\aleph_\omega$ many such $f(\beta)(n)$, and so such an $\alpha$ exists. Thus, $g:\omega\to \aleph_\omega$. But notice that for any particular $\alpha\lt\aleph_\omega$, we have $\alpha\lt\aleph_n$ for some $n$ and consequently $g(n)\neq f(\alpha)(n)$, and thus $g\neq f(\alpha)$. Thus, $f$ was not surjective to $(\aleph_\omega)^\omega$ after all, a contradiction. This is just a standard proof of Konig's theorem (that $\aleph_\omega^\omega\gt\aleph_\omega$), and the point is that it doesn't use AC. - This argument is essentially the same as the argument that Andres gave in the comments above, but it seems to answer the whole question, and not just the issue about the reals being a countable union of countable sets. The proof shows more generally that the continuum, if well-orderable, cannot have countable cofinality, even in ZF. – Joel David Hamkins Oct 20 2011 at 3:41 Thanks a lot, Joel. I swear I read somewhere that in the Feferman-Levy model $2^\omega=\omega_1$. I can't find it, and the proofs you and Andres gave convince me that it was probably in a dream. – Asaf Karagila Oct 20 2011 at 8:33 2 Just for the record, the standard proof of the more general Konig's theorem ($\kappa^{cof(\kappa)}>\kappa$ for $\kappa$ any aleph) doesn´t use AC (it is essentially the same proof that Joel already gave). – Ramiro de la Vega Oct 20 2011 at 13:00
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http://mathoverflow.net/questions/49724?sort=newest
## The functional equation $f(f(x))=x+f(x)^2$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'd like to gather information and references on the following functional equation for power series $$f(f(x))=x+f(x)^2,$$$$f(x)=\sum_{k=1}^\infty c_k x^k$$ (so $c_0=0$ is imposed). First things that can be established quickly: • it has a unique solution in $\mathbb{R}[[x]]$, as the coefficients are recursively determined; • its formal inverse is $f^{-1}(x)=-f(-x)$ , as both solve uniquely the same functional equation; • since the equation may be rewritten $f(x)=f^{-1}(x)+x^2$, it also follows that $f(x)+f(-x)=x^2$, the even part of $f$ is just $x^2/2$, and $c_2$ is the only non-zero coefficient of even degree; • from the recursive formula for the coefficients, they appear to be integer multiples of negative powers of $2$ (see below the recursive formmula). Rmk. It seems (but I did not try to prove it) that $2^{k-1}c_k$ is an integer for all $k$, and that $(-1)^k c_{2k-1} > 0$ for all $k\geq 2$. Question : how to see in a quick way that this series has a positive radius of convergence, and possibly to compute or to evaluate it? [updated] A more reasonable question, after the numeric results and various comments, seems to be, rather: how to prove that this series does not converge. Note that the radius of convergence has to be finite, otherwise $f$ would be an automorphism of $\mathbb{C}$. Yes, of course I did evaluate the first coefficients and put them in OEIS, getting the sequence of numerators A107700; unfortunately, it has no further information. Motivation. I want to understand a simple discrete dynamical system on $\mathbb{R}^2$, namely the diffeomorphism $\phi: (x,y)\mapsto (y, x+y^2)$, which has a unique fixed point at the origin. It is not hard to show that the stable manifold and the unstable manifold of $\phi$ are $$W^s(\phi)=\mathrm{graph}\big( g_{|(-\infty,\ 0]}\big)$$ $$W^u(\phi)=\mathrm{graph}\big( g_{|[0, \ +\infty)}\big)$$ for a certain continuous, strictly increasing function $g:\mathbb{R}\to\mathbb{R}$, that solves the above functional equation. Therefore, knowing that the power series solution has a positive radius of convergence immediately implies that it coincides locally with $g$ (indeed, if $f$ converges we have $f(x)=x+x^2/2+o(x^2)$ at $x=0$ so its graph on $x\le0$ is included in $W^s$, and its graph on $x\ge0$ is included in $W^u$: therefore the whole graph of $f$ would be included in the graph of $g$,implying that $f$ coincides locally with $g$). If this is the case, $g$ is then analytic everywhere, for suitable iterates of $\phi$ give analytic diffeomorphism of any large portion of the graph of $g$ with a small portion close to the origin. One may also argue the other way, showing directly that $g$ is analytic, which would imply the convergence of $f$. Although it seems feasible, the latter argument would look a bit indirect way, and in that case I'd like to make sure there is no easy direct way of working on the coefficients (of course, it may happen that $g$ is not analytic and $f$ is not convergent). Details: equating the coefficients in both sided of the equation for $f$ one has, for the 2-Jet $$c_1^2x+(c_1c_2+c_2c_1^2)x^2 =x + c_1^2x^2,$$ whence $c_1=1$ and $c_2=\frac 1 2;$ and for $n>2$ $$2c_n=\sum_{1\le j\le n-1}c_jc_{n-j}\ \ -\sum_{1 < r < n \atop \|k\|_1=n}c_rc_{k_1}\dots c_{k_r}.$$ More details: since it may be of interest, let me add the argument to see $W^s(\phi)$ and $W^u(\phi)$ as graphs. Since $\phi$ is conjugate to $\phi^{-1}=J\phi J$ by the linear involution $J:(x,y)\mapsto (-y,-x)$, we have $W^u(\phi):=W^s(\phi^{-1})=J\ W^s(\phi)$, and it suffices to study $\Gamma:=W^s(\phi)$. For any $(a,b)\in\mathbb{R}^2$ we have $\phi^n(a,b)=(x_n,x_{n+1})$, with $x_0=a$, $x_1=b$, and $x_{n+1}=x_n^2+x_{n-1}$ for all $n\in\mathbb{N}$. From this it is easy to see that $x_{2n}$ and $x_{2n+1}$ are both increasing; moreover, $x_{2n}$ is bounded above iff $x_{2n+1}$ is bounded above, iff $x_{2n}$ converges, iff $x_n\to 0$, iff $x_n\le 0$ for all $n\in\mathbb{N}$. As a consequence $(a,b)\in \Gamma$ iff $\phi^n(a,b)\in Q:=(-\infty,0]\times(-\infty,0]$, whence $\Gamma=\cap_{ n\in\mathbb{N}} \phi^{-n}(Q)$. The latter is a nested intersection of connected unbounded closed sets containing the origin, therefore such is $\Gamma$ too. In particular, for any $a\leq 0$ there exists at least a $b\leq 0$ such that $(a,b)\in \Gamma$: to prove that $b$ is unique, that is, that $\Gamma$ is a graph over $(\infty,0]$, the argument is as follows. Consider the function $V:\Gamma\times\Gamma\to\mathbb{R}$ such that $V(p,p'):=(a-a')(b-b')$ for all $p:=(a,b)$ and $p':=(a',b')$ in $\Gamma$. Showing that $\Gamma$ is the graph of a strictly increasing function is equivalent to show that $V(p,p')>0$ for all pair of distinct points $p\neq p'$ in $\Gamma$. By direct computation we have $V\big(\phi(p),\phi(p')\big)\leq V(p,p')$ and $\big(\phi(p)-\phi(p')\big)^2\geq \|p-p'\|^2+2V(p,p')(b+b')$. Now, if a pair $(p,p')\in\Gamma\times\Gamma$ has $V(p,p')\le0$, then also by induction $V\big(\phi^n(p),\phi^n(p')\big)\leq 0$ and $\big(\phi^n(p)-\phi^n(p')\big)^2\geq \|p-p'\|^2$ for all $n$, so $p=p'$ since both $\phi^n(p)$ and $\phi^n(p')$ tend to $0$. This proves that $\Gamma$ is a graph of a strictly increasing function $g:\mathbb{R}\to\mathbb{R}$: since it is connected, $g$ is also continuous. Of course the fact that $\Gamma$ is $\phi$-invariant implies that $g$ solves the functional equation. - why it has unique solution, not two solutions (one for $c_1=1$, one for $c_1=-1$)? – Fedor Petrov Dec 17 2010 at 14:46 This is strongly reminiscent of problems 163 and 165 in the currnet (December 15 version) of EC1, 2nd edition: math.mit.edu/~rstan/ec/ec1.pdf (pp. 147-8 and 204 onward). Maybe some of the ideas or references there will be helpful. (Warning: this is a massive pdf.) – JBL Dec 17 2010 at 16:32 @JBL: you meant massive by content right? :-) – S. Sra Dec 17 2010 at 18:08 2 In case of usefulness, I had already put a number of references on functional equations at: zakuski.math.utsa.edu/~jagy/other.html emphasizing early work of Irvine Noel Baker. That is not to say that your problem can be written in his framework. But from what I have seen, it is easy for a functional equation solution to behave as $e^{-1/x^2},$ that is $C^\infty$ on the real line, analytic except at $0,$ but when extended to the complex plane (if that is even possible) the origin becomes an essential singularity. – Will Jagy Dec 17 2010 at 23:35 3 I also put three early items by Kuczma at zakuski.math.utsa.edu/~jagy/other.html one being a survey, see page 29, formula (82), as well as two of his articles about $$\phi ( \phi(x)) = g(x, \phi(x))$$ which includes your example. What I am not seeing is anything about smoothness, either at a fixpoint, or away from it. – Will Jagy Dec 20 2010 at 22:36 show 10 more comments ## 10 Answers Having thought about this question some more, including making some plots of the trajectories of points under iterated applications of $f$ (see Gottfried's circular trajectories), it is possible to devise a numerical test which should show that the series expansion diverges (if indeed it does). Whether it is a practical test or not depends on how badly it fails. My initial rough attempt wasn't conclusive, so I haven't determined the answer to this yet. This answer still needs working through the details, but I'll post what I have so far. Also, I think that much of what I have to say is already well known. As the thread is now community wiki and anyone can edit this post, feel free to add any references or further details. The main ideas are as follows, and should apply much more generally to analytic functions defined near a fixed point via an iterative formula, such as $f(f(z))=\exp(z)-1$ in this MO question. 1. There are two overlapping open regions bounding the origin from the right and left respectively, and whose union is a neighbourhood of the origin (with the origin removed). The functional equation $f(f(z))=z+f(z)^2$ with $f(z)=z+O(z^2)$ can be uniquely solved in each of these regions, on which it is an analytic function. 2. The solution $f$ on each of the regions has the given power series as an asymptotic expansion at zero. Furthermore, it is possible to explicitly calculate a bound for the error terms in the (truncated) expansion. 3. The functional equation has a solution in a neighbourhood of the origin (equivalently, the power series has a positive radius of convergence) if and only if the solutions on the two regions agree on their overlap. One way to verify that $f$ cannot be extended to an analytic function in a neighbourhood of the origin would be to accurately evaluate the solutions on the two domains mentioned above at some point in their overlap, and see if they differ. Another way, which could be more practical, is to use the observation that after the second order term, the only nonzero coefficients in our series expansion are for the odd terms, and the signs are alternating [Edit: this has not been shown to be true though and, in any case, I give a proof that this implies zero radius of convergence below]. Consequently, if we evaluate it at a point on the imaginary axis, truncating after a finite number of terms, we still get a lower bound for $\vert f\vert$. If it does indeed diverge, then this will eventually exceed the upper bound we can calculate as mentioned above, proving divergence. Looking at the first 34 terms from OEIS A107700 was not conclusive though. Choose a point $z_0$ close to (and just to the right of) the origin. Using the power series to low order, we approximate $z_1=f(z_0)$. Then, the functional equation can be used to calculate $z_n=f(z_{n-1})=z_{n-2}+z_{n-1}^2$. Similarly, choosing points just to the left of the origin, we can calculate good approximations for the iterates of $f^{-1}$. Doing this for a selection of initial points gives a plot as follows. Concentrating on a small region about the origin, the iterates of $f$ give clearly defined trajectories - the plot includes a region of radius 0.26 about the origin (much larger, and the paths do start to go wild). As can be seen, those paths leaving the origin do one of two things. Either they move to the right, curving up or down, until they exit the region. Or, they bend round in a circle and re-enter the origin from the left. The iterates of $f^{-1}$ leaving the origin from the left behave similarly, but reflected about the imaginary axis. This should not be too surprising, and is behaviour displayed by any analytic function of the form $f(z)=z+\lambda z^2+O(z^3)$ where $\lambda > 0$. Consider approximating to second order by the Mobius function $f(z)\approx g(z)\equiv z/(1-\lambda z)$. Then, $g$ preserves circles centered on the imaginary axis and passing through the origin, and will move points on these circles in a counterclockwise direction above the origin and clockwise below the origin. A second order approximation to $g$ should behave similarly. In our case, we have $\lambda=1/2$ and $g$ actually agrees with $f$ to third order, so it is not surprising that we get such accurate looking circles (performing a similar exercise with $f(z)=\exp(z)-1$ gave rather more lopsided looking 'circles'). One thing to note from the plot above: the circles of diameter 0.25 above and below the origin are still very well-defined. So, if $f$ does define an analytic function, then its radius of convergence appears to be at least 0.25, and $f(\pm0.25i)$ is not much larger than 0.25 in magnitude. I wonder if summing a few hundred terms of the power series (as computed by Gottfried) will give a larger number? If it does, then this numerical evidence would point at $f$ not being analytic, and a more precise calculation should make this rigorous. To understand the trajectories, it is perhaps easiest to consider the coordinate transform $z\mapsto -1/z$. In fact, setting $u(z)=-1/z$, then the Mobius transform above satisfies $g(u(z))=u(z+\lambda)$. More generally, we can calculate the trajectories exiting and entering the origin for a function $f(z)=z+\lambda z^2+O(z^3)$ as follows $$\begin{align} &u_1(z)=\lim_{n\to\infty}f^{n}(-1/(z-n\lambda)),\\ &u_2(z)=\lim_{n\to\infty}f^{-n}(-1/(z+n\lambda)). \end{align}\qquad\qquad{\rm(1)}$$ Then, $u_1$ and $u_2$ map lines parallel to the real axis onto the trajectories of $f$ and $f^{-1}$ respectively and, after reading this answer, I gather are inverses of Abel functions. We can do a similar thing for our function, using the iterative formula in place of $f^{n}$. Then, we can define $f_i$ according to $f_i(z)=u_i(\lambda+u_i^{-1}(z))$, which will be well-defined analytic functions on the trajectories of $f$ (resp. $f^{-1}$) before they go too far from the origin (after which $u_i$ might not be one-to-one). Then $f_i$ will automatically satisfy the functional equation, and will give an analytic function in a neighbourhood of the origin if they agree on the intersection of their domain (consisting of the circular trajectories exiting and re-entering the origin). [Maybe add: Calculate error bounds for $u_i$ and the asymptotic expansion.] Update: Calculating trajectories of larger radius than plotted above gives the following. The trajectories leaving the origin from the right and entering from the left do not agree with each other, and intersect. This is inconsistent with the existence of a function $f$ in a neighborhood of the origin solving the functional equation, as the two solutions $f_1,f_2$ defined on trajectories respectively leaving and entering the origin do not agree. And, if the solutions $f_1,f_2$ do not agree on the larger trajectories then, by analytic continuation, they cannot agree close to the origin. So, if it can be confirmed that this behaviour is real (and not numerical inaccuracies), then the radius of convergence is zero. Update 2: It was noted in the original question that, for $n\ge3$, all coefficients $c_n$ in the power series expansion of $f$ are zero for even $n$, and that the odd degree coefficients are alternating in sign, so that $(-1)^kc_{2k-1}\ge0$ for $k\ge2$. This latter observation has not been proven, although Gottfried has calculated at least 128 coefficients (and I believe that this observation still holds true for all these terms). I'll give a proof of the following: if the odd degree coefficients $c_n$ are alternating in sign for $n\ge3$, then the radius of convergence is zero. To obtain a contradiction, let us suppose a positive radius of convergence $\rho$, and that the odd degree coefficients are alternating in sign after the 3rd term. This would imply that $$f(it)=-t^2/2 + it(1-t^2/4 - h(t^2))\qquad\qquad{\rm(2)}$$ where $h$ has nonnegative coefficients, so $h\ge0$ for real $t$. Also, $h(t)\to\infty$ as $t\to\rho_-$. For small $t$, $f(it)=it + o(t)$ has positive imaginary part so, by continuity, there will be some $0 < t_0 < \rho$ such that $\Im[f(it_0)]=0$. Choose $t_0$ minimal. If $\rho > 2$ then (2) gives $\Im[f(2i)]\le2(1-2^2/4)=0$ so, in any case, $t_0\le2$. Then, for $0 \le t \le t_0$, the imaginary part of $f(it)$ is bounded by $t(1-t^2/4)$ and (2) gives $$\begin{align} \vert f(it)\vert^2 &\le t^4/4 + t^2(1-t^2/4)^2\\ &=t^2(1-t^2(4-t^2)/16)\\ &\le t^2 < \rho^2. \end{align}$$ So, $f(it)$ is within the radius of convergence for $0 \le t \le t_0$. Also, by construction, the functional equation $f(f(it))=it+f(it)^2$ holds for $\vert t\vert$ small. Then, by analytic continuation the functional equation holds on $0\le t \le t_0$ so, $$\Im[f(f(it_0))]=\Im[it_0+f(it_0)^2]=t_0\not=0.$$ However, $f(it_0)$ and the power series coefficients are all real, so $f(f(it_0))$ is real, giving the contradiction. - The existence of these circular trajectories seems close to a contradiction situation for the following reason too. It looks like (though I didn't try to prove it) that the odd degree coefficients of the formal power series have alternate sign, that in terms of the formal series reads: f(it)=−t^2/2+it(1−h(t^2)), with h a series with positive coefficients. Thus, assuming R>0, h(t)→+∞ for t→R−, and Im(f(it))→−∞ as t→R, in contrast with the existence of circle orbits (at least, if they exist up to the height iR). – Pietro Majer Dec 22 2010 at 16:19 @Pietro: yes, if the odd degree coefficients have alternating sign, it is possible to show that the radius of convergence is zero using the fact that f(it) will pass through the real axis. – George Lowther Dec 22 2010 at 19:33 Concerning the remark, that the function for diameters 0.25 and below could be convergent. I've not yet new definitive results. But another heuristic: if you parallel the coefficients of the powerseries with the bernoulli-numbers (shifted by an index-difference of 1) and divide $c_k / b_{k-1}$ the sequence of ratios gives a very civilized picture. The ratios seem slowly to increase - but now without much up&down. So I take this as another strong suggestion for the rate-of-increase as hypergeometric and thus the convergence-radius of the powerseries of f as zero – Gottfried Helms Dec 22 2010 at 23:37 @George: do you mean you see how to prove the implication "$(-1)^k c_{2k-1} > 0$ for all $k\ge 3$ $\Rightarrow R=0$"? It seems to me that one needs to prove the existence of Gottfried's circles up to the quote $iR$. – Pietro Majer Dec 23 2010 at 9:48 @Pietro: Yes. I added this to the answer. – George Lowther Dec 23 2010 at 16:40 ### 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. It seems interesting to complexify the map you are studying. Thus we can study now $$F:\binom{x}{y} \mapsto \binom{y}{x+y^2}$$ as a map from $\mathbb{C}^2$ to itself ( the so-called "complex Henon map"). Now near the origin, the second iterate $F \circ F$ is tangent to the identity. There is now a whole body of results concerning germs of maps of $\mathbb{C}^N$ that are tangent to the identity, that can be applied here. In particular, such germs can have "parabolic curves" (analytic images of a disk $\Delta$ in $\mathbb{C}^2$ such that $(0,0) \in \partial \Delta$), included in the stable manifold of the origin. Such curves should be seen as generalizations of the "Leau-Fatou flowers" appearing in the study of parabolic fixed points in $\mathbb{C}$. A very nice survey about local dynamics in $\mathbb{C}^N$ has been written by M. Abate (in LNM 1998, Springer "Holomorphic dynamical systems"). When studying such Henon maps in $\mathbb{C}^2$ it is also usually helpful to plot slices of the set $K^+$ (points with bounded forward orbits): here is a horizontal slice y=0 of $K^+$. In blue, you see the set of points with unbounded forward orbit. . - This answer should go as a comment at George's analysis, but I had an error in the comments and also I couldn't format them properly. So here it goes. To force a symmetry around the imaginary axis means to introduce a definition independent of the ansatz by the formal powerseries. We'll have to check, whether they finally match. Assume some circle of radius $c$ over the origin with center at $0+c*î$ Then we choose $z_0 = 2*c*î$, just the top of the circle. Using the functional equation $z_1 = z_0^2+z_{-1}$ together with the symmetry-assumtion and the assumtion, that $z_1$,$z_0$,$z_{-1}$ are on the circumference of the circle we can uniquely determine all needed coordinates and have thus a "germ" for the iteration. What we get is the following. denote $s= \sqrt{1-4*c^2}$ Then we get $z_{-1} = 2 c^2 + i*c*(1+s)$ $z_0=i*2*c$ $z_1 = - 2 c^2+i*c*(1+s)$ and the list of the trajectory counterclockwise through the second quadrant $z_0 = i* 2*c$ $z_1 = -2*c^2 + i*(s + 1)*c$ $z_2 = 4*c^4 - (s+1)^2*c^2 + i*(-4*(s+1)*c^3 + 2*c)$ $z_3 = 16*c^8 - 24*(s+1)^2*c^6 + (s^4 + 4*s^3 + 6*s^2 + 20*s + 17)*c^4 - 6*c^2$ $+ i*((-32*s - 32)*c^7 + (8*s^3 + 24*s^2 + 24*s + 24)*c^5 + (-4*s^2 - 8*s - 4)*c^3 + (s + 1)*c)$ (The output can be much more simplified) The resulting trajectory is very near to that which was computed using the truncated powerseries. It is symmetric by construction, however it leaves the circumference of the circle already at $z_2$ [update] here is a comparision of the "circularity" of the trajectories (as it was already computed) using the powerseries and that using the above ansatz with the assumtion of a symmetric and circular initializing around the center $(0,i*c)$ [end update] - Look also at these pages which describe the technique of solving such equations: First look here: http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf And then look here: http://eqworld.ipmnet.ru/en/solutions/fe/fe2314.pdf - Let me try and redeem myself. I again claim that there is no positive radius of convergence of the power series (which seems to go along with the numerical evidence provided). Well I failed to redeem myself. Some Observations: First of all formally one has: $f^{-1}(x)=-f(-x)$ then it follows that for $g(x)=-f(x)$ one has $g(g(x))=-f(-f(x))=f^{-1}(f(x))=x$. Notice that (fomally) $g(x)=-x-\frac{1}{2} x^2+O(x^3)$ and that if $f$ had a positive radius of convergence then so would $g$. Lemma: Let $g$ be real analytic on some interval $I=(-\epsilon,\epsilon)$ and have an expansion $g(x)=x+O(x^2)$ about $0$. If $g(g(x))=x$ for all $x\in (-\delta, \delta)$ then $g(x)=x$ for all $x\in I$. Remark: I realized after posting this that as stated i.e $g=x+O(x^2)$ that there aren't even (non-trivial) formal solutions. However, if $g(x)=-x+O(x^2)$ there are formal solutions but but I don't know if there are actual analytic solutions Proof: Let $r>0$ be the radius of convergence of $g$ at $0$ notice that $g$ extends to a holomorphic function (which we continue to denote by $g$) on $$D_r=\lbrace z: |z|< r \rbrace \subset \mathbb{C}.$$ Notice that $g(z)=z+O(z^2)$ and $g'(z)=1+O(z)$. In particular, by the inverse function theorem there is a $r>\rho>0$ so that 1) $g(D_\rho)\subset D_r$ is open. 2) $g$ is an conformal isomorphism between $D_\rho$ and $g(D_\rho)$. Notice, that on $D_\rho$ one has $g(g(z))=z$ by analytic continuation. Now let $U=D_\rho\cap g(D_\rho)$ notice that $U$ is open and $0\in U$ by shrinking $\rho$ if needed we can ensure that $U$ is convex and hence simply-connected, we can also ensure $g(g(z))=z$ on $U$. We claim that $g(U)=U$. To see this note that if $p\in U$ then $p\in D_\rho$ and $p=g(p')$ for $p'\in D_\rho$. But $p'=g(g(p'))=g(p)\in g(D_\rho)$ so $p'\in U$ so $p\in g(U)$. Hence $U\subset g(U)$. On the other hand, if $q\in g(U)$ then $q=g(q')$ with $q'\in U$ but as we just showed $q'=g(q'')$ for $q''\in U$ and so $q=g(g(q''))=q''\in U$. By the Riemann mapping theorem there is a conformal isomorphism $\psi: U\to D_1$ so that $\psi(0)=0$ and $\psi'(0)=\lambda$ for $\lambda>0$. Now consider the map: $$G=\psi \circ g \circ \psi^{-1}.