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H: Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a>0$, show that: Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a>0$, show that: $$\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)-f(0)}{x}dx=0$$ This question came from:How can one show that $ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$? AI: Supposing that $b > a$ - otherwise interchange the roles - you have $$\begin{align} \left\lvert \int_{\epsilon a}^{\epsilon b}\frac{f(x)-f(0)}{x}\,dx\right\rvert &\leqslant \int_{\epsilon a}^{\epsilon b} \frac{\lvert f(x) - f(0)\rvert}{x}\,dx\\ &\leqslant \sup_{0 \leqslant \xi \leqslant \epsilon b} \lvert f(\xi) - f(0)\rvert \cdot\int_{\epsilon a}^{\epsilon b} \frac{dx}{x}\\ &= \log \frac{b}{a}\cdot \sup_{0 \leqslant \xi \leqslant \epsilon b} \lvert f(\xi) - f(0)\rvert. \end{align}$$ Since $f$ is assumed continuous in $0$, the supremum tends to $0$ for $\epsilon\to 0$.
H: Poincare Inequality implies Equivalent Norms I am currently working through the subject of Sobolev Spaces using the book 'Partial Differential Equations' by Lawrence Evans. After the result proving the Poincare Inequality it says the following in the book(page 266.) "In view of the Poincare Inequality, on $W_{0}^{1,p}(U)$ the norm $\|\|DU\|\|_{L^{p}}$ is equivalent to $\|\|u\|\|_{W^{1,p}(U)}$, if $U$ is bounded." Do you know the argument behind this statement? AI: This applies to $1 \leq p < n$. You want constants $B_{1}$ and $B_{2}$ such that $B_{1}\|\|u\|\|_{W_{0}^{1,p}(U)} \leq \|\|Du\|\|_{L^{p}(U)} \leq B_{2}\|\|u\|\|_{W_{0}^{1,p}}$. The inequality $\|\|Du\|\|_{L^{p}(U)} \leq B_{2}\|\|u\|\|_{W_{0}^{1,p}}$ is trivial. From Gagliardo-Nirenberg-Sobolev Inequality we have: $\|\|u\|\|_{L^{p}(U)} \leq C\|\|Du\|\|_{L^{p}(U)}$ where $p := \frac{np}{n-p}$ $p* > p$. We also have $\|\|u\|\|_{L^{q}(U)} \leq C\|\|u\|\|_{L^{p}(U)}$ if $1\leq q \leq p$ by Generalized Holder Inequality. If we take $A = \frac{1}{c^{2}}$ and $q=p$ then we have $A\|\|u\|\|_{L^{p}} \leq \|\|Du\|\|_{L^{p}}$ then since $\|\|u\|\|_{W_{0}^{1,p}} \leq A\|\|u\|\|_{L^{p}} + A\|\|Du\|\|_{L^{p}} \leq (1+A)\|\|Du\|\|_{L^{p}}$ it follows that $\frac{A}{A+1}\|\|u\|\|_{W_{0}^{1,p}} \leq \|\|Du\|\|_{L^{p}}$ Take $B_{1} := \frac{A}{A+1}$. The result follows from combining the two inequalities.
H: A function which is continuous at everywhere in its domain, but diff only at one point Suppose a real valued function $f:\mathbb R \to \mathbb R$ is continuous everywhere. Is it possible to construct $f$ that is differentiable at only one point? If possible give an example also. AI: It is possible. Start with a function $g$ that is contiuous everywhere and nowhere differentiable, like the Weierstraß function. Then choose the point where you want it differentiable, say $a$, and set $$f(x) = (x-a)\cdot g(x).$$ Then $f$ is differentiable in $a$, with $f'(a) = g(a)$, and nowhere else.
H: Showing that a set is open/closed $\def\R{\mathbb R}$ Is the set $$S=\{(x_1,x_2,x_3) \in \R^3 \mid e^{x_1} + x_2^2 <x_3 \} \subset \R^3$$ open or closed? My attempt: Let $f:\R^3 \to \R$, $f(x_1,x_2,x_3)$ =$e^x_1 + x_2^2-x_3$. Then $S=\{(x_1,x_2,x_3) \in \R^3 \mid f(x_1,x_2,x_3)<0 \}$. Also $f(x_1,x_2,x_3)$ is continuous in $\R^3$ and the set $S=\{(x_1,x_2,x_3) \in \R^3 \mid f(x_1,x_2,x_3)<0\}$ is the inverse image of $(-\infty,0)$,which is open in $\R$ and therefore $S$ is open in $\R^3$. Am i correct ? AI: You are right if you mean by $R = \mathbb{R}$ the real numbers.
H: Are my answers correct? (finding intercepts, asymptotes, and extrema) Are my answers correct? a) (0, 4/3) and (2,0) and (-2, 0) b) Horizontal asymptote: $y = 3$, Vertical asymptotes: $x = 3, x = -3$ c) Extremum is at $(0, 4/3)$, maximum. AI: Yes, all of the answers are correct. In future, maybe try and show your work as well, just in case there is a mistake? (This is an answer because I am answering the original question.)
H: Supremum over dense subset of banach space Let $\{x_n\}$ be a countable dense subset of a Banach space $X$. How can I show that $$\sup_{x \in X}f(x) = \sup_{n \in \mathbb{N}}f(x_n)$$ where $f$ is continuous and real-valued?? AI: For a continuous function, you always have $f(\overline{A}) \subset \overline{f(A)}$. So in your situation, you have $$f(X) = f\left( \overline{\left\lbrace x_n : n \in \mathbb{N} \right\rbrace}\right) \subset \overline{f\left(\lbrace x_n : n \in \mathbb{N}\rbrace \right)}.$$ Now use the fact that $$\sup \overline{B} = \sup B$$ for all $B\subset \mathbb{R}$.
H: When is round-robin scheduling possible and with in minimal time? Suppose that you have six teams $x_0, x_1, x_2, x_3, x_4, x_5$. Can you schedule round-robin games between them so that if one game is played each day, the series of games can be completed in five weeks? A naive attempt fails: $$(x_0,x_1), (x_2, x_3), (x_4, x_5)$$ $$(x_0,x_2), (x_1, x_3) .. oops$$ Some care is needed to get this to work. A solution: $$(x_0,x_3), (x_1, x_5), (x_2, x_4)$$ $$(x_1,x_4), (x_2, x_0), (x_3, x_5)$$ $$(x_2,x_5), (x_3, x_1), (x_4, x_6)$$ $$(x_0,x_1), (x_2, x_3), (x_4, x_5)$$ $$(x_0,x_5), (x_2, x_1), (x_4, x_3)$$ This solution certainly is not easy to generalize. Questions: Is this in fact always possible for an even number of teams? Is there a convenient algorithm for producing the schedule? Can this problem be easily expressed in terms of graph theory, combinatorics, or abstract algebra? AI: Let me explain the algorithm with 6 teams. You start x1 vs x2 x3 vs x4 x5 vs x6 Then you keep x1 fixed and move the other teams (counter-)clockwise x1 vs x4 x2 vs x6 x3 vs x5 and proceed doing this until every team has played against each other. For an odd number of teams you define x1 to be the bye. x1 vs x6 x4 vs x5 x2 vs x3 then x1 vs x5 x6 vs x3 x4 vs x2 and finally x1 vs x3 x5 vs x2 x6 vs x4
H: Confirming proofs of properties of preimages I do not think that I made terrible mistakes, nevertheless a conforming word would be good for me. Thank you! Let $f\colon X\to Y$ be a map and $(B_i)_{i\in I}$ a family of subsets of $Y$ ($I$ is any index set). Overmore consider any $A,B\subset Y$. Show: \begin{align} &(1)~f^{-1}(\bigcup_{i\in I}B_i)=\bigcup_{i\in I}f^{-1}(B_i)\\ &(2)~f^{-1}(\bigcap_{i\in I} B_i)=\bigcap_{i\in I}f^{-1}(B_i)\\ &(3)~(f^{-1}(A))^C=f^{-1}(A^C)\\ &(4)~f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)\\ &(5)~f^{-1}(A\Delta B)=f^{-1}(A)\Delta f^{-1}(B) \end{align} My proofs: (1) "$\subset$": \begin{align} x\in f^{-1}(\bigcup_{i\in I}B_i) &\Rightarrow\exists~j\in I: f(x)\in B_j\\&\Rightarrow x\in f^{-1}(B_j)\\&\Rightarrow x\in\bigcup_{i\in I}f^{-1}(B_i) \end{align} "$\supset$": \begin{align} x\in\bigcup_{i\in I}f^{-1}(B_i) &\Rightarrow\exists j\in I: x\in f^{-1}(B_j)\\&\Rightarrow f(x)\in B_j\\&\Rightarrow f(x)\in\bigcup_{i\in I}B_i\\&\Rightarrow x\in f^{-1}(\bigcup_{i\in I}B_i)~~~\Box\end{align} (2) "$\subset$": \begin{align} x\in f^{-1}(\bigcap_{i\in I} &\Rightarrow f(x)\in\bigcap_{i\in I}B_i\\&\Rightarrow f(x)\in B_i~\forall~i\in I\\&\Rightarrow x\in f^{-1}(B)~\forall~i\in I\\&\Rightarrow x\in\bigcap_{i\in I}f^{-1}(B_i) \end{align} "$\supset$": \begin{align} x\in\bigcap_{i\in I}f^{-1}(B_i) &\Rightarrow x\in f^{-1}(B_i)~\forall~i\in I\\&\Rightarrow f(x)\in B_i~\forall~i\in I\\&\Rightarrow f(x)\in\bigcap_{i\in I}B_i\\&\Rightarrow x\in f^{-1}(\bigcap_{i\in I}B_i)~~~\Box \end{align} (3) "$\subset$": \begin{align} x\in (f^{-1}(A))^C=X\setminus f^{-1}(A) &\Rightarrow x\in X\wedge x\notin f^{-1}(A)\\&\Rightarrow x\in X\wedge f(x)\notin A\\&\Rightarrow x\in X\wedge T(x)\in A^C\\&\Rightarrow x\in X\wedge x\in f^{-1}(A^C)\\&\Rightarrow x\in X\cap f^{-1}(A^C)=f^{-1}(A^C) \end{align} "$\supset$": \begin{align} x\in f^{-1}(A^C)=Y\ &\Rightarrow f(x)\in A^C=Y\setminus A\\&\Rightarrow f(x)\in Y\wedge f(x)\notin A\\&\Rightarrow x\in f^{-1}(Y)=X\wedge x\notin f^{-1}(A)\\&\Rightarrow x\in X\setminus f^{-1}(A)=(f^{-1}(A))^C~~~\Box \end{align} (4) "$\subset$": \begin{align} x\in f^{-1}(A\setminus B) &\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow x\in f^{-1}(A)\wedge x\notin f^{-1}(B)\\&\Rightarrow x\in f^{-1}(A)\setminus f^{-1}(B) \end{align} "$\supset$": \begin{align} x\in f^{-1}(A)\setminus f^{-1}(B) &\Rightarrow x\in f^{-1}(A)\wedge f(x)\notin f^{-1}(B)\\&\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow f(x)\in A\setminus B\\&\Rightarrow x\in f^{-1}(A\setminus B)~~~\Box \end{align} (5) "$\subset$": \begin{align} x\in f^{-1}(A\Delta B) &\Rightarrow f(x)\in (A\setminus B)\cup (B\setminus A)\\&\Rightarrow f(x)\in (A\setminus B)\vee f(x)\in B\setminus A \end{align} I. $f(x)\in A\setminus B$: \begin{align} &\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow x\in f^{-1}(A)\wedge x\notin f^{-1}(B)\\&\Rightarrow x\in f^{-1}(A)\setminus f^{-1}(B)\\&\Rightarrow x\in (f^{-1}(A)\setminus f^{-1}(B))\cup (f^{-1}(B)\setminus f^{-1}(A))=f^{-1}(A)\Delta f^{-1}(B) \end{align} II. $f(x)\in B\setminus A$: \begin{align} &\Rightarrow f(x)\in B\wedge f(x)\notin A\\&\Rightarrow x\in f^{-1}(B)\setminus f^{-1}(A)\\&\Rightarrow x\in f^{-1}(A)\Delta f^{-1}(B) \end{align} "$\supset$": \begin{align} x\in f^{-1}(A)\Delta f^{-1}(B) &\Rightarrow x\in (f^{-1}(A)\setminus f^{-1}(B))\vee (f^{-1}(B)\setminus f^{-1}(A)) \end{align} I. $x\in f^{-1}(A)\setminus f^{-1}(B)$: \begin{align} &\Rightarrow x\in f^{-1}(A)\wedge x\notin f^{-1}(B)\\&\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow f(x)\in A\setminus B\\&\Rightarrow x\in f^{-1}(A\setminus B)\cup f^{-1}(B\setminus A)\overset{\text{with}}{\underset{\text{(1)}}{=}}f^{-1}((A\setminus B)\cup (B\setminus A))=f^{-1}(A\Delta B) \end{align} II. $x\in f^{-1}(B)\setminus f^{-1}(A)$: \begin{align} &\Rightarrow x\in f^{-1}(B)\wedge x\notin f^{-1}(A)\\&\Rightarrow f(x)\in B\wedge f(x)\notin A\\&\Rightarrow f(x)\in B\setminus A\\&\Rightarrow x\in f^{-1}(B\setminus A)\\&\Rightarrow x\in f^{-1}(B\setminus A)\cup f^{-1}(A\setminus B)\overset{\text{with}}{\underset{\text{(1)}}{=}}f^{-1}((B\setminus A)\cup (A\setminus B))=f^{-1}(A\Delta B)~~~\Box \end{align} AI: Nothing blatantly wrong that I can see (apart from a few typos, like missing "$B_i)$" in the $\subset$ direction of part 2, "$T(x)$" in the $\subset$ direction of part 3, and a random "$=Y$" in the $\supset$ direction of part 3), but you're making it a bit too complicated in your proofs of the last two. Indeed, $$f^{-1}(A\setminus B)=f^{-1}(A\cap B^c)\overset{(2)}{=}f^{-1}(A)\cap f^{-1}(B^c)\overset{(3)}{=}f^{-1}(A)\cap(f^{-1}(B))^c=f^{-1}(A)\setminus f^{-1}(B),$$ and $$\begin{align}f^{-1}(A\triangle B) &= f^{-1}\bigl((A\setminus B)\cup(B\setminus A)\bigr)\\ &\overset{(1)}{=}f^{-1}(A\setminus B)\cup f^{-1}(B\setminus A)\\ &\overset{(4)}{=}\bigl(f^{-1}(A)\setminus f^{-1}(B)\bigr)\cup\bigl(f^{-1}(B)\setminus f^{-1}(A)\bigr)\\ &= f^{-1}(A)\triangle f^{-1}(B),\end{align}$$ with no need for element-chasing, double-inclusion, or breaking into cases. Also, you could do parts 1 through 3 as $\iff$ all the way, rather than a double-inclusion proof.
