qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,151,937 | <p>Let $A$ be an $m\times n$ real matrix, $x$ an $n\times 1$ vector and $b$ an $m\times 1$ vector. I want to compute
\begin{equation}
\dfrac{\partial }{\partial x} \Vert Ax+b\Vert^{2}.
\end{equation}
First, I expanded
\begin{equation}
\Vert Ax+b\Vert^{2}=(Ax+b)^{T}(Ax+b)=x^{T}A^{T}Ax+2x^{T}A^{T}b+b^{T}b
\end{equation}
then I computed
\begin{eqnarray}
\dfrac{\partial }{\partial x}(x^{T}A^{T}Ax+2x^{T}A^{T}b+b^{T}b)=A^{T}Ax+x^{T}A^{T}A+2A^{T}b
\end{eqnarray}
but I know the above is wrong since $A^{T}Ax$ and $x^{T}A^{T}A$ does not have the same dimention. Thanks for the help.</p>
| Logan Stapleton | 399,630 | <p>A general way of finding the gradient <span class="math-container">$\nabla_x f(x)$</span> of any vector-valued function <span class="math-container">$f:\mathbb{R}^n \rightarrow \mathbb{R}$</span> is by using Taylor's Theorem, which can be expressed as such:
<span class="math-container">$$f(x+\delta) = f(x) + [\nabla_x f(x)]^\intercal \delta + o(||\delta||_2),$$</span>
where <span class="math-container">$\delta\in \mathbb{R}^n$</span> is small.</p>
<p>For your example, let <span class="math-container">$f(x) := ||Ax+b||_2^2$</span>. Applying Taylor's Theorem (above),
we have that:
<span class="math-container">\begin{align}
f(x + \delta) &= ||A(x+\delta)+b||_2^2 \\
&= ||Ax+b + A\delta||_2^2\\
&= ||Ax + b||_2^2 + 2(Ax+b)^\intercal A\delta + ||A\delta||_2^2\\
&= f(x) + [2A^\intercal(Ax+b)]^\intercal \delta + o(||\delta||_2)\\
&= f(x) + [\nabla_x f(x)]^\intercal \delta + o(||\delta||_2).
\end{align}</span></p>
<p>Thus, the gradient <span class="math-container">$\nabla_x f(x)=2A^\intercal(Ax+b)$</span>.</p>
<p>(The third line in the equations above is given by the equality
<span class="math-container">$||U+V||_2^2 = ||U||_2^2 + 2U^\intercal V + ||V||_2^2$</span>, for any vectors <span class="math-container">$U, V \in \mathbb{R}^n$</span>.)</p>
|
2,939,163 | <p>I want to find a certain <span class="math-container">$x$</span> that belongs to <span class="math-container">$\mathbb R$</span> so that </p>
<p><span class="math-container">$$\left|\begin{array}{r}1&x&1\\x&1&0\\0&1&x\end{array}\right|=1$$</span></p>
<p>This should be easy enough. I apply the Laplace extension on the third row so I get</p>
<p><span class="math-container">$$0-\left|\begin{array}{a}1 & 1\\x&0\end{array}\right|+x\left|\begin{array}{r}1&x\\x &1\end{array}\right|=1$$</span></p>
<p>So we have</p>
<p><span class="math-container">$$-(0-x)+x(1-x^2)=1\implies x+x-x^3=1\implies x^3-2x+1=0$$</span></p>
<p>I'm kind of stuck because I'm not entirely familiar with solving cubic functions. I don't think there's a way to refactor this. Perhaps I should have found another way to solve this. <span class="math-container">$x=1$</span> is definitely a solution, but there's another one that I'm missing. Any hints?</p>
| Peter Szilas | 408,605 | <p>Polynomial long division:</p>
<p><span class="math-container">$\small{(x^3-2x+1)÷(x-1)= x^2 +x-1}$</span>;</p>
<p><span class="math-container">$\small{ -(x^3-x^2)}$</span></p>
<p><span class="math-container">$-----$</span></p>
<p><span class="math-container">$\small{x^2-2x +1}$</span></p>
<p><span class="math-container">$\small{ -(x^2-x)}$</span></p>
<p><span class="math-container">$------$</span></p>
<p><span class="math-container">$\small{-x +1}$</span></p>
<p><span class="math-container">$\small{ -(-x+1)}$</span></p>
<p><span class="math-container">$------$</span></p>
<p><span class="math-container">$\small{0}$</span></p>
|
3,715,475 | <p>About a year ago I asked <a href="https://math.stackexchange.com/questions/3258617/alaoglu-theorem-over-the-p-adics">here</a> whether the Banach-Alaoglu Theorem works over the <span class="math-container">$p$</span>-adics. The satisfactory answer I got is that the "usual" proof only uses local compactness, and so the Banach-Alaoglu Theorem holds for any local field.</p>
<p>Now I would like to look at other, more general non-Archimedean fields. I know that Hahn-Banach holds for all spherically complete such fields, and so I was wondering if it is possible to prove Banach-Alaoglu for such fields as well? Because Hahn-Banach works, a related question is whether in the complex setting there is a proof of Banach-Alaoglu that uses Hahn-Banach, but not local compactness of <span class="math-container">$\mathbb{R}$</span> or <span class="math-container">$\mathbb{C}$</span>.</p>
| Robert Furber | 184,596 | <p>Chilote has pointed to the right notion for the general case. I will answer the literal question.</p>
<p>The Banach-Alaoglu theorem (using the usual topological notion of compactness) cannot hold for normed spaces over a field with valuation <span class="math-container">$(k,|\cdot|)$</span> if the unit ball of <span class="math-container">$k$</span> is noncompact. The reason is that the weak-* dual of <span class="math-container">$k$</span> is isomorphic to <span class="math-container">$k$</span> with its original topology.</p>
<p>The spherical completion of the algebraic closure of <span class="math-container">$\mathbb{Q}_p$</span>, for <span class="math-container">$p$</span> a prime, is a spherically complete field whose unit ball is noncompact.</p>
|
1,554,603 | <p>Let $\theta \in \mathbb R$, and let $T\in\mathcal L(\mathbb C^2)$ have canoncial matrix</p>
<p>$M(T)$ = $$
\left(
\begin{matrix}
1 & e^{i\theta} \\
e^{-i\theta} & -1 \\
\end{matrix}
\right)
$$
(a) Find the eigenvalues of $T$.</p>
<p>(b) Find an orthonormal basis for $\mathbb C^2$ that consists of eigenvectors for $T$.</p>
<p>I can get the eigenvalues of T and they are $\sqrt 2$ and $-\sqrt 2$. However, I cannot get each eigenvector respect to each eigenvalue. I know how to get eigenvectors by calculating the null space of $(T - \lambda I)$, but it looks like this is not a proper method to solve this problem. So, anyone can help? Thank you! </p>
| lhf | 589 | <p>The ideals of $\mathbb Z/60$ correspond to the ideals of $\mathbb Z$ that contain $60\mathbb Z$ and so correspond to the divisors of $60$. Since $60=2^2\cdot 3 \cdot 5$, it has $(2+1)\cdot(1+1)\cdot(1+1)=12$ divisors.</p>
|
3,989,921 | <p>Answered (I think!):</p>
<p>The triple product's purpose is to find a direction to the origin, perpendicular to the baseline, which is super trivial in 2D as there is only two perpendicular orientations, but the "cylinder" distinction is made in 3D because there are infinite perpendicular orientations - hence the triple product. Diagram in one of the answers shows this nicely!</p>
<p>Original Question:</p>
<p>I was looking at a video explaining the G-J-K algorithm for finding the two closest features on two convex shapes - when it came to a bit about, given a line, what is the direction in 3D space, from a point on the line, which points to the origin. Seems easy enough: just invert the position vector to point backwards to the origin - and that's exactly what the code does in an earlier step of the algorithm. Bizarrely, they say that as "the origin could be in a cylinder around the line in 3D space", the direction to the origin is given by the following:</p>
<p>Let X be the vector between points A and B on the line (from A to B), and Y be the inverse position vector from A to the origin. The direction to the origin, apparently, is (X cross Y) cross X! I expanded this out, saw an interesting pattern but could not link it to the problem:</p>
<p>For two vectors A, B; (A cross B) cross A I found to be equal to:</p>
<p>Ax(AyBy + AzBz) - Bx(AyAy + AzAz)</p>
<p>Ay(AxBx + AzBz) - By(AxAx + AzAz)</p>
<p>Az(AxBx + AyBy) - Bz(AxAx + AyAy)</p>
<p>Notated as a column vector, with "Vn" meaning the Nth coordinate of vector V. By "cross" I mean the 3D cross product.</p>
<p>I know this isn't the computer science stack exchange - I would like the mathematical meaning of "(A cross B) cross A", and a reason why the code does this, and not just another inverse position vector as it did previously, to find a direction to the origin. Many thanks for any help, and I will repost this on a different stack if it's too offensive here!</p>
| Z Ahmed | 671,540 | <p>Vector trple product is defined as <span class="math-container">$$\vec P \times (\vec Q \times \vec R)= (\vec P \cdot \vec R)\vec Q-(\vec P\cdot \vec Q) \vec R.$$</span> Then
<span class="math-container">$$\vec V=(\vec A \times \vec B) \times \vec A=-\vec A \times (\vec A \times \vec B)=-(\vec A \cdot \vec B)\vec A+(\vec A \cdot \vec A)\vec B.$$</span>
If <span class="math-container">$\vec A$</span> is unit vector then,
<span class="math-container">$$\vec V=\vec B-(\vec B \cdot \hat A)\hat A.$$</span>
Then <span class="math-container">$\vec V$</span> is the component of <span class="math-container">$\vec B$</span> perpendicular to <span class="math-container">$\vec A$</span>.</p>
<p>Otherwise ingeneral <span class="math-container">$\vec V$</span> is perpendicular to <span class="math-container">$\vec A$</span> and coplanar with <span class="math-container">$\vec A$</span> anf <span class="math-container">$\vec B$</span>.</p>
<p><strong>Edit:</strong>
<span class="math-container">$\vec V$</span> is also the direction of projection of a line (with direction as <span class="math-container">$\vec B$</span>) in the plane <span class="math-container">$ \vec A. \vec r=d$</span>.</p>
|
46,905 | <p>I need to draw a set of curves on one graph (characteristics equations). As you can see they have exchanged x and y axes. My goal is to plot all those curves on one graph. Are there ways to do that? </p>
<pre><code>f[t_, t0_] := -(2 - 4/Pi*ArcTan[2])*Exp[-t]*(t - t0);
g[x_, x0_] := (x - x0)/(-(2 - 4/Pi*ArcTan[x + 2]));
Show[Table[Plot[f[t, t0], {t, 0, 1},
PlotRange -> {0, -0.3},
AxesLabel -> {t, x}], {t0, 0, 1, 0.1}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/6BDw9.jpg" alt="enter image description here"></p>
<pre><code>Show[
Table[
Plot[g[x, x0], {x, 0, -0.3}, PlotRange -> {0, 1}, AxesLabel -> {x, t}],
{x0, 0, -0.3, -0.05}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/F9h7Q.jpg" alt="enter image description here"></p>
| kglr | 125 | <p>Kuba's answer using <code>ParametricPlot</code> is the most convenient way to get the result you need. Alternatively, you can use a geometric transformation function that rotates and then reflects <code>s2</code> around the vertical axis:</p>
<pre><code>f[t_, t0_] := -(2 - 4/Pi*ArcTan[2])*Exp[-t]*(t - t0);
g[x_, x0_] := (x - x0)/(-(2 - 4/Pi*ArcTan[x + 2]));
s1 = Show[Table[Plot[f[t, t0], {t, 0, 1}, PlotRange -> {0, -0.3},
AxesLabel -> {"t", "x"}, PlotStyle -> Red], {t0, 0, 1, 0.1}]];
s2 = Show[Table[Plot[g[x, x0], {x, 0, -0.3}, PlotRange -> {0, 1},
AxesLabel -> {"x", "t"}], {x0, 0, -0.3, -0.05}]];
</code></pre>
<p>Define </p>
<pre><code>trF = GeometricTransformation[#,
Composition[ReflectionTransform[{-1, 0}], RotationTransform[Pi/2]]] &;
</code></pre>
<p>This transformation can be <code>Apply</code>'ed to <code>s2</code></p>
<pre><code>Show[s1, Graphics@(trF @@ s2)]
</code></pre>
<p>or <code>MapAt</code>'ed at position <code>{1}</code> of <code>s2</code></p>
<pre><code>Show[s1, MapAt[trF, s2, {1}]]
</code></pre>
<p>to get</p>
<p><img src="https://i.stack.imgur.com/iwYTY.png" alt="enter image description here"></p>
|
3,113,083 | <p>Why does every CNF formula for <span class="math-container">$(x_{1} \vee y_{1}) \wedge (x_{2} \vee y_{2})\wedge \ldots \wedge (x_{n} \vee y_{n})$</span> have at least <span class="math-container">$2^{n}$</span> terms?</p>
<p>This statement is on the Wikipedia page for DNF form here: <a href="https://en.wikipedia.org/wiki/Disjunctive_normal_form" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Disjunctive_normal_form</a></p>
<p>But, I don't understand why it's true. Can someone please clarify? As shown above, the CNF has <span class="math-container">$2n$</span> terms. I'm not sure why the DNF has <span class="math-container">$2^{n}$</span> at least terms though.</p>
| dan_fulea | 550,003 | <p>The combinatorial problem is (relatively) too complicated, and writing some code is (really) too simple, so here is the code and the count:</p>
<pre><code>S = SymmetricGroup(6)
R = Zmod(6)
cansPermutations = \
[ s for s in S
if not {R(0), R(1), R(-1)}.intersection(
{R(s(k)-1) - R(k-1) for k in [1..6]}) ]
</code></pre>
<p>(This is "all the code".)</p>
<p>Then we ask for the variables.</p>
<pre><code>sage: len(cansPermutations)
20
sage: cansPermutations
[(1,5,3)(2,6,4),
(1,4)(2,5)(3,6),
(1,3,5)(2,4,6),
(1,5,2,4)(3,6),
(1,3,6,4)(2,5),
(1,4,6,3)(2,5),
(1,5,2,4,6,3),
(1,3)(2,5)(4,6),
(1,3,5,2,6,4),
(1,4)(2,6,3,5),
(1,4,2,5)(3,6),
(1,5)(2,4)(3,6),
(1,3,6,4,2,5),
(1,3,5)(2,6,4),
(1,4,2,6,3,5),
(1,4)(2,5,3,6),
(1,5,3,6,2,4),
(1,4,6,2,5,3),
(1,5,3)(2,4,6),
(1,4)(2,6)(3,5)]
</code></pre>
<p>There are <span class="math-container">$20$</span> "good permutations".</p>
<hr>
<p>The question does not make it clear, if we need to compute the probability, or only to show it is not <span class="math-container">$(3/6)^6$</span>. Here, i will compute explicitly the set of the "good" permutations of the cans. (I.e. of those that move each can <span class="math-container">$k\in R=\Bbb Z/6$</span> to some place which is not among <span class="math-container">$k$</span> and <span class="math-container">$k\pm 1$</span>.)</p>
<p>Note that each permutation can be uniquely written as a product of disjoint cycles. We do this with each of the good permutations. No element is fixed, so the lengths of the cycles that appear are at least <span class="math-container">$2$</span>. We have only the following possibilities to split the total length <span class="math-container">$6$</span> in pieces <span class="math-container">$\ge 2$</span>:
<span class="math-container">$$
\begin{aligned}
6 &= 2+2+2\\
&=2+4\\
&=3+3\\
&=6\ .