$$ This is a conformal automorphism of $D_1$ and hence is a Mobius transformation that also satisfies $G(G(z))=z$. Now one checks that the only fractional linear transformations that that satisfy $G(G(z))=z$ are $G(z)=z$ and $G(z)=-z$. The latter can't occur as it would contradict $g(z)=z+O(z^2)$. In particular, one has $G(z)=z$. But then $g(z)=\psi^{-1}( G( \psi(z)))=\psi^{-1}( \psi(z))=z$ as claimed. - How did you switch from $g(x)=-x+O(x^2)$ to $g(x)=x+O(x^2)$? – George Lowther Dec 21 2010 at 2:55 You're right. I should stay away from this problem it seems... – Rbega Dec 21 2010 at 3:19 Here is a plot for the trajectories starting at a couple of initial values in the complex plane. Here I computed the initial coordinates using the powerseries (including Noerlund-summation) and used then the functional equation to produce the trajectories for the iterates. [Update] I tried to find more information about the deviation of the trajectories from the circular shape. It seems, that these deviations are systematic though maybe small. Here is a plot of the trajectories from $z_1= 0.2 î$; the center of the perfect circle was assumed at $0.1 î$ I modified the value of $z_0$ for best fit (visually) of all $z_{k+1} = z_k^2+z_{k-1}$ and with some examples it seemed, that indeed the powerseries-solutions are the best choice for $z_0$ Here is the plot for the near circular trajectories in positive and negative directions (grey-shaded, but difficult to recognize the "perfect circle") The following plot shows the radial deviation, just the differences of $|z_k - center | - radius$ which is here $dev(z_k) = | z_k - 0.1 î| - 0.1$ The increasing wobbling at the left and right borders seems due to accumulation of numerical errors over iterations (I used Excel for the manual optimization) Note: for purely imaginary $z_1$ the Noerlund-summation for the powerseries does not work because we get a divergent series (of imaginary values) with non-alternating signs beginning at some small index. [end update] - I also tried my own simulation, which was consistent with your results. i53.tinypic.com/16kau02.png – George Lowther Dec 19 2010 at 19:11 I have to say, it does look like $f$ is a well defined analytic function in a neighbourhood of the origin. And, you might be able to put together a proof by showing that these circles are well-defined, so $f$ is well defined. Then apply the functional equation to $f^\prime$ to show that $f$ is differentiable. – George Lowther Dec 19 2010 at 19:15 However, I get a very similar result for $f(f(z))=e^z-1$. i53.tinypic.com/1zd2gcl.png. Yet, that does not give an analytic function in a neighbourhood of the origin (according to this answer: mathoverflow.net/questions/4347/…). I'm not sure what goes wrong in that case. I'll have to try reading the linked paper of Baker, although I don't read german, so it could be tricky. – George Lowther Dec 19 2010 at 19:48 I think the issue could be the following: consider calculating the path taken by the iterates of f starting in the upper-right quadrant, which will go around the circles in an anticlockwise direction, eventually approaching the origin from the upper-left quadrant. Try the same exercise, taking iterates of $f^{-1}$, which go around the circles in a clockwise direction. Do they agree? If they do, then it looks like you have an analytic function. Otherwise, you only get an analytic function if you exclude either the positive or negative real line. – George Lowther Dec 19 2010 at 20:23 Hmm, two observations. First: iterating $f$ from an initially real value goes to real infinity. Second: I tried also $z=0.05+0.001*î$ but got an overflow in the exponent at some iterate (didn't investigate this further) - which is remarkable when using Pari/GP. Don't know how things behave/how the radius of the circle/ellipse/or whatever "explodes" for z0 even nearer to the real axis. – Gottfried Helms Dec 19 2010 at 21:07 show 8 more comments Note: this should go as an answer to the comment of J.C.Ottern but I want to show data and thus need the entry for answers to the original questions. Hi J.C. - For the following I took each second coefficient. To adapt the signs I multiply by powers of $î =\sqrt{-1}$ to get coefficients say $d_k$. Then I show the quotients of subsequent $d_k$: $q_k = d_k/d_{k-1} = - c_{2k}/c_{2k-2}$ ````q_k -1.00000000000 -0.250000000000 0.500000000000 1.62500000000 2.78846153846 4.01379310345 5.34621993127 ... < some 50 coefficients ignored > ... 350.224481320 362.230770820 374.439712641 386.851306244 399.465551130 // at k=62 412.282446834 // at k=63 ```` To have convergence-radius >0 that ratios must converge to a constant value. But even the differences of the ratios increase: ````-1.00000000000 0.750000000000 0.750000000000 1.12500000000 1.16346153846 1.22533156499 ... 11.3983288747 11.6009830751 11.8036366001 12.0062894999 12.2089418203 12.4115936031 12.6142448861 // q_62-q_61 12.8168957040 // q_63-q_62 ... ```` If that behaviour continues the generated powerseries must have convergence-radius zero. [update] Here are the first 24 terms of the powerseries for $f(x)$ as I got them. left column in float, middle column in most-cancelled rational format,right column normalized rational format (numerators can be found in OEIS): ```` 0 0 . 1.00000000000 1 1/2 0.500000000000 1/2 1/4 0.250000000000 1/4 1/8 0 0 . -0.125000000000 -1/8 -2/32 0 0 . 0.203125000000 13/64 13/128 0 0 . -0.566406250000 -145/256 -145/512 0 0 . 2.27343750000 291/128 2328/2048 0 0 . -12.1542968750 -6223/512 -49784/8192 0 0 . 82.9446411133 1358965/16384 1358965/32768 0 0 . -703.072265625 -359973/512 -46076544/131072 0 0 . 7256.32673264 1902202515/262144 1902202515/524288 0 0 . -89745.2179527 -23526170415/262144 -94104681660/2097152 0 0 . 1312224.19186 42998962319/32768 5503867176832/8388608 ```` - Thanks for taking the trouble, Gottfried. I'm a bit confused about the terms though, could you maybe post the first terms in your Taylor series? – J.C. Ottem Dec 18 2010 at 20:05 @J.C. : I've included that terms in the previous answer-form. Please tell me if you got something else/if I've made some error here. – Gottfried Helms Dec 18 2010 at 20:24 No, this is fine. Thanks! – J.C. Ottem Dec 18 2010 at 20:31 I computed the coefficients of the formal powerseries for $f(x)$ to $n=128$ terms. As it was already mentioned in other answers/comments, the coefficients seem to form a formal powerseries of concergence-radius zero; equivalent to that the rate of growth of absolute values of the coefficients is hypergeometric. To get a visual impression of the characteristics of the function I evaluated $f(x)$ for some $0 < x <2$ Method 1 : I used (an experimental version of) Noerlund-summation with which I could estimate approximations for that interval of $0 < x < 2$ . Method 2, just to crosscheck that results: I repeated the evaluation for that range of x using the functional equation. I computed the "primary interval" of $x_0=0.1 \text{ to } 0.105\ldots -> y_0=0.105\ldots \text{ to } 0.111\ldots$ which defines one unit-interval of iteration height where the Noerlund sum seems to converge very good (I assume error $<1e-28$ using 128 terms of the powerseries). Then the functional equation allows to extend that interval to higher x, depending on whether we have enough accuracy in the computation of the primary interval $x_0 \text{ to } y_0$. Result: both computations gave similar and meaningful results with an accuracy of at least 10 digits - but all this requires a more rigorous analysis after we got some likelihood that we are on the right track here. Here is a plot for the estimated function $f(x)$ for the interval $0 < x < 4.5$ (the increase of the upper-bound of the interval was done using the functional equation) and here a plot which includes the inverse function to see the characteristic in terms of $x^2 = f(x)-f^{o-1}(x)$ [update] Based on the second plot I find this representation interesting which focuses on squares. It suggests to try to use integrals to find the required coordinates for the positioning of the squares. Does the sequence of green-framed squares has some direct representation which allows to determine the coordinates independently from recursion? The "partition" of the big green square into that red ones alludes to something like the golden ratio...[end update] - 1 Thank you, very clear pictures. Another possibility should be iterating the image of the positive half-axis $\\{(0,y): y\ge0\\}$ via the map $\phi:(x,y)\mapsto(y,x+y^2)$. This way we get a sequence of graphs that converge to the graph of $f$, alternately above and below. – Pietro Majer Dec 18 2010 at 15:38 Why do you think that the concergence-radius is zero? From the plot oeis.org/A107699/graph it seems that the numerators of the coefficients grow exponentially. – J.C. Ottem Dec 18 2010 at 17:57 Very nice! I'm wondering if its possible to do something similar, using the functional equation to extend $f$ to the right half-plane $\Re[z] > 0$? Just a wild stab in the dark, but maybe it does extend to an analytic function on that region, and explodes as you approach the imaginary axis? – George Lowther Dec 18 2010 at 23:47 @George: I plotted a couple of trajectories for RE(z) = 0.05 and small imaginary values. Couriosly - they look like perfect circles. Is this something what you were asking for? – Gottfried Helms Dec 19 2010 at 17:02 @Gottfried: Yes! I was expecting that the trajectories might go astray. Or maybe that different trajectories cross each other. But what you observed seems perfectly consistent with an analytic function. – George Lowther Dec 19 2010 at 17:42 show 4 more comments Your equation can be re-written as a difference equation, if g(t) is the flow of f(x) then $$g(t+2)=g(t)+g(t+1)^2$$ or $$\Delta^2 g-(\Delta g+g)^2+2\Delta g-2g=0$$ It is second order non-linear ordinary difference equation. It looks weird but at least you can try to apply methods of discrete calculus to this equation (with not much hope though). - I think there can't be a positive radius of convergence. I don't know what happens now... My reasoning is was as follows: Basically suppose there was a positive radius of convergence and call it $R>0$. Now for $1>\lambda> 0$ set $g_\lambda(x)=\frac{1}{\lambda} f(\lambda x)$. Clearly, $g_\lambda$ is analytic on $|x|<\frac{R}{\lambda}$. Moreover, as $\lambda\to 0$ we see that $g_\lambda\to x$ (NOT $g_\lambda\to c_1=1$) uniformly on compact subsets (by Arzela-Ascoli). Now it is clear that $g_\lambda(g_\lambda(x))=\frac{1}{\lambda} f(f(\lambda x))$ but $\frac{1}{\lambda} f(f(\lambda x))=\frac{1}{\lambda} (\lambda x+f(\lambda x) ^2)=x+\lambda g_\lambda(x)^2$. That is $g_\lambda$ satisfies the equation $g_\lambda(g_\lambda(x))=x+\lambda(g_\lambda(x))^2$ Now let $\lambda\to 0$. For any $x$ with $|x|<2$ the left hand side tends to $x$ (NOT $c_1=1$) while the right hand side tends to $x$. This is clearly impossible. So everything is consistent. I think this works. Obviously not. - Thinking about it some more I guess the above argument and Taylor's theorem actually shows that $f$ can't even be $C^1$ in a neighborhood of $0$. – Rbega Dec 17 2010 at 23:49 2 $g_\lambda(x)$ tends to $x$, not to 1 – Fedor Petrov Dec 18 2010 at 0:01 Good point. I knew something was wrong... – Rbega Dec 18 2010 at 0:14 2 But there is a good hint. Thanks to the rescaling argument the equation $f(f(x))=x+f(x)^2$ for analytic germs is equivalent to $g(g(x))=x+\lambda g(x)^2$,for $\lambda>0$, even for a small one, and the latter may be treated as aperturbation of $g(g(x))=x$ by means of the implicit funciton theorem. – Pietro Majer Dec 18 2010 at 1:08
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http://math.stackexchange.com/questions/155589/schur-weyl-duality-for-sl-2-and-s-n
# Schur -Weyl duality for $sl_2$ and $S_n$ $V$ is an $m$ dimensional vector space having a structure of $sl_2(\mathbb{C})$-module, where $sl_2(\mathbb{C})$ is the Lie algebra of the Lie group $SL_2(\mathbb{C})$. The symmetric group $S_n$ acts on the tensor product $V^{\otimes n}$. What does Schur-Weyl duality say in this case? What is the irreducible decomposition of $V^{\otimes n}$? If we have $S_n$ irreducible decomposition can we get $sl_2(\mathbb{C})$ decomposition and vice versa? I would be very grateful if someone could give a detailed answer. Thanking you in advance. - – Marc van Leeuwen Jun 8 '12 at 13:46 ## 1 Answer Here is an answer to your question about the decomposition of $V^{\otimes n}$ as an $\mathfrak{sl}_2(\mathbb{C})$-module (I assume $V$ is the standard $2$-dimensional irreducible module). The irreducible $\mathfrak{sl}_2(\mathbb{C})$-modules are indexed by non-negative integers and the one corresponding to the integer $m$ will be denoted $L(m)$ (it has dimension $m+1$ and we know exactly what it looks like, see for example Humphrey's Introduction to Lie Algebras and Representation Theory chapter 7). So to see how to decompose $V^{\otimes n}$ we need to know what $V\otimes L(m)$ is for some integer $m$. This is easiest to see if we look at these as $\mathfrak{gl}_2(\mathbb{C})$-modules where the decomposition of tensor products is given by the Littlewood-Richardson rule. In this case, since $V = L(1)$, we simply get that $V\otimes L(m) = L(m+1)\oplus L(m-1)$. Now we want to apply this to see how many times $L(m)$ appears as a summand in $V^{\otimes n}$. Let us denote this multiplicity by $a_{m,n}$. We see from the above that $a_{m,n} = a_{m-1,n-1} + a_{m+1,n-1}$. One can check that $$a_{m,n} = \binom{n}{\frac{m+n}{2}} - \binom{n}{\frac{n - m - 2}{2}}$$ satisfies the above recursive formula when $n$ and $m$ have the same parity (and when they don't, $a_{m,n} = 0$). Note that $a_{0,2k} = a_{1,2k-1}$ is the $k$'th Catalan number. - Thank you for your answer, i was trying to get some information about $V$ not necessarily an irreducible module, having dimension m... – Jacob Feb 20 at 6:42 If $V$ is not irreducible, it will be the direct sum of some irreducible modules. It is possible to calculate the tensor powers of an arbitrary module in the same way as above, but the combinatorics become very complicated in the general case. – Tobias Kildetoft Feb 20 at 13:48
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http://mathoverflow.net/questions/40161/why-do-we-care-about-the-hilbert-scheme-of-points/40180
## Why do we care about the Hilbert scheme of points? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If $X$ is a scheme, the Hilbert scheme of points $X^{[n]}$ parameterizes zero dimensional subschemes of $X$ of degree $n$. Why do we care about it? Of course, there are lots of "in subject" reasons, which I summarize by saying that $X^{[n]}$ is maybe the simplest modern moduli space, and as such is an extremely fertile testing ground for ideas in moduli theory. But it is not clear that this would be very convincing to someone who was not already interested in $X^{[n]}$. The question I am really asking is: Why would someone who does not study moduli care about $X^{[n]}$? The main reason I ask is for the sake of having some relevant motivation sections in talks. But an answer to the following version of the question would be extremely valuable as well: What can someone who knows a lot something* about $X^{[n]}$ contribute to other areas of algebraic geometry, or mathematics more generally, or even other subjects? *reworded in light of the answer of Nakajima - ## 11 Answers What can someone who knows a lot about $X^{[n]}$ contribute to other areas of algebraic geometry, or mathematics more generally, or even other subjects? I know a little about $X^{[n]}$. And I have no contribution to mathematics nor other areas of algebraic geometry. But I find study of Hilbert schemes is very interesting. Isn't it enough to motivate to study Hilbert schemes ? - 27 If I am asked this question by a politician, I will answer differently. – Hiraku Nakajima Sep 29 2010 at 12:43 6 -1 mark for excessive modesty – Richard Thomas Oct 20 2010 at 13:38 ### 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. One of the reasons is the following. If $X$ is a smooth curve the symmetric product $X^{(n)}$ is smooth. If $X$ is a smooth surface, $X^{(n)}$ is singular, but it is a theorem of Fogarty that $X^{[n]}$ is smooth. So it is a (rather natural) resolution of singularities of $X^{(n)}$. If moreover $X$ is a symplectic surface (i.e. either a $K3$ or an abelian surface), $X^{[n]}$ has a symplectic form. As Beauville showed, in the $K3$ case it is even an irreducible symplectic variety; in the abelian case a certain subvariety $K_{n-1}(X) \subset X^{[n]}$ is irreducible symplectic. There are really few known examples of irreducible symplectic varieties in higher dimension: up to deformation, I have listed them all, except for two sporadic examples constructed by O'Grady! This shows that the Hilbert schemes $X^{[n]}$ are indeed very relevant to people who study holomorphic symplectic geometry. - 2 Andrea- I assume you mean irreducible projective symplectic varieties? There are loads of quasi-projective ones that don't fit in your class. – Ben Webster♦ Sep 29 2010 at 3:03 @Ben: thank you for pointing out; of course I am talking about complete examples. – Andrea Ferretti Sep 30 2010 at 19:40 Mark Haiman used Hilbert schemes of surfacesto prove the Macdonald positivity conjecture about Macdonald polynomials: see Hilbert schemes, polygraphs, and the Macdonald positivity conjecture - The connection between the geometry of the Hilbert scheme of points on $\mathbb{C}^d$ and the combinatorics of $d$-dimensional partitions (piles of $d$-dimensional boxes in the corner of an $d$-dimensional room) may resonate with some audiences. In dimension 3 especially, geometric knowledge has led to beautiful combinatorial results about 3 dimensional box counting (I have in mind Young arXiv:0802.3948 and Okounkov-Reshetikhin-Vafa hep-th/0309208). The connection that Allen mentions with quivers is nice, and to counter to his dimension 2 bias, I think the quiver story is arguably even better in dimension 3. The Hilbert scheme of points on $\mathbb{C}^3$ (or $\mathbb{C}^3/G$) is given by representations of a quiver with super-potential. Unlike the surface case, the relations on the quiver are given by the critical locus of a single function --- a phenomenon special to dimension 3. - I believe everyone so far has said "Hilbert schemes of points on surfaces" are worth studying. Personally, I take this as a sign that Hilbert schemes of points (on general $X$) are not that interesting, and that it's merely a historical accident that these glittering jewels were first found within the many tons of worthless Hilbert scheme ore. In particular, note that the Hilbert scheme of $n$ points in the plane (or on other $\widetilde{{\mathbb C}^2/\Gamma}$, for $\Gamma \leq SU(2)$) can be alternately seen as a Grojnowski-Nakajima quiver variety, and these are smooth holomorphic symplectic manifolds in general. - 4 Have you heard of any type of process that gives you the Hilbert scheme of points if you input a surface but gives you something better behaved than the Hilbert scheme of points if you plug in something higher-dimensional? – Peter Samuelson Sep 29 2010 at 5:07 Hilbert schemes of points in the plane come up in representation theory. More specifically in the representation theory of rational Cherednik algebras. Rational Cherednik algebras, and more generally symplectic reflection algebras have a close connection with symplectic resolutions, algebraic combinatorics, integrable systems and more and are thus very interesting. See for example: This paper and This paper - Slogan: the Hilbert scheme of $n$ points on the minimal resolution of an ADE singularity (associated to a finite subgroup $G \subseteq SL_2(\mathbb{C})$) is to the symplectic reflection algebra of type $G^n \rtimes S_n$ as the flag variety $G/B$ is to the universal enveloping algebra of $\mathrm{Lie}(G)$. So far this is not completely understood. – Sheikraisinrollbank Sep 28 2010 at 9:43 er... typo. Of course I meant, cotangent bundle to $G/B$. – Sheikraisinrollbank Sep 29 2010 at 13:42 One can use Hilbert schemes of points on a surface $S$ to detect sigularities of embedded curves $C \subset S$. For instance let $\xi \in S^{[3]}$ be a "fat point", i.e. a subscheme of $S$ isomorphic to $Spec(\mathbb{C}[x,y]/(x^2,xy,y^2))$, then $C$ contains $\xi$ if and only if $C$ has a singularity at $supp(\xi)={ p }$. In a recent preprint Thomas, Kool and Shende (http://front.math.ucdavis.edu/1010.3211) use this method to proof Goettsche's conjecture about counting singular members of linear systems on surfaces. The basic idea is to write the counting problem as an intersection product of class on $S \times S^{[n]}$ and use results of Ellingsrud-Goettsche-Lehn (building on work of Nakajima) which describe the intersection theory on $S^{[n]}$. REMARK: I just got aware of the fact, that V. Schende who coauthored the above paper is actually asking the question. So he probably knows about this application. - 4 +1 for the remark :-) – Kevin Lin Oct 19 2010 at 6:47 Someone answered, when asked by braids and braid groups were important, answered that they are important because they are histories of permutations and permutations are important (or something along that line; I'd love a precise quotation!) The Hilbert scheme is, in a way, the collection of all possible states in those histories (and all the possible ways you can mess up when permuting points...) - 3 I would really love any more precise information about these ideas... – Vivek Shende Sep 27 2010 at 17:48 1 See the work of Nakajima and Grojnowski, where for a fixed algebraic surface $X,$ the points of the Hilbert schemes $X^{[n]}$ for all $n$ are interpreted as instantons on $X.$ – Victor Protsak Sep 28 2010 at 3:28 As a student I once asked John Mather for advice about dealing with spaces of ordered $n$-tuples of distinct points in a manifold -- how to get organized about limiting cases where the points come together. He said I needed the Hilbert scheme, and he was right. Well, I wasn't doing algebraic geometry -- I was doing something with smooth manifolds -- and he knew it, so "scheme" was the wrong word. But some kind of essence of the Hilbert scheme idea, adapted to what would have been real semi-algebraic geometry if I had been systematic about it, did the trick. - One context in which Hilbert schemes provide a useful conceptual picture is in the Fourier expansion of partition functions that are associated to certain Siegel modular forms of genus 2. - If I understand things right, the Hilbert scheme of points in $\mathbb{A}^2$ is hyperkahler manifold and the $n^\textrm{th}$ Calogero-Moser $CM_n$ space can be obtained by deforming the complex structure. The disjoint union of the $CM_n$ parametrizes isomorphism classes of (right) ideals in the first Weyl algebra (polynomial differential operators in 1 variable). This classification is related to Hyugen's principle in differential equations. -
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http://tcsmath.wordpress.com/2010/06/15/majorizing-measures/
# tcs math – some mathematics of theoretical computer science ## June 15, 2010 ### Majorizing measures Filed under: Math — Tags: chaining, cover time, Gaussian process, majorizing measures, Talagrand — James Lee @ 12:44 am At STOC 2010 last week, Talagrand gave a presentation on some of his favorite open problems, which included a quick review of Gaussian processes and the majorizing measures theory. In joint work with Jian Ding and Yuval Peres, we recently showed how the cover time of graphs can be characterized by majorizing measures. While I’ll eventually try to give an overview of this connection, I first wanted to discuss how majorizing measures are used to control Gaussian processes. In the next few posts, I’ll attempt to give an idea of how this works. I have very little new to offer over what Talagrand has already written; in particular, I will be borrowing quite heavily from Talagrand’s book (which you might read instead). Gaussian processes Consider a Gaussian process ${\{X_t\}_{t \in T}}$ for some index set ${T}$. This is a collection of jointly Gaussian random variables, meaning that every finite linear combination of the variables has a Gaussian distribution. We will additionally assume that the process is centered, i.e. ${\mathbb E(X_t) = 0}$ for all ${t \in T}$. It is well-known that such a process is completely characterized by the covariances ${\{\mathop{\mathbb E}(X_s X_t)\}_{s,t \in T}}$. For ${s,t \in T}$, consider the canonical distance, $\displaystyle d(s,t) = \sqrt{\mathbb E\,|X_s-X_t|^2},$ which forms a metric on ${T}$. (Strictly speaking, this is only a pseudometric since possibly ${d(s,t)=0}$ even though ${X_s}$ and ${X_t}$ are distinct random variables, but we’ll ignore this.) Since the process is centered, it is completely specified by the distance ${d(s,t)}$, up to translation by a Gaussian (e.g. the process ${\{X_t + X_{t_0}\}_{t \in T}}$ will induce the same distance for any ${t_0 \in T}$). A concrete perspective If the index set ${T}$ is countable, one can describe every such process in the following way. Let ${\{g_i\}_{i=1}^{\infty}}$ be a sequence of i.i.d. standard Gaussians, let ${T \subseteq \ell^2}$, and put $\displaystyle X_t = \sum_{i \geq 1} g_i t_i.$ In this case, it is easy to check that ${d(s,t) = \|s-t\|_2}$ for ${s,t \in T}$. (That this construction is universal follows from the fact that every two separable Hilbert spaces are isomorphic.) Random projections If ${T}$ is finite, then we can think of ${T \subseteq \mathbb R^n}$ for some ${n \in \mathbb N}$. In this case, if ${g}$ is a standard ${n}$-dimensional Gaussian, then $\displaystyle X_t = \langle g, t \rangle,$ and we can envision the process as the projection of ${T}$ onto a uniformly random direction. Studying the maxima We will be concerned primarily with the value, $\displaystyle \mathbb E \sup_{t \in T} X_t.$ (I.e. the expected value of the extremal node circled above.) One may assume that ${T}$ is finite without losing any essential ingredient of the theory, in which case the supremum can be replaced by a maximum. Note that we are studying the tails of the process. Dealing with these extremal values is what makes understanding the above quantity somewhat difficult. As some motivation for the classical study of this quantity, one has the following. Theorem 1 For a separable Gaussian process ${\{X_t\}_{t \in T}}$, the following two assertions are equivalent. 1. The map ${t \mapsto X_t(\omega)}$ is uniformly continuous (as a map from ${(T,d)}$ to ${\mathbb R)}$ with probability one. 2. As ${\varepsilon \rightarrow 0}$, $\displaystyle \mathop{\mathbb E}\sup_{d(s,t) \leq \varepsilon} |X_s-X_t| \rightarrow 0.$ However, from our viewpoint, the quantitative study of ${\mathbb E\sup_{t \in T} X_t}$ in terms of the geometry of ${(T,d)}$ will play the fundamental role. Bounding the sup We will concentrate first on finding good upper bounds for ${\mathop{\mathbb E}\sup_{t \in T} X_t}$. Toward this end, fix some ${t_0 \in T}$, and observe that $\displaystyle \mathop{\mathbb E}\sup_{t \in T} X_t = \mathop{\mathbb E}\sup_{t \in T} (X_t - X_{t_0}).$ Since ${\sup_{t \in T} (X_t - X_{t_0})}$ is a non-negative random variable, we can write $\displaystyle \mathop{\mathbb E} \sup_{t \in T} (X_t - X_{t_0}) = \int_{0}^{\infty} \mathop{\mathbb P}\left(\sup_{t \in T} X_t - X_{t_0} > u\right)\,du,$ and concentrate on finding upper bounds on the latter probabilities. Improving the union bound As a first step, we might write $\displaystyle \mathop{\mathbb P}\left(\sup_{t \in T} X_t - X_{t_0} > u\right) \leq \sum_{t \in T} \mathop{\mathbb P}\left(X_t - X_{t_0} > u\right).$ While this bound is decent if the variables ${\{X_t-X_{t_0}\}_{t \in T}}$ are somewhat independent, it is rather abysmal if the variables are clustered. Since the variables in, e.g. ${S_1}$, are highly correlated (in the “geometric” language, they tend to project close together on a randomly chosen direction), the union bound is overkill. It is natural to choose representatives ${t_1 \in S_1}$ and ${t_2 \in S_2}$. We can first bound ${X_{t_1}-X_{t_0}}$ and ${X_{t_2}-X_{t_0}}$, and then bound the intra-cluster values ${\{X_t - X_{t_1}\}_{t \in S_1}}$ and ${\{X_t - X_{t_2}\}_{t \in S_2}}$. This should yield better bounds as the diameter of ${S_1}$ and ${S_2}$ are hopefully significantly smaller than the diameter of ${T}$. Formally, we have $\displaystyle \mathop{\mathbb P}\left(\sup_{t \in T} X_t - X_{t_0} > u\right) \leq$ $\displaystyle \mathop{\mathbb P}(X_{t_1} - X_{t_0} > u/2) + \sum_{t \in S_1} \mathop{\mathbb P}(X_t - X_{t_1} > u/2)$ $\displaystyle + \mathop{\mathbb P}(X_{t_2} - X_{t_0} > u/2) + \sum_{t \in S_2} \mathop{\mathbb P}(X_t - X_{t_2} > u/2).$ Of course, there is no reason to stop at one level of clustering, and there is no reason that we should split the contribution ${u = u/2 + u/2}$ evenly. In the next post, we’ll see the “generic chaining” method which generalizes and formalizes our intuition about improving the union bound. In particular, we’ll show that every hierarchical clustering of our points offers some upper bound on $\mathbb E \sup_{t \in T} X_t$. ## 3 Comments » 1. [...] Processes in Theory Jump to Comments James Lee has a new post on his tcsmath blog about Gaussian Processes, a topic I’ve been enamored with for the last while. I love the graphics he includes in his [...] Pingback by — June 16, 2010 @ 11:58 am 2. Thank you! Even though I knew about the chaining technique, but still your post helped me to have a better intuition. I liked the cluster diagram! Comment by — June 16, 2010 @ 9:29 pm
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http://unapologetic.wordpress.com/2007/03/page/3/
