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 |
|---|---|---|---|---|
2,589,155 | <blockquote>
<p>Let $d$ be the standard metric on $\mathbb{R}$. i.e $d(x,y) = |x-y|$. Is there a distance $d'$ with $d'(x,y) \geq d(x,y)$ so that $(\mathbb{R},d')$ is compact?</p>
</blockquote>
<p>Is my proof correct?
Let $B_{\epsilon}^{d'}(x)$ denote a ball of radius $\epsilon$ centered on $x$ with respect to $d'$.... | Guy Fsone | 385,707 | <p>Well your proof is correct but it could be shortened as follow:</p>
<p>
<blockquote>
<p>Indeed, if such metric exists we have that the $Id:(\Bbb R, d')\to(\Bbb R, d)$ is continuous since we assume $$d(x,y)\le d'(x,y)$$
Then $(\Bbb R, d)= Id((\Bbb R, d'))$ is also compact as continuous image of compact set... |
2,840,091 | <blockquote>
<p>Consider the linear map$:T:\mathbb{R}^3 → \mathbb{R}$ with
$$T\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right)=x-2y-3z$$
Find the basis of its kernel.</p>
</blockquote>
<p><strong>My try</strong></p>
<p>Since the plane is the nullspace of the matrix
$$A=\begin{bmatrix} 1 & -2 & -3 ... | user247327 | 247,327 | <p>The "kernel" of a linear transformation, $A$, is defined as the subspace, $\{v\}$, of its domain such that $Av= 0$. Here that is $A(x, y, z)= x- y+ 3z= 0$. We can solve for $x$: for any $y, z$, $x= y- 3z$. A vector in the kernel is of the form $(x, y, z)=(y- 3z, y, z)= (y, y, 0)+ (-3z, 0, z)= y(1, 1, 0)+ z(-3, 0,... |
90,006 | <p>I have two problem collections I am currently working through, the "Berkeley Problems in Mathematics" book, and the first of the three volumes of Putnam problems compiled by the MAA. These both contain many problems on basic differential equations.</p>
<p>Unfortunately, I never had a course in differential equation... | Will Jagy | 10,400 | <p>Yes, the value of the fraction is always below 1/2. So it stays far away from 3, indeed never gets within 5/2 of it. Note that it is not necessary to know that there is any limit to do answer your question. Anyway, take $\varepsilon = 2$ and no $n$ ever gets you within $\varepsilon$ of 3.</p>
|
2,497,799 | <h2>Question</h2>
<p>While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is </p>
<blockquote>
<p>$$\sum_{j=0}^{n-2}2^j (n-j)$$</p>
</blockquote>
<h2>My Approach</h2>
<blockquote>
<p>$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2}
2^... | Prajwal Kansakar | 49,781 | <p>$$\sum_{j=1}^{n-2}j\;2^j=2+2\times 2^2+3\times 2^3+\cdots+(n-2)2^{n-2}$$$$=[2+2^2+2^3+\cdots+2^{n-2}]+[2^2+2^3+\cdots+2^{n-2}]+[2^3+\cdots+2^{n-2}]+\cdots+2^{n-2}$$$$=2(2^{n-2}-1)+2^2(2^{n-3}-1)+2^3(2^{n-4}-1)+\cdots+2^{n-2}(2^{n-(n-1)}-1)$$ $$=(2^{n-1}-2)+(2^{n-1}-2^2)+(2^{n-1}-2^3)+\cdots+(2^{n-1}-2^{n-2})$$ $$=(n... |
2,831,469 | <p>How can we solve $$y'''(t)+a(y''(t))^2+b(y'(t))^3=0$$</p>
<p>Could one make some kind of least common denominator argument to decide possible substitutions? Since the chain rule will come into play, I suppose a substitution both for the variable $t$ and the function $y$ could be useful. Possibly some powers of them... | Alex Jones | 350,433 | <p>As a more general solution, if you have an equation of the form
$$ x''(t) + a(x(t))x'(t)^2+b(x(t)) = 0 $$
then you can make the substitution $f(x) = x'(t)^2$ to arrive at the equation
$$ \frac{1}{2}f'(x) + a(x)f(x)+b(x)= 0 $$
Letting $\mu(x) = \exp\left[\int a(x) dx\right]$, we can solve for $f(x)$:
$$ f(x) = \mu(x... |
107,653 | <p>Suppose you have the set of all possible $n$ x $n$ square adjacency matrices where $n$={1,2,3,4...}. For each matrix, compute the logarithm of the largest eigenvalue. Is it true that the set of logarithms you obtain is dense in $\mathbb{R}$? How do you begin to prove/disprove this?</p>
| Nikita Sidorov | 8,131 | <p>In addition to Doug's nice answer above: it is probably even easier to show that the set of simple Parry numbers is dense in $(1,\infty)$. More precisely, let $\beta>1$ and let $(d_n)_{n=1}^\infty$ be the <em>greedy $\beta$-expansion</em> of 1, i.e.,
$$
1=\sum_{n=1}^\infty d_n\beta^{-n},
$$
where $d_1=\lfloor \b... |
4,230,409 | <p>I recently stumbled upon Edwin Moise's proof of the area of a square, but there is something that is bothering me in his proof - he claims that since the inequalities <span class="math-container">$(1)$</span> and <span class="math-container">$(5)$</span> are true, then <span class="math-container">$a =\sqrt{S_a}$</s... | Sayan Dutta | 943,723 | <p>Here <span class="math-container">$\alpha S$</span> represents the area of <span class="math-container">$S$</span>.</p>
<p>Now, your question is why do <span class="math-container">$(1)$</span> and <span class="math-container">$(5)$</span> imply <span class="math-container">$a=\sqrt{\alpha S_a}$</span>. This is a ve... |
2,797,419 | <p>Calculating the moment of inertia is fairly simple, however, how would I proceed to calculate it about an arbitrary axis?
The question asks the moment of inertia of $C=\{(x,y,z)|0\leq x,0\leq y, 0 \leq z,x+y+z\leq 1\}$, so, if I'm not wrong about the boundaries, the moment of inertia about the "usual" Z axis would b... | Ted Shifrin | 71,348 | <p>Just integrate the function $f(x,y,z)$ that equals the square of the distance from $\vec x=(x,y,z)$ to your axis. If $\vec a$ is a unit vector in the direction of the axis, then $$f(x,y,z) = \|\vec x - (\vec x\cdot\vec a)\vec a\|^2 = \|\vec x\|^2 - (\vec x\cdot\vec a)^2.$$</p>
|
1,607,108 | <p>Using the ratio test:
$$\frac{1}{3}\lim\limits_{n\to\infty}\left|\frac{(n+1)^3(\sqrt{2}+(-1)^{n+1})^{n+1}}{n^3(\sqrt{2}+(-1)^n)^n}\right|$$</p>
<p>Without evaluating the limit, the numerator is greater then denominator and the series is divergent. Is there an easier method for checking the convergence of this seque... | Clive Newstead | 19,542 | <p>If you were to apply the result you quote directly, what you'd actually obtain is that if $2 \mid x^2-1$ then $8 \mid (x^2-1)^2$. Specifically, you're taking $a=2$, $c=4$ and $b=d=x^2-1$ in the result you mention.</p>
<p>However, it <em>is</em> true that if $2 \mid x^2-1$ then $8 \mid x^2-1$; it's just true for dif... |
2,822,176 | <p>I'm solving a list of exercises of double integrals and they normally have a range for $x$ and $y$, but in this case it says that $y = x^2$ and $y = 4$, so I thought that $x$ would be $\sqrt{y}$, but my answer was wrong. The answer should be 25,60).</p>
<p>A thin metal plate occupies a shadow below the figure below... | A.Γ. | 253,273 | <p>Step 1: You need to translate information "the region <strong>is limited</strong>" by two curves into inequalities for $x$ and $y$ to describe the shadowed $D$.</p>
<blockquote class="spoiler">
<p> For example, $D=\{(x,y)\colon -2\le x\le 2,\,x^2\le y\le 4\}$.</p>
</blockquote>
<p>Step 2: After that the problem ... |
1,842,448 | <p>$F(x)= 2x^2 -3x$. </p>
<p>find the range of $x$ to check whether the function the is strictly increasing and strictly decreasing.</p>
| Community | -1 | <p>Apply a small positive increment $h$ to $x$ and compare</p>
<p>$$2(x+h)^2-3(x+h)><2x^2-3x.$$</p>
<p>This can be rewritten</p>
<p>$$4xh+2h^2-3h><0,$$ or</p>
<p>$$4x-3+2h><0.$$</p>
<p>Ignoring $h$, which can be made smaller and smaller, the LHS is positive when </p>
<p>$$x>\frac34$$ and conv... |
588,228 | <p>A,B,C are distinct digits of a three digit number such that </p>
<pre><code> A B C
B A C
+ C A B
____________
A B B C
</code></pre>
<p>Find the value of A+B+C.</p>
<p>a) 16 b) 17 c) 18 d) 19</p>
<... | M.R. Yegan | 134,463 | <p>$ABC+BAC+CAB=ABBC \\
C+10B+100A+C+10A+100B+B+10A+100C=C+110B+1000A$</p>
<p>and this yields to $B+C=880A-100C$,
and this in turn implies the number 20 divides $B+C$. But $B$ and $C$ are digits. So $2\leq{B+C}\leq18$. Clearly a contradiction.</p>
|
1,600,266 | <p>I'm trying to work out sum of this series $$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \ldots$$</p>
<p>I know one method is to do substitutions and getting the series into a form of a known series. So far I've converted the series into $$ 1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \ldots $$ where $x=... | CommonerG | 74,357 | <p>I think it's likely a mistake, although I don't see any mention in the <a href="http://infolab.stanford.edu/~ullman/ialc/errata.html" rel="nofollow">errata</a>.</p>
<p>Perhaps he means every prime except 2 when expressed as a binary digit ends with a 1? </p>
|
835,454 | <p>I was struggling with the following problem (from linear algebra):</p>
<blockquote>
<p>Let $V$ be the vector space of the $2 \times 2$ matrices with real coefficients. Consider the action of the group $SL_2(\mathbb{R})$ on $V$, namely for any matrix $T \in SL_2(\mathbb{R})$ and for every matrx $A \in V$ we define... | user1551 | 1,551 | <p>The question setter probably mean this. Consider the following ordered basis $\mathcal B$ of $V=M_2(\mathbb R)$.
