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 |
|---|---|---|---|---|
383,063 | <p>I need some help with this exercise:</p>
<p>Suppose $A\subseteq{G}$ is abelian, and $|G:A|$ is a prime power. Show that $G'\lt{G}$</p>
<p>Thank you very much in advance.</p>
| 包殿斌 | 324,668 | <p>If we assume that <span class="math-container">$A$</span> is a normal abelian subgroup of <span class="math-container">$G$</span>. Then we can use the idea in the above comments,i.e., show that there is a nontrivial degree 1 representation of <span class="math-container">$G$</span>. </p>
<p>First we have <span clas... |
104,875 | <p>I've been looking for a solution to this problem for other applications too, for some time, but haven't come up with a solution that does not involve <code>Animate</code> or similar (and it never works).</p>
<p>Take this example:
plot a function (say <code>f=a/x</code>) for different <code>a</code>. The y-axis plot... | Edmund | 19,542 | <p>You can avoid the <code>Which</code> by dynamically setting <code>PlotRange</code> and <code>FrameStyle</code> as a function of <code>f</code></p>
<pre><code>Manipulate[
Plot[f[x], {x, 0, 10},
PlotRange -> {Automatic, {0, Max[axisStep Quotient[f[1], axisStep], 2]}},
Frame -> True,
FrameStyle ->
... |
204,150 | <p>If I had a list of let's say 20 elements, how could I split it into two separate lists that contain every other 5 elements of the initial list?</p>
<p>For example:</p>
<pre><code>list={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}
function[list]
(*
{1,2,3,4,5,11,12,13,14,15}
{6,7,8,9,10,16,17,18,19,20}
*)
</... | Coolwater | 9,754 | <p>You could use:</p>
<pre><code>function = Flatten[Transpose[Fold[Partition, #, {5, 2}]], {{1}, {2, 3}}] &;
</code></pre>
<p>If the length of the input is not a multiple of <code>10</code>, it will effectively cut off the remainder at the end.</p>
|
44,552 | <p>I was pushing my way through a physics book when the author separated the variables of the Schrödinger equation and I lost the plot:</p>
<p>$$\Psi (x, t) = \psi (x) T(t)$$</p>
<p>can someone please explain how this technique works and is used? It can be in general maths or in the context of this problem. Thanks </... | Ross Millikan | 1,827 | <p>Some functions (not all) $\psi (x,t)$ can be written as a product of a function of $x$ and another function of $t$. For example, $\psi (x,t)=xt$ can be, while $\psi_2 (x,t)=x^2+t^2$ cannot. The author is guessing that this will yield a solution to the problem and will go on to show that it does. After some manipu... |
2,480,528 | <blockquote>
<p>Find a formula for $\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)$ then prove it. </p>
</blockquote>
<p>I assumed that $\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)=\frac{2n}{2n-1}$ after doing a few cases from above then I tried to prove it with induction would this be a fair approach or ... | xpaul | 66,420 | <p>Let $z=x+yi$ and then from
$$ |1+z|=|1-i\bar{z}| $$
it is easy to obtain
$$ (x+1)^2+y^2=(-y+1)^2+x^2, $$
So
$$ x=-y $$
and hence $z=(1-i)x$.</p>
|
2,480,528 | <blockquote>
<p>Find a formula for $\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)$ then prove it. </p>
</blockquote>
<p>I assumed that $\prod_{i=1}^{2n-1} \left(1-\frac{(-1)^i}{i}\right)=\frac{2n}{2n-1}$ after doing a few cases from above then I tried to prove it with induction would this be a fair approach or ... | ivanculet | 493,187 | <p>Take this
$$|1+z|=|1-i\bar{z}|$$
then square this babies
$$(1+z)(1+\bar{z})=(1-i\bar{z})(1+iz)$$
$$1+z+\bar{z}+z\bar{z}=1-i\bar{z}+iz+z\bar{z} $$
so
$$ z+\bar{z}=i(z-\bar{z})$$
$$2Re(z)=i(i2Im(z)$$
therefore $$Re(z)=-Im(z)$$</p>
|
3,452,493 | <p>I remember hearing / reading about a scenario during WW2 where the US / Western Powers were thinking about attacking Japan via an overland route from India. The problem involved how to leapfrog the supplies from India over to China and there begin the fight with the Japanese. There’s a logic to it as the distance i... | user284331 | 284,331 | <p><span class="math-container">\begin{align*}
\left(1+\dfrac{a}{n}\right)^{n}\geq 1+n\cdot\dfrac{a}{n}=1+a>a,
\end{align*}</span>
so <span class="math-container">$1+a/n>a^{1/n}$</span>.</p>
<p><span class="math-container">\begin{align*}
\left(\dfrac{na}{1+na}\right)^{n}&=\left(\dfrac{1}{1+\dfrac{1}{na}}\rig... |
4,287,733 | <blockquote>
<p>How can we show <span class="math-container">$$\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $$</span> for all <span class="math-container">$n \in \mathbb{N}$</span>, <span class="math-container">$0 \leq c \leq x
\leq 1, c \neq 1$</span>?</p>
</blockquote>
<p>The context is <a href="ht... | MathematicsStudent1122 | 238,417 | <p>Using the idea proposed by <strong>Ivan Kaznacheyeu</strong>, we can generalize this to the following inequality:</p>
<blockquote>
<p>For any <span class="math-container">$n \in \mathbb{N}$</span>, suppose <span class="math-container">$(x_i)_{i=1}^{n} \in [0,1]^n$</span>, <span class="math-container">$(c_i)_{i=1}^{n... |
78,143 | <p>I don't know the meaning of geometrically injective morphism f of schemes. </p>
<p>What's the definition of "geometrically injective"?</p>
<p>I can't find it. I hope your answer.</p>
<p>Thanks.</p>
| Leo Alonso | 6,348 | <p>A map of schemes $f \colon X \to Y$ is <em>geometrically injective</em> if it is injective on <em>geometric points</em>, i.e. points with values in an algebraic closed field. In more detail, let $K$ be an algebraically closed field. For all pairs of maps ($K$-valued points) $x, y \colon \operatorname{Spec}(K) \to X$... |
2,800,015 | <p>Prove $p(x)=\frac{6}{(\pi x)^2}$ for $x=1,2,...$where $p$ is a probability function. and $E[X]$ doesn't exists.</p>
<p><b> My work </b></p>
<p>I know $\sum _{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$</p>
<p>Moreover,</p>
<p>$p(1)=\frac{6}{\pi^2}$<br>
$p(2)=\frac{6}{\pi^24}$<br>
$p(3)=\frac{6}{\pi^29}$<br>
$p(4)... | Ted Shifrin | 71,348 | <p><strong>HINT</strong>: What can you say about $\sum\limits_{n=1}^\infty n\cdot \dfrac 6{(\pi n)^2}$?</p>
|
3,831,387 | <p><span class="math-container">$X,Y\sim N(0,1)$</span> and are independent, consider <span class="math-container">$X+Y$</span> and <span class="math-container">$X-Y$</span>.</p>
<p>I can see why <span class="math-container">$X+Y$</span> and <span class="math-container">$X-Y$</span> are independent based on the fact th... | John Dawkins | 189,130 | <p>Intuitively, it's because the joint density of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is rotation invariant, and the transformation from <span class="math-container">$(X,Y)$</span> to <span class="math-container">$((X+Y)/\sqrt{2},(X-Y)/\sqrt{2})$</span> is a rotation. The... |
3,597,172 | <h2>The problem</h2>
<p>Let <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span></p>
<p>Determine <span class="math-container">$f(x)$</span> knowing that </p>
<p><span class="math-container">$ 3f(x) + 2 = 2f(\left \lfloor{x}\right \rfloor) + 2f(\{x\}) + 5x $</span>, where <span class="math-container">$... | José Carlos Santos | 446,262 | <p>Yes, this is the matrix of a surjective linear map. Look at the first and the fourth columns: <span class="math-container">$\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$</span> and <span class="math-container">$\left[\begin{smallmatrix}0\\1\end{smallmatrix}\right]$</span> respectively. It follows from this ... |
271,255 | <p>I'm reading <em>Mathematica Programming</em> by Leonid Shifrin. And it said</p>
<blockquote>
<p><code>ClearAll</code> serves to clear all definitions (including attributes) for a given symbol (or symbols), and not to clear definitions of all global symbols in the system (it is a common mistake to mix these two thing... | user293787 | 85,954 | <p>Let me rewrite the paragraph that you quote:</p>
<p><em>In <code>ClearAll</code> the "all" stands for "all definitions". It does not stand for "all symbols". Therefore calling <code>ClearAll[f]</code> means to "clear all definitions associated to the symbol <code>f</code>".</e... |
2,403,608 | <p>I was asked to solve for the <span class="math-container">$\theta$</span> shown in the figure below.</p>
<p><a href="https://i.stack.imgur.com/3Yxqv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Yxqv.png" alt="enter image description here" /></a></p>
<p>My work:</p>
<p>The <span class="math-con... | haqnatural | 247,767 | <p>Hope it will help,ask if it will not clear
<a href="https://i.stack.imgur.com/C8xli.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C8xli.png" alt="enter image description here"></a></p>
|
627,815 | <p>What is the best way to write 'exclusively divisible by' a given number in terms of set notation? eg: the set of natural numbers that are divisible by $2$ and only $2$; the set of natural numbers that are divisible by $3$ and only $3$; $\dots 5$, $7\dots$ etc.</p>
