qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
148,002 | <p>I'm trying to get my head round the following calculation of the fundamental group of the torus, using Seifert Van-Kampen (I know it's easier to do this by considering covering spaces, but I'm trying to learn the Seifert Van-Kampen method). </p>
<p>Consider the torus $T$ as the unit square in $\mathbb R^2$ with opp... | John Engbers | 31,778 | <p>The key here is that you want to analyze the image of $l_*$. The generator of $\pi_1(A\cap B)$ is one loop around the 'circle' $A \cap B$. When this generator of $\pi_1(A \cap B)$ is mapped into $\pi_1(B)$, it maps onto the boundary of the circle, and so it maps onto $u.v.\bar{u}.\bar{v}$. In other words, pushing... |
425,713 | <blockquote>
<p>We say that a metric space <span class="math-container">$M$</span> is <em>totally bounded</em> if for every <span class="math-container">$\epsilon>0$</span>, there exist <span class="math-container">$x_1,\ldots,x_n\in M$</span> such that <span class="math-container">$M=B_\epsilon(x_1)\cup\ldots\cup B... | Brian M. Scott | 12,042 | <p>HINT: Let $\langle x_k:k\in\Bbb N\rangle$ be any sequence in $M$, and suppose that $n\in\Bbb Z^+$. There is a finite $F\subseteq M$ such that $M=\bigcup_{x\in F}B\left(x,\frac1n\right)$, so there is a $y_n\in M$ such that $\left\{k\in\Bbb N:x_k\in B\left(y_n,\frac1{2n}\right)\right\}$ is infinite. Use this observati... |
194,642 | <p>Given a poset $(P,\leq)$ the <em>interval topology</em> on $P$ is generated by
$$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$
where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.</p>
<p>Let $\{P_i : i\in I\}$ be a family of posets such that the interval... | Todd Trimble | 2,926 | <p>The statement in the answer by Dominic that the interval topology of the product poset is the product topology of the interval topologies is incorrect. The argument that the product topology contains the interval topology is correct, but the one for the opposite containment is not, as shown in this <a href="https://... |
2,410,655 | <p><em>Okay so I'm asking this quesion knowing a thing or two about sequences and general terms</em></p>
<p>What is the sum of the series :
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$</p>
<blockquote>
<p><strong>My Try:</strong> <em>I tried calculating the general term $T_{n}$ for the sequence but ... | Martin Sleziak | 8,297 | <p>Let me start with some steps. Perhaps you'll be able to finish.</p>
<p>$$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 &-2 &-1 &-3\\
-1 & 5 & 2 & 9\\
3 & \alp... |
282,889 | <p>I am trying to show that if $f$ is continuous on the interval $[a,b]$ and its upper derivative $\overline{D}f$ is such that $ \overline{D}f \geq 0$ on $(a,b)$, then $f$ is increasing on the entire interval. Here $\overline{D}f$ is defined by
$$
\overline{D}f(x) = \lim\limits_{h \to 0} \sup\limits_{h, 0 < |t| \leq... | Arin Chaudhuri | 404 | <p>This is based on David's comment above. </p>
<p>Choose $u,v, \epsilon$ such that $ a < u < v < b$ and $ \epsilon > 0 $. </p>
<p>Let $$ S = \{ x \in [u,v] : f(x) + \epsilon x \geq f(u) + \epsilon u \}.$$ $S$ is not empty as $ u \in S$ and $S$ is closed as $f$ is continuous. </p>
<p>Let $\sup S = t$, $t... |
282,889 | <p>I am trying to show that if $f$ is continuous on the interval $[a,b]$ and its upper derivative $\overline{D}f$ is such that $ \overline{D}f \geq 0$ on $(a,b)$, then $f$ is increasing on the entire interval. Here $\overline{D}f$ is defined by
$$
\overline{D}f(x) = \lim\limits_{h \to 0} \sup\limits_{h, 0 < |t| \leq... | Mark Perlman | 74,779 | <p>The statement is not true. The Weierstrass function has upper derivative greater than zero everywhere, it is continuous, and it is not an increasing function. This question comes as an error in Royden, and should read that the lower derivative of $f$ is greater than or equal to zero, per this errata.</p>
<p><a href... |
95,176 | <p>Does the octic,</p>
<p>$\tag{1} x^8+3x^7-15x^6-29x^5+79x^4+61x^3+29x+16 = nx^2$</p>
<p>for any constant <em>n</em> have Galois group of order 1344? Its discriminant <em>D</em> is a perfect square,</p>
<p>$D = (1728n^4-341901n^3-11560361n^2+3383044089n+28121497213)^2$</p>
<p>Surely (1) is not an isolated result. ... | Community | -1 | <p>Maple gives Galois group $H$ of your polynomial $f=f_n$ as the subgroup of $S_8$ generated by these permutations: $$(1 2)(5 6), (1 2 3)(4 6 5), (1 2 6 3 4 5 7), (1 8)(2 3)(4 5)(6 7), (2 8)(1 3)(4 6)(5 7), (4 8)(1 5)(2 6)(3 7).$$ That can be easily proved by using the standard technique of Galois theory assuming that... |
3,255,737 | <blockquote>
<p>If <span class="math-container">$\phi$</span> is the solution of the integral equation <span class="math-container">$$\phi(x)=1-2x-4x^2+\int_0^x[3+6(x-t)-4(x-t)^2]\phi(t)dt$$</span></p>
<p>Then the value of <span class="math-container">$\phi(\log 2)$</span> is</p>
<p>(a). 2</p>
<p>(b). ... | Ali Shadhar | 432,085 | <p>in the body, I reached
<span class="math-container">\begin{align}
S&=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1-x^2)}{x(1-x^2)}\ dx\\
&=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\left[\ln(1-x)+\ln(1+x)\right]}{x(1-x)(1+x)}\ dx,\qquad 1-x=y\\
&=-2\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln x+\ln(2-x)... |
2,337,856 | <p>Is there an easy way to show, that the only solution of this system of non-linear equations has only the solution a=0, b=0, c=0</p>
<p>$a+b+c=0$</p>
<p>$ax+by+cz=0$</p>
<p>$ax^2+by^2+cz^2=0$</p>
<p>For $x,y,z\neq 0$ and different.</p>
<p>Solving this gets really ugly.
Is there an elegant way?
It is obvious that... | Claude Leibovici | 82,404 | <p>As said in comments, using $$x=\sqrt t\implies dx=\frac{dt}{2 \sqrt{t}}$$ you have
$$\int x^k e^{-x^2} \,dx=\frac{1}{2}\int t^{\frac{k-1}{2}}e^{-t}\,dt=-\frac{1}{2} \Gamma \left(\frac{k+1}{2},t\right)$$ where appears the incomplete gamma function.</p>
<p>For the case of $$\int_{-\infty}^\infty x^k e^{-x^2} \,dx=... |
2,337,856 | <p>Is there an easy way to show, that the only solution of this system of non-linear equations has only the solution a=0, b=0, c=0</p>
<p>$a+b+c=0$</p>
<p>$ax+by+cz=0$</p>
<p>$ax^2+by^2+cz^2=0$</p>
<p>For $x,y,z\neq 0$ and different.</p>
<p>Solving this gets really ugly.
Is there an elegant way?
It is obvious that... | zipirovich | 127,842 | <p>If $k$ is odd, $k=2n+1$, then we can substitute $t=x^2$ to obtain
$$\int x^ke^{-x^2}\,dx=\int x^{2n+1}e^{-x^2}\,dx=\frac{1}{2}\int t^ne^{-t}\,dt,$$
which then can be integrated by parts $n$ times. It's not too difficult to figure out an inductive formula or maybe even an explicit formula for this case from here.</p>... |
1,118,398 | <p>So i have $ABCD$ is a convex quadrlateral and $E$ is the intersection point of diagonals. Given that $AE=2,BE= 5, CE = 6, DE =10$ and side $BC = 5$. I know the formula $A=\frac{1}{2}d_1 d_2 \sin \theta$ but I don't have the angle and there is no figure so Im not able to find the sides I was thinking to get all the s... | DeepSea | 101,504 | <p><strong>Hint:</strong> $BC^2 = BE^2+EC^2-2\cdot BE\cdot EC\cos \angle BEC\Rightarrow 5^2=5^2+6^2 - 2\cdot 5\cdot 6\cos \angle BEC$</p>
|
1,118,398 | <p>So i have $ABCD$ is a convex quadrlateral and $E$ is the intersection point of diagonals. Given that $AE=2,BE= 5, CE = 6, DE =10$ and side $BC = 5$. I know the formula $A=\frac{1}{2}d_1 d_2 \sin \theta$ but I don't have the angle and there is no figure so Im not able to find the sides I was thinking to get all the s... | JimT | 409,742 | <p>You don't need to bother with exact value of the angle. Since — see above — $\cos\angle BEC = 3/5$ then $\sin\angle BEC = 4/5$ and thus area equals $AC\cdot BD\cdot\sin\angle BEC/2 = 8\cdot 15 \cdot (3/5)/2 = 36$</p>
|
4,438,491 | <p><span class="math-container">$$\sum _{k=2}^{\infty }\:\frac{1}{\sqrt{k}\left(\ln k\right)^{\ln k}}$$</span></p>
<p>I've tried limit comparison with <span class="math-container">$\frac{1}{\sqrt{n}}$</span> and <span class="math-container">$\frac{1}{n^2}$</span> and also Cauchy condensation test but nothing seems to w... | RRL | 148,510 | <p>Note that for <span class="math-container">$k > e^{e^2}$</span> we have <span class="math-container">$\ln \ln k > 2$</span> and</p>
<p><span class="math-container">$$(\ln k )^{\ln k} = e^{\ln \ln k \cdot \ln k}= k ^{\ln\ln k}> k^2$$</span></p>
<p>Thus, for all sufficiently large <span class="math-container"... |
772,997 | <p>My question is pretty much what it says in the headline.<br>
Is $A = f\mathbb{R^2}$ complete, where $f:\mathbb{R^2}\to\mathbb{R^3}$, $f(x,y) = (x,y,x^2-y)$, $(x,y)\in\mathbb{R^2}$.</p>
<p>$f\mathbb{R^2}=$$\{ f(x,y), (x,y) \in \mathbb{R^2} \}$</p>
<p>My initial thought is that as $f\mathbb{R^2}$ is $f$:s image (i.e... | souf | 135,820 | <p>Thanks for your help, I'm pretty sure I got it now. And I think I understood the idea of homeomorphisms better.
