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 |
|---|---|---|---|---|
1,002,719 | <p>If we have</p>
<p>$f: \{1, 2, 3\} \to \{1, 2, 3\}$</p>
<p>and</p>
<p>$f \circ f = id_{\{1,2,3\}}$</p>
<p>is the following then always true for every function?</p>
<p>$f = id_{\{1,2,3\}}$</p>
| amWhy | 9,003 | <p>Note that we have $f^{-1}\circ f = f\circ f^{-1}= id\{1, 2, 3\}\;\forall \text{ permutations on } \{1, 2, 3\}$.</p>
<p>By definition, all permutations on a set are bijective functions, so every permutation has an inverse.</p>
<p>You need only find a single bijective function $f$ which <em>is its own inverse,</em>... |
201,576 | <p>Struggling with something basic. Suppose when defining f[x_] the outcome depends on Sign[x]. When calling this function, how do I tell Mathematica the sign of the argument?
My attempt:</p>
<pre><code>f[x_]=x Sign[x];
Assuming[a>0, f[a]]
</code></pre>
<p>The output I get is </p>
<pre><code>a Sign[a]
</code></p... | Bob Hanlon | 9,362 | <pre><code>$Version
(* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *)
Clear["Global`*"]
p = x /. Solve[(a - x)/y^2 == 1, x];
q = ((3 a - 2 p)/(4 y^2));
f[x_] = x Sign[x];
Assuming[q > 0, FullSimplify[f[q]]]
(* {Abs[a + 2 y^2]/(4 y^2)} *)
</code></pre>
<p>Note the presence of <a href="https://reference... |
458,088 | <p>I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!</p>
<p>Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.</p>
... | Daniel Franke | 86,152 | <p>When I saw your question I was immediately reminded of this:</p>
<p><a href="http://logitext.mit.edu/logitext.fcgi/tutorial" rel="nofollow">http://logitext.mit.edu/logitext.fcgi/tutorial</a></p>
<p>See also this related blog post by the author:</p>
<p><a href="http://blog.ezyang.com/2012/05/an-interactive-tutoria... |
458,088 | <p>I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!</p>
<p>Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.</p>
... | hardmath | 3,111 | <p>Graph theory, particularly for planar graphs, seems a good area for the sort of recreational math puzzles to keep club members interested.</p>
<p>Here are two that I learned about recently:</p>
<blockquote>
<p><a href="https://en.wikipedia.org/wiki/Planarity" rel="nofollow noreferrer">Planarity</a> The goal is to mo... |
1,329,398 | <p>So, I've posted a question regarding Wikipedia's quartic page. This was from the first question.</p>
<blockquote>
<p>I'm trying to implement the general quartic solution for use in a ray tracer, but I'm having some trouble. The solvers I've found do cause some strange false negatives leaving holes in the tori I'm t... | asmeurer | 781 | <p>If you are interested in the various ways of representing the quartic solution symbolically, the <a href="https://github.com/sympy/sympy/blob/acb005ce9971ea7714b785383414258279878fd4/sympy/polys/polyroots.py#L244" rel="nofollow">SymPy code</a> goes through a few methods and tries to pick the best one for the given p... |
2,881,673 | <p>I've searched all over the internet and cannot seem to factorise this polynomial.</p>
<p>$x^4 - 2x^3 + 8x^2 - 14x + 7$</p>
<p>The result should be $(x − 1)(x^3 − x^2 + 7x − 7)$</p>
<p>What are the steps to get to that result?
I've tried grouping but doesn't seem to work...</p>
| mfl | 148,513 | <p>$x-a$ is a factor of $x^4 - 2x^3 + 8x^2 - 14x + 7$ if and only if $a^4 - 2a^3 + 8a^2 - 14a + 7=0.$ Integers that can work are the divisors of $7$ (the independent term). That is: $\pm 1,\pm 7.$ If we check with $x=1$ we get</p>
<p>$$1^4 - 2\cdot 1^3 + 8\cdot 1^2 - 14\cdot1 + 7=0.$$ So $x-1$ is a factor. That is, th... |
1,415,752 | <p>I test my answer using wolfram alpha pro but it gets a different result to what I am getting. This is homework.</p>
<p>My result is z= 2(y-1)</p>
<p>partial derivative with respect to y is </p>
<pre><code> x.y^x-1
</code></pre>
<p>partial derivative with respect to x is
ln(y).y^x</p>
<p>ln(1) is zero... | Yes | 155,328 | <p>In fact, the square root $\sqrt{1-x}$ is meaningless for all $x > 1$. So, in particular, the limit $\lim_{x \to 1+}\sqrt{1-x}$ does not exist at all.</p>
|
375,372 | <p>Using the $\epsilon-M $ definition of the limit, calculate
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$</p>
<p>Working so far: </p>
<p>$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$</p>
<p>Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$</p>
<p>$$\left|\fr... | Santosh Linkha | 2,199 | <p>use the fact that $ \left | \frac 1 x - 0\right| < \delta $
$$\left| \frac{-3x - 17}{x^2 + x + 8}\right| < \left| \frac{-3x - 17}{x^2 }\right| \le 3\left |\frac 1 x \right | + 17\left |\frac 1 {x^2} \right | < 20 \delta = \epsilon $$</p>
|
1,918,435 | <p>Assume I have a set $S=\{1,\ldots,N\}$, where $N$ is an integer. In my case, $N$ can be zero or non-zero, depending on situation. </p>
<p>If $N=0$, does it automatically mean that $S=\emptyset$? I feel like it should, but would like to be sure. Thanks in advance for replying.</p>
| Mees de Vries | 75,429 | <p>Typically, if $S$ is informally defined by $S = \{1,\ldots,N\}$, then if $N = 1$ this means $S = \{1\}$ and if $N = 0$ this means $S = \emptyset$.</p>
|
1,918,435 | <p>Assume I have a set $S=\{1,\ldots,N\}$, where $N$ is an integer. In my case, $N$ can be zero or non-zero, depending on situation. </p>
<p>If $N=0$, does it automatically mean that $S=\emptyset$? I feel like it should, but would like to be sure. Thanks in advance for replying.</p>
| Asaf Karagila | 622 | <p>Yes. In the context of the natural numbers, if we write $S=\{1,\ldots,N\}$, then we mean really $\{x\in\Bbb N\mid 1\leq x\leq N\}$. If $N<1$, then there are no such $x$ and the set is empty.</p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argumen... | Per Alexandersson | 1,056 | <p>Read more papers, related to your problem. You can get inspiration, or stumble upon a proof of exactly what you are looking for. It could also be that you learn some technique in a proof.</p>
<p>I recently realized that a proof of something I needed was already proved in an earlier paper of mine, but it was not actu... |
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argumen... | მამუკა ჯიბლაძე | 41,291 | <p>My personal experience - it helps to listen to some talks, not necessarily very closely related to your problem, ideally not online, so you can discuss it with somebody after the talk, etc. Try to find analogies, maybe even superficial ones, the more unexpected the better. It is not only about thinking out of the bo... |
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square r... | GPerez | 118,574 | <p>To be honest, I would refrain from saying that $$\sqrt{-1} = i$$
It's not even like the real numbers, where one can agree that the square root function takes on the positive values, because what is positive, in the complex plane? Even if you did establish the identity as convention, how would you extend the conventi... |
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square r... | Espen Nielsen | 45,874 | <p>This question can be approached from one of two angles. Let's do one at a time.</p>
<ol>
<li>Can $\sqrt{-1}$ be both $i$ and $-i$?</li>
</ol>
<p>The answer is in fact yes. People realized early on that some functions may take any of a number of values. This is true already in the real number system with the square... |
1,869,119 | <p>Show that the Monotone Convergence Theorem may not hold for decreasing sequences of functions.</p>
<p>Suppose $\left\{f_{n}\right\}$ is a sequence of nonnegative decreasing functions converging to $f$ pointwise. I know that if $f_{1}$ is finite,we can construct the sequence say $\left\{f_{1}-f_{n}\right\}$ which is... | Martín Vacas Vignolo | 297,060 | <p>Put $f_n=\infty\chi_{(0,1/n)}$</p>
|
209,856 | <p>Of course, I can use Stirling's approximation, but for me it is quite interesting, that, if we define $k = (n-1)!$, then the left function will be $(nk)!$, and the right one will be $k! k^{n!}$. I don't think that it is a coincidence. It seems, that there should be smarter solution for this, other than Stirling's ap... | platinor | 93,324 | <p>Take $\log$ on both sides and use the $\log {n!} = \Theta(n\log n)$.