\end{aligned}
$$</span>
Now we count for each split the number of the possibilities to fill in the places.</p>
<ul>
<li><p>For the type <span class="math-container">$2+2+2$</span> we search for permutations of the shape <span class="math-container">$(1a)(bc)(de)$</span>. If we map <span class="math-container">$1\to 3$</span>, i.e. <span class="math-container">$a=3$</span>, then for <span class="math-container">$5$</span> we have only the possibility <span class="math-container">$2$</span>, and there is only one case, <span class="math-container">$(13)(25)(46)$</span>. Symmetrically, if we decide to map <span class="math-container">$1\to 5$</span>, then there is only one possibility for the <span class="math-container">$3$</span>, we get the only case <span class="math-container">$(15)(36)(24)$</span>. For <span class="math-container">$1\to 4$</span> we have after some chase of possibilities only <span class="math-container">$(14)(25)(36)$</span> and <span class="math-container">$(14)(26)(35)$</span>. We have thus <span class="math-container">$1+1+2=4$</span> cases in this shape. This combinatorial problem was simple. </p></li>
<li><p>For the case <span class="math-container">$3+3$</span> we have still an easy count. A good permutation of the shape <span class="math-container">$(1bc)(def)$</span> must have odd values for <span class="math-container">$b,c$</span>, else the <span class="math-container">$4$</span> is either <span class="math-container">$b$</span> or <span class="math-container">$c$</span>. But the neighbors of <span class="math-container">$1,4$</span> cover all possible remaining values, so there is no "good place" for <span class="math-container">$c$</span>. So <span class="math-container">$1,b,c$</span> are the numbers <span class="math-container">$1,3,5$</span> in some order (for <span class="math-container">$3,5$</span>), the remaining <span class="math-container">$2,e,f$</span> are <span class="math-container">$2,4,6$</span> in some order (of <span class="math-container">$4,6$</span>). We have thus <span class="math-container">$2\times 2=4$</span> cases in this shape. Again a simple case.</p></li>
<li><p>The next case is <span class="math-container">$2+4$</span>. Let <span class="math-container">$(ab)(cdef)$</span> be such a good permutation, <span class="math-container">$a<b$</span>. We are conjugating it cyclically, i.e. permute cyclically the letters till <span class="math-container">$a$</span> becomes <span class="math-container">$1$</span>. In this was it is enough to study the good permutations of the shape <span class="math-container">$(1B)(CDEF)$</span>, then produce all conjugates of it, taking care to not repeat solutions. Since <span class="math-container">$B\ne 2$</span>, the <span class="math-container">$2$</span> is among <span class="math-container">$C,D,E,F$</span>, and we can and do assume <span class="math-container">$C=2$</span>. Also, <span class="math-container">$5$</span> is among <span class="math-container">$D,E,F$</span>. The first observation is that <span class="math-container">$B=4$</span>. Else we may assume (by changing the orientation of the circle) <span class="math-container">$B=3$</span>, then <span class="math-container">$2,D,E,F$</span> are <span class="math-container">$2,4,5,6$</span> in some order. So <span class="math-container">$5\to 2$</span>, but then which number goes to <span class="math-container">$5$</span>, we have only neighbors left, contradiction. Our permutation is thus <span class="math-container">$(14)(2DEF)$</span>, and a quick chase validates only <span class="math-container">$(14)(2536)$</span> and its invers <span class="math-container">$(14)(2635)$</span>. Now back to <span class="math-container">$(ab)(cdef)$</span>. It is clear that <span class="math-container">$(ab)$</span> may be only among <span class="math-container">$(14)$</span>, and/or <span class="math-container">$(25)$</span> and/or <span class="math-container">$(36)$</span> and in each case we obtain two solutions. So there are <span class="math-container">$3\times 2$</span> solutions of this shape.</p></li>
<li><p>It remains to get the good permutations of shape <span class="math-container">$6$</span>, i.e. looking like <span class="math-container">$(1bcdef)$</span>. Maybe the simplest way to proceed is to fix the possible place of the <span class="math-container">$4$</span>. It cannot be <span class="math-container">$c$</span> (and symmetrically in the cycle <span class="math-container">$e$</span>) because we have no good option for the letter in between, <span class="math-container">$b$</span> (and respctively <span class="math-container">$f$</span>). We can place thus the <span class="math-container">$4$</span> on <span class="math-container">$b$</span>, getting finally only the cases <span class="math-container">$(142635)$</span> and <span class="math-container">$(146253)$</span>, the reversed cycles in the pattern <span class="math-container">$(1bcde4)$</span>, and two more solutions for the pattern <span class="math-container">$(1bc4ef)$</span>, which are <span class="math-container">$(152463)$</span> and its inverse <span class="math-container">$(136425)$</span>. So we have <span class="math-container">$2+2+2$</span> more solutions.</p></li>
</ul>
<p>Totally, there are <span class="math-container">$4+4+6+6=20$</span> solutions. </p>
|
2,994,296 | <p>I'm trying to figure out how to prove, that <span class="math-container">$$\lim_{n\to \infty} \frac{n^{4n}}{(4n)!} = 0$$</span>
The problem is, that <span class="math-container">$$\lim_{n\to \infty} \frac{n^{n}}{n!} = \infty$$</span>
and I have no idea how to prove the first limit equals <span class="math-container">$0$</span>. </p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Starting like other two answers:
<span class="math-container">$$a_n = \frac{n^{4n}}{(4n)!}\implies
\frac{a_{n+1}}{a_n} = \left(\left(1+\frac1n\right)^n\right)^4\cdot\frac{(n+1)^4}{(4n+1)(4n+2)(4n+3)(4n+4)}\to\frac{e^4}{4^4} < 1,
$$</span>
and by the quotient test <span class="math-container">$\sum a_n$</span> converges, but this implies <span class="math-container">$a_n\to 0$</span>.</p>
|
2,860,360 | <p>It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book <em>Algebraic Number Theory</em>:</p>
<p>"<em>Example 7.20</em> For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."</p>
<p>The question is why does it suffice to look at the principal ideal generated by 2?</p>
| Mark Bennet | 2,906 | <p>The probabilities work because there is a chance that more than one person is successful at the same time, even though there is also a chance that none are successful. The average number of successes for six people is six times the average for one person, but this average covers the case where all succeed at the same time (for example) as well as the cases where two out of the six succeed.</p>
|
2,860,360 | <p>It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book <em>Algebraic Number Theory</em>:</p>
<p>"<em>Example 7.20</em> For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."</p>
<p>The question is why does it suffice to look at the principal ideal generated by 2?</p>
| Martin Argerami | 22,857 | <p>I think the situation is easier to see in a simpler example. Drop a coin, "heads" is success. The probability of success is 50%. This does <strong>not</strong> mean that if you drop the coin three times your probability of getting one head is 150%.</p>
<p>The probability of getting at least one head is one minus the probability of getting three tails, so it's $1-1/8=7/8$. In percentage, that would be around 87%. </p>
|
2,860,360 | <p>It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book <em>Algebraic Number Theory</em>:</p>
<p>"<em>Example 7.20</em> For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."</p>
<p>The question is why does it suffice to look at the principal ideal generated by 2?</p>
| JKreft | 457,797 | <p>The problem is your third paragraph, where you've confused the expected number of successes in 6 tries (1.2) with a percentage chance (120%). This is one reason whey probability students are encouraged to work in decimals/fractions instead of percentages.</p>
<p>If you let six people try it, you can expect 1.2 successes on average. If you're looking for the probability that at least one succeeds out of 6, you have (as noted elsewhere) {1-P(all six fail)} which would be {1-0.8^6}.</p>
|
2,860,360 | <p>It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book <em>Algebraic Number Theory</em>:</p>
<p>"<em>Example 7.20</em> For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."</p>
<p>The question is why does it suffice to look at the principal ideal generated by 2?</p>
| Acccumulation | 476,070 | <p>You've currently phrased this is terms of whether the probability is multiplicative (does having six times the number of trials give six times the probability of success), but we can equivalently ask whether it's additive (is the probability of success over two trials equal to the sum of probabilities for each individual trial).</p>
<p>The reason it's not additive is that on the first flip, everyone is eligible to get a success. If you have 20 people flipping coins, then all of them could get heads, and on average you're going to have 10 of them getting a success.</p>
<p>But on the second flip, while everyone has a chance of getting heads, only the ones who got tails are eligible to be <em>new</em> successes. If someone gets heads the first time, then you've already counted them, and counting them if they get heads the second time is double-counting.</p>
<p>For the number of successes to be additive, you would have to have the same number of <em>new</em> successes each trial. But you don't: to get the number of new successes, you have to multiply the probability of getting a success on that trial by the number of people who haven't gotten a success yet. </p>
<p>In your case, after one trial, 20% will get a success on the first trial, so 80% won't have a success after one trial. During the second trial, 20% will get a success, but only 80% of them will be <em>new</em> successes. So after the first trial, you'll have 20% success from the first trial, plus 20%*80% from the second trial, giving a total of 36% success, rather than the 40% you would expect from taking 20% and multiplying it by 2. The missing 4% represents people who got successes both times, but should be counted only once.</p>
<p>If you work through the math, you'll find that the percentage of people who don't have any successes after $n$ trials will be $(1-p)^n$, and thus the percentage with at least one success will be $1-(1-p)^n$.</p>
|
2,880,830 | <p><a href="https://i.stack.imgur.com/a4Wh2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a4Wh2.png" alt="enter image description here"></a></p>
<p>Not sure how to do this one. If <span class="math-container">$S$</span> is a field, then I was considering that <span class="math-container">$\exists r_1,\ldots, r_n\in R$</span> s.t. <span class="math-container">$1 = r_1s_1+\cdots+r_ns_n$</span> so for <span class="math-container">$r = rr_1s_1+\cdots+rr_ns_n$</span>. Maybe that is somehow useful for taking inverses of elements. </p>
<p>The assumption that <span class="math-container">$S$</span> is an integral domain is necessary because otherwise we could have <span class="math-container">$S = \mathbb{Z}_p[x]/f(x)$</span> where <span class="math-container">$f(x)$</span> is not irreducible. This is still a finitely generated <span class="math-container">$\mathbb{Z}_p$</span>-module, but its not a field. </p>
<p>Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple</p>
<p>Source: <a href="https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf" rel="nofollow noreferrer">https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf</a></p>
| Alex Wertheim | 73,817 | <p>Uncover the spoilers for solutions completing the hints below:</p>
<ul>
<li><p>Suppose <span class="math-container">$R$</span> is a field, and let <span class="math-container">$s \in S$</span> be a nonzero element. Then multiplication by <span class="math-container">$s$</span> is an <span class="math-container">$R$</span>-linear endomorphism of <span class="math-container">$S$</span>, which is injective since <span class="math-container">$s$</span> is nonzero and <span class="math-container">$S$</span> is a domain. </p>
<blockquote class="spoiler">
<p> Since <span class="math-container">$S$</span> is a finite-dimensional <span class="math-container">$R$</span>-vector space, it follows that multiplication by <span class="math-container">$s$</span> is also surjective, and so <span class="math-container">$1$</span> is in the image of this map. </p>
</blockquote></li>
<li><p>Suppose <span class="math-container">$S$</span> is a field. This solution I have in mind for this direction is a bit trickier. Let <span class="math-container">$r \in R$</span> be nonzero, and let <span class="math-container">$s$</span> be the inverse to <span class="math-container">$r$</span> in <span class="math-container">$S$</span>. As before, consider the <span class="math-container">$R$</span>-linear map <span class="math-container">$\varphi_{s} \colon S \to S$</span> corresponding to multiplication by <span class="math-container">$s$</span>. Since <span class="math-container">$S$</span> is a finitely generated <span class="math-container">$R$</span>-module, <span class="math-container">$\varphi_{s}$</span> satisfies a monic polynomial relation with coefficients in <span class="math-container">$R$</span> by Cayley-Hamilton. That is, there exist <span class="math-container">$r_{1}, \ldots, r_{n} \in R$</span> such that multiplication by <span class="math-container">$s^{n}+r_{1}s^{n-1} + \cdots +r_{n}$</span> is the zero element of <span class="math-container">$\mathrm{End}_{R}(S)$</span>. </p>
<blockquote class="spoiler">
<p>Since <span class="math-container">$S$</span> is a faithful <span class="math-container">$R$</span>-module (<span class="math-container">$R$</span> is a subring of <span class="math-container">$S$</span>,and so contains <span class="math-container">$1$</span>), this implies that <span class="math-container">$s^{n}+r_{1}s^{n-1} + \cdots +r_{n} = 0$</span>. Now multiply both sides by <span class="math-container">$r^{n-1}$</span> to conclude that <span class="math-container">$s \in R$</span>.</p>
</blockquote></li>
</ul>
|
3,844,448 | <p>Find all values of <span class="math-container">$h$</span> such that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span>.</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
3 & -1 & 0\\
-4 & 1 & 3
\end{bmatrix} $</span></p>
<p>I used row transformations to get</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
0 & -1-3h & 3\\
0 & 1+4h & -1
\end{bmatrix} $</span></p>
<p>But how do I solve to get the rank? I know the general idea is that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span> when dim(col(<span class="math-container">$A$</span>)) = dim(row(<span class="math-container">$A$</span>)) = <span class="math-container">$2$</span></p>
| Matthew H. | 801,306 | <p>Notice the plane <span class="math-container">$9x+y+3z=0$</span> is precisely the span of the first and third column of your matrix. All you need to do is find <span class="math-container">$h$</span> so that <span class="math-container">$(h,-1,1)$</span> resides on this plane; this will guarantee that you matrix has rank two. So <span class="math-container">$9h-1+3=0$</span> implies <span class="math-container">$h=-2/9$</span> as desired.</p>
|
79,084 | <p>Let $X$ be a topological space (say a manifold). A result of R. Thom states that the pushforwards of fundamental classes of closed, smooth manifolds generate the rational homology of $X$. This work of Thom predates the development of bordism. Is there now a more elementary proof of this result that does not rely on spectral sequence techniques?</p>
| Sergey Melikhov | 10,819 | <p>A nice, direct combinatorial construction was given by Gaifullin, see his <a href="https://arxiv.org/abs/0712.1709" rel="nofollow noreferrer">papers</a> <a href="https://arxiv.org/abs/0806.3580" rel="nofollow noreferrer">on the</a> <a href="https://arxiv.org/abs/0912.3933" rel="nofollow noreferrer">arXiv</a> (equivalently: <em><a href="https://arxiv.org/abs/0712.1709" rel="nofollow noreferrer">Explicit construction of manifolds realizing the prescribed homology classes</a></em>, <em><a href="https://doi.org/10.4213/rm9198" rel="nofollow noreferrer">Realisation of cycles by aspherical manifolds</a></em> and <em><a href="https://doi.org/10.1134/S0081543810010074" rel="nofollow noreferrer">
Configuration spaces, bistellar moves, and combinatorial formulae for the first Pontryagin class</a></em>). A drawback of this approach is that if you think of it as realizing some multiple <span class="math-container">$m\alpha$</span> of the given integral homology class <span class="math-container">$\alpha$</span> by an oriented smooth manifold, then <span class="math-container">$m$</span> is not bounded in terms of the dimension of <span class="math-container">$\alpha$</span>.</p>
<p>There has also been another geometric approach.
Thom also proved that <span class="math-container">$\bmod2$</span> homology classes are representable by maps of smooth (possibly unorientable) manifolds. This was reproved geometrically in</p>
<blockquote>
<p>S. Buoncristiano and D. Hacon, <em>An elementary geometric proof of two theorems of Thom,</em> Topology 20 (1981), no. 1, 97–99 (<a href="https://core.ac.uk/download/pdf/82811869.pdf" rel="nofollow noreferrer">Core pdf</a>)</p>
</blockquote>
<p>The other theorem of their title is that unoriented bordism is determined by Stiefel-Whitney numbers, and it is used in their proof that mod 2 homology classes are representable by smooth manifolds.</p>
<p>I believe the same geometric argument should also work to show that rational homology classes are representable by oriented smooth manifolds - <em>modulo</em> the fact that Pontryagin numbers determine oriented bordism tensored by <span class="math-container">$\Bbb Q$</span>. This fact I'm afraid I don't know how to prove geometrically (for some proof, see e.g. the Milnor-Stasheff book). But note that in a subsequent paper Buouncristiano and Hacon also gave a geometric proof that Chern numbers determine complex bordism (Ann. of Math., 118 (1983), 1-7 <a href="https://doi.org/10.2307/2006950" rel="nofollow noreferrer">https://doi.org/10.2307/2006950</a>). Their other papers may also be of interest if you care about geometric proofs of classical results on bordism.</p>
|
1,967,847 | <blockquote>
<p>A vector space $V$ is called <strong>finite-dimensional</strong> if there is a finite subset of $V$ that is a basis for $V$. If there is no such finite subset of $V$, then $V$ is called <strong>infinite-dimensional</strong>.</p>
<hr>
<p>We now establish some results about finite-dimensional vector spaces that will tell about the number of vectors in a basis, compare two different bases, and give properties of bases. First, we observe that if $\{\mathbf{v}_1, \mathbf v_2,\dotsc, \mathbf v_k\}$ is a basis for a vector space $V$, then $\{c\mathbf v_1, \mathbf v_2, \dotsc, \mathbf v_k\}$ is also a basis when $c\neq 0$ (Exercise 35). Thus a basis for a nonzero vector space is never unique.<br>
<a href="https://i.stack.imgur.com/ghEh9.png" rel="nofollow noreferrer">Image.</a></p>
</blockquote>
<p>I am confused if $\mathbb R^2$ is a finite dimensional vector space. $[1 \ 0],[0 \ 1]$ will be the standard basis of $\mathbb R^2$. However, there are also $[2 \ 0],[0 \ 1]$ and I can find infinite many to be the basis of $\mathbb R^2$. So, $\mathbb R^2$ is an infinite dimensional vector space? </p>
| copper.hat | 27,978 | <p>For each $n$, the following is a basis for $\mathbb{R}^2$, ${\cal B}_n = \{ (0,1), (n,1) \}$.</p>
<p>So, each ${\cal B}_n$ has exactly two elements, but there is an infinite number of bases.</p>
|
650,866 | <p>So I have the function $$ e^{-2x} $$ and if I derive this I thought that I should get $$ -2xe^{-2x} $$ But the $x$ disappears, why? Is it an inner derivative and because of that, I also have to differentiate the expression $-2x$ when I put it in front of $e$? If that is the case, then $x$ would be 1 and -2 is the only character left.. Am I right?</p>
| amWhy | 9,003 | <p>Yes, you are correct in your thoughts in the last paragraph. </p>
<p>We use the chain rule:</p>
<p>Let $u = -2x.\;$ Then $\;\dfrac {du}{dx} = -2.$</p>
<p>This gives us $$\frac{d}{dx}(e^{-2x}) = \frac d{dx}(e^u) = e^u\left(\dfrac{du}{dx}\right) = e^{-2x}(-2) = -2e^{-2x}$$</p>
|
888,319 | <p><span class="math-container">$ABC$</span> is an acute angled triangle, where <span class="math-container">$P$</span> is the orthocenter, and <span class="math-container">$R$</span> is the circumradius. I want to show that <span class="math-container">$PA+PB+PC\le 3R$</span> geometrically, that is without using trigonometry. I have a trig solution, but I want to know whether we can do it by pure geometry.</p>
<p><img src="https://i.stack.imgur.com/jqIKE.png" alt="enter image description here" /></p>
<p>Note: In the image, the direction of the inequalities should be the opposite.</p>
| Jack D'Aurizio | 44,121 | <p>Yes, there is a way. Due to the Euler theorem, $O,G,H$ are collinear and $HO=3\,GO$. </p>
<p>This implies that if we take $O$ as the origin, the vector equation:
$$ H = A+B+C $$
holds. By applying the triangular inequality:
$$ OH = |H| \leq |A|+|B|+|C| = OA+OB+OC = 3R.$$</p>
|
1,572,593 | <p>Let $G$ a group, $H$ and $K$ two subgroups of G of finite order such that $H \cap K = \{1_G\}$. </p>
<p>I already show the first exercise which says that the cardinal of $HK$ is $|H||K|$.</p>
<p>The second exercise ask to deduce that if $|G|=pm$ where $p$ is a prime number and $p>m$, then $G$ has at most one subgroup of order $p$. Show that if the subgroup exist, it is normal in $G$.</p>
<p>I think a can use the Cauchy and Sylow theorem, but I'm a bit block off on this problem. Is anyone is able to give me a hint?</p>
| Justpassingby | 293,332 | <p>No need for sophistication. If $G$ has two distinct subgroups of order $p$ then those subgroups satisfy the conditions on $H$ and $K.$</p>
<p>Now since the subgroup is the only one of its order, it is identical to all of its conjugates and therefore normal.</p>
|
1,335,640 | <p>1) A disease has hit a city. The percentage of the population infected $t$ days after the disease arrives is approximated by $$p(t) = 12te^{\frac{-t}{7}} \qquad \mbox{for} \qquad0\leq t \leq 35.$$ </p>
<p>After how many days is the percentage of infected people a maximum? What is the maximum percent of the population infected?</p>
<p>The maximum percent of the population infected is ______ %</p>
<p>2) A container contains 12 diesel engines. The company chooses 5 engines at random and will not ship the container if any of the engines chosen are defective. Find the probability that a container will be shipped even though it contains 2 defectives if the sample size is 5.</p>
<p>For the first problem, the number of days at which the percentage is at maximum is 7. Clearly, if I substitute this to $p(t)$ I will get the maximum percentage. My problem is how did they get the answer of 7 days? How do I deal with this kind of problem? Is there a specific formula? I'm trying to figure it out but can't.</p>
<p>Also for the second problem I made use of the hypergeometric formula that is $$p(x) = \frac{\left[C(k,x) \cdot C(N-k, n-x)\right]}{C(N,n)}$$ where $N$ is the size of population, $k$ is the number of successes in the population, $x$ is the number of successes in the sample and $n$ is the sample size. I used this and I got a different answer. The answer should have been 0.318 but I got a different one. Please help. </p>
| BruceET | 221,800 | <p><strong>Mode of gamma distribution.</strong> For (1), recognize that $p(x)$ is the PDF of a gamma distribution with shape parameter 2 and scale parameter 7. Look at the Wikipedia article. Its mode is $x = (2-1)7 = 7.$ (The bound 35 is so far out as to be irrelevant to the discussion.)</p>
<p><strong>Application of hypergeometric to acceptance sampling.</strong> For (2), I used the hypergeometric PDF programmed into R as follows:</p>
<pre><code> dhyper(0, 2, 10, 5)
## 0.3181818
</code></pre>
<p>So it seems you have made a mistake. </p>
<p>Hint: What is ${10 \choose 5}$
and what is ${12 \choose 5}?$</p>
|
201,999 | <p>Prove that the given sequence ${a_n}$ diverges to infinity.</p>
<p>$a_n=\frac{n^3+5}{-n^2+8n}$</p>
<p>I believe that the sequence diverges to -infinity. And I have this for my proof so far:</p>
<p>Let $M>0$ and let $N=$ ?. Then $n>N$ implies... I am confused on how to solve for the N. I believe I have to make the numerator larger and denominator smaller, but I have a hard time visualizing this. </p>
| André Nicolas | 6,312 | <p>We have
$$a_n=-\left(\frac{n^3+5}{n^2-8n}\right).$$
Informally, we now want to show that that if $n$ is "large" then $\dfrac{n^3+5}{n^2-8n}$ is large. We do not want to worry about negative values of $n^2-8n$, so suppose that $n\gt 8$, making $n^2-8n\gt 0$. For $n \gt 8$, we have
$$\frac{n^3+5}{n^2-8n} \gt \frac{n^3}{n^2-8n},\tag{$1$}$$
for on the right we have made the numerator smaller than on the left.</p>
<p>Also, for $n\gt 8$, we have
$$\frac{n^3}{n^2-8n}\gt \frac{n^3}{n^2},\tag{$2$}$$
for on the right we have made the denominator bigger than on the left.</p>
<p>Combining Inequalities $(1)$ and $(2)$, we get that if $n\gt 8$ then
$$\frac{n^3+5}{n^2-8}\gt \frac{n^3}{n^2}=n,$$
and therefore
$$a_n=-\left(\frac{n^3+5}{n^2-8}\right)\lt -n.\tag{$3$}$$ </p>
<p>Finally, we want to show that for any number $K$, so matter how large <strong>negative</strong> it is, we can find $N$ such that if $n\gt N$, then $a_n\lt K$.
