Split (1)

train · 1.46M rows

text stringlengths 83 79.5k
H: $f$ measurable, $g$ either 1 or 0 if $f$ rational or irrational I have trouble advancing with this problem, though it seemed quite straight forward at glance. $f$ is a measurable function, and let $$g(x)= \left\{ \begin{array}{ll} 0 & \mbox{if } f(x)&\mbox{is rational} \\ 1 & \mbox{if } f(x)&\mbox{is irrational} \end{array} \right.$$ show g is measurable I know the definition of measurability with preimage and such, but I don't have the space or sigma algebras given. Isn't that necessary? I struggle to apply the definition with only the above given information. I'm probably missing something. any tip or hint would be great! Thanks AI: I am guessing you are trying to show that $g$ is measurable. I am assuming that single points are measurable. Since $\mathbb{Q}$ is countable, it is measurable, as is its complement $\mathbb{Q}^c$. Hence $A= f^{-1} \mathbb{Q}^c$ is measurable. Hence $g= 1_A$ is measurable.
H: Open sets in $\mathbb{R}^2$ as countable union of disjoint open rectangles From this question I realize that there exists an open set in $\mathbb{R}^2$ that is not a disjoint union of open rectangles. The example given is the set of points lying below the line $y=-x$. However, I can't quite see how one would prove that that particular set is not a disjoint union of open rectangles. What contradiction can one derive if that set were to be a disjoint union of open rectangles? AI: For one thing, a disjoint union of a family of open sets can only be connected if at most one member of the family is non-empty. The half-plane $\{y < -x\}$ is connected and not an open rectangle, hence it cannot be the disjoint union of open rectangles. Another way to see it is to note that in a disjoint union of open rectangles, no boundary point of any rectangle can be covered. For any open set containing the boundary point must intersect the rectangle of which it is a boundary point.
H: Finding the limit of $\left( 1-\frac{1}{n} \right)^{n}$ How would one compute the following limit? $$\lim_{n \to \infty} \left( 1 - \frac{1}{n} \right)^{n}$$ I know $$\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n} = e$$ but right there is a minus keeping that limit from being used. Another problem I am questioning is finding the limit of $$\frac{n!}{2n}$$ Of course, $\frac{x^n}{n!}$ has zero as a limit but here it is the opposite. AI: You can obtain the limit of $$\left(1-\frac1n\right)^n$$ easily from the one you know, $\left(1 + \frac1n\right)^n \to e$, by noting $$1 - \frac1n = \frac{n-1}{n} = \frac{1}{\left(\frac{n}{n-1}\right)} = \frac{1}{\left(1 + \frac{1}{n-1}\right)}.$$ Then you can write $$\left(1 - \frac1n\right)^n = \left(1-\frac1n\right)\frac{1}{\left(1+\frac{1}{n-1}\right)^{n-1}},$$ where the first factor obviously converges to $1$, and the second one converges to $\frac{1}{e}$ by what you already know.
H: stuck in finding $\frac{d(\sin^{-1}(x))}{dx}$ what is wrong with the following reasoning? $$\sin(\sin^{-1}(x))=x$$ This is of the form $f(g(x))=h(x)$. Then, by implicit differentiation: $$(f(g(x)))'=f'(g(x))\cdot g'(x)=h'(x) $$ Or in this case: $$\cos(\sin^{-1}(x))\frac{d(\sin^{-1}(x))}{dx} = 1$$ Then $$\frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\cos(\sin^{-1}(x))}$$ However I do not think this is correct. I wonder if I made a conceptual mistake in the implicit differentiation. Perhaps there are some extra steps needed to get to the result $$\frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\sqrt{1-x^2}}$$ If anyone could help me out that would be awesome! Thanks in advance. AI: Note we are making the usual assumptions on $x$ here. Note that since $\sin(\arcsin x)=x$ and $\cos^2t+\sin^2t=1$ we have:$$\cos(\arcsin x)^2+\sin(\arcsin x)^2=1\\\cos(\arcsin x)^2=1-\sin(\arcsin x)^2\\\cos(\arcsin x)^2=1-x^2\\\cos(\arcsin x)=\sqrt{1-x^2}$$ Can you see now?
H: Tate's Thesis: in what sense is Tate's Theorem 4.2.1 the Riemann-Roch theorem for curves? I am reading Tate's Thesis. Tate derives a theorem which he calls "the number-theoretic analogue of the Riemann-Roch theorem" from an abstract Poisson summation formula. I am accustomed to thinking of the Riemann-Roch theorem as a statement about the dimension of global sections of invertible sheaves over a nonsingular projective curve. Can you help me understand the connection between Tate's theorem and the theorem for curves? The theorem for curves: Let $k$ be an algebraically closed field. Let $C$ be a nonsingular projective curve over $k$. Let $\mathcal{L}$ be an invertible sheaf on $C$. Let $\Omega^1$ be the invertible sheaf of 1-forms on $C$. Then $$h^0(C,\mathcal{L}) - h^0(C,\Omega^1\otimes \check{\mathcal{L}}) = d-g+1$$ where $d$ is the degree of $\mathcal{L}$, and $g$ is the genus of $C$, defined as $h^0(C,\Omega^1)$. Tate's theorem: Let $k$ be a number field and let $V$ be its adele ring. Let $U$ be the idele group (i.e. the units of $V$). Let $D$ be a fundamental domain of $V$ for the discrete action of $k$. Let $f:V\rightarrow\mathbb{C}$ be a continuous, $L^1$ function. Let $\hat f$ be its fourier transform. Let $\|\cdot\|$ be the canonical absolute value on $U$, which is the product of the local absolute values, each appropriately normalized so the product is trivial on $k$. If $f$ satisfies $\sum_{\xi\in k}f(a(x+\xi))$ is convergent for all $a\in U, x\in V$, and convergence is uniform for $x\in D$ $\sum_{\xi\in k} \|\hat f(a\xi)\|$ converges for all $a\in U$ then $$\frac{1}{\|a\|}\sum_{\xi\in k} \hat f(\xi / a) = \sum_{\xi\in k} f(a\xi)$$ How are these two theorems related? Thoughts: Is there a way to think of an invertible sheaf as a continuous $L^1$ function on the adele ring of the curve? If so, I guess the fourier transform is related to the dual sheaf? AI: If you work in the function field case, i.e. replace the number field $K$ and its places $v$ by a finite extension of $k[x]$ for some finite field $k$, and work with its places, then this statement becomes the Riemann--Roch theorem. The appearance of $1-g$ in the RR thm., and the role of the canonical bundle in forming the correct kind of dual, will here be absorbed into the the definition of the self-dual measure on the adeles. To be a little more precise, you should imagine that your line bundle is of the form $\mathcal L(D)$ for some divisor $D$ (it always is, after all!); then $a$ will play the role of $D$ (or maybe $-D$). And you should take $f$ to the characteristic function of the integral adeles. Then one side of the equality will count rational functions $\xi$ for which $\xi a^{-1}$ has no demoninators (so global sections of $\mathcal L(D)$), and the other side will count rational functions $\xi$ such, roughly, $\xi a^{-1}$ is integral, which is the global sections of $\mathcal L(-D)$; except that $f$ is not quite self-dual. In the number field case the different comes in, and in the function field case we are considering here, the canonical bundle will come in (as well as a factor related to $1-g$). Finally, to get the familiar statement about dimensions, take log of both sides and divide by $\log q$ (where $q = \|k\|$). (You can check then that the $\|a\|^{-1}$ on the LHS, after taking logs, gives the $\deg D$ term.)
H: Complex Numbers.... Suppose a is a complex number such that: $$a^2+a+\frac{1}{a}+\frac{1}{a^2}+1=0$$ If m is a positive integer, find the value of: $$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}$$ My Approach: After I could not solve it using the usual methods I tried a bit crazier approach. I thought that as the question suggests that the value of the expression does not depend upon the value of m, provided it is positive, hence the graph of the expression on Y-axis and m on X-axis would be parallel to X-axis and thus the slope be zero. So I differentiated it with respect to m and equated it to zero and after factorizing and solving I got $a^m=-1$ or $a^m=1$ or $a^m=\left(\frac{-1}{4}+i\frac{\sqrt15}{4}\right)$ or $a^m=\left(\frac{-1}{4}-i\frac{\sqrt15}{4}\right)$. But if $a^m=1$ then $a=1$ which does not sattisfy the first equaion. Thus the value of $$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=\frac{-9}{4}$$ OR $$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}=0$$ What other approach would you suggest? What are the flaws in my approach (if any)? AI: If we multiply $$a^2 + a + \frac1a + \frac{1}{a^2} + 1 = 0$$ with $a^2$, we obtain (since evidently $a \neq 1$) $$0 = a^4 + a^3 + a^2 + a + 1 = \frac{a^5-1}{a-1},$$ so $a^5 = 1$, $$a = e^{(2\pi ik)/5},\quad k \in \{1,2,3,4\}.$$ Then, if $m$ is a multiple of $5$, we have $$(a^2)^m + a^m + \frac1{a^m} + \frac{1}{(a^2)^m} = 1+1+1+1 = 4,$$ and if $m$ is not a multiple of $5$, the four numbers $$a^{2m},\, a^m,\, a^{-m}\, a^{-2m}$$ are the numbers $a^2,\, a,\, a^{-1},\,a^{-2}$, possibly in a different order, then the sum is $-1$.
H: Diagonal sequence trick: Uniform bound necessary? There is a treatment of the "diagonal sequence trick" in Reed and Simon (Functional Analysis Vol.1) stated there as follows: Let $f_n(m)$ be a sequence of functions on the positive integers which is uniformly bounded, i.e. $\|f_n(m)\| \le C$ for all $n,m$. Then there is a subsequence $\{f_{\hat n(i)}(m)\}^\infty_{i = 1}$ so that for each fixed $m$, the sequence $f_{\hat n(i)}(m)$ converges as $i \to \infty$. I was wondering whether the uniform bound is actually necessary. I am not sure where it is essential in the proof, outlined below: Consider the sequence $f_n(1)$. It is a bounded set of numbers, so we can find a subsequence $f_{n(i)}$ such that $f_{n_1(i)}(1) \to f_\infty(1)$, for some number $f_\infty(1)$. Now consider the sequence $f_{n_1(i)}(2)$. We can find a subsequence $f_{n_2(i)}(2) \to f_\infty(2)$ as $i \to \infty$. Proceeding inductively, we find successive subsequences $f_{n_k(i)}$ so that $f_{n_{k+1}(i)}$ is a subsequence of $f_{n_k(i)}$, and $f_{n_k(i)} \to f_\infty(k)$ as $i \to \infty$. Thus, in particular, $f_{n_k(i)}(j) \to f_\infty(j)$ as $i \to \infty$ for $j = 1,2,\dots,k$. To get a subsequence $f_{\hat n(i)}$ converging for each $j$, take the diagonal sequence $\hat n(k) = n_k(k)$. Then $f_{\hat n(k)}, f_{\hat n(k+1)}, \dots$ is a subsequence of $f_{n_k(i)}$ so $f_{\hat n(i)}(k) \to f_\infty(k)$ as $i \to \infty$ for each $k$. AI: The proof needs that $n\mapsto f_n(m)$ is bounded for each $m$ in order to find a convergent subsequence. But it is indeed not necessary that the bound is uniform in $m$ as well. For example, you might have something like $f_n(m) = \sin(nm) e^m$ and the argument still works. Intuitively, we don't need to obtain convergence of the diagonal sequence afresh but obtain it already from it being essentially a subsequence of all subsequences introduced.
H: Let $\mu$ be the measure constructed in the Riesz representation theorem. Is $\mu(\partial A)=0$? I am currently self-studying Rudin's real complex analysis. The Riesz representation theorem in the book states: Let $X$ be a locally compact Hausdorff topological space. Let $T:C_c(C)\rightarrow \mathbb{C}$ be a positive linear functional. Then there exists a $\sigma$-algebra $\mathscr{M}$ on $X$ that contains all the Borel sets and there exists a unique measure on the measurable space $(X,\mathscr{M})$ such that the following conditions hold (I will show the first condition only to be brief): 1) $T(f)=\int_X f\,d\mu$ for every $f\in C_c(X)$ 2)... 3)....4).... Question: Let $\mu$ be the measure constructed in the proof of the theorem. Is it true that for every $A\subseteq X$ that $\mu(\partial A)=0$? AI: One example of a functional $T$ is given by choosing $x \in X$ and defining $T(f) := f(x)$. Then the measure $\mu$ is simply the $\delta$-function supported at $x$, whose value on a set $A$ is $1$ if $x \in A$ and $0$ otherwise. So $\mu(\partial A)$ will be $0$ if $x \not\in \partial A$, but will be $1$ otherwise. As a general remark, it's a good idea to keep $\delta$-functions in mind as an example of the kind of measure that can come up in the Riesz rep'n theorem, just to have examples in mind of a different nature than Lebesgue measure (although that is not necessary for answering your particular question). Another class of examples is given by letting $Y$ be a closed subset of $X$, and $\nu$ be a Borel measure on $Y$. We can then define $T(f)$ by $T(f) = \int_Y f d\nu$. (Note that $f$ restricted to $Y$ is continuous with compact support, so the integral makes sense.) The resulting measure $\mu$ on $X$ will be supported on $Y$. (Delta-functions are the special case when $Y$ is a point and $\nu$ gives the point a measure of $1$.) So you could now take e.g. $A$ to be a ball in $X = \mathbb R^n$, and $Y = \partial A$ to be its boundary. Then for any non-zero Borel measure $\nu$ on $Y$, you would get a counterexample.
H: Equation of a line parallel to $5x-3y=7$ That goes through the point (3,-1) This is a study question in preparation for my midterm. It's multiple choice. The answers are: A) $y=(5/3)x-(7/3)$ B) $y=(3/5)x-(14/5)$ C) $y=(5/3)x-6$ D) $y=-(3/5)x+(4/5)$ Here is my process: $5x-3y=7$ Move the 3x over and change the sign. $5x=3y+7$ Move the 7 over and switch it to negative. $3y=5x-7$ Divide entire formula by 3. $y=(5x/3)-(7/3)$ Which is answer A), so I know that answer A is not correct, since this is not a parallel line, it is the same line. Now I use the point slope equation: $y-y_1=(5(x-x_1))/3$ $y+1=(5(x-3))/3$ $y=(5x-15-1)/3$ $y=(5x-16)/3$ So I know only (C) can be the correfct answer, since it has the same slope as the original equation, but my y intercept is off in my equation. I would appreciate some help as to how to arrive at the $-6$ in answer C. Also, I don't know how to create fractions on this forum, any correction of my equations for clarification would be appreciated. AI: You just made an arithmetic error trying to do two things at once. $$y=\frac{5(x-3)}3-1=\frac{5x-15}{3}-1=\frac53x-5-1=\frac53x-6$$
H: How to prove algorithm for solving a square congruence when p ≡ 5 (mod 8) I'm having trouble understanding why this algorithm works and where it comes from: "Suppose p ≡ 5 (mod 8) is a prime and y is a square (mod p); that is, for some $ x, x^2 ≡ y\ (mod\ p)$. This can be solved by the following algorithm Compute $d ≡ y^{(p−1)/4} \pmod {p}$. Show that $d^4 ≡ 1 \pmod {p}$ Hence show $d ≡ ±1 \pmod {p}$ or $d^2 ≡ −1 \pmod {p}$. If $d ≡ 1 \pmod {p}$ compute $r ≡ y^{(p+3)/8} \pmod {p}$; if $d ≡ −1 \pmod {p}$ compute $r ≡ 2y(4y)^{(p−5)/8}\pmod {p}$; Then (r, −r) are the square roots of y. [We will return to the case $d^2 ≡ −1 \pmod {p}$ later.]" Here are my attempts: Since $ p \equiv 5 \pmod {8}$ then $p = 8k + 5$ for some integer $k$. With some algebra I found that $\frac{p-1}{2} = 4k +2 = 2(2k+1)$. We can let $d \equiv y^{2k+1} \pmod {p}$. So $d^2 \equiv y^{2(2k+1)} \pmod {p}$. Then $d \equiv y^\frac{p-1}{4} \pmod {p}$. And $d^4 \equiv 1 \pmod {p}$ by Fermat's theorem. For the second step, since $d^2 \equiv y^\frac{p-1}{2} \equiv (\frac yp) \equiv1 \pmod {p}$ by Legendre's theorem. So $d \equiv \pm 1 \pmod {p}$. I'm confused why $d^2$ would equal $-1$ though. I'm even more confused about the third step. After playing around with it, I found that if we set $r \equiv y^{k+1} \pmod {p}$ and substitute $k = \frac{p-5}{8}$ (since $p = 8k + 5$) then we get the first r, but I have no idea where the $2y$ and the $4y$ came from in the second one. Any advice, hints, or clarifications? Thanks Edit: For the second step, since $d^4 \equiv 1 \pmod {p}$, then $d^2 \equiv \pm 1 \pmod {p}$. So either $d^2 \equiv 1 \pmod {p}$ which implies that $d \equiv \pm 1 \pmod {p}$ or $d^2 \equiv -1 \pmod {p}$. AI: The $d^2 \equiv -1 \pmod{p}$ part is wrong. Presumably, the author mixed some parts of the $p \equiv 1 \pmod{8}$ case in, where one computes $d = y^{(p-1)/8}$. For $p \equiv 5 \pmod{8}$, we always have $1 = \left(\frac{y}{p}\right) \equiv d^2 \pmod{p}$. So if $d \equiv 1 \pmod{p}$, we choose $r \equiv y^{(p+3)/8}\pmod{p}$, and that yields $$r^2 \equiv y^{(p+3)/4} \equiv y\cdot y^{(p-1)/4} \equiv y\cdot d \equiv y \pmod{p}.$$ For $d \equiv -1\pmod{p}$, choosing $r \equiv y^{(p+3)/8} \pmod{p}$ would lead to $r^2 \equiv y\cdot d \equiv -y \pmod{p}$ by the same computation as above, so we mix a square root of $-1$ in to obtain a square root of $y$. For $p \equiv 5 \pmod{8}$, we have $\left(\frac2p\right) = -1$, so $2^{(p-1)/4}$ is a square root of $-1$ modulo $p$. The author just wrote it in an obscure way, we have $2\cdot 4^{(p-5)/8} = 2\cdot 2^{(p-5)/4} = 2^{(p-1)/4}$.