The first terms becomes $\Theta(n!\log{(n!)})$, the second one becomes $\Theta((n-1)!\log {(n-1)!}) + \Theta(n!\log{(n-1)!})$. So it's obvious that the first terms grows faster than the second one.</p>
|
424,694 | <p>Let <span class="math-container">$p$</span> be a prime, and consider <span class="math-container">$$S_p(a)=\sum_{\substack{1\le j\le a-1\\(p-1)\mid j}}\binom{a}{j}\;.$$</span>
I have a rather complicated (15 lines) proof that <span class="math-container">$S_p(a)\equiv0\pmod{p}$</span>. This must be
extremely classic... | Ira Gessel | 10,744 | <p>Here's a straightforward proof, using generating functions, though it's not as elegant as Ofir's.
We have
<span class="math-container">$$
\sum_{a=0}^\infty\binom{a}{j}x^a =\frac{x^j}{(1-x)^{j+1}}.
$$</span>
Setting <span class="math-container">$j=(p-1)k$</span> and summing on <span class="math-container">$k$</span> ... |
1,492,027 | <p>Defining R to be the relationship on real numbers given by xRy iff x-y is rational, I've been asked to find the equivalence class of $\sqrt2$. My instincts say that the equivalence class of $\sqrt2$ would just be the empty set. But after a riveting conversation on a similar subject <a href="https://math.stackexchang... | DRF | 176,997 | <p><strong>Hint</strong> first thing you need to do is figure out what an equivalence class <strong>is</strong>. Once you have an understanding of what it is that result should be in general terms (real number, rational number, set of real numbers, set of rational numbers etc.) you can continue.</p>
|
107,399 | <p>Let's say we have a set a\of associations:</p>
<pre><code>dataset = {
<|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "a", "subtype" -> "II", "value" -> 2|>,
<|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "b", "subtype" -> ... | WReach | 142 | <p>One approach is to employ a helper function that unwraps singleton lists:</p>
<pre><code>{delist[v_]} ^:= v
</code></pre>
<p>With this, the <code>GroupBy</code> expression is fairly succinct:</p>
<pre><code>dataset // GroupBy[{#type&, #subtype& -> delist}]
(*
<| "a" -> <| "I" -> <|"ty... |
107,399 | <p>Let's say we have a set a\of associations:</p>
<pre><code>dataset = {
<|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "a", "subtype" -> "II", "value" -> 2|>,
<|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "b", "subtype" -> ... | Ronald Monson | 2,249 | <p>If you need to retain the original keys, I'd be inclined to follow Edmund's answer; alternatively, if you are happy enough to throw away the keys (also given that "every entry is unique") one flexible approach follows a <a href="https://mathematica.stackexchange.com/a/102483/2249">nice solution of your own</a>.</p>
... |
1,613,185 | <p>There are five red balls and five green balls in a bag. Two balls are taken out at random. What is the probability that both the balls are of the same colour</p>
| Em. | 290,196 | <p>Notice that you are interested in the event $GG$ or $RR$.
Let's consider the $GG$ case. The probability that you get first $G$
is $$\frac{5}{10}.$$ Then there are $4$ $G$ left and a total of $9$ total. Now, in the second draw, what is the probability that you get another $G$? That is
$$\frac{4}{9}.$$
Thus the probab... |
97,393 | <p>The polynomial</p>
<p>$F(x) = x^5-9x^4+24x^3-24x^2+23x-15$</p>
<p>has roots $x=1$ and $x=j$. Calculate all the roots of the polynomial.</p>
<p>I was told I had to use radicals or similar to solve this but after reading up on it I'm still confused about how to solve it.</p>
| yoyo | 6,925 | <p>if $F(a)=0$, then $(x-a)|F(x)$. also, if a polynomial with real cofficients has a complex root, then the complex conjugate is also a root. so $F$ is divisible by $(x-1)$ and $x^2+1=(x-i)(x+i)$ (according to you). after dividing by these, you will have a polynomial of degree 2, which you can easily factor:
$$
F/(... |
97,393 | <p>The polynomial</p>
<p>$F(x) = x^5-9x^4+24x^3-24x^2+23x-15$</p>
<p>has roots $x=1$ and $x=j$. Calculate all the roots of the polynomial.</p>
<p>I was told I had to use radicals or similar to solve this but after reading up on it I'm still confused about how to solve it.</p>
| guy | 231,779 | <p>As you know that x=1 is a solution, factor (x-1) out of the equation then solve the resulting quartic equation with the formula. You can see the formula at <a href="http://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg" rel="nofollow">http://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formul... |
3,172,485 | <p>Consider the familiar trigonometric identity: <span class="math-container">$\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$</span></p>
<p>Show that the identity above can be interpreted as Fourier series expansion.</p>
<p>so we know that cos is periodic between <span class="math-container">$\pi$</span> and... | Mostafa Ayaz | 518,023 | <p><strong>Hint</strong></p>
<p>Simply show that<span class="math-container">$$\int_{0}^{2\pi}\cos^3 x\cos nx=\int_{0}^{2\pi}\cos^3 x\sin mx=0$$</span>for <span class="math-container">$n\ne 1,3$</span>. What about <span class="math-container">$n=1$</span> or <span class="math-container">$n=3$</span>?</p>
|
2,055,559 | <blockquote>
<p>Let <span class="math-container">$a,b,c$</span> be the length of sides of a triangle then prove that:</p>
<p><span class="math-container">$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$</span></p>
</blockquote>
<p>Please help me!!!</p>
| Lelouch vi Britannia | 395,510 | <p>let</p>
<p>$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=0$ (1)</p>
<p>let</p>
<p>$x=-a+b+c; y=a-b+c; z=a+b-c$</p>
<p>$z,y,x$ are twice the length of the segments between the vertices and the touching point of the incircles. so</p>
<p>$a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}$</p>
<p>substitute them to (1) and ... |
2,055,559 | <blockquote>
<p>Let <span class="math-container">$a,b,c$</span> be the length of sides of a triangle then prove that:</p>
<p><span class="math-container">$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$</span></p>
</blockquote>
<p>Please help me!!!</p>
| Michael Rozenberg | 190,319 | <p>Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$ and $u\geq0$, $v\geq0$.</p>
<p>Hence, $\sum\limits_{cyc}(a^3b-a^2b^2)=(u^2-uv+v^2)x^2+(u^3+2u^2v-uv^2+v^3)x+2u^3v\geq0$.</p>
<p>Done!</p>
|
1,581,161 | <p>Let the triangle $ABC$ and the angle $\widehat{ BAC}<90^\circ$ </p>
<p>Let the perpendicular to $AB$ passing by the point $C$ and the perpendicular to $AC$ passing by $B$ intersect the circumscribed circle of $ABC$ on $D$ and $E$ respectively .
