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 |
|---|---|---|---|---|
4,259,561 | <p>I am looking for the derivation of the closed form along any given diagonal <span class="math-container">$a$</span> of Pascal's triangle,<br />
<span class="math-container">$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$</span>
Numbered observations follow. As for the limit proposed in the title given by:</p>
<p><stron... | Robert Shore | 640,080 | <p>You have <span class="math-container">$25$</span> possible choices for the clockwise member of the pair that's sitting together. Once you've selected that pair, (and assuming the question asks for <em>exactly</em> two members to be adjacent to one another), you can choose any of the <span class="math-container">$21... |
1,218,582 | <p>I was presented with the function $max (|x|,|y|)$ which should output a maximum value of given two.... I can only suppose this one creates some body in $\mathbb{R}^3$ but how do you sketch it and what does it mean in $\mathbb{R}^3$? for that matter in $\mathbb{R}^2$ I cant really imagine it also.. </p>
| MPW | 113,214 | <p>The most revealing approach would probably be to draw the set of level curves in the plane. This projects the slices through the graph by horizontal planes back down onto the domain plane, like a topographic map.</p>
<p>The equation of the level curve corresponding to the planar section at height $c$ is
$$\max(|x|,... |
3,105,664 | <p><span class="math-container">$$
I_n=\int_{0}^{1}\frac{(1-x)^n}{n!}e^x\,dx
$$</span></p>
<blockquote>
<p>Prove that
<span class="math-container">$$
I_n=\frac{1}{(n+1)!}+I_{n+1}
$$</span></p>
</blockquote>
<p>I tried integration by parts and still can't prove it, I appreciate any hint/answer. </p>
| Zacky | 515,527 | <p>I used <span class="math-container">$n!$</span> as in the original image.<span class="math-container">$$I_n=\int_{0}^{1}\frac{(1-x)^n}{n!}e^x\,dx=\frac{1}{n!}\int_0^1 \left(-\frac{(1-x)^{n+1}}{n+1}\right)'e^x\,dx$$</span>
<span class="math-container">$$=\underbrace{-\frac{1}{n!}\frac{(1-x)^{n+1}}{n+1} e^x\bigg|_0^1... |
2,267,165 | <p>Find the distance between the two parallel planes?</p>
<p>$$a: x-2y+3z-2=0$$
$$b: 2x-4y+6z-1=0$$</p>
<p>The given answer is: $\dfrac{3}{\sqrt{56}}$</p>
| hamam_Abdallah | 369,188 | <p>Take a point in the plane $(a) $.</p>
<p>for example $A (2,0,0) $.</p>
<p>the distance to the plane $(b) $ is</p>
<p>$$\frac {|2.2- 4.0 +6.0-1|}{ \sqrt {2^2+4^2+6^2 } }=$$
$$\frac {3}{\sqrt {56}} $$</p>
|
1,832,080 | <p>The converse is pretty obvious. If G is a cycle, then it is isomorphic to it's line graph.
How to prove that if L(G) is isomorphic to G, then G is a cycle...?</p>
<p><strong>P.S.</strong>- Assume G is connected</p>
| Ashwin Ganesan | 157,927 | <p>Let $G$ be a connected graph on $n$ vertices and $m$ edges and suppose $G$ is isomorphic to its line graph $L(G)$. Then, $m=n$ and so $G$ is a connected graph having $n$ edges. This implies $G$ is of the form $T+e$, where $T$ is a tree and $e$ is an edge not in $T$. If $G=T+e$ is the $n$-cycle graph, then we are ... |
948,329 | <p>I have come across this trig identity and I want to understand how it was derived. I have never seen it before, nor have I seen it in any of the online resources including the many trig identity cheat sheets that can be found on the internet.</p>
<p>$A\cdot\sin(\theta) + B\cdot\cos(\theta) = C\cdot\sin(\theta + \Ph... | Jasser | 170,011 | <p>Hint : Just divide and multiply LHS by C and consider $\frac AC =\cos \phi$ and similarly $ \frac BC =\sin \phi$ and then try to simplify.</p>
<p>To get the proof consider</p>
<p>\begin{align}
\sin (a-b) &= \cos (\pi/2-(a-b)) \\
&=\cos ((\pi/2-a)+b)
\end{align}
following on from this, one can then apply th... |
3,933,296 | <p>What I already have,</p>
<ol>
<li>Palindrome in form XYZYX, where X can’t be 0.</li>
<li>Divisibility rule of 9: sum of digits is divisible by 9. So, we have 2(X+Y)+Z = 9M.</li>
<li>The first part is divisible by 9 if and only if X+Y is divisible by 9. So, we have 10 pairs out of 90. And each such pair the total sum... | Measure me | 854,564 | <p>The answer is <span class="math-container">$100$</span>, my argument is as follows.</p>
<p><span class="math-container">$X$</span> is choosen as a digit from <span class="math-container">$1$</span> to <span class="math-container">$9$</span>, and <span class="math-container">$Y$</span> from <span class="math-containe... |
3,506,091 | <p>Solve <span class="math-container">$2^m=7n^2+1$</span> with <span class="math-container">$(m,n)\in \mathbb{N}^2$</span></p>
<p>Here is what I did:
First try, I have seen first that the obvious solutions are <span class="math-container">$n=1$</span> and <span class="math-container">$m=3$</span> , and <span class="ma... | Barry Cipra | 86,747 | <p>If <span class="math-container">$2^m=7n^2+1$</span> with <span class="math-container">$m=3k$</span>, as the OP found must be the case (since <span class="math-container">$2^m\equiv1$</span> mod <span class="math-container">$7$</span>), we have</p>
<p><span class="math-container">$$7n^2=2^{3k}-1=(2^k-1)(2^{2k}+2^k+1... |
1,059,939 | <blockquote>
<p>Can a function exist which is both $o(g(n))$ and $\omega(g(n))$?</p>
</blockquote>
<p>Wouldn't this imply
$$m |g(n)| \le |f(n)| \le k |g(n)| $$</p>
<p>If $f(n) = g(n)$ then wouldn't an arbitrary integer $m$ be greater than $f(n)$?</p>
<p>If $f(n) \ne g(n)$ wouldn't for $n$ sufficiently large the... | alexjo | 103,399 | <p>The distribution of $V$ is $\chi^2(n)$. The square root of a $\chi^2(n)$ random variable is a <a href="http://mathworld.wolfram.com/ChiDistribution.html" rel="noreferrer">$\chi(n)$ random variable</a>.</p>
<p>Let the random variable $V$ have the chi-square distribution with $n$ degrees of freedom
with probability d... |
1,059,939 | <blockquote>
<p>Can a function exist which is both $o(g(n))$ and $\omega(g(n))$?</p>
</blockquote>
<p>Wouldn't this imply
$$m |g(n)| \le |f(n)| \le k |g(n)| $$</p>
<p>If $f(n) = g(n)$ then wouldn't an arbitrary integer $m$ be greater than $f(n)$?</p>
<p>If $f(n) \ne g(n)$ wouldn't for $n$ sufficiently large the... | Kamster | 159,813 | <p>First note that each $X^{2}_i\sim Gamma(\frac{1}{2},\frac{1}{2})$ for each $i$ (i.e. chi square with df=1) now to find distribution of $\sum_{i=1}^{n}X^2_{i}$ we see that using moment generating functions and fact that each are independent of each other that
$$M_{\sum_{i=1}^{n}X^2_{i}}=M(t)^{n}=\left(\frac{\frac{1}... |
1,189,216 | <p>Wikipedia and other sources claim that </p>
<p>$PA +\neg G_{PA}$</p>
<p>can be consistent, where $\neg G_{PA}$ is the Gödel statement for PA.</p>
<p>So what is the error in my reasoning?</p>
<p>$G_{PA}$ = "$G_{PA}$ is unprovable in PA"</p>
<p>$\neg G_{PA} $</p>
<p>$\implies$ $\neg$ "$G_{PA}$ is unprovable in P... | Asaf Karagila | 622 | <p>The point is that $G_{PA}$ is neither provable nor refutable in $\sf PA$. But it is a concrete sentence, and in a given model of $\sf PA$ it has a concrete truth value.</p>
<p>But if in all models $\sf PA$ it would have the same truth value, then the completeness theorem tells us that it is provable from $\sf PA$. ... |
