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H: Group structure from involutions, exercise devised by Richard Brauer. The following is an idea originally communicated by Richard Brauer. I'm having difficulty following some of the combinatorial elements. Let $G$ be a finite group containing exactly two conjugacy classes of involutions. Let $u_1$ and $u_2$ be nonconjugate involutions in $G$. Let $c_i=\|C(u_i)\|$, the respective centralizers of the $u_i$. Let $S_i$ be the set of ordered pairs $(x,y)$ with $x$ conjugate to $u_1$, $y$ conjugate to $u_2$, and $(xy)^n=u_i$ for some $n$. Let $s_i=\|S_i\|$. Then $\|G\|=c_1s_2+c_2s_1$. The idea is to count the ordered pairs $(x,y)$ with $x$ conjugate to $u_1$ and $y$ conjugate to $u_2$ in two different ways. Is it somehow clear that this number is $(\|G\|/c_1)(\|G\|/c_2)$ right off the bat? I believe this follows since the possible choices of the $x$ is the cardinality of the orbit of $u_1$ under the conjugation action, which equals the index of the stabilizer in $G$, but the stabilizer is just the centralizer in this case. Likewise for the choices of $y$. On the other hand, I know that if the product $uv$ of two involutions $u$ and $v$ has odd order, then $u$ and $v$ are conjugate. So if $x$ and $y$ are not conjugate here, their product $xy$ has even order. Then $(xy)^n$ for $n=o(xy)/2$ is an involution, and thus conjugate to either $u_1$ or $u_2$. This exhausts all the pairs in question, but why does counting in this way give $(\|G\|/c_1)s_1+(\|G\|/c_2)s_2$? I only know that $(xy)^n$ is conjugate to some $u_i$ for some $n$, not that $(xy)^n=u_i$ as in the definition of the $s_i$. After that, the result will follow from the equality $$ (\|G\|/c_1)(\|G\|/c_2)=(\|G\|/c_1)s_1+(\|G\|/c_2)s_2. $$ Thanks. AI: $G$ acts on the conjugacy class of $u_i$ by conjugation, so the number of elements conjugate to $u_i$ is the index of the stabilizer of $u_i$ in $G$. The stabilizer is $C(u_i)$, so the number of involutions conjugate to $u_i$ is $\|G\|/c_i$. For your second question: Let $S$ be the set of pairs $(x,y)$ such that $x$ is conjugate to $u_1$ and $y$ is conjugate to $u_2$. Let $T_i$ be the set of pairs $(x,y) \in S$ such that $(xy)^n$ (for $n=o(xy)/2$) is conjugate to $u_i$. You want to prove that $\|T_i\| = \|S_i\| (\|G\|/c_i)$. For each involution $u$, let $S_u$ be the set of pairs $(x,y) \in S$ such that $(xy)^n$ (for $n=o(xy)/2$) equals $u$. We then have $\|T_i\| = \sum \|S_u\|$, where the sum takes place over the conjugacy class of $u_i$. But the size of $S_u$ is independent of the choice of $u$ within the conjugacy class of $u_i$, because if $u^g = v$ then $(x,y) \mapsto (x^g, y^g)$ is a bijection between $S_u$ and $S_v$. Hence $\|S_u\| = \|S_i\|$ for all $u$ in the conjugacy class of $u_i$. There are $\|G\|/c_i$ conjugates of $u_i$, so $\|T_i\| = \|S_i\| (\|G\|/c_i)$.
H: Recurrence relation telescoping Hi there I am trying to solve the following recurrence relation using telescoping. How would I go about doing it? $$T(n) = \frac 2n \Big(T(0) + T(1) + \ldots+ T(n-1)\Big) + 5n$$ Assuming $n\ge 1$ AI: Rearrange $$T(n) = \frac 2n \Big(T(0) + T(1) + \ldots+ T(n-1)\Big) + 5n\tag{1}$$ to get $$T(0)+T(1)+\ldots+T(n-1)=\frac12\Big(nT(n)-5n^2\Big)\;,$$ and hence $$T(0)+T(1)+\ldots+T(n-2)=\frac12\Big((n-1)T(n-1)-5(n-1)^2\Big)\;.$$ Then substitute this into $(1)$: $$\begin{align} T(n)&=\frac2n\left(\frac12\left((n-1)T(n-1)-5(n-1)^2\right)+T(n-1)\right)+5n\\ &=\frac{n-1}nT(n-1)-\frac{5(n-1)^2}n+\frac2nT(n-1)+5n\\ &=\frac{n+1}nT(n-1)-\frac5n(n^2-2n+1)+5n\\ &=\frac{n+1}nT(n-1)+10-\frac5n\;.\tag{2} \end{align}$$ Now let $a=T(0)$, and calculate a few values of $T$: $$\begin{array}{c\|l} n&T(n)\\ \hline 0&a\\ 1&2a+5\\ 2&3a+15\\ 3&4a+\frac{85}3\\ 4&5a+\frac{265}6\\ 5&6a+62 \end{array}$$ This suggests that $T(n)=c_na+b_n$, where $c_n$ is probably $n+1$. Indeed, if $$T(n-1)=na+b_{n-1}\;,$$ then from $(2)$ we find that $$\begin{align} T(n)&=(n+1)a+\left(1+\frac1n\right)b_{n-1}+10-\frac5n\\ &=(n+1)a+b_{n-1}+10+\frac1n(b_{n-1}-5)\;, \end{align}$$ confirming that $c_n=n+1$. Thus, the problem reduces to solving the recurrence $$b_n=b_{n-1}+10+\frac1n(b_{n-1}-5)$$ with initial condition $b_0=0$. It appears to be convenient to let $u_n=b_n-5$, so that $u_0=-5$, and $$u_n=u_{n-1}+10+\frac1nu_{n-1}=\frac{n+1}nu_{n-1}+10\;.$$ Then $$\begin{align} u_n&=\frac{n+1}nu_{n-1}+10\\ &=\frac{n+1}n\left(\frac{n}{n-1}u_{n-2}+10\right)+10\\ &=\frac{n+1}{n-1}u_{n-2}+10\left(1+\frac{n+1}n\right)\\ &=\frac{n+1}{n-1}\left(\frac{n-1}{n-2}u_{n-3}+10\right)+10\left(1+\frac{n+1}n\right)\\ &=\frac{n+1}{n-2}u_{n-3}+10\left(1+\frac{n+1}n+\frac{n+1}{n-1}\right)\\ &\qquad\qquad\qquad\vdots\\ &=(n+1)u_0+10\left(\frac{n+1}{n+1}+\frac{n+1}n+\frac{n+1}{n-1}+\ldots+\frac{n+1}2\right)\\ &=-5(n+1)+10(n+1)(H_{n+1}-1)\\ &=-5-5n+10(n+1)(H_{n+1}-1)\;, \end{align}$$ where $H_n$ is the $n$-th harmonic number. (Properly speaking, this should now be proved by induction on $n$.) Finally, then, $b_n=u_n+5=10(n+1)(H_{n+1}-1)-5n$, and $$\begin{align} T(n)&=(n+1)\Big(T(0)+10(H_{n+1}-1)\Big)-5n\\ &=(n+1)\Big(T(0)+10H_{n+1}\Big)-10(n+1)-5n\\ &=(n+1)\Big(T(0)+10H_{n+1}\Big)-15n-10\;. \end{align}$$
H: Cofinal subset of ordinals Let $\alpha$ be an ordinal and suppose this ordinal has a cofinal subset. Let the cofinality of $\alpha$ be $\beta$. Thus $\beta$ is isomorphic with $B$, cofinal subset of $\alpha$. Then $\beta$ is a cofinal subset of itself. Now, let the cofinality of $\beta$ be $\gamma$. Here, how to prove that $\alpha$ has a cofinal subset similar to $\gamma$? I see this is completely different from proving that cofinal subset of cofinal subset of $A$ is a cofinal subset of $A$, since $\beta$ described above is not a cofinal subset of $\alpha$. (Actually if $\alpha ≠ \beta$, then $\beta$ is not a cofinal subset of $\alpha$.) AI: Since the cofinality of $\alpha$ is $\beta$, there is an increasing $\beta$ sequence $(a_{\eta})_{\eta < \beta}$ where each $a_\eta < \alpha$ and $\alpha = \lim_{\eta < \beta} a_\eta$. Since the cofinality of $\beta$ is $\gamma$, there is a increasing $\gamma$-sequence $(b_\zeta)_{\zeta < \gamma}$ such that each $b_\zeta < \beta$ and $\lim_{\zeta < \gamma} b_\zeta = \beta$. Then $a_{b_{\zeta}}$ is a $\gamma$ sequence of elements in $\alpha$ such that $\lim_{\zeta < \gamma} a_{b_{\zeta}} = \alpha$.
H: Journals of math history? In a related question to this one, in what journals do math historians publish their article in? Brian M. Scott provided a link to Judy Grabiner's, who is a math historian, home page and it seems that she publishes in general mathematical journals for the most part, such as the American Mathematical Monthly. However there is mention of a journal that seems to be exclusively for math history called "Historia Mathematica". Are there any other journals exclusively for math history? AI: BSHM Bulletin. Journal of the British Society for the History of Mathematics. Taylor & Francis, 4 Park Square, Milton Park, Abingdon 0X14 4RN England. ISSN: 1749-8430. E-ISSN: 1749-8341. 3 issues/vol./yr. Revista Brasileira de História da Matemática. An International Journal on the History of Mathematics, Sociedade Brasileira de História da Matemática Caixa postal 1631 CEP 59.078-970 Campus Universitario Natal RN Brazil. ISSN: 1519-955X. Gaṇita Bhāratī. Bulletin of the Indian Society for the History of Mathematics. Formerly Gaṇita-Bhāratī. MD Publ. Pvt., Ltd. "MD House'' 11 Darya Ganj, New Delhi 110 002 India. ISSN: 0970-0307. 2 issues/vol./yr.
H: combinations and probablity A committee of 5 members is selected randomly from 10 parents, 16 students and 4 teachers. how can one find the probability that a teacher will be the chairman if the first person selected is to be made chairman. AI: For a teacher to be Chair, we need to select a teacher on the first pick. How many people are there? How many teachers? (The size of the committee is irrelevant.)
H: Where is $f(x)={(x^2+2x-48)}/{x^2}$ increasing? decreasing? The question is to find where the graph is increasing or decreasing. The original function is $f(x)={(x^2+2x-48)}/{x^2}$ I know I need to find the prime of this function and I think it is this after using the quotient rule: $2(-x^2-x+49)/x^3$ Finally, in order to draw a graph I need to find the points on the x axis that are either undefined and/or the slope equals 0. Immediately I know that 0 is an undefined point because it makes the prime graph undefined. It's when I take the numerator and set it equal to 0 that trips me up. $2(-x^2-x+49)=0$ $-x^2-x+49=0$ $(x)(x)=0$ I'm guessing I did something wrong finding this prime because the answers are all including the number 48. I'm stumped because I've been at this question for about 1 hour. Thanks! AI: You computation of derivative is incorrect. It is easier to separate out each term and compute the derivative. $$f(x) = \dfrac{x^2 + 2x - 48}{x^2} = 1 + \dfrac2{x} - \dfrac{48}{x^2}$$ Hence, $$f'(x) = 0 - \dfrac2{x^2} + \dfrac{2 \times 48}{x^3} = \dfrac{96-2x}{x^3}$$ Now setting, $f'(x) = 0$ gives us $96-2x = 0 \implies x = 48$. EDIT As FrankScience rightly points out in his comment, you do not really need to differentiate to figure out the behavior of this function. We have that \begin{align} f(x) & = \dfrac{x^2 + 2x - 48}{x^2}\\ & = -48 \left( \left(\dfrac1x \right)^2 - \dfrac1{24} \dfrac1x\right) + 1\\ & = -48 \left( \left(\dfrac1x \right)^2 - 2 \dfrac1x\dfrac1{48} + \left(\dfrac1{48} \right)^2 - \left(\dfrac1{48} \right)^2\right) + 1\\ & = -48 \left( \dfrac1x - \dfrac1{48}\right)^2 + \dfrac1{48} + 1\\ & = \dfrac{49}{48} -48 \left( \dfrac1x - \dfrac1{48}\right)^2 \end{align} Hence, note that the functions hits the maximum at $x=48$, since for all $x \neq 48$, you always subtract a positive quantity from $\dfrac{49}{48}$. Analyze what happens when $x \to \infty$, $x \to - \infty$, $x \to 0^{\pm}$, $x \to 48^{\pm}$ to draw conclusions.
H: Clarification on bounded convergence theorem. For the proof of Bounded Convergence Theorem, I see how to get most all the information, but I don't see exactly why $f$ is measurable. I assume I am missing something completely obvious. Here is the theorem as I understand it so far. Theorem: Let $\{ f_n \}$ be a sequence of functions bounded by $M$, supported on a set $E$ of finite measure, and $f_n(x) \to f(x)$ a.e. $x$. Then $f$ is bounded measurable function supported on $E$ for a.e. $x$ and $\int \|f_n(x) - f(x)\|dx \to 0$ as $n \to 0$. Proof: Almost everywhere we have, $$\|f(x)\| - \|f_n(x)\| \le \|f(x) - f_n(x)\| < \epsilon.$$ Therefore, $\|f(x)\| \le M$ almost everywhere, and $f(x) = 0$ for $x \in E$ almost everywhere. Let $\epsilon > 0$. By Egorov's theorem we can find a measurable set $A \subset E$ such that $f_n \to f$ uniformly on $A$ and $m(E - A) < \epsilon$. Thus $\|f_n(x) - f(x)\| < \epsilon$ for large $n$ for all $x \in A$. $$\int \|f_n(x) - f(x)\|dx \le \int_A \|f_n(x) - f(x)\| dx + \int_{E - A} \|f_n(x) - f(x)\| dx \le \epsilon m(A) + 2M \epsilon.$$ As $\epsilon$ is arbitrary, this completes the proof. Is measurability something that I just get automatically? AI: For any sequence of measurable functions $\{f_n\}$, both their lim sup and lim inf as $n\rightarrow \infty$ are measurable. Because the limit function $f$ here is equal to the lim sup (actually both) it is measurable. Proof: I will use the fact that a function is measurable if the subset $$f^{-1}((a, \infty])$$ is measurable. First we prove that $\sup_n$ $f_n$ is measurable. Note that $\{\sup f_n>a\}=\cup_n\{f_n>a\}$, and as a countable union of measurable sets, it is measurable. As $\inf_n f_n$ is $-\sup (-f_n(x))$, we get the same result for $\inf f_n$. Now note that $\lim$ $\sup$ as $n\rightarrow \infty$ of $f_n(x)$ is just $$\inf_k\{\sup_{n\ge k} f_n\}$$ and you're done.
H: Angle of a triangle inscribed in a square Say we have a square $ABCD$. Put points $E$ and $F$ on sides $AB$ and $BC$ respectively, so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. What is $\angle DNF$? I'm inclined to say that it's a right angle because that's what it looks like from what I've drawn, but I have no idea how to proceed. AI: $\angle DNF = 90^\circ \Longleftrightarrow \angle BNF = \angle CND$. It suffics to prove that $\triangle BNF \sim \triangle CND$. Well, it's trivial: $\angle NBF = \angle BEC = \angle NCD$ and $\displaystyle \frac{NB}{BF} = \frac{NB}{BE} = \sin\angle NEB = \cos\angle NCB = \frac{NC}{CB} = \frac{NC}{CD}$. Q.E.D
H: probability of a horse winning a race. Lets suppose ten horses are participating in a race and each horse has equal chance of winning the race. I am required to find the following: (a) the probability that horse A wins the race followed by horse B. (b) the probability that horse C becomes either first or second in the race. I know there are $10 \cdot 9 \cdot 8 $ ways of having first, second or third. Since each horse has an equal chance of winning, each has probability of 1/10. Would I be right in saying that the probability that A wins followed by B is $\frac{1}{10} \cdot \frac{1}{10} $? Is it okay if I do this for (b)? $\frac{1}{10} +\frac{1}{10} $? AI: You can double-check yourself by counting outcomes. There are $10!$ possible finish orders, and it’s implied that they’re all equally likely. If $A$ wins and $B$ finishes second, there are $8!$ equally likely possible orders for the remaining $8$ horses. Thus, there are $8!$ outcomes in which $A$ wins and $B$ comes second out of a total of $10!$ possible outcomes, for a probability of $$\frac{8!}{10!}=\frac{8!}{10\cdot9\cdot8!}=\frac1{10\cdot9}=\frac1{90}\;,$$ just as Andrew Salmon argued directly from the probabilities. For the second question, if $C$ finishes first, there are $9!$ possible finish orders for the other $9$ horses. The same is true if $C$ finishes second. Since $C$ can’t finish both first and second, these $9!+9!$ outcomes are all distinct, and the probability that $C$ finishes first or second is therefore $$\frac{9!+9!}{10!}=\frac{2\cdot9!}{10\cdot9!}=\frac2{10}=\frac15\;,$$ again just as Andrew argued directly from the probabilities.
H: Multiple roots of a polynomial over a field of characteristic $p$ I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$ have a root of multiplicity $>1$ over the field of characteristic $p$. $ p=2, 3, 5, 7 $ Please help. For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ and $f'(x) = x^3+1$. Hence, $f^′(x)= 0$ for $x=2$. Therefore in an algebraically closed field of characteristic $3$, $f(x)$ has multiple roots. AI: The hint is to use the following result : Let $f(x) \in K[x]$ where $K$ is some field. $\alpha$ is a multiple root of $f(x)$ if and only if $\alpha$ is a root of $f'(x)$, the formal derivative of $f(x)$. In the algebraic closure, $\bar{F}$ of a field of characteristic $2$, $f(x) = x^4 + x = x(x^3 + 1)$ and $f'(x) = 1$. $f(x)$ has four roots in $\bar{F}$, counting multiplicity. However, $f'(x) = 1$ which has no roots. Therefore in an algebraically closed field of characteric 2, $f(x)$ does not have multiple roots. Do something similar for the others.
H: a sequence question, 4 options $x_n$, $y_n$ be two sequence satisfying $x_n\le y_n\le x_{n+2}$, then which are correct below? $y_n$ is increasing $x_n$ and $y_n$ converge together. $x_n$ decreasing $y_n$ bounded. Well, I can see $x_1\le y_1\le x_3$, $x_2\le y_2\le x_4\dots$ by putting $n=1,2..$ but how to compare seperately oftheir individula terms? AI: From $x_n\le y_n\le x_{n+2}$ for all $n\in\Bbb N$ you can infer that $$x_0\le y_0\le x_2\le y_2\le x_4\le y_4\le\ldots\tag{1}$$ and $$x_1\le y_1\le x_3\le y_3\le x_5\le y_5\le\ldots\;,\tag{2}$$ but there’s nothing to tie the odd and even subsequences together. In particular, it does not follow that $\langle y_n:n\in\Bbb N\rangle$ is increasing: it might be, for instance, that $$x_n=y_n=\begin{cases} n,&\text{if }n\text{ is even}\\ 1-\frac1n,&\text{if }n\text{ is odd}\;. \end{cases}$$ Then the odd and even subsequences are both increasing, but $\langle y_n:n\in\Bbb N\rangle$ is bouncing up and down like crazy. This example also shows that $\langle y_n:n\in\Bbb N\rangle$ need not be bounded (though it can be), and that $\langle x_n:n\in \Bbb N\rangle$ need not be either increasing or decreasing. I’ll just give you a hint for the remaining one: you can use $(1)$ and $(2)$ to show that if one of $\langle x_n:n\in\Bbb N\rangle$ $\langle y_n:n\in\Bbb N\rangle$ converges, then so does the other, though it’s quite possible that neither converges. Remember, a sequence $\sigma$ converges to some limit $L$ if and only if every subsequence of $\sigma$ converges to $L$.