That will show that $\lim_{n\to\infty}a_n=-\infty$.</p>
<p>So let $K$ be negative, and let $N$ be any integer greater than the larger of $8$ and $|K|$. By Inequality $(3)$, if $n\gt N$, we have
$$a_n\lt -n \lt -N\lt -|K|=K.$$ </p>
<p><strong>Remark:</strong> It is useful to work as long as possible with positive quantities, since there our intuition is likely to be more accurate.</p>
<p>There were additional items of unpleasantness introduced because of the fact that the limit is $-\infty$. It would be more comfortable to let $b_n=-a_n$, and show that $\lim_{n\to\infty} b_n=\infty$, and then refer to a little theorem somewhere to the effect that a sequence $(x_n)$ has limit $p$ (which may be a real number or one of $\pm\infty$) iff the sequence $(-x_n)$ has limit $-p$. </p>
|
2,915,685 | <blockquote>
<p>Let $X$ be Banach space and $Y$ be a normed vector space and suppose that I find a linear map $T \in L(X,Y)$. With the aid of $T$, I wonder under what condition I can conclude that $Y$ is also a Banach space.</p>
</blockquote>
<p>Is $T$ a linear isomorphism enough? I encounter this question when think of an example like $\ell^p$-space equipped with norm
$$
\|a\|_p := \left(\sum_{n=1}^\infty |a_n|^p\right)^{1/p}
$$
and $\ell^p(w)$, a weighted $\ell^p$-space, equipped with norm
$$
\|a\|_{p,w}:=\left(\sum_{n=1}^\infty |a_n|^p w(n)\right)^{1/p}
$$
for $p \geq 1$ and $w: \mathbb{N} \to [0,\infty)$. It seems to me these two guys are isomorphic, and I was trying to argue $\ell^p(w)$-space is Bananch space without using the standard Cauchy sequence argument. Does this thought make sense? or it is completely off-track? Any comment or feedback is appreciated.</p>
<p><strong>Edit</strong> What if $w(n) > 0$ is forced. Does the idea above alive?</p>
| detnvvp | 85,818 | <p>If $X$ is a Banach space, $Y$ is a linear space and $T:X\to Y$ is a linear isomorphism, then $Y$ is a Banach space. To show this let $(y_n)$ be a Cauchy sequence in $Y$, then the sequence $(T^{-1}y_n)$ is Cauchy in $X$, so it converges to some $x$, and then $y_n\to Tx$. Therefore $Y$ is complete.</p>
<p>Now, with the definition of $\ell^p(w)$, choose $w(n)=\frac{1}{n}$. Then $T:\ell^2\to\ell^2(w)$ sending $x$ to $x$ is well defined, injective and continuous. However, $T$ is not surjective: note that $y=\left(1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\dots \frac{1}{\sqrt{n}},\dots\right)\in \ell^2(w)$, but there is no $x\in\ell^2$ such that $Tx=y$. Therefore $T$ cannot be an isomorphism.</p>
|
38,439 | <p>I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define integrals.</p>
<p>This all seems very ad hoc, however. Not natural. I'm aware this is a pretty trivial question for a lot of you (which is why I'm asking it!), but what is the "correct" natural definition we should think of when we think of integrals?</p>
<p>I know there's some relation to a perfect pairing of homology and cohomology, somehow relating to Poincare duality (is that right?). And there's also something about chern classes? My geometry, as you can see, is pretty confused (being many years in my past).</p>
<p>If you can come up with a natural framework that doesn't have to do with the keywords I mentioned, that would also be very welcome.</p>
| Theo Johnson-Freyd | 78 | <p>If you are interested in manifolds, then you might be interested in various related notions of <em>measures</em> and <em>distributions</em>. Let $M$ be a smooth manifold with algebra of functions $C^\infty(M)$. There is a very general notion of a <em>distribution</em>, which is that of any linear function $C^\infty(M) \to \mathbb R$. In this framework, a distribution is a <em>measure</em> if it satisfies a positivity condition, namely that it takes everywhere-nonnegative functions to nonnegative numbers.</p>
<p>Another, different definition of "distribution" corresponds to the <em>distribution bundle on $M$</em>, which is a canonical trivializable line bundle on $M$. It can be presented by gluing data and transition amplitudes as follows. Let $U,V \subseteq M$ with $\phi: U \to \mathbb R^n$ and $\psi: V \to \mathbb R^n$ be coordinate charts, and consider the trivial bundles one-dimensional bundles over $U,V$. We glue them together by giving transition data: if $f$ is a section over $U$ of the trivial bundle, on $U\cap V$ we identify it with the section $f \cdot \left| \det \frac{\partial \phi}{\partial \psi}\right|$ of the trivial bundle over $V$. (When $M$ is oriented, this bundle is the same as the <em>determinant bundle</em> $\wedge^{\operatorname{top}} {\rm T}^*M$; the determinant bundle is always a line bundle, and so its square is trivializable, and has a trivializable square root, which is the distribution line bundle whether $M$ is oriented or not.) Note that the transition functions preserve positivity of the sections, and so the notion of "positive distribution" and so on are well-defined.</p>
<p>Finally, if you are interested in a totally algebraic notion of integration for $\mathbb R^n$, you might be interested in the following observation, which in some form is older but nevertheless deserves to be called an observation of Berezin. Namely, the <em>integral</em>, as a linear map $C^\infty_{\operatorname{compact}}(\mathbb R^n) \to \mathbb R$, is uniquely defined up to scalar multiple by the fact that it vanishes on the images of $\frac{\partial}{\partial x_i} : C^\infty_{\operatorname{compact}}(\mathbb R^n) \to C^\infty_{\operatorname{compact}}(\mathbb R^n)$. Here $C^\infty_{\operatorname{compact}}(\mathbb R^n)$ is the algebra of smooth functions with compact support, and $x_1,\dots,x_n$ are the usual coordinate functions on $\mathbb R^n$. There are many situations in which by naming an algebra "of functions" and some "partial derivatives" you can uniquely (up to scalar) determine an "integral". An example is the algebra of de Rham differential forms on an oriented manifolds $M$, and the "partial derivatives" are the de Rham $d$ and the Lie derivatives for all vector fields on $M$. This <strike>uniquely</strike> picks out the integral that is zero on non-top forms and integrates top forms over $M$ as a canonical "measure" on the "space" whose "algebra of functions" is the differential forms. This is an example of a "superintegral", and it was to motivate a definition of superintegrals that Berezin made the above observation.</p>
|
1,749,909 | <p>I need a help with somthing:
I need to tell if these two integrals are Convergence\Absolute convergence:</p>
<p>$\int _1^{\infty }\frac{\ln x}{\sqrt{x^3-x^2-x+1}}dx$, $\int _0^{\infty \:}\:\frac{\left(x^{\frac{1}{4}}+1\right)\cdot \sin\left(2\sqrt{x}\right)}{x}dx$
Now I compute this and I find that both converge. But I don't know how to check
if it is also Absolute convergence.
Thanks.</p>
| jst345 | 327,162 | <p>Regarding your proof, to be honest, I think the student is right to be sceptical as there are a couple of issues with it! Firstly your suggestion that it is 'clear visually' that $\int^1_0 f(x) dx -\int^1_0 g(x) dx$ represents the area between the curves may very well be the case, but unfortunately when it comes to doing arithmetic with infinite quantities what is clear visually is often not actually the case! Secondly, I assume that the expression I've written above was what you were really thinking of when you made the statement, but by writing it as $\int^1_0 f(x) - g(x) dx$, you've assumed that this new definition of integral distributes over addition (or subtraction), which is wrong — in fact, the contradiction you reach arguably works better as a refutation of this latter assumption than what you were originally aiming to do.</p>
<p>Really, I don't think you should be trying to 'disprove' the student's understanding of 'area under the curve' at all, as to an extent they're making a valid point: the points below the x-axis <strong>are</strong> 'under the curve'. The main problem with using that to make a definition is that it would always be infinite and therefore of limited use or interest. Also, the fact that everyone else is taking the integral as 'the area under the curve but above the x-axis' means that to do anything else would be doomed to cause major confusion.</p>
<p>If I were you, when he questions the meaning of area under the curve, I would simply admit that the phrase is a bit ambiguous and clarify that you mean only down to the axis.</p>
|
2,278,431 | <p>"Apply Green's Theorem to evaluate the line integral of F around positively oriented boundary"</p>
<p>$$F(x,y)=x^2yi+xyj$$</p>
<p>C: The region bounded by y=$x^2$ and y=4x+5</p>
| Community | -1 | <p>Use the fact that $(1+\frac 1 n)^n$ <a href="https://math.stackexchange.com/questions/167843/i-have-to-show-1-frac1nn-is-monotonically-increasing-sequence"> is increasing</a> and $\lim (1+\frac1n)^n = e$ therefore $(1+\frac1n)^n \lt e$ and $\frac{u_{n+1}}{u_n} \lt 1$</p>
|
2,278,431 | <p>"Apply Green's Theorem to evaluate the line integral of F around positively oriented boundary"</p>
<p>$$F(x,y)=x^2yi+xyj$$</p>
<p>C: The region bounded by y=$x^2$ and y=4x+5</p>
| Clement C. | 75,808 | <p>The sequence is <strong>non-increasing</strong>. Before proving it, first: <em>compute the first terms, it will help you check whether your intuition is correct!</em></p>
<p>On <code>Mathematica</code> or <a href="http://www.wolframalpha.com/input/?i=DiscretePlot%5B+n%5En%2F(n!+E%5En),+%7Bn,+1,+100%7D%5D&t=crmtb01" rel="nofollow noreferrer">Wolfram</a>:</p>
<blockquote>
<p>DiscretePlot[ n^n/(n! E^n), {n, 1, 100}][<img src="https://i.stack.imgur.com/FFgKA.png" alt="enter image description here">]<a href="https://i.stack.imgur.com/FFgKA.png" rel="nofollow noreferrer">1</a></p>
</blockquote>
<p>Then, once you have done that and gotten a sense of what you have to prove (here: the sequence, if anything, does look decreasing) you can attempt to prove it.</p>
<p>Let $a_n \stackrel{\rm def}{=} \frac{n^n}{n!e^n}$. Then it is easy to see that
$$
\frac{a_n}{a_{n+1}} = e\left(\frac{n}{n+1}\right)^n
= \frac{e}{(1+\frac{1}{n})^n}
$$
Now, from $$
(1+\frac{1}{n})^n = e^{n\ln(1+\frac{1}{n})} \leq e^{n\cdot \frac{1}{n}} =e^1 =e
$$
(which follows from the standard inequality $\ln(1+x)\leq x$) we get
$$
\frac{a_n}{a_{n+1}} \geq 1
$$
for all $n\geq 1$.</p>
|
207,807 | <p>Is there an explicit example of a non-commutative monoid $M$ such that for all elements $m,n \in M$ and $p \in \mathbb{N}$ we have $(m \cdot n)^p=m^p \cdot n^p$?</p>
<p>It suffices to construct a semigroup $H$ with an absorbing element $0$ such that $a^2=0$ for all $a$, because then $M := H \cup \{e\}$ will do the job. This sounds easy at first glance, but I cannot find an example.</p>
| Berci | 41,488 | <p>Let $M:=\{1,a,b\}$ with the following multiplication:
$$\begin{align} a\cdot b:=a & \quad b\cdot a:=b \\
a\cdot a:= a & \quad b\cdot b:= b \end{align} $$
(In other words, $xy=x$ holds for all $x\ne 1$, so associativity comes easily.)</p>
|
2,713,201 | <p>How would you work something like this out? </p>
<p>Are there similar problems to
$$\frac{d\left( (\cos(x))^{\cos(x)}\right)}{dx}$$
which could be worked out the same way?</p>
| Community | -1 | <p>The go-to way for the derivative of $f^g$ is $(f(x)^{g(x)})'=(e^{g(x)\ln f(x)})'$ and then use chain rule. Or knowing by heart the formula that follows directly from this procedure. Of course, when $f(x)>0$, because for some reason we appear to be in a moment where people seem to be very adamant about saying stuff like "$\left(\frac1x\right)'=-\frac{1}{x^2}$ holds only if $x\ne 0$".</p>
|
2,713,201 | <p>How would you work something like this out? </p>
<p>Are there similar problems to
$$\frac{d\left( (\cos(x))^{\cos(x)}\right)}{dx}$$
which could be worked out the same way?</p>
| The Integrator | 538,397 | <p>For any function of the type $u(x) ^{v(x)} $, use the logarithmic rule</p>
<p>ie, let $y =u^v$</p>
<p>$\implies ln(y) = vln(u) $</p>
<p>Differentiate wrt x</p>
<p>$\frac{dy}{dx} \frac 1y = v'ln(u) +\frac vu u'$</p>
<p>$\frac{dy}{dx} =y\big(v'ln(u) +\frac vu u'\big) $</p>
<p>For $ cos(x)^{cos(x)} $</p>
<p>$\frac{dy}{dx} = \small cos(x)^{cos(x)} (-sin(x)ln(cos(x)) - sin(x) ) $</p>
|
2,713,201 | <p>How would you work something like this out? </p>
<p>Are there similar problems to
$$\frac{d\left( (\cos(x))^{\cos(x)}\right)}{dx}$$
which could be worked out the same way?</p>
| Latin Wolf | 536,550 | <p>So $\cos(x)^{\cos(x)}=e^{\cos(x)ln(\cos(x))}$ and $$\frac{d}{dx} e^{\cos(x)ln(\cos(x))}=e^{\cos(x)ln(\cos(x)} \cdot (-\sin(x) ln(\cos(x))+ -\sin(x)=\cos(x)^{\cos(x)} \cdot -\sin(x) ln(\cos(x))-\sin(x)$$</p>
|
3,443,094 | <blockquote>
<p>If <span class="math-container">$$\lim_{x\to 0}\frac{ae^x-b}{x}=2$$</span> the find <span class="math-container">$a,b$</span></p>
</blockquote>
<p><span class="math-container">$$
\lim_{x\to 0}\frac{ae^x-b}{x}=\lim_{x\to 0}\frac{a(e^x-1)+a-b}{x}=\lim_{x\to 0}\frac{a(e^x-1)}{x}+\lim_{x\to 0}\frac{a-b}{x}=\boxed{a+\lim_{x\to 0}\frac{a-b}{x}=2}\\
\lim_{x\to 0}\frac{a-b}{x} \text{ must be finite}\implies \boxed{a=b}\\
$$</span>
Now I think I am stuck, how do I proceed ?</p>
| trancelocation | 467,003 | <p>If you plug <span class="math-container">$a=b$</span> into your starting expression you get</p>
<p><span class="math-container">$$\lim_{x\to 0}a\frac{e^x-1}{x} = a \left.\left(e^x \right)'\right|_{x=0}= a \stackrel{!}{=} 2$$</span></p>
<p>Btw, you can show <span class="math-container">$a=b$</span> a bit easier by noting that</p>
<p><span class="math-container">$$\lim_{x\to 0}\frac{ae^x-b}{x}= 2 \Rightarrow \lim_{x\to 0}\left( \frac{ae^x-b}{x}\cdot x\right) = \lim_{x\to 0}\left(ae^x-b\right) = 0$$</span></p>
<p><span class="math-container">$$\Rightarrow a-b = 0$$</span></p>
|
3,408,846 | <p>This is an example in Serge Lang "Introduction to Linear Algebra", page 48. I try to multiply these two <span class="math-container">$2$</span>x<span class="math-container">$3$</span> and <span class="math-container">$3$</span>x<span class="math-container">$2$</span> matrices but fail to obtain the result as mentioned in the text.</p>
<p>I have:
<span class="math-container">$
\left( \begin{array}{cc}
2 & 1 & 5\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array} \right)
=\left( \begin{array}{c}
15 & 30 \\
4 & 2
\end{array} \right)
$</span></p>
<p>Serge's result is, however:</p>
<p><span class="math-container">$
\left( \begin{array}{cc}
2 & 1 & 5\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array} \right)
=\left( \begin{array}{c}
15 & 15 \\
4 & 2
\end{array} \right)
$</span></p>
<p>Where did I do wrong?</p>
| NoChance | 15,180 | <p>My answer is a different. Notice the entry <span class="math-container">$a_{22}$</span> in the result.