Yes, thank you for pointing out the two different concepts. I still think it is <em>image</em> I was trying to refer to. I think my course book uses unorthodox notations every now and then.
According to m... |
3,758,763 | <p>Does there exist a symmetric matrix <span class="math-container">$A$</span> such that <span class="math-container">$2^{\sqrt{n}}\le |\operatorname{Tr}(A^n)|\le2020 \cdot2^{\sqrt{n}}$</span> for all <span class="math-container">$n$</span>?</p>
<p>I think no. The trace of <span class="math-container">$A^n$</span> equ... | Dmitry | 743,044 | <p>As you said, for a symmetric matrix <span class="math-container">$A$</span> we have <span class="math-container">$Tr(A^n) = \sum_i \lambda_i^n$</span>. Now consider two cases for <span class="math-container">$\lambda_\max = \max_i |\lambda_i|$</span>:</p>
<ul>
<li><span class="math-container">$\lambda_\max \le 1$</s... |
3,758,763 | <p>Does there exist a symmetric matrix <span class="math-container">$A$</span> such that <span class="math-container">$2^{\sqrt{n}}\le |\operatorname{Tr}(A^n)|\le2020 \cdot2^{\sqrt{n}}$</span> for all <span class="math-container">$n$</span>?</p>
<p>I think no. The trace of <span class="math-container">$A^n$</span> equ... | Driss Alami Louati | 336,891 | <p>Let <span class="math-container">$A\in\mathbb{R}^{d\times d}$</span> and <span class="math-container">$A$</span> is symmetric. The eigenvalues <span class="math-container">$\lambda_1,\ldots,\lambda_d$</span> of <span class="math-container">$A$</span> are real.
For all <span class="math-container">$n\in\mathbb{N}$</s... |
194,813 | <p>Please help me solving $\displaystyle\lim_{x\to a}\frac{a^{a^{x}}-{a^{x^{a}}}}{a^x-x^a}$</p>
| Madrit Zhaku | 34,867 | <p>$\displaystyle\lim_{x\to a}\frac{a^{a^{x}}-{a^{x^{a}}}}{a^x-x^a}$=$\displaystyle\lim_{x\to a}\frac{a^{x^{a}}({a^{a^{x}-x^a}}-1)}{a^x-x^a}$=$a^{a^{a}}\displaystyle\lim_{x\to a}\frac{{a^{a^{x}-x^a}-1}}{a^x-x^a}$=$|\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\ln a|$=$a^{x^{a}}\ln a$.</p>
|
395,678 | <p>Please, I need to know the proof that </p>
<p>$$\left(\sum_{k=0}^{\infty }\frac{n^{k+1}}{k+1}\frac{x^k}{k!}\right)\left(\sum_{\ell=0}^{\infty }B_\ell\frac{x^\ell}{\ell!}\right)=\sum_{k=0}^{\infty }\left(\sum_{i=0}^{k}\frac{1}{k+1-i}\binom{k}{i}B_in^{k+1-i}\right)\frac{x^k}{k!}$$</p>
<p>where $B_\ell$, $B_i$ are Be... | Community | -1 | <p>$$\left(\sum_{k=0}^{\infty} \dfrac{n^{k+1}}{k+1} \dfrac{x^k}{k!} \right) \left(\sum_{l=0}^{\infty} B_l \dfrac{x^l}{l!}\right) = \sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l}$$</p>
<p>$$\sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l} = \sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{n^{m-l+1}}{m-l+... |
1,037,316 | <p>I want to write this polynomial in factored form:</p>
<p>$$x^4+4$$</p>
<p>The reason I want to do this is to be able to make <strong>partial-fraction decomposition</strong> on it to make an integrand easier to integrate. <em>What's the general method?</em></p>
<p>In addition to this, I searched on how to figure o... | Community | -1 | <p><strong>Hint</strong></p>
<p>$$x^4+4=(x^4\color{red}{+4x^2}+4) \color{red}{-4x^2}$$
and notice that
$$a^2-b^2=(a-b)(a+b)$$</p>
|
338,625 | <p>P-adic numbers are complete in one sense and incomplete in another sense. Is it so?</p>
<p>Firstly, does not complete mean connected? I read somewhere that there is not intermediate value theorem for p-adics because they are not connected. (if I am correct).</p>
<p>It seems I need elaboration of this "It can be sh... | Andy | 1,415 | <p>After re-reading your question I think I understand what you want explained: take $\mathbb{Q}$, define the $p$-adic absolute value on it as $\left| x \right|_p = p^{-n}$, when $p$ is a factor in the unique factorization of $x$ ($ x = p^n a/b$, and $p$ does not divide $a$ or $b$), if $p$ is not part of the unique pri... |
1,518,804 | <p>In real analysis we showed that if $\displaystyle \lim_{x\to x_0}|f(x)|=|L|$, then not necessarily $\displaystyle \lim_{x\to x_0}f(x)=L$ (the converse is true). </p>
<p>I want to find a counter example in complex analysis, i.e, if $\displaystyle \lim_{z\to z_0}|f(z)|=|L|$, then not necessarily $\displaystyle \lim_{... | Neeraj Kumar | 263,885 | <p>If $\displaystyle \lim_{z\to z_0}|f(z)|=|-L|=|L|$. So which one to converge to $L $ or $-L$?</p>
|
3,441,225 | <p>Let <span class="math-container">$S=1-1/3+1/5-1/7+\cdots$</span>. As each term in the series is decreasing and tends to <span class="math-container">$0$</span>, it is known that their sum exists and is finite by alternating series test. And by considering <span class="math-container">$\int_0^11/(1+x^2)dx$</span>, it... | user | 505,767 | <p>We have that for any <span class="math-container">$k$</span></p>
<p><span class="math-container">$$\frac1{2k-1}-\frac1{2k+1}=\frac2{4k^2-1}\ge\frac1{2k^2}$$</span></p>
<p>therefore for any <span class="math-container">$N\ge 1$</span></p>
<p><span class="math-container">$$S_{2N}=\sum_{k=1}^{2N} \frac{(-1)^{k+1}}{2... |
1,142,568 | <p>Let $A$ be the ring of germs of real analytic functions in $0\in\mathbb R$. Let $x\in A$ be the identity map on $\mathbb R$. How can I show that the map $f\mapsto \sum_n(f^{(n)}(0)/n!)T^n$ from $A$ to $\mathbb R[[T]]$ is injective, and induces an isomorphism from the $xA$-adic completition of $A$ onto $\mathbb R[[T]... | Mariano Suárez-Álvarez | 274 | <p>For the first question: the map is injective simply because the kernel is the set of germs of analytic functions all of whose derivatives at $0$ are zero, and there is clearly only one such germ. To show that the map is surjective on the completion it is enough to show that it is surjective modulo powers of the maxi... |
1,142,568 | <p>Let $A$ be the ring of germs of real analytic functions in $0\in\mathbb R$. Let $x\in A$ be the identity map on $\mathbb R$. How can I show that the map $f\mapsto \sum_n(f^{(n)}(0)/n!)T^n$ from $A$ to $\mathbb R[[T]]$ is injective, and induces an isomorphism from the $xA$-adic completition of $A$ onto $\mathbb R[[T]... | Olórin | 187,521 | <p>The injectivity is trivial : if a germ maps to zero, then it corresponds to a function analytic in a neighbourhood of $0 $ that has all it's derivatives vanishing at zero, which implies by analyticity (expend the function at $0$ !) that the function is zero, and same for the germ. To see that it induces and isomorph... |
267,442 | <p>Suppose $a,b\in\Bbb N$ are odd coprime with $a,b>1$ then is it true that if all four of $$x_1a+x_2b,\mbox{ }x_2a-x_1b,\mbox{ }x_1\frac{(a+b)}2+x_2\frac{(a-b)}2,\mbox{ }x_2\frac{(a+b)}2-x_1\frac{(a-b)}2$$ are in $\Bbb Z$ for some $x_1,x_2\in\Bbb R$ then $x_1,x_2\in\Bbb Z$ should hold?</p>
| Cherng-tiao Perng | 104,791 | <p>Yes, it holds. Here is a different proof. The given conditions are equivalent to $$(a+bi)(x_1-x_2i)\in {\mathbb Z}[i]$$ and $$\left(\frac{a+b}2+\frac{a-b}2i\right)(x_1-x_2i)\in {\mathbb Z}[i].$$ To show that $x_1-x_2i\in {\mathbb Z}[i],$ it suffices to show that $$\gcd\left(a+bi,\frac{a+b}2+\frac{a-b}2i\right)=1~~{\... |
267,442 | <p>Suppose $a,b\in\Bbb N$ are odd coprime with $a,b>1$ then is it true that if all four of $$x_1a+x_2b,\mbox{ }x_2a-x_1b,\mbox{ }x_1\frac{(a+b)}2+x_2\frac{(a-b)}2,\mbox{ }x_2\frac{(a+b)}2-x_1\frac{(a-b)}2$$ are in $\Bbb Z$ for some $x_1,x_2\in\Bbb R$ then $x_1,x_2\in\Bbb Z$ should hold?</p>