H: are $f$ and $g$ bounded function? $f$ and $g$ are entire function such that $\|f^2+g^2\|=1$ Then which of the following are correct? $f$ and $g$ are constant. $f$ and $g$ are bounded. $f$ and $g$ have no zeroes on unit circle. $ff'+gg'=0$ Well What I do is let $h(z)= f^2(z) + g^2(z)$ then clearly $h(z)$ is bounded entire hence constant by Liouville theorem, hence (4) is correct right? AI: The first option is wrong since for instance, $f(z) = \cos(z)$ and $g(z) = \sin(z)$. The second option is wrong since for instance, $f(z) = \cos(z)$ and $g(z) = \sin(z)$. The third option is wrong since for instance, $f(z) = \cos(\pi z)$ and $g(z) = \sin(\pi z)$. Note that $g(1) = 0$. The fourth option is correct since $f^2 + g^2$ is again entire and by Liouville's theorem, we have that $f^2 + g^2 = c$, where $c$ is a constant such that $\lvert c \rvert = 1$. Hence, taking the derivative gives us $ff' + gg' = 0$.
H: Solve an absolute value equation simultaneously My question is : Solve simultaneously $$\left\{\begin{align}&\|x-1\|-\|y-2\|=1\\&y = 3-\|x-1\|\end{align}\right.$$ What I did : $y=3 - \|x-1\|$ is given. Thus $y = 3-(x-1)$ or $y = 3-\left(-(x-1)\right),$ and so $$y = 4-x\qquad\mbox{ or } \qquad y = 2+x.$$ If $y = 2+x$, then $$x - 1 = 4-y \quad(1)\qquad\mbox{ or } \qquad y - 2 = x \quad(2).$$ Substituting 1), we get $$\|4-y\| - \|y-2\| = 1 \qquad\mbox{ or } \qquad \|y-2-1\|-\|y-2\| = 1.$$ I got here but I am not getting how to get the final solution. Any help would be greatly appreciated. AI: You have that $\lvert x - 1 \rvert - \lvert y - 2 \rvert = 1$. This gives us that $$\lvert x - 1 \rvert = 1 + \lvert y-2 \rvert.$$Plugging this into the second equation gives us $$y = 3 - \left( 1 + \lvert y-2 \rvert \right) = 2 - \lvert y - 2 \rvert$$ This gives us that $$y + \lvert y - 2 \rvert = 2.$$ If $y \geq 2$, then we get that $$y + y- 2 =2 \implies y = 2$$ If $y < 2$, then we get that $$y + 2 - y =2$$ which is true for all $y <2$. Hence we get that $$y \leq 2$$ From the second equation, we get that $$\lvert x - 1 \rvert = 3 - y$$ Note that since $y \leq 2$, $3-y > 0$ always. If $x >1$, then $$x - 1 = 3 - y \implies x = 4-y$$Note that $4-y > 1$, since $y \leq 2$. If $x \leq 1$, then $$x - 1 = y-3 \implies x = y-2$$Note that $y-2 \leq 1$, since $y \leq 2$. Hence, the solution set is given as follows. $$y \leq 2 \\ \text{ and }\\ x = 4-y \text{ or } y-2$$
H: What is the smallest integer which is $1\mod8$, $2\mod25$ and $7\mod11$? Like the question says, what is the smallest integer which is $1\mod8$, $2\mod25$ and $7\mod11$? I've worked out a number which is $1\mod8$ and $2\mod25$ by using that $$25 - 3 \times 8 = 1$$ so the number is $25 - 2\times 3 \times 8 = -23$ which also equals $177\mod200$. But now I'm stuck, as I can't think of integers $a$ and $b$ such that $11a + 200b = 1$? AI: Note that $200\equiv 2\pmod{11}$, and $-23\equiv-1\pmod{11}$. Thus, $$-23+4\cdot200\equiv-1+4\cdot2=7\pmod{11}\;,$$ and it follows that $800-23=777$ is a solution to your system of congruences. All you need do now is decide whether it’s the smallest positive solution, and if not, reduce it appropriately.
H: Subring of polynomials Let $k$ be a field and $A=k[X^3,X^5] \subseteq k[X]$. Prove that: a. $A$ is a Noetherian domain. b. $A$ is not integrally closed. c. $dim(A)=?$ (the Krull dimension). I suppose that the first follows from $A$ being a subring of $k[X]$, but I don't know about the rest. Thank you in advance. AI: a) Not every subring of a noetherian ring is noetherian (there are plenty of counterexamples), so this doesn't work here. Instead, use Hilbert's Basis Theorem. b) The element $X^2 = \frac{X^5}{X^3}$ is in $\mathrm{Quot}(A)$. Try to show that it is integral over $A$, but not in $A$. c) The dimension is the transcendence degree of $\mathrm{Quot}(A)$ over $k$. But this field is easy to compute.
H: Laurent Series $\exp(1/z)/(1-z)$ I need some help finding the Laurent expansion and residue of $$\dfrac{\exp \left(\frac1z \right)}{(1-z)}$$ So far I've done $$\sum_{j=0}^\infty \frac{z^{-j}}{j!} \sum_{k=0}^\infty z^k = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}$$ but don't know where to go from here. And is it also possible to use Cauchy product when one of the powers is $<0$ and the other is $>0$? AI: You can rewrite the series in the form: $$\sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}=\sum_{n=1}^{\infty}\left(\sum_{j=n}^{\infty}\frac{1}{j!}\right)z^{-n}+\sum_{n=0}^{\infty}\left(\sum_{j=0}^{\infty}\frac{1}{j!}\right)z^n=\sum_{n=1}^{\infty}\left(e-\sum_{j=0}^{n-1}\frac{1}{j!}\right)z^{-n}+\sum_{n=0}^{\infty}ez^n$$ So the residue at $0$ is the coefficient of $z^{-1}$, which is $\sum_{j=1}^\infty\frac{1}{j!}=e-1$.
H: Remembering Taylor series Could anyone suggest a good way of memorizing Taylor series for common functions? I have tried to remember them but never seem to be able to commit them to permanent memory. AI: As have already been said there is no golden trick here. However it is sometimes useful to know some ways to manipulate series and functions. Suppose we know $$\frac{1}{1-x}=\sum_{k=0}^\infty x^k,\qquad\|x\|<1\tag{1}$$ and $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}, \qquad\text{for all $x$}\tag{2}$$ Formally if we integrate (1) we get $$-\log(1-x)=\sum_{k=0}^\infty \frac{x^{k+1}}{k+1}=\sum_{k=1}^\infty \frac{x^{k}}{k},\qquad\|x\|<1$$ and then $$\log(1+x)=-\sum_{k=1}^\infty \frac{(-1)^kx^{k}}{k},\qquad\|x\|<1$$ Computing $\arctan x$ is similar since by (1) $$(\arctan x)'=\frac{1}{1+x^2}=\sum_{k\geq0} (-1)^kx^{2k}$$ Also if we use $\sin x= (e^{ix}-e^{-ix})/2i$ in conjunction to (2) we get $$2i\sin x = \sum_{k=0}^\infty \frac{(ix)^k}{k!}- \sum_{k=0}^\infty \frac{(-ix)^k}{k!} = \sum_{k=0}^\infty \frac{i^k(1-(-1)^k)x^k}{k!}$$ now $1-(-1)^{k} = 0 $ for even $k$, and $1-(-1)^{k} = 2$ otherwise. So $$2i\sin x = 2\sum_{k=0}^\infty \frac{i^{2k+1}x^{2k+1}}{(2k+1)!} = 2i\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!}$$ where in the last step we used $i^{2k+1}=i\cdot i^{2k}=i\cdot (-1)^k$, and we reach the expansion $$\sin x=\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{(2k+1)!} $$ Cosine is similar to sine. Note that these are formal derivations - proofs needs justifications of the steps done.
H: How many elements of order $10$ are there in the symmetric group $S_7$? I have a feeling you use Sylow's Theorems but I'm not sure where to start, any hints? AI: If you write a permutation in disjoint cycle notation: $(\alpha_1 \alpha_2 ... \alpha_{n_1})(\beta_1 ... \beta_{n_2})...$ then the order of the permutation is the lowest common multiple of the $n_i$. So it is clear that elements of order $10$ in $S_7$ must have cycle type $(a b)(c d e f g)$. How many of these are there? Well there are $7$ choices for $a$, then for each choice there are $6$ choices for $b$ etc. We get $7!$ choices for $a,b,c,d,e,f,g$. Divide by $2$ to account for counting $(a b)$ and $(b a)$ as the same. Divide by $5$ to do the same for the $5$-cycle $(c d e f g)$. Thus there are $7!/10 = 504$ elements of order $10$ in $S_7$.
H: Proof: $\alpha_1>\alpha_2$ in a triangle I have to proof in a triangle, that $\alpha_1>\alpha_2$ holds. The inner point P (from where I draw the smaller triangle) is set randomly. Does anyone have a suggestion where I have to start? Greetings AI: Given that the origin triangle is $\triangle ABC$, and the inner point is $P$, where $\alpha_1 = \angle BPC$, $\alpha_2 = \angle BAC$. Supposing that $D$ is the intersection of $AP$ and $BC$, we have $\angle BPC = \angle BPD + \angle DPC = \angle BAP + \angle ABP + \angle CAP + \angle ACP$ $= \angle BAC + \angle ABP + \angle ACP > \angle BAC$, Q.E.D.
H: Why can neither GeoGebra, nor MathWay solve this simple math problem? $\frac{e^{0.75}}{-0.5^e+10000}$ This doesn't work either: $\frac{2.72^{0.75}}{-0.5^{2.72}+10000}$ I can even solve it with a calculator. AI: First of all, $(-0.5)^e$ is not the same as $-0.5^e$. Secondly, $x^y$ is not well defined (at least not as a real number) if $x < 0$ and $y$ is not an integer.
H: The limit of a sum with elements from 2 sequences Let's consider the sequence of real numbers $(a_{n}),n\geq1,a_{1}>0$ that satisfies the following recurrence: $$\frac{n(n+2)}{(n+1)}a_{n}-\frac{n^2-1}{n}a_{n+1}=n(n+1)a_{n}a_{n+1}$$ I'm supposed to calculate: $$\lim_{n\rightarrow\infty}\frac{\sum_{i=1}^{n} a_{i}b_{i}}{n}$$ where $(b_{n})$ is a convergent sequence of real numbers. AI: First, supposing $a_{n+1} = 0$ for some $n$, then $a_n = 0$, and if $n > 1$, we get $a_{n-1} = 0$. Repeatedly, we have $a_n = 0$, so $a_n \neq 0$ whenever $n > 0$. \[ \frac{n(n+2)a_n}{n+1} - \frac{(n-1)(n+1)a_{n+1}}{n} = n(n+1)a_na_{n+1} \] is equivalent to \[ \frac{n(n+2)}{(n+1)a_{n+1}} - \frac{(n-1)(n+1)}{na_n} = n(n+1) \] We obtain that \[ \frac{(n-1)(n+1)}{na_n} = \frac13(n+1)n(n-1), \qquad n > 1 \] Therefore $a_n = 3n^{-2}$ whenever $n > 1$. Given that $\|b_n\| \le M$ over all positive integers $n$, thus whenever $n > 1$, \begin{align} \left\|\sum_{k=1}^n a_kb_k\right\| &\le \|a_1b_1\| + \sum_{k=2}^n \frac{3\|b_n\|}{n^2} \\ &\le \|a_1b_1\| + \sum_{k=2}^n \frac{3M}{(n-1)n} \\ &= \|a_1b_1\| + 3M \sum_{k=2}^n\left(\frac 1 {n-1} - \frac 1 {n}\right) \\ &< \|a_1b_1\| + 3M \end{align} So \[ \lim_{n\to\infty}\frac{\sum_{k=1}^n a_kb_k}n = 0 \] Q.E.D.
H: Relationship between smooth function and ordinary derivative. The function $F(x)$ is said to be smooth at the point $x$, if $$\lim_{h\to 0} \frac{F(x+h)+F(x-h)-2F(x)}{h}=0.$$ My question is that if function $F$ is differentiable at some point then how can we show that function $F$ is smooth? AI: For once this is the weirdest definition of smooth I have seen so far, but working with the definition just observe $$\frac{F(x+h)+F(x-h)-2F(x)}{h}=\frac{F(x+h)-F(x)}{h}+\frac{F(x-h)-F(x)}{h}$$ If $F$ is differentiable the first summand converges to $F'(x)$ and for the second summand we get by substituting $h\mapsto -h$ $$-\frac{F(x+h)-F(x)}{h} .$$ Taking the limit gives the desired result. We may change the sign of $h$ above since $h\to 0\Leftrightarrow -h\to 0$.
H: Running in the rain, good or bad idea? I originally wanted to ask this on the Skeptiks site, but while having a look around I came across this site which seems to have some complicated maths, but not a reliable answer. The conclusion it attempted to arrive at: So here we have it - more mathematical advice to avoid getting wet. Because we divide by VP in this equation, maximising our velocity now emerges as a good idea, assuming there is a shelter available. However, this mentions that the longer you stay in the rain the wetter you get. So, my question is: Is there a point where if the nearest shelter is at a certain distance it is better to walk to that shelter rather than run? or Is it always better to run no matter the distance of the shelter? AI: Let's assume that the cow is spherical .. sorry, assume that the walker is a box with height $h$, width $w$ and length $l$. The box needs to cover a horizontal distance $D$ while being hit by a minimal amount of water. Further assume that the air is uniformly filled with $\rho$ kilograms of water per cubic meter, moving straight downwards at a uniform speed $v_r$ (since we imagine that the raindrops have reached terminal velocity). If the box is walking at speed $v_h$, how much water hits him? In each infinitesimal moment $\Delta t$, the water in the air falls a vertical distance of $\Delta t v_r$, and $\Delta t v_r w l\rho$ amount of water hits the top of the box. The total amount of water to hit the top during the trip is $\frac{D}{v_h} v_r w l \rho$. Similarly, during each infinitesimal moment $\Delta t$ the front of the box pushes into a volume $\Delta t v_h h w$ of rain-filled air. The rain is moving downwards, but hits the box nevertheless, so the total amount of water to hit the front is $\frac{D}{v_h} v_h h w \rho = D h w \rho$ -- in other words this amount is independent of $v_h$. The only influence $v_h$ has is that the larger $v_h$ is, the less water hits the top of the box. So under the above simplifying assumptions one should indeed attempt to maximize speed. Things get rather more interesting if the rain is falling at an angle -- then the amount of rail to hit the front/back of the box does depend on $v_h$ and matters are not so simple anymore. If I remember correctly, then if there is a tailwind, there's an optimal walking/running speed above which the amount of water to hit the front increases with speed faster than the amount of water to hit the top decreases. (But I don't have time to derive that right now). More specifically, if there's so much tailwind that, standing still, you would receive more water on your back than on your top, then you should run exactly at the wind speed in order to minimize the amount of water hitting you.
H: Easy questions about Linear equations Prove that Each field of characteristic zero contains a copy of the rational number field. For an $n$ by $n$ matrix $A,$ if it is not invertible, then there exists an $n$ by $n$ matrix $B$ such that $AB=0$ but $B\ne0.$ For (1), I think I have to use the fact that each subfield of the complex number field contains every rational number. Am I right? For (2), I have no idea what to do first. Thanks. AI: For the first question you can start right at the definition and build your way up from there: a) Any field contains $1\neq 0$ b) Since the characteristic is zero it contains $1\neq 1+1\neq 1+1+1\neq \cdots$ hence a copy of the natural numbers. c) It contains inverses wrt addition and multiplication, thus a copy of the integers and the rationals. For your second question: There is a vector in the kernel of $A$, make this vector a column of $B$ and fill with zeros.
H: Golden parallelepiped Define a golden parallelepiped as a $d$-dimensional box with side lengths $(1, \phi, \phi^2, \ldots, \phi^{d-1})$, where $\phi$ is the golden ratio: The volume of the parallelepiped is $\phi^{d(d-1)/2}$, e.g., $\phi^3$ for $d=3$. I wonder if there is a natural geometric explanation of the relationship $\phi^n = \phi^{n-1} + \phi^{n-2}$, e.g., $\phi^3 = \phi^2 + \phi$ for $d=3$? I am imagining some type of partition or dissection of the volume of the golden parallelepiped that corresponds to the $\phi$ equation. AI: Since $\phi^2=\phi+1$, we can break up the edges of side length $\phi^2$ into parts of length $\phi$ and $1$, and breaking up the solid along that plane divides it into pieces of volume $\phi\times \phi\times 1=\phi^2$ and $1\times\phi\times1=\phi$, thereby demonstrating $\phi^3=\phi^2+\phi$. The same idea should work for any $d$: given the box with lengths $1$, $\phi$, ..., $\phi^d$, dividing the sides of length $\phi^k$ (for any choice of $k$) into parts of length $\phi^{k-1}$ and $\phi^{k-2}$ and then dividing the solid along that hyperplane gives pieces of volume $\phi^{\frac{d(d-1)}{2}-1}$ and $\phi^{\frac{d(d-1)}{2}-2}$.
H: complex analysis multiple choice question Suppose $\langle z_n \rangle_{n\ge0}$ is a sequence of complex numbers such that $\sum_{n=0}^\infty z_n$ converges,Given that $f$ is an entire function such that $f(z_n)=n$, then $f\equiv 0$. $f$ is unbounded. No such $f$ exists. $f$ has no zeroes. AI: As Gerry pointed out, $z_n \rightarrow 0$. Entire functions are continuous, so think about what we can say about $f(0)$. Also, what you have written about $g(z)$ doesn't make any sense. You can't say $g(z) = f(z) - n$ has zeros at all of the $z_k$ because for the definition of $g(z)$ to make sense, $n$ must be fixed. So it would have a zero at $z_n$ but not at $z_k$ if $k \neq n$.
H: About a continuous function I'm trying to solve this problem, but I don't have any idea. Can you help me? Let X a compact metric space and $f:X\times\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function. Consider $m(t_0)=\max_x (f(x,t_0))$. Show that $m$ is continuous. Thanks. AI: Fix $t\in\Bbb R$. We just have to show sequential continuity, since we are working in a metric space. Let $\{t_n\}\subset\Bbb R$ a sequence which converges to $t$. Since $X$ is compact, we can find $x_n$ such that $m(t_n)=f(x_n,t_n)$. We show that for each subsequence of $\{t_n\}$ we can find a further subsequence $\{t_{n_k}\}$ such that $m(t_{n_k})\to m(t)$. It will show that $m(t_n)\to m(t)$. Let $\{t_{n'}\}$ a subsequence of $\{t_n\}$. The sequence $\{x_{n'}\}$ admits a converging subsequence $\{x_{n_k}\}$, say to $x$. Then $(t_{n_k},x_{n_k})\to (t,x)$ and we conclude using the continuity of $f$ (and the fact that $f(x_n,t_n)\geq f(y,t_n)$ for all $y\in Y$, to show that $f(x,t)=u(t)$.