H: Prove that $x_{n+1}=\frac{2}{9}(x_n^3+3)$ converges Let $x_1=1/2$ and $x_{n+1}=\frac{2}{9}(x_n^3+3)$ for $n\geq 1$. We want to prove that the sequence $(x_n)$ converges to real number $r\in (0,1)$ satisfying the equation $2r^3-9r+6=0$. First part For the sequence to converge, it must be bounded. A sequence $X:=(x_n)$ of real numbers is said to be bounded if there exists a real number $M>0$ such that $\|x_n\| \leq M$ for all $n\in \mathbb{N}$. First, we show that $(x_n)$ converges. This is where I struggle I tried showing by induction that $(x_n)$ < 1 for all $n$** but it didn't work I also tried the ratio test... Second part Claim The real number $r$ satisfies the equation $2r^3-9+6=0$ Proof - Since $x_n$ and $x_{n+1}$ convergence to the same real number $r$, we can substitute $r$ in the equation for $x_{n+1}$ when $n$ gets sufficiently large : $$ \begin{aligned} x_{n+1}&=\frac{2}{9}(x_n^3+3)\\ r&=\frac{2}{9}(r^3+3))\\ r&=\frac{2}{9}r^3+\frac{6}{9}\\ 0&=\frac{2}{9}r^3-r+\frac{6}{9}\\ 0&=2r^3-9r+6 \quad \text{multiply by 9} \end{aligned} $$ AI: Induction works fine. $x_n<1$ implies that $x_n^3<1$. Hence $$x_{n+1}<\frac{2}{9}(1+3)=\frac{8}{9}<1$$ In fact the same proof shows $\frac{8}{9}$ as an upper bound.
H: $\mathbb{Q}[X,Y]/(X,Y^{2}-1)$ is this a maximal ideal or a prime ideal? $\mathbb{Q}[X,Y]/(X,Y^{2}-1)$ is this a maximal ideal or a prime ideal? So far i got: $\mathbb{Q}[X,Y]/(X,Y^{2}-1) = \left(\mathbb{Q}[Y]\right)X/(X,Y^{2}-1)$ is isomorphic to $\mathbb{Q}[Y]/(Y^{2}-1)$ = $\mathbb{Q}[Y]/(Y+1)(Y-1)$. Because we now have zero divisors this can't be a domain, so its cant be a prime/maximal ideal. I don't trust my proof because i don't know why the second step would be valid. Please help :) Kees AI: You are right. The polynomials $Y - 1$ and $Y + 1$ are both outside the ideal $(X, Y^2 - 1)$, but their product is contained in it, thus it cannot be a prime ideal. Since any maximal ideal is prime, it cannot be maximal either.
H: Let $R$ be the set of all integers with alternative ring operations defined below. Show that $\Bbb Z$ is isomorphic to $R$. For any integers $a,b$, define $a\oplus b=a + b - 1$ and $a\odot b=a + b - ab.$ Let $R$ be the ring of integers with these alternative operations. Show that $\Bbb Z$ is isomorphic to $R$. What I thought about doing was showing $f(a + b) = a + b - 1$, meaning it isn't isomorphic since it doesn't equal $f(a)f(b)$. True? AI: To establish an isomorphism, you would have to find a mapping $f:\Bbb Z \to R$ satisfying the definition of an isomorphism: $f(a+b)=f(a)\oplus f(b)$ $f(ab)=f(a)\odot f(b)$ $f$ one-to-one and onto (you could just find an inverse homomorphism) You're going to have to come up with a candidate for $f$. My first hint would be to look at $a\odot 1$ and $a\odot 0$ and $a\oplus 1$. $0$ and $1$ are "special" in $R$, and if you sort out what they are doing in the new operation, you can deduce the right $f$ to pick.
H: The $n\times n$ matrix $(a_{ij})$ with $\sum_{j=1}^{n} a_{ij} = 1$ has an eigenvalue 1. The Problem Let $A = (a_{ij})$ be an $n\times n$ matrix such that $$ \sum_{j=1}^n a_{ij} = 1 $$ for all $i = 1,2,\dots,n$. Show that $A$ has an eigenvalue $1$. I tried to figure out from $A - \lambda I v = \mathbf{0}$ but now I'm stuck. AI: Take $v = (1,1,\ldots, 1)$. What is $A(v)$?
H: Lebesgue integral over discrete set Let $X=\{x_1,x_2,\ldots\}$, and for any $A\subseteq X$ let $$\mu(A)=\sum_{x_n\in A}2^{-n}$$ Suppose $f$ is given by $f(x_n)=1/n!$. Find $$\int_X fd\mu$$ I tried to estimate from below by simple functions $s_k$ where $s_k(x_n)=1/n!$ for $n=1,2,\ldots,k$ and $s_k(x_n)=0$ otherwise. Then $\int_Xs_kd\mu=\sum_{i=1}^k2^{-i}/i!$. I'm quite sure that $\int_Xfd\mu=\sum_{i=1}^\infty 2^{-i}/i!$. But since there could be other simple functions that are $\leq f$, how can I prove it rigorously? AI: Your sequence $\{s_k\}$ converges to $f$ monotonically. The monotone convergence theorem gives you the value of the integral.
H: Series $\sum_{i=1}^\infty2^{-i}/i!$ The series $\sum_{i=1}^\infty2^{-i}/i!$ is clearly convergent by the ratio test with $\sum_{i=1}^{\infty}2^{-i}$, but is it possible to calculate the exact sum? AI: More generally, $$e^x=\sum_{i=0}^\infty \frac{1}{i!}x^i$$ Applying this with $x=2^{-1}$ gives the series you want, but starting at $i=0$. That is an extra $2^0/0!=1$ in the sum. Subtract this off and you get $\sqrt{e}-1$, as @Daniel points out in the comments.
H: The existence of inequalities between the sum of a sequence and the sum of its members Let $(a_n)_{n=m}^\infty$ be a sequence of positive real numbers. Let $I$ denote some finite subset of $M := \{m, m+1, \cdots \}$, i.e., $I$ is the index of some points of $(a_n)_{n=m}^\infty$. Does there exist a real number $r$ such that for any valid $I$, $\sum_{n=m}^\infty a_n > r$ and $\sum_{i \in I}a_i \leq r$? It seems that if $\sum_{n=m}^\infty a_n$ is finite, let's denoted by $S$. Such a $r$ can be written as $S - \epsilon$ where $\epsilon$ is a positive real number. Assume that for any $\epsilon > 0$, $\sum_{i \in I} a_i > S - \epsilon$, which means that the sequence $(\sum_{i \in I}a_i)_{I \in 2^M}$ is bounded by $S$. Then how to do next? Thanks a lot! AI: No. Since each $a_n$ is positive you have $$\sum_{n=m}^\infty a_n = \sup_I \sum_{i \in I} a_i$$ where the supremum is taken over all finite subsets $I$ of $M$. This is a consequence of the monotone convergence theorem for series.
H: Moment generating function of a function I need to find MGF of $Z = \mu + \frac{1}{\lambda}X$ ($\mu, \lambda$ are constants). How do I do that? UPDATE If it helps: $$f(x) = \frac{1}{\sqrt{2\pi}}\exp{ \{ -\frac{x^2}{2} \} }$$ So ... $$M_X(t) = te^{t^2/2}$$ AI: The general rule is this: $$ M_Z(t) = E[e^{tZ}] = E[e^{t(\mu + \frac{1}{\lambda})X}] = e^{t\mu} E[e^{(t \frac{1}{\lambda})X}] $$ $$ M_Z(t) = e^{t\mu} M_X(t \frac{1}{\lambda}) $$ So if you know the moment generating function of $X$, you have the moment generating function of $Z$. In any event the "additional information" in the Question now specifies that $X$ is a normally distributed random variable, so its moment generating function is well-known.
H: Balancing objects of varying length in a collection of set length while maintaining order I have an object each with a length associated with it. I can then have multiple of these objects and I want to put them into another collection/array with a certain set count. Order matters and I need every element in the top level list to have at least one object in it. example: object 1 called x1 with a count of 4 object 2 called x2 with a count of 3 object 3 called x3 with a count of 1 object 4 called x4 with a count of 20 put all those into an array with 3 elements (columns) and balance it much as possible. the good outcome would be [x1] [x2,x3] [x4] Is there an algorithm or pattern that can give me a balanced (as much as possible) collection. Thoughts so far: My original solution was to take the target length (totalCount/numberOfColums) of all objects, then any item over that target length would/should get its own columm. Next I would re-calculate the target length with the remaining objects and with the remaining columns and break those up. This solution breaks down in some situations. example: [10] [2] [2] [3] [3] [9] [2] [10] this breaks it out into \| [10] \| [2] [2] [3] \|* [3] \| [9] [2] [10] \| the last column is filled so much because as I get to the end I have to still put the items in the collection. The third column just has the one item of count 3, because the 9 would throw it over the target length. (I later added a tolerance to account for this, but this added many new issues when list are perfectly balance and you don't want the tolerance) AI: Update: misunderstood question. I advise a quick, greedy algorithm. First choose your target bin size, then fill the bins greedily (as much as possible without going over). When you're done, count the number of bins; if it's not too many you're done. If there are too many, then increase the target and start over. The outcome described in the comments will be achieved for a target of 11. For a target of 10, the 9 and 2 will be split into separate bins, so that won't work.
H: Learning roadmap for Non-commutative Geometry I am interested in learning Non-commutative geometry and K-theory of operator algebras. Please suggest a learning roadmap for this subject. My present knowledge of Measure theory & Functional Analysis is very little.Please advise which topics in Functional Analysis and Operator Theory should I learn before starting Alain Connes' book 'Non Commutative Geometry' and references for the same. What are other prerequisites for reading this book ? AI: I would recommend that you learn some Operator Algbras first, from say Murphy's Operator Algebras and Operator Theory. Then you can learn some K-theory from Rordam/Larsen/Laustsen's book Introduction to K-theory for Operator Algebras. After that, you should be knowledgeable enough to find your own way around.
H: Fatou's lemma on bounded domain and bounded range Fatou's lemma states that if $f_1,f_2,\ldots$ are nonnegative measurable functions, and $f=\liminf f_i$, then $$\int_E fd\mu\leq \liminf\int_E f_id\mu$$ The two examples for strict inequality given in the link are where $f_n$ takes on a nonzero value on an arbitrarily large domain, and where $f_n$ gets arbitrarily large. If we restrict the domain to $[0,1]$ and require $0\leq f_i\leq 1$, does equality always hold? AI: No, even with bounded functions on a space of finite measure we can find examples for the strict inequality, if the space has two disjoint sets $A,B$ with positive measure. Then one can simply oscillate $$f_i = \begin{cases}\chi_A &, i \equiv 0 \pmod{2}\\ \chi_B &, i \equiv 1 \pmod{2} \end{cases}$$ and one has $f = \liminf f_i = 0$, but $\liminf \int f_i = \min \{\mu(A),\,\mu(B)\} > 0$.
H: Probability of drawing 3 balls A box contains 8 red, 3 white, and 9 blue balls. If 3 balls are drawn at random determine the probability that all 3 balls are red all 3 balls are white 2 are red and 1 is white at least 1 is white What I tried for the first one is 3/20 * 2/19 * 1/18 = 6/6840 But they said it was wrong. AI: There are $8$ balls that are red. So you should have $$(1)\quad \dfrac {8}{20}\cdot \frac 7{19}\cdot \frac 6{18} = \frac{336}{6840} = \frac{14}{285}$$ What you actually calculated is the probability that all three balls are white: $$(2)\quad \dfrac 3{20}\cdot \dfrac 2{19}\cdot \frac 1{18} = \dfrac{6}{6840} = \dfrac 1{1140}$$ For $(3)$, there are three ways to get draw 2 red balls and 1 white ball: RRW, RWR, WRR. Add the probabilities of each of the three sequences of draws to find the overall probability of drawing $2$ red balls and one white ball. For $(4)$, compute the probability of getting zero white balls, and subtract it from $1$ to obtain the probability of getting at least one white ball.
H: What is the $\lVert v\rVert$ sign mean in the context of vectors? Suppose $V$ a inner product space, $u, v \in V$. I need to prove this identity: $$\lVert u+v\rVert^2 +\lVert u-v\rVert^2 = 2\left(\lVert u\rVert^2 +\lVert v\rVert^2\right) $$ what is the $\lVert v\rVert$ ("two lines") sign mean? thanks. AI: It is the norm. In this case, $$ \\|u\\| = \sqrt{\langle u, u \rangle} $$
H: $f(z)=f(iz)\forall z$, my question is : is there such $f$ exists? $f$ be an entire function such that $\|f(z)\|\le c\|z\|^3\forall \|z\|\ge 3,f(z)=f(iz)\forall z$, my question is : is there such $f$ exists? $f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4+a_5z^5+\dots=a_0+a_1iz-a_2z^2-ia_3z^3+a_4z^4+ia_5z^5\dots$ $\|g(z)\|=\|{f(z)\over z^3}\|\le c\forall \|z\|\ge 3$ so by Liuvilles $f(z)=kz^3\forall \|z\|\ge 3$ also $kz^3=-ki z^3$ so $k=0$, so $f(z)=0\forall \|z\|\ge 3$? so $f\equiv 0$? am I wrong anywhere? AI: Since you only have the estimate $$\lvert f(z)\rvert \leqslant c\lvert z\rvert^3\tag{1}$$ for $\lvert z\rvert \geqslant 3$, you can't assume that $$g(z) = \frac{f(z)}{z^3}$$ is an entire function. It may have a pole in $0$. However, $(1)$ says, by the Cauchy-estimates, that $f$ is a polynomial of degree at most $3$. Then $$a_0 + a_1 z + a_2 z^2 + a_3 z^3 \equiv a_0 + a_1 iz - a_2 z^2 - a_3 i z^3$$ tells you that $a_1 = a_2 = a_3 = 0$, so the only functions with these properties are the constant functions.
H: Is $GL(n;R)$ closed as a subset of $M_n(R)$? Let $M_n(R)$ denote the space of all $n×n$ matrices with real entries. The general linear group over real numbers,denoted $GL(n,R)$, is given by $GL(n,R)=${$A∈M_n(R)\|det(A)\neq0$}. Is $GL(n,R)$ closed as a subset of $M_n(R)$? Some thought of mine:Obviously,there exists a sequence $A_n$ in $GL(n,R)$ such that $A_n$ tends to $A$ as $n$ to infinity where $A$ is not invertible.For instance,considering $n=2$,let $A_n=\begin{pmatrix} 1 && 0\\ 0 && \frac{1}{n}\end{pmatrix}$.Then $limA_n=A=\begin{pmatrix} 1 && 0\\ 0 && 0\end{pmatrix} ,n\to{+\infty}$,where $det(A)=0$.The example tells us that $GL(n,R)$ is not closed in $M_n(R)$. How can we prove $GL(n,R)$ is open in $M_n(R)$ directly and strictly?And how do you define the topology in $M_n(R)$? AI: The topology of $M_n({\mathbb R})$ is the standard topology of ${\mathbb R}^{n^2}$, Hint: $\det$ is continuous.
H: Prove that a nonzero homomorphic image of a local ring is a local ring The problem says that prove that a nonzero homomorphic image of a local ring is a local ring. Could you give me a scratch of a proof for this or maybe a full answer if you don't mind? AI: The reason is that for every ring $R$ and a proper ideal $I$ there's a one on one correspondence between ideals of $R$ containing $I$ and ideals of $R/I$. Since for every homomorphism $f\colon R \to S$ we have that $\text{Im} f \cong R/\ker f$, we have that the ring $\text{Im}f$ have to have as much maximal ideals as the maximal ideals of $R$ which contain $I=\ker f$. Since $R$ has just one maximal ideal also $\text{Im}f$ have a unique maximal ideal, hence it's local.