1,520,028 | <p>I'm struggling to figure out how to find a bound on my error for this problem:</p>
<p>Let T_{6}(x) be the Taylor polynomial of degree 6 based at a = 0 for the function f(x)=\cos(x). Suppose you approximate f(x) by T_{6}(x). If |x|\leq 1, find a bound on the error in your approximation by using the alternating serie... | John Molokach | 90,422 | <p>Your denominators are incorrect. The factorials should be the same as the exponents of $x$. Also the alternating series error bound says that your $b_n$ term is an upper bound for the error of the 6th degree polynomial. </p>
|
1,248,331 | <p>That's the question :</p>
<p>Let $a$ be a cardinality such that this following statment is true :</p>
<p>For every $A, C$, if $ A \subseteq C$, $|A| = a$ and $|C| > a$, then $|C \setminus A| > |A|$.</p>
<p>Without using cardinality arithmethics, prove that $a + a = a$.</p>
<p>This is how the question is wr... | Hagen von Eitzen | 39,174 | <p>You are missing the "for every". With $A=\{42,666\}$, $C=\{13,42,666\}$ we have $A\subseteq C$, $|A|=a$ and $|C|>a$, but not $|C\setminus A|>|A|$. Hence your $a=2$ does not have the required property.</p>
|
2,146,929 | <p>Let $f:S^n \to S^n$ be a homeomorphism. I know the result that a rigid motion in $\mathbb R^{n+1}$ is always <a href="https://math.stackexchange.com/a/866471/185631">linear</a>, but can we get more information from the assumption that $f:S^n \to S^n$ is a homeomorphism?</p>
| Andreas Caranti | 58,401 | <p>$\newcommand{\C}{\mathbb{C}}$$\renewcommand{\theta}{\vartheta}$$\newcommand{\R}{\mathbb{R}}$Consider $S^{1} \subseteq \R^{2}$. The map
$$
f(e^{i \theta}) = e^{i \theta^{2}/2 \pi}
$$
is a homeomorphism of $S^{1}$, as $\theta \mapsto \dfrac{\theta^{2}}{2 \pi}$ is a homeomorphism on the interval $[0, 2 \pi]$. </p>
<p>... |
1,785,633 | <p><em>(First, I am very aware of the fact that Brownian motion is actually probably more difficult to understand than at least basic complex analysis, so the pedagogical merits of such an approach would be questionable for anyone besides a probabilist wanting to refresh or reshape their already existing complex analys... | Redundant Aunt | 109,899 | <p>You could consider two linearly independent vectors $a,b$ and then pose $c=a+b$.</p>
|
1,250,020 | <blockquote>
<p>Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$</p>
</blockquote>
<p>I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.</p>
<p>Any hints on how to do it... | Nicolas | 213,738 | <p>We have
$$\int\frac{1+\sin x\cos x}{1-5\sin^2x}\mathrm{d}x
=\int\frac{\mathrm{d}x}{1-5\sin^2x}+\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x$$
and
$$\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x
=-\frac{1}{10}\int\frac{\mathrm{d}\left(1-5\sin^2x\right)}{1-5\sin^2x}
=-\frac{1}{10}\ln\left(1-5\sin^2x\right).$$
Then,... |
3,006,022 | <p>I have a radioactive decay system to solve for <span class="math-container">$x(t)$</span> and <span class="math-container">$y(t)$</span> (no need for <span class="math-container">$z(t)$</span>):
<span class="math-container">$$\begin{cases}x'=-\lambda x\\
y'=\lambda x-\mu y\\
z'=\mu y\end{cases}$$</span>
with the ini... | Offlaw | 571,888 | <p><span class="math-container">$$y'+\mu y=\lambda e^{-\lambda t}$$</span></p>
<p><span class="math-container">$$d(e^{\mu t}y)=\lambda e^{(\mu - \lambda)t} dt \text{ }[\text{I.F.} = e^{\mu t}]$$</span></p>
<p><span class="math-container">$$\text{Integrating, } e^{\mu t}y=\frac{\lambda e^{(\mu - \lambda)t}}{\mu - \lam... |
3,012,416 | <p>I know the answer is obvious: In <span class="math-container">$\mathbb{Z}$</span> the only solutions of <span class="math-container">$xy=-1$</span> are <span class="math-container">$x=-y=1$</span> and <span class="math-container">$x=-y=-1$</span>.
My problem is that I want to formally prove it and I don't know how t... | Siong Thye Goh | 306,553 | <p>if <span class="math-container">$xy=-1$</span>, then we we have <span class="math-container">$|x||y|=1$</span>, that is we must have <span class="math-container">$|x|=1$</span> and <span class="math-container">$|y|=1$</span>.</p>
<p>Also, determinining <span class="math-container">$x$</span> would completely determ... |
3,012,416 | <p>I know the answer is obvious: In <span class="math-container">$\mathbb{Z}$</span> the only solutions of <span class="math-container">$xy=-1$</span> are <span class="math-container">$x=-y=1$</span> and <span class="math-container">$x=-y=-1$</span>.
My problem is that I want to formally prove it and I don't know how t... | fleablood | 280,126 | <p>Alternatively.</p>
<p>If <span class="math-container">$x $</span> or <span class="math-container">$y $</span> is <span class="math-container">$0$</span>, <span class="math-container">$xy=0$</span>.</p>
<p>Otherwise, <span class="math-container">$|x|\ge 1$</span> and <span class="math-container">$|y|\ge 1$</span>. ... |
3,471,292 | <p>I need to find the value of the series <span class="math-container">$\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$</span>.I've computed its radius of convergence which comes out to be zero.</p>
<p>I'm not getting how to make adjustments in the general terms of the series to get the desired result...</p>
| E.H.E | 187,799 | <p><span class="math-container">$$y=\sum_{n=0}^{\infty}\frac{nx^n}{n!}+\sum_{n=0}^{\infty}\frac{x^n}{n!}$$</span>
<span class="math-container">$$y=e^x+\sum_{n=0}^{\infty}\frac{nx^n}{n!}$$</span>
<span class="math-container">$$\frac{y}{x}=\frac{e^x}{x}+\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}$$</span>
<span class="math-co... |
2,355,579 | <blockquote>
<p><strong>Problem:</strong> James has a pile of n stones for some positive integer n ≥ 2. At each step, he
chooses one pile of stones and splits it into two smaller piles and writes the product
of the new pile sizes on the board. He repeats this process until every pile is exactly
one stone.</p>
... | user64742 | 289,789 | <p>The answer is <a href="https://chat.stackexchange.com/transcript/message/38721939#38721939">4 8 15 16 23 42</a></p>
<hr>
<p>Joke aside what you actually need is iterated induction (sorta).</p>
<p>Some setup:</p>
<ul>
<li>The ordering of two piles is irrelevant. A pile of 2 and a pile of 3 is the same as a pile o... |
1,212,262 | <p>The statement goes as follow: </p>
<p>$ B ∩ C ⊆ A ⇒ (C − A) ∩ (B − A) = ∅. $</p>
<p>First, the sign "=>" represents a tautology, no? ( apparently I get it confuse with the 3 bar sign, if you know what I mean).</p>
<p>Second, the fact that it equals to no solution, how do I prove that? Seems to contradict itself, ... | Christian Blatter | 1,303 | <p>From your handling of cases I have got the impression that the boxes are labeled, as are the colors, but balls of the same color are undistinguishable.</p>
<p>If there were enough balls of all colors we could just assign a color to each of the boxes. This can be done in $4^5=1024$ ways.</p>
<p>But assignments wher... |
2,806,432 | <p>Let $(\mathbb{R}^N,\tau)$ a topological space, where $\tau$ is the usual topology.