H: Maximize volume of a box in first octant So the question is Find the volume of the largest rectangular box with 3 faces on the co-ordinate planes and one vertex in the plane $x+2y+3z=6$. When I started typing this question, I didn't know what to do but as I got into it I tried some things and now actually have an answer. I'm not 100% sure it's right so all I'm asking now is confirmation of my ideas. The first thing I tried is to deal with fewer variables, so I get $x$ in terms of $y$ and $z$: $$x = 6-2y-3z.$$ This then gives the formula for volume: $$V= x\cdot y\cdot z = (6-2y-3z)\cdot y\cdot z = 6yz -2y^2z - 3z^2.$$ Then I get the gradient $\Delta V = \left( \begin{array}{ccc} 6z - 4yz - 3z^2 \\ 6y - 2y^2 -6yz \end{array} \right).$ Setting $\Delta V = 0$ gives $$z(6-4y-3z) = 0 \implies z=0 \ \text{ or } \ (6-4y-3z) = 0,$$ $$y(6-2y-6z) = 0 \implies y=0 \ \text{ or } \ (6-2y-6z) = 0.$$ Now $z=0$, $y=0$ surely can't be maximums, so I'm concerned with $$(6-4y-3z) = 0 \ \text{ and } \ (6-2y-6z) = 0$$ Solving gives me $y = 1$, $z = \frac{2}{3}$. $\quad()$ To then check if it is a maximum, we compute the Hessian which will be $$\text{Hessian} = \left( \begin{array}{ccc} -4z & 6-4y-6z \\ 6-4y-6z & -6y \end{array} \right),$$ and the determinant of this is then $12$ (using $()$). To the best of my knowledge, because $12>0$, this implies that $(*)$ gives a maximum or minimum ... but now how do I check this is a max and not a min? When I tried reading it up I get terms thrown at me like "positive semi-definite" and I'm not really sure what that means. Anyway, I hope I'm right about my thoughts and any of your thoughts will be appreciated. AI: As the comment suggests you don't really have to go through the hassle to check whether you have a maximum or a minimum. The function is non-negative and takes on the value $0$. If you have only one critical point it has to be a maximum. But to give a complete answer: The usual criterion for functions $\mathbb R\to\mathbb R$ that you have a minimum if the second derivative is positive and a maximum if it is negative translates to the following: The critical point is a maximum if the Hessian is negative definite and a minimum if it is positive definite. To check whether a (symmetric) matrix is positive definite is relatively easy: All leading principal minors have to be positive. In other words the determinants of all upper left $n\times n$ submatrices have to be positive (here you just have the $1\times 1$ and the $2\times 2$ case). To check whether a matrix is negative definite you have a condition that the signs of the principal minors must be alternating. However it is easier to multiply the matrix by $-1$ and check for positive definiteness. Note that there are more possibilities than postive or negative definite, so this characterisation is not exhaustive. In you example the principal minors are $-4z=-\frac 83<0$ and $12>0$ (alternating). For minus the Hessian you have $4z>0$ and $12>0$ (all positive). Thus the Hessian is negative definite and you have a maximum.
H: Perimeter of a triangle A question states: The length of each side of a certain triangle is an even number.If no two sides have the same length what is the smallest perimeter the triangle could have ? According to the author the answer is 18 but i think the answer is 12. (Scalene triangle with 2+4+6 as sides). Any suggestions please? AI: The sum of the lengths of any two sides of a triangle must be greater than the third side.
H: "Lock" a direction vector by preventing motion along a second vector I have 2 unit vectors, o and v. o is the orientation of a cylinder and v is a direction I wish to move inside this cylinder. However, I want to allow v to only move perpendicular to the cylinder, so it cannot move along the axis defined as moving along o. I've been banging my head together and trying various equations I can apply to v to get a vector that will move perpendicular to the cylinder in the direction of v, but not along that axis. My current idea is to rotate o and v such that o points along one of the cardinal axes (I was thinking Z for visualization purposes) and then setting the Z component of v to 0, before rotating back, but there surely must be a more condensed way of doing this. I think part of the problem is that I just cannot think of the correct search terms for this. Anybody got any ideas? Thanks AI: $\frac{o\cdot v}{\|o\|}=\|v\|\cos(\theta)$ is the length of the component of $v$ that lies along $o$. Subtract that component from $v$ to get the perpendicular component. This is a part of what's known as the Gram-Schmidt orthogonalization process. So your formula for the perpendicular component is $$v-\frac{o\cdot v}{\|o\|^2}o$$
H: Show that $\left( \lnot a \implies a \right) \stackrel{?}{\implies} a$ Consider the following axiom: $$\lnot a \implies a$$ Intuitively, this seems like a contradiction. But all implications hold if the LHS is false. Does this mean that: $$a$$ is a valid conclusion? Or is there a contradiction in the axiom? AI: This is a tautology, and since there is only one free variable, it is easy to check. However, you can simplify the expression symbolically before. $\lnot a \implies a$ is logically equivalent to $\lnot(\lnot a)\lor a$, which is $a$. Hence your formula is just $a \implies a$, which is more obviously a tautology.
H: Max size of the problem that can be solved in 2 hours if the algorithm takes n^2 microseconds I have been going through my textbook trying to solve this problem but I can't seem to completely understand. It seems so simple yet I can't figure it out. How do I go about solving this? AI: 2 hours is equal to $2\times 3600\times 1000000$ microseconds which is $7200000000$ microseconds. So you want to find $n$ such that $n^2 = 7200000000$.
H: Can we distinguish $\aleph_0$ from $\aleph_1$ in Nature? Can we even find examples of infinity in nature? AI: If by nature mean physical universe, note that there is no proof that an infinite set exists. While some of us would like to believe that universe is infinite, this cannot be proved. You also have to remember that while we wish to perceive reality as a continuous process, we cannot measure it in a precise accuracy. Given two pieces of metal of length 1m (by the same ruler) they might differ by a few atoms, which means an actual increase in length since while very very small atoms still have a physical size. So we cannot really point out an infinite object in the universe, but we also fail pointing out something which is "exactly half" of another thing, etc. If, on the other hand, you mean "in mathematics outside of set theory", then $\aleph_1$ is somewhat hidden. The reason is that in non-set theoretic constructions there is often little importance to cardinalities but rather countability; the cardinality of the continuum; and perhaps the power set of that. However there are still mild appearances of $\aleph_1$. For example Shelah proved in 1980 that there is a group of size $\aleph_1$ such that every proper subgroup of it is countable. Another example (although slightly more set theoretic in nature) is the solution for Whitehead problem's for $\aleph_1$. If we assume in addition to ZFC a certain set theoretical assumption then every $W$-group of size $\aleph_1$ is free; if we prove a different assumption (again, with ZFC) then there is a $W$-group of size $\aleph_1$ which is not free. Somewhat related: Proofs given in undergrad degree that need Continuum hypothesis? Examples of sets whose cardinalities are $\aleph_{n}$, or any large cardinal. (not assuming GCH) Does infinity and zero really exist?
H: Numerical solution of the Laplace equation on circular domain I was solving Laplace equation in MATLAB numerically. However I have problems when the domain is not rectangular. The equation is as follows: $$ \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 $$ domain is circular $$ x^2 + y^2 < 16 $$ and boundary condition $$ u(x,y)= x^2y^2 $$ How should I start with solving this equation numerically ? AI: Perhaps begin by rewriting the problem in polar coordintates: $$\frac{\partial^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}u}{\partial\theta^{2}}=0$$ $$r^2<16$$ $$\left. u(r,\theta)\right\|_D=r^4\cos^2\theta\sin^2\theta$$
H: Upper bound for complex polynomial I have a polynomial $p$ of degree $n$ satisfying $\lvert p(z) \lvert \leq c\ \ \forall z\in\partial B_1(0)$. (Isn't this true for any polynomial?) Show $\lvert p(z)\lvert \leq c \lvert z\lvert^n \ \ \forall z\in \mathbb{C}\backslash B_1(0)$. The obvious attempt would be $\|p(z)\|=\|p(\lvert z\lvert\frac{z}{\lvert z\lvert})\|=\|\sum_{i=0}^n a_i \|z\|^i (\frac{z}{\lvert z\lvert})^i\|\leq \|z\|^n \sum_{i=0}^n \|a_i (\frac{z}{\lvert z\lvert})^i\|$ which doesn't lead anywhere. I guess I have to apply some maximum principle but don't know how. AI: Let $\Omega:=\{z\in\Bbb C, \|z\|<1\}$. It's a domain. Define $g(z):=z^np\left(\frac 1z\right)$: it's a holomorphic function, and by maximum modulus principle the modulus is reached at the boundary, and is $\leq c$. Now, for $\|z\|>1$, work with $\frac 1z$ in the previous argument.
H: Strategies to find the set of functions $f:\mathbb R\to\mathbb R$ satisfing the functional equation: $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$ My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following: Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation: $$f\big(x^3\big)+f\big(y^3\big)=(x+y)\Big(f\big(x^2\big)+f\big(y^2\big)-f(xy)\Big):\forall x, y\in\mathbb R$$ I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis. Thanks in advance. AI: Here is a typical start. Put $x=y=0$. The right side is $0$, so $f(0)=0$. Now set $y=0$, and let $x$ roam freely. Since $f(0)=0$, we get $f(x^3)=xf(x^2)$. Set $y=-x$. We get $f(x^3)+f(-x^3)=0$. Since everything is a cube, we have $f(-u)=-f(u)$ for all $u$. Now explore $x=y$. Can we learn anything from setting $x$ and/or $y$ qual to $1$? A little playing has gotten us a lot of information, enough that we should be able to complete things.
H: #(cardinality of irreps Lie algebras) > #(irreps of ssociative algebras). Proof? I know that irreducible representations of associative $$-algebras are fairly restricted: any $$-algebra $A$ is isomorphic to a finite sum of simple algebras $A\cong\oplus_{i=0}^{N}M_{n_i}(\mathbb{C})$ What's the cardinality of the irreps of a Lie algebra? AI: The smallest non-zero Lie algebra of all is the one-dimensional Lie algebra. If you only look at finite dimensional representations, its irreps are in bijection with pairs $(\lambda,n)$ with $\lambda\in\mathbb C$ and $n\in\mathbb N$: this fact is an immediate consequence of the theorem of Jordan Canonical forms. If you want to consider infinite dimensional representations, then things are much more complicated. A representation of this algebra is roughly the same as an endomorphism of a vector space, so in a sense the study of its representions is the same thing as operator theory: whole books have been written about this.
H: Simple clarification - deduction using big-O notation A set of lecture notes I'm reading on Halasz's theorem makes the following statement in a proof, which I can't quite follow - I was hoping someone might be able to clear up what I'm missing: Therefore, $Re(\sum_p p^{-s} = \log \log x + O_M(1)$, where $O_M$ denotes that the constant involved in the big O notation depends on $M$, and $s:= 1 + \frac{1}{\log x} +it$. Recall that $Re(\sum_p p^{-s}) = \log \|\zeta(s)\| +O(1)$; hence $\|\zeta(s)\| \geq c_M\log x$, some $c_M$. I hopefully won't need to give any more context about the proof itself as I don't think it's relevant, but I can if needs be. The zeta referred to is the usual zeta function, and $x$ is usually some real $\geq 1$. What I can't understand is the $\geq$ in the final line; it appears both in my handwritten notes and the lecturer's typed notes, so I'm fairly sure it's meant to be there. However, the deduction which seems to be being made is to equate $ \log \log x + O_M(1)= \log \|\zeta(s)\| +O(1)$ and then deduce the inequality; but wouldnt this give us (taking exponents) $\|\zeta(s)\|=\log x \cdot \exp(O_M(1))$? So then if anything surely that would give us an upper bound for $\|\zeta\|$, not a lower bound? If it helps, I can try to provide more information to clarify. The lecture notes are online at http://www.dpmms.cam.ac.uk/~bjg23/primenumbers-2012.html courtesy of Ben Green (who I hope doesn't mind me linking to his page), namely on the penultimate page of chapter 2, though I'm not sure whether the notes are accessible outside the Cambridge domain. Many thanks in advance for your help. AI: Taking exponents you can just as easily arrive at $\log x = \|\zeta(s)\| \cdot \exp(O_M(1))$ which should provide the lower bound on $\|\zeta(s)\|$, I think
H: Why would the author ask if I used the Associative Law to prove + is not equiv. to ? I just started reading An Introduction to Mathematical Analysis by H.S. Bear and problem 1 goes as follows: Problem 1: Show that + and are necessarily different operations. That is, for any system (F, +, ) satisfying Axioms I, II, and III, it cannot happen that x + y = x y for all x, y. Hint: You do not know there are any numbers other than 0 and 1, so that your argument should probably involve only these numbers. Did you use Axiom II? If not, state explicitly the stronger result that you actually proved. In this book, Axiom I is commutativity of + and , Axiom II is associativity of + and , and Axiom III is existence of identities (x+0=x, x1=x, 0 does not equal 1). My question: Simply why would the author specifically ask the reader if he/she used Axiom II (associativity) and what exactly do they mean by "If not, state explicitly the stronger result you actually proved"? Why not not include those last two sentences? FWIW, here is my solution: To prove: Restated: And I justified 5 by citing Axiom III since Axiom III includes the statement that 0 does not equal 1. AI: The basic idea is as follows. From the neutral Axioms III, and commutativity of addition we have $$\begin{eqnarray}\rm x = 0 + x &\rm \\ \rm y \: 1 &=\rm y \end{eqnarray}$$ If $\ +\, =\, \ $ then aligned terms are unified for $\rm\:y = 0,\ x = 1,\:$ yielding $$\rm\ 1 = 0 + 1 = 0 1 = 0 $$ contra hypothesis $\rm\:1 \ne 0.\:$ Thus $\rm\: +\: \ne\: *\:$ because they take different values at the point $\rm\:(0,1)$. Note that the proof does not use associativity, and doesn't use commutativity if you state the neutral axioms as above. In any case, only one of the commutative axioms is needed, so that the neutral axioms can be ordered so the above unification is possible. In particular, the inference works in noncomutative rings, i.e. rings where multiplication is not necessarily commutative. Further, because the proof did not use associativity, it will also work in nonassociative rings. Note $\ $ This method of deriving consequences by unifying terms in identities is a basic method in equational reasoning (term rewriting), e.g. google Knuth-Bendix or Grobner basis algorithms.
H: Reflections generating isometry group I was reading an article and it states that every isometry of the upper half plane model of the hyperbolic plane is a composition of reflections in hyperbolic lines, but does not seem to explain why this is true. Could anyone offer any insight? Thanks. AI: An isometry $\phi:M\to N$ between connected Riemannian manifolds $M$ and $N$ is completely determined by its value at a single point $p$ and its differential at $d\phi_p$. Take any isometry $\phi$ of $\mathbb{H}^2$. Connect $i$ and $\phi(i)$ by a (unique) shortest geodesic and let $C$ be a perpendicular bisector of the connecting geodesic. Then the reflection $r_C$ across $C$ maps $i$ to $\phi(i)$. Now take an orthonormal basis $e_j$ at $i$. It is mapped to $d\phi e_j\in T_{\phi(i)}\mathbb{H}^2$. Linear algebra tells us that in $T_{\phi(i)}\mathbb{H}^2$ we can map $d\phi e_j$ to $dr_C e_j$ by a reflection across a line (if $\phi$ is orientation-preserving) or by a rotation (if $\phi$ is orientation-reversing). A two-dimensional rotation can be written as a composition of two reflections. By the exponential map, the lines across which we're reflecting map to geodesics in $\mathbb{H}^2$ and the reflections extend to reflections of $\mathbb{H}^2$ across those geodesics. We have therefore written $\phi$ as a composition of reflections. In some cases, e.g. in Thurston, Three-Dimensional Geometry and Topology, the isometry group of $\mathbb{H}^2$ is defined as the group generated by reflections across circles (e.g., in the Poincare model). Chapter two of that book has a discussion of hyperbolic geometry and exercises comparing the various perspectives on isometries, e.g., as $$Sl(2;\mathbb{R}) = SO(1,1) = \mbox{Möbius transformations with real coefficients} = \langle\mbox{refl. across circles}\rangle.$$You might also check Chapter B (I think) of Benedetti and Petronio, Lectures on Hyperbolic Geometry.
H: "Any one of" or "either"? First there was this question: Q. Find the number of numbers from 100 to 400 which are divisible by either 2, 3, 5, and 7 And the obvious solution was to find numbers divisible only by 2 i.e 100 102 104 ... 400 numbers divisible only by 3 Now we are left with only odd numbers i.e 101 103 105 ... 399 we'll check for only multiples of 3 here, because even multiples of 3 have been taken cared by 2 the numbers will be 105 111 ... 399 numbers divisible only by 5 The numbers left will be 101 103 107 109 113 115 ... 395 We'll check for only multiples of 5 115 125 ... 395 numbers divisible only by 7 Now the numbers left we'll be 101 103 ... 397 And once again we'll check for only multiples of 7 The sum of multiples individually checked in 1, 2, 3, and 4 will give the answer But then came this question Q. Find the number of numbers from 100 to 400 which are divisible by any one of 2, 3, 5, or 7 I can't seem to understand is there any difference between the first and the second question, and if there is, then wont the second question have a different answer for each of 2, 3, 5, and 7 Example question by the author, followed by the first question AI: The word "either" typically denotes exclusivity of a logical disjunction. In other words, given statements $x_1,...,x_n$, the following are equivalent: (1) Either $x_1$ or ... or $x_n$ holds. (2) Exactly one of the statements $x_1,...,x_n$ holds. (3) (At least / No less than) one of the statements $x_1,...,x_n$ holds, and (at most / no more than) one of them holds. Unfortunately, the word "either" is often misused, even by people (e.g.: English-speaking mathematicians) who really ought to know better! Some use it simply as a lead-off word for a list of options, and in that case, it is effectively a throwaway word, meaning nothing at all, and so allowing that any non-null subcollection of the list of statements may hold simultaneously. As a further complication, the phrase "any one of" seems to carry with it the connotation of "exactly one of"! It would be better to say "any of" instead--as that suggests that more than one may hold, but at least one must hold--but again, this is a common error, even among those who should know better. Ultimately, this is just a frustrating reality. The best advice I can offer is to try to determine which meaning is intended by the questioner. If it is a textbook, try to find an early example where "either" is used, and see if it means "exactly" or if it is a throwaway. Likewise with "any one of". It is probable that they are intended to have different meanings, so if one is exclusive, then the other is almost surely inclusive. Edit: Based on what you've told me, it looks as though the author is using "either" as a throwaway word, which suggests that "any one of" is intended to mean "exactly one of" in this text. Your approach to the first problem is then the right one to take. I am going to break things down as follows, to give a slightly more general approach that should help out in both problems and more, if you're familiar with the basic concepts of set theory (specifically: union, intersection, complement, universe of discourse) and how we may use venn diagrams to help sort out the elements (or cardinalities of sets) of a universe of discourse. Here, the universe of discourse ($U$) will be taken to be the integers from $100$ to $400$ (inclusive). $A$ will be taken to be the set those of elements of $U$ that are divisible by $2$; $B$, by $3$; $C$, by $5$; $D$, by $7$. First, we will determine the cardinalities of (number of elements in) each of the sets $A,B,C,D$ (using the technique of common differences described by the author): $$\|A\|=151,\: \|B\|=100,\: \|C\|=61,\: \|D\|=43.$$ Next, we determine the cardinalities of the intersections of any pair of the sets $A,B,C,D$ (again with common differences): $$\|A\cap B\|=50,\: \|A\cap C\|=31,\: \|A\cap D\|=21,\: \|B\cap C\|=20,\: \|B\cap D\|=15,\: \|C\cap D\|=9.$$ Then we look at intersections of any trio of the sets $A,B,C,D$, similarly: $$\|A\cap B\cap C\|=10,\: \|A\cap B\cap D\|=7,\: \|A\cap C\cap D\|=4,\: \|B\cap C\cap D\|=3.$$ Now the intersection of all of them has only $210$ as an element, so $\|A\cap B\cap C\cap D\|=1$. Thus, there is only one number on the list divisible by all four of $2,3,5,7$. To find how many are divisible by exactly three, we take those divisible by at least three (in this case, members of intersections of trios) and toss out the one divisible by all four. In particular: $$\|A\cap B\cap C\cap(\neg D)\|=9,$$ $$\|A\cap B\cap(\neg C)\cap D\|=6,$$ $$\|A\cap(\neg B)\cap C\cap D\|=3,$$ $$\|(\neg A)\cap B\cap C\cap D\|=2,$$ so there are a total of $9+6+3+2=20$ numbers on the list divisible by exactly three of $2,3,5,7$. To determine how many are divisible by exactly two, we take those divisible by at least two (members of intersections of pairs) and toss out those divisible by all four and those divisible by exactly three. For example, $\|A\cap B\cap(\neg C)\cap(\neg D)\|=50-1-9-6=34$, and we similarly find $$\|A\cap(\neg B)\cap C\cap(\neg D)\|=18,$$ $$\|(\neg A)\cap B\cap C\cap(\neg D)\|=8,$$ $$\|A\cap(\neg B)\cap(\neg C)\cap D\|=11,$$ $$\|(\neg A)\cap B\cap(\neg C)\cap D\|=6,$$ $$\|(\neg A)\cap(\neg B)\cap C\cap D\|=3,$$ giving us a total of $80$ numbers on the list divisible by exactly two of $2,3,5,7$. For those divisible by exactly one, start with those divisible by at least one, and toss those divisible by more. So, $\|A\cap(\neg B)\cap(\neg C)\cap(\neg D)\|=151-1-6-3-9-34-18-11=69$, and simliarly, $$\|(\neg A)\cap B\cap(\neg C)\cap(\neg D)\|=34,$$ $$\|(\neg A)\cap(\neg B)\cap C\cap(\neg D)\|=17,$$ $$\|(\neg A)\cap(\neg B)\cap(\neg C)\cap D\|=11,$$ giving us $131$ numbers on the list divisible by exactly one of $2,3,5,7$. Unless I made a mistake, $131$ should be the answer to the second problem you encountered, if "either" and "any one of" are, in fact, intended to mean different things and "either" is a throwaway word. In total, then, there are $1+20+80+131=232$ numbers on the list that are divisible by at least one of $2,3,5,7$, and since there were only $301$ on the list in the first place, then there are $69$ that are divisible by none of $2,3,5,7$. While time-consuming, the advantage of proceeding as above is that, at this point, you can answer any question along this line, since everything is nicely sorted out.