| Tom Collinge | 98,230 | <p>The natural numbers are an example of a “ Euclidean Ring”. A ring is almost a "field" but doesn't have a multiplicative inverse, and the Euclidean property as applied to natural numbers means that any number a can be written as a = q.b + r. The Euclidean property has a corresponding but slightly more complicated def... |
3,492,155 | <p>I am looking for the values <span class="math-container">$ x \in R $</span> which satisfy the following equation :</p>
<p><span class="math-container">$ e^{-\alpha x} = \frac{a}{x - c} $</span></p>
<p>Where <span class="math-container">$ \alpha $</span>, <span class="math-container">$ a $</span> and <span class="m... | Claude Leibovici | 82,404 | <p>Doing what @user721481 suggested, you should arrive at
<span class="math-container">$$x=c-\frac{W\left(-a \alpha e^{\alpha c}\right)}{\alpha }$$</span></p>
|
4,614,853 | <p>I'm trying to prove that the BinPacking problem is NP hard granted the partition problem <em>is NP hard</em>. If I have E a set of positive integers, can I split it into two subsets such that the sums of the integers in both subsets are equal?</p>
<p>The polynomial reduction I found would be the following:</p>
<ul>
... | Apass.Jack | 580,448 | <p>It is pleasantly surprising that you got a simple reduction from the partition problem to the BinPacking problem. Instead of being skeptic, you are supposed to celebrate your elegant proof.</p>
<p>Here is a little lemma for this occasion.</p>
<p>Let us represent <a href="https://en.wikipedia.org/wiki/Decision_proble... |
89,810 | <p>I have defined a table </p>
<pre><code>Table[Table[
Graphics3D[
Cuboid[radijDensity[[j, i]] {-Sin[kotiDensity[[j, i]]],
1 - Cos[kotiDensity[[j, i]]], 0}, {radijDensity[[j, i]]*
Sin[kotiDensity[[j, i]]],
radijDensity[[j, i]] (1 - Cos[kotiDensity[[j, i]]]) +
visina[[j, i + 1]], 0}]], {i... | Szabolcs | 12 | <blockquote>
<p>Something like ListPlot3D just that I want it to show those cuboids.</p>
</blockquote>
<p>If you need to place the same shape at multiple points either in 2D or 3D, the best solution is <a href="http://reference.wolfram.com/mathematica/ref/Translate.html" rel="nofollow noreferrer"><code>Translate</co... |
4,398,207 | <p>I have the following exercise:</p>
<blockquote>
<p>Prove if the functor that sends an abelian group to it's <span class="math-container">$n$</span>-torsion subgroup for <span class="math-container">$n\geq 2$</span> is exact.</p>
</blockquote>
<p>I know that I need to take <span class="math-container">$f\colon M\to N... | Andreas Blass | 48,510 | <p>Let <span class="math-container">$f:\mathbb Z\to\mathbb Z$</span> be the function <span class="math-container">$n\mapsto2n$</span>. Then the short exact sequence
<span class="math-container">$$
0\to\mathbb Z\overset{f}\to\mathbb Z\to\mathbb Z/2\to0
$$</span>
is sent by the "torsion part" functor to
<span c... |
2,110,561 | <p>so I want to find the volume of the body D defined as the region under a sphere with radius 1 with the center in (0, 0, 1) and above the cone given by $z = \sqrt{x^2+y^2}$. The answer should be $\pi$. A hint is included that you should use spherical coordinates. I've started by making a equation for the sphere, $x^2... | Kuifje | 273,220 | <p>In spherical coordinates:
$$
E=\{(\rho,\theta,\phi)|0\le \theta\le 2\pi, 0\le \phi \le \pi/4, 0\le \rho \le 2\cos \phi \}
$$
It follows that</p>
<p>$$
\boxed{
V= \iiint_E dV = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\cos\phi}\rho^2\sin\phi \;d\rho d\phi d\theta = \pi
}
$$</p>
<p><strong>Note</strong>: to find your bou... |
1,600,054 | <p>The graph of $y=x^x$ looks like this:</p>
<p><a href="https://i.stack.imgur.com/JdbSv.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/JdbSv.gif" alt="Graph of y=x^x."></a></p>
<p>As we can see, the graph has a minimum value at a turning point. According to WolframAlpha, this point is at $x=1/e$.</p>
<p>... | Narasimham | 95,860 | <p>Try to minimize logarithm of $ y= x^x,$ i.e., $y=x \, \log x$ </p>
<p>Its derivative is</p>
<p>$$ 1 + \log(x) $$</p>
<p>When equated to zero, it solves to that the minimum of $y(x)$ ocuurs at </p>
<p>$$ x= \dfrac{1}{e}= 0.36788$$</p>
<p>and the minimum value is</p>
<p>$$ y_{min}= \dfrac{1}{{e}^{\frac{1}{e}}} ... |
187,197 | <p>I have a logic expression:
<code>f0[a0_, a1_, a2_, a3_] := a0 And Not a1 And Not a2 And a3 Or Not a0 And a2 And a3 Or Not a0 And a1 And a3</code>, I know I should use <code>BooleanTable</code>, but it cannot generate a table like below.</p>
<p>How to generate a truth table in mathematica like below?</p>
<p><a hre... | Conor Cosnett | 36,681 | <pre><code>truthTableFormattor[rawData_] := Insert[Insert[
Grid[rawData /. {0 -> 0,
1 -> Item[1, Background -> Lighter[Magenta]]},
FrameStyle -> Gray,
Frame -> All], {Background -> {None, {GrayLevel[0.7], {White}}},
Dividers -> {Black, {2 -> Black}}, Frame -> True,
Spacings -> ... |
424,445 | <p>I'm studying Pattern recognition and statistics and almost every book I open on the subject I bump into the concept of <strong>Mahanalobis distance</strong>. The books give sort of intuitive explanations, but still not good enough ones for me to actually really understand what is going on. If someone would ask me "W... | Adnan Baysal | 953,976 | <p>I found <a href="https://www.machinelearningplus.com/statistics/mahalanobis-distance/" rel="nofollow noreferrer">this link</a> useful for understanding what Mahalanobis distance measures actually. The following image captures the essence very well: <a href="https://i.stack.imgur.com/LECVD.jpg" rel="nofollow noreferr... |
903,049 | <p>I have to find the series expansion and interval of convergence for the function $\ln(1 - x)$.</p>
<p>For the expansion, I have gone through the process and obtained the series:</p>
<p>$-x - (x^2/2) - (x^3/3) - . . . - (-1)^k((-x)^k)/k$</p>
<p>I know that the interval of convergence will be $(-1,1)$, but am havin... | amWhy | 9,003 | <p>Don't let the $(-1)^k$ or $(-x)^k = (-1)^kx^k$ trouble you. They have the effect of canceling each other out for odd $k$, and besides, for the ratio test, we apply it taking the absolute value of the general term $|a_k|$.</p>
<p>$$|a_k| = \frac{(x)^k }{k}$$</p>
<p>$$\frac{a_{k+1}}{a_k} = \frac{\frac{(x)^{k+1}}{k+1... |
2,502,711 | <p>If the cross ratio $Z_1, Z_2, Z_3$ and $Z_4$ is real, then</p>
<p>which of the following statement is true? </p>
<p>1)$Z_1, Z_2$ and $Z_3$ are collinear</p>
<p>2)$Z_1, Z_2$ and $Z_3$ are concyclic</p>
<p>3)$Z_1, Z_2$ and $Z_3$ are collinear when atleast one $Z_1, Z_2$ or $Z_3$ is real </p>
<p>My attempt : By ... | aleden | 468,742 | <p>$$\int_{0}^\infty e^{-azx}e^{-sx}dx=\int_{0}^\infty e^{-x(az+s)}dx=-\frac{e^{-x(az+s)}}{az+s}|_{0}^\infty=\frac{1}{az+s}$$</p>
|
1,840,159 | <blockquote>
<p>Question: Prove that a group of order 12 must have an element of order 2.</p>
</blockquote>
<p>I believe I've made great stride in my attempt.</p>
<p>By corollary to Lagrange's theorem, the order of any element $g$ in a group $G$ divides the order of a group $G$.</p>
<p>So, $ \left | g \right | \mi... | N. S. | 9,176 | <p><strong>Hint:</strong> Here is a simple proof idea that every group of even order must have an element of order $2$.</p>
<p>Pair every element in $G \backslash \{ e \}$ with its inverse. If all pairs consist of two different elements then $G \backslash \{ e \}$ would have an even number of elements. </p>
<p>What d... |
1,527,137 | <p>Usually one has the matrix and wishes to estimate the eigenvalues, but here it's the other way around: I have the positive eigenvalues of an unknown real positive definite matrix and I would like to say something about it's diagonal elements.</p>
<p>The only result I was able to find is that the sum of the eigenval... | Ilya | 5,887 | <p>For any fixed partition $a=x_0\leq\dots\leq x_n =b$ we have
$$
S^f(b) -S^f(a) + f(b) - f(a) \geq \sum_{i = 0}^{n-1} |f(x_{i+1}) - f(x_i)| + \sum_{i = 0}^{n-1} (f(x_{i+1}) - f(x_i)) \geq 0
$$
since $y + |y| \geq 0$ for all $y$. Hence, when you take $\sup$ over partitions, the inequality still holds.</p>
|
104,297 | <p>How would I go about solving</p>
<p>$(1+i)^n = (1+\sqrt{3}i)^m$ for integer $m$ and $n$?</p>
<p>I have tried </p>
<pre><code>Solve[(1+I)^n == (1+Sqrt[3] I)^m && n ∈ Integers && m ∈ Integers, {n, m}]
</code></pre>
<p>but this does not give the answer in the 'correct' form.</p>
| Eric Towers | 16,237 | <p>One can get a truly ridiculous solution by <code>ComplexExpand[]</code>ing the real and imaginary parts:</p>
<pre><code>Reduce[ComplexExpand[{Re /@ #, Im /@ #}] &[
(1 + I)^n == (1 + Sqrt[3] I)^m],
{m, n}, Integers]
(* ... an astonishing mess involving 14 integer parameters ... *)
</code></pre>
<p>Howev... |
705,829 | <p>I'm trying to solve a problem here.</p>
<p>It says: "Prove that a triangle is isoceles if $\large b=2a\sin\left(\frac{\beta}{2}\right)$."