<a href="https://i.stack.imgur.com/vq2aS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vq2aS.png" alt="enter image description here"></a></p>
|
3,811,154 | <p>Prove that this integral is less than infinity. If <span class="math-container">$0<a<c$</span> and <span class="math-container">$0<b$</span>: <span class="math-container">$$\int_0^\infty \frac{|x|^a}{(x+b)^{c+1}} dx.$$</span></p>
<p>From inspection, because <span class="math-container">$a<c$</span> and <span class="math-container">$|x| < |x+b|$</span>, if this I was looking at the absolute convergence, then the numerator is smaller than the denominator, so it would converge. But, I am not used to trying to find convergence of integrals, I am used to series.</p>
<p>I tried doing integration by parts and substituting the original integral back in, but that did not work. I can show that here if someone wants to see.</p>
<p>Any suggestions on how to proceed?</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>Near <span class="math-container">$ +\infty $</span>, the integrand function satisfies</p>
<p><span class="math-container">$$\frac{x^a}{(x+b)^{c+1}}\sim \frac{1}{x^{c+1-a}}$$</span></p>
<p>So, if</p>
<p><span class="math-container">$$c+1-a>1$$</span>
or
<span class="math-container">$$a<c,$$</span></p>
<p>it will converge.</p>
|
619,890 | <p>I have a question.There is a group of 5 men and a group of 7 women.With how many ways can each of the 5 men get married with one of the 7 women?</p>
| miniBill | 33,238 | <p>The projection of a point $x$ onto $L(S)$ is the intersection of $x + W^\bot$ with $L(S)$, where $W$ is the linear part of $L(S)$ and $W^\bot$ is it's orthogonal space, that is, the linear space of vector orthogonal to $W$.</p>
|
984,558 | <p>Is it possible to find a representation of the infinitesimal generators of the special unitary group SU(3) that contains 4 by 4 matrices, by say taking a Kronecker product of its irreducible representation(s) with itself?</p>
<p>I know this is possible for SU(2), where one can express the three 4 by 4 matrices spanning the unit quaternion group in terms of some of the generators of SU(4), which are 4 by 4 matrices as well.</p>
<p>I am trying to do the same for SU(3). This question is motivated by the investigation of higher dimensional gauge theories. Thanks.</p>
| Stephen | 146,439 | <p>You might have a look at Fulton's book <em>Young tableaux</em> for the combinatorics relevant for computing these dimensions. Here is the relevant fact: the irreducible representations of $SU(n)$ may be indexed by partitions with at most $n-1$ parts in such a way that the dimension of the irreducible $L(\lambda)$ corresponding to a partition $\lambda$ is the number of column-strict Young tableaux (in the alphabet $\{1,2,\dots,n\}$) on $\lambda$. This gives a practical way of computing the dimensions that is a bit faster than using Weyl's character formula (the classical formula that can be expressed in a root-system uniform fashion) or more recent tools such as the Littelmann path model. The point being, for $SU(n)$ you don't need to know about root systems to operate a machine that will compute the dimensions. (In fact, things are even better: Schur functions give you the characters of the irreps and not just their dimensions).</p>
<p>In your example, the irreps would therefore be indexed by the partitions (in roughly increasing order of size) $$(0),(1),(1,1),(2),(2,1),(3),(2,2),(3,1),(4),\dots$$ of dimensions
$$1,3,3,6,8,10,6,15,15,... $$ as noted in Jyrki's comment above. The upshot for your problem is that a four dimensional representation is one of three things: the sum of four copies of the trivial irrep, or the sum of one copy of the trivial irrep with one of the two $3$ dimensional irreps. </p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| linksideal | 171,582 | <p>Note that you don`t want to prove or contradict $P$ or $R$ themselves. What you want to prove is the implication $P\rightarrow\neg R$. So you pick the assumption of this implication, in this case $P$, and use it for further argumentation.<br>
Now you can start your argumentation with another assumption like $R$ or with anything else. Important difference: now you can look for contradictions.</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| MattClarke | 118,792 | <p>Let me give a brief and less formal explanation than the other answers.</p>
<p>If you start with a set of assumptions {A<sub>1</sub>, A<sub>2</sub>, ... A<sub>n</sub>} and derive a contradiction, it proves that the set of assumptions is internally inconsistent (or as @J-Marcos phrased it, "jointly incompatible") -- i.e. they can't <em>all</em> be true.</p>
<p>So you can pick any one of the assumptions, let's say A<sub>i</sub>, and conclude that on the basis of the <em>other</em> assumptions, A<sub>i</sub> must be false.</p>
|
357,520 | <p>How can I show that
$$\lim_{a\to{0}}\frac{{\pi}a\ \coth{\mathrm{{\pi}a}-1}}{2a^2}=\frac{\pi^2}{6}$$
I think the limit is in $\frac{0}{0}$ form, so I am using L'Hospital's rule, and then I cannot solve further, Please Help.</p>
<p>Thanks!</p>
| Community | -1 | <p>Let $t = \pi a$. We then get that
$$\dfrac{\pi a \coth(\pi a)-1}{2a^2} = \pi^2 \dfrac{t \coth(t)-1}{2t^2}$$
Now let us look at $f(t) = \dfrac{t \coth(t)-1}{2t^2}$. We then have
$$f(t) = \dfrac{\coth(t)}{2t} - \dfrac1{2t^2}$$
Now use the Laurent series expansion. $$\coth(t) = \dfrac1t + \dfrac{t}3 + \mathcal{O}(t^3)$$
Hence, we have
$$f(t) = \dfrac{\dfrac1t + \dfrac{t}3 + \mathcal{O}(t^3)}{2t} - \dfrac1{2t^2} = \dfrac16 + \mathcal{O}(t^2)$$
Hence, we have
$$\lim_{t \to 0}f(t) =\dfrac16$$
Hence, we conclude that
$$\lim_{a \to 0}\dfrac{\pi a \coth(\pi a)-1}{2a^2} = \pi^2 \lim_{t \to 0}\dfrac{t \coth(t)-1}{2t^2} = \dfrac{\pi^2}6$$</p>
|
183,749 | <p>Considering two functions <span class="math-container">$\psi_{1}(u,v)$</span> and <span class="math-container">$\psi_{4}(u,v)$</span>. we have these two parial differential equation for them</p>
<p><span class="math-container">$(-2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{4}}{\partial u}+2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{4}}{\partial v})+(2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{1}}{\partial u}+2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{1}}{\partial v}-m\psi_{1})=0$</span></p>
<p><span class="math-container">$(2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{1}}{\partial u}-2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{1}}{\partial v})+(-2 i Sech[\frac{u}{\alpha}] \frac{\partial\psi_{4}}{\partial u}-2 i Sech[\frac{v}{\alpha}] \frac{\partial\psi_{4}}{\partial v}-m\psi_{4})=0$</span></p>
<p>I need to find <span class="math-container">$\psi_{1}(u,v)$</span> and <span class="math-container">$\psi_{4}(u,v)$</span>
I wrote these code for them but I don't know why is the problem and why it doesn't give the answer.</p>
<pre><code>DSolve[{((-2 I Sech[u/α] D[ψ4[u, v], u]+2 ISech[v/α] D[ψ4[u, v], v]) + (-m ψ1 [u, v] + 2 I Sech[u/α] D[ψ1[u, v], u] +2 I Sech[v/α] D[ψ1[u, v], v])) ==0, ((2I Sech[u/α] D[ψ1[u, v], u]-2 I Sech[v/α] D[ψ1[u, v], v]) + (-m ψ4[u, v] - 2 I Sech[u/α] D[ψ4[u, v], u]-2 I Sech[v/α] D[ψ4[u, v], v])) == 0}, {ψ1[u,v],ψ4[u, v]}, {u, v}]
</code></pre>
<p>I also tried <code>NDSolve</code> but still didn't get the answer</p>
| bbgodfrey | 1,063 | <p>This system of equations also can be solved symbolically, although not with <code>DSolve</code>. Beginning with </p>
<pre><code>eq = {((-2 I Sech[u/α] D[ψ4[u, v], u] + 2 I Sech[v/α] D[ψ4[u, v], v]) +
(-m ψ1[u, v] + 2 I Sech[u/α] D[ψ1[u, v], u] + 2 I Sech[v/α] D[ψ1[u, v], v])),
(( 2 I Sech[u/α] D[ψ1[u, v], u] - 2 I Sech[v/α] D[ψ1[u, v], v]) +
(-m ψ4[u, v] - 2 I Sech[u/α] D[ψ4[u, v], u] - 2 I Sech[v/α] D[ψ4[u, v], v]))};
</code></pre>
<p>make the transformation,</p>
<pre><code>rulet = {Sinh[u/α] -> ut, Sinh[v/α] -> vt};
</code></pre>
<p>with corresponding transformation of derivatives,</p>
<pre><code>ruled = {Derivative[1, 0][ψ_][u, v] -> Derivative[1, 0][ψ][ut, vt] Cosh[u/α]/α,
Derivative[0, 1][ψ_][u, v] -> Derivative[0, 1][ψ][ut, vt] Cosh[v/α]/α,
ψ_[u, v] -> ψ[ut, vt]}
</code></pre>
<p>Applying this transformation then eliminates the independent variables from the coefficients of the equations.</p>
<pre><code>Simplify[α eq /. ruled]
(* {((-2 I D[ψ4[ut, vt], ut] + 2 I D[ψ4[ut, vt], vt]) +
(-m α ψ1[ut, vt] + 2 I D[ψ1[ut, vt], ut] + 2 I D[ψ1[ut, vt], vt])),
(( 2 I D[ψ1[ut, vt], ut] - 2 I D[ψ1[ut, vt], vt]) +
(-m ψ4[ut, vt] - 2 I D[ψ4[ut, vt], ut] - 2 I D[ψ4[ut, vt], vt]))} *)
</code></pre>
<p>It follows, therefore, that the solution can be decomposed into Fourier modes,</p>
<pre><code>{ψ1 -> Function[{ut, vt}, ψ10 Exp[I ku ut + I kv vt]],
ψ4 -> Function[{ut, vt}, ψ40 Exp[I ku ut + I kv vt]]}
</code></pre>
<p>or, in terms of the original variables,</p>
<pre><code>{ψ1 -> Function[{u, v}, ψ10 Exp[I ku Sinh[u/α] + I kv Sinh[v/α]]],
ψ4 -> Function[{u, v}, ψ40 Exp[I ku Sinh[u/α] + I kv Sinh[v/α]]]}
</code></pre>
<p>The wavenumbers, <code>{ku, kv}</code> can be determined in the usual manner,</p>
<pre><code>Simplify[eq Exp[-I ku Sinh[u/α] - I kv Sinh[v/α]]/.%];
CoefficientArrays[% I α/4, {ψ10, ψ40}] // Last // Normal // Det
(* ku kv - (m^2 α^2)/16 *)
</code></pre>
|
4,039,424 | <p>I would like to calculate the elements of <span class="math-container">$\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$</span>.
I know that the elements of <span class="math-container">$\mathbb{Q}(\sqrt[3]{2})$</span> have the form of <span class="math-container">${a+b\sqrt[3]{2}+c\sqrt[3]{4}}$</span>, where a,b,c <span class="math-container">$\in \mathbb{Q}$</span> and the elements of <span class="math-container">$\mathbb{Q}(\sqrt{3})$</span> have the form of <span class="math-container">$a+b\sqrt{3}$</span>, where a,b <span class="math-container">$\in \mathbb{Q}$</span>.
I also calculated the minimal polynomial of <span class="math-container">$(\sqrt[3]{2}+\sqrt{3})$</span> over <span class="math-container">$\mathbb{Q}$</span> , which is: <span class="math-container">$x^6−9x^4−4x^3+27x^2−36x−23$</span>.</p>
<p>Can you help me to calculate the form of the elements of <span class="math-container">$\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$</span>? I have to find the elements of the linear combinations which form the higher powers of <span class="math-container">$(\sqrt[3]{2}+\sqrt{3})$</span> which are also powers of <span class="math-container">$(\sqrt[3]{2}+\sqrt{3})$</span>?
Could you give me a proper method to find the solution? Also, could you write down your calculation in your answer? Thank you for helping me!</p>
| reuns | 276,986 | <p><span class="math-container">$$\sum_{p|n} 1/p \le \sum_{p| \prod_{q\le k_n} q}1/p= \sum_{p\le O(\log n)} 1/p= O(\log\log\log n)$$</span></p>
<p>where <span class="math-container">$\prod_{q\le k_n} q$</span> is the least primorial <span class="math-container">$\ge n$</span>:</p>
<p><span class="math-container">$\prod_{q\le k_n} q\ge \exp(k_n/10)$</span> gives <span class="math-container">$k_n\le 10\log(n)$</span>.</p>
|
3,478,098 | <p><a href="https://i.stack.imgur.com/dcxhi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dcxhi.jpg" alt="enter image description here" /></a></p>
<p>Guys, Why is this weird statement true? It seems counterintuitive to me I cannot understand or lack creativity understanding it can you help me explain it? Guys please if possible make it visual.</p>
| Eric Wofsey | 86,856 | <p>This is immediate from the definition of <span class="math-container">$\delta$</span>. If <span class="math-container">$x$</span> is a rational number with denominator at most <span class="math-container">$N$</span>, that means <span class="math-container">$x\in Q_N$</span>, so by definition <span class="math-container">$|a-x|\geq\delta$</span> and thus <span class="math-container">$x\not\in (a-\delta,a+\delta)$</span>. So, any rational number which is in <span class="math-container">$(a-\delta,a+\delta)$</span> must have denominator greater than <span class="math-container">$N$</span>.</p>
|
33,993 | <p>I am given the parameters for a bivariate normal distribution ($\mu_x, \mu_y, \sigma_x, \sigma_y,$ and $\rho$). How would I go about finding the Var($Y|X=x$)? I was able to find E[$Y|X=x$] by writing $X$ and $Y$ in terms of two standard normal variables and finding the expectation in such a manner. I am unsure how to do this for the variance.</p>
<p>Also, how do I find the probability that both $X$ and $Y$ exceed their mean values (i.e., $P(X>\mu_x, Y > \mu_y)$)?</p>
<p>Thanks for the help!</p>
| Did | 6,179 | <p>Rather than embarking on some pretty involved computations of conditional distributions, one should rely on one of the main assets of Gaussian families, namely, the... </p>
<blockquote>
<p><strong>Key feature:</strong> In Gaussian families, conditioning acts as a linear projection.</p>
</blockquote>
<p>Hence, as the OP suggested, one could do worse than to start from a representation of $(X,Y)$ by standard i.i.d. Gaussian random variables $U$ and $V$, for example,
$$
X=\mu_x+\sigma_xU\qquad
Y=\mu_y+\sigma_y(\rho U+\tau V)$$ where the parameter $\tau$ is $$\tau=\sqrt{1-\rho^2}
$$
Since $\sigma_x\ne0$, the sigma-algebra generated by $X$ is also the sigma-algebra generated by $U$ hence conditioning by $X$ or by $U$ is the same. Furthermore, constants and functions of $X$ or $U$ are all $U$-measurable while functions of $V$ are independent on $U$, thus,
$$
\mathrm E(Y\mid X)=\mu_y+\sigma_y(\rho U+\tau \mathrm E(V))=\mu_y+\sigma_y \rho U
$$
which is equivalent to
$$
\color{red}{\mathrm E(Y\mid X)=\mu_y+\rho\frac{\sigma_y}{\sigma_x}(X-\mu_x)}
$$
Likewise, when computing conditional variances conditionally on $X$, deterministic functions of $X$ or $U$ should be considered as constants, hence their conditional variance is zero, and functions of $V$ are independent on $X$, hence their conditional variance is their variance. Thus,
$$
\mbox{Var}(Y\mid X)=\mbox{Var}(\sigma_y\tau V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V\mid X)=\sigma_y^2\tau^2\mbox{Var}(V)
$$
that is,
$$
\color{red}{\mbox{Var}(Y\mid X)=\sigma_y^2(1-\rho^2)}
$$
Finally, the event $$A=[X>\mu_x,Y>\mu_y]$$ is also
$$
A=[U>0,\rho U+\tau V>0].