H: Functional Analysis, Why this statement true? Why this statement true? If $f \in C^0([0,1], W^{2,2}(K)) $ then $ f \in C^0 ([0,1] \times K)$. $ K \subset R^n $ , W : Sobolev Space. AI: I don't think this is true. The Sobolev embedding theorem suggests this should only work for small $n$; in this case I think it should be $n < 4$ if I have the parameters right. To try for a counterexample, let's take, say, $K$ to be the unit ball of $\mathbb{R}^6$, and let $f(t,x) = \|x\|^t$. One should be able to show that $$f(t, \cdot), \partial_i f(t, \cdot), \partial_i \partial_j f(t,\cdot)$$ are continuous as functions $[0,1] \to L^2(K)$, which would make $f$ continuous as a map from $[0,1]$ to $W^{2,2}(K)$. However, as a function on $[0,1] \times K$, $f$ is clearly not continuous at $(0,0)$.
H: Prove: If $L \leq X$, $L$ has finite dimension, $M\leq X$ Then $L+M$ is closed. Prove: If $X$ is a locally convex space, $L \leq X$, $L$ has finite dimension, $M\leq X$ Then $L+M$ is closed. What I know: If $L$ is a finite dimensional subspace, then $L$ is closed. AI: If $M$ is not closed then this is not true in general. Indeed, consider $$ X=\ell_1(\mathbb{N}),\quad L=\mathrm{span}\{e_1\},\quad M=\mathrm{span}\{e_n:n>1\} $$ Obviously, $$ \mathrm{dim}(L)=1,\quad \overline{M}=\{x\in X: x(1)=0\} $$ $$ L+M=\mathrm{span}\{e_n:n\in\mathbb{N}\} $$ $$ \overline{L+M}=\ell_1(\mathbb{N}) $$ If we additionally assume that $M$ is closed then this statement is true. This proof is taken from Rudin's Functional analysis theorem 1.42. Since $M$ is closed then $X/M$ is Hausdorff locally convex space. Consider quotient map $$ \pi:X\mapsto X/M:x\mapsto x+M $$ Since $L$ is finite dimensional, then does $\pi(L)$. Since $\pi(L)$ is finite dimensional then it is closed. Since $\pi$ is continuous then $\pi^{-1}(\pi(L))$ is closed. It is remains to note that $M+L=\pi^{-1}(\pi(L))$
H: Combinatorics: Things always included together Find the number of permutations of $n$ different things taken $r$ at a time so that two particular things are always included and are together? Including two things initially, i have $(n-2)$ things from which I can choose $(r-2)$ things. Hence $\,^{(n-2)}C_{(r-2)}$ denote the combinations which can be arranged in $(r-2)!$ The two things can be interchanged in 2! ways within themselves. Furthermore, when $r$ things are selected, I have $r+1$ ways where I can insert two things which have to remain together. So I get $\,^{(n-2)}C_{(r-2)} \cdot (r-2)! \cdot 2! \cdot (r+1)$ Am I doing the right thing? AI: You have started off in the correct way, that is, recognizing that you now have an increased restriction upon the number of items you need to choose and the number of items to chose from. You now need to permute $(r-2)$ things from $(n-2)$ things as you identified, so we have $\,^{n-2}P_{r-2}$ ways of doing this. However, we have to consider two more factors, firstly, the fact that we can arrange the two items that must be next to each other in $2!=2$ ways (as you already stated), and secondly that we can put this pair of items in any of the $(r-2+1)=(r-1)$ places in the set. Therefore the number of ways of arranging the items as described in your problem can be found by: $$\,^{n-2}P_{r-2} \cdot 2! \cdot (r-1)$$ N.B: $$\,^{n}P_{r}\equiv {(n)}_{r}\equiv \frac{n!}{(n-r)!}, \forall r\le n$$ And: $$\,^{n}P_{r}\equiv0, \forall r \gt n$$
H: How to draw an ellipse using its center and two points on its perimeter Suppose I have an ellipse centered at the origin, and two points on its perimeter which are not antipodal to one another (i.e. not negative to each other as vectors in $\mathbb R^2$). How can I draw the whole ellipse? AI: The Wikipedia animation you point to show an ellipse being generated by two vectors $\mathbf a$ and $\mathbf b$ via the parametric equation $$ t \mapsto \cos(t)\cdot \mathbf a + \sin(t)\cdot \mathbf b $$ with $t$ ranging from $0$ to $2\pi$, say. Getting an equation for the ellipse is not as simple to describe. What you need there is to express an arbitrary point in the (non-rectangular) coordinate system that has $\mathbf a$ and $\mathbf b$ as basis vectors. Then the equation for (say) the closed ellipse is $$ \{ v \mid p^2+q^2\le 1 \text{ where } v=p\mathbf a+q\mathbf b \}$$
H: Counting ways of sitting in adjacent seats In how many ways can $m$ people entering a theatre be seated in two rows, each containing $n$ seats with the condition that no two sit in adjacent seats in the first row? AI: First calculate how many ways $j$ seats can be occupied from $n$: this is ${n \choose j}$ if $n\ge j$. Then calculate how many ways $k$ seats can be occupied from $n$ with none adjacent: this is ${n+1-k \choose k}$ if $n+1 \ge 2k$ So the answer to the original question is $$\sum_{k= \max(0,m-n)}^{\min \left(m ,\lfloor(n+1)/2\rfloor \right)} {n+1-k \choose k}{n \choose m-k} $$ if $2m \ge 3n+1$, and $0$ otherwise. Multiply by $m!$ if order of the individuals matters.
H: Is the dual graph simple? According to the book Topological Graph Theory by Gross and Tucker, given a cellular embedding of a graph on a surface (by 'surface' I mean here a sphere with $n\geq 0$ handles), one can define a dual multigraph by treating the faces of the original graph embedding as vertices and adding an edge between two vertices for every edge the corresponding faces have in common in the original graph. My question is this: Let $G$ be cellularly embedded in some surface and let $G'$ be the dual of this embedding. If $\delta (G)\geq 3$, can I be sure that $G'$ is simple (there is at most one edge between two vertices)? This should be simple, but I am not comfortable enough with proofs in topological graph theory. Note: $\delta (G)$ denotes the minimum degree of vertices in $G$. Edit: Follow-up to this question: When is the dual graph simple? AI: I am not strong in topological graph theory, but I am pretty sure this should hold on all surfaces. Create a graph $G$ in the following way. Create two diamonds (graph on 4 vertices, where two triangles are sharing an edge). Now create an edge joining a vertex of degree 2 in one diamond to a vertex of degree 2 in the other diamond. At last join the two remaining vertices of degree 2. Now the dual graph $G'$ has 6 vertices, where two of them $v,u$ corresponds to faces of degree 6 in $G$. There is 2 edges from $u$ to $v$, thus $G'$ is a multigraph. This graph is shown as the second to last graph here: http://mathworld.wolfram.com/CubicGraph.html
H: Eigenvalues of block matrices Let $K$ be a field of characteristic 0, and consider the following block matrix $$M=\left(\begin{array}{cc} A & B\\ -B&D\end{array}\right),$$ where each block is an $n\times n$ matrix with coefficients in $K$. I am looking for a relation between the eigenvalues of $M$ and those of $A$ and $D$. Context: Here, I'm assuming that $M$ is invertible and semisimple. I was wondering if there is a way to show that both $A$ and $D$ are invertible and semisimple as well. Moreover, we can also assume that $B$ is $2\times2$ and has one of the following forms: $$\left\{\left(\begin{array}{cc} 1 & 0\\ 0&1\end{array}\right),\left(\begin{array}{cc} 1 & 0\\ 0&0\end{array}\right),\left(\begin{array}{cc} 0 & 0\\ 0&0\end{array}\right)\right\}.$$ AI: It's certainly not necessary for $A$ and $D$ to be invertible, e.g. with $B = \pmatrix{1 & 0\cr 0 & 1\cr}$ you could have $A = D = \pmatrix{0 & 0\cr 0 & 0\cr}$, or with $B = \pmatrix{1 & 0\cr 0 & 0\cr}$ you could have $A = \pmatrix{0 & 0\cr 0 & 1\cr}$ and $D = \pmatrix{1 & 1\cr 1 & 1\cr}$. Of course with $B = \pmatrix{0 & 0\cr 0 & 0\cr}$ the eigenvalues of $M$ are the union of the eigenvalues of $A$ and of $D$. In all cases $\text{Tr}(M) = \text{Tr}(A) + \text{Tr}(D)$, so the sum of the eigenvalues of $M$ is the sum for $A$ plus the sum for $D$. In the case $B = \pmatrix{1 & 0\cr 0 & 1\cr}$, the coefficient of $\lambda^2$ in the characteristic polynomial of $M$ (which is $\sum_{i < j} \lambda_i \lambda_j$ where $\lambda_i$ are the eigenvalues of $M$) is $a_{{1}}a_{{2}}+a_{{1}}d_{{1}}+a_{{1}}d_{{2}}+a_{{2}}d_{{1}}+a_{{2}}d_{ {2}}+d_{{1}}d_{{2}}+2$, where $a_i$ and $d_i$ are the eigenvalues of $A$ and $D$ respectively. In the case $B = \pmatrix{1 & 0\cr 0 & 0\cr}$, that coefficient would be $a_{{1}}a_{{2}}+a_{{1}}d_{{1}}+a_{{1}}d_{{2}}+a_{{2}}d_{{1}}+a_{{2}}d_{ {2}}+d_{{1}}d_{{2}}+1$. In the case $B = \pmatrix{1 & 0\cr 0 & 0\cr}$, I think these are the only equations linking the eigenvalues of $M$ with those of $A$ and $D$: you can choose $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ arbitrarily satisfying the two constraints on $\sum_i \lambda_i$ and $\sum_{i<j} \lambda_i \lambda_j$ and find a suitable $A$ and $D$ with eigenvalues $a_i$ and $d_j$ that works. In the case $B = \pmatrix{1 & 0\cr 0 & 1\cr}$, it looks to me like there is an additional constraint: you can only choose one eigenvalue (say $\lambda_1$) arbitrarily, and then the other three will satisfy $$\eqalign{{\lambda}^{3}&+ \left( -a_{{1}}-a_{{2}}-d_{{1}}-d_{{2}}+\lambda_{{1}} \right) {\lambda}^{2}\cr + \left( a_{{1}}a_{{2}} \right. & \left. +a_{{1}}d_{{1}}+a_{{1}}d_ {{2}}-a_{{1}}\lambda_{{1}}+a_{{2}}d_{{1}}+a_{{2}}d_{{2}}-a_{{2}} \lambda_{{1}}+d_{{1}}d_{{2}} -d_{{1}}\lambda_{{1}} -d_{{2}}\lambda_{{1}} +{\lambda_{{1}}}^{2}+2 \right) \lambda \cr +a_{{2}}d_{{2}}\lambda_{{1}} &-a_{ {1}}a_{{2}}d_{{2}}-a_{{2}}d_{{1}}d_{{2}}-a_{{1}}a_{{2}}d_{{1}}-a_{{1}} d_{{1}}d_{{2}}-a_{{1}}-a_{{2}}+{\lambda_{{1}}}^{3}+2\,\lambda_{{1}}-d_ {{1}}-d_{{2}}-d_{{1}}{\lambda_{{1}}}^{2}\cr -d_{{2}}{\lambda_{{1}}}^{2} & +a_ {{1}}a_{{2}}\lambda_{{1}}+a_{{1}}d_{{2}}\lambda_{{1}}+a_{{2}}d_{{1}} \lambda_{{1}}+a_{{1}}d_{{1}}\lambda_{{1}}+d_{{1}}d_{{2}}\lambda_{{1}}- a_{{2}}{\lambda_{{1}}}^{2}-a_{{1}}{\lambda_{{1}}}^{2}=0\cr} $$
H: Subspaces of all real-valued continuous functions on $\mathbb{R}^1$ I'll go ahead and give you the problem first, and then explain my trouble with it. Which of the following subsets are subspaces of the vector space C(-$\infty$,$\infty$) defined as follows: Let V be the set of all real-valued continuous functions defnined on $\mathbb{R}^1$. If $f$ and $g$ are in $V$, we define $(f + g)(t) = f(t) + g(t)$. If $f$ is in $V$ and $c$ is a scalar, we define $c$ · $f$ by $(c · f)(t) = cf(t)$. Then $V$ is a vector space, which is donatedby C(-$\infty$,$\infty$). (a) All nonnegative functions (b) All constant functions There are more, but I have a feeling that after the (a) I'll be able to get the rest myself. Anyay, my problem is that the problem says it required calculus, and I don't understand how calculus even comes into it. As far as I can tell, (b) would be a subspace (because it's are closed under operations of V), but (a) would not because V could make a negative value positive. I may be completely wrong though; any suggestions? AI: The set of constant functions makes a subspace since (1) each constant function is continuous, (2) if $f$ and $g$ are constant functions and $c$ is a scalar, then $f+g$ and $c\cdot f$ are also constant functions. The set of non-negative functions is not a subspace for various reasons: (1) not every non-negative function is continuous, (2) if $f$ is non-negative and non-zero, then $-1\cdot f$ takes a negative value so the set of non-negative functions is not closed under scalar multiplication. (It is closed under addition).
H: Moment of inertia - formula derivation: Missing $\frac{1}{2}$ I'm trying to deduce the formula of the moment of inertia of an object of rotation. The general formula for the moment of inertia is declared as: $$J=mr^2 =\sum{m_i r_i^2}$$ If I replace $m_i$ of the $\sum{m_i * r_i^2}$ with $\int dm$ (where dm are the masses) and $r_i^2$ with the $\int(y)^2dx$, I get $J=\int y^2 dm = \int y^2(\rho\space dV)$ (remember: $(\rho\space dV)$ since $\rho=\frac{m}{V}$) Furthermore $V=\int{\piy^2}dx$ leads me to $J = \int y^2(\rho\space dV) = \int y^2 \rho\piy^2 dx = $ $$J=\pi\rho\int y^4 dx$$ Now my question: If I compare my formula with the formula it should be, I perceive that there is $\frac{1}{2}$ missing. $$J=\frac{\pi\sigma}{{\color{red}2}}\int y^4 dx$$ What mistake did I make? AI: The problem is that you didn't trace the meaning of $y$ through the formulas. The $y$ in $$J=\int y^2\,\mathrm dm$$ is the $y$ coordinate of a mass element in the body, whereas the $y$ in $$V=\int\pi y^2\,\mathrm dx$$ and in $$J=\frac{\pi\rho}2\int y^4\,\mathrm dx$$ is the $y$ coordinate of the curve forming the boundary of the body. In replacing $\mathrm dm$ by $\rho\,\mathrm d V$ with $\mathrm dV$ representing an entire infinitesimal disk of the body of rotation, you'd have to average $y^2$ over the disk. However, I get a factor of $1/4$ from that, not $1/2$, so there might be another problem, too.
H: Impossible to prove vs neither true nor false First off I am not a logician, so I probably won't use the correct terms. Sorry ! I have heard, like most mathematicians, about questions like the continuum hypothesis, or the independance of the axiom of choice from ZF. These statements (continuum hypothesis or axiom of choice) were referred to as "neither true nor false", because you could add them or their negation to form two different sets of axioms that would be equally self-consistent. On the other hand, I have also heard about statements that "were true but could not be proven", i.e. could not be proven in a finite number of applications of axioms. For instance, it could be that the Goldbach conjecture is true, but that there is no other way to "prove" it than to verify it for all integers, which is not really a proof of course. Is that distinction correct ? (and what is the real terminology?) I fail to understand what the problem would be, for example, if you were to add the negation of a "true but impossible to prove" statement as an axiom. There would be a contradiction, but you could never find it, so... AI: First we need to assert the general framework. We have a language with relation symbols and function symbols and constants, etc. With this language we can write sentences and formulas. We say that $T$ is a theory if it is a collection of sentences in a certain language, often we require that $T$ is consistent. If $T$ is a first-order theory, whatever that means, then we can apply Goedel's completeness theorem and we know that $T$ is consistent if and only if it has a model, that is an interpretation of the language in such way that all the sentences in $T$ are true in a specific interpretation. The same theorem also tells us that if we have some sentence $\varphi$ in the same language, then $T\cup\{\varphi\}$ is consistent if and only if it has a model. We go further to notice that if we can prove $\varphi$ from $T$ then $\varphi$ is true in every model of $T$. On the other hand we know that if $T$ is consistent it cannot prove a contradiction. In particular if it proves $\varphi$ it cannot prove $\lnot\varphi$, and if both $T\cup\{\varphi\}$ and $T\cup\{\lnot\varphi\}$ are consistent then neither $\varphi$ nor $\lnot\varphi$ can be proved from $T$. When we say that CH is unprovable from ZFC we mean that there exists a model of ZFC+CH and there exists a model of ZFC+$\lnot$CH [1]. Similarly AC with ZF, there are models of ZF+AC and models of ZF+$\lnot$AC. Now we can consider a specific model of $T$. In such model there are things which are true, for example in a given model of ZF the axiom of choice is either true or false, and similarly the continuum hypothesis. Both these assertions are true (or false) in a given model, but cannot be proved from ZF itself. Some theories, like Peano Axioms treated as the theory of the natural numbers, have a canonical model. It is possible that the Goldbach conjecture is true in the canonical model, and therefore we can regard it as true in some aspects, but the sentence itself is false in a different, non-canonical model. This would cause the Goldbach conjecture to become unprovable from PA, while still being true in the canonical model. Footnotes: Of course this is all relative to the consistency of ZFC, namely we have to assume that ZFC is consistent to begin with, but if it is then both ZFC+CH and ZFC+$\lnot$CH are consistent as well.