H: What are the properties of the roots of the incomplete/finite exponential series? Playing around with the incomplete/finite exponential series $$f_N(x) := \sum_{k=0}^N \frac{z^k}{k!} \stackrel{N\to\infty}\longrightarrow e^z$$ for some values on alpha (e.g. solve sum_(k=0)^19 z^k/(k!) = 0 for z), I made a few observations: The sum of the roots of $f_N$ are $-N$ The product of the roots of $f_N$ are $(-1)^N\cdot N!$ Their imaginary part seems to lie between $\pm10$ The zeros seem to form an interesting shape: Patterns for $N=17, 18, 19$ Now the sum and product part are clear, since $$\begin{align} f_N(x) &= \frac1{N!}\left(z^N + N z^{N-1} + N(N-1)z^{N-2} + ... + N!\right) \\ &= \frac1{N!}(z-z_{N0})(z-z_{N1})\cdots(z-z_{NN}) \\ &= \frac1{N!}\left(z^N - \left(\sum_{k=0}^Nz_{Nk}\right) z^{N-1} + ... + (-1)^N\prod_{k=0}^N z_{NK}\right) \end{align}$$ and since $e^z=0 \Leftarrow \Re z\to-\infty$ it is clear that the roots tend towards real parts with negative infinity, but I'm still intrigued by the questions what ($N$-dependent) curve do the zeros of $f_N(z)$ lie on, does that curve maintain its shape for varying $N$ and merely translate or also deform; and what other properties of the zeros (e.g. absolute value) can be derived? AI: The zeros of the scaled functions $f_N(Nz)$ do converge to an airfoil-like curve. See an animation here. See also these: Zeros of truncated Taylor series by Jonas (see the references at the end) (2013) Zeroes of the partial sums of the exponential function by Zemke (2009) On the zeroes of the nth partial sum of the exponential series by Zemyan (2005) The zeros of the partial sums of the exponential series by Walker (2003)
H: I have five eggs to color for Easter. I can color them red, yellow, or blue. How many ways are there to do this? Not sure if my thinking is correct on this problem. I have five eggs to color for Easter. I can color them red, yellow, or blue. How many ways are there to do this? I was thinking 5 * 5 choose 3 since you would have 3 choices for color for each egg. Not sure if this would be correct though. Let me know what you think. AI: Consider a simpler problem. Suppose that you have just one egg. How many ways are there to color that egg using red, yellow or blue? The answer is: $3$, right? Now, instead of one egg, you have two. How many ways can you color the second one? The answer is again $3$, right? But, recall that you had $3$ ways to color the first egg. Thus, for every possible color choice for the first egg, you can color the second one in $3$ ways. Therefore, the two eggs can be colored in $3 * 3$ ways. So, the final answer is ...?
H: Dot product over complex vectors: Conjugate first or second? Does there exist a truly "standard" dot product over complex vectors? Wikipedia and Wolfram's MathWorld indicate directly or indirectly that the second argument is conjugated. Matlab's dot product is the opposite. "When A and B are both column vectors, dot(A,B) is the same as A'B." This has attractiveness from the perspective that it is more consistent with the vector definition for real numbers. I.e., if dot conjugated the second argument it would be B'A I am confused about whether there is a "standard" approach. Did Matlab just go rogue on this one? Or were others before them using a dot product that implied linearity in the second argument? This ambiguity seems more dangerous and error-prone than a simple notational preference. E.g., how mathematicians use "i" to represent the square root of negative one, vs. engineers who use "j". AI: In fact I've got a book in which the dot product is considered antilinear in the first argument despite other books having antilinearity in the second one. However this ambiguity is in no way dangerous, because you may make a composition of operators to have linearity in the argument that you want. The book in question is Richtmyer, Morton. Difference Methods for Initial-Value Problems
H: O(p,q; C) isomorphic to the usual orthogonal group O(p + q; C) for complex field I'm trying to make sense of this statement that appears on wiki: "The group O(p,q) is defined for vector spaces over the reals. For complex spaces, all groups O(p,q; C) are isomorphic to the usual orthogonal group O(p + q; C), since the transform $z_j \mapsto iz_j$ changes the signature of a form." I don't quite get it. Can anybody give me a proof? And also, does it work for SO(p,q) as well? Thanks AI: Let $$Q_{p,q}(z,w) = \sum_{i = 1}^{p} z_i w_i - \sum_{i=p+1}^{p+q} z_i w_i.$$ This is the standard symmetric bilinear form of signature $(p,q)$ on $\Bbb C^{p+q}$ (note that the definition of indefinite orthogonal groups uses bilinear forms, not sesquilinear). The complex indefinite orthogonal group is then defined by $$\mathrm{O}(p,q; \Bbb C) = \{A \in \mathrm{GL}_{p+q}(\Bbb C) : Q_{p,q}(Az, Aw) = Q_{p,q}(z,w)\}.$$ If we take $\{e_i\}$ to be the standard basis of $\Bbb C^{p+q}$ and define $$\varphi(e_i) = \begin{cases} e_i, & 1 \leq i \leq p, \\ ie_i, & p+1 \leq i \leq q \end{cases}$$ then we have that $$Q_{p,q}(\varphi(z), \varphi(w)) = \sum_{i = 1}^p z_i w_i + \sum_{i = p+1}^{p+q} z_i w_i = \sum_{i = 1}^{p+q} z_i w_i = Q_{p+q,0}(z,w).$$ Hence $\varphi$ is an isomorphism between the symmetric bilinear forms $Q_{p,q}$ and $Q_{p+q,0}$. Then we get an induced isomorphism $$\mathrm{O}(p,q; \Bbb C) \xrightarrow{~\cong~} \mathrm{O}(p+q; \Bbb C),$$ $$A \mapsto A \circ \varphi.$$ This of course works for $\mathrm{SO}(p,q; \Bbb C)$ as well since we just need to take the determinant $1$ elements for that.
H: If $Y \subseteq X$ is open, and $S \subseteq Y$ is nowhere dense in $Y$, then $S$ is nowhere dense in $X$ Let $X$ be a topological space, let $Y$ be an open subspace of $X$. Suppose that $S \subseteq Y$ is nowhere dense in $Y$, that is, $\operatorname{int}_Y(\operatorname{cl}_Y(S)) = \emptyset$. How to prove that $S$ is nowhere dense in $X$? I have not made much progress so far. As a starting point, the assumption $\operatorname{int}_Y(\operatorname{cl}_Y(S)) = \emptyset$ can be written in terms of closures and interiors in $X$, but this does not seem to be of much help so far.. AI: I recommend an alternate characterization of nowhere dense. The following is a good exercise to prove. Lemma: Given a subset $S$ of a topological space $Z,$ the following are equivalent: $S$ is nowhere dense in $Z$. For every non-empty open $V\subseteq Z$, there is a non-empty open $W\subseteq V$ such that $S\cap W=\emptyset.$ That makes the result almost trivial. Take any non-empty open subset $U$ of $X$. If $U\cap Y=\emptyset,$ then $U\cap S=\emptyset,$ so suppose not. Then $V=U\cap Y$ is a non-empty open subset of $Y$, and so, bearing in mind that $S$ is nowhere dense in $Y$ and that $Y$-open sets are $X$-open, what can we conclude?
H: What does it mean to "determine" an equivalence relation? I don't understand the following problem: What does it mean exactly that a number of pairs can "determine" an equivalence relation? Say if I have the following set: {1, 2, 3}, and a relation R that's true for (a, b) if a=b. Then would the pairs {1, 1}, {2, 2}, and {3, 3} "determine" this relation? How then, does the author arrive at n/2 for this solution? This problem is from this problem set. AI: This is explained, although in a slightly roundabout way, at the beginning of Problem 4 on the problem set. Rephrasing the definition, hopefully slightly more clearly: if $R$ is any set of pairs, then the equivalence relation it determines is a relation $E_R$ on the set $\bigcup R$ (i.e. the set of all elements that appear in any pair in $R$); $E_R$ is defined as the smallest equivalence relation on $\bigcup R$ containing $R$. Exercise: for $(x,y) \in \bigcup R$, you have $(x,y) \in E_R$ if and only if there is some sequence $(x_0,x_1,\ldots,x_n)$ such that $x_0 = x$, $x_n = y$, and for each $0 \leq i < n$, either $(x_i,x_{i+1}) \in R$ or $(x_{i+1},x_i) \in R$. The slightly nonstandard part of the definition is taking the domain of the equivalence relation to be also determined by $R$, rather than as given separately. A more common version of the definition is: suppose $X$ is a set, and $R \subseteq X \times X$; then the equivalence relation on $X$ generated by $R$ is the smallest equivalence relation on $X$ containing $R$. (And again, one can give an explicit definition of this by sequences of elements that are related in $R$.)
H: Prove that $[0,1]$ is not a compact subset of $\mathbb{R}$ with the lower limit topology, i.e. open sets are of the form $[a,b)$. Prove that $[0,1]$ is not a compact subset of $\mathbb{R}$ with the lower limit topology, i.e. open sets are of the form $[a,b)$. My question is will different topology affect compactness of a set? If this is so, why? At first, when I see this question, I thought something is wrong with this question because I know that $[0,1]$ is compact by using Heine-Borel theorem. AI: In that case you can't use the Heine-Borel theorem because this theorem only apply to the case of normal topology. To prove this proposition, you just find the open cover of $[0,1]$ such that every finite subcover does not cover $[0,1]$. Let $\mathcal{A}$ be a set of open sets defined as $$\mathcal{A}=\{[0,r):0<r<1\}\cup \{[1,2)\}$$ then $\mathcal{A}$ cover $[0,1]$. However every finite subcover of $\mathcal{A}$ does not cover $[0,1]$.
H: Balls and boxes probability problem Here is another question from the book of V. Rohatgi and A. Saleh. I would like to ask help again. Here it goes: Let A, B, and C be three boxes with three, four, and five cells, respectively. There are three yellow balls numbered 1 to 3, four green balls numbered 1 to 4, and five red balls numbered 1 to 5. The yellow balls are placed at random in box A, the green in B, and the red in C, with no cell receiving more than one ball. Find the probability that only one of the boxes will show no matches. My question is more on how to interpret the problem. I actually cannot understand what is being asked and how was the experiment performed. Can anyone help me please? Also, if you have an answer, please explain the solution as well. Thanks AI: The only interpretation I can think of is that the cells are also numbered: in Box A they are $1,2,3$, in Box B $1,2,3,4$, in Box C $1,2,3,4,5$. Then a match is a ball numbered $i$ landing in a cell numbered $i$. For a related single box problem, look up derangements. The probability that Box A has no match is easy to find. Box B is a little harder, and for Box C the theory of derangements will greatly simplify the counting.
H: How do I ascertain whether an equation defines a function? Asides from graphing an equation are there any other ways of determining if an equation is a function? AI: If you can re-write the equation as a dependent variable and a independent variable, so Y = an equation only involving X, then verify that each X determines only 1 value for Y (i.e, there are no $\pm$ or you don't need to take even roots of Y to isolate it etc.)
H: Explanation on a "different" proof that $C_c(\Omega)$ is dense in $L^p(\Omega)$. Theorem: Let $\Omega\subset \mathbb{R}^n$ be an open set and $1\leq p < \infty$. The space $C_c(\Omega)$ is dense in $L^p(\Omega)$. Haim Brezis has a French book called "Analyse fonctionnelle: theorie et applications" (my version, Análysis Functional: Teoría y aplicaciones, is Spanish) that says: We know that $C_c(\Omega)$ is dense in $L^1(\Omega)$. So, we can suppose $1<p< \infty$. To prove that $C_c(\Omega)$ is dense in $L^p(\Omega)$ it's enough to show that if $h\in L^{p'}(\Omega)$ satisfies $$\int_\Omega hu=0\,\textit{ for all }\,u\in C_c(\Omega),$$ then $h=0$. Here $p'$ is a number such that $1/p+1/p'=1$. Could someone explain me why is it enough to prove it? Note: The rest of the proof is to show that $h$ is locally integrable to conclude that $$\int_\Omega hu=0\text{ for all }u\in C_c(\Omega)\Rightarrow h=0.$$ This proof seems "different" of other proofs (like this one) that uses, for example, approximation by simple functions. Thanks. AI: I suspect this is what is going on : Let $X$ denote the closure of $C_c(\Omega)$ in $L^p(\Omega)$, you want to show that $X = L^p(\Omega)$. Suppose not, then by the Hahn-Banach theorem, there must exist a continuous linear function $\varphi : L^p(\Omega) \to \mathbb{C}$ such that $$ \varphi(u) = 0 \quad\forall u \in C_c(\Omega) $$ but $\varphi \neq 0$. Now, the dual of $L^p$ is $L^{p'}$, so that means that there should exist $h \in L^{p'}$ such that $$ \int hu = 0 \quad\forall u \in C_c(\Omega), \text{ but } h \neq 0 $$ This is what you are trying to contradict. (ie. if no such $h$ exists, then it must happen that $X = L^p(\Omega)$)
H: How to calculate this series/with factorial in the numerator? I'm wondering if anyone could help me out with figuring out this series? $$\sum_{k=1}^{\infty }\frac{2^{2k}}{(2k)!}$$ with the factorial in the bottom, what trick we need here to calculate it? Thank you!! AI: This is $\sum_{k=1}^{\infty }\frac{2^{2k}}{(2k)!}=\cosh 2 -1$ as shown by Wolfram Alpha where the $-1$ comes because your sum starts from $k=1$ instead of $k=0$
H: Integrable function via Fubini Suppose $f(x,y)\geq 0$ is a measurable function on $\mathbb{R}\times\mathbb{R}$, and $$\int_\mathbb{R}\left(\int_\mathbb{R}f(x,y)d\mu(x)\right)d\mu(y)<\infty$$ Do we have that $\dfrac{xyf(x,y)}{x^2+y^2}$ is integrable on $\mathbb{R}\times\mathbb{R}$. By Fubini, we have that $$\int_{\mathbb{R}\times\mathbb{R}}f(x,y)d(\mu\times\mu)<\infty$$but I don't know about the factor $\dfrac{xy}{x^2+y^2}$. AI: It would help to show $\dfrac{xy}{x^2+y^2}$ is bounded --- say $\left\|\dfrac{xy}{x^2+y^2}\right\| \le C$ for all $x,y$. Then $$ \left\|f(x,y)\dfrac{xy}{x^2+y^2}\right\| \le C\|f(x,y)\|, $$ so the integral of that can easily be shown to be finite. The function $\dfrac{xy}{x^2+y^2}$ is homogeneous of degree $0$, i.e. multiplying $(x,y)$ by a constant has the effect of multiplying the whole fraction by $1$. Therefore it's enough to show that the function is bounded on the unit circle $x^2+y^2=1$. It's a continuous function on a compact set (the circle is compact), so it's bounded.