$$
To evaluate $\mathrm P(A)$, one can turn to the planar representation of couples of independent standard Gaussian random variables, which says in particular that the distribution of $(U,V)$ is invariant by rotations. The event $A$ means that the direction of the vector $(U,V)$ is between the angle $\vartheta$ in $(-\pi/2,\pi/2)$ such that $$\tan(\vartheta)=-\rho/\tau$$ and the angle $\pi/2$. Thus,
$$\mathrm P(A)=\frac{\pi/2-\vartheta}{2\pi}$$
that is,
$$
\color{red}{\mathrm P(X>\mu_x,Y>\mu_y)=\frac14+\frac1{2\pi}\arcsin\rho}
$$
<strong>Numerical application:</strong> If $\mu_x=2$, $\mu_y=-1$, $\sigma_x=2$, $\sigma_y=1$ and $\rho=-\sqrt3/2$, then
$$
\mathrm E(Y\mid X)=-1+\sqrt3/2-(\sqrt3/4)X\qquad
\mbox{Var}(Y\mid X)=1/4
$$
and $\tau=1/2$, hence $\vartheta=\pi/3$ and $$\mathrm P(A)=1/12$$</p>
|
3,185,226 | <p>I have a simple question that confuses me for a while:</p>
<blockquote>
<p><span class="math-container">$$f(X) = \text{tr} \left( [ \log(X) ]^2 \right)$$</span></p>
<p>where <span class="math-container">$X$</span> is an <span class="math-container">$m \times m$</span> symmetric positive definite (SPD) matrix and <span class="math-container">$\log(X)$</span> is the matrix logarithm of matrix <span class="math-container">$X$</span>. What is <span class="math-container">$\frac{\partial f}{\partial X}$</span>?</p>
</blockquote>
<p>Using the chain rule, I have</p>
<p><span class="math-container">$df = \text{tr}(2ZdZ)$</span>,</p>
<p>where <span class="math-container">$Z=\log (X)$</span>. I think we should have <span class="math-container">$dZ = X^{-1}dX$</span> as a scalor function, so we will have</p>
<p><span class="math-container">$\frac{\partial f}{\partial X} = 2\log(X)X^{-1}$</span>,</p>
<p>but I haven't found any related reference.</p>
<p>Any comment or hint will be appreciated!</p>
| Robert Israel | 8,508 | <p>Careful: most of the standard calculus formulas for differentiation require things to commute. Matrices don't. So <span class="math-container">$d(Z^2)$</span> is not <span class="math-container">$2 Z \; dZ$</span>, it's <span class="math-container">$Z \; dZ + (dZ)\; Z$</span>. And I don't think there is a closed-form formula for <span class="math-container">$d(\log Z)$</span>.</p>
<p>However, if <span class="math-container">$X$</span> is symmetric, we can assume wlog that it is diagonal. Then you can easily compute
<span class="math-container">$f(X + dX)$</span> for <span class="math-container">$dX$</span> with a single matrix element.</p>
|
2,637,337 | <p>$ABC$ is a triangle and $A_1, B_1, C_1$ are points on $BC, CA, AB$ such that $$\frac{BA_1}{A_1C}=\frac{CB_1}{B_1A}=\frac{AC_1}{C_1B}=\lambda$$</p>
<p>If $A_2, B_2, C_2$ are points on $B_1C_1, C_1A_1$, and $A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1}=\frac{C_1B_2}{B_2A_1}=\frac{A_1C_2}{C_2B_1}=\frac{1}{\lambda}$$</p>
<p>prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of simlitude.</p>
<p>I'm missing something obvious, but I don't know what.</p>
| Jack D'Aurizio | 44,121 | <p>We have $B_1 = \frac{1}{1+\lambda} A +\frac{\lambda}{1+\lambda} C$ and cyclic identities.<br>
We have $B_2 = \frac{\lambda}{1+\lambda} A_1 +\frac{1}{1+\lambda} C_1$ and cyclic identities.<br>
By combining them
$$ B_2 = \frac{\lambda}{1+\lambda}\left(\frac{1}{1+\lambda}C+\frac{\lambda}{1+\lambda}B\right)+\frac{1}{1+\lambda}\left(\frac{1}{1+\lambda}B+\frac{\lambda}{1+\lambda}A\right)$$
$$ B_2 = \frac{\lambda}{(1+\lambda)^2}(A+B+C)+\color{red}{\frac{\lambda^2-\lambda+1}{(\lambda+1)^2}} B $$
where the red term is the wanted ratio of similitude.</p>
|
325,588 | <p>Is there an analytic function $f$ in $\mathbb{C}\backslash \{0\}$ s.t. for every $z\ne0$: $$|f(z)|\ge\frac{1}{\sqrt{|z|}}\, ?$$</p>
| MBM | 64,739 | <p>How about this:</p>
<p>Since $f(z)$ is analytic on $\mathbb{C}-\{0\}$, $g(z) = \frac{1}{(f(z))^2}$ is analytic on
$\mathbb{C}-\{0\}$. Also $\bigg|\frac{g(z)}{z}\bigg| \leq 1$. </p>
<p>I am sure you will finish the rest (think about the order of the pole at $0$ and use Liouville's Theorem).</p>
|
252,820 | <p><code>Sound[]</code> generates a visual representation of notes. I would like to extract that image, only notes, without controls and borders. How can I do it?</p>
<p>Take this example</p>
<pre><code>Sound[SoundNote @@@ Transpose @ {
{"E5", "D5", "F#4", "G#4", "C#5", "B4", "D4", "E4", "B4", "A4", "C#4", "E4", "A4"},
{1/8, 1/8, 1/4, 1/4, 1/8, 1/8, 1/4, 1/4, 1/8, 1/8, 1/4, 1/4, 1/2 }
}]
</code></pre>
<p><a href="https://i.stack.imgur.com/v3qNP.png" rel="noreferrer"><img src="https://i.stack.imgur.com/v3qNP.png" alt="enter image description here" /></a></p>
<p>So far, I found that I can <code>Rasterize</code> the whole output.
Also, if I press <code>Ctrl Shift E</code> on the output it seems like there is something, but the code is not simple and very low level. I'm not sure how to translate that into graphics primitives. This method requires manual intervention anyway, and I would like to get the image automatically.</p>
| creidhne | 41,569 | <p>Use <a href="http://reference.wolfram.com/language/ref/SparseArray.html" rel="nofollow noreferrer"><code>SparseArray</code></a> and <a href="http://reference.wolfram.com/language/ref/Band.html" rel="nofollow noreferrer"><code>Band</code></a> to reproduce your example 6-by-6 matrix:</p>
<pre><code>m = SparseArray[
{Band[{1,2},{-1,-1}]->{1,0}, Band[{2,1},{-1,-1}]->{-1,0}}, {6,6}];
AntisymmetricMatrixQ@m (* True *)
</code></pre>
<p>This simple change from <code>{6,6}</code> to <code>{36,36}</code> makes a 36-by-36 matrix:</p>
<pre><code>m = SparseArray[
{Band[{1,2},{-1,-1}]->{1,0}, Band[{2,1},{-1,-1}]->{-1,0}}, {36,36}];
AntisymmetricMatrixQ@m (* True *)
</code></pre>
<p>The matrix is a SparseArray. Use <code>Normal@m</code> to create an ordinary list.</p>
|
2,804,495 | <p>I was asked to solve this double integral:
Compute the area between $y=2x^2$ and $y=x^2$ and the hyperbolae $xy=1$ and $xy=2$ in </p>
<p>$$ \iint dx \,dy$$</p>
<p>I tried to solve it starting with considering that </p>
<p>$$x^2 \leq y \leq 2x^2 $$</p>
<p>suitabile for integration interval in $y$, obtaining the incomplete form</p>
<p>$$ \int^{x^2}_{2x^2} \int_\ldots^\ldots dx \,dy$$</p>
<p>but I also have
$$1 \leq xy \leq 2$$ and I would obtain a result in which I still have one independent variabile. </p>
<p>Please, can anyone help me? Thanks in advance.</p>
| mechanodroid | 144,766 | <p>$y \in [x^2, 2x^2]$ can be interpreted as $\frac{y}{x^2} \in [1,2].$</p>
<p>Consider the change of variables</p>
<ul>
<li><p>$u = \frac{y}{x^2} \in [1,2]$</p></li>
<li><p>$v = xy \in [1,2]$</p></li>
</ul>
<p>The Jacobian is given by</p>
<p>$$\frac1J = \begin{vmatrix} u_x & u_y \\ v_x & v_y
\end{vmatrix} = \begin{vmatrix} -\frac{2y}{x^3} & \frac1{x^2} \\ y & x
\end{vmatrix} = -\frac{3y}{x^3} = -3u$$</p>
<p>so $$dx\,dy = |J| \,du\,dv = \frac1{3u} \, du\,dv$$ </p>
<p>Therefore you need to calculate</p>
<p>$$\int_{[1,2]^2} \frac{1}{3u}\,du\,dv= \frac13\ln 2$$</p>
|
3,100,957 | <p>A fair coin is tossed until one of the patterns show up: TTH or THT.
Let A be the event that TTH shows up before THT.</p>
<p>What is P(A)?</p>
<p>Here is my solution but I am not sure if it is correct or there is a better solution.</p>
<p>Let <span class="math-container">$p=P(A)$</span>. Define </p>
<p><span class="math-container">$A_1=$</span> the event that the first toss is H</p>
<p><span class="math-container">$A_2=$</span> the event that the first two tosses are TT</p>
<p><span class="math-container">$A_3=$</span> the event that the first three tosses are THT</p>
<p><span class="math-container">$A_4=$</span> the event that the first three tosses are THH</p>
<p>Then this is a partition for the sample space. </p>
<p><span class="math-container">$p=P(A|A_1)P(A_1)+P(A|A_2)P(A_2)+P(A|A_3)P(A_3)+P(A|A_4)P(A_4)$</span>.</p>
<p>Then</p>
<p><span class="math-container">$p=p\frac{1}{2}+1\frac{1}{4}+0\frac{1}{8}+p\frac{1}{8}$</span>
which implies that <span class="math-container">$p=\frac{2}{3}.$</span></p>
| Kavi Rama Murthy | 142,385 | <p>Convergence in <span class="math-container">$\ell^{1}$</span> implies convergence of the coordinates. If a subsequence of <span class="math-container">$\{f_n\}$</span> converges, say to <span class="math-container">$f$</span>, then then we must have <span class="math-container">$f(k)=\frac 1 k$</span> for each <span class="math-container">$k$</span> but then <span class="math-container">$f \notin \ell^{1}$</span>. </p>
|
2,995,327 | <p>Suppose a,b ∈ Z. If 4 | <span class="math-container">$(a^2 + b^2)$</span> then a and b are not both odd.</p>
<p>So, assuming that 4 | <span class="math-container">$(a^2 + b^2)$</span> and <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are odd</p>
<p>this gives <span class="math-container">$4k=(2l+1)^2+(2u+1)^2$</span> for some <span class="math-container">$k,l,u\in z$</span></p>
<p>eventually leading to <span class="math-container">$4k=4(l^2+l+u)+2(u^2+1)$</span></p>
<p>The RHS is not a multiple of 4 when <span class="math-container">$u=2$</span> contradiction.</p>
<p>Is this valid, thanks.</p>
| 5xum | 112,884 | <p>It's not yet valid, because you haven't shown why the RHS cannot be a multiple of <span class="math-container">$4$</span>. You cannot simply set <span class="math-container">$u=2$</span>, because the <span class="math-container">$u$</span> you have is already determined by <span class="math-container">$b$</span>, since <span class="math-container">$b=2u+1$</span>.</p>
<hr>
<p>To correct your proof, re-think how you got from</p>
<p><span class="math-container">$$4k=(2l+1)^2 + (2u+1)^2$$</span></p>
<p>to </p>
<p><span class="math-container">$$4k = 4(l^2+l+u) + 2(u^2+1)$$</span></p>
<p>because I think you were a bit sloppy here.</p>
|
2,653,483 | <p>Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.</p>
<p>Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.</p>
<p>Without using a calculator, evaluate $\sqrt{a − b}$.</p>
| Ng Chung Tak | 299,599 | <p>\begin{align}
a &= \frac{10^{2018}-1}{9} \\
b &= \frac{2(10^{1009}-1)}{9} \\
a-b &= \frac{10^{2018}-2\times10^{1009}+1}{9} \\
&= \frac{(10^{1009}-1)^{2}}{9} \\
\end{align}</p>
<p><strong>Can you proceed?</strong></p>
|
2,545,516 | <p>So I have to assess the convergence of $$\displaystyle\sum_{n=1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right).$$</p>
<p>I'm told that it diverges, but can't really see why.</p>
<p>The divergence test doesn't really help, because
$\lim\limits_{x\to\infty}\displaystyle\frac{1}{\sqrt{n}}=0$, so</p>
<p>$\lim\limits_{x\to\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right)=0$, which doesn't conclude its divergence.</p>
<p>I doubt the ratio test would be much of use in this situation.</p>
<p>I can't imagine using the integral comparison test, as I wouldn't know where to start with $\displaystyle\int_{1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right) \mathrm dx$.</p>
| Randall | 464,495 | <p>Limit-compare to $\sum_n \frac{1}{\sqrt{n}}$. You need the limit
$$
\lim_{n \to \infty} \frac{\sin(1/\sqrt{n})}{1/\sqrt{n}}
$$
which is $1$ by a change of variables, making use of
$$
\lim_{t \to 0^+} \frac{\sin t}{t}=1.
$$
To get from point A to point B, change variables by $t=\frac{1}{\sqrt{n}}$. As $n \to \infty$, $t \to 0^+$.</p>
<p>Since $\sum_n \frac{1}{\sqrt{n}}$ diverges, so does your series.</p>
<p>The terms in your series are positive, so this is legal. </p>
|
52,079 | <p>I'm in doubt about the topology of maps between fibres of vector bundles.</p>
<p>Consider $E$ and $F$ vector bundles and the set of all linear maps from a fibre of $E$ to a fibre of $F$, ie, the set of all linear maps $T:E_x \rightarrow F_y$, where $E_x$ is the fiber over $x$ and $F_y$ is the fiber over $y$.</p>
<p>I want to know how to define the topology of this set.</p>
<p>I need this topology for this question: Consider $f: E \rightarrow F$, a map that preserves each fiber and its restriction to each fibre, $f_x : E_x \rightarrow F_y$, is differentiable. The differential of $f_x$ calculated in a vector $v \in E_x$ is the linear maps $df_x(v):E_x \rightarrow F_y$. I want to say that $f$ is a $C^1$ map if the function $v \in E \rightarrow df_{\pi(v)} (v)$ is continuous. And for this I need a topology for the set defined above.</p>
<p>Does anybody know how to define the topology? What does it mean that two of these maps are close to each other?</p>
<p>Note that $df_x (v)$ is not necessarily a homomorphism, because this is only defined over the fiber that contains $v$.</p>
| Johannes Ebert | 9,928 | <p>Let $G_i$, $i=0,1$, be topological groups and $P_i \to X_i$ be $G_i$-principal bundles. Then $P_0 \times P_1 \to X_0 \times X_1$ is a $G_0 \times G_1$-principal bundle. Let $V_i$ be topological vector spaces with continuous $G_i$-actions. The group $G_0 \times G_1$ acts on $Hom(V_0,V_1)$. If the topology on the Hom-space is appropriately chosen, the action is continuous and you can form the bundle $(P_0 \times P_1) \times_{G_0 \times G_1 } Hom (V_0,V_1)$ on $X_0 \times X_1$. A point in the total space is precisely a linear map between two fibres. In the finite-dimensional case, you can take $P_i$ to be the frame bundle of $E_i$. All the topologies on linear groups and Hom-spaces are unique and you get a unique answer. In the infinite-dimensional case, you need to be very careful about the topology on the groups $G_i$.</p>
|
2,365,933 | <p>I'm aware of how we can simplify functions which have $Arc$ as an argument . For example $\sin(\cos^{-1}(x)) = \sqrt{1-x^2}$ but what about cases which $Arc$ is out of the parentheses ? For instance consider this : $\sin^{-1}(\tan x)$ . Is there any way for simplification ? </p>
| Sidharth Ghoshal | 58,294 | <p>You can squeeze some more out of this.</p>
<p>Consider <span class="math-container">$$ f = \ln ( \sec x + \tan x)$$</span></p>
<p>It’s easy to see that <span class="math-container">$f’ = \sec x$</span>. Now consider <span class="math-container">$g = sin^{-1} (i \tan(x))$</span></p>
<p>We have that <span class="math-container">$g’ = \sec(x)$</span></p>
<p>So there is some constant <span class="math-container">$K$</span> such that: </p>
<p><span class="math-container">$$ \ln ( \sec x + \tan x) + K = \sin^{-1} (i \tan(x))$$</span></p>
<p><span class="math-container">$$ \ln ( ( (\frac{i \tan x }{i}) ^2 + 1) ^{1/2}+ \frac{i \tan x }{i} )+ K = sin^{-1} (i \tan(x))$$</span></p>
<p>So we can conclude: </p>
<p><span class="math-container">$$ \ln ( \sqrt{ 1 - \tan^2} - i \tan(x) ) = \sin^{-1} \tan x $$</span></p>
<p>Which might be a little more insightful. </p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Ian Agol | 1,345 | <p>This is not what you're looking for, but I always remember Milnor's diagram in Chapter 9 of his book on <a href="https://www.maths.ed.ac.uk/%7Ev1ranick/papers/milnmors.pdf" rel="noreferrer">Morse Theory</a> describing the symmetries of the curvature tensor.</p>
<p><a href="https://i.stack.imgur.com/WwXkX.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WwXkX.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/sOP0Z.png" rel="noreferrer"><img src="https://i.stack.imgur.com/sOP0Z.png" alt="enter image description here" /></a></p>
|
1,368,988 | <p>I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.</p>
<p><img src="https://i.stack.imgur.com/vueMQ.png" alt="An example for n=8."></p>
<p>Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$</p>
<p>We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit:
$$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$</p>
<p>Now for my question: <strong>How would you solve the opposite problem?</strong> To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?</p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice, $$\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}$$
$$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(1-\cos\left(\frac{2\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(2\sin^2\left(\frac{\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{2n}{2}\sin\left(\frac{\pi}{n}\right)$$ $$=\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\frac{1}{n}}$$ $$=\pi\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\left(\frac{\pi}{n}\right)}$$ Let $\frac{\pi}{n}=t\implies t\to 0\ as\ n\to \infty$ $$=\pi \lim_{t\to 0}\frac{\sin t}{t}$$ $$=\pi\times 1=\pi$$</p>
|
200,903 | <p>My teacher was explaining quadratics in my class and it was a little bit unclear to me. The problem was <br> <br>
Suppose $at^2 + 5t + 4 > 0$, show that $a > 25/16$ . <br> <br></p>
<p>My teacher said that there are no solutions for this function when it is greater than $0$ and used $b^2-4ac \lt 0$, and this is the part that confused me. I understand why he used $b^2-4ac \lt 0$ but I cannot understand why there are no solutions by just looking at the function. Could someone explain this to me?</p>
| i. m. soloveichik | 32,940 | <p>In order for $f(t)=at^2+5t+4>0$ all $t$, then the discriminant must be negative, else there are real roots to $f(t)$. The discriminant is $25-16a$, so $25-16a<0$ and thus $a>25/16$.</p>
|
103,675 | <p>I have defined a recursive sequence</p>
