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H: basis functions do not lie in the space they form For example, any continuous function in $\mathbb{L}^2(-\infty,\infty)$ space can be expanded by delta functions $\delta(x-a)$ or Fourier basis $e^{ikx}$. However, the basis functions, both $\delta(x-a)$ and $e^{ikx}$, are not square-integrable, thus not in $\mathbb{L}^2(-\infty,\infty)$ space. It's hard to understand there is a space whose has basis that are not in the space. Is there an intuitive explanation? AI: I think there is a confusion between two similar-sounding statements: every element of space $X$ can be written in terms of functions in the set $S$ the set $S$ spans $X$ The second statement has a precise meaning, and it does require $S$ to be a subset of $X$. The meaning of the first is determined from context: "written in terms of" means whatever the writer wants it to mean. Here is a simple example without exotic things like delta functions: Fact. Every function $f\in C[0,1]$ is the sum of a uniformly convergent series $f=\sum_{k=1}^\infty c_k g_k$, where $g_1,g_2,\dots$ are characteristic functions of the dyadic intervals $[(j-1)/2^m,j/2^m]$ enumerated in some way. Note that the topology of $C[0,1]$ is indeed the topology of uniform convergence, so the infinite sum is almost like an expansion of $f$ with respect to a spanning set of $C[0,1]$. But it's not, precisely because $g_k$ do not belong to $C[0,1]$. Not being contained in $C[0,1]$, the set $\{g_k\}$ does not satisfy the definition of a spanning set or of a basis.
H: number of possible matrix entries swaps Let us say that there is a $n \times n$ matrix with entries defined/given as some natural number. (duplicates are fine.) What would be the number of possible matrices that result from swapping/switching the entries in the original matrix that respects the entries in each row and column (order of rows and columns does not matter.)? (By respecting entries in each row and column, I mean that if there were some defined numbers in each row, the matrix must contain these numbers in some row.) AI: You would be selecting a permutation of columns and a permutation of the rows. The largest possible number is $(n!)^2$, if all entries are distinct: choose a permutation of the rows, $n!$; the resulting matrix satisfies the conditions. Then choose a permutation of the columns, $n!$ possibilities, and the matrix still satisfies the conditions. If there are repeated entries, then it depends on exactly how you have repeats. For example, if $n=2$, you get $4$ different matrices: $$\left(\begin{array}{cc} a&b\\c&d\end{array}\right),\quad\left(\begin{array}{cc}b&a\\d&c\end{array}\right),\quad \left(\begin{array}{cc}c&d\\a&b\end{array}\right),\quad\left(\begin{array}{cc}d&c\\b&a\end{array}\right).$$ If $n=3$, you have $36$ possibilities ( which I will not write in extenso); here is a way of verifying this by counting in a different way: if the original matrix is $$\left(\begin{array}{ccc} a&b&c\\ d&e&f\\ g&h&k \end{array}\right)$$ then you have $9$ possible positions for $a$; once $a$ is selected, you get two possibilities for placing $b$ in the same row, and two for placing $d$ in the same row. That determines uniquely the positions of $c$ and $g$; Once $c$ and $d$ are determined, that uniquely determines the position of $f$ (must be in the same row as $d$, same column as $c$), hence of $e$, and likewise the positions of $g$, $h$, and $k$ are now fixed. So in total you get $9\times2\times2=36 = 6\times 6 = 3!\times 3! = (3!)^2$ ways of doing it.
H: Introductory Treatment of Differential Geometry I'm going to be taking a graduate course in differential geometry, this coming fall, but I am not prepared for it. Can anyone recommend a good introductory treatment of the background materials? The list of topics in the course is: Manifolds, Local Study of Manifolds, Vector bundles, Submanifolds, Vector Fields, Lie Groups (brief treatment), Differential forms, Orientation and Integration, Statement of the Hodge Theorem, The Kähler condition My calculus background (particularly advanced calculus) is not strong. I had a three semester coverage of calculus (the typical Calc 1, Calc 2, and Calc 3) and this was almost a decade ago. Since then my experience has been almost exclusively with pure math -- applied math courses always made me uncomfortable. I have about a month to prepare for this course so I'd like to make as much of this time as possible; and arbitrarily choosing books on the topic is a great way to waste time I've found. AI: Spivak's Calculus on Manifolds is a great little book.
H: Solve $ \left( \log_3 x \right)^2 + \log_3 (x^2) + 1 = 0$ I'm new to logarithms and I am having trouble solving this equation $$ \left( \log_3 x \right)^2 + \log_3 (x^2) + 1 = 0.$$ How would I solve this? A step-by-step response would be appreciated. Also, I know how to solve it with assigning $\log_3 (x)$ as $x$ and solving $x^2 + x + 1 = 0$, and getting the answer from there. I am looking to see how I would do it with just logarithms and no quadratics. Thanks AI: Hint: The second term is equal to $2\log_3(x)$. Let $w=\log_3(x)$. We are looking at the equation $w^2+2w+1=0$.
H: How to solve this logarithm system? I am new to logarithms and I am having trouble with this logarithm system. \begin{align} \log_9(x) + \log_y(8) & = 2, \\ \log_x(9) + \log_8(y) & = 8/3. \end{align} A step-by-step procedure would be highly appreciated. Thanks in advance. AI: Use the fact that $$\log_b(a) = \dfrac1{\log_a(b)}$$ Hence, if we denote $\log_9(x) = a$ and $\log_y(8) = b$, we get that \begin{align} a+b & = 2\\ \dfrac1a + \dfrac1b & = \dfrac83 \implies \dfrac{a+b}{ab} = \dfrac83 \implies ab = \dfrac34 \end{align} Now solve for $a$ and $b$ and hence $x$ and $y$.
H: Roots of a polynomial mod $n$ Let $n=n_1n_2\ldots n_k$ where $n_i$ are pairwise relatively prime. Prove for any polynomial $f$ the number of roots of the equation $f(x)\equiv 0\pmod n$ is equal to the product of the number of roots of each of the equations $f(x)\equiv 0\pmod{n_1}$, $f(x)\equiv 0\pmod{n_2}$, $\ldots$, $f(x)\equiv 0\pmod{n_k}$. I encountered this problem while reading a chapter on RSA encryption. It has something to do with the Chinese remainder theorem but I can't see precisely how to use it. AI: The Chinese Remainder Theorem says that if $n_1,\ldots,n_k$ are pairwise coprime, and $a_1,\ldots,a_k$ are any integers, then there is a solution to the system of congruences $$\begin{align} x &\equiv a_1\pmod{n_1}\\ x&\equiv a_2\pmod{n_2}\\ &\cdots\\ x&\equiv a_k\pmod{n_k}, \end{align}$$ and moreover, the solution is unique modulo $n=n_1\cdots n_k$. We show that there is a bijection between the set of integers $$A=\{a\in\mathbb{Z}\mid 0\leq a\lt n, f(a)\equiv 0\pmod{n}\}$$ and the set $$B=\{(a_1,\ldots,a_k)\in\mathbb{Z}^k\mid 0\leq a_i\lt n_i, f(a_i)\equiv 0\pmod{n_i},\text{ for }i=1,\ldots,k\}.$$ Suppose first that $a$ is a solution to $f(x)\equiv 0\pmod{n}$. Since $n_i\|n$ for each $i$, then $f(a)\equiv 0\pmod{n}$ implies $f(a)\equiv 0\pmod{n_i}$. Hence, $(a\bmod n_1,a\bmod n_2,\ldots,a\bmod n_k)$ is a $k$-tuple of integers whose $i$th entry is a solution to $f(x)\equiv 0\pmod{n_i}$, and satisfies $0\leq a_i\lt n_i$. Conversely, suppose that $(a_1,\ldots,a_k)$ is a tuple such that $f(a_i)\equiv 0\pmod{n_i}$ for each $i$. Then, by the Chinese Remainder Theorem, there is a unique $a$, $0\leq a\lt n$, such that $a\equiv a_i\pmod{n_i}$ for each $i$. Therefore, $f(a)\equiv f(a_i)\equiv 0\pmod{n_i}$ for $i=1,\ldots,n$. Since the $n_i$ are pairwise coprime and $n_i\|f(a)$ for all $i$, then $n=n_1\cdots n_k\|f(a)$, so $f(a)\equiv 0\pmod{n}$. That is, each element of $A$ yields an element of $B$. It is now easy to verify that the maps we have defined $A\to B$ and $B\to A$ are inverses of each other (by the uniqueness clause of the Chinese Reainder Theorem), yielding the desired bijection: $\|A\|=\|B\|$. Note that $A$ is the set of all solutions to $f(x)\equiv0\pmod{n}$, whereas $B$ is the cartesian product of the sets of solutions of $f(x)\equiv 0\pmod{n_i}$; so $\|A\|$, the number of solutions to $f(x)\equiv 0\pmod{n}$, is the same as the product of the number of solutions of $f(x)\equiv 0\pmod{n_i}$ for $i=1,\ldots,k$.
H: What is the meaning of this analysis problem and give some hint please? What is the meaning of this analysis problem and give some hint please? This problem was founded on Analysis 1 by Herbert Amann and Joachim Escher on page 100. Determine the following subsets of $\Bbb R^2$ by drawing: $$A = \{(x,y) \in \Bbb R^2 : \|x-1\| + \|y+1\| \leq 1\},$$ $$B = \{(x,y) \in \Bbb R^2 : 2x^2+y^2>1, \|x\| \leq \|y\|\},$$ $$C = \{(x,y) \in \Bbb R^2 : x^2-y^2>1, x-2y<1, y-2x<1\}.$$ AI: For example C, there are three regions of the plane defined. To find each one, change the inequality to an equals sign and plot the graph, then decide which of the graph is the region of interest. The first is a hyperbola opening toward the $+x$ and $-x$ axes. You want the regions that do not contain the origin. The second is a line of slope $\frac 12$ going through $(1,0)$ and you want the half-plane above it. The third is a line of slope $2$ going through $(0,1)$ and you want the half-plane below it. If you plot all three on the same axes and shade the regions of interest, your result is the area shaded all three ways, the intersection of the regions of each inequality. Since the inequalities are strict, the dividing lines are not included.
H: Can the Coast Guard catch the thief, given speed and distance between them? The problem is: Given two boats, one, the coast-guard boat, stands on the point zero, the thief's boat, on the point $D$, the coast-guard boat travels in a speed of $Vg$ knots and the thief's boat, in $Vf$ knots, Given $D$, $Vf$ and $Vg$, say if it's possible to the coast guard boat reach the thiefs boat before the thief's boat reach $12$ miles away from the point $0$ I've started with an Inequalitie $0 + (iVg) <= D + (iVf)$ with $D + (i*Vf) < 12 miles$ being $i$ a moment. but know I'm stuck on how to continue, and how to get a formula to solve it. Thanks EDIT: Sorry for the inconvenience, I've just misunderstood the statement, they aren't perpenticular to the cost when they leave, imagine as a cartesian plane, the coast-guard boat is in the point $(0,0)$ and the thief, in the point $(d, 0)$ and the objective of the thief is reach the point $(d, 12)$ before the coast guard boat reach him, it gives an all new view to to problem, that's why my example in the comments isn't wrong.. Sorry for the incovenience. AI: Presumably the thief will go directly away from the origin, so it takes $\frac {12-D}{Vf}$ hours to reach the point $12$ miles away. In that time the Coast Guard boat can go $\frac {(12-D)Vg}{Vf}$ miles. If this is at least $12$, it will catch the thief. So your condition is $\frac {(12-D)Vg}{Vf} \ge 12$. The question did not seem to ask when the Coast Guard catches the thief, but if you want the time $i$, you are close in that it comes when $iVg=D+iVf$, which gives $i=\frac D{Vg-Vf}$
H: Proof if $a \vec v = 0$ then $a = 0$ or $\vec v = 0$ I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right: for a vector space $V$ with an underlying field $F$: Take an element $a$ from $F$ and $\vec{v}$ from $V$. $a\vec{v}=\vec{0}\implies a=0 $ or $ \vec{v}=\vec{0}$ After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive $a\neq 0 \wedge \vec{v} \neq \vec{0} \implies a\vec{v}\neq \vec{0}$ Say $a\vec{v}=\vec{u}$. Since $a$ is non-zero, we can divide both sides by $a$. $$\vec{v}=\frac 1 a \vec{u}$$ If $\vec{u}$ were $\vec{0}$ then by $$\frac 1 a \vec{0}=\frac 1 a (\vec{0}+\vec{0})\implies\frac 1 a \vec{0}=\vec{0}$$ $v$ would be $0$ as well. Since it isn't by assumption, $\frac 1 a \vec{u}$ cannot be zero and so $\vec{u}$ cannot be as well. Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of $\frac 1 a \vec{u}\neq 0 \implies \vec{u}\neq 0$ doesn't seem obvious. I think I need to use the $1\vec{v}=\vec{v}$ axiom, but I'm not sure how. Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple. AI: Let $a\in F,\vec v\in V$ and suppose $a\vec v=0$. If $a\neq 0$, then $a^{-1}\in F$ so $$\vec v=(a^{-1}a)\vec v=a^{-1}(a\vec v)=a^{-1}\vec 0=\vec 0$$ thus either $a=0$ or $\vec v=0$.
H: A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing. I would like to prove the statement in the title. Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x < y$. And that's pretty much how far I got. Help will be appreciated. AI: Prove the contrapositive instead: if $f$ is not strictly increasing and not strictly decreasing, then it is not one-to-one. For example, say there are points $a\lt b\lt c$ such that $f(a)\lt f(b)$ and $f(b)\gt f(c)$. Either $f(a)=f(c)$ (in which case $f$ is not one-to-one), or $f(a)\lt f(c)$, or $f(c)\lt f(a)$. If $f(a)\lt f(c)\lt f(b)$, then by the Intermediate Value Theorem there exists $d\in (a,b)$ such that $f(d)=f(c)$; hence $f$ is not one-to-one. Now, there are other possibilities (I made certain assumptions along the way, and you should check what the alternatives are if they are not met).
H: Covariance question: A non squared matrix possible? I have an academic economic paper that says the following: $$q_r = \operatorname{Cov}(rx,v')\lambda$$ $$(14 \times 1)=(14 \times 4)(4 \times 1)$$ My vector $q_r$ is of size $14 \times 1$, my matrix $rx$ is size $T \times 14$, and my matrix $v$ is of size $T \times 4$. I don't have the value of $\lambda$. How can I get a covariance matrix of size 14 $\times$ 4? AI: The $(i,j)$ entry of the matrix is the covariance of the $i$'th column of $rx$ and the $j$'th column of $v$.
H: Homework question regarding inverse function with a cosine I'm given $f(x)= 3-4\cos(x-2)$ I've gotten to $(-x+3)/4 = \cos(2)\cos(y)+\sin(s)\sin(y)$ But I can't get the $y$ out to create an inverse ... AI: Don't expand the $\cos(x-2)$. Solve the whole thing as you normally would, to get $x-2 = \cos^{-1}($something involving $f(x))$, and then just add 2 to both sides. Spoiler: $$\begin{eqnarray} f(x) & = & 3 - 4 \cos(x-2) \\ \frac14(3-f(x)) & = & \cos(x-2) \\ \cos^{-1}\left(\frac14(3-f(x))\right) &=& x-2 \\ 2+\cos^{-1}\left(\frac14(3-f(x))\right) &=& x\end{eqnarray}$$
H: What is the volume of this 3d shape? I'm wondering if there is an equation that represents the volume of an arbitrary 3d primitive matching this description: 1.) Point at center of sphere 2.) Each edge is the length of the radius 3.) 3 flat sides, 1 arc side Image: So it's kind of a sector of a sphere, but instead of a conical shape it's more of a tetrahedral shape, but with a curved end. AI: The volume of your figure is $\frac13rA$, where $r$ is the radius of the sphere, and $A$ is the area of the curved end, which is a spherical triangle. A simple formula for $A$ is $r^2(\alpha+\beta+\gamma-\pi)$, where $\alpha$, $\beta$, and $\gamma$ are the angles of the spherical triangle in radians, which are the same as the dihedral angles between the flat faces of your figure. Beyond that, the formula you get will depend on how the three faces or three edges of your figure are specified.
H: Proving Inequalities using Induction I'm pretty new to writing proofs. I've recently been trying to tackle proofs by induction. I'm having a hard time applying my knowledge of how induction works to other types of problems (divisibility, inequalities, etc). I've been checking out the other induction questions on this website, but they either move too fast or don't explain their reasoning behind their steps enough and I end up not being able to follow the logic. I do understand how to tackle a problem which involves a summation. This is the one I just did (the classic "little gauss" proof): Prove $1+2+3+\dots+n = n(n+1)/2$ I. Basis $1=(1+1)/2$ $1=1$ II. Induction Assume the expression holds for an arbitrary $n=k$ such that $1+2+3+\dots+k = k(k+1)/2$ Show that the expression holds for $n=k+1$ $1+2+3+...+n+k+(k+1) = k+1[(k+1)+1]/2$ And this is done mainly by observing that we already have a formula for 1 through k on the LHS, so the equation can be rewritten as $k(k+1)/2 + (k+1) = k+1[(k+1)+1]/2$ NOTE: I believe this is using the inductive hypothesis. Please correct me if I'm wrong. Anyway, finding common denominators on the left hand side and distributing on the right, you eventually show that it's true. This (so far) has worked for every proof I've attempted that involves a summation on the left hand side. Now, I start losing it when the format changes. For example, this inequality proof I'm trying to write. I'll post what I have here: $n^2 \ge 2n$ for all $n>1$ I. Basis $2^{2} \ge 2(2)$ $4 \ge 4$ II. Induction Assume the inequality holds for an arbitrary $n=k$, such that $k^2 \ge 2(k)$ Show that the expression holds for $n=k+1$ such that $(k+1)^2 \ge 2(k+1)$ This is where I get lost andI know I'm supposed to invoke the IH somewhere in the expression. But unlike the summation problem earlier, I'm not sure where to begin. Could anyone point me in the right direction? AI: Induction hypothesis is not $2^k\geq 2k$ but $k^2\geq 2k$. Then, for $P(k+1)$, we have to prove $(k+1)^2\geq 2(k+1)$. Proof: $(k+1)^2=k^2+2k+1$ but $k^2 \geq 2k$ (by IH) $\implies k^2+2k+1\geq (2k+2k+1=4k+1)\geq 2k+2$ as $k\geq 1\implies (k+1)^2\geq 2(k+1)$. Hence ,$P(k+1)$ is true whenever $P(k)$ is true. Since $P(1)$ is true $\implies P(n)$ is true $\forall n\in \Bbb Z^+$.
H: Given $(a_n)$ and $(b_n) \in \mathbb{R}$,if we have $\|a_n - b_n\| < \frac{1}{n} \forall n \in \mathbb{N}$ Given $(a_n)$ and $(b_n) \in \mathbb{R}$,if we have $\|a_n - b_n\| < \frac{1}{n} \forall n \in \mathbb{N}$, I think it is possible to find an $N$ such that $\forall n \ge N$, we have $\|a_n-b_n\| < \frac{\epsilon}{2}$. I think we can definitely find that $N$ where we pick $N= \frac{2}{\epsilon}$. Then we will have for $n>N$, we will have $N> \frac{2}{\epsilon} \Rightarrow \|a_n-b_n\| <\frac{1}{n}< \frac{1}{N}< \frac{\epsilon}{2}$ I think my argument is correct though, point out if there is any flaw. I need to use it as a lemma for other proof. AI: This has nothing to do with $a_n$ or $b_n$; you're just saying that you can always find an $N$ such that $\frac1n\lt\frac\epsilon2$ for all $n\gt N$. This is the Archimedean property of the real numbers.
H: Fourth roots Complex analysis I am trying to find the fourth roots of $8\sqrt2(1+i)$. So then, I was deciding to convert $1+i$ to an $re^{i\theta}$, where $r = \sqrt2$ and $\theta = 45^\circ$ or $\pi/4$. then; $z = \sqrt2e^{i\pi/4}$, but then we need the fourth roots so then, take $\sqrt2e^{i\pi/4}$ and raise it to the fourth power? AI: You noticed that $1+i=\sqrt{2}e^{\pi i/4}$. That implies that $$8\sqrt{2}(1+i)=8\cdot\sqrt{2}\cdot\sqrt{2}\cdot e^{\pi i/4}=16e^{\pi i/4}.$$ If $z=re^{i\theta}$, where $r>0$ and $\theta\in[0,2\pi)$, has the property that $z^4=8\sqrt{2}(1+i)$, then we must have that $$z^4=(re^{i\theta})^4=r^4e^{4i\theta}=16e^{\pi i/4}.$$ Because $r$ is a non-negative real number, we know that $r^4=16$ implies that $r=2$. However, solving for $\theta$ is slightly trickier because angles "wrap around" after $2\pi$. To be precise, $$e^{4i\theta}=e^{\pi i/4}$$ implies that $$4i\theta=\frac{\pi i}{4}+2\pi ik\text{ for some }k\in\mathbb{Z}$$ and therefore, $\theta\in[0,2\pi)$ is one of the four values $\frac{\pi }{16}+k\frac{\pi}{2}$ where $k\in\{0,1,2,3\}$.