H: Interesting Medieval Mathematics Lecture/Activity Ideas? Recently I have been invited to give a talk about Medieval Mathematics or mathematics in the 500 AD - 1500 AD time frame. I have been researching the time frame for the past week and have found interesting snippets of Arabian mathematicians and Fibonacci. I know it was called the Dark Ages, but I didn't think it would be this dark... The lecture or(inclusive) activity is 45-90 min long and my audience are students at a boarding school I know (well behaved, funny, smart students). So here is what I was thinking: Talk about pre-Arabic number systems and arithmetic, talk some about Arabian mathematicians, and then talk about Fibonacci/Golden Ratio and conclusion. For the activity I thought of doing Abacus races or medieval mathematics trivia with prizes (they would appreciate this). What do you guys think? Any ideas/suggestions? AI: You really ought to mention some of the mathematical achievements of the $14$th century philosopher Nicolas Oresme: he worked with fractional exponents; he was the first to prove that the harmonic series diverges; he gave in essence a formula for the sum of a geometric series with arbitrary first term and ratio $\frac1n$ for integers $n\ge 2$; and he came close to inventing Cartesian coordinates, using his version to prove that the distance travelled in a given period by an object moving under constant acceleration is equal to the distance travelled in the same period by an object moving at a constant speed equal to that of the first object at the midpoint of the period. There’s a fair bit of information available on the web; the summary here should be helpful. I’d replace the abacus, which in the form in which we think of it was little used in mediæval Europe, with its mediæval equivalent, the counting board or counting table.
H: Nonlinear difference equation Maybe this is a trivial question, but how to find the general solution to the following first order difference equation? $$ y_{t+1}=a+\frac{b}{y_{t}} $$ Also, could someone recommend a reference textbook on difference (and possibly differential) equations? I would need a quite comprehensive text, at an intermediate level. Thank you. AI: I would write $y_t = \frac{p_t}{q_t}$, so the general step becomes $$ \frac{p}{q} \rightsquigarrow a+\frac{bq}{p} = \frac{ap+bq}{p}$$ and the recurrence splits into two coupled linear ones: $$ p_{t+1} = ap_t + bq_t $$ $$ q_{t+1} = p_t $$ Substituting the second of these into the first we get $$ p_{t+1} = ap_t + bp_{t-1} $$ which can then be solved by standard methods.
H: Uncountable open cover of $\mathbb{R}$ The question posed to me is to find an uncountable open subcover of $\mathbb{R}$ such that it has no finite subcover, but I can't even think of a way to define an uncountable open cover. AI: Consider the collection $\{(a,b)\mid a<b\in\Bbb R\}$, the collection of non-trivial finite length intervals. Note that while there is no finite subcover, there is always a countable subcover. Spaces with this property, $\Bbb R$ included, are known as Lindelof spaces.
H: The Decomposition VS. The Partition of a set The book that I am reading says that the decomposition of a set $A$ is any representation of $A$ as the union of a disjoint family of set, $A=\sum_{i\in I}A_i$ (if a family of set is pairwise disjoint the author writes $\bigcup_{i\in I}A_i$ as $\sum_{i\in I}A_i$). The family $\{A_i:i\in I\}$ is referred to as the partiton of the set $A$. Does this imply that all of the elements of a partition on the set $A$ are pairwise disjoint? Also, is it true that the union over every element of $\{A_i:i\in I\}$ is equal to $A$? Also, a quick question on notation, when we write a union in the form $\bigcup_{i\in I}A_i$ do we mean that we take the union for all $i$ in $I$? Or, we take the union for any $i$ in $I$ not necessarily every $i$? AI: Yes: a partition of $A$ is a family $\mathscr{P}$ of pairwise disjoint non-empty subsets of $A$ whose union is $A$. (Sometimes the requirement that the members of the partition be non-empty is relaxed.) Thus, if $\mathscr{P}=\{A_i:i\in I\}$, then $$\bigcup\mathscr{P}=\bigcup_{i\in I}A_i=\bigcup\{A_i:i\in I\}=A$$ And yes, all of the union notations above mean the union of all of the sets in the collection.
H: A strange characterization of connected spaces Let $X$ be a topological space. Recall that an open cover $\mathcal U$ of $X$ is point finite if every point of $X$ is contained in finitely many elements of $\mathcal U$. Let us say that a point finite open cover $\mathcal U$ is fair if any two points of $X$ are contained in the same number of elements of $\mathcal U$. Let us call a space unfair, if the only fair covers of $X$ are the trivial covers $\mathcal U = \{X\}$ and $\mathcal U = \{X, \emptyset\}$. Question: Is it true that $X$ is connected if and only if it is unfair? If $X$ is not connected, say $X= U \cup V$, with $U, V$ disjoint and nonempty, then $\{U, V\}$ is a nontrivial fair cover of $X$. This establishes one side of the implication. What about the other implication? AI: Suppose that $\mathscr{U}$ is a point finite open cover of $X$ such that each $x\in X$ belongs to exactly $n$ members of $\mathscr{U}$. For each $x\in X$ let $V(x)=\bigcap\{U\in\mathscr{U}:x\in U\}$; clearly $\{V(x):x\in X\}$ is an open cover of $X$. Now suppose that $y\in V(x)$; then if $x\in U\in\mathscr{U}$, we must have $y\in U$, so the $n$ members of $\mathscr{U}$ containing $x$ all contain $y$, and it follows that $V(y)=V(x)$. Thus, $\{V(x):x\in X\}$ is a partition of $X$ into open (and therefore clopen) sets. If this partition has only one element, then $V(x)=X$ for all $x\in X$, $\mathscr{U}=\{X\}$, and $n=1$. Otherwise, the partition is non-trivial, and $X$ is not connected.
H: How many possible deals in straight poker. Each player is dealt 5 cards. 5 players how many deals are possible? I know that for one player there is 2598960 possible outcomes i.e. 52 C 5, I need to know how I can do this for the next four players and why if possible AI: Hint: having chosen five cards for the first player, how many cards are left? Now choose the second player's hand, multiply that by the number of first player hands and you have the number of first and second player hands. Continue.
H: The convergence of a sequence of sets A sequence $\{A_n : n=0,1,2,...\}$ is said to be monotone nondecreasing if we have $$A_0\subseteq A_1\subseteq \cdot \cdot \cdot \subseteq A_n \subseteq \cdot \cdot \cdot $$ The same sequence is said to be monotone nonincreasing if we have $$A_0\supseteq A_1\supseteq \cdot \cdot \cdot \supseteq A_n \supseteq \cdot \cdot \cdot$$ To specify the type of convergence, for a monotone nondecreasing sequence of set we write we write $\{A_{n}\}\uparrow A$, likewise for a monotone nonincreasing sequence we write $\{A_{n}\}\downarrow A$. I am confused by the notation, does the latter statement about convergence imply that $\bigcup \{A_n\}=A$? Or, does it only mean that the as $n\to\infty$ the $n$-th element of $\{A_n\}$ will get closer and closer to the some set $A$ (not necessarily $\bigcup\{A_n\}$) but it will never be equal to $A$ because the sequence is infinite? AI: The down- and up-arrows imply the monotonicity of the sequence; the $A$ is the actual limit, which is the intersection for a non-increasing sequence and the union for a non-decreasing sequence. If $\langle A_n:n\in\Bbb N\rangle$ is monotone non-increasing, then its limit is $$A=\bigcap_{n\in\Bbb N}A_n\;,$$ and we write $$\langle A_n:n\in\Bbb N\rangle\downarrow A\;.$$ If the sequence is monotone non-decreasing, then its limit is $$A=\bigcup_{n\in\Bbb N}A_n\;,$$ and we write $$\langle A_n:n\in\Bbb N\rangle\uparrow A\;.$$
H: Counting the number of integer sequences Count the number of sequences of integers, a(1), a(2), .... a(n), containing n positive integers such that1<=a(i)<=m, for all 1<=i<=n and max_value - min_value = q, where max_value means the highest integer in the above sequence and similarly for min_value. Note that repetition is allowed. NOTE1 : Just to avoid confusion, sequence 1,2,3 and 3,2,1 are different. NOTE2 : In case someone computes the count to be equal to nC2 * 2! * (m-q) * (q+1)^(n-2) [nC2 = n choose 2] which, bytheway, seems pretty logical at the first glance, please note that it is incorrect as we are overcounting; many cases are counted more than once. AI: Suppose $1\le q\le m-1$. We can pick the minimum value in $m-q$ ways, i.e., any element of the set $\{1,2,\dots,m-q\}$. Now we count the number of sequences of length $n$ with minimum value $x$ and maximum value $x+q$. There are $(q+1)^n$ sequences with values in the interval $[x,x+q]$. Among these there are $q^n$ that miss $x$, there are $q^n$ that miss $x+q$, and there are $(q-1)^n$ that miss both $x$ and $x+q$. By the in-and-out principle, the number of sequences that hit both endpoints is $(q+1)^n-2q^n+(q-1)^n$. Hence the number of sequences satisfying the conditions stated in the question is $[m-q][(q+1)^n-2q^n+(q-1)^n]$ provided $0\lt q\le m$. (The formula is obviously correct in the case $q=m$ which was not discussed.)
H: Show $P(A \subset B) = (\frac{3}{4})^n$ Let $S = \{1, 2, \dots, n \}$ and suppose that $A$ and $B$ are, independently, equally likely to be any of the $2^n$ subsets of $S$. Show that $P(A \subset B) = (\frac{3}{4})^n$ and show that $P(A \cap B = \emptyset) = (\frac{3}{4})^n$. Applying the law of total probability , $$P(A \subset B) =\sum_{k=0}^n P(A \subset B \Large \mid \normalsize \|B\| = k)P(\|B\| = k) $$ If $k = 0$ then $B = \varnothing $. Therefore $P(A \subset B \Large \mid \normalsize \|B\| = k) = 0$ because there do not exist any sets $A$ such that $A \subset B$. If $k = 1$ and $A \subset B$, then there exists precisely one set $A$ such that $A \subset B$, and this $A = \varnothing$. Therefore $$\begin{align} &P(A \subset B \mid \normalsize \|B\| = k) = \frac{1}{2^n} = \frac{2^1 - 1}{2^n} = \frac{2^k - 1}{2^n} \\ &P(\|B\| = k) = \frac{\binom{n}{k}}{2^n} \end{align}$$ If $A \subset B$, then $A \in \mathcal{P}(B) - B$. Thus, there are $\|\mathcal{P}(B) - B\| = 2^{k} - 1$ possibilities for $A \subset B$ out of a possible $2^{\|S\|} = 2^n$ sample space. Therefore I have $$\begin{align} &P(A \subset B) = 0 + \sum_{k=1}^n P(A \subset B \Large \mid \normalsize \|B\| = k)P(\|B\| = k) = \sum_{k=1}^n \frac{2^k - 1}{2^n}\frac{\binom{n}{k}}{2^n} \\ &\mathsf{but} \ \ \ \sum_{k=1}^n \frac{2^k - 1}{2^n}\frac{\binom{n}{k}}{2^n} \ne (\frac{3}{4})^n \end{align}$$ Where is my mistake? Hints only please. AI: It is easy to see that $P(A\subseteq B)=(\frac34)^n$ so the first part of what you're trying to show can't be true. Namely, suppose we choose for each of the $n$ possible elements first whether it is in $A$ and then whether it is in $B$. All of the $2n$ choices are independent and uniformly distributed between yes and no. $A\subseteq B$ means that each pair of answers is one of (yes, yes), (no, yes) or (no, no). For each pair the probability of that happening is $\frac 34$, so the probabilty of that happening for all $n$ pairs is $(\frac{3}{4})^n$. Similarly $A\cap B=\varnothing$ means that each pair of answers is one of (yes,no), (no, yes), or (no,no). Again, this has probability $\frac 34$ and must happen $n$ times, so the probability of disjoint $A$ and $B$ is $(\frac 34)^n$. Trying to count possibilities, grouped by the size of the sets is definitely the long way around to an answer here.
H: CFG - whose words contain exactly twice as many b's as a's. I am trying to built a CFG for the language that accepts all words that have twice as many b's as a's. The only idea I could come up with is: Start -> S S-> SaSbSbS \| SbSaSbS \| SbSbSaS \| $\epsilon$ But obviously it will not be able to parse the word aaabbbbbb Because it doesn't matter what combination I pick, there will still be an S which can not be parsed further as it will contain either only a's or b's (if I am wrong please guide me how do I parse this word using the above CFG). The worst part is that googling brought either the same solution or something similar although written is some other manner, but still unable to parse the above word. The question is: is there a CFG for the language that accepts twice as many b's as a's (being able to parse the given word) and if yes, what is it? AI: You’re forgetting the $\epsilon$-productions. Your grammar does generate $aaabbbbbb$: $$S\Rightarrow SaSbSbS\Rightarrow^* aSbb\Rightarrow aSaSbSbSbb\Rightarrow^* aaSbbbb\Rightarrow^* aaabbbbbb\;.$$ The first and second $\Rightarrow^*$ use $S\to\epsilon$ three times each; the last abbreviates a longer sequence of applications of productions, but at that point it should be clear what they are.
H: Can a directed hamiltonian path be found in polynomial time? I was discussing a programming competition problem with one of my math professors in Linear Algebra that reads as follows: A matrix is an $r\times c$ array of numbers, where $r$ is the number of rows and $c$ is the number of columns. Given two matrices $M_1$ and $M_2$ with dimensions $r_1\times c_1$ and $r_1\times c_2$, respectively, their multiplication, $M_1M_2$, is defined only if the number of columns, $c_1$, in $M_1$ is equal to the number of rows, $r_2$ , in $M_2$. The matrix resulting from $M_1M_1$ will have $r_1$ rows and $c_2$ columns. Similarly, $M_2M_1$ is defined only if $c_2=r_1$. In this case, the resulting matrix will have dimensions $r_2\times c_1$. Given a list of matrix dimensions, your job is to determine whether or not some combination of all the matrices can be successfully multiplied together. The actual solution presented after the competition made use of dynamic programming techniques to solve it in polynomial time... but this professor of mine, having not seen the problem before, thought that it could be converted into a graph problem where each node in the graph represented a matrix, and there was a directed edge between two nodes if their matrix product was defined. He then posited that you just needed to find a Hamiltonian path through the graph... After mentioning to him that the Hamiltonian path problem is NP-Hard, he says "Not if it's a digraph". Is he mistaken, or am I missing something? AI: It sounds like the professor is confused. Finding Hamiltonians in a directed graph is certainly just as NP-hard as finding them in an undirected one. If you have an efficient solver for digraph Hamiltonians and an undirected graph, just replace edge with a pair of edges going in both directions, and feed the resulting graph to your assumed digraph solver. The obvious graph representation of the problem is to construct a (directed multi)graph where there is one vertex for each dimension, and an edge for each matrix, connecting its horizontal dimension with its vertical one. The problem is then just to find out whether this directed graph has an Eulerian path, which is quite a bit easier than finding a Hamiltonian one. Dynamic programming doesn't seem to be needed here -- just check that (1) the graph is connected, and (2) every vertex has indegree equal to outdegree, except possibly for two vertices having a degree unbalance of $\pm 1$.
H: A question about a sequence. I'm preparing for the subject exam in November. This is a question that I thought I had the correct answer to. Let $\left \{a_n \right \}_{n=1}^\infty$ be defined recursively by $a_1=1$ and $a_{n+1}=\left( \frac{n+2}{n} \right) a_n$ for $n\geq 1$. Then what is $a_{30}$ equal to? So I had no idea how to approach this, so I sought to find a pattern. I noticed $a_2=\binom{3}{1}$, $a_3=\binom{4}{2}$, $a_4=\binom{5}{3}$, $a_5=\binom{6}{4}$, etc., so naturally I thought that $a_{30}=\binom{31}{29}$. However, I checked the answer in the back of my practice test and it says I'm incorrect! Can someone tell me how to do this problem? Thank you! AI: Look at some small ones, and you’ll see what’s happening: $$a_4=\frac53a_3=\frac53\cdot\frac42a_2=\frac{5\cdot4\cdot3}{3!}=\frac{5\cdot4}2=\binom52\;,$$ and $$a_5=\frac64a_4=\frac64\cdot\frac53a_3=\frac64\cdot\frac53\cdot\frac42a_2=\frac{6\cdot5\cdot4\cdot3}{4!}=\frac{6\cdot5}2=\binom62\;.$$ Now state and prove the general result. (Note that this is equivalent to your answer; if this is the answer in the key, both it and your answer are correct, since they’re the same answer in different form. If the key has some other answer, it’s probably wrong.)
H: A question about Riesz - Fischer theorem's proof In Riesz-Fischer theorem's proof, when we put $$ g_k =\|f_{n_1}\|+\|f_{n_2}-f_{n_1}\|+ \cdots + \|f_{n_k}-f_{n_{k-1}}\| $$ it is easy to get (by Minkowski's inequality) $$ \left \\| g_k \right \\|_p \leq \left\\|f_{n_1}\right\\|_p+\left\\|f_{n_2}-f_{n_1}\right\\|_p+ \cdots + \left\\|f_{n_k}-f_{n_{k-1}}\right\\|_p. $$ Can you show me how to formally pass to the limit in order to obtain $$ \left \\| g \right \\|_p \leq \sum_{k=1}^\infty \left \\| f_{n_k}-f_{n_{k-1}} \right \\|_p $$ ? Thanks AI: Since for all $k$ we have $$\lVert g_k\rVert_p \leqslant \sum_{m=1}^\infty \lVert f_{n_m} - f_{n_{m-1}}\rVert_p$$ (with $f_{n_0} = 0$), and $\lvert g_k\rvert^p \uparrow \lvert g\rvert^p$ monotonically, the monotone convergence theorem tells us $$\lVert g\rVert_p \leqslant \sum_{m=1}^\infty \lVert f_{n_m} - f_{n_{m-1}}\rVert_p.$$
H: answer check for a calculus 3 variable chain rule problem yo. find $\frac{\partial{V}}{\partial{t}} = ?$ $V =\frac{1}{3}{x^2}h$ $x = \frac{t}{t+1}$ $h = \frac{1}{t+1}$ $$\frac{\partial{V}}{\partial{t}} = \frac{2t-t^2}{3(t+1)^4}$$ Does this look correct? AI: The way I solve this problem is to plug $x, h$ into $V$. Then calculate the derivative. So $V=\frac{1}{3}\cdot \frac{t^2}{(t+1)^2}\cdot \frac{1}{(t+1)}$ Using quotient rule: $\frac{dV}{dt} = \frac{2t\cdot 3(t+1)^3-9(t+1)^2\cdot t^2}{9(t+1)^6} = \frac{2t-t^2}{3(t+1)^4}$ So your answer is correct.
H: Prove that $x_n$ converges, and find $\lim_{n\rightarrow \infty} x_n$ I need some critique and suggestions on how to go over this question: Prove that $x_n$ converges, and find $\lim_{n\rightarrow \infty} x_n$: $x_1 > 0 $, $x_{n+1} = \frac{1}{2}(x_n + \frac{5}{x_n}) \ \forall n \geq 1 $. Here's how I solved it: Let $x_1 =1 \Rightarrow x_2 = 3 \Rightarrow x_3 = \frac{28}{12} < x_2$ and $x_4 < x_3 $. So the sequence is actually decreasing and therefore it is bounded below by $x_1$. Let $\lim_{n\rightarrow \infty} x_{n+1} = \lim_{n\rightarrow \infty} x_{n} = A$. $A =\frac{1}{2} (A+\frac{5}{A}) \Rightarrow A = \sqrt5 $. AI: You have the right idea. You have a recursive formula and to show it has a limit, first show it is bounded--this can be done by induction, then show it is monotone, and then yes it correct to assert that $lim$ $a_{n+1}$ = $lim$ $a_n$.