H: Is $f(t)=t^\alpha$ for $\alpha\in(0,1)$ a sub-additive function? Possible Duplicate: Does $\|x\|^p$ with $0<p<1$ satisfy the triangular inequality on $\mathbb{R}$? Is the function $$f(t)= t^{\alpha},\quad \alpha\in (0,1)$$ a subadditive function? My teacher said categorically that this is true. But I'm not so sure. EDIT: $~~~~~0\leq t\leq 1$ AI: By homogeneity, in the inequality $(x+y)^{\alpha}\leq x^{\alpha}+y^{\alpha}$ just deal with the case $y=1$. Let $f(t):=t^{\alpha}+1-(t+1)^{\alpha}$. The derivative has the sign of $t^{\alpha-1}-(t+1)^{\alpha-1}$ which is non-negative since $\alpha<1$. Hence $f(t)\geq f(0)=0$ and we are done.
H: Prove $\int 1-\prod_{i=1}^n (1- \mathbb{I}_{A_i}) d \mu= \mu ( \bigcup_{i=1}^n A_i )$ Let $(\Omega, \mathcal{A}, \mu)$ be a measurable space. $A_1, A_2,...,A_n \in \mathcal{A}$ are sets with finite measure. I have to prove $\int 1-\prod_{i=1}^n (1- \mathbb{I}_{A_i}) d \mu= \mu ( \bigcup_{i=1}^n A_i )$. But I am once again totaly puzzeld how to start. Result: I solved the problem, using the hint by Weltschmerz. By now I found an elegant way of the prove, which simply is $$\int 1 - \prod_{i=1}^n \left( 1- \mathbb{I}_{A_i} \right) d \mu= \int 1 - \prod_{i=1}^n \left( \mathbb{I}_{A_i^C} \right) d \mu$$ $$ = \int 1 - \mathbb{I}_{\bigcap_{i=1}^n A_i^C} d \mu = \int \mathbb{I}_{\left( \bigcap_{i=1}^n A_i^C \right)^C} d \mu = \int \mathbb{I}_{\bigcup_{i=1}^n A_i} d \mu= \mu \left( \bigcup_{i=1}^n A_i \right)$$ AI: Giveaway hint: $1-\prod_{i=1}^n (1-\mathbb{I}_{A_i})$ is $1$ if $x\in$ ... and $0$ if $x\in$ ..., and that $=\mathbb{I}_{\textrm{something}}$.
H: An equality in a group Let $G$ be a non-abelian group of automorphisms, where composition of automorphisms is the group operator. We have the following notation $$g = g_1 \quad x_0 = g_1^{-1}g_2 \quad x_i = g^{-i}x_0g^i$$ where $g_1,g_2 \in G$. Now I want to show that $$g_1^{-j}g_2^j=x_{j-1} x_{j-2}\cdots x_0$$ I have looked at it for some time now, and I can not get it to work. It is part of a proof in "Algebraic Graph Theory" by Biggs 1974. AI: Note that $g_1^{-1}x_kg_1 = x_{k+1}$ for all $k$: indeed, $g_1^{-1}x_kg_1 = g_1^{-1}g_1^{-k}x_0g_1^kg_1 = g_1^{-(k+1)} x_0 g_1^{k+1} = x_{k+1}$. We now proceed by induction on $j$. If $j=1$, then the left hand side is $g_1^{-1}g_2$, and the right hand side is $x_0$, so we have equality. Assume the asserted equality holds for $j$. Then $$\begin{align} g_1^{-j-1}g_2^{j+1} &= g_1^{-1}\Bigl( g_1^{-j}g_2^j\Bigr)g_2\\ &= g_1^{-1}\Bigl( x_{j-1}x_{j-2}\cdots x_0\Bigr) g_2\\ &= g_1^{-1}x_{j-1}1x_{j-2}1\cdots 1x_01g_2\\ &= g_1^{-1}x_{j-1}(g_1g_1^{-1})x_{j-2}(g_1g_1^{-1})\cdots (g_1g_1^{-1})x_0(g_1g_1^{-1})g_2\\ &= (g_1^{-1}x_{j-1}g_1) (g_1^{-1}x_{j-2}g_1) \cdots (g_1^{-1}x_0g_1) g_1^{-1}g_2\\ &= x_{j}x_{j-1}x_{j-2}\cdots x_1(g_1^{-1}g_2)\\ &= x_jx_{j-1}\cdots x_1x_0, \end{align}$$ as desired. Note. It doesn't matter that the elements of $G$ are automorphisms, just that you have a group. The fact that $G$ is nonabelian is also irrelevant (though if $G$ is abelian then $x_i = x_0$ for all $i$, and the equation reduces to $(g_1^{-1}g_2)^j = x_0^j$, which is trivially true). And we don't actually need to rename $g_1$ (though it was probably done for some reason in the proof you were reading).
H: calculating (nPr/q!) % m for calculating the value of choosing r items from n items where q are of same kind, and we should take %m , i used the following relation (nPr/q!) %m where m is prime For calculating this i calculated n! n!%m then, i calculated (n-r)! and multiplied it with q!, i.e temp = (n-r)!q!; Then i mulitplied modular mulitplicative inverse of temp with n! and took mod of result but am not getting the correct answer..E.g if n= 3 ; r = 2; q = 2 then the expected result is (3P2/2!)%1000000007 = 3 but am getting 250000004 ..I can't understand my mistake here..Thanks. AI: The method you've described is correct, so there must be a bug in your implementation. In this case (n=3, r=2, q=2, m=1000000007): n! = 6 temp = (n-r)!q! = 2 Multiplicative inverse of 2 (mod m) = 500000004 result = (6*500000004) % 1000000007 = 3, the expected result.
H: How to show that $x \mapsto d(x,f(x))$ is continuous for a continuous $f$? I'm busy with a topology course, but the following question has me somewhat stumped. The entire question is, let $(X,d)$ be a metric space and $f:X \rightarrow X$ be continuous. Show that $X \rightarrow R$, $x \mapsto d(x,f(x))$ is continuous. I've tried proving it with the old-fashioned $\epsilon$-$\delta$ definition, but I quickly run out of options with that approach. If someone could give me a nudge in the right direction, it would be much appreciated! AI: The distance function $d: X\times X\longrightarrow \mathbb{R}^+ \cup \{0\}$ is continuous (this is an important fact on its own and you can prove it by using the triangle inequality). So is the function $i:X\longrightarrow X\times X$ defined by $i(x)=(x,f(x))$, because its projections are continuous. The composition of continuous functions is continuous.
H: Polynomial irreducibility criterion Given $f \in \mathbb{Z}[X]$ and $2\deg(f)+1$ distinct $a_i \in \mathbb{Z}$ such that $f(a_i)$ are prime numbers. Then $f$ is irreducible. I'm trying to prove that and I am stuck. Any hints for a good starting point? AI: Hint: Suppose $f=gh$ were a factorization where $g,h\in\mathbb{Z}[x]$ are not units (i.e. not equal to $\pm1$). Because $f(a_i)=g(a_i)h(a_i)$ is a prime number for each $i$, we must have that either $g(a_i)=\pm1$ or $h(a_i)=\pm1$ for each $i$ (the signs might be different for different $i$). Further hint (mouse over to reveal): Let $n=2\deg(f)+1$. WLOG, suppose $a_1,\ldots,a_k$ are the $a_i$ such that $g(a_i)=\pm1$, and $a_{k+1},\ldots,a_n$ are the $a_i$ such that $h(a_i)=\pm1$. Since $$2\deg(f)+1=2\deg(g)+2\deg(h)+1=n,$$ we have $$(2\deg(g)-k)+(2\deg(h)-(n-k))=-1$$ so that either $2\deg(g)< k$ or $2\deg(h)< n-k$.
H: Antiderivative $1/z$ on $\mathbb C$ Let $\Omega \subseteq \mathbb{C}$ be open and $\gamma:[\alpha,\beta]\rightarrow \Omega$ be a piecewise continuously differentiable and closed path. Why does $z^{-1}$ have no antiderivative on $\mathbb{C}\setminus0$? Why is $\int\limits_\gamma z^{-1} dz = 2\pi i \cdot \text{ind}_\gamma (0)$, where $\text{ind}_\gamma$ denotes the winding number. AI: Suppose, for the sake of contradiction, that $1/z$ has an antiderivative $F$ on $\mathbb{C}-0$. Then if $\gamma:[a,b]\rightarrow \mathbb{C}-0$ is a path, the integral over this path is just $F(\gamma(a))-F(\gamma(b))$, by the fundamental theorem of calculus. Now consider a closed loop that winds around 0. The integral should be $F(\gamma(a))-F(\gamma(a))=0$ because the starting and ending points are the same. But if you directly compute the integral of $1/z$ over the unit circle, you get $2\pi i$. This is a contradiction, and there is no antiderivative. Edit: I see you have edited your question, so I will edit my answer. Your second equation, about winding numbers, is true because of the generalized residue theorem. You can find this in any good book on complex analysis.
H: Need help with finding matrix inverse in $\mathbb{Z}/26\mathbb{Z}$ I am trying to learn the Hill Cipher and I am facing difficulties understanding how to find the inverse of a matrix in Modulo 26. What I've learnt so far is that I need to apply elementary row operations and apply modulo 26 after each operation until my matrix becomes in row echelon reduced form. I've taken the example matrix found in the upper link and tried to find its inverse following the algorithm, hoping that I would reach the result showed in wikipedia. Despite my effors no avail so far. Below are my steps, I've spent few nights banging my head against this. Please note that modulo 26 is applied after each elementary row operation: Start: \begin{array}{ccc\|ccc} 6 & 24 & 1 & 1 & 0 & 0\\ 13 & 16 & 10 & 0 & 1 & 0\\ 20 & 17 & 15 & 0 & 0 & 1\\ \end{array} R1 + R2 = \begin{array}{ccc\|ccc} 19 & 14 & 11 & 1 & 1 & 0\\ 13 & 16 & 10 & 0 & 1 & 0\\ 20 & 17 & 15 & 0 & 0 & 1\\ \end{array} 11R1,2R2 = \begin{array}{ccc\|ccc} 1 & 24 & 17 & 11 & 11 & 0\\ 0 & 6 & 20 & 0 & 2 & 0\\ 20 & 17 & 15 & 0 & 0 & 1\\ \end{array} 1/2R2 and then 9R2= \begin{array}{ccc\|ccc} 1 & 24 & 17 & 11 & 11 & 0\\ 0 & 1 & 12 & 0 & 9 & 0\\ 20 & 17 & 15 & 0 & 0 & 1\\ \end{array} 6R1 + R3= \begin{array}{ccc\|ccc} 1 & 24 & 17 & 11 & 11 & 0\\ 0 & 1 & 12 & 0 & 9 & 0\\ 0 & 5 & 13 &14 & 14 & 1\\ \end{array} 21R2 + R3= \begin{array}{ccc\|ccc} 1 & 24 & 17 & 11 & 11 & 0\\ 0 & 1 & 12 & 0 & 9 & 0\\ 0 & 0 & 5 & 14 & 21 & 1\\ \end{array} 21R3= \begin{array}{ccc\|ccc} 1 & 24 & 17 & 11 & 11 & 0\\ 0 & 1 & 12 & 0 & 9 & 0\\ 0 & 0 & 1 & 8 & 25 & 21\\ \end{array} As you can see I've 8 24 21 in the third row, whereas according to wikipedia's example there should have been 21 12 8. I've checked multiple times my calculations and I cant see anything wrong. Can anyone shed some light on what I am doing wrong? AI: It must be observed that determinant is initially $441= 25 \mod 26$ and it is a unit because $25\cdot 25=1 \mod 26$. And in the second step you multiply by $2$ that is a zero divisor in $\mathbb{Z}/~26\mathbb{Z}$. From this step the matrix has not inverse.
H: Are seminorms convex and a question on local base? How can I prove that seminorms are convex? Another question is that, when we talk about topological vector spaces, why do we emphasise neighborhoods about $0$? As far as I know, if we know the neighborhoods about $0$ then we can find all the open sets by translation. But can I not talk about open sets about any other point? I think I am missing something here. Thank you for helping. AI: Just use the axioms of a semi-norm: if $x,y$ are elements of the space and $\alpha\in [0,1]$, use triangular inequality and the fact that $\alpha$, $1-\alpha$ are non-negative. If $V$ is a neighborhood of $0$, $x+V:=\{x+v,v\in V\}$ is a neighborhood of $x$. Since in the context of topological vector spaces translations are supposed to be continuous, we just focus on open sets which contain $0$ (we can deduce the other by translation).
H: $3x^3 = 24$ quadratic equation Completing the square I know by factoring $$x^3 - 8 = 0\\ x-2 = 0$$ that one of the solutions is 2. but the other solutions is $1 ± i \sqrt 3$. Can someone explain to me how to get that? AI: May be you know this already but it's not clear to me from your post that you have the progression from the original problem statement to the final two factors exactly correct. $3x^3 = 24 \\ \Rightarrow 3x^3 - 24 = 0 \\ \Rightarrow 3\left(x^3 - 8\right) = 0 \\ \Rightarrow x^3 - 8 = 0 \\ \Rightarrow \left(x-2\right)\left(x^2+2x+4\right) = 0 \\ \Rightarrow x-2 = 0, \ \ x^2+2x+4 = 0$ The first factor results in $x = 2$. The second factor must be solved using the quadratic formula or by completing the square since it is not factorable (Quadratic Formula on Wikipedia).
H: Euler's summation by parts formula I'm beginning analytic number theory and I see this formula in Apostol's book : If $f$ has a continuous derivative $f'$ on the interval $[y,x]$, where $0 < y < x$, then $$ \sum_{y < n \le x} f(n) = \int_y^x f(t) \, dt + \int_y^x (t-[t]) f'(t) \, dt + f(x) ([x] - x) - f(y) ([y]-y). $$ The proof in Apostol's can be followed easily if one uses Riemann Integration. But since I meet with number theorists often I see more this kind of notation : $$ \sum_{y < n \le x} f(n) = \int_y^x f(t) d[t] = \text{something here I don't recall} - \int_y^x [t]f'(t) dt $$ because for some reason they can "integrate $d[t]$ and it gives $[t]$", which I don't understand, and I also don't really understand precisely what $d[t]$ stands for. I have done a measure theory course ; what I'm saying is that I don't understand all the details ; I understand that they "integrate by parts with the measure $d[t]$" which makes the proof quite simpler, but I don't understand the assumptions they make and how the details work out. I think that $d[t]$ could be a measure such that for $E \subseteq \mathbb R$ or $\mathbb C$, $d[t](E) = \| \mathbb N \cap E \|$, but I'm not sure. Here's what I'm looking for : I don't want an intuitive point of view with plots or summations ; I want a formal proof from the viewpoint of a measure theorist, with details. Is there anyway this can be made clear? The reason why I want this is because I don't have much faith in the "integration by parts with $d[t]$" version of the proof, but number theorists seem to love it so much and they all sketch it ; I never managed to do it myself formally, even though I did a measure theory course. Thanks for the help, AI: Here's what I'm looking for : I don't want an intuitive point of view with plots or summations ; I want a formal proof from the viewpoint of a measure theorist, with details. Find the book Montgomery and Vaughan's "Multiplicative Number Theory I. Classical Theory". This book is an excellent reference for many different subjects analytic number theory. Appendix A, "The Riemann Stieltjes integral," deals with precisely your question. It is $8$ pages long, and should answer everything. Edit: Zev kindly added links to Appendix A, so no need to check your library, or ask your prof.
H: A function and its Fourier transform cannot both be compactly supported I am stuck on the following problem from Stein and Shakarchi's third book. I can't figure out how to use the hint productively. Once I know $f$ is a trigonometric polynomial, I see how to finish the problem, but I don't know how to conclude that $f$ is a trigonometric polynomial. I tried substituting in the Fourier transform in the formula for Fourier coefficients and switching the order of integration, but I couldn't get that to work. I can't think of any more ideas. Problem: Suppose $f$ is continuous on $\mathbb{R}$. Prove $f$ and $\hat f$ cannot both be compactly supported unless $f=0$. Hint: Suppose $f$ is supported in $[0, 1/2]$ and expand it in a Fourier series in $[0,1]$. Show $f$ must be a trigonometric polynomial. This question was asked before, but with different hypotheses and in the context of complex analysis. Please do not close as a duplicate. AI: Further hint: let $c_n$ the $n$-th Fourier coefficient of $f$. We can write $\widehat f(n)=c_n$ and since $\widehat f$ is compactly supported, $c_n$ vanishes for $n$ large enough. It implies that $f$ is a trigonometric polynomial. The hypothesis of the hint is not restrictive: using a substitution in the integral defining the Fourier transform, we can assume that the support of $f$ is contained in $[0,a]$ for some $a>0$, then define $g(x):=f\left(\frac x{2a}\right)$.