$B-\beta$
I've tried to prove it but I can't</p>
<p>Can anyone help me?</p>
| sirfoga | 83,083 | <p><strong>Hint</strong> : for the law of sines you have $$ \frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{2a\sin\left(\frac{\beta}{2}\right)}{\sin\beta} \Rightarrow ...$$ </p>
|
3,226,320 | <p>I have some data points that need to be fit to the curve defined by</p>
<p><span class="math-container">$$y(x)=\frac{k}{(x+a)^2} - b$$</span></p>
<p>I have considered that it can be done by the least squares method. However, the analytical solution gives me a negative <span class="math-container">$a$</span>, so it... | JJacquelin | 108,514 | <p>I cannot see any difficulty with the your data points x = {0, 1, 2, 3}, y = {-23, -32, -38, -40}.</p>
<p>With least squares fitting my result is shown on the figure below. The computed value of <span class="math-container">$a$</span> is positive as expected.</p>
<p>If you obtain a negative <span class="math-conta... |
3,226,320 | <p>I have some data points that need to be fit to the curve defined by</p>
<p><span class="math-container">$$y(x)=\frac{k}{(x+a)^2} - b$$</span></p>
<p>I have considered that it can be done by the least squares method. However, the analytical solution gives me a negative <span class="math-container">$a$</span>, so it... | JJacquelin | 108,514 | <p>Supposing that the OP is looking for a conventional method of regression, the present answer would be not convenient. This is why I post it as a distinct answer.</p>
<p>The calculus below is ultra simple since there is no iteration and no need for initial guessed values. </p>
<p><a href="https://i.stack.imgur.com/... |
4,181,524 | <p><span class="math-container">$$\text{I need to prove the following lemma : }\frac{\zeta'(s)}{\zeta(s)} = - \sum_{n=1}^{\infty} \frac{\Lambda(n)}{n^s}$$</span></p>
<p><strong>My attempt:</strong></p>
<p><span class="math-container">$$\text{We know that }\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} $$</span><span clas... | Kenta S | 404,616 | <p>HINT:<span class="math-container">$$\frac{\zeta'(s)}{\zeta(s)}=\frac{d}{ds}\ln\zeta(s)=-\sum_p\frac d{ds}\ln(1-p^{-s}).$$</span></p>
|
4,539,637 | <blockquote>
<p>If the digits <span class="math-container">$7,7,3,2$</span>, and 1 are randomly arranged from left to right, what is the probability both of the 7 digits are to the left of the 1 digit?</p>
</blockquote>
<p>The answer is <span class="math-container">$1/3$</span> because <span class="math-container">$1 7... | user97357329 | 630,243 | <p>You may find a simple solution in <strong>(Almost) Impossible Integrals, Sums, and Series</strong>, Sect. <strong>6.60</strong>, page <span class="math-container">$533$</span>, showing that
<span class="math-container">$$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{\Gamma(i) \Gamma(j) \Gamma(x)}{\Gamma(i+j+x)}=\frac{1}... |
89,000 | <p>Let $f:I \rightarrow \mathbb{R}$, where $I\subset \mathbb{R}$ is an interval, be midconvex, that is
$$f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$$ for all $x,y \in I$.
Assume that for some $x_0, y_0 \in \mathbb{R}$ such that $x_0 < y_0$ holds equality
$$f\left(\frac{x_0+y_0}{2}\right)= \frac{f(x_0)+f(... | Robert Israel | 8,508 | <p>Consider the following example. Let $B$ be a Hamel basis of $\mathbb R$ over the rationals $\mathbb Q$. Thus each $x \in \mathbb R$ can be written uniquely as
$x = \sum_{b \in B} c_b(x) b$, where all but finitely many of the $c_b(x)$ are $0$ for any particular $x$. Let $f(x) = c_{b_0}(x)^2$ for some given $b_0 \i... |
3,571,047 | <p>Here's what I have so far:</p>
<p><span class="math-container">$$\frac{\partial f}{\partial y}|_{(a,b)} = \lim\limits_{t\to 0} \frac{\sin(a^2 + b^2 + 2tb + t^2) - \sin(a^2 + b^2)}{t} = \lim\limits_{t\to 0} \frac{\sin(a^2 + b^2)[\cos(2tb + t^2) - 1] + \cos(a^2 + b^2)\sin(2tb + t^2)}{t}$$</span>
I can see that side l... | Hadi | 645,692 | <p>We can play around with the substitution choice to ensure that the integral is expressed purely in terms of <span class="math-container">$t$</span>.</p>
<p>From the substitution choice we infer:
<span class="math-container">$$t=\sqrt{x}+1$$</span>
<span class="math-container">$$\implies t-1=\sqrt{x}$$</span>
<span... |
3,001,747 | <p>I'd need some help evaluating this limit:</p>
<p><span class="math-container">$$\lim_{x \to 0} \frac{\ln\sin mx}{\ln \sin x}$$</span></p>
<p>I know it's supposed to equal 1 but I'm not sure how to get there.</p>
| J.G. | 56,861 | <p>You want <span class="math-container">$1+\lim_{x\to 0}\dfrac{\ln\frac{\sin mx}{\sin x}}{\ln\sin x}$</span>. The numerator <span class="math-container">$\to \ln m$</span>, the denominator <span class="math-container">$\to -\infty$</span>. The result is therefore <span class="math-container">$1+0=1$</span>.</p>
|
2,703,639 | <p>a) $f: L^1(0,3) \rightarrow \mathbb{R}$</p>
<p>b) $f: C[0,3] \rightarrow \mathbb{R}$</p>
<p>for part a I got $\|f\| = 1$ because $\|f(x)\|=|\int_0^2x(t)dt| \leq \int_0^2|x(t)|dt \leq\int_0^3|x(t)|dt = \|x(t)\|_1$ so $\|f\|=1$</p>
<p>for b, I think its similar: $\|f(x)\|=|\int_0^2x(t)dt| \leq \int_0^2|x(t)|dt \le... | angryavian | 43,949 | <p>For part a), you have only shown $\|f\| \le 1$. But it should not be hard to find some $x$ such that $|f(x)| = \|x\|$.</p>
<p>For part b), note $|f(x)| \le \int_0^2 |x(t)|\,dt \le 2 \|x\|_{\max}$, so $\|f\| \le 2$. If you take some $x \in C[0,3]$ with $x(t)=1$ for $t \in [0,2]$ and $\|x\|_{\max}=1$, then $|f(x)|=2=... |
57,213 | <p>Let <span class="math-container">$A \in \mathbb{Q}^{6 \times 6}$</span> be the block matrix below:</p>
<p><span class="math-container">$$A=\left(\begin{array}{rrrr|rr}
-3 &3 &2 &2 & 0 & 0\\
-1 &0 &1 &1 & 0 & 0\\
-1&0 &0 &1 & 0 & 0\\
-4&6 ... | Gerry Myerson | 8,269 | <p>Expanding on the comment: </p>
<p>If $A$ has eigenvalue $\lambda$, then $f(A)$ has eigenvalue $f(\lambda)$. So $f(A)$ is not invertible if $f(\lambda)=0$. </p>
|
2,372,171 | <p>Let $T$ be a bounded linear operator on a Hilbert space $H$. I have to show that the following are equivalent:</p>
<p>(i) $T$ is unitary</p>
<p>(ii) For every orthonormal basis $\{u_{\alpha}:\alpha\in \Lambda\}$, $\{T(u_{\alpha}):\alpha\in \Lambda\}$ is an orthonormal basis.</p>