H: Are the only $b$ such that $b^2 + 4c$ and $b^2 - 4c$ are both perfect squares primes of the form $4k + 1$? And of course, multiples of primes of the form $4k + 1$ $b, c, k$ are positive integers. Ran into a math problem essentially involving this, just curious. I observed it to hold for the first $50$ integers, but not sure past that nor how to prove it. Does it have to do with Fermat's "Christmas" theorem ($p = x^2 + y^2$ iff $p \equiv 1 \pmod 4$)? AI: The number $b$ need not be prime. We have $65^2+3696=89^2$ and $65^2-3696=23^2$. A boring kind of example is $10^2-96=4$, $10^2+96=14^2$. Remark: One can generate arbitrarily many non-boring examples using $(r,s,b)$ such that $(r,s,b)$ is a primitive Pythagorean triple and $b$ is not prime. Any such example requires $b$ to have at least two not necessarily distinct prime divisors of the form $4k+1$.
H: If a commutative ring with identity is the sum of two ideals, then their product is equal to their intersection. My problem is to prove exactly as the title says; particularly if I+J=R for some commutative ring R with identity and ideals I and J of the ring R, then IJ = I ∩ J. I know already that IJ is an ideal in I ∩ J. I tried supposing that there is some element k of I ∩ J which is not in IJ, and from the hypothesis it's clear there's some x in I and y in J so that x + y = k. I know that k is in IJ if some set of products $x_i y_i$ for $i = 1, 2, 3, ..., n$ sum to k, but I don't quite know how to arrive at this contradiction from the rest. AI: It is immediate that $IJ \subset I \cap J$. For the reverse direction, take $x \in I \cap J$. Since $I+J=R$, there exist elements $a \in I$ and $b \in J$ such that $1=a+b$. Now, $x = x 1 = xa+xb$ and $xa, xb \in I J$. Hence $x \in IJ$ and this proves that $IJ = I\cap J$.
H: $n$ points on a circle Choose $n$ points on a circle so that no three of the $\binom{n}{2}$ chords have a common point inside the circle. Let $a_{n}$ be the number of regions formed inside the circle by drawing the cords. Obtain the recurrence relatin $a_{n}=a_{n-1}+f(n)$ for $n\geq 1$ where $f(n)=n-1+\sum_{i=1}^{n-1}(i-1)(n-1-i)$. It is clear where the $a_{n-1}$ comes from and the $n-1$ in $f(n)$ but what does this product mean? We have $(i-1)(n-1-i)$ extra regions for each $i$? And what would the $i$ be indexing? The points on the circle? Any hints on this problem would be nice. I've been stumped for a few days now. AI: Fix the $n$th point that you introduced. The chords which do not involve this point will yield $a_{n-1} $ regions. How many regions are introduced by adding chords involving the $n$th point? Hint: The number of regions is equal to the number of points of intersection of each chord with the remanding chord, plus 1. Hint: If we connect it to the point that is $i$ vertices away, then chords intersect it if one of the endpoints is the $i$ vertices, and another of the endpoints is the other $n-1 - i$ vertices. Do you see how to arrive at $f(n)$ now?
H: How can I find the maximum velocity if I've already found when it occurs? I've been working on this problem for awhile and got the second half figured out, but I can't seem to get what the maximum actually is. Here's the question: A bullet is fired in the air vertically from ground level with an initial velocity 274 m/s. Find the bullet's maximum velocity and maximum height. I found that a maximum occurs at time t = $\frac{274}{9.8}$ $\approx 27.959$ So I plugged s ($\frac{274}{9.8}$) in and got that the maximum height $\approx$ 3830.408 I tried looking up how to find it and found stuff telling me to do "the first derivative test." I tried using the examples to solve my problem, but can't get the answer. Can someone please help me figure out how to get the maximum velocity? Thanks! I can upload a picture of my work if necessary. AI: I think you need some physical reasoning here. Once you fire your bullet vertically, it's only slowed by gravity until it's velocity is zero and then it starts to fall back. So, in absolute value, velocity is maximum in the moment of firing and when it returns to the surface. If you think about the vertical direction as positive, then the maximum occurs precisely in the firing moment. The reason why the derivative test doesn't work here is the following. Mathematically, it's stated as the proposition: "If $f$ has a relative maximum in an open interval, then $f'$ is zero in this point" The problem here is that velocity is a decreasing linear function of time, say $v(t) = v_{0} - gt$, where $v_{0}$ is the firing velocity and $g$ is gravity. Also, you're working on a closed interval $[0, t_{max} ]$, where $t_{max}$ is the instant corresponding to maximum height, calculated as above by you. Notice that $v(t)$ has no relative minimum in $(0, t_{max} )$, and that's why the test doesn't work (indeed, it's derivative is a constant). On the other hand, Weierstrass Theorem assures you that the continuous function $v(t)$ attains a maximum in the compact interval $[0, t_{max} ]$. Since you know it's not in the interior, it must be one end-point, and it's clear that it's actually $0$.
H: A function of $f\circ g$ This is studying for my midterm. Let $f(x)=x^2/(x+1)$ and $g(x)=2x-3$ A function of $f\circ g$ is: So I begin with the equation: $$x^2/(x+1) \cdot 2x-3$$ Add one to the denominator of the second equation. $$x^2/(x+1)\cdot(2x-3)/1$$ Multiplying i get: $$(2x^2-3x^2)/(x+1)$$ Although this isn't right. Could somebody help me out? AI: Here $f(x)=\frac{x^2}{x+1}$ and $g(x)=2x-3$. So $f(g)= \frac{(2x-3)^2}{(2x-3)+1}=\frac{(2x-3)^2}{2x-2}$.
H: inf A of $n^22^{-n}$ Let $A=\{n^22^{-n}, n \in \mathbb{N}\}$. Find $\inf A$, $\sup A$. I tried starting by proving that $\frac{n^2}{2^n} \leq 1/n$ by induction. After, I showed that $\frac{n^2}{2^n} \geq 0$. By the squeeze theorem, $$0 \geq \frac{n^2}{2^n} \geq \frac{1}{n}.$$ But my solution is awfully complicated : I used strong induction with $n+5$... Is there an easier way? AI: $$ \begin{array}{r\|l} n & n^2 2^{-n} \\ \hline 1 & \frac12 \\[6pt] 2 & 1 \\[6pt] 3 & 1 + \frac18 \\[6pt] 4 & 1 \\[6pt] 5 & \frac{25}{32} \\[6pt] 6 & {}\ \ \vdots \end{array} $$ If you can show that it keeps getting smaller after that, then you've got the supremum. If you can show the limit is $0$ and every term is positive, then you've got the infimum. If $n\ge 4$, then every time $n$ is incremented by $1$, the numerator is multiplied by $\left(\frac{n+1}{n}\right)^2=\left(1+\frac1n\right)^2 \le \frac{25}{16} = 1.5625$ and the denominator is multiplied by $2$. That is certainly enough to show that the sequence decreases after that, so the sup is the max, $1+\frac18$. At every step one multiplies by something $\le \frac{25}{32}$, and the problem is to show that the inf is $0$. If, not then the inf is some positive number $c$, and $0<c\le\left(\frac{25}{32}\right)^n$ for all positive integers $n$. So $1/c>\left(\frac{32}{25}\right)^n$ for all $n$. Hence the following exists: $$ d=\sup\left\{ \left(\frac{32}{25}\right)^n : n=1,2,3,\ldots \right\}. $$ Since $d\cdot\frac{25}{32}<d$, then $d\cdot\frac{25}{32}$ must not be an upper bound of that set. Consequently for some $n$ $d\cdot\frac{25}{32}\le \left(\frac{32}{25}\right)^n$. But then $d\le\left(\frac{32}{25}\right)^{n+1}$ and we get a contradiction. The assumption that the inf is $>0$ led to a contradiction.
H: Is the set consisting of $0$ and all polynomials with coefficients in $\mathbb{F}$ and with degree $m$ a subspace of $P(\mathbb{F})$? Is the set consisting of $0$ and all polynomials with coefficients in $\mathbb{F}$ and with degree $m$ a subspace of $P(\mathbb{F})$? I want to say no, since it seems that it is not closed under addition. Consider $p$,$q$ with $$\begin{align} p(z) = a_0+a_1z+ \cdots + a_mz^m,\\q(z) = b_0+b_1z+\cdots-a_mz^m,\\ \\ \text{where } a_m \neq 0. \end{align}$$ Then $p(z)+q(z) = (a_0+b_0)+(a_1+b_1)z+\cdots +(a_{m-1}+b_{m-1})z^{m-1}$, whose degree is less than or equal to $m-1$. AI: You are correct. The space of all polynomials with coefficients in a field $\mathbb{F}$ is a vector space (better still, it is an algebra). For a subset of this vector space to be a subspace, it needs to contain the zero vector, be closed under multiplication by scalars (elements of $\mathbb{F}$), and be closed under addition. The proposed subset satisfies the first two conditions but not the third, as you have shown, so it is not a subspace. However, if we consider the space of all polynomials with coefficients in $\mathbb{F}$ and degree less than or equal to $m$, we do obtain a subspace - in general $\operatorname{deg}(f + g) \leq \max\{\operatorname{deg}(f), \operatorname{deg}(g)\}$.
H: Flipping a coing probability Given that we have already tossed a balanced coin ten times and obtained zero heads, what is the probability that we must toss it at least two more times to obtain the first head? One thing I know is that tossing it initially 10 times is of no use for the answer. However, how can I give an exact value when the first head could come 2+n time? Where n could be [0, infinity). AI: This is equal to the Probability of getting no heads in the first toss. This can be formed as an infinite series as $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\dots=1-\frac{1}{2}$
H: How to prove this assertion? If $a<b$, then there exists a positive integer $n$ such that $a \leq b - 1/n$. I was thinking I could use the Archimedean Property, but I don't know how since I need to get a non-strict inequality $\leq$. I'd appreciate any hints or ideas. Thank you. AI: This is really equivalent to showing that if $\epsilon > 0$, then there is an integer $n$ for which $$0 < \frac{1}{n} < \epsilon$$ (Try setting $\epsilon = b - a$). On the other hand, this is equivalent to showing that there is an integer $n$ for which $$n > \frac{1}{\epsilon} > 0$$ The Archimedean property may be useful here. Note that if you find $n = \frac{1}{\epsilon}$, you can always conclude that $n + 1 > \frac{1}{\epsilon}$. This removes the issue about $\le$ versus $<$.
H: Formally prove: $\lim_{n\to\infty}x_n=L_1\Longrightarrow\lim_{n\to\infty}x_{n+k}=L_1,\forall k\in\mathbb{N}$ OK, so I'm given the following: $$\lim_{n\to\infty}x_n=L_1\iff\forall\epsilon>0,\exists N(\epsilon)\in\mathbb{N}\ni\forall n>N(\epsilon),\ \left\|x_n-L_1\right\|<\epsilon$$ I just have no idea how to use that to prove $$\lim_{n\to\infty}x_{n+k}=L_1\iff\forall\epsilon>0,\exists N(\epsilon)\in\mathbb{N}\ni\forall n>N(\epsilon),\ \left\|x_{n+k}-L_1\right\|<\epsilon$$ What can I possibly do with $\left\|x_{n+k}-L_1\right\|$ to make it so I can make use of the given information? AI: You know $$\forall\varepsilon>0\;\exists N(\varepsilon)\in\mathbb{N}\;\forall n>N(\varepsilon)\ \left\|x_n-L_1\right\|<\varepsilon$$ and want to prove $$\forall\varepsilon>0\;\exists N'(\varepsilon)\in\mathbb{N}\;\forall n>N'(\varepsilon)\ \left\|x_{n+k}-L_1\right\|<\varepsilon$$ Just set $N'(\varepsilon)=N(\varepsilon) - k$. If $n>N'(\varepsilon)$ then $n+k > N'(\varepsilon)+k=N(\varepsilon)$ and you can use the first statement.
H: Why must we account for the domain of an original function when we reduce? Why are reduced rational functions not always "equal to" the original function? Why must we account for the domain of an original function when we reduce? For example, $$g(x) = \dfrac{x^2 + 3x - 4}{x - 1}$$ has a limited domain, because $x = 1$ will cause division by zero. Thus, there will be a hole in the graph at $x = 1$. However, the reduced function, $$f(x) = x + 4$$ does not have a limited domain, and therefore will not have a hole, but we are supposed to put one in anyway. I can't really understand why this is the case. It sort of makes sense, but I would appreciate an explanation for it. I tried to rationalize it in my brain by thinking of the following situation: If I have 8 one-quarter pieces of pie, I have two whole pies because $ 8 \times (1/4) = 2$. However, it is not the same as having two whole pies because they have slices in them. Is this an "okay" analogy to think about what is going on here? AI: Since one of them has a hole and the other doesn't, they're not equal. If $x\ne1$, then $f(x)=g(x)$. If $x=1$, then $f(x)=f(1)=5$ and $g(x)$ reduces to $0/0$, so it's undefined. That's why the reduced rational function is not equal to the original rational function. You say the reduced function does not have a hole. That is precisely why it is not equal to the original function. But if you "put one in anyway", then what you get is equal to the original function. One of the most important reasons for reducing is precisely to remove the hole. If a car goes $60$ miles in $2$ hours, then dividing, you conclude that its average speed during that time is $60/2=30$ miles per hour. If the car goes $0$ miles in $0$ hours, what is its speed at that time? Is it $0/0$ miles per hour? No matter what the car's speed is at that instant, you get $0/0$. But reducing and thereby getting rid of the hole is the way to find the speed at an instant.
H: Analytic function bounded by polynomial Prove that a function which is analytic in the whole plane and satisfies the inequality $\|f(z)\|<\|z\|^n$ for some $n$ and all sufficiently large $\|z\|$ reduces to a polynomial. The function is analytic, so $f^{n}(z)$ exists for all $n$, all $z$. We have the Cauchy's integral formula for higher derivatives $$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}dz.$$ So $$\int_C\frac{f(z)}{(z-a)^{n+1}}dz<\|z\|^n$$ for all large $\|z\|$. How does that help? AI: Take $C$ to be a circle of radius $R$ around the origin, large enough to contain $a$. Parameterize via $z = Re^{it}$. Then \begin{align} \|f^{(n)}(a)\| &= \frac{n!}{2\pi} \left\|\int_C \frac{f(z)}{(z - a)^{n + 1}} dz \right\| \\ &= \frac{n!}{2\pi} \left\|\int_0^{2\pi} \frac{f(Re^{it})}{(Re^{it} - a)^{n + 1}} Rie^{it} dt\right\| \\ &\le \frac{n!}{2\pi} \int_0^{2\pi} \frac{\|f(Re^{it})\|}{\|Re^{it} - a\|^{n + 1}} R dt \\ &\le \frac{n!}{2\pi} \int_0^{2\pi} \frac{R^n}{\|Re^{it} - a\|^{n + 1}} R dt \end{align} The denominator can be bounded below by, say, $\frac{1}{2} R^{n + 1}$ by choosing $R$ sufficiently large, and so we see that $$\|f^{(n)}(a)\| \le n! \frac{R^{n + 1}}{\frac{1}{2} R^{n + 1}} = 2n!$$ Hence, $f^{(n)}$ is a bounded entire function, and so is constant.
H: Is the function $ f(x,y)=\frac{xy}{x^{2}+y^{2}}$ where $f(0,0)$ is defined to be $0$ continuous? Is the function $ f(x,y)=\frac{xy}{x^{2}+y^{2}}$ where $f(0,0)$ is defined to be $0$ continuous? I don't think it is and I am trying to either show this by the definition or by showing that maybe a close set in $\mathbb{R}$ has an inverse set that is not closed in $ \mathbb{R} ^{2}$. I tried the point $0$ but this is open in $\mathbb{R}$. Any hints or ideas? Thanks! AI: It is not continuous at $(0,0)$. Because $f(t,t)=1/2$ but $f(t,0)=0$, so if we approach to the origin along the line $y=x$ then $f(x,y)\to 1/2$ but if we approach to the origin along the x-axis then $f(x,y)\to 0$.
H: In which spaces,$F$ is irreducible iff $F=\overline{\{x\}}$, for all closed $F$? I'm wondering which spaces $X$ meet the following condition: For any closed $F\subseteq X$, $F$ is irreducible if and only if $F=\overline{\{x\}}$ for some $x\in X$. Where $F$ is irreducible if and only if for all closed sets $F_1,F_2$ such that $F\subseteq F_1\cup F_2$, we have that $F\subseteq F_1$ or $F\subseteq F_2$. Thanks. AI: A point $x\in F$ whose closure is $F$ is called a generic point of $F$ (note that the closure of an irreducible subspace of topological space is always irreducible). A sober topological space is one in which every irreducible closed subset has a unique generic point. Any such space has the property you ask about (the existence of a generic point for each irreducible closed subset). Schemes (e.g. spectra of rings) are sober. The uniqueness in the definition of sobriety is equivalent to requiring that for distinct points $x\neq y$ in the space, there is either a neighborhood of $x$ not containing $y$, or a neighborhood of $y$ not containing $x$ (I believe this is called $T_0$ or Kolmogorov); more succinctly, $\overline{\{x\}}\neq\overline{\{y\}}$. The indiscrete topology on a set with two elements $X=\{x,y\}$ is such that every irreducible closed set has a generic point, but the generic point is not unique (the space is not $T_0$). So this is another example of a space having the property you ask about, but which is not sober. I'm not sure what can be said more generally about spaces satisfying your condition.