H: Finding the Derivative Of $f(x) = 7\ln(5xe^{-x})$ The original question is $f(x) = 7\ln(5xe^{-x})$ I'm not sure if I have to use the chain rule to figure out $\ln(5xe^{-x})$ because $5xe^{-x}$ is one term within ln. My guess is that it's like this: $$7(-(e^{-x-1})/e^{-x})$$ or just simply $-7$. I'm specifically unsure with how to find the derivative of $5xe^{-x}$. I know that $e^x$'s derivative is simply $e^x(1)$ because the derivative of $x = 1$ so when I find the derivative of $e^{-x}$ I'd expect it to be $-1e^x$ and in my case $-5xe^{-x}$ AI: Hint:$$\ln\left(5xe^{-x}\right)=\ln 5+\ln x+\ln e^{-x}\ldots$$ Now just remember that the natural log. and the exponential function are inverse to each other and also $\,\left(\ln x\right)'=\frac{1}{x}\,$
H: Calculating formula to store location of Lower Triangular Matrix I am struggling with a problem from this textbook. The question is as follows: Determine a formula h = f(i,j) to store location MATRIX[i][j] in h. Ensure to only store nonzero elements. Then it asks how they can be stored in a single dimensional array. I'm unsure of how to go about solving this. Can someone help me better understand this? AI: Lets consider the matrix: \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} You can memorize it as \begin{pmatrix} a & b & c & d \end{pmatrix} You can do it with every matrix, and in fact the computer do it every time. If you have a $m\times n$ matrix you can access to the entry $a_{ij}$ with the formulae $(i-1)\times n + j$ (I start the array with $1$). Let's consider a lower triangular matrix. You can ignore the upper part of the matrix. You can actually use three major methods: store it by row, store it by column, proceed by diagonal entry. I will show you the first and the third method by you have to find the formulae (because it is an homework). Let's consider the matrix: \begin{pmatrix} a_{11} & 0 \\ a_{21} & a_{22} \\ \end{pmatrix} You save it as \begin{pmatrix} a_{11} & a_{21} & a_{22} \end{pmatrix}. In the third method you memorize it as \begin{pmatrix} a_{11} & a_{22} & a_{21} \end{pmatrix}
H: Why are there $3$ conjugacy classes of involutions in this centralizer? I'm trying to work through a sketch proof attributed to Walter Feit on characterizing $S_5$. Suppose $G$ is a finite group with exactly two conjugacy classes of involutions, with $u_1$ and $u_2$ being representatives. Suppose $C_1=C(u_1)\simeq \langle u_1\rangle\times S_3$ and $C_2=C(u_2)$ be a dihedral group of order $8$. The eventual result is that $G\simeq S_5$. Also, $C(u)$ denotes the centralizer of $u$ in $G$. The sketch says "$C_2$ contains three classes of involutions. If $x$ is an involution in $C_2$, $x\neq u_2$ then $x$ is conjugate to $xu_2$. Why does $C_2$ contain three classes of involutions, and why is any involution $x\in C_2$ such that $x\neq u_2$ conjugate to $xu_2$? My thoughts: From the answer here, I know that $C_2$ is a Sylow $2$-subgroup. Since $u_1$ is contained in a Sylow $2$-subgroup, and all the Sylow $2$-subgroups are conjugate to each other, I can replace $u_1$ by some conjugate, and without loss of generality assume that $u_1\in C_2$. Then $u_2\in C_1$ as well. Then I know $C_2$ contains at least two classes for $u_1$ and $u_2$. They can't be conjugate in $C_2$ lest they be conjugate in $G$. Thus $C_2$ has at least $2$ classes of involutions, but why are there exactly $3$? My guess is $u_1u_2$ might be the third class, but I can't make progress with this guess. Thanks. AI: The dihedral group of order $8$, $D_8 = \langle r,s \mid r^4 = s^2 = 1, sr = r^3s\rangle$, has exactly five elements of order $2$: $r^2$, which is central, and $s$, $rs$, $r^2s$, and $r^3s$, which are not central. Since $u_2\in C(u_2)$ and is central (since it commutes with everything in its centralizer), $u_2$ is a conjugacy class by itself, and corresponds to the element $r^2$ in $D_8$. Now notice that the conjugates of $s$ are just $s$ and $r^2s$: conjugating by a power of $r$ we get $r^isr^{-i} = r^{2i}s$, so we either get $r^2s$ (when $i$ is odd) or $s$ (when $i$ is even). And conjugating by an element of the form $r^is$ we get $(r^is)s(sr^{-i}) = r^isr^{-i} = r^{2i}s$ again. Similarly, the conjugates of $rs$ are $rs$ and $r^3s$. Now, notice that $r^3s = (rs)r^2$, $rs = (r^3s)r^2$, $s=(r^2s)r^2$, and $r^2s = sr^2$. So, in each case, if $x\in D_8$ is an involution different from $r^2$, then the other conjugate of $x$ is $xr^2$. So this happens simply because you have a group isomorphic to the dihedral group of order $8$, and nothing to do with the fact that it is a centralizer of an involution of a group with exactly two conjugacy classes of involutions. Note also that you will not necessarily have $u_1u_2\in C_2$: that would require $c_1$ and $c_2$ to centralize each other, and this will depend on the representatives you choose (since, in $S_5$, the two classes of involutions are the transpositions $(i,j)$, and the products of two transpositions $(r,s)(t,u)$; these may commute, e.g., $(1,2)$ commutes with $(1,2)(3,4)$; but they need not commute, as for example $(1,2)$ does not commute with $(1,3)(4,5)$ ).
H: Cauchy integral formula for convex sets Again, I'm struggling with a proof. Cauchy integral theorem for convex sets (Preliminary lemma) Let $\Omega\subseteq \mathbb{C}$ open and convex, $ p\in\Omega,\ f$ : $\Omega\rightarrow \mathbb{C}$ continuous, $f\in H(\Omega\backslash \{p\})$ ($f$ analytic on $\Omega \backslash \{p\}$. Then there is a $F\in H(\Omega)$ (F analytic on $\Omega$) with $F'(z)=f(z)$ for all $ z\in\Omega$. It also folows that $\displaystyle \int\limits_{\gamma}f(z)\mathrm{d}z=0$ for every piecewise continuously differentiable and closed path $\gamma\in\Omega$. Cauchy integral formula Let $\Omega\subseteq \mathbb{C}$ be open and convex, $f\in H(\Omega)$ ($f$ analytic on $\Omega$) and $\gamma$ a piecewise continuously differentiable and closed path in $\Omega$. Then, we have: $$ f(z)\cdot \mathrm{ind}_{\gamma}(z)=\frac{1}{2\pi i}\int\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in\Omega\backslash \gamma^{}) $$ where $\mathrm{ind}_\gamma$ is the winding number and $\gamma^ = \gamma([a,b])$ for $\gamma:[a,b]\rightarrow\mathbb{C}$. Proof Fix $z\in\Omega\backslash \gamma^{}$ and define $g:\Omega\rightarrow \mathbb{C}$ by: $$ g(\xi):=\left\{\begin{array}{l} \frac{f(\xi)-f(z)}{\xi-z},\ \xi\in\Omega\backslash \{z\}\\ f'(z),\ \xi=z \end{array}\right. $$ Then we have $g$ continuous on $\Omega$ and $g \in H(\Omega\backslash \{z\})$. $\color{red}{\text{(1) Why is g continuous on } \Omega \text{ and analytical on } \Omega\backslash \{z\}}$ From the Cauchy integral theorem for convex sets we have $\displaystyle \frac{1}{2\pi i}\int\limits_{\gamma}g(\xi)\mathrm{d}\xi=0$, so: $\color{red}{\text{(2) Where does that } \frac{1}{2\pi i} \text{ come from? There is no factor in the mentioned Cauchy integral theorem}}$ $$ \frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)-f(z)}{\xi-z}\mathrm{d}\xi\ =0\ (z\in\Omega\backslash \gamma^{}) $$ This completes the proof. AI: (1) $g$ is analytic on $\Omega\backslash \{z\}$ because both the function $\xi\mapsto f(\xi)-f(z)$ and the function $\xi \mapsto \frac{1}{\xi-z}$ are analytic there, hence the product is analytic. $g$ is continuous on $\Omega\backslash \{z\}$ since it is analytic there, the only point in question is at $z$, but this follows since $f$ is differentiable there. (2) The factor doesn't matter, and is a convenience so that you end up with $f(z)\cdot \mathrm{ind}_{\gamma}(z)$, but it will work out anyway, even if you don't multiply it.
H: For a reduced ring $A$, must $A[x]\setminus A$ be multiplicatively closed? Suppose $A$ is a reduced commutative ring. Is $A[x]\setminus A$ is multiplicatively closed? AI: Let $A$ be any reduced commutative ring with nonzero zero divisors, say $ab=0$ for some nonzero $a$ and nonzero $b$. Since $(ax)(bx)=0$, $A[x]\setminus A$ is not multiplicatively closed. So, you need to assume $A$ is at least a domain. Once that is true, then $A[x]\setminus A$ is obviously multiplicatively closed: just consider the degrees of polynomials in question!
H: Triangle related question My question is: In $\Delta ABC$, let $AE$ be the angle bisector of $\angle A$. If $\frac{1}{AE} = \frac{1}{AC} + \frac{1}{AB}$ then prove that $\angle A = 120^\circ$. What I tried: I extended side $AB$ and took a point $M$ on it such that $AC$ is congruent to $AM$. Then I proved that $AE$ is parallel to $MC$. I was trying to prove that $\triangle AMC$ is equilateral so that I get $\angle MAC=60^\circ$. But I am not able to prove it. Any help to solve this question would be greatly appreciated! AI: Mark point $D$ on side $AC$ such that $AE=AD$. Rewrite the given relation as follows: $$\frac{1}{AE}-\frac{1}{AC}=\frac{1}{AB}$$ $$\frac{AC-AE}{AE\cdot AC}=\frac{1}{AB}$$ $$\frac{AC-AD}{AD\cdot AC}=\frac{1}{AB}$$ $$\frac{DC}{AD}=\frac{AC}{AB}$$ Now by the angle bisector theorem: $$\frac{BE}{EC}=\frac{AB}{AC}$$ Combining we obtain: $$\frac{DC}{AD}=\frac{EC}{BE}$$ Therefore, $$\frac{AC}{DC}=\frac{AD+DC}{DC}=1+\frac{AD}{DC}=1+\frac{BE}{EC}=\frac{BE+EC}{BE}=\frac{BC}{EC}$$ Hence $\Delta ABC$ is isometric to $\Delta CDE$. Hence $\angle CDE=\angle CAB$. Now $AE=AD$ hence in $\Delta AED$ we have $\angle AED = \angle ADE=\beta$. $\angle DAE = \frac{1}{2}\angle CAB = \frac{1}{2} \angle CDE = \alpha$. We obtain: $$\angle DAE + \angle ADE + \angle AED = \alpha + 2\beta = \pi$$ $$\angle CDE +\angle ADE = 2\alpha + \beta = \pi$$ Solving we find $\alpha = \beta$, hence $\angle DAE = \angle ADE = \angle AED =\frac{\pi}{3}$, hence $\angle CAB = \frac{2\pi}{3}$.
H: Power series expansion for analytic functions Theorem Let $\Omega\subseteq \mathbb{C}$ be open and $f\in H(\Omega)$ ($f$ analytic on $\Omega$). If $ C(z_{0},R)\subseteq\Omega$ (where $C(z_0,R)$ is the circle with origin $z_0$ and radius $R$), then we can represent $f$ on $C(z_{0},R)$ as a power series with convergence radius $\geq R$. Proof Let $0<r<R$ and $\gamma:[0,2\pi]\rightarrow\Omega$ be the path $\gamma(t)=z_{0}+re^{it}$. Because $C(z_{0},R)$ is convex and $\gamma$ is in this circle, we have the following from the Cauchy integral formula: $$ f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in C(z_{0},\ r)) $$ $\color{red}{\text{(1) Why not directly on } C(z_{0},\ R)? }$ Because $C(z_{0},\ r)\subseteq \mathbb{C}\backslash \gamma^{}$ it follows that $f(z)=\displaystyle \sum_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ for a power series with convergence radius $\geq r$. Because $ r\in (0,\ R)$ is chosen arbitrarily and because the the coefficients $c_{n}$ are determined by the differentials $f^{(n)}(z_{0})$, the power series $\displaystyle \Sigma_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ has the convergence radius $\geq R$. $\color{red}{\text{(2) Why are the coefficients calculated via }f^{(n)}(z_{0})? }$ $\color{red}{\text{(3) Why does it follow that the convergence radius is } \geq R? }$ AI: For (1), we choose $r<R$ because the path $\gamma_R(t)=z_0+Re^{it}$ for $t\in[0,2\pi]$ may not lie inside the region $\Omega$. This is because while $C(z_o,R)\subset\Omega$, its boundary, which is $\gamma_R^$, may not be contained within $\Omega$. Now, to see why the integral formula only holds when $z\in C(z_0,r)$, and not on its boundary (which is $\gamma^*$ in your statement), this is simply due to the statement of the Cauchy Integral Formula. That is, it only holds for points that do not lie on $\gamma$. For (2), the coefficients $c_n$ are determined by the differentials $f^{(n)}(z_0)$ because of the Taylor formula (which we can obtain from the finite Taylor formula if need be). For (3), we have shown that the power series has radius of convergence $r$ where $r$ is an arbitrary positive number less than $R$. Thus, the radius of convergence is at least $\sup_{0<r<R} r= R$. So the radius of convergence is greater than or equal to $R$.
H: $\lim _{(x,y)\rightarrow (0,0)} \ln(\sin(x^2+y^2))=?$ I am working on this practice question for an upcoming exam. I am not sure if I am oversimplifying this here (as the question is worth 3 marks): In polar coordinates I have: $\ln(\sin(r^2))$, so on any path, as $(x,y)\rightarrow(0,0)$, $r\rightarrow 0$. For $r$ sufficiently close to 0, lim $r\rightarrow 0$ r$\sin(r^2) = 0$. Therefore, $\ln(\sin(r^2))\rightarrow -\infty$. We are allowed to take $-\infty$ as a a valid limit. Any help/hints would be greatly appreciated! AI: You intuition is correct but we need to get your justification a little more solid. If you want to establish the limit is $-\infty$, you can argue that for every $n\in\mathbb{N}$, you can find an $r\in(0,\infty)$ such that $\ln(\sin(r^2))<-n$. Added: To be absolutely clear, this works here because $\ln(\sin(r^2))$ is monotonically increasing between 0 and a small positive number. You might be more comfortable arguing $\epsilon$-$\delta$ style too.
H: Parametric Plot for Ellipse I have two vectors $a$ and $b$ and want to perform a parametric plot that generates an ellipse out of it. Normally you would do that for: $$t \mapsto \cos(t) \cdot a + \sin(t) \cdot b$$ But when I look up parametric plot, I only find examples like: $\sin(t), \cos(t)$ or $t \cdot \sin(t), 5 \cdot \cos(t)$ These plots are defined through commas, one value for $x$ and one for $y$. But how do I know map the definition above to this comma notation? AI: a and b are vectors (over $\mathbb{R}$), so one can write them as $a=(a_1,a_2)$ and $b = (b_1,b_2)$. Thus your parameterization is $t \rightarrow (a_1 \cos(t) + b_1 \sin(t), a_2 \cos(t)+b_2 \sin(t)) $ for $t \in [0,2\pi]$. That should do the trick. Have a play with the numbers $a_1,..,b_2$ and see what the results are when you plot them!
H: Derivative of a function with variable range Suppose for example that I have a function $g_y(x)$ such that $g_y(x) = \begin{cases} y+x &\mbox{if } -x<y<1-x \\ 1 & \mbox{if } y>1-x \\0 &\mbox{if } y<-x \end{cases}$ How, if possible, would I find $ \frac{dg}{dx} $? Edit: Also, I need to have that $ x\in (0,1)$ AI: $$g_y(x) = \begin{cases}0 & \text{if } x < -y\\ x+y & \text{if }x \in (-y,1-y)\\ 1 & \text{if }x > 1-y\end{cases}$$ You also need $x \in (0,1)$. We will split it into cases depending on where $y$ lies. If $y \leq -1$, then we get that $g_y(x) = 0$ for all $x \in (0,1)$. If $-1 \leq y \leq 0$, then we get that $$g_y(x) = \begin{cases}0 & \text{if } x \in (0,-y)\\ x+y & \text{if }x \in (-y,1)\end{cases}$$ for all $x \in (0,1)$. If $0 \leq y \leq 1$, then we get that $$g_y(x) = \begin{cases} x+y & \text{if }x \in (0,1-y)\\ 1 & \text{if }x \in(1-y,1)\end{cases}$$ for all $x \in (0,1)$. If $y \geq 1$, then we get that $g_y(x) = 1$ for all $x \in (0,1)$. Now you should be able to work out the derivative with relative ease. Note that if $y \in [-1,0]$, then the function is not differentiable at $-y$ and for $y \in [0,1]$, the function is not differentiable at $1-y$.
H: Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing $4\sin(A)\sin(B)\sin(C)$ to $$4\sin(B+C)\sin(A+C)\sin(A+B)$$ by making substitutions by reorganizing $A+B+C=\pi$. I then did the same thing to the other side to get $$-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))$$ and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. I am pretty sure there is some simpler way of proving the equality, but I can't seem to figure it out. Maybe there is a geometric interpretation or maybe it can be done using just algebra and trig. Any hint's would be appreciated (I would prefer an algebraic approach, but it would be nice to see some geometric proofs as well) AI: Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities \begin{align} \sin(\pi-\theta) & = \sin(\theta)\\ \cos(\pi-\theta) & = -\cos(\theta)\\ \sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\ \sin(2\theta) & = 2\sin(\theta) \cos(\theta)\\ \cos(\theta - \phi) - \cos(\theta + \phi) & = 2 \sin(\theta) \sin(\phi) \end{align} We have that \begin{align} \sin(2A) + \sin(2B) & = 2 \sin(A+B) \cos(A-B)\\ & = 2 \sin(\pi-C) \cos(A-B)\\ & = 2 \sin(C) \cos(A-B) \end{align} Hence, \begin{align} \sin(2A) + \sin(2B) + \sin(2C) & = 2 \sin(C) \cos(A-B) + 2 \sin(C) \cos(C)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(C) \right)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(\pi-(A+B)) \right)\\ & = 2 \sin(C) \left(\cos(A-B) - \cos(A+B) \right)\\ & = 2 \sin(C) \times 2 \sin(A) \sin(B)\\ & = 4 \sin(A) \sin(B) \sin(C) \end{align}
H: Simpson's rule failure on $\cos\pi x$ I think I know why but I can't represent this accurately with a graph. I am supposed to show why Simpson's Rule is so far off for the integral $$\int_0^{20} \cos \pi x$$ I know that the answer should be zero because it repeats on that interval, 10 up and 10 down evenly. I know that the antiderivative $\sin \pi x$ will be zero for any 0 or pi value so that evaluates to zero. When I use Simpson's Rule I get $$\frac{2}{2} (1 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 +1) = 9$$ I am pretty certain that the reason the numbers are so far off is because the interval is 2 and that will cover an up and down which Simpson's Rule will overestimate but I can not show this on a graph to equal 9. AI: I am an idiot and I did the graph wrong. It overestimates because it makes a rectange on each subinterval.