H: How are these two equal? Which of these terms is greater ? $2x-6y+1$ or $1$ if $x^4 + 3y^2=0$ According to the text they are equal ?How is that ? AI: But since $x^2$ and $3y^2$ are positive, (assuming $x,y$ real) we have that $x^2+3y^2=0$ forces $x=y=0$ so the text is correct!
H: Uncountable disjoint union of $\mathbb{R}$ I'm doing 1.2 in Lee's Introduction to smooth manifolds: Prove that the disjoint union of uncountably many copies of $\mathbb{R}$ is not second countable. So first, let $I$ be the set over which we are unioning. Then I believe the disjoint union is just $\mathbb{R}\times I$. Then I believe that sets of the form $\cup_{x\in A}(x,i)$ is open if and only if $A\subset\mathbb{R}$ is open (I know the open sets are defined with the canonical injection though). At first I thought if I let $I=\mathbb{R}$, then $\mathbb{R}\times I=\mathbb{R^2}$, but now I am thinking maybe they just have the same elements, but the topology is different, and this is why $\mathbb{R}\times I$ is not second countable? AI: It will be easier if we talk about $(0,1)$ which is homeomorphic to $\mathbb R$, but its diameter is $1$. First note that the disjoint union is metrizable, by setting the distance between any two points coming from two different copies to be $2$. For metric spaces second countability implies separability. However a countable subset of the disjoint union cannot meet all the copies, only countably many of them. Therefore the disjoint union is not separable and thus not second countable.
H: Does there exist continuous map from $S^1$ to $\mathbb{R}$ such that $f(x)=f(y)$ for uncountably many $x,y$? Does there exist continuous map from $S^1$ to $\mathbb{R}$ such that $f(x)=f(y)$ for uncountably many $x,y\in S^1$? By the Borsuk-Ulam theorem, I know there is no injective map from $S^1\rightarrow \mathbb{R}^1$. If we consider some map like $g(x)=f(x)-f(-x)$ for $x\in S^1$, I can say by the intermediate value theorem that there is at least one point where $g$ vanishes; is that vanishing set uncountable? AI: Any constant map $f:\mathbb{S}^1\to\mathbb{R}$ is continuous and would satisfy $f(x)=f(y)$ for every $x,y\in\mathbb{S}^1$, of which there are certainly uncountably many. In fact, a continuous map $f$ that is only constant on a non-empty open set of $\mathbb{S}^1$ would have $f(x)=f(y)$ for uncountably many $x,y\in\mathbb{S}^1$, because any non-empty open set of $\mathbb{S}^1$ is uncountable.
H: RSA probabilistic decryption problem I encountered this problem in an algorithms book and could not see how multiplicative could be used as a probabilistic algorithm. Using the fact that RSA is multplicative: $$P_A(M_1) P_A(M_2)\equiv P_A(M_1M_2)\pmod n$$ if someone could efficiently decrypt 1 percent of messages from $Z_n$ encrypted with $P_A$, then he could employ a probabilistic algorithm to decrypt every message encrypted with $P_A$ with high probability AI: Ok, so the attacker has a way of calculating $M$ from $P_A(M)$, if $P_A(M)$ is in a set $S$ that covers about 1 per cent of the residue classes modulo $n$. The attacker, facing the task of calculating $M_1$, given $P_A(M_1)$ can then generate a few hundred random $(M_2,P_A(M_2))$ pairs. S/he can then check, whether $P_A(M_1)P_A(M_2)$ is in the set $S$ for any $M_2$. If that happens, the attacker will know $$M_1M_2=P_A^{-1}(P_A(M_1)P_A(M_2))$$ AND s/he will know $M_2$, so figuring out $M_1$ is then easy. If the choices for $M_2$ were truly random, the probabilities of failure with each $M_2$ are independent from each other, and all about $0.99$. So with, say, $200$ trials, the probability of failure is $0.99^{200}\approx e^{-2}$ or about $13$ per cent. Make four hundred attempts, if that is not good enough. Observe that there is no need to have a good description of the set $S$. The attacker can just attempt to decode any $P_A(M_1)P_A(M_2)$ that is generated.
H: Nth derivative of $\tan^m x$ $m$ is positive integer, $n$ is non-negative integer. $$f_n(x)=\frac {d^n}{dx^n} (\tan ^m(x))$$ $P_n(x)=f_n(\arctan(x))$ I would like to find the polynomials that are defined as above $P_0(x)=x^m$ $P_1(x)=mx^{m+1}+mx^{m-1}$ $P_2(x)=m(m+1)x^{m+2}+2m^2x^{m}+m(m-1)x^{m-2}$ $P_3(x)=(m^3+3m^2+2m)x^{m+3}+(3m^3+3m^2+2m)x^{m+1}+(3m^3-3m^2+2m)x^{m-1}+(m^3-3m^2+2m)x^{m-3}$ I wonder how to find general formula of $P_n(x)$? I also wish to know if any orthogonal relation can be found for that polynomials or not? Thanks for answers EDIT: I proved Robert Isreal's generating function. I would like to share it. $$ g(x,z) = \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) = \tan^m(x+z) $$ $$ \frac {d}{dz} (\tan^m(x+z))=m \tan^{m-1}(x+z)+m \tan^{m+1}(x+z)=m \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^{m-1}(x)+m \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^{m+1}(x)= \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} (m\tan^{m-1}(x)+m\tan^{m+1}(x))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} (\dfrac{d}{dx}(\tan^{m}(x)))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^{n+1}}{dx^{n+1}} (\tan^{m}(x))=\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^{n+1}}{dx^{n+1}} (\tan^{m}(x))$$ $$ \frac {d}{dz} ( \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) )= \sum_{n=1}^\infty \dfrac{z^{n-1}}{(n-1)!} \dfrac{d^n}{dx^n} \tan^m(x) = \sum_{n=1}^\infty \dfrac{z^{n-1}}{(n-1)!} \dfrac{d^n}{dx^n} \tan^m(x)=\sum_{k=0}^\infty \dfrac{z^{k}}{k!} \dfrac{d^{k+1}}{dx^{k+1}} \tan^m(x)$$ I also understood that it can be written for any function as shown below .(Thanks a lot to Robert Isreal) $$ \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} h^m(x) = h^m(x+z) $$ I also wrote $P_n(x)$ as the closed form shown below by using Robert Israel's answer. $$P_n(x)=\frac{n!}{2 \pi i}\int_0^{2 \pi i} e^{nz}\left(\dfrac{x+\tan(e^{-z})}{1-x \tan(e^{-z})}\right)^m dz$$ I do not know next step how to find if any orthogonal relation exist between the polynomials or not. Maybe second order differential equation can be found by using the relations above. Thanks for advice. AI: I don't know if this will help: The exponential generating function of $f_n(x)$ is $$ g(x,z) = \sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} \tan^m(x) = \tan^m(x+z) = \left(\dfrac{\tan(x)+\tan(z)}{1-\tan(x)\tan(z)}\right)^m $$ So the exponential generating function of $P_n(x)$ is $$ G(x,z) = g(\arctan(x),z) = \left(\dfrac{x+\tan(z)}{1-x \tan(z)}\right)^m $$
H: Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. First approach. $\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$ From this relationship, I get: $2\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$ Then: $\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{1}{2}\left[\frac{x}{1+x^2}+\arctan x\right]+C$ This is a recursive solution. Second approach. $x=\tan t$ in $t\in (- \pi/2, \pi/2)$, i.e. $t=\arctan x$, then $dx=(1+x^2) dt$. $\int \frac{1}{\left(1+x^2\right)^2}dx=\int \frac{1}{1+x^2}dt=\int \frac{\cos^2t}{\sin^2t+\cos^2t}dt=\int \cos^2t dt=\frac{1}{2}\int \left(1+\cos 2t \right) dt=\frac{t}{2}+\frac{1}{4}\sin 2t$ This result can be rewritten (using trigonometric formulas): $\frac{t}{2}+\frac{1}{4}\sin 2t=\frac{t}{2}+\frac{1}{2}\sin t \cos t$ From $\cos^2 t=\frac{1}{1+x^2}$, I have: $\|\cos t\|=\sqrt{\frac{1}{1+x^2}}$ but in $t\in (- \pi/2, \pi/2)$, $\|\cos t\|=\cos t$. So: $\cos t=\sqrt{\frac{1}{1+x^2}}$. Now I have a problem: $\|\sin t\|=\sqrt{\frac{1}{1+x^2}}$, but $\|\sin t\|\neq \sin t$ for $t\in (- \pi/2, \pi/2)$. Any suggestions, please? This integral can be solved in other ways? Thanks. AI: One way to make the trigonometric solution substitution solution end nicely, without worries about signs, is to note that $$\sin t \cos t =\frac{\sin t}{\cos t} \cos^2 t.$$ Since $\cos^2 t=\frac{1}{\sec^2 t}=\frac{1}{1+\tan^2 t}$, we get that $$\sin t \cos t =\tan t\frac{1}{1+\tan^2 t}=\frac{x}{1+x^2}.$$ Another way: The following is closer to your calculation. Use the fact that $\sin t\cos t$ has the same sign as $\tan t$ to resolve ambiguities of sign. We know from your calculation that $\sin t \cos t =\pm \frac{x}{1+x^2}$. But this has the same sign as $\tan t$, that is, the same sign as $x$. That resolves the $\pm$ problem in favour of $\frac{x}{1+x^2}$, which has the same sign as $x$.
H: quadratic polynomial investigation in my mathematics textbook,i have found one interesting problem and i have one question.textbook asks following problem deduce all possible value of $a$,for which equation $4x^2-2x+a=0$ has roots in given interval $(-1;1)$ textbook used following method.it found axis of symmetry $x=-b/2a$(here $a$ denotes coefficient before $x^2$,so in this case $a=4$) and it got $x=1/4$ because $1/4$ is located in $(-1;1)$,we have to find maximum root which is less then $1$,and minimum one,which is greater then $-1$,also consider all $a$ for which discriminant is non negative and finally we would have set of equation $1)1-4a>=0$ $2)(1-\sqrt{1-4a})/4>-1$ $3)(1+\sqrt{1-4a})/4<1$ we will get solution $-2<a<=1/4$ my question is what if axis of symmetry is not located in given interval?let us suppose we have equation $5x^2-17x+a=0$ here axis of symmetry $x=-b/2*a=17/10=1.7$,but $1.7$ is outside of given interval $(-1;1)$,in this case what i have to do? AI: $4x^2-2x+a=0$ have solutions $x=\frac{2\pm \sqrt{4-16a}}{8}$. Since $x\in (-1,1)\implies -1\lt\frac{2\pm \sqrt{4-16a}}{8}\lt 1\implies -8\lt {2\pm \sqrt{4-16a}}\lt 8\implies -10\lt {\pm \sqrt{4-16a}}\lt 6$. Solve for these inequalities, you will get the required values for $a$.
H: Is the following proof to Hölders inequality correct ? Hölders inequality is $\int \|fg\|dx \le\|\|f\|\|_p\|\|g\|\|_q$ Define $F(x)= \frac{f(x)}{(g(x))^{q/p}}$ and $\nu dx =g(x)^q$ and $\Phi(t)= \|t\|^p$, $p\in (1,\infty)$ Now lets find $\Phi(\int F(x) d\nu)= \frac{(\int fg)^p}{\|\|g\|\|_q^{qp}}$ and similarly find $\int \Phi (F(x))$ and apply jensens inequality . My doubt is in the setting of $F(x)$ and $\nu dx $ Can anyone help me to make this proof correct . Thanks AI: We assume WLOG that $f$ and $g$ are non-negative. We should take $F(x):=\frac{f(x)}{(g(x))^{q/p}}\chi_{g\neq 0}$ and we have to make the change of measure, if $\mu$ is the initial one $\nu=\frac{g(x)^q}{\lVert g\rVert^q_q}\mu$, that is, if $h$ is integrable $$\int h(x)\nu(dx)=\int h(x)g(x)^qd\mu.$$ We can apply Jensen's inequality since $\nu$ is a probability measure. We have \begin{align} \Phi(\int F(x)d\nu(x))&=\left\|\int F(x)d\nu(x)\right\|^p\\ &=\left\|\int F(x)\frac{g(x)^q}{\lVert g\rVert^q_q}d\mu(x)\right\|^p\\ &=\frac 1{\lVert g\rVert^{qp}_q}\left\|\int\frac{f(x)}{(g(x))^{q/p}}\frac{g(x)^q}{\lVert g\rVert^q_q}\chi_{g\neq 0}d\mu(x)\right\|^p\\ &=\frac 1{\lVert g\rVert^{2qp}_q}\left\|\int fgd\mu(x)\right\|^p, \end{align} since $q-q/p=q/q=1$. We also have \begin{align} \int\Phi(F(x))d\nu(x)&=\int\|F(x)^p\|d\nu\\ &=\int \|F(x)\|^p\frac{g(x)^q}{\lVert g\rVert^q_q}d\mu(x)\\ &=\int \left\|\frac{f(x)}{(g(x))^{q/p}}\chi_{g\neq 0}\right\|^p\frac{g(x)^q}{\lVert g\rVert^q_q}d\mu(x)\\ &=\frac 1{\lVert g\rVert_q^q}\int \|f(x)\|^pd\mu(x). \end{align}
H: Need help with a differential equation -like problem. $\forall y \in \mathbb{R}, \int_{-\infty}^{\infty} f(x)f(x-y)dx=f(y)$ I also know that $\int_{-\infty}^\infty f(x) dx$ converges and that $f$ is symmetric about the origin. What does $f$ look like? Is it possible to identify a parametric set of solutions for $f$? How do you even go about solving a problem like this? AI: A Fourier transformation leads to $\hat f^2=\hat f$, which is solved by any Fourier transform $\hat f$ that takes only the values $0$ and $1$. For instance, $\hat f=\chi_{[-a,a]}$, the characteristic function on an interval centred on the origin, leads to a $\operatorname{sinc}$-like function $f$.
H: Example of a function continuous at only one point. Possible Duplicate: Find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous at precisely one point? I want to know some example of a continuous function which is continuous at exactly one point. We know that $f(x)=\frac{1}{x}$ is continuous everywhere except at $x=0$. But i think this in reverse manner but i dont get any example. So please help me out! AI: One standard example is the function $$f(x)=\begin{cases} x,&\text{if }x\in\Bbb Q\\ 0,&\text{if }x\in\Bbb R\setminus\Bbb Q\;. \end{cases}$$ That is, $f(x)=x$ if $x$ is rational, and $f(x)=0$ if $x$ is irrational. This function is continuous only at $x=0$. Added: The same basic idea can be used to build a function that is continuous at any single specified point. With a little more ingenuity, you can use it to get, for instance, a function that is continuous just at the integers: $$f(x)=\begin{cases} \sin\pi x,&\text{if }x\in\Bbb Q\\ 0,&\text{if }x\in\Bbb R\setminus\Bbb Q\;. \end{cases}$$ This works because $\sin\pi x=0$ if and only if $x\in\Bbb Z$.
H: Find a prime number $p$ so that $f = \overline{3}x^3+ \overline{2}x^2 - \overline{5}x + \overline{1}$ is divided by $x-\overline{2}$ in $\mathbb Z_p$ Let $f = \overline{3}x^3+ \overline{2}x^2 - \overline{5}x + \overline{1}$ be defined in $\mathbb Z_p$. Find a prime number $p$ so that $f$ can be divided by $g = x-\overline{2}$, then factorize $f$ as product of irreducible factors in the $\mathbb Z_p$ found. I am really stuck because I can't figure out on which principle should I rely on to solve this exercise, a push in the right direction is definitely greatly appreciated. AI: $x-a$ divides a polynomial $f(x)$ iff $f(a)=0\implies (x-2)$ divides $f(x)=3x^3+2x^2-5x+1$ iff $f(2)=24+8-10+1\equiv 0\pmod p\implies 23\equiv 0 \pmod p$. The only prime for which $23\equiv 0\pmod p$ is $23$ itself. So, the answer is $23$. Now you can easily factorize $3x^3+2x^2-5x+1$, since after finding one root, it becomes quadratic which is easy to factorize.
H: Insight of some concepts in commutative algebra I really enjoyed the basic algebra course and wanted to teach myself a little more. So I am trying to learn commutative algebra from Atiyah-MacDonald and Eisenbud. The department in our university is very good for analysis based subjects. I have enjoyed courses like basic analysis, measure theory and probability theory. I have a clean insight of what a theorem is trying to say, in these subjects, and I can make a mental picture before I formulate and prove propositions rigorously. In algebra, I dont seem to have this insight. I wish I could speak to people proficient in commutative algebra to get an insight and find the right style of thinking in this field. By the way, I have read this thread. I will make the question specific: 1) I do not understand the idea of an ideal quotient. I know it's definition and I can prove the properties listed in Atiyah-MacDonald's book. But yet I have no idea of the big picture. What is it's purpose? How can I spot an ideal quotient? 2) I came across an idea called 'exact sequences' in Eisenbud's intro to modules. In two swift examples, he constructs exact sequences. I can verify that the second example is indeed an exact sequence. But I could not figure out how he constructed such an example!! The second example was the exact sequence: Given a ring $R$, an ideal $I \subset R$, $a \in R$ $0 \to \dfrac{R}{(I : a)} \to \dfrac{R}{I} \to \dfrac{R}{I + (a)} \to 0$ Is there an insight to this construction that, sort of, lets me guess the 'exact sequence' relation between the objects? P.S: I will have more specific questions as I read along. Thank you for your answers. AI: To answer your first question: I never really got the feel for ideal quotients until I realised it can be used to prove the following two facts: The set of zero divisors in a commutative ring $A$ is a union of prime ideals. The way I proved this at first was using Krull's Lemma, but you can also use ideal quotients as in here: A ring in which every prime ideal is finitely generated is Noetherian. If you would like general advice for commutative algebra, I am not an expert but here's what I can say. When I first started commutative algebra I had 0 intuition. One thing I realised that was helpful as I was going along was to draw a commutative diagram. For example, you should be aware that if $A \subset B$ is a finite ring extension, then given any $Q \subset A$ a prime ideal there is always a prime ideal $P \subset B$ lying over $A$. If you draw a commutative diagram, the proof of this fact is not hard. Also, since you mentioned Eisenbud, I remember proving some isomorphism in there using just a big diagram chase. So, draw a diagram if you can!
H: stuck on a differential equation let be the differential equation $ x^{2} y''(x)+ y(x)(a^{2}+k^{2} _{n})=0 $ the boundary conditions are $ \int_{0}^{\infty}dx \|y(x)\|^{2} < \infty $ and $ y(0) $ must be finite (regular solutions near the origin ) here $ a^{2} >0 $ and $ k^{2} _{n}>0 $ these $ k_{n} $ are a discrete set of eigenvalues my question is how can i transform my differential equation into a more well-known differential equation so i can get the eigenvalues AI: It is an homogeneous Euler equation. Its solution is $y(x)=C_1x^{r_1}+C_2x^{r_2}$ where $r_1,r_2$ are the solutions of the indicial equation $$ r(r-1)+a^2+k_n^2=0. $$ None of these solutions is square integrable on $(0,\infty)$ unless $C_1=C_2=0$.
H: Let $S_n=\sum_{k=1}^{n}\frac{\sin\frac{k\pi}{25}}{k}$,how many positive $S_n$? Let $S_n=\sum_{k=1}^{n}\frac{\sin\frac{k\pi}{25}}{k}$,how many positive $S_n$ are in $S_1,S_2,...S_{100}$ AI: I think it is positive for all values of n.As Prasad G mentioned $sin(\frac{k\pi}{25})$ is positive for the values between 25-50 and 50-75. However, we will get positive sum when n is 55 for example, because of symmetry of sin function(consider the sin on the unit circle and observe that it is symmetric wrt x-axis) and the term k on the denominator. Negative terms will be divided by bigger k so sum turns out to be positive eventually.
H: General nonatomic measure that cannot be expressed as an integral I read in a paper (Kingman — Poisson Processes, 2005) that: In most cases the mean [of an inhomogenous Poisson process on a set $A$] is given in terms of the rate function $\lambda(x)$ on $S$ by \begin{align} \mu(A) &= \int_A \lambda(x)dx\qquad\textrm{for }A > \subseteq S \end{align} but it can be a general (nonatomic) measure on $S$. I know this is a silly question, but what kinds of nonatomic measures cannot be expressed as in the equation? AI: The Cantor set has Lebesgue measure zero, so it is enough to find an atomless measure that puts positive measure on the Cantor set and nowhere else. Now one can identify the Cantor set with the space $\{0,1\}^\mathbb{N}$ in a measurably isomorphic way and the latter space can be endowed with the coin flipping probability measure.