H: Calculate $\lim_{n\to\infty}\binom{2n}{n}2^{-n}$ I would like to show that: $$\lim_{n\to\infty}\binom{2n}{n}2^{-n} = \infty$$ I have gotten as far as: $$ \binom{2n}{n}={(2n)!\over (n!)^2}=\left({n\over1}+1\right)\left({n\over2}+1\right)(\dots)\left({n\over n}+1\right)\ge2^n $$ But the $2^{⁻n}$ factor defeats that attempt, any suggestion on how to continue would be most appreciated. AI: Hint Keep the first bracket $({n\over1}+1)$ unchanged: $$\binom{2n}{n}={(2n)!\over (n!)^2}=({n\over1}+1)({n\over2}+1)(\dots)({n\over n}+1)\ge2^{n-1}(n+1)$$
H: How many ways are there to distribute 8 teachers to 4 schools where each school must get at least 1 teacher? Additional details: the teachers are considered distinct from one another. So here is what I thought: 1) Choose four teachers to go to each one of the schools: $\binom{8}{4}\cdot4!$ 2) For each of those situations, distribute the other 4 teachers to the 4 schools: $4^4$ So total: $\binom{8}{4}\cdot4!\cdot4^4$ However, I am almost 100% sure that I am over-counting but can't quite put my finger on it. Any help would be appreciated. AI: We can use Inclusion/Exclusion. There are $4^8$ ways to assign the $8$ teachers, with no restrictions. We need to remove the bad assignments, where some school(s) get no teacher. Let the schools be A, B, C, D. There are $3^8$ ways to assign the teachers, avoiding school $A$. There are also $3^8$ ways to avoid schools B, C, D, for a total of $\binom{4}{1}3^8$. However, this overcounts the bad assignments. For example, $\binom{4}{1}3^8$ counts twice the assignments that avoid schools A and B. The same is true for all the $\binom{4}{2}$ pairs of schools. So to count the bad assignments, we must subtract $\binom{4}{2}2^8$. But we have subtracted too much, for we have subtracted one too many times the assignments that avoid all schools but A, also the ones that avoid all schoola but B, or all but C, or all but D. So we must add back $\binom{4}{3}1^8$. The total number of good assignments is therefore $$4^8-\binom{4}{1}3^8+\binom{4}{2}2^8-\binom{4}{3}1^8.$$ Remark: The counting is presumably not being done by a School Board, since a Board is likely to consider teachers indistinguishable.
H: A linear transformation defined by a system of equations carries $\mathbb{R}^n$ onto $\mathbb{R}^m$ iff the rank of the coefficient matrix is $m$. How do I show that a linear transformation defined by a system of equation carries $\mathbb{R}^n$ onto $\mathbb{R}^m$ if and only if the rank of the coefficient matrix of the system is $m$. So I understand that we can define $T:\mathbb{R}^n \to \mathbb{R}^m$ by $T(v)=Av$, where $A$ is an $m \times n$ matrix with entries from $\mathbb{R}$. But it seems so trivial though. Is there a way that I can explain this well? AI: Hint: $\operatorname{rank}(A) = \dim\,\operatorname{Col}(A)$. Using this you should be able to determine the image of $T$.
H: Want to check analyticity of a series on a open disk. How do we check the analyticity of a any power series? For example: How will we show that $$f(z):= \sum_{n=1}^\infty z^{n!}= z^1+z^2+z^6+z^{24}......+z^{n!}......$$ is anaytic on disk {$z : \|z\|<1$} Thanks. AI: The coefficients $a_n$ of your power series are $0$ or $1$, and $a_n=1$ for infinitely many $n$, so $\limsup_{n\to\infty}\|a_n\|^{1/n}=1$. Thus, your power series converges on the open unit disk by Cauchy-Hadamard theorem, and by standard results it is analytic on such disk.
H: Radius of Convergence of $\sum ( \sin n) x^n$. Thank you very much in advance for any assistance/advice on solving this problem. I am fairly new to power series and determining the radius of convergence. Determine, with proof, the radius of convergence of $\sum ( \sin n) x^n$. My understanding of this problem is that we cannot employ the ratio test since it will not really give us something we can use to determine the convergence (actually, I don't think it even applies). One approach that I can think of is to compare this series with another one, but could it involve the integral test since $\sin n$ is continuous? Also, my book does not include the lower and upper limits of the sum, but I am assuming they go from 0 to $\infty$. I am using the textbook Introduction to Analysis by Arthur Mattuck. AI: Since $\|\sin n\|\le 1$ for all $n$, our series converges absolutely for any $x$ with $\|x\|\lt 1$, by comparison with the geometric series $\sum x^n$. Thus the radius of convergence is $\ge 1$. It remains to show that the radius is not $\gt 1$. To do this, it is enough to show that for any $\delta\gt 0$, the series $\sum \sin n (1+\delta)^n$ does not converge. We do this by showing that the terms $\sin n (1+\delta)^n$ cannot have limit $0$. The numbers $\sin n$ are dense in the interval $[-1,1]$. In particular, there are infinitely many $n$ such that $\|\sin n\|\gt 1/2$. This completes the proof. Remark: We can prove the result without referring to the fact that the numbers $\sin n$ are dense in $[-1,1]$. We show that $\sin n$ cannot have limit $0$. For suppose to the contrary that $\|\sin n\|$ is close to $0$ for all large enough $n$. Suppose for instance that for all $n\gt N$, we have $\|\sin n\|\lt 1/10$. Consider $\sin(n+1)=\sin n \cos 1+\cos n\sin 1$. The term $\sin n \cos 1$ has absolute value $\lt 1/10$. The term $\cos n\sin 1$ has absolute value $\ge \sqrt{1-1/100}\sin 1\gt 0.8$, and therefore $\|\sin(n+1)\|\gt 0.7$.
H: Uniform continuity on $\mathbb{C}$ If a continuous function $f: \mathbb{C} \rightarrow \mathbb{C}$ satisfies $f(z) \rightarrow 0$ as $\|z\| \rightarrow \infty$, then $f$ is uniformly continuous on $\mathbb{C}$. Should I be thinking about the Riemann sphere here? I have no clue what my intuition should I have be. Any helpful comments are appreciated. Just trying to brush up on my complex analysis, but I have forgotten everything. AI: $f(z)\to 0$ as $\|z\|\to\infty$ implies for each $\varepsilon>0$ there exists $R>0$ such that $\|f(z)\|<\varepsilon/2$ for $\|z\|>R$. Since the disk $D_R=\{z\in\Bbb{C}:\|z\|\le 2R\}$ is compact, $f$ is uniformly continuous on $D_R$. Therefore, for each $\varepsilon>0$ there is $\delta>0$ such that $\|f(z)-f(w)\|<\varepsilon$ for all $z,w\in D_R$ satisfy that $\|z-w\|<\delta$. For each $\varepsilon>0$, let $R$ and $\delta$ are positive numbers given above. Let $z$ and $w$ are complex numbers satisfy that $\|z-w\|<\min(\delta,R/2)$. If $\|z\|,\|w\|\le 2R$ then $\|f(z)-f(w)\|<\varepsilon$ because of the definition of $\delta$. If not, then $\|z\|>2R$ or $\|w\|>2R$. Without loss of generality we assume that $\|z\|>2R$. Since $\|z-w\|<R/2$, we get $\|w\|>3R/2$ and $\|f(z)-f(w)\|\le \|f(z)\|+\|f(w)\|<\varepsilon$.
H: Infinitely many moons, or one ring to bring them all, a limit to bind it? The Kanagy clan makes its home on a distant planet of mass $M_p$ with $k$ moons. Suppose the moons are identical with mass $m$. Furthermore, these moons share a common circular orbit on an orbital plane. The circular orbits are at distance $L$ from the center of the planet and the moons are evenly spaced. What is the speed $v(k)$ of the lunar orbits for a fixed, but finite, value of $k$? Suppose we hold $km=M_m$ fixed as $k \rightarrow \infty$, what is the limiting value of $\lim_{k \rightarrow \infty}v(k)$ in this context? My ideal solution to this problem addresses $(1.)$ by vector analysis paired with the equation of motion for constant speed circular motion. Often cases $k=2$ or $k=3$ are given as homework problems in first semester university physics. I've found an expression for this in a previous attempt, but I'd rather not include it here for fear of biasing the reader. Next, the solution continues to $(2.)$, when I attempted to compute the limit directly it was rather involved. However, by intuition, I know the answer should easily derive from Newton's Law of Gravitation as follows: $$ \frac{M_mv^2}{L}= \frac{GM_mM_p}{L^2} \qquad \Rightarrow \qquad v = \sqrt{\frac{GM_p}{L}} $$ Given the preceding discussion, show show $v(k) \rightarrow \sqrt{\frac{GM_p}{L}}$ as $k \rightarrow \infty$. Alternatively, prove my intuition is incorrect. Incidentally, I gave this as a bonus problem on a final exam in my university physics course. I had a student pretty well solve $(1.)$, but $(2.)$ I've not yet cracked. It is assumed that classical mechanics applies to this problem and any relativistic effects may be neglected. AI: The centre of mass of the system is at the planet, which we can thus assume is stationary. If we number the moons $0$ to $k-1$, moon $0$ sees an angle $\theta_j = \dfrac{\pi}{2} - \dfrac{j \pi}{k}$ between moon $j$ and the planet, and the distance from moon $0$ to moon $j$ is $2 L \sin(j \pi/k)$. So the component in the direction of the planet of the gravitational acceleration of moon $0$ due to moon $j$ is $\cos(\theta_j) G m/(2 L \sin(j \pi/k))^2 = G m/(4 L^2 \sin(j \pi/k))$. The net gravitational acceleration (directed toward the planet) due to all the other moons is then $$ \sum_{j=1}^{k-1} \dfrac{G m}{4 L^2 \sin(j \pi/k)} = \sum_{j=1}^{k-1} \dfrac{G M_m}{4 k L^2 \sin(j \pi/k)} > \sum_{j=1}^{k-1} \dfrac{G M_m}{4 L^2 \pi j}$$ But that goes to $+\infty$ as $k \to \infty$. So we must also have $v(k) \to \infty$.
H: Sequence of Ordinals and Ordinal Definability in Levy Collapse Extensions Let $\kappa$ be an inaccessible cardinal. Let $G$ be generic for $Col(\omega, < \kappa)$, the Levy Collapse. If $f\in \text{ }^\omega \text{Ord}^{V[G]}$, is $f \in OD_{\text{ }^\omega\omega}^{V[G]}$? A possibly useful fact is that if $f\in \text{ }^\omega \text{Ord}^{V[G]}$, then there exists a $\lambda < \kappa$ such that $f \in V[G \| \lambda]$. However, I do not know how to use this, if relevant at all. I am interested in this since some people define the Solovay model as $\text{HOD}^{V[G]}_{\text{ }^\omega \text{ORD}}$ and other people use $\text{HOD}_{\text{ }^\omega\omega}^{V[G]}$. I think if the above question has a positive answer, then the two are the same. Thanks for any help. $x \in \text{OD}_A$ if $x$ is definable with parameters in $OD \cup A$. $x \in \text{HOD}_A$ if $tc(\{x\}) \subset OD_A$, where $tc$ is the transitive closure. AI: The answer is negative, and those two models are not necessarily the same. The basic problem is that one might have an $\omega$-sequence of ordinals very high up, above $\kappa$, and if this sequence isn't sufficiently definable in $V$, then it will not be in $\text{HOD}^{V[G]}_{{}^\omega\omega}$, but of course it is in $\text{HOD}^{V[G]}_{{}^\omega\text{Ord}}$. For example, start with a model where there is a measurable cardinal $\delta$ above $\kappa$ in $V_0$, and let $V=V_0[s]$ be obtained by Prikry forcing at $\delta$ to add a new cofinal $\omega$-sequence $s$ in $\delta$. The forcing to $V[G]$ will be small relative to $\delta$, and we can view $V[G]$ as $V_0[G][s]$, and we can in effect interchange the order of forcing, because the old measure generates a unique new measure, and so $s$ is $V_0[G]$-generic. One can show that after Prikry forcing, the generic sequence is not definable from reals and ordinal parameters, and so $s$ is not in $\text{HOD}^{V[G]}_{{}^\omega\omega}$. But it is an $\omega$-sequence of ordinals, and so it shows the two models are different in this case.
H: Limit of $n$-th derivative over factorial and exponential function Suppose $f(z)$ is analytic on the disk $\|z\|<1$. Prove that $\lim_{n\rightarrow\infty}\dfrac{f^{(n)}(0)}{n!n^n}=0$. When I see the $n$-th derivative, I think of the Cauchy's formula: $$\dfrac{f^{n}(0)}{n!n^n}=\dfrac{1}{2\pi in^n}\int_C\dfrac{f(z)}{z^{n+1}}dz$$ whenever the closed curve $C$ is contained in the disk $\|z\|<1$ and doesn't contain the origin. As $n$ grows large, $n^n$ in the denominator certainly grows very large, but I don't know how to bound the term $\int_C\dfrac{f(z)}{z^{n+1}}dz$. AI: Take $C$ to be the circle with radius $\frac12$ centered at $0$. (Your statement of Cauchy's formula isn't right for all closed curves, but it applies for this one. $C$ should be simple, oriented positively, and wind around $0$.) Then $$\left\|\int_C\dfrac{f(z)}{z^{n+1}}dz\right\|\leq 2\pi\cdot \frac12\dfrac{\max\{\|f(z)\|:\|z\|=\frac12\}}{\left(\frac12\right)^{n+1}}=2^{n+1}\pi\max\limits_{\|z\|=1/2}\|f(z)\|.$$ The max exists because $C$ is compact and $f$ is continuous. The result follows because $$\lim\limits_{n\to\infty}\dfrac{2^n}{n^n}=0.$$ Alternatively, $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ where the power series has radius of convergence $R\geq 1$. Thus $$1\geq\frac{1}{R}=\limsup\limits_{n\to\infty}\sqrt[n]{\|a_n\|} =\limsup\limits_{n\to\infty}n\cdot\sqrt[n]{\left\|\dfrac{f^{(n)}(0)}{n!n^n}\right\|}.$$ This implies that $\lim\limits_{n\to\infty}\sqrt[n]{\left\|\dfrac{f^{(n)}(0)}{n!n^n}\right\|}=0$, which in turn implies that $\lim\limits_{n\to\infty}\left\|\dfrac{f^{(n)}(0)}{n!n^n}\right\|=0$.
H: Question on Discrete metric space Let $X = \{1,1/2,1/4,...,1/2^n,...\} \cup \{0\}$ and $Y = \{X\} - \{0\}$. Is $Y$ dense in $X$? The metric is the usual. If yes, why a separable discrete metric space is then countable? In this setting $X$ is not discrete? We cannot have accumulation points in discrete spaces? AI: Yes, $Y$ is dense in $X$. I don't understand why this answer leads to your other question, but I will answer it. The only dense subset of a discrete metric space is the whole space. Therefore if the space is separable and discrete, it must be countable.
H: Matrices as Functions A friend of mine was criticized in undergrad by a Professor for saying that a matrix is a function. Now, a matrix can be represented by a linear transformation, and linear transformations by definition are functions. Is there any theoretical reason as to why a matrix can't be dubbed a function? From my understanding, they are functions. Am I missing something? AI: A linear transformation is a function. Given a linear transformation $T : \mathbb{R}^n \to \mathbb{R}^m$, and a choice of bases on $\mathbb{R}^n$ and $\mathbb{R}^m$, there is an $m\times n$ matrix $A$ such that $T(x) = Ax$ for all $x \in \mathbb{R}^n$. Note, choices were made to write $T$ in this form. Note, $A$ is not the function, it is just used in the definition of the function. In the other direction, given an $n\times m$ matrix $A$, and a choice of bases on $\mathbb{R}^n$ and $\mathbb{R}^m$, you can define a linear transformation $T : \mathbb{R}^n \to \mathbb{R}^m$ by $T(x) = Ax$ but again, $A$ is not the function, $T$ is. Consider the analogous situation of a function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x$. Would you say that $2$ is a function? Here $2$ plays the role of the matrix $A$. This isn't so much an analogy as it is the special case $n = m = 1$.
H: How prove this $\lim_{x\to+\infty}(f'(x)+f(x))=l$ let $f(x)$ is continous and $f'(x)$ is continous on $[0,\infty)$,show that $$\lim_{x\to+\infty}(f'(x)+f(x))=l$$ if and only if: $\displaystyle\lim_{x\to+\infty}f(x)=l$ and $f'(x)$ is uniformly continuous on $[0,+\infty)$. How prove this it? Thank you. I can prove this if $$\lim_{x\to+\infty}(f'(x)+f(x))=l$$ then we have $\displaystyle\lim_{x\to+\infty}f(x)=l$ My Part of the Solution: without loss of we let $l=0$,Give $\epsilon>0$,let $a>0$ be such that $\|f(x)+f'(x)\|<\epsilon$ for $x\ge a$ Then by the generalized mean value theorem there is $\xi\in (a,x)$ such that $$\dfrac{e^x f(x)-e^af(a)}{e^x-e^a}=f(\xi)+f'(\xi)$$ Thus $$\|f(x)-f(a)e^{a-x}\|<\epsilon\|1-e^{a-x}\|$$ so $$\|f(x)\|<\|f(a)\|e^{a-x}+\epsilon\|1-e^{a-x}\|$$ so $$\|f(x)\|<2\epsilon$$ for sufficiently large $x$. But How can prove $f'(x)$ is uniformly continuous on $[0,+\infty)$ and other question: How prove if$\displaystyle\lim_{x\to+\infty}f(x)=l$ and $f'(x)$ is uniformly continuous on $[0,+\infty)$ then we have $$\lim_{x\to+\infty}(f'(x)+f(x))=l$$ AI: Hint 1 If $\lim_{x \to \infty} f(x)=l$ then $\lim_{x \to \infty} f'(x)=0$. Thus, for each $\epsilon >0$ there exists an $a$ so that $\|f'(x)\|< \epsilon$ for $x \in (a, \infty)$. You also know that $f'(x)$ is uniformly continuous on $[0,a]$ (why)? Hint 2 For the other question, try to apply the generalized mean value Theorem for $(x,x+a)$ when $a$ is very small and $x$ very large.
H: Left and Right Cosets of a Subgroup of Index 2 Show that if $H$ is a subgroup of index $2$ in a finite group $G$, then every left coset of $H$ is also a right coset of $H$. AI: $H$ is a coset, and the only other possible coset is $G \setminus H$ by order considerations.