We suppose that $DE=BC$</p>
<p>What is the angle $\widehat{BAC}$ ... | user24142 | 208,255 | <p><a href="https://i.stack.imgur.com/r1Yal.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r1Yal.png" alt="A diagram of the setup"></a></p>
<p>The above is basically what you're given. I've labeled a lot of the angles with $\alpha, \beta, \gamma, \delta$ and $\epsilon$. Two inscribed angles are equ... |
356,530 | <p>I'm a really confused as to how to start this question, would really appreciate any help you guys could give me!</p>
| Paul Gustafson | 66,345 | <p>There are $2^{17}$ total strings. There are ${17 \choose {n}}$ ways of getting a string with $n$ ones in it. What's next?</p>
|
356,530 | <p>I'm a really confused as to how to start this question, would really appreciate any help you guys could give me!</p>
| Cameron Buie | 28,900 | <p>How many total bit strings of that length are there? There are $\binom{17}{k}$ bit strings of length $17$ with exactly $k$ ones, since choosing the positions of all the ones determines the string completely. It will be quicker to find the number of bit strings that <strong>don't</strong> meet your criterion, then su... |
273,499 | <blockquote>
<p>Show that every group $G$ of order 175 is abelian and list all isomorphism types of these groups. [HINT: Look at Sylow $p$-subgroups and use the fact that every group of order $p^2$ for a prime number $p$ is abelian.]</p>
</blockquote>
<p>What I did was this. $|G| = 175$. Splitting 175 gives us $175 ... | amWhy | 9,003 | <p>The fact that $P, Q $ are both <em>normal</em> tells you that $G$ is a direct product of $P$ and $Q$. $P$ is abelian because of the hint given in the problem statement: $|P| = 25 = 5^2$, and $5$ is prime.$|Q| = 7$ with $7$ prime. All groups of prime order are cyclic, and all cyclic groups are abelian, $Q$ is therefo... |
2,140,192 | <p>I want to show that $C^1[0,1]$ isn't a Banach Space with the norm:</p>
<p>$$||f||=\max\limits_{y\in[0,1]}|f(y)|$$</p>
<p>Therefore, I want to show that the sequence $\left \{ |x-\frac{1}{2}|^{1+\frac{1}{n}} \right \}$ converges to $|x-\frac{1}{2}|$, but I can't find $N$ in the definition of convergence. Could anyo... | Jonas Meyer | 1,424 | <p>To show $\left|x-\frac12\right|^{1+1/n}=\left|x-\frac12\right|\left|x-\frac12\right|^{1/n}\to \left|x-\frac12\right|$ pointwise, which is clear when $x=\frac12$, it suffices to show that when $0<t<1$, $t^{1/n}\to 1$. Note that $(t^{1/n})$ is bounded above by $1$ and monotone increasing, so it has a limit $L$,... |
2,794,704 | <p>Can the following sum be further simplified? $${1\over 20}\sum_{n=1}^{\infty}\left(n^2+n\right)\left(\frac45\right)^{n-1}$$
(It's part of a probability problem)</p>
| nonuser | 463,553 | <p>$$f(x)=\sum_{n=1}^{\infty}n(n+1)x^{n-1}= (\sum_{n=1}^{\infty}(n+1)x^{n})'=(\sum_{n=1}^{\infty}x^{n+1})''$$</p>
<p>$$=\Big({x^2\over 1-x}\Big)''=\Big({2x-x^2\over (1-x)^2}\Big)'= {2\over (1-x)^3}$$ </p>
<p>So $${1\over 20}f\Big({4\over 5}\Big) = 12,5$$</p>
|
754,012 | <p>Is it possible to show that the harmonic series is divergent by showing that the sequence of partial sums is a monotone increasing sequence that is unbounded?</p>
| Steven Gubkin | 34,287 | <p>Perhaps you have seen the proof by the integral test, and you would like a more "hands on" proof, where you get a bound you can feel. Here is such a proof:</p>
<p>$$
\begin{align}1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{2^{n}-1} &> \frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+... |
548,776 | <p>I am new in Topos theory. I have actually just started learning. I am reading MacLane-Moerdijk's book, as it was suggested to me as the best introduction to the subject. Unfortunately I can not make sense of the following.</p>
<p>In section 5 of Chapter I, (page 41) they build what is needed to prove that in the pr... | Slade | 33,433 | <p>Yes, that's what they mean.</p>
<p>Writing $up$ or $u(p)$ would seem to make more intuitive sense, as this feels more like $u$ is "acting" on $p$ (and since we only have one gadget, $P$, for turning $u$ into a function on sets, your definition is really the only way), but since we are using contravariant functors, ... |
2,528,716 | <p>How can I prove </p>
<blockquote>
<p>$$x^2+y^2-x-y-xy+1≥0$$</p>
</blockquote>
<p>I tried $(x+y)^2-3xy-(x+y)+1≥0 \rightarrow(x+y-1)(x-y)-3xy+1≥0$ I can not continue</p>
| lab bhattacharjee | 33,337 | <p>Let $u=x^2+y^2-x-y-xy\iff x^2-x(1+y)+y^2-y-u=0$</p>
<p>As $x$ is real, the discriminant must be $\ge0$</p>
<p>$$\implies(1+y)^2-4(y^2-y-u)\ge0$$</p>
<p>$$4u\ge3y^2-6y-1=3(y-1)^2-4\ge-4$$ </p>
|
2,431,375 | <p>A continuous function $f$ on $[a,b]$, differentiable in $(a,b)$, has only 1 point where its derivative vanishes. What is true about this function?</p>
<p>A. $f$ cannot have an even number of extrema.</p>
<p>B. $f$ cannot have a maximum at one endpoint and minimum at the other.</p>
<p>C. $f$ might be monotonically... | zwim | 399,263 | <p>If $f$ is differentiable on $[a,b]$ then the example below shows only C works</p>
<p><a href="https://i.stack.imgur.com/vBaCX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vBaCX.png" alt="enter image description here"></a></p>
<p>$\bbox[5px,border:2px solid]{f(x)=-x^2\text{ on interval }[-1,0]... |
1,611,506 | <blockquote>
<p>$$\int (2x^2+1)e^{x^2} \, dx$$</p>
</blockquote>
<p>It's part of my homework, and I have tried a few things but it seems to lead to more difficult integrals. I'd appreciate a hint more than an answer but all help is valued.</p>
| user304682 | 304,682 | <p>Here is a hint; you can not analytically evaluate $\int \exp(x^2) dx$.
Read more about that on <a href="https://en.wikipedia.org/wiki/Error_function" rel="nofollow">wiki</a>.
So your final answer should contain terms with $\int \exp(x^2) dx$.</p>
|
58,024 | <p><img src="https://i.stack.imgur.com/cTpA2.jpg" alt="Show that..."></p>
<p>The picture says it all. "Vis at" means "show that". My first thought was that h is 2x, which is not correct. Maybe the formulas for area size is useful? </p>
<p>EDIT: (To make the question less dependent from the <a href="https://math.meta.... | Américo Tavares | 752 | <p>From the geometry of the problem (see figure and identify two <a href="http://en.wikipedia.org/wiki/Similarity_%28geometry%29#Similar_triangles" rel="nofollow noreferrer">similar triangles (Wikipedia)</a> or <a href="http://www.gcseguide.co.uk/similar_triangles.htm" rel="nofollow noreferrer">here</a>), we can get th... |
3,444,673 | <p>How to evaluate this double integral
<span class="math-container">$$\int_{0}^1\int_{x^2}^x \frac{x}{y}e^{-\frac{x^2}{y}}dydx.$$</span></p>
<p>It seems like I am evaluating the double integral of a non-elementary function. I tried substitutions but the integral is growing. </p>
<hr>
<p>Following Fred's suggestion,... | Kavi Rama Murthy | 142,385 | <p>The second one is correct. For the first one consider <span class="math-container">$a+b\int |f|d\mu =\int (a+b|f|)d\mu \leq \int \sqrt {a^{2}+b^{2}} \sqrt {1+|f|^{2}} d\mu$</span>. If <span class="math-container">$a^{2}+b^{2} \leq 1$</span> this gives <span class="math-container">$a+b\int |f|d\mu \leq \int \sqrt {1... |
19,880 | <p>I want to write down $\ln(\cos(x))$ Maclaurin polynomial of degree 6. I'm having trouble understanding what I need to do, let alone explain why it's true rigorously.</p>
<p>The known expansions of $\ln(1+x)$ and $\cos(x)$ gives:</p>
<p>$$\forall x \gt -1,\ \ln(1+x)=\sum_{n=1}^{k} (-1)^{n-1}\frac{x^n}{n} + R_{k}(x... | Agustí Roig | 664 | <p>I think Joe Johnson's is a good idea:</p>
<p>$$
\frac{d}{dx} \ln\cos x = -\tan x
$$</p>
<p>plus the knowledge of the <a href="http://en.wikipedia.org/wiki/Taylor_series" rel="nofollow">Taylor series for the tangent</a>,</p>
<p>$$
\tan x = \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} \ , \qquad ... |
666,297 | <p>Find the value of $x$, what is the value of $x$ in this equation, step by step solution will be great.