H: "Connection Space" Can we distill the idea of "connectivity" away from their topological context and study abstract properties of "connectivity"? I define a connective space to be a set $X$ together with a collection $\gamma$ of subsets of $X$, which we define as "connected". $\gamma$ contains every singleton subset of $X$, and for all $A, B \in \gamma$ such that $A \cap B \neq \emptyset$ we have $A \cup B \in \gamma$. It might be interesting to study functions between connective spaces that preserve connected sets. Or, more suggestively, perhaps functions such that every pre-image of a connected set is connected... Does this exist in literature? AI: The answer is yes. See this article (J. Muscat & D. Buhagiar - Connective Spaces). Added: I did some googling and found also this preprint (S. Dugowson - On Connectivity Spaces) which examines related ideas. The references here suggest that such spaces have been studied already by R. Börger in 1983.
H: Coherent Sheaves on Projective Space I am having trouble proving the following claim and would be glad if someone could help me out. Claim: Let $\mathbb P$ denote n-dimensional projective space, and let $F$ be a coherent sheaf on $\mathbb P$. Then there exists some integer k such that $F\otimes O(k)$ is generated by a finite number of global sections. I think I know how to begin: Let $\mathbb P^n=\bigcup U_i$ be the usual cover by open affines, and observe that $F\restriction U_i$ is finitely generated (as a coherent sheaf on an affine variety). Now I would like to choose generators of all the $F\restriction U_i$ and tensor them with elements of $O(k)(U_i)$ (for some k) to obtain sections of $F\otimes O(k)$ which lift to global sections generating this sheaf. The problem is that I don't understand well enough what sections of $F\restriction U_i$ look like, so I don't know what to tensor with. Thanks in advance! Roy AI: The key requirement is to to do things in an orderly fashion! 1) First choose for each $i$ finitely many sections $s_i^\alpha \in \Gamma(U_i,F)$ which generate every fiber $F_x$ $(x\in U_i)$. This is possible by Theorem A for affine schemes. [You write "The problem is that I don't understand well enough what sections of $F\restriction U_i$ look like". Actually, you don't have to !] 2) Now we'll consider a variable integer $k$ and the key observation is that a global section $s\in \Gamma(X,F(k))$ is a collection of sections $s_i \in \Gamma(U_i,F)$ satisfying $s_i=\frac {x_j^k}{x_i^k}s_j $ on $U_i\cap U_j$. The global sections $\Gamma(X,F(k))$ will generate all the fibers $F_x$ if we can find sections $s^\alpha\in \Gamma(X,F(k))$ whose restrictions to $U_i$ are $s_i^\alpha$. 3) Finally prove that given $s_i \in \Gamma(U_i,F)$ there exists $k_0$ such that for all $k\geq k_0$ there exists $s\in \Gamma(X,F(k))$ given on $U_i$ by $s_i$. This a little technical. The main ingredient for that last result is 4) Reminder: Given an affine scheme $X$ , a coherent sheaf $F$ on $X$ and $f\in \mathcal O(X)$, then for any global section $s\in\Gamma(X, F)$, zero on $D(f)$, there exists $N$ such that $f^N\cdot s=0\in \Gamma(X,F)$. I've learned all this in Serre's FAC. Here is an opinion on that paper .
H: Why aren't these loops homotopic? Let $S^1 = \{z \in \mathbb{C} : \|z\| = 1\}$. Take the loops $f,g : [0,1] \rightarrow S^1$, $f(t) = 1$, $g(t) = e^{2\pi it}$. I know these represent different elements in $\pi_1(S^1, 1)$, but I don't see why $F(t,s) = e^{2\pi its}$ isn't a homotopy between $f$ and $g$. AI: Loops are paths for which initial and end point coincide (in other words: closed paths), see here. The reason is simply that $F(\cdot,s)$ is not a closed path if $s\neq 0,1$.
H: How do the $L^p$ spaces lie in each other? Let $(S,\mathcal{B},\mu)$ be a measure space, $Y$ be a banach space and for $1\le p <\infty$ let $L^p(\mu;Y)$ be the set of all maps $f:S\rightarrow Y$ that are measurable and for which $\|f\|^p$ is integrable. Let $L^{\infty}(\mu;Y)$ be the of essentially bounded maps. I wonder if it is known for which measure spaces and for which $p,p'$ these function spaces lie in each other, i.e. for which setting there exists a (reasonably well behaved) injection $L^p(\mu;Y)\hookrightarrow L^{p'}(\mu;Y)$. Are there known results in this generality? What if the measure space is simply a subset of $\mathbb{R}^n$ (compact, convex, or with any other property). Or are there other restrictions one can make to get an injection? Thank you in advance! I'd be glad for any pointers. AI: If $S$ has finite total measure, than Hölders inequality shows that $L^p$ embeds into $L^q$ if $p>q$ (I memorize: $L^1(S)$ is the largest one, $L^\infty(S)$ the smallest). If $S$ has infinite total measure and free of atoms, then no $L^p$ embeds into another $L^q$. One always construct functions which are in $L^p$ but not in $L^q$ and vice versa. If $S$ consists of atoms only (e.g. $S=\mathbb{N}$ with the counting measure), than $L^p$ embeds into $L^q$ if $p<q$ (I memorize: $\ell^\infty$ is the largest one, $\ell^1$ is the smallest).
H: What is $d(y dx)$? Let $x$, $y$ be 0-forms, thus $dx$, $dy$ are 1-forms. Since 1-forms compose an algebra over 0-forms ring, expressions like $$y dx$$ make perfect sense. Now I ask what is $$d(y dx)$$ I suggest it to be $y d(dx) = 0$, since $d$ is linear, however I feel it is likely to be wrong. Is there any other meaningful product except $\land$ between forms that would generalize the situation above? AI: $dƒ$ is the differential of $ƒ$ for smooth functions $ƒ$. $d(dƒ) = 0$ for any smooth function $ƒ$. $d(α∧β) = dα∧β + (−1)^p(α∧dβ) $ where $\alpha$ is a p-form. Take $\alpha=y$ and $\beta=dx$ in this last formula.
H: Unambiguous Way of Stating a Biconditional in Plain English I am having a hard time understanding this section in Wikipedia's article on Logical biconditionals: Colloquial usage One unambiguous way of stating a biconditional in plain English is of the form "b if a and a if b". Another is "a if and only if b". Slightly more formally, one could say "b implies a and a implies b". The plain English "if'" may sometimes be used as a biconditional. One must weigh context heavily. For example, "I'll buy you a new wallet if you need one" may be meant as a biconditional, since the speaker doesn't intend a valid outcome to be buying the wallet whether or not the wallet is needed (as in a conditional). However, "it is cloudy if it is raining" is not meant as a biconditional, since it can be cloudy while not raining. Should the example read: "I'll buy you a new wallet if you need one" may be meant as a biconditional, since the speaker does intend a valid outcome to be buying the wallet whether or not the wallet is needed (as in a biconditional). My question is how can the plain English "if'" sometimes be used as a biconditional? I'm OK with the word "biconditional." I don't understand how the reader is to know the "speaker doesn't intend a valid outcome to be buying the wallet whether or not the wallet is needed (as in a conditional)" especially how this amounts to "(as in a conditional)". AI: No. Let $Q$ be the statement "I will buy you a wallet" and $P$ the statement "You need a wallet. "I'll buy you a wallet if you need one" could conceivably be translated as $P \Rightarrow Q.$ However, I hope you would agree that buying a wallet for someone who doesn't need one is quite silly - thus buying someone a wallet implies that they had needed one, ie) $Q \Rightarrow P.$ Hence $(Q\Rightarrow P ) \wedge (P \Rightarrow Q),$ or $P \iff Q$
H: How to calculate $\pi\int^1_0\sqrt{x(1-x)} \mathrm \, dx$ =? I need to find the value of the following integral $$\pi\int^1_0\sqrt{x(1-x)} \mathrm \, dx = ? $$ I tried parts integration, $a\sin x$ substitution and none of them seems work. Can someone get me on the right path? Thank you very much! AI: Put $x=\sin^2\theta\implies dx= \sin2\theta d\theta$.Therefore, your integral becomes, $$ \int_0^{\pi/2}\sin\theta \cos\theta \sin2\theta d\theta=(1/2)\int_0^{\pi/2}\sin^22\theta d\theta=(1/4)\int_0^{\pi/2}(1-\cos4\theta) d\theta=\pi/8$$
H: Finding $\sup$ of the given function in Ball of radius $1$ centered at origin. I want to find the the supremum of $\dfrac{\|x\|^{2/3}-\|y\|^{2/3}}{\|x-y\|^{2/3}} $ in the unit ball centered at the origin . Here $x\neq y$, $x,y \in \mathbb R^n$. How do I proceed ? Thank you for your help. AI: Hint: By a modification of the triangular inequality you see that $$\|\|x\|^{2/3}-\|y\|^{2/3}\|\leq \|x-y\|^{2/3}$$ for all $x\neq y$. So we have$$\dfrac{\|x\|^{2/3}-\|y\|^{2/3}}{\|x-y\|^{2/3}}\leq 1$$ Is the value of the function $=1$ at some point?
H: Solving $f(2011)=2012$, $f(4xy)=2yf(x+y)+f(x-y)$ How to find the all functions $f$ :$ \mathbb{R}\longrightarrow\mathbb{R}$ such that $f(2011)=2012$,for every $x,y\in\mathbb{R}$ then: $$f(4xy)=2yf(x+y)+f(x-y)$$ AI: If such a function exists, then $f(x)=f(0) \ne 0$ for every $x$, therefore $f(0)=2yf(0)+f(0)$ for every $y$, i.e. $y=0$ for every $y$. Thus such a function does not exist.
H: Chi Squared Distribution with $\mu = 0$, $\sigma^2 \neq 1$ Let $X_i$ be independent normally distributed random variables with zero mean and variance $\sigma^2 \neq 1$. What is the probability density function of the random variable formed by the sum of their squares? Here is my attempt: Let $Y = \sum_{i=1}^{k}X_i^2$. Then $$\begin{align}Y &= \sigma^2\sum_{i=1}^{k}(\frac{X_i}{\sigma^2})^2\\ &\sim \sigma^2\chi^2_{k}\\ &\sim \sigma^2\Gamma_{k/2}(\theta = 2)\\ &\sim \Gamma_{k/2}(\theta = 2\sigma^2) \end{align} $$ where $\chi^2_{k}$ is the central chi-squared distribution, $\Gamma_{k}(\theta)$ is the Gamma distribution with Scaling parameter $\theta$. Therefore, the mean of $Y$ is $\frac{k}{2}2\sigma^2 = k \sigma^2$ and variance $\frac{k}{2}(2\sigma^2)^2 = 2k\sigma^4$. Also, is it possible to derive the mean and variance without going back to the Gamma distribution? AI: Since $Y$ is the sum of $k$ i.i.d. random variables distributed like $Z$, $\mathrm E(Y)=k\cdot\mathrm E(Z)$ and $\mathrm{Var}(Y)=k\cdot\mathrm{Var}(Z)$. Here, $Z$ is distributed like $\sigma^2 X_0^2$ where $X_0$ is standard normal hence $\mathrm E(Z)=\mathrm E(X_0^2)\cdot\sigma^2=\sigma^2$ and $\mathrm{Var}(Z)=\left(\mathrm E(X_1^4)-\mathrm E(X_1^2)^2\right)\cdot\sigma^4=(3-1)\cdot\sigma^4=2\sigma^4$. To prove the first assertions, since $Y=Z_1+\cdots+Z_k$, one can use the linearity of the expectation for $\mathrm E(Y)$, and, for $\mathrm{Var}(Y)$, the expansion $$ \mathrm E(Y^2)=\sum_{i=1}^k\mathrm E(Z_i^2)+\sum_{i\ne j}\mathrm E(Z_i)\cdot\mathrm E(Z_j)=k\cdot\mathrm E(Z^2)+k(k-1)\cdot\mathrm E(Z)^2, $$ which yields $$ \mathrm{Var}(Y)=\mathrm E(Y^2)-\mathrm E(Y)^2=\mathrm E(Y^2)-k^2\cdot\mathrm E(Z)^2=k\cdot\mathrm{Var}(Z). $$
H: Application of Pell's equation Need to find $n,r$ (if any) for $121^r-2n^2=1$ where $n,r$ are natural numbers. Observed that $n$ is odd then $n=2m+1$ (say). But on replacement of $n$ by $2m+1$, increases the complexity of the problem. AI: All solutions to $x^2-2y^2=1$ are given implicitly by $$ x_k+y_k\sqrt{2} = (3+2\sqrt{2})^k $$ To solve your equation would require $x_k=11^r$ for some $k,r$. But looking at solutions modulo 11 and 3 we find a cycle of 12 possibilities mod 11 $(x_{13},y_{13})\equiv (3,2)\pmod{11}$ and a cycle of four possibilities mod 3, $(x_5,y_5)\equiv (0,2) \pmod{3}$. Considering divisibility of $x$ we find $11\mid x_k$ only when $k\equiv 3 \pmod{6}$, and $3\mid x_k$ whenever $k$ is odd. Thus $x_k$ cannot be a power of $11$ and your equation has no solutions.
H: Why is $\mathbb{C}^{g}$ the universal cover of any connected compact complex Lie group of dimension g? This question came up while I was studying the book "Complex Abelian Varieties" by Lange/Birkenhake. More precisely, the authors prove in Lemma 1.1 that every connected compact complex Lie group of dimension $g$, $g$ a positive integer, is a complex torus. At one point they are using the fact that the universal cover of any such Lie group is a complex vector space of dimension $g$. They even give a reference (Theorem 18.4.1, "The Structure of Lie Groups" by Gerhard Hochschild). However, Theorem 18.4.1 (which is instead listed as Proposition 18.4.1 in the cited book) reads "If a semisimple analytic group has a faithful finite-dimensional continuous representation then its center is finite" I really don't have a clue how this should apply to my problem, because a complex torus is abelian, hence in particular not semisimple. On the other hand my knowledge of Lie groups is very very rudimentary so I might overlook something. Can anyone show me how Proposition 18.4.1 in Hochschild's book helps me, or instead give me a reference to a proof of the statement in the title of my question? Of course any outline of a proof would also be appreciated, however I'd rather favor a reference. A reference to a direct proof of the above cited Lemma 1.1 would be very helpful too. Thank you in advance AI: The following steps lead to a solution. Let me first define standard notation. If $G$ is a Lie group and if $g\in G$, then the map $\phi_{g}:G\to G$ defined by the rule $\phi_g(x)=gxg^{-1}$ for $x\in G$ is referred to as conjugation by $g$. If $f:M\to N$ is a smooth map, then we denote by $T_{p}(f):T_{p}(M)\to T_{f(p)}(N)$ the differential of $f$ at $p\in M$. Exercise 1: Let $G$ be a complex Lie group. The adjoint representation of $G$ is the map $\rho:G\to \text{GL}(\mathfrak{g})$ defined by the rule $\rho(g)=T_{e}(\phi_{g})\in \text{GL}(\mathfrak{g})$ where $e\in G$ is the identity element. Prove that $\rho:G\to \text{GL}(\mathfrak{g})$ is a complex (holomorphic) representation of $G$. Exercise 2: In the context of Exercise 1, assume that $G$ is in addition compact and connected. Prove that the adjoint representation $\rho:G\to \text{GL}(\mathfrak{g})$ is trivial. (Hint: use the maximum modulus principle.) Conclude that $G$ is an abelian group. We have now set the stage to prove that the exponential map $\text{exp}:\mathfrak{g}\to G$ is a covering map. Exercise 3: Let $G$ be a Lie group. If $X,Y\in \mathfrak{g}$ commute, i.e., $[X,Y]=0$, then prove that $e^{X}e^{Y}=e^{X+Y}=e^{Y}e^{X}$. In the context of Exercise 2, conclude that $\text{exp}:\mathfrak{g}\to G$ is a homomorphism. Exercise 4: Use the Hopf-Rinow theorem (or look it up if necessary) to prove that $\text{exp}:\mathfrak{g}\to G$ is surjective in the context of Exercise 2. Exercise 5: Prove that $\text{exp}:\mathfrak{g}\to G$ is a covering map in the context of Exercise 2. (Hint: use Exercise 3 and Exercise 4. Recall that $\text{exp}:\mathfrak{g}\to G$ is a local homeomorphism at the origin of $\mathfrak{g}$ because its differential at the origin of $\mathfrak{g}$ is the identity map.) I hope this helps!
H: Oblique asymptotes When we find oblique asymptotes, we divide the numerator by the denominator and take only the polynomial portion of the expression as the equation of the slant asymptote. Why? What is the proof that this equation is the slant asymptote? AI: Because you're interested in the behaviour of the function as $\|x\|$ grows large. You can split the function up. I'll take an example: $$\frac{3x^2+1}{x+2}$$ $$\frac{3x(x+2)-6x+1}{x+2}$$ $$3x+\frac{-6x+1}{x+2}$$ $$3x+\frac{-6(x+2)+13}{x+2}$$ $$3x-6+\frac{13}{x+2}$$ at each step I'm rewriting the highest term in a way that involves $x+2$, and then adjusting the lower terms so that the expansion comes out right. Now look what happens to the fraction as $x$ grows large: $$\frac{13}{102},\frac{13}{1002},\frac{13}{10002}, \cdots$$ It gets smaller and smaller and goes to $0$. You can show the same thing for the other direction as well (negative numbers that get closer and closer to zero). So an asymptote is concerned with behaviour as $x$ grows large, and as $x$ grows large the fractional portion drops out to $0$. It only influences the function near the origin, which the asymptote isn't concerned with.
H: How to prove $p$ divides $a^{p - 2} + a^{p - 3} b + a^{p - 4} b^2 + \cdots + b^{p - 2}$ when $p$ is prime, $a, b \in \mathbb{Z}$ and $a,b \lt p$? If $p$ is a prime number and $a, b \in \mathbb{Z}$ such that $a,b \lt p$, then how could we prove that $p$ divides $\left(a^{p - 2} + a^{p - 3} b + a^{p - 4} b^2 + \cdots + b^{p - 2}\right)$? AI: We have $$(a-b)(a^{p-2} + a^{p-3}b + \ldots + ab^{p-3} + b^{p-2}) = a^{p-1} - b^{p-1}.$$ If $p\nmid a$ and $p\nmid b$ then the right side is congruent to $1-1 = 0$ by Fermat's little theorem, i.e $p$ divides the product on the left. If furthermore $a \not\equiv b \pmod p$ then $p$ must already divide $a^{p-2} + a^{p-3}b + \ldots + ab^{p-3} + b^{p-2}$. The statement becomes false if you do not assume $a \not\equiv b \pmod p$, e.g. for $a=b=1$ the sum equals $p-1$ which is not divisible by $p$. Also, if $p\| a$ (w.l.o.g.) the sum is congruent to $b^{p-2}$ which is only divisible by $p$ if $p\|b$.
H: Less than infinity or Less or Equal to infinity What is the difference between Less than infinity or Less or Equal to infinity? We cannot substitute infinity anyway, and have to use limits. So does it make sense to write less and equal to infinity? AI: In the context of the real numbers $\infty$ denotes "larger than any $x\in\mathbb R$". It is often useful to have this as a possible index, to indicate this what happens "after". For example in the context of measure theory, a positive measure is a function from sets into $[0,\infty]$. That is, some sets have infinite measure. In another related context we have $p$-norms for $p\in[1,\infty]$ which is also a well-defined notion. When we then say that $p\leq\infty$ we mean that the following proposition would hold for any value of $p$, either finite or infinite. For example, $\\|f+g\\|_p\leq\\|f\\|_p+\\|g\\|_p$ holds for finite $p$ and for $p=\infty$. On the other hand sometimes we would like to indicate something holds only for the finite values of $p$, e.g. if $1\leq p<\infty$ then $\ell_p$ is separable. It is not true anymore for $\ell_\infty$, despite the space itself is a well-defined object whose index is $\infty$. Or for example if $A\subseteq\mathbb R$ and the $0<m(A)\leq\infty$ then there is a subset of $A$ which is not Lebesgue measurable. On the other hand, if $m(A)<\infty$ then there is some $n\in\mathbb N$ such that $m(A\cap[-n,n])=m(A)$.