H: What is $O(\sqrt{2^n}n^2)$? What is $O(\sqrt{2^n}n^2)$? Is it $O(2^n)$, or does the square root cause it to be reduced? I'm trying to analyze an algorithm that I came up with, and if it still has exponential time cost, I'm going to have to try harder. Thanks! AI: $\sqrt {2^n}=2^{\frac n2}$, still exponential in the sense that it will dominate $n^k$ for any $k$
H: showing that there is no simple group of order 48 I tried to solve this prob. by using Sylow theorems. But i can't solve it because it is not seen in the text as a example or exercise. I even use a diagram of sets and consider all possible cases, but it is not effective.... AI: Suppose G is simple. Since G is simple, we must have that $n_2 = 3$. Let P be a Sylow 2-subgroup of G. Then $\|G:N_G(P)\|=3$. Theorem: If G is a simple group, $\|G\|\neq 2$, and if $H<G, \|G:H\|=n > 1$ then $G\cong$ subgroup of $A_n$. So in our case we have that $G\cong$ subgroup of $A_3$ but $\|A_3\|=3$ and $\|G\|=48$, which is a contradiction. Thus G cannot be simple. Proof of Theorem: Let G be a simple group, $\|G\|>2$. Let $H<G$ and $\|G:H\|=n>1$. There exists a homomorphism $\theta :G \to S_n$ with $ker(\theta)\subseteq H$ (the permutation representation of the G acting on cosets H of G by left multiplication). Since G is simple, we must have that $ker(\theta) = 1$. So $\theta (G) \cong G$ by the isomorphism theorem. Suppose that the result is not true, that $\theta(G) \nsubseteq A_n$. Then we must have $A_n\theta(G) = S_n$. Since $\frac{S_n}{A_n} = \frac{A_n\theta(G)}{A_n}\cong \frac{\theta(G)}{A_n\cap\theta(G)}$ by the second isomorphism theorem, and G is simple, we must have $A_n\cap \theta(G)=1$ or $\theta(G)$. If $A_n\cap \theta(G)= \theta(G)$, then $S_n/A_n=1$, which is a contradiction. In the case $A_n\cap \theta(G)=1$ we have that $S_n/A_n \cong \mathbb{Z}/2\mathbb{Z}\cong \theta(G)\cong G$, i.e. that $\|G\|=2$, which is a contradiction. Thus we must have that $\theta(G)\subseteq A_n$
H: Complex integral of product of two polynomial powers Compute $$\int_{\|z\|=1}z^n\left(z-\dfrac{1}{2}\right)^mdz$$ where $m,n$ are integers. If $m,n\geq 0$, the function is entire, and so the integral is $0$. If $n<0$ and $m\geq 0$, the function becomes $\dfrac{\left(z-\dfrac12\right)^m}{z^{\|n\|}}$. This can be handled by the Cauchy formula. Let $f(z)=\left(z-\dfrac12\right)^m$. Then the integral is $\dfrac{2\pi i}{\|n-1\|!}\cdot f^{(\|n-1\|)}(0)$. If $m<0$ and $n\geq 0$, it should be similar. But what if $m,n<0$? Now there are two values for which the denominator vanish. How can we use Cauchy's theorem or otherwise? Edit: Since this exercise appears in Ahlfors many sections before the Residue theorem, I would like to see a solution that doesn't use Residue theorem. AI: $\displaystyle{% \int_{\left\vert\, z\,\right\vert\ =\ 1} z^{n}\left(z- {1 \over2}\right)^{m}\,{\rm d}z:\ {\large ?}\\[5mm] }$ In general, we have two expansion of the integrand: $$ z^{n}\left(z- {1 \over2}\right)^{m} = \left(-1\right)^{m} \sum_{\ell = 0}^{\infty}{m \choose \ell}{1 \over 2^{m - \ell}}\, {\left(-1\right)^{\ell} \over z^{-n - \ell}} $$ $$ z^{n}\left(z- {1 \over2}\right)^{m} = \left[{1 \over 2} + \left(z - {1 \over 2}\right)\right]^{n}\left(z- {1 \over2}\right)^{m} = \sum_{\ell = 0}^{\infty}{n \choose \ell}{1 \over 2^{n - \ell}}\, {1 \over \left(z - 1/2\right)^{-m - \ell}} $$ where $\displaystyle{x \choose y}$ is the $\it\underline{\mbox{generalized binomial coefficient}}$ as explained here. When $m \leq -1$ and $n \leq -1$ ( $m$ and $n$ integers ) we have to consider both poles: At $z = 0$ and at $z = 1/2$. The result turns out to be: \begin{align} \int_{\left\vert\, z\,\right\vert\ =\ 1} z^{n}\left(z- {1 \over2}\right)^{m}\,{\rm d}z &= 2\pi{\rm i}\left[% {m \choose -n - 1}\left(-\,{1 \over 2}\right)^{m + n + 1} + {n \choose -m - 1}{1 \over 2^{n + m + 1}}\right] \\[3mm]&= 2\pi{\rm i}\left[% {n \choose -m - 1} + \left(-1\right)^{m + n + 1}{m \choose -n - 1} \right]\,{1 \over 2^{m + n + 1}} \\[3mm]&= 2\pi{\rm i}\left[% \left(-1\right)^{m + 1}{-m - n - 2 \choose -m - 1} + \left(-1\right)^{m}{-m - n - 2\choose -n - 1} \right]\,{1 \over 2^{m + n + 1}} \\& \phantom{% \left[% {n \choose -m - 1} + \left(-1\right)^{m + n + 1}{m \choose -n - 1} \right]} m, n \in {\mathbb Z}\,,\quad m \leq -1\,,\quad n \leq -1. \end{align}
H: Problem with $\text{Tor}$ functor Please explain to me about small $\text{Tor}$ functor problem. I use $\text{Tor(A,B)}$ define at http://en.wikipedia.org/wiki/Tor_functor. we take a projective resolution: $\cdots\rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow A\rightarrow 0$ (1) then remove the A term and tensor the projective resolution with B to get the complex: $\cdots \rightarrow P_2\otimes_R B \rightarrow P_1\otimes_R B \rightarrow P_0\otimes_R B \rightarrow 0$ (2) and take the homology of this complex. Clearly, since right exactness of $\otimes$-functor, so from (1) we have (2) is right exact sequence. It's mean, homology $H_n(x)=0$. But it's impossible! I'm really misunderstand! Thanks for regarding! AI: Tensoring with $B$ doesn't take you from (1) to (2). It takes you from (1) to $$\cdots P_2 \otimes_R B \to P_1\otimes_R B \rightarrow P_0 \otimes_R B \rightarrow A\otimes_R B \rightarrow 0$$ which is indeed exact at the last few terms.
H: Limit at infinity $\lim_{x\rightarrow \infty}\left(x-\lfloor x\rfloor\right) = $ $\displaystyle \lim_{x\rightarrow \infty}\left(x-\lfloor x\rfloor\right) = $ I have Tried like this way:: Let $x = k\rightarrow \infty$. Then Limit Convert into $\lim_{x\rightarrow k}(x-\lfloor x \rfloor)$ Now When $x = k\rightarrow \infty$. Then $x=\lfloor x \rfloor $. So $\displaystyle \lim_{x\rightarrow \infty}\left(x-\lfloor x\rfloor\right) = 0$ But Answer given is Limit does not exists. AI: The limit does not exist. Note that $x\mapsto x-\lfloor x\rfloor$ traces out all numbers in $[0,1)$ indefinitely many times as $x\to+\infty$ (to convince yourself, you may want to draw a graph; it's gonna be a sequence of segments pointing to the northeast). That is, there is not a particular number to which this function gets ever closer as $x$ tends to infinity (which, intuitively, is the very essence of the definition of convergence). In your example, what you really show is that $$\lim_{\substack{x\to+\infty\\\text{$x$ is an integer}}}\left(x-\lfloor x\rfloor\right)=0.$$ This is elusive, because you have chosen only a particular sequence of $x$'s (namely, the sequence of integers) and shown that the associated sequence of function values converges to zero. However, by the definition of convergence, you ought to show this result for all sequences of $x$'s tending to infinity. But then again, you can't show this, because convergence does not occur.
H: Why is the modulus of a complex number $a^2+b^2$? Why is the modulus not $\sqrt{a^2-b^2}$? Carrying out standard multiplication this would be the result-why is this not the case? I know viewing the complex plane you can easily define the sum as being the distance to the points, but what meaning does $\sqrt{a^2-b^2}$ have? AI: The geometric meaning of the modulus is the length of the vector from the origin to the point in the complex plane. So it's just defined like the Euclidean distance to a point $(x, y)$, $\sqrt{x^2 + y^2}$. It just happens that "y component" is the imaginary part of the complex number $a + i b$, so you use $\sqrt{a^2 + b^2}$.
H: Help Solving First Order nonlinear ODE $e^y + [t e^y -sin(y)] \frac{dy}{dt}=0 \quad y(2)=1.5$ Is it possible to find an exact solution to this ODE using some method? Following Adriano's hint: (second edit) $d( e^y t + cos(y) ) =0$ or $e^y t + cos(y)=c_1 $ for some constant $c_1$ and then using the initial value condition $c_1 = e^{1.5} 2 + cos(1.5)$ So we have the relation $e^y t + cos(y) = e^2 1.5 + cos(2)$ which satisfies the ODE. or $ t = e^{-y} [-cos(y)+e^{1.5} 2 + cos(1.5)]$ AI: Hint: Yes. Note that this ODE is exact, since by rewriting it as: $$ \underbrace{[e^y]}_{M}dt + \underbrace{[t e^y - \sin(y)]}_N dy=0 $$ we observe that it has the form $M(t,y)~dt + N(t,y)~dy = 0$, where: $$ M_y = e^y = N_t $$ Now try to find a function $\psi(t,y)$ such that $\psi_t = M$ and $\psi_y = N$. Once you have done this, you can apply product rule in reverse.
H: How can we prove that a Fourier Series exists? How does one show that an arbitrary periodic function, so long as it is reasonably well behaved, can always be represented as a sum of sine and cosine functions? It sounds like the first thing you would learn when it comes to Fourier Series but my text simply claims that this is true and goes on to tell me how to work out the coefficients. Can anyone help? AI: As alluded to in the comments, it is probably too difficult to prove given what the text assumes you know. The basic idea is that $B = \{1\}\cup\{\cos(nx)\mid n \in \mathbb{N}\}\cup\{\sin(nx)\mid n\in\mathbb{N}\}$ forms a basis for the space you're interested in; it's $L^2[-\pi, \pi]$, but I won't explain what that means. Note, a lot of care needs to be taken here as the space is infinite dimensional; in particular, the definition of basis is not clear. As $B$ is a basis, we can write any element of $L^2[-\pi, \pi]$ as a linear combination of the elements in $B$. Furthermore, the space $L^2[-\pi, \pi]$ has an inner product. With respect to this inner product, the elements of $B$, up to a constant factor, form an orthonormal basis for $L^2[-\pi, \pi]$. This allows us to write out explicitly the coefficients of the basis vectors in the linear combination (this result is true in finite and infinite dimensional vector spaces, see here); these are precisely the formulae you have for $a_n$ and $b_n$. Note, some of what I've said is not strictly true, but the ideas roughly match up with the theory. Further details can be found on Wikipedia.
H: Can Prims and Kruskals algorithm yield different min spanning tree? In this problem I am trying to find the min weight using the Prims and Kruskals and list the edges in the order they are chosen. For Prims I am getting order (A,E), (E,F), (F,C), (C,D), (C,B) with a weight of 21 With Kruskals I get (C,F), (A,E), (C,D), (B,C) with a weight of 14 I feel like I shoudnt be getting two different min weights. Could the problem be that I am forced to start at Vertex A when using Prims? If that is not my problem what is? AI: You messed up with Kruskal's algorithm, since the answer you give isn't even a tree: there is, for example, no path between A and B.
H: Invertible polynomials and power series Consider the polynomial ring $A[x]$ and $f(x)=\sum_{i=0}^{n}a_ix^i\in A[x]$, where $A$ is a commutative ring with unit. Show that $f$ is a unit in $A[x]$ if and only if $a_0$ is a unit in $A$ and $a_1,\dots, a_n$ are nilpotent. Consider the formal power series ring $A[[x]]$ and $f(x)=\sum_{i=0}^{\infty}a_ix^i\in A[[x]]$, where $A$ is a commutative ring with unit. Show that $f$ is a unit in $A[[x]]$ if and only if $a_0$ is a unit in $A$. How to prove these two questions? AI: For the second one, the only if is clear from $f(x)g(x)=1$ by comparing coefficients of the terms of degree zero. For the if just do long division for series (remember that it is like long division for polynomials but organizing the terms by increasing degrees). For the first one, for the if part, consider a large power of $a_0^{-1}f(x)$. For the only if it is clear $a_0$ is going to be a unit by comparing coefficients in $f(x)g(x)=1$. For the only if part we can assume $a_0=1$ (by considering $a_0^{-1}f(x)$) to have to write less. We use induction in the number of terms. If $B$ is a nilpotent term and $f(x)$ is a unit then $f(x)-B$ is also a unit. Begin from polynomials of at most degree one (we can always assume the inverse with the same number of terms by adding if necessary terms with zero coefficients). $f(x)=1+a_1x+a_2x^2+...+a_nx^n$ and $g(x)=1+b_1x+b_2x^2+...+b_nx^n$. From $f(x)g(x)=1$ you can solve, comparing coefficients, for the coefficients of $g$. For example, $b_1+a_1=0$. Then $b_1=-a_1$. But then $0=b_1a_1+a_2+b_2=-a_1^2$. So, $a_1$ is nilpotent. ...
H: Condition on spectrum of T Let $T$ $\in \mathfrak{B}(\mathbb{H})$ be normal. Let $A$ be the closed subalgebra generated by $T$, $T^{}$ and $I$. Suppose $T$ can be approximated in norm by finite linear combinations of projections in $A$. What is the necessary and sufficient condition on $\sigma(T)$ for this to happen? I am not sure how to approach this question. Do we use the spectral theorem for normal operators? Does it follow that since $T$ can be norm-approximated by linear combinations of projections, its spectrum must be finite? I would be grateful for some hints on how to start. AI: Have you seen that $A\cong C(\sigma(T))$? This is equivalent to asking for conditions on $\sigma(T)$ such that the function $z\mapsto z$ can be uniformly approximated by linear combinations of continuous characteristic functions on $\sigma(T)$, which is asking for a lot of clopen subsets of $\sigma(T)$ to exist. For example, this would hold if $\sigma(T)$ were the Cantor set. (I noticed the problem asks for hints so I turned this comment into an answer. Related: Is a von Neumann algebra just a C-algebra which is generated by its projections?)
H: ways to fill n places with fixed values in a given range and avoid diplicates formed. what are the number of ways in which we can fill n places with 2 fixed values and rest places with values between the 2 selected ones such that we get no duplicates? example : n=4 , fixed values 1 and 3 case 1 : fix 1 on position 1 and 3 on position 2 1 3 1 1 1 3 1 2 1 3 2 1 1 3 2 2 1 3 1 3 1 3 3 1 1 3 3 3 1 3 2 3 1 3 3 2 case 2 : fix 1 on position 1 and 3 on position 3 1 1 3 1 1 1 3 2 1 2 3 1 1 2 3 2 1 3 3 2 1 2 3 3 1 3 3 1 1 1 3 3 1 3 3 3 now in case 1 and 2 : 1 3 3 3 and 1 3 3 1 and 1 3 3 2 are repeating case 3 : ....similarly other cases follow what i have done till now is :nC2 * POWER( (max-min+1) , n-2 ) - duplicates but not able to subtract duplicates. AI: Use the in-and-out principle, also known as the inclusion-exclusion principle. Let's say $n\ge2$ and call the fixed values $x$ and $x+q$ where $q\ge1$. (In your example, $n=4,x=1,q=2$.) There are $(q+1)^n$ ways to fill the $n$ positions with integers in the range $[x,x+q]$. This is an overcount, because some of the ways we counted miss $x$ or $x+q$ or both. Of the $(q+1)^n$ things we counted, there are $q^n$ that miss $x$; these are the ways of filling the $n$ positions with integers in $[x+1,q]$. Count them out: that leaves $(q+1)^n-q^n$ ways. By the same token, our original overcount included $q^n$ things that miss $x+q$; count those out too. That leaves $(q+1)^n-2q^n$. But now we have undercounted, because things that miss both $x$ and $x+q$ have been counted out twice. There are $(q-1)^n$ of those, so count them back in to get the correct final answer:$$(q+1)^n-2q^n+(q-1)^n.$$ In your example with $n=4,x=1,q=2$ we get $81-2\cdot16+1=50$. Is that what you got? To check my answer, let's count them by brute force. There are $19$ ways starting with a $1$, namely: $1113$, $1123$, $1131$, $1132$, $1133$, $1213$, $1223$, $1231$, $1232$, $1233$, $1311$, $1312$, $1313$, $1321$, $1322$, $1323$, $1331$, $1332$, $1333$. There are $12$ ways starting with a $2$, namely: $2113$, $2123$, $2131$, $2132$, $2133$, $2213$, $2231$, $2311$, $2312$, $2313$, $2321$, $2331$. There are $19$ ways starting with a $3$. I'm getting tired of typing, so just take the list of $19$ ways starting with a $1$ and change all the $1$s to $3$s and the $3$s to $1$s. The grand total (no duplicates!) is $19+12+19=50$. P.S. To explain the solution more formally: If $S$ is a set, $\|S\|$ denotes the number of elements in $S$. Let $S$ be the set of all sequences of length $n$ taking (integer) values in the closed interval $[x,x+q]$; then $\|S\|=(q+1)^n$. Let $A$ be the subset of $S$ consisting of all sequences in $S$ in which $x$ is missing; in other words, $A$ is the set of all sequences taking values in the interval $[x+1,x+q]$; so $\|A\|=q^n$. Let $B$ be the set of all sequences in $S$ in which $x+q$ is missing; so $\|B\|=q^n$. The union $A\cup B$ is the set of all sequences in $S$ missing $x$ or $x+q$ or both. The inclusion-exclusion principle says that $\|A\cup B\|=\|A\|+\|B\|-\|A\cap B\|$. The intersection $A\cap B$ is the set of all sequences in $S$ missing both $x$ and $x+q$, in other words, the set of all sequences taking values in the interval $[x+1,x+q-1]$, and so $\|A\cap B\|=(q-1)^n$, and $\|A\cup B\|=\|A\|+\|B\|-\|A\cap B\|=q^n+q^n-(q-1)^n$=2q^n-(q-1)^n. That's the number of elements in $S$ which do not contain both an $x$ and an $x+q$. Since there are a total of $(q+1)^n$ of sequences in $S$, the number of sequences which do contain an $x$ and a $q$ is equal to $\|S\|-\|A\cup B\|=(q+1)^n-[2q^n-(q-1)^n]=(q+1)^n-2q^n+(q-1)^n$.
H: General question on parameter functions I have always wondered why considering paths, $\gamma(t)$, there is inevitably a condition that $\gamma'(t) \neq 0$ associated within the same sentence. Can someone please give me a motivational reasoning behind this; whether by examples or goals etc. I have intentionally left this vague as I have seen this in Calculus, real/Complex analysis, geometry and so forth. Any comments will be appreciated. AI: I think it's often because we're mostly interested in considering smooth curves (i.e. ones that don't have sharp corners in them). If $\gamma'(t_0) = 0$, then there might be a sharp corner at $t=t_0$ even though the function $\gamma$ is infinitely differentiable. For example, consider the curve $\gamma(t) = (t^2, t^3)$ at $t=0$. Another way of saying the same thing ... at places where $\gamma'(t) = 0$, the naive computation of the unit tangent (by unitizing the first derivative vector) fails, so, without some extra fuss, you can't even get started doing simple differential geometry. Best to exclude such places.