H: Greatest common divisor, no prime factor without power If I try to get the GCD of 272 and 210 I end up with the following $272 = 272/2=136/2=68/2=34/2=17$ $2^4 \times 17$ $210 = 210/2=105/5=21/3=7$ $2 \times 5 \times 3 \times 7$ Normally if there are 2 equal numbers with a different force, the one with the highest force is scrapped. But that would mean there are no equal numbers left. So am I to understand the scrapping rule is ignored in these situations and the GCD is 2 or did I make a mistake? AI: Scrapping rule is not ignored. Rather, scrapping means taking away the extra force of the larger number. So 2^4 is scrapped down to 2, because it has 3 extra force. Hope this helped. (I assumed force means power/exponent. Also, the word "scrap" is not really used to describe what is being done here.)
H: The closed form of of the Dirac Delta Function I know that $\delta(x)$ is the Dirac Delta function - satisfying -$$\int^{\infty}_{- \infty}\delta(x)dx=1$$Can anyone tell me a closed form of $\delta(x)$ . I guess it might be in a form of $e^{a(x)}$ for some $a(x)$ AI: The Dirac delta function is not a function. Every function which is zero everywhere and "not zero" at zero has integral zero - changing the value of the function at one point shouldn't change the integral (since you can make the partition of the domain of integration arbitrarily small around that bad point). Think about the other part of the definition $\int \delta \phi = \phi(0)$. This is closer to how the Dirac Delta "function" is defined - it is actually a linear functional (a linear function going to the real numbers) on a space of "test functions," which is the vector space of all functions (typically $R^n \to C$) that are infinitely differentiable and which are zero outside of some bounded set (i.e. compactly supported). In this case, the dirac delta functional is the linear map that takes a test function and sends it to its value at zero. It is easy to check that this is linear. It is also true that this is a continuous linear functional when you put the appropriate topology on the space of test functions, but communicating the construction of that topology few words is exceedingly difficult (without using phrases like inductive limit of Frechet spaces, at least). If you take the space of all of these continuous linear functionals on the test functions you get what are called distributions. Intuitively, you can think of distributions as the family of functions that associate each test function to some number in a way that is linear and also continuous with respect to the maximum values of the partial derivatives of each test function. (So that if two test functions have similar partial derivatives of all orders, then they are sent to nearby numbers.) Many of these distributions have no closed form (in the sense that you want), including the dirac delta function. (You can prove that there is no function $f$ that satisfies $\int f \phi = \phi(0)$ for all test functions $\phi$.) However, many familiar functions can be thought of as distributions, just by taking sending $f$ to the functional $\phi \to \int f \phi$ - at least if you are willing to ignore differences between functions that occur on "measure zero sets". (Which is to say, {the constant function 1} and {the function which is constantly 1 except when it is zero when it is 12} are the same as distributions. A point is a measure zero set. Any countable set is a measure zero set. The middle thirds cantor set is also a measure zero set. Fat cantor sets are not measure zero sets. Functions which are identical except on measure zero sets give rise to the same distribution because integration ignores measure zero sets.) What is cool about this is that if you think about a function as a distribution instead of as a "function", then placing it inside the vector space of distributions actually allows you to assign meaningful partial derivatives of every order. These are called its distributional derivatives. For example, the function $H$, which is zero when $x \leq 0$ and 1 when $x > 0$, has a distributional derivative equal to the dirac delta functional. The way this works is that we define the distributional derivative to be a formal differentiation by parts (this is not nonsense - it is the transposition of the differentiation operator from the space of test functions to itself up to the space of distributions, the idea being that you can take a differentiation of a distribution and pull it down to a differentiation on the test functions it is acting on. This is like taking the transpose of a matrix, but in an infinite dimensional vector space.) $\int H' \phi = - \int H \phi' = - \phi (\infty) + \phi(0) = \phi(0)$, where the $\phi(\infty)$ term disappears because each of the functions $\phi$ are zero outside of some bounded set (which may vary for each $\phi$.) If you draw pictures of these functions it makes some intuitive sense that the derivative of $H$ is the dirac delta function - there is a spike at zero, and it is zero everywhere else. Maybe this gives you some of the intuition you are looking for.
H: How to solve system of 4 equations with four unknowns? I have got system of 4 equations as shown below and I am considering if there is any other method than brute force to solve them. B + C + D = S1 A + C + D = S2 A + B + D = S3 A + B + C = S4 Values of S1-S4 are given: S1 = 70 S2 = 75 S3 = 80 S4 = 75 I tried to to solve that, but always I have infinite equation what means to solve A I need B, to solve B I need C, to solve C I need D and to solve D I need.. A. I know the answer, but need the solution how to solve it. Answer: A = 30 B = 25 C = 20 D = 25 AI: If you subtract the last from the third, you get $D-C=-1$ or $D=C-1$. Now you can substitute this into three equations and eliminate $D$. Similarly subtracting the first two allows you to eliminate $B$ You will then have two equations in two unknowns, and again can solve one equation for one unknown in terms of the other. This is a standard technique for simultaneous equations, which works very well with this set because of the structure.
H: Alternating series problem My mind is blanking completely on how to do this one. the base string of numbers i have to pull a series out of is $$ \frac{2}{3} -\frac{2}{5} + \frac{2}{7} - \frac{2}{9} + \frac{2}{11} - ...$$ which i found to be $$\sum_{n=1}^\infty(-1)^n( \frac{2}{2n+1} ) $$ I have to do an An to Bn, and if Bn is >= 0 and Bn is decreasing it's convergent right? so i'm thinking my Bn is $$ \frac{2}{2n+1} .$$ ? which Is greater than 0 so I think that that's Divergent. Am I on the right track? AI: The Alternating Series Test says that $$ \sum_{k=0}^\infty (-1)^ka_k $$ converges as long as $a_{k+1}\le a_k$ and $\lim\limits_{k\to\infty}a_k=0$. Comparing to the Leibniz Series, your series converges to $2-\pi/2$.
H: If $f: [0,1]\rightarrow \mathbb{R}$ is continuous function positive If $f: [0,1]\rightarrow \mathbb{R}$ is continuous function positive, so $$\int_{0}^{1} \frac{f(x)}{f(x)+f(1-x)}dx=\frac{1}{2}$$??? all examples that I tested have worked. AI: Since $$ \int_{1/2}^1\frac{f(x)}{f(x)+f(1-x)}\,dx\stackrel{y=1-x}{=}\int_0^{1/2}\frac{f(1-y)}{f(1-y)+f(y)}\,dy=\int_0^{1/2}\frac{f(1-x)}{f(x)+f(1-x)}\,dx, $$ it follows that \begin{eqnarray} \int_0^1\frac{f(x)}{f(x)+f(1-x)}\,dx&=&\int_0^{1/2}\frac{f(x)}{f(x)+f(1-x)}\,dx+\int_{1/2}^1\frac{f(x)}{f(x)+f(1-x)}\,dx\\ &=&\int_0^{1/2}\frac{f(x)}{f(x)+f(1-x)}\,dx+\int_0^{1/2}\frac{f(1-x)}{f(x)+f(1-x)}\,dx\\ &=&\int_0^{1/2}\frac{f(x)+f(1-x)}{f(x)+f(1-x)}\,dx=\int_0^{1/2}\,dx=\frac12. \end{eqnarray}
H: Solving a system of equations with matrices I have \begin{align} x_1 + 2x_2 & = 3 \\[0.5ex] 4x_1 + 5x_2 & = 6 \\[0.5ex] 9x_1 + 12x_2 & = 14 \end{align} I get the reduced row echelon form of $$\left[\begin{array}{cc\|c} 1 & 0 & 0 \\[0.55ex] 0 & 1 & 0 \\[0.55ex] 0 & 0 & 1 \end{array}\right]$$ This is obviously inconsistent but the book is saying that it isn't and gives nothing further, is this right or wrong? AI: You are absolutely correct: the system is indeed inconsistent. As you know, there is no possible way that $0x_1 + 0x_2 = 1$, hence the inconsistency. Since there are three equations and two unknowns, the system is "overdetermined", and the only way the system can be consistent is if one (or more) of the rows in the $RREF$ matrix were strictly a row of zeros.
H: how to find center/radius of a sphere Say you have an irregular tetrahedron, but you know the (x,y,z) coordinates of the four vertices; is there a simple formula for finding a sphere whose center exists within the tetrahedron formed by the four points and on whose surface the four points lie? AI: Simple formula, maybe not. Take any three out of four points. The sphere in question must contain the circle through the three points within the plane of the three points. Which is to say, take three points, circumscribe a circle around that triangle. That circle has a center. If you draw a line through that center, orthogonal to that plane, the center of the sphere is somewhere along that line. Now take out one of the points and put in the fourth, you now have a different three points in a different plane. Do the same things. The two lines must intersect in the center of the sphere. If any three of the points are nearly collinear, or the four are nearly coplanar, there are any number of ways to make the method more robust
H: What is the CPLX button on a calculator? On my cheap "dollar store" scientific calculator, it has a 2nd function button named "CPLX". When you press it, the calculator displays some text similar to the "DEG, RAD, GRAD" that says "CPLX". When in this mode, you can't add, subtract, multiply, or divide. The equals button doesn't even work. What is this and how is it used? I'm wondering if maybe it's for complex numbers, seeing that there's also a "a" and a "b" button next to it. AI: This is for complex numbers, ones of the form $a+bi$ where $a,b$ are real and $i$ satisfies $i^2 = -1$. Which calculator is it - you need the manual to learn to manipulate it.
H: Taking power of a simple function with disjoint domains I have a question in mind: Suppose I have a nonnegative simple function $$\sum_{i=1}^{N}{a_i}{\chi_{E_i}}$$ where the $E_i$s are pairwise disjoint. Then consider $$(\sum_{i=1}^{N}{a_i}{\chi_{E_i}})^p$$ where $1<p<\infty$. Is it true that $$(\sum_{i=1}^{N}{a_i}{\chi_{E_i}})^p=\sum_{i=1}^{N}{{a_i}^p}{\chi_{E_i}}$$? If yes, is it due to the pairwise disjoint $E_i$s involved? If no, kindly provide a counterexample. Thank you. AI: Yes, and it is due to the pairwise disjoint $E_i$. Suppose $E_1=[0,2]$ and $E_2=[1,3]$. Let $f=2\chi_{E_1}+3\chi_{E_2}$. Then $f(1)^2=25$, since $f(1)=5$. But if we assume $f(1)^2=2^2\chi_{E_1}+3^2\chi_{E_2}$, we get $13$. This is because the sum of squares is not the square of the sum.
H: Definition: finite type vs finitely generated The mathematical term "finite type" appears more and more in the modern articles nowadays. But it is still hard to be found in the standard textbooks. I learned the definition of it from Stacks Project http://stacks.math.columbia.edu/tag/00F2 , it is defining on the ring maps. What is the right point of view when we say "ring of finite type", "group of finite type", "module of finite type"? Would the definition in each case be exactly equivalent to the definition of "finitely generated"? AI: I think that "finite type" and "finitely generated" ring homomorphisms are really just synonyms. But "finitely generated" is also used for modules and in fact arbitrary algebraic structures (see below), so that one often prefers "finite type" in the setting of ring homomorphisms. In order to differentiate these notions even more, one says "finite" if the corresponding module is finitely generated. Similarly, for schemes, one can define finite morphisms and morphisms (locally) of finite type. See here for the relations between these two notions. If $C$ is a variety in the sense of universal algebra, then an object $M \in C$ is called finitely generated if there are elements $a_1,\dotsc,a_n$ such that $M = \langle a_1,\dotsc,a_n \rangle$, where the right hand side is the smallest subobject of $M$ containing the $a_1,\dotsc,a_n$. This yields the usual notion when $C=\mathsf{Set},\, \mathsf{Grp},\,R\mathsf{-Mod},\, R\mathsf{-CAlg}$ etc. Even more generally, an object $M$ of an arbitrary category $C$ is called finitely generated if for every directed diagram $\{N_i\}$ of objects whose transition maps are monomorphisms the canonical map $\varinjlim_i \hom(M,N_i) \to \hom(M,\varinjlim_i N_i)$ is bijective. This coincides with the definition above if $C$ is a variety (easy exercise).
H: How can i solve this probability problem fast? We have a box and we have on it 6 balls with numbers from 0 to 5. We push out 3 balls in the way that after pushing out a ball we turn it back again in the box. What is the probability that the sum of the numbers will be equal to 5. Thanks in advance. AI: $x_1+x_2+x_3=5,~0 \le x \le 5~,x \in \mathbb{N}$ solution number is $\binom{5+2}{2}=21$ Answer is $\frac{21}{6.6.6}$.
H: Prove that image and kernel of a matrix are invariant subspaces A subspace $\mathcal{S}$ of $\mathbb{R}^n$ is called invariant for an $n \times n$ matrix $A$ if $Av$ lies in $\mathcal{S}$ whenever $v \in \mathcal{S}$. The image and kernel of a matrix are invariant under the matrix. If $A$ and $B$ are commuting matrices, then the kernel of $A$ is invariant under $B$ and the kernel of $B$ is invariant under $A$. Likewise, the image of $A$ is invariant under $B$ and the image of $B$ is invariant under $A$. How would you start proving these assertions? AI: Hint: Take $v \in KerA$. Then $A(Bv)=(AB)v=(BA)v=B(Av)=B\vec{0}=\vec{0}$ (here $AB=BA$, since they are commuting; $Av=\vec{0}$, since $v \in KerA$). What does that mean?
H: Prove that $\text{argmin}_x f(x) \leq \text{argmin}_x (f(x) - mx)$ for any $f(x)$ and $m\geq0$ How to prove that $\text{argmin}_x f(x) \leq \text{argmin}_x (f(x) - mx)$ for any function $f(x)$ and $m\geq0$? AI: Hint: Let $y = \mathrm{argmin}_x f(x)$ and $z = \mathrm{argmin}_x (f(x) - mx)$. Write inequalities for $f(y)$ and $f(z)$ and for $f(y)-my$ and $f(z)-mz$.