<p>(iii) For some orthonormal basi... | Prahlad Vaidyanathan | 89,789 | <p>For $(iii) \Rightarrow (i)$, you want to show that
$$
\langle Tx,Ty\rangle = \langle x,y\rangle \qquad (\ast)
$$
for all $x,y\in H$. First note that $(\ast)$ is true if $x,y\in S := \{u_{\alpha}\}$. By sesqui-linearity, $(\ast)$ is true if $x,y\in \text{span}(S)$. However, $\text{span}(S)$ is dense in $H$, so if $x\... |
283,747 | <p>Let $BG$ denote the classifying space of a finite group $G$. For which group cohomology classes $c\in H^2(G;\mathbb{Z}/2)$ does there exist a real vector bundle $E$ over $BG$ such that $w_2(E)=c$?</p>
| Mark Grant | 8,103 | <p>You can set this up as an obstruction theory problem. Your cohomology class is represented by a map $c: BG\to K(\mathbb{Z}/2,2)$, and the question is whether this map lifts through the universal Stiefel-Whitney class $w_2:BO\to K(\mathbb{Z}/2,2)$. </p>
<p>The primary obstruction to such a lift turns out to be the c... |
2,647,123 | <p>I'm asked to to find a $3\times3$ matrix, in which no entry is $0$ but $A^2=0$. </p>
<p>The problem is if I I brute force it, I am left with a system of 6 equations (Not all of which are linear...) and 6 unknowns. Whilst I could in theory solve that, is there more intuitive way of solving this problem or am I going... | M. Winter | 415,941 | <p>Here is a geometric approach. Think about a matrix $P$ which orthogonally projects all of $\Bbb R^3$ onto a one-dimensional subspace spanned by a vector $n$:</p>
<p>$$P_{ij}=n_i n_j$$</p>
<p>After projection, rotate your space by $90^\circ$ around some axes $v$ orthogonal to $n$ by using a rotation matrix $R$. You... |
1,441,349 | <p>Given is that $a∈ℤ_n^*$ and $d|ord(a)$.</p>
<p>I need to show that $ord(a^d)= ord(a)/d$.</p>
<p>I started with the following:</p>
<p>$ord(a^d) = e$, such that $(a^d)^e\equiv 1\pmod n$</p>
<p>$ord(a)/d =f/d$ where $ord(a)=f$, such that $a^f\equiv 1\pmod n$</p>
<p>Now I want to prove that $e=f/d$. </p>
<p>I have... | Noah Schweber | 28,111 | <p>I think you're asking, "When do we have '$\emptyset\models\psi$'?"</p>
<p>If this is the case, the answer is: only when $\psi$ is a tautology. By definition, $\emptyset\models\psi$ iff every valuation making every formula in $\emptyset$ true, makes $\psi$ true. However, every valuation <em>at all</em> makes every f... |
366,654 | <p>Find all values of real number p or which the series converges:</p>
<p>$$\sum \limits_{k=2}^{\infty} \frac{1}{\sqrt{k} (k^{p} - 1)}$$ </p>
<p>I tried using the root test and the ratio test, but I got stuck on both. </p>
| preferred_anon | 27,150 | <p>$$\frac{dy}{dx}+\frac{y}{x-2}=5(x-2)\sqrt{y}$$
Multiply through by $\frac{1}{2}\sqrt \frac{x-2}{y}$ to get
$$\frac{\sqrt{x-2}}{2\sqrt{y}}\frac{dy}{dx}+\frac{\sqrt{y}}{2\sqrt{x-2}}=\frac{5}{2}(x-2)^{3/2}$$
The LHS is the derivative of $\sqrt{x-2}\sqrt{y}$, so we can integrate:
$$\sqrt{x-2}\sqrt{y}=\int\frac{5}{2}(x-2... |
2,218,716 | <p>Explain why 1/i is − i. </p>
<p>(That is: explain why the multiplicative inverse of i is the complex number − i.)</p>
<p>And then the hint that I was given was, what property defines the multiplicative inverse?</p>
<p>I know how to algebraically prove 1/i = -i, but need help writing the proof.</p>
| Stella Biderman | 123,230 | <p>By the definition of multiplicative inverse, $1/i$ satisfies $(1/i)i=1$. Presumably you know that $i^2=-1$. Then you have that $$(-i)i=-i^2=1=(1/i)i$$ By cancellation, $-i=1/i$.</p>
|
3,249,735 | <p>My question relates to this problem:</p>
<p>Prove by induction that 54 divides <span class="math-container">$2^{2k+1}-9k^2+3k-2$</span>. </p>
<p>My solving so far gives this answer: (after all calculations)</p>
<p><span class="math-container">$2^{2(k+1)+1}-9(k+1)^2+3(k+1)-2= 54 \cdot2^2k+27k^2-27k$</span></p>
<p... | Deepak | 151,732 | <p><span class="math-container">$27k^2-27k = 27k(k-1)$</span>.</p>
<p>Exactly one of <span class="math-container">$k$</span> and <span class="math-container">$k-1$</span> is even, so... </p>
|
819,830 | <p>Is the idea of a proof by contradiction to prove that the desired conclusion is both true and false or can it be any derived statement that is true and false (not necessarily relating to the conclusion)? Or can it simply be an absurdity that you know is false but through your derivation comes out true?</p>
| Kaj Hansen | 138,538 | <p>If revolving around the $x$-axis makes more sense to you, then we can consider revolving the inverse function of $f(x) = \sqrt{x}$ around the $x$-axis with the appropriate bounds.</p>
<p>Of course, $f^{-1}(x) = x^2$, and so revolving $y = x^2$ around the $x$-axis from $x = 0$ to $x = 2$ will yield the same result. ... |
3,579,346 | <p>I've been learning some introductory analysis on manifolds and have had a small issue ever since the notion of tangent spaces at points on a differentiable manifold was introduced.</p>
<p>In our lectures, we began with the definition using equivalence classes of curves. But it is also possible to define tangent spa... | painday | 433,808 | <p>I think what OP wants is a definition using the concept of functors. Below is the repost of my answer <a href="https://mathoverflow.net/q/429683">https://mathoverflow.net/q/429683</a> to the question "<strong>An easy way to to explain the equivalence definitions of tangent spaces?</strong>"</p>
<hr />
<p><... |
611,788 | <p>I'm here to ask you guys if my logic is correct.
I have to calculate limit of this:
$$\lim_{n\rightarrow\infty}\sqrt[n]{\sum_{k=1}^n (k^{999} + \frac{1}{\sqrt k})}$$
At first point. I see it's some limit of $$\lim_{n\rightarrow\infty}\sqrt[n]{1^{999} + \frac{1}{\sqrt 1} + 2^{999} + \frac{1}{\sqrt 2} \dots + n^{99... | hmakholm left over Monica | 14,366 | <p>Yes -- in fact the 3-sphere can be <em>completely</em> covered by <em>disjoint</em> great circles, the <a href="http://en.wikipedia.org/wiki/Hopf_fibration">Hopf fibration</a>.</p>
<p>(Terminology nitpick: The subset $\{ (x,y,z,w)\in\mathbb R^4\mid x^2+y^2+z^2+w^2=1\}$ is usually known as the <strong>three</strong>... |
611,788 | <p>I'm here to ask you guys if my logic is correct.