\begin{equation}
0.4x+15=x
\end{equation}</p>
| dato datuashvili | 3,196 | <p>So we have,</p>
<p>$15=x-0.4x$</p>
<p>$15=0.6x$</p>
<p>$x=\frac{15}{0.6}=25$</p>
<p>Because, </p>
<p>$$\frac{15}{0.6}=\frac{15}{\frac{6}{10}}=\frac{(15\cdot 10)}{6}=\frac{150}{6}=25$$</p>
|
2,663,537 | <p>Suppose G is a group with x and y as elements. Show that $(xy)^2 = x^2 y^2$ if and only if x and y commute.</p>
<p>My very basic thought is that we expand such that $xxyy = xxyy$, then multiply each side by $x^{-1}$ and $y^{-1}$, such that $x^{-1} y^{-1} xxyy = xxyy x^{-1}$ , and therefore $xy=xy$.</p>
<p>I realiz... | mfl | 148,513 | <p><strong>Hint</strong></p>
<p>$$(xy)^2=x^2y^2\iff xyxy=xxyy\iff x^{-1}xyxy=x^{-1}xxyy\iff yxy=xyy.$$</p>
<p>Can you finish?</p>
|
481,086 | <blockquote>
<p>Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by
$\{(x,y,z)\ | \ x=y, f(x)=z\}$.</p>
</blockquote>
<p>How will I be able to do this problem? </p>
<p>I know that a regular parametrization of a curve then the curvature a... | FireGarden | 87,896 | <p>For a unit speed curve, the curvature can be simply $$ \kappa = || \ddot\gamma || $$
Or if it is regular, then $$ \kappa = \dfrac{||\ddot\gamma \times \dot\gamma ||}{||\dot\gamma||^3} $$</p>
|
481,086 | <blockquote>
<p>Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by
$\{(x,y,z)\ | \ x=y, f(x)=z\}$.</p>
</blockquote>
<p>How will I be able to do this problem? </p>
<p>I know that a regular parametrization of a curve then the curvature a... | Christian Blatter | 1,303 | <p>Your curve lies in the plane $x=y$; therefore it is a plane curve. Denoting the length along the first main diagonal in the $(x,y)$-plane by $s$ the curve can be written as a graph in the form
$$z(s)=f\left({s\over\sqrt{2}}\right)\qquad (s_0\leq s\leq s_1)\ .$$
For such graphs the formula for the curvature reads
$$... |
87,963 | <p>Assume that $L/K$ is an extension of fields and $[L:K]=n$, with $n$ composite. Assume that $p\mid n$, can we always produce a subextension of degree $p$ and if not under what conditions can it be done? I would guess this is very false, but I couldn't come up with any trivial counterexamples.</p>
| Gerry Myerson | 8,269 | <p>We know that equations of degree 2 and 3 have solutions in radicals, but most equations of degree 6 (or any degree 5 or greater) don't. So let $f$ be a polynomial over $K$ of degree 6 not solvable in radicals, let $\alpha$ be a root of $f$ in some extension, let $L=K(\alpha)$. If there were an intermediate field $E$... |
87,963 | <p>Assume that $L/K$ is an extension of fields and $[L:K]=n$, with $n$ composite. Assume that $p\mid n$, can we always produce a subextension of degree $p$ and if not under what conditions can it be done? I would guess this is very false, but I couldn't come up with any trivial counterexamples.</p>
| Georges Elencwajg | 3,217 | <p>You are very right when you write "I would guess this is very false": here is a precise statement. </p>
<p><strong>Proposition 1</strong><br>
For any $n\gt 1$ there exists a field extension $\mathbb Q\subset K$ of degree $[K:\mathbb Q]=n \:$ with no intermediate extension $\mathbb Q \subsetneq k\subsetneq ... |
512,037 | <p>This is a question from our reviewer for our exam for linear algebra. I just want to have some ideas how to tackle the problem.</p>
<p>If $A$ is an $n\times n$ matrix with integer coefficients, such that the sum of each row's elements is equal to $m$, show that $m$ divides the determinant.</p>
| DanielV | 97,045 | <p>The other answer is probably better, but I offer this as an alternative.</p>
<p>Let B be an upper triangular matrix of 1s and 0s. For example, 4x4:</p>
<p>$B = \begin{array} {c c c c }
1 & 1 & 1 & 1\\
0 & 1 & 1 & 1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & 1\\
\end{array}$</p>
... |
268,778 | <p>Consider a random process where integers are sampled uniformly with replacement from $\{1...n\}$. Let $X$ be a random variable that represents the number of samples until either a duplicate is found or both the values $1$ and $2$ have been found. So if the samples where $1,6,3,5,1$ then $X=5$ and if it was $1,6,3,... | cardinal | 7,003 | <p>First, note that by the <a href="http://en.wikipedia.org/wiki/Pigeonhole_principle" rel="nofollow">pigeonhole principle</a>, $\renewcommand{\Pr}{\mathbb P}\Pr(X > n) = 0$.</p>
<p>Next let
$$
A_m := \{\text{no duplicates in first $m$ trials}\}
$$
and
$$
B_{m,i} := \{\text{The value $i$ has been seen in the fir... |
268,778 | <p>Consider a random process where integers are sampled uniformly with replacement from $\{1...n\}$. Let $X$ be a random variable that represents the number of samples until either a duplicate is found or both the values $1$ and $2$ have been found. So if the samples where $1,6,3,5,1$ then $X=5$ and if it was $1,6,3,... | Did | 6,179 | <p>Let $x\geqslant1$. There are $n^x$ samples of length $x$. Amongst these, $(n)_x=\frac{n!}{(n-x)!}$ samples have no duplicate. </p>
<p>A sample of length $x$ without duplicate where two distinct given results $i$ and $j$ appear is uniquely described by a sample of length $x-2$ without duplicate where neither result ... |
721,514 | <p>Let $T$ be a normal operator and $f$ be a bounded borel function on ${\sigma}(T)$. If $E_{T}$ and $E_{f(T)}$ are the spectral decompositions of $T$ and $f(T)$ respectively, prove that for any borel set $w$ we have that $E_{f(T)}(w)=E_{T}(f^{-1}(w))$ for any borel subset $w$ of ${\sigma}(T)$</p>
| Cm7F7Bb | 23,249 | <p>As mentioned in the comments, we need to show that the sequence is monotonic and bounded.</p>
<p>First, we observe that
$$
x_n-x_{n+1}=x_n-\frac12\Bigl(x_n+\frac a{x_n}\Bigr)=\frac1{2x_n}(x_n^2-a).