$$
\mathcal B=\left\{
B_1=\pmatrix{1&0\\ 0&0},
\ B_2=\pmatrix{0&0\\ 1&0},
\ B_3=\pmatrix{0&1\\ 0&0},
\ B_4=\pmatrix{0&0\\ 0&1}
\right\}.
$$
We can identify every $B=\pmatri... |
180,122 | <p>Could any one give me hint how to show that the zero section of any smooth vector bundle is smooth?</p>
<p>Zero section is a map $\xi:M\rightarrow E$ defined by $$\xi(p)=0\qquad\forall p\in M.$$</p>
| Community | -1 | <p>Say $E\to M$ is a smooth vector bundle, and let $p\in M$. Then there is a neighborhood $U\subseteq M$ of $p$ and a smooth local trivialization $\Phi:\pi^{-1}(U)\to U\times\mathbb R^k$ of $E$ over $U$. What can you say about $(\Phi\circ\zeta)(q)$ where $q\in U$? Can you conclude that $\Phi\circ\zeta$ is smooth? Now r... |
2,598,848 | <p>I have a project I'm doing in desmos and want to know if this is possible.</p>
<p>I want to be able to set some parameter, a, equal to the solution of f(x) = 0. I plugged my specific equation into Maple and it could not arrive at an explicit solution for x.</p>
<p>However, when I type in the equation f(x) = 0 into... | Emilio Novati | 187,568 | <p>I suppose $a<b$. The intersection points of the two ellipses are (see the figure where $a=2$ and $b=4$) $A,B,C,D =(\pm k,\pm k)$ with
$$k=\frac{ab}{\sqrt{a^2+b^2}}$$ </p>
<p><a href="https://i.stack.imgur.com/ZpHMj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZpHMj.jpg" alt="enter image de... |
1,583,540 | <p>Is it possible to find an example of an one-sided inverse of a function? other than matrix?</p>
<p>I am trying to find such an example but having no luck. Anybody got an idea about it?</p>
| boxotimbits | 299,738 | <p>Think of the function $f:\{0,1\}\to \{1\}$ that is constant. It has a right inverse $g:\{1\}\to\{0,1\}$ which maps $1\mapsto 1$. This is because $f\circ g=Id_{\{1\}}$, the identity function on $\{1\}$. But it doesn't have a left inverse since it is not injective (1 to 1). </p>
<p>Also you should notice that the rig... |
1,583,540 | <p>Is it possible to find an example of an one-sided inverse of a function? other than matrix?</p>
<p>I am trying to find such an example but having no luck. Anybody got an idea about it?</p>
| Federico Poloni | 65,548 | <p>Another classical example is the shift maps. Let $S$ be the space of real sequences $S = \{(a_0,a_1,a_2,\dots) : a_i\in\mathbb{R} \, \text{for all $i$}\}$ (which, essentially, is the space of functions $\mathbb{N}\to\mathbb{R}$).</p>
<p>Define the "left shift map" $f:S\to S$ as
$$
f((a_0,a_1,a_2,a_3,\dots)) = (a_1,... |
267,963 | <p>Why <code>HoldForm</code> do this?</p>
<p><a href="https://i.stack.imgur.com/DDFDT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DDFDT.png" alt="enter image description here" /></a></p>
<pre><code>HoldForm[a/b \[Integral]f[x] \[DifferentialD]x]
</code></pre>
<p>Output is this ugly looking formula:
<span ... | Alexey Popkov | 280 | <p><sup>(The basic idea is from <a href="https://mathematica.stackexchange.com/a/186026/280">this</a> answer by <a href="https://mathematica.stackexchange.com/users/1871/xzczd">xzczd</a>.)</sup></p>
<p>Try this:</p>
<pre><code>MakeBoxes[Times[a_, b_], StandardForm] := RowBox[{MakeBoxes@a, MakeBoxes@b}]
MakeBoxes[Times[... |
116,494 | <p>I am trying to plot contour plots and some of them show a white patch because of a change in sign of the points plotted. How can I make the contour plot show this change in sign without having to take the absolute of the values?</p>
<p>The data is the following:</p>
<pre><code>axialP1plot= {{30, 0, 0.509185}, {60,... | dantopa | 29,452 | <p>Use <code>Contours -> {0}</code> to force the contour line at zero value.<a href="https://i.stack.imgur.com/x7MgK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x7MgK.jpg" alt="enter image description here"></a></p>
<pre><code>ListContourPlot[axialP1plot, ColorFunction -> "Rainbow",
Cont... |
2,161,913 | <p>Consider the angle $\theta$ with $$0<\theta<\frac{\pi}{2}$$ Suppose that $$cos\theta=\frac{l}{n} \space and \space \space sin\theta=\frac{m}{n}$$
are rational numbers, with postive integers $l,m$ and $n$. Show that $l$ and $m$ cannot both odd integers.<br>
MY ATTEMPT: Assume that $l$ and $m$ are both odd int... | RRL | 148,510 | <p>You got the right answer for the wrong reason, eg. $(2^n)^{1/n} \to 2 \neq 1$.</p>
<p>Cauchy's second limit theorem states</p>
<p>$$\lim_{n \to \infty} (a_n)^{1/n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n},$$</p>
<p>if the limit on the RHS exists.</p>
<p>This case reduces to </p>
<p>$$\lim_{n \to \infty} \frac{... |
602,353 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be given by $$f(x)= \begin{cases}
-x-1, &\text{if }x≥0\\
1, &\text{if }x<0.