I have to calculate limit of this:
$$\lim_{n\rightarrow\infty}\sqrt[n]{\sum_{k=1}^n (k^{999} + \frac{1}{\sqrt k})}$$
At first point. I see it's some limit of $$\lim_{n\rightarrow\infty}\sqrt[n]{1^{999} + \frac{1}{\sqrt 1} + 2^{999} + \frac{1}{\sqrt 2} \dots + n^{99... | Jeremy Daniel | 115,164 | <p>Firstly, you should speak of the $3$-dimensional sphere, viewed in the $4$-dimensional euclidian space. </p>
<p>Secondly, the term equator in this situation should certainly refer to the intersection in $\mathbb{R}^4$ of the sphere $S^3$ and a hyperplane, that is a $2$-dimensional object.</p>
<p>For your question,... |
2,361,336 | <p>Note: This is <strong>not a duplicate</strong> as I am asking for a proof, not a criteria, and this is a specific proof, not just any proof – <strong>please treat like any other question on a specific math problem.</strong> Please do not close. thanks!</p>
<p><a href="https://i.stack.imgur.com/5R2aE.png" rel="nofol... | Bernard | 202,857 | <p>We don't really need induction here (or, at a pinch, finite induction).</p>
<p>Multiplying by $(x-x_0)^{n-k}$, we deduce that $\dfrac{Q(x)}{(x-x_0)^k}\to0$ for all $k=0,1,\dots n$.</p>
<p>In particular, $Q(0)=0$, whence $a_0=0$.</p>
<p>Next $\dfrac{Q(x)}{x-x_0}=a_1+a_2(x-x_0)+\dots+a_n(x-x_0)^{n-1}\to 0$, so $a_... |
4,248,766 | <p>How many functions <span class="math-container">$f: \{1,...,n_1\} \to \{1,...,n_2\}$</span> are there such that if <span class="math-container">$f(k)=f(l)$</span> for some <span class="math-container">$k,l \in \{1,...,n_1\}$</span>, then <span class="math-container">$k=l$</span>?</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Assuming that <span class="math-container">$n_2\ge n_1$</span> we have <span class="math-container">$$n_1!\binom{n_2}{n_1}$$</span> functions.</p>
<p>Note that <span class="math-container">$f(1)$</span> has <span class="math-container">$n_2$</span> choices and <span class="math-container">$f(2)$</span> has <span cla... |
4,639,011 | <p>I'm interested in finding the asymptotic at <span class="math-container">$n\to\infty$</span> of
<span class="math-container">$$b_n:= \frac{e^{-n}}{(n-1)!}\int_0^\infty\prod_{k=1}^{n-1}(x+k)\,e^{-x}dx=e^{-n}\int_0^\infty\frac{e^{-x}}{x\,B(n;x)}dx$$</span>
Using a consecutive application of Laplace' method, I managed ... | Gary | 83,800 | <p><strong>First approach.</strong> We have
<span class="math-container">\begin{align*}
b_n & = \frac{{{\rm e}^{ - n} }}{{\Gamma (n)}}\int_0^{ + \infty } {\frac{{\Gamma (x + n)}}{{\Gamma (x + 1)}}{\rm e}^{ - x} {\rm d}x}
\\ & = \frac{{{\rm e}^{ - n} }}{{\Gamma (n)}}\int_0^{ + \infty } {\frac{1}{{\Gamma (x + 1)... |
2,589,837 | <p>Let $\Omega \subset \mathbb R^d$ smooth, bounded and connected domain. Let $A\in \mathbb R^{d\times d}$ symetric and uniformly elliptic, i.e. there is $C>0$ such that $$C^{-1}\|x\|^2\leq Ax\cdot x\leq C\|x\|^2.$$</p>
<p>How can I prove that $$a(u,v)=\int_\Omega A \nabla u\cdot \nabla v,$$
continuous ? I know tha... | gerw | 58,577 | <p>If $A$ is a symmetric matrix, then $Ax \cdot x \le C \|x\|^2$ implies that all eigenvalues of $A$ are bounded from above by $C$. Hence, $\|A x\| \le C \|x\|$.</p>
|
2,589,837 | <p>Let $\Omega \subset \mathbb R^d$ smooth, bounded and connected domain. Let $A\in \mathbb R^{d\times d}$ symetric and uniformly elliptic, i.e. there is $C>0$ such that $$C^{-1}\|x\|^2\leq Ax\cdot x\leq C\|x\|^2.$$</p>
<p>How can I prove that $$a(u,v)=\int_\Omega A \nabla u\cdot \nabla v,$$
continuous ? I know tha... | Community | -1 | <p>Note that for a symmetric matrix A we have </p>
<p>$$ \|A\| =\sup_{\|x\|_2=1}\{\|Ax\|_2\}= \sup_{\|x\|_2=1}\{|\langle Ax,x\rangle_2|\}$$
Therefore, since for all $x\in \Bbb R^d$
$$|Ax\cdot x| =\left|\sum_{i,j=1}^{d}A_{ij}x_ix_j \right|\le C\|x\|_2$$</p>
<p>We automatically for all $x\in \Bbb R^d$ we get</p>
<p>... |
119,876 | <pre><code>Module[{x},
f@x_ = x;
p@x_ := x;
{x, x_, x_ -> x, x_ :> x}
]
?f
?p
</code></pre>
<p>gives</p>
<pre><code>{x$17312, x$17312_, x_ -> x, x_ :> x}
f[x_]=x
p[x_]:=x
</code></pre>
<p>but I'd like to get</p>
<pre><code>{x$17312, x$17312_, x$17312_ -> x$17312, x$17312_ :> x$17312}
f[x$17312... | masterxilo | 6,804 | <p>I think a third kind of behaviour would also be possible: it is debatable whether <code>f@x_ = x, x_ -> x</code> within <code>Module[{x}, ...</code> should not become <code>f@x_ = x$123, x_ -> x$123</code> because then the sequence</p>
<pre><code>ClearAll[Global`x];
x = 1;
0 /. x_ -> x
</code></pre>
<p>wo... |
2,971,143 | <p>Let me choose <span class="math-container">$n=1$</span> for my induction basis: <span class="math-container">$2 > 1$</span>, true.</p>
<p>Induction Step : <span class="math-container">$2^n > n^2 \rightarrow 2^{n+1} > (n+1)^2 $</span></p>
<p><span class="math-container">$2^{n+1} > (n+1)^2 \iff$</span></... | Mohammad Riazi-Kermani | 514,496 | <p>Obviously your inequality is not true for <span class="math-container">$n=4$</span> </p>
<p>Thus you better start at <span class="math-container">$n\ge 5 $</span> which is proved the same way. </p>
|
2,971,143 | <p>Let me choose <span class="math-container">$n=1$</span> for my induction basis: <span class="math-container">$2 > 1$</span>, true.</p>
<p>Induction Step : <span class="math-container">$2^n > n^2 \rightarrow 2^{n+1} > (n+1)^2 $</span></p>
<p><span class="math-container">$2^{n+1} > (n+1)^2 \iff$</span></... | Kolja | 386,635 | <p>You chose <span class="math-container">$n=1$</span> as induction base, but the induction step works only for <span class="math-container">$n\geq 3$</span>, i.e. you showed that <span class="math-container">$2^n > n^2$</span> implies <span class="math-container">$2^{n+1}>(n+1)^2$</span> only when <span class="m... |
2,179,317 | <p>We know that, if $\mathcal D$ is a domain containing the origin $(0,0,0)$, then</p>
<p>$$\int_{\mathcal D} \delta(\vec r) d \vec r= \int_{\mathcal D} \delta(x) \delta(y) \delta(z) dx dy dz=1$$</p>
<p>However, <a href="http://mathworld.wolfram.com/DeltaFunction.html" rel="nofollow noreferrer">we also know that</a> ... | Ruslan | 64,206 | <p>First, your integral identity doesn't make any sense. An example of an asymmetric nascent delta function, which will have another result is</p>
<p>$$\delta_s(x)=\frac1{s\sqrt\pi}\exp\left(-\frac{(x-s)^2}{s^2}\right).$$</p>
<p>For it we'll have</p>
<p>$$\int\limits_0^\infty \delta_s(x)\,dx=\frac12+\frac{\operatorn... |
4,220,972 | <p>I'm studying a for the GRE and a practice test problem is, "For all real numbers x and y, if x#y=x(x-y), then x#(x#y) =?</p>
<p>I do not know what the # sign means. This is apparently an algebra function but I cannot find any such in several searches. I'm an older student and haven't had basic algebra in over 4... | zd_ | 381,421 | <p>By Chebyshev's inequality,
<span class="math-container">$$
\mathbb{P}(|X| > k) \leq \frac{\mathbb{E}(|X|^p)}{k^p},
$$</span>
for any <span class="math-container">$p>0$</span>.</p>
<p>Since we have <span class="math-container">$\mathbb{E}(X^2)$</span>,
<span class="math-container">$$
\mathbb{P}(|X| > k) \leq... |
1,179,497 | <p>Let $(F,+,\cdot)$ be a field. </p>
<p>Then to prove that $(F,+)$ and $(F-\{0\},\cdot)$ are not isomorphic as groups.</p>
<p>I am facing difficulty in finding the map to bring a contradiction!!</p>
| quid | 85,306 | <p>Show: </p>
<ul>
<li><p>if the additive group contains an element of order $2$ then the multiplicative does not. </p></li>
<li><p>if the additive group contains no element of order $2$ then the multiplicative does. </p></li>
</ul>
|
2,111,402 | <p>Simple exercise 6.2 in Hammack's Book of Proof. "Use proof by contradiction to prove"</p>
<p>"Suppose $n$ is an integer. If $n^2$ is odd, then $n$ is odd"</p>
<p>So my approach was:</p>
<p>Suppose instead, IF $n^2$ is odd THEN $n$ is even</p>
<p>Alternatively, then you have the contrapositive, IF $n$ is not even... | jupiterd | 409,586 | <p>Your proof looks correct to me, but I would like to share with you my strategy for proof by contradiction.</p>
<p>Consider the if/then statement $p\Rightarrow q$. In your case, $p$ represents "$n^{2}$ is odd" and $q$ represents "$n$ is odd". To achieve the proof by contradiction, we want to show that when $p$ is tr... |
1,982,102 | <p>If I wanted to figure out for example, how many tutorial exercises I completed today.</p>
<p>And the first question I do is <strong>question $45$</strong>, </p>
<p>And the last question I do is <strong>question $55$</strong></p>
<p>If I do $55-45$ I get $10$.</p>
<p>But I have actually done $11$ questions:<br>
$... | dxiv | 291,201 | <p>It depends on whether the range includes or not its endpoints. For example, let $n \in \mathbb{N}$. Then:</p>
<ul>
<li>$45 \lt n \lt 55\quad$ has $\;55-45-1=9$ solutions;</li>
<li>$45 \le n \lt 55\;$ and $\;45 \lt n \le 55\quad$ both have has $\;55-45=10$ solutions;</li>
<li>$45 \le n \le 55\quad$ has $\;55-45+1=11... |
1,982,102 | <p>If I wanted to figure out for example, how many tutorial exercises I completed today.</p>
<p>And the first question I do is <strong>question $45$</strong>, </p>
<p>And the last question I do is <strong>question $55$</strong></p>
<p>If I do $55-45$ I get $10$.</p>
<p>But I have actually done $11$ questions:<br>
$... | Deusovi | 256,930 | <p>If you want to use pure subtraction, here's how you'd do it:</p>
<p>At the beginning, you were at the start of problem <strong>45</strong>.</p>
<p>At the end, you were at the end of problem 55, <em>which is the start of problem <strong>56</em></strong>.</p>
<p>And $56-45=11$. You finished $11$ problems.</p>
<hr>... |
2,781,827 | <p>I need to find a symmetric matrix of real values (not the zero matrix) of any order that is orthogonal to any diagonal matrix of real values.