$$
Secondly, we obtain that
\begin{align*}
x_n^2-a
&=\frac14\Bigl(x_{n-1}+\frac a{x_{n-1}}\Bigr)^2-a\\
&=\fr... |
721,514 | <p>Let $T$ be a normal operator and $f$ be a bounded borel function on ${\sigma}(T)$. If $E_{T}$ and $E_{f(T)}$ are the spectral decompositions of $T$ and $f(T)$ respectively, prove that for any borel set $w$ we have that $E_{f(T)}(w)=E_{T}(f^{-1}(w))$ for any borel subset $w$ of ${\sigma}(T)$</p>
| bobbym | 77,276 | <p>That looks a lot like the well known method for computing square roots. It is derived by using Newton's on</p>
<p>$$ x^2 - a = 0 $$</p>
<p>$$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $$</p>
<p>$$=x_n-\frac{x_n^2-a}{2x_n}$$</p>
<p>$$=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right) $$</p>
<p>If it converges it will converge ... |
2,290,395 | <p>What if in Graham’s Number every “3” was replaced by “tree(3)” instead? How big is this number? Greater than Rayo’s number? Greater than every current named number?</p>
| Simply Beautiful Art | 272,831 | <p>No. If you replaced all the $3$'s in the construction of Graham's number with $\operatorname{TREE}(3)$, the resulting number would be smaller than $g_{\operatorname{TREE(3)}}$ where $g_n$ denotes the $n$th number in Graham's sequence with $g_{64}$ being Graham's number. This is much much smaller than $\operatorname{... |
3,280,095 | <p>Given the function
<span class="math-container">$$\int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$</span>
You would need to have <span class="math-container">$u=\sqrt{x}-3$</span> and <span class="math-container">$du=\frac{1}{2 \sqrt{x}}$</span>, when I use a online calculator it suggests to rewrite the numerator as
<span clas... | cmk | 671,645 | <p>If <span class="math-container">$u=\sqrt{x}-3,$</span> then <span class="math-container">$\sqrt{x}=u+3,$</span> and <span class="math-container">$dx=2\sqrt{x} du=2(u+3)du.$</span> Making these substitutions, <span class="math-container">$$\int\frac{\sqrt{x}}{\sqrt{x}-3}\, dx=\int\frac{u+3}{u}2(u+3)\, du=\int\frac{2(... |
3,280,095 | <p>Given the function
<span class="math-container">$$\int \frac{\sqrt{x}}{\sqrt{x}-3}dx $$</span>
You would need to have <span class="math-container">$u=\sqrt{x}-3$</span> and <span class="math-container">$du=\frac{1}{2 \sqrt{x}}$</span>, when I use a online calculator it suggests to rewrite the numerator as
<span clas... | Bernard | 202,857 | <p>Simple: 1) <span class="math-container">$\;\sqrt x=u+3$</span>, and 2) <span class="math-container">$\;\mathrm d u=\dfrac{\mathrm dx}{2\sqrt x}$</span>, so
<span class="math-container">$$\mathrm dx=2\sqrt x\,\mathrm d u=2(u+3)\,\mathrm d u, \quad\text{so }\quad\sqrt x\,\mathrm dx=2(u+3)^2\mathrm du.$$</span></p>
|
1,154,592 | <p>I was doing some basic Number Theory problems and came across this problem :</p>
<blockquote>
<p>Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime</p>
</blockquote>
<p><strong>My Solution : (Sloppy)</strong></p>
<blockquote>
<ul>
<li>$a^{n}-... | Agni Chakraborty | 521,925 | <p>Since $a$ and $n$ are both positive integers so $a>0$ and $n>0$
then taking $a=1$ putting in $a^n-1$ we get $1^n-1=0$ which is not prime hence take $a=2$ where $2^n-1$ is a prime if and only if $n$ is a prime</p>
|
1,089,078 | <p>Suppose we have a deck of cards, shuffled in a random configuration. We would like to find a $k$-bit code in which we explain the current order of the cards. This would be easy to do for $k=51 \cdot 6=306$, since we could encode our deck card-by-card, using $2$ bits for the coloring and $4$ bits for the number on ea... | Arthur | 15,500 | <p>Enumerate all the cards in the deck, say by value and suit in a simple manner, e.g. let diamonds be numbered 1 to 13, ace to king, then clubs 14 to 26, then hearts 27 to 39 and lastly spades 40 to 52.</p>
<p>As you encode cards one by one, decrease the value of every card above it. So the sequence $1,2,3,4$ means t... |
4,542,985 | <p>I want to fully understand the probabilistic interpretation. As in, I know once we have a probabilistic model, we differentiate for maximum likelihood and find the weights/regressors but what i really find difficult to grasp is how exactly are we developing a probabilistic model for linear regression.
I have see th... | Snoop | 915,356 | <p><span class="math-container">$(\subseteq)$</span>. Note <span class="math-container">$V$</span> is real valued. Also
<span class="math-container">$$\{\omega\in \Omega:V(\omega)\leq x\}=\bigcup_{k\leq \lfloor x\rfloor }B_k\in G,\,\quad \forall x\geq 1$$</span>
<span class="math-container">$$\{\omega\in \Omega:V(\omeg... |
4,050,893 | <p>Given a linear transformation <span class="math-container">$T: V \rightarrow W$</span> where <span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite dimensional, then is it true that nullity(<span class="math-container">$T$</span>) = nullity(<span class="math-container">$[T]_\b... | RobPratt | 683,666 | <p>You can formulate this as an integer linear programming problem, and here is a feasible solution:
<span class="math-container">\begin{align}
A&= \begin{pmatrix}
0 &0 &0 &1 \\
1 &0 &1 &0 \\
1 &1 &1 &1 \\
1 &0 &0 &1 \\
\end{pmatrix} \\
B&= \begin{pmatrix}
1 &... |
8 | <p>Contexts have backticks, which conflict with the normal way to enter inline code. How do I enter an inline context, since the initial approach:</p>
<pre><code>`System``
</code></pre>
<p>doesn't work ( `System`` ).</p>
| J. M.'s persistent exhaustion | 50 | <p>Backslashes before backticks seem to work too. For example,</p>
<pre><code>`?Global\`*`
</code></pre>
<p>in the comment below:</p>
|
8 | <p>Contexts have backticks, which conflict with the normal way to enter inline code. How do I enter an inline context, since the initial approach:</p>
<pre><code>`System``
</code></pre>
<p>doesn't work ( `System`` ).</p>
| Szabolcs | 12 | <p>This is so common in Mathematica that I suggest this should be specially included in the editor help or the FAQ on this site. Let me list the most common usages:</p>
<ul>
<li><p>Contexts, for example, <code>Global`</code>. </p>
<p>Markdown: <code>``Global` ``</code> (note the space before the closing <code>``</co... |
3,965,455 | <p>Find <span class="math-container">$E(X^3)$</span> given <span class="math-container">$X$</span> is in <span class="math-container">$Exp(2)$</span></p>
<p>My idea is that we can use <span class="math-container">$f_X(x)=2e^{-2x}$</span> and integrate <span class="math-container">$\int_{0}^{a}x^3f_X(x)dx$</span>. But f... | Community | -1 | <p><span class="math-container">$f(x) = \dfrac{10x}{x - 10}$</span><br><br>
Let <span class="math-container">$x - 10 = k.$</span> So, <span class="math-container">$x = 10+k$</span></p>
<p>Therefore, <span class="math-container">$\dfrac{10x}{x-10} = \dfrac{10(k+10)}{k} = 10 + \dfrac{100}{k}$</span><br><br></p>
<p>So... |
43,611 | <p>I posted this on Stack Exchange and got a lot of interest, but no answer.</p>
<p>A recent <a href="http://people.missouristate.edu/lesreid/POW12_0910.html" rel="nofollow">Missouri State problem</a> stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing i... | Jeremy West | 10,204 | <p>Start with the collection of half-open intervals of the form $[a,a+1) \times 0$ where $a \geq 0$ is an integer. This decomposes the positive $x$-axis into half-open intervals. Now, for every value of $0 < \theta < 2\pi$, decompose the ray whose angle with the positive $x$-axis is $\theta$ into half-open interv... |
4,031,476 | <p>I recently completed a variation of a problem I found from a mathematical olympiad which is as follows:</p>
<p>Prove that, for all <span class="math-container">$n \in \mathbb{Z}^+$</span>, <span class="math-container">$n \geq 1$</span>, <span class="math-container">$$\sum_{k=1}^n \frac{k}{2^k} < 2 $$</span></p>
<... | Brian M. Scott | 12,042 | <p>Note that no calculus is needed:</p>