<pre><code>a[0] := 1
a[n_] := Sqrt[3] + 1/2 a[n - 1]
</code></pre>
<p>because I want to calculate the <code>Limit</code> for this sequence when n tends towards infinity.</p>
<p>Unfortunately I get a <code>recursion exceeded</code> error when doing:</p>
<pre><code>Limit[a[n], n -> Infinity]
</code></pre>
<p>How can I calculate the <code>Limit</code> for this sequence using Mathematica?</p>
| Daniel Lichtblau | 51 | <p>[Possibly this is too cheap and should just be a comment, I'm not sure.]</p>
<p>One way to go about such problems is to realize that "in the limit", <code>a[n]==a[n-1]</code>. So just solve an algebraic equation. If there are multiple solutions then you need to figure out which is correct based on initial values and some other reasoning.</p>
<pre><code>Solve[x == Sqrt[3] + x/2, x]
(* Out[130]= {{x -> 2 Sqrt[3]}} *)
</code></pre>
|
103,675 | <p>I have defined a recursive sequence</p>
<pre><code>a[0] := 1
a[n_] := Sqrt[3] + 1/2 a[n - 1]
</code></pre>
<p>because I want to calculate the <code>Limit</code> for this sequence when n tends towards infinity.</p>
<p>Unfortunately I get a <code>recursion exceeded</code> error when doing:</p>
<pre><code>Limit[a[n], n -> Infinity]
</code></pre>
<p>How can I calculate the <code>Limit</code> for this sequence using Mathematica?</p>
| J. M.'s persistent exhaustion | 50 | <p>As of version 11.2, this can be dealt with by <code>RSolveValue[]</code>:</p>
<pre><code>RSolveValue[{a[n] == Sqrt[3] + 1/2 a[n - 1], a[0] == 1}, a[∞], n]
2 Sqrt[3]
</code></pre>
|
3,497,420 | <p>Consider the function <span class="math-container">$$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$</span> The point <span class="math-container">$(0,0)$</span> is a critical point. Observe,
<span class="math-container">\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\
f_{yy} & = 4, f_{yy}=4
\end{align*}</span></p>
<p>So, in order to determine the nature of the above critical point, we need to check the Hessian at <span class="math-container">$(0,0)$</span> which is <span class="math-container">$0$</span> and hence the test is inconclusive. <span class="math-container">$$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$</span>So, I tried to see the function on slices like <span class="math-container">$y=0$</span> and <span class="math-container">$y=x$</span> but nothing worked. So please suggest me how do I find the nature of the critical point in this case?</p>
| Luca Goldoni Ph.D. | 264,269 | <p>You have that
<span class="math-container">$$
g_a(x)=f(x,ax^2 ) = 2\left( {a^2 - a} \right)x^4 + \left( {1 - a} \right)x^6
$$</span>
With <span class="math-container">$0<a<1$</span> the function <span class="math-container">$g_a(x)$</span> has a local maximum. With <span class="math-container">$a>1$</span> the function has a local minimum. This means that <span class="math-container">$(0,0)$</span> is a saddle point.</p>
|
3,826,994 | <p>I would like to find <span class="math-container">$z$</span> which minimizes the below, when <span class="math-container">$x$</span> is held at a specific value.</p>
<p><span class="math-container">$f(x,z) =\sqrt{\sqrt{x^2 + z^2} - 0.25}$</span></p>
<p>For example; I would like to find the value of <span class="math-container">$z$</span> which minimizes the function when <span class="math-container">$x = 0.5$</span></p>
| Harshit Raj | 782,789 | <p>As you said solving gets you here:
<span class="math-container">$$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$</span></p>
<p>and finally, substituting gets u here:
<span class="math-container">$$x = \frac{-2b +6b^2}{1+b^2}$$</span> and <span class="math-container">$$y=\frac{-6b +2}{1+b^2}$$</span></p>
<p>Just divide above two equations you get</p>
<p><span class="math-container">$$x = \frac{-2b + 6b^2}{1 + b^2} = -\frac{2 - 6b}{1 + b^2}\cdot b = -by$$</span>
<span class="math-container">$$\implies x+by=0$$</span></p>
|
1,572,045 | <p>This is maybe a stupid question, but I want to find the roots of:</p>
<blockquote>
<p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$</p>
</blockquote>
<p>What that I did:</p>
<p>$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$</p>
<p>So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ </p>
<p>My questions:</p>
<p>$1)$ Is there an easy way to see that $x=-8$ is a root too?</p>
<p>$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$</p>
| Laurent Duval | 257,503 | <p>First advice: when you are looking for roots of a polynomial, factorize as much as you can. You have $(x-1)^2$ and $(x-1)^3$, so you can factorize $(x-1)^2$. You have $(x+2)$ and $(x+2)^2$, so you can factorize $(x+2)$. Thus you get:
$$(x+2)(x-1)^2 (2(x-1)-3(x+2))\,. $$ So:</p>
<ol>
<li>For question 1, the $-8$ root should be evident (develop the last term)</li>
<li>For question 2: look at powers, $(x-1)^2$ means that $1$ is a root, twice. So your roots are $-8$, $-2$ and $1$ and $1$ again.</li>
</ol>
|
38,659 | <p>I know how to use Matrix Exponentiation to solve problems having linear Recurrence relations (for example Fibonacci sequence). I would like to know, can we use it for linear recurrence in more than one variable too? For example can we use matrix exponentiation for calculating ${}_n C_r$ which follows the recurrence C(n,k) = C(n-1,k) + C(n-1,k-1). Also how do we get the required matrix for a general recurrence relation in more than one variable?</p>
| Patrick Da Silva | 10,704 | <p>You shouldn't be so dramatically holding on to those rules ; notice that they are not called theorems, but rules ; that is because they are principles that we did not prove but stated so that students could help themselves to count stuff. I am usually comfortable with basing myself on axioms to prove theorems, but what you are trying to do here is not to prove a theorem, but to count stuff. To do so, use whatever you feel comfortable with, and do not bother about "rules" whose name seem to govern your mind... because you can get confused very quickly.</p>
<p>Once you have understood that to choose $k$ items amongst $n$ possible items, you have $n \choose k$ ways of doing so, it doesn't matter the rule you used to understand how to do so, because once you did, this becomes a new "rule" you can apply as free as you want. What I'm saying is, do not try to "fit in the name of the rules" ; understanding why $10 \choose 2$ finds the correct number is a good thing, but trying to fit in the name of the rules in the calculations you do is not something important. </p>
<p>A purely product rule can be applied if you still wish to "fit in the names of the rules" in your calculation" : you have to choose 2 chapters among 10, now there are 10 choices for your first chapter, and 9 choices for your second, since everytime you're about to choose your second chapter, one chapter's gone. Hence there is 90 choices of first and second chapter by the product rule (The number of choices in each case is independent of each other, even though the choices available are dependent, but the counting does not differ.). Since you do not wish to distinguish a pair of chapters by whether one of the two is first or second, you've noticed you've counted the double of what you wished to count, hence you divide 90 by 2 to get ${10 \choose 2} = 45$ and you never "added" stuff in here.</p>
<p>I hope this helped.</p>
|
691,494 | <p>Suppose I am given a state space $S=\{0,1,2,3\}$ with transition probability matrix </p>
<p>$\mathbf{P}= \begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3} & 0\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3}\\[0.3em]
0 & 0 & 0 & 1
\end{bmatrix}$</p>
<p>and I want the expected number of steps from states $0 \rightarrow 3$ which I will denote $E_0(N(3))$.</p>
<p>Attempt at solving: First I write the transient states $\{0,1,2\}$ and recurrent state $\{3\}$ which I got from drawing the chain. I now want to write $\mathbf{P}$ in canonical form, i.e. with state space $S=\{3,0,1,2\}$ as so:</p>
<p>$\mathbf{P}=\begin{bmatrix}
1 & 0 & 0 & 0\\[0.3em]
0 & \frac{2}{3} & \frac{1}{3} & 0 \\[0.3em]
0 & \frac{2}{3} & 0 & \frac{1}{3}\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3}
\end{bmatrix}$</p>
<p>It's clear that the transient matrix is</p>
<p>$\mathbf{Q}=
\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3}\\[0.3em]
0 & 0 & \frac{1}{3}
\end{bmatrix}$</p>
<p>Now I can get the matrix I want for computing expected steps (calculated with Mathematica):</p>
<p>$\mathbf{M}=(\mathbf{I}-\mathbf{Q})^{-1}=\begin{bmatrix}
9 & 3 & 3\\[0.3em]
6 & 3 & 3\\[0.3em]
0 & 0 & \frac{3}{2}
\end{bmatrix}$</p>
<p>From this, we get $E_0(N(3))=9+3+3=15$. Is this correct? I am sort of weak in finding the "canonical form" of a matrix.
Note: although this looks like a homework question, it's simply a preparation problem for an upcoming exam, so a complete solution/correction of my work is appreciated.</p>
| Gareth | 79,908 | <p>The distribution for the number of time steps to move between marked states in a discrete time Markov chain is the <a href="https://en.wikipedia.org/wiki/Discrete_phase-type_distribution" rel="noreferrer">discrete phase-type distribution</a>. You made a mistake in reorganising the row and column vectors and your transient matrix should be
$$\mathbf{Q}=
\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{2}{3} & 0 & \frac{1}{3}\\
\frac{2}{3} & 0 & 0
\end{bmatrix}$$
which you can then continue to find
$$\mathbf M = (\mathbf I - \mathbf Q)^{-1} =
\begin{bmatrix}
27 & 9 & 3\\
24 & 9 & 3 \\
18 & 6 & 3\end{bmatrix}$$
and summing the first row gives you (as you require) 39.</p>
<p>Generally when I have seen these distributions written the 'exit' state is put last, which would mean your matrix with elements in the order $\{0,1,2,3\}$ was already in canonical form and you needed to extract the top left portion as your transient matrix,</p>
<p>$$\mathbf P =
\left(
\begin{array}{ccc|c}
\frac{2}{3} & \frac{1}{3} & 0 & 0\\
\frac{2}{3} & 0 & \frac{1}{3} & 0 \\
\frac{2}{3} & 0 & 0 & \frac{1}{3}\\
\hline\\
0 & 0 & 0 & 1
\end{array}
\right)
$$
For further moments you might be interested in this paper:</p>
<ul>
<li>Dayar, Tuǧrul. "On moments of discrete phase-type distributions." Formal Techniques for Computer Systems and Business Processes. Springer Berlin Heidelberg, 2005. 51-63. <a href="http://dx.doi.org/10.1007/11549970_5" rel="noreferrer">doi:10.1007/11549970_5</a>.</li>
</ul>
|
3,490,329 | <blockquote>
<p>Show that a 2-dimensional subspace of the space of <span class="math-container">$2\times2$</span> matrices contains a non-zero symmetric matrix. </p>
</blockquote>
<p>I don't know if it should be written like the addition of two symmetric and skew-symmetric matrix or there is another way to show it. </p>
| Josh Messing | 579,368 | <p>If <span class="math-container">$V$</span> is the linear subspace in question then we can write <span class="math-container">$V$</span> as
<span class="math-container">$$
V = \{ \lambda_1A + \lambda_2B : \lambda_1,\lambda_2 \in \mathbb{R}\}
$$</span>
for two linearly independent matrices <span class="math-container">$A \text{ and }B$</span>. An element of <span class="math-container">$V$</span> is symmetric if an only if
<span class="math-container">$$
\lambda_1A_{1,2} + \lambda_2B_{1,2} = \lambda_1A_{2,1} + \lambda_2B_{2,1}
\iff
\begin{pmatrix} A_{1,2} & B_{1,2} \end{pmatrix}\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix}
=
\begin{pmatrix} A_{2,1} & B_{2,1} \end{pmatrix}\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix}
$$</span>
<span class="math-container">$$
\iff
\begin{pmatrix} A_{1,2}-A_{2,1} & B_{1,2}-B_{2,1} \end{pmatrix}\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix} = 0
$$</span>
There must be a non zero solution <span class="math-container">\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix}</span> to the equation above since the matrix <span class="math-container">\begin{pmatrix} A_{1,2}-A_{2,1} & B_{1,2}-B_{2,1} \end{pmatrix}</span> can't be 1-1(it is from <span class="math-container">$\mathbb{R}^2$</span> to <span class="math-container">$\mathbb{R}$</span>). The fact that the solution vector is non zero implies that the resultant symmetric matrix <span class="math-container">$\lambda_1A+\lambda_2B$</span> is non zero since <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are linearly independent. </p>
|
1,066,484 | <p>Given a simple connected bipartite graph $G$ with degree of vertices equal to $k$, where $k\ge 2$. Prove that there is no cut vertex exist in $G$. </p>
<p>Cut vertex $v$ here is a vertex which make the graph induced have number of connected component $>1$ when $v$ is removed.</p>
<p>I have tried to prove by contradiction but i have no clue about what contradiction can be obtained. I am quite curious about what the point the graph has to be bipartite is here. I have not come across any place to adopt the property of a bipartite graph so far. Any hints on tackling this problem would be appreciated. Last but not least, thanks for reading.</p>
<p>Edited: i have found something </p>
| Rebecca J. Stones | 91,818 | <p>Let $x$ be a cut vertex and properly $2$-color the graph red and green. If we delete $x$ we get some connected components: in one of these components, let the red vertices belong to $A$ and green vertices belong to $B$.</p>
<p>The graph induced by $\{x\} \cup A \cup B$ looks as follows:</p>
<p><img src="https://i.stack.imgur.com/ALIGo.png" alt="The graph induced by $\{x\} \cup A \cup B$"></p>
<p>We reach a contradiction if we compare modulo $k$ (a) the number of edges coming out of $A$ and (b) the number of edges going into $A$.</p>
|
2,996,920 | <p>Please recommend me a good book to study interpolation techniques such as polynomial interpolation, cubic, spline interpolations, if possible tell me the branch of mathematics that deals with this subject. I want to go in depth with this topic.</p>
| TurlocTheRed | 397,318 | <p>Numerical Recipes is a highly recommended classic. I personally know working physicists who use it whenever they need some numerical work done:</p>
<p><a href="http://numerical.recipes/oldverswitcher.html" rel="nofollow noreferrer">http://numerical.recipes/oldverswitcher.html</a></p>
|
211,427 | <p>Can't seem to figure this one out. Could anyone help me out and explain it to me?<br>
Thank you.</p>
<p>Let $P$ and $Q$ be relations on $Z$ by x$P$y iff x + 1 <= y and a$Q$b iff a + 2 <= b. Prove that P $\circ$ Q = {(p,q) belonging to ZxZ | p + 3 <= q} </p>
| Michael Albanese | 39,599 | <p>I will outline how to go from a symmetric bilinear form to a linear operator.</p>
<p>Given a symmetric bilinear form $\sigma : V \times V \to F$, consider the map $L' : V \to V^*$ given by $L'(v) = \sigma(v, \cdot)$. It is not a linear operator as the codomain is $V^*$ not $V$. However, as $V$ is finite dimensional, $V^*$ is isomorphic to $V$; let $\phi : V^* \to V$ be an isomorphism. Now it is not hard to check that $L = \phi\circ L'$ defines a linear operator on $V$.</p>
<hr>
<p>It is worth noting that the isomorphism between $V$ and $V^*$ is not canonical, you need to choose a basis for $V$ in order to define one.</p>
|
3,411,081 | <p>A group of 12 people are going out to a concert on Saturday night. The group will take three cars with four people in each car. If they distribute themselves at random, what is the probability that A and B will be in the same car?</p>
<p>I tried (12C2*10C2*8C4*4C4)/(12C4*8C4*4C4) because you're choosing two first and then sorting the rest. This gave me a more than 100% probability.</p>
<p>The answer at the back is 0.273 assuming the cars are non-distinct.
The closest I came to it was 4C2/12C3 but I'm not sure why this works.</p>
| Alexander Geldhof | 560,477 | <p>Let the number be <span class="math-container">$z + 1$</span>. Jot down <span class="math-container">$1$</span> on your paper.</p>
<p>Now decide whether you want to add <span class="math-container">$1$</span> to the <em>last</em> number you already wrote down (i.e. make it <span class="math-container">$2$</span>) or proceed to the next factor in the sum (<span class="math-container">$1+1$</span>). You have two choices here.</p>
<p>Continue like this and make the decision to add one to the last number or proceed <span class="math-container">$z$</span> times. This leads to <span class="math-container">$2^z$</span> distinct configurations, depending on which series of two-option choices you made.</p>
<p>You still have to prove that you can achieve any desired factorization with this process, and that different choices made lead to different factorizations, but neither are particularly hard.</p>
|
3,411,081 | <p>A group of 12 people are going out to a concert on Saturday night. The group will take three cars with four people in each car. If they distribute themselves at random, what is the probability that A and B will be in the same car?</p>
<p>I tried (12C2*10C2*8C4*4C4)/(12C4*8C4*4C4) because you're choosing two first and then sorting the rest. This gave me a more than 100% probability.</p>
<p>The answer at the back is 0.273 assuming the cars are non-distinct.