| GH from MO | 11,919 | <p>This is a simplification of Cehrng-tiao Perng's proof. As observed in that proof, it suffices to show that the Gaussian integers
$$ \gamma:=a+bi\qquad\text{and}\qquad \delta:=\frac{1+i}{2}(a-bi) $$
are coprime. We know that some $\mathbb{Z}$-linear combination of $a$ and $b$ equals $1$, while also
$$\gamma+(1-i)\del... |
2,707,941 | <p>$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$</p>
<p>I can change the form of this limit saying that
$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2(1+\frac{2}{n}+\frac{1}{n^2})})=n^{\alpha+\frac{2}{5}} (\sqrt[5]... | user | 505,767 | <p>By binomial expansion</p>
<ul>
<li>$\sqrt[5]{n^2+n}=\sqrt[5]{n^2}\sqrt[5]{1+1/n}=\sqrt[5]{n^2}(1+\frac1{5n}+o(1/n))$</li>
<li>$\sqrt[5]{n^2+2n+1}=\sqrt[5]{n^2}\sqrt[5]{1+2/n+1/n^2}=\sqrt[5]{n^2}(1+\frac2{5n}+o(1/n))$</li>
</ul>
<p>then</p>
<p>$$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=-\frac{n^\alpha\sqrt[5]... |
35,014 | <p>I want to display a rational number in <em>Mathematica</em> in periodic style.
<code>PeriodicForm</code> isn't working anymore. It worked in <em>Mathematica</em> 5 and now I'm using <em>Mathematica</em> 9.</p>
<p>I want to display the number $3.13678989898989898989\ldots$, where the repeating $89$ part should be di... | bill s | 1,783 | <p>Or, without the package, you can use </p>
<pre><code>RealDigits[19/7]
{{2, {7, 1, 4, 2, 8, 5}}, 1}
</code></pre>
<p>which shows the repeated decimal portion in the second (list) element of the answer. This tells you that the answer is <code>2</code> followed by repeating <code>714285</code>. The final 1 is the exp... |
4,101,139 | <blockquote>
<p>Calculate the volume bounded by the surface <span class="math-container">$x^n + y^n + z^n = a^n$</span> <span class="math-container">$(x>0,y>0,z>0)$</span>.</p>
</blockquote>
<p><span class="math-container">$$\iiint\limits_{x^n+y^n+z^n \le a^n \\ \ \ \ \ \ \ x,y,z > 0}\mathrm dx~ \mathrm dy ... | metamorphy | 543,769 | <p>Considering the <span class="math-container">$d$</span>-dimensional generalisation <span class="math-container">$\{x\in\mathbb{R}_+^d : \lVert x\lVert_n\leqslant a\}$</span>, where <span class="math-container">$a>0$</span> and <span class="math-container">$$x=(x_1,\dots,x_d)\in\mathbb{R}^d\implies\lVert x\lVert_n... |
1,040,442 | <p>How do I find one value of $x$ in these equations?
$$
\begin{cases}
x \equiv 3 \pmod{5}\\
x \equiv 4 \pmod{7}
\end{cases}
$$</p>
| Deepak | 151,732 | <p>First ascertain that the moduli are coprime, guaranteeing a solution by the CRT.</p>
<p>Then solve one equation first (doesn't matter which):</p>
<p>$x \equiv 4 \pmod 7$</p>
<p>$x = 7k + 4$</p>
<p>Substitute this into second equation.</p>
<p>$7k + 4 \equiv 3 \pmod 5$</p>
<p>$7k \equiv -1 \pmod 5$</p>
<p>$k \e... |
1,620,526 | <p>In Mathematical Analysis by Apostol he mentions that the "Intersection of all open intervals of the form $(-\frac{1}{n}, \frac{1}{n})$ is $\{0\}$"</p>
<p>Obviously this is a super basic question but I thought that an open interval does not include the endpoints, so from the limit as $n\to\infty$ on both sides we ge... | jimbo | 115,363 | <p><span class="math-container">$\{0\}$</span> is a subset of <span class="math-container">$\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$</span>.</p>
<p>Since <span class="math-container">$0$</span> is an element of the open interval <span class="math-container">$\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)... |
3,319,010 | <p>I would like to learn more about combinatorics of finite sets (including theorems such as Sperner, Erdos-Ko-Rado theorems, LYM inequality). Is there any good book or article for this topic (if possible with problems and exercises)?</p>
| Mark | 470,733 | <p>They are linearly independent as functions of variable <span class="math-container">$x$</span>. Yes, you are right that given a specific value of <span class="math-container">$x$</span> you can find non-zero <span class="math-container">$a,b\in\mathbb{R}$</span> such that <span class="math-container">$ax+bx^2=0$</sp... |
2,584,968 | <p>I know that if X is locally compact and Hausdorff, then any non-empty open set $S$ contains a non-empty closed set. I know this to be the case because a locally compact space is a regular space, in which the claim holds.</p>
<p>But why does any open $S$ contain a <em>non-empty open set whose closure is compact and ... | DanielWainfleet | 254,665 | <p>The closure bar denotes closure in $X$.</p>
<p>Take $p\in S.$ Let $T$ be an open set containing $p$ such that $\overline T$ is compact. Let $U=S\cap T.$</p>
<p>Method (I). </p>
<p>(i). If $\overline U=U$ then $U$ is a non-empty open subset of $S$ whose closure is compact (because $U=\overline U$ is closed in th... |
1,800,821 | <p>Consider an $n \times n$ matrix of the form
$$
A = \begin{bmatrix}
a_1 & a_2 & \ldots & a_{n-1} & a_n \\
1 \\
& 1 \\
& & \ddots \\
& & & 1
\end{bmatrix}
$$
for certain $a_1, \ldots, a_n \in \mathbb{R}$ (and zeroes in all blank spaces). Can this matrix ever (for some $n \in \ma... | Marc van Leeuwen | 18,880 | <p>To expand upon the answer by loup blanc, the <em>minimal</em> polynomial of$~A$ is of degree$~n$, namely it is $X^n-a_1X^{n-1}-a_2X^{n-2}-\cdots-a_n$. If some eigenvalue$~\lambda$ had geometric multiplicity${}>1$ then there would to the contrary exist a monic polynomial of degree less than$~n$ that annihilates$~A... |
658,815 | <p>I'm looking for a good way to remember/understand part of the <a href="https://en.wikipedia.org/wiki/Convergence_of_measures" rel="nofollow noreferrer">Portmanteau theorem</a>. Specifically, let <span class="math-container">$X$</span> be a metric space. The part of the Portmanteau theorem I'm asking about says that ... | Davide Giraudo | 9,849 | <p>Consider a sequence $(x_n)_{n\geqslant 0}$ of elements of $X$ which converges to $x$ with $x_i\neq x$ for each $i$. Then define $\mu_n$ as the Dirac measure at $x_n$ and $\mu$ at $x$ and $F:=\{x\}$. Hence if you already know that in the general statement<br>
$$\limsup_n\mu(F)\overset{\large{\leqslant}}{\geqslant}\mu... |
5,877 | <p>I've recently run across a series of problems that didn't reflect reality. </p>
<p>For example - </p>
<ul>
<li>An algebra problem with two teens on bicycles. The resulting times showed the bike was moving at 120MPH. </li>
<li>A quadratic equation, "The football follows a path of....." but the equation didn't refl... | guest | 8,745 | <p>Physically (or even economically) unreasonable answers are confusing. The student needs to be shown that math is a tool that can solve practical problems. Answers that are unreasonable send the wrong message. (That math is a logic game versus a commercially relevant skill.) Why do that, when there are so many ea... |
1,369,669 | <p>We are given the following:</p>
<p>$$\int \sin(xy)dy$$</p>
<p>We start by assigning anything algebraic into our first variable, $u$. Recalling LIATE (Logarithmic, Inverse-Trig., Algebra, Trig., Exponential) we start with algebra.</p>
<p>If I assign $$u=xy$$ then</p>
<p>$$\int \sin(u)\frac{du}{x}dy$$</p>
<p>Henc... | wythagoras | 236,048 | <p><strong>Hint.</strong> $\dfrac{1}{x}$ is a constant, since $x$ is a constant, so you can take it out to get $$\frac{1}{x} \int \sin(u) du$$</p>
<p>which you should be able to integrate. </p>
|
1,369,669 | <p>We are given the following:</p>