H: Expressing the negation of $[\neg(p\land\neg q)]\land\neg r$ without $\neg,\land,\lor$ Negate $[\neg(p\land\neg q)]\land\neg r$ and replace the resulting formula by an equivalent which does not involve $\neg, \land$ or $\lor$. Can someone tell me how to get through this question? Help! I start with $$\neg[[\neg(p\land\neg q)]\land\neg r] = (p\land\neg q)\lor(r).$$ Is that right so far? AI: By DeMorgan's $$\lnot\Big(\lnot(p \land \lnot q) \land \lnot r\Big) \equiv (p \land \lnot q) \lor r$$ Now use the equivalency of $\lnot a \lor b \equiv a \rightarrow b\tag{1}$ and so $a \lor b \equiv \lnot a \rightarrow b\tag{2}$ $$\begin{align} (p \land \lnot q) \lor r & \equiv \lnot (p \land \lnot q) \rightarrow r \tag{2}\\ \\ & \equiv (\lnot p \lor q )\rightarrow r \tag{DeMorgan}\\ \\ & \equiv (p\rightarrow q)\rightarrow r\tag{1}\end{align}$$
H: Rate of change: area and perimeter The side of rectangle $x = 20m$ increases at the rate of $5m/s$, the other side $y=30m$ decreases at $4m/sec$. What is the rate of change o the perimeter and area of the retangle? If we put this rectangle in the cartesian plan (starting on the origin) , we can calculate it's perimeter and area by $P = 2x + 2y$ and $A=x.y$(maybe I should use $\triangle x$ and $\triangle y$ instead of x and y ). Since my perimeter is lnearly increasing in x and y, we can find $\frac{dP}{dt} = \frac{\partial P}{\partial x} \frac{\partial x}{ \partial t} + \frac{ \partial P }{\partial y} \frac{\partial y}{\partial t} $ = $2 m/sec$ I couldn't do the same with my area (i don't even know if with I did with the perimeter is correct). Thanks for your help! AI: We have $\displaystyle\frac{dx}{dt}=5$ and $\displaystyle\frac{dy}{dt}=-4$ Now the rate of change of area $$=\frac{d(xy)}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}$$ Now the rate of change of perimeter $$=\frac{d\{2(x+y)\}}{dt}=\cdots$$
H: Morphism from a proper irreducible scheme into an affine scheme of finite type Let $K$ be a field and $X$ be a proper irreducible $K$-scheme. Show that the image of any $K$-morphism $X \rightarrow Y$ into an affine $K$-scheme $Y$ of finite type consists of a single point. I came across this exercise in my reading, and was wondering how you might prove it. Any help would be appreciated! AI: The set-theoretic image of the $K$-morphism $f:X\rightarrow Y$ is irreducible. Since $f$ is quasi-compact, the underlying topological space of the scheme-theoretic image of $f$ is equal to the closure of the set-theoretic image, which will again be irreducible. Since closed subschemes of affine finite type $K$-schemes are also affine and of finite type over $K$, replacing $Y$ by the scheme-theoretic image of $f$, we may assume $f:X\rightarrow Y$ is surjective. Since $Y$ is separated and of finite type over $K$, properness of $X$ implies that in fact $Y$ is also proper over $S$ (see part (f) of Proposition 3.16 on page 104 of Liu's book). An affine $K$-scheme which is also proper over $K$ is finite over $K$. But then $Y$ is the spectrum of a finite $K$-algebra, which means its underlying topological space is discrete. Since we know that $Y$ is irreducible, $Y$ must be a single point.
H: Find $\int_0^{2\pi}\frac{1-\frac{1}{4}\cos\theta}{1+\frac{1}{16}\cos^2\theta}d\theta$ How to evaluate the following integral? $$\int_0^{2\pi}\frac{1-\frac{1}{4}\cos\theta}{1+\frac{1}{16}\cos^2\theta}d\theta$$ This is an exercise in complex analysis. It looks like a holomorphic function $f(z)$ that we integrate along a circle and take the real part, but I can't see which function could produce that. More precisely, by Cauchy Integral formula, $$f(z)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}-z}ie^{i\theta}d\theta$$ so I tried to find a holomorphic function and a $z\in\mathbb{C}$ such that $$\Re\left(\frac{f(e^{i\theta})}{e^{i\theta}-z}ie^{i\theta}\right)=\frac{1-\frac{1}{4}\cos\theta}{1+\frac{1}{16}\cos^2\theta}$$ but this attempt failed. AI: The standard way to attack integrals of this sort (i.e., rational functions of sine and cosine over a complete cycle) is to sub $z=e^{i \theta}$; then $d\theta=-i dz/z$ and $\cos{\theta}=\frac12 (z+z^{-1})$, and the integral becomes $$-i \oint_{\|z\|=1} \frac{dz}{z} \frac{1-\frac18 (z+z^{-1})}{1+\frac{1}{64} (z+z^{-1})^2}$$ which is equal to $$i 8 \oint_{\|z\|=1} dz \frac{ z^2-8z+1}{z^4+66 z^2+1}$$ You may now use the residue theorem to evaluate the integral: find the poles of the integrand and evaluate the residues of those poles inside the unit circle. The poles are where $$z^4+66z^2+1=0 \implies z^2=-33\pm 8\sqrt{17} = -(\sqrt{17}\pm 4)^2$$ which means that either $z=\pm i \left (\sqrt{17}-4\right )$ or $z=\pm i \left (\sqrt{17}+4\right )$. Of these, only the former pair are within the unit circle, so we need only take residues at those poles. The integral is $i 2 \pi$ times the sum of these residues, which may be taken to be twice the real part of one of the residues. The expression for the residue is $$i 8 \frac{z^2-8 z+1}{4 z^3+132 z}$$ Plug in $z= i \left (\sqrt{17}-4\right )$, take the real part, and multiply by two. The arithmetic is messy but straightforward; the result is $$\int_0^{2 \pi} d\theta \frac{1-\frac14 \cos{\theta}}{1+\frac{1}{16} \cos^2{\theta}} = \frac{8 \pi}{\sqrt{17}}$$
H: Length of vector How do you find the length of this vector? $(5 e^{20}) \mathbf{ \hat i}-e^{-4}\mathbf{ \hat j}+\mathbf{\hat k}$ Every time I try to do this, it turns out very ugly. I need to find the length to solve for the unit tangent vector of $(e^5\mathrm t)\mathbf{ \hat i}+(e^{-1}\mathrm t)\mathbf{ \hat j}+t\space \mathbf{ \hat k}$ at $\mathrm t(4)$. The above vector is the derivative of this one with 4 already plugged. From what i see in my math book, $T(t)$ is equal to $\frac{the \space derivative}{the\space length\space of\space the\space derivative}$. I'm pretty sure this is how you do it. If anyone could help, it would be very much appreciated... Thank you so much!! AI: Just use the pythagorean theorem. $5e^{20}\textbf i-e^{-4}\textbf j+\textbf k$ has length $\sqrt{(5e^{20})^2+(-e^{-4})^2+1^2}$. It might not be super pretty, but it's not so horrible either: you can expand it out with just a few minutes of extra effort, and then you can even pull the denominator out of the radical.
H: Does $d(x_{n+1},x_n) Let $(X,d)$ be a complete metric space, $(x_n)_{n\in\mathbb{N}}\subset X$ such that $d(x_{n+1},x_n)<d(x_n,x_{n-1})$ for all $n\in\mathbb{N}$. Since I cannot construct such sequence which is not convergent, I wonder if every such sequence is convergent? I first thought of defining a sequence $y_n:=d(x_{n+1},x_n)\in[0,\infty)$ and checking if it converges to zero, but it does not seem fruitful. (Of course, I am aware of the definition of convergence, which is much stronger than this.) AI: Let $x_0=1$, $x_n=x_{n-1}+1+\frac{1}{n}$ for $n\ge 1$. This satisfies your condition but is not convergent.
H: Geometry: angle relative to x-axis for a tangent of a circle (see picture) Excuse my poor description in the title, I think a picture is needed to explain my question: Theta is the angle to the x-axis. So my question is: given the radius of the circle, theta, and beta, what is theta'? The problem arises when simulating a car (with heading theta) that is turning. Beta can be assumed to be small. Where I saw this problem it was stated: theta' = theta+beta % 2*pi. But I want to know why. AI: This is about the rotation of the tangent line and horizontal line by $\beta$. If you rotate the first tangent line by $\beta$ you get the second tangent line. Then the first horizontal line is rotated by $\beta$ and still makes the angle $\theta$ with the rotated tangent line. However the angle between the rotated horizontal line and the new horizontal line is $\beta$ and hence the whole angle is $\theta+\beta$.
H: Understanding Limit Points I am doing some analysis and trying to understand the idea of a limit point. This is the way I am thinking of it. I would just like someone to confirm if I have the right idea, or to correct me if I am mistaken. Thanks. The way I'm thinking of a limit point of $p$: If there is a $q\neq p$, such that $q\in E$, then $q$ is a limit point. So, if I imagined an epsilon neighborhood, the limit points of that neighborhood, will be all points in that neighborhood besides the center, $p$, and the "boundary" points. (I don't think the "boundary" points are going to be limit points, because the definition of a neighborhood says that the metric $d(p,q)<r).$ Can someone confirm if I'm thinking about this correctly? Edit: After consulting a few analysis books, I've changed the way I think of it as follows: A point $p$ is a limit point of $E\subset X$, where $X$ is a metric space, if the open ball with radius $\epsilon$, $\forall\epsilon>0$, centered at $p$ intersects a point in $E$ other than $p$. AI: You've worded it very strangely. Try: let $X$ be a topological space (or metric space). Then a limit point $x \in X$ of a subset $S \subset X$, is a point such that for any open neighborhood $U$ (or open ball) about $x$, there exists $y \in U \cap S, y \neq x$.
H: Ideals in a Noetherian ring Let $R$ be a ring, let $\mathfrak{i}$ be an ideal of $R$, let $\{x_{\alpha}\}_{\alpha\in\mathcal{A}}$ be a set of generators for $\mathfrak{i}$, suppose $\mathcal{A}$ has infinitely many elements. Now, assume $R$ Noetherian. Can I extract from $\{x_{\alpha}\}$ a finite set of generators for $\mathfrak{i}$? I know since $R$ is Noetherian there are finitely many generators for $\mathfrak{i}$. But can I find this set extracting finitely many elements from a set of generators arbitrarily fixed at the beginning? AI: Yes, you can. Start with one element among the set $\{x_\alpha\}$, call it $y_0$. If $\left<y_0\right> = \mathfrak i$ then you are done. Otherwise, there is a $y_1 \in \mathfrak \{x_\alpha\}$ such that $y_1 \notin \left<y_0\right>$. If $\left<y_0, y_1\right> = \mathfrak i$ then you are done. Otherwise, there is a $y_2$... If this process never terminates, then you get a strictly increasing chain of ideals, which is impossible. Therefore at some point you have obtained a complete generating set.
H: Prove that $A\cap (B\setminus C)=(A \cap B)\setminus(A \cap C)$. Problem: Prove that $A\cap (B\setminus C)=(A \cap B)\setminus(A \cap C)$. I've tried it on my own: \begin{align} x&\in A\cap (B\setminus C) \\ &\Leftrightarrow (x\in A) \wedge (x\in B\setminus C) \\ &\Leftrightarrow (x\in A) \wedge (x\in B \wedge x\notin C) \\ &\Leftrightarrow (x\in A \wedge x\in B) \wedge (x\notin C) \\ &\Leftrightarrow x\in (A\cap B)\setminus C\\ &\Leftrightarrow \dots \end{align} What would the next step be? I've no idea how to get from $x\in (A\cap B)\setminus C$ to $x\in(A \cap B)\setminus(A \cap C)$. If I'm trying to do the right side, we get \begin{align} x&\in(A \cap B)\setminus(A \cap C)\\ &\Leftrightarrow (x\in A \wedge x\in B)\wedge x\notin (A \cap C)\\ &\Leftrightarrow (x\in A \wedge x\in B)\wedge (x\notin A \wedge x\notin C)\\ &\Leftrightarrow \dots \end{align} How do I make it simpler? AI: You replaced $x\notin A\cap C$ with $(x\notin A)\land(x\notin C)$, which is incorrect. The correct one would be: $$x\notin A\cap C \\ \Leftrightarrow \neg (x\in A\cap C)\\ \Leftrightarrow \neg (x\in A \land x\in C)\\ \Leftrightarrow (x\notin A)\lor (x\notin C) $$ Try, whether you can finish the problem using this.
H: What is the purpose of sets? Why do we use them? All is in the title. Why sets? Why do we need them and where are they important? AI: The purpose of sets is to house a collection of related objects. They are important everywhere in mathematics because every field of mathematics uses or refers to sets in some way. They are important for building more complex mathematical structure. For example the ordered set $(x_1, x_2, x_3)$ of three points forms a vector in $\Bbb{R}^3$ if $x_i \in \Bbb{R}$. I couldn't get through that definition without referring to another set $\Bbb{R}$. Sets can come with algebraic or topological properties that are useful.
H: simplify the summation over i It's a question on the 1000 Exercise in probability. Let $X_1, X_2, X_3$ be independent random variables talking values on positive integers, and having mass function given by $P(X_i=x)=(1-p_i)p_i^{x-1}$, for x =1,2,3,..., i=1,2,3 Show: $$P(X_1<X_2<X_3)=\frac{(1-p_1)(1-p_2) p_2 p_3^2}{(1-p_2 p_3)(1-p_1p_2 p_3)}$$ The solution on the book is as following My question is from first line to second line in the solution, index "k" disappears, I think it's by making k=j+1, but I don't know where $(1-p_3)$ goes, Also, from 2nd to 3rd line of the solution, I don't know how the term $(1-p_2p_3) $ in the denominator coming from. Thanks for your time!! AI: Hint: $\sum_{i<j<k} = \sum_{i<j}\sum_{k=j+1}^\infty$. What is $\sum_{k=j+1}^\infty p_3^{k-1}$? Key is that $1-p_3$ is in the denominator of that sum. The same thing for the next step. What is $$\sum_{j=i+1}^\infty (p_2p_3)^{j-1}?$$
H: What's the integral of a constant? If the derivative of a constant is $0$ then what is the integral of a constant? What is the integral of $0$? AI: The integral of a constant $C$ with respect to $x$ is $Cx + A$, $A$ constant. Applying this rule to the constant function $y(x) = 0$, $\int {0}dx = 0+A = A$.
H: Proving mutual orthogonality of vectors Let three vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ in three-space be such that: $$ a_ia_j + b_ib_j + c_ic_j = \delta_{ij} $$ where a vector name with subscript represents a component of the vector (the subscript can take the value 1,2 or 3) and $ \delta_{ij} $ is the Kronecker delta. Is it true that $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are therefore mutually orthogonal? AI: If $P$ is the square matrix with one vector in each column, you are saying $P P^T = I.$ What do you think you get for $P^T P?$
H: An interesting (unknown) property of prime numbers. I don't know if this is the right place to ask this question. Please excuse my ignorance if it is not. I like to play with integers. I have been doing this since my childhood. I spend a lot of time looking up new integer sequences on OEIS. Last week I stumbled upon a unique property of prime numbers. I have been searching the internet since then to find if there are any papers talking about this property and haven't found any. I want to publish it. I am not a mathematician but I am an engineer and I can write a decent paper to clearly express the property. I haven't attempted to prove this property. It is just an observation that I verified upto the largest known prime number (http://primes.utm.edu/). It holds! The problem is I do not want to divulge this property for fear of being denied credit to its discovery. How should I disseminate this finding? Again, I apologize if this is not the right place for this question. AI: If you feel you really have something, you could always type up a paper and post it on the preprint archive... http://arxiv.org/ That would establish precedence and probably get you a host of feedback as well.