\end{cases}$$</p>
<p>Denote by</p>
<ul>
<li>$U$ the <em>usual</em> topology on $\mathbb{R}$.</li>
<li>$H$ the <em>half-open interval</em> topology on $\mathbb{R}$.</li>
<li>$C$ is the <em>open ... | Randomstop | 113,784 | <p>Here is a breakdown of my attempt for each one of my questions:</p>
<ol>
<li><p>U-U Continuous: $(-2,1)$ is open in the U topology. $f^{-1}((-2,1))=[0,1)$. $[0,1)$ is not open in the U topology and therefore f is not U-U continuous.</p></li>
<li><p>U-H Continuous: $(-2,1)$ is also H-open, which implies that f is... |
4,516,006 | <p>Let <span class="math-container">$a,~b\in\Bbb Q$</span> and suppose <span class="math-container">$\sqrt{a},~\sqrt{b}$</span> is irrational and <span class="math-container">$\sqrt{a}-\sqrt{b}\in\Bbb Q$</span>. I want to prove that <span class="math-container">$\sqrt{a}-\sqrt{b}=0$</span>; that is, <span class="math-c... | Alessandro Cigna | 641,678 | <p>Observe that <span class="math-container">$(\sqrt a-\sqrt b)(\sqrt a +\sqrt b)=a-b\in \Bbb Q$</span>, if <span class="math-container">$\sqrt a\ne\sqrt b$</span> then <span class="math-container">$\sqrt a+\sqrt b=\frac {a-b}{\sqrt a-\sqrt b}\in\Bbb Q$</span>. But so <span class="math-container">$\sqrt a=\frac 12(\sqr... |
4,516,006 | <p>Let <span class="math-container">$a,~b\in\Bbb Q$</span> and suppose <span class="math-container">$\sqrt{a},~\sqrt{b}$</span> is irrational and <span class="math-container">$\sqrt{a}-\sqrt{b}\in\Bbb Q$</span>. I want to prove that <span class="math-container">$\sqrt{a}-\sqrt{b}=0$</span>; that is, <span class="math-c... | Andrea Mori | 688 | <p>This is a <strong>hint</strong></p>
<p>Suppose <span class="math-container">$\sqrt a-\sqrt b\in\mathbb{Q}$</span>. Then <span class="math-container">$(\sqrt a-\sqrt b)^2\in\mathbb{Q}$</span> and so
<span class="math-container">$$
\sqrt{ab}=\frac{a+b}2\in\mathbb{Q}
$$</span>
with <span class="math-container">$ab\in\m... |
3,827,302 | <p>In part of Kenneth Falconer's proof of the countable stability of Hausdorff dimension on p. 49, sect 3.2 of <a href="https://www.wiley.com/en-us/Fractal+Geometry%3A+Mathematical+Foundations+and+Applications%2C+3rd+Edition-p-9781119942399" rel="nofollow noreferrer">Fractal Geometry</a>, I understand him to say that
<... | usr0192 | 275,654 | <p>Based on Kevin Arlin's answer here is my revised understanding (decided to post as an answer because not enough space for a comment)</p>
<p>For composition fill, in the <span class="math-container">$\Lambda^3_1$</span>-horn
<a href="https://i.stack.imgur.com/pBIXQ.jpg" rel="nofollow noreferrer"><img src="https://i.s... |
2,111,579 | <p>A question:</p>
<p>If $\displaystyle{m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n},{\quad\text{and}\quad}{n}\ne{m};$</p>
<p>Proof that $\displaystyle{m}^{7}+{n}^{7}+{30}={1}$</p>
<p>Without finding the roots of equation $\displaystyle{x}^{2}+{x}-{1}={0}$.</p>
<p>Is there such a shortcut solution?</p>
| Community | -1 | <p>Yes, your inverse is correct. Let me show you an alternative approach of finding this inverse without the use of the adjugate matrix.
<hr>
Denote $A_n$ as the $n \times n$ matrix that has elements $a_{ij} = 1$ if $i \geq j$, and $0$ otherwise. Then, $A_n$ can be constructed by applying the elementary row operations... |
3,190,781 | <blockquote>
<p>Suppose the polynomial <span class="math-container">$P(x)$</span> with integer coefficients satisfies the following conditions:<br>
(A) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 − 4x + 3$</span>, the remainder is <span class="math-container">$65x −... | know dont | 663,454 | <p>We notied that: <span class="math-container">$x^2-4x+3=(x-1)(x-3)$</span> and <span class="math-container">$x^2+6x-7=(x-1)(x+7)$</span>.</p>
<blockquote>
<p>So, we have: <span class="math-container">$P(1)=65*1-68=-5*(1)+a$</span>.
<span class="math-container">$\implies -3=-5+a\iff a=2$</span>.</p>
</blockquote>... |
2,014,806 | <p>I was hoping to get some help on a review problem from my text book. The problem reads: </p>
<p>Let $A$ be $m\times{n}$ matrix. Prove that</p>
<p>a) $A^*A$ is self-adjoint.</p>
<p>b) All eigenvalues of $A^*A$ are non-negative.</p>
<p>c) $A^*A + I$ is invertible.</p>
<p>For part a), it seems to follow very clear... | Gyu Eun Lee | 52,450 | <p>For part (b), you are on the right track but you want to adopt a different approach. The question is about the eigenvalues of $A^*A$, not those of $A$. So suppose $x$ is an eigenvector of $A^*A$, i.e. $A^*Ax = qx$ for some scalar $q$. Now you can use your approach (use the inner product) to deduce things about $q$.<... |
2,876,040 | <p>So i have the following problem: I need to cover the whole interval from $1$ to $10^{12}$ on a computer program. But i can´t do it on a for loop because it would be too slow. So can i iterate over a smaller amount ($\sqrt{n}$ for instance), and still cover the whole interval? </p>
<p>What i've done so far:</p>
<p>... | Deepak gupta | 450,988 | <p>The optimization depends on what you want to do with numbers in the interval. Example:
1. If you want them to write them to a file then there is no optimization you can do.
2. If you want to find out multiples of 1000, then it is possible to optimize the loop to 10^9 iterations.</p>
<p>The approach you tried will o... |
2,588,225 | <p>To start off, I'm very new to the complex plane and complex numbers.</p>
<p>I'm trying to determine if the series $\sum_{n=1}^\infty\frac{1}{n^2 + i}$ is convergent or divergent.</p>
<p>However, I'm not exactly sure how to express the value of the function. Could I just treat the denominator of the expression as t... | max_zorn | 506,961 | <p>$$\left|\frac{1}{n^2+i}\right| = \frac{1}{\sqrt{n^4+1}} < \frac{1}{\sqrt{n^4}} = \frac{1}{n^2};$$
thus, by the comparison test and $p$-test (with $p=2$) the series converges absolutely. </p>
|
3,772 | <p>In <a href="https://math.stackexchange.com/questions/102764/non-polynomial-functions-on-fields-of-finite-characteristic">this</a> question, there are two answers, both of which I up-voted, but I didn't "accept" either since there didn't appear to be a reason why one stands out as clearly preferable to the other. Ap... | hmakholm left over Monica | 14,366 | <p>If your only problem is that there isn't a clear reason to prefer one answer to the other, but you still feel that <em>your question has been answered satisfactorily</em>, then I'd say to go ahead, flip a coin and accept one of them. Or choose the one whose author needs the rep the most, or whatever.</p>
<p>Getting... |
1,591,990 | <p>A while ago I saw this question <a href="https://math.stackexchange.com/questions/1561877/quartic-diophantine-equation-16r4112r3200r2-112r16-s2">Quartic diophantine equation: $16r^4+112r^3+200r^2-112r+16=s^2$</a> which was very relevant to a undergraduate research paper I am currently working on. The answer given fo... | Allan MacLeod | 110,165 | <p>The following shows how to transform the quartic
\begin{equation}
y^2=ax^4+bx^3+cx^2+dx+e ,
\end{equation}
with $a,b,c,d,e \in \mathbb{Q}$, into an equivalent elliptic curve.</p>
<blockquote>
<p><strong>Case 1:</strong></p>
</blockquote>
<p>We first consider the case when $a=1$, which is very common. If $a=\alph... |
330,677 | <p>I was thinking last days about the following problem - the sum of Darboux is a Darboux function?</p>
<p>Do You know a proof or counter example?</p>
| MathematicsStudent1122 | 238,417 | <p>There's a simpler example. Let $g,h: \mathbb{R} \to \mathbb{R}$ be given by</p>
<p>$$g(x)=\begin{cases}
{\sin x^{-1}} & x \neq 0 \\
1 & x=0\\
\end{cases}$$</p>
<p>$$h(x)=\begin{cases}
{-\sin x^{-1}} & x \neq 0 \\
0 & x=0\\
\end{cases}$$</p>
<p>These are Darboux. However, </p>
<p>$$g(x) + h(x) =\b... |
1,016,585 | <p>I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$</p>
<p>How do I do this?</p>
<p>It looks like it is close to being an $\operatorname{arcsin}$?</p>
<hr>
<p>I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{??... | Community | -1 | <p>The trick is to use trig substitution:</p>
<p>HINT:</p>
<p>Let $x = \sqrt{c} \sec u \implies x^2 = c \sec^2u \implies x^2 - c = c ( \sec^2 u - 1 ) = c \tan^2 u$</p>
<p>and</p>
<p>$$ dx = \sqrt{c} \sec u \tan u $$</p>
<p>Hence, your integral looks like</p>
<p>$$ \int \frac{\sqrt{c} \sec u \tan u}{\sqrt{c } \... |
30,970 | <p>I try to filter sublists of a list which match a pattern. </p>
<pre><code>test = {{"String1", "a"}, {"String2", "b"}, {"String3",
"a"}, {"String4", "a"}};
</code></pre>
<p>The result should be:</p>
<pre><code>result = {{"String1", "a"}, {"String3", "a"}, {"String4", "a"}}
</code></pre>
<p>That means the first e... | user1066 | 106 | <p>Another possibility:</p>
<pre><code>Pick[#, (Thread@#)[[2]], "a"] &@test
</code></pre>
<blockquote>
<pre><code>{{"String1", "a"}, {"String3", "a"}, {"String4", "a"}}
</code></pre>
</blockquote>
|
1,531,525 | <p>I would like to compute the following sum:</p>
<p>$$\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}$$</p>
<p>I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $\cos \theta = {e^{i\theta} + e^{-i\theta}\over 2}$. I'm not sure how to proc... | Anurag A | 68,092 | <p>Consider the series
$$S=\sum_{n=0}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^n.$$
This is a geometric series whose sum is
$$S=\frac{2}{2-e^{i\theta}}.$$
Now the real part of $S$ is the sum you are looking for.</p>
|
1,531,525 | <p>I would like to compute the following sum:</p>
<p>$$\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}$$</p>
<p>I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $\cos \theta = {e^{i\theta} + e^{-i\theta}\over 2}$. I'm not sure how to proc... | Caleb Stanford | 68,107 | <p>\begin{align*}
\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}
&= \sum_{n=0}^\infty \frac{e^{i n \theta} + e^{- i n \theta}}{2^{n+1}} \\
&= \sum_{n=0}^\infty \frac{(e^{i \theta})^n + (e^{- i \theta})^n}{2^{n+1}} \\
&= \sum_{n=0}^\infty \frac{(e^{i \theta})^n}{2^{n+1}} + \sum_{n=0}^\infty \frac{(e^{- i \the... |
209,259 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/16374/universal-chord-theorem">Universal Chord Theorem</a> </p>
</blockquote>
<p>I am having a problem with this exercise. Could someone help?</p>
<p>Suppose $a \in (0,1)$ is a real number which is not of... | Hagen von Eitzen | 39,174 | <p><strong>Hint:</strong>
Since $\frac1a\notin\mathbb N$, there is some $n\in\mathbb N_0$ such that $n<\frac1a<n+1$. Let $r= 1-na$. Then $0<r<a$.