<p>$$\int \sin(xy)dy$$</p>
<p>We start by assigning anything algebraic into our first variable, $u$. Recalling LIATE (Logarithmic, Inverse-Trig., Algebra, Trig., Exponential) we start with algebra.</p>
<p>If I assign $$u=xy$$ then</p>
<p>$$\int \sin(u)\frac{du}{x}dy$$</p>
<p>Henc... | John_dydx | 82,134 | <p>$$ \int \sin xy \ dy$$</p>
<p>As you've done: let $u = xy$, $\frac{du}{dy}=x $</p>
<p>$$\int \sin xy \ dy = \int \frac{\sin u}{x} du$$</p>
<p>$$ \frac{-\cos u}{x} + c = \frac{-\cos xy}{x} + c$$</p>
<p>Hint: when integrating wrt y, x becomes a constant so you don't need to worry about it ($\sin xy $ is a functi... |
301,176 | <p>What on Earth do Russian <a href="https://en.wikipedia.org/wiki/Matryoshka_doll" rel="noreferrer">Matryoshka dolls</a> have in common with large cardinal axioms?! Well, the answer lies in Jónsson algebras! Here is how:</p>
<p>As illustrated in the pictures, a Matryoshka set is a self-replicating <em>container</em> i... | Jochen Glueck | 102,946 | <p>The answer is <strong>yes</strong>, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on <a href="https://en.wikipedia.org/wiki/Uniform_boundedness... |
301,176 | <p>What on Earth do Russian <a href="https://en.wikipedia.org/wiki/Matryoshka_doll" rel="noreferrer">Matryoshka dolls</a> have in common with large cardinal axioms?! Well, the answer lies in Jónsson algebras! Here is how:</p>
<p>As illustrated in the pictures, a Matryoshka set is a self-replicating <em>container</em> i... | Jochen Wengenroth | 21,051 | <p>Instead of inspecting the Banach-Steinhaus proof as in Jochen Glueck's answer one can <em>apply</em> Banach-Steinhaus to the Banach space $[A]$ (the linear hull of $A$) endowed with the Minkowski functional $\|x\|_A=\inf\{t>0: x\in tA\}$ (completeness of this norm follows from completeness of the Banach space $X$... |
4,294,950 | <p>I am interested in knowing how often (in terms of percentage) you would expect the total of 3 fair rolled dice to exceed the total of 2 fair rolled dice.</p>
<p>Thanks</p>
| Lorenzo Pompili | 884,561 | <p>Numerically, it happens about 77,85% of the times. Precisely, 6054 times every 7776. Below the C++ code (I just happened to have it written for calculating probabilities in Risiko).</p>
<hr />
<pre><code>#include<stdio.h>
#include<stdlib.h>
using namespace std;
void main(){
int i,j,k,a,b;
int win=0... |
3,020,024 | <blockquote>
<p>Let <span class="math-container">$f: \Bbb R \to \Bbb R$</span> be a function such that <span class="math-container">$f'(x)$</span> exists and is continuous over <span class="math-container">$\Bbb R$</span>. Moreover, let there be a <span class="math-container">$T > 0$</span> such that <span class="... | lulu | 252,071 | <p>suppose <span class="math-container">$f(x)$</span> is negative for all values of <span class="math-container">$x$</span>. Periodicity tells us that there must be solutions to <span class="math-container">$f'(x)=0$</span> and for such <span class="math-container">$x$</span> we would then have <span class="math-contai... |
1,610,798 | <p>If two numbers are less than a given number, how can we algebraically show that their difference is also less than the given number .
Both numbers are greater than zero and in $\mathbb{Q}$.</p>
| Archis Welankar | 275,884 | <p>Counterexample (only 1) giving so take $a=-7,b=-200,c=-5$ now take difference $a-b=-7+200>c$ also you can get eg where $a-b,b-a>c$</p>
|
4,283,075 | <p>I have a vector <span class="math-container">$[d] = [d_1, d_2, ..., d_i]$</span>. All elements of <span class="math-container">$d$</span> are always <span class="math-container">$0$</span> except for one of them which can be either <span class="math-container">$+1$</span> or <span class="math-container">$-1$</span>.... | DanielV | 97,045 | <blockquote>
<p>You just did. Why use confusing symbols when you can plainly and quickly say what you want to say in words? – Alex Kruckman</p>
</blockquote>
<p>Comment is correct. But if you insist on first order language:</p>
<p><span class="math-container">$$\exists k ~.~ (d_k = -1 \lor d_k = 1) \land \forall j ~.~... |
1,302,738 | <p>I'm trying to evaluate this sum </p>
<p>$$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{(n^2+n-1)(k^2+k-1)}$$</p>
<p>I have no idea how to deal with it. </p>
<p>With one sum I can, with partial-fraction decomposition, express it as a function of Digamma function and I stuck there.</p>
| Jack D'Aurizio | 44,121 | <p>As already pointed by the other answers, we just have to compute:
$$ S_1=\sum_{n\geq 1}\frac{1}{n^2+n-1},\qquad S_2=\sum_{n\geq 1}\frac{1}{(n^2+n-1)^2}.\tag{1}$$
and by exploiting the logarithmic derivative of the Weierstrass product for the cosine function we have, for any $\alpha\in\mathbb{R}^+\setminus\mathbb{N}$... |
517,904 | <blockquote>
<p>Let $A,B$ be non-empty sets and $f:A\to B$ a function. Proof that $f$ is injective, iff $f\circ g=f\circ h$ implies that $g=h$ for all functions $g,h:Y\to A$, for every set $Y$?</p>
</blockquote>
<p>I can see why this is. But how do I prove this? I get confused by the if and only if part.</p>
| abiessu | 86,846 | <p>It appears that you have all the pieces, so here's how I would approach the problem:</p>
<p>Given rational $\frac xy$ such that $0\lt \frac xy \lt 1$, we have $0\lt x \lt y$, or $1 \lt \frac yx$. There must exist a positive integer $n$ such that $n\le \frac yx\le n+1$, which means that $\frac 1n\ge \frac xy\ge \fr... |
753,231 | <p>Solve $$z^2+2iz-1+2i$$</p>
<p>I tried:</p>
<p>$(z+i)^2-1-1+2i$</p>
<p>$(z+i)^2 = 2-2i$</p>
<p>Which gives me $a^2-b^2 = 2, 2ab = -2, a^2+b^2 = \sqrt(8)$ And this I cannot solve.</p>
| Ron Gordon | 53,268 | <p>$$z^2+2 i z-1 = (z+i)^2$$</p>
<p>So the equation for the roots is
$$(z+i)^2=-2 i = 2 \, e^{i 3\pi/2} \implies z+i = \begin{cases} \sqrt{2} \, e^{i 3 \pi/4} \\ \sqrt{2} \, e^{-i \pi/4}\end{cases} $$</p>
<p>The roots are $z=-1$, or $z=1-2 i$.</p>
|
2,005,981 | <p>Let
$f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that:
$$f(x+y) = f(x) + f(y)$$
$$f(1) = 1$$ Show that $$f(x) = x$$ I have been having trouble approaching this problem. I have shown, through a system of equations, that $f(x+y) = x + y$, but that's about as far as I can get. Appreciate an... | Kaynex | 296,320 | <p>If we let $y = 1$:
$$f(x + 1) = f(x) + 1$$
This is a recurrence relation and can be solved with simple techniques. However, we already know what the solution is supposed to be. Proving that $f(x) = x$ is a solution:</p>
<p>LS:
$$f(x + 1) = x + 1$$</p>
<p>RS:
$$f(x) + 1 = x + 1$$</p>
<p>Because LS = RS, $f(x) = x$... |
1,236,603 | <p>Sorry, this may be a stupid question, but I am just beginning to learn about this and cannot find the answer anywhere I have looked so far. Clearly if we have any polynomial $P(z)$, then it is easy to show that the order is $0$.</p>
<p>Clearly though not all entire functions that grow at roughly this rate are polyn... | mrf | 19,440 | <p>There are many non-polynomial entire functions of order zero. You can show that the order $\rho$ can be computed as
$$
\rho = \limsup_{n\to\infty} \frac{n\log n}{\log(1/|a_n|)}
$$
where $a_n$ are the Maclaurin coefficients of $f$. To get a function of order $0$, you only have to make sure that $|a_n|$ tends sufficie... |
40,181 | <p>Currently, my math training includes Calc 1-3, linear algebra, and some introduction to set theory/discrete math.