The Unapologetic Mathematician Flag varieties As another part of preparing for the digestion of the $E_8$ result, I need to talk about flag vareties. You’ll need at least some linear algebra to follow from this point. A flag in a vector space is a chain of nested subspaces of specified dimensions. In three-dimensional space, for instance, one kind of flag is a choice of some plane through the origin and a line through the origin sitting inside that plane. Another kind is just a choice of a plane through the origin. The space of all flags of a given kind in a vector space can be described by solving a certain collection of polynomial equations, which makes it a “variety”. It’s sort of like a manifold, but there can be places where a variety intersects itself, or comes to a point, or has a sharp kink. In those places it doesn’t look like $n$-dimensional space. Flag varieties and Lie groups have a really interesting interaction. I’ll try to do the simplest example justice, and the rest are sort of similar. We take a vector space $V$ and consider the group $SL(V)$ of linear transformations $T:V\rightarrow V$ with $\det(T)=1$. Clearly this group acts on $V$. If we pick a basis $\{b_1,b_2,...,b_n\}$ of $V$ we can represent each transformation as an $n\times n$ matrix. Then there’s a subgroup of “upper triangular” matrices of the form $\left(\begin{array}{cccc}1&a_{1,2}&\cdots&a_{1,n}\\ 0&1&\cdots&a_{2,n}\\\vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1\end{array}\right)$ check that the product of two such matrices is again of this form, and that their determinants are always $1$. Of course if we choose a different basis, the transformations in this subgroup are no longer in this upper triangular form. We’ll have a different subgroup of upper triangular matrices. The subgroups corresponding to different bases are related, though — they’re conjugate! Corresponding to each basis we also have a flag. It consists of the line spanned by the first basis element contained in the plane spanned by the first two elements, contained in… and so on. So why do we care about this flag? Because the subgroup of upper triangular matrices with respect to this basis fixes this flag! The special line is sent back into itself, the special plane back into itself, and so on. In fact, the group $SL(V)$ acts on the flag variety “transitively” (there’s only one orbit) and the stabilizer of the flag corresponding to a basis is the subgroup of upper triangular matrices with respect to that basis! The upshot is that we can describe the flag variety from the manifold of $SL(V)$ by picking a basis, getting the subgroup of upper triangular matrices $U$, and identifying elements of $SL(V)$ that “differ” by an element of $U$. The subgroup $U$ is not normal in $SL(V)$, so we can’t form a quotient group, but there’s still a space of cosets: $SL(V)/U$. So studying the flag variety in $V$ ends up telling us about the relationship between the group $SL(V)$ and its subgroup $U$. In general if we have a Lie group $G$ and a subgroup $B$ satisfying a certain condition we can study the relation between these two by studying a certain related variety of flags. Posted by John Armstrong | Atlas of Lie Groups | 2 Comments Lie groups, Lie algebras, and representations I’m eventually going to get comments from Adams, Vogan, and Zuckerman about Kazhdan-Lusztig-Vogan polynomials, but I want to give a brief overview of some of the things I know will be involved. Most of this stuff I’ll cover more thoroughly at some later point. One side note: I’m not going to link to the popular press articles. I’m glad they’re there, but they’re awful as far as the math goes. Science writers have this delusion that people are just incapable of understanding mathematics, and completely dumb things down to the point of breaking them. Either that or they don’t bother to put in even the minimum time to get a sensible handle on the concepts like they do for physics. Okay, so a Lie group is a group of continuous transformations, like rotations of an object in space. The important thing is that the underlying set has the structure of a manifold, which is a space that “locally looks like” regular $n$-dimensional space. The surface of the Earth is curved into a sphere, as we know, but close up it looks flat. That’s what being a manifold is all about. The group structure — the composition and the inversion — have to behave “smoothly” to preserve the manifold structure. One important thing you can do with a Lie group is find subgroups with nice structures. Some of the nicest are the one-dimensional subgroups passing through the identity element. Since close up the group looks like $n$-dimensional space, let’s get really close and stand on the identity. Now we can pick a direction and start walking that way, never turning. As we go, we trace out a path in the group. Let’s say that after $t$ minutes of elapsed time we’re at $g(t)$. If we’ve done things right, we have the extremely nice property that $g(t_1)g(t_2)=g(t_1+t_2)$. That is, we can multiply group elements along our path by adding the time parameters. We call this sort of thing a “1-parameter subgroup”, and there’s one of them for each choice of direction and speed we leave the origin with. So what if we start combining these subgroups? Let’s pick two and call them $g_1(t)$ and $g_2(t)$. In general they won’t commute with each other. To see this, get a ball and put it on the table. Mark the point at the top of the ball so you can keep track of it. Now, roll the ball to the right by 90°, then away from you by 90°, then to your left by 90°, then towards you by 90°. The point isn’t back where it started, it’s pointing right at you! Try it again but make each turn 45°. Again, the point isn’t back at the top of the ball. If you do this for all different angles, you’ll trace out a curve of rotations, which is another 1-parameter subgroup! We can measure how much two subgroups fail to commute by getting a third subgroup out of them. And since 1-parameter subgroups correspond to vectors (directions and speeds) at the identity of the group, we can just calculate on those vectors. The set of vectors equipped with this structure is called a Lie algebra. Given two vectors $\mathbf{v}$ and $\mathbf{w}$ we write the resulting vector as $\left[\mathbf{v},\mathbf{w}\right]$. This satisfies a few properties. • $\left[c\mathbf{v}_1+\mathbf{v}_2,\mathbf{w}\right]=c\left[\mathbf{v}_1,\mathbf{w}\right]+\left[\mathbf{v}_2,\mathbf{w}\right]$ • $\left[\mathbf{v},\mathbf{w}\right]=-\left[\mathbf{w},\mathbf{v}\right]$ • $\left[\mathbf{u},\left[\mathbf{v},\mathbf{w}\right]\right]=\left[\left[\mathbf{u},\mathbf{v}\right],\mathbf{w}\right]+\left[\mathbf{v},\left[\mathbf{u},\mathbf{w}\right]\right]$ Lie algebras are what we really want to understand. So now I’m going to skip a bunch and just say that we can put Lie algebras together like we make direct products of groups, only now we call them direct sums. In fact, for many purposes all Lie algebras can be broken into a finite direct sum of a bunch of “simple” Lie algebras that can’t be broken up any more. Think about breaking a number into its prime factors. If we understand all the simple Lie algebras, then (in theory) we understand all the “semisimple” Lie algebras, which are sums of simple ones. And amazingly, we do know all the simple Lie algebras! I’m not remotely going to go into this now, but at a cursory glance the Wikipedia article on root systems seems to be not completely insane. The upshot is that we’ve got four infinite families of Lie algebras and five weird ones. Of the five weird ones, $E_8$ is the biggest. Its root system (see the Wikipedia article) consists of 240 vectors living in an eight-dimensional space. This is the thing that, projected onto a plane, you’ve probably seen in all of the popular press articles looking like a big lace circle. So we already understood $E_8$? What’s the big deal now? Well, it’s one thing to know it’s there, and another thing entirely to know how to work with such a thing. What we’d really like to know is how $E_8$ can act on other mathematical structures. In particular, we’d like to know how it can act as linear transformations on a vector space. Any vector space $V$ comes equipped with a Lie algebra $\mathfrak{end}(V)$: take the vector space of all linear transformations from $V$ to itself and make a bracket by $\left[S,T\right]=ST-TS$ (verify for yourself that this satisfies the requirements of being a Lie algebra). So what we’re interested in is functions from $E_8$ to $\mathfrak{end}(V)$ that preserve all the Lie algebra structure. And this, as I understand it, is where the Kazhdan-Lusztig-Vogan polynomials come in. Root systems and Dynkin diagrams are the essential tools for classifying Lie algebras. These polynomials are essential for understanding the structures of representations of Lie algebras. I’m not ready to go into how they do that, and when I am it will probably be at a somewhat higher level than I usually use here, but hopefully lower than the technical accounts available in Adams’ paper and Baez’ explanation. Posted by John Armstrong | Atlas of Lie Groups | 3 Comments Quandles At long last I come to quandles. I know there are some readers who have been waiting for this, but I wanted to at least get through a bunch of group theory before I introduced them, because they tend to feel a bit weirder so it’s good to warm up before jumping into them. The story of quandles really begins back in the late ’50s and early ’60s when John Conway and Gavin Wraith considered the wrack and ruin that’s left when you violently rip away the composition from a group and just leave behind its conjugation action. This is a set with an operation $x\triangleright y=xyx^{-1}$, and it’s already a quandle. The part of the structure they considered, though, has lost its ‘w’ and become known as a “rack”. In 1982, David Joyce independently discovered these structures while a student under Peter Freyd working on knot theory with a very categorical flavor (hmm.. sounds familiar). He called them “quandles” because he wanted a word that didn’t mean anything else already, and when the term popped into his head he just liked the sound of it. There are other names for similar structures, but “quandle” is the one that really took hold, partially because there are a lot of unusual algebraic structures that aren’t good for much but their own interest, but “quandle” was the term the knot theorists picked up and ran with. Actually, after hearing one of my talks Dr. Freyd mentioned that Joyce had come up with a lot of good things while a student, but he (Freyd) never thought much would come of quandles. In the end quandles have become the biggest thing to come out of his (Joyce’s) thesis. Okay, so let’s get down to work. There are three axioms for the structure of a quandle, and I’ll go through them in the reverse of the usual order for reasons that will become apparent. We start with a set with two operations, written $x\triangleright y$ and $y\triangleleft x$. The third and most important axiom is that $x\triangleright y$ distributes over itself: $x\triangleright (y\triangleright z)=(x\triangleright y)\triangleright(x\triangleright z)$. A set with one operation satisfying this property is called a “shelf”, leading to Alissa Crans’ calling the property “shelf-distributivity”. No, I’m not going to let her live down making such an awful pun, mostly because she beat me to it. We can verify that conjugation in a group satisfies this property: $x\triangleright (y\triangleright z)=x(yzy^{-1})x^{-1}=(xyx^{-1})(xzx^{-1})(xy^{-1}x^{-1})=$ $=(xyx^{-1})(xzx^{-1})(xyx^{-1})^{-1}=(x\triangleright y)\triangleright(x\triangleright z)$ The second axiom is that the two operations undo each other: $(x\triangleright y)\triangleleft x=y=x\triangleright(y\triangleleft x)$. Some authors just focus on the one operation and insist that for every $a$ and $b$ the equation $a\triangleright x=b$ have a unique solution. Our second operation is just what gives you back that solution. A shelf satisfying (either form of) this axiom is called a rack. We again verify this for conjugation, using conjugation by the inverse as our second operation: $(x\triangleright y)\triangleleft x=x^{-1}(xyx^{-1})x=y=x(x^{-1}yx)x^{-1}=x\triangleright(y\triangleleft x)$. Finally, the first axiom is that $x\triangleright x=x$. Indeed, for a group we have $x\triangleright x=xxx^{-1}=x$. This axiom makes a rack into a quandle. One more specialization comes in handy: we call a quandle “involutory” if $x\triangleright y=y\triangleleft x$. Equivalently (by the second axiom), $x\triangleright(x\triangleright y)=y$. That is, $x$ acts on $y$ by some sort of reflection, and acting twice restores the original. As a bit of practice, check that in a rack the second operation is also self-distributive. That is, $(z\triangleleft y)\triangleleft x=(z\triangleleft x)\triangleleft(y\triangleleft x)$. Also verify that if we start with an abelian group $G$ (writing group composition as addition), the operation $g\triangleright h=2g-h$ makes the set of elements of $G$ into an involutory quandle. Progress from the Atlas A team of 18 mathematicians working on the Atlas of Lie Groups and Representations have completed computing the Kazhdan-Lusztig-Vogan polynomial for the exceptional Lie group $E_8$. There’s a good explanation by John Baez at The n-Category Café (warning: technical). I feel a sort of connection to this project, mostly socially. For one thing, it may seem odd but my advisor does a Lie algebra representations — not knot theory like I do — which is very closely related to the theory of Lie groups. His first graduate student, Jeff Adams, led the charge for $E_8$. Dr. Adams was one of the best professors I had as an undergrad, and I probably owe my category-friendliness to his style in teaching the year-long graduate abstract algebra course I took, as well as his course on the classical groups. That approach of his probably has something to do with his being a student of Zuckerman’s. And around we go. Posted by John Armstrong | Atlas of Lie Groups | 2 Comments Integers I’m back from Ohio at the College Perk again. The place looks a lot different in daylight. Anyhow, since the last few days have been a little short on the exposition, I thought I’d cover integers. Okay, we’ve covered that the natural numbers are a commutative ordered monoid. We can add numbers, but we’re used to subtracting numbers too. The problem is that we can’t subtract with just the natural numbers — they aren’t a group. What could we do with $2-3$? Well, let’s just throw it in. In fact, let’s just throw in a new element for every possible subtraction of natural numbers. And since we can get back any natural number by subtracting zero from it, let’s just throw out all the original numbers and just work with these differences. We’re looking at the set of all pairs $(a,b)$ of natural numbers. Oops, now we’ve overdone it. Clearly some of these differences should be the same. In particular, $(S(a),S(b))$ should be the same as $(a,b)$. If we repeat this relation we can see that $(a+c,b+c)$ should be the same as $(a,b)$ where we’re using the definition of addition of natural numbers from last time. We can clean this up and write all of these in one fell swoop by defining the equivalence relation: $(a,b)\sim(a',b')\Leftrightarrow a+b'=b+a'$. After checking that this is indeed an equivalence relation, we can pass to the set of equivalence classes and call these the integers $\mathbb{Z}$. Now we have to add structure to this set. We define an order on the integers by $(a,b)\leq(c,d)\Leftrightarrow a+d\leq b+c$. The caveat here is that we have to check that if we replace a pair with an equivalent pair we get the same answer. Let’s say $(a,b)\sim(a',b')$, $(c,d)\sim(c',d')$, and $(a,b)\leq(c,d)$. Then $a'+b+c+d'=a+b'+c'+d\leq b+b'+c'+c$ so $a'+d'\leq b'+c'$. The first equality uses the equivalences we assumed and the second uses the inequality. Throughout we’re using the associativity and commutativity. That the first inequality implies the second follows because addition of natural numbers preserves order. We get an addition as well. We define $(a,b)+(c,d)=(a+c,b+d)$. It’s important to note here that the addition on the left is how we’re defining the sum of two pairs, and those on the right are additions of natural numbers we already know how to do. Now if $(a,b)\sim(a',b')$ and $(c,d)\sim(c',d')$ we see $(a+c)+(b'+d')=(a+b')+(c+d')=(b+a')+(d+c')=(a'+c')+(b+d)$ so $(a+c,b+d)\sim(a'+c',b'+d')$. Addition of integers doesn’t depend on which representative pairs we use. It’s easy now to check that this addition is associative and commutative, that $(0,0)$ is an additive identity, that $(b,a)+(a,b)\sim(0,0)$ (giving additive inverses), and that addition preserves the order structure. All this together makes $\mathbb{Z}$ into an ordered abelian group. Now we can relate the integers back to the natural numbers. Since the integers are a group, they’re also a monoid. We can give a monoid homomorphism embedding $\mathbb{N}\rightarrow\mathbb{Z}$. Send the natural number $a$ to the integer represented by $(a,0)$. We call the nonzero integers of this form “positive”, and their inverses of the form $(0,a)$ “negative”. We can verify that $(a,0)\geq(0,0)$ and $(0,a)\leq(0,0)$. Check that every integer has a unique representative pair with ${}0$ on one side or the other, so each is either positive, negative, or zero. From now on we’ll just write $a$ for the integer represented by $(a,0)$ and $-a$ for the one represented by $(0,a)$, as we’re used to. Slides for Bracket Extension talk I’ve posted the slides for my talk. There are a few typos that I noticed as I was speaking, but nothing that makes it incomprehensible. I actually started the lecture on page 10 to save time, and because the run-up is pretty standard material. Posted by John Armstrong | Knot theory | 1 Comment AMS Sectional, Day 2 Still no wireless, so I’ll again jot a little something about the noteworthy talks. Louis Kauffman gave a talk about an invariant of “virtual” knots and links, which are described in his paper that I linked to the other day. The invariant is an extension of the Kauffman bracket that I extend to tangles. An obvious question is how to do both extensions, getting functors on the category of virtual tangles. Heather Dye, Kauffman’s former student, then spoke on virtual homotopy. This may well be related to Allison Henrich’s work on Legendrian virtual knots. It’s all tangled (har) up together. I gave my talk after that. I’ll make a separate post with the link to my slides. After lunch, Alexander Shumakovitch gave a very clear (though not yet complete) combinatorial categorification of HOMFLY evaluations. There’s a parameter $n$ in his theory, and setting it to 2 gives back the combinatorial version of Khovanov homology. Setting it to higher values should correspond to what Josh Sussan — currently finishing his Ph.D. here at Yale — has done in the representation-theoretic picture for $U_q({\mathfrak sl}_n)$. The last talk that really grabbed me was Michael “Cap” Khoury’s explanation of a new definition for the Alexander-Conway polynomial. The really interesting thing here is that it really looks like he’s realizing it as some sort of representable functor on some sort of category. Almost, but not quite. We talked for a bit about what’s missing, and I don’t think it’s impossible to push it a bit and get that last lousy point. Joyce on Geometry Beannachtai na Feile Padraig. In honor of the day, I’d like to post a passage of Finnegans Wake. There’s a beautiful section from pages 293 to 299 “explaining” Euclid I.1. It’s as good a place as any to start in on “the book of Doublends Jined”, so if this sort of thing intrigues you I hope you’ll get a copy and go from here. If nothing else I hope that my own exegeses are somewhat easier to follow. This area of the text is particularly… texty. I’ll do my best to match the original as exactly as possible. Individual pages will be separated by hard rules. The passage itself follows the jump. Read more » New result Never believe anyone who says that drinking never helps anything, for tonight over many a pint o’ Guinness I hit upon the solution to a problem that’s been nagging at me for some time. The answer is so incredibly simple that I’m feeling stupid for not thinking of it before. So here it is: Cospans in the comma category of quandles over a given quandle Q from the free quandle on $m$ letters to the free quandle on $n$ letters categorify the extension of link colorings by Q to tangles. This gives me a new marker to aim at. I can explain this, but it will require more preliminaries before it’s really accessible to the GILA (Generally Interested Lay Audience) as yet. Those who know quandles — a topic I’ll be covering early next week — and category theory and some knot theory should be able to piece together the meaning now. For the rest of you.. stay tuned. Posted by John Armstrong | Knot theory | 1 Comment AMS Sectional, Day 1 I don’t have wireless access here in the lecture hall, so I can’t “live blog”. I’m writing notes on the lectures I find noteworthy. David Radford spoke about something called the Hennings Invariant of a finite-dimensional Hopf algebra. I’ve always liked his style, since he manages to boil down a lot of complicated algebraic structure to what’s essential for the application at hand. He also describes it incredibly clearly. His lectures are very accessible to a grad student who has a basic background in algebra, which is more than I can say about many algebraists. I think it should be clear why I think this is a Good Thing. I’ll get to Hopf algebras in more depth eventually, but for now let me say this: they’re very much like groups, but using somewhat heavier machinery. In the long run, groups and Hopf algebras both work off of a very similar structure. Pat Gilmer gave a talk on “congruence” and “similarity” of 3-manifolds. A 3-manifold is a space that looks close-up like three-dimensional space, like the surface of the Earth looks flat since we’re so close to it. These two concepts he’s pushing are equivalence relations. Two 3-manifolds may be different, but might still be “similar” or “congruent” if they’re related by certain modifications, called “surgeries”. Congruence was evidently studied about ten years ago and Gilmer reinvented it himself along with similarity. He’s particularly interested in how certain well-known invariants of 3-manifolds change as you apply these surgeries. One interesting thing this brings to mind is the fact that we can get any 3-manifold from the 3-sphere (the surface of the Earth is a *2*-sphere) by cutting out a bunch of bagel-shaped regions that might be knotted, twisting up the boundaries of the parts we cut out, and putting them back in. This means that there’s a connection between knot theory and 3-manifold theory. In fact, a very large portion of mathematicians calling themselves knot theorists are really more interested in 3-manifolds and just use knot theory as a tool. After lunch, Carmen Caprau gave her talk about an ${\mathfrak sl}_2$ tangle homology. It manages to fix a big problem I’ve had with Khovanov’s homology theory — it tends to screw up the signs. Knot homology theories are really big business these days. Most people I know who are on the job market and work directly with these sorts of things have jobs nailed down, and Carmen is no exception. Good luck to her. Scott Carter talked about cohomology in symmetric monoidal categories with products and coproducts. This extends the stuff he has done with Alissa Crans, Mohammed Elhamdadi, Pedro Lopes, and Masahico Saito. The last version of this talk I saw at last Spring’s Knots In Washington was some of the nicest theory I’ve seen. Now they’re taking this abstract setup describing Hoschschild homology theories (which try to capture the underlying essence of associativity), and “dualize” all the diagrams to get some sort of topological invariant. I always love mixing up notation and subject matter, and this is very much in that Kauffman-esque spirit. Hopefully there will be an updated version of their paper on the arXiv soon. Maciej Niebrzydowski spoke on homology of dihedral quandles, which he worked on with his advisor, Jozef Przytycki. I’ll leave this alone since I’m almost ready to talk about quandles in full detail. 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/57951?sort=newest