H: Prove $T_n(x)$ of Chebyshev Polynomial given the recurrence relation Using the recursion formula for Chebyshev polynomials, show that $T_n(x)$ can be written as $$T_n(x)=2^{n-1}(x-x_1)(x-x_2)...(x-x_n)$$ where $x_i$ are the $n$ roots of $T_n$ The recurrence relation: $T_0(x)=1$,$T_1(x)=x$, and $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$ I have an intuition that I need to use induction,but what is my inductive hypothesis? Here is my work: Assume $T_n(x)=2^{n-1}(x-x_1)(x-x_2)...(x-x_n)$ Prove $T_{n+1}(x)=2^{n}(x-x_1)(x-x_2)...(x-x_n)(x-x_{n+1})$ Use the definition of Chebyshev polynomial, $T_n(x) = cos(n\theta),x =cos(\theta)$ $T_0(x)=1$ and $T_1(x)=x$ And then use the recursion formula $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$, I get $T_{n+1}(x) = 2x[2^{n-1}(x-x_1)(x-x_2)...(x-x_n)]-[2^{n-2}(x-x_1)(x-x_2)...(x-x_{n-1})]$ Then $T_{n+1}(x) = 2x(2^{-1})(x-x_n)[2^n(x-x_1)(x-x_2)...(x-x_{n-1})]-2^{-2}[2^n(x-x_1)(x-x_2)...(x-x_{n-1})]$ Rearrange I get $T_{n+1}(x) = [2x(2^{-1})(x-x_n)-2^{-2}][2^n(x-x_1)(x-x_2)...(x-x_{n-1})]$ What should I do next(assume what I did above is correct) so that I can get $T_{n+1}(x)=2^{n}(x-x_1)(x-x_2)...(x-x_n)(x-x_{n+1})$ ?? AI: An inductive hypothesis is: $T_n=2^{n-1}x^n+f_n(x)$ where $\deg f\le n-1$. Addendum: From $T_n=2^{n-1}x^n+f_n(x)$ we have $T_{n+1}=2^nx^{n+1} +2xf_n(x)-2^{n-2}x^{n-1}-f_{n-1}(x)$ and evidently $\deg f_{n+1}(x)\le n$ where $f_{n+1}(x)=2xf_n(x)-2^{n-2}x^{n-1}-f_{n-1}(x)$. Further if $x_1,\ldots,x_n$ are the roots of $T_n$, we get $T_n=2^{n-1}(x-x_1)\ldots(x-x_n)$.
H: Are these bases for a topology? I have the following topology : $$\tau= \Bigl\{U\subseteq \mathbb{R}^2: (\forall(a,b) \in U) (\exists \epsilon >0) \bigl([a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq U\bigr)\Bigr\}$$ Are these a basis for the previous topology: $\beta_1= \{[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}$ $\beta_2= \{[a,a+\epsilon) \times [b-\epsilon, b+\epsilon)\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}$ The first one is obviously a basis for $\tau$ because of the definition of $\tau$ and I would say that the second is also a basis, because $[a,a+\epsilon) \times [b-\epsilon, b+\epsilon) \subseteq [a,a+\epsilon] \times [b-\epsilon, b+\epsilon] $ Is it correct? what do you think? AI: Note that no element in $\beta_1$ is actually an open set in the topology $\tau$. Consider the element $A$ in $\beta_1$ with $a=b=0$ and $\epsilon=1$. The point $(1,1)\in A$ but there exists no $\epsilon>0$ such that $[1,1+\epsilon]\times[1-\epsilon,1+\epsilon]\subset A$ because $(1,1)$ is on the boundary of $A$. For $\beta_2$, consider the same $a,b,\epsilon$ but look at the point $(0,0)$.
H: Solving a system of equations with 3 unknows X/Y = 5 X/Z = 2.5 Y/Z = 0.5 I don't want to check if this system is correct, but rather get actual values of X, Y, Z which could actually fit into this. From my head I can think of examples like x= 5, y = 1, z = 2 and 50, 10, 20 would work as well here. But what would be a formula to actually get numbers which would fit as x, y, z. DOesn't matter if its 5, 1, 2 or 50, 10, 20. Any ideas? AI: I suggest converting your system of equations into a 3 by 4 matrix and using Gaussian elimination (google it to find out how). You find that your system has an infinite number of solutions (one for each value of z), with $y = \frac{2z}{5}$ and $x = 5y = 2z$.
H: Laurent series of $z^{-3}$ at $z_0 = i$. Is there a way to do this by hand or is the question just evil? I have to find the two Laurent series expansions of $\frac{1}{z^3}$ about $i$. The only approach I can think of is to do: $$\frac{1}{z^3} = \frac{1}{(z-i)^3} \left( \frac{z-i}{z} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{i}{i+(z-i)} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{1-i(z-i)} \right) ^3 $$ This expansion will work in the disk $\|z-i\|<1:$ $$=\frac{1}{(z-i)^3} \left( 1 - \sum_{n=0}^{\infty}i^n(z-i)^n \right) ^3 =i \left( \sum_{n=0}^{\infty}i^{n}(z-i)^{n} \right) ^3 .$$ But this is nasty, because I don't know how to cube the series! And I don't think anything nice would come out of it, anyway. The second expansion is in the annulus $\|z-i\|>1:$ $$\frac{1}{z^3} = ... = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{z-i}\frac{1}{ \frac{1}{z-i}-i} \right) ^3 = \frac{1}{(z-i)^3} \left( 1 - \frac{1}{z-i}\frac{i}{ 1-\frac{-i}{z-i}} \right) ^3 =$$ $$= \left( \frac{1}{z-i} - i \sum_{n=0}^{\infty} \frac{(-i)^n}{(z-i)^{n-2}} \right) ^3$$ Same problem, except now it's even worse... How do I deal with this? AI: I'm sure you have been taught about this before. $\frac{1}{z^3}$ is regular at $z = i$, so the Laurent series there is simply the ordinary Taylor series expansion (in the vicinity of $i$). For any $\alpha > 0$ and $a \in \mathbb{C}$ not on the negative real axis, we have $$\begin{align} \frac{1}{z^\alpha} = & \sum_{k=0}^{\infty} \frac{1}{k!} \left( \left.\frac{d^k}{dz^k} \frac{1}{z^{\alpha}}\right\|_{z=a}\right) (z-a)^k\\ = & \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{(-\alpha)(-\alpha-1)\cdots(-\alpha-k+1)}{a^{\alpha+k}}\right) (z-a)^k\\ = & \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left( \frac{\alpha(\alpha+1)\cdots(\alpha+k-1)}{a^{\alpha+k}}\right) (z-a)^k\\ \end{align}$$ When $\alpha$ is a positive integer $n$, this reduces to $$ \frac{1}{z^{n}} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} \frac{(z-a)^k}{a^{n+k}} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{n-1} \frac{(z-a)^k}{a^{n+k}} \tag{1}$$ In particular, when $n = 3$ and $a = i$, one has $$\frac{1}{z^3} = \sum_{k=0}^{\infty} (-1)^k \binom{k+2}{2} \frac{(z-i)^k}{i^{3+k}} =\sum_{k=0}^{\infty} i^{k+1}\frac{(k+1)(k+2)}{2} (z-i)^k \tag{2}$$ Update To derive the other Laurent series (i.e, the one in the vicinity of $\infty$), one can let $u = \frac{1}{z-i}$ and rewrite $$\frac{1}{z^3} \quad\text{ as }\quad\frac{1}{(i + \frac{1}{u})^3} = \frac{i u^3}{(u-i)^3}.$$ One then consider the Taylor series expansion of it at $u = 0$. Compare this with $(1)$ and $(2)$, one find $u$ is now taking the role of $z-i$ and $-i$ is taking the role of $a$ there. As a result, $$\frac{1}{z^3} = i u^3 \sum_{k=0}^{\infty}(-1)^k\binom{k+2}{k} \frac{u^k}{(-i)^{3+k}} = \sum_{k=0}^{\infty}(-i)^k \frac{(k+1)(k+2)}{2}\frac{1}{(z-i)^{k+3}} $$
H: Why does the google calculator give $\tan 90^{\circ} = 1.6331779e^{+16}$? I typed in $\tan 90^{\circ}$ in Google and it gave $1.6331779\mathrm{E}16$. How did it come to this answer? Limits? Some magic? AI: The closest IEEE-754 double value to $\pi/2$ is $1.5707963267948965579989817342720925807952880859375$. The cosine of that, on standard x86_64 hardware evaluates to $6.123233995736766 \times 10^{-17}$. The reciprocal of that is $1.633123935319537 \times 10^{16}$.
H: Cross product of vector functions I was trying to make sense of a problem when I stumbled upon this on yahoo answers. I was just wondering if it was correct. If it is, can you please maybe explain why? ${\bf r}'(t) = \langle -5 \cos t, -5 \cos t, -4 \sin t \rangle$ ${\bf r}''(t) = \langle 5 \sin t, 5 \sin t, -4 \cos t \rangle$. ${\bf r}'(t) \times {\bf r}''(t) = \langle 20, -20, 0 \rangle$. AI: $${\bf r}'(t) \times {\bf r}''(t) = \langle -5 \cos t, -5 \cos t, -4 \sin t \rangle \times \langle 5 \sin t, 5 \sin t, -4 \cos t \rangle = \begin{vmatrix} {\bf i} & {\bf j} & {\bf k} \\ -5 \cos t & -5 \cos t & -4 \sin t \\ 5 \sin t & 5 \sin t & -4 \cos t \end{vmatrix}$$ $$= \begin{vmatrix} -5 \cos t & -4 \sin t \\ 5 \sin t & -4 \cos t \end{vmatrix} {\bf i}+ \begin{vmatrix} -5 \cos t & -4 \sin t \\ 5 \sin t & -4 \cos t \end{vmatrix} {\bf j}+ \begin{vmatrix} -5 \cos t & -5 \cos t \\ 5 \sin t & 5 \sin t \end{vmatrix} {\bf k} = \langle (-5\cos t)(-4\cos t)-(5\sin t)(-4\sin t), -\left( (-5\cos t)(-4 \cos t)-(5\sin t)(-4\sin t) \right), (-5\cos t)(5\sin t)-(5\sin t)(-5\cos t) \rangle$$ $$=\langle 20\cos^2 t +20\sin^2 t, -20\cos^2 t-20\sin^2 t, -25\sin t\cos t+25\sin t \cos t\rangle = \langle 20, -20, 0 \rangle$$ Using $20\cos^2 t +20\sin^2 t = 20(\cos^2 t+\sin^2 t) = 20(1) = 20$ etc.
H: Probability of a pair of vertices being selected in a random subset of k vertices Let $G$ be a graph of $n$ vertices. I select $k$ vertices uniformly at random. What is the probability that a specific pair of vertices are in the subset? AI: There are $\binom{n}{k}$ equally likely ways to choose $k$ vertices. There are $\binom{n-2}{k-2}$ ways to choose $k$ vertices that include a specific $2$, say Alicia and Beti, since we have to choose $k-2$ vertices from $n-2$ to accompany the two friends. Divide. The resulting expression $\dfrac{\binom{n-2}{k-2}}{\binom{n}{k}}$ can be considerably simplified. Another way: The probability that the vertex Alicia is chosen is $\frac{k}{n}$. Given that Alicia is chosen, the probability that vertex Beti is chosen is $\frac{k-1}{n-1}$. Thus the probability they are both chosen is $\frac{k}{n}\cdot\frac{k-1}{n-1}$.
H: Is this space a banach space? Hi I want to find out whether $l^1$ with the norm $\|\|x\|\|:=sup_n \|\sum_{i=1}^{n} x_i\|$ is a Banach space. In case that you think that it is a Banach space, just say: It's a Banach space(and then I will first try to prove this), but in case that it is not, I would be grateful to you, if you could give me a sequence, that destroys the Banach space question. AI: Let $e^i$ be the sequence with a $1$ in the $i$-th place and zeros elsewhere. Let $x^{k} = \sum\limits_{n=1}^k \dfrac{(-1)^{n-1}}{n}e^n$. Then, for $k < m$ $$\lVert x^k - x^m\rVert = \frac{1}{k+1},$$ since the partial sums of the difference $\sum_{n=k+1}^m \dfrac{(-1)^{n-1}}{n}e^n$ are bounded by the first nonzero partial sum, since the nonzero terms have alternating sign and decreasing magnitude. So $(x^k)$ is a Cauchy sequence, but its "limit", the alternating harmonic series, is not in $l^1$.
H: When does $xxyy = xyxy$ not imply $xy = yx$ in a ring? This is a follow-up to a graded question on some homework. In attempting to prove that a ring R such that $x^2 = x$ for all $x \in R$ is commutative, I tried using the fact that $xy = x^2y^2 = (xy)^2$ for all $x, y\in R$, so $xxyy = xyxy$. I thought this was enough information to know for sure that $xy=yx$, but apparently it isn't. What is the problem with this conclusion? AI: You cannot simply cancel off $x$ and $y$. To get $xy=yx$ from $xxyy=xyxy$. Here's a counter-example: $x = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $y = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$. Notice that $xy = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ so $xyxy = 0$. Also, $x^2=0$ so $x^2y^2=0=xyxy$. But notice that $yx=0 \not= xy$. How did this happen? Well, cancelation doesn't work when we have zero divisors hanging around. Notice that $yx=0$ (and with both $x$ and $y$ not zero) says that $y$ and $x$ are zero divisors. But in the ring of $2 \times 2$ matrices, things that multiply to be zero in one order don't necessarily work that way in the other order as well.
H: How should subscripts be used when evaluating midpoint, trapezoid, and simpsons rule? I conceptually understand how these rules work, and I can visually see what I need to calculate, but the subscripts sometimes confuse me. Some start at $x_{0}$ and others start at $x_{1}$ If I'm looking at evaluating $\int f(x)dx$ using the midpoint rule, well in my mind I see my first interval beginning at $x_{0}$ and the next interval (rectangle) beginning at $x_{1}$ (where the other interval ends). I then know I have to calculate the midpoint between those two intervals. So would that mean the initial midpoint is $x_{0}$ or $x_{1}$ or something else? It makes sense to me when I'm evaluating using specific endpoints like $L$ or $R$, but generally speaking, when considering the Midpoint, Trapezoid, and Simpsons rule, how should one treat the interval subscripts? AI: Yes, the midpoint you would have to calculate is the midpoint of $x_0$ and $x_1$, i.e. $\frac{1}{2}(x_0 + x_1)$. This is sometimes denoted $x_{1/2}$; likewise $x_{(2k-1)/2} := \frac{1}{2}(x_k +x_{k+1})$. This notation is very helpful when using Simpson's rule for example.
H: What is probability of independent events? I have N similar hard drives. There is a probability f% that any one of them will crash. What is the probability that at least two of them will crash? AI: With $p=\frac f{100}$, the probability of no crash at all is $(1-p)^N$ and the probability of exactly one crash is $Np(1-p)^{N-1}$ (picke one, let it crash, let the others not crash), so the probability of two or more crashs is $$1-(1-p)^N-Np(1-p)^{N-1} = 1-(1-(N+1)p)(1-p)^{N-1}. $$
H: The roots of $z^4+z^3+1=0$ are in $\frac{3}{4}<\|z\|< \frac{3}{2}$. How can we show that all the roots of the complex polynomial $$p(z)=z^4+z^3+1$$ lie in $\frac{3}{4}<\|z\|< \frac{3}{2}$? This is an exercise in complex analysis. AI: Broad hints: if $\|z\|\leq \frac34$, what can you say about $\|z^4+z^3\|$? (Try using the triangle inequality) Similarly, if $\|z\|\geq \frac32$ you should be able to compare $\|z^4\|$ with $\|z^3+1\|$.