Any hints?</p>
| max_zorn | 506,961 | <p>Put zeros on the diagonal and ones on the off-diagonal. </p>
|
4,524,554 | <p>Consider we have an <span class="math-container">$n \times n$</span> matrix, <span class="math-container">$A$</span>. This matrix represents a linear function from <span class="math-container">$\Bbb R^n$</span> to <span class="math-container">$\Bbb R^n$</span>. Let's say we found a sub-space spanned by the vectors <... | Anne Bauval | 386,889 | <p>No, <span class="math-container">$f$</span> is not necessarily of this form. It is given by: for a fixed nonempty set <span class="math-container">$A$</span> of primes,</p>
<p><span class="math-container">$f(n)=1$</span> if some <span class="math-container">$p\in A$</span> divides <span class="math-container">$n$</s... |
2,377,816 | <p>I was solving problems based on Bayes theorem from the book "A First Course in Probability by Sheldon Ross". The problem reads as follows:</p>
<blockquote>
<p>An insurance company believes that there are two types of people: accident prone and not accident prone. Company statistics states that accident prone pers... | Satish Ramanathan | 99,745 | <p>Another rationale for the answer is to get the cue from the statements:</p>
<p>1) Probability that an accident prone drive will have an accident on a given year is 0.4</p>
<p>2) Probability that an non accident prone driver will have an accident on a given year is 0.2</p>
<p>3) Probability that a person is accide... |
3,905,197 | <p>Stirling's Formula states that <span class="math-container">$\Gamma(z+1) \sim \sqrt{2 \pi z} (\frac{z}{\mathbb{e}})^{z}$</span> as <span class="math-container">$z \rightarrow \infty$</span>. I need to prove the following identity using Stirling's formula:</p>
<p><span class="math-container">$$ (2n)! \sim \frac{2^{2n... | Claude Leibovici | 82,404 | <p>Hoping that you enjoy hypergeometric functions
<span class="math-container">$$I_n=\int_0^\infty
\frac{1}{x^{2n+3}}
\left (
\sin x -
\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}
\right )
\,dx$$</span></p>
<p><span class="math-container">$$\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}=\frac{(-1)^n x^{2 n+3} \,
... |
2,245,408 | <blockquote>
<p>How is the following result of a parabola with focus <span class="math-container">$F(0,0)$</span> and directrix <span class="math-container">$y=-p$</span>, for <span class="math-container">$p \gt 0$</span> reached? It is said to be <span class="math-container">$$r(\theta)=\frac{p}{1-\sin \theta} $$</spa... | Chappers | 221,811 | <p>No, that can't work, because you've shown that these parabolae have different foci.</p>
<p>Since there are lots of parabolae ($x^2=4ay$, $y^2=4ax$, any rotation and translation of one of these), it's easier to work backwards. So let's do that: $\sin{\theta} = y/r$, so
$$ p= r(1-\sin{\theta}) = r-y, $$
so
$$ x^2+y^2... |
2,756,686 | <p>I have a second derivative that I need to use to find inflection points to create a graph. The second derivative is $$f^{\prime\prime}(x)=-4\pi^2\cos(\pi(x-1))$$</p>
<p>So I set the equation to $0$ and solve for $x$</p>
<p>$$-4\pi^2\cos(\pi(x-1))=0$$</p>
<p>I divide by the constant $-4\pi^2$ and get </p>
<block... | Phil H | 554,494 | <p>$\cos\frac{\pi}{2} = \cos\frac{3\pi}{2}$. There is a solution in another region of the 4 quadrants.</p>
|
3,410,150 | <p>If we try solving it by finding <span class="math-container">$f''(x)$</span> then it is very long and difficult to do, so my teacher suggested a way of doing it, he said find nature of all the roots of <span class="math-container">$f(x) =f'(x)$</span>, and on finding nature of the roots we got them to be real(but no... | Community | -1 | <p><strong>Proof the roots are real and distinct</strong></p>
<p>We can scale the axes and translate the curve so that, without loss of generality, the function is <span class="math-container">$f(x)=(x-1)^3(x+1)^3$</span>.</p>
<p>Then <span class="math-container">$f'(x)=6x(x-1)^2(x+1)^2, f''(x)=6(x-1)(x+1)(5x^2-1)$<... |
528,591 | <p>I need to prove there are zero divisors in $\mathbb{Z}_n$ if and only if $n$ is not prime.
What should I consider first? </p>
| Cameron Buie | 28,900 | <p><strong>Hint</strong>: If $n$ is not prime, then either $n=1$ (in which case this is trivial) or there are some integers $k,m\in\{2,...,n-1\}$ such that $n=km$ (as you've correctly deduced). What can we conclude from there about certain elements of $\Bbb Z_n$?</p>
<p>If $n$ is prime, and $[km]=[k][m]=[0]$ for some ... |
1,172,893 | <p>My textbook says I should solve the following integral by first making a substitution, and then using integration by parts:</p>
<p>$$\int cos\sqrt x \ dx$$</p>
<p>The problem is, after staring at it for a while I'm still not sure what substitution I should make, and hence I'm stuck at the first step. I thought abo... | John Hughes | 114,036 | <p>Try $x = u^2$, and $dx = 2u ~ du$. </p>
|
1,172,893 | <p>My textbook says I should solve the following integral by first making a substitution, and then using integration by parts:</p>
<p>$$\int cos\sqrt x \ dx$$</p>
<p>The problem is, after staring at it for a while I'm still not sure what substitution I should make, and hence I'm stuck at the first step. I thought abo... | Frank Lu | 41,622 | <p>Let $x=t^2$, then $dx=2tdt$, so
$$\int\cos\sqrt{x}dx=\int 2t\cos tdt$$</p>
<p>Then use integration by parts.</p>
|
713,098 | <p>The answer to my question might be obvious to you, but I have difficulty with it. </p>
<p>Which equations are correct:</p>
<p>$\sqrt{9} = 3$</p>
<p>$\sqrt{9} = \pm3$</p>
<p>$\sqrt{x^2} = |x|$</p>
<p>$\sqrt{x^2} = \pm x$</p>
<p>I'm confused. When it's right to take an absolute value? When do we have only one va... | PossibilityZero | 135,599 | <p>The mathematical symbol
√
refers to positive number of the two possible square roots.</p>
<p>If the question is written as "What is the square root of 9?", then the answer is both 3 and -3. </p>
<p>However, if the question is "Evaluate √9," the answer would only be 3. Consequentially, -√9 = -3</p>
|
2,715,374 | <p>We know that \begin{equation*}
a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ddots+\cfrac{1}{a_n}}}}}=[a_0,a_1, \cdots, a_n]
\end{equation*}</p>
<p>If $\frac{p_n}{q_n}=[a_0,a_1, \cdots, a_n]$.</p>
<blockquote>
<p>How to prove that $$
\begin{pmatrix}
p_n & p_{n-1} \\
q_n & q_{n-1} \... | Henno Brandsma | 4,280 | <p>The false implication $x \neq y \to ax \neq ay$ is used. This is true iff $a$ is non-zero (in a field), because then the equivalent (by contraposition) $ax = ay \to x=y$ can be shown by division by $a$. So indeed it's a division by $0$ error implicitly.</p>
|
676,573 | <p>Exercise: Write the polynomial $1 + 2x -x^2 + 5x^3 - x^4$ at powers of $(x-1)$.</p>
<p>I presume this exercise is solved using Taylor Series, since it belongs to that chapter, but have no idea how to solve it. Otherwise, it's very straightforward.</p>
<p>Note: The above exercise is <strong>not</strong> homework.</... | Mark Bennet | 2,906 | <p>Here is a method. Let $t=x-1$ so that $x=t+1$.</p>
<p>$$p(t)=1+2(t+1)-(t+1)^2+5(t+1)^3-(t+1)^4$$</p>
<p>Then it is just arithmetic to get the answer.</p>
<p>You can also use Taylor Series - sometimes these problems are set to solve by Taylor Series to show the method works.</p>