## Lifting of projective representations of a torus over a valuation ring ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $R$ be a valuation ring. We don't assume it to be discrete or have a finite residue field. Let $T$ be a split torus over $R$, so $T\cong {\mathbb G}_m^r$ for some $r$. Left there be given a homomorphism of group schemes $T\to {\rm PGL}_n$, defined over $R$ ,does there exist a lift to a morphism $T\to {\rm GL}_n$ defined over $R$? I expect this to be somewhere in the literature, but I couldn't locate it. - ## 1 Answer The obstruction to lifting is given by a central extension $E$ of $T$ by $\mathbb{G}_m$. Such an extension must be commutative because the commutator induces a bi-homomorphism $T\times T\to\mathbb{G}_m$, and every such map is trivial. So $E$ is a torus, and the extension is dual to an extension of the constant group scheme $\underline{\mathbb{Z}}$ by $\underline{\mathbb{Z}}^r$, hence trivial. The argument works over any base scheme $S$ satisfying $H^1(S,\underline{\mathbb{Z}})=0$. -
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http://te.wikipedia.org/wiki/%E0%B0%A6%E0%B1%83%E0%B0%B7%E0%B1%8D%E0%B0%9F%E0%B0%BF_%E0%B0%95%E0%B1%8B%E0%B0%A3%E0%B0%82
# దృష్టి కోణం వికీపీడియా నుండి ఇక్కడికి గెంతు: పేజీకి సంబంధించిన లింకులు, వెతుకు ఈ వ్యాసాన్ని పూర్తిగా తరువాత ఈ మూసను తీసివేయండి. అనువాదం చేయాల్సిన వ్యాస భాగం ఒకవే‌ళ ప్రధాన పేరుబరిలో వున్నట్లయితే పాఠ్యం సవరించు నొక్కినప్పుడు కనబడవచ్చు. దృష్టి కోణం(Angle of view) A camera's angle of view can be measured horizontally, vertically, or diagonally. In photography, angle of view describes the angular extent of a given scene that is imaged by a camera. It parallels, and may be used interchangeably with, the more general visual term field of view. It is important to distinguish the angle of view from the angle of coverage, which describes the angle of projection by the lens onto the focal plane. For most cameras, it may be assumed that the image circle produced by the lens is large enough to cover the film or sensor completely.[1] If the angle of view exceeds the angle of coverage, however, then vignetting will be present in the resulting photograph. For an example of this, see below. ## కెమెరా దృష్టికోణాన్ని లెక్కకట్టడం In 1916, Northey showed how to calculate the angle of view using ordinary carpenter's tools.[2] The angle that he labels as the angle of view is the half-angle or "the angle that a straight line would take from the extreme outside of the field of view to the center of the lens;" he notes that manufacturers of lenses use twice this angle. For lenses projecting rectilinear (non-spatially-distorted) images of distant objects, the effective focal length and the image format dimensions completely define the angle of view. Calculations for lenses producing non-rectilinear images are much more complex and in the end not very useful in most practical applications. Angle of view may be measured horizontally (from the left to right edge of the frame), vertically (from the top to bottom of the frame), or diagonally (from one corner of the frame to its opposite corner). For a lens projecting a rectilinear image, the angle of view (α) can be calculated from the chosen dimension (d), and effective focal length (f) as follows:[3] $\alpha = 2 \arctan \frac {d} {2 f}$ Because this is a trigonometric function, the angle of view does not vary quite linearly with the reciprocal of the focal length. However, except for wide-angle lenses, it is reasonable to approximate $\alpha\approx \frac{d}{f}$ radians or $\frac{180d}{\pi f}$ degrees. The effective focal length is nearly equal to the stated focal length of the lens (F), except in macro photography where the lens-to-object distance is comparable to the focal length. In this case, the magnification factor (m) must be taken into account: $f = F \cdot ( 1 + m )$ (In photography $m$ is usually defined to be positive, despite the inverted image.) For example, with a magnification ratio of 1:2, we find $f = 1.5 \cdot F$ and thus the angle of view is reduced by 33% compared to focusing on a distant object with the same lens. ### ఉదాహరణ Consider a 35 mm camera with a normal lens having a focal length of F=50 mm. The dimensions of the 35 mm image format are 24 mm (vertically) × 36 mm (horizontal), giving a diagonal of about 43.3 mm. Now the angles of view are: • horizontally, $\alpha_h = 2\arctan\; h/2f =$ 39.6° • vertically, $\alpha_v = 2\arctan\; v/2f =$ 27.0° • diagonally, $\alpha_d = 2\arctan\; d/2f =$ 46.7° ### Derivation of the angle-of-view formula Consider a lens in a camera being used to photograph an object at a distance $S_1$, and forming an image that just barely fits in the dimension $d$ of the frame (the film or image sensor); treat the lens as if it were a pinhole at distance $S_2$ from the image plane: Now $\alpha/2$ is the angle between the optical axis of the lens and the ray joining its optical center to the edge of the film. Here $\alpha$ is defined to be the angle-of-view, since it is the angle enclosing the largest object whose image can fit on the film. We want to find the relationship between: the angle $\alpha$ (half of the angle-of-view) the "opposite" side of the right triangle, $d/2$ (half the film-format dimension) the "adjacent" side, $S_2$ (distance from the lens to the image plane) Using basic trigonometry, we find: $\tan ( \alpha / 2 ) = \frac {d/2} {S_2} .$ which we can solve for α, giving: $\alpha = 2 \arctan \frac {d} {2 S_2}$ To project a sharp image of distant objects, $S_2$ needs to be equal to the focal length $F$, which is attained by setting the lens for infinity focus. Then the angle of view is given by: $\alpha = 2 \arctan \frac {d} {2 f}$ where $f=F$ #### సూక్ష్మ(మాక్రో)ఫోటోగ్రఫి For macro photography, we cannot neglect the difference between $S_2$ and $F$ From the thin lens formula, $\frac{1}{F} = \frac{1}{S_1} + \frac{1}{S_2}$. We substitute for the magnification, $m = S_2/S_1$, and with some algebra find: $S_2 = F\cdot(1+m)$ Defining $f=S_2$ as the "effective focal length", we get the formula presented above: $\alpha = 2 \arctan \frac {d} {2 f}$ where $f=F\cdot(1+m)$. ## కాటకాల రకాలు వాటి ప్రభావం Lenses are often referred to by terms that express their angle of view: • Ultra wide-angle lenses, also known as fisheye lenses, cover up to 180° (or even wider in special cases) • Wide-angle lenses generally cover between 100° and 60° • Normal, or Standard lenses generally cover between 50° and 25° • Telephoto lenses generally cover between 15° and 10° • Super Telephoto lenses generally cover between 8° through less than 1° Zoom lenses are a special case wherein the focal length, and hence angle of view, of the lens can be altered mechanically without removing the lens from the camera. Longer lenses magnify the subject more, apparently compressing distance and (when focused on the foreground) blurring the background because of their shallower depth of field. Wider lenses tend to magnify distance between objects while allowing greater depth of field. Another result of using a wide angle lens is a greater apparent perspective distortion when the camera is not aligned perpendicularly to the subject: parallel lines converge at the same rate as with a normal lens, but converge more due to the wider total field. For example, buildings appear to be falling backwards much more severely when the camera is pointed upward from ground level than they would if photographed with a normal lens at the same distance from the subject, because more of the subject building is visible in the wide-angle shot. Because different lenses generally require a different camera–subject distance to preserve the size of a subject, changing the angle of view can indirectly distort perspective, changing the apparent relative size of the subject and foreground. 28 mm lens, 65.5° × 46.4° 50 mm lens, 39.6° × 27.0° 70 mm lens, 28.9° × 19.5° 210 mm lens, 9.8° × 6.5° ### గుండ్రటి చేపకన్ను దృష్టి కోణం(circular fisheye) A circular fisheye lens (as opposed to a full-frame fisheye) is an example of a lens where the angle of coverage is less than the angle of view. The image projected onto the film is circular because the diameter of the image projected is narrower than that needed to cover the widest portion of the film. ## సాదారణ కటక దృష్టి కోణం This table shows the diagonal, horizontal, and vertical angles of view, in degrees, for lenses producing rectilinear images, when used with 36 mm × 24 mm format (that is, 135 film or full-frame 35mm digital using width 36 mm, height 24 mm, and diagonal 43.3 mm for d in the formula above[4]). | | | | | | | | | | | | | | | | Focal Length (mm) | Diagonal (°) | Vertical (°) | Horizontal (°) | |-------|-------|-------|------|------|------|------|------|------|-------|-------|--------|--------|--------|--------|---------------------|----------------|----------------|------------------| | 13 | 15 | 18 | 21 | 24 | 28 | 35 | 50 | 85 | 105 | 135 | 180 | 210 | 300 | 400 | 500 | 600 | 830 | 1200 | | 118 | 111 | 100 | 91.7 | 84.1 | 75.4 | 63.4 | 46.8 | 28.6 | 23.3 | 18.2 | 13.7 | 11.8 | 8.25 | 6.19 | 4.96 | 4.13 | 2.99 | 2.07 | | 85.4 | 77.3 | 67.4 | 59.5 | 53.1 | 46.4 | 37.8 | 27 | 16.1 | 13 | 10.2 | 7.63 | 6.54 | 4.58 | 3.44 | 2.75 | 2.29 | 1.66 | 1.15 | | 108 | 100.4 | 90 | 81.2 | 73.7 | 65.5 | 54.4 | 39.6 | 23.9 | 19.5 | 15.2 | 11.4 | 9.8 | 6.87 | 5.15 | 4.12 | 3.44 | 2.48 | 1.72 | ## Three-dimensional digital art Displaying 3d graphics requires 3d projection of the models onto a 2d surface, and uses a series of mathematical calculations to render the scene. The angle of view of the scene is thus readily set and changed; some renderers even measure the angle of view as the focal length of an imaginary lens. The angle of view can also be projected onto the surface at an angle greater than 90°, effectively creating a fish eye lens effect. ## సినిమాటోగ్రఫి Modifying the angle of view over time, or zooming, is a frequently used cinematic technique. ### Video games As an effect, some first person games, especially racing games, widen the angle of view beyond 90° to exaggerate the distance the player is travelling, thus exaggerating the player's perceived speed. This effect can be done progressively, or upon the activation of some sort of "turbo boost." An interesting visual effect in itself, it also provides a way for game developers to suggest speeds faster than the game engine or computer hardware is capable of displaying. Some examples include Burnout 3: Takedown and Grand Theft Auto: San Andreas. Players of first-person shooter games sometimes set the angle of view of the game, widening it in an unnatural way (a difference of 20 or 30 degrees from normal), in order to see more peripherally. ## References and notes 1. Neil Wayne Northey (Sept. 1916). "The Angle of View of your Lens". The Camera 20 (9). 2. Ernest McCollough (1893). "Photographic Topography". Industry: A Monthly Magazine Devoted to Science, Engineering and Mechanic Arts: 399–406. 3. However, most interchangeable-lens digital cameras do not use 24x36 mm image sensors and therefore produce narrower angles of view than set out in the table. See crop factor and the subtopic digital camera issues in the article on wide-angle lenses for further discussion. ## See also "http://te.wikipedia.org/w/index.php?title=దృష్టి_కోణం&oldid=810446" నుండి వెలికితీశారు
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http://math.stackexchange.com/questions/263496/properties-of-fourier-transform-of-distributions?answertab=votes
# Properties of Fourier transform of distributions For distributions the scaling property, $f(ax) = \frac{1}{|a|} \mathcal{F(\frac{u}{a})}$, of the Fourier transform is no longer true. Is there a source that lists which properties of the Fourier transform remain true even for distributions and which are false? - 3 Are you sure about the failure of the scaling property? – Giuseppe Negro Dec 22 '12 at 1:30 I agree with Guiseppe. The scaling property is commonly exploited with Dirac delta functions. – Ron Gordon Dec 22 '12 at 4:23
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http://physics.stackexchange.com/questions/54549/how-deep-is-the-well-given-this-data?answertab=oldest
# How deep is the well, given this data? A rock is dropped into a well. A timer starts when the rock is dropped and is stopped when the noise of the rock hitting the ground of the well is heard. How deep is the well? Here's what I have so far: $t = \left(\left(\frac{2\times h}{g}\right)^{\frac{1}{2}}+\left(\frac{h}{340}\right)\right)$ when I try to use the quadratic formula, I can express $h$, but when substitute different amounts for $t$, it doesn't seem right. I get a very small and a very large value for $h$. For example for $t=30\:\mathrm{s}$ I get $3\:\mathrm{m}$ and $42\:\mathrm{km}$, which is quite odd. I also tried it with the following set of equations: $$\frac{1}{2}\left(g\times t_1^2\right)=h$$ $$t_2\times 340 \:\mathrm{m/s}=h$$ $$t_1+t_2=t$$ - 1 – Qmechanic♦ Feb 20 at 19:57 1 – Emilio Pisanty Feb 20 at 21:38 ## 1 Answer Let $v$ denote the speed of sound, and let's use your equations. Combine the second and third equations to get $$\frac{h}{v} = t-t_1$$ so that $$t_1 = t-\frac{h}{v}$$ and therefore using the first equation $$h = \frac{1}{2}g\left(t-\frac{h}{v}\right)^2$$ Putting this in standard form to use the quadratic equation gives $$h^2 - 2v\left(t+\frac{v}{g}\right)h + (vt)^2 = 0$$ which gives $$h = \frac{gtv+v^2\pm v\sqrt{v(2gt+v)}}{g}$$ for $g=9.8\,\mathrm{m}/\mathrm{s}^2$, $t=30\,\mathrm{s}$, adn $v=340\,\mathrm{m}/\mathrm s$, I get $2.5\,\mathrm {km}$ and $41.5\,\mathrm{km}$. The first solution is the correct one, for a 30 second fall, I think it seems quite reasonable; if you were freely falling for that amount of time without air resistance, you would travel about 4.4 kilometers. - Thanks, but the timer is stopped, when I hear the sound coming back from the well, how do you account for that? if you look carefully, at the end of my question, i also have this function. I need to somehow account for the time that the sound of the rock took to come back to me. – Andris Feb 20 at 21:25 Oh sorry misread the question; solution edited. – joshphysics Feb 20 at 21:53 thanks, upvote, but can you unroll the square? because I did 7 (!) different ways of these starting equations and im always left with something different, could you take a look at Qmechanics solutions in the other question? how does that works out for him? – Andris Feb 20 at 22:37 Ok I included another step. – joshphysics Feb 20 at 22:52 thank you, i had a sign error, yours work, accepted – Andris Feb 21 at 0:18
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http://math.stackexchange.com/questions/842/why-is-compactness-in-logic-called-compactness
Why is compactness in logic called compactness? In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set. Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn't seem strong enough to justify the terminology. Is there more to the choice of the terminology in logic than this analogy? - – Qiaochu Yuan Jul 27 '10 at 23:26 6 Answers The Compactness Theorem is equivalent to the compactness of the Stone space of the Lindenbaum–Tarski algebra of the first-order language L. (This is also the space of 0-types over the empty theory.) A point in the Stone space SL is a complete theory T in the language L. That is, T is a set of sentences of L which is closed under logical deduction and contains exactly one of σ or ¬σ for every sentence σ of the language. The topology on the set of types has for basis the open sets U(σ) = {T: σ ∈ T} for every sentence σ of L. Note that these are all clopen sets since U(¬σ) is complementary to U(σ). To see how the Compactness Theorem implies the compactness of SL, suppose the basic open sets U(σi), i ∈ I, form a cover of SL. This means that every complete theory T contains at least one of the sentences σi. I claim that this cover has a finite subcover. If not, then the set {¬σi: i ∈ I} is finitely consistent. By the Compactness Theorem, the set consistent and hence (by Zorn's Lemma) is contained in a maximally consistent set T. This theory T is a point of the Stone space which is not contained in any U(σi), which contradicts our hypothesis that the U(σi), i ∈ I, form a cover of the space. To see how the compactness of SL implies the Compactness Theorem, suppose that {σi: i ∈ I} is an inconsistent set of sentences in L. Then U(¬σi), i ∈ I, forms a cover of SL. This cover has a finite subcover, which corresponds to a finite inconsistent subset of {σi: i ∈ I}. Therefore, every inconsistent set has a finite inconsistent subset, which is the contrapositive of the Compactness Theorem. - Did you find this question interesting? Try our newsletter email address The analogy for the compactness theorem for propositional calculus is as follows. Let $p_i$ be propositional variables; together, they take values in the product space $2^{\mathbb{N}}$. Suppose we have a collection of statements $S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can prove that they are all simultaneously satisfiable by using a compactness argument. Let $F$ be a finite set. Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) which satisfy $S_t$ for $t \in F$ is a closed set $V_F$ of assignments satisfying the sentences in $F$. The intersection of any finitely many of the $V_F$ is nonempty, so by the finite intersection property, the intersection of all of them is nonempty (since the product space is compact), whence any truth in this intersection satisfies all the statements. I don't know how this works in predicate logic. - A motto which is related (and sometimes true) is "proofs are finite". In most systems of logic under consideration, the statement and a proof of the statement are finite strings of symbols. One can think of them as "compactly" represented. How nice then that certain systems will always allow a proof to be found if there is one. While this does not give a tight connection to topology, it suggests (and you need to work this through on your own to be convinced) that certain infinite abnormalities (like an infinite proof) will not occur. A similar topological claim is that, for compact sets, an infinite inclusive chain of certain subsets will have nonempty intersection, such a claim easily seen to be false for some non-compact sets, such as the collection { C_n | C_n = {x | x >= n and x is a real number } for n a non-negative integer } . - As far as I know, the link comes from the syntactic theory. You are given a set of symbols of sentence F = {f_i} and you are allowed allowed to form complex statements using them. You can combine the elementary statements with AND, OR, NOT operators and parentheses, in the usual way. So you get a set X of composed sentences, like (f AND g) OR (NOT h) or something like that. A syntactic version of the compactness theorem states the following. Assume that for every finite subset Y of X you can assign truth values to the f_i in such a way that all sentences in Y are true. Than you can do the same for X. Proof Consider the topological space A obtained by taking the product of {0, 1} over the set F. The topology on A is the product topology. By Tychonoff's theorem, A is compact. For every composed statement s, the set of truth values which make s true is a finite intersection of cylinders, hence it is a closed set of A. The hypothesis say that every finite intersection of such closed sets, for s ranging in X, is not empty. Hence the intersection of all such closed set is not empty, which means one can make all statements in X true. - Why the downvote?!? This is the same answer as Akhil, even if in a different language. I used as elementary language as possible since I did not know whether the reader knew about propositional calculus. I would like very much if the downvoter could explain what is wrong in this answer. – Andrea Ferretti Aug 20 '10 at 10:14 Lemma: A topological space $X$ is compact if and only if for every collection $\mathcal{C}$ of closed sets with the finite intersection property has nonempty intersection over the collection. Proposition: $\mathbb{M}(\mathcal{L})$ is compact if and only if every finitely satisfiable $\mathcal{L}$-theory is satisfiable. Proof: Consider the space $\mathbb{M}(\mathcal{L})=\{\Phi \; | \; \Phi \; \text{is a maximal} \; \mathcal{L} \text{-theory}\}$. For each $\mathcal{L}$-sentence $\varphi$ let $[(\varphi)]=\{ \Phi \in \mathbb{M}(\mathcal{L}) \; | \; \varphi \in \Phi\}$. A subbase $\cal{B}$ for a topology on $\mathbb{M}(\mathcal{L})$ is given by the sets $[(\varphi)]$. That is, the open subsets of $\mathbb{M}(\mathcal{L})$ are the union of the finite intersections of elements of $\mathcal{B}$. To see that $\mathcal{B}$ defines a topology on note that $[(\forall x(x\neq x))]=\emptyset$ and $[(\forall x(x=x))]=\mathbb{M}(\mathcal{L})$. Furthermore, arbitrary unions are already defined to be in the topology and finite intersections are in the topology since $[(\varphi)]\cap [(\theta)]=[(\varphi \wedge \theta)]$, which is defined to be open. Assume logical compactness. Note that $\bigcap_{i\in I} [(\varphi_i)]=\emptyset$ if and only if it is not satisfiable. Let $\mathcal{C}$ be a subcollection of the subbasis of the topology of $\mathbb{M}(\mathcal{L})$ with the finite intersection property. Then every finite subset of $\mathcal{C}$ is satisfiable and by the compactness theorem this implies $\bigcap [(\varphi)]$ which ranges over all the elements of $\cal{C}$ is satisfiable, hence $$\bigcap_{[(\varphi)] \in \cal{B}} [(\varphi)]\neq \emptyset.$$ Thus $\mathbb{M}(\mathcal{L})$ is compact. Assume $\mathbb{M}(\mathcal{L})$ is compact. Let $\Phi$ be an $\mathcal{L}$-theory in which is finitely satisfiable. Let $\mathcal{C}_\Phi=\{[(\varphi)] \; | \; \varphi \in \Phi\}$. Every element of $\mathcal{C}_\Phi$ is closed and every finite subset of $\mathcal{C}_\Phi$ has nonempty intersection since $\Phi$ is finitely satisfiable. Since $\mathbb{M}(\mathcal{L})$ is compact, $\bigcap_{\varphi \in \Phi} [(\varphi)]\neq\emptyset$, hence it is satisfiable. $\square$ - This isn't a complete answer, partly because I've had discussions with other grad students and we weren't able to work it out satisfactorily, but I've been told that you can actually put a topology on logical statements such that the compactness theorem translates to "The set of true statements is a compact subset of the set of all statements" or something similar. But as I said, a few of my friends and I weren't able to work out the topology. -
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http://mathhelpforum.com/differential-equations/126512-determine-whether-function-solution-de.html
# Thread: 1. ## Determine whether the function is a solution of the DE? Determine whether the function is a solution of the differential equation y''' - 8y = 0 (a) y = cos(2x) (b) y = e^(2x) What exactly is one suppose to do here? Is this suppose to be the third degree because we only covered first degree DE? 2. Find $y'''$ for each possibility and sub them back into the equation to check. Sound good? 3. Well, you could differentiate the given functions and see if they satisfy the equation, i.e., you want to know if $y''' = 8y$. 4. $cos(2x)-8y=0$ I end up with y = 1/8 cos(2 x) + C What should i do with this solution? Originally Posted by pickslides Find $y'''$ for each possibility and sub them back into the equation to check. Sound good? 5. Your derivative is not correct. $y'= -2\cdot \sin(2x)$. 6. So i am suppose to replace a&b with Y''' or with 8(y) and what does the 3 ''' mean? Originally Posted by lvleph Your derivative is not correct. $y'= -2\cdot \sin(2x)$. 7. $y'''$ means the third derivative. You are suppose to use the solutions in a) and b) and plug them in the original DEQ. If $y$ from a) or b) satisfies $y''' + 8y =0$ then it is a solution to the DEQ. From a) $y''' = 8\cos(2x)$ and so $8\cos(2x) - 8cos(2x) = 0$ So, a) is a solution to the DEQ. Now, check if b) works.