H: How do we validate the equality of function while reducing the indeterminate form to simpler function? To evaluate the limit of an indeterminate form, we need to reduce the function to its simplest form in order to get a meaningful answer. For example: $f(x_1)$=${x^3-1\over x-1}$, evaluate $lim_{x->1} {x^3-1\over x-1}$; putting $x=1$ in the equation we get $\frac 00$. We reduce the equation to the simpler form, $f(x_1)$=${x^3-1\over x-1}$ = ${(x-1)(x^2+x+1)\over (x-1)}$= $(x^2+x+1)$$=$$f(x_2)$ Therefore, $lim_{x->1} (x^2+x+1)$; putting $x=1$ we get, $lim_{x->1} f(x_2)$$=3$ It suggests from the above that $f(x_1)=f(x_2)=3$ ; the functions are equal. But, the condition of equality of function: For $f(x_1)=f(x_2)$, $dom f(x_1)=dom f(x_2)$ and $range f(x_1)=range f(x_2)$ We see, the domain of $f(x_1) = {R}-(1)$ and domain of $f(x_2)=R$ It implies, that $dom f(x_1)dom≠f(x_2)$ contradiction! The functions seem unequal. How does the validity of equality of function holds, not only here for general case as well? AI: It is true that: $$ {x^3-1\over x-1} \neq x^2+x+1 $$ since the domains of these two functions are not equal. However, these two functions are equal everywhere else other than at $x = 1$. Hence, since evaluating the limit of a function $f$ as $x \to a$ does not depend on the actual value of $f(a)$, we are allowed to use the equality: $$ \lim_{x\to1}{x^3-1\over x-1} = \lim_{x\to1}(x^2+x+1) $$
H: Calculate the Factor Group $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$ I am attempting to understand and compute: $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$ I know $(0,2)$ generates $H = \{(0,0),(0,2),(0,4)\}$, which has order 3 because there are 3 elements. Now, I must find all the cosets which there should be 8. The reason there should be 8 is because $4\cdot 6/3 = 8$. This is where I run into the issue. I have a hard time generating the cosets, but furthermore, once I actually do get the cosets, I have a hard time analyzing them and knowing what it means. If someone could walk me through a step-by-step solution that would be great. The answer is that the factor group is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$. Reference: Fraleigh, A First Course in Abstract Algebra, p. 147, Example 15.10. AI: You are right, the order of the quotient is 8. You want to enumerate all the cosets of $H$. Each coset is of the form $(a,b)H$ for some $(a,b) \in G:= \mathbb{Z}_4\times \mathbb{Z}_6$. The question is, when are two cosets the same? Now $$ (a,b)H = (c,d)H \Leftrightarrow (a-c, b-d) \in H \Leftrightarrow a=c, 2\mid (b-d) $$ So the distinct left cosets would correspond to the elements of the set $$ \{(0,1), (0,2), (1,1), (1,2), (2,1), (2,2), (3,1), (3,2)\} $$ Now the next question is the group structure on $G/H$ : This is given by multiplication of cosets $$ (a,b)H \cdot (c,d)H = (a+c,b+d)H $$ Consider the multiplication $$ (0,1)H \cdot (0,2)H = (0,3)H = (0,1)H $$ since $(0,3)-(0,1) = (0,2) \in H$. Similarly, $$ (1,2)H \cdot (3,2)H = (4,4)H = (0,0)H $$ since $4 \equiv 0$ in $\mathbb{Z}_4$ and $(0,4 \in H$; and so on. Try and see if you can write down the multiplication table for $G/H$. Once you have done that, explain why (i) $G/H$ is abelian (ii) $G/H$ has a cyclic subgroup $H$ of order 4 and a cyclic subgroup $K$ of order 2 which are disjoint. (iii) Define the "natural" map $f : H\times K \to G$ and show that it is an isomorphism. Conclude that $G/H \cong \mathbb{Z}_4\times \mathbb{Z}_2$
H: Probability of 4 or fewer errors in 100,000 messages The probability of an error occurring in a message is 10^-5. The probability is independent for different messages. There are 100,000 messages sent. What is the probability that 4 or fewer errors occur? AI: In principle, the number $X$ of errors in $100000$ messages has binomial distribution. But in this kind of situation (probability $p$ of an "error" small, number $n$ of trials large, $np$ of moderate size) it is standard to approximate the distribution of $X$ by using the Poisson distribution with parameter $\lambda=np$. In our case we have $\lambda=np=(10^{-5})(100000)=1$. The probability of $4$ or fewer errors is approximately $$\sum_{k=0}^4 e^{-1} \frac{1^k}{k!}.$$
H: Complete separable metric space Let $K$ be an algebraically closed field of cardinality continuum containing the rationals. Is there always a (canonical if possible) way to equipp $K$ with a topology which makes it into a separable complete metric space? AI: If $K$ is an algebraically closed field of cardinality $2^{\aleph_0}$ containing the rationals, then there is a field isomorphism between $K$ and $\Bbb C$. Since $\Bbb C$ is a separable complete metric space, we can just transport the structure to $K$ via that isomorphism.
H: Real random variable $X$ such that $ \lim_{n\to \infty}{nP(\|X\|>n)}=0$ what about $ E\|X\|<\infty$? Let $X\colon (\Omega,\mathcal F,\mathbb P)\to (\mathbb R,\mathcal B(\mathbb R))$ be a random variable from a probability space to the real numbers with the Borel sets. I proved that if $\mathbb E\|X\|<\infty$ then $ \lim_{n\to \infty}{n\mathbb P(\|X\|>n)}=0$. I want to know if the reciprocal is true or not. If not, I'd like to know if there are some extra hypothesis that can assure me that it's true. AI: The short answer is NO. There exists a probability distribution on $\mathbb{R}^+$ such that $\lim_{n \to \infty} n \mathbb{P}[X > n] = 0$ but $\mathbb{E}[X] = \infty$. (The basic observation is simply the fact that $\int_{n_0}^{\infty} \frac{1}{x \ln(x)}$ is divergent.) Define the distribution such that $\mathbb{P}[X \leq n_0] = 0$ for some $n_0 > 0$, and $\mathbb{P}[X > n] = \frac{C}{n \ln(n)}$ for all $n > n_0$. For such $\mathbb{P}$ it holds that $\mathbb{E}[X] = \int_{n_0}^{\infty} \mathbb{P}[X > n] dn = \int_{n_0}^{\infty} \frac{C}{n \ln(n)} = \ln\ln(n)\|_{n_0}^\infty = \infty$.
H: Question About Dedekind Cuts Rudin gives the definition of a Dedekind Cut to be: A set of rational numbers is said to be a cut if (I) $\alpha$ contains at least one rational, but not every rational; (II) if $p\in\alpha$ and $q<p$ (q rational), then $q\in\alpha$; (III) $\alpha$ contains no largest rational. I'm confused as to how a set of rationals can contain no largest rational, yet not contain every rational. AI: This is very possible. Consider $\{r \in \mathbb Q \mid r < 0\}$. Suppose this set contained some largest rational $p$. But $p$ is negative, so $\frac p2$ is also negative and greater than $p$ (i.e. "less negative"). This can easily be adapted into an argument that for all $q \in \mathbb Q$ the set $L_q = \{r \in \mathbb Q \mid r < q\}$ is a Dedekind cut. But it's not all of them as the standard "$\sqrt2$" example shows.
H: Why is orthogonal basis important? Lets take the $\mathbb{R}^3$ space as example. Any point in the $\mathbb{R}^3$ space can be represented by 3 linearly independent vectors that need not be orthogonal to each other. What is that special quality of orthogonal basis (extending to orthonormal) that we choose them over non-orthogonal basis? AI: If $\{v_1, v_2, v_3\}$ is a basis for $\mathbb{R}^3$, we can write any $v \in \mathbb{R}^3$ as a linear combination of $v_1, v_2,$ and $v_3$ in a unique way; that is $v = x_1v_2 + x_2v_2+x_3v_3$ where $x_1, x_2, x_3 \in \mathbb{R}$. While we know that $x_1, x_2, x_3$ are unique, we don't have a way of finding them without doing some explicit calculations. If $\{w_1, w_2, w_3\}$ is an orthonormal basis for $\mathbb{R}^3$, we can write any $v \in \mathbb{R}^3$ as $$v = (v\cdot w_1)w_1 + (v\cdot w_2)w_2 + (v\cdot w_3)w_3.$$ In this case, we have an explicit formula for the unique coefficients in the linear combination. Furthermore, the above formula is very useful when dealing with projections onto subspaces. Added Later: Note, if you have an orthogonal basis, you can divide each vector by its length and the basis becomes orthonormal. If you have a basis, and you want to turn it into an orthonormal basis, you need to use the Gram-Schmidt process (which follows from the above formula). By the way, none of this is restricted to $\mathbb{R}^3$, it works for any $\mathbb{R}^n$, you just need to have $n$ vectors in a basis. More generally still, it applies to any inner product space.
H: In topological group left and right multiplication are homeomorphic From Wikipedia (http://en.wikipedia.org/wiki/Topological_group): The inversion operation on a topological group $G$ is a homeomorphism from $G$ to itself. Likewise, if $a$ is any element of $G$, then left or right multiplication by $a$ yields a homeomorphism $G \rightarrow G$. Maybe it is trivial, but I am not able to find argments, why left and right multiplication are homeomorphic. AI: One of the axioms for a topological group says that, for every $a \in G$, the left-multiplication map $f_a : G \to G$ given by $f_a(x) = ax$ is continuous. By definition, a homeomophism of $G$ is a continuous map $f: G \to G$ with a continuous inverse. Can you give the continuos inverse for $f_a$?
H: When does the limit of a sum become an integral? In many maths and physics texts and courses, I've been told in many cases that the limit of a sum becomes an integral, i.e. (very roughly): $$\lim_{n\to\infty} \sum_{x=0}^n f(x) = \int_0^\infty f(x) dx$$ However I know that this equation can't be true in every case, otherwise series would not exist. So I'd like to ask: what characteristics does $f(x)$ have for the above equation to be true? More details: Yesterday I had to resolve a series like this: $$\lim_{n\to\infty} \sum_{x=0}^n (1 - e^{1/n}) e^{(n-x)/n}$$ This comes from a real-life problem. I decided to find the value of that limit using the following integral: $$\lim_{n\to\infty} \int_0^n (1 - e^{1/n}) e^{(n-x)/n} dx$$ The reasons why I decided to go with the integral are two: habit and because I needed the sum to be continuous (meaning: every value summed must be infinitesimal). Also, I've always been told that sums are for discrete values, integrals are for continuous/infinitesimal ones. The problem is that I do not have a mathematical justification for that. And while testing shows that the result matches reality, I don't know whether it is mathematically correct, or it is just a coincidence. So I asked myself: why should my limit be an integral and not a series? I decided to calculate first the expression of $\sum_{x=0}^n (1-e^{1/n}) e^{(n-x)/n}$ (which is $1 - e^{1/n+1}$) and afterwards the limit for $n$ to infinity ($1 - e$). The result is the same of the integral, and I'm a bit surprised by that. The reason why I'm surprised can be explained looking at the picture on this answer. My sum is not an area, however it it were, I'd know for sure that what I need is the picture on the left rather than the one on the right (being a real-life problem, I know in advance the nature of the operands I'm dealing with). What's going on here? Which of the following equations is the most correct? $$ \lim_{n\to\infty} \sum_{x=0}^n (1 - e^{1/n}) e^{(n-x)/n} = \lim_{n\to\infty} \int_0^n (1 - e^{1/n}) e^{(n-x)/n} dx \\ \lim_{n\to\infty} \sum_{x=0}^n (1 - e^{1/n}) e^{(n-x)/n} = \sum_{x=0}^n (1-e^{1/n}) e^{(n-x)/n} $$ What I'm really interested in knowing: when do I need to transform a limit of a sum into an integral? And: does it happens every time that, when a limit of a sum is equal to an integral, the corresponding series have the same result? AI: When you look up the definition of the Riemann integral (http://en.wikipedia.org/wiki/Riemann_integral#Riemann_sums), you will find that it is the limit of a sequence of sums. The same holds for the Lebesgue integral. Update Now I looked properly at the remainder of your question. If you think about your sum, you are actually computing an integral over [0,1], approximated by points $x/n$ for $x = 0, ..., n$. Also note that, for large $n$, $(1-e^{1/n}) \approx -1/n$. So let's rewrite \begin{align} \sum_{x=0}^n (1-e^{1/n}) e^{(n-x)/n} & \approx - \sum_{x=0}^n e^{1-x/n} \frac 1 n \approx - \int_0^1 e^{1-\xi} \ d \xi = 1 - e \end{align}
H: A quick question on scalars and commutators. I just had a quick question about the factoring of scalars within the commutator. Say we have operators $\hat{A}$ and $\hat{B}$, and scalars $a$ and $b$. If we take the commutator: $$[a\hat{A}, b\hat{B}]$$That is just equivalent to $ab(\hat{A}\hat{B} - \hat{B}\hat{A})$, right? Probably a dumb question, but I just wanted some clarification. AI: We have \begin{align} [\def\A{\hat A}\def\B{\hat B}a\A, b\B] &= (a\A)(b\B) - (b\B)(a\A)\\ &= ab\A\B - ba\B\A \\ &= ab(\A\B - \B\A)\\ &= ab[\A, \B] \end{align} for a commutative ring of scalars. So the answer is: Yes.
H: Prove that flow is a linear combination of flow cycles and flow paths Let $D=(N,A)$ be a directed graph, and for an arc $e=xy$ define $h(e)=x$ and $t(e)=y$. A flow is $\mathbf{x}=(x(e_1),\dots,x(e_k))$ with $\sum_{e:t(e)=v}x(e)=\sum_ {e:h(e)=v}x(e)$ for all $v\in A\backslash \{s,t\}$. A flow cycle is a flow $\mathbf{y}$ on a directed cycle $C$ given by $y(e)=\epsilon(C)$ if $e\in E(C)$ and $y(e)=0$ otherwise, for some positive number $\epsilon(C)$. An $s-t$ flow path is a flow $\mathbf{z}$ on a directed $s-t$ path $P$ with $z(e)=\epsilon(P)$ for $e\in E(P)$ and $y(e)=0$ otherwise, for some positive number $\epsilon(P)$. Show that $\mathbf{x}$ can always be written as a sum of flow paths and flow cycles. The first part of the problem was to show that a circulation, which is just a flow with no source or sink, can be written as a sum of flow cycles. I did this by way of the incidence matrix $A$ of $D$, showing that a circulation must be in the nullspace of $A$, which means we can form a basis of cycles, and since this basis is made of vectors with values either 0, -1 or 1, they are flow cycles. I can't figure out how to apply this approach in this case though since we are no longer dealing with the nullspace of the indicence matrix. Can the approach be adapted? Otherwise should I do something else? I'm really stuck here. AI: First, your notation is a bit confusing, you have multiple $t$-s (tail and target) and multiple $x$-s and $y$-s (vertices and a flow), etc. (even using different fonts would help, like \mathrm and \mathbf, e.g. see the equations below). As for the second part, the flow condition guarantees, that influx and outflux is the same for all vertices $v$ but for the source $s$ and target $t$ $$\sum_{e:\ \mathrm{t}(e)=v}\mathbf{x}(e)=\sum_{e:\ \mathrm{h}(e)=v}\mathbf{x}(e).$$ However, it also guarantees that the surplus of outflux in $s$ is the surplus of influx in $t$ (and given your definition also the other way around). $$+\sum_{e:\ \mathrm{t}(e)=s}\mathbf{x}(e)-\sum_{e:\ \mathrm{h}(e)=s}\mathbf{x}(e) = -\sum_{e:\ \mathrm{t}(e)=t}\mathbf{x}(e)+\sum_{e:\ \mathrm{h}(e)=t}\mathbf{x}(e) .$$ So you can add another vertex $m$ that would connect $t \to s$, direct appropriate flow through it, then use the previous lemma to obtain cycle decomposition, and finally remove $m$ to break the cycle going through $m$ into $s-t$ path. I hope this helps $\ddot\smile$
H: Can someone help me find the sum of the following series? I am working on one of the fractals and finding its convergent area. $$\begin{align} S & = 1+3\left(\frac{1}{9}+4(\frac{1}{9^2})+4^2(\frac{1}{9^3})+...\right)\\ & = 1+3*\sum_{i=0}^{\infty} \left[\frac{4^i}{9^{i+1}}\right] \end{align}$$ I want to say that I am looking at a geometric series, but I'm not 100% sure. AI: It is a geometric series. Pull out a factor $\frac19$ from the sum and you should see that the general term under the sum is $\frac{4^i}{9^i}$.
H: How does the law of cosines help with this problem? I've gotten stuck on this problem from Stewart Calculus, 7th edition: A runner sprints around a circular track of radius 100 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m? From the hints provided on the website, I know that the law of cosines is somehow involved in finding the solution. So far I think I've figured out that the angle of his position on the track is changing by 7/200pi radians/second, as he's moving at 7 meters and the circumference of the track is 200pi meters. I recognize that two sides of the triangle are not changing; the runner's position is a constant 100 m from the track's center and his friend's position is a constant 200 m from the track's center. I cannot figure out how to combine this information to get any closer to the answer. Help, please? Edit: I'm now not even sure I've set up this for correct application of the law of cosines. Given side lengths of 100, 200, and 200, I tried this: 200^2 = 200^2 + 100^2 - 2200100cos(x), where x is presumably the angle formed between the friend, the center of the track, and the runner. This gives 40,000 = 40,000 + 10,000 - 40000cos(x). For this to work, cos(x) would have to be 1/4, but that seems impossible. AI: Let $C$ be the distance between the runner and his friend. Let $A$ be the distance between the runner and the center of the circle. Let $B$ be the distance between the friend and the center of the circle. Now you have a triangle. Let $\theta$ be the angle opposite the side of length $C$. You can see that the change in $\theta$ is a constant. We know $C^2 = A^2 + B^2 - 2ABcos(\theta)$. So $2C\frac{dC}{dt} = 0 + 0 - 2ABcos(\theta + \frac \pi 2)\frac{d\theta}{dt}$. You are trying to solve for $\frac{dC}{dt}$, and you have already figured out $\frac{d\theta}{dt}$. Edit: oops I forgot to take the derivative of the cosine, added that in.