Let $A\subset\mathbb{R}^N$ a compact. If $(A_n)_n$ is a family of open such that
\begin{equation}
\bigcup_nA_n\supset A,
\end{equation}
then, from compact definition
\begin{equation}
\bigcup_{i=1}^{k}A_i\supset A
\end{equation}
Now, ... | lulu | 252,071 | <p>Your method works fine, but you have to remember the continuity correction. You are approximating a discrete distribution with a continuous one. Thus, for instance, your method gives a non-zero chance of getting $224$ points (say) though that is in fact impossible. </p>
<p>As you are working with the scores, the... |
1,736,098 | <p>Wrote some Python code to verify if my Vectors are parallel and/or orthogonal. Parallel seems to be alright, orthogonal however misses out in one case. I thought that if the dotproduct of two vectors == 0, the vectors are orthogonal? Can someone tell me what's wrong with my code?</p>
<pre><code>def isParallel(self,... | bluppfisk | 330,237 | <pre><code>def isOrthogonal(self,v,tolerance=1e-10):
if abs(self.dotProduct(v)) < tolerance:
return True
return False
</code></pre>
<p>the above code runs fine, as there are chances of rounding errors in my original code. Hans Lundmark got is straight: floating point rounding errors can cause the c... |
251,818 | <p>In other words if a graph is $3$-regular does it need to have $4$ vertices? I ask because I have been asked to prove that if $n$ is an odd number and $G$ is an $n$-regular graph then $G$ must have an even number of vertices.</p>
| joriki | 6,622 | <p>It seems from your last sentence that you're asking whether an $n$-regular graph must have <em>exactly</em> $n+1$ vertices (rather than <em>at least</em> $n+1$ vertices). If so, as Gregor commented, the answer is no.</p>
<p>For the proof you're trying to find, try counting the number of incidences in two different ... |
200,278 | <p>Say I have two TimeSeries:</p>
<pre><code>x = TimeSeries[{2, 4, 1, 10}, {{1, 2, 4, 5}}]
y = TimeSeries[{6, 2, 6, 3, 9}, {{1, 2, 3, 4, 5}}]
</code></pre>
<p>x has a value at times: 1,2,4,5</p>
<p>y has a value at times: 1,2,3,4,5</p>
<p>I would like to build a list of pairs {<span class="math-container">$x_i$</sp... | user42582 | 42,582 | <p>One possible solution is to use <a href="https://reference.wolfram.com/language/ref/TimeSeriesResample.html?q=TimeSeriesResample" rel="noreferrer"><code>TimeSeriesResample</code></a>:</p>
<pre><code>td = TimeSeriesResample[TemporalData[{x, y}], "Intersection"];
</code></pre>
<p>Using <code>"Intersection"</code> in... |
3,525,488 | <p>So I have the polar curve </p>
<p><span class="math-container">$r=\sqrt{|\sin(n\theta)|}$</span></p>
<p>Which I am trying to evaluate between <span class="math-container">$0$</span> and <span class="math-container">$2\pi$</span>. By smashing it into wolfram it returns a constant value 4 for any <span class="math-c... | Quanto | 686,284 | <p>Use the variable change <span class="math-container">$t=n\theta$</span> to rewrite the integral as</p>
<p><span class="math-container">$$\textrm{I}=\frac{1}{2}\int_{0}^{2\pi}|\sin(n\theta)|d\theta
= \frac1{2n} \int_{0}^{2\pi n}|\sin t|dt$$</span></p>
<p>Due to the periodicity of the sine function </p>
<p><span cl... |
9,484 | <p>Let <span class="math-container">$F(k,n)$</span> be the number of permutations of an n-element set that fix exactly <span class="math-container">$k$</span> elements.</p>
<p>We know:</p>
<ol>
<li><p><span class="math-container">$F(n,n) = 1$</span></p>
</li>
<li><p><span class="math-container">$F(n-1,n) = 0$</span></p... | Michael Lugo | 143 | <p>The "semi-exponential" generating function for these is</p>
<p><span class="math-container">$\sum_{n=0}^\infty \sum_{k=0}^n {F(k,n) z^n u^k \over n!} = {\exp((u-1)z) \over 1-z}$</span></p>
<p>which follows from the exponential formula.</p>
<p>These numbers are apparently called the <a href="https://oeis.or... |
9,484 | <p>Let <span class="math-container">$F(k,n)$</span> be the number of permutations of an n-element set that fix exactly <span class="math-container">$k$</span> elements.</p>
<p>We know:</p>
<ol>
<li><p><span class="math-container">$F(n,n) = 1$</span></p>
</li>
<li><p><span class="math-container">$F(n-1,n) = 0$</span></p... | Joshua Baehring | 490,064 | <p>Let <span class="math-container">$S_n$</span> be the set of all permutations of <span class="math-container">$X$</span> \(i.e. <span class="math-container">$S_n = \{ f$</span> <span class="math-container">$|$</span> <span class="math-container">$f : X \rightarrow X\}$</span>). Now consider the set of permutations, <... |
346,950 | <p>The equation $$3\sin^2 x - 3\cos x -6\sin x + 2\sin 2x + 3=0$$ has a solution $x = 0$. That is mean it has a factor $\cos x - 1$. I tried write the given equation has the form
$$(\cos x - 1)P(x)=0.$$ I am looking for the factor $P(x)$. How to do that?</p>
| chenbai | 59,487 | <p>LHS$=sinx(3sinx-2)+4sinx(cox-1)-3(cosx-1)=\sqrt{1-cos^2 x}(3sinx-2)+4sinx(cox-1)-3(cosx-1)=\sqrt{|cosx-1||cosx+1|}(3sinx-2)+4sinx(cox-1)-3(cosx-1)$</p>
<p>if you insist $cosx-1$ as a factor, it has $\sqrt{|cosx-1|}$,but not $cosx-1$.</p>
<p>if you simply want to solve it, $1-cosx=2sin^2 \dfrac{x}{2}$,$sinx=2sin\df... |
138,658 | <p>Suppose $X$ is a topological space, and $\mu$ is a Borel measure on $X$. Also suppose we have an $n$-dimensional vector bundle $E \to X$, with an inner product $\langle \cdot,\cdot \rangle_x$ on the fibre $E_x$ for all $x \in X$, in such a way that each $E_x$ is complete and such that there exists a vector bundle tr... | johndoe | 36,502 | <p>The existence of a finite trivialising cover is a less stringent condition than one would expect: see <a href="https://mathoverflow.net/questions/94479/does-every-vector-bundle-allow-a-finite-trivialization-cover">Does every vector bundle allow a finite trivialization cover?</a></p>
<p>(Sorry for the commentlike an... |
399,934 | <p>How to derive the specific case of the generating element of a group given its generating set. For example, when
$$G=\langle a,b,c|a^3=b^3=c^2=1,ab=ba,ca=a^2c,cb=b^2c\rangle$$
we can let $G\subset S_3\times S_3$, and let
$$a=((123),1),b=(1,(123)),c=((12),(12))$$
to get the desired result. However, when $3$ is repla... | Mariano Suárez-Álvarez | 274 | <p>It is not true that if a linear map $f:V\to W$ has a one sided inverse, then $\dim V=\dim W$ nor that $f$ is an isomorphism.</p>
|
399,934 | <p>How to derive the specific case of the generating element of a group given its generating set. For example, when
$$G=\langle a,b,c|a^3=b^3=c^2=1,ab=ba,ca=a^2c,cb=b^2c\rangle$$
we can let $G\subset S_3\times S_3$, and let
$$a=((123),1),b=(1,(123)),c=((12),(12))$$
to get the desired result. However, when $3$ is repla... | Ink | 34,881 | <p>It is a standard fact from elementary set theory that a map $f$ is a bijection $\iff$ $f$ has a two-sided inverse. Remember, $df_x$ is a linear map, and therefore, an isomorphism. This shows $k = l$ since isomorphic vector spaces have equal dimension.</p>
|
2,618,804 | <p>Let $V$ be a vector space of dimension $m\geq 2$ and $ T: V\to V$ be a linear transformation such that $T^{n+1}=0$ and $T^{n}\neq 0$ for some $n\geq1$ .Then choose the correct statement(s):</p>
<p>$(1)$ $rank(T^n)\leq nullity(T^n)$</p>