|
1,383,956 | <p>Having two points <span class="math-container">$A(xa, ya)$</span> and <span class="math-container">$B(xb, yb)$</span> and knowing a value <span class="math-container">$k$</span> representing the length of a perpendicular segment in the middle of <span class="math-container">$[AB]$</span>, how can I find the other po... | Hetebrij | 252,750 | <p>About part (4), you have $P ( X \ge 0.5 \mid X \ge 0.3) = \frac{ P( X \ge 0.5 \textrm{ and } X \ge 0.3)}{ P( X \ge 0.3) }$. And $ P( X \ge 0.5 \textrm{ and } X \ge 0.3) \neq P( X \ge 0.15)$. So you must divide by $P( X \ge 0.3)$ instead of $P(X \ge 0.5)$. And we have $P(X \ge 0.5 \textrm{ and } X \ge 0.3) = P(X \... |
367,204 | <p>I'm trying to prove that $\mathbb Z_p^*$ ($p$ prime) is a group using the Fermat's little theorem to show that every element is invertible.</p>
<p>Thus using the Fermat's little theorem, for each $a\in Z_p^*$, we have $a^{p-1}\equiv1$ (mod p). The problem is to prove that p-1 is the least positive integer which $a^... | Vittorino Mandujano Cornejo | 1,013,589 | <p>In my case I made an Isomorphism between <span class="math-container">$\mathbb{Z}^*_p = \{1,2,3,...,p-1\}$</span> and all the Automorphisms of a Group G of order p which is of the form <span class="math-container">$\varphi_{x}(g)=g^{x}$</span> (Where <span class="math-container">$x \in \mathbb{Z}^*_p$</span> and <sp... |
675,718 | <p>I have a non-linear system of equations, $$\left\{ \begin{array}{rcl} x^2 - xy + 8 = 0 \\ x^2 - 8x + y = 0 \\ \end{array} \right.$$
I have tried equating the expressions (because both equal 0), which tells me: $$x^2 - xy + 8 = x^2 - 8x + y$$
Moving all expressions to the right yields: $$0 = xy - 8x + y - 8$$
Factori... | Acrobat | 262,162 | <p>Выражаешь из второго $y$. Потом подставляешь в первое вместо $y$, далее решаешь квадратное уравнение относительно $x$. Находишь два корня. Потом для каждого из корней находишь $y$. В итоге получится два корня $x$ и два корня $y$. Всё банально просто. </p>
<p>Quoth google translate:</p>
<blockquote>
<p>Expresses... |
1,282,843 | <p>I'm having trouble proving the following statement:</p>
<blockquote>
<p>$x(u, v) = (u − u^ 3/ 3
+ uv^2 , v − v^ 3/ 3
+ u^ 2 v, u^2 − v^ 2 )$ is a minimal surface and x is not injective</p>
</blockquote>
<p>Proving that $x(u,v)$, which is also known as the Enneper surface, is minimal is not a problem. However, ... | Community | -1 | <p>One way is to look for symmetries. The second and third components are even functions of $u$, while the first is an odd function of $x$. So, if $(u_0, v_0)$ is a point such that the first component of $x$ is zero, then $x(u_0,v_0)=x(-u_0,v_0)$.</p>
<p>It's not hard to find a solution of $u−u^3/3+uv^2=0$ with nonzer... |
1,837,807 | <p>Let $\mathbb{N}$ denote the set of natural numbers, then a subbasis on $\mathbb{N}$ is </p>
<p>$$S = \{(-\infty, b), b \in \mathbb{N}\} \cup \{(a,\infty), a \in \mathbb{N}\}$$</p>
<p>Let $\leq$ be the relation on $\mathbb{N}$ identified with "less or equal to"</p>
<p>Then I saw a claim that says: (the order topol... | Henno Brandsma | 4,280 | <p>The subbase is a union of two families: the family of left intervals, and the family of right intervals. So the proof is correct.</p>
<p>If your interpretation would hold, you would write it as </p>
<p>$$S = \{(-\infty,b) \cup (a, \infty): a,b \in \mathbb{N} \}$$</p>
<p>which is different: a family of unions, par... |
1,714,902 | <p>(Question edited to shorten and clarify it, see the history for the original)</p>
<p>Suppose we are given two $n\times n$ matrices $A$ and $B$. I am interested in finding the closest matrix to $B$ that can be achieved by multiplying $A$ with orthogonal matrices. To be precise, the problem is</p>
<p>$$\begin{align}... | reuns | 276,986 | <p>you want to minimize over $U$ an unitary matrix :</p>
<p>$$\|U A-B\|_F^2 = \sum_{n=1}^N \|U A_n - B_n\|^2$$</p>
<p>where $A_n,B_n$ are the rows of $A,B$. since any permutation matrix is unitary, you can also consider minimizing : </p>
<p>$$\sum_{n=1}^N \|U A_{\phi_n} - B_n\|^2$$
over $U$ and a permutation $\phi$.... |
3,451,301 | <p>The following classical generalization</p>
<blockquote>
<p><span class="math-container">$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$</span>
where <span class="math-container">$\eta(a)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^... | Ali Shadhar | 432,085 | <p>A different proof with a big bonus:</p>
<p>By the definition of the skew harmonic number we have
<span class="math-container">\begin{gather}
\sum_{n=1}^\infty\frac{(-1)^n \overline{H}_n}{n^{2q}}=\sum_{n=1}^\infty \frac{(-1)^n}{n^{2q}}\left(\ln(2)-\int_0^1 \frac{(-x)^n}{1+x}\mathrm{d}x\right)\\
=\ln(2)\sum_{n=1}^\inf... |
2,189,445 | <p>I try to solve this:
$$
\frac{\partial^{2} I}{\partial b \partial a} = I.
$$
I guessed $ I = C e^{a+b} $, but it's not the general solution. So, how to find the last one?</p>
| Bram28 | 256,001 | <p>The Barber Paradox is a straight-forward contradiction, and your logical analysis shows exactly that. Good job!</p>
<p>OK, but then what about the professor's claim?</p>
<p>The professor used the Barber's paradox to illustrate the self-reference or diagonalization method that is behind Godel's proof of the incompl... |
314,238 | <p>Let $R$ be a ring and $\mathfrak{m},\mathfrak{m'}$ two ideals of $R$.</p>
<p>Suppose that $\frac{R}{\mathfrak{m}}$ and $\frac{R}{\mathfrak{m'}}$ are isomorphic. Can i san say that $\mathfrak{m}$ and $\mathfrak{m'}$ are isomorphic too?</p>
| Tom Oldfield | 45,760 | <p>Unfortunately not!</p>
<p>For a particularly interesting example, consider $R = C[0,1]$, the ring of continuous functions from $[0,1] \rightarrow R$. Take the ideal $I = \{f \in R : f(x) = 0 \space \forall x \in [0,\frac12]\}$. The quotient $R/I$ is then isomorphic to $C[\frac12, 1]$, which is in turn isomorphic to... |
3,121,361 | <p>Given <span class="math-container">$G$</span> has elements in the interval <span class="math-container">$(-c, c)$</span>. Group operation is defined as:
<span class="math-container">$$x\cdot y = \frac{x + y}{1 + \frac{xy}{c^2}}$$</span></p>
<p>How to prove closure property to prove that G is a group?</p>
| trancelocation | 467,003 | <p>I would also use Binomial theorem but would estimate a bit differently using a telescoping sum:</p>
<p><span class="math-container">\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^{n}
& = & 1+1 +\sum_{k=2}^n\frac{n(n-1)\cdots (n-k+1)}{n^k}\cdot\frac{1}{k!}\\
& < & 2 +\sum_{k=2}^n\frac{1}{(k-1)k}\\
&... |
2,555,499 | <p>Let $v_1=(1,1)$ and $v_2=(-1,1)$ vectors in $\mathbb{R}^2$. They are <strong>clearly linearly independent</strong> since each is not an scalar multiple of the other. The following information about a linear transformation $f: \mathbb{R}^2 \to \mathbb{R}^2$ is given: $$f(v_1)=10 \cdot v_1 \text{ and } f(v_2)=4 \cdot ... | user334639 | 221,027 | <p>Wring $x \div y \times z$ should be avoided as it is just a bad idea.</p>
<p>It doesn't matter how many people will say it has one correct interpretation, the number of people who will be confused or misinterpret that is just too big.</p>
<p>If you mean $(x \div y) \times z$, then just write $x \times z \div y$.