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http://mathhelpforum.com/discrete-math/50779-disproving-prime-numbers.html
# Thread: 1. ## disproving prime numbers how do you prove that number is not Prime? the example in my book says prove that 2 to the power of 66 is not prime and then says the keys lies in this simple algebra $x squared-1=(x-1)(x+1)$ it says to use that to write a proof that the number is not prime so what am supposed to do with that equation? 2. Hello, Originally Posted by andrewuk how do you prove that number is not Prime? the example in my book says prove that 2 to the power of 66 is not prime and then says the keys lies in this simple algebra $x squared-1=(x-1)(x+1)$ it says to use that to write a proof that the number is not prime so what am supposed to do with that equation? huh 2 divides $2^{66}$, therefore it is not prime Another way : $2^{66}-1=(2^{33}-1)(2^{33}+1)$ $\implies 2^{66}=(2^{33}-1)(2^{33}+1)+1$ But since $2^{33}$ is an even number, $2^{33}-1 \text{ and } 2^{33}+1$ are odd numbers. Thus their product is an odd number. If we substract 1 to it, we get an even number. Hence $2^{66}$ is an even number (and different from 2), so it is not prime. But that's finding the most complicated way... 3. Ok thanks for that, don't know why the book couldn't of included an example first to show how the equation is used. How would i split it if the power is an odd number? e.g. 2 to the power of 67 4. Originally Posted by Moo Hello, huh 2 divides $2^{66}$, therefore it is not prime Another way : $2^{66}-1=(2^{33}-1)(2^{33}+1)$ $\implies 2^{66}=(2^{33}-1)(2^{33}+1)+1$ But since $2^{33}$ is an even number, $2^{33}-1 \text{ and } 2^{33}+1$ are odd numbers. Thus their product is an odd number. If we substract 1 to it, we get an even number. Hence $2^{66}$ is an even number (and different from 2), so it is not prime. But that's finding the most complicated way... This is ridiculus, the very definition of $2^{66}$ makes it composite and hence not prime as it has at least three distinct divisors. So any proof that $2^{66}$ is not prime that uses any of the laws of powers is using this, and so is just a complicated version of the simple proof. RonL 5. Ye the example does seem stupid but i still need to know how to use that equation when the power is an odd number, can you have a decimal power? 6. Its considered rude to post the same question on two forums. Prime numbers - The Student Room Bobak 7. Given any number A with prime factorization $x_0^{n_0} \cdot x_1^{n_1} \cdot \dots \cdot x_k^{n_k}$ is composite with $(n_0 + 1) \cdot (n_1 + 1) \cdot \dots \cdot (n_k + 1)$ factors. So, this works for any prime factorization, whether it is to an odd power or not. I do not quite understand what you mean by decimal power. Do you mean can you have $2^{0.3}$? If so, that number does exist, however you will not find factors for it. 8. Hello, Here is my opinion: someone made an error when copying. The real question in the book should be "Prove that $2^{66}-1$ is a prime." Bye. 9. Originally Posted by wisterville Hello, Here is my opinion: someone made an error when copying. The real question in the book should be "Prove that $2^{66}-1$ is a prime." Bye. I'm sorry now I'm lost. $2^{66}-1$ is not prime as: $2^{66}-1=(2^{33}+1)(2^{33}-1)$ RonL 10. Hello, Sorry, I wanted to say: The real question in the book should be "Prove that is NOT a prime." Bye.
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http://math.stackexchange.com/questions/101969/calculate-the-maximum-number-of-linearly-independent-subsets
# calculate the maximum number of linearly independent subsets I am not sure if this question has been asked before, apologies if I am repeating this question again. My question is as follows: given a set i know how to find all the subsets of the set. But what I would really like to know is how do i find all linearly independent subsets that can be obtained from this super set. Thanks, Bhavya - 2 What do you mean by linear independence? A set in itself has no linear structure, and the phrase "linear independence" is meaningless. – Willie Wong♦ Jan 24 '12 at 13:35 1 @bhavya: If your set $S$ is a finite subset of a say $3$-dimensional vector space, then you need only worry about subsets of $s$ that have $3$ or fewer elements, for the other subsets are not linearly independent. Then you can test all such "small" subsets of $S$ for linear independence. Testing a single subset of $S$, say with $3$ elements, is easy, you probably have had to do it. I do not know any efficient way of testing a large number of subsets of $S$, except for the tedious process of testing one subset, then another, and so on. – André Nicolas Jan 24 '12 at 13:47 1 @bhavya: As for calculating the maximum, it could easily be that all the subsets of $S$ with $3$ or fewer elements are linearly independent. So if $S$ has $n$ elements drawn from a $3$-dimensional vector space, the number of linearly independent subsets of $S$ could be as large as $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}$. – André Nicolas Jan 24 '12 at 13:53 @Andre thank you for the explanation it makes a lot of sense – bhavya Jan 24 '12 at 15:07
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http://en.wikipedia.org/wiki/Faraday's_law_of_induction
# Electromagnetic induction (Redirected from Faraday's law of induction) For the relationship between a time-varying magnetic field and an induced electric field, see Maxwell's equations. Electromagnetism • Electromagnetic induction Scientists Electromagnetic induction is the production of a potential difference (voltage) across a conductor when it is exposed to a varying magnetic field. Michael Faraday is generally credited with the discovery of induction in 1831 though it may have been anticipated by the work of Francesco Zantedeschi in 1829.[1] Around 1830[2] to 1832,[3] Joseph Henry made a similar discovery, but did not publish his findings until later. Faraday's law of induction is a basic law of electromagnetism that predicts how a magnetic field will interact with an electric circuit to produce an electromotive force (EMF). It is the fundamental operating principle of transformers, inductors, and many types of electrical motors, generators and solenoids.[4][5] The Maxwell–Faraday equation is a generalisation of Faraday's law, and forms one of Maxwell's equations. ## History A diagram of Faraday's iron ring apparatus. Change in the magnetic flux of the left coil induces a current in the right coil.[6] Faraday's disk (see homopolar generator) Electromagnetic induction was discovered independently by Michael Faraday and Joseph Henry in 1831; however, Faraday was the first to publish the results of his experiments.[7][8] In Faraday's first experimental demonstration of electromagnetic induction (August 29, 1831[9]), he wrapped two wires around opposite sides of an iron ring or "torus" (an arrangement similar to a modern toroidal transformer). Based on his assessment of recently discovered properties of electromagnets, he expected that when current started to flow in one wire, a sort of wave would travel through the ring and cause some electrical effect on the opposite side. He plugged one wire into a galvanometer, and watched it as he connected the other wire to a battery. Indeed, he saw a transient current (which he called a "wave of electricity") when he connected the wire to the battery, and another when he disconnected it.[10] This induction was due to the change in magnetic flux that occurred when the battery was connected and disconnected.[6] Within two months, Faraday had found several other manifestations of electromagnetic induction. For example, he saw transient currents when he quickly slid a bar magnet in and out of a coil of wires, and he generated a steady (DC) current by rotating a copper disk near the bar magnet with a sliding electrical lead ("Faraday's disk").[11] Faraday explained electromagnetic induction using a concept he called lines of force. However, scientists at the time widely rejected his theoretical ideas, mainly because they were not formulated mathematically.[12] An exception was Maxwell, who used Faraday's ideas as the basis of his quantitative electromagnetic theory.[12][13][14] In Maxwell's papers, the time varying aspect of electromagnetic induction is expressed as a differential equation which Oliver Heaviside referred to as Faraday's law even though it is slightly different in form from the original version of Faraday's law, and does not describe motional EMF. Heaviside's version (see Maxwell–Faraday equation below) is the form recognized today in the group of equations known as Maxwell's equations. Lenz's law, formulated by Heinrich Lenz in 1834, describes "flux through the circuit", and gives the direction of the induced EMF and current resulting from electromagnetic induction (elaborated upon in the examples below). Faraday's experiment showing induction between coils of wire: The liquid battery (right) provides a current which flows through the small coil (A), creating a magnetic field. When the coils are stationary, no current is induced. But when the small coil is moved in or out of the large coil (B), the magnetic flux through the large coil changes, inducing a current which is detected by the galvanometer (G).[15] ## Faraday's Law ### Qualitative statement The most widespread version of Faraday's law states: The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. This version of Faraday's law strictly holds only when the closed circuit is a loop of infinitely thin wire,[16] and is invalid in other circumstances as discussed below. A different version, the Maxwell–Faraday equation (discussed below), is valid in all circumstances. ### Quantitative The wire loop (red) forms the boundary of a surface Σ (blue). The black arrows denote any vector field F(r, t) defined throughout space; in the case of Faraday's law, the relevant vector field is the magnetic flux density B, and it is integrated over the blue surface. The red arrow represents the fact that the wire loop may be moving and/or deforming. The definition of surface integral relies on splitting the surface Σ into small surface elements. Each element is associated with a vector dA of magnitude equal to the area of the element and with direction normal to the element and pointing “outward” (with respect to the orientation of the surface). Faraday's law of induction makes use of the magnetic flux ΦB through a hypothetical surface Σ whose boundary is a wire loop. Since the wire loop may be moving, we write Σ(t) for the surface. The magnetic flux is defined by a surface integral: $\Phi_B = \iint\limits_{\Sigma(t)} \mathbf{B}(\mathbf{r}, t) \cdot d \mathbf{A}\ ,$ where dA is an element of surface area of the moving surface Σ(t), B is the magnetic field, and B·dA is a vector dot product (the infinitesimal amount of magnetic flux). In more visual terms, the magnetic flux through the wire loop is proportional to the number of magnetic flux lines that pass through the loop. When the flux changes—because B changes, or because the wire loop is moved or deformed, or both—Faraday's law of induction says that the wire loop acquires an EMF $\mathcal{E}$, defined as the energy available per unit charge that travels once around the wire loop (the unit of EMF is the volt).[16][17][18][19] Equivalently, it is the voltage that would be measured by cutting the wire to create an open circuit, and attaching a voltmeter to the leads. According to the Lorentz force law (in SI units), $\mathbf{F} = q \left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right)$ the EMF on a wire loop is: $\mathcal{E} = \frac{1}{q} \oint_{\mathrm{wire}}\mathbf{F}\cdot d\boldsymbol{\ell} = \oint_{\mathrm{wire}} \left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right)\cdot d\boldsymbol{\ell}$ where E is the electric field, B is the magnetic field (aka magnetic flux density, magnetic induction), dℓ is an infinitesimal arc length along the wire, and the line integral is evaluated along the wire (along the curve the conincident with the shape of the wire). The EMF is also given by the rate of change of the magnetic flux: $\mathcal{E} = -{{d\Phi_B} \over dt} \ ,$ where $\mathcal{E}$ is the electromotive force (EMF) in volts and ΦB is the magnetic flux in webers. The direction of the electromotive force is given by Lenz's law. For a tightly wound coil of wire, composed of N identical loops, each with the same ΦB, Faraday's law of induction states that[20][21] $\mathcal{E} = -N {{d\Phi_B} \over dt}$ where N is the number of turns of wire and ΦB is the magnetic flux in webers through a single loop. ## Maxwell–Faraday equation An illustration of Kelvin-Stokes theorem with surface Σ its boundary ∂Σ and orientation n set by the right-hand rule. The Maxwell–Faraday equation is a generalisation of Faraday's law that states that a time-varying magnetic field is always accompanied by a spatially-varying, non-conservative electric field, and vice-versa. The Maxwell–Faraday equation is $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$ (in SI units) where $\nabla\times$ is the curl operator and again E(r, t) is the electric field and B(r, t) is the magnetic field. These fields can generally be functions of position r and time t. The Maxwell–Faraday equation is one of the four Maxwell's equations, and therefore plays a fundamental role in the theory of classical electromagnetism. It can also be written in an integral form by the Kelvin-Stokes theorem:[22] $\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$ where, as indicated in the figure: Σ is a surface bounded by the closed contour ∂Σ, E is the electric field, B is the magnetic field. dℓ is an infinitesimal vector element of the contour ∂Σ, dA is an infinitesimal vector element of surface Σ. If its direction is orthogonal to that surface patch, the magnitude is the area of an infinitesimal patch of surface. Both dℓ and dA have a sign ambiguity; to get the correct sign, the right-hand rule is used, as explained in the article Kelvin-Stokes theorem. For a planar surface Σ, a positive path element dℓ of curve ∂Σ is defined by the right-hand rule as one that points with the fingers of the right hand when the thumb points in the direction of the normal n to the surface Σ. The integral around ∂Σ is called a path integral or line integral. Notice that a nonzero path integral for E is different from the behavior of the electric field generated by charges. A charge-generated E-field can be expressed as the gradient of a scalar field that is a solution to Poisson's equation, and has a zero path integral. See gradient theorem. The integral equation is true for any path ∂Σ through space, and any surface Σ for which that path is a boundary. If the path Σ is not changing in time, the equation can be rewritten: $\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \frac{d}{dt} \int_{\Sigma} \mathbf{B} \cdot d\mathbf{A}.$ The surface integral at the right-hand side is the explicit expression for the magnetic flux ΦB through Σ. ## Proof of Faraday's law The four Maxwell's equations (including the Maxwell–Faraday equation), along with the Lorentz force law, are a sufficient foundation to derive everything in classical electromagnetism.[16][17] Therefore it is possible to "prove" Faraday's law starting with these equations.[23][24] Click "show" in the box below for an outline of this proof. (In an alternative approach, not shown here but equally valid, Faraday's law could be taken as the starting point and used to "prove" the Maxwell–Faraday equation and/or other laws.) Outline of proof of Faraday's law from Maxwell's equations and the Lorentz force law. Consider the time-derivative of flux through a possibly moving loop, with area $\Sigma(t)$: $\frac{d\Phi_B}{dt} = \frac{d}{dt}\int_{\Sigma(t)} \mathbf{B}(t)\cdot d\mathbf{A}$ The integral can change over time for two reasons: The integrand can change, or the integration region can change. These add linearly, therefore: $\left. \frac{d\Phi_B}{dt}\right|_{t=t_0} = \left( \int_{\Sigma(t_0)} \left. \frac{\partial\mathbf{B}}{\partial t}\right|_{t=t_0} \cdot d\mathbf{A}\right) + \left( \frac{d}{dt} \int_{\Sigma(t)} \mathbf{B}(t_0) \cdot d\mathbf{A} \right)$ where t0 is any given fixed time. We will show that the first term on the right-hand side corresponds to transformer EMF, the second to motional EMF (see above). The first term on the right-hand side can be rewritten using the integral form of the Maxwell–Faraday equation: $\int_{\Sigma(t_0)} \left. \frac{\partial \mathbf{B}}{\partial t}\right|_{t=t_0} \cdot d\mathbf{A} = - \oint_{\partial \Sigma(t_0)} \mathbf{E}(t_0) \cdot d\boldsymbol{\ell}$ Area swept out by vector element dℓ of curve ∂Σ in time dt when moving with velocity v. Next, we analyze the second term on the right-hand side: $\frac{d}{dt} \int_{\Sigma(t)} \mathbf{B}(t_0) \cdot d\mathbf{A}$ This is the most difficult part of the proof; more details and alternate approaches can be found in references.[23][24][25] As the loop moves and/or deforms, it sweeps out a surface (see figure on right). The magnetic flux through this swept-out surface corresponds to the magnetic flux that is either entering or exiting the loop, and therefore this is the magnetic flux that contributes to the time-derivative. (This step implicitly uses Gauss's law for magnetism: Since the flux lines have no beginning or end, they can only get into the loop by getting cut through by the wire.) As a small part of the loop $d\boldsymbol{\ell}$ moves with velocity v for a short time $dt$, it sweeps out a vector area vector $d\mathbf{A}=\mathbf{v} \, dt \times d\boldsymbol{\ell}$. Therefore, the change in magnetic flux through the loop here is $\mathbf{B} \cdot (\mathbf{v} \, dt \times d\boldsymbol{\ell}) = -dt \, d\boldsymbol{\ell} \cdot (\mathbf{v}\times\mathbf{B})$ Therefore: $\frac{d}{dt} \int_{\Sigma(t)} \mathbf{B}(t_0) \cdot d\mathbf{A} = -\oint_{\partial \Sigma(t_0)} (\mathbf{v}(t_0)\times \mathbf{B}(t_0))\cdot d\boldsymbol{\ell}$ where v is the velocity of a point on the loop $\partial \Sigma$. Putting these together, $\left. \frac{d\Phi_B}{dt}\right|_{t=t_0} = \left(- \oint_{\partial \Sigma(t_0)} \mathbf{E}(t_0) \cdot d\boldsymbol{\ell}\right) + \left(- \oint_{\partial \Sigma(t_0)} (\mathbf{v}(t_0)\times \mathbf{B}(t_0))\cdot d\boldsymbol{\ell} \right)$ Meanwhile, EMF is defined as the energy available per unit charge that travels once around the wire loop. Therefore, by the Lorentz force law, $EMF = \oint \left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right) \cdot \text{d}\boldsymbol{\ell}$ Combining these, $\frac{d\Phi_B}{dt} = -EMF$ ## "Counterexamples" to Faraday's law • Faraday's disc electric generator. The disc rotates with angular rate ω, sweeping the conducting radius circularly in the static magnetic field B. The magnetic Lorentz force v × B drives the current along the conducting radius to the conducting rim, and from there the circuit completes through the lower brush and the axle supporting the disc. Thus, current is generated from mechanical motion. • A counterexample to Faraday's Law when over-broadly interpreted. A wire (solid red lines) connects to two touching metal plates (silver) to form a circuit. The whole system sits in a uniform magnetic field, normal to the page. If the word "circuit" is interpreted as "primary path of current flow" (marked in red), then the magnetic flux through the "circuit" changes dramatically as the plates are rotated, yet the EMF is almost zero, which contradicts Faraday's Law. After Feynman Lectures on Physics Vol. II page 17-3 Although Faraday's law is always true for loops of thin wire, it can give the wrong result if naively extrapolated to other contexts.[16] One example is the homopolar generator (above left): A spinning circular metal disc in a homogeneous magnetic field generates a DC (constant in time) EMF. In Faraday's law, EMF is the time-derivative of flux, so a DC EMF is only possible if the magnetic flux is getting uniformly larger and larger perpetually. But in the generator, the magnetic field is constant and the disc stays in the same position, so no magnetic fluxes are growing larger and larger. So this example cannot be analyzed directly with Faraday's law. Another example, due to Feynman,[16] has a dramatic change in flux through a circuit, even though the EMF is arbitrarily small. See figure and caption above right. In both these examples, the changes in the current path are different from the motion of the material making up the circuit. The electrons in a material tend to follow the motion of the atoms that make up the material, due to scattering in the bulk and work function confinement at the edges. Therefore, motional EMF is generated when a material's atoms are moving through a magnetic field, dragging the electrons with them, thus subjecting the electrons to the Lorentz force. In the homopolar generator, the material's atoms are moving, even though the overall geometry of the circuit is staying the same. In the second example, the material's atoms are almost stationary, even though the overall geometry of the circuit is changing dramatically. On the other hand, Faraday's law always holds for thin wires, because there the geometry of the circuit always changes in a direct relationship to the motion of the material's atoms. Although Faraday's law does not apply to all situations, the Maxwell–Faraday equation and Lorentz force law are always correct and can always be used directly.[16] Both of the above examples can be correctly worked by choosing the appropriate path of integration for Faraday's Law. Outside of context of thin wires, the path must never be chosen to go through the conductor in the shortest direct path. This is explained in detail in "The Electromagnetodynamics of Fluid" by W. F. Hughes and F. J. Young, John Wiley Inc. (1965) ## Applications The principles of electromagnetic induction are applied in many devices and systems, including: ### Electrical generator Rectangular wire loop rotating at angular velocity ω in radially outward pointing magnetic field B of fixed magnitude. The circuit is completed by brushes making sliding contact with top and bottom discs, which have conducting rims. This is a simplified version of the drum generator Main article: electrical generator The EMF generated by Faraday's law of induction due to relative movement of a circuit and a magnetic field is the phenomenon underlying electrical generators. When a permanent magnet is moved relative to a conductor, or vice versa, an electromotive force is created. If the wire is connected through an electrical load, current will flow, and thus electrical energy is generated, converting the mechanical energy of motion to electrical energy. For example, the drum generator is based upon the figure to the right. A different implementation of this idea is the Faraday's disc, shown in simplified form on the right. In the Faraday's disc example, the disc is rotated in a uniform magnetic field perpendicular to the disc, causing a current to flow in the radial arm due to the Lorentz force. It is interesting to understand how it arises that mechanical work is necessary to drive this current. When the generated current flows through the conducting rim, a magnetic field is generated by this current through Ampère's circuital law (labeled "induced B" in the figure). The rim thus becomes an electromagnet that resists rotation of the disc (an example of Lenz's law). On the far side of the figure, the return current flows from the rotating arm through the far side of the rim to the bottom brush. The B-field induced by this return current opposes the applied B-field, tending to decrease the flux through that side of the circuit, opposing the increase in flux due to rotation. On the near side of the figure, the return current flows from the rotating arm through the near side of the rim to the bottom brush. The induced B-field increases the flux on this side of the circuit, opposing the decrease in flux due to rotation. Thus, both sides of the circuit generate an emf opposing the rotation. The energy required to keep the disc moving, despite this reactive force, is exactly equal to the electrical energy generated (plus energy wasted due to friction, Joule heating, and other inefficiencies). This behavior is common to all generators converting mechanical energy to electrical energy. ### Electrical transformer Main article: transformer The EMF predicted by Faraday's law is also responsible for electrical transformers. When the electric current in a loop of wire changes, the changing current creates a changing magnetic field. A second wire in reach of this magnetic field will experience this change in magnetic field as a change in its coupled magnetic flux, d ΦB / d t. Therefore, an electromotive force is set up in the second loop called the induced EMF or transformer EMF. If the two ends of this loop are connected through an electrical load, current will flow. ### Magnetic flow meter Main article: magnetic flow meter Faraday's law is used for measuring the flow of electrically conductive liquids and slurries. Such instruments are called magnetic flow meters. The induced voltage ℇ generated in the magnetic field B due to a conductive liquid moving at velocity v is thus given by: $\mathcal{E}= - B \ell v,$ where ℓ is the distance between electrodes in the magnetic flow meter. ## Eddy currents Main article: Eddy current Conductors (of finite dimensions) moving through a uniform magnetic field, or stationary within a changing magnetic field, will have currents induced within them. These induced eddy currents can be undesirable, since they dissipate energy in the resistance of the conductor. There are a number of methods employed to control these undesirable inductive effects. • Electromagnets in electric motors, generators, and transformers do not use solid metal, but instead use thin sheets of metal plate, called laminations. These thin plates reduce the parasitic eddy currents, as described below. • Inductive coils in electronics typically use magnetic cores to minimize parasitic current flow. They are a mixture of metal powder plus a resin binder that can hold any shape. The binder prevents parasitic current flow through the powdered metal. ### Electromagnet laminations Eddy currents occur when a solid metallic mass is rotated in a magnetic field, because the outer portion of the metal cuts more lines of force than the inner portion, hence the induced electromotive force not being uniform, tends to set up currents between the points of greatest and least potential. Eddy currents consume a considerable amount of energy and often cause a harmful rise in temperature.[26] Only five laminations or plates are shown in this example, so as to show the subdivision of the eddy currents. In practical use, the number of laminations or punchings ranges from 40 to 66 per inch, and brings the eddy current loss down to about one percent. While the plates can be separated by insulation, the voltage is so low that the natural rust/oxide coating of the plates is enough to prevent current flow across the laminations.[26] This is a rotor approximately 20mm in diameter from a DC motor used in a CD player. Note the laminations of the electromagnet pole pieces, used to limit parasitic inductive losses. ### Parasitic induction within inductors In this illustration, a solid copper bar inductor on a rotating armature is just passing under the tip of the pole piece N of the field magnet. Note the uneven distribution of the lines of force across the bar inductor. The magnetic field is more concentrated and thus stronger on the left edge of the copper bar (a,b) while the field is weaker on the right edge (c,d). Since the two edges of the bar move with the same velocity, this difference in field strength across the bar creates whorls or current eddies within the copper bar.[27] High current power-frequency devices such as electric motors, generators and transformers use multiple small conductors in parallel to break up the eddy flows that can form within large solid conductors. The same principle is applied to transformers used at higher than power frequency, for example, those used in switch-mode power supplies and the intermediate frequency coupling transformers of radio receivers. ## Faraday's law and relativity ### Two phenomena Some physicists have remarked that Faraday's law is a single equation describing two different phenomena: the motional EMF generated by a magnetic force on a moving wire (see Lorentz force), and the transformer EMF generated by an electric force due to a changing magnetic field (due to the Maxwell–Faraday equation). James Clerk Maxwell drew attention to this fact in his 1861 paper On Physical Lines of Force. In the latter half of part II of that paper, Maxwell gives a separate physical explanation for each of the two phenomena.[citation needed] A reference to these two aspects of electromagnetic induction is made in some modern textbooks.[28] As Richard Feynman states:[16] So the "flux rule" that the emf in a circuit is equal to the rate of change of the magnetic flux through the circuit applies whether the flux changes because the field changes or because the circuit moves (or both).... Yet in our explanation of the rule we have used two completely distinct laws for the two cases  –    $\stackrel{\mathbf{v\times B}}{}$  for "circuit moves" and   $\stackrel{\mathbf{\nabla \times E = - \part_t B}}{}$   for "field changes". We know of no other place in physics where such a simple and accurate general principle requires for its real understanding an analysis in terms of two different phenomena. — Richard P. Feynman, The Feynman Lectures on Physics ### Einstein's view Reflection on this apparent dichotomy was one of the principal paths that led Einstein to develop special relativity: It is known that Maxwell's electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena. Take, for example, the reciprocal electrodynamic action of a magnet and a conductor. The observable phenomenon here depends only on the relative motion of the conductor and the magnet, whereas the customary view draws a sharp distinction between the two cases in which either the one or the other of these bodies is in motion. For if the magnet is in motion and the conductor at rest, there arises in the neighbourhood of the magnet an electric field with a certain definite energy, producing a current at the places where parts of the conductor are situated. But if the magnet is stationary and the conductor in motion, no electric field arises in the neighbourhood of the magnet. In the conductor, however, we find an electromotive force, to which in itself there is no corresponding energy, but which gives rise—assuming equality of relative motion in the two cases discussed—to electric currents of the same path and intensity as those produced by the electric forces in the former case. Examples of this sort, together with unsuccessful attempts to discover any motion of the earth relative to the "light medium," suggest that the phenomena of electrodynamics as well as of mechanics possess no properties corresponding to the idea of absolute rest. — Albert Einstein, On the Electrodynamics of Moving Bodies[29] ## References 1. S M Dhir (2007). "§6 Other posive results and criticism". Hans Christian Ørsted and the Romantic Legacy in Science:Ideas, Disciplines, Practices. Springer. ISBN 978-1-4020-2987-5. 2. "Magnets". ThinkQuest. Retrieved 2009-11-06. 3. "Joseph Henry". Notable Names Database. Retrieved 2009-11-06. 4. Sadiku, M. N. O. (2007). Elements of Electromagnetics (fourth ed.). New York (USA)/Oxford (UK): Oxford University Press. p. 386. ISBN 0-19-530048-3. 5. 6. ^ a b Giancoli, Douglas C. (1998). Physics: Principles with Applications (Fifth ed.). pp. 623–624. 7. Ulaby, Fawwaz (2007). Fundamentals of applied electromagnetics (5th ed.). Pearson:Prentice Hall. p. 255. ISBN 0-13-241326-4. 8. "Joseph Henry". Distinguished Members Gallery, National Academy of Sciences. Retrieved 2006-11-30. 9. Faraday, Michael; Day, P. (1999-02-01). The philosopher's tree: a selection of Michael Faraday's writings. CRC Press. p. 71. ISBN 978-0-7503-0570-9. Retrieved 28 August 2011. 10. Michael Faraday, by L. Pearce Williams, p. 182-3 11. Michael Faraday, by L. Pearce Williams, p. 191–5 12. ^ a b Michael Faraday, by L. Pearce Williams, p. 510 13. Maxwell, James Clerk (1904), A Treatise on Electricity and Magnetism, Vol. II, Third Edition. Oxford University Press, pp. 178–9 and 189. 14. "The flux rule" is the terminology that Feynman uses to refer to the law relating magnetic flux to EMF.Richard Phillips Feynman, Leighton R B & Sands M L (2006). The Feynman Lectures on Physics. San Francisco: Pearson/Addison-Wesley. Vol. II, pp. 17-2. ISBN 0-8053-9049-9. 15. ^ a b Griffiths, David J. (1999). Introduction to Electrodynamics (Third ed.). Upper Saddle River NJ: Prentice Hall. pp. 301–303. ISBN 0-13-805326-X. 16. Tipler and Mosca, Physics for Scientists and Engineers, p795, google books link 17. Note that different textbooks may give different definitions. The set of equations used throughout the text was chosen to be compatible with the special relativity theory. 18. Nave, Carl R. "Faraday's Law". HyperPhysics. Georgia State University. Retrieved 29 August 2011. 19. Roger F Harrington (2003). Introduction to electromagnetic engineering. Mineola, NY: Dover Publications. p. 56. ISBN 0-486-43241-6. 20. ^ a b Davison, M. E. (1973). "A Simple Proof that the Lorentz Force, Law Implied Faraday's Law of Induction, when B is Time Independent". American Journal of Physics 41 (5): 713–711. doi:10.1119/1.1987339. 21. ^ a b 22. K. Simonyi, Theoretische Elektrotechnik, 5th edition, VEB Deutscher Verlag der Wissenschaften, Berlin 1973, equation 20, page 47 23. ^ a b 24. Griffiths, David J. (1999). Introduction to Electrodynamics (Third ed.). Upper Saddle River NJ: Prentice Hall. pp. 301–3. ISBN 0-13-805326-X.  Note that the law relating flux to EMF, which this article calls "Faraday's law", is referred to in Griffiths' terminology as the "universal flux rule". Griffiths uses the term "Faraday's law" to refer to what article calls the "Maxwell–Faraday equation". So in fact, in the textbook, Griffiths' statement is about the "universal flux rule". ## Further reading • Maxwell, James Clerk (1881), A treatise on electricity and magnetism, Vol. II, Chapter III, §530, p. 178. Oxford, UK: Clarendon Press. ISBN 0-486-60637-6.