Let us try to find a function such that $f(x+a)-f(x)=1$ for all $x$ with $0\le x \le 1-a$.
Thus we have by induction $f(x)=k+f(t)$ if $x=ka+t$ with $k\in\mathbb ... |
224,474 | <p>I want to generate the following matrix for any n:</p>
<pre><code> Table[A[i, j], {i, 0, n}, {j, 0, n}];
</code></pre>
<p>Where,</p>
<pre><code> A ={{0,1,0},{0,0,2},{0,0,0}} when n=2;
A={{0,1,0,0},{0,0,2,0},{0,0,0,3},{0,0,0,0}} when n=3;
A={{0,1,0,0,0},{0,0,2,0,0},{0,0,0,3,0},{0,0,0,0,4},{0,... | Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
A[n_Integer?Positive] := DiagonalMatrix[Range[n], 1, n + 1]
A /@ Range[2, 4]//Column
</code></pre>
<p><a href="https://i.stack.imgur.com/M6CK8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/M6CK8.png" alt="enter image description here" /></a></p>
|
199,946 | <p>I came across this step in an inductive proof, but my algebra skills seem a bit rusty..</p>
<p>$7(7^{k+2}) + 64(8^{2k+1}) = 7(7^{k+2}+8^{2k+1})+57(8^{2k+1})$</p>
<p>How did they do this?</p>
<p>Note: The point was to show that the expressions are divisible by 57.</p>
| Chris Eagle | 5,203 | <p>$64=7+57$, so $64\cdot 8^{2k+1}=(7+57)\cdot 8^{2k+1}=7\cdot 8^{2k+1}+57 \cdot 8^{2k+1}$.</p>
|
1,256,688 | <p>Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of 2015 distinct real numbers such that after one initial move is applied to the sequence — no matter what move — there is always a way to continue with a finite se... | mweiss | 124,095 | <ol>
<li>If you assume $A,B,C$ and are able to use them to not only deduce $Z$ but also a contradiction to $Z$, then the assumptions you began with are <em>inconsistent</em>, and then you <em>really</em> have a problem, because literally anything can be deduced from a set of inconsistent hypotheses. So what you seem t... |
1,826,094 | <p>$i\cdot\bar{z} = 2+2i$</p>
<p>I know that $\bar{z} = a-bi$ so then i get $i(a-bi)=2+2i$
Then $ai+b=2+2i$ (because $i^2=-1$)</p>
<p>When 2 complex numbers are equal you usually can equal their parts
Ex: $2+2i=a+bi$ so $a=2$ and $b=2$</p>
<p>But in this case I aint got $a+bi$ I have $ai+b$</p>
<p>The answer is $z=... | levap | 32,262 | <p>It might be useful also to think of the problem geometrically. The operation $z \mapsto i\overline{z}$ takes the vector $z$, reflects it across the $x$-axis (this is $z \mapsto \overline{z}$) and then rotates it by $90$ degrees counterclockwise (this is $w \mapsto iw$). Solving $i\overline{z} = 2 + 2i$ amounts to fi... |
1,826,094 | <p>$i\cdot\bar{z} = 2+2i$</p>
<p>I know that $\bar{z} = a-bi$ so then i get $i(a-bi)=2+2i$
Then $ai+b=2+2i$ (because $i^2=-1$)</p>
<p>When 2 complex numbers are equal you usually can equal their parts
Ex: $2+2i=a+bi$ so $a=2$ and $b=2$</p>
<p>But in this case I aint got $a+bi$ I have $ai+b$</p>
<p>The answer is $z=... | Tazwar Sikder | 347,302 | <p>We have: $i\cdot\bar{z}=2+2i$</p>
<p>Let $z=a+bi \Rightarrow \bar{z}=a-bi$:</p>
<p>$\Rightarrow i\cdot(a-bi)=2+2i$</p>
<p>$\Rightarrow ai-bi^{2}=2+2i$</p>
<p>$\Rightarrow ai+b=2+2i$</p>
<p>$\Rightarrow b+ai=2+2i$</p>
<p>Comparing real and imaginary parts:</p>
<p>$\therefore\hspace{5 mm}b=2$ and $a=2$</p>
|
1,381,017 | <p>Could you have me to find the ferivative of $$f(t)=\frac {1}{\rho}\log (1+\rho t)$$ with repsect to $t$?</p>
<p>And Is it
$$\lim_{t\to \infty} \frac {f'(t)}{t}=1?$$
<em>Update:</em>
Based on Hint of Surb:</p>
<p>$$f(t)'=\frac {1}{1+\rho t}$$</p>
<p>Then
$$\lim_{t\to \infty} \frac {f'(t)}{t}=0$$</p>
<p>Is it co... | Lucian | 93,448 | <blockquote>
<p><em>I have been working on days to find the integral of the following question</em></p>
</blockquote>
<p>Even if you were to have kept on working on it for <em>years</em>, you still would not have been able to reach any conclusion, since it cannot be expressed in terms of elementary functions. See <a... |
957,604 | <p>Let <strong>X</strong> and <strong>Y</strong> have the joint probability density function given by:</p>
<p>f(x,y)=$\frac{1}{4}exp{\frac{-(x+y)}{2}}$, x>0 and y>0</p>
<p>(a) Find Pr(x<1,y>1)</p>
<p>(b) Find Pr(y<$x^2$)</p>
<p>This is how I tackled (a):</p>
<p>Pr(x<1,y>1)=$\int\int exp{\frac{-(x+y)}{2}}$... | Platehead | 29,459 | <p>Hint for (a): use the law of total probability, conditioning over which two chips are transferred.</p>
<p>Then for (b), Bayes' theorem gives
\begin{equation}
P(\text{2 transferred} \mid \text{3 selected}) = \frac{P(\text{3 selected} \mid \text{2 transferred})P(\text{2 transferred})}{P(\text{3 selected})}.