<p>$(2)$ $rank(T^n)\leq nullity(T^{n+1})$</p>
<p><strong>Try:</strong></p>
<... | Andrea Marino | 177,070 | <p>Note that $f=T^n$ is such that $f^2=0$. Thus $Im f \subseteq \ker f$ implies $rank(f) \le nullity(f) $, which is (1).</p>
|
1,784,679 | <p>if $p,q,r$ are three positive integers prove that</p>
<p>$$LCM(p,q,r)=\frac{pqr \times HCF(p,q,r)}{HCF(p,q) \times HCF(q,r) \times HCF(r,p)}$$</p>
<p>I tried in this way;</p>
<p>Let $HCF(p,q)=x$ hence $p=xm$ and $q=xn$ where $m$ and $n$ are relatively prime.</p>
<p>similarly let $HCF(q,r)=y$ hence $q=ym_1$ and $... | Mayank Bomb | 256,329 | <p>To get a start on the problem first let us try to understand it intuitively. We need to show that for three numbers <span class="math-container">$a,b,c$</span>, <span class="math-container">$$abc=\frac{LCM(a,b,c) \times HCF(a,b) \times HCF(b,c) \times HCF(c,a)}{HCF(a,b,c)}$$</span>
On the LHS- the product of three n... |
3,522,867 | <p>Consider the sequence <span class="math-container">$\{x_n\}_{n\ge1}$</span> defined by <span class="math-container">$$x_n=\sum_{k=1}^n\frac{1}{\sqrt{k+1}+\sqrt{k}}, \forall n\in\mathbb{N}.$$</span> Is <span class="math-container">$\{x_n\}_{n\ge 1}$</span> bounded or unbounded. </p>
<p>I solved the problem as stated... | Eduline | 743,749 | <p>Here goes the solution I found out. Let <span class="math-container">$a_k$</span> be defined as <span class="math-container">$$a_k=\frac{1}{\sqrt{k}+\sqrt{k+1}}, \forall k\in\mathbb{N}.$$</span> Then, <span class="math-container">$$x_n=\sum_{k=1}^n a_k, \forall n\in\mathbb{N}.$$</span> Now <span class="math-containe... |
3,313,603 | <p>I am assigned with a question which states the rate of a microbial growth is exponential at a rate of (15/100) per hour. where y(0)=500, how many will there be in 15 hours?</p>
<p>I know this question is generally modelled as: </p>
<p><span class="math-container">$y=y_0*e^{kt}$</span></p>
<p>However, my solution ... | Amy Ngo | 692,535 | <p><span class="math-container">$e^{\ln 500}$</span> is equal to 500. Notice that if <span class="math-container">$y = e^x$</span>, taking the natural log of both sides gives you <span class="math-container">$\ln y = \ln (e^x) = x$</span>. Thus, to undo this operation, take each side as the power of <span class="math-c... |
4,086,995 | <p>Let <span class="math-container">$ABCD$</span> be a rectangle.</p>
<p>Given:</p>
<p><span class="math-container">$A(2;1)$</span></p>
<p><span class="math-container">$C(5;7)$</span></p>
<p><span class="math-container">$\overline{BC}=2\overline{AB}$</span>.</p>
<p>I tried to solve it, but after using the Pythagoras th... | José Carlos Santos | 446,262 | <p>If <span class="math-container">$f$</span> is an automorphism of <span class="math-container">$\Bbb Z$</span>, then there is some <span class="math-container">$m\in\Bbb Z$</span> such that <span class="math-container">$f(m)=1$</span>, and therefore <span class="math-container">$mf(1)=1$</span>. But this is possibly ... |
3,251,337 | <p>Be <span class="math-container">$E,F,K , L,$</span> points in the sides <span class="math-container">$AB,BC,CD,DA$</span> of a square <span class="math-container">$ABCD$</span>, respectively. Show that if <span class="math-container">$EK$</span> <span class="math-container">$\perp$</span> <span class="math-containe... | Community | -1 | <p>After William Elliot's feedback on your proof and <a href="https://math.stackexchange.com/questions/3251331/boundary-points-and-metric-space#comment6686712_3251331">this comment</a> of yours, I don't think there is much that needs to be clarified. Still if you have anything specific regarding your proof to ask me, I... |
1,925,245 | <p>Find the eigenvalues of
$$
\left(\begin{matrix}
C_1 & C_1 & C_1&\cdots&C_1 \\
C_2 & C_2 & C_2&\cdots &C_2 \\
C_3 & C_3 & C_3&\cdots&C_3 \\
\vdots&\ & \ & \ & \vdots\\
C_n & C_n & C_n&\cdots&C_n \\
\end{matrix}\right)
$$</p>
<p>My approa... | Community | -1 | <p>Since $rank(A)=1$, $A$ admits $n-1$ times the eigenvalues $0$. The last eigenvalue is $trace(A)=C_1+\cdots+C_n$.</p>
|
3,995,046 | <p>Refer to <a href="https://oeis.org/A340800" rel="nofollow noreferrer">https://oeis.org/A340800</a> to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase. Is there an explanation for this? How can the size of primes have any influence on th... | Empy2 | 81,790 | <p>Supposing the last digits of primes form a random sequence, from the set 1,3,7,9.<br />
Let the prime we want end in a 1. The next <span class="math-container">$n-1$</span> primes must end with something else, and the <span class="math-container">$n$</span>th end with a 1. This happens with probability <span class... |
2,811,825 | <p>I'm trying to solve the following problem (under which is my attempt at it)</p>
<p><a href="https://i.stack.imgur.com/qxHGI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qxHGI.png" alt="enter image description here"></a></p>
<p>I'm confused on how to solve for the expectation here in a non-con... | Sungjin Kim | 67,070 | <p>Consider the following very bad scenario: </p>
<p>$$A_0=p^{-1}(A)=[0,1/2], \ \ B_0=p^{-1}(B)=[3/4,1].$$
Then your process may give
$$
I_0=[0,1],
$$
$$
I_1=[1/2,1],
$$
$$
I_2=[1/2,3/4], \ \ I_3=[1/2,5/8], \ \ I_4=[1/2, 9/16], \cdots
$$</p>
<p>Then the intersection of the nested intervals is $\{1/2\}$, and it belon... |
2,811,825 | <p>I'm trying to solve the following problem (under which is my attempt at it)</p>
<p><a href="https://i.stack.imgur.com/qxHGI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qxHGI.png" alt="enter image description here"></a></p>
<p>I'm confused on how to solve for the expectation here in a non-con... | William Elliot | 426,203 | <p>P = p([0,1]) is connected.<br>
P $\cap$ A and P $\cap$ B are separated and not empty.<br>
P $\cap$ (A $\cup$ B) is disconnected.<br>
Thus P /= P $\cap$ (A $\cup$ B).<br>
Whereupon be the beseached t in [0,1].</p>
|
784,673 | <p>Consider stationary autoregression AR(1):
$$u_t=\beta u_{t-1}+ \varepsilon_t, \quad t \in \mathbb{Z}.$$ </p>
<p>$\{\varepsilon_t\}$ - i.i.d. $N(0,\sigma^2)$ random variables.</p>
<p>I know that $\mathbf{E}u_t =0$ and $\mathbf{E}u_t^2=\sigma^2/(1-\beta^2).$</p>
<p>The question is: what is the distribution of $u_t$... | square_one | 148,176 | <p>That is one of the objectives of the whole modeling exercise. </p>
<p>We desire to obtain the distribution of the solution to a time series. Thus, we need to obtain the distribution of the error terms since the AR(1) equation is determined simply by past observations of the random variable X and the error terms. So... |
784,673 | <p>Consider stationary autoregression AR(1):
$$u_t=\beta u_{t-1}+ \varepsilon_t, \quad t \in \mathbb{Z}.$$ </p>
<p>$\{\varepsilon_t\}$ - i.i.d. $N(0,\sigma^2)$ random variables.</p>
<p>I know that $\mathbf{E}u_t =0$ and $\mathbf{E}u_t^2=\sigma^2/(1-\beta^2).$</p>
<p>The question is: what is the distribution of $u_t$... | Did | 6,179 | <p>Iterating the recursion, one sees that, for every $t$,
$$u_t=\sum_{s\geqslant0}\beta^s\varepsilon_{t-s},
$$
where the family $(\varepsilon_s)$ is i.i.d. normal $(0,\sigma^2)$, hence each $u_t$ is normal $(0,\tau^2)$, where $\tau^2$ is indeed
$$
\tau^2=\sum_{s\geqslant0}\beta^{2s}\sigma^2=\frac{\sigma^2}{1-\beta^2}.