H: Evaluating integral similar to $\sin^{-1}(3x)$ I was just doing revision for an upcoming exam and I came across a question I do not know how to answer. It seems pretty simple, however I am having a blank moment. If anyone knows how to solve that I would really appreciate the help. The question: It is given that y = $\sin^{-1}(ax)$, where a is a positive constant and $-\frac{1}{a}\leq x\leq \frac{1}{x}$. a) Find $\frac{dy}{dx}$ in terms of y and a. b) Hence find $\frac{dy}{dx}$ in terms of x and a. c) Hence, or otherwise, find $\int \frac{1}{\sqrt{1-9x^{2}}}dx$. d) Find the exact value of $\int_{0}^{1/6}\sqrt{1-9x^{2}} dx$ I have finished parts a to c (workings below) but cannot seem to solve part d (I do realize it's probably relatively simple, however I have just spent good 45 minutes trying to solve it or find the answer and I can't seem to solve it) $$ y=\sin^{-1}(ax)\\\sin(y)=ax\\\frac{dy}{dx}\cos(y)=a\\\Rightarrow\frac{dy}{dx}=\frac{a}{\cos(y)}=\frac{a}{\sqrt{1-\sin^{2}(y)}}=\frac{a}{\sqrt{1-a^{2}x^{2}}}\\\int\frac{1}{\sqrt{1-9x^{2}}}dx = \frac{1}{3}\int{\frac{3}{\sqrt{1-9x^{2}}}}dx=\frac{1}{3}\sin^{-1}(3x)+c$$ Any help would be greatly appreciated. AI: For the last part, let $x = \dfrac{\sin(\theta)}{3}$. As $x$ goes from $0$ to $1/6$, we have that $\theta$ goes from $0$ to $\pi/6$. Also, $dx = \dfrac{\cos(\theta)}{3} d \theta$. Hence, $$I = \int_0^{1/6} \sqrt{1-9x^2} dx = \int_0^{\pi/6} \sqrt{1 - \sin^2(\theta)} \dfrac{\cos(\theta)}{3} d \theta = \int_0^{\pi/6} \dfrac{\cos^2(\theta)}{3} d \theta$$ Hence, $$I = \dfrac16 \int_0^{\pi/6} 2 \cos^2(\theta) d \theta = \dfrac16 \int_0^{\pi/6} (1 + \cos(2 \theta)) d \theta = \dfrac16 \left( \theta + \dfrac{\sin(2 \theta)}{2} \right)_{0}^{\pi/6}$$ Hence, $$I = \dfrac16 \left(\dfrac{\pi}{6} + \dfrac{\sin(\pi/3)}{2}\right) = \dfrac16 \left(\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{4}\right) = \dfrac1{72} \left(2 \pi + 3 \sqrt{3} \right)$$
H: The theorem on ordinary differential equations stated in the appendix of Kobayashi-Nomizu The following theorem is stated in the appendix I of Foundation of differential geometry by Kobayashi-Nomizu. They say the proof will be found in various text books on differential equations. I checked several books, but could not find it. Theorem Let $E$ and $F$ be finite dimensional normed spaces over $\mathbb R$. Let $U$ and $V$ be non-empty open subsets of $E$ and $F$ respectively. Let $J$ be an open interval of $\mathbb R$ containing $0$. Let $f:J×U×V → E$ be a map. Suppose $f$ is differentiable of class $C^p, 0 ≦ p ≦ ω$ in $J$ and of class $C^q, 1 ≦ q ≦ ω$ in $U$ and $V$. Then there exist open subinterval $J_0$ of $J$ containing $0$, non-empty open connected subsets $U_0, V_0$ of $U, V$ respectively and a unique map $g:J_0×U_0×V_0 → U$ which satisfy the following properties. (1) $g$ is differentiable of class $C^{p+1}$ in $J_0$ and of class $C^q$ in $U_0$ and $V_0$. (2) $D_tg(t, x, s) = f(t, g(t, x, s)$, s) for all $(t, x, s) ∈ J_0×U_0×V_0$ (3) $g(0, x, s) = x$ for all $(x, s) ∈ U_0×V_0$ AI: This is a combination of the existence and uniqueness theorem and the smooth dependence on parameters. See e.g. Solomon Lefschetz, "Differential Equations: Geometric Theory", II.9.1.
H: The spectra of weighted shifts Since weighted shifts are like the model-operators in operator theory and people have been studying them for so long, I think there should be quite a large literature on the spectra of such operators. However, after some search I hardly found anyone which gives a complete picture of what the spectra of a weighted shift. Most of the papers I found deal with some specific properties of their spectra, but the question I have in mind is what the spectra are, exactly. I wonder whether there is some good reference on this. Thanks! AI: I've found that some good material on this subject can be found in Paul Halmos' A Hilbert Space Problem Book. Chapter 11, entitled Spectral Radius, contains some of the basic results in operator theory along with some specific problems about weighted shifts. In particular, the exercises work through the norm, spectral radius, point spectrum and the approximate point spectrum of a weighted shift. Some other exercises in this section utilize weighted shifts to construct examples and counterexamples to various questions presented. One nice fact about weighted shifts is that given weights, $a_n,b_n$, if $\|a_n\|=\|b_n\|$ for all $n$ (either in $\mathbb{N}$ or $\mathbb{Z}$ depending on whether it is the unilateral or bilateral shift), then the weighted shifts determined by $a_n$ and $b_n$ are unitarily equivalent (This is exercise 89). A direct corollary to this is that the spectrum of a weighted shift is radially symmetric about zero. The questions contained here don't answer in its entirety the question: "what is the spectrum of a weighted shift." However, they do contain some particularly illuminating facts and examples. A Hilbert Space Problem Book contains exercises, hints, and solutions. I highly recommend it for a better understanding of many aspects of operator theory. Edit: After doing a little light reading, I came across an article that gives a much more complete picture of the the spectra of weighted shifts. The paper is entitled Approximate Point Spectrum of a Weighted Shift by William C. Ridge. It gives a very explicit breakdown of the parts of the spectra of weighted shifts based on different conditions on their weights. But, for the big picture, the spectrum of a unilateral weighted shift is always a closed disk centered at the origin; the spectrum of a bilateral weighted shift is always connected.
H: Rudin's Principles of Analysis Theorem 1.11 I'm having a problem with this theorem. What if set $B$ is all $x$ such that $\sqrt{2} < x \le 2$, and $S$ is the set of all $y$ such that $\sqrt{2} < y \le 3$. $\sup L = \sqrt{2}$, which does not exist in $S$. Does this prove the theorem wrong by contradiction? 1.11 Theorem Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$. AI: No, your example does not meet the hypotheses of the theorem, because $B$ is not bounded below in $S$. Note that $L=\{a\in S:\text{for all }b\in B, a\leq b\}=\emptyset$.
H: Lemma on bounding one metric by a multiple of another - clarification My book (Principles of Topology by Fred Croom) states the following lemma: "Let $d_1$ and $d_2$ be two metrics for the set $X$ and suppose that there is a positive number $c$ such that $d_1(x,y) \le cd_2(x,y)$ for all $x,y\in X$. Then the identity function $i: (X,d_2) \to (X,d_1)$ is continuous." My question is simple: must $c$ be a constant independent of $x$ and $y$, or is it a not-necessarily-constant positive number? Experience suggests the former, but the lack of the word 'constant' and two possible interpretations of the wording make me question it. I worked out a proof that every metric space is homeomorphic to a bounded metric space in which it works to set $c=d(x,y)$, and this is the reason that I want to get this straight. AI: The constant $c$ is independent of the points $x$ and $y$. To see why this is important, let $d_1$ be the usual metric on $\Bbb R$, and let $d_2$ be the discrete metric given by $$d_2(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\;.\end{cases}$$ Let $f:\Bbb R\to\Bbb R:x\mapsto x$ be the identity map. It’s easy to see that $f$ is not continuous as a function from $\langle\Bbb R,d_1\rangle$ to $\langle\Bbb R,d_2\rangle$. On the other hand, if we set $c(x,y)=\max\{1,\|x-y\|\}$ for each $x,y\in\Bbb R$, it’s easy to check that $$d_1(x,y)=\|x-y\|\le c(x,y)\cdot1=c(x,y)d_2(x,y)$$ for all $x,y\in\Bbb R$.
H: A (non-artificial) example of a ring without maximal ideals As a brief overview of the below, I am asking for: An example of a ring with no maximal ideals that is not a zero ring. A proof (or counterexample) that $R:=C_0(\mathbb{R})/C_c(\mathbb{R})$ is a ring with no maximal ideals. A homework question in my algebra class earlier this year asked to exhibit a ring (necessarily without identity) without any maximal (proper) ideals. For solutions to this, it suffices to exhibit an abelian group $(G,+)$ without maximal (proper) subgroups. For, given such a group, define multiplication to be constantly zero. In this case, $G$ becomes a zero ring without maximal ideals (because ideals correspond to subgroups). It is not particularly difficult to construct examples of the above. For example, consider $(\mathbb{Q},+)$. Another interesting example is $P=\{z\in\mathbb{C}\mid \exists n\in\mathbb{N}, z^{p^n}=1\}$ with standard complex number multiplication as "addition" (that is, as the abelian group operation). However, any example constructed in this manner is a zero ring, and as such seems "artificial," which I admit, is not a rigorous term. I would like to find a somewhat less artificial example of a ring without maximal ideals. For a definition of "less artificial," let us start with "not a zero ring." I have a candidate in mind, but I am having trouble explicitly proving that it has no maximal ideals. Let $C_0(\mathbb{R})$ denote the ring of continuous real-valued function on $\mathbb{R}$ vanishing at infinity. Let $C_c(\mathbb{R})$ denote the (two-sided) ideal of compactly supported functions. I believe that the ring $R:=C_0(\mathbb{R})/C_c(\mathbb{R})$ contains no maximal ideals, but I am having trouble showing it. My intuition for this problem is as follows. Given a function $f\in C_0(\mathbb{R})$, $f(x)$ approaches zero at some "rate" as $x\to\pm\infty$ (possibly different based on $\pm$). Furthermore, for any given rate, we can find a function with a larger rate, in the sense that we can find a $g\in C_0(\mathbb{R})$ such that $f(x)=o(g(x))$ ($f$ is little-$o$ of $g$). Now, even if $f$ is non-vanishing, there is no $h\in C_0(\mathbb{R})$ such that $fh=g$, for any $h$ could not vanish at infinity. Thus the principal ideal generated by $f$ does not contain $g$. By iterating this process we could construct a strictly ascending chain of principal ideals. Now, the idea is that the ring $R$ consists of these "rates" as described above. I know that this is not precise, or necessarily even correct. But it's my intuition. The previous paragraph shows that we can find an ascending chain of "rates," but a lot of work still needs to be done. If anyone can clean this up, it would be much appreciated. AI: In this paper Mel Henriksen shows that a commutative ring $R$ has no maximal ideals iff (a) $J(R)=R$, where $J(R)$ is the Jacobson radical of $R$, and (b) $R^2+pR=R$ for every prime $p\in\Bbb Z$. He then gives three examples. One starts with a field $F$ of characteristic $0$ and forms the integral domain $$S(F)=\left\{h(x)=\frac{f(x)}{g(x)}\in F(x):f(x),g(x)\in F[x]\text{ and }g(0)\ne 0\right\}\;;$$ its unique maximal ideal is $R(F)=xS(F)$, which has no maximal ideals. This paper by Patrick J. Morandi also constructs some examples.
H: Evaluate the partial derivatives of the following function: The function $f: \mathbb{R} \longrightarrow \mathbb{R}$ is defined by the rule $$f(x,y) = \begin{cases} \frac{x^5y}{x^4+y^2}, & (x,y) \neq (0,0), \\ 0, & (x,y)=(0,0). \end{cases}$$ Evaluate $f_x(0,y)$, $f_x(0,0)$, $f_y(x,0)$, and $f_y(0,0)$. (The definition of partial derivative as a limit is the recommended method.) My question is really a conceptual question, that I am struggling to understand. At the point $(0,0)$, the function is defined to be $0$. So, finding the partial derivatives at those points, why would you not use $f(x,y)=0$? My instincts tell me that no, you would use the other part of the function (partly because the question suggests using the limit definition): $\dfrac{x^5y}{x^4+y^2}$, $(x,y) \neq (0,0)$. But why is this? Secondly, in general, why would you use the limit definition of a derivative, as opposed to simply taking the derivative mechanically? What is the difference? Thanks for any help! AI: First: Your instinct is correct. Derivatives are "local information." This means that the derivative of a function at any point is computed by using information about the function's nearby values. For example, consider the limit definition of a derivative of a single-variable function: $$f'(x) = \lim_{h\to x}\frac{f(h) - f(x)}{h - x}.$$ In order to compute this, we need to understand all of the values $f$ takes at the $h$ values near $x$. Second: Taking the derivative "mechanically" is just cranking through a set of rules that you have learned for some particularly nice functions (products, sums, quotients of differentiable functions; exponentials, trig functions, polynomials, etc). You can't take the derivative at $0$ mechanically here because, at zero, your function is not one of those particularly nice functions. Since you can't use the differentiation rules, you're stuck with computing the derivative straight from the limit definition.
H: Dedekind Cut Proof I am greatly confused with Dedekind cuts... I am trying to prove that this is a Dedekind cut: If $D$ and $E$ are in $\mathbb{Q}$ and are Dedekind cuts, then prove that $$D*E=(-\infty, 0] \cup \{r_1r_2\mid 0 < r_1 \in D, 0 < r_2 \in E\}$$ is a Dedekind cut as well. My three propositions of a Dedekind cut are: 1.) If $r\in D$ and $s < r$, then $s \in D$. 2.) There is a number $x \in \mathbb{Q}$ so that $r\leq x$ for all $r \in D$. 3.) If $r \in D$, then there is a number $s \in D$ so that $r < s$. After looking at many sources, my concept of a Dedekind cut is falling short... and so this proof is. I would be greatly appreciative for a simple definition and example of a Dedekind cut, and/ or help on this proof. Thx! AI: Let’s take a closer look at the conditions defining a Dedekind cut. A subset $D$ of $\Bbb Q$ is a Dedekind cut if: whenever $r\in D$ and $s<r$, then $s\in D$; there is a number $q\in\Bbb Q$ such that $r\le q$ for all $r\in D$; and for each $r\in D$ there is an $s\in D$ such that $r<s$. Think of the rationals in the usual pictorial fashion, laid out as a line extending infinitely far in both directions, negative rationals to the left and positive rationals to the right. Condition (1) says that if $D$ contains some rational $r$, it contains every rational to the left of $r$ as well. One set that satisfies this condition is $\Bbb Q$ itself. Another is $(\leftarrow,0)$, the set of negative rationals, containing every rational strictly to the left of $0$. Yet another is $(\leftarrow,1]$, the set of rationals at or to the left of $1$. On the other hand, the set $(\leftarrow,0)\cup(2,4)$ does not satisfy condition (1): it contains $3$, but it doesn’t contain $1$ even though $1<3$. It’s not an initial segment of $\Bbb Q$. Condition (2) is the simplest: it just says that $D$ cannot extend infinitely far to the right. There must be some rational number $q$ such that every member of $D$ is at or to the left of $q$. Almost all of the sets that satisfy condition (1) satisfy condition (2) as well; the only exception is $\Bbb Q$ itself, and condition (2) is designed specifically to rule out $\Bbb Q$. Condition (3) is perhaps the hardest to get a grip on, but what it says is actually quite simple: it says that $D$ must not have a largest element. No matter what $r$ you choose in $D$, $D$ contains some bigger number $s$. This rules out sets like $(\leftarrow,1]=\{q\in\Bbb Q:q\le 1\}$, that have a maximum element. A Dedekind cut, therefore, is a subset of the rational numbers that looks more or less like $(\leftarrow,2)$, say: it’s an initial segment of the rational number line, it’s not the whole line, and it has no largest element. In fact, every $r\in\Bbb Q$ defines a Dedekind cut in just this way, namely, the cut $$(\leftarrow,r)=\{q\in\Bbb Q:q<r\}$$ consisting of every rational to the left of (smaller than) $r$. However, these aren’t the only Dedekind cuts. For example, let $$D=\{q\in\Bbb Q:q<0\text{ or }q^2<2\}\;.$$ It’s not hard to check that this $D$ satisfies conditions (1)-(3); if you’ve not already seen this, you probably will soon. But $D$ is not $(\leftarrow,r)$ for any rational number $r$, because if it were, we could show that $r^2=2$, when in fact we know that $\sqrt2$ is irrational. It’s these ‘extra’ Dedekind cuts, the ones not of the form $(\leftarrow,r)$ for any rational number $r$, that make Dedekind cuts useful and interesting: they ‘fill in’ the gaps in $\Bbb Q$ corresponding to the irrational numbers and allow us to construct the real numbers rigorously starting with just the rationals. In the end the cuts $(\leftarrow,r)$ for $r\in\Bbb Q$ are going to correspond to the rationals, and the ‘extra’ Dedekind cuts are going to correspond to the irrationals. But in order to make that work, we have to be able to define the various arithmetic operations on these Dedekind cuts in such a way that they behave the way they should. Your exercise here is part of showing how to define multiplication of Dedekind cuts. I’ll say nothing about the exercise itself, as breeden has already covered that in some detail. Added: In view of the comments, I think that I should emphasize that in what I’ve written above, $(\leftarrow,q)$ and so forth are to be understood as subsets of $\Bbb Q$. That is, I’m writing $(\leftarrow,q)$ as an abbreviation for $\{r\in\Bbb Q:r<q\}$, not for $\{r\in\Bbb R:r<q\}$. As another example to show how a set can be bounded above (as is required by (2)) and still have no largest element, let $D$ be the set of negative rationals. Every member of $D$ is less than $0$ (and hence less than every positive rational as well), but $D$ has no largest element. If $m$ and $n$ are positive integers, so that $-\frac{m}n\in D$, then $-\frac{m}n<-\frac{m}{2n}<0$, so $-\frac{m}{2n}$ is a member of $D$ that is bigger than $-\frac{m}n$. This shows that $D$ has no largest element: give me any element of $D$, and I’ve just shown you how to find a bigger one.
H: Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where $r = 2a\cos\theta + 2b\sin\theta$ Now I forgot how to derive this. So I tried using the standard form of a circle. $$(x-a)^2 + (y - b)^2 = a^2 + b^2$$ $$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$ $$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) = a^2 + b^2$$ $$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$ Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought? AI: I think your substitutions from the first line to the second aren't quite right. It looks like you used $x=a\cos\theta$ and $y=b\sin\theta$, but you probably wanted $$\begin{align}x&=r\cos\theta\\y&=r\sin\theta.\end{align}$$ Using those, $$\begin{align} (x-a)^2 + (y - b)^2 &= a^2 + b^2 \\ (r\cos\theta-a)^2+(r\sin\theta-b)^2&=a^2+b^2 \\ r^2\cos^2\theta+a^2-2ar\cos\theta+r^2\sin^2\theta+b^2-2br\sin\theta&=a^2+b^2 \\ r^2(\sin^2\theta+\cos^2\theta)+(a^2+b^2)-2r(a\cos\theta+b\sin\theta)&=a^2+b^2 \\ r^2-2r(a\cos\theta+b\sin\theta)&=0 \\ r^2=2r(a\cos\theta+b\sin\theta) \\ r=0\;\;\text{ or }\;\;r=2a\cos\theta+2b\sin\theta \end{align}$$ and since $r=0$ describes only the pole, which is also contained in the graph of $r=2a\cos\theta+2b\sin\theta$, $$r=2a\cos\theta+2b\sin\theta$$ is sufficient to describe the circle.
H: Question on some arithmetic calculations When $6272$ is multiplied by $0.94$ the answer is $5895.68$. When it is divided by $1.06$ the answer is $\approx 5916.9811$. Why is it so? Just as a little background, I am using the default Microsoft calculator for this calculation. I haven't pulled out Mathematica yet. Could this be because of some weird variable conversion (int to double)? Any help would be much appreciated. AI: You may be doing rounding too much and too early. You may figure, correctly, that multiplying by $0.94$ is equivalent to dividing by $\frac{1}{0.94}$. Calculate $\frac{1}{0.94}$. It is approximately $1.0638298$. You may have thrown away the part after the $1.06$, figuring it wouldn't make much difference. Well, percentagewise it is not much difference, but in dollars the difference is significant. There is no mysterious fault in the Windows calculator, it is doing the calculation correctly, to good accuracy. To get good performance out of a calculator, let it do its thing, only round off at the end. Remark: In general, if $a\ne 0$, multiplying by $a$ is equivalent to dividing by $\frac{1}{a}$.
H: How can we solve: $\sqrt{x} + \ln(x) -1 = 0$? How could we solve $$\sqrt{x} + \ln(x) -1 = 0$$ without using Mathematica? Obviously a solution is $x = 1$, but what are the other exact solutions? AI: Both $\sqrt x$ and $\ln x$ are increasing functions of $x$, so $\sqrt x+\ln x=1$ can have at most one solution. As you note, it does have one, namely $x=1$, but that must be the only one: $\sqrt x+\ln x<1$ when $0<x<1$, and $\sqrt x+\ln x>1$ when $x>1$.