\end{equa... |
3,912,076 | <p>A particle of mass m slides down on the curve <span class="math-container">$z=1+\frac{x^2}{2}$</span> in the <span class="math-container">$xz$</span>-plane without friction under the action of constant gravity.Suppose <span class="math-container">$z$</span>-axis points vertically upward.Find the Lagrangian of given ... | Cesareo | 397,348 | <p>The movement is restricted to the manifold <span class="math-container">$f(x,z) = 1+\frac 12 x^2-z=0$</span>. The potential energy is taken having a referential plane as for instance, the plane <span class="math-container">$XY$</span> so the potential energy is given by <span class="math-container">$m g z$</span>. T... |
978,202 | <p>When I use $\sin x \sim x$ , the answer is $1$ , is the answer correct?</p>
| Shine Mic | 182,335 | <p>We can use $sinx \sim x$ only when $x \to 0$ , when $x \to \infty $, we cannot use this substitution.</p>
<p>We may solve this problem in this way:
$$0 \leftarrow - \mathop {\lim }\limits_{x \to \infty } {1 \over {{x^2} + x}} \le \mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}} \le \mathop {... |
680,501 | <blockquote>
<p>Let $M$ be an $R$-module and let $F$ be a free $R$-module of finite rank. Let $\phi : M \to F$ be an epimorphism. Then show that $M$ has a submodule $F' \cong F $ such that $M=F' \oplus \ker\phi$.</p>
</blockquote>
<p>I am new to Module theory. If I apply fundamental theorem of homomorphism then $M/\... | Did | 6,179 | <p>Hint: Each $\mathbf 1_{A_n}$ is measurable with respect to $X=\sum\limits_k\frac1{3^k}\mathbf 1_{A_k}$.</p>
|
3,312,959 | <p>If the distance from the point <span class="math-container">$(K, 4)$</span> to <span class="math-container">$(2, 4)$</span> and y axis are same find k. </p>
<p>I can't understand what to calculate. According to my calculation the answer is 1, but the given answer is 0 and 4.</p>
| Michael Rozenberg | 190,319 | <p>I think we need to change the word "to" on "and". It's just a typo. </p>
<p>If so we have:
<span class="math-container">$$|K-2|=2$$</span>
Can you end it now?</p>
|
217,711 | <p>Set
$$
g(x)=\sum_{k=0}^{\infty}\frac{1}{x^{2k+1}+1} \quad \text{for} \quad x>1.
$$</p>
<p>Is it true that
$$
\frac{x^{2}+1}{x(x^{2}-1)}+\frac{g'(x)}{g(x)}>0 \quad \text{for}\quad x>1?
$$
The answer seems to be positive. I spent several hours in proving this statement but I did not come up with anyth... | Max Alekseyev | 7,076 | <p>The inequality is equivalent to
$$S := (x^2+1)g(x) + x(x^2-1)g'(x) > 0.$$
The left hand side here can be expanded to
$$S = \sum_{k\geq 0} \frac{(x^2+1)(x^{2k+1}+1) - (2k+1)x^{2k+1}(x^2-1)}{(x^{2k+1}+1)^2} $$
$$= \sum_{k\geq 0} \frac{(x^2+1) - (2k+1)(x^2-1)}{x^{2k+1}+1} + \sum_{k\geq 0}\frac{(2k+1)(x^2-1)}{(x^{2... |
3,005,591 | <p>I wish to find the integers of <span class="math-container">$a,b,c$</span> and <span class="math-container">$d$</span> such that:
<span class="math-container">$$225a + 360b +432c +480d = 3$$</span>
which is equal to:
<span class="math-container">$$75a + 120b +144c+ 160d =1$$</span></p>
<p>I know I have to use the... | Will Jagy | 10,400 | <p><span class="math-container">$$ 144 \cdot 12 - 75 \cdot 23 = 3 $$</span>
<span class="math-container">$$ 160 \cdot 1 - 3 \cdot 53 = 1 $$</span>
<span class="math-container">$$ 160 \cdot 1 - 53 ( 144 \cdot 12 - 75 \cdot 23 ) =1 $$</span>
<span class="math-container">$$ 160 \cdot 1 - 636 \cdot 144 + 1219 \cdot 75 =... |
3,005,591 | <p>I wish to find the integers of <span class="math-container">$a,b,c$</span> and <span class="math-container">$d$</span> such that:
<span class="math-container">$$225a + 360b +432c +480d = 3$$</span>
which is equal to:
<span class="math-container">$$75a + 120b +144c+ 160d =1$$</span></p>
<p>I know I have to use the... | Bill Dubuque | 242 | <p><span class="math-container">$\color{#c00}6 = 2(75)\!-\!144,\, $</span> <span class="math-container">$\color{#0a0}{80} = 2(120)\!-\!160\ $</span> so <span class="math-container">$\,\bbox[5px,border:1px solid red]{1 = 75\!+\!\color{#c00}6\!-\!\color{#0a0}{80} = 3(75)-2(120) -144 + 160}$</span></p>
<p><strong>Remark<... |
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you a... | Jonas Meyer | 1,119 | <p>The Jones polynomial: a knot invariant that originally came from subfactor theory.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you a... | Jose Brox | 1,234 | <p>Root systems and Dynkin diagrams for classification matters (see both <a href="http://en.wikipedia.org/wiki/Root_system" rel="nofollow">Root system</a> and <a href="http://en.wikipedia.org/wiki/ADE_classification" rel="nofollow">ADE classification</a> at Wikipedia).</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you a... | Qiaochu Yuan | 290 | <p>Suppose you are interested in random walks on an extremely structured graph such as a hypercube graph or a cycle graph. If your graph happens to be the Cayley graph of an abelian group $G$, as in both of the above examples, then it is easy to describe the behavior of random walks on it because the eigenvectors of t... |
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you a... | GMRA | 135 | <p>Well for a long time there was no proof of the Burnside theorem avoiding Representation theory. Now there are methods to proof it without Representation theory, but they are still a lot harder then the original representation theoretic one.</p>
<p><a href="http://en.wikipedia.org/wiki/Burnside_theorem" rel="nofollo... |
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you a... | Martin Brandenburg | 2,841 | <p>Sometimes you want to understand a group $G$, but the only thing you know is that there is an extension $1 \to A \to G \to E \to 1$. If everything is abelian, $G$ corresponds to an element in $Ext^1(E,A)$. If at least $A$ is abelian, then $E$ acts on $N$ by conjugation and $G$ corresponds to an element in $H^2(E,A)$... |
1,072,427 | <p>Prove using contradiction that any prime number greater than $3$ is of the form $6n \pm 1$.</p>
<p>Thanks for any help</p>
| barak manos | 131,263 | <p>Assume that $N$ is a prime larger than $3$ and <strong>not</strong> of the form $6n\pm1$:</p>
<ul>
<li>$N\equiv0\pmod6\implies N$ is divisible by $6\implies N$ is not a prime</li>
<li>$N\equiv2\pmod6\implies N$ is divisible by $2\implies N$ is not a prime (or $N=2$)</li>
<li>$N\equiv3\pmod6\implies N$ is divisible ... |
3,069,369 | <p>I know how to perform polynomial regression. But is there any method to use for estimating the degree of the polynomial that is best suited? Some kind of meta-regression.</p>
<p>With best suited I mean the grade that has the highest probability of being the true degree of the source for the data.</p>
<p>For exampl... | Gnumbertester | 628,028 | <p>In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.</p>
<p>You can study the rational root theorem here <a href="https://en.wikipedia.org/wiki/Rational_root_theorem" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Ratio... |
3,069,369 | <p>I know how to perform polynomial regression. But is there any method to use for estimating the degree of the polynomial that is best suited? Some kind of meta-regression.</p>
<p>With best suited I mean the grade that has the highest probability of being the true degree of the source for the data.</p>
<p>For exampl... | Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>This polynoial is divisible by <span class="math-container">$x^2-1$</span>:
<span class="math-container">\begin{align}
2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\\
&=2x^2(x^2-1)+ x^2(x-5)-x+5 \\
&= (x^2-1)(2x^2+x-5).
\end{align}</span>+</p>
|
1,562,907 | <p>I want to know whether $$\mathbb{Q}[x]/(x^3-1)$$ is a field or not. Is it as simple as determining if $x^3-1$ is irreducible in $\Bbb Q$?</p>
<p>But since it has roots , $x=1$ wouldn't this imply it is reducible and hence not a field? Or is there much more to it I am not considering?</p>
| fleablood | 280,126 | <p>The thing is $g(g^{-1}(x)) = x$. So we solve for $g^{-1}(x)$ in terms of $x$. Somehow, it is easy and less confusing if we use the notation $y = g^{-1}(x)$ and solve $x = g(g^{-1}(g(x)) = g(y)$ and solve for $y$ (which is actually $g^{-1}(x))$.</p>
<p>So $x = g(y)$; what is $y$?</p>
<p>$x = g(y) = f(y) + 1$. (w... |
4,581,481 | <p>How would one go about ALGEBRAICALLY finding the range of a semicircle?</p>
<p>eg. <span class="math-container">$y = \sqrt{4-x^2}$</span></p>
<p>Since there is no plus-minus sign, we know that <span class="math-container">$y$</span> must at least have a range of <span class="math-container">$[0, \infty)$</span>. Yet... | Ogglie Ostrich | 1,119,801 | <p>Assuming we are concerned with real <span class="math-container">$y$</span> we have:
<span class="math-container">$$y=\sqrt{4-x^2}\in\mathbb{R} \implies 4-x^2\geq0\implies x^2\leq4$$</span>
also<span class="math-container">$$ x\in\mathbb{R} \implies x^2\geq0 \implies 0\leq x^2\leq 4$$</span>
Hence:
<span class="math... |
1,117,924 | <p>How can I show with linear approximation that $y \approx x$ for small x? I know the rule $$f(x) \approx f(a) + f^{\prime}(a) (x-a),$$ but I don't know how to put it to use in this case.</p>
| randomgirl | 209,647 | <p>So we know the tangent line will look like $y-y_1=m(x-x_1)$
and we want $y-0=1\cdot(x-0)$.<br>
Since we want our line to go through $(0,0)$, I say find the tangent line at
$x=0$.</p>
|
1,580,796 | <p>In a game players take turns saying up to 3 numbers (starting at 1 and working their way up) and who every says 21 is eliminated. So we may have a situation like the following for 3 players:</p>
<blockquote>
<p>Player 1: 1,2</p>
<p>Player 2: 3,4,5</p>
<p>Player 3: 6</p>
<p>Player 1: 7,8,9</p>
<p>Player 2:10,11</p>
<... | joriki | 6,622 | <p>Community wiki answer so the question can be marked as answered:</p>
<p>As Abstraction pointed out in a comment, there is no such strategy (except in extreme cases) because the other players can cooperate against you; they have a wider range of options, collectively, than you do, so you can't use the sort of strate... |
1,923,226 | <p>I have been trying to prove this with some continuity theorems but haven't put together a good proof yet. </p>
| operatorerror | 210,391 | <p>Fix any $\epsilon$ you wish. Setting $\delta=1/2$ insures that $|x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$, since if $|x-y|<1/2$ means that $f(x)=f(y)$, since there is only one point in a radius of 1/2 on the integer lattice.</p>
<p>Might also be worth looking up the open preimage definition of continui... |
4,113,744 | <p>The number of real values <span class="math-container">$(x,y)$</span> for which <span class="math-container">$$2^{x+1}+3^y=3^{ y+2}-2^x$$</span> is ?