<p><span class="math-container">$$\begin{align*}
\sum_{k=1}^n\frac{k}{2^k}&=\sum_{k=1}^n\sum_{\ell=1}^k\frac1{2^k}\\
&=\sum_{\ell=1}^n\sum_{k=\ell}^n\frac1{2^k}\\
&=\sum_{\ell=1}^n\frac{\frac1{2^{\ell}}-\frac1{2^{n+1}}}{1-\frac12}\\
&=\sum_{\ell=1}^n\left(\frac1{2... |
2,279,281 | <blockquote>
<p>$\displaystyle F(x)=\int_0^x (t-2)f(t)\; dt$ with $f(0)=1$, $f(1)=0$ has an extremum in $(0,3)$?</p>
</blockquote>
<p>The title explains a lot. Given
$$
\displaystyle F(x)=\int_0^x (t-2)f(t)\,dt
$$
with $f(0)=1$, $f(1)=0$ and $f:\mathbb{R}\to\mathbb{R}$ is a strictly decreasing differentiable functio... | Jay Zha | 379,853 | <p>$$F'(x)=(x-2)f(x)$$</p>
<p>$f$ is strictly decreasing and differentiable (thus continuous). So $f(x) > 0$ on $[0,1)$, $f(1)=0$, and $f(x) <0$ on $(1,3]$.</p>
<p>$F'(x)<0$ on $[0,1)$, $F'(1) =0$, $F'(x) >0$ on $(1,2)$, $F'(2)=0$, and $F'(x)<0$ on $(2,3]$</p>
<p>So $1$ is the minimum point, $2$ is th... |
2,279,281 | <blockquote>
<p>$\displaystyle F(x)=\int_0^x (t-2)f(t)\; dt$ with $f(0)=1$, $f(1)=0$ has an extremum in $(0,3)$?</p>
</blockquote>
<p>The title explains a lot. Given
$$
\displaystyle F(x)=\int_0^x (t-2)f(t)\,dt
$$
with $f(0)=1$, $f(1)=0$ and $f:\mathbb{R}\to\mathbb{R}$ is a strictly decreasing differentiable functio... | Kevin Limanta | 152,696 | <p>Since $F'(x) = (x-2)f(x)$, then if $F'(x) = 0$, it implies that either $x = 2$ or $f(x) = 0$. We also know that $f$ is strictly decreasing, so $f$ has at most one root. Given that $f(1) = 0$, then the only solution for $f(x) = 0$ is $x = 1$. Hence $F'(x) = 0$ implies that either $x = 1$ or $x = 2$.</p>
<p>Now, on $... |
3,358,449 | <blockquote>
<p>I have 8 variables; <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, <span class="math-container">$E$</span>, <span class="math-container">$F$</span>, <span class="math-container">$G$</span... | gandalf61 | 424,513 | <p>Some hints:</p>
<p>Since <span class="math-container">$8$</span> cannot occur on the left hand side of an equation, we know that either <span class="math-container">$A=8$</span> or <span class="math-container">$F=8$</span>.</p>
<p>From equations <span class="math-container">$1$</span> and <span class="math-contain... |
25,488 | <p>I have noticed a common pattern followed by many students in crisis:</p>
<ul>
<li>They experience a crisis or setback (injury, illness, tragedy, etc)</li>
<li>This causes them to miss a lot of class.</li>
<li>They may stay away from class longer than they "need to" because of shame: they feel that since t... | guest troll on a phone | 20,297 | <p>Policies differ a lot. My dad died December 1 of my second year at USNA. I got 3 days off and zero allowances for turning in final papers or doing final exams. My sister got sent home from a civilian school, total forgiveness on finals and assignments, and credit to move on. Both semester system. Neither one seemed... |
438,263 | <p>Is there a concrete example of a <span class="math-container">$4$</span> tensor <span class="math-container">$R_{ijkl}$</span> with the same symmetries as the Riemannian curvature tensor, i.e.
<span class="math-container">\begin{gather*}
R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} = R_{klij}, \\
R... | Sam Hopkins | 25,028 | <p>I am one of the moderators of the <a href="https://realopacblog.wordpress.com/" rel="noreferrer">Open Problems in Algebraic Combinatorics blog</a>.</p>
<p>First of all, we welcome submissions from anyone who has a good open problem in algebraic combinatorics that they want to advertise (see the top post for info abo... |
143,070 | <p>Suppose whole square and the left square in the diagram below are pullbacks, then we may wonder whether the right square is a pullback. It is usually not the case. </p>
<p><img src="https://i.stack.imgur.com/yhrcd.jpg" alt="square"></p>
<p>Now we seek some addition condition on $X\to Y$ that forces the right squar... | Peter LeFanu Lumsdaine | 2,273 | <p><strong>No, epic is not sufficient;</strong> there is a counterexample in <strong>Pos</strong>, the category of posets. Take <em>A</em>, <em>B</em>, and <em>X</em> each to be the discrete two-element poset $\{0,1\}$; take <em>C</em>, <em>Y</em>, and <em>Z</em> to be the same elements with their natural ordering. $\... |
754,583 | <p>Write <span class="math-container">$$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$</span> so that <span class="math-container">$\phi_n=n+\frac{n}{\phi_n},$</span> which gives <span class="math-container">$\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$</span> We know <span class="math-container">$\phi_1=\phi$</span... | mercio | 17,445 | <p>If $f_n(x) = n+n/x$, I'm not sure about your notation, but I think $(1)$ defines $\phi_n$ as the limit (provided it exists) of $f_n \circ f_n \circ \ldots \circ f_n (n)$.</p>
<p>Such a limit would be a fixpoint of $f_n$, which means one of the two complex numbers $y_n,z_n = \frac {n \pm \sqrt {n(n+4)}}2$.</p>
<p>T... |
36,774 | <p>Do asymmetric random walks also return to the origin infinitely?</p>
| Yuval Filmus | 1,277 | <p>Proof sketch: let $P(x,y)$ be the generating function of all walks which end up at the origin for the first time, with $x$ meaning left and $y$ meaning right. You can write a recurrence relation for the walks and deduce an expression for $P$ by solving a quadratic. Now substitute $pt$ for $x$ and $1-p$ for $y$.</p>
|
36,774 | <p>Do asymmetric random walks also return to the origin infinitely?</p>
| Charles | 1,778 | <p>No. Heuristic: If the walk goes right with probability $1/2+\alpha/2>1/2$ then the expected position after $n$ steps is $\alpha n,$ while the expected variation is only $O(\sqrt n).$ Thus the walk crosses the origin only finitely often.</p>
|
680,364 | <p>I want some verification for my proof to a homework problem. (Is it correct? Is there a simpler way to do this?)</p>
<p>Let $G$ be a finite group of odd order and suppose there is an element $g$ that is conjugate to its own inverse. In other words, there is $h \neq e$ such that $h^{-1}gh = g^{-1}$. We will show $g=... | zyy | 1,019,350 | <p>Assume there exists <span class="math-container">$x \neq 1$</span> in <span class="math-container">$G$</span> with <span class="math-container">$g^{-1}xg=x^{-1}$</span> for some <span class="math-container">$g \in G$</span>. Then <span class="math-container">$g^{-1}x^{m}g = x^{-m}$</span> for any <span class="math-c... |
3,062,597 | <p>For the statement of Rouché's theorem, I've always seen that both <span class="math-container">$f$</span> and <span class="math-container">$g$</span> have to be holomorphic on and inside a simple closed curve <span class="math-container">$ C $</span>. However, I am solving a problem which seems to suggest that I sho... | mathcounterexamples.net | 187,663 | <p>Yes that also looks strange to me. Rouché’s theorem hypothesis is to have a simply connected open subset <span class="math-container">$U\subseteq \mathbb C$</span> and a compact <span class="math-container">$K \subset U$</span> whose boundary is a closed simple curve positively oriented.</p>
|
134,407 | <p>Some shapes, such as the disk or the <a href="http://en.wikipedia.org/wiki/Reuleaux_triangle" rel="nofollow noreferrer">Releaux triangle</a> can be used as manholes,
that is, it is a curve of constant width.
(The width between two parallel tangents to the curve are independent of the orientation of the curve.)</p>
... | Włodzimierz Holsztyński | 8,385 | <p>(<strong>EDIT (remarks after the actual answer below)</strong> Technically word <em>cover</em> allows for arbitrary overlaps. However, despite the so familiar definition of notion <em>cover</em>, I was still under impression that the Question meant <em>holes</em> which had pairwise disjoint interiors. Thus m... |
150,809 | <p>An Iwasawa manifold is a compact quotient of a 3-dimensional complex Heisenberg group by a cocompact, discrete subgroup.