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http://unapologetic.wordpress.com/2008/05/page/2/
# The Unapologetic Mathematician ## Commutativity in Series II We’ve seen that commutativity fails for conditionally convergent series. It turns out, though, that things are much nicer for absolutely convergent series. Any rearrangement of an absolutely convergent series is again absolutely convergent, and to the same limit. Let $\sum_{k=0}^\infty a_k$ be an absolutely convergent series, and let $p:\mathbb{N}\rightarrow\mathbb{N}$ be a bijection. Define the rearrangement $b_k=a_{p(k)}$. Now given an $\epsilon>0$, absolute convergence tells us we can pick an $N$ so that any tail of the series of absolute values past that point is small. That is, for any $n\geq N$ we have $\displaystyle\sum\limits_{k=n+1}^\infty\left|a_k\right|<\frac{\epsilon}{2}$ Now for $0\leq n\leq N$, the function $p^{-1}$ takes only a finite number of values (the inverse function exists because $p$ is a bijection). Let $M$ be the largest such value. Thus if $m>M$ we will know that $p(m)>N$. Then for any such $m$ we have $\displaystyle\sum\limits_{j=m+1}^{m+d}\left|b_k\right|=\sum\limits_{j=m+1}^{m+d}\left|a_{p(k)}\right|\leq\sum\limits_{k=N+1}^\infty\left|a_k\right|$ and we know that the sum on the right is finite by the assumption of absolute convergence. Thus the tail of the series of $b_j$ — and thus the series itself — must converge. Now a similar argument to the one we used when we talked about associativity for absolutely convergent series shows that the rearranged series has the same sum as the original. This is well and good, but it still misses something. We can’t handle reorderings that break up the order structure. For example, we might ask to add up all the odd terms, and then all the even terms. There is no bijection $p$ that handles this situation. And yet we can still make it work. Unfortunately, I arrive in Maryland having left my references back in New Orleans. For now, I’ll simply assert that for absolutely convergent series we can perform these more general rearrangements, though I’ll patch this sometime. Posted by John Armstrong | Analysis, Calculus | 3 Comments ## Commutativity in Series I We’ve seen that associativity may or may not hold for infinite sums, but it can be improved with extra assumptions. As it happens, commutativity breaks down as well, though the story is a bit clearer here. First we should be clear about what we’re doing. When we add up a finite list of real numbers, we can reorder the list in many ways. In fact, reorderings of $n$ numbers form the symmetric group $S_n$. If we look back at our group theory, we see that we can write any element in this group as a product of transpositions which swap neighboring entries in the list. Thus since the sum of two numbers is invariant under such a swap — $a+b=b+a$ — we can then rearrange any finite list of numbers and get the same sum every time. Now we’re not concerned about finite sums, but about infinite sums. As such, we consider all possible rearrangements — bijections $p:\mathbb{N}\rightarrow\mathbb{N}$ — which make up the “infinity symmetric group $S_\infty$. Now we might not be able to effect every rearrangement by a finite number of transpositions, and commutativity might break down. If we have a series with terms $a_k$ and a bijection $p$, then we say that the series with terms $b_k=a_{p(k)}$ is a rearrangement of the first series. If, on the other hand, $p$ is merely injective, then we say that the new series is a subseries of the first one. Now, if $\sum_{k=0}^\infty a_k$ is only conditionally convergent, I say that we can rearrange the series to give any value we want! In fact, given $x\leq y$ (where these could also be $\pm\infty$) there will be a rearrangement $b_k=a_{p(k)}$ so that $\displaystyle\liminf\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=x$ $\displaystyle\limsup\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nb_k=y$ First we throw away any zero terms in the series, since those won’t affect questions of convergence, or the value of the series if it does converge. Then let $p_n$ be the $n$th positive term in the sequence $a_k$, and let $-q_n$ be the $n$th negative term. The two series with positive terms $\sum_{k=0}^\infty p_k$ and $\sum_{k=0}^\infty q_k$ both diverge. Indeed, if one converged but the other did not, then the original series $\sum_{k=0}^\infty a_k$ would diverge. On the other hand, if they both converged then the original series would converge absolutely. Conditional convergence happens when the subseries of positive terms and the subseries of negative terms just manage to balance each other out. Now we take two sequences $x_n$ and $y_n$ converging to $x$ and $y$ respectively. Since the series of positive terms diverges, they’ll eventually exceed any positive number. We can take just enough of them (say $k_1$ so that $\displaystyle\sum_{k=0}^{k_1}p_k>y_1$ Similarly, we can then take just enough negative terms so that $\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l<x_1$ Now take just enough of the remaining positive terms so that $\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k>y_2$ and enough negatives so that $\displaystyle\sum\limits_{k=0}^{k_1}p_k-\sum\limits_{l=0}^{l_1}q_l+\sum\limits_{k=k_1+1}^{k_2}p_k-\sum\limits_{l=l_1+1}^{l_2}q_l<x_2$ and so on and so forth. This gives us a rearrangement of the terms of the series. Each time we add positive terms we come within $p_{k_j}$ of $y_j$, and each time we add negative terms we come within $q_{l_j}$ of $x_j$. But since the original sequence $a_n$ must be converging to zero (otherwise the series couldn’t converge), so must the $p_{k_j}$ and $q_{l_j}$ be converging to zero. And the sequences $x_j$ and $y_j$ are converging to $x$ and $y$. It’s straightforward from here to show that the limits superior and inferior of the partial sums of the rearranged series are as we claim. In particular, we can set them both equal to the same number and get that number as the sum of the rearranged series. So for conditionally convergent series, the commutativity property falls apart most drastically. Posted by John Armstrong | Analysis, Calculus | 4 Comments ## Quantum Knot Mosaics Today, Sam Lomonaco and Louis Kauffman posted to the arXiv a paper on “Quantum Knots and Mosaics”. I had the pleasure of a sneak preview back in March. Here’s what I said then (I haven’t had a chance to read the paper as posted, so some of this may be addressed): About half the paper consists of setting up definitions of a mosaic and the Reidemeister moves. This concludes with the conjecture that before you allow superpositions the mosaic framework captures all of knot theory. The grading by the size of the mosaic leads to an obvious conjecture: there exist mosaic knots which are mosaic equivalent, but which require arbitrarily many expansions. This is analogous to the same fact about crossing numbers. Obviously, I’d write these combinatorial frameworks as categories with the mosaics as objects and the morphisms generated by the mosaic moves. Superpositions just seem to be the usual passage from a set to the vector space on that basis. See my new paper for how I say this for regular knots and Reidemeister moves. Then (like I say in the paper) we want to talk about mosaic “covariants”. I think this ends up giving your notion of invariant after we decategorify (identify isomorphic outputs). The only thing I’m wondering about (stopping shy of saying you two are “wrong”) is the quantum moves. The natural thing would be to go from the “group” (really its a groupoid like I said before) of moves to its linearization. That is, we should allow the “sum” of two moves as a move. This splits a basis mosaic input into a superposition. In particular, the “surprising” result you state that one quantum mosaic is not quantum equivalent to the other must be altered. There is clearly a move in my view taking the left to the right. “Equivalence” is then the statement that two quantum mosaics are connected by an *invertible* move. I’m not sure that the move from left to right is invertible yet, but I think it is. ## Associativity in Series II I’m leaving for DC soon, and may not have internet access all day. So you get this now! We’ve seen that associativity doesn’t hold for infinite sums the way it does for finite sums. We can always “add parentheses” to a convergent sequence, but we can’t always remove them. The first example we mentioned last time. Consider the series with terms $a_k=(-1)^k$: $\displaystyle\sum\limits_{k=0}^\infty a_k=1+(-1)+1+(-1)+...$ Now let’s add parentheses using the sequence $d(j)=2j+1$. Then $b_j=(-1)^{(2(j-1)+1)+1}+(-1)^{2j+1}=1+(-1)=0$. That is, we now have the sequence $\displaystyle\sum\limits_{j=0}^\infty b_j=(1+(-1))+(1+(-1))+...=0+0+...=0$ So the resulting series does converge. However, the original series can’t converge. The obvious fault is that the terms $a_k$ don’t get smaller. And we know that $\lim\limits_{k\rightarrow\infty}a_k$ must be zero, or else we’ll have trouble with Cauchy’s condition. With the parentheses in place the terms $b_j$ go to zero, but when we remove them this condition can fail. And it turns out there’s just one more condition we need so that we can remove parentheses. So let’s consider the two series with terms $a_k$ and $b_j$, where the first is obtained from the second by removing parentheses using the function $d(j)$. Assume that $\lim_{k\rightarrow\infty}a_k=0$, and also that there is some $M>0$ so that each of the $b_j$ is a sum of fewer than $M$ of the $a_k$. That is, $d(j+1)-d(j)<M$. Then the series either both diverge or both converge, and if they converge they have the same sum. We set up the sequences of partial sums $\displaystyle s_n=\sum\limits_{k=0}^na_k$ $\displaystyle t_m=\sum\limits_{j=0}^mb_j$ We know from last time that $t_m=s_{d(m)}$, and so if the first series converges then the second one must as well. We need to show that if $t=\lim\limits_{m\rightarrow\infty}t_m$ exists, then we also have $\lim\limits_{n\rightarrow\infty}s_n=t$. To this end, pick an $\epsilon>0$. Since the sequence of $t_m$ converge to $t$, we can choose some $N$ so that $\left|t_m-t\right|<\frac{\epsilon}{2}$ for all $m>N$. Since the sequence of terms $a_k$ converges to zero, we can increase $N$ until we also have $\left|a_k\right|<\frac{\epsilon}{2M}$ for all $k>N$. Now take any $n>d(N)$. Then $n$ falls between $d(m)$ and $d(m+1)$ for some $m$. We can see that $m\geq N$, and that $n$ is definitely above $N$. So the partial sum $s_n$ is the sum of all the $a_k$ up through $k=d(m+1)$, minus those terms past $k=n$. That is $\displaystyle s_n=\sum\limits_{k=0}^na_k=\sum\limits_{k=0}^{d(m+1)}a_k-\sum\limits_{n+1}^{d(m+1)}a_k$ But this first sum is just the partial sum $t_{m+1}$, while each term of the second sum is bounded in size by our assumptions above. We check $\displaystyle\left|s_n-t\right|=\left|(t_{m+1}-t)-\sum\limits_{n+1}^{d(m+1)}a_k\right|\leq\left|t_{m+1}-t\right|+\sum\limits_{n+1}^{d(m+1)}\left|a_k\right|$ But since $n$ is between $d(m)$ and $d(m+1)$, there must be fewer than $M$ terms in this last sum, all of which are bounded by $\frac{\epsilon}{2M}$. So we see $\displaystyle\left|s_n-t\right|<\frac{\epsilon}{2}+M\frac{\epsilon}{2M}=\epsilon$ and thus we have established the limit. Posted by John Armstrong | Analysis, Calculus | 4 Comments ## Associativity in Series I As we’ve said before, the real numbers are a topological field. The fact that it’s a field means, among other things, that it comes equipped with an associative notion of addition. That is, for any finite sum we can change the order in which we perform the additions (though not the order of the terms themselves — that’s commutativity). The topology of the real numbers means we can set up sums of longer and longer sequences of terms and talk sensibly about whether these sums — these series — converge or not. Unfortunately, this topological concept ends up breaking the algebraic structure in some cases. We no longer have the same freedom to change the order of summations. When we write down a series, we’re implicitly including parentheses all the way to the left. Consider the partial sums: $\displaystyle s_n=\sum\limits_{k=0}^na_k=((...(((a_0+a_1)+a_2)+a_3)...+a_{n-1})+a_n)$ But what if we wanted to add up the terms in a different order? Say we want to write $\displaystyle s_6=(((a_0+a_1)+(a_2+a_3))+((a_4+a_5)+a_6))$ Well this is still a left-parenthesized expression, it’s just that the terms are not the ones we looked at before. If we write $b_0=a_0+a_1$, $b_1=a_2+a_3$, and $b_2=a_4+a_5+a_6$ then we have $\displaystyle s_6=((b_0+b_1)+b_2)=\sum\limits_{j=0}^2b_j=t_2$ So this is actually a partial sum of a different (though related) series whose terms are finite sums of terms from the first series. More specifically, let’s choose a sequence of stopping points: an increasing sequence of natural numbers $d(j)$. In the example above we have $d(0)=1$, $d(1)=3$, and $d(3)=6$. Now we can define a new sequence $\displaystyle b_0=\sum\limits_{k=0}^{d(0)}a_k$ $\displaystyle b_j=\sum\limits_{k=d(j-1)+1}^{d(j)}a_k$ Then the sequence of partial sums $t_m$ of this series is a subsequence of the $s_n$. Specifically $\displaystyle t_m=\sum\limits_{j=0}^mb_j=\sum\limits_{k=0}^{d(0)}a_k+\sum\limits_{j=1}^m\left(\sum\limits_{k=d(j-1)+1}^{d(j)}a_k\right)=\sum\limits_{k=0}^{d(m)}a_k=s_{d(m)}$ We say that the sequence $t_m$ is obtained from the sequence $s_n$ by “adding parentheses” (most clearly notable in the above expression for $t_m$). Alternately, we say that $s_n$ is obtained from $t_m$ by “removing parentheses”. If the sequence $s_n$ converges, so must the subsequence $t_m=s_{d(m)}$, and moreover to the same limit. That is, if the series $\sum_{k=0}^\infty a_k$ converges to $s$, then any series $\sum_{j=0}^\infty b_j$ obtained by adding parentheses also converges to $s$. However, convergence of a subsequence doesn’t imply convergence of the sequence. For example, consider $a_k=(-1)^k$ and use $d(j)=2j+1$. Then $s_n$ jumps back and forth between zero and one, but $t_m$ is identically zero. So just because a series converges, another one obtained by removing parentheses may not converge. Posted by John Armstrong | Analysis, Calculus | 2 Comments ## The Ratio and Root Tests Now I want to bring out with two tests that will tell us about absolute convergence or (unconditional) divergence of an infinite series $\sum_{k=0}^\infty a_k$. As such they’ll tell us nothing about conditionally convergent series. First is the ratio test. We take the ratio of one term in the series to the previous one and define the limits superior and inferior $\displaystyle R=\limsup\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$ $\displaystyle r=\liminf\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$ Now if $R<1$ then the series converges absolutely. If $r>1$ then the series diverges. But if $r\leq1\leq R$ the test fails and we get no result. In the first case, pick $x$ to be a number so that $R<x<1$. Then there is some $N$ so that $x$ is an upper bound for the sequence of ratios past $N$. For large enough $n$, this means $\displaystyle\left|\frac{a_{n+1}}{a_n}\right|<x=\frac{x^{n+1}}{x^n}$ and so $\displaystyle\frac{\left|a_{n+1}\right|}{x^{n+1}}<\frac{\left|a_n\right|}{x^n}\leq\frac{\left|a_N\right|}{x^N}$ Now if we set $c=\frac{\left|a_N\right|}{x^N}$, this tells us that $|a_n|\leq cx^n$. Then the comparison test with the geometric series tells us that $\sum_{k=0}^\infty\left|a_k\right|$ converges. On the other hand, if $r>1$ then eventually $\left|a_{n+1}\right|>\left|a_n\right|$, so the terms of the series are getting bigger and bigger and bigger. But this would throw a monkey wrench into Cauchy’s condition for convergence of the series. As for the root test, we will consider the sequence $\sqrt[n]{\left|a_n\right|}$ and define $\displaystyle\rho=\limsup\limits_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}$ If $\rho<1$ then the series converges absolutely. If $\rho>1$ then the series diverges. And if $\rho=1$ the test is inconclusive. In the first case, as we did for the ratio test, pick $x$ so that $\rho<x<1$. Then above some $N$ we have $\left|a_n\right|<x^n$ and the comparison test works straight away. On the other hand, if $\rho>1$ then $\left|a_n\right|>1$ infinitely often, and Cauchy’s criterion falls apart again. Posted by John Armstrong | Analysis, Calculus | 5 Comments ## Limits Superior and Inferior As we look at sequences (and nets) of real numbers (and more general ordered spaces) a little more closely, we’ll occasionally need the finer notion of a “limit superior” (“limit inferior”). This is essentially the largest (smallest) value that a sequence takes in its tail. In general, let $x_\alpha$ be a net (indexed by $\alpha\in A$) in some ordered space $X$. Then we can consider the “tail” $A_\alpha=\{\beta\in A|\beta\geq\alpha\}$ of the index set consisting of all indices above a given index $\alpha$. We then ask what the least upper bound of the net is on this tail: $\sup\limits_{\beta\geq\alpha}x_\beta$. Alternately, we consider the greatest lower bound on the tail: $\inf\limits_{\beta\geq\alpha}x_\beta$. Now as we move to tails further and further out in the net, the least upper bound (greatest lower bound) may drop (rise) as we pass maxima (minima). That is, the supremum (infimum) of a set bounds the suprema (infima) of its subsets. So? So if we pass such a maximum it clearly doesn’t affect the long-run behavior of the net, and we want to forget it. So we’ll take the lowest of the suprema of tails (the highest of the infima of tails). Thus we finally come to defining the limit superior $\displaystyle\limsup x_\alpha=\inf\limits_{\alpha\in A}\sup\limits_{\beta\geq\alpha}x_\alpha$ and the limit inferior $\displaystyle\liminf x_\alpha=\sup\limits_{\alpha\in A}\inf\limits_{\beta\geq\alpha}x_\alpha$ Now these are related to our usual concept of a limit. First of all, $\displaystyle\liminf x_\alpha\leq\limsup x_\alpha$ and the limit converges if and only if these two are both finite and equal. In this case, the limit is this common finite value. If they both go to infinity, the limit diverges to infinity, and similarly for negative infinity. If they’re not equal, then the limit bounces around between the two values. If we’re considering a sequence of real numbers, then we’re taking a bunch of infima and suprema, all of which are guaranteed to exist. Thus the limits superior and inferior of any sequence must always exist. As an illustrative example, work out the limits superior and inferior of the sequence $(-1)^n(1+\frac{1}{n})$. Show that this sequence diverges, but does so by oscillating rather than by blowing up. Finally, note that we can consider a function $f(x)$ defined on a ray to be a net on that ray, considered as a directed subset of real numbers. Then we get limits superior and inferior as $x$ goes to infinity, just as for sequences. Posted by John Armstrong | Analysis, Calculus | 1 Comment ## Dirichlet’s and Abel’s Tests We can now use Abel’s partial summation formula to establish a couple other convergence tests. If $a_n$ is a sequence whose sequence $A_n$ of partial sums form a bounded sequence, and if $B_n$ is a decreasing sequence converging to zero, then the series $\sum_{k=0}^\infty a_kB_k$ converges. Indeed, then the sequence $A_nB_{n+1}$ also decreases to zero, so we just need to consider the series $\sum_{k=0}^\infty A_kb_{k+1}$. The bound on $|A_k|$ and the fact that $B_k$ is decreasing imply that $|A_k(B_k-B_{k+1})|\leq M(B_k-B_{k+1})$, and the series $\sum_{k=0}^\infty M(B_k-B_{k+1})$ clearly converges. Thus by the comparison test, the series $\sum_{k=0}^\infty A_kb_{k+1}$ converges absolutely, and our result follows. This is called Dirichlet’s test for convergence. Let’s impose a bit more of a restriction on the $A_n$ and insist that this sequence actually converge. Correspondingly, we can weaken our restriction on $B_n$ and require that it be monotonic and convergent, but not specifically decreasing to zero. These two changes balance out and we still find that $\sum_{k=0}^na_kB_k$ converges. Indeed, the sequence $A_nB_{n+1}$ converges automatically as the product of two convergent sequences, and the rest is similar to the proof in Dirichlet’s test. We call this Abel’s test for convergence. Posted by John Armstrong | Analysis, Calculus | 2 Comments ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://gilkalai.wordpress.com/2008/09/07/diameter-problem-3/?like=1&source=post_flair&_wpnonce=b19f116cf4
Gil Kalai’s blog ## Diameter Problem (3) Posted on September 7, 2008 by ### 3. What we will do in this post and and in future posts We will now try all sorts of ideas to give good upper bounds for the abstract diameter problem that we described. As we explained, such bounds apply to the diameter of graphs of simple d-polytopes. All the methods I am aware of for providing upper bounds are fairly simple. (1) You think about a strategy from moving from one set to another, (2) You use this strategy to get a recursive bound, (3) You solve the recursion and hope for the best. What I would like you to think about, along with reading these posts, is the following questions: (a) Can I come up with a different/better strategy for moving from one set to the other? (b) Can I think about a mathematically more sophisticated way to get an upper bound for the diameter? (c) Can this process of finding a strategy/writing the associated recurrence/solving the recurrence be automatized? The type of proofs we will describe are very simple and this looks like a nice example for a “quasi-automatic” proof process. Let me repeat the problem and prove to you a nice upper bound: ### Reminder: Our Diameter problem for families of sets Consider a family $\cal F$ of subsets of size d of the set N={1,2,…,n}. Associate to $\cal F$ a graph $G({\cal F})$ as follows: The vertices of  $G({\cal F})$ are simply the sets in $\cal F$. Two vertices $S$ and $T$ are adjacent if $|S \cap T|=d-1$. For a subset $A \subset N$ let ${\cal F}[A]$ denote the subfamily of all subsets of $\cal F$ which contain $A$. MAIN ASSUMPTION: Suppose that for every $A$ for which ${\cal F}[A]$ is not empty $G({\cal F}[A])$ is connected. MAIN QUESTION:   How large can the diameter of $G({\cal F})$ be in terms of $d$ and $n$. Let us denote the answer by $F(d,n)$. ### 4. A one line observation. What the upper bound $F(d,n)$ tells us about the diameter of  $G({\cal F}[A])$? Let ${\cal F'}[A]$ be the family obtained from ${\cal F}[A]$ by removing the elements of A from every set. Note that $G({\cal F}[A]) = G({\cal F}' [A])$. Therefore, the diameter of  $G({\cal F}[A])$ is at most $F(d',n')$, where $d'= d-|A|$ and $n'$ is the number of elements in the union of all the sets in $G({\cal F}'[A])$. ### 5. Linear bounds for a fixed dimension Let’s use the following strategy to move from one set to the other. Given two sets S and T in $\cal F$ we first try to move from S to T using a different type of path. $S_0, S_1, S_2, \dots, S_t$,  where this time $|S_i \cap S_{i+1}| \ge1$.  