What would you recommend that I study over summer in preparation for the Putnam? Real analysis, topology, abstract algebra (all of the above)? What would be the most pertinent? Thanks!</p>
| daOnlyBG | 173,397 | <p>If you want to do well on the Putnam, I think you'd do well to look over the books that Graphth suggested. However, to dismiss real analysis, abstract algebra, and topology, could be a pretty grave mistake- there are often at least 2-3 questions that cover those topics. </p>
<p>Fortunately, such questions really on... |
318,754 | <p>This is more of a philosophical or historical question, and I can be totally wrong in what I am about to write next.</p>
<p>It looks to me, that complex-analytic geometry has lost its relative positions since 50's, especially if we compare it to scheme theory. <em>Are there internal mathematical reasons for why tha... | Donu Arapura | 4,144 | <p>I'm not sure this question is a good fit form mathoverlow, but here are a few thoughts. I'll probably delete this answer in a while.</p>
<ol>
<li>Let me elaborate my comment concerning sociology a bit further. A (sub)field (not necessarily in mathematics) becomes "hot" when exciting new techniques are introduced th... |
3,163,123 | <p>I am trying to prove that the determinant of a magic square, where all rows, columns and diagonal add to the same amount, is divisible by 3. </p>
<p>I proved it for magic squares which have entries <span class="math-container">$1,\ldots, 9$</span>, but it turns out I need to show it for magic squares which can have... | Arthur | 15,500 | <p>Collecting from all the comments above:</p>
<p>Let <span class="math-container">$s$</span> be the sum of one row. This <span class="math-container">$s$</span> must be divisible by <span class="math-container">$3$</span>: Since the sum of the two diagonals along with the row and the column going through the center e... |
1,883,047 | <p>I have this logical statement</p>
<p>$$\neg x\lor (x \wedge y)$$</p>
<p>However I do not know what is considered a valid transformation. Normally if there is an $\wedge$ in the middle I treat it like multiplication and pull out some "shared" piece but here I don't know how to use distributive properties. </p>
| Community | -1 | <p>If you write the truth table, you see a familiar pattern. Note that your expression is just $x\to y$ . We know that $x\to y\equiv \neg x \vee y$. </p>
<p>You can also use the distributive properties as given in the other answer.</p>
|
2,460,155 | <p>I solved it by myself, but I don't know I did it right.
I hope you guys to check this out.</p>
<p>Suppose $(AB-BA)^{T}=BA-AB$</p>
<p>$(AB-BA)^{T}=B^{T}A^{T}-A^{T}B^{T}$</p>
<p>then,</p>
<p>$B^{T}=B$</p>
<p>$A^{T}=A$</p>
<p>therefore,</p>
<p>$B^{T}A^{T}-A^{T}B^{T}=BA-AB$</p>
<p>It's my first time to ask quest... | Ziad Fakhoury | 295,839 | <p>What you showed was that if $AB - BA$ was skew symmetric then $A,B$ are symmetric. However what they asked us to show was the opposite, if $A,B$ are symmetric then show $AB - BA$ is skew symmetric.</p>
<p>To prove $AB -BA$ is skew symmetric you must show that $(AB - BA)^T = BA - AB.$
$$(AB-BA)^T = (AB)^T - (BA)^T ... |
546,809 | <p>The question is :Find the derivative of $f(x)=e^c + c^x$. Assume that c is a constant.</p>
<p>Wouldn't $f'(x)= ce^{c-1} + xc^{x-1}$. It keeps saying this answer is incorrect, What am i doing wrong?</p>
| 1233dfv | 102,540 | <p>Consider $$f(x)=e^{c}+c^{x}$$ where $c$ is a constant. We know that since $c$ is a constant, $e^c$ is also a constant making ${d\over dx}(e^c)=0$. Also, ${d\over dx}(c^x)=c^x\ln c$. The reason for this is because ${d\over dx}(c^x)={d\over dx}{(e^{\ln c})^x}={d\over dx}({e^{{(\ln c}){x}})}=e^{({\ln c)} x}\cdot {d\ove... |
1,910,910 | <p>I am dealing with the following exercise:</p>
<p>Let $u_n$ bounded in $L^\infty[0,1]$ such that, for any continuous function $f: [0,1]\times R$ to $R$
$$\lim_n \int_0^1 f(x, u_n(x))=\int_0^1 f(x, u(x)).$$</p>
<p>Prove that $u_n$ converges to $u$ in $L^1.$
There is a hint saying to prove it first for $u\in C^0$.</p... | GiantTortoise1729 | 219,849 | <p>On a finite measure space $(X,\mu)$, $L^q(X) \subset L^p(X)$ for $q\geq p$. This is proved by Holder's inequality. Apply this to be able to approximate your $L^\infty$ functions.</p>
|
114,645 | <p>How can I make a moving point on the circle with control?</p>
<pre><code>Manipulate[
ParametricPlot[Sqrt[50]{Cos[x],Sin[x]},{x,0,10Pi},
Epilog->{PointSize[Large], Point[Table[{2,0}]]},
PlotRange->{{0,10},{0,10}}],{{Sqrt[50],2,"Play"}, 1, 10}
]
</code></pre>
| kglr | 125 | <p><strong>Update:</strong></p>
<pre><code>Manipulate[ ParametricPlot[radius {Sin[x], Cos[x]}, {x, 0, 2 Pi},
PlotRange -> {{-10, 10}, {-10, 10}}, MeshFunctions -> {#3 &},
Axes -> True, Mesh -> {{{t, {Red, PointSize[.05]}}}}],
{t, 0, 2 Pi}, {{radius, 2, "Play"}, 1, 10}]
</code></pre>
<p><a ... |
1,822,811 | <p>Suppose a function $f$ is defined as follows: </p>
<p>$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$</p>
<p>Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I... | JimmyK4542 | 155,509 | <p>Since $x^4-2x^2y^2+y^4= (x^2-y^2)^2 \ge 0$, we have $2x^2y^2 \le x^4+y^4$. </p>
<p>Therefore, $\dfrac{x^2y^2}{x^4+y^4} \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$. </p>
<p>Also, $\dfrac{x^2y^2}{x^4+y^4} \ge 0$ for all $(x,y) \neq (0,0)$. </p>
<p>From the above inequalities, we have that $\left|\dfrac{x^2y^2}{x^4... |
1,822,811 | <p>Suppose a function $f$ is defined as follows: </p>
<p>$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$</p>
<p>Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I... | Will Kwon | 311,215 | <p>Hint: use arithmetic-geometric mean inequality to show the function is continuous at $(0,0)$: for $a,b>0$,
$$ ab \leq \frac{a^2 + b^2}{2} $$ </p>
|
1,822,811 | <p>Suppose a function $f$ is defined as follows: </p>
<p>$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$</p>
<p>Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I... | Mark Viola | 218,419 | <p>A fairly efficient way to approach this problem is to transform to polar coordinates and write</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{x^3y^2}{x^4+y^4}=\lim_{r\to \infty}\left(r\,\,\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right)$$</p>
<p>Noting that we can write </p>
<p>$$\begin{align}
\left|\frac{\... |
3,886,173 | <p>Please help me to prove this inequality</p>
<p><span class="math-container">$$\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} < \ln n < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1}$$</span></p>
| TravorLZH | 748,964 | <p>Whenever <span class="math-container">$n$</span> is not a prime power, we have <span class="math-container">$\Lambda(n)=0$</span>, but when <span class="math-container">$n=p^k$</span>, we have</p>
<p><span class="math-container">$$
{\Lambda(n)\over\log n}={\log p\over\log p^k}=\frac1k
$$</span></p>
<p>Hence, we have... |
3,956,963 | <p>What is the difference between countable infinity and uncountable infinity? Are there any examples? How can I imagine it?