H: Integration problem in matrix calculus Let $\mathbf{A}=\begin{bmatrix} f(x_1,x_1), & \ldots,& f(x_1,x_n)\\ \vdots&\ddots& \vdots \\f(x_n,x_1),&\ldots, &f(x_n,x_n) \end{bmatrix} $, where $f:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$. I want to calculate $\int \mathbf{A}\mathrm{d}\mathbf{x}$, where $\mathbf{x}=\begin{bmatrix}x_1\\ \vdots\\x_n \end{bmatrix}$. It would be nice if you can show some references. Thanks :) AI: Consider the change $$ {\rm d} {\boldsymbol y} = {\boldsymbol A}\, {\rm d} {\boldsymbol x} $$ $$ = \left[ \begin{matrix} f(x_1,x_1)\,{\rm d}x_1 + f(x_1,x_2)\,{\rm d}x_2 + \ldots \\ f(x_2,x_1)\,{\rm d}x_1 + f(x_2,x_2)\,{\rm d}x_2 + \ldots \\ \vdots \end{matrix} \right] $$ and its integral $$ {\boldsymbol y} = \int {\boldsymbol A}\, {\rm d} {\boldsymbol x} $$ $$ = \left[ \begin{matrix} \int f(x_1,x_1)\,{\rm d}x_1 + \int f(x_1,x_2)\,{\rm d}x_2 + \ldots \\ \int f(x_2,x_1)\,{\rm d}x_1 + \int f(x_2,x_2)\,{\rm d}x_2 + \ldots \\ \vdots \end{matrix} \right] $$
H: How many triangles with integral side lengths are possible, provided their perimeter is $36$ units? How many triangles with integral side lengths are possible, provided their perimeter is $36$ units? My approach: Let the side lengths be $a, b, c$; now, $$a + b + c = 36$$ Now, $1 \leq a, b, c \leq 18$. Applying multinomial theorem, I'm getting $187$ which is wrong. Please help. AI: The number of triangles with perimeter $n$ and integer side lengths is given by Alcuin's sequence $T(n)$. The generating function for $T(n)$ is $\dfrac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$. Alcuin's sequence can be expressed as $$T(n)=\begin{cases}\left[\frac{n^2}{48}\right]&n\text{ even}\\\left[\frac{(n+3)^2}{48}\right]&n\text{ odd}\end{cases}$$ where $[x]$ is the nearest integer function, and thus $T(36)=27$. See this article by Krier and Manvel for more details. See also Andrews, Jordan/Walch/Wisner, these two by Hirschhorn, and Bindner/Erickson.
H: Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$ What is the number of Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$? AI: When $q$ is a power of $2,$ we have ${\rm PSL}(2,q) = {\rm SL}(2,q)$ and a Sylow $2$-normalizer is a Borel subgroup of order $q(q-1).$ Hence there are $q+1$ Sylow $2$-subgroups as ${\rm SL}(2,q)$ has order $(q-1)q(q+1)$. When $q$ is odd, the order of ${\rm PSL}(2,q)$ is $\frac{q(q-1)(q+1)}{2}.$ A Sylow $2$-subgroup of ${\rm SL}(2,q)$ is (quaternion or) generalized quaternion and a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is either a Klein $4$-group or a dihedral $2$-group with $8$ or more elements. In all these cases, a Sylow $2$-subgroup of ${\rm SL}(2,q)$ contains its centralizer, and some elementary group theory allows us to conclude that the same is true in ${\rm PSL}(2,q).$ The outer automorphism group of a dihedral $2$-group with $8$ or more elements is a $2$-group. Hence a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is self-normalizing when $q \equiv \pm 1$ (mod 8), and in that case the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(q^{2}-1)_{2^{\prime}}$ where $n_{2^{\prime}}$ denotes the largest positive odd divisor of the positive integer $n.$ When $q \equiv \pm 3$ (mod 8), then a Sylow $2$-normalizer of ${\rm PSL}(2,q)$ must have order $12$ ( a Sylow $2$-subgroup is a self-centralizing Klein $4$-group, but there must be an element of order $3$ in its normalizer by Burnside's transfer theorem). In this case, the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(\frac{q^{2}-1}{24})$
H: What's the precise meaning of imaginary number? The same to the title,what's the precise meaning of imaginary number? And on the other hand,how can the imaginary number be reflected in Physics? AI: I'm unsure what you mean by the "precise" meaning of an imaginary number, but to me it seems best to talk about the prototypical example of an imaginary number, the imaginary unit $i$, and then work from there. You can think of the imaginary unit as being a solution of the equation $X^2+1=0$ i.e. you define $i$ to be a number such that $i^2=-1$ (note that I cannot say "the" number since it is not unique, as $-i$ also has this property). To me, this is the precise meaning of the imaginary unit. One then defines the complex numbers $\mathbb{C}$ in terms of this imaginary unit as $$\mathbb{C}=\{a+bi \; \|\; a,b\in \mathbb{R}\}.$$ More formally, the complex numbers are obtained by adjoining a root of $X^2+1$ to $\mathbb{R}$ by taking the quotient of $\mathbb{R}[X]$ by the maximal ideal $(X^2+1)$. Let $C= \mathbb{R}[X]/(X^2+1)$. Note that $\{1,X\}$ is a basis for $C$ and $X$ has the property that $X^2=-1$. From this it is seen that the map $\varphi:C \rightarrow \mathbb{C}$ defined by $\varphi(a+bX)=a+bi$ is an isomorphism of fields. Another way of viewing the complex numbers is as follows: The set $R$ of matrices of the form $$\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix},$$ where $a \in \mathbb{R}$, behaves exactly like the real numbers with respect to the operations of matrix addition and multiplication i.e. they are isomorphic as fields ($R \cong \mathbb{R}$ ). Consider the matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Notice that this matrix has the property that $$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.$$ Setting $i=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and using our identification above, we see that $i$ is a solution of the equation $X^2+1=0$. So with this one could reasonably believe that the precise meaning of the imaginary unit $i$ is $i=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. One then notices that the set of matrices $C$ of the form $$\begin{pmatrix} a & -b \\ b & a \end{pmatrix},$$ where $a,b \in \mathbb{R}$ behave precisely like the complex number $\mathbb{C}$ i.e. they are isomorphic as fields ($C\cong \mathbb{C}$), and takes this to be the precise meaning of the complex numbers. Additionally, one could formally develop the complex numbers by defining $C$ to be the set of pairs $(a,b) \in \mathbb{R}^2$ such that additions and multiplication are defined by $$(a,b)+(c,d)=(a+c,b+d)$$ and $$(a,b)(c,d)=(ac-bd,ad+bc).$$ First note that subset $D$ of $C$ consisting of elements of the form $(a,0)$ behaves just like elements of $\mathbb{R}$, so, in particular, $(1,0)$ is the multiplicative identity of $C$. One then notices that the element $(0,1) \in C$ has the property that $(0,1)(0,1)=(-1,0)$ and so it is a solutions to the equation $X^2+1=0$.
H: Spectrum Of The Laplacian Question I'm given the following question (from Davies' book - Spectral theory of differential operators): Use the theorem () below to prove that if $\Omega$ is a convex region in $ \mathbb{R}^2 $ , then: $ \frac{1}{4} \int _\Omega \frac{\|f\|^2}{d^2} d^2 x \leq \int _\Omega \| \bigtriangledown f \| ^2 d^2 x $ for all $f \in C_c ^\infty (\Omega ) $ . Deduce that $ Spec(H) \subseteq [\lambda, \infty ) $ , where $ \lambda^{-1} := 4Inradius(\Omega)^2 $. Theorem : Let $\Omega $ be a region in $ \mathbb{R}^2 $ and let $f \in C_c ^\infty (\Omega ) $. Then : $ \frac{1}{2} \int _\Omega \frac{\|f\|^2}{m(x)^2} d^2 x \leq \int _\Omega \| \bigtriangledown f \| ^2 d^2 x $ for all $f \in C_c ^\infty (\Omega ) $ where the pseudodistance $m(x)$ is defined by $ \frac{1}{m(x)^2 } := \frac{1}{2\pi } \int_{-\pi}^{\pi} \frac{d \Theta }{d_\Theta (x)^2 }$ and $ d_\Theta : \Omega \to (0,+\infty ] $ is defined by $ d_\Theta (x) = min \{ \|s\| : x+se^{i \Theta} \notin \Omega \} $ . It's also known that if $\Omega $ is regular ( i.e.- there exists a constant $c$ such that $d(x)\leq m(x) \leq cd(x) $ ) , then $ Spec (H) \leq [ \frac{1}{2c^2 Inradius^2}, \infty ) $ . How can I use these facts in order to answer Davies' question? Any help will be greatfully acknowledged ! Thanks in advance $ d(x):=min \{ \|x-y\| : y \notin \Omega \} $ . AI: To use the theorem and the fact, it suffices to prove that $\Omega$ is regular, and that $d(x) \leq m(x) \leq \sqrt{2} d(x)$. The first inequality is evident (since the average must be greater than the minimum). It suffices to prove the second inequality, which we re-write equivalently as $$ \frac{1}{2\pi}\int_0^{2\pi}\frac{1}{d_\theta(x)^2} \mathrm{d}\theta = \frac{1}{m^2(x)} \geq \frac{1}{2d^2(x)} \tag{} $$ Now, since $\Omega$ is assumed to be convex, its boundary "lies to one side of the tangent line(s)" (I use plural since a convex boundary need not be differentiable, but can have multiple supporting tangent lines). Furthermore, if $\theta_0$ is such that $d_{\theta_0}(x) = \inf_\theta d_\theta(x) = d(x)$ we must have that the boundary is a differentiable function of $\theta$ there, and that the tangent line is perpendicular to the direction $\theta_0$ (this is elementary convex analysis [see below the cut]). We thus have [see below second cut] that $d_\theta(0) \leq d_{\theta_0}(x) \sec (\theta - \theta_0) $ by convexity. This implies that $$ \frac{1}{2\pi} \int_0^{2\pi} \frac{1}{d_\theta(x)^2}\mathrm{d}\theta \geq \frac{1}{2\pi}\frac{1}{d(x)^2} \int_0^{2\pi} \cos^2(\theta - \theta_0) \mathrm{d}\theta = \frac{1}{2d(x)^2} $$ which is precisely (*). If $\Omega$ is convex, it has non-vanishing tangent support. That is for every point $x\in \partial\Omega$ there exists at least one line $\ell_x$ through $x$ such that $\ell_x$ does not intersect the interior of $\Omega$. Now let $y$ be a point in the interior of $\Omega$. Look at the line $xy$. Suppose $\ell_x$ is not orthogonal to $xy$. Then you can choose a point $z$ on $\ell_x$ such that $\|z-y\|$ is less than $\|x-y\|$ strictly (using that the angle $\angle zyx$ is acute). Since along the line $zy$, the distance from $y$ to the boundary is $\leq \|z-y\|$ (by virtue of $\ell_x$ not intersecting the interior of $\Omega$), we arrive at the conclusion If there exists a supporting line $\ell_x$ which is not perpendicular to $xy$, the direction $xy$ cannot minimise $d_\theta(y)$. The contrapositive of that statement states that if $x$ is the point such that the direction $xy$ minimises $d_\theta(y)$, there exists a supporting line (automatically true by convexity) and that supporting line can only be the perpendicular to $xy$. This tells you that the function defining the boundary of $\Omega$ has a unique tangent at $x$, and hence is differentiable there. Recall that the formula in radial coordinates for a line $\ell$ which comes closest to the origin at the point $(r_0,\theta_0)$ is precisely $$ r = r_0 \sec (\theta - \theta_0) $$ where $\theta \in (\theta_0 - \pi/2, \theta_0+\pi/2)$.
H: Pointwise convergence implies $L^p$ convergence? Let $f: X \to [0, \infty) \subset \mathbb R$ measurable where $X$ is a measure space. Let $f_n : X \to [0, \infty) $ be simple functions (i.e. linear combinations of characteristic functions of measurable sets) such that for each $x \in X$, $f_n(x) \leq f_{n+1}(x)$ and $f_n(x)$ converges to $f(x)$. How can I prove that $$ \\|f_n - f \\|_p = \left ( \int_X \|f - f_n\|^p d \mu\right )^{1/p} \xrightarrow{n \to \infty} 0$$ I don't think this is right but if for $n > N_x$, $\|f_n(x) - f(x)\| \leq \varepsilon$, we can let $N = \sup_{x \in X} N_x$ to get $\\|f_n - f\\|_\infty \leq \varepsilon$ and then $$ \\|f_n - f \\|_p = \left ( \int_X \\|f - f_n\\|^p d \mu\right )^{1/p} \leq \left ( \int_X \varepsilon^p d \mu\right )^{1/p} = \mu(X)^{1/p} \varepsilon $$ But $\mu(X)$ could be infinite so I'm not sure what to do. Thanks. Edit What assumptions do I need to make this true? AI: The answer to the question in the title is: No, even on finite measured spaces. For an example, consider $X=(0,1)$ endowed with the Lebesgue measure, and $f_n=2^n\cdot\mathbf 1_{(0,1/n)}$.
H: Compute $\lim\limits_{n\to\infty} \int_{0}^{2\pi} \cos x \cos 2x\cdots \cos nx \space{dx}$ Compute the following limit: $$\lim_{n \to \infty}\int_{0}^{2\pi}\cos\left(x\right)\cos\left(2x\right)\ldots \cos\left(nx\right)\,{\rm d}x$$ Today I was working on a W. L. Putnam competition's problem containing this integral and wondered how I may compute its value when $n$ goes to $\infty$. So far I've found no answer. Could you give me some suggestions about the way I should go? AI: Let $u = e^{ix}$. Then $\cos x \cos 2x ... \cos nx = \frac{1}{2^n}\prod_{k=1}^n (u^k + u^{-k})$. Write $$\prod_{k=1}^n (z^k + z^{-k}) = \sum_{j=-N}^N a_j z^j$$ where $N=\frac{n(n+1)}2$. Then the integral above is just $\frac{2\pi}{2^n}a_0$. Combinatorially, we can see that $a_0$ is the number of ways of picking a subset of $\{1,2,...,n\}$ which adds up to $\frac{n(n+1)}{4}$. Now, the set of subsets of $\{1,...,n\}$ that add up to a particular value is an antichain, so we have a bound on the size of $a_0$, namely, $$0\leq a_0 \leq \binom {n}{\lfloor n/2\rfloor}$$. And we know that $$\lim_{n\to\infty}\frac{1}{2^n}\binom {n}{\lfloor n/2\rfloor} =0$$ So we know your limit is zero. (The maximum size of an antichain in the Boolean poset is Sperner's theorem.) Note that when $n\equiv 1,2\pmod 4$, $\frac{n(n+1)}4$ is not an integer, so in those cases, $a_0=0$. Anybody have an "interesting" lower bound on $a_0$ for the other cases? [In comments below, Robert Israel shows that when $n\equiv 0,3\pmod 4$, $a_0\geq 2^{\lfloor (n+1)/4\rfloor}$. I suspect that there is a polynomial, $q$, such that $a_0\geq \frac{2^n}{q(n)}$.] This proof generalizes. Given any sequence of positive integers, $\{b_1,...,b_n\}$, we have the same result: $$\int_{0}^{2\pi} \cos b_1 x \cos b_2 x ... \cos b_n x dx= \frac{2\pi}{2^n}A$$ where $A$ is the count of an antichain, so $0\leq A\leq \binom {n}{\lfloor n/2\rfloor}$ If $b_i=1$ for all $i$, then we get exact equality when $n$ is even, which is to say that: $$\int_{0}^{2\pi} \cos^{2m} x dx = \frac{2\pi}{2^{2m}} \binom {2m}{m}$$
H: proving convergence of a sequence and then finding its limit For every $n$ in $\mathbb{N}$, let: $$a_{n}=n\sum_{k=n}^{\infty }\frac{1}{k^{2}}$$ Show that the sequence $\left \{ a_{n} \right \}$ is convergent and then calculate its limit. To prove it is convergent, I was thinking of using theorems like the monotone convergence theorem. Obviously, all the terms $a_{n}$ are positive. So, if I prove that the sequence is decreasing, then by the monotone convergence theorem it follows that the sequence itself is convergent. $a_{n+1}-a_{n}=-\frac{1}{n}+\sum_{k=n+1}^{\infty }\frac{1}{k^{2}}$. But, I can't tell from this that the difference $a_{n+1}-a_{n}$ is negative. If anybody knows how to solve this problem, please share. AI: [Edit: the first proof or convergence wasn't quite right, so I removed it.] It is useful to find some estimates first (valid for $n>1$): $$\sum_{k=n}^{\infty }\frac{1}{k^{2}}<\sum_{k=n}^{\infty }\frac{1}{k(k-1)}=\sum_{k=n}^{\infty }\left(\frac1{k-1}-\frac1k\right)=\frac1{n-1}\\\sum_{k=n}^{\infty }\frac{1}{k^{2}}>\sum_{k=n}^{\infty }\frac{1}{k(k+1)}=\sum_{k=n}^{\infty }\left(\frac1{k}-\frac1{k+1}\right)=\frac1{n}$$ The last equality in each of these lines holds because those are telescoping series. This gives us the estimate: $1<a_n<\frac{n}{n-1}$. By the squeeze theorem, we can conclude that our sequence converges and $\lim_{n\to\infty}a_n=1$.
H: Could someone please confirm the contradiction in this equation? I'm following an equation from a published paper in order to calculate probabilities using a Markov Chain. The equation says: Construct a set $U$ that consists of all items that appear in the top-$k$ in at least one list. For each pair of items $i$ and $j$ in $U$, let the preference for $i$ over $j$, $m_i{_j}$, equal $1$ if the majority of the lists ($>=50$%) that rank both $i$ and $j$ rank $j$ above $i$ and $0$ otherwise. Let $m_i{_j} = m_j{_i} = 0.5$ if items $i$ and $j$ are never directly compared in any list. My problem with the above is: $U$ is composed of the top-k items in at least one list. Wouldn't this mean that $m_i{_j} = 0.5$ would never be possible because if $U$ is composed of items that must all appear in one list, each item is directly compared in at least one list. I just want someone to confirm my reading of this. AI: If there are at least two lists, the convention is necessary if item $i$ appears in some lists, item $j$ appears in some other lists, and no list contains both items $i$ and $j$.
H: Problem 18.1 in I. Martin Isaacs' Algebra I am trying to prove the following: Let $E/F$ be an arbitrary extensions. Show that $E/F$ is normal if and only if $E$ is the union of all those intermediate fields $K$ such that $K$ is the splitting field for some $f(X) \in F[X]$. My attempt at the solution is the following: Let $A$ be the union of all those intermediate fields $K$ such that $K$ is the splitting field for some $f(X) \in F[X]$. Assume $E/F$ is normal. Clearly $A \subseteq E$. To prove the opposite containment let $\alpha \in E$ and $f(X)=\text{min}_F(\alpha)$. Since $E/F$ is normal we know that $f(X)$ splits over $E$, and so $E$ contains the splitting field of $f(X)$, call it $K_f$. Since $f(X) \in F[X]$ we know $K_f \in A$ and thus $\alpha \in A$. Therefore $E\subseteq A$. Assume $E=A$ and let $\alpha \in E$ and $f(X)=\text{min}_F(\alpha)$. The splitting field of $f(X)$ is contained in $E$ by definition, so $f(X)$ splits completely over $E$. Thus $E/F$ is normal. One immediate problem with my proof is that fact that I implicitly assume $E/F$ is algebraic when I start talking about the minimal polynomial of an element of $E$. My question is: Is there any way to fix this proof so as to not assume $E/F$ is algebraic? AI: What is a normal extension? Lang, for example, in his "Algebra", makes it crystal clear there are (at least) three equivalent conditions for this, one of which is "$\,E/F\,$ is a normal extension iff $\,E\,$ is the splitting field of a family of polynomials $\,\mathcal P\,$ in$\,F[x]\,$" Now, since "the splitting field" carries a meaning of minimality (i.e., the minimal extension containing all the roots of all the polynomials in some such family $\,\mathcal P\,$) , it seems clear that we can "assume" or "agree" on the extension being algebraic, otherwise we could simply take $\,F(\{\alpha\; :\;\alpha\,\,\text{is a root of some element in}\,\,\mathcal P\}) $ , which would be strictly contained in $\,E\,$ if $\,E/F\,$ weren't algebraic, thus contradicting the above mentioned minimality...
H: Weak convergence and weak star convergence. If region $\Omega$ is bounded and $u_n$ has weak star convergence in $L^\infty ( \Omega)$ to some $u\in L^\infty(\Omega)$ , does it imply that $u_n$ converges weakly in any $L^p(\Omega) $ ? I think i got it : If $sup$ of a function is finite then integral over a bounded region is finite with any $p$ norm . is it right ? AI: $\{u_n\}\subset L^\infty(\Omega)$ converges in the weak star topology to $u\in L^\infty(\Omega)$ if $$ \lim_{n\to\infty}\int_\Omega u_n\phi\,dx=\int_\Omega u\,\phi\,dx\quad\forall\phi\in L^1(\Omega). $$ Since $\Omega$ is bounded, $L^\infty(\Omega)\subset L^p(\Omega)\subset L^1(\Omega)$ for all $p\ge1$. It follows that $u_n$ converges weakly to $u$ in $L^p(\Omega)$ for all $p\in[1,\infty)$.