T... |
1,849,588 | <p>The exercise asks if exists a definition of $Nat(x)$ such that $Nat(x) \Rightarrow Nat(S(x))$, and $\exists x $such that $ Nat(x)$ is false without the use of axiom of infinity. Here $Nat(x) \Leftrightarrow x$ is a natural number, and $S(x)$ is the successor of $x$. </p>
<p>I tried to define $$S(a)=\{a\}.$$ Then I ... | Eric Wofsey | 86,856 | <p>I don't understand your attempt. In particular, you haven't actually defined what "$C_a$" is. You wrote "$a \in C_a \Leftrightarrow (a\in C_a \Rightarrow S(s) \in C_a)$", but this is circular when taken as a definition (since $C_a$ appears on the right-hand side) and even if it weren't, it would only define when $... |
1,849,588 | <p>The exercise asks if exists a definition of $Nat(x)$ such that $Nat(x) \Rightarrow Nat(S(x))$, and $\exists x $such that $ Nat(x)$ is false without the use of axiom of infinity. Here $Nat(x) \Leftrightarrow x$ is a natural number, and $S(x)$ is the successor of $x$. </p>
<p>I tried to define $$S(a)=\{a\}.$$ Then I ... | Kaa1el | 95,485 | <p>Yes, all you need is some kind of type theory with induction principle.</p>
<p>You just define it like this:</p>
<pre><code>N : Type
0 : N
s : N -> N
</code></pre>
<p>It works in many proof verifier and maybe some functional programming languages.</p>
|
1,071,564 | <p>Let's a call a directed simple graph $G$ on $n$ labelled vertices <strong>good</strong> if every vertex has outdegree 1 and, when considered as if it were undirected, it is connected. How many good graphs of size $n$ are there?</p>
<p>Here's my work so far. Let's call this number $T(n)$. Clearly, $T(2) = 1$: there'... | David | 119,775 | <p>My solution does not agree with your answer for $T(5)$, but let's give it a try anyway . . .</p>
<p>To construct such a graph on $n$ vertices, consider the vertices with indegree $0$. If there are none of these then the graph is a (directed) cycle, and there are $(n-1)!$ possibilities. If there are $k$ specified ... |
1,832,320 | <p>I know there are n linearly independent and n + 1 affinely independent vectors in $\mathbb{R}^n$. But how many convexly independent there are?</p>
<p>I think there are infinity number of them because if I have a convex polytope I can always add another point that is "outside" of said polytope. </p>
<p>But I'm not ... | Luísa Borsato | 296,426 | <p>You can use that those metrics are equivalent (because they derived from norms). Then, you just have to verify if the sets $B_i$ are open or not in the $d_i$ metric. </p>
|
1,010 | <p>For periodic/symmetric tilings, it seems somewhat "obvious" to me that it just comes down to working out the right group of symmetries for each of the relevant shapes/tiles, but its not clear to me if that carries over in any nice algebraic way for more complicated objects such as a <a href="http://en.wikipedia.org/... | Danny Calegari | 1,672 | <p>Aperiodic tilings can be thought of (in a sometimes useful way) as leaves of laminations; the groupoid in question (as in Emily's answer) is then the holonomy groupoid of the lamination.</p>
<p>There is a standard description of the Penrose tiles in this way; think of an irrational plane (i.e. an $R^2$) in $R^n$ fo... |
2,418,954 | <p>Using Vieta's formulas, I can get $$\begin{align} \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\&= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\ &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{\left (-\frac{d}{a} \ri... | Andreas | 317,854 | <p>Let $y = 1/x$. So you ask for the value of $f(y) = y_1^3 + y_2^3 + y_3^3$ where the $y_i$ are solutions of $dy^3 + cy^2 + by + a = 0$. With this equation, write
$$- d f(y) = c (y_1^2 + y_2^2 + y_3^2) + b (y_1 + y_2 + y_3) + 3 a.$$</p>
<p>Note the identity $$y_1^2 + y_2^2 + y_3^2=(y_1+y_2+y_3)^2-2(y_1y_2+y_1y_3+... |
1,477,325 | <p>It is known that an integrable function is a.e. finite. Is an a.e. finite function integrable? What if the measure is finite?</p>
| Giovanni | 263,115 | <p>No, just consider the constant function $1$. It is not integrable on the real line. </p>
<p>You don't even need an unbounded domain. Let $f(x) = \frac 1x$ and integrate over $[0,1]$ to find a counterexample to your statement.</p>
|
2,261,500 | <p>I try to prove that statement using only Bachet-Bézout theorem (I know that it's not the best technique). So I get $k$ useful equations with $n_1$ then $(k-1)$ useful equations with $n_2$ ... then $1$ useful equation with $n_{k-1}$. I multiply all these equations to obtain $1$ for one side. For the other side I'm lo... | Tengu | 58,951 | <p>The condition $\forall i \ne j \in \{1,2 \ldots, k \}, \gcd (n_i,n_j)=1$ is too strong. Only <strong>two</strong> $i \ne j$ so $\gcd(n_i,n_j)=1$ is enough to imply $\gcd (n_1, \ldots, n_k)=1$.</p>
<p>Indeed, let $d=\gcd (n_1, \ldots, n_k)$ then $d \mid n_i, d \mid n_j$ so $d \mid \gcd (n_i,n_j)=1$ so $d=1$. Thus, $... |
1,637,748 | <p>I was given the following thing to prove:</p>
<p>$$\lim_{n \to \infty} {d(n) \over n} = 0$$
where $d(n)$ is the number of divisors of n.</p>
<p>I'm so sure how to approach this question. One way I thought of is to use the UFT to turn the expression to:</p>
<p>$$\lim_{n \to \infty} {\prod (x_i + 1) \over \prod p_i... | wythagoras | 236,048 | <p><strong>Hint:</strong> For every divisior $p$ greater than $\sqrt{n}$ of $n$, there is a divisor $q$ smaller than $\sqrt{n}$ such that $n=pq$. It follows that we can divide the divisiors in pairs where one of the elements is smaller than $\sqrt{n}$, and hence $d(n) \leq 2\sqrt{n}$. Can you use this to evaluate the l... |
3,368,402 | <p>I am utilizing set identities to prove (A-C)-(B-C).</p>
<p><span class="math-container">$\begin{array}{|l}(A−B)− C = \{ x | x \in ((x\in (A \cap \bar{B})) \cap \bar{C}\} \quad \text{Def. of Set Minus}
\\
\quad \quad \quad \quad \quad =\{ x | ((x\in A) \wedge (x\in\bar{B})) \wedge (x\in\bar{C})\} \quad \text{Def. o... | Matthew Leingang | 2,785 | <p>You might also just try to prove this in words. Suppose <span class="math-container">$x \in (A-B)-C$</span>.</p>
<ul>
<li>Then <span class="math-container">$x \in (A-B)$</span> and <span class="math-container">$x \notin C$</span>.</li>
<li>Since <span class="math-container">$x \in (A-B)$</span>, we know <span clas... |
440,615 | <p>Let $R$ be the region in the first quadrant bounded above by the circle $(x-1)^2 + y^2 = 1$ and below by the line $y = x$ . Sketch the region $R$ and evaluate the double integral $\iint 2y \;\mathrm dA$ . </p>
| Y.H. Chan | 71,563 | <p>First you need to find the region: you can see it <a href="http://www.wolframalpha.com/input/?i=%28x-1%29%5E2%2By%5E2%3D1+and+y%3Dx+and+x%3D0+and+x%3D1" rel="nofollow">here</a>. The minor segment is your required region and bounded by
$$x=0$$$$x=1$$$$y=x$$$$y=\sqrt{1-(x-1)^2}=\sqrt{2x-x^2}$$
so all you need to do is... |
2,137,706 | <p>$$\sum_{k=0}^m {n \choose k} {n-k \choose m-k} = 2^m {n \choose m}, m<n$$</p>
<p>I know that $2^m$ represents the number of subsets of a set of length $m$, which I can see there being a connection to the ${n \choose k}$ term, but I can't see how the combination it's multiplied by affects this.</p>
| Nicolás Vilches | 413,494 | <h1>A "double-counting" proof</h1>
<p>Let $m<n$ be two positive integers. We have $n$ objects, and we want to choose $m$ of them to paint them, then choose some of the painted objects (possibly none of them) to store them on a box. We have two ways to do this</p>
<p>First, we can select the $m$ objects to be paint... |
1,310,233 | <p>Fundamental theorem of calculus states that the derivative of the
integral is the original function, meaning that
$$
f(x)=\frac{d}{dx}\int_{a}^{x}f(y)dy.\tag{*}
$$
To motivate the statement of the Lebesgue differentiation theorem, observe
that (*) may be written in terms of symmetric differences as
$$
f(x)=\lim_{r\t... | aexl | 33,764 | <p>Let $g: \Bbb R \to \Bbb R$ be differentiable at a point $x \in \Bbb R$, i.e. the limit
$$ g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} $$
exists.
So it follows, that
$$\begin{align*} \frac{g(x+h) - g(x-h)}{2h}
&= \frac{\left[ g(x+h) - g(x) \right] + \left[ g(x) - g(x-h) \right]}{2h} \\
&= \frac 1 2 \fra... |
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