H: Prove that $\|\cos(\sin(x_1)) - \cos(\sin(x_2))\| \leq \|x_1 - x_2\|, \forall x_1, x_2 \in \mathbb R$. I asked this question without any limitation on methods that might be used. I believe it's turned out to be interesting to see a variety of different approaches. It turns out that the aim of the exercise that my students were given was to provide an example of the application of the Mean Value Theorem. I really like the MVT solution as it's a nice little application showing how powerful elementary calculus can be. None the less, I've enjoyed following other lines of thought. I'm ticking Christian Blatter's solution as it's a particularly nice application of precalculus knowledge. It would be a good challenge question for brighter maths students in the final year of high school after introducing the various trigonometric formulae. It's a neat application of converting a difference to a product. Finally, it's a good example of the contrast between pre and post calculus. Thanks to everyone who has contributed to this page. AI: One has $$\|\sin x\|\leq\|x\|\qquad(x\in{\mathbb R})$$ and consequently $$\|\cos\alpha-\cos\beta\|=2\left\|\sin{\alpha+\beta\over2}\right\|\ \left\|\sin{\alpha-\beta\over2}\right\|\leq \|\alpha-\beta\|\ .$$ Similarly one proves $\|\sin\alpha-\sin\beta\|\leq\|\alpha-\beta\|$. Putting it together we obtain $$\|\cos(\sin x_1)-\cos(\sin x_2)\|\leq\|\sin x_1-\sin x_2\|\leq\|x_1-x_2\|\ .$$
H: $z$-transform of $1/n$ How can one calculate the $z$-transform of: $x(n) = \frac{1}{n}$ , where $n \geq 1$? I have searched for table entries, then got stuck while trying to do it with the definition of $z$-transform (summation). AI: From what I can gather, by definition, the $z$-transform is the sum $$ \sum_{n = 1}^\infty\frac{z^{-n}}{n} $$ which just so happens to be the power series of $-\ln(1 - 1/z)$, converging as long as $\|z\| > 1$.
H: if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ $\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get. $\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$ again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get $\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$ Now I did not understand how can I solve This $8^{th}$ Degree equation, So Help Required, Thanks AI: $f(f(x)) =f^{-1}(1)$ $$f(x) = x-\frac{1}{x}$$ $$f^{-1}(x) = \frac{x+\sqrt{x^2+4}}{2}$$ or $$f^{-1}(x) = \frac{x-\sqrt{x^2+4}}{2}$$ It means we have 2 possible values of $f^{-1}(1)$ $f^{-1}(1)=\phi_1=\frac{1+\sqrt{5}}{2}$ or $f^{-1}(1)=\phi_2=\frac{1-\sqrt{5}}{2}$ For $\phi_1=\frac{1+\sqrt{5}}{2}$ $f(f(x)) =\phi_1=\frac{1+\sqrt{5}}{2}$ As you pointed in question we know that $f(f(x)) = x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} $ $x-\frac{1}{x}=P$ then $f(f(x)) = P-\cfrac{1}{P}= \phi_1$ You can solve $P$ $P^2-\phi_1P-1=0$ You will find 2 values from here and you will use in it $x-\frac{1}{x}=P_1$ and $x-\frac{1}{x}=P_2 $ then you will find 4 different x values. Then you will do the same things for For $\phi_2=\frac{1-\sqrt{5}}{2}$, $f(f(x)) = P-\cfrac{1}{P}= \phi_2$ and you will find other 4 different x values Totally 8 different x , you will find.
H: remainder on division by 9 for a tricky number What will be the remainder when $ 32^{32^{32}} $ is divided by 9? I was able to solve this by using the cyclicity of remainders when $2^{2^n}$ is divided by 9. For $n$ =even it gave remainder 7 and for $n$ =odd it gave remainder 4. So using this I worked out the above to get answer as 4. However, can there be a more concise way possible using some theorem like Fermat's or something else? AI: Euler's theorem says that any number coprime to $9$, raised to the sixth power is congruent to $1$ mod $9$. We have that $32$ is coprime to $9$, so $32^6\equiv 1 \mod 9$. Now, we need to know what the exponent $32^{32}$ is mod $6$. By the chinese remainder theorem, we can reduce to mod $2$ and mod $3$. The exponent is obviously even, so it's congruent to $0$ mod $2$. It then remains to check congruency modulo $3$. We have (working mod $3$): $$ 32^{32} \equiv 2^{32} \equiv (2^2)^{16} \equiv 1^{16} \equiv 1 $$ so that finally we get $32^{32}\equiv 4$ mod $6$. Back to the original question, this means that for some natural number $n$ $$ 32^{32^{32}} = 32^{6n + 4} \equiv 32^4 \equiv 5^4 \mod 9 $$ This is not too difficult to calculate, since $5^2 = 25 \equiv -2$ mod $9$, so $5^4 \equiv (-2)^2 \equiv 4$.
H: Invariant Subrepresentation of Induced Representation In what follows we let $G$ be a compact Lie group and $H\lneqq G$ a closed subgroup. We denote by $\mu_{G},\mu_{H}$ the respective Haar measures. Let $(\mathcal{H},\langle,\cdot\,,\cdot\rangle_{\mathcal{H}})$ be a Hilbert space. A unitary representation $\Phi$ of a group $G$ on $\mathcal{H}$ is a group homomorphsim $\Phi:G\to\mathfrak{U}(\mathcal{H})$, where $\mathfrak{U}(\mathcal{H})$ is the group of unitary operators on $\mathcal{H}$, such that for each $v\in\mathcal{H}$ the map $g\mapsto\Phi(g)v$ is continuous. In what follows, we will write $gv$ for $\Phi(g)v$, whenever $\mathcal{H}$ is a Hilbert space, $\Phi:G\to\mathfrak{U}(\mathcal{H})$ a unitary representation of $G$ over $\mathcal{H}$, $g\in G$ and $v\in\mathcal{H}$. I was trying to prove the following: Question: Let $\mathcal{H}$ be a finite-dimensional, unitary representation of $H\lneqq G$, then $\operatorname{Ind}_{H}^{G}\mathcal{H}$ contains a non-trivial, finite-dimensional, unitary subrepresentation of $G$. I will first spend a few lines on defining all the symbols used above -- just to make sure we talk about the same objects and for you to find out whether I have in fact misunderstood the basics --, then I want to explain the context in which the question popped up and finally I want to outline what I intended to do. Definition: Let $\mathcal{H}$ be a unitary representation of $G$. The left-regular representation of $G$ (over $\mathcal{H}$) is the unitary representation $G\to L^{2}(G,\mathcal{H})$ defined by: $$(gf)(x):=f(g^{-1}x)\quad\forall g,x\in G,\forall f\in\mathscr{C}(G,\mathcal{H})$$ Definition: Let $H\lneqq G$ be a closed subgroup, $\mathcal{H}$ a unitary representation of $H$. Set $$\mathcal{H}_{0}:=\{f\in\mathscr{C}(G,\mathcal{H});f(gh)=h^{-1}f(g)\forall g\in G,\forall h\in H\}$$ The induced representation $\operatorname{Ind}_{H}^{G}\mathcal{H}$ of $G$ is the closure of $\mathcal{H}_{0}\subseteq L^{2}(G,V)$ as a subrepresentation of the left-regular representation over $\mathcal{H}$. Definition: Given a representation $\mathcal{H}$ of $G$ and $H\lneqq G$ a closed subgroup, we denote by $\operatorname{Res}_{G}^{H}W$ the representation of $H$ on $W$ obtained by restricting the map $G\to\mathfrak{U}(\mathcal{H})$ to $H$. The question popped up as such a finite-dimensional, non-trivial unitary subrepresentation of the induced representation is chosen in an article. In particular, at the beginning of the proof of lemma 2.3 in Jinpeng An, Jiu-Kang Yu & Jun Yu: \emph{On the Dimension Datum of a Subgroup and its Application to Isospectral Manifolds}, Journal of Differential Geometry, \textbf{94} (2013), pp. 55-85 a non-trivial, finite-dimensional subepresentation of $\operatorname{Ind}_{H}^{H'}\mathbf{1}$ with $H\lneqq H'\leq G$ closed subgroups and $\mathbf{1}$ the trivial representation of $H$ is selected. I guessed that we could do this for any given finite-dimensional representation. 1st approach: I intended to use the Frobenius reciprocity together with the Peter-Weyl decomposition of $\operatorname{Ind}_{H}^{G}\mathcal{H}$, i.e. we note that for any representation $W$ of $G$ holds: $$\operatorname{Hom}_{H}(\operatorname{Res}_{G}^{H}W,\mathcal{H})\cong \operatorname{Hom}_{G}(W,\operatorname{Ind}_{H}^{G}\mathcal{H})$$ Now assume otherwise, i.e. all the finite dimensional subrepresentations of $\operatorname{Ind}_{H}^{G}\mathcal{H}$ are trivial, then application of Peter-Weyl implies that for all irreducible, finite-dimensional representations $W$ of $G$ holds: $$\operatorname{Hom}_{H}(\operatorname{Res}_{G}^{H}W,\mathcal{H})\cong\begin{cases} \mathbb{C}^{n} & \text{if }W\cong{\mathbf{1}_{G}}\\ \{0\} & \text{else} \end{cases}$$ where $n$ is the multiplicity of the trivial representation $\mathbf{1}_{G}$ in $\operatorname{Ind}_{H}^{G}\mathcal{H}$ (possibly infinity). I do not see how this can be pushed further to end up with a contradiction. All I get is that the components of $\mathcal{H}$ are all trivial which does not yield a contradiction by itself. 2nd approach: I intended to construct an element which spans a finite dimensional subspace which is not $G$-invariant. The idea was to use normality of Lie groups and compactness of $H$: we can construct a function $\phi:G\to\mathbb{C}$ such that $\phi\big\|_{H}\equiv 1$ and $\phi\big\|_{gH}\equiv 0$ for some $g\in G\setminus H$ and $\phi(G)\subseteq [0,1]$. Let $v\in\mathcal{H}\setminus\{0\}$, then the map $f:\hat{g}\mapsto \phi(\hat{g})v$ is continuous non-zero, hence for each $\hat{g}\in G$ we obtain a well-defined vector $\operatorname{Ind}f(\hat{g})$ by: $$\langle\operatorname{Ind}f(\hat{g}),w\rangle_{\mathcal{H}}:=\int_{H}\langle hf(\hat{g}h),w\rangle\operatorname{d}\mu_{H}(h)\quad \forall w\in\mathcal{H}$$ Using uniform continuity of $f$ it is not hard to deduce (uniform) continuity of $\operatorname{Ind} f$ and one easily sees $\operatorname{Ind} f\in\mathcal{H}_{0}$ thanks to the many invariance properties of $\mu_{H}$. The idea now was that clearly $\operatorname{Ind} f(g)=0$ and: $$\langle\operatorname{Ind} f(1),w\rangle_{\mathcal{H}}=\int_{H}\langle hv,w\rangle_{\mathcal{H}}\operatorname{d}\mu_{H}(h)\quad\forall w\in\mathcal{H}$$ So the question reduces to: Is there a way to guarantee that for some $v\in\mathcal{H}$ the average $\int_{H}hv\operatorname{d}\mu_{H}$ defined by: $$\langle\int_{H}hv\operatorname{d}\mu_{H},w\rangle_{\mathcal{H}}=\int_{H}\langle hv,w\rangle_{\mathcal{H}}\operatorname{d}\mu_{H}(h)$$ is non-trivial? Once the question is answered with yes, I can approximate the vector by matrix coefficients for irreducible representations of $G$ to obtain a non-zero vector in a finite-dimensional, non-trivial subrepresentation with non-trivial orbit under $G$. Note that with respect to the context the question popped up in, this construction indeed answers the question: if we induce the trivial representation, we integrate over $\lVert v\rVert_{\mathcal{H}}^{2}$, which is non-zero for $v\neq 0$. AI: Edit: I see now that you want to rule out that $Ind_H^G \sigma$ is a multiple of the trivial representation. If $\sigma$ is nontrivial, then you are of course done, because $Res_H Ind_H^G \sigma$ contains $\sigma$. If $\sigma$ is trivial, and $H$ is normal you have that $Ind_H^G 1$ is the regular representation of $H\backslash G$, and if $H$ is a proper subgroup, there is at least one non-trivial representation of $H\backslash G$, conseqently in $Ind_H^G 1$. The following addresses the discreteness of the induced representation, which I thought was the question at first reading. The quotient of $H\backslash G$ is compact and $\sigma$ is unitary and finite-dimensional, hence the induced representation $(\pi, V_\pi)= Ind_H^G \sigma$ decomposes into a Hilbert direct sum of irreducible representations of $G$ with finite multiplicity. One way how to prove that an induced representation decomposes discretely in the above sense is to show that $$ \pi (\phi): v \mapsto \int\limits_{G} \phi(g) \pi(g) v d g$$ is a compact operator for $\phi \in C_c^\infty(G)$. In fact, you obtain a Hilbert-Schmidt operator: Consider $V_\pi$ as the space of functions $f:G \rightarrow V_\sigma$ with $f(hg) = \sigma(h) f(g)$ and $\pi(g) f(x) = f(xg)$, then $$ \pi(\phi) : v \mapsto \int\limits_{G} \phi(x^{-1}g) f(g) d g,$$ so $\pi(\phi)$ is a kernel transformation with kernel $k(x,g) = \phi(x^{-1}g)$. The kernel is continuous, hence lies in $L^2(G \times G)$, which implies Hilbert Schmidt. Details can be found in "Principles of Harmonic Analysis" By Deitmar-Echterhoff (Chapter 9). The operator is actually trace class, but this requires the Dixmier-Malliavin theorem, which lies deeper. This strategy works, whenever $H$ is cocompact in $G$. It fails for example for $SL_n(\mathbb{Z}) \subset SL_n(\mathbb{R})$, which is a cofinite embedding for $n \geq 2$.
H: Bounding the integral of the tails of a random variable. I found an argument like this in a book, but I couldn't understand how we got this bound. Suppose $X_n$ is a sequence of random variables. For some $\delta > 0$ and all $n \geq 1$, $$ \int_{\|X_n\| \geq M} \|X_n\| \, dP \leq \frac{1}{M^\delta} E[\|X_n\|^{1 + \delta}]. $$ How can I justify this bound? AI: We don't need a sequence, only the fact that if $X$ is a non-negative random variable, then $$\int_{\{X\geqslant M\}}X\mathrm d\mathbb P\leqslant \frac 1{M^\delta}\mathbb E[X^{1+\delta}].$$ To see that, notice that $$X\cdot\chi_{\{X\geqslant M\}}\leqslant \frac{X^{1+\delta}}{M^\delta},$$ then integrate.
H: Derivation of weak form of Euler Lagrange Equation In Giaquinta's and Giusti's 1982 paper entitled "On the regularity of the minima of variational integrals", they look at the following quadratic functional: \begin{equation} F(u)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}u^i\ \mathrm{d}x \end{equation}where the coefficients are differentiable, uniformly bounded and $D_{\alpha}\equiv\frac{\partial}{\partial x_{\alpha}}$. Here, $\Omega\subset\mathbb{R}^n$ and the integrand is the mapping \begin{equation} \Omega\times \mathbb{R}^N\times\mathbb{R}^{nN}\ni(x, u(x), Du(x))\mapsto \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}u^i \end{equation} They say that the corresponding system of Euler equations (satisfied by every bounded local minimum $u$) is: \begin{equation} \int_{\Omega} \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}-\frac{1}{2}A^{\alpha\beta}_{u^i}(x, u)\frac{\partial u^h}{\partial x_{\alpha}}\frac{\partial u^h}{\partial x_{\beta}}\varphi^i\ \mathrm{d}x \end{equation} for all $\varphi\in L^{\infty}(\Omega, \mathbb{R}^N)\cap H_{0}^{1}(\Omega, \mathbb{R}^N)$. I am trying to deduce the above equation for $\varphi\in C_{c}^{\infty}(\Omega, \mathbb{R}^N)$ since I suppose it follows by approximation that it is true for $\varphi\in L^{\infty}(\Omega, \mathbb{R}^N)\cap H_{0}^{1}(\Omega, \mathbb{R}^N)$. However, I'm having difficulty in showing this. Below is my working. Let $\varphi\in C_{c}^{\infty}(\Omega)$. Define, \begin{equation} i(\tau)\equiv F(u+\tau\varphi)\quad\tau\in\mathbb{R}. \end{equation}For $\tau\neq 0$ we have \begin{align} i'(\tau)&=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u+\tau\varphi)\Big[D_{\alpha}\varphi^iD_{\beta}[u^i+\tau\varphi^i]+D_{\beta}\varphi^iD_{\alpha}[u^i+\tau\varphi^i]\Big]\\ &+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u+\tau\varphi)\varphi^h\Big[D_{\alpha}[u^i+\tau\varphi^i]D_{\beta}[u^i+\tau\varphi^i]\Big]. \end{align}Consequently, \begin{equation} i'(0)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big]+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^h\Big[D_{\alpha}u^iD_{\beta}u^i\Big]. \end{equation}After integrating by parts and equating to zero (since $i'(0)=0$), we obtain: \begin{equation} \int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}D_{\alpha}\Big[A^{\alpha\beta}(x, u)D_{\beta}u^i\Big]\varphi^i+D_{\beta}\Big[A^{\alpha\beta}(x, u)D_{\alpha}u^i\Big]\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^hD_{\alpha}u^iD_{\beta}u^i\ \mathrm{d}x. \end{equation} So this is not the system I was supposed to end up with. What am I doing wrong? AI: Why have you integrated by parts? The author of your paper didn't do that, starting from your line \begin{equation} i'(0)=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big]+\sum_{h=1}^{N}A^{\alpha\beta}_{u^h}(x, u)\varphi^h\Big[D_{\alpha}u^iD_{\beta}u^i\Big]\tag{$*$} \end{equation} we note that the first sum is symmetric in $\alpha$, $\beta$, hence $$\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)\Big[D_{\alpha}\varphi^iD_{\beta}u^i+D_{\beta}\varphi^iD_{\alpha}u^i\Big] = 2 \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i $$ Now divide by 2, rename some of the indices (exchange $h$ and $i$), and we arrive at \begin{equation} \int_{\Omega} \sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}A^{\alpha\beta}(x, u)D_{\alpha}u^iD_{\beta}\varphi^i\ \mathrm{d}x=\int_{\Omega}\sum_{i=1}^{N}\sum_{\alpha, \beta=1}^{n}\sum_{h=1}^{N}-\frac{1}{2}A^{\alpha\beta}_{u^i}(x, u)\frac{\partial u^h}{\partial x_{\alpha}}\frac{\partial u^h}{\partial x_{\beta}}\varphi^i\ \mathrm{d}x \end{equation}
H: Does $i^T=-i$ or $i^T = i$ ?(T is transpose) Assume $A$ is a skew-symmetric matrix(its eigenvalues is zero or purely imaginary). If $x$ is its eigenvector, we have $$x^TAx=\lambda\|x\|^2$$, take transpose on both sides. we have $$-x^TAx=(\lambda\|x\|^2)^T$$ If $\lambda \in \Bbb{R}$,then $\lambda = 0$. If $\lambda$ is purely imaginary, the equation hold if and only if $i^T = -i$. but $i$ is a number, its dimension is $1$. So I think its transpose is itself. what's wrong? AI: The formula $x^T A x=\lambda \|x\|^2$ holds for real matrix $A$ and real eigenvector $x$. In that case, there is no reason that $x$ is a real vector. So this formula does not hold. In the complex case, $x^* Ax=\lambda \|x\|^2$ holds (if $Ax=\lambda x$ holds,) where $\|x\|^2=x^x$ and $x^$ is conjugate transpose of the vector $x$.