... |
3,124,285 | <p>I have written a proof, and I would appreciate verification. The problem is picked from "Set Theory and Matrices" by I. Kaplansky.</p>
<hr>
<p><em>Proof</em>. Let <span class="math-container">$a_1=f(x)$</span> and <span class="math-container">$a_2=f(y)$</span>, then</p>
<p><span class="math-container">$g(a_1)=g(a... | Hector Blandin | 170,571 | <p>Let <span class="math-container">$a_1,a_2\in\mathrm{dom}(g)$</span>. Since <span class="math-container">$f$</span> is surjective for each pair <span class="math-container">$a_1,a_2\mathrm{dom}(g)$</span> there exists <span class="math-container">$x,y\in\mathrm{dom}(f)$</span> such that <span class="math-container">$... |
2,804,716 | <p>Given this Maclaurin series:</p>
<p>$$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$</p>
<p>And the following Catenary curve, assuming that $a=1$:</p>
<p>$$g(x)=\frac{a(e^\frac{x}{a}+e^{-\frac{x}{a}})}{2}$$</p>
<p>Why does $f(x)=g(x)$ seem to hold true (at least when graphed)?</p>
<p>I'm looking for a purely al... | user | 505,767 | <p>Let consider the function</p>
<p>$$C(x)=\frac{e^x+e^{-x}}{2}$$</p>
<p>and let define also</p>
<p>$$S(x)=C'(x)=\frac{e^x-e^{-x}}{2} \implies S'(x)=C(x)$$</p>
<p>thus since</p>
<ul>
<li>$C(0)=1$</li>
<li>$C'(0)=S(0)=0$</li>
<li>$C''(0)=S'(0)=C(0)=1$</li>
<li>...</li>
</ul>
<p>by Taylor's expansion at $x=0$, that... |
2,804,716 | <p>Given this Maclaurin series:</p>
<p>$$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$</p>
<p>And the following Catenary curve, assuming that $a=1$:</p>
<p>$$g(x)=\frac{a(e^\frac{x}{a}+e^{-\frac{x}{a}})}{2}$$</p>
<p>Why does $f(x)=g(x)$ seem to hold true (at least when graphed)?</p>
<p>I'm looking for a purely al... | Emilio Novati | 187,568 | <p>The starting point is the series expansion of the exponential function:
$$
e^x=\sum_{k=0}^\infty \frac{x^k}{k!}
$$
substituting in $g(x)$ with $a=1$ we have:
$$
g(x)=\frac{e^x+e^{-x}}{2} = \frac{1}{2}\left(\sum_{k=0}^\infty \frac{x^k}{k!}+\sum_{k=0}^\infty \frac{(-x)^k}{k!} \right)=\frac{1}{2}\left[\sum_{k=0}^\inft... |
3,950,808 | <p><em>(note: this is very similar to <a href="https://math.stackexchange.com/questions/188252/spivaks-calculus-exercise-4-a-of-2nd-chapter">a related question</a> but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)</em></p>
<p>I'm self-learning math, and an <a href="h... | Christoph | 86,801 | <p>Instead of setting <span class="math-container">$x=1$</span> in
<span class="math-container">$$
\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k,
$$</span>
expand the LHS and collect terms of degree <span class="math-container">$k$</span> toge... |
1,945,116 | <p>I need a simple definition of Disjoint cycles in Symmetric Groups.I already understand what cycles and Transpositions are. I need a simple definition and if possible,give a clear example.Thanks in advance Mathematician</p>
| P Vanchinathan | 28,915 | <p>Consider a permutation of a set that has the property that, for example, elements of two disjoint subsets are permuted within each subset as cycles in each subset.</p>
<p>$S=\{1,2,3,4,5,6,7,8,9\};\ A=\{1,4,9\};\ B={2,3,5,6,7,8}$</p>
<p>The permutation of $S$ sending 1 to 4, 4 to 9, 9 to 1, 2 to 3, 3 to 5, 5 to ... |
3,308,291 | <p>I have an array of numbers (a column in excel). I calculated the half of the set's total and now I need the minimum number of set's values that the sum of them would be greater or equal to the half of the total. </p>
<p>Example:</p>
<pre><code>The set: 5, 5, 3, 3, 2, 1, 1, 1, 1
Half of the total is: 11
The least a... | Cameron Buie | 28,900 | <p>It is not closed. <span class="math-container">$1+1=2,$</span> which is not in <span class="math-container">$\{0,1,4\}.$</span></p>
|
2,582,046 | <p>My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form <span class="math-container">$\rho e^{i\theta}$</span>, so the proof begins the... | José Carlos Santos | 446,262 | <p>The error lies in assuming that $(\forall a,b\in\mathbb{C}):e^{ab}=(e^a)^b$. </p>
<p>It's worse than wrong; it doesn't make sense. The reason why it doesn't make sense is because $e^a$ can be an arbitrary complex number (except that it can't be $0$). And what is $z^w$, where $z,w\in\mathbb C$? A reasonable definiti... |
386,073 | <p>For which values of a do the following vectors for a <strong><em>linearly dependent</em></strong> set in $R^3$?</p>
<p>$$V_1= \left(a,\, \frac{-1}{2}, \,\frac{-1}{2}\right),\;\; V_2= \left(\frac{-1}{2},\, a, \,\frac{-1}{2}\right),\; \;V_3= \left(\frac{-1}{2}, \,\frac{-1}{2},\, a\right)$$</p>
<p>Please would it be ... | amWhy | 9,003 | <p>We want to find all (only) those value(s) that will make the vectors linearly <em>dependent</em>. </p>
<p>Can you see, for example, why $\,a = -\frac 12\,$ <em>is a problem</em>? Why would $\bf \,a = -\frac 12\,$ make the vectors linearly dependent? And why would $\bf\, a = 1\,$ make the vectors linearly dependent... |
386,073 | <p>For which values of a do the following vectors for a <strong><em>linearly dependent</em></strong> set in $R^3$?</p>
<p>$$V_1= \left(a,\, \frac{-1}{2}, \,\frac{-1}{2}\right),\;\; V_2= \left(\frac{-1}{2},\, a, \,\frac{-1}{2}\right),\; \;V_3= \left(\frac{-1}{2}, \,\frac{-1}{2},\, a\right)$$</p>
<p>Please would it be ... | egreg | 62,967 | <p>Row reduction has little to do with eigenvalues, but it has <em>much</em> to do with linear dependence. Actually it's the method of choice.</p>
<p>It's better to change the order into $v_3,v_2,v_1$</p>
<p>$$
\begin{bmatrix}
-\frac{1}{2} & -\frac{1}{2} & a \\
-\frac{1}{2} & a & -\frac{1}{2} \\
a &am... |
357,101 | <p>There exists a minimal subshift <span class="math-container">$X$</span> with a point <span class="math-container">$x \in X$</span> such that <span class="math-container">$x_{(-\infty,0)}.x_0x_0x_{(0,\infty)} \in X$</span>?</p>
| Ilkka Törmä | 66,104 | <p>We can produce such a subshift by a standard hierarchical construction.
Let <span class="math-container">$w_{0,0} = 01$</span> and <span class="math-container">$w_{0,1} = 011$</span>.
For each <span class="math-container">$k \geq 0$</span>, define <span class="math-container">$w_{k+1,0} = w_{k,0} w_{k,0} w_{k,1}$</s... |
4,537,685 | <p>Let <span class="math-container">$H$</span> be a Hilbert space, and let
<span class="math-container">$$H_n = \otimes_n H = \Big\{\sum_{i_1,\ldots, i_n} \alpha_{i_1, \ldots, i_n} \big(e_{i_1} \otimes \cdots \otimes e_{i_n}\big) : \sum_{i_1,\ldots, i_n} |\alpha_{i_1, \ldots, i_n}|^2<\infty \Big\}$$</span>
denote th... | Sammy Black | 6,509 | <p>A (small) example to get the feel of these objects:
<span class="math-container">$$
2 \, e_3 \otimes e_{17} - 5 \, e_4 \otimes e_4
$$</span>
is an arbitrary element of the two-fold tensor product space <span class="math-container">$H_2$</span>. Here the multi-index <span class="math-container">$(i_1, i_2) = (3, 17)... |
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Joel Dodge | 493 | <p>I never think about groupoids in any technical sense, but my favorite easy example of one can be built out of a separable field extension K/k. It is the category whose points are the subfields of the algebraic closure of k which are k-isomorphic to K. The morphisms between two objects are just the k-isomorphisms b... |
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Alfonso Gracia-Saz | 3,065 | <p>Contra dance (or square dance) gives us a nice example of a groupoid. The objects are the formations (i.e. the positions of the dancers) and the morphisms are the calls up to homotopy. A choreography (or a dance, if you wish) if a set of composable calls whose product is a morphism between two specific objects.</p... |
3,223,732 | <p>Let <span class="math-container">$X=U\cup V$</span> where <span class="math-container">$U,V$</span> are simply-connected open sets and <span class="math-container">$U\cap V$</span> is the disjoint union of two simply connected sets. We also have the condition that any subspace <span class="math-container">$S$</span>... | Ariel Serranoni | 253,958 | <p>By definition, if <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are vector spaces and <span class="math-container">$T\colon V\to W$</span> is a linear transformation, then </p>
<p><span class="math-container">$$\text{Null}(T)=\{x\in V\,\colon T(x)=0\}.$$</span></p>
<p>Therefor... |
2,002,385 | <blockquote>
<p>Prove $|e^{i\theta_1}-e^{i\theta_2}|\geq|e^{i\theta_1/2}-e^{i\theta_2/2}|$
where $ \theta_1, \theta_2 \in (0,\pi]$.</p>
</blockquote>
<p>Even though geometrically it is an obvious fact, somehow I couldn't prove it elegant way (it's really frustrating), and I'm sure some of you guys know how to prov... | Ewan Delanoy | 15,381 | <p>Hint :
$\bigg|
\frac
{e^{i\theta_1}-e^{i\theta_2}}
{e^{\frac{i\theta_1}{2}}-e^{\frac{i\theta_2}{2}}}\bigg|=
\bigg|e^{\frac{i\theta_1}{2}}+e^{\frac{i\theta_2}{2}}\bigg|=\bigg|e^{i\frac{\theta_1-\theta_2}{4}}+e^{i\frac{\theta_2-\theta_1}{4}}\bigg|=2|\cos(\frac{\theta_2-\theta_1}{4})|$.</p>
|
3,238,914 | <p>When is the <a href="https://en.wikipedia.org/wiki/Euler_line" rel="nofollow noreferrer">Euler line</a> parallel with a triangle's side?</p>
<p>I have found that a triangle with angles <span class="math-container">$45^\circ$</span> and <span class="math-container">$\arctan2$</span> is a case.</p>
<p>Is there any o... | Blue | 409 | <p>Here's an approach that's more satisfying than <a href="https://math.stackexchange.com/a/3239223/409">my previous attempt</a>.</p>
<hr>
<p>Let <span class="math-container">$\triangle ABC$</span> have midpoints <span class="math-container">$D$</span>, <span class="math-container">$E$</span>, <span class="math-conta... |
2,845,085 | <p>Find $f(5)$, if the graph of the quadratic function $f(x)=ax^2+bx+c$ intersects the ordinate axis at point $(0;3)$ and its vertex is at point $(2;0)$</p>
<p>So I used the vertex form, $y=(x-2)^2+3$, got the quadratic equation and then put $5$ instead of $x$ to get the answer, but it's wrong. I think I shouldn't hav... | Community | -1 | <p>As we know:
$2ax + b=0$ when $x=2$</p>
<p>This means $4a + b=0$ or $b=-4a$</p>
<p>Now you can substitute </p>
<p>$a×2^2 - 4a×2 +3=0$</p>
<p>Or $4a - 8a +3=0$</p>
<p>Or $a=\frac{3}{4}$</p>
<p>Now you can get $b$ and you can get $f(5)$</p>
|
265,949 | <p>Consider the set of all $n \times n$ matrices with real entries as the space $ \mathbb{R^{n^2}}$ .