The closest I came to it was 4C2/12C3 but I'm not sure why this works.</p>
| Community | -1 | <p>Let <span class="math-container">$f(n)$</span> be the number you're looking for for a given <span class="math-container">$n$</span>. Let's work by strong induction.</p>
<p>Consider <span class="math-container">$f(n+1)$</span>. It could either have one term (which happens in one way) or more than one term. In this second case, the second term can be any number from <span class="math-container">$1$</span> through <span class="math-container">$n$</span>. If the first term is <span class="math-container">$k$</span>, then there are <span class="math-container">$f(n+1-k)$</span> ways to complete that sum. Adding all that together with our induction hypothesis that <span class="math-container">$f(k)=2^{k-1}$</span>, we get</p>
<p><span class="math-container">$$f(k+1)=1+2^0+2^1+...+2^{k-1}$$</span></p>
<p>which is obviously <span class="math-container">$f(k+1)=2^k$</span>.</p>
|
1,270,042 | <p>$$(a+5)(b-1)=ab-a+5b-5=20-5=15.$$</p>
<p>So, both $a + 5$ and $b-1$ divide $15$. </p>
<p>Then, $a + 5$ is one of $15, -15, 3, -3, 5, -5, 1, -1$, so $a$ is one of $10, -20, -2, -8, 0, -10, -4, -6$ and $b – 1$ is one of $15, -15, 3, -3, 5, -5, 1, -1$, so $b = 14, -14, 4, -2, 6, -4, 2, 0$.</p>
<p>Could all possibilities for $a, b$ found by considering $(a+5)(b-1)$ be just random(not in a probability sense) and not connected to $ab = a - 5b + 20$ at all? In other words, could it be that if some of the possible $a, b$ found this way happen to satisfy $ab = a - 5b + 20$, then it's just a coincidence? </p>
| DeepSea | 101,504 | <p>Or write $b = \dfrac{a+20}{a+5} = 1 + \dfrac{15}{a+5}$, thus $a+5 \mid 15$, and $a+5 = 1, 3, 5, 15$ are the possibilities. But since we consider only positive values, we have:$a+5 = 15 \to a = 10 \to b = 1+1 = 2$.</p>
|
2,135,191 | <p>In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence? </p>
<p>The answer given is $\frac{100}{101}$, but I am not sure how.</p>
<p>So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$</p>
<p>The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator. </p>
| Arnaldo | 391,612 | <p>$$a_1+a_2+a_3+...+a_{100}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{100}-\frac{1}{101}\right)=1-\frac{1}{101}=\frac{100}{101}$$</p>
<p>Note that $-\frac{1}{2}$ cancel $\frac{1}{2}$, $-\frac{1}{3}$ cancel $\frac{1}{3}$ and go on until $-\frac{1}{100}$ cancel $\frac{1}{100}$.</p>
|
2,135,191 | <p>In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence? </p>
<p>The answer given is $\frac{100}{101}$, but I am not sure how.</p>
<p>So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$</p>
<p>The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator. </p>
| Mark Fischler | 150,362 | <p>$$a_{100} = \left( \frac11 - \frac12 \right)+\left( \frac12 - \frac13 \right)+
\left( \frac13 - \frac14 \right)+\cdots + \left( \frac1{100} - \frac1{101} \right)
\\= \frac11 + \left( -\frac12+\frac12\right)+\left( -\frac13+\frac13\right)
+\cdots +\left( -\frac1{100}+\frac1{100}\right) -\frac1{101} = 1 - \frac1{101}
$$</p>
|
635,077 | <p>$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$</p>
<p>How can I prove this statement?</p>
| DanielWainfleet | 254,665 | <p>Here is a modern proof using sleight of hand.(1)An isometry of the Euclidean plane $E$ is a function $f:E\to E$ that preserves the distance $d$ between points: $d(A,B)=d(f(A),f(B))$ for all $A,B\in E$. We have (2):An isometry $f$ is onto.Because, for $Q\in E$, let $A_1,A_2,A_3\in E$ be non-co-linear with $Q\not \in \{f(A_1),f(A_2),f(A_3)\}$.The unique $P$ such that $d(P,A_i)=d(Q,f(A_i))$ for $i=1,2,3$ must satisfy $f(P)=Q$.We have (3):An isometry that fixes 3 non-co-linear points $A_1,A_2,A_3$ is the identity function.Because,as in (2),for any $P$, the only $Q$ that can satisfy $ d(P,A_i)=d(Q,f(A_i))=d(Q,A_i)$ for $i=1,2,3$ is $Q=P$.We have (4):If isometries $ f,g$ agree on 3 non-co-linear points then $f=g$. Because $g^{-1}:E\to E$ exists by (2), and hence is also an isometry (obviously), and by (3) we have $g^{-1}(f(P))=P$ for all $ P$, so $$g(P)=g(g^{-1}(f(P))=f(P).$$ Now to use all this :(5): Choose a pair of orthogonal co-ordinate axes for $E$ with origin $O$. Consider two types of isometry :First, $R_b$ is a counter-clockwise rotation about $O$ through angle $b$. Second, $M_b$ sends the point with co-ordinates $(x,y)$ to $$(x\cos b-y\sin b,x\sin b+y\cos b).$$The fact that $M_b$ is an isometry relies only on the theorem of Pythagoras, or in other words, $\cos^2 b+\sin^2 b=1$. Now for any $b$, observe that $R_b$ and $M_b$ agree on the 3 non-co-linear points $(0,0),(1,0),(0,1)$.Therefore by (4) we have $$R_b=M_b\text { for all b }.$$Finally, since $R_b$ and $R_c$ are rotations about $O$ we have$$R_bR_c=R_{b+c} \text{ for all b,c }.$$ Therefore we have $$M_{b+c}=R_{b+c}=R_bR_c=M_bM_c\text{ for all b,c}.$$ By comparing the co-ordinates of $M_{b+c}(1,0)$ with the co-ordinates of $M_bM_c(1,0)$, we have the angle-sum formulas for $\cos$ and $\sin$.</p>
|
226,551 | <p><strong>(1)</strong> Which graph classes are extremely tough to test for graph non-isomorphic pairs from isomorphic pairs?</p>
<p><strong>(2)</strong> Is there a repository of adjacencies from such classes?</p>
| John Machacek | 51,668 | <p>In <a href="http://people.mpi-inf.mpg.de/~pascal/docs/KutzSchw_ScrewBox.pdf" rel="noreferrer">ScrewBox: a Randomized Certifying Graph-Non-Isomorphism Algorithm</a> by Kutz and Schweitzer it is stated that "It is generally accepted that the incidence graphs of finite projective planes confront graph isomorphism algorithms with great challenges."</p>
<p>Also the <a href="https://people.mpi-inf.mpg.de/~pascal/docs/thesis_pascal_schweitzer.pdf" rel="noreferrer">PhD Thesis</a> of Schweitzer discusses difficult graphs (including incidence graph of projective planes) in section 2.8.</p>
<p>I do not know any repository of adjacencies of these graph, but Moorhouse has <a href="http://www.uwyo.edu/moorhouse/pub/planes27/" rel="noreferrer">data</a> online for known projective planes of order 27 online. Also the paper of Kutz and Schweitzer as well as the thesis of Schweitzer both cite data that Royle has on projective planes on order 16 but that link seems to broken for me right now.</p>
|
226,551 | <p><strong>(1)</strong> Which graph classes are extremely tough to test for graph non-isomorphic pairs from isomorphic pairs?</p>
<p><strong>(2)</strong> Is there a repository of adjacencies from such classes?</p>
| Brendan McKay | 9,025 | <p>There is a move towards creating a "standard" benchmark set for graph isomorphism, but it didn't happen yet. Meanwhile, the largest collection that includes hard graphs as well as easy graphs is <a href="http://pallini.di.uniroma1.it/Graphs.html" rel="noreferrer">here</a>. To make examples of difficult nonisomorphic pairs, take two graphs with the same parameters and randomly label them. For difficult isomorphic pairs take two random labellings of the same graph.</p>
<p>The graphs here that will cause trouble to most programs are in the classes ag, cfi, had, latin-sw, pg, pp, f-lex.</p>
|
2,068,951 | <p>I'm interested in proving the following claim:</p>
<p>There exists a sequence of natural numbers $\left(a_{n}\right)_{n=1}^{\infty}$ such that
$$\lim_{n\to\infty}\left(1-\frac{1}{2^{n}}\right)^{a_{n}}=\frac{1}{2}
$$</p>
<p>I've studied a fair amount of calculus and algebra, yet I've never encountered such a problem before.
How should I approach this claim, or rather what tools should I read about?</p>
<p>Thanks!</p>
| John Hughes | 114,036 | <p>Write $\ln 2$ as a binary number with bits $b_1, b_2, \ldots$. </p>
<p>As others have suggested in briefly-present answers, taking logs is the secret: you want
$$
\lim_{n\to+\infty}\frac{a_n}{2^n}=\ln(2)
$$</p>
<p>and hence
$$a_n=2^n\ln(2).$$
but this choice of $a_n$ is not an integer. On the other hand, the
floor of it is. So we pick</p>
<p>$$a_n=\lfloor 2^n\ln(2) \rfloor$$</p>
<p>which is just the binary "whole number" whose bits (before the decimal point, so to speak) are $b_1, b_2, \ldots, b_n$. </p>
<p>Then the question becomes why the limit in this case is the same as in the case without the floors. </p>
<p>Letting
$$q_n=2^n\ln(2),$$
I'm claiming that
$$
\lim \left(1-\frac{1}{2^{n}}\right)^{a_{n}}=\lim \left(1-\frac{1}{2^{n}}\right)^{q_{n}}.
$$</p>
<p>Taking log of both sides, this is the same as
$$
\lim a_n \log \left(1-\frac{1}{2^{n}}\right)= \lim q_n \log\left(1-\frac{1}{2^{n}}\right)^{q_{n}}
$$
or
$$
\lim (a_n-q_n) \log \left(1-\frac{1}{2^{n}}\right)= 0.
$$
Now $a_n - q_n$ is between 0 and 1, and $\log \left(1-\frac{1}{2^{n}}\right) \approx -\frac{1}{2^{n}}$, so it's clear that the limit is bounded below by $$-2\frac{1}{2^{n}},$$ which goes to 0, so the squeeze lemma applies and the limit is indeed 0. </p>
|
886,070 | <p>This is a follow-up question to <a href="https://math.stackexchange.com/questions/884642/an-equation-of-the-form-a-b-c-abc">An equation of the form A + B + C = ABC</a> . I totally messed up with making the equation from the question specification . Actually the question was $$ \arctan(\frac{1}{A}) = \arctan(\frac{1}{B}) + \arctan(\frac{1}{C})$$</p>
<p>So the equation after rearranging becomes $$\frac{1}{A} = \frac{B+C}{BC-1}.$$ Now we have to find the $\min(B+C)$ where A is given to be a fixed positive integer and $B$ and $C$ and are some positive integers which satisfy the equation. I tried some values which satisfy the equation but I don't have enough mathematical background on solving diophantine equations. </p>
<p>PS : Sorry for the mistake in the previous post </p>
| Robert Israel | 8,508 | <p>Write your equation as $$B = \dfrac{AC+1}{C-A}$$
Let $C = A + x$, so this becomes
$$ B =A + \dfrac{A^2+1}{x}$$
Thus $x$ must divide $A^2+1$, and
$B + C = 2A + x + \dfrac{A^2+1}{x}$.
You'll want a divisor of $A^2+1$ that is closest to $\sqrt{A^2+1}$ (on one side or the other).</p>
|
886,070 | <p>This is a follow-up question to <a href="https://math.stackexchange.com/questions/884642/an-equation-of-the-form-a-b-c-abc">An equation of the form A + B + C = ABC</a> . I totally messed up with making the equation from the question specification . Actually the question was $$ \arctan(\frac{1}{A}) = \arctan(\frac{1}{B}) + \arctan(\frac{1}{C})$$</p>
<p>So the equation after rearranging becomes $$\frac{1}{A} = \frac{B+C}{BC-1}.$$ Now we have to find the $\min(B+C)$ where A is given to be a fixed positive integer and $B$ and $C$ and are some positive integers which satisfy the equation. I tried some values which satisfy the equation but I don't have enough mathematical background on solving diophantine equations. </p>
<p>PS : Sorry for the mistake in the previous post </p>
| The Great Seo | 166,806 | <p>$$\begin{align}
\frac1A={B+C\over BC-1} &\iff BC-AB-AC-1=0 \\&\iff (B-A)(C-A)=BC-BA-BC+A^2=A^2+1.
\end{align}$$
Thus as Robert Isreal pointed out, to minimize $B+C$, you have to find a divisor of $A^2+1$ that is closest to $\sqrt{A^2+1}$, and let it be $B-A$.</p>
|
671,407 | <p>I have problem with equation: $4^x-3^x=1$. </p>
<p>So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions? </p>
| Barry Cipra | 86,747 | <p>Let $f(x)=1-\left({3\over4}\right)^x-\left({1\over4}\right)^x$. The functions $(3/4)^x$ and $(1/4)^x$ are strictly decreasing functions of $x$, so the minus signs make $f(x)$ strictly increasing. </p>
|
279,520 | <p>So I have to find the integral of $$ \int \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx$$</p>
<p>I think I have to do this using the integration by parts..so I will take $f = \sin^{-1}(x)$ and $ \sqrt {1+x}=g' $...what about now? </p>
| Stefan Hansen | 25,632 | <p>Do integration by parts with $f(x)=\arcsin(x)$ and $g(x)=\frac{1}{\sqrt{1+x}}$. Then
$$
I=\int\frac{\arcsin x}{\sqrt{1+x}}\,\mathrm dx=\int f(x)g(x)\,\mathrm dx=f(x)G(x)-\int f'(x)G(x)\,\mathrm dx\\
=2\arcsin (x)\sqrt{1+x}-\int \frac{1}{\sqrt{1-x^2}}\cdot2\sqrt{1+x}\,\mathrm dx,
$$
but
$$
\frac{1}{\sqrt{1-x^2}}\cdot2\sqrt{1+x}=\frac{2\sqrt{1+x}}{\sqrt{1+x}\sqrt{1-x}}=\frac{2}{\sqrt{1-x}},
$$
so
$$
I=2\arcsin (x)\sqrt{1+x}-\int \frac{2}{\sqrt{1-x}}\,\mathrm dx=2\arcsin (x)\sqrt{1+x}+4\sqrt{1-x}.