H: Convergence of $a_{n}=\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}$? For $n$ in $\mathbb{N}$, consider the sequence $\left \{ a_{n} \right \}$ defined by: $$a_{n}=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$$ I would like to prove whether this sequence is convergent, and if so what its limit is. I can prove by induction that $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}\geqslant \sqrt{n}$ For any $n$ in $\mathbb{N}$. Hence, $a_{n}\geqslant 1$. I wanted to prove that the sequence is decreasing and then use the monotone convergence theorem to prove it is convergent. However, I couldn't come up with a proof for this part. Anyone know how to prove convergence and find the limit? I had another proof based on using Riemann sums, but I am looking for another proof using onne of the tricks used to prove convergence of sequences. Here is my proof: $$ a_{n}=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{\frac{k}{n}}}. $$ Hence, $$\lim_{n \to \infty }a_{n}=\int_{0}^{1}\frac{dx}{\sqrt{x}}=2$$ AI: Using only elementary inequalities and no (improper) integral: For every $k$, $\sqrt{k+1}-\sqrt{k}=\dfrac1{\sqrt{k+1}+\sqrt{k}}$ and $2\sqrt{k}\lt\sqrt{k+1}+\sqrt{k}\lt2\sqrt{k+1}$, hence $$ \sqrt{k+1}-\sqrt{k}\lt\dfrac1{2\sqrt{k}}\lt\sqrt{k}-\sqrt{k-1}. $$ Summing the rightmost inequality yields $$ a_n\lt\frac2{\sqrt{n}}\sum_{k=1}^n\sqrt{k}-\sqrt{k-1}=2. $$ Likewise, summing the leftmost inequality yields $$ a_n\gt\frac2{\sqrt{n}}\sum_{k=1}^n\sqrt{k+1}-\sqrt{k}=2\frac{\sqrt{n+1}-1}{\sqrt{n}}\gt2-\frac2{\sqrt{n}}. $$ Since $\dfrac2{\sqrt{n}}\to0$, this proves that $\lim\limits_{n\to\infty}a_n=2$.
H: Solving a set of "circular" quadratic equations $x_a'$ and $y_a'$ are unknown. What's the simplest way to solve it? Every time I tried, it grew into tremendous size or was unable to think out in reasonable amount of time due to it's complexity. $\begin{align} (x_f-x_a')^2+(y_f-y_a')^2&=r^2\\ (x_a-x_a')^2+(y_a-y_a')^2&=4r^2\sin^2(\frac12\alpha) \end{align}$ P.S. I don't need to specify real value, that is, I want to get a function $x_a'$ and $y_a'$ of known values. AI: Expand the squares on the left-hand sides and subtract the two equations. The $x_a'^2$ and $y_a'^2$ terms will cancel out, and what is left is a linear equation in $x_a'$ and $y_a'$ which you can use to express $y_a'$ in terms of $x_a'$ (or vice versa -- one of these may be impossible depending on the constants). Stick the expression for $y_a'$ into one of the original equations and simplify. The result is a quadratic equation in $x_a'$ which you can solve with the quadratic formula. Finally use the linear equation again to find the corresponding value of $y_'a$. If you try to unfold this as a single closed formula in $x_f$, $y_f$, $r$ $x_a$, $y_a$, and $\alpha$, the resulting formula is going to be huge alright. But if you give names to some of the intermediate values and split the whole thing into steps, each step is individually simple enough.
H: Different types of continuity for operators on Hilbert spaces In chapter one of K-theory and $C^$-algebras, a Friendly Approach, the author gives a very brief discussion about several types of continuity of operators between Hilbert spaces. Let $T: \mathcal{H}_1\to\mathcal{H}_2$ be a linear operator between Hilbert spaces, we say $T$ is $\tau_1-\tau_2$ continuous if $T$ is continuous when $\mathcal{H}_1$ is equipped with topology $\tau_1$ and $\mathcal{H}_2$ is equipped with $\tau_2$. The author says $T$ is norm-norm continuous iff $T$ is norm-weak continuous iff $T$ is weak-weak continuous, and that $T$ is weak-norm continuous iff $T$ is of finite rank. I know that norm-norm countinuous implies norm-weak continuous, and weak-weak continuous implies norm-weak continuous. But I cannot figure out the rest of the assertion. Can somebody give a hint? Also, before I continue to read the later parts of the book, I wonder whether these different types of continuity plays a crucial part in the theory. Thanks! AI: If $T$ is norm-weak continuous, apply the Principle of Uniform Boundedness to the family of linear functionals $x \to \langle y, Tx \rangle$ for $\\|y\\| \le 1$ to conclude that they are uniformly bounded, and therefore that $T$ is a bounded linear operator (i.e. norm-norm continuous). To show $T$ is weak-weak continuous, note that $\langle y, Tx \rangle = \langle T^ y, x \rangle$. If $T$ is weak-norm continuous, there is a finite set $\{y_1, \ldots, y_n\}$ such that $\\|Tx\\| \le 1$ whenever all $\|\langle y_j, Tx \rangle\| \le 1$. In particular, if all $\langle y_j, Tx\rangle = 0$ we must have $Tx=0$. But in any vector space of dimension $>n$, the intersection of $n$ hyperplanes contains a nonzero vector. So $\text{Ran}(T)$ has dimension at most $n$.
H: Holoedric isomorphism? While trying to read the following article Schottenfells, Ida May. Two Non-Isomorphic Simple Groups Of The Same Order 20,160. I found the term "holoedrically isomorphic". In an abstract for the article, I also came across the claim that "Holoedric isomorphism is the only isomorphism that can exist between two simple groups." I haven't been able to find a definition for this term anywhere. Presumably it is a stronger notion than that of a normal isomorphism of groups? What does it mean? AI: Isomorphism did not used to mean 1-1, just onto. Holoedric isomorphisms are both 1-1 and onto. See page 381 of Burnside's Theory of Groups (1ed). The standard english term was "simply isomorphic", the French term was isomorphisme holoédrique. The corresponding term for non-injective epimorphism (so onto, not 1-1) was "multiply isomorphic" in English, and isomorphisme meriédrique in French. "édrique" appears to be about the same as "edral" as in "dihedral". Schottenfels, Ida May. "Two non-isomorphic simple groups of the same order 20,160." Ann. of Math. (2) 1 (1899/00), no. 1-4, 147–152. MR1502265 DOI:10.2307/1967281
H: Is the Kleene/Brouwer ordering dense? This question was motivated by a statement in Simpson's Subsystems of Second Order Arithmetic (second edition), p. 168. It is straightforward to verify (in $\mathsf{RCA}_0$ for instance) that $\leq_{KB}$ is a dense liner [sic] ordering with no left endpoint and with the empty sequence $\langle \rangle$ as its right endpoint. (My emphasis.) That it's a linear ordering with no left endpoint and $\langle \rangle$ as its right endpoint is indeed straightforward to verify. So is the Kleene/Brouwer ordering dense, too? Let $Seq$ be the set of codes for finite sequences of natural numbers, and let $KB$ be the set of all pairs $(\sigma, \tau) \in Seq \times Seq$ such that either $\sigma \supseteq \tau$ or $$\exists{j < \min(lh(\sigma), lh(\tau))} \, [ \sigma(j) < \tau(j) \wedge \forall{i < j} (\sigma(i) = \tau(i)) ].$$ Now consider the following: $\langle 0 \rangle <_{KB} \langle \rangle$, so if the order is dense then there must be some $\upsilon$ such that $\langle 0 \rangle <_{KB} \upsilon <_{KB} \langle \rangle$. Clearly it can't be the case that $\langle 0 \rangle \supset \upsilon \supset \langle \rangle$. And there is no point that $\upsilon$ can diverge from the empty sequence at. So it must be the case that $\upsilon \supset \langle \rangle$ and there exists some $j$ such that $\upsilon(j) < 0$. But since there is no natural number $n < 0$ this can't be the case either. So we have a counterexample to the density of the Kleene/Brouwer ordering. AI: Proof of density: Suppose that $\sigma <_{KB} \tau$. The first case is that $\tau \prec \sigma$ (where $\prec$ is proper initial segment relation). Then $\|\tau\| < \|\sigma\|$. Define $\gamma$ or length $\|\tau\| + 1$ as follows $\gamma(n) = \begin{cases} \sigma(n) & \quad n < \|\tau\| \\ \sigma(\|\tau\|) + 1 & \quad n = \|\tau\| \end{cases}$ $\sigma <_{KB} \gamma$ since for all $n < \|\tau\|$, $\sigma(n) = \tau(n)$ but $\sigma(\|\tau\|) < \sigma(\|\tau\|) + 1 = \gamma(\|\tau\|)$. $\gamma <_{KB} \tau$ since $\tau \prec \gamma$. The second case is that there exists a $j < \min\{\|\sigma\|, \|\tau\|\}$ such that for all $n< j$, $\sigma(n) = \tau(n)$ and $\sigma(j) < \tau(j)$. Then define $\gamma$ to be the string of length $\|\tau\| + 1$ define by $\tau0$, i.e. $\tau$ concatenated by $0$. Than $\sigma <_{KB} \gamma$ since $j$ witnesses this property again. $\gamma <_{KB} \tau$ since $\tau \prec \gamma$.
H: How to compute the min-cost joint assignment to a variable set when checking the cost of a single joint assignment is high? I want to compute the min-cost joint assignment to a set of variables. I have 50 variables, and each can take on 5 different values. So, there are 550 (a huge number) possible joint assignments. Finding a good one can be hard! Now, the problem is that computing the cost of any assignment takes about 15-20 minutes. Finding an approximation to the min-cost assignment is also okay, doesn't have to be the global solution. But with this large computational load, what is a logical approach to finding a low-cost joint assignment? AI: In general, if the costs of different assignments are completely arbitrary, there may be no better solution than a brute force search through the assignment space. Any improvements on that will have to come from exploiting some statistical structure in the costs that gives us a better-than-random chance of picking low-cost assignments to try. Assuming that the costs of similar assignments are at least somewhat correlated, I'd give some variant of shotgun hill climbing a try. Alternatively, depending on the shape of the cost landscape, simulated annealing or genetic algorithms might work better, but I'd try hill climbing first just to get a baseline to compare against. Of course, this all depends on what the cost landscape looks like. Without more detail on how the costs are calculated, you're unlikely to get very specific answers. If you don't even know what the cost landscape looks like yourself, well, then you'll just have to experiment with different search heuristics and see what works best.
H: derivative at a given point I often see the following: $$ \left. \frac{\partial q}{\partial \alpha} \right\|_{\alpha = 0} $$ Where $q$ is a function $q(q', t, \alpha)$. Is that just the same as that? $$ \frac{\partial}{\partial \alpha} q(\alpha=0) $$ If they are the same, why do people write the first one? I find the second easier. AI: I don't know if I can add much that is helpful to what Dirk has written, but: You understanding is correct. The two expressions mean the same thing. It is indeed just a matter of taste. The vertical line $\bigg\lvert_{\alpha = 0}$ simply means that you take the expression before the line and evaluate it at $\alpha = 0$. The notation is also used in other cases than when evaluating derivatives. Usually I would interpret $\frac{\partial f}{\partial x}$ as a function again, and so if I wanted to use a notation similar to your second option I would probably write $$ \frac{\partial q}{\partial \alpha}(q',t,0). $$ In my opinion it seems a bit more messy when you put in the $(\alpha = 0)$. So if for example $f(x) = 2x + 1$, I wouldn't write $f(x=1) = 3$ (but some might do that). If you have a function of several variables, like $f(x,y) = xy + 2$, I would write $f(0,y)$ if I wanted to out the variable $x$ equal to zero. Also, I would say that the notation that you have in your comment to your question isn't good since it isn't clear if you are putting $\alpha = 0$ or if it is $q'$ or $t$ that is zero.
H: contour integral computations Let $C$ be the boundary of the square whose vertices are $1+i$, $1-i$, $-1 + i$ and $-1 -i$. Suppose that $C$ is oriented counterclockwise. How to compute a) $$\int_C \frac{e^z}{z-1/2} \, dz$$ b) $$\int_C \ln(z+3) \, dz$$ c) $$\int_C \bar{z} \, dz$$ I seee that we can use $f'(z)z'$ and maybe we can proceed this using a definition of Cauchy Riemann? Also, can we use the fundamental theorem of calculus on this? AI: In your first integral, the numerator is an entire function that has no zeros. Since it's an entire function, it has no poles, so the fraction can have a pole only if the denominator has a zero. Since the numerator has no zeros, every place where the denominator has a zero will be a pole of the function defined by the fraction. The point where the denominator has a zero is INSIDE the square. The value of the numerator at that point is $e^{1/2}$. So the integral is equal to $$ e^{1/2}\int_C \frac{dz}{z-1/2}. $$ In general, the integral $\displaystyle\int_B \frac{dz}{z-a}$ depends only on whether $a$ is inside or outside of the curve $B$, if $B$ winds once around every point that it surrounds. Is that enough for you to figure out the first one? (If not, I can add more.) The second one has a branch point at $-3$ but behaves well within the square you've describe, i.e. it's holomorphic on the curve and in the region it winds around. That gets you the value of the integral almost instantly, by citing a well known theorem. For the third one, the integral of the complex conjugate of a function is just the complex conjugate of the integral. So find $\displaystyle \int_C z~dz$ by citing the theorem mentioned in the paragraph above, and take its conjugate. Later note: @mary : Your comment above about the residue theorem suggests you might have some difficulties with what I wrote above. The $2\pi i$ that you mention is indeed the value of the integral $\displaystyle\int_C\frac{dz}{z-1/2}$, and if you're arare of that "$2\pi i$", you can probably figure that out. However, the residue theorem is not limit to finding integrals whose value is $2\pi i$. Rather, the residue theorem says that the integral along a curve that winds once around all the points it surrounds is $2\pi i$ times the sum of the residues at all of the points surrounded by the curve where a singularity occurs. (At least if one assumes only finitely many such points, and that's enough for the present occasion.) Where you integrate $z~dz$, you're integrating a function that has no singularities, hence that sum is $0$. When you integrate $\ln(z+3)~dz$, there are no sigularities at points that the curve winds around, so again that sum is $0$.
H: Solving a Maximum Likelihood Estimation with an exponential distribution I need someone's insight on applying a MLE for an exponential distribution. In a finance paper, I have the following: $\displaystyle d_i \sim \frac{\epsilon_i}{\lambda_i}$ where $\epsilon_i$ is i.i.d. exponentially distributed with parameter $= 1$. and $i=1,\ldots,n$. $d_i$ are duration time values like time between two events. The $\epsilon$ are not observed. $\lambda_i$ are not observed and must be replaced with estimates from an optimal filter under a $2^k$ states where $k$ can take value $2 \ldots 10$. Conditional on $\lambda_i$ the $d_i$ have an exponential distribution of $\lambda_i$ with density $p(d_i\|\lambda_i) = \lambda_i \exp[-\lambda_i d_i]$ The $\epsilon_i$ in $\displaystyle d_i \sim \frac{\epsilon_i}{\lambda_i}$ confuses me in the MLE application. First, is the $\epsilon_i$ relevant in the MLE computation? If yes, how does it influence the likelihood fucntion below: $$ \mathcal{L}(\lambda,d_1,\dots,d_n)=\prod_{i=1}^n f(d_i,\lambda)=\prod_{i=1}^n \lambda e^{-\lambda d}=\lambda^ne^{-\lambda\sum_{i=1}^nd_i} $$ AI: You seem to be mixing up two different ideas. The $n$ random variables are independent but not identically distributed. If your observation is $(d_1, d_2, \ldots, d_n)$, then the likelihood function is the product $$\prod_{i=1}^n \lambda_i \exp(-\lambda_i d_i)$$ which is a function of $n$ unknown parameters $\lambda_1, \lambda_2, \ldots, \lambda_n$. This likelihood function has maximum value at $(\lambda_1, \lambda_2, \ldots, \lambda_n) = (d_1^{-1}, d_2^{-1}, \ldots, d_n^{-1})$. More likely, your model is that of $n$ independent samples of an exponential random variable with unknown parameter $\lambda$. If your observation is $(d_1, d_2, \ldots, d_n)$, then the likelihood function is the product $$\prod_{i=1}^n \lambda \exp(-\lambda d_i) = \lambda^n\exp\left (-\lambda \sum_{i=1}^n d_i \right)$$ which is a function of the single unknown parameter $\lambda$. The maximum value of this likelihood occurs at $\displaystyle \lambda = \frac{n}{\sum_{i=1}^n d_i } = \frac{1}{\bar{d}}$ where $\bar{d}$ is the sample mean $\displaystyle \frac{1}{n}\sum_{i=1}^n d_i$.
H: Does every set have a well ordering with greatest element? Does every non-empty set have a well ordering with greatest element? It is well known that every set has a well ordering. But can we also assume that this well ordering has greatest element? [Edited to remove an ambiguity revealed by the answers and comments] AI: Yes. Let $\langle S,\le\rangle$ be a well-order. If $S$ already has a greatest element with respect to $\le$, we’re done. If not, let $s_0$ be the $\le$-least element of $S$, and define a new well-ordering $\preceq$ of $S$ as follows: if $s,t\in S\setminus\{s_0\}$, then $s\preceq t$ iff $s\le t$; if $s\in S$, then $s\preceq s_0$. This produces a well-ordering of $S$ with $s_0$ as the largest element: it simply moves $s_0$ from the beginning of the order to the end.
H: Why can/do we multiply all terms of a divisor with polynomial long division? I'm trying to understand why polynomial long division works and I've hit a wall when trying to understand why we multiply all terms of the divisor by the partial quotient. Consider: $$\frac{x^2 + 3x + 2}{x + 2}$$ During the first step we divide ${x^2}$ by ${x}$ giving us a partial quotient of ${x}$. Next, we multiply the partial quotient by the first term of the divisor, giving us ${x^2}$. So far so good, but this next step is what I don't understand: why do we then multiply the next term of the divisor by the partial quotient (i.e. ${x}$ * ${2}$)? It seems as though we're testing to see if ${x}$ can be divided into the first term of the dividend, and if it can, then we distribute the result over the whole divisor which is then subtracted. I don't understand how we can do that when we're only testing the divisibility of those first terms. I tried a slightly different example to see what happens which highlights what I mean: $$\frac{x^2}{x + 1}$$ First, we test the divisibilty of ${x^2}$ by ${x}$. Obviously it goes ${x}$ times, but then I wanted to see if I could carry on and multiply ${x}$ * ${1}$. I carried on and the quotient becomes ${x + \frac{1}{x + 1} - 1}$. Checking $({x + 1})({x + \frac{1}{x + 1} - 1})$ does give the original dividend of ${x^2}$. Having worked through that problem I just can't see why we're able to do the multiplication of the second term of the divisor, subtract it and get everything to hold true. Essentially, if I subtract ${x^2}$ from ${x^2}$ I end up with nothing. So the second multiplication, to me at least, seems unclear as to why it works and what the purpose is. I know I'm missing something simple here but I can't seem to make the connection. Could someone explain this to me please? AI: Think about polynomial long division in much the same way that you think about division of real numbers. For instance, say you want to divide 48820 by 28. In a sense, doing division in this way is "greedy" -- you divide by the largest possible parts first, then work down from there. Now, , when you long divide by 28, you first see if 28 goes into the leading term, 4. It does not, so then you check if it goes into 48 (which it does exactly once). Then, when you "bring down" the next term, you multiply the "partial quotient" -- 1 -- by 28. Right? Now think of 28 as 2*10+8. Same operations hold. Now, replace 10 with x. In order to keep the algorithm the same, you have to multiply by the whole denominator. Edit: To expand a little... Consider the previous example. 48820/28 = 1743+16/28. Now, let's compute $\frac{4x^4+8x^3+8x^2+2x+0}{2x+8}$. By doing polynomial long division, we obtain: $2x^3-4x^2+20x-79+\frac{632}{2x+8}$. Now you might say, "wait a minute, that has a leading coefficient of 2, but obviously 28 goes into 48 only once!" Yes, this is true, but notice that we have minus signs here. That is key. Now, let's set $x = 10$ and see what we compute: $2000-400+200-79+\frac{632}{28}$. Doing the arithmetic, we see this is exactly our desired result.
H: no simple group of order $945$ I need to show that there are no simple groups of order $945$. I've tried the regular method using the Sylow theorems. $$\|G\|=945=3^3\cdot5\cdot7 $$ If $G$ is simple then there should be 7 Sylow-3 groups ; 21 Sylow-5 groups and 15 Sylow-7 groups. Even if they would all intersect trivially, there will still be no contradiction. Any ideas? AI: If as you say there are 7 Sylow 3-subgroups, then the normalizer $N$ of one of these subgroups has index 7. If $G$ is simple, then there is an injective permutation representation $G \hookrightarrow S_7$, and so $\|G\| = 945$ must divide $7!$, but this is not the case.