Which of the following sets are compact?</p>
<ol>
<li>The set of all orthogonal matrices.</li>
<li>The set of all matrices with determinant equal to unity.</li>
<li><p>The set of all invertible matrices.</p>
<p>I am ... | anonymous | 50,284 | <p>Let's think about $1)$. A matrix is orthogonal if it consists of rows and columns of orthogonal unit vectors (in this case, of $\mathbb{R}^n$). Let $A \in \mathbb{R}^{n^2}$ be such a matrix. Pick a row or column from $A$, say $(a_1, ..., a_n)$. By assumption, $|(a_1, ..., a_n)| = 1$, from which we have that each... |
879,886 | <p>If one number is thrice the other and their sum is $16$, find the numbers.</p>
<p>I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question </p>
<p>$$
\begin{align}
x&=3y &\iff x-3y=0 &&(1)\\
x&=16-3y&&&(2)
\end{align}
$$</p>
| evinda | 75,843 | <p>$$x=3y$$
$$y+x=16 \Rightarrow y+3y=16 \Rightarrow 4y=16 \Rightarrow y=4$$</p>
<p>So, $x=12$.</p>
|
2,918,091 | <p>Suppose I want to find the locus of the point $z$ satisfying $|z+1| = |z-1|$</p>
<p>Let $z = x+iy$</p>
<p>$\Rightarrow \sqrt{(x+1)^2 + y^2} = \sqrt{(x-1)^2 + y^2}$ <br/>
$\Rightarrow (x+1)^2 = (x-1)^2$ <br/>
$\Rightarrow x+1 = x-1$ <br/>
$\Rightarrow 1= -1$ <br/>
$\Rightarrow$ Loucus does not exist</p>
<p>Is my a... | Siong Thye Goh | 306,553 | <p>As you remove the square root sign, there is another possible solution</p>
<p>$$x+1 = -(x-1)$$</p>
<p>Hence $x=0$ which is the $y$-axis.</p>
<p>A faster way is to recognize that this means the distance from $1$ and $-1$ are equal and hence the perpendicular bisector is the locus.</p>
|
3,014,670 | <p>I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors. </p>
<p><em>I've done some initial research, and reviews these Stack questions: <a href="https://math.stackexchange.com/quest... | Adrian Keister | 30,813 | <p>Your initial translation is correct, though in standard form I would write </p>
<p>All non-rational people are lakers.</p>
<p>This is an Aristotelian A type statement, and A type statements have valid obverses and valid contrapositives. They do not have valid converses.</p>
<p>Obverse: No non-rational people are ... |
1,885,068 | <p>Prove $$\int_0^1 \frac{x-1}{(x+1)\log{x}} \text{d}x = \log{\frac{\pi}{2}}$$</p>
<p>Tried contouring but couldn't get anywhere with a keyhole contour.</p>
<p>Geometric Series Expansion does not look very promising either.</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. One may set
$$
f(s):=\int_0^1 \frac{x^s-1}{(x+1)\log{x}}\: \text{d}x, \quad s>-1, \tag1
$$ then one is allowed to differentiate under the integral sign, getting
$$
f'(s)=\int_{0}^{1}\frac{x^s}{x+1}\:dx=\frac12\psi\left(\frac{s}2+\frac12\right)-\frac12\psi\left(\frac{s}2+1\right), \quad s>... |
3,873,433 | <p>Is it true that for all square, complex matrices A, B
<span class="math-container">$$
\left\|AB\right\|_p\leq\left\|A\right\|\left\|B\right\|_p$$</span></p>
<p>where <span class="math-container">$\left\| .\right\|_p$</span> refers to the Schatten p-norm and <span class="math-container">$\left\| .\right\|$</span> ref... | user8675309 | 735,806 | <p>You can also prove this (and most Schatten norm results) via the theory of majorization</p>
<p>It's worth noting that p norms for real non-negative vectors and Schatten p norms for diagonal Positive (semi)definite matrices are essentially the same thing. In both cases the norms are homogenous with respect to re-sca... |
3,720,272 | <p>My textbook employs a brute force method: add the number of committees that could be formed with one woman, two women, and three women in them. Then, the total number of such committees will be:
<span class="math-container">$$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 ... | Shiv Tavker | 687,825 | <p>You are recounting! For simplicity consider there are only three people. <span class="math-container">$W_1, W_2, M_1$</span>. Suppose you count the same way. You will get number way as <span class="math-container">$\left(\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right) \times \left(\begin{smallmatrix} 2 \\ 2 \end... |
2,377,946 | <blockquote>
<p>The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$</p>
</blockquote>
<p>I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?</p>
| Doug M | 317,162 | <p>$\displaystyle\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$</p>
<p>Where do we get with the substitution you have suggested?</p>
<p>$x = a\tan\theta\\
dx = a\sec^2\theta\\
\displaystyle\int_0^{\frac \pi 4} \frac{(a^4\tan^4\theta)(a\sec^2\theta)}{(a^2\tan^2\theta+a^2)^4}d\theta\\
$</p>
<p>Looks promising:
Keep simplifying</... |
1,419,483 | <p>Can anyone please help me in solving this integration problem $\int \frac{e^x}{1+ x^2}dx \, $?</p>
<p>Actually, I am getting stuck at one point while solving this problem via integration by parts.</p>
| Claude Leibovici | 82,404 | <p>Since $x^2+1=(x+i)(x-i)$, partial fraction decomposition leads to $$\frac 1{x^2+1}=\frac 1{2i}\Big(\frac{1}{x-i}-\frac{1}{x+i}\Big)=-\frac i{2}\Big(\frac{1}{x-i}-\frac{1}{x+i}\Big)$$ So $$I=\int \frac{e^x}{1+ x^2}\,dx =-\frac i{2}\int\Big(\frac{e^x}{x-i}-\frac{e^x}{x+i}\Big)\,dx=-\frac i{2}\int\Big(\frac{e^i\,e^{x-i... |
4,411,247 | <blockquote>
<p>If <span class="math-container">$G$</span> is finite group, how to prove that <span class="math-container">$f(g)=ag$</span>, <span class="math-container">$a \in G$</span>, is a bijection for all <span class="math-container">$g \in G$</span>? Here <span class="math-container">$ag$</span> is <span class="... | Community | -1 | <p>Though the claim holds true for any group, you can take advantage of the assumed finiteness of <span class="math-container">$G$</span> to get immediately:</p>
<ol>
<li>the injectivity holds by the left cancellation law;</li>
<li>since <span class="math-container">$G$</span> is finite, the surjectivity follows from 1... |
4,547,480 | <p>I am working with some data for which I am interested in calculating some physical parameters. I have a system of linear equations, which I can write in matrix form as:</p>
<p><span class="math-container">$$
\textbf{A} \textbf{x} = \textbf{b}
$$</span></p>
<p>where <span class="math-container">$\textbf{A}$</span> is... | llorente | 810,541 | <p>I figured out the solution to my question, so I am posting it here in case others are interested.</p>
<p>The main problem is how to apply bounds on the unknown parameters <span class="math-container">$\textbf{x}$</span>. The key is to substitute the <span class="math-container">$x_i$</span> using a transformation th... |
92,967 | <p>Let <span class="math-container">$d(n)$</span> be the number of divisors function, i.e., <span class="math-container">$d(n)=\sum_{k\mid n} 1$</span> of the positive integer <span class="math-container">$n$</span>. The following estimate is well known
<span class="math-container">$$
\sum_{n\leq x} d(n)=x \log x + (2 ... | Dr. Pi | 9,232 | <p>The $n=ab$ trick is very effective. There is a similar trick of writing in a unique way $n=ab$ with $n$ squarefree, $b$ a square and I was wondering whether this could also work here.</p>
|
2,401,281 | <blockquote>
<p>Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square. </p>
</blockquote>