H: Finding two different matrices that have the same product with another one? (AB=AC) Say you have a matrix, $A$, and you want to find two matrices $B, C$ for which $AB=AC$, but being $B\neq C$. I know this is quite possible due to matrices' nature, but I can't quite find a method to find them except for just random search. i.e. Say $A = \begin{pmatrix} 1&1&1&1\\ 1&2&3&4\\ 2&4&4&6 \end{pmatrix}$ You can probably use some linear combination trick, either with rows or columns; or by creating a template matrix with a parameter that doesn't intervene on the final result, and hence you would be able to find infinite examples; but I'm unable to find a method for either of these ways. I'm not looking for a particular solution, but rather for a method that would allow you to find them, I don't mind the result as long as I learn how to do it. Anyway, thanks in advance. Any help appreciated. AI: You need to find a matrix whose column vectors are in the nullspace of $A$. Say that $\{v_1,\dots,v_n\}$ span the nullspace of $A$. Then, choose any matrix $D$ and construct any matrix $E$ whose column vectors are linear combinations of $\{v_1,\dots,v_n\}$ (you need to pick $D$ and $E$ to be of appropriate dimension). Then, $$A(D+E)=AD+AE=AD.$$ This gives you your "infinite source" of $B$s and $C$s. You can compute $v_1,\dots,v_n$ simply solving a system of linear equations, see here.
H: A Better Way to Solve this Factorial Problem? I had a problem that asked me to find which of the following is larger: ${2013 \choose 500}$ or ${2013 \choose 1500}$ Beneath is my proof. I think it is correct (though your verification and suggestions would nonetheless be appreciated). I haven't worked with factorials all that much, and am curious to see if it could be done any more swiftly. Are there easier ways to prove this proposition? ${\bf Proof:}$ $${n \choose k} = \frac{n!}{k!(n-k)!}$$ $${2013 \choose 1500} = \frac{2013!}{1500!(2013-1500)!} = \frac{2013!}{1500!\cdot 513!}$$ $${2013 \choose 500} = \frac{2013!}{500!(2013-500)!} = \frac{2013!}{500!\cdot 1513!}$$ Obviously, if $500!1513! > 1500!513!$, then ${2013 \choose 1500} > {2013 \choose 500}$ and vice-versa. Note that $\frac{500!1513!}{500!} = 1513!$, and $\frac{513!1500!}{500!} = 1500!\cdot(501\cdot 502 \cdot ... \cdot 513)$. We can rewrite $1513!$ as $1500 \cdot (1501 \cdot 1502 \cdot ... \cdot 1513)$. As both expansions $A = (501 \cdot ... \cdot 513), B = (1501 \cdot ... \cdot 1513)$ have 13 terms each, and $\forall a \in A, a < b$ for any $b \in B$, it must be that $\prod\limits_{a\in A} a < \prod\limits_{b \in B} b$. So $1513! > 1500!(501\cdot ... \cdot 513)$, so $500!1513! > 1500!513!$, so $\frac{2013!}{500!1513!}<\frac{2013!}{1500!513!}$, so ${2013 \choose 1500} > {2013 \choose 500}$. $\square$ AI: The sequence $$\left\langle\binom{2013}k:k=0,1,\ldots,2013\right\rangle$$ is centrally symmetric and single-peaked, increasing from $\binom{2013}0=1$ to a maximum at $\binom{2013}{1006}=\binom{2013}{1007}$ and then falling to $\binom{2013}{2013}=1$. Since $500<513\le 1006$, we must have $$\binom{2013}{500}<\binom{2013}{513}=\binom{2013}{1500}\;.$$
H: A finite sum with cosines I'm not able to compute the following sum : $$\sum_{k=0}^n\frac{\cos(kx)}{(\cos(x))^k}$$ AI: You have $\cos(kx) = \Re(e^{ikx})$. Can you use a geometric series argument to do the calculation?
H: $\mod 4$ properties of Fermat number Let $k\in\Bbb N$. Let $2^{2^k}+1$ be a composite Fermat number. Let $p$ be a prime factor of $2^{2^k}+1$. Then is $p\mod 4\equiv 1$? AI: If $p$ is a prime divisor of $F_k = 2^{2^k}+1$, then we have $$2^{2^k} \equiv -1 \pmod{p},$$ hence $2^{2^{k+1}} \equiv 1 \pmod{p}$ and the order of $2$ modulo $p$ is $2^{k+1}$. That directly yields $2^{k+1} \mid (p-1)$ or, expressed differently, $p \equiv 1 \pmod{2^{k+1}}$. But, since $k > 4$, as we assumed $F_k$ to be composite, that means in particular $p \equiv 1 \pmod{8}$, and hence $2$ is a quadratic residue modulo $p$. Let $x^2\equiv 2 \pmod{p}$. Then the order of $x$ modulo $p$ is $2^{k+2}$, and hence $p \equiv 1 \pmod{2^{k+2}}$.
H: What is the underlying structure that makes this analogy so good? In "Linear Algebra Done Right", the author draws (in my opinion) a fantastic parallel between $\mathbb{C}$ and $\mathcal{L}(V)$ (where $V$ is an $\mathbb{F}$-inner product space). In this analogy, he establishes: A complex number $z$ corresponds to an operator $T$, The conjugate $\overline{z}$ corresponds to the adjoint $T^$, The complex number $z$ is real corresponds to $T$ being self-adjoint, The complex number $z$ is non-negative corresponds to $T$ being positive-semidefinite, The complex number $z$ satisfies $\|z\| = 1$ corresponds to $T$ being an isometry ($TT^ = I$), among others. I'm curious though if there is some underlying structure linking $\mathbb{C}$ and $\mathcal{L}(V)$ that makes the parallel so great (I know they are both vector spaces), or is this just a coincidental observation by the author? AI: It is not a coincidence. $\mathbb C$ can be seen as $\mathcal L(V)$ in the case where $V$ is of dimension one. If you want, you can see the notions of adjoint, positive, etc., for operators as the right generalizations of the corresponding notions of $\mathbb C$.
H: Do the algebraic properties of the exponential and log functions specify them uniquely in probability theory? I come from a physics background and in classical mechanics, we construct a Hamiltonian function whose partial derivatives generates a vector field, two independent systems are assigned a total Hamiltonian which is the algebraic sum of the Hamiltonians for the isolated systems. In statistical mechanics, Liouville's theorem requires that in equilibrium, the probability that a system's dynamical variables have specific values is a function of the value Hamiltonian for those specific values. So the standard probability theory argument that the probability to measure state a in system A and state b in system B for independent systems A and B is the product of the individual probabilities: $Pr(a \text{ & } b) = Pr(a)Pr(b)$. This is essentially a quick derivation of the maxwell-boltzmann distribution. I've seen this argument about probability theory in several places, and can easily see that the exponential satisfies the requirement. So I think my question is essentially this, does $$ f(a+b) = f(a)f(b) \\ f(x) > 0 \text{ for all real x} \\ $$ uniquely specify that $f(x)=Ae^{cx}$? Or is there some other consideration that I must be overlooking? AI: If $f$ is a differentiable function, taking a derivative with respect to $b$ means that $$f'(a+b)=f(a)f'(b).$$ Now substitute $b=0$ to get the differential equation $f'(a)=f(a) f'(0)$ which has the solution $f(x) = A e^{f'(0) x}$. Thus $f'(x) = f'(0) A e^{f'(0) x}$. Solving for $f'(0)$ gives $f'(0)=Af'(0)$ which is only true if $A=1$ or $f'(0)=0$. In the latter case we end up with the constant function $f=1$ (since $A=A^2$). Otherwise, the solution is $f(x) = A e^{c x}$ where choosing $c$ chooses $f'(0)$.
H: If $\lfloor a \rfloor \le b$, what is $a$ less than? I know that $\lfloor a \rfloor \le b$. Can I then conclude that $a \le b + 1$? Is that the most precise way to describe a? AI: $\lfloor a\rfloor\le b$ if and only if $a<b+1$. Note the strict inequality: if $a$ were equal to $b+1$, its floor would be $b+1$, not $b$ or less. In more detail, $\lfloor a\rfloor=n$ if and only if $n\le a<n+1$, i.e., if and only if $a\in[n,n+1)$. Thus, $$\{a\in\Bbb R:\lfloor a\rfloor\le b\}=\bigcup_{n\le b}[n,n+1)=(\leftarrow,b+1)\;,$$ the set of all real numbers less than $b+1$.
H: What would be the algebraic representation to this verbal function? You are exploring a hidden cave when suddenly, your flashlight goes out. Luckily, you have a 10-inch wax candle with you... Whew! You light the candle, but you must get out of the cave quickly, because every two minutes, one inch of the wax candle melts. Also, once the height of the candle reaches 2 inches or less, you will no longer be able to hold it(otherwise you would be burned). How long do you have to get out of the cave safely? That is the word problem. I'm having some trouble coming up with an equation to it. Can anyone tell me what the algebraic representation of this function would be? This is part of my homework that is due 10/23/13. AI: Go through the word problem carefully, noting each piece of (possibly-) relevant information at a time. Then try to build up a function/equation/whatever you need from the information. For example, in this problem you might make the following quick observations: - The candle is $10$ inches tall. - The length of the candle decreases by $1$ inch every $2$ minutes - You're trying to figure out how long until the candle is $2$ inches tall (i.e., how long you have until you can no longer hold the candle). You could definitely come up with an equation to solve here algebraically, but in this case it's probably going to be easier for you to just make a table of values giving the length of the candle after it's been lit for $t$ minutes (where, say, $t = 2, 4, 6, \ldots$). How many minutes until the candle is $2$ inches long? Edit: If you're instructed to come up with a function $f(t)$ for the length (in inches) of the candle $t$ minutes after being lit, first assume that $f(t)$ is linear (which is reasonable here). Then you know $f(0) = 10$. You can compute $f(2)$ from the above observations. Now you have two points $(0,10)$, $(2,f(2))$ on this line, and so you can solve for the general formula. (Since it's worth being nitpicky here, you should really have a piecewise function if you want to define $f(t)$ for all $t \geq 0$ (since otherwise you'd be saying that eventually the length of the candle is negative. But that's a secondary concern.)
H: Open immersion from a proper scheme to a separated, irreducible scheme. Fix a scheme $S$ and let $X$ and $Y$ be $S$-schemes. Assume that $X$ is proper over $S$ and $Y$ is separated over $S$. Let $f: X \rightarrow Y$ be an open immersion of $S$-schemes. If $Y$ is connected, show that $f$ is an isomorphism. Here's another exercise I encountered. I'm not exactly sure how to use the properties ascribed to $Y$ to prove this. Any help would be appreciated! AI: Separatedness of $Y$ over $S$ together with properness of $X$ over $S$ implies that the set-theoretic image of $f$ is closed (see my answer here The image of a proper scheme is closed). The image is open by assumption, so (assuming $X$ is non-empty) connectedness of $Y$ forces it to be all of $Y$. An open immersion which is surjective is an isomorphism.
H: Functions and Mapping question? Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the condition $$2f(x) = f(x + y) + f(x + 2y)$$ for all real numbers $x$ and all non-negative real numbers $y$. I just don't know where to start. AI: HINT: If you set $x=0$, the condition requires that $$2f(0)=f(y)+f(2y)$$ for all real $y\ge 0$. Thus, $$f(y)+f(2y)=2f(0)=f(2y)+f(4y)\;,$$ so $f(y)=f(4y)$ for all $y\ge 0$. This doesn’t prove that $f$ has to be a constant function, but it certainly points in that direction. This suggests that one should suppose that $f$ isn’t constant and see whether it’s possible to derive a contradiction. Even before that one might examine a numerical example. First let’s rewrite the condition: $$f(x)=\frac12\Big(f(x+y)+f(x+2y)\Big)\tag{1}$$ whenever $y\ge 0$. Now suppose, just to get started, that some $f(x)=2$ and $f(x+y)=1$, where $y>0$. We can use $(1)$ to find that $f(x+2y)=3$. Replacing $x$ by $x+y$ in $(1)$, we see that $$f(x+y)=\frac12\Big(f(x+2y)+f(x+3y)\Big)\;,\tag{2}$$ and we can solve to find that $f(x+3y)=-1$. Replacing $x$ by $x+2y$ in $(1)$ gives us $$f(x+2y)=\frac12\Big(f(x+3y)+f(x+4y)\Big)\;,\tag{3}$$ which in turn can be solved to show that $f(x+4y)=7$. But now we have a problem: replacing $y$ by $2y$ in $(1)$, we find that $$f(x)=\frac12\Big(f(x+2y)+f(x+4y)\Big)\;,\tag{4}$$ which in our numerical example says that $2=\frac12(3+7)=5$, which is impossible. Now generalize this example: show in similar fashion that if $f$ is not constant, then it cannot satisfy $(1)$, because then it must also satisfy $(2)$-$(4)$, and we can always derive a contradiction.
H: How to invert number of days since 1 JAN 2000 Given the day of a year $d_y$ and the year $y$, it is straightforward to calculate the number of days since 1 JAN 2000: $D = d_y-1+(y-2000)*365+floor((y-2000)/4)-floor((y-2000)/100)+floor((y-2000)/400) $ But now, given $D$, I would like to invert and calculate $d_y$ and $y$. Anyone know of a way to do this that does not boil down to counting and comparing? Thanks. AI: Every 400-year period contains exactly 146,097 days, so by dividing and taking remainder you can reduce the problem to finding a date between 2000 and 2399. First, if $D\le 366$, it's in 2000. That's reasonable to treat as a special case, because 2000 is the only multiple of 100 within our period that's a leap year. Otherwise subtract $366$ from $D$; we're now looking for a date between 2001 and 2399 and counting January 1 2001 as day 0. Any 100-year period in this interval contains 36,524 days; by dividing with remainder once more we can reduce the problem to find a date between 2001 and 2100. Now, any 4-year period between 2001 and 2100, except the last, contains 1461 days. The last one has only 1460 days, but since $D$ at this stage can never be 36525 we can ignore that difference. Just divide by 1461 and the remainder will point to a date between 2001 and 2004. Finally divide by 365; if the quotient is 4 we're at December 31 2004; otherwise the quotient plus 2001 is the year and the remainder is the day within the year (which is 0-based, so add one if you want an 1-based day number).
H: Finding the equation of the tangent line at point P Find an equation of the tangent line, at the point with coordinates (2,1), to the curve described by $2x^2y+xy^3-3x=4$ I'm not too sure how to solve for this question as I can't find anything in my textbook about it. I realize I need to find the derivative of the curve, which would be the following I believe, and after that I am lost. $$4xy + y^3 - 3= 0$$ AI: Apply the $\,\def\dd{{\rm d}} \dd\,$ operator, which satisfies the following rules: $\dd(c)=0,\,$ if $c$ is a constant. $\dd(c\cdot f+g)=c\cdot\dd f+\dd g$ $\dd(f\cdot g)=\dd f\cdot g+f\cdot \dd g$ So that, for example $\dd(x^2)= \dd x\cdot x + x\cdot\dd x=2x\,\dd x$. Thus, we get $$4xy\,\dd x+2x^2\dd y + y^3\dd x+3xy^2\dd y -3\dd x = 0 \,.$$ Now we can express $\displaystyle\frac{\dd y}{\dd x}$ and substitute the given coordinates in the given expression to obtain the slope of the tangent. (You should also check that the given point lies in the curve at all..)
H: Probability that one out of four players gets all four aces What is the probability that exactly one player gets all four aces if a randomly shuffled deck is dealt to four players? My attempted working: ${4\choose1}\cdot\frac{1}{4!}\cdot\frac{1}{52\cdot51\cdot50\cdot49}\cdot{48\choose 9, 13, 13, 13}/{52\choose 13, 13, 13, 13}$. However this probability seems to be too small. What am I doing wrong? AI: There are ${52\choose 13,13,13,13}$ ways to deal the cards to players N,S, E, and W. There are 4 ways to choose a player who gets all of the aces. Say it's N. He can have ${48\choose 9}$ ways to get the rest of his cards. Now deal the remaining 39 cards to E, S and W.