I went like, <span class="math-container">$$2^{x+1}+2^x=3^{y+2}-3^y$$</span>
after solving which I got; <span class="math-container">$(x,y)=(3,1)$</span>
Is there any ... | Mostafa Ayaz | 518,023 | <p>Note that
<span class="math-container">$$2^{y}+2^{y+1}=3^{y+2}-3^y\iff 3\times 2^y=8\times3^y\iff 2^{x-3}=3^{y-1}.$$</span>
Now, an investigation on tuples <span class="math-container">$(x,y)$</span> (whether real or integer) should be easy.</p>
|
364,059 | <p>While reading up on "Glivenko Cantelli Theorem" from Probability Models by K.B Athreya, the author used 2 lemmae to prove it. One was called Scheffe's lemma, the other Polya's theorem.</p>
<p>Scheffe's Lemma is stated as follows:</p>
<p>Let $f_n, f$ be non negative $\mu$ integrable functions. If $f_n \to f$ a.e an... | Sangchul Lee | 9,340 | <p>Let</p>
<p>$$u_n = \max\{ f, f_n\} \quad \text{and} \quad l_n = \min \{ f, f_n \} $$</p>
<p>so that both $(u_n)$ and $(l_n)$ converge pointwise to $f$, $l_n \leq f \leq u_n$ and $|f - f_n| = u_n - l_n$. By DCT, it is clear that $\int l_n \, d\mu \to \int f \, d\mu$. thus from</p>
<p>$$ \int u_n \, d\mu = \int (f ... |
364,059 | <p>While reading up on "Glivenko Cantelli Theorem" from Probability Models by K.B Athreya, the author used 2 lemmae to prove it. One was called Scheffe's lemma, the other Polya's theorem.</p>
<p>Scheffe's Lemma is stated as follows:</p>
<p>Let $f_n, f$ be non negative $\mu$ integrable functions. If $f_n \to f$ a.e an... | Xiao | 131,137 | <p>Observe $|f| + |f_n| - |f-f_n|\geq 0$, with Fatou's lemma
$$\liminf \int |f| + |f_n| - |f-f_n| \geq 2 \int |f| $$
by the assumption $\lim \int |f_n| = \int |f|$, the left-hand-side of above inequality also equals
$$\liminf \int |f| + |f_n| - |f-f_n| = 2\int |f| -\limsup \int |f-f_n|,$$
combine them we get
$$2\int ... |
100,801 | <p>When doing calculations in notebook, I modify the code a lot, and there are several versions of similar codes in the same notebook. But I found myself easily forgot which version is the newest, and I have to confirm which cell is the newest.</p>
<p>So if mathematica can provide a menu options like "cell latest modi... | Albert Retey | 169 | <p>From the OPs comments I deduce that the problem is only the dialog that opens when one opens with the menu entry "Cell" -> "Notebook History" and which shows a relatively involved visualization of the history and defaults to analyze the whole notebook. Obviously that is too slow for really large notebooks. But if on... |
200,414 | <p>Some people may carelessly say that you need calculus to find such a thing as a local maximum of $f(x) = x^3 - 20x^2 + 96x$. Certainly calculus is <em>sufficient</em>, but whether it's <em>necessary</em> is another question.</p>
<p>There's a global maximum if you restrict the domain to $[0,8]$, and $f$ is $0$ at t... | André Nicolas | 6,312 | <p>We do an analysis using the Arithmetic Mean Geometric Mean Inequality.</p>
<p>Make the usual change of variable <span class="math-container">$x=t+\frac{20}{3}$</span> to eliminate the term in the
square of the variable. We get the cubic polynomial
<span class="math-container">$$t^3-\frac{112}{3}t+k,$$</span>
where ... |
17,398 | <p>I am teaching my daughter, who is currently about <span class="math-container">$46$</span> months old, additions. She is very curious and asks a lot of <em>good</em> questions. For example, when I told her that <span class="math-container">$2+6=8$</span> and <span class="math-container">$4+4=8$</span>, she asked me ... | Selene Auckland | 12,899 | <p>None of the answers so far seem to give a name for the kind of concept in the inquiry. What you're asking is "how to make 8" or more generally "<strong>making numbers</strong>". See <a href="https://www.amazon.co.uk/Making-Numbers-Using-manipulatives-arithmetic/dp/0198375611" rel="nofollow noreferrer">here</a> and <... |
3,131,791 | <p>I read this:</p>
<blockquote>
<p><strong>Definition 1.1</strong>. <em>The</em> complex projective line <span class="math-container">$\mathbb{CP}^1$</span> <em>(or just</em> <span class="math-container">$\mathbb{P}^1$</span><em>) is the set of all ordered pairs of complex numbers</em> <span class="math-container">... | Servaes | 30,382 | <p>The coordinates of different representatives of the same point differ by a scalar multiple; if <span class="math-container">$(x_0,x_1)$</span> and <span class="math-container">$(x_0',x_1')$</span> represent the same point then there exists some nonzero scalar <span class="math-container">$\lambda$</span> such that <... |
3,423,138 | <p>A) Always positive </p>
<p>B) Always negative </p>
<p>C)Always non-negative </p>
<p>D) Always zero </p>
<p>Obviously, this isn’t a solving question, so I can’t really show any working here (I would have otherwise). I didn’t really understand the question, so help would be greatly appreciated </p>
<p>Thanks!</p>... | b00n heT | 119,285 | <p>Hint: Show that</p>
<p><span class="math-container">$$\frac{d}{dx}\text{Li}_n(x)=\frac{1}{x}\text{Li}_{n-1}(x)$$</span></p>
<p>then by induction if <span class="math-container">$\text{Li}_n(x)$</span> is rational, so is its derivative and hence <span class="math-container">$\text{Li}_{n-1}(x)$</span>.</p>
|
235,425 | <p>Ok, here is what I have for the proof of this conjecture. Let me know if I'm on the right path? all input appreciated.</p>
<p>There exist integers $j$, $k$, and $m$, such that,
$b = aj $ and $ c = ajk.$ Then $c = ajk $ (substituting $aj$ for $b$)
let $m = jk$, then $c = ma, => a|c.$ </p>
| Mr Q | 221,240 | <p>Since you given a/b, this means b=ka for some k-interger. Also b/c implies that c=lb for some interger l. now, substituting b-ka into c=lb gives</p>
<pre><code> c=lka.....>>>c=lk(a)>>>>we conclude a can divide c
</code></pre>
<p>Hence a/c
N:B_ the / is not a mathematical symbol fo... |
591,014 | <p>I need help with this proof:</p>
<p>$f: X\rightarrow Y$</p>
<p>$C,D\subseteq Y$</p>
<p>$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$</p>
<p>Thanks.</p>
| Riccardo | 74,013 | <p>Just pick an element in $f^{-1}( C \cap D) $ and verify the first inclusion. The same for the second</p>
<p>Maybe this definition will help
$ f^{-1}( C) := \{ x \in X \text { such that } f (x) \in C \} $</p>
<p>Just apply it</p>
|
2,327,181 | <p>I'm a bit confused about the definition of finite sets/intervals. I know that a set S is called finite when it has a finite number of elements, or formally, when there exists a bijection $f:S\to\{1,...,n\}$ for some natural number n. </p>
<p>However, the interval $(1,2)$ is called finite. I don't understand why; $(... | Andre | 303,581 | <p>Yes, we agreed to do so. A "finite interval" is an interval of finite length, i.e, the number $b-a$ is finite. A finite set, on the other hand, is a set of finite cardinality, so consisting of only finitely many elements. Maybe the terminology is a bit unfortunate, but since EVERY nonempty interval possesses uncount... |