H: Solve the limits by alternative method We know, that $\displaystyle\lim_{n\to ∞}$ $(1+2+3+...+n)\over n^2$ = $\displaystyle\lim_{n\to ∞}$ $1\over n^{2}$.${n(n+1)\over 2}$ = $\displaystyle\lim_{n\to ∞}$ $1\over 2$$({1}+{1\over n})$=${1\over 2}$$({1}+{1\over ∞})$=$1\over 2$ But, doing it in alternate way by the property of algebra of limits: $\displaystyle\lim_{x\to a}$ $(f±g)(x)$ = $\displaystyle\lim_{x\to a}$ $f(x)$ $±$ $\displaystyle\lim_{x\to a}$ $g(x)$ The equation goes: $\displaystyle\lim_{n\to ∞}$ $(1+2+3+...+n)\over n^2$ $\displaystyle\lim_{n\to ∞}$$1\over n^2$ + $\displaystyle\lim_{n\to ∞}$$2\over n^2$ + $\displaystyle\lim_{n\to ∞}$$3\over n^2$ +....+$\displaystyle\lim_{n\to ∞}$$n\over n^2$ =$\displaystyle\lim_{n\to ∞}$$1\over n^2$ + $\displaystyle\lim_{n\to ∞}$$2\over n^2$ + $\displaystyle\lim_{n\to ∞}$$3\over n^2$ +....+$\displaystyle\lim_{n\to ∞}$$n\over n^2$ = $0+0+0+...+0 = 0$ not $1\over 2$ [$\displaystyle\lim_{x\to ∞}$ $c\over x^N$ = $0$] Why does it contradict on solving alternately? If it is incorrect to solve it by the method? AI: Hint: The separation of limits work only for the finite sum. In other words if $\{a^i_n\}$ are convergent sequences for $i=1,..,k$ such that $k$ is finite, then you can have the following: $$ \displaystyle\lim_{n\to\infty}(a^1_n+\dots+a^k_n)=\displaystyle\lim_{n\to\infty}a^1_n+\dots+\displaystyle\lim_{n\to\infty}a^k_n $$ In your case, $\displaystyle\lim_{n\to\infty}(\frac{1}{n^2}+\dots+\frac{n}{n^2})$ is indeed sum of infinite number of sequences and you cannot separate them.
H: Asymptotic behaviour of real sequences Let's say we have two real sequences $(a_n)_{n\in\mathbb{N}}$ and $(c_n)_{n\in\mathbb{N}}$ with $c_n\in o(\frac1n)$ (i.e. $c_n(\frac1n)^{-1}\xrightarrow{n\rightarrow\infty}0$). And for all $\epsilon>0$ holds that $$\frac{1-\epsilon}n-\frac1\epsilon \|c_n\|\leq a_n\leq \frac{1+\epsilon}n+\frac1\epsilon \|c_n\|.$$Is it true that $a_n\sim \frac1n$ i.e. $a_n(\frac1n)^{-1}\xrightarrow{n\rightarrow\infty}1$? I think it should hold but I'm really confused by the $\epsilon$'s because I can't just consider the limit for $\epsilon\rightarrow0$. Thanks and I'm sorry if it's a really stupid question! AI: Fix $\epsilon\gt0$. Then $n\|c_n\|\to0$ and, for every $n$, $$ 1-\epsilon-\epsilon^{-1}n\|c_n\|\leqslant na_n\leqslant1+\epsilon+\epsilon^{-1}n\|c_n\|, $$ hence $$ 1-\epsilon\leqslant\liminf_{n\to\infty}\,na_n\leqslant\limsup_{n\to\infty}\,na_n\leqslant1+\epsilon. $$ This holds for every $\epsilon\gt0$ hence $$ 1=\liminf_{n\to\infty}\,na_n\leqslant\limsup_{n\to\infty}\,na_n=1, $$ which implies $$ \liminf_{n\to\infty}\,na_n=\limsup_{n\to\infty}\,na_n=1, $$ that is $$ \lim_{n\to\infty}na_n=1. $$
H: Given $n \in \mathbb{N}$ prove that a polynomial result gives a natural number. A friend asked me this question: Prove that for every $n\in \Bbb N$ the next equation result: $\dfrac{n^3}{6}+\dfrac{n^2} {2}+\dfrac{n}{3}$ would be a natural number. My instincts were that i need to use induction, Though i'm not sure how. So i've started with this: $$\frac{n^3+3n^2+2n}{6}\to\frac{n(n^2+3n+2)}{6}\to\frac{n(n+1)(n+2)}{6}$$ As i see it, I need to prove that $n(n+1)(n+2)$ divides $6$, with no reminder, Or even just prove that every $3$ consecutive numbers divides $6$ without reminder. But, I can't seem to understand how to do it. I guess i'm missing something. AI: Hints: (1) By uniqueness of prime decomposition, to be a multiple of 6 it is necessary and sufficient to be a multiple of 2 and 3. (2) Can you show that for any natural $n$ either $n$ or $n+1$ is even, that is a multiple of 2? (3) For 3, argue as in (2) for 2, but now using $n+2$ also.
H: How do I prove that $\lim_{x\to0^+} x^{1/x}=0$ How do I prove that $\lim_{x\to 0^+} x^{1/x}=0$ It looks like on of those situations in which l'hospital will come in handy but that doesn't seem to work as well. AI: Hint: Try to find the limit of logarithm of the function and use $\ln (x^{1/x})=\frac{\ln(x)}{x}$. Then you get $\ln L=\lim_{x\to0^+}\frac{\ln(x)}{x}$ where $L$ is the limit.
H: Video lectures on Group Theory The web is full of video lectures these days but, try as I might, I can find very little for Introduction to Group Theory. The closest I found was http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra . Are they any online introductory group theory lectures people would recommend? AI: http://www.youtube.com/user/LadislauFernandes Has uploaded quite a few videos on introductory group theory. I have not watched that much of any of them though, so I can't say much about the quality (the bit I have watched seemed ok. A bit slow, but that might just be because this is not new stuff to me).
H: Closed form for $\int \frac{1}{x^7 -1} dx$? I want to calculate: $$\int \frac{1}{x^7 -1} dx$$ Since $\displaystyle \frac{1}{x^7 -1} = - \sum_{i=0}^\infty x^{7i} $, we have $\displaystyle(-x)\sum_{i=0}^\infty \frac{x^{7i}}{7i +1} $. Is there another solution? That is, can this integral be written in terms of elementary functions? AI: Factorise $x^{7}-1$ and use partial fractions - I do not expect your answer to be nice. You should get 3 unfactorable quadratics which will contribute $\tan^{-1}$ terms, and one factor $x-1$ which will give some multiple of $\ln(x-1)$. To perform the actual factorisation, note that the roots of $x^{7}=1$ are $e^{\frac{2ni\pi}{7}}=\cos(2n\pi/7)+i\sin(2n\pi/7)$ for, say, $n\in\{-3,-2,-1,0,1,2,3\}$. Thus we can factorise as follows: $$x^{7}-1=(x-e^{-6i\pi/7})(x-e^{-4i\pi/7})(x-e^{-2i\pi/7})(x-1)(x-e^{2i\pi/7})(x-e^{4i\pi/7})(x-e^{6i\pi/7})$$ Now note that $(x-e^{i\theta})(x-e^{-i\theta})=x^{2}-(e^{i\theta}+e^{-i\theta})x+1=x^{2}-2x\cos\theta+1$ is a real-valued quadratic, and hence if we group the linear complex factors into pairs like this, we can get 3 real quadratics as we wanted. The factorisation becomes $$x^{7}-1=(x^{2}-2c_{1}x+1)(x^{2}-2c_{2}x+1)(x^{2}-2c_{3}x+1)(x-1)$$ where $c_{k}=\cos(2k\pi/7)$. Now try partial fractions - you need to find constants $A_{1}\ldots A_{7}$ such that $$\frac{1}{x^{7}-1}=\frac{A_{1}x+A_{2}}{x^{2}-2c_{1}x+1}+\frac{A_{3}x+A_{4}}{x^{2}-2c_{2}x+1}+\frac{A_{5}x+A_{6}}{x^{2}-2c_{3}x+1}+\frac{A_{7}}{x-1}$$ By multiplying by each factor, you can get a set of seven linear equations for your seven constants, which you can solve by any method you choose. This will be very unpleasant. Integrating the last term is trivial (simply $A_{7}\ln(x-1)$) and the quadratic terms are less so. Taking the first term as an example, note $$x^{2}-2c_{1}x+1=(x-c_{1})^{2}+(1-c_{1}^{2})=(1-c_{1}^{2})\left[\left(\frac{x-c_{1}}{\sqrt{1-c_{1}^{2}}}\right)^{2}+1\right]$$ So, making the substution $u=\frac{x-c_{1}}{\sqrt{1-c_{1}^{2}}}$ will give you some terms of the form $\frac{2au+b}{u^{2}+1}$ which trivially integrates to $a\ln(u^{2}+1)+b\tan^{-1}(u)$. The arithemtic of finding these constants is left as an exercise for you.
H: How to prove that $x/y$ is continuous in R $f:R^2$ \{y=0} $\Rightarrow R$ , $f:(x,y)\Rightarrow x/y$. Prove (formally) that $f$ is continuous. I think what I should show is that any point that belongs to an open ball of radius $\epsilon$ of image, has a pre-image that belongs to an open ball around (x,y), and since image and pre-image are both open, then $f$ is continuous. But this doesn't seem correct to me. Any help is appreciated. AI: Yes, to show that such a function $f$ is continuous you want to show that the preimage of an open ball of radius $\epsilon$ contains an open ball around $(x,y)$. Here's why you should believe this is true. If $(x,y)$ is close to $(x',y')$, say $$\sqrt{(x' - x)^2 + (y'-y)^2} <\delta$$ then we know that $\|x-x'\|<\delta$ and $\|y-y'\| < \delta$ for some small $\delta>0$. Try and compute $$\left\| \frac{x}{y} - \frac{x + \eta}{y +\rho} \right\|$$ For $\|\eta\|,\|\rho\| < \|\delta\|$. Can you show that this can be made less than $\epsilon$ by making $\delta$ small enough?
H: Is $\Gamma L(r,q^t)$ a subgroup of $\Gamma L(rt,q)$? I know that $GL(r,q^t)$ can be seen as a subgroup of $GL(rt,q)$ since every linear transformation on the vector space $V(r,q^t)$ extends in a unique way to a linear transformation on the vector space $V(rt,q)$. For $\Gamma L(r,q^t)$ however, I run into trouble if I want to embed the group in $\Gamma L(rt,q)$ in a direct way because several semilinear mappings on $V(r,q^t)$ correspond to the same semilinear mapping on $V(rt,q)$, several elements of $Aut(F_{q^t})$ have the same restriction to $F_q$. Is there a way of embedding $\Gamma L(r,q^t)$ in $\Gamma L(rt,q)$? Or is this just not (always) possible? AI: The answer must be yes, although I can only show this indirectly, and I cannot see how to construct a canonical embedding - perhaps there isn't one! If $q=p^e$ with $p$ prime, then we have embeddings $A \to B \to C$ with $A={\rm GL}(r,q^t)$, $B={\rm GL}(rt,q)$, $C= {\rm GL}(rte,p)$, which are essentially unique (i.e. images of any two such embeddings are conjugate). The normalizers in $C$ of $A$ and $B$ are respectively ${\rm \Gamma L}(r,q^t)$ and ${\rm \Gamma L}(rt,q)$, whereas the normalizer of $A$ in $B$ is the extension of $A$ by a field automorphism of order $t$, induced by the map $x \mapsto x^q$ of the field of order $q^t$. By the uniqueness of the embedding $A \to B$, we have $\|N_{N_C(B)}(A)\| = \|N_B(A)\| \times \|N_C(B):B\|$, which is equal to $\|N_C(A)\|$, so $N_C(A) < N_C(B)$, which is the required embedding.
H: Map from $\mathbb {R}$ to $\mathbb {R^2}$ Is there a way to construct an injective function that map from $\mathbb{R}$ to $\mathbb{R^2}$? If yes, please give me an example. Thank you! AI: Injective means that for $(x,y) \in \mathbb R^2$ and $a,b\in\mathbb R$ we have $f(a)=(x,y) = f(b) \Rightarrow a=b$ thus all we need is every $\mathbb R$ to go to a different point in $\mathbb R^2$. Any line through $\mathbb R^2$ will satisfy this (amongst others) Thus $$f(a) = \left(2a-0.5,1.7-a\right)$$ is injective, for example.
H: Singular values in SVD I have recently started reading about SVD. If factorization of a matrix $A$ is required, we calculate the eigenvectors of $AA^T$ and $A^TA$ and they become the column vectors of $U$ and $V$ matrices correspondingly. The $\Sigma$ matrix is filled with square roots of the eigenvalues. Now we find that the eigenvalues of $AA^T$ and $A^TA$ are the same. We also find that the obtained eigenvectors are orthogonal to each other. Question 1: Why are the eigenvalues of $AA^T$ and $A^TA$ the same? Question 2: Are eigenvectors for a matrix always orthogonal? If yes, why? If not, why are they orthogonal here? PS - I'm new to Lin-Al, so it would be really helpful if the explanations are intuitive. AI: (1) Let $A$, $B$ any two matrices, and $\lambda \ne 0$ an eigenvalue of $AB$, with eigenvector $x$, say. Then $ABx = \lambda x$, and $Bx \ne 0$ (as $\lambda \ne 0$). We have $BA(Bx) = B(ABx) = \lambda Bx$, so $Bx$ is an eigenvector for $BA$ with eigenvalue $\lambda$, that is, if we denote by $\sigma(AB)$ the set of eigenvalues of $AB$: $$ \sigma(AB) -\{0\} \subseteq \sigma(BA) $$ and by symmetry of our argument $$ \sigma(AB) - \{0\} = \sigma(BA) - \{0\} $$ Note that for $0$ this need not hold: For $A = B^t = (0 \, 1)$, we have $AB = 1$, so $AB$ does not have $0$ as an eigenvalue, but $BA = \binom{0\,0}{0\, 1}$ has. So in your case, the non-zero eigenvalues of $AA^t$ and $A^t A$ are the same. (2) No, eigenvectors of general matrices needn't be orthogonal, but eigenvectors to different eigenvalues of real, symmetric matrices must. So let $B = B^t$ be a real, symmetric matrix, $Bx = \lambda x$, $By = \mu y$, with $\lambda, \mu \in \mathbb R$ eigenvalues and $x,y \ne 0$ eigenvectors of $B$. Then we have \begin{align} \def\sp#1{\left<#1\right>}\lambda\sp{x,y} &= \sp{\lambda x,y} \\ &= \sp{Bx,y}\\ &= \sp{x,B^ty}\\ &= \sp{x,By}\\ &= \mu\sp{x,y}\\ \iff (\lambda-\mu)\sp{x,y} &= 0 \end{align} As $\lambda \ne \mu$, we have $\sp{x,y} = 0$.
H: Analytical solution to Poisson’s equation 1D I need some help in finding u(x) analytically where equation and the boundary conditions are satisfied AI: Anything wrong with integrating twice? $$u(x) = -\frac{x^2}{2} + A x + B$$ $$u(0) = 1 \implies B = 1$$ $$u(1) = 2 \implies -\frac12 + A + B = 2 \implies A=\frac{3}{2}$$
H: Why “Syracuse” in “Syracuse problem” Is “Syracuse” in “Syracuse problem” (a variant name of Collatz conjecture) a reference to the city of Syracuse in Sicily, to one of several Syracuses in USA or something else (a person's name, for instance)? AI: "The name Syracuse problem was proposed by Hasse during a visit to Syracuse University [in Syracuse, New York] in the 1950s" according to Jeff Lagarias, on page 32 of his book about the problem.
H: prove uniqueness of a measure If $(X,\mathcal{M},\mu)$ is a measure space and $\mathcal{\overline{M}}:=\{E\cup F:E\in\mathcal{M}\text{ and }F\subset N\text{ for some }N\in \mathcal{N}\}$ is a completion of $\mathcal{M}$ with respect to $\mu$ where $\mathcal{N}:=\{N\in\mathcal{M}:\mu(N)=0\},$ then $\mathcal{\overline{M}}$ is a $\sigma$-algebra. If now $\overline{\mu}(E\cup F):=\mu(E)$ then $\overline{\mu}$ is a complete measure. Need to prove the uniqueness of the measure, but I am not following the proof 100%. Suppose $\mu^{\prime}$ is another measure on $\mathcal{\overline{M}}$ that extends $\mu$ to a complete measure. So I guess we need to show that $\mu^{\prime}=\overline{\mu}$ let $E\cup F\in\overline{\mathcal{M}}$ with $E\in\mathcal{M}$ and $F\subset N\in\mathcal{N}$ we have $\mu^{\prime}(E)\le \mu^{\prime}(E\cup F)\le \mu^{\prime}(E\cup N)=\mu(E\cup N)\le \mu(E)+\mu(N)=\mu(E)=\mu^{\prime}(E)$ (since $E\in\mathcal{M}$) so $\mu^{\prime}(E\cup F)=\mu(E)=\overline{\mu}(E\cup F)$ hence $\mu^{\prime}=\mu$ It is the line with the inequalities that I dont' get. 1)$\mu^{\prime}(E)\le \mu^{\prime}(E\cup F)$ this I think follows from $E\subset E\cup F$ 2)$\mu^{\prime}(E\cup F)\le \mu^{\prime}(E\cup N)$ as $E\cup F\subset E\cup N$ 3) $\mu^{\prime}(E\cup N)=\mu(E\cup N)$ where does this one come from? 4) $\mu(E\cup N)\le \mu(E)+\mu(N)$ subaddtivity of the measure 5)$\mu(E)=\mu^{\prime}(E)$ since $N$ is a null-set but then the equality? AI: (1) and (2) exactly, this follows from monotonicity of $\mu'$, for (3) note, that $\mu'$ extends $\mu$, and both $E, N \in \mathcal M$, so from $\mu'\|_{\mathcal M} = \mu$ and $E \cup N \in \mathcal M$ we have $$ \mu'(E \cup N) = \mu(E \cup N). $$
H: Irrationality of a unique positive root of $\sin{x} = x^2$ The equation $\sin{x} = x^2$ has a unique positive real root. I wonder if there is any standard technique how to show that this number is irrational (rational), preferably a technique which works also in other similar scenarios. I tried an inverse symbolic calculator on the numerical approximation of the root, but it didn´t find anything whatsoever. AI: In fact $x$ is irrational. See Corollary 2.7 in Niven's book. If the root $x$ in question was rational, then $\sin(x)$ is irrational and so $\sin(x)-x^2$ could not be zero (since this would be an irrational minus a rational).

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.