H: An equicontinous headache (problem) There's this family of functions: $$\mathcal H=\{ f_k\in\mathcal C^0([-1,1],\Bbb R) : f_k (x)=\begin{cases} -1 \;\text{if $x\in[-1,-\frac1k]$} \\ kx \;\text{if $x\in[-\frac1k,\frac1k]$} \\ 1 \;\text{if $x\in[\frac1k,1]$}\end{cases} \text{with $k\in\Bbb N$} \}$$ We have to show that $\mathcal H$ is not equicontinous in $0$. What I've donde is this: let $\varepsilon=\frac12>0$. For any $\delta>0$ exists a $k\in\Bbb N$ such that $[-\frac1k,\frac1k]\in(-\delta,\delta)$, ie, $\frac1k=\|\frac1k\|<\delta$, so for the same $k$, there's a $f_k\in\mathcal H$, such that $\|f_k(\frac1k)\|=1>\frac12=\varepsilon$. However I don't feel this is totally correct, so I tried looking at other properties of the functions, and I got that: if $m,n\in\Bbb N$ such that $n\le m$ then $d_{\infty}(f_n,f_m)=\max_{x\in[-1,1]} \|f_m(x)-f_n(x)\|=\|1-f_n(\frac1m)\|=\|1-n\frac1m\|=\|\frac{m-n}m\|$, but this seems pointless. What am I doing wrong? This problem seems very innocous, I don't understand why I can't get it... AI: You can find a difference of $1$ on an arbitrarily small interval for one of the functions in the sequence. This violates equicontinuity.
H: Prove: Dividing an odd number by 2 always produces a remainder of 1 How would I go about proving that for all n belonging to the natural numbers, if any given odd number n is divided by 2, then the remainder is at least 1? I got a hint: Try to reduce the number of n, but I have no idea how that would help. I was thinking along the lines of induction, but what would be the best approach to this? I just want hints, please. I want to solve this myself, just need a heads up on where to get started. AI: We can show that it's exactly one. Let $n$ be our number, such that $n = 2m + r$, m and r whole numbers. If $r < 1$, then it has to be zero. In which case we just have $n = 2m$ and n isn't odd any more. If $r>1$, then if it's even r is divisible by two so $2\|(2m+r)$, meaning n isn't odd anymore. If r is odd, then we can write it as $s+1$, s is even, and $n = 2(m + s/2) + 1$, meaning 1 is the new remainder.
H: Do I need different constant names for different levels of integration? I'm just getting my feet wet in integration, so pardon me if I misuse a term. Let's take the anti-triple-derivative (I'm not sure if that's what it's actually called) of $8x$. $y'''=8x$ $y''=4x^2+c$ $y'=\frac43x^3+cx+d$ $y=\frac13x^4+cx^2+dx+e$ Now, could (should) I have used $c$ for all of the constants because they could be anything? For example, $y'=\frac43x^3+cx+c$ and $y=\frac13x^4+cx^2+cx+c$. Or are the different constants necessary because the constants are not necessarily the same? AI: Different constants are necessary because the constants are not necessarily the same. Well done: $y = \frac 13x^4 + cx^2 + dx + e$ is correct.
H: How to show that a sequence is positive, monotonically decreasing and converges to 0 I have the following sequence $$ y_n = \int_0^1 \frac{x^n}{x+5}\,dx, n = 0,1,\dots $$ So first question is, how to show that it's always positive? Second is, how to show that it's monotonically decreasing. I think I have to prove something like this, but how?: $$ \int_0^1 \frac{x^{n+1}}{x+5}\,dx \leq \int_0^1 \frac{x^n}{x+5}\,dx $$ Third question is, how to prove that it converges to zero? AI: For your first question: integrals of positive functions are positive. For your second: if $f \leq g$ then $\int f \leq \int g$. Note that $x^{n+1} \leq x^{n}$ on $[0,1]$ (note that this answers your first question too, take $f=0$). For your third question: We have shown the sequence is monotonically decreasing, and it is positive so it is bounded below by 0. Thus it converges to its infimum, which corresponds to its limit. Note that $\frac{1}{x+5}$ is bounded on $[0,1]$ so you may say $$ \frac{x^{n}}{x+5} \leq C x^{n} $$ for some big constant $C>0$. Hence $$\int_0^1\frac{x^{n}}{x+5}dx \leq C\int_0^1 x^ndx = \frac{C}{n+1} \to 0$$ as $n \to \infty$. By the squeeze theorem it follows that $$\int_0^1 \frac{x^n}{x+5}dx \to 0.$$
H: Base 10 notation Why are the ten digits in base 10 noted $0, 1, 2 ,3 ,4 ,5 ,6 ,7 ,8 ,9$ ? I heard it has to do with angles, but since I can't see how $9$ has $9$ angles, this may not be the historical reason ? AI: It has nothing to do with angles. The digits we have now are alterations of the Arabic versions, which in turn are modified versions of the original Hindi symbols in a script called Devanagari: (I nabbed this diagram from the Wikipedia article on digit symbols.) Florian Cajori History of Mathematical Notations Vol. 1 (Open Court, 1928) has an extensive explanation of the development of digit symbols on pages 45–70. According to Cajori, the Hindi system dates to sometime between 1100 and 1300 years ago; the symbols were adopted in Europe only about 800 years ago, although their earliest use in Europe dates as far back as the year 976. This table of early numeral forms is from Cajori p. 50: If you look at the letter shapes we use now, you can see similarly drastic alterations over the course of two thousand years, as the letters were adopted by the Phoenicians from Egyptian hieroglyphic, then by the Greeks from the Phoenicians, then by the Romans from the Greeks, and then by medieval Europeans from the Roman forms.
H: How are these definitions of Frechet derivatives related to each other? First def. for f exists the fréchet-derivative if there is a continuous operator $T \in L(X,Y)$, such that $\lim_{h \rightarrow 0} \frac{f(x_0 + hv) - f(x_0)}{h} = Tv$ converges uniformly in $v$, vs. Second def. $f(x_0+h)=f(x_0)+ Th + o(h)$. How do get the uniform convergence from the second one? I guess it is pretty obvious, but I do not see it. My doubt is, that this definition might not be true for every normed vector space, since not every norm is equivalent to the sup-norm that refers to uniform convergence. Or does uniform convergence for a abstract normed vector space refer to something different than convergence in the sup-norm? AI: The "uniform convergence" of the first definition is not uniform convergence on all of $X$. If $f$ is defined only on a proper subset of $X$, it cannot possibly mean that since $f(x_0 + hv)$ isn't defined for all $v$, and even if it is defined on all of $X$, the difference quotient $$\frac{f(x_0 + hv) - f(x_0)}{h}$$ has no reason whatsoever to be close to $Tv$ when $\lVert v\rVert$ is large, really really large. What is meant is uniform convergence on bounded subsets of $X$. It is enough to pick one special bounded subset of $X$, the (closed) unit ball $B_X = \{ v\in X : \lVert v\rVert \leqslant 1\}$. Then the first definition means that the function $$q(h) := \sup_{v \in B_X} \left\lVert \frac{f(x_0 + hv) - f(x_0)}{h} - Tv\right\rVert\tag{1}$$ has the property that $\lim\limits_{h\downarrow 0} q(h) = 0$. The second definition means that $$r(w) := f(x_0+w) - f(x_0) - Aw$$ belongs to $o(w)$, which in turn means $$\lim_{w\to 0} \frac{r(w)}{\lVert w\rVert} = 0,$$ and that means that for every $\varepsilon > 0$, there is a $\delta > 0$ such that $$\lVert w\rVert < \delta \Rightarrow \lVert r(w)\rVert \leqslant \varepsilon\cdot \lVert w\rVert,$$ or, with $w_0 = \frac{1}{\lVert w\rVert}\cdot w$ for $w \neq 0$, $$\sup_{\lVert w_0\rVert = 1,\, h < \delta} \left\lVert \frac{f(x_0 + hw_0) - f(x_0)}{h} - Aw_0\right\rVert \leqslant \varepsilon.\tag{2}$$ Now, if we write $v = t\cdot v_0$ with $\lVert v_0\rVert = 1$ and $t\in [0,1]$ in $(1)$, we see that $$\begin{align} \left\lVert \frac{f(x_0 + htv_0)-f(x_0)}{h} - T(tv_0)\right\rVert &= t\cdot \left\lVert \frac{f(x_0+htv_0) - f(x_0)}{th} - Tv_0\right\rVert\\ &\leqslant \left\lVert \frac{f(x_0+htv_0) - f(x_0)}{th} - Tv_0\right\rVert, \end{align}$$ so we have $$\sup_{h < \delta} \sup_{\lVert v\rVert \leqslant 1}\left\lVert \frac{f(x_0+hv) - f(x_0)}{h} - Tv\right\rVert = \sup_{h < \delta} \sup_{\lVert v\rVert = 1}\left\lVert \frac{f(x_0+hv) - f(x_0)}{h} - Tv\right\rVert,\tag{3}$$ and $\lim\limits_{h\downarrow 0} q(h) = 0$ is equivalent to the second definition, since with regard to $(3)$, it reduces to $(2)$.
H: Proof by Induction for a recursive sequence and a formula So I have a homework assignment that has brought me great strain over the past 2 days. No video or online example have been able to help me with this issue either and I don't know where to turn. I’m given $a_0=0$ $a_n=2a_{n-1}+1$ After writing the first 6 terms of the series: 0, 1, 3, 7, 15, 31, 63 I come up with an alternate formula of $a_n=2^n-1$ I then have to prove these formulas are the same using Induction in 3 parts: Proving the base case Stating my Inductive Hypothesis Showing the Inductive Step I have done Inductive proofs before but I don’t know how to show cases or do manipulations on a recursive formula. I don’t know how to represent when n = k then n = k + 1 or showing the approach by using n = k – 1 then n = k. Any ideas? AI: For the setup, we need to assume that $a_n = 2^n - 1$ for some $n$, and then show that the formula holds for $n + 1$ instead. That is, we need to show that $$a_{n + 1} = 2^{n + 1} - 1$$ Let's just compute directly: \begin{align} a_{n + 1} &= 2a_n + 1 \hspace{1.55in}\text{// recursion relation} \\ &= 2 \cdot (2^n - 1) + 1 \hspace{1in} \text{// induction hypothesis} \\ &= 2^{n + 1} - 2 + 1 \hspace{1.15in} \text{// arithmetic} \\ &= 2^{n + 1} - 1 \end{align} which is exactly what we wanted to be true.
H: Proving there are at least $N$ surjective functions from $A$ to $B$ Let $A = \{1,2,\ldots,m\}$; $B = \{1,2,\ldots,n\}$. I have to prove that there are at least $\frac{m!}{(m-n+1)!}$ surjective functions from $A$ to $B$. I've given it some thought, but I don't know how to work out the proof. I've looked at some similar answers, but I'm not finding them helpful. Could anyone please give me a hint or tell me how I need to attack this? I understand that there are $m!$ ways to arrange the elements of $A$, and that there are $n!$ ways to arrange the elements of $B$. I'm not sure how to draw the possible correspondences between them. I also know that the collection of all $n$-element subsets of $m$ can be expressed as ${m \choose n} = {m \choose m-n}$. AI: HINT: Suppose that $f:A\to B$ is a surjection. For $k=1,2,\ldots,n-1$ let $$g_f(k)=\min\{a\in A:f(a)=k\}\;.$$ Show that if $f$ and $h$ are surjections from $A$ to $B$, and $$\langle g_f(1),\ldots,g_f(n-1)\rangle\ne\langle g_h(1),\ldots,g_h(n-1)\rangle\;,$$ then $f\ne h$. How many possibilities are there for the $(n-1)$-tuple $\langle g_f(1),\ldots,g_f(n-1)\rangle$?
H: If $V$ is finite-dimensional and exist $\beta$ basis of $V$ such that $T(\beta)$ is a basis for $W$, then $T$ is a isomorphism? Let $V$ and $W$ vector space over $F$ and $T : V \rightarrow W$ lineal. The statement is false, but I can't find a counterexample. AI: Consider $V=\Bbb{R}^2$ and $W=\Bbb{R}^1$, and let $\pi:V\to W$, $\pi(x,y)=x$. Let $\beta=\{(1,0),(1,1)\}$, then $\pi(\beta)=\{(1)\}$, which is basis of $W$.
H: Sample size from population? This is probably very rudimentary maths, but given a strict population size ($N = 20$ for example), is the sample size any number $<N$? For use in calculation of confidence intervals using a population size, all of the formulas use $n$ and not $N$, meaning I need a sample size rather than population size. $$N = 20;\qquad n = 1, 2, 3, \dots , 19$$ $$N = 10;\qquad n = 1, 2, 3, \dots , 9$$ Is this right? This isn't really a maths question, but moreso a semantics question; is a sample size just taken from a population? A sample size of $n = 5$ can come from a population size of $N = 10$? Basically: for determining confidence intervals of a population, is the sample size used in calculations an arbitrary number as long as it fits within the amount of the population? edit; according to the textbook (and online sources), a sample should be an approximation of a population, so $n = N - 1$? Math is my weakest area, so I apologize if this is silly. AI: I'm guessing what you're looking for is a confidence interval for the population size $N$. Your question says "using a population size" rather than "for a population size", so you're not making it altogether clear. The formulas for confidence intervals found in elementary textbooks are usually based samples in which the observations are mutually independent. If it's done that way, then the sample size can exceed the population size. The sequence $2,1,5,3,3,5,1$ of size $7$ could be taken, and in this case, not all of the members of the population are included in the sample ($4$ is not). One can do better by sampling without replacement, so that no member of the population is used more than once in the sample. In that case, the sample size cannot exceed the population size. But the formulas for confidence intervals are messier and harder to derive in that case.
H: why is one point set in a first countable $T_1$ space a $G_\delta$? why is one point set in a first countable $T_1$ space a $G_\delta$ set? I thought, in first countable space, there may be a point $x$ such that a local basis at x does not contain one point set ${x}$.. Thank you for attention in advance. AI: Let $X$ be a first countable $T_1$ space, and let $x\in X$. Since $X$ is first countable, there is a countable local base $\mathscr{B}= \left\{B_n:n\in \Bbb N \right\}$ at $x$. Let $$ G=\bigcap_{B \in \mathscr{B}} B =\bigcap_{n\in\Bbb N}B_n; $$ clearly $G$ is a $G_\delta$ set in $X$, and I claim that $G=\{x\}$. Certainly $x\in G$. Suppose that $y\in X$ and $y \neq x$. Let $U=X\setminus\{y\}$; $X$ is $T_1$, so $U$ is open, and clearly $x\in U$. Thus, there is some $B_n\in\mathscr{B}$ such that $$ x\in B_n\subseteq U. $$ But then $$ G \subseteq B_n\subseteq U, $$ so $y\notin G$. This shows that $x$ is the only point of $X$ lying in $G$ and hence that $G=\{x\}$, so that $\{x\}$ is indeed a $G_\delta$ set in $X$.

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