118,985 | <p>Here's a question I came across:</p>
<blockquote>
<p>count the number of permutations of 10 men, 10 women and one child,
with the limitation that a man will not sit next to another man, and a
woman will not sit next to another woman</p>
</blockquote>
<p>What I tried was using the inclusion-exclusion principl... | André Nicolas | 6,312 | <p>Look at the power series expansion of $e^x$. The $x^4$ term is $\dfrac{x^4}{4!}$, so for positive $x$, we have $e^x>\frac{x^4}{4!}$ and therefore
$$\frac{x^2}{e^x} <\frac{4!}{x^2}.$$
We know that
$$\int_2^\infty \frac{dx}{x^2}$$
converges, and the $4!$ on top makes no difference. Note that the same idea can... |
2,797,039 | <p><strong>1)</strong> Let $\{f_n\}$ be a sequence of nonnegative measurable functions of $\mathbb R$ that converges pointwise on $\mathbb R$ to $f$ integrable. Show that</p>
<p>$$\int_{\mathbb R} f = \lim_{n\to \infty}\int_{\mathbb R}f_n \Rightarrow \int_{E} f = \lim_{n\to \infty}\int_{E}f_n $$</p>
<p>for any measu... | Atif Farooq | 451,530 | <p><em>Solution</em>. Let $\mathcal{J}$ denote the event that exactly two jacks are selected in addition let $D_i$ denote the event that the die roll yielded $i\in\{1,2,3,4,5,6\}$. We may compute $\mathbf{P(\mathcal{J})}$ by making use of bayes's formula if you observe that
$$\mathbf{P}(\mathcal{J}) = \sum_{k=2}^{6}\m... |
4,083,588 | <p>Can somebody help me how to solve</p>
<p><span class="math-container">$x* 23 \equiv_{60} 1 $</span> with <span class="math-container">$ x \in \mathbb{N} $</span> and <span class="math-container">$x > 100$</span></p>
<p>What would be a good approach?</p>
<p>I know that x = 107 would be a solution. However how can ... | Roddy MacPhee | 903,195 | <p><span class="math-container">$x\cdot a\equiv_{b} 1\implies x=\frac{-(b^{-1})\cdot b+1}{a}\bmod b$</span></p>
|
4,083,588 | <p>Can somebody help me how to solve</p>
<p><span class="math-container">$x* 23 \equiv_{60} 1 $</span> with <span class="math-container">$ x \in \mathbb{N} $</span> and <span class="math-container">$x > 100$</span></p>
<p>What would be a good approach?</p>
<p>I know that x = 107 would be a solution. However how can ... | Joffan | 206,402 | <p><a href="https://en.wikipedia.org/wiki/Euler%27s_theorem" rel="nofollow noreferrer">Euler's theorem</a> tells you that for <span class="math-container">$n$</span> and <span class="math-container">$a$</span> coprime, you have <span class="math-container">$$a^{\phi(n)} \equiv 1 \bmod n$$</span> where <span class="math... |
2,866,399 | <p>I was studying Ring and modules and now I am studying Group theory. I feel very confuse regarding generating set .
In Ring Theory if we have an Ideal $ \ I = < x,y > $
Then x doesn't contained in that Ideal, however in Group theory if we have $ \ G = < x, y >$ then x is in the group.
Can any one please... | egreg | 62,967 | <p>The confusion stems from the definition of “ideal generated by $x$” (let's keep it simple, for the moment, with $R$ a commutative ring).</p>
<p>There are two possible definitions:</p>
<ol>
<li>the ideal generated by $x$ is $Rx$</li>
<li>the ideal generated by $x$ is the intersection of all the ideals containing $x... |
2,897,374 | <p>So today in my Algebraic Topology class we were trying to construct a homotopy and at one point we needed to basically construct a (continuous) bijection between some intervals $[a, b]$ and $[c, d]$ sending $a \mapsto c$ and $b \mapsto d$. </p>
<p>My professor said that a bijection $\varphi : [a, b] \to [c, d]$ wou... | amd | 265,466 | <p>Since $P^2=P$, you have a perfectly good projection matrix. However, the definition of “projective matrix” that you have also requires that $P^T=P$, which makes it specifically an <em>orthogonal</em> projection. For that, the column space of the matrix must be the orthogonal complement of the null space. </p>
<p>T... |
267,274 | <p>Suppose <span class="math-container">$\left(A,\leq_{A}\right)$</span>
and <span class="math-container">$\left(B,\leq_{B}\right)$</span>
are <a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="nofollow noreferrer">posets</a> and <span class="math-container">$f:A\to B$</span>
is an order-preserving bije... | Brian M. Scott | 12,042 | <p>Let $A=\{0,1\}\times\Bbb N$, and define $\langle i,m\rangle\le_A\langle j,n\rangle$ iff $i=j$ and $m\le n$. Let $B=\Bbb N$ with the usual order. Finally, let</p>
<p>$$f:A\to B:\langle i,n\rangle\mapsto 2n+i\;.$$</p>
<p>Then $f$ is an order-preserving bijection, but $f^{-1}$ is not order-preserving for exactly the ... |
267,274 | <p>Suppose <span class="math-container">$\left(A,\leq_{A}\right)$</span>
and <span class="math-container">$\left(B,\leq_{B}\right)$</span>
are <a href="https://en.wikipedia.org/wiki/Partially_ordered_set" rel="nofollow noreferrer">posets</a> and <span class="math-container">$f:A\to B$</span>
is an order-preserving bije... | Asaf Karagila | 622 | <p>To drive Brian's point further, let's assume for simplicity that $f$ is the identity. This simply means that $\leq_A\subseteq\leq_B$ as sets of ordered pairs.</p>
<p>For example, if $(A,R)$ is any partially ordered set, and $(A,\leq_R)$ is a linearlization of $R$ then the identity is an order-preserving bijection, ... |
1,825,392 | <p>Let $f(z)$ be an entire function so that,</p>
<p>$$ \int \frac{|f(z)|}{1 + |z|^3} dA(z) < \infty$$</p>
<p>where the integral is taken over the entire complex plane. Show that $f$ is a constant.</p>
<p>I believe that the idea is to use the mean value property; that is:</p>
<p>$$f(z) = \frac{1}{\pi\delta^2}\in... | zhw. | 228,045 | <p>Suppose $f$ is nonconstant. Then there is a constant $c>0$ such that</p>
<p>$$\tag 1 \frac{1}{2\pi}\int_0^{2\pi} |f(re^{it})|\, dt > cr$$</p>
<p>for large $r.$</p>
<p>Proof: We can write $f(z) = f(0) + z^ng(z)$ for some $n\in \mathbb N$ and some entire $g$ such that $g(0) \ne 0.$ We then have $(1)$ bounded ... |
2,904,432 | <blockquote>
<p>Let the subspace <span class="math-container">$S=\{(x, y, z) \in R^3; x = -2z \}$</span></p>
<p>a) Determine its dimension and a basis B of S;</p>
<p>b) Complete B such that it becomes a basis M of <span class="math-container">$R^3$</span>.</p>
</blockquote>
<p><span class="math-container">$\operatornam... | Dan Sp. | 539,506 | <p>Yes, $dim(S)=2$. Here is a basis for $B$:</p>
<p>$\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}2\\0\\-1\end{bmatrix}$</p>
<p>The basis does not span $R^3$ but it is a basis for $S$. To complete $B$ for $R^3$ just find one more vector that is orthogonal to the above two.</p>
|
3,520,893 | <p>As in group theory, there is a concept of isomorphism between metric spaces called isometry.