H: Quantative Comparison - Which is bigger A quantitative comparison question states: if $r<s<t$ and the average arithmetic mean of r , s and t is 90 . Which of the following is bigger a)The average of s and t or b)$90$. The answer is a. However I cant quiet figure out how they got this: I know that : $\frac{r+s+t}{3}=90$ so Average of s and t would be $\frac{270-r}{2}$. How can I precisely say that this average is definitely greater than 90 ? Edit: After reviewing some of the pointers here is what I got. Please Let me know if this is correct since $s+t = 270 -r$ so $\frac{270-r}{2}> 90$ thus $270-r > 180$ so $90-r >0$ Now since r cant be negative which is impossible (even if it is) the $90-r > 0$ Is this correct ? AI: Suppose that the average of $s$ and $t$ is $90$ or more; then $\frac{s+t}2\ge 90$, so $s+t\ge 180$, and therefore $r+s+t\ge r+180$. On the other hand, you know that $r+s+t=270$, so $270\ge r+180$, and therefore $90\ge r$. Is it possible to have $90\le r<s<t$ and $r+s+t=270$? Added: You ask whether this argument is correct: since $s+t = 270 -r$ so $\frac{270-r}{2}> 90$ thus $270-r > 180$ so $90-r >0$ Now since $r$ cant be negative which is impossible (even if it is) the $90-r > 0$ It needs a bit of work. The fact that $s+t=270-r$ does not imply that $\frac{270-r}2>90$, so your first so is inappropriate. What you want to say is this: Suppose that $\frac{270-r}2>90$; then $270-r>180$, so $90-r>0$. Next, it’s true that $r$ can’t be negative, but this is irrelevant. What you can conclude here is that $r<90$. Unfortunately, none of this really helps: it just shows that there is no immediate problem with the assumption that the average of $s$ and $t$ is more than $90$. In order to show that this actually is the case, you want to suppose that it isn’t, and derive a contradiction. Thus, you really ought to start out: Suppose that $\frac{270-r}2\le 90$; then $270-r\le 180$, so $90-r\le 0$, and therefore $90\le r$. Now you can go on to argue as follows: By hypothesis, $r<s<t$, so we have $90\le r<s<t$. This implies that $$r+s+t>r+r+r\ge 3\cdot 90=270$$ and hence that $$\frac{r+s+t}3>\frac{270}3=90\;.$$ This, however, contradicts the hypothesis that the arithmetic mean of $r,s$, and $t$ is $90$. Thus, the supposition that $\frac{270-r}2\le 90$ must be false, i.e., it must be true that $\frac{270-r}2>90$ and therefore that the arithmetic mean of $s$ and $t$ is more than $90$.
H: Verify trigonometric equation $\frac{(\sec{A}-\csc{A})}{(\sec A+\csc A)}=\frac{(\tan A-1)}{(\tan A+1)}$ How Would I verify the following identity. $$\frac{(\sec{A}-\csc{A})}{(\sec A+\csc A)}=\frac{(\tan A-1)}{(\tan A+1)}$$ I simplified it to $$\frac{(\sin{A}-\cos{A})}{(\sin{A} \cos{A})}\div\frac{(\sin{A}+\cos{A})}{(\sin{A}\cos{A})}$$ AI: Hint: Start by multiplying top and bottom on the left by $\sin A$.
H: Differentiating the posterior distribution function I am learning about Bayesian statistics and I'm currently doing loss functions. Let $f(\theta \| \mathbf{x} ) $ be a posterior pdf . Let $F(\theta \| \mathbf{x} ) $ be the associated distribution function. I want to differentiate $F(a - D\| \mathbf{x} )$ with respect to $D$. In this case $D$ is the "decision" which is to be optimised and $a$ is constant. I am having a problem here: $$\frac{d}{d D}F(a - D\| \mathbf{x} ) = \frac{d}{d D} \left( \int_{-\infty}^{a - D} f(\theta\|\mathbf{x}) d \theta \right)$$ $$=f(a - D \| \mathbf{x})$$ Is this correct ? I think it might be wrong. Should it be $=-f(a - D \| \mathbf{x})$ because I have to apply the chain rule somewhere ?? Or is something else wrong ? I know there are some issues about integration under differentiation but my teacher said I don't need to worry about that now, and just use this: $$\frac{d}{dy} \int_{-\infty}^y f(t) dt=f(y)$$ I'm doing self-study (with a bit of teacher guidance in his own time so I don't like to ask him too much) but I feel a bit out of depth now and school breaks up for the holidays next week ! AI: Think of this as a special case of differentiating $h(f(D))$ with respect to D: $h(f(D)) = \int_{-\infty}^{f(D)}g(x)\text{d}x$ where $\text{d}h/\text{d}D = h'(f(D))f'(D)$ via the chain rule, as you thought. In your case, $f(D) = a-D$, so $h'(f(D)) = g(a-D)$, as you know, and $f'(D) = -1$.
H: In Taxicab Geometry, what is the solution to d(P, A) = 2 d(P, B) for two points, A and B? Taxicab and Euclidean geometry differ a great deal, due to the modified metric function: $$d_T(A,B)=\|x_a-x_b\|+\|y_a-y_b\|$$ (Note that this means when measuring distance, it is not the length of the hypotenuse, but the sum of the legs of the same right triangle.) My Main Problem In Euclidean geometry, the answer to the question "Find the locus of points $X$ such that: $d(X, A) = 2 * d(X, B)$" yields a regular, Euclidean circle. A little bit of algebra makes this very trivial. But what is the answer to the same problem, but for $d_T$? What I Know So Far This kind of geometry actually has a very interesting property, namely that as things rotate, their measures change. Consider the cases where points share either one of their coordinates. Many times, those situations yield the same answers as do their Euclidean counterparts. Some things are noticeably different, though. For instance, a circle, as defined as the set of points a fixed distance from one point, actually comes out as a square, rotated 45 degrees. It is also trivial to illustrate that. It did occur to me that the answer to this problem could be analogous to Euclidean geometry, and the solution may simply be a Taxicab circle (a square). But this didn't seem to work out. Plus, I worked out the solution for the points sharing an x or y coordinate, I end up with two mirror-image line segments. But the general case, where the two points are corners of any rectangle still eludes me. My second educated guess was that the solution could be a Euclidean circle, but this didn't work out either. Lastly, some constructions seemed to differ depending on whether the points I chose formed the opposite diagonal corners of a general rectangle, or a square. E.g. (0,0) and (3, 3) seem to be a yet different type of exception. Any thoughts on this problem would be greatly appreciated! AI: EDIT: alright, got it. There is a special case, when $AB$ are both on a horizontal or a vertical line. Then the shape asked for is a convex kite shape, two orthogonal edges of equal length with slopes $\pm 1,$ the other two edges with slopes out of $3,-3, \frac{1}{3},\frac{-1}{3}, $ the whole figure symmetric across the $AB$ line, as one would expect. Otherwise, draw the rectangle with $AB$ as the endpoints of one diagonal. If this rectangle is a square, or the longer edge is no more than twice the length of the other, the shape is an isosceles trapezoid, as described below. If the longer edge is more than twice as long as the other, the shape is a nonconvex hexagon, as in Rahul's image. Furthermore, if we fix the longer edge and call it length $1,$ as the shorter edge goes to length $0$ the resulting hexagon shape comes to resemble the kite shape, which is its limit. Anyway, there are just the three possibilities. Note that, in the cases when $AB$ are not in a line parallel to the $x$ or $y$ axis, we can color the nine regions of the tic-tac-toe board as a chess or checkerboard; then regions the same color as the rectangle have segments (if any) of slope $\pm 1,$ while any segments in the regions that are the other color have slope among $3,-3, \frac{1}{3},\frac{-1}{3}. $ It is all pretty rigid. ORIGINAL: A little to add to what Ross and Rahul pointed out. One segment is automatically there, the segment inside the rectangle passing through the 2/3 point along the $AB$ diagonal, but with slope $\pm 1,$ and closer to $B.$ This segment is part of a "circle" around $A$ as well as a "circle" around $B.$ There is another such with the same slope, passing through a point we might as well call $A',$ which is the reflection of $A$ along the line $AB,$ the same distance from $B$ as $A$ is but on the other side. This can be the longest edge involved, as it continues through one of Ross's tic-tac-toe regions. There may also be common "circle" segments rotated $90^\circ$ from those, as in Rahul's diagram. So one ought to check for those first, in the nine regions, any segments with slope $\pm 1$ that are the overlap of a circle around $A$ and circle around $B.$ Rahul has shown that you can get three such segments, and there are automatically at least two, but I do not see how one could get four such. So I think the diagram you are asking about is likely to be either a quadrilateral (an isosceles trapezoid) or a non-convex hexagon, being an isosceles trapezoid with one vertex replaced by an extra triangle, one edge of which is orthogonal to the two parallel edges of the trapezoid part. For that matter, there are really only two genuinely distinct slopes allowed, $\pm 1$ and $3,-3, \frac{1}{3},\frac{-1}{3}. $
H: Finding the derivative of a function using the Product Rule I'm home teaching myself calculus because I'm 16 and therefore too young to take an actual class with a teacher, so I apologise if this seems simple. I understand the definition of the Product Rule and its formula: "If a function $h(x)=f(x)\times g(x)$ is the product of two differentiable functions $f(x)$ and $g(x)$, then $h'(x) = f(x)\times g'(x)+f'(x)\times g(x)$". I did a question to find the derivative of $g(x) = (2x+1)(x+4)$ using the Product Rule. Now on the solutions sheet it says I must begin by writing: $g'(x)=(2x+1){\bf (1)}+{\bf (2)}(x+4)$ What confuses me are the terms that I have put in bold. (the terms $(1)$ and $(2)$). I believe the term $(1)$ is $g'(x)$ from the formula and the term $(2)$ is $f'(x)$ from the formula. How am I supposed to know these 2 terms? Am I supposed to find the derivative of $(2x+1)$ and $(x+4)$ before going on to the question? I also apologise if this is quite messy. AI: I don't really understand what the stars are suppose to mean, but I can try to explain the product rules. As you pointed out, $h'(x) = f'(x)g(x) + f(x)g'(x)$. In your example, the function $h(x) = (2x +1)(x + 4)$. First, you need to recognize that $h$ is the product of two functions: $f(x) = 2x + 1$ $g(x) = x + 4$ In order to apply the product rule formula, you need to find $f'(x)$ and $g'(x)$. I assume that you know how to do this for the two functions above. Hence you should obtain $f'(x) = 2$ $g'(x) = 1$ Sticking all of these terms into the product the rule $h'(x) = f'(x)g(x) + f(x)g'(x)$ $h'(x) = (2)(x + 4) + (2x + 1)(1) = 2x + 8 + 2x + 1 = 4x + 9$
H: Compute: $\lim_{n\to\infty} n^{p+1} \int_{0}^{1} e^{-nx} \ln (1+x^p) \space dx $ Find the following limit for any $p$ natural number: $$\lim_{n\to\infty} n^{p+1} \int_{0}^{1} e^{-nx} \ln (1+x^p) \space dx $$ If i'm not wrong, without much effort one may see that this integral may be rewritten as Gamma function and the limit is $p!$ . I'm just curious about more different ways one might go here. AI: The limit is $\Gamma(p+1)$. To see that, first integrate by parts. We have \begin{align} I_n&:=n^{p+1}\int_0^1e^{—nx}\ln(1+x^p)dx\\ &=n^{p+1}\left(\left[\frac{-e^{-nx}}n\ln(1+x^p)\right]_0^1-\int_0^1\frac{-e^{-nx}}n\frac{px^{p-1}}{1+x^p}dx\right)\\ &=-n^pe^{-n}+pn^p\int_0^1e^{-nx}\frac{x^{p-1}}{1+x^p}dx, \end{align} so we need to compute the limit of $J_n:=n^p\int_0^1e^{-nx}\frac{x^{p-1}}{1+x^p}dx$. We use the substitution $t=nx$ (then $dt=ndx$) to get \begin{align}J_n&=n^p\int_0^ne^{-t}\frac{t^{p-1}}{n^{p-1}\left(1+\left(\frac tn\right)^p\right)}dt\frac 1n \\ &=\int_0^ne^{-t}\frac{t^{p-1}}{1+\left(\frac tn\right)^p}dt. \end{align} This quantity converges to $\Gamma(p)$. Indeed, $\int_0^ne^{—t}t^{p-1}dt\to \Gamma(p)$ and \begin{align} \left\|J_n-\int_0^ne^{—t}t^{p-1}dt\right\|&\leq \int_0^ne^{-t}t^{p-1}\frac{\left(\frac tn\right)^p}{1+\left(\frac tn\right)^p}dt\\ &=\int_0^ne^{—t}t^{p-1}\frac{t^p}{n^p+t^p}dt\\ &\leq \int_0^{+\infty}e^{—t}t^{p-1}\frac{t^p}{n^p+t^p}dt, \end{align} and we conclude by the monotone convergence theorem.
H: $<$ in a preorder The author of the book I am studying defines $<$ for a poset as If $x, y \in X$, where $X$ is a poset, then we shall write $x < y$ to mean that $x \le y$ and $x \ne y$. From this, I can conceive of two definitions for $<$ for a preorder: 1) If $x, y \in X$, where $X$ is a preorder, then we shall write $x < y$ to mean that $x \le y$ and $x \ne y$. or 2) If $x, y \in X$, where $X$ is a preorder, then we shall write $x < y$ to mean that $x \le y$ and $y \not\le x$. Which of these is more appropriate? AI: You may know by now that if you have a preorder then you can take a quotient by the equivalent relation $x\sim y\iff x\leq y\land y\leq x$ and have a poset. The definition should be such that it carries over to the quotient, so the second definition is more appropriate. In the first one we can have $x\neq y$ and $x\leq y\land y\leq x$, but $x\sim y$ so in the induced poset $[x]=[y]$.
H: Embedding elliptic curves into the general linear group Is it possible to embedd an elliptic curve $E:\;\; y^2=x^3+ax+b$, defined over an algebraically closed field $k$, into some $GL_n(k)$ ? AI: If $E\hookrightarrow\mathrm{GL}_n$ is a closed immersion of $k$-schemes ($k$ any field) inducing an isomorphism with the closed subscheme $Z$ of $\mathrm{GL}_n$, then $Z$ is necessarily affine, and, being isomorphic over $k$ to $E$, is also proper over $k$. It is therefore finite over $k$. But this implies that $E$ is finite over $k$, which is not the case.
H: What is the relationship between integrals and areas? Possible Duplicate: Why is the area under a curve the integral? Why does calculating an integral or an anti-derivatives represent an area? How do they figure it out? What is the relationship between them? AI: This is basic level stuff covered in first course of calculus. You may also refer to this: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition
H: Distance Formula for n-dim Barycentric Coordinates Assume that we are given every distance between each pair of points from a $n$-simplex $\triangle$. Given $n$-dimensional barycentric coordinates (measured with respect to $\triangle$) of two points, how do we compute the distance between the two points? A more detailed explanation of the question is given below: Given $n+1$ vectors $\{\mathbf{a}_1,\cdots,\mathbf{a}_{n+1}\}$ from $\mathbb{R}^n$, we can give homogeneous coordinates to another vector $\mathbf{x}\in\mathbb{R}^{n+1}$ as $(\lambda x_1:\lambda x_2:\cdots:\lambda x_n)$ where $\lambda\in\mathbb{R}-\{0\}$ and $\mathbf{x}={(\sum x_i \mathbf{a}_i)}/{(\sum x_i)}$. $x_i$ can be alternatively defined as the volume of the simplex formed by vectors $\{\mathbf{x},\mathbf{a}_1, \cdots, \mathbf{a}_{i-1},\mathbf{a}_{i+1},\cdots,\mathbf{a}_{n+1}\}$ (which can be computed via Cayley-Menger determinant, I believe.) Assume that we are given matrix $M$ such that $M_{ij}=\\|\mathbf{a}_i-\mathbf{a}_j\\|^2$, i.e. we are given the distances between arbitrary two points from $\{\mathbf{a}_1,\cdots,\mathbf{a}_{n+1}\}$. If we are given barycentric coordinates $P,Q\in\mathbb{R}^{n+1}$ of two vectors $\mathbf{p},\mathbf{q}\in\mathbb{R}^n$, can we express $\\|\mathbf{p}-\mathbf{q}\\|^2$ in terms of $P,Q,M$? This is already answered for $n=1,2$. For $n=1$ it is trivial. For $n=2$, assuming that $P$ and $Q$ are normalized (that is, their coordinate sums are equal to 1), and $P=\{p_1:p_2:p_3\},Q=\{q_1:q_2:q_3\}$, distance $\\|\mathbf{p}-\mathbf{q}\\|^2=\sum_{cyc}\frac12(-a^2+b^2+c^2)(p_1-q_1)^2$ where $a,b,c$ are sidelengths of the simplex. AI: Suppose $p = \sum_i \pi_i a_i$, $q = \sum_i \zeta_i a_i$, with $\sum_i \pi_i = \sum_i \zeta_i = 1$. Then $\\|p-q\\| = \\| \sum_i (\pi_i - \zeta_i) a_i\\|$. Then we can express the norm as $\\|p-q\\|^2 = \sum_{i,j} (\pi_i - \zeta_i)(\pi_j-\zeta_j) \langle a_i, a_j \rangle$. So we just need to figure out the $\langle a_i, a_j \rangle$. Suppose we let $\hat{a}_i = a_i -a_1$. Then since $\sum_i (\pi_i - \zeta_i) a_1 = 0$, we have that $p-q = \sum_i (\pi_i - \zeta_i) \hat{a}_i$. Furthermore, since $a_i-a_j = \hat{a}_i - \hat{a}_j$, the matrix $M$ gives us the distances $\\|\hat{a}_i - \hat{a}_j\\|$. Since $\hat{a}_1 = 0$, we have $\\|\hat{a}_i\\|^2 = M_{1i}$ We need to compute $\\|p-q\\|^2 = \sum_{i,j} (\pi_i - \zeta_i)(\pi_j-\zeta_j) \langle \hat{a}_i, \hat{a}_j \rangle$. The polarization identity gives the inner product as (note the minus sign) $$-\langle x , y \rangle = \frac{1}{2} ( \\|x-y\\|^2-\\|x\\|^2-\\|y\\|^2),$$ from which we have $\langle \hat{a}_i, \hat{a}_j \rangle = -\frac{1}{2}(M_{ij} - M_{i1}-M_{j1})$. Consequently, $\\|p-q\\|^2 = - \frac{1}{2} \sum_{i,j} (\pi_i - \zeta_i)(\pi_j-\zeta_j) (M_{ij} - M_{i1}-M_{j1})$. This can be written as $- \frac{1}{2} \langle \pi-\zeta, B(\pi-\zeta) \rangle$, where $B = M - M_{\cdot 1} 1^T - 1^T M_{1 \cdot}$, where $1$ is a vector of $1$'s, $M_{\cdot 1}$ represents the first column of $M$, and $M_{1 \cdot}$ represents the first row of $M$ (these last two vectors are transposes of each other, of course).
H: A problem with definitions of rotation/reflection matrix/operator I am a math undergraduate student taking a course called "Geometry and symmetry" and I have something I don't understand with the definition the lecture gave in class. Definition: $T\,:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is called a linear rotation operator if there exist an orthonormal basis $B$ s.t $$\begin{pmatrix}cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta)\\ & & 1\\ & & & 1\\ & & & & 1\\ & & & & & & .\\ & & & & & & & .\\ & & & & & & & & .\\ & & & & & & & & 1\\ & & & & & & & & & 1 \end{pmatrix}$$. Definition: $T\,:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is called a linear reflection operator if there exist an orthonormal basis $B$ s.t $$\begin{pmatrix}1\\ & 1\\ & & .\\ & & & .\\ & & & & .\\ & & & & & 1\\ & & & & & & -1 \end{pmatrix}$$. The definitions for matrix $P$ are that a matrix is called linear reflection matrix if there is an orthogonal matrix $P$ s.t $P^{-1}AP$ is in one of the above forms, accordingly. Later on, there is a theorem that says that is $T\,:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is orthogonal then there exist an orthonormal basis $B$ s.t $$[T]_{B}=\begin{pmatrix}R_{\theta_{1}}\\ & R_{\theta_{2}}\\ & & .\\ & & & .\\ & & & & .\\ & & & & & R_{\theta_{k}}\\ & & & & & & 1\\ & & & & & & & 1\\ & & & & & & & & .\\ & & & & & & & & & 1\\ & & & & & & & & & & -1\\ & & & & & & & & & & & .\\ & & & & & & & & & & & & .\\ & & & & & & & & & & & & & -1 \end{pmatrix};R_{\theta}:=\begin{pmatrix}cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta) \end{pmatrix}$$ My question is: if $T$ is orthogonal then if $A$ is a matrix s.t $Tv=Av$ then $A$ is orthogonal hence $\|A\|=1$ or $\|A\|=-1$. in the first case I learned that by $A$ is a linear rotation matrix, in the second case I learned that $A$ is linear reflection operator, so I deduce that there is an orthonormal basis in which $A$ have the form of the two matrix above - i.e rotating around some $2$ dimensional space fixing the other $n-2$ subspace of $\mathbb{R}^{n}$ or rotating around some vector fixing the other $n-1$ subspace of $\mathbb{R}^{n}$. It seems that the matrix in the theorem is very different than the first two matrix in the question since it has $k$ (that can be $>1$) rotations, and it can reflects around more then $1$ vector... Can someone please explain whats going on here? I expected the theorem to be the $[T]_{B}$ is one of the first two matrices in the question, instead I have in front of me a strange looking matrix that I can not classify as a rotation or a reflection matrix, so where in my reasoning I am wrong (I determined that $[T]_{B}$ is the first matrix in the question if $A$ above have det $1$ and the second if $A$ have det -1) ? AI: I think your confusion stems from the mistaken belief that every orthogonal matrix with positive determinant is what you call a "linear rotation operator". That happens to be true in three dimensions, but not more. However, every orthogonal matrix with positive determinant is a composition of such linear rotation operators, also known as simple rotations. So you can think of the latter as the set of "atomic" rotations that all other rotations are composed of.