We can also refer to Griffiths and Harris's Principles of Algebraic Geometry p. 444 for simpler description.</p>
<p>I want to compute the automorphism of Iwasawa manifold,i.e.the group of biholo... | abx | 40,297 | <p>Here are some easy remarks to start with. The group $\mathrm{Aut}(X)$ is a complex Lie group, its Lie algebra is $H^0(X,T_X)$. Since $X=G/\Gamma $, the tangent bundle $T_X$ is trivial, so $H^0(X,T_X)$ has dimension 3, and is naturally identified with the Lie algebra of $G$; this implies that the neutral component of... |
4,620,319 | <p>Let's assume that for <span class="math-container">$0<\beta<\alpha<\frac{\pi}{2}$</span>, <span class="math-container">$\sin(\alpha+\beta) = \frac{4}{5}$</span>, and <span class="math-container">$\sin(\alpha-\beta) = \frac{3}{5}$</span>. Then, how could we find <span class="math-container">$\cot(\beta)$</sp... | mrtechtroid | 960,957 | <p><span class="math-container">$\displaystyle sin( a+b) =4/5\ \ \ sin( a-b) =3/5.$</span><br />
<span class="math-container">$\displaystyle So\ cos( a+b) \ =\ 3/5$</span> but that means <span class="math-container">$\displaystyle sin\left(\frac{\pi }{2} -( a+b)\right) =sin( a-b) =3/5$</span><br />
So <span class="math... |
2,977,645 | <p>I'm trying to prove that <span class="math-container">$\sqrt[n]{\frac{s}{t}}$</span> is irrational unless both s and t are perfect nth powers. I have found plenty of proofs for nth root of an integer but cannot find anything for rationals. Also trying to work up from the proofs I have found is rather difficult.</p>
... | Calum Gilhooley | 213,690 | <p>This is more of a comment than an answer, because it doesn't use the uniqueness of prime factorisation. Instead, it uses the more basic result that if a positive integer <span class="math-container">$c$</span> divides the product of two positive integers <span class="math-container">$a$</span> and <span class="math-... |
234,866 | <p>Problem: when I draw a rectangle and put a coloured edge around it, the displayed edge is centred along the nominal edge and if it follows the same course as one of the axes then it does not show up. For example:</p>
<pre><code>Graphics[{EdgeForm[Red], FaceForm[], Rectangle[{0, 0}, {4, 3}]}, Axes -> True]
</code>... | kglr | 125 | <p>We can mimic thick edges overlaying a white scaled rectangle over a colored rectangle:</p>
<pre><code>t = .1;
Graphics[Table[
{EdgeForm[], If[EvenQ[i + j], Red, Blue], Rectangle[{i, j}, 1 + {i, j}],
FaceForm[White], Scale[Rectangle[{i, j}, 1 + {i, j}], 1 - 2 t]},
{i, 7}, {j, 5}],
ImageSize -> Large]
<... |
2,423,569 | <p>I am asked to show that if $T(z) = \dfrac{az+b}{cz+d}$ is a mobius transformation such that $T(\mathbb{R})=\mathbb{R}$ and that $ad-bc=1$ then $a,b,c,d$ are all real numbers or they all are purely imaginary numbers. </p>
<p>So far I've tried multiplying by the conjugate of $cz+d$ numerator and denominator and see i... | Angina Seng | 436,618 | <p>Prove first that
$$T(z)=\frac{a'z+b'}{c'z+d'}$$
for real $a'$, $b'$, $c'$ and $d'$ and that then $a=\pm a'/\sqrt{a'd'-b'c'}$ etc.</p>
|
234,851 | <p>Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.</p>
<p>My attempt involves finding $\frac {dy}{dx}$ of that function first, which leaves me with a massive equation.</p>
<p>Next, I used this formula, </p>
<p>$$\int_1^2\sqrt{1+(\frac{dy}{dx})^2}$$</p>
<p>this attempt leave... | Community | -1 | <p>You are asked to find the length of the curve from $y=1$ to $y=2$ and $x$ is a function of $y$, so you need to use this equation (page 585 of 5th edition of Stewart's calculus):</p>
<p>$$L=\int_1^2\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)^2}\mathrm{d}y$$</p>
|
234,851 | <p>Find the length of the curve $x=0.5y\sqrt{y^2-1}-0.5\ln(y+\sqrt{y^2-1})$ from y=1 to y=2.</p>
<p>My attempt involves finding $\frac {dy}{dx}$ of that function first, which leaves me with a massive equation.</p>
<p>Next, I used this formula, </p>
<p>$$\int_1^2\sqrt{1+(\frac{dy}{dx})^2}$$</p>
<p>this attempt leave... | Dilawar | 1,674 | <p>Make the following substitution,
$ y = \sec w $ (since y is between 1 and 2, it should be fine) and change the limits of integration accordingly.</p>
<p>Use the trigonometric identities to simplify intermediate expressions. </p>
|
811,856 | <p>Let $X,Y,H$ be the standard base for the Lie algebra $\mathrm{sl}_2({\mathbb{C}})$, i.e. $H=\begin{pmatrix} 1 & 0\\ 0 &-1\end{pmatrix}$, $X=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $Y=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$. Let $V$ be some finite dimensional irreducible representatio... | Rene Schipperus | 149,912 | <p>There are two ways to see this one is the fact that the Jordan decomposition is preserved under representations of semisimple Lie algebras. The second is Weyl's unitary trick which is based on the fact that all representations of compact groups are completely reducible.
You can find more on both in Fulton and Harris... |
811,856 | <p>Let $X,Y,H$ be the standard base for the Lie algebra $\mathrm{sl}_2({\mathbb{C}})$, i.e. $H=\begin{pmatrix} 1 & 0\\ 0 &-1\end{pmatrix}$, $X=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$, $Y=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$. Let $V$ be some finite dimensional irreducible representatio... | Olivier Bégassat | 11,258 | <p>An elementary proof relies on the folllowing observation:</p>
<blockquote>
<p>Let $\lambda\in\Bbb C$ be an eigenvalue of $H$ acting on $V$ (<em>any</em> finite dimensional representation of $\mathfrak{sl}_2(\Bbb C)$). Then the following sum of eigenspaces of $H$
$$W=\bigoplus_{n\in\Bbb Z}E_{\lambda+2n}(H)$$
i... |
2,416,071 | <p>I have this integral: $\displaystyle \int^{\infty}_0 kx e^{-kx} dx$.</p>
<p>I tried integrating it by parts:</p>
<p>$\dfrac{1}{k}\displaystyle \int^{\infty}_0 kx e^{-kx} dx = ... $. But I'm stuck </p>
<p>now. Can you help me please?</p>
| DeepSea | 101,504 | <p><strong>hint:</strong> $\displaystyle \int kxe^{-kx}dx = -\displaystyle \int xd(e^{-kx})$</p>
|
572,276 | <p>I am trying to write the definition of greatest common divisor using symbolic notation. Here is my current attempt: </p>
<p>$d = gcd(m,n) \Leftrightarrow d \in Z \wedge max(d | m \wedge d | n)$</p>
<p>Any help or hints are greatly appreciated! Thanks!</p>
| amWhy | 9,003 | <p>Let's try to say what you are saying on the right-hand side: $$d = \gcd(m, n) \iff \Big((d\in \mathbb Z \land d\mid m \land d\mid n) \land \forall d' \in \mathbb Z\left((d'\mid m \land d'\mid n) \rightarrow d' \leq d\right)\Big)$$</p>
|
572,276 | <p>I am trying to write the definition of greatest common divisor using symbolic notation. Here is my current attempt: </p>
<p>$d = gcd(m,n) \Leftrightarrow d \in Z \wedge max(d | m \wedge d | n)$</p>
<p>Any help or hints are greatly appreciated! Thanks!</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>Here is an alternative definition: $\;\gcd(m,n)\;$ is the (unique) non-negative number that satisfies
$$
\langle \forall d \in \mathbb Z :: d \mid \gcd(m,n) \;\equiv\; d \mid m \:\land\: d \mid n \rangle
$$
See <a href="https://math.stackexchange.com/search?q=is%3Aanswer+user%3A11994+gcd">some answers of mine which ... |
4,247,268 | <p><strong>Q:</strong></p>
<blockquote>
<p>If <span class="math-container">$f\left(x\right)=-\frac{x\left|x\right|}{1+x^{2}}$</span> then find <span class="math-container">$f^{-1}\left(x\right)$</span></p>
</blockquote>
<p>My approach:</p>
<ol>
<li>Dividing the cases when <span class="math-container">$x\ge0$</span> and... | Alessio K | 702,692 | <p>The sign function is given by</p>
<p><span class="math-container">$$\operatorname{sgn}(x)=\begin{cases}-1, \space\text{if}\space x<0\\ 0, \space\space\text{if}\space x=0\\1, \space\text{if}\space x>0\end{cases}$$</span></p>
<p>and the modulus of <span class="math-container">$x$</span> is given by</p>
<p><span ... |
5,528 | <p>Let H be a subgroup of G. (We can assume G finite if it helps.) A complement of H in G is a subgroup K of G such that HK = G and |H∩K|=1. Equivalently, a complement is a transversal of H (a set containing one representative from each coset of H) that happens to be a group.</p>
<p>Contrary to my initial naive... | William DeMeo | 9,124 | <p>(This is a follow up question rather than an answer.) </p>
<p>Wouldn't it make more sense if the complement of a subgroup $H \leq G$ were defined to be a subgroup $K\leq G$ such that $H \cap K = 1$ and $\langle H, K \rangle = G$? That is, shouldn't we allow for the possibility that the set $HK$, consisting of all... |
2,118,931 | <p>If $A$ is a $4\times2$ matrix and $B$ is a $2\times 3$ matrix, what are the possible values of $\operatorname*{rank}(AB)$?</p>
<p>Construct examples of $A$ and $B$ exhibiting each possible value of $\operatorname*{rank}(AB)$ and explain your reasoning.</p>
| Fernando Revilla | 401,424 | <p><strong>Hint.</strong> Use that in general, $$\text{rank}(AB)\leq \min\{\text{rank }A, \text{rank }B\},$$ and in our case, $$0\le \text{rank }A \le 2,\quad 0\le \text{rank }B \le 2.$$</p>
|
3,886,523 | <p>Each of <span class="math-container">$3$</span> urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is ... | Théophile | 26,091 | <p>Any polygon can be decomposed into triangles, giving <span class="math-container">$A = A_1 + \cdots + A_n$</span>.</p>
<p>Thus,
<span class="math-container">$$\begin{align}
V &= V_1 + \cdots + V_n\\
&=\frac13A_1h + \cdots + \frac13A_nh\\
&=\frac13(A_1 + \cdots + A_n)h\\
&=\frac13Ah.