H: How can we solve: $\sqrt{x} - \ln(x) -1 = 0$? How could we solve $$\sqrt{x} - \ln(x) -1 = 0$$ without using Mathematica? Obviously a solution is $x = 1$, but what are the other exact solutions? This question is inspired by my first question How can we solve: $\sqrt{x} + \ln(x) -1 = 0$?. Here the situation is not clear. Any ideas? AI: The other real solution, according to Maple, is $4 \text{LambertW}(-1, -e^{-1/2}/2)^2$, which is approximately $12.34020237$. Note that if $f(x) = \sqrt{x} - \ln(x)$, $f'(x) = \dfrac{1}{2\sqrt{x}} - \dfrac{1}{x} = \dfrac{\sqrt{x}-2}{x}$, so $f(x)$ is decreasing on $(0,4]$ and increasing on $[4,\infty)$. Since $f(4) = 2 - 2 \ln(2) < 0$ while $f(x) \to +\infty$ as $x \to 0+$ and as $x \to +\infty$, there are two real solutions, one in $(0,4)$ and one in $(4,\infty)$. To explain that LambertW solution: if $x = 4 t^2$ with $t > 0$, the equation says $$4 t^2 = x = e^{\sqrt{x}-1} = e^{2 t-1}$$ and thus $$ -t e^{-t} = -e^{-1/2}/2 $$ Now $\text{LambertW}(s)$ is defined to be a solution $w$ of $w e^w = s$. If $e^{-1} < s < 0$, as in this case $s = -e^{-1/2}/2$, there are two real solutions, the "principal branch" (which in this case gives $t=1/2$ and thus $x=1$), and the "$-1$ branch". There are infinitely many other branches, which are all complex.
H: If a line bundle and its dual both have a section (on a projective variety) does this imply that the bundle is trivial? Is there any reason that, on a projective variety X, if a line bundle L has a (non-zero) section and also its dual has a section then this implies that L is the trivial line bundle? AI: Yes, there is a reason for $L$ to be trivial and here it is: Let $0\neq s\in \Gamma(X,L)$ and $0\neq \sigma\in \Gamma(X,L^)$ be two non zero sections. Then $s\otimes \sigma\in \Gamma(X,L\otimes L^)=\Gamma(X,\mathcal O)$ is a constant since $X$ is complete: $s\otimes \sigma =c\in k$ (the base field). Now, since $s$ and $\sigma$ are non-zero there is a non-empty open subset $U\subset X$ on which both do not vanish and on which $s\otimes \sigma=c $ does not vanish either: in other words $c\neq0\in k$ . Since $s\otimes \sigma =c\neq 0$, a non-zero constant, vanishes nowhere we conclude that a fortiori $s$ vanishes nowhere, so that $L$ is trivial, as announced, since $ s\in \Gamma(X,L)$ .
H: lower bound for probability of no 2 balls per bin. There are $n$ balls and $m$ bins and every ball is placed independently and uniformly at random into a bin. I'm trying to show that there exists a constant $c$ such that, if $m=c\sqrt{n}$ then with probability $\geqslant 1/2$, no two balls will collide into the same bin, for sufficiently large $n$. (In textbooks I found the solution for upper bounding this probability.) Let $A_i$ be the event that the $i$-th ball does not collide with balls $1,\dots,i-1$. So essentially I want to lower bound the probability $$ p = P\left[\ \bigcap_{i=2}^m A_i\ \right] = \prod_{i=2}^m\left( 1 - \frac{i-1}{n}\right). $$ Assuming that $n \geqslant 2m$, we can use the bound $1-x\geqslant e^{-x-x^2}$ (for $x\leqslant \frac{1}{2}$). This gives $$ p\geqslant \exp\left(\sum_{i=2}^m\left(-\frac{(i-1)^2}{n^2}-\frac{i-1}{n}\right)\right) = \exp\left( \frac{-(m-1) m (2 m+3 n-1)}{6 n^2} \right). $$ But now I'm stuck, because setting $m=\Theta(\sqrt{n})$, doesn't seem to yield constant probability. What am I missing? AI: @André Nicolas: $m$ and $n$ are transposed in the first sentence (notice that the product of $A_i$ over all balls goes up to $m$). @somebody: Your strategy of using the lower bound $e^{-x-x^2}$ is fatally flawed, since the $x^2$ term gives rise to the undesirable cubic polynomial in $m$ that is causing a problem later. What you want is to find a lower bound for $1-x$ which looks more like $e^{-cx}$. For instance, it is easy to check that $$1-x \ge e^{-2x},$$ for $0\le x\le \tfrac12$. Now you should have no difficulty following through the argument to find a value of $m = c\sqrt{n}$ which does the job.
H: Question about distribution Let $(f_k)_{1\le k\le \infty}\in L_{1}^\mathrm{loc}(\mathbb{R}^n)$ be a sequence of real valued functions such that $\operatorname{supp} f_k \subset \{\|x\|\le k^{-1}\}$, $$\int f_k (x)\,dx=1,k\in \{1,2,\ldots,\infty\}$$ Show that the sequence $(f_k^2)_{1\le k\le\infty}$ does not converge in $\mathcal{D}'(\mathbb{R}^{n})$ as $k\rightarrow \infty$. AI: Take $\psi$ a bump function: smooth function, with support contained in $B(0,2)$, non-negative, and $\psi=1$ on $B(0,1)$. If $\{f_k^2\}$ where convergent in $\mathcal D'(\Bbb R^n)$, in particular the sequence $\{a_k\}:=\{\int_{\Bbb R^n}f_k^2(x)\psi(x)dx\}$ would be bounded. Since the support of $f_k^2$ is contained in $B(0,k^{-1})$, we have $a_k=\int_{\|x\|\leq k^{-1}}f_k^2(x)dx$. Denoting $M:=\sup_{j\in\Bbb N}\int_{\|x\|\leq j^{-1}}f_j^2(x)dx$ $$1=\int_{B(0,k^{-1})}f_k(x)dx\leq \sqrt{\|B(0,k^{-1})\|}\sqrt{\int_{\|x\|\leq k^{-1}}f_k^2(x)dx}\leq \sqrt{\|B(0,k^{-1})\|}\sqrt M$$ hence $\frac 1M\leq \|B(0,k^{-1}\|$, which is a contradiction, because the volume of the ball of radius $k^{-1}$ converges to $0$ as $k\to +\infty$.
H: Citable Reference for Picard's Theorem in Banach Space I was wondering if anyone knew of a legitimate citable reference where Picard's Theorem on the existence of solutions to ODEs in Banach space is proven? For some reason I can only find proofs for the case of $\mathbb{R}^n$ in any of the books I have looked at. I realize that the proof is the same, but I would like to have a reference where I do not have to mention this fact. Thanks in advance AI: Try Theorem 6, pp.117-118 of the survey paper Lobanov, S. G. & Smolyanov, O. G. "Ordinary differential equations in locally convex spaces". Uspekhi Mat. Nauk, 1994, 49, 93-168 (link to paper). It being partially a survey paper, the theorem is stated without proof; if you want the proofs, you'd have to follow the references therein.
H: Fields as a reflective subcategory of integral domains? A subcategory $\mathbf A$ is reflective subcategory of $\mathbf B$ if for every $B\in\mathbf B$ there exists an $A_B\in\mathbf A$ and a $\mathbf B$-morphism $r_B \colon B \to A_B$ such that: for any $A\in\mathbf A$ and any $\mathbf B$-morphism $f \colon B \to A$ there exists unique $\mathbf A$-morphism $\overline f \colon A_B \to A$ such that $\overline f\circ r_B=f$. This is equivalent to saying that the inclusion functor $E \colon \mathbf A \to \mathbf B$ has a left adjoint. Current revision of the Wikipedia article on reflective subcategories claims that The category of fields is a reflective subcategory of the category of integral domains. The reflector is the functor which sends each integral domain to its field of fractions. I don't think this is true - an easy counterexample is the homomorphism $f \colon \mathbb Z \to\mathbb Z_2$ given by $f(n)=n\mod 2$. This homorphism obviously cannot be extended to $\overline f \colon \mathbb Q \to \mathbb Z_2$. The above claim would be true if we considered the categories of integral domains only with injective ring homomorhpisms. Am I correct? Did I miss something there? The same problem has already been mentioned on the talk page of this Wikipedia article, so this is more-or-less just a sanity check. AI: Yes, you are correct... and the category of integral domains is defined to only have the injective homomorphisms as arrows. Unfortunately, this part of the definition sometimes seems to be left out when the category is introduced....
H: Derivative of $\sec^{-1}(\frac{x}{3})$ Derivative of $\sec^{-1}(\frac{x}{3})$ I have tried these types of problems with two different approaches and keep getting the same answer which seems to be wrong. I suspect I am doing something obvious incorrectly; however, I can't seem to figure it out. First method: $y = \sec^{-1}(\frac{x}{3})$ therefore, $\sec(y) = \frac{x}{3}$ $\frac{dy}{dx}\sec(y)\tan(y) = \frac{1}{3}$ $\frac{dy}{dx} = \frac{1}{3\sec(y)\tan(y)}$ Since $\sec(y) = \frac{x}{3}$ and $\tan(y) = \sqrt{\sec^2(y) -1} = \sqrt{\frac{x^2}{9} - 1}$ $\frac{dy}{dx} = \frac{1}{x\sqrt{\frac{x^2}{9} - 1}}$ Method two: $(f')^{-1}(\frac{x}{3}) = \frac{1}{f'(\sec^{-1}(\frac{x}{3}))}$ $ = \frac{1}{3\sec(\sec^{-1}{(\frac{x}{3})})\tan(\sec^{-1}(\frac{x}{3}))}$ $= \frac{1}{x\sqrt{\sec^2(\sec^{-1}(\frac{x}{3})) - 1}}$ $= \frac{1}{x\sqrt{\frac{x^2}{9} -1}}$ AI: WA says $\frac{d}{dx}\sec^{-1}(x/3)=3/\left(x^2\sqrt{1-\frac{9}{x^2}}\right)$, which confirms your results: $$ \begin{eqnarray} \frac{dy}{dx} &=& \frac{1}{x\sqrt{\frac{x^2}{9} - 1}} \text{ take out $\frac{x^2}{9}$ from the $\sqrt{\cdot}$ }\\ &=&\frac{1}{x\frac{x}{3}\sqrt{1-\frac{9}{x^2} }}\\ &=&\frac{3}{x^2\sqrt{1-\frac{9}{x^2} }} \end{eqnarray} $$
H: A problem of compactness and connectedness Consider the subset $A$ and $B$ of $\mathbb{R}^2$ defined by $A =\{(x, x\sin\frac{1}{x}) :x\in(0,1]\}$ $B = A\cup \{(0,0)\}$ I have to check for compactness and connectedness of $A$ and $B$. Here is my attempt. $A$ is bounded but not closed as 0 is the limit point of set $A$ but it doesn't belongs to $A$. Hence $A$ is not compact. $B$ is compact since it is closed and bounded subset of $\mathbb{R}^2$. I am not able to figure out connectedness of given sets. Am I correct? Is there any other way to tackle this problem? I need help with this. Thank you very much AI: I think you mean that $A$ is not closed because $(0,0)$ is a limit point of $A$ that is not in $A$, not $0$. That’s correct, and $B$ is closed and therefore compact, though you haven’t really justified the assertion that it’s closed. Note that the function $$f:(0,1]\to\Bbb R^2:x\mapsto\left(x,x\sin\frac1x\right)$$ is continuous, and $(0,1]$ is connected; what does that tell you about the connectedness of $A$? Can you show that the function $$f:[0,1]\to\Bbb R^2:x\mapsto\begin{cases}f(x),&\text{if }x\in(0,1]\\0,&\text{if }x=0\end{cases}$$ is also continuous? If so, that gives you an easy way to see that $B$ is both compact and connected, because $[0,1]$ is both compact and connected.
H: Equating coefficients in a Fourier series Suppose, for example, using Fourier series techniques to solve a differential equation leads to the following: $a_0 + \sum_{n=1}^{\infty}a_n\sin(nx)+b_n\cos(nx)=4\sin x$ At this point, why can you equate the coefficients of $\cos(nx)$ and $\sin(nx)$ on the two sides the equation. I don't understand why this step works. With very many thanks, Froskoy. AI: Informally you can recover the coefficient by averaging both sides against $\sin mx$: $$\frac{1}{\pi}\int_{-\pi}^{\pi}\left( a_0 + \sum_{n=1}^\infty a_n\sin(nx) + b_n\cos(nx)\right)\sin(mx)\, dx =\frac{1}{\pi}\int_{-\pi}^\pi 4\sin x \sin (mx)\, dx, $$ because the left hand side becomes then $$a_0 \frac{1}{\pi}\int_{-\pi}^\pi\sin(mx)\, dx+ \sum_{n=1}^\infty a_n\frac{1}{\pi}\int_{-\pi}^\pi\sin(nx)\sin(mx)\, dx+b_n\frac{1}{\pi}\int_{-\pi}^\pi\cos(nx)\sin(mx)\, dx =a_m $$ while the left hand side is $4\delta_{1m}$. This procedure, similar to the original reasoning by Fourier, requires the interchange of integral and sum in left hand side, which can be done if the series converges uniformly. However, that request is excessively restrictive: turns out that convergence in $L^2$ sense will suffice. This can be proven by means of Hilbert space methods.
H: Evaluating $f_x(0,0)$ and $f_y(0,0)$ using the limit definition of the partial derivative Possible Duplicate: Evaluate the partial derivatives of the following function: The function $f:\mathbb R^2 \to \mathbb R$, defined as: $$\left\{\begin{align}&\frac{x^5y}{x^4+y^2}&&(x,y) \neq 0\\&0&&(x,y)=(0,0)\end{align}\right.$$ Using the limit definition I get: $$f_x(0,0) = \frac{(0+h)^2 (0)}{h(0+h)^4+(0)} - \frac{(0)^5(0)}{h(0)^4+h(0)^2}$$ for both $f_x$ and $f_y$, I get $\frac00$ terms for the last term. I know that the answer to both is $0$, but how do I deal with the $\frac00$ term (I can't seem to get rid of it). Is this allowed? Thanks for any insight! AI: For $h\ne 0$ you should have $$\frac{f(0+h,0)-f(0,0)}h=\frac{f(h,0)-f(0,0)}h=\frac{h^5\cdot0}{h(h^4+0^2)}-0=\frac0{h^5}=0\;.$$ As a function of $h$ this is the constant $0$, so its limit as $h\to 0$ is $0$.
H: Summing the series $ \sum_{n=0}^{\infty} (-1)^n\int_0^1 \cos(nt^2)\mathrm dt$ I would like to find the sum of the series $$ \sum u_{n}$$ where $$ u_{n}=(-1)^n\int_0^1 \cos(nt^2)\mathrm dt$$ Using the change of variable $t\rightarrow \sqrt{n}t$: $$ u_{n}=\frac{(-1)^n}{\sqrt{n}} \int_0^{\sqrt{n}} \cos(t^2)\mathrm dt\sim_{n\rightarrow \infty} \frac{(-1)^n}{2}\sqrt{\frac{\pi}{2n}}$$ So $\sum u_{n}$ is convergent. What about $$ \sum_{n=0}^{\infty} (-1)^n\int_0^1 \cos(nt^2)\mathrm dt$$ ? AI: The m-th partial sum of your sum is the real part of $$ \sum_{n=0}^{m} (-1)^n \int^1_0 \exp(int^2) dt = \int^1_0 \sum_{n=0}^{m} \left(-\exp(it^2)\right)^n dt= \int^1_0 \frac{1-(-\exp(it^2))^{m+1}}{1+ \exp(it^2) } dt.$$ The integral $\displaystyle I_m = \int^1_0 \frac{\exp(imt^2)}{1+\exp(it^2)} dt $ tends to $0$ by the Riemann Lebesgue Lemma. Thus $$\sum_{n=0}^{\infty} (-1)^n \int^1_0 \exp(int^2) dt = \int^1_0 \frac{1}{1+\exp(it^2)} dt.$$ We can compute $$ \Re\left(\frac{1}{1+\exp(it^2)}\right) = \frac{1+\cos(t^2)}{\sin^2(t^2)+(\cos(t^2)+1)^2} =1/2$$ so $$\sum_{n=0}^{\infty} (-1)^n \int^1_0 \cos(nt^2) dt = 1/2.$$
H: Integrating Reciprocals of Polynomials I have seen integrals of the form $$\int \frac{1}{ax+b}dx$$ and $$\int \frac{1}{ax^{2}+bx+c} dx$$ But I cannot see how to integrate reciprocals of higher degree - does there exist a general solution to the integrals of reciprocals of cubics, quartics, and higher? AI: Reciprocals of higher degree can have their denominators factored into linear and/or irreducible quadratic terms, and from there, our result can be obtained through partial fraction decomposition. For more details, see Arturo's excellent answer to this question.
H: A basic question related with compactness I have to check for compactness of given subets of $\mathbb{R}^2$. $A =\{(x, y) :xy = 1\}$ $B =\{(x, y) :x^2y^2 = 1\}$ $C =\{(x, y) :e^x = \cos y\}$ $D =\{(x, y) :\mid x\mid +\mid y \mid \leq 10^{100}\}$ The purpose of asking above question is not just to get answers. I need concepts to deal with these kind of problems. Let me explain where I face difficulties. Take set $A$; Intuitively this is clear to me that $A$ is not a compact subset of $\mathbb{R}^2$ as it is closed but unbounded. My problem is I am having trouble with checking boundedness or unboundedness of given subsets. Here I know that set $A$ consists of points which lies on rectangular hyperbola. So I have no difficulty in judging that set $A$ is unbounded. But I am not sure about others. Since I am not able to figure out set them. Edit: I don't want graphical approach to solve these problems. Because quite often i face problems where I find myself unable to visualize graph of given functions. I think there must be available some mathematical tool to deal with this. I need help to understand this. I would be very much thankful to all of you. AI: Draw pictures, or at least visualize. In set $B$, $x$ can be arbitrarily big. If you pick any $x\ne 0$, there is a $y$ such that $x^2y^2=1$. So there is no disk with centre the origin that contains all of $B$. Also in set $C$, $x$ can be arbitrarily large negative. Just choose $y$ close to $\pi/2$, but a tiny bit below. Or else note that $y$ can be anything that makes $\cos y$ positive, and there are arbitrarily large $y$ with this property, such as $y=2n\pi$ where $n$ is any positive integer. The set $D$ is clearly bounded, it is inside the disk with centre the origin and radius $10^{100}$. So you need to check whether or not $D$ is closed.
H: Working with Linear Operators on Vector Space Let $\phi: V \rightarrow V$ be a linear operator on a vector space V over a field F Prove that $V = \phi(V)\bigoplus NS (\phi )$ if and only if $\phi(V ) = \phi^2(V)$ AI: I'll use $\,\ker\phi\,$ instead of $NS$: Suppose $\,V=\phi(V)\oplus\ker\phi\,$ and let$\,x\in\phi(V)\Longrightarrow\,\exists y\in V\,\,s.t.\,\,x=\phi y$ , but:$$y\in V\Longrightarrow\,\exists!\,v=\phi t\in\phi(V)\,,\,u\in\ker\phi\,\,s.t.\,\,y=\phi t+u\Longrightarrow$$$$\Longrightarrow x=\phi y=\phi^2t+\phi u=\phi^2t\Longrightarrow x\in\phi^2(V)$$Now you try the other direction

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