We will choose such a path with t being as small as possible. Let $w_i \in S_{i-1} \cap S_{i}, i=1,2,\dots ,t$. We will consider the families ${\cal F}_i = {\cal F}'[w_i]$. The one line observation tells us that the diameter of $G({\cal F}_i)$ is bounded from above by $F(d-1,n_i),$ where $n_i$ the number of elements in the union $X_i$ of all the sets in ${\cal F}_i$. We want to prove an upper bound on $F(d,n)$ of the form $c_d \cdot d$. For this purpose, let us have a closer look at these $n$ sets $X_1, X_2, \dots, X_n$. Claim: $X_i \cap X_j = \emptyset$ if $j-i > 2$. Proof: Suppose that  $y \in (X_i \cap X_j)$ and $j-i>2$. So there is a set $R \in {\cal F}$ which contains $w_i$ and $y$, and there is a set $U \in {\cal F}$  which contains both $w_j$ and $y$. Now we can shortcut! We replace the segment $S_i, S_{i+1}, \dots S_j$ by $S_{i-1},R,U,S_j$. This will give us a shorter path of the peculiar type we consider here. The claim implies that every element of N is included in at most three $S_i$s. We are done! If $F(d-1,n) \le c_{d-1}n$ then we get that the distance between $S$ and $T$ in $G({\cal F})$ is at most $\sum F(d-1,n_i) \le c_{d-1} \sum n_i \le c_{d-1} 3n$. This gives us $F(d,n) \le 3^d n$. This argument is due to Barnette. ### Reminder: The connection with Hirsch’s Conjecture The Hirsch Conjecture asserts that the diameter of the graph G(P) of a d-polytope P with n facets is at most n-d. Not even a polynomial upper bound for the diameter in terms of d and n is known. Finding good upper bounds for the diameter of graphs of d-polytopes is one of the central open problems in the study of convex polytopes. If d is fixed then a linear bound in n is known, and the best bound in terms of d and n is $n^{\log d+1}$. We will come back to these results later. One basic fact to remember is that for every d-polytope P, G(P) is a connected graph. As a matter of fact, a theorem of Balinski asserts that G(P)\$ is d-connected. The combinatorial diameter problem I mentioned in an earlier post (and which is repeated below) is closely related. Let me now explain the connection. Let P be a simple d-polytope. Suppose that P is determined by n inequalities, and that each inequality describes a facet of P. Now we can define a family $\cal F$ of subsets of {1,2,…,n} as follows. Let $E_1,E_2,\dots,E_n$ be the n inequalities defining the polytopeP, and let $F_1,F_2,\dots, F_n$ be the n corresponding facets. Every vertex v of P belongs to precisely d facets (this is equivalent to P being a simple polytope). Let $S_v$ be the indices of the facets containing v, or, equivalently, the indices of the inequalities which are satisfied as equalities at v. Now, let $\cal F$ be the family of all sets $S_v$ for all vertices of the polytope P. The following observations are easy. (1) Two vertices v and w of P are adjacent in the graph of P if and only if $|S_v \cap S_w|=d-1$. Therefore, $G(P)=G({\cal F})$. (2)  If A is a set of indices. The vertices v of P such that $A \subset S_v$ are precisely the set of vertices of a lower dimensional face of P. This face is described by all the vertices of P which satisfies all the inequalities indexed by $i \in A$, or equivalently all vertices in P which belong to the intersection of the facets $F_i$ for $i \in A$. Therefore, for every $A \subset N$ if ${\cal F}[A]$ is not empty the graph $G({\cal F}[A])$ is connected – this graph is just the graph of some lower dimensional polytope. This was the main assumption in our abstract problem. Remark: It is known that the assertion of the Hirsch Conjecture fails for the abstract setting. There are examples of families where the diameter is as large as n-(4/5)d.
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http://www.physicsforums.com/showthread.php?t=337830
Physics Forums ## electrical potential difference given velocity 1. The problem statement, all variables and given/known data An electron moving parallel to the x axis has in initial speed of 3.7x10^6m/s at the origin. Its speed is reduced to 1.4x10^5m/s at the point x=2.0 cm. What is the potential difference between the origin and this point? Which point is at the higher potential? 2. Relevant equations I am not sure what equations are useful with this equation. 3. The attempt at a solution I have not been able to attempt this problem, I have no idea how i would go about starting it PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug I'm a little unsure, but perhaps you could use the fact that Uq = 0,5 m v^2, at least for the first questions. I don't quite know what is meant by the second question, but electrons are negative. So try that. Anden, examine the equation you posted carefully, what happens if the charge is negative? Does that make sense physically? What is the electron's kinetic energy at the origin? What is its kinetic energy at point x? What does this say about the work done on the electron? What is the relationship between the work done on a mass moving it from point A to point B, and the potential difference between those two points? ## electrical potential difference given velocity Is it really that simple? there is nothing special about it being in the electrical potential section? If its Uq then is q the charge of the electron? Or do i simply have to find the kinetic energy? Ups, yes of course you're right, the electron has negative charge. But I think if you forget that and count the electrons charge as positive, you can use the equation to get directly to the potential difference. EDIT: A thought: If you get a negative potential difference, doesn't that describe whats going on here? I mean, you get a general idea of the direction of the acceleration of the electron. Quote by Anden Ups, yes of course you're right, the electron has negative charge. But I think if you forget that and count the electrons charge as positive, you can use the equation to get directly to the potential difference. EDIT: A thought: If you get a negative potential difference, doesn't that describe whats going on here? I mean, you get a general idea of the direction of the acceleration of the electron. Oh, so since it comes out nonsensical (A square equaling a negative) we just ignore the problem? :P I haven't brushed up on the subject in a while, but I'm pretty sure that $$U_{AB}\cdot q=W_{AB}$$ What do you mean by a square equaling a negative? And why is it so non-sensical? ;) And what happens with a negative charge in Uq = W? Thread Tools | | | | |---------------------------------------------------------------------|-------------------------------|---------| | Similar Threads for: electrical potential difference given velocity | | | | Thread | Forum | Replies | | | Introductory Physics Homework | 8 | | | Classical Physics | 5 | | | Introductory Physics Homework | 4 | | | Introductory Physics Homework | 4 | | | Introductory Physics Homework | 3 |
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http://www.physicsforums.com/showthread.php?s=86e0cba3bcf2c774e27f1eb960bc99b0&p=4083425
Physics Forums ## Finding a basis for a set of polynomials and functions Find a basis for and the dimension of the subspaces defined for each of the following sets of conditions: {p $\in$ P3(R) | p(2) = p(-1) = 0 } { f$\in$Span{ex, e2x, e3x} | f(0) = f'(0) = 0} Attempt: Having trouble getting started... So I think my issue is interpreting what those sets are and setting it up. So I think the sets are: i) the set of all polynomials s.t P(2) = p(-1) = 0 and ii) the set of exp functions where at 0 equal 0. So how do I put these each into a matrix form to find the basis and dimension? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 1 Recognitions: Gold Member Homework Help Science Advisor Quote by trap101 Find a basis for and the dimension of the subspaces defined for each of the following sets of conditions: {p $\in$ P3(R) | p(2) = p(-1) = 0 } { f$\in$Span{ex, e2x, e3x} | f(0) = f'(0) = 0} Attempt: Having trouble getting started... So I think my issue is interpreting what those sets are and setting it up. So I think the sets are: i) the set of all polynomials s.t P(2) = p(-1) = 0 I doubt this. It is probably the set of all polynomials with degree <= 3, such that p(2) = p(-1) = 0. and ii) the set of exp functions where at 0 equal 0. Not just any exp functions. They have to be in $\text{span}(e^x, e^{2x}, e^{3x})$. I suggest that you start by finding the dimension of these two spaces: $P_3(\mathbb{R})$ and $\text{span}(e^x, e^{2x}, e^{3x})$. Also, what is the form of a general element for each of these two spaces? Recognitions: Gold Member Science Advisor Staff Emeritus Personally, I wouldn't use a matrix, I would use the basic definition. First, I am going to assume that P3 is the vector space of polynomials of degree 3 or less, which has dimension 4 (some texts use that to mean the space of polynomials 2 or less which has dimension 3- the same ideas will apply but it is simpler). Any such polynomial canbe written $p(x)= ax^3+ bx^2+ cx+ d$. The condition that p(2)= 0 means that we must have $p(2)= 8a+ 4b+ 2c+ d= 0$ or $d= -(8a+ 4b+ 2c)$. The condition that p(-1)= 0 means that $-a+ b- 2c+ d= 0[/tex] or [tex]d= -(a- b+ 2c)[/tex]. Then d= -(8a+ 4b+ 2c)= -(a- b+ 2c) so that 8a+ 4b+ 2c= a- b+ 2c which reduces to 7a= -5b. Sp we can replace a by -5b/7 which means d= -(-(5/7)b- b+ 2c)= -(12/7)b- 2c. Using those, [itex]ax^3+ bx^2+ cx+ d= -(5/7)bx^3+ bx^2+ cx- (12/7)b- 2c= (-(5/7)x^3+ x^2- 12/7)b+ c(x- 2)$ Now, do you see what a basis is and what the dimension is? (You could have made a quick "guess" at what the dimension is by the fact that the basic space has dimension 4 and there are 2 conditions put on it.) For the second one, any f in the span of ex, e2x, and e3x, can be written as f(x)= aex+ be2x+ ce3x, and f'(x)= aex+ 2bex+ 3cex. The condition that f(0)= 0 gives a+ b+ c= 0 and f'(0)= 0 gives a+ 2b+ 3c= 0. We can subtract the first equations from the second to get b+ 2c= 0 or b= -2c. Putting that into the first equation a- 2c+ c= a- c= 0 so a= c. That is, we can write aex+ bex+ cex= ce[sup]x[/sup- 2ce2x+ ce3x= c(ex- 2e2x+ e3x. Now, what is the dimension and what is a basis? (Here, the basic space has dimension three and there are two conditions.) ## Finding a basis for a set of polynomials and functions Quote by jbunniii I doubt this. It is probably the set of all polynomials with degree <= 3, such that p(2) = p(-1) = 0. Not just any exp functions. They have to be in $\text{span}(e^x, e^{2x}, e^{3x})$. I suggest that you start by finding the dimension of these two spaces: $P_3(\mathbb{R})$ and $\text{span}(e^x, e^{2x}, e^{3x})$. Also, what is the form of a general element for each of these two spaces? That's what I intend on doing, but my issue is setting it up in order ot find those dimensions. So here's how I'm trying to piece it together: I know the general form for P3(R) is: ax3+bx2+cx+d, now the condition is that p(2) = P(-1) = 0. So I have to some how write out a set of vectors that satisfy that form. As for ii) a function would be f(x) = ex-2e2x+e3x, but I'm utterly clueless as to how this is line independent and how I could even find this vector if I set up a matrix Thread Tools Similar Threads for: Finding a basis for a set of polynomials and functions Thread Forum Replies General Math 5 Precalculus Mathematics Homework 6 Precalculus Mathematics Homework 1 Linear & Abstract Algebra 1 Calculus & Beyond Homework 2
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http://mathoverflow.net/questions/38943?sort=oldest
## An orthogonal companion matrix ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $P\in{\mathbb R}[X]$ be a monic polynomial with roots on the unit circle. For the problem below, we may assume wlog that the roots are simple and distinct from $\pm1$. It can be shown that there exists a matrix $M\in{\bf SO}_n({\mathbb R})$, whose characteristic polynomial is $P$ (an orthogonal companion matrix of $P$, in short OCM). See for instance Exercise 99 on my list http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf . Regretfully, this exercise uses the square root of Hermitian positive definite matrices, which cannot be computed in finitely many operations. Does there exist a construction of an OCM that uses only finitely many elementary operations (including the square root of complex numbers) ? Thanks to the reduction to Hessenberg form, which can be done in finite time and which preserves the orthogonal group, we may restrict our attention to a Hessenberg orthogonal matrix $M$. It writes \left( \begin{array}{ccccc} c_1 & s_1c_2 & s_1s_2c_3 & s_1s_2s_3c_4 & \ldots \\ -s_1 & c_1c_2 & c_1s_2c_3 & c_1s_2s_3c_4 & \ldots \\ 0 & -s_2 & c_2c_3 & s_2s_3c_4 & \ldots \\ 0 & 0 & -s_3 & c_3c_4 & \ldots \\ 0 & 0 & 0 & -s_4 & \ldots \end{array} \right) where $(c_j,s_j)$ are cosine/sine pairs. - 1 This sounds like the "inverse unitary eigenproblem"; see dx.doi.org/10.1016/0024-3795(93)00188-6 for instance. – J. M. Sep 16 2010 at 10:07 Nice paper, but my question is different: $n$ is even and the data consists of $n/2$ real numbers, the arguments of the roots (associated pairwise by c.c.). Whereas there are $n$ parameters left (the angles of the cosines). The solution exists but must be highly non-unique. I wish an algorithm, involving elementary operations and square roots of complex numbers, giving such a matrix. Such a trick is well-known in the Inverse Symmetric Eigenvalue Problem. – Denis Serre Sep 16 2010 at 16:11 ## 2 Answers I'd do this in three steps: 1. Find any $2n \times 2n$ matrix $A$ whose eigenvalues are $e^{\pm i \theta}$. 2. Find a positive definite quadratic form preserved by $A$. In equations, we want $A P A^T = P$. 3. Find an orthonormal basis for $P$, using the Gram-Schimdt algorithm. In equations, we want $S P S^T = \mathrm{Id}$. Then $S A S^{-1}$ is orthogonal and has the required eigenvalues. I can think of two ways to do step 2. The first is more purely algebraic, the second I think would be much easier to implement. Algebra: Let $f(x) = \prod_{j=1}^{n} (x-e^{i \theta_j}) (x - e^{-i \theta_j}) = \prod (x^2 - 2 \cos \theta_j + 1)$ be your characteristic polynomial. Let $V$ be the ring $\mathbb{R}[x]/f(x)$. Note $1$, $x$, ..., $x^{2n-1}$ is a basis for this ring, in which multiplication can be written down algebraically in terms of the coefficients of $f$. Also, multiplication by $x$ has the desired eigenvalues, so that accomplishes part 1. For $y \in T$, let $T(y)$ be the trace of multiplication by $y$. Also, let $y \mapsto \overline{y}$ be the automorphism of $V$ induced by $x \mapsto x^{-1}$. Again, both of these can be written down, in the monomial basis, algebraically in terms of the coefficients of $f$. Then `$\langle y,z \rangle = T(y*\overline{z})$` is the desired positive definite quadratic form. Namely, observe that the ring $V$ is isomorphic to $\mathbb{C}^{\oplus n}$. In terms of this isomorphism, $\langle (z_1, \ldots, z_n), (z_1, \ldots, z_n) \rangle = 2 \sum |z_i|^2$. In practice: The condition that $APA^T = P$ is a linear condition on $P$. Let $W$ be the subspace of the vector space of symmetric matrices where this condition is satisfied; finding $W$ is just algebra. Now, our goal is to find a positive definite element of $W$. For large $N$, $(1/N) \left( \mathrm{Id} + AA^T + A^2 (A^{T})^2 + \cdots + A^{N-1} (A^T)^{N-1} \right)$ is positive definite and is near $W$. I would guess that the orthogonal projection of this matrix onto $W$ would probably be positive definite for large $N$. - In second paragraph of Algebra, did you mean For $y \in V$, let $T(y)$ ... automorphism of $V$ ... ? Is this a fair summary: Let $A_0$ be the companion matrix of $f$ (in the usual sense). Let $SA_0$ be the induced action $Q \to A_0 Q A_0^t$ on the vector space of quadratic forms. Then $1$ is an eigenvalue of multiplicity $n$. The eigenspace $Q_0$ has a basis consisting of symmetric integer matrices. The set of positive definite quadratic forms form an open convex cone within $Q_0$ defined by positivity of various determinants, so a positive definite integer matrix can be readily found. – Bill Thurston Sep 17 2010 at 1:21 Thanks for catching those typos. And, yes, that's a very fair summary. – David Speyer Sep 17 2010 at 11:11 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. You can use the Cayley transform and reduce this problem to generating a skew-symmetric matrix with a prescribed characteristic polynomial. For example, this works in the $3\times 3$ case (although when $n$ is odd, $1$ is always an eigenvalue). I have not thought about it thoroughly, but presumably, methods used in Inverse Symmetric Eigenvalue Problem should apply in the skew-symmetric case. - You can go from the skew-symmetric to the self-adjoint problem by multiplication by $i$. I think this approach would work, one just needs to carefully check that one always stays in the right set... – Helge Sep 16 2010 at 22:08 I didn't understand the comment about multiplication by $i:$ both symmetry and skew symmetry are linear conditions, therefore, they are invariant under scalar multiplication. – Victor Protsak Sep 16 2010 at 23:37 Self-adjointness isn't, since it's $a_{ij} = \overline{a_{ij}}$. It's only REAL linear. – Helge Sep 17 2010 at 11:30 And it's "well-known" that self-adjointness is the "correct" condition for complex matrices ... (There is some room for debate here, that's why the "). – Helge Sep 17 2010 at 11:31 The problem is not about "correct conditions", it is manifestly about reconstructing REAL orthogonal matrices with prescribed eigenvalues, which become real skew-symmetric after Cayley transform. Replacing "real skew-symmetric" with "complex skew-hermitian" corresponds to replacing special orthogonal group with special unitary group. There are other ways to generalize the question, too, but is there any evidence that these generalizations can be solved more easily than the original question? No matter how many times you say "halva", your mouth wouldn't become sweet. – Victor Protsak Sep 17 2010 at 15:07
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http://mathoverflow.net/questions/22065?sort=oldest
## Algebraic properties of the algebra of continuous functions on a manifold. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Does the algebra of continuous functions from a compact manifold to $\mathbb{C}$ satisfy any specific algebraic property? I'm not sure what kind of algebraic property I expect, but I feel that because of the Gel'fand transform, it may not be unreasonable to expect something. We can drop the compactness condition if we switch to continuous functions to $\mathbb{C}$ that vanish at infinity. I'm really hoping for some necessary and sufficient condition, but if anybody knows of any sort of condition, that would be appreciated. - Check out: mathoverflow.net/questions/5344/… – Steven Gubkin Apr 21 2010 at 15:27 @Steven: so far, on MO only $C^{\infty}(M)$ has been discussed, not $C(M)$, as in my question mathoverflow.net/questions/21168/… – Martin Brandenburg Apr 21 2010 at 15:41 1 @Eric: so you're asking for a condition of a commutative $C^*$-algebra (in their language) whose spectrum is a manifold? – Martin Brandenburg Apr 21 2010 at 15:42 @Martin: yes, that's right. – Eric A. Bunch Apr 21 2010 at 16:02 @Steven and Martin: thanks for the references. I thought I remembered similar questions, but I couldn't find them for the life of me. I feel like it may be a bit simpler to get some condition on the C^*-algebra of continuous functions to C rather than the R-algebra of smooth functions to R simply because of the Gel'fand transform, but this may be a misplaced suspicion. – Eric A. Bunch Apr 21 2010 at 16:13 show 10 more comments ## 1 Answer I found a reference for a necessary property that might be called algebraic. Browder proved a theorem relating the number of generators of a complex commutative Banach algebra to the Čech cohomology with complex coefficients of the maximal ideal space, and as a corollary concluded that if $M$ is a compact orientable $n$-dimensional manifold, then $C(M)$ cannot be generated as a Banach algebra by fewer than $n+1$ elements. The paper is very short, but for an even shorter summary here's the MR review. Just some comments, added later: One obtains the compact Hausdorff space $X$ (up to homeomorphism) from $C(X)$ by considering the maximal ideal space of $C(X)$ with Gelfand topology, but clearly you want something less tautological than "the maximal ideal space is a manifold." A small step in this direction would be to try to formulate the topological properties of $X$ in terms of the closed ideals of $C(X)$. As alluded to in Qiaochu's comment, there is an analogue of Nullstellensatz: each closed ideal in $C(X)$ consists of all functions vanishing on a (uniquely determined) closed subset of $X$. So for example, the locally Euclidean property could be reformulated for a commutative C*-algebra $A$ as follows: There is an $n$ such that for every maximal ideal $M$ of $A$ there is a closed ideal $I$ of $A$ such that $I$ is not contained in $M$ and $I$ is $*$-isomorphic to $C_0(\mathbb{R}^n)$. Second countability of the maximal ideal space is equivalent to $A$ being separable in the norm topology; that's not algebraic, but might be considered more intrinsic to the C*-algebra. But this only leads to another, more specific question: Is there a useful or interesting (C*-)algebraic characterization of $C_0(\mathbb{R}^n)$? - If I remember rightly from having skimmed over that paper en passant a few years ago, an important point is that one gets $n$ rather than $2n$? Or have I misremembered? – Yemon Choi Apr 22 2010 at 18:02 Browder emphasizes that if we were talking about real-valued generators then the corresponding result would be trivial, but for some compact Hausdorff spaces fewer generators are required in the complex case. So the point seems to be that the lower bound of $n+1$ had previously not been known, even in the case of $S^2$ as Browder (and Rudin in the review) point out. I'm not sure exactly what you mean about $2n$, but I might just be slow. – Jonas Meyer Apr 22 2010 at 18:44 Oh, I'm sure you're not being slow and I'm just going prematurely senile;) What you describe in your comment sounds familiar -- I just remembered that there was something to do with posssibly, a priori, having different answers in the real and complex cases. – Yemon Choi Apr 22 2010 at 21:50 @Jonas: Thanks for the paper, it was interesting. While it fits my requirements of an algebraic property (well, C^*-algebraic, but we have no other choice in that), I think I'm going to wait and see what other interesting things crop up :). – Eric A. Bunch Apr 23 2010 at 1:41 @Eric: For what it's worth, it's better than C*-algebraic; it's Banach algebraic. E.g., $C(S^1)$ is singly generated as a C*-algebra, but requires 2 generators as a Banach algebra. I was tentative about posting, hence the "might be called algebraic" and the community wiki, but I'm glad you find it interesting. – Jonas Meyer Apr 23 2010 at 4:48 show 4 more comments
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http://mathoverflow.net/questions/63580/is-a-double-centralizer-type-theorem-encountered-in-semisimple-algebras-true