H: Which is the biggest integer that divides all integers that are the product of three consecutive odd numbers? I read this problem from a high-school-math-problems-calendar, and I'm solving them in my spare time just for the fun of it (what in math is not about the fun? =) ), but this little one it's been hard for me (maybe is a silly problem for mathematicians). Again, this is the problem statement: Which is the biggest integer that divides all integers that are the product of three consecutive odd numbers? So far, I did this: $$ (m - 1)\cdot(m+1)\cdot(m + 3)\tag{1} $$ where $m = 2k$, expanding: $$ m^3+3m^2-m-3=m(m^2-1)+3(m^2-1) $$ (Of course, if I keep simplifying the equation I'll get (1) again). The right term is divisible by $3$, but the left term is divisible by $2$, don't know what else to do... Any ideas? I know that the answer is $3$ because I made a little program that test the mod $3$ of the product of first $1000$ consequtive odd numbers and it prints $0$ for all of them, but don't know how to prove it python -c "for n in xrange(1, 1000, 2): print (n * (n + 2) * (n + 4)) % 3;" Any help would be appreciated (I'd appreciate if somebody edit/adds the correct tags for this question) AI: $-3=(-1)\cdot1\cdot3$ is the product of three consecutive odd integers, so any integer that divides all such products must divide $-3$. The divisors of $-3$ are $-1,1,-3$, and $3$, so these are the only candidates. To finish the proof, we need only show that $3$ always does divide the product of three consecutive odd integers. Let $n$ be any odd integer, and consider the product $m=n(n+2)(n+4)$. There are three possibilities. If $n$ is a multiple of $3$, then certainly so is $m$. If $n=3k+1$ for some integer $k$, then $n+2=3k+3=3(k+1)$ is a multiple of $3$, and therefore so is $m$. If $n=3k+2$ for some integer $k$, then $n+4=3k+6=3(k+2)$ is a multiple of $3$, and therefore so is $m$. In every case, therefore, $m$ is a multiple of $3$, and $3$ is therefore the largest integer that divides every product of three consecutive odd integers.
H: Relation between Diameter and Tangent of circle A comparative question states: One side of rectangle is the diameter of a circle. The opposite side of rectangle is tangent to the circle.Which is bigger a)The perimeter of rectangle or b)The circumference of the circle (Ans=$b$) Now I know a tangent is perpendicular to the circle. However I can't figure the rest out.How did the text conclude that a) is bigger? AI: Let $r$ be the radius of the circle. The long sides of the rectangle have length $2r$, and the short sides have length $r$, so the perimeter of the rectangle is $2(2r+r)=6r$. The circumference of the circle is $2\pi r$, and $2\pi>6$, so the circumference of the circle is greater than the perimeter of the rectangle. This picture may help:
H: The probabilty of a new arrangement of 52 cards deck? I just read this article, it claims that if you just shuffle a 52 card deck, you will mostly be creating an arrangement that no human have ever seen before. But this doesn't seem right since every time we create a new arrangement, this one is now a candidate to happen again, so every time we fail, the odds increase. How can we mathematically review this claim? AI: The number of arrangements is $52!$. Using a tool like Wolfram Alpha, or by using the Stirling approximation,, or even just a scientific calculator, we find that $52!$ is larger than $8\times 10^{67}$. Assume, outrageously, that there have been $10$ billion people on Earth for $10000$ years, each shuffling and dealing a deck of cards every second. In $10000$ years, there are fewer than $3.2\times 10^{11}$ seconds. Multiplying by $10$ billion gets us $3.2\times 10^{21}$ shuffled decks, far short of $8\times 10^{67}$, which is a seriously large number. So a very tiny fraction of the possible hands has been dealt. It is now probably intuitively reasonable that if all orderings are equally likely, then with probability close to $1$, all orderings have been different. But the intuition can be unreliable (witness the Birthday Problem). So we make a more detailed estimate. It turns out that the relevant fact is that the square of $3.2\times 10^{21}$, though huge, is a very tiny fraction of the number of possible deals. Mathematical details: Suppose that there are $N$ different arrangements of cards, and that we shuffle and deal the cards out independently $n$ times, where $n$ is smaller than $N$. Then the probability that all the results are different is $$\frac{N(N-1)(N-2)\cdots(N-n+1)}{N^n}.$$ The top is bigger than $N-n$, so the probability is bigger than $$\left(1-\frac{n}{N}\right)^n.$$ Using the fact that $\left(1-\frac{x}{k}\right)^k$ is approximately $e^{-x}$, we find that the probability the results are all different is $\gt e^{-n^2/N}$. Let $N=52!$, and let $n$ be our absurdly high number of $3.2\times 10^{21}$ shuffles. If $x$ is close to $0$, then $e^{-x}$ is approximately $1-x$. Thus the probability the shuffles are all different is greater than $1-\frac{n^2}{N}$. (A better estimate gives that it is approximately $1-\frac{n^2}{2N}$.) This is a number which is very close to $1$. The probability there has been one or more repetition, even with $n$ grossly overestimated, is less than $10^{-25}$. The probability of at least one match grows rapidly as $n$ grows. Already at $n=10^{33}$, it is roughly $6\times 10^{-3}$, not large but certainly not negligible.
H: Convergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$? I need to prove the convergence/divergence of the series $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$ based on the convergence/divergence of the series $\sum_{n=1}^{\infty }a_{n}$. It is given that $a_{n}> 0$, $\forall n\in \mathbb{N}$ If the series $\sum_{n=1}^{\infty }a_{n}$ is convergent, then from $\frac{a_{n}}{1+na_{n}}< a_{n}$ and the comparison test, we conclude that $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$ is convergent. However, if the series $\sum_{n=1}^{\infty }a_{n}$ is divergent, I have no idea how to prove the convergence/divergence of $\sum_{n=1}^{\infty }\frac{a_{n}}{1+na_{n}}$. It is definitely divergent (just take $a_{n}=\frac{1}{n}$), but I have no clue how to prove it. AI: As you noted, if $a_n=\frac{1}{n}$, then $\sum_{n=1}^\infty \frac{a_n}{1+na_n}=\sum_{n=1}^\infty \frac{1}{2n}$ diverges. On the other hand, say $a_{2^k}=\frac{1}{2}$ and $a_{n}=2^{-n}$ for $n$ not a power of two. Then $$\sum_{n=1}^\infty \frac{a_n}{1+na_n} < \sum_{n=1}^\infty \frac{2^{-n}}{1+n2^{-n}} + \sum_{k=1}^\infty \frac{1}{2+2^k} \, ;$$ since both right-hand sums are convergent and every term of the left-hand sum is positive, the left-hand sum also converges. In both cases, $\sum a_n$ diverges. So merely knowing that $\sum a_n$ diverges is insufficient to tell you about the convergence behavior of $\sum \frac{a_n}{1+na_n}$. On the other hand, if $a_n$ is decreasing and $\sum a_n$ is divergent, then $\frac{a_n}{1+na_n}>\frac{a_n}{1+s_n}>\frac{a_n}{2s_n}$ for $n$ sufficiently large, where $s_n=\sum_{k=1}^n a_k$. Moreover, I claim that if $\sum a_n$ diverges, then so does $\sum \frac{a_n}{s_n}$. Hence, if you add the condition that $a_n$ is decreasing to the original problem, it will in fact be true that $\sum a_n$ converges if and only if $\sum \frac{a_n}{1+na_n}$ does. To establish the claim, fix some positive integer $N$ and look at the tail of $\sum \frac{a_n}{s_n}$ beginning at the $N^{\rm th}$ term. For any positive integer $k$, we have $$\sum_{i=1}^k \frac{a_{N+i}}{s_{N+i}} > \sum_{i=1}^k \frac{a_{N+i}}{s_{N+k}}=\frac{s_{N+k}-s_N}{s_{N+k}}=1-\frac{s_N}{s_{N+k}} \, .$$ Since the sequence $s_n$ is divergent, we can choose $k$ such that $s_{N+k}>2s_N$, and thus $1-\frac{s_N}{s_{N+k}}>\frac{1}{2}$. But $N$ was arbitrary; hence the tails of $\sum \frac{a_n}{s_n}$ don't decrease to zero, and so the sum does not converge.
H: PDE with series Consider this PDE The solution to the PDE is So what I am having trouble is solving it using this method. I am going to say that my $u(x,t) = \sum_{n=1}^{\infty} u_n(t) \sin(nx)$ and $x \sin(t) = \sum_{n=1}^{\infty}h_n(t)\sin(nx)$ The reason I chose sine for my inhomogeneous term is because my book recommends it. But I think it is because if I use cosine, I would get a $\frac{a_0}{2}$ term and it would be difficult. To solve for the coefficients of $h_n(t)$, I get $h_n(t) = \frac{2}{\pi}\int_{0}^{\pi} x\sin(t) \sin(nx) dx = \frac{2\sin(t)(-1)^n}{n}$ Substituting everything into $u_{tt} = u_{xx} + x\sin(t)$ gives me $$ \sum_{n=1}^{\infty}u''_n(t) \sin(nx) + \sum_{n=1}^{\infty}u_n(t)n^2\sin(nx) = \sum_{n=1}^{\infty}\frac{2\sin(t)(-1)^n}{n}\sin(nx)$$ Dividing out that sine, I'll get $$ \sum_{n=1}^{\infty}u''_n(t) + \sum_{n=1}^{\infty}u_n(t)n^2 = \sum_{n=1}^{\infty}\frac{2\sin(t)(-1)^n}{n}$$ Here is where I am stuck, can someone tell me what value of n to use? Thank you very much AI: I might do it this way, using Duhamel's principle. Start by looking for the solution $\phi(x,t,s)$ to the homogeneous partial differential equation (the ordinary wave equation $\dfrac{\partial^2 \phi}{\partial t^2} = \dfrac{\partial^2 \phi}{\partial x^2}$, depending on the parameter $s$, with boundary conditions $\phi(0,t,s)=\phi(\pi,t,s) = 0$ and initial conditions $\phi(x,0,s)=0$, $\dfrac{\partial \phi}{\partial t}(x,0,s) = x \sin(s)$. Then a solution of your nonhomogeneous wave equation is $u(x,t) = \int_0^t \phi(x,t-s,s)\ ds$. EDIT: Here's another way, if you don't want to use Duhamel's principle. Let $u(x,t)$ be expanded in a Fourier series $u(x,t) = \sum_{n=1}^\infty u_n(t) \sin(n x)$ (we use sines rather than cosines here because of the boundary conditions $u(0,t) = u(\pi,t) = 0$). Correspondingly, the same type of Fourier series for the inhomogeneous term is $x \sin(t)= \sum_{n=1}^\infty a_n \sin(n x) \sin(n t)$ where $$a_n = \dfrac{2}{\pi} \int_0^\pi x \sin(n x)\ dx = \dfrac{2}{n} (-1)^{n+1}$$ Plugging these into the pde and the initial condition, for the coefficients of $\sin(n x)$ to balance we need $$ u_n''(t) = -n^2 u_n(t) + a_n \sin(t), \ u_n(0) = 0, \ u_n'(0) = 0 $$ For $n \ne 1$ the solution of this (which you can get using the method of Undetermined Coefficients) is $$u_n(t) = \dfrac{a_n}{n(n^2-1)}(n \sin(t) - \sin(nt)) = \dfrac{2 (-1)^{n+1}}{n^2 (n^2-1)} (n \sin(t) - \sin(nt))$$ For $n=1$ that solution doesn't work (it's the case of "resonance") and the solution is $$u_1(t) = \dfrac{a_1}{2} (\sin(t) - t \cos(t)) = \sin(t) - t \cos(t)$$ Thus putting it all together, $$ u(x,t) = \sin(x) (\sin(t) - t \cos(t)) + \sum_{n=2}^\infty \dfrac{2 (-1)^{n+1}}{n^2 (n^2-1)} \sin(nx) (n \sin(t) - \sin(nt))$$
H: Calculating Percentile Rank Using Relative Strength Ranking I have a spreadsheet of stock quotes that contains Relative Strength Ranking (RSR) ranging from -45 to 65 and the count of ranks is 1500. Could someone please explain how I can calculate percentile ranks using these numbers? Is there a specific formula? AI: If the total number of stocks is exactly 1500, the percentile rankis $\frac {(\text {number below + number tied/2)}100}{1500}$. So if a stock has 579 stocks with lower RSR and 44 with the same RSR, its percentile rank is $\frac {(579 + 44/2)100}{1500}=\frac{60100}{1500}\approx 40.06$. Some people then round to the nearest whole number, or truncate, or something like that.
H: Is $(x^3-x^2+2x-1)$ prime in $\mathbb{Z}/(3)[x]$? This is somewhat of a follow up on this question: Why is $(3,x^3-x^2+2x-1)$ not principal in $\mathbb{Z}[x]$? I'm curious, is $\mathbb{Z}[x]/I$ a domain, with $I=(3,x^3-x^2+2x-1)$? I know $I$ is not principal. Also, I took the sequence of epimorphisms $$ \mathbb{Z}[x]\stackrel{\varphi}{\longrightarrow}\bar{\mathbb{Z}}[x]\stackrel{\pi}{\longrightarrow}\bar{\mathbb{Z}}[x]/\bar{I} $$ where $\bar{I}=(\bar{x}^3-\bar{x}^2+\bar{2}\bar{x}-\bar{1})$ is the image of $I$ in $\bar{\mathbb{Z}}:=\mathbb{Z}/(3)$. Since the kernel of $\pi\circ\varphi=I$, I know $\mathbb{Z}[x]/I\simeq\bar{\mathbb{Z}}[x]/\bar{I}$. The latter is a ring of $27$ elements, but I don't want to go through and verify by hand that it is a domain. I know it's a domain iff $\bar{I}$ is prime, but I can't tell if it is or isn't. How can I efficiently tell if $\bar{\mathbb{Z}}[x]/\bar{I}$ is a domain? Thank you. AI: Let $F$ be the field with $3$ elements. Since $F[x]$ is a unique factorization domain, a principal ideal $(f)$ being prime is equivalent to $f$ being irreducible. In your case $f$ is a cubic polynomial, so being irreducible is equivalent to not having any roots in $F$. Just check the three values: $f(0)=-1$, $f(1)=1$, and $f(2)=1$ (all modulo $3$). Thus $f$ is irreducible and the ideal it generates is prime.
H: What role does $\mathfrak D$ play in the definition of the union and intersection of the complements of a collection of sets? I'm puzzled by something that might be complete silly. Halmos writes in his "Naive Set Theory": If $\mathcal C$ is a collection of subsets of a set $E$ (that is, $\mathcal C$ is a subcollection of $\mathcal P(E)$), then write $$\mathfrak D = \{X \in \mathcal P(E):X^\prime \in \mathcal C\}$$ It is customary to denote the union and intersection of $\mathfrak D$ by the symbols: $$\bigcup_{X \in \mathcal C}X' \;\;\text{ and } \;\;\bigcap_{X \in \mathcal C}X'$$ He is defining the union and intersection of the complements of the sets $X \in \mathcal C$, which are subsets of the set $E$. $E$ is the "everything" set, which contains all sets considered in the current section. I'm a little confused by why he is considering the set $\mathfrak D$. Why not just consdier the set of complements? $\mathfrak D$ is apparently, as I understand, the set of all subsets of $E$ such that the complement of $X$, namely $X'$ is in the collection of subsets of $E$, $\mathcal C$. By definition of the powerset of $E$, $$\mathcal C \subset \mathcal P(E)$$ so $\mathfrak D$ is the set of subsets of $E$, $X$ such that $X'$ is also a subset of $E$. I still can't see how $\mathfrak D$ enters in the picture here. AI: $\mathfrak{D}=\{X\in\mathcal{P}(E):X'\in\mathcal{C}\}$ is the collection of subsets of $E$ whose complements are in $\mathcal{C}$. For instance, if $\mathcal{C}$ is the collection of all finite subsets of $E$, then $\mathfrak{D}$ is the collection of all cofinite subsets of $E$, i.e., those subsets of $E$ whose complements are finite. Note that for subsets $X$ of $E$ we have $X\in\mathfrak{D}$ iff $X'\in\mathcal{C}$. Take complements: this is equivalent to saying that $X'\in\mathfrak{D}$ iff $X''\in\mathcal{C}$. But $X''=X$, so $X'\in\mathfrak{D}$ iff $X\in\mathcal{C}$, and $\mathfrak{D}=\{C':C\in\mathcal{C}\}$. Thus, we can just as well describe $\mathfrak{D}$ as the set of complements (relative to $E$) of members of $\mathcal{C}$. And this is what justifies writing $$\bigcup\mathfrak{D}=\bigcup_{C\in\mathcal{C}}C'\quad\text{ and }\quad\bigcap\mathfrak{D}=\bigcap_{C\in\mathcal{C}}C'\;.\tag{1}$$ If it weren’t the case that $\mathfrak{D}=\{C':C\in\mathcal{C}\}$, $(1)$ would not be justifiable.
H: Find undetermined coefficients in polynomial quotient and remainder When a polynomial $$P(x)=x^4- 6x^3 +16x^2 -25x + 10$$ is divided by another polynomial $$Q(x)=x^2 - 2x +k,$$ then the remainder is $$x+a.$$ I have to find the values of $a$ and $k$. Can somebody tell me shortest way to get these values? Which theorem should be applied here? AI: Hint: Perform a long division between $P(x)$ and $Q(x)$ and you will get a remainder of degree $1$ of the form $Ax + B,$ where both $A$ and $B$ are expressions in terms of $k.$ Equate with the given $x + a.$ First, equate $A = 1$ to get the value of $k.$ Then equate $B = a$ to get the value of $a.$ For example, you will get something like (NOT the actual answer): $(k+1) x + (2k-1).$ This means $1 = k+1$ and $a = 2k-1.$ Two equations in two variables.