Two metric spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are isometric if there is a function that perserves the distance of two elements. That function is called isometry.</p>
<... | Community | -1 | <p>I don't think so, but two metrics are equivalent iff there are <span class="math-container">$m, M$</span> such that <span class="math-container">$1/md'(x,y)\le d(x,y)\le Md'(x,y)$</span> for all <span class="math-container">$x,y\in X$</span>.</p>
|
2,975,647 | <p>Is this following statement valid (where both <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are positive)?</p>
<blockquote>
<p>If <span class="math-container">$x>y$</span>, then <span class="math-container">$\dfrac{1}{x} < \dfrac{1}{y}$</span>. </p>
</blockquote>
| Robert Israel | 8,508 | <p>No. Note that if <span class="math-container">$a$</span> or <span class="math-container">$b$</span> is a negative integer (and <span class="math-container">$c$</span> is not) the series has only finitely many nonzero terms.</p>
<p>EDIT: If <span class="math-container">$a,b,c$</span> are positive integers,
<span cl... |
1,360,910 | <p>My book (Introduction to Ring Theory, Paul Cohn) states this as a theorem and gives a proof. The book usually skips over trivial/easy proofs, so I don't really understand why this is in here.</p>
<p>Isn't the statement absolutely obvious for sets $A,B,C$ with $A\subseteq C$? You just need to draw a diagram and it b... | Eoin | 163,691 | <p>I'm not sure if the statement is obvious to you but in general $+$ and $\cap$ don't distribute over each other. I wouldn't expect them to either since $\cap$ is a set theoretic operation and $+$ is only defined for modules/ideals/abelian groups. However, when $A\subset C$ they do distribute over one another.</p>
<p... |
3,467,203 | <p><span class="math-container">$$\frac{(-1)^{k+1}k}{3^k}$$</span></p>
<p>Correct me if I am wrong but with <span class="math-container">$(-1)^{k+1}$</span> this makes it an alternating series test therefore i think I need to show that <span class="math-container">$\frac{k}{3^k}$</span> is decreasing and non-negative.... | user284331 | 284,331 | <p>It is clear that <span class="math-container">$k/3^{k}\rightarrow 0$</span> as <span class="math-container">$k\rightarrow\infty$</span>. Now by elementary calculus applied to the derivative of <span class="math-container">$x/3^{x}$</span> for <span class="math-container">$x>0$</span>, we conclude that <span class... |
3,592,368 | <p>I managed to solve this question using trigonometry. But I wondered if there'd be anyway of doing it using only synthetic geometry. Here it is.</p>
<blockquote>
<p>Let <span class="math-container">$ABC$</span> be a right isosceles triangle of hypotenuse <span class="math-container">$AB$</span>. Let also <span class=... | ole | 725,731 | <p>Let us find the altitude of point <span class="math-container">$P$</span>, a is a leg. Intersection of the circles <span class="math-container">$x^2+h^2=k^2, x^2+(h-\frac{a}{2})^2=(\frac{a}{2})^2 $</span></p>
<p><span class="math-container">$h=\frac{k^2}{a} $</span></p>
<p><span class="math-container">$A\; PBC=\fr... |
2,073,794 | <p>Let's consider the following variant of Collatz (3n+1) : </p>
<p>if $n$ is odd then $n \to 3n+1$</p>
<p>if $n$ is even then you can choose : $n \to n/2$ or $n \to 3n+1$</p>
<p>With this definition, is it possible to construct a cycle other than the trivial one, i.e., $1\to 4 \to 2 \to 1$?</p>
<p>Best regards</p>... | Sjoerd | 6,688 | <p>$$8 \rightarrow 25 \rightarrow 76 \rightarrow 38 \rightarrow 19 \rightarrow 58 \rightarrow 29 \rightarrow 88 \rightarrow 44 \rightarrow 22 \rightarrow 11 \rightarrow 34 \rightarrow 17 \rightarrow 52 \rightarrow 26 \rightarrow 13 \rightarrow 40 \rightarrow 20 \rightarrow 10 \rightarrow 5 \rightarrow 16 \rightarrow 8$... |
1,218,912 | <p>My problem is following:
$$\binom{n}{r} + \binom{n+1}{r+1} + \binom{n+2}{r+2} + \dots + \binom{n+M}{r+M}$$</p>
<p>how can we reduce it to a more short solution</p>
<p>Here $\dbinom{n}{r} = \dfrac{n!}{r! (n-r)!}$ and thus same as regular.Please help me in solving the above expression</p>
| Elaqqad | 204,937 | <p><strong>Hint</strong> what happens if you add :$\dbinom{n}{r-1}$ to your sum, can you prove that it will be simplified recursively using $$\dbinom{n}{k}+\dbinom{n}{k+1}=\dbinom{n+1}{k+1}$$</p>
<p><strong>Answer</strong> $\dbinom{n+M+1}{r+M}-\dbinom{n}{r-1}$</p>
|
4,474,803 | <p>Let <span class="math-container">$f$</span> be a differentiate function and consider two operators
<span class="math-container">$$
A(f(x))=\int_0^1 \frac{ d}{dx}f(x t^\mu) dt,\\
B(f(x))=\mu xf(x)+\int_0^x f(t) dt,
$$</span>
where <span class="math-container">$\mu $</span> is a parameter.</p>
<p>I need to prove that<... | Dark Rebellion | 858,891 | <p>In this very particular case, NO, <span class="math-container">$\phi_1$</span> is NOT equivalent to <span class="math-container">$\phi_2$</span>, but the overall "intuition" is correct, except for one edge case that was overlooked.</p>
<p>Let me elaborate. It is indeed the case (assuming no name collisions... |
1,671,933 | <p>In the question <a href="https://math.stackexchange.com/questions/1502484/the-application-of-nimbers-to-nim-strategy">on nimbers</a>, the original poster asks for the meaning of Nimber <a href="https://en.wikipedia.org/wiki/Nimber#Multiplication" rel="nofollow noreferrer">multiplication</a> in the context of imparti... | Oscar Lanzi | 248,217 | <p>The "game" in <a href="https://www.youtube.com/watch?v=9JN5f7_3YmQ" rel="nofollow noreferrer">this video</a> may be analyzed in terms of a limited form of nimber multiplication, specifically involving the nimbers <span class="math-container">$1,2,3$</span>.</p>
<p>The rules of the game are as follows:</p>
... |
1,085,620 | <p>Which number between the interval 7902 & 7918 can be divided without remainder only to itself and to the number 1?</p>
| Batman | 127,428 | <p>You want to find a prime between 7902 and 7918. One way to do this is to use a <a href="http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes" rel="nofollow">number sieve</a>, or since there are only a few small numbers, brute force by testing divisors between 1 and 88 ($\lfloor \sqrt{7918}\rfloor$). From this we see 7... |
2,107,192 | <p>I am trying to find the total number of signals that can be created from $3$ pink, $3$ white and $2$ black flags when arranged in a straight line. But, only $5$ flags are allowed in a signal.</p>
<p>I know how to find permutations in this situation when all are taken at a time i.e. a signal has $8$ flags, which wou... | Sunil | 249,557 | <p>I was able to solve this problem but not sure if there is a better and shorter method. I used a $case$ approach to solving this by listing all possible groups of 5 flags using the original collection of 8 flags and then within each group of 5 flags we apply the formula for permutations of 5 objects at a time.</p>
<... |
2,107,192 | <p>I am trying to find the total number of signals that can be created from $3$ pink, $3$ white and $2$ black flags when arranged in a straight line. But, only $5$ flags are allowed in a signal.</p>
<p>I know how to find permutations in this situation when all are taken at a time i.e. a signal has $8$ flags, which wou... | awkward | 76,172 | <p>Let's generalize the problem slightly and find the number of signals with $r$ flags, say $a_r$. Let $f(x)$ be the exponential generating function for $a_r$, i.e.
$$f(x) = \sum_{r=0}^{\infty} \frac{1}{r!} a_r x^r$$
Since there are 3 pink, 3 white, and 2 black flags,
$$f(x) = \left( 1 + x + \frac{1}{2!} x^2 + \frac{1... |
1,309,572 | <p>The main idea of Russel's paradox is that, in Naive Set Theory, if we define $R = \{x\ |\ x \not\in x \}$, then $R \in R \Leftrightarrow R \not \in R$.</p>
<p>ZFC deals with this by making unrestricted set comprehension illegal, while type theory creates a hierarchy of sets such that there is no place in it for $R$... | Asaf Karagila | 622 | <p>If $R$ is a set such that $x\in R\leftrightarrow x\notin x$, and the law of excluded middle holds, then either $R\in R$ or $R\notin R$, but not both since it is impossible that both a statement and its negation are true.</p>
<p>But now if $R\in R$ then $R\notin R$; and if $R\notin R$ then $R\in R$.</p>
<p>So it's ... |
1,309,572 | <p>The main idea of Russel's paradox is that, in Naive Set Theory, if we define $R = \{x\ |\ x \not\in x \}$, then $R \in R \Leftrightarrow R \not \in R$.</p>
<p>ZFC deals with this by making unrestricted set comprehension illegal, while type theory creates a hierarchy of sets such that there is no place in it for $R$... | hmakholm left over Monica | 14,366 | <p>This wouldn't be possible in ordinary first-order logic, because no matter what the axioms say <em>about</em> $R\in R$, it is necessarily a <em>well-formed</em> formula.</p>
<p>But there's a more conceptual problem here: If you're defining $R$ as "the set that consists of all sets that don't contain themselves", yo... |
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