<p>This question is from a math olympiad contest. </p>
<p>I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful... | farruhota | 425,072 | <p>Denote:
$$a+b+c=0; ab+ac+bc=k; abc=t$$
Then $a,b,c$ are the roots of:
$$x^3+kx+t=0$$
Note:
$$a^3+ka+t=0 \Rightarrow a^4+ka^2+ta=0,$$
$$b^3+kb+t=0 \Rightarrow b^4+kb^2+tb=0,$$
$$c^3+kc+t=0 \Rightarrow c^4+kc^2+tc=0.$$
Add and multiply by $2$:
$$2(a^4+b^4+c^4)=-2k(a^2+b^2+c^2)-t(a+b+c)=-2k((a+b+c)^2-2k)-0=(2k)^2.$$</p... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Scott Carter | 36,108 | <p><a href="http://www.youtube.com/user/ProfessorElvisZap" rel="nofollow noreferrer">Elvis's youtube link </a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Ferran V. | 892 | <p>My personal favorite in Dimensions, that was mentioned before by Gerald Edgar. For a neat and clear exposition the Geom.of 3 manifolds, Poincaré conjecture, etc I recommend <a href="http://athome.harvard.edu/threemanifolds/index.html">this</a> lecture by C.McMullen. Or Das Schöne denken (hosted at the HIM in Bonn), ... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | JoeG | 3,457 | <p>This <a href="https://www.youtube.com/watch?v=SccDUpIPXM0" rel="nofollow">video</a> about Andrew Wiles and the proof of Fermat's Last Theorem is the only time I've seen the real excitement of mathematics presented accurately. </p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Paolo | 1,831 | <p>GRASP is a new lecture series at the University of Texas at Austin, which is aimed at bringing some of the fundamental concepts and big picture of the GRASP areas (Geometry, Representation, and Some Physics) to a wider audience (the intended target audience are beginning graduate students).</p>
<p><a href="http://w... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Valerio Talamanca | 11,018 | <p><a href="http://www.youtube.com/watch?v=VdPrCWr9Ruk&feature=player_embedded#" rel="nofollow">http://www.youtube.com/watch?v=VdPrCWr9Ruk&feature=player_embedded#</a>!</p>
<p>is a video made by a student in the school of arichitecture using pov-ray
is about algebraic surfaces and how they "deform"</p>
<p>th... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | Justin Hilburn | 333 | <p>I am surprised no one has mentioned that the <a href="http://online.kitp.ucsb.edu/online/" rel="nofollow">Kavli Institute for Theoretical Physics</a>, the <a href="http://www.scgp.stonybrook.edu/" rel="nofollow">Simons Center for Geometry and Physics</a>, and the <a href="http://www.perimeterinstitute.ca/" rel="nofo... |
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.yo... | rnegrinho | 29,818 | <p>I know of some youtube channels with good content. The last two link are not strictly pure math, but still worth a look.</p>
<p>Institut Henri Poincaré:
<a href="https://www.youtube.com/channel/UCrKGv5WY5ryaIXEmnxKVxOQ" rel="nofollow">https://www.youtube.com/channel/UCrKGv5WY5ryaIXEmnxKVxOQ</a></p>
<p>princetonmat... |
361,755 | <p>Let $S$ be a multiplicatively closed subset of a commutative noetherian ring $A$. Let $M$ and $N$ be finitely generated $A$-modules. If $M_S$ is isomorphic to $N_S$, show that $M_t$ is isomorphic to $N_t$ for some $t \in S.$</p>
| Martin Brandenburg | 1,650 | <p>There is a more general geometric statement whose proof I find a little bit clearer.</p>
<p>Let $F,G$ sheaves of modules over some ringed space $X$. If $F$ is of finite presentation, then $\underline{\hom}_{\mathcal{O}_X}(F,G)_x \to \hom_{\mathcal{O}_{X,x}}(F_x,G_x)$ is an isomorphism (one easily reduces to the cas... |
480,828 | <p>We can suppose that we will create a new number system with essentially two imaginaries that do not interact. (Besides this, all quantities are taken to be integers) For example, we have an $i_1$ and an $i_2$. Then we could say</p>
<p>$$(a+b i_1)(c+d i_1) = ac + (ad + bc)i_1-bd$$</p>
<p>and, similarly for $i_2$... | Anixx | 2,513 | <p>If you drop the rule that <span class="math-container">$i_1 i_2=0$</span> but maintain commutativity, you will get <a href="https://en.wikipedia.org/wiki/Bicomplex_number" rel="nofollow noreferrer">bicomplex numbers</a>.</p>
|
4,099,649 | <p>I’m trying to solve two (in my opinion, tough) integrals which appear in part of my problem. I tried different ways but in the end I failed. See them below, please.</p>
<p><span class="math-container">$${\rm{integral}}\,1 = \int {{{\left( {\frac{A}{{{x^\alpha }}}\, + \sqrt {B + \frac{C}{{{x^{2\alpha }}}}\,} } \right... | Tyma Gaidash | 905,886 | <p>Here will be an evaluation of:</p>
<blockquote>
<p><span class="math-container">$${ \rm{integral}}\,1,2 = \int {{{\left( {\frac{A}{{{x^\alpha }}}\, + \sqrt {B + \frac{C}{{{x^{2\alpha }}}}\,} } \right)}^{\pm\frac{1}{3}}}} dx ,$$</span></p>
</blockquote>
<p>Note that if <span class="math-container">$$(a+b)^v=a^v\lef... |
51,341 | <p>I have a function that is a summation of several Gaussians. Working with a 1D Gaussian here, there are 3 variables for each Gaussian: <code>A</code>, <code>mx</code>, and <code>sigma</code>:</p>
<p>$A \exp \left ( - \frac{\left ( x - mx \right )^{2}}{2 \times sigma^{2}} \right )$</p>
<pre><code>A*Exp[-((x - mx)^2/... | Apple | 10,193 | <pre><code>f[data_] := Total[#1*Exp[-((x - #2)^2/(2 #3^2))] & @@@ data];
f[{{A, mx, sigma}}]
f[{{A, mx, sigma}, {A2, mx2, sigma2}}]
</code></pre>
<p>Use Function, Apply and Total.</p>
<p><img src="https://i.stack.imgur.com/rfPbd.jpg" alt="enter image description here"></p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?<... | Michael Hardy | 11,667 | <p>Sets have members, and two sets are the same set if, and only if, they have the same members.</p>
<p>That is not quite enough to characterize what sets are.</p>
<p>For example, is the set of all sets that are not members of themselves a member of itself? If so, you get a contradition, and if not, you get a contra... |
1,186,517 | <p>I do some ex for preparing discrete mathematics exam, i get stuck in one problem, anyone could help me?</p>
<blockquote>
<p>How many ways we can partition set {1,2,...,9} into subsets of size 2
and 5?</p>
</blockquote>
<p>anyway, some tutorials for solving such a question...</p>
<p>Edit: Like always Scott is ... | Brian M. Scott | 12,042 | <p>A partition of $\{1,\ldots,9\}$ into sets of sizes $2$ and $5$ must contain two sets of size $2$ and one of size $5$: no other combination of $2$’s and $5$’s adds up to $9$. Thus, the question boils down to determining how many ways there are to pick a $5$-element subset of $\{1,\ldots,9\}$ and then split the remain... |
1,722,948 | <blockquote>
<p>$$\frac{1}{x}-1>0$$</p>
</blockquote>
<p>$$\therefore \frac{1}{x} > 1$$</p>
<p>$$\therefore 1 > x$$</p>
<p>However, as evident from the graph (as well as common sense), the right answer should be $1>x>0$. Typically, I wouldn't multiple the x on both sides as I don't know its sign, bu... | Michael Hoppe | 93,935 | <p>Consider the function $f(x)=1/x-1=(1-x)/x$ defined for $x\neq0$. As $ f$ is continuous it can only change sign where it is zero or undefined, i.e., at $x=1$ or $ x=0$. Hence the sign of $f$ is constant on $]-\infty, 0[$, $]0,1[$, and $]1,\infty[$.</p>
<p>Finally compute the sign of say $f(-1)$, $f(1/2)$, and $f(2)... |