H: Intuition about the Pixley-Roy topology Let $R$ be Real line and let $F[R]$ be $\{x\subset R:\text{is finite}\}$ with Pixley-Roy topology. Definition of Pixley-Roy topology is this: Basic neighborhoods of $F\in F[R]$ are the sets $$[F,V]=\{H\in F[R]; F\subseteq H\subseteq V\}$$ for open sets $V\supseteq F$, see e.g. here. I'm not very clear to the topology of this space. I'm try to have an intuitional impress of the space. Could someone draw a picture for me such that help me to familar with this special toplogical space? Thanks ahead:) AI: Consider the very rough sketch below: p s q t x r u ---------(-\|--------\|-------\|--\|----)-----\|------(-----\|---\|---)------ U W Let $V=U\cup W$, the union of the two open intervals shown in the sketch, and let $F=\{p,q,r\}$. Then $[F,V]$ is the collection of all finite subsets $H$ of $\Bbb R$ such that $F\subseteq H$ and $H\subseteq V$. For example, let $H=\{p,q,r,s,t,u\}$; then $F\subseteq H\subseteq V$, so $H\in[F,V]$. On the other hand, $K=\{p,q,s,t,u\}$ is not in $[F,V]$, because $F\nsubseteq K$ (since $r\notin K$), and $G=\{p,q,r,s,t,u,x\}\notin[F,V]$ because $G\nsubseteq V$ (since $x\notin V$). Added: It’s not easy to give a good global picture of $X=\mathscr{F}[\Bbb R]$, but perhaps these observations will help. For $n\in\Bbb Z^+$ let $X_n=\{F\in\mathscr{F}[\Bbb R]:\|F\|=n\}$. Each $X_n$ is a discrete subspace of $X$. Suppose that $F\in X_n$, and let $V$ be an open set in $\Bbb R$. If $F\nsubseteq V$, then of course $[F,V]=\varnothing$, so assume that $F\subseteq V$. If $G\in[F,V]\cap X_n$, then $F\subseteq G\subseteq V$; but $\|F\|=\|G\|=n$, so $F\subseteq G$ iff $F=G$, and therefore $[F,V]=\{F\}$. For each $n\in\Bbb Z^+$ let $Y_n=\bigcup_{k\le n}X_k$, the set of points of $X$ that contain at most $n$ points of $\Bbb R$; then $Y_n$ is closed in $X$. For suppose that $F\in X\setminus Y_n$; then $\|F\|>n$, and $[F,\Bbb R]\cap Y_n=\varnothing$, since every member of $[F,\Bbb R]$ contains $F$ and therefore has more than $n$ points of $\Bbb R$. For each $n\in\Bbb Z^+$, $X_n$ is a dense set of isolated points of $Y_n$. That the points of $X_n$ are isolated in $Y_n$ follows from (1) and (2). To see that they are dense in $Y_n$, let $F\in Y_n$ be arbitrary, and let $V$ be any open subset of $\Bbb R$ containing $F$. Clearly $\|F\|\le n$. If $\|F\|=n$, $F\in[F,V]\cap X_n$. If $\|F\|<n$, let $G$ be a set of $n-\|F\|$ points of $V\setminus F$; clearly $F\cup G\in[F,V]\cap X_n$. An easy extension of (3) is that the Cantor-Bendixson rank of $Y_n$ is $n$, and that in fact $Y_n^{(k)}=Y_{n-k}$ for $k\le n$, where $Y_0=\varnothing$. Let $[F,V]$ and $[G,W]$ be basic open sets in $X$. Then $$[F,V]\cap[G,W]=[F\cup G,V\cap W]\;,$$ which is empty iff $F\cup G\nsubseteq V\cap W$. $[F,V]\cup[G,W]$ doesn’t have a really simple description. It’s not too hard to use (5) and the second countability of $\Bbb R$ to show that $X$ is ccc: every collection of pairwise disjoint non-empty subsets of $X$ is countable. $X$ is metacompact. To see this, let $\mathscr{U}$ be an open cover of $X$. Without loss of generality assume that the members of $\mathscr{U}$ are basic open sets. For each $F\in X$ there is a $U_F=[G_F,V_F]\in\mathscr{U}$ such that $F\in U_F$. Let $R_F=[F,V_F]$; $G_F\subseteq F$, so $[F,V_F]\subseteq U_F$, and $\mathscr{R}=\{R_F:F\in X\}$ is an open refinement of $\mathscr{U}$. Let $G\in X$ be arbitrary; if $G\in R_F\in\mathscr{R}$, then $F\subseteq G$. $G$ has only finitely many subsets, so $G$ belongs to at most finitely many members of $\mathscr{R}$. Thus, $\mathscr{R}$ is a point-finite open refinement of $\mathscr{U}$, and $X$ is metacompact. (In fact $X$ is hereditarily metacompact, by a similar argument.)
H: What is the relation between the average rate of change and the derivative? A value in the range for any base polynomial function with a y-intercept of zero can be expressed as: $$f\left(x\right) = px$$ where $p$ is the average rate of change between $0$ and $x$. The average rate of change can be in turn expressed as $$p = \frac{ f'\left(0\right) + f'\left(x\right)}{2}$$ where $f'\left(0\right)$ is the instantaneous rate of change at $0$ (or in other words the derivative evaluated at zero) and $f'\left(x\right)$ is the instantaneous rate of change evaluated at $x$ (or in other words the derivative evaluated at the specific range value). In a base polynomial equation (or in other words a polynomial with one term of degree $d$ and leading coefficient $1$) with degree greater than 1 the derivative evaluated at zero can be further simplified to $0$. This leads to a final simplified equation: $$f(x) = \frac{f'(x)x}{2}$$ This final equation however fails to provide the range value for a polynomial above degree $2$. According to the power rule, if: $$f(x) = x^3$$ then: $$f'(x) = 3x^2$$ According to the above derived equation: $$f(3) = \frac{f'(3)(3)}{2} = \frac{(3(3)^2)3}{2} = \frac{81}{2}$$ which is clearly the wrong answer. What is the flaw in the equation relating the average change of rate and the derivative? AI: Unfortunately, $$p = \frac{ f'\left(0\right) + f'\left(x\right)}{2}$$ does not give the average rate of change. For example, try $f(x)=1-\cos x$. Your formula gives the average rate of change from $0$ to $\pi$ as $0$, when instead it should clearly be positive from examining the graph. This is like saying that the average value of a function $f(x)$ on an interval $[a,b]$ is $\frac{f(a)+f(b)}{2}$. This is untrue for most functions. It happens to be true for linear functions, which are exactly what you get when taking the derivative of a quadratic polynomial, explaining why it works in that case.
H: If $\|f(z)\| \geq \|g(z)\|$ for $z \in D$ and $E = \{ z \in D : \|f(z)\| =\|g(z)\| \}$ has a limit point, then $E=D$. This is my problem: Let $D := \{ z \in \mathbb{C} : \|z\| <1 \}$. Let $f$ and $g$ be analytic functions on $D$. Suppose $\|f(z)\| \geq \|g(z)\|$ for all $z \in D$. Define $E = \{ z \in D : \|f(z)\| =\|g(z)\| \}$. Show that if $E$ has a limit point in $D$, then $E=D$. This looks like it should be related to the identity theorem for analytic functions, but I can't seem to get it to work out. Going for a proof by contradiction, I assume that $E \;$ has a limit point in $D\;$ and that $E \neq D\;$. Then there is a sequence $z_n$ of points in $E$ which converges to a limit point $z_0$ in $D$. By continuity, $z_0 \in E$; that is, $f(z) = re^{i\theta}$ and $g(z) = re^{i\phi}$ for some $r$, $\theta$, and $\phi$. And here I stop. Any advice on how to proceed? Thanks. AI: I think that you can use the maximum modulus principle to get what you want. Suppose first that $f$ has no zeros in the unit disk $D$. Then that means that $\dfrac{g(z)}{f(z)}$ is analytic in $D$. Then your condition that $\|f(z)\| \geq \|g(z)\|$ for every $z \in D$ translates to $$\left \| \frac{g(z)}{f(z)} \right \| \leq 1 \quad \text{for every $z \in D$}$$ Then since $E$ has a limit point in $D$, there is a sequence of points $z_n \in E$ such that $z_n \to a$ for some $a \in D$ and since for every $z_n \in E$ we have $\left \| \dfrac{g(z_n)}{f(z_n)} \right \| = 1$ then by continuity also $$\left \| \frac{g(z_n)}{f(z_n)} \right \| \to \left \| \frac{g(a)}{f(a)} \right \| $$ so $\left \| \dfrac{g(a)}{f(a)} \right \| = 1$. Thus we have that $$\left \| \frac{g(z)}{f(z)} \right \| \leq \left \| \frac{g(a)}{f(a)} \right \| \quad \text{for every $z \in D$}$$ so by the maximum modulus principle we conclude that $g/f$ is constant, say $g/f = c$ and $\|c\| = 1$. Then this implies that $\|f(z)\| = \|g(z)\|$ for all $z \in D$ and this implies that $E = D$ as you wanted to show. Now, what happens if $f$ has zeros in $D$? In this case you can use the inequality $\|f(z)\| \geq \|g(z)\|$ to conclude that if $f$ has a zero of order $n$ at $a \in D$, then $a$ is also a zero of $g$ of order $m$ and actually $m \geq n$. This implies that the quotient $g/f$ has removable singularities at the zeros of $f$ and then we can redefine the function at those points and again we would have $g/f$ analytic in $D$ and the previous argument applies.
H: Ergodicity of the First Return Map I was looking for some results on Infinite Ergodic Theory and I found this proposition. Do you guys know how to prove the last item (iii)? I managed to prove (i) and (ii) but I can't do (iii). Let $(X,\Sigma,\mu,T)$ be a $\sigma$-finite space with $T$ presearving the measure $\mu$, $Y\in\Sigma$ sweep-out s.t. $0<\mu(Y)<\infty$. Making $$\varphi(x)= \operatorname{min}\{n\geq0; \ T^n(x)\in Y\}$$ and also $$T_Y(x) = T^{\varphi(x)}(x)$$ if $T$ is conservative then (i) $\mu\|_{Y\cap\Sigma}$ under the action of $T_Y$ on $(Y,Y\cap\Sigma,\mu\|_{Y\cap\Sigma})$; (ii) $T_Y$ is conservative; (iii) If $T$ is ergodic, then $T_Y$ is ergodic on $(Y,Y\cap\Sigma,\mu\|_{Y\cap\Sigma})$. Any ideas? Thank you guys in advance!!! AI: Let be $B \subset Y$. To prove the invariance of $\mu_A$ it is sucient to prove that, $$\mu(T_Y^{-1}B)=\mu(B)$$ First, $$\mu(T_Y^{-1}B)=\sum_{n=1}^{\infty}\mu(Y\cap\{\varphi_Y =n \}\cap T^{-n}B)$$ Now, $$\{ \varphi_Y\leq n\}\cap T^{-n-1}B=T^{-1}( \{\varphi_Y\leq n-1\}\cap T^{-n}B)\cup T^{-1}(Y \cap\{\varphi_Y =n \}\cap T^{-n}B ) $$ This gives by invariance of the measure $$\mu(Y\cap \cap\{\varphi_Y =n \}\cap T^{-n}B)=\mu(B_n)-\mu(B_{n-1}) $$ where $B_n=\{ \varphi_Y\leq n\}\cap T^{-n-1}B.$ We have $\mu(B_n)\to\mu(B)$ as $n\to \infty$, thus $$\mu(T_Y^{-1} B)=\lim\mu(B_n)=\mu(B).~~~~ :)$$ Let us assume now the ergodicity of the original system. Let $B\subset Y$ be a measurable $T_Y$-invariant subset. For any $x \in B$ , the first iterate $T^n x~~~(n\geq 1)$ that belongs to $Y$ also belongs to $B$ , which means that $\varphi_B=\varphi_Y$ on $B.$ But if, $\mu(B)\neq 0$ , Kac's lemma gives that $$\int_{B}\varphi_B d\mu=1=\int_{Y}\varphi_{Y} d\mu$$ which implies that $\mu(B \setminus A) = 0$, proving ergodicity. :)
H: Sum of the thirteenth power of the roots of given polynomial Find the sum of the thirteenth powers of the roots of $x^{13} + x - 2\geq 0$. Any solution for this question would be greatly appreciated. AI: Any root $r_i$ of $x^{13} + x - 2 = 0$ satisfies $r_i^{13} + r_i - 2 = 0,$ or $r_i^{13} = 2 - r_i.$ A polynomial of degree $13$ has $13$ roots (counting repititons). See here and here. Sum them up: $$ \sum_{i = 1}^{13} r_i^{13} = 26 - \sum_{i = 1}^{13} r_{i} .$$ Also observe that the $x^{n-k}$th coefficient of a polynomial is the $k$th symmetric polynomials in the roots (see this), with $$\text{ceoff of } x^{12} = r_1 + r_2 + \ldots + r_{13} = 0.$$ (to convince yourself of the latter fact, expand a smaller example: $(x-r_1)(x-r_2)(x-r_3),$ and observe the coefficient of $x^2$.)
H: Why is the product of all units of a finite field equal to $-1$? Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by $a$. Then the product in question can be written as $$ a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}. $$ However, I'm having trouble jumping from that product to $-1$. What is the key observation to make here? Thanks! AI: The only elements of a field so that $x^2=1$ are $1$ and $-1$. Therefore, when you multiply all the non-zero elements together, each element but $1$ and $-1$ will be paired with its inverse. Thus, the product is $-1$. Note for Fields with Characteristic 2: In a field of characteristic $2$, $-1=1$, so there is only one element so that $x^2=1$. Thus, the product is still $-1=1$.
H: Local boundedness of continuous functions on a Banach space Let $X$ be an infinite-dimensional Banach space and $f : X \to \mathbb{R}$ continuous (not necessarily linear). Can $f$ be unbounded on the unit ball? Of course, in a locally compact space these are impossible. Since $X$ is not locally compact one would guess these are possible, but I cannot think of an example. Are such examples possible even when $X$ is, say, separable Hilbert space? AI: For the first question, let $X=\ell_\infty$. For $n\in\Bbb N$ let $x_n\in X$ be defined by $$x_n(k)=\begin{cases}1,&\text{if }k=n\\0,&\text{otherwise}\;,\end{cases}$$ and let $$V_n=\left\{y\in X:\\|y-x_n\\|<\frac14\right\}\;.$$ Let $y\in X$ and $B=\{z\in X:\\|y-z\\|<1/4\}$, and suppose that $B\cap V_n\ne\varnothing$. Then there is a $z\in X$ such that $\\|z-y\\|<1/4$ and $\\|z-x_n\\|<1/4$, so $\\|y-x_n\\|<1/2$. If $\\|y-x_m\\|<1/2$ as well, then $\\|x_n-x_m\\|<1$, and therefore $m=n$. Thus, $\mathscr{V}=\{V_n:n\in\Bbb N\}$ is a locally finite family of open sets. Let $f_n:X\to\Bbb R$ be any continuous function such that $f(x_n)=n$ and $f(y)=0$ for $y\in X\setminus V_n$. Finally, let $f=\sum_{n\in\Bbb N}f_n$. Since $\mathscr{V}$ is locally finite, $f$ is a continuous function from $X$ to $\Bbb R$, and clearly $f$ is unbounded on the unit ball.
H: Working with phi function with larger numbers? I've been recently learning about Phi function(Euler's totient function). I am attempting to efficiently find the $\phi(n)$ of higher numbers. What I wanted to asked about was, if I have say: $$\frac{30}{\phi(30)} = 3.75,$$ would I be able to now know that for every multiple of $30$? $$\frac{180}{\phi(180)} = 3.75,$$ $$\frac{999990}{\phi(999990)} = 3.75.$$ I have done some research but, with me being quite the novice I am still unsure? AI: We will prove that $\frac{a}{\varphi(a)}=\frac{b}{\varphi(b)}$ iff the prime factorizations of $a$ and $b$ involve the same primes. If $p$ is a prime, then $\varphi(p^k)=(p-1)p^{k-1}$ for any positive integer $k$. So $\frac{p^m}{\varphi(p^m)}=\frac{p^n}{\varphi(p^n)}$ for all positive integers $m$ and $n$. Since $\varphi$ is multiplicative, it follows that if the prime factorizations of $a$ and $b$ involve the same primes, then $\frac{a}{\varphi(a)}=\frac{b}{\varphi(b)}$. Conversely, suppose that $\frac{a}{\varphi(a)}=\frac{b}{\varphi(b)}$. Then the same primes divide $a$ and $b$. This is trickier to prove. Let $p_1, p_2,\dots,p_k$ be the (distinct) primes that divide $a$, listed in decreasing order, and $q_1,q_2, \dots,q_l$ be the primes that divide $b$, again listed in decreasing order. From the fact that $\frac{a}{\varphi(a)}=\frac{b}{\varphi(b)}$, we can fairly easily conclude that $$(q_1-1)\cdots(q_l-1)p_1\cdots p_k=(p_1-1)\cdots(p_k-1)q_1\cdots q_l.\tag{$1$}$$ Suppose that $q_1 \ge p_1$. Since $q_1$ divides the right-hand side of $(1)$, it must divide the left-hand side. It cannot divide any $q_i-1$, and the only $p_i$ it can possibly divide is $p_1$, since $q_1 \ge p_1$. It follows that $q_1=p_1$. Now in Equation $(1)$, cancel the terms $q_1$ and $p_1$, also $q_1-1$ and $p_1-1$. (If $p_1 \ge q_1$, use the same argument.) We obtain an equation of the same type as $(1)$. Continue in his way, from the largest primes down. We conclude hat $k=l$, and $p_i=q_i$ for all $i$. Remark: Your post says you are interested in efficiently finding $\varphi(n)$ for large $n$. Let $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, where the $p_i$ are distinct pimes, and the $a_i$ are $\ge 1$. Then $$\varphi(n)=(p_1-1)p_1^{a_1-1} (p_2-1)p_2^{a_2-1}\cdots (p_k-1)p_k^{a_k-1}.$$ For very large $n$, the above formula is not efficient, since it involves factoring $n$, which seems to be a computationally difficult problem. It is known that if $n$ is the product of two primes, and we only know $n$ and $\varphi(n)$, we can efficiently find the two primes. So if there is an efficient way to find $\varphi(n)$ for large $n$, then the RSA encryption scheme, which is thought to be secure, is in fact not at all secure. There has been a huge amount of effort expended in trying to "break" RSA. One can probably safely say that there is known efficient way to compute $\varphi(n)$ for very large $n$.
H: Segregation of countable sequence into two countable subsequences This is I think a very simple question about infinite sequences. I thought I knew the answer but the manipulation described below worries me. Suppose I divide the interval $(0,\frac{1}{4})$ into infinitely many subintervals $S_n = (\frac{1}{(n+1)^2},\frac{1}{(n)^2})$, to wit: $S_1 = (\frac{1}{9},\frac{1}{4}),S_2 = (\frac{1}{16},\frac{1}{9})$, etc. Suppose there is a countably infinite subsequence $\{S_{n_i} \}$ of these intervals that interests me. I wish to segregate this subsequence by moving it to the left of the interval, so that the two countable sequences $\{S_n\} \setminus \{S_{n_i}\}$ and $\{S_{n_i}\}$ are segregated and remain in length-order, respectively. So we would have, $0,...S_{n_2},S_{n_1},0,...,S_2,S_1$. Are any special assumptions needed to justify this manipulation (and is the situation clear)? Thanks for any help. Edit, example: We have the line segment s: 0 _______ 1/4 I divide it into subintervals as described. Now suppose I want to take the subset of intervals indexed by odd n, and move them to the left of the segment. From the right at x = 1/4, I have a subsequence of intervals whose length approaches zero near (let us say) x = s, and then a new subsequence that begins at s, whose lengths approach 0 as they move towards x = 0. Does this help? AI: Revised: Let $a_n$ be the length of $S_n$, so that $\sum_{n\ge 1}a_n=\frac14$. This is an absolutely convergent series, so we may rearrange the terms as we wish. Let $\langle n_k:k\in\Bbb Z^+\rangle$ be an increasing enumeration of the subscripts of the intervals that are to be ‘moved’ to the left, and let $\langle m_k:k\in\Bbb Z^+\rangle$ be an increasing enumeration of the subscripts of the remaining intervals. For $k\in\Bbb Z^+$ let $s_k=\sum_{i=1}^ka_{m_i}$; $s_k$ is the total length of the first $k$ of the remaining intervals. Let $x_0=\frac14$, and for $k\in\Bbb Z^+$ let $x_k=\frac14-s_k$ and $T_k=(x_k,x_{k-1})$. Let $s=\sum_{i\ge 1}a_{m_i}$; then $0<s<\frac14$, and $$\bigcup_{k\ge 1}T_k=\left(s,\frac14\right)\setminus\{x_k:k\in\Bbb Z^+\}\;.$$ Now repeat the process with the $S_{n_k}$ in the interval $(0,s)$. For $k\in\Bbb Z^+$ let $u_k=\sum_{i=1}^ka_{n_k}$. Let $y_0=s$, and for $k\in\Bbb Z^+$ let $y_k=s-u_k$ and $U_k=(y_k,y_{k-1})$. Then $\sum_{k\ge 1}a_{n_k}=\frac14-s$, since $\sum_{k\ge 1}a_k=\frac14$, and $$\bigcup_{k\ge 1}U_k=(0,s)\setminus\{y_k:k\in\Bbb Z^+\}\;.$$ The intervals $U_k$ and $T_k$ are now arranged as you want them.
H: How to simplify [3a(b-c)+5][-3a(b-c)-5] by using special product? In simplifying $$[3a(b-c)+5][-3a(b-c)-5],$$ I used $$4(au+bv)(cu+dv)=acu^2+(ad+bc)uv+bdv^2.$$ I failed to apply the formula to the equation because $a=3a$, $b=-3a$, $c=5$, $d=-5$, $u=(b-c)$, $v=?$ There's no value of $v$ so I tried to find other special product but it's only $$(au+bv)(cu+dv)=acu^2+(ad+bc)uv+bdv^2$$ fit to the equation. AI: Hint: Let $x=3a(b-c)$ and $y=5$. Then \begin{align}[3a(b-c)+5][-3a(b-c)-5]&=(x+y)(-x-y)\\\\ &=(x+y)\cdot(-1)(x+y)\\\\ &=-(x+y)^2\\\\ &=-x^2-2xy-y^2. \end{align}
H: How to simplify $(3a-b^2-a)^3$ by using special product? When I simplify $(3a-b^2-a)^3$, I used $(u±v)^3=u^3±3u^2v+3uv^2±v^3$ but I'm confused with $(b^2-a)$ AI: First of all, your cited identity is incorrect; the exponent on the $v$ was wrong in the last term. The correct version is $$(u\pm v)^3=u^3\pm 3u^2v + 3uv^2\pm v^3.$$ Also, note that $$3a-b^2-a=2a-b^2$$ so that $$(3a-b^2-a)^3=(2a-b^2)^3,$$ and you can then let $u=2a$ and $v=-b^2$, making the problem somewhat more straightforward.

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