## Is a double centralizer type theorem ( encountered in semisimple algebras) true for algebraic groups ? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let, $G(k^{al})$ be an algebraic group, over an algebraically closed field, and $\Gamma_{G}$ is the set of all closed subgroups of $G(k^{al})$. Then is the map $Z_{G}: \Gamma_{G} \rightarrow \Gamma_{G}$ which takes a closed subgroup to its centralizer in $G$, an involution? (probably not true) If we now assume that $G(k^{al})$ is reductive or semisimple is there a characterization of all such closed subgroups for which $Z_{G}$ is an involution? More generally if $G_{k}$ is an algebraic group scheme (now $k$ is no longer algebraically closed ) and $\Gamma_{G}$ is the set of closed group subschemes of $G_{k}$, do the previous two questions have a meaningful answer? - ## 2 Answers When `$G=\mathrm{GL}_n$`, then the centraliser $C$ of a subgroup scheme $H$ form the invertible elements of the algebra $M$ of matrices commuting with $H$. The group of such elements is Zariski dense in $M$ so $C$ and $M$ determine each other. Hence, the image of $Z_{\mathrm{GL}_n}$ and the question of involutivity on that image is completely reduced to double centraliser results for algebras. - ### 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. Not necessarily. For cases when this does happen see the question 28354 -
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http://math.stackexchange.com/questions/15707/for-what-functions-fx-is-fxfy-convex
# For what functions $f(x)$ is $f(x)f(y)$ convex? For which functions $f\colon [0,1] \to [0,1]$ is the function $g(x,y)=f(x)f(y)$ convex over $(x,y) \in [0,1]\times [0,1]$ ? Is there a nice characterization of such functions $f$? The obvious examples are exponentials of the form $e^{ax+b}$ and their convex combinations. Anything else? EDIT: This is a simple observation summarizing the status of this question so far. The class of such functions f includes all log-convex functions, and is included in the class of convex functions. So now, the question becomes: are there any functions $f$ that are not log-convex yet $g(x,y)=f(x)f(y)$ is convex? EDIT: Jonas Meyer observed that, by setting $x=y$, the determinant of the hessian of $g(x,y)$ is positive if and only if $f$ is a log-convex. This resolves the problem for twice continuously differentiable $f$. Namely: if $f$ is $C^2$, then $g(x,y)$ is convex if and only if $f$ is log-convex. - 1 In the $C^2$ case you can use the criterion of positivity of the Hessian, which leads you to look for nonnegative convex $f$ such that $f'(x)^2f'(y)^2\leq f''(x)f(x)f''(y)f(y)$ for all $x$ and $y$. – Jonas Meyer Dec 28 '10 at 3:06 3 log-convexity might be a helpful search term. – user1709 Dec 28 '10 at 8:37 Good point Slowsolver. This gives more examples: log convexity of $f(x)$ is a sufficient condition for $g(x,y)$ to be convex. Now, is it also necessary? – shaddin Dec 28 '10 at 20:45 2 @user5159: In the $C^2$ case it is necessary, as can be seen by letting $x=y$ in the inequality in my previous comment, yielding $f'^2\leq f''f$, which is equivalent to $(\log f)''\geq 0$. – Jonas Meyer Dec 30 '10 at 9:03 @Jonas Meyer Nice! As far as I'm concerned this resolves the question -- I'm only really interested in $C^2$ functions. Moreover, I suspect something similar is probably true in general. Feel free to post an answer and I'll vote / accept it. – shaddin Dec 31 '10 at 0:05 ## 2 Answers Suppose $f$ is $C^2$. First of all, because $g$ is convex in each variable, it follows that $f$ is convex, and hence $f''\geq0$. I did not initially have Slowsolver's insight that log convexity would be a criterion to look for, but naïvely checking for positivity of the Hessian of $g$ leads to the inequalities $$f''(x)f(y)+f(x)f''(y)\geq0$$ and $$f'(x)^2f'(y)^2\leq f''(x)f(x)f''(y)f(y)$$ for all $x$ and $y$, coming from the fact that a real symmetric $2$-by-$2$ matrix is positive semidefinite if and only if its trace and determinant are nonnegative. The first inequality follows from nonnegativity of $f$ and $f''$. The second inequality is equivalent to $f'^2\leq f''f$. To see the equivalence in one direction, just set $x=y$ and take square roots; in the other direction, multiply the inequalities at $x$ and $y$. Since $\log(f)''=\frac{f''f-f'^2}{f^2}$, this condition is equivalent to $\log(f)''\geq0$, meaning that $\log(f)$ is convex. - This may just be a silly observation. But, let $\alpha=(a_1,a_2)$ and $\beta=(b_1,b_2)$. If $f$ is convex and the sum of $g$ at the four corners of the square formed by $\alpha$ and $\beta$ is non-positive, then $g(t\alpha + (1-t)\beta) \leq tg(\alpha)+(1-t)g(\beta)$, where $t \in [0,1]$ -
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http://mathhelpforum.com/advanced-algebra/135480-operation.html
# Thread: 1. ## Operation Hi, $_* \$is a binary operation in E, C is a set defined as: $C={ y \in E/ (\forall x \in E): y_*x=x_*y }$ I must determine the specifications of $(C,_*)$: associativity, commutativity, neutral element, inverse element... I found that $_*$ is commutative but for the others i don't know how to do it. 2. Originally Posted by bhitroofen01 Hi, $_* \$is a binary operation in E, C is a set defined as: $C={ y \in E/ (\forall x \in E): y_*x=x_*y }$ I must determine the specifications of $(C,_*)$: associativity, commutativity, neutral element, inverse element... I found that $_*$ is commutative but for the others i don't know how to do it. If E has an identity element (is that what you mean by "neutral element"?) then that identity element commutes with everything in E. If * is associative on E, then it's clearly associative on any subset of E. If y is in C and x is in E, then $y^{-1}x=(x^{-1}y)^{-1}=(yx^{-1})^{-1}=xy^{-1}$.
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http://mathhelpforum.com/calculus/135796-find-centroid.html
Thread: 1. Find the centroid $x=y^2, y=x-2$ I have the formulas $<br /> \bar{x}=\frac{1}{A}\int_{a}^b x[f(x)-g(x)]\,dx<br />$ and $<br /> \bar{y}=\frac{1}{A}\int_{a}^b \frac{1}{2}[(f(x))^{2}-(g(x))^{2}]\,dx$ I found A= 9/2 Now I do not know how to write $x=y^2$ as f(x) function, since $x=y^2$would be $f(x) = y = +-\sqrt{x}$ $g(x) = x-2$ Thanks for any help. 2. suggestion Originally Posted by DBA $x=y^2, y=x-2$ I have the formulas $<br /> \bar{x}=\frac{1}{A}\int_{a}^b x[f(x)-g(x)]\,dx<br />$ and $<br /> \bar{y}=\frac{1}{A}\int_{a}^b \frac{1}{2}[(f(x))^{2}-(g(x))^{2}]\,dx$ I found A= 9/2 Now I do not know how to write $x=y^2$ as f(x) function, since $x=y^2$would be $f(x) = y = +-\sqrt{x}$ $g(x) = x-2$ Thanks for any help. i would substitute the obtained value and calculate for both the positive and the negative values. 3. Sorry, but I do not understand what you mean. Calculate two times $\bar {x}$ and two times $\bar {y}$ ? How do I end up with one centroid point then? 4. Originally Posted by DBA Sorry, but I do not understand what you mean. Calculate two times $\bar {x}$ and two times $\bar {y}$ ? How do I end up with one centroid point then? i drew the graph. one of the points lie on the complex plane . your triangle has the following points (0,0),(2,sqrt(2)),(2,i.sqrt(2)). so you cannot find the centroid on the real plane. or if you want to find out then some imaginary part will stay behind. 5. Thanks for your answer. We did not have this type of questions with imaginary numbers. So, I am not familiar how graph this. Is there a possibility to express the formulas for $\bar{x}$ and $\bar{y}$ in terms of y (so instead of dx, we would use dy)?
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http://math.stackexchange.com/questions/281413/effective-well-ordering-of-reals
# Effective Well ordering of reals Is there an effective (constructive) well order on reals ? I know several questions were already asked on this topic, and the answers were very good to this well known problem. My question is more specific : Is there some axiom $A$ ($A$ is equivalent to "there exists some large cardinal with such properties...") such that one can effectively build a well order on reals in ZFC+$A$ ? I think there is such $A$ but I did not find any articles or relevant links about it. So if someone can give me a link OR assure me I'm wrong about it and nothing was published about that in the last years OR assure me I'm totally wrong and nothing like that is possible, and I'm a guy who must stop maths asap. Thanks ! - I think this answer is 'No', since ZFC+'there is no definable definable well-ordering of the reals' is consistent if ZFC is consistent. – tetori Jan 18 at 14:05 Yeah, I deleted my comment. But saying there is a consistent axiom that says "there is no definable well ordering" is not enough (no ?) to say there is no consistent axiom that says "there is one way to define..." – Xoff Jan 18 at 14:11 2 – Amit Kumar Gupta Jan 18 at 14:14 No, but it does mean that, in ZFC, there is no definable function which is provably a well-ordering of the reals. @Xoff – Thomas Andrews Jan 18 at 14:23 Thanks for the math overflow link, this is the answer I needed, and there are links to nice articles. Thank you ! – Xoff Jan 18 at 14:29 ## 1 Answer I'm glad you found the links to the MO question useful. Let me add a couple of complementary remarks (that I should probably add to the answer over there at some point): I. Recently, Woodin has been studying an axiom he calls "Ultimate $L$", see for example this MO question, and the links I provide there. The intention of the axiom is to provide a framework that should be consistent with all known large cardinals, and implies that the universe admits a fine structure akin to that of $L$ or, more precisely, to that of the core model. It is a consequence of the axiom (appropriately formulated) that $\mathsf{GCH}$ holds (in a strong sense, we even have definability). In fact, one can provide an upper bound for the complexity of a well-ordering of the reals, using a predicate for the universally Baire sets. The presence of this predicate can be explained, loosely speaking, as follows: The axiom postulates that the universe is approximated by "mice", and a well-ordering of reals can be traced back to the existence of strategies for successfully comparing these mice, but these strategies can be coded via universally Baire sets. One is left with the question of whether this is a "reasonable" axiom to add to $\mathsf{ZFC}$. In fact, one can think of it as a "completion." The point is that we should be able to show the consistency of any "natural" theory by showing it holds in some appropriate inner model of a forcing extension of a rank initial segment of $V$. This is expected to be a consequence of an appropriate version of the so-called $\Omega$-conjecture, see for example this paper by Bagaria-Castells-Larson for an introduction. The philosophy advanced here is that all we should care about is the interpretability power of a (set) theory. Two theories that are bi-interpretable are considered just as good, and preferring one over another is a matter of aesthetics more than anything else. II. Curiously, I didn't mention in the forcing-axioms part of my answer to the question first linked above that there is some expectation that Martin's maximum, $\mathsf{MM}$, or at least a natural strengthening, should provide us with a (lightface) definable well-ordering of the reals(!). There's been a steady amount of work towards settling this question, by several first rate set theorists, but we are not there yet. - I'm not a first rate set theorist, nor even second or third (omegath perhaps). But thank you for your answer, it is very useful to me :) – Xoff Jan 18 at 20:05
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http://physics.stackexchange.com/questions/38077/quark-compositions-in-pi-to-pi0-pion-decay?answertab=votes
# Quark compositions in $\pi^+$ to $\pi^0$ pion decay Pions can undergo a rare beta-like decay into leptons: Pion beta decay (with probability of about $10^{−8}$) into a neutral pion plus an electron and electron antineutrino (or for positive pions, a neutral pion, positron, and electron neutrino). • Why is the quark composition of the neutral pion is so different to the charged pion after pion decay? $$\pi^{+}(\overline u,d) \to \pi^{o}(\frac {\overline u,u-\overline d, d}{\sqrt 2})+e^{+}+ \nu_e$$ $$\pi^{-}(\overline d,u) \to \pi^{o}(\frac {\overline u,u-\overline d, d}{\sqrt 2})+e^{-}+\overline \nu_e$$ • Why is the quark composition of the neutral pion is so different with neutral hadron (like neutron)? - ## 2 Answers I think I know where your misconception lies-- you appear to think that the individual quarks are the real thing, so you wonder why the $\pi^{+}$, made up of ''pure'' states of $|{\bar u}\rangle$ and $|d\rangle$, could decay into a ''mixed'' state. The problem here is that the physical thing is not the quarks, themselves, but the quark field. And this field is spanned by four basis states (so long as you don't allow higher generations of quarks): $|{\bar u}\rangle$, $|u\rangle$, $|{\bar d}\rangle$, and $|d\rangle$. Since we know that free states cannot carry color charge, and we are interested in the lightest possible strongly interacting particles, we are restricted to build states out of quark/antiquark pairs. It turns out that the three low energy eigenstates to the hamiltonian are $|u{\bar d}\rangle$, $|d{\bar u}\rangle$ and $\frac{1}{\sqrt{2}}\left(|u{\bar u}\rangle - |d{\bar d}\rangle\right)$, but any other linear combination of quark/antiquark states would be, in the abstract, an eqally good value for the quark field. - Because your formulae for the pion wave functions are written in a confusing way. The quark composition of the positively charged pion is $$|\pi^+\rangle = |u\bar d\rangle$$ while the neutral pion is $$|\pi^0\rangle = \frac{|u\bar u\rangle - |d\bar d\rangle}{\sqrt{2}}.$$ These two composite formulae are completely analogous: these pions are mesons, i.e. quark-antiquark bound states, and they just differ in which quarks from the double $(u,d)$ are used. The neutral pion is a superposition of two pieces: superpositions of these types are omnipresent everywhere in quantum theory. The charged pions only contain one term because there is only one quark-antiquark combination involving $u,d$ quarks and antiquarks whose total charge is $\pm 1$. To get $Q=0$, one may either combine $u\bar u$ or $d\bar d$ and the relative complex coefficient between these two terms is a priori arbitrary and determines whether the superposition is a mass eigenstate and if it is, whether it's lighter or heavier. Of course that the detailed formulae for the two pions must be different in some respects, otherwise they would be an identical particle. During the decay, all the conservation laws and other laws of physics are satisfied. - why baryon beta decay does not works in this way – Neo Sep 23 '12 at 9:29 1 Which way? The fundamental processes behind beta decays are always the same. The wave functions of the decaying particles and final particles are different because they're different particles. But it's always the case that the beta decay transforms a $d$ quark to $u e^+ \nu$ or some rearrangement from the left to right hand side and/or some replacement of particles by antiparticles. Both quarks in the charged pion may get transformed by this elementary beta process at the level of quarks and leptons which is why the superposition is possible on the right hand side. – Luboš Motl Sep 23 '12 at 10:05 Why is the quark composition of the neutral pion is so different with neutral hadron – Neo Sep 23 '12 at 10:10 2 It is not "so different". The neutral pion is a bound state of a light quark and its antiparticle; neutron is a $udd$ bound state. They are different composite particles which means that they use different combinations of the elementary particles but all of the light hadrons use $u,d$ quarks and antiquarks. – Luboš Motl Sep 23 '12 at 15:33 1 Dear FrankH, because the final state is a superposition, there are actually two diagrams that contribute to the decay amplitude. In one of them, for $\pi^+$ decay, $\bar u$ is just running horizontally from the initial to final state while $d$ converts to a $u$ by emitting a virtual $W$ which later decays to $e^+$ and $\nu$ by another vertex. In the other diagram, producing the second term, $d$ runs from the initial to final state while $\bar u$ becomes $\bar d$ by emitting a virtual $W$ which also splits to $e^+\nu$ on the right side. – Luboš Motl Sep 23 '12 at 17:02 show 2 more comments
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http://math.stackexchange.com/questions/221153/find-the-directions-in-which-the-directional-derivative-has-the-value-1
# Find the directions in which the directional derivative has the value 1 Can anyone show me how to adjust my work below so that it is a correct answer? This is question number 14.6.28 in the 7th edition of Stewart Calculus. Find the directions in which the directional derivative of $f(x,y)=ye^{-xy}$ at the point $(0,2)$ has the value 1. My work is: $\nabla f(x,y) = <-y^2 e^{-xy}, e^{-xy}(1-xy)>$ $\nabla f(0,2) = <-4,1>$ $|\nabla f(0,2)| = \sqrt{17}$ $D_v f(x,y)=|\nabla f|\cos{\theta}=\sqrt{17}\cos{\theta}$ $\sqrt{17} \cos{\theta}=1$ when $\cos{\theta}={{\sqrt{17}}/17}$ $\theta =\arccos{{\sqrt{17}}/17}\approx +1.326$ and $-1.326$ EDIT: I tried the following, based on suggestions below: $\vec{u}=<\cos{\theta},\sin{\theta}>$ $D_u f(x,y)=\nabla f(x,y)\cdot \vec{u}$ $D_u f(0,2)=\nabla f(0,2)\cdot \vec{u}=-4\cos{\theta}+\sin{\theta}=1$ I then plugged this into a spreadsheet and found that $-4\cos{\theta}+\sin{\theta}=1$ when $\theta = \pi/2 , 5\pi/2 , 9\pi/2 , ...$ and when $\theta = 4\pi/3 , 10\pi/3 , 16\pi/3 , ...$ Can anyone check the correctness of this approach? Also, I found this result experimentally. If it is correct, I would rather be able to find it using calculus. - 1 Directional derivative is "del f dot u" where u is a unit vector in the given direction. If $u=(x,y)$ then in your case the "del f dot u" is -4x+y, which you set to 1, along with $x^2+y^2=1$ to make u a unit vector. I don't follow your method... – coffeemath Oct 25 '12 at 22:14 @TMS: check your math. – robjohn♦ Oct 25 '12 at 22:21 @coffeemath Can you elaborate? I am trying to work on this now. If you post it as an answer, and if I can follow/check it, I will mark it as the answer. – CodeMed Oct 26 '12 at 20:46 Just entered an "answer"-- turns out there are two different directions! – coffeemath Oct 27 '12 at 2:14 But $-4\cos(4\pi/3)+\sin(4\pi/3)=2-\sqrt{3}/2$ which is not 1. – coffeemath Oct 27 '12 at 9:50 ## 2 Answers The directional derivative is $\nabla f \bullet u$, where $u$ is a unit vector which points in the direction desired. What you want is the unit vector $u=(x,y)$; your del $f$ is $(-4,1)$ as you say, and then $\nabla f \bullet u$ is simply $-4x+1y$. Since it should be 1 you know that $-4x+y=1$, i.e. $y=1+4x$. Since $(x,y)$ is a unit vector you also know that $x^2+y^2=1$. So plugging in we have $x^2+(1+4x)^2=1$, which when you move the 1 over and expand gives the equation $17x^2+8x=0$. This factors as $x(17x+8)=0$. So either $x=0$ or else $x=-8/17$. Then plugging these into $y=1+4x$ gives the two unit vectors $(0,1)$ and $(-8/17,-15/17)$. [note the question said "find the directions" rather than "direction"; I think in general for a desired value of the directional derivative strictly between the gradient and the negative of the gradient, one usually has two directions. If you're on a hill not pointing straight up, and you find one way to walk so you're going up at a lesser rate than straight up, there should be another such direction...] - Thanks. +1 for giving a working answer. Sorry for the delay in response, but it took me this long to have the time to sit down to really think through what you wrote. Eric Angle also seems to have given the start of an answer. In his case, x and y are both functions of $\theta$. I just did not know how to get from $<\cos{\theta}, \sin{\theta}>$ to the answer, so I am marking yours as the answer. In theory, we should be able to use both methods to check each other. – CodeMed Oct 28 '12 at 19:48 Actually, from $cos^2\theta+\sin^2\theta=1$, the other approach is equivalent. However I don't myself immediately see how the formula for the cosine of the sum could help in this case. The two angles $\pi/2$ and $\arctan(15/8)+\pi$ [one is in quadrant 3] don't seem related enough for addition trig formulas to be of use... – coffeemath Oct 28 '12 at 20:15 Let $u$ be the unit vector along which you would like to take the directional derivative of $f$. With $\theta$ the angle that $u$ makes with the $x$ axis, $$u = \left(\cos\theta,\sin\theta\right).$$ You can check that $u$ is a unit vector. Now, the directional derivative of $f$ in the direction of $u$, at the point $\left(0,2\right)$, is $$\nabla f \Big|_{\left(0,2\right)} \bullet u = \left(-4,1\right) \bullet \left(\cos\theta,\sin\theta\right) = -4 \cos \theta + \sin \theta.$$ The problem states that this should be 1, so $$-4 \cos \theta + \sin \theta = 1.$$ Now it remains to solve for $\theta$. You can probably do this with $$- \cos\left(a+b\right) = - \cos a \cos b + \sin a \sin b.$$ - Can you please elaborate? I am not sure that I understand. – CodeMed Oct 26 '12 at 0:09 I tried your approach in a revision to my post above. Not sure if it is correct. Can you elaborate with something that I can verify? – CodeMed Oct 26 '12 at 22:21 What do I use for a and b? – CodeMed Oct 26 '12 at 23:23
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