1,893,280 | <p>How to show $\frac{c}{n} \leq \log(1+\frac{c}{n-c})$ for any positive constant $c$ such that $0 < c < n$?</p>
<p>I'm considering the Taylor expansion, but it does not work...</p>
| Mark Viola | 218,419 | <blockquote>
<p><strong>I thought it would be instructive to present a way forward that does not rely on calculus, but rather an elementary inequality. To that end, we proceed.</strong></p>
<p>I showed in <a href="https://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263">THIS AN... |
3,029,708 | <p>Suppose there is a vector <span class="math-container">$U \in \mathbb{R}^n$</span>. How would you find the derivative of:</p>
<p><span class="math-container">$$
F(U)=trace\left(diag(U) A\ diag(U) \right)
$$</span>
where <span class="math-container">$A \in \mathbb{R}^{n \times n} \succ 0 $</span> and where <span cl... | p32fr4 | 586,851 | <p>Posting the solution I identified.</p>
<p>Due to the trace operator evaluating the above is equivalent to evaluating:</p>
<p><span class="math-container">$$
{\partial \left(\sum\limits_{i=1}^{n} u_i \ A_{(i,i)}u_i\right)\over \partial U }=
\left( \begin{align}
\begin{array}
{\partial \left(\sum\limits_{i=1}^{n} u... |
4,252,428 | <p>I encountered this system of nonlinear equations:
<span class="math-container">$$\begin{cases}
x+xy^4=y+x^4y\\
x+xy^2=y+x^2y
\end{cases} $$</span></p>
<p>My ultimate goal is to show that this has only solutions when <span class="math-container">$x=y$</span>. I didn't find any straight forward method to solving this.... | herb steinberg | 501,262 | <p>Proof by contradiction:</p>
<p>Take the difference of the two equations and divide out common factors to get <span class="math-container">$y^3-y=x^3-x$</span>. This is a cubic in either variable in terms of the other, giving three solutions in each case, possible duplicates (x=y will appear in both sets). Use syn... |
1,464,747 | <p>I am trying to solve this question:</p>
<blockquote>
<p>How many ways are there to pack eight identical DVDs into five indistinguishable boxes so that each box contains at least one DVD?</p>
</blockquote>
<p>I am very lost at trying to solve this one. My attempt to start this problem involved drawing 5 boxes, an... | soctiggs | 276,328 | <p>5 boxes 8 dvds ...
firstly you put one dvd in each box .
and now you solve no. of ways of placing 3 dvds in 5 boxes.
which is same as no of solution to the equation <br>
b1 + b2 + b3 + b4 + b5 = 3 <br> i.e.,
$ (5+3-1)\choose (3)$ = 35 . .... [solution to the equation a1+a2+a3+...an = r is $ (n+r-1) \choose n ... |
194,954 | <p>Is there a reason to use <code>Hold*</code> attributes for functional code (e.g. no intention to mutate input)? I'd expect performance gains as in pass by value vs pass by reference. </p>
<p>E.g. </p>
<pre><code>data = RandomReal[1, 10^8];
data // Function[x, x[[1]]] // RepeatedTiming
</code></pre>
<blockquote>
... | b3m2a1 | 38,205 | <p>In general, as Henrik and I note in the comments, Mathematica makes sure only to copy data when it has undergone some sort of change internally. An easy way to see this is to set a flag like <code>Valid</code> and see when it disappears:</p>
<pre><code>myBigData = RandomReal[{}, {500, 800}];
myBigData // System`Pri... |
3,399,195 | <p>So I've seen various questions with the limit 'equal' to <span class="math-container">$\infty$</span> or that the limit doesn't exist in a case where the function tends to <span class="math-container">$\infty$</span>.</p>
<p>For example, the limit of <span class="math-container">$\sqrt{x}$</span> as <span class="mat... | user | 505,767 | <p>Usually we say that limit exists when it is finite or finite. In the first case we say that the function converges to <span class="math-container">$L$</span>, in the second case we say that the function diverges to plus or minus infinity.</p>
<p>We say that the limit doesn’t exist in all the remaining cases, for ex... |
3,009,362 | <p>I need to find
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$</span></p>
<p>Looking at the graph, I know the answer should be <span class="math-container">$\frac{20}{17}$</span>, but when I tried solving it, I reached <span class="math-container">$0$</span>.</p>
<p>Here are the... | Mefitico | 534,516 | <p><strong>Hint:</strong> Try factorization!</p>
<p><span class="math-container">$$
\frac{2x^2-50}{2x^2+3x-35}=\frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=\frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
$$</span></p>
|
2,358,385 | <p>I had a test and I couldn't solve this problem:<br></p>
<p>Given $f: \mathbb R^2 \rightarrow \mathbb R$.<br>For every constant $y_0$, $f(x,y_0)$ is known to be continuous.<Br>Also, $\frac{\partial f}{\partial y}(x,y)$ is defined and bounded for all $(x,y)$.
<br><br>I needed to prove that $f$ is continuous for all $... | StuartMN | 439,545 | <p>f(x,y)-f($x_0$,$y_0$) = [f(x,y)-f(x,$y_0$) ] + [ f(x,$y_0$) - f($x_0$,$y_0$) ] .</p>
<p>On the first bracketed term use the mean value theorem and the boundedness of the partial derivative : on the second use the given continuity . </p>
|
2,622,092 | <p>I want to study the convergence of the improper integral $$ \int_0^{\infty} \frac{e^{-x^2}-e^{-3x^2}}{x^a}$$To do so I used the comparison test with $\frac{1}{x^a}$ separating $\int_0^{\infty}$ into $\int_0^{1} + \int_1^{\infty}$.</p>
<p>For the first part, $\int_0^{1}$, I did $$\lim_{x\to0} \frac{\frac{e^{-x^2}-e^... | Jack D'Aurizio | 44,121 | <p>Your problem boils down to computing
$$ I(\alpha)=\int_{0}^{+\infty}\frac{1-e^{-z^2}}{z^\alpha}\,dz = \frac{1}{2}\int_{0}^{+\infty}\frac{1-e^{-z}}{z^{\frac{\alpha+1}{2}}}\,dz. $$
In a right neighbourhood of the origin $\frac{1-e^{-z}}{z^{\frac{\alpha+1}{2}}}$ behaves like $z^{\frac{1-\alpha}{2}}$ and in a left neig... |
2,972,957 | <p><strong>Artin's Theorem-</strong> Let <span class="math-container">$E$</span> be a field and <span class="math-container">$G$</span> be a group of automorphisms of <span class="math-container">$E$</span> and <span class="math-container">$k$</span> be the set of elements of <span class="math-container">$E$</span> fix... | Tsemo Aristide | 280,301 | <p><span class="math-container">$A$</span> is not itself a group, think about the set of transpositions which generated <span class="math-container">$S_n$</span>, (a permutation whose signature is even is not a transposition). The Artin theoem's allows to say that and <span class="math-container">$[E:F]=|G|\geq |A|$</... |
2,252,579 | <p>$$ \lim_{n\to\infty}\left (\frac n {n+1} \right )^{2n} = \lim_{n\to\infty}\left (\frac{n+1}{n} \right )^{-2n} =\lim_{n\to\infty} \left (1 + \frac 1n \right )^{-2n}= \left (\lim_{n\to\infty}\left (1 + \frac 1n \right )^{n} \right )^{-2} = e^{-2}$$</p>
<p>What I don't understand is why is it a -2 and not +2? Also, ... | Alex Jones | 350,433 | <p>$\lim (\frac{n}{n+1})^{2n} = \lim e^{\log(\frac{n}{n+1})^{2n}} = \lim e^{2n\log(\frac{n}{n+1})} = e^{\lim 2n\log(\frac{n}{n+1})}$.</p>
<p>$\lim 2n \log (\frac{n}{n+1}) = \lim \frac{\log (\frac{n}{n+1})}{\frac{1}{2n}} = \lim \frac{\frac{1}{\frac{n}{n+1}(n+2)^2}}{\frac{-1}{2n^2}} = \lim -2 \frac{n}{n+1} = -2(1) = -2.... |
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