Can you offer some assistance? please.</p>
| FFjet | 597,771 | <p>Intuitively speaking, if <span class="math-container">$A$</span> is countable, you have a way to <strong>list out</strong> the items in <span class="math-container">$A$</span>. If <span class="math-container">$A$</span> is uncountable, you will not have such ways, and if you try to make a list of <span class="math-c... |
147,425 | <p>So I've come across the following inequality for probability measures:</p>
<p>$$
P(X \cap Y) \ge P(X) + P(Y) - 1
$$</p>
<p>I'm trying to work out why it should be true. I'm sure I'm missing something obvious.</p>
<p>I have the following:</p>
<p>$$
P(X \cap Y) = P(X) +P(Y) - P(X \cup Y) \le P(X) +P(Y) - 1
$$</p>
... | Michael Greinecker | 21,674 | <p>$P(X\cup Y)\leq 1$. Hence $-P(X\cup Y)\geq -1$.</p>
|
132,936 | <p>In propositional logic, a <strong>theory</strong> <span class="math-container">$T$</span> consists of a set of logical symbols and statements which we call axioms. A logical statement <strong><span class="math-container">$A$</span> can be proven by <span class="math-container">$T$</span></strong> if there is a proof... | Doug Spoonwood | 11,300 | <p>My impression of this goes that in propositional calculus, with only proof theory, there exist ways to derive formulas which intuitively have meanings which either surpass a person's understanding or come as ridiculously difficult to see. This implies that more mathematics can get formed than possible with proofs j... |
821,654 | <p>I have a Taylor series problem, well more precisely a Maclaurin series.</p>
<p>I am trying to find convergence of: $f(x) = e^{x^3} + e^{{2x}^3}$</p>
<p>Okay here goes:</p>
<p>$$f'(x) = 3xe^{x^3} + 6x e^{{2x}^3}$$
$$f''(x) = 9x^2e^{x^3} + 3e^{x^3} + 36x^2e^{{2x}^3} + 6e^{{2x}^3}=e^{x^3}(9x^2+3) + e^{{2x}^3}(36x^2+... | ABC | 91,270 | <p>At the moment, you have $x^2-x=380$, which is the same as $x^2-x-380=0$. In general, a function with degree two can be written as $ax^2+bx+c=0$, so in this case, $a=1$, $b=-1$ and $c=-380$.</p>
<p>Now you can compute the discriminant $D$ as $D=b^2-4ac=(-1)^2-4*1*(-380) = 1 + 4*380 = 1521$.</p>
<p>Because the value... |
821,654 | <p>I have a Taylor series problem, well more precisely a Maclaurin series.</p>
<p>I am trying to find convergence of: $f(x) = e^{x^3} + e^{{2x}^3}$</p>
<p>Okay here goes:</p>
<p>$$f'(x) = 3xe^{x^3} + 6x e^{{2x}^3}$$
$$f''(x) = 9x^2e^{x^3} + 3e^{x^3} + 36x^2e^{{2x}^3} + 6e^{{2x}^3}=e^{x^3}(9x^2+3) + e^{{2x}^3}(36x^2+... | lhf | 589 | <p>You may recognize that $x^2-x=x(x-1)$ and that $380=20\cdot 19$, which will give you $x=20$.</p>
<p>Writing $380=(-20)\cdot (-19)$ will give you $x=-19$.</p>
|
185,295 | <p>I would like to solve the following equation <span class="math-container">$y^2=x^2+ax^2y^2+by^2x^3+cy^3x^2$</span> where <span class="math-container">$a,b,c$</span> are small, so <span class="math-container">$y\approx x+O(x^3)$</span>. I would like to have a series approximation of the solution rather than an exact ... | Daniel Lichtblau | 51 | <p>Could use implicit differentiation, solve for all derivatives at <code>x==0</code>.</p>
<pre><code>ee = y^2 - (x^2 + a x^2 y^2 + b y^2 x^3 + c y^3 x^2) /. y -> y[x];
dpolys = Table[D[ee, {x, j}], {j, 0, 5}] /. x -> 0
(* {y[0]^2, 2*y[0]*Derivative[1][y][0], -2 - 2*a*y[0]^2 - 2*c*y[0]^3 +
2*Derivative[1]... |
18,772 | <p>Suppose $G$ is a semisimple group, and $V_{\lambda}$ is an irreducible finite-dimensional representation of highest weight $\lambda$. Suppose $H \subset G$ is a semisimple subgroup. What is the multiplicity of the trivial representation in $V_{\lambda}|_{H}$? Is there a simple way to read this off from $\lambda$ ... | Allen Knutson | 391 | <p>Consider the case that $H$ is a maximal torus of $G'$, and your $G = G' \times H$. (Well, you said $G,H$ semisimple, but I'm going to pretend you meant reductive, because really you should have.)
Then your question is answered by the Kostant multiplicity formula.</p>
<p>If you're willing to take that formula as "si... |
18,772 | <p>Suppose $G$ is a semisimple group, and $V_{\lambda}$ is an irreducible finite-dimensional representation of highest weight $\lambda$. Suppose $H \subset G$ is a semisimple subgroup. What is the multiplicity of the trivial representation in $V_{\lambda}|_{H}$? Is there a simple way to read this off from $\lambda$ ... | David Bar Moshe | 1,059 | <p>I want to refer you to the following <a href="http://www.emis.ams.org/journals/MV/084/mv08407.pdf" rel="nofollow">article</a> by B. Cahen, treating the case of compact Lie groups, where an approach based on the realization of the irreducible representations of G according to the Borel-Weil theorem as reproducing ker... |
1,118,269 | <p>I'm reading: <em>Mathematical thought from ancient to modern times by Kline</em>. My question is about this pasasge:</p>
<blockquote>
<p>Beyond its achievements in subject matter, the nineteenth century
reintroduced rigorous proof. No matter what individual mathematicians may
have thought about the soundness ... | Edward Porcella | 403,946 | <p>The proof of "Ptolemy's Theorem", in <em>Almagest I, 10</em> is as rigorous as Euclid's <em>Elements</em>, which Ptolemy is clearly acquainted with and relying on, ca. 150 A.D. Even though Kline qualifies his claim by saying "<em>almost</em> all of mathematics..." it is still going too far to say "the concept of ded... |
1,118,269 | <p>I'm reading: <em>Mathematical thought from ancient to modern times by Kline</em>. My question is about this pasasge:</p>
<blockquote>
<p>Beyond its achievements in subject matter, the nineteenth century
reintroduced rigorous proof. No matter what individual mathematicians may
have thought about the soundness ... | CopyPasteIt | 432,081 | <p>Below I copy and paste wikipedia segments as well as a quote from Richard Dedekind. We concentrate here on just the nineteenth century, but see the last section where we pay tribute to Gottfried Wilhelm Leibniz.</p>
<p>I don't agree with the author. Indeed, a couple of axioms needed to be pinned down, and set theory... |
263,349 | <p>If secant and the tangent of a circle intersect at a point outside the circle then <strong>prove that the area of the rectangle formed by the two line segments corresponding to the secant is equal to the area of the square formed by the line segment corresponding to the tangent</strong><br>
I find this question hig... | Nate Eldredge | 822 | <p>Note first that contrary to the claim in comments, this set (call it $LC(X)$) need not be open.</p>
<p>Let $X \subset \mathbb{R}^2$ be the union of all the lines through the origin that have rational slope, with the Euclidean metric. $X$ is locally connected at the origin 0; indeed any open ball in $X$ centered at... |
1,951,733 | <blockquote>
<p>Can any polynomial $P\in \mathbb C[X]$ be written as $P=Q+R$ where $Q,R\in \mathbb C[X]$ have all their roots on the unit circle (that is to say with magnitude exactly $1$) ? </p>
</blockquote>
<p>I don't think it's even trivial with degree-1 polynomials... In this supposedly simple case, with $P(X)=... | fleablood | 280,126 | <p>Two rules that work for any base $b$.</p>
<p>If $n$ divides $b-1$, then if the sum of the digits is a multiple of $n$ then the number is a multiple of $n$. (That's why the $3$ and $9$ rule work).</p>
<p>If the sum of the even place digits is a multiple of $b+1$ more or less than the sum of the odd place digits th... |
239,120 | <blockquote>
<p>The line graph <span class="math-container">$L(G)$</span> of a graph <span class="math-container">$G$</span> is defined in the following way: the vertices of <span class="math-container">$L(G)$</span> are the edges of <span class="math-container">$G$</span>, <span class="math-container">$V(L(G)) = E(G)$... | Chamataka | 80,562 | <p>The complement of the line graph of $K_5$ can be constructed as follows. Label the vertices of $K_5$ as $1,2,\ldots,5$. The 10 edges of this graph are the ${5 \choose 2}$ 2-subsets of $\{1,\ldots,5\}$. The line graph $L(K_5)$ thus has 10 vertices, labeled by these 10 2-subsets $\{i,j\}$. Two vertices $\{i,j\}, \{k... |
3,917,129 | <p>I have some problems trying to solve this exercise: <br />
Consider the following Cauchy problem: <span class="math-container">$\left\{\begin{matrix}x'=t+\frac{y}{1+x^2}
\\ y'=txe^{-ty^2}
\\ x(0)=y(0)=1
\end{matrix}\right.$</span><br />
Discuss the existence of a solution and its uniqeness. Then I'm asked to prove o... | Lutz Lehmann | 115,115 | <p>If the function is sub-linear in state-space direction,
<span class="math-container">$$
\|F(t,u)\|\le a(t)\|u\|+b(t)
$$</span>
then the solution is bounded per Grönwall lemma by the solution of the linear ODE
<span class="math-container">$$
v'(t)=a(t)v(t)+b(t),~~ v(0)=\|u(0)\|.
$$</span>
As solutions of linear DE ex... |
1,959,131 | <p>Hallo :) I am hopeless with this exercise:</p>
<blockquote>
<p>Solve the system of equations over the positive real numbers</p>
<p><span class="math-container">$$\sqrt{xy}+\sqrt{xz}-x=a$$</span></p>
<p><span class="math-container">$$\sqrt{zy}+\sqrt{xy}-y=b$$</span></p>
<p><span class="math-container">$$\sqrt{xz}+\sq... | Andreas | 317,854 | <p>Here are some ideas which, with some hindsight, save you the long computations.</p>
<p>Notice that your system of equations is cyclic in the variables (x,y,z) and (a,b,c). I.e. the next equation follows from the previous one by shifting all variables by one position ($x \to y$ and simultaneoulsy $a\to b$ etc.), wh... |
383,478 | <p>I'm trying to develop a reduction formaula for the integral - $\int \sin^n x dx$.