\end{align}$$</sp... |
1,482,644 | <p>I am trying to find the shortest equivalent expression of the following:</p>
<p>((C → D) $\wedge$ (D → C)) $↔$ (C $\wedge$ D ∨ ¬C $\wedge$ ¬D)</p>
<p>I have "simplified" the expression into the following:</p>
<p>(($\neg$C $\vee$ D) $\wedge$ ($\neg$D $\vee$ C)) $↔$ ((C $\wedge$ D) ∨ (¬C $\wedge$ ¬D))</p>
<p>I am ... | Jeevan Devaranjan | 220,567 | <p>Your simplification is correct. One key quality of the material conditional(if) in classical logic is that $C \rightarrow D$ is equivalent to $\lnot C \lor D$. You can read more about the material conditional <a href="https://en.wikipedia.org/wiki/Material_conditional" rel="nofollow">here</a>
Since your simplication... |
2,155,180 | <p>Let $f,g$ be analytic on some domain $\Omega \subset \mathbb{C}$. By Cauchy's formula, we have
$$
\frac{1}{2\pi i} \oint_{\partial\Omega}
\frac{f(z) \, g(z)}{z - z_0}
\, dz
=
f(z_0) \, g(z_0)
=
-\frac{1}{4\pi^2}
\oint_{\partial\Omega}
\frac{f(u)}{u - z_0}
\, du
\,
\oint_{\partial\Omega}
\frac{g(v)}{v - z_0}
\, d... | Mark Fischler | 150,362 | <p>For this specific case, of course it is easiest to notice that $20\equiv -3$ and $49 \equiv +3$, as in the comments.</p>
<p>In the general case, it is useful to first apply Euclid's algorithm on the multiplier of $x$ (here, $23$ and the modulus. So let's say we needed to solve $23x=2\pmod {79}$; we would have
$$
7... |
499,840 | <p>I have three points $A=(2,3), B=(6,4)$ and $C=(6,6).$
Given $\vec{AB}=\vec v$ and $\vec{BC}={0 \choose 2}$. I have also that for every $t\in [0,1]$ there is a point $D$ given as $\vec{AD}=t\vec{v}.$ </p>
<p>My question is determine $t$ such that the area of triangle $ADC$ equals area of the triangle $DBC$.</p>
<p... | njguliyev | 90,209 | <p>Yes, $t=\frac12$. Hint: The triangles $ADC$ and $DBC$ have the same altitude.</p>
|
343,281 | <p>Consider the following note written by Gerhard Gentzen in early 1932, on the onset of his work on a consistency proof for arithmetic:</p>
<blockquote>
<p>The axioms of arithmetic are obviously correct, and the principles of proof obviously preserve correctness. Why cannot one simply conclude consistency, i.e., w... | Noah Schweber | 8,133 | <p>Your question seems to boil down to <em>(after fixing an error)</em> the following:</p>
<blockquote>
<p>Any model <span class="math-container">$\mathfrak{M}$</span> of ACA<span class="math-container">$_0$</span> has a first-order part <span class="math-container">$Num(\mathfrak{M})$</span>, which satisfies PA; wh... |
1,119,634 | <p>Find the point on the curve $y=x^2+2$ where the tangent is parallel to the line $2x+y-1=0$</p>
<p>I understand the answer is $(-1,3)$ but I can't find a way to get there... Thanks </p>
| abel | 9,252 | <p>pick a point $(a, a^2 + 2)$ on the graph of $y = x^2 + 2.$ the tangent at this point has slope $2a$ so the tangent has the equation $y = a^2 + 2 + 2a(x-a) = 2ax + 2 - a^2$ if this line is to match the line $2x + y - 1 = 0$ then $a = -1$ and the point at which this line is tangent is $(-1, 3).$</p>
|
3,285,036 | <p>Obviously this cannot happen in a right rectangle, but otherwise - as Sin(0) or 180 or 360 equals 0, I guess there is no way to find out what the original angle was?</p>
| fleablood | 280,126 | <p>In general if <span class="math-container">$\sin x = M$</span> and <span class="math-container">$-180 < x \le 180$</span> degrees there are two possible answers to what <span class="math-container">$x$</span> is. </p>
<p>There is one <span class="math-container">$k$</span> where <span class="math-container">$0 ... |
2,083,460 | <p>While trying to answer <a href="https://stackoverflow.com/questions/41464753/generate-random-numbers-from-lognormal-distribution-in-python/41465013#41465013">this SO question</a> I got stuck on a messy bit of algebra: given</p>
<p>$$
\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)
$$</p>
<p>I n... | Arnaldo | 391,612 | <p>$$\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)=\log n+\log \biggl( 1 + \frac{v}{m^2}\biggr)^{3/2}=\log n\biggl( 1 + \frac{v}{m^2}\biggr)^{3/2}$$</p>
<p>Then</p>
<p>$$m=n\biggl( 1 + \frac{v}{m^2}\biggr)^{3/2} \rightarrow m^{2}=\frac{n^2(m^2+v)^3}{m^6}\rightarrow m^8=n^2(m^2+v)^3$$</p>
<p>I j... |
233,238 | <p>I am just practicing making some new designs with Mathematica and I thought of this recently. I want to make a tear drop shape (doesn't matter the orientation) constructed of mini cubes. I am familiar with the preliminary material, I am just having some difficulty getting it to work.</p>
| NonDairyNeutrino | 46,490 | <p>You were <em>really</em> close. You just need to add another underscore to <code>y_</code> as</p>
<pre><code>list = {
{Position, {Code}},
{1, {0000, 0001}},
{2, {0100, 0011}},
{3, {0110, 0111}},
{4, {1000, 1001}},
{5, {1100,1011}},
{6, {1110, 1111}}
};
list /. {x_, {y__}} :> {x, y}
</code></pre>
<b... |
257,567 | <p>Assuming <a href="http://www.springer.com/gp/book/9783642649059" rel="nofollow noreferrer">Bishop's</a> constructive mathematics, is it true that any real-valued square matrix with <strong>distinct</strong> roots of the characteristic polynomial can be diagonalized? By distinct, I mean <strong>apart</strong>: $x \ne... | Suvrit | 8,430 | <p>As suspected, the desired inequality actually holds for all hyperbolic polynomials; the inequality in the OP follows as corollary (Corollary 1) to Theorem 2 (which seems to be new). </p>
<p>We will need the following remarkable theorem to obtain our result.</p>
<blockquote>
<blockquote>
<p><strong>Theorem 1 ... |
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