I've successfully developed a formula which is depended on two elements jumps, which is more or less:</p>
<p>$$\int \sin^k x d x = \frac{k-1}{k} \int \sin^{k-2}x d x
- \frac{1}{k} \cos x \sin^{k-1} x
$$</p>
... | vonbrand | 43,946 | <p>I'm afraid it won't work. $\sin$ and $\cos$ are intimately interwoven, you should look at $\int \sin^n x dx$ and $\int \cos^n x dx$ in parallel. You'll see each leading to the other, and back (your "two steps").</p>
|
4,416,238 | <p>From Serge Lang's Introduction to Linear Algebra, page 152:</p>
<blockquote>
<p>Let <span class="math-container">$L:V\rightarrow V$</span> be a linear map [<span class="math-container">$V$</span> is a vector space]. Suppose that there exists a
basis <span class="math-container">$\{v_1,...,v_n\}$</span> and numbers <... | SomeCallMeTim | 767,376 | <p>Note that as <span class="math-container">$v_i\in V$</span> is a vector in a vector space, in particular we can multiply by scalars from the base field, so <span class="math-container">$c_i v_i\in V$</span>. The statement <span class="math-container">$L(v_i)=c_i v_i$</span> says that <span class="math-container">$L$... |
2,202,473 | <p>Can you please help me for this problem?</p>
<p>Show that the map $x:M\rightarrow \mathbb{R}$, $\mathbb{R}$ has group structures under addition, is defined by $x\left( \left[ \begin{matrix} a& b\\ o& c\end{matrix} \right] \right)$ =$ \log \left( \dfrac {a} {c}\right) $ is a group homomorphism. </p>
<p>Also... | Alan Wang | 165,867 | <p>I assume that $M \subseteq GL(2,\Bbb{R}).$
\begin{align*}
x\left( \begin{bmatrix} a& b\\ 0& c\end{bmatrix} \begin{bmatrix} p& q\\ 0& r\end{bmatrix}\right) &= x\left( \begin{bmatrix} ap& aq+br\\ 0& cr\end{bmatrix}\right)\\ &= \log \left( \dfrac {ap} {cr}\right) \\ &=\log\left(\frac... |
1,733,721 | <p>Determine $\sup E$, $\inf E$, and (where possible) $\max E$, $\min E$ for the set $E = \{ \sqrt[n]{n}: n \in \mathbb{N}\}$.</p>
<p><strong>Attempt:</strong> I've written that $\inf E = 1 = \min E$.</p>
<p>When it comes to finding $\sup E$, I've noticed punching in increasing values of n on my calculator, the eleme... | carmichael561 | 314,708 | <p>The function $f(x)=x^{\frac{1}{x}}$ has derivative
$$ f^{\prime}(x)=x^{\frac{1}{x}}\frac{1-\log x}{x^2}$$
Therefore $f$ has its global maximum on $[1,\infty)$ at $x=e$, and is increasing on $[1,e)$ and decreasing on $(e,\infty)$. Therefore the only values of $n$ you need to check are $n=2$ and $n=3$. And $3^{\frac{1... |
1,397,991 | <p>Consider the real interval $[0,1)$, this is partially ordered set (totally ordered actually). This set has a upper bound like $1$, and according to Zorn's Lemma each partially ordered set with a upper bound should have at least one maximal element. However, in this set there is no maximal element, i.e., element that... | Asaf Karagila | 622 | <p>As a partial order $[0,1)$ has no upper bound. Sure $1$ is an upper bound of $[0,1)$ in $[0,1]$ or in $\Bbb R$. But that is not the same partial order. You are not allowed to go to larger partial orders when you apply Zorn's lemma.</p>
<p>So $[0,1)$ has many chains without upper bounds. E.g. $[0,1)$ itself.</p>
|
1,543,722 | <p>We are learning about inequalities. I originally assumed it would be the same as equations, except with a different sign. And so far, it has been - except for this.</p>
<p>Take the simple inequality:
$-5m>25$ To solve it, we divide by $-5$ on both sides, as expected.
$m>-5$.</p>
<p>But, I have been told that... | talrnu | 293,089 | <p>For those who benefit from imagery, consider this number line:</p>
<p>$<----- (-5) ---- 0 ------ (7) --->$</p>
<p>You can see $7$ is farther to the right than $-5$, so $7 > -5$.</p>
<p>Multiply both of those values by $-1$, and you flip the number line:</p>
<p>$<--- (-7) ------ 0 ---- (5) ----->$<... |
2,248,754 | <p>Take $b>a>1$ By considering $x^{-y}$ over $(1,\infty)\times (a,b)$, show that $$\int_{1}^{\infty}\frac{x^{-a}-x^{-b}}{\log(x)}dx$$ exists and find its value</p>
<p>I've assumed they want me to write the intagral as $$\int_{1}^{\infty}\int_{a}^{b}\frac{yx^{-y-1}}{\log(x)}dxdy$$ and use Tonelli's Theorem to jus... | Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\n... |
1,705,736 | <p>Is this $$\sqrt{e^{ix}}=e^{\frac{ix}{2}}=\cos\frac{x}{2}+i\sin\frac{x}{2}$$ true?</p>
<p>Or
$$\sqrt{e^{ix}}=\sqrt{\cos x+i\sin x}$$</p>
<p>How do I express square root of $e^{ix}$ as a non-square root expression?</p>
| Lee Mosher | 26,501 | <p>Although "the" square root of $e^{ix}$ is not well-defined, every nonzero complex number $c$ has exactly two square roots in the sense that there are exactly two solutions of the equation $z^2=c$.</p>
<p>To find both square roots of $e^{ix}$ you must remember that $e^{ix}$ may also be written as
$$e^{i x} = e^{i (... |
4,323,344 | <p>Let <span class="math-container">$\phi$</span> be Euler's totient function and <span class="math-container">$n$</span> be a positive integer. Let <span class="math-container">$\phi^k(n)$</span> denote <span class="math-container">$k$</span> sucessive applications of the totient function.</p>
<p>Since <span class="ma... | Thomas Andrews | 7,933 | <p><span class="math-container">$\phi(2m)\leq m,$</span> and <span class="math-container">$\phi(n)$</span> is even except when <span class="math-container">$\phi(n)=1.$</span> So you get <span class="math-container">$k\leq 1+\log_2(n-1).$</span></p>
<p>This upper bound can be achieved when <span class="math-container">... |
3,777,856 | <p>I have the following equation that I need to solve:</p>
<p><span class="math-container">$$1000.00116=\frac{1000}{\left(1+x\right)^{16}}+\frac{1-\left(1+x\right)^{-16}}{x}$$</span></p>
<p>However, software I use is refusing to do it. Which software/web is capable of solving it? Or could you please show me the answer?... | Claude Leibovici | 82,404 | <p>This looks like a finance problem. Consider</p>
<p><span class="math-container">$$y=\frac{1000}{\left(1+x\right)^{16}}+\frac{1-\left(1+x\right)^{-16}}{x}$$</span> and expand as a series around <span class="math-container">$x=0$</span> to get</p>
<p><span class="math-container">$$y=1016-16136 x+136816 x^2-819876 x^3+... |
1,089,193 | <p>The theorem is that every reflection $R_{S}$ in an affine subspace $S$ of $\mathbb{E}^{n}$ is an isometry:</p>
<p>$R_S:\ \mathbb{E}^{n} \rightarrow \mathbb{E}^{n}:\ x \mapsto R_{S}(x) = x + 2 \overrightarrow{x\pi_{S}(x)}$</p>
<p>I'm horrendously stuck with the proof.
I get that I'm trying to prove that $R_{S}$ pre... | Squirtle | 29,507 | <p>Here's a partial answer which is suitable for Euclidean space (not quite as general as an affine space). Given a vector $a$ in Euclidean space $\mathbb{R}^n$, the formula for the reflection in the hyperplane through the origin, orthogonal to $a$, is given by</p>
<p>$$\text{Ref}_a(v) = v - 2\frac{v\cdot a}{a\cdot a... |
22,078 | <p>It is a standard and important fact that any smooth affine group scheme $G$ over a field $k$ is a closed $k$-subgroup of ${\rm{GL}}_n$ for some $n > 0$. (Smoothness can be relaxed to finite type, but assume smoothness for what follows.) The proof makes essential use of $k$ being a field, insofar as it uses freen... | Torsten Ekedahl | 4,008 | <p>I don't understand why the usual proof over a field base doesn't work over a
(local) artinian base $R$ for a flat finite type group scheme over $R$: Let $A$
be the affine algebra of $G$. Take any basis for $A$ modulo the maximal ideal of
$R$ and lift it to $A$. As $A$ is $R$-flat it is a basis $\{e_i\}$ of $A$.</p>
... |
22,078 | <p>It is a standard and important fact that any smooth affine group scheme $G$ over a field $k$ is a closed $k$-subgroup of ${\rm{GL}}_n$ for some $n > 0$. (Smoothness can be relaxed to finite type, but assume smoothness for what follows.) The proof makes essential use of $k$ being a field, insofar as it uses freen... | Philipp Gross | 4,101 | <p>This is not a direct answer to the question for a general group scheme $G \to S$ and I am not an expert in this area. However, I would like to point out that the resolution property of stacks is a natural condition that appears in this context of Hilbert's 14th problem by work of R. W. Thomason:</p>
<p><em>Equivari... |
168,040 | <p>Is there an algorithm which on input "$(a,p)$" (where $0\leq a<p$ are integers) takes time polynomial in $\log p$ and outputs "NOT PRIME" if $p$ is not prime and otherwise outputs the Legendre symbol $(a/p)$?</p>
<p>By the <a href="http://en.wikipedia.org/wiki/AKS_primality_test" rel="nofollow noreferrer">AKS pr... | user130204 | 130,204 | <p>You can take a look on this talk by Brent and Zimmermann in 2010.</p>
<p><a href="https://maths-people.anu.edu.au/~brent/pd/Brent_ACCMCC_10.pdf" rel="nofollow noreferrer">https://maths-people.anu.edu.au/~brent/pd/Brent_ACCMCC_10.pdf</a></p>
|
4,128,110 | <p>Find the <span class="math-container">$\max$</span> and the <span class="math-container">$\min$</span> with Lagrange multipliers, given <span class="math-container">$$f(x,y,z)=xyz^2,$$</span> <span class="math-container">$$g(x,y,z)=x^2+y^2+z^2-1=0.$$</span></p>
<p><a href="https://i.stack.imgur.com/Nr8G8.jpg" rel="... | user247327 | 247,327 | <p>I would do this: imagine 5000 such scenarios. "10 out of every 50" is 1/5 so in 1000 such situations, the bus arrives at the station late, in 4000, the bus is on time. In 1/5 of the times the bus is late, so 200, the train is also late so he can take it anyway. There are 4000+ 200= 4200 times out of 500... |
369,722 | <p>Let <span class="math-container">$\mathcal{H}$</span> denote a Hilbert space and <span class="math-container">$B(\mathcal{H})$</span> denote the algebra of all bounded operators on <span class="math-container">$\mathcal{H}$</span>. By recognizing the (Banach) dual of <span class="math-container">$B(\mathcal{H})$</sp... | Jochen Glueck | 102,946 | <p>The answer to the first question is <strong>yes</strong>. This follows from the following more general result.</p>
<p><strong>Terminology I: Ordered Banach spaces.</strong> By a <em>pre-ordered Banach space</em> I mean a pair <span class="math-container">$(X,X_+)$</span> where <span class="math-container">$X$</span>... |
1,290,444 | <p>It seems pretty clear to me that both of these are at least uncountable (which I think I could prove with some work). It also seems that you should be able to make some diagonal argument about the two, but I'm not really sure how to make that. I've been trying to think of functions between groups and rings and ways ... | lhf | 589 | <p>There are more groups than rings in the sense that the forgetful functor from <strong>Rings</strong> to <strong>Groups</strong> which selects the additive group of a ring is not surjective, even if you restrict it to <strong>Abelian Groups</strong>. In other words, <a href="https://math.stackexchange.com/questions/9... |
2,041,534 | <p>I think I have a method to solve the problem. I am aware that its NP complete and its so tempting to solve. I am aware that I can be wrong 99.99% but I wanted to give a shot at it. I want to put it to test.</p>
<p>Given 2 Graphs : A, B (No Self loops / No Multi-edges)<a href="https://i.stack.imgur.com/PZUUN.png" re... | Simon | 377,049 | <p>Actually, the graph isomorphism problem is not np hard. In fact there is a kinda recent paper describing a quasipolynomial algorithm (<a href="https://arxiv.org/abs/1512.03547" rel="nofollow noreferrer">https://arxiv.org/abs/1512.03547</a>). Their method starts at similar ideas to yours (looking at degree of nodes),... |
3,065,659 | <p>I have been looking online and on lecure notes and I have observed that there are 2 definitions for the completeness axiom and I cannot relate them together.</p>
<p>These are:</p>
<p>1) Every non-empty set of real numbers that is bounded above has
a supremum. Every non-empty set of real numbers that is bounded
bel... | jmerry | 619,637 | <p>There are a lot more than just those two; it's a fairly common exercise/lecture topic in an analysis course to prove that various definitions of completeness are equivalent, by building a directed graph of implications that allows us to reach anywhere from anywhere. For these two:</p>
<p>(1) implies (2): Take a seq... |
99,239 | <p>Suppose $\mathcal{H}$ is a separable Hilbert space over $\mathbb{C}$ (countable dimensions) with inner product $\langle,\rangle$. Let $A$ be a bounded linear operator on $\mathcal{H}$, i.e, in $B(\mathcal{H}$). Suppose further that $A$ is not a multiple of the identity
operator. Then is it true that there exist t... | wildildildlife | 10,729 | <p>Nothing new compared to Andreas's answer, just wanted to stress the polarization idea:</p>
<p><strong>Notation</strong>: For $H$ a Hilbert space, and $A\in B(H)$ (bounded linear operator), write $q_A$ for the quadratic form $x\mapsto \langle Ax,x\rangle$. </p>
<p><strong>Lemma</strong> ('polarization'): If $H$ is ... |
1,131,475 | <p>If $f(x)$, is a twice differentiable function, and $f"(x)=0$ at $x=c$, then $f(x)$ has an inflection point at $x=c$. </p>
<p>Does the above statement always apply? It seems so to me, because if the second derivative is set equal to zero, and there is a solution, then there must be an inflection point. Can anyone pr... | Hagen von Eitzen | 39,174 | <p>Given sufficient differentiability, an inflection point of $f$ is a locla extremum of $f'$; and a local extremum of $f'$ is a point where $f''$ changes signs. Does a function <em>necessarily</em> switch signs at its zeroes?</p>
|
2,624,703 | <p>I'm having trouble proving these, can anyone help?</p>
<p>The question is as follows:</p>
<p>For each $n \in \mathbb{N}$, let $A_n = \lbrace{ k \in \mathbb{Z} ; k^2 <= n}\rbrace$ </p>
<p>Prove that:</p>
<p>1) $\bigcap A_n$ = $\lbrace 0, 1, -1 \rbrace$</p>
<p>2) $\bigcup A_n = \lbrace ...,-2,-1,0,1,2,...\rbra... | Robert Z | 299,698 | <p>The integrand function has just ONE singularity at $z=-2$. For any non-zero integer $n$ then by letting $z=\frac{1}{n \pi}-2$, we have that
$$\sin\left(\frac{1}{z+2}\right)=\sin(n\pi)=0.$$</p>
<p>So, for $R>2$, the singularity $z=-2$ is inside the circle $|z|=R$, and, by the <a href="https://en.wikipedia.org/wik... |
656,791 | <p>For all $n\geq 1$, prove with mathematical induction </p>
<p>$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$</p>
<p>So far.. I have substituted 1 and saw that the statement is true and I have plugged in n+1 to show that the proof is true for all integers but I don't know how to g... | Semsem | 117,040 | <p>you simply have
$$2-\frac{1}{k}+\frac{1}{(k+1)^2} \\
=2-\{\frac{1}{k}-\frac{1}{(k+1)^2}\} \\
=2-\{\frac{(k+1)^2-k}{k(k+1)^2}\}\\
=2-\{\frac{k^2+k+1}{k(k+1)^2}\}\\
\leq 2-\frac{k(k+1)}{k(k+1)^2}$$</p>
<p>Since
$$k^2+k+1 \gt k^2+k\\
\frac{k^2+k+1}{k(k+1)^2} \gt \frac{k^2+k}{k(k+1)^2}\\
-\frac{k^2+k+1}{k(k+1)^2} \le ... |
754,392 | <p>Consider the following integrals in variables $x,y$ over the whole $\mathbb{R}$, where $a,b\in\mathbb{R}/0$ are constants:</p>
<p>$$\int dx \int dy ~\delta(x-a)\delta(y-b\,x)=\int dy ~\delta(y-b\,a)=1$$</p>
<p>In the following we will evaluate the above integrals in a slightly different way and obtain a completely... | druckermanly | 102,424 | <p>The power series for $F'(X)$ is $$F'(x) = \sum_0^\infty \frac{(-x)^{2n}}{n!} = \sum_0^\infty \frac{(-1)^nx^{2n}}{n!}$$
We now integrate each side and get:
$$
\begin{align}
F(X) &= \sum_0^\infty \frac{(-1)^n (x)^{2n+1}}{(2n+1)(n!)}
\end{align}$$</p>
|
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