$$</p>
|
3,057,198 | <p><span class="math-container">$$\iint_{G}\!x^2\,\mathrm{d}x\mathrm{d}y$$</span></p>
<p>where <span class="math-container">$G := \left\{(x,y)\in\mathbb{R}^{2}\,;\,|x|+|y| \le 1\right\}$</span></p>
<p>How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).</p>
<p>Oh and also Happy New Year in advance!</p>
| Greg Martin | 16,078 | <p>In double integrals of the form <span class="math-container">$\int (\int f(x,y)\,dx)\,dy$</span>: the limits of the outer integral are the largest possible values of <span class="math-container">$y$</span> over the entire domain of integration; then, for every fixed <span class="math-container">$y$</span>, the limits of the inner integral are the largest and smallest values of <span class="math-container">$x$</span> that can occur in conjunction with that particular value of <span class="math-container">$y$</span> (in particular, these are often functions of <span class="math-container">$y$</span>, whereas the outer limits are always constant).</p>
<p>How big and small can <span class="math-container">$y$</span> get in the region <span class="math-container">$G$</span>?</p>
<p>Given a fixed value of <span class="math-container">$y$</span>, how big and small can <span class="math-container">$x$</span> get (as a function of <span class="math-container">$y$</span>) so that the point <span class="math-container">$(x,y)$</span> is still in <span class="math-container">$G$</span>?</p>
|
4,651,596 | <p>I know the proof of the "<a href="https://en.wikipedia.org/wiki/Doubling_the_cube" rel="nofollow noreferrer">Doubling the cube problem</a>". What is used there is the fact that if a number <span class="math-container">$a$</span> is constructible then <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]$</span> is a power of <span class="math-container">$2$</span>.</p>
<p>I found in a German textbook the remark that the inversion is not correct:
If <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]$</span> is a power of <span class="math-container">$2$</span>, then <span class="math-container">$a$</span> is not necessarily constructible.</p>
<p>Do you know an example for an <span class="math-container">$a$</span> where <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]$</span> is a power of <span class="math-container">$2$</span> and which is not constructible – or a textbook with an example?</p>
| coudy | 716,791 | <p>The <a href="https://en.wikipedia.org/wiki/Quartic_function" rel="nofollow noreferrer">formulas</a> solving the quartic involve taking cube roots in general. Some <a href="https://en.wikipedia.org/wiki/Resolvent_cubic" rel="nofollow noreferrer">auxiliary cubic</a> equation is needed in the resolution of a quartic in general.</p>
|
4,177,639 | <p>I have an object with known coordinates in in 3D but on the ground (<code>z=0</code>). The object has a direction vector. My goal is to move this object on the ground (so <code>z</code> stays <code>0</code>) using its direction vector and via randomly-generated velocity vectors with one condition: I want to ensure that the generate velocity vector can only move the object within a "valid arc", defined in degrees with respect to the direction vector of the object. More specifically, the way I determine valid range is by ensuring that the new position is within 45 degrees of the old object's direction vector (-45 degrees to the left and +45 degrees to the right). Can someone write a pseudocode on how I can achieve this?</p>
<p>Here's my attempt to do this but this doesn't seem to be the correct way to help me achieve what I want:</p>
<pre><code>object_dir = object_position # the direction could be the same as the object's coortinates
while True:
vel_vec=[uniform(-max_vel, max_vel), uniform(-max_vel, max_vel)] # generate a random velocity vector
new_pos = object_dir + vel_vec # compute a new position (and/or object direction vector) for the object
if (compute_angle(new_pos, object_dir) < 45 or compute_angle(new_pos, object_dir) > 315):
break
</code></pre>
| Tyma Gaidash | 905,886 | <p>Unfortunately, there is no closed form so far, at least without using special functions, but here it is using this <a href="https://www.desmos.com/calculator/eaqmpcq4t7" rel="nofollow noreferrer">graph</a>. You can also think of plugging in some value of x into the final answer at the top of the power tower like answer or just imagine it goes away and just the value of x is left in terms of a power tower like answer of a and b. The inequality at the end is from @ThomasAndrews:</p>
<p><span class="math-container">$$a^x+b^x=1\implies a^x=1-b^x\implies c_0=x=\log_a(1-b^x)=\log_b(1-a^x)=c_1\implies c_{n+1}= \log_a(1-b^{c_n})=\log_b(1-a^{c_n}) \implies x=\lim_{n \to \infty} c_n= \log_a\left(1-b^{\log_a(1-b^{…})}\right)= \quad\log_b\left(1-a^{\log_b(1-a^{…})}\right) \implies x=c_\infty,0\not<a\not\le 1\not\le b\ \mathrm{or} \ 0 \not <b \not\le 1 \not\le a$$</span></p>
<p>Here are two other special case solutions I have. Just set the c-value to 1 in one of my answers</p>
<blockquote>
<p><a href="https://math.stackexchange.com/questions/4109555/how-to-isolate-x-in-ax-bx-c-for-use-in-medical-statistics/4109634#4109634"><em><strong>here</strong></em></a></p>
</blockquote>
<p>I will try to find a closed form or alternate non-closed form if possible. In the graph, the solution curve turns into y=k or x=k as the solution as x crosses the curve at (k,k). Please correct me and give me feedback!</p>
|
3,371,922 | <p>The definition of the limit states that limit of <span class="math-container">$f(x)$</span> when <span class="math-container">$x$</span> approaches <span class="math-container">$c$</span> is <span class="math-container">$L$</span> iff for every <span class="math-container">$\epsilon > 0$</span> there exists <span class="math-container">$\delta > 0$</span> such that <span class="math-container">$|f(x) - L | < \epsilon$</span> and <span class="math-container">$0 < |x - c| < δ )$</span>.</p>
<p>This states that <span class="math-container">$f(x)$</span> can reach <span class="math-container">$L ( L- ε < f(x) < L + \epsilon)$</span>, while <span class="math-container">$x$</span> cannot reach <span class="math-container">$c ( 0 < |x-c|)$</span>.</p>
<p>The informal definition says that limit means the value the function approaches as the input approaches some value. (They use the same word.) Why can one reach its corespondent value (<span class="math-container">$L$</span>) while the other can't (<span class="math-container">$x$</span> to equal <span class="math-container">$c$</span>)? Why f(x) can equal L , and x can't equal c. What is the intuitive answer to this question?</p>
| Theo Bendit | 248,286 | <p>The definition is not quite what you wrote. It should be:</p>
<blockquote>
<p>For all <span class="math-container">$\varepsilon > 0$</span>, there exists <span class="math-container">$\delta > 0$</span> such that, <strong>if</strong> <span class="math-container">$0 < |x - c| < \delta$</span>, <strong>then</strong> <span class="math-container">$|f(x) - L| < \varepsilon$</span>.</p>
</blockquote>
<p>Putting the <span class="math-container">$0 < $</span> next to the <span class="math-container">$|x - c| < \delta$</span> actually makes the definition <em>weaker</em>, in the sense that it's easier to satisfy. By putting the <span class="math-container">$0 < $</span> there, it means that we don't care what happens when <span class="math-container">$x = c$</span>. The if...then statement imposes something to be true whenever <span class="math-container">$0 < |x - c| < \delta$</span> is true. When <span class="math-container">$0 < |x - c| < \delta$</span> is false (e.g. if <span class="math-container">$x = c$</span>), the premise is false, so the conclusion <span class="math-container">$|f(x) - L| < \varepsilon$</span> can be true or false.</p>
<p>If we were to get rid of the <span class="math-container">$0 < $</span>, we actually recover the concept of continuity, i.e. we don't just get a limit, we actually force the limit <span class="math-container">$L$</span> to be <span class="math-container">$f(c)$</span>. Why? Because we can always choose <span class="math-container">$x = c$</span>, and knowing that
<span class="math-container">$$|x - c| < \delta \implies |f(x) - L| < \varepsilon$$</span>
implies that, in the case of <span class="math-container">$x = c$</span>,
<span class="math-container">$$0 < \delta \implies |f(c) - L| < \varepsilon.$$</span>
Note that <span class="math-container">$0 < \delta$</span> is always true, hence the conclusion <span class="math-container">$|f(c) - L| < \varepsilon$</span> must also be always true, regardless of <span class="math-container">$\varepsilon$</span>. But, there's only one number <span class="math-container">$L$</span> so that <span class="math-container">$|f(c) - L| < \varepsilon$</span> for all <span class="math-container">$\varepsilon > 0$</span>, and that's <span class="math-container">$L = f(c)$</span>.</p>
<p>On the other hand, if we added the <span class="math-container">$0 <$</span> to the conclusion <span class="math-container">$|f(x) - L| < \varepsilon$</span>, this becomes a stronger definition, in the sense that it's harder to satisfy. Not only do we require that, on the interval <span class="math-container">$(c - \delta, c + \delta)$</span>, the function to be <span class="math-container">$\varepsilon$</span>-close to <span class="math-container">$c$</span>, we also require it to be different from <span class="math-container">$c$</span>. This would exclude constant functions from having limits, as on every <span class="math-container">$\delta$</span>-neighbourhood, they take their limit as a function value at every point!</p>
|
106,126 | <blockquote>
<p><strong>Problem</strong> Prove that $n! > \sqrt{n^n}, n \geq 3$. </p>
</blockquote>
<p>I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. However, both methods don't seem to get close to the answer. I wonder is there another method to prove this problem that I'm not aware of? Any suggestion would be greatly appreciated.</p>
| André Nicolas | 6,312 | <p>There can be no method prettier than Henry's pairing argument. So let's work in the opposite direction, and look for an ugly argument.</p>
<p>When we go from $n$ to $n+1$, the factorial function grows by a factor of $n+1$. What about the other function? It grows by the factor
$$\frac{\sqrt{(n+1)^{n+1}}}{\sqrt{n^n}},$$
which simplifies to
$$(n+1)^{1/2}\left(1+\frac{1}{n}\right)^{n/2}.$$
Recall that $(1+1/n)^n<3$. So if $n+1>3$, the second function grows by a factor that is less than the growth factor of the factorial function. Now we hunt for a place >$2$ at which $n!>\sqrt{n^n}$. Beyond it, everything will be OK. </p>
<p><strong>Another way:</strong> Next we look at estimates, in an attempt to carry our your other proposed argument. It is easier to work with logarithms. We want to show that if $n$ is large enough, then $\log(n!)>
\frac{1}{2}n\log n$.</p>
<p>Draw the curve $y=\log x$. Draw the rectangle with base $[1,2]$ and height $\log 2$, the rectangle with base $[2,3]$ and height $\log 3$, and so on up to the rectangle with base $[n-1,n]$ and height $\log n$. The sum of the areas of these rectangles is $\log 1+\log 2+\cdots+\log n$. By comparison with the area under the curve $y=\log x$ from $1$ to $n$, we conclude that
$$\log(n!) > \int_1^n \log x\,dx =n\log n -n +1.$$
(The integration is not hard, one does it by parts, letting $u=\log x$ and $dv=dx$.)</p>
<p>Now it is enough to show that $n\log n -n+1 >\frac{1}{2}n\log n$, or equivalently that $\frac{1}{2}n\log n -n+1>0$, if $n$ is large enough. Making sure that $\frac{1}{2}\log n >0$ will do the job, but we can do better because we get some help from the $1$. </p>
<p>The integral estimate we got for $\log(n!)$ can be refined quite a bit. After all the refinements are done, we obtain the <em><a href="http://en.wikipedia.org/wiki/Stirling_approximation">Stirling approximation</a></em> to $n!$, a result of great importance.</p>
|
106,126 | <blockquote>
<p><strong>Problem</strong> Prove that $n! > \sqrt{n^n}, n \geq 3$. </p>
</blockquote>
<p>I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. However, both methods don't seem to get close to the answer. I wonder is there another method to prove this problem that I'm not aware of? Any suggestion would be greatly appreciated.</p>
| robjohn | 13,854 | <p>$\log$ is strictly concave $\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\log(x)=-\frac{1}{x^2}<0\right)$, so for $1\le k\le n$, we have
$$
\log(k)\ge\frac{(k-1)\log(n)+(n-k)\log(1)}{n-1}\tag{1}
$$
with equality only when $k=1$ or $k=n$. Summing $(1)$ yields
$$
\begin{align}
\log(n!)
&\ge\frac{n}{2}(\log(n)+\log(1))\\
&=\log(\sqrt{n^n})\tag{2}
\end{align}
$$
If $n\ge3$, then for at least one of the summands, equality fails; therefore,
$$
n!>\sqrt{n^n}\tag{3}
$$</p>
|
222,596 | <p>I would like to find a temperature by knowing the enthalpy, is this possible?
This is what i've tried so far:</p>
<pre><code>V1 = 150;
V2 = 4;
T1 = 15 + 273;
Enthalpy
h[T_] := QuantityMagnitude[
ThermodynamicData["Air",
"Enthalpy", {"Temperature" -> Quantity[T, "Kelvins"]}]]
h1 = h[T1]
sol = Solve[h1 + V1^2/2 == h2 + V2^2/2]
{{h2 -> 425466.}}
FindRoot[h[T2] == h2 /. sol, {T2, 300}]
During evaluation of ThermodynamicData::quant: T2 is not a real number.
During evaluation of FindRoot::jsing: Encountered a singular Jacobian at the point {T2} = {300.}. Try perturbing the initial point(s).
{T2 -> 300.}
</code></pre>
| Tim Laska | 61,809 | <p>I would explicitly state the reference pressure (assuming 1 Bar). Here is an alternative way:</p>
<pre><code>Clear[h]
h[t_?NumericQ] :=
QuantityMagnitude@
ThermodynamicData["Air",
"Enthalpy", {"Pressure" -> Quantity[1, "Bars"],
"Temperature" -> Quantity[t, "Kelvins"]}]
V1 = 150;
V2 = 4;
T1 = 15 + 273;
h1 = h[T1];
sol = First@Solve[h1 + V1^2/2 == h2 + V2^2/2, h2];
t /. FindRoot[h[t] == h2 /. sol, {t, 273}]
(* 299.173 *)
</code></pre>
|
787,926 | <p>I need some help to solve this integral:</p>
<p>$$\int_0^1 dy\int_0^{1-y} \cos \left(\frac{x-y}{x+y} \right) \mathrm dx$$</p>
<p>Thank you.</p>
| 2'5 9'2 | 11,123 | <p>Every $5$-smooth number is of the form $2^x3^y5^z$. Let's call the $1000$th one $M$. That means there are $1000$ nonnegative integer solutions to $$2^x3^y5^z<M$$ This is the same as having $1000$ nonnegative integer solutions to $$\log(2)x+\log(3)y+\log(5)z<\log(M)$$</p>
<p>There are approximately as many nonnegative integer solutions as the volume of the tetrahedron that this linear inequality defines: $\frac{1}{6}\frac{(\log(M))^3}{\log(2)\log(3)\log(5)}$. So we can immediately zoom in on the order of magnitude of $M$ by solving $$\begin{align}
\frac{1}{6}\frac{(\log(M))^3}{\log(2)\log(3)\log(5)}
&\approx1000\\
\implies M
&\approx \exp\left(\sqrt[3]{6000\log(2)\log(3)\log(5)}\right)\approx 278817463=:\tilde{M}
\end{align}$$</p>
<p>Now define $S_0=\{1\}$. Since each $5$-smooth number will yield three more through multiplication by $2$, $3$, and $5$, generate more $5$-smooth numbers this way. We have
$S_1=\{2,3,5\}$. And then $S_2=\{4,6,10,6,9,15,10,15,25\}$, which we cull down to $\{4,6,10,9,15,25\}$. </p>
<p>Keep creating these $S_i$, culling away duplicates and <em>also</em> culling away any results that take you too far past our ballpark for $\tilde{M}$. For instance, if you reach $2\tilde{M}$, throw that number out. </p>
<p>Eventually you will have run out of options (after about $\log_2(2\tilde{M})\approx28$ iterations). This will leave you with a collection $S=S_0\cup S_1\cup \cdots S_n$ of all of the $5$-smooth numbers less than $2\tilde{M}$, of which there ought to be $K>1000$. Apply a sorting algorithm (at worst ${\cal O}(K\log(K))$, but probably quicker since there is some natural ordering already from this process) and you can pick out the $1000$th term.</p>
<p>Note: it might be faster to get $S$ just to iterate through it like</p>
<pre><code>for i = 0..log_2(2M)
for j = 0..log_3(2M/2^i)
for k = 0..log_5(2M/2^i/3^j)
append 2^i*3^j*5^k to S
</code></pre>
|
4,177,829 | <p>Given angles <span class="math-container">$0<\theta_{ij}<\pi$</span> for <span class="math-container">$1\leq i<j\leq k$</span>, what conditions are there on the angles to ensure that there exists <span class="math-container">$k$</span> unit vector <span class="math-container">$v_i\in \mathbb R^k$</span> so that the angle between <span class="math-container">$v_i$</span> and <span class="math-container">$v_j$</span> is <span class="math-container">$\theta_{ij}$</span>?</p>
<p>There are clearly problems when <span class="math-container">$\theta_{12},\theta_{13},\theta_{23}$</span> are all close to <span class="math-container">$\pi$</span>. What if I can ensure, for a given <span class="math-container">$\epsilon>0$</span> that <span class="math-container">$|\theta_{ij}-\frac{\pi}2|<\epsilon?$</span></p>
<p>Intuitively, this is true for <span class="math-container">$k=3$</span> and the angles close to <span class="math-container">$\frac{\pi}2$</span>. We can easily pick <span class="math-container">$v_1,v_2$</span> and the locus of points for <span class="math-container">$v_3$</span> meeting the angle requirement with <span class="math-container">$v_1$</span> is a near-great circle in the unit sphere. With <span class="math-container">$v_2$</span>, the same. And these two near-great circles are near-perpendicular. We need them to intersect for <span class="math-container">$v_3$</span> to be found.</p>
<p>But I can’t prove it, and my intuition for <span class="math-container">$k>3$</span> spheres is negligible.</p>
<p>This is a possible solution for <a href="https://math.stackexchange.com/q/4177715">this question</a>. (In fact, I only need <span class="math-container">$k>3$</span> to complete that question - I’ve got an entirely different solution for <span class="math-container">$k=3$</span> in that question.)</p>
<hr />
<p>I think a minimum necessary condition is, for all distinct <span class="math-container">$i,j,k$</span>:
<span class="math-container">$$\theta_{ij}+\theta_{jk}\geq \theta_{ik}.\tag 1$$</span> (Here we need <span class="math-container">$\theta_{ij}=\theta_{ji}$</span> for <span class="math-container">$i>j$</span> to include all the correct cases in (1).)</p>
<p>Also, probably:
<span class="math-container">$$\theta_{ij}+\theta_{jk}+\theta_{ik}\leq 2\pi\tag 2$$</span></p>
<p>It seems like, when <span class="math-container">$k=3$</span>, (1) and (2) should be enough.</p>
| Ѕᴀᴀᴅ | 302,797 | <p><span class="math-container">$\def\vec{\boldsymbol}\def\v{\vec{v}}\def\x{\vec{x}}\def\R{\mathbb{R}}\def\Ω{{\mit Ω}}\def\<{\langle}\def\>{\rangle}\DeclareMathOperator{\diag}{diag}$</span>On the one hand, if there exists such <span class="math-container">$\v_1, \cdots, \v_n$</span>, define <span class="math-container">$V = [\v_1, \cdots, \v_n] \in \R^{n × n}$</span>, then<span class="math-container">$$
V^T V = \begin{bmatrix}
\<\v_1, \v_1\> & \cdots & \<\v_1, \v_n\>\\
\vdots & \ddots & \vdots\\
\<\v_n, \v_1\> & \cdots & \<\v_n, \v_n\>
\end{bmatrix} = \begin{bmatrix}
1 & \cos θ_{1, 2} & \cdots & \cos θ_{1, n}\\
\cos θ_{2, 1} & 1 & \ddots & \cos θ_{2, n}\\
\vdots & \ddots & \ddots & \vdots\\
\cos θ_{n, 1} & \cos θ_{n, 2} & \cdots & 1
\end{bmatrix} =: A
$$</span>
is a positive semidefinite matrix.</p>
<p>On the other hand, if <span class="math-container">$A \geqslant 0$</span>, then there exists an orthogonal matrix <span class="math-container">$\Ω$</span> and <span class="math-container">$D = \diag(λ_1, \cdots, λ_n)$</span> such that <span class="math-container">$λ_1, \cdots, λ_n \geqslant 0$</span> and <span class="math-container">$A = \Ω^T D\Ω$</span>. Take <span class="math-container">$V = \sqrt{D}\Ω$</span> and <span class="math-container">$\v_k$</span> as the <span class="math-container">$k$</span>-th column of <span class="math-container">$V$</span>, then such <span class="math-container">$\v_k$</span>'s satisfies <span class="math-container">$\dfrac{\v_i · \v_j}{\|\v_i\| \|\v_j\|} = \cos θ_{i, j}$</span> for any <span class="math-container">$i ≠ j$</span>. (When <span class="math-container">$A > 0$</span>, <span class="math-container">$V$</span> can also be computed by Cholesky decomposition.) Therefore the given condition is equivalent to <span class="math-container">$A \geqslant 0$</span>.</p>
<p>Now, an easy-to-check condition to ensure <span class="math-container">$A \geqslant 0$</span> is<span class="math-container">\begin{gather*}
\sum_{\substack{1 \leqslant j \leqslant n\\j ≠ i}} |\cos θ_{i, j}| \leqslant 1.\quad \forall 1 \leqslant i \leqslant n \tag{1}
\end{gather*}</span>
In fact, when (1) is true, for any <span class="math-container">$\vec{x} \in \R^n$</span>,<span class="math-container">\begin{gather*}
\x^T A \x = \sum_{k = 1}^n x_k^2 + 2 \sum_{i < j} \cos θ_{i, j} x_i x_j\\
\geqslant \sum_{k = 1}^n x_k^2 - \sum_{i < j} |\cos θ_{i, j}| (x_i^2 + x_j^2) = \sum_{i = 1}^n \biggl( 1 - \sum_{\substack{1 \leqslant j \leqslant n\\j ≠ i}} |\cos θ_{i, j}| \biggr) x_i^2 \geqslant 0.
\end{gather*}</span>
An even simpler condition implied by (1) is<span class="math-container">\begin{gather*}
\max_{1 \leqslant i < j \leqslant n} |\cos θ_{i, j}| \leqslant \frac{1}{n - 1}, \tag{2}
\end{gather*}</span>
or<span class="math-container">\begin{gather*}
\max_{1 \leqslant i < j \leqslant n} \left| θ_{i, j} -\frac{π}{2} \right| \leqslant \arcsin\frac{1}{n - 1}. \tag{2$'$}
\end{gather*}</span></p>
|
3,045,899 | <p>In the lecture notes we have a fact:</p>
<blockquote>
<p>If <span class="math-container">$A$</span> has orthonormal columns then <span class="math-container">$||Ax||^2_2 = ||x||^2_2$</span></p>
</blockquote>
<p>Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?</p>
<p>Thank you</p>
| C. Ding | 320,080 | <p>Since <span class="math-container">$A$</span> has orthonormal columns, <span class="math-container">$A^TA=I$</span>.</p>
|
4,407,210 | <p><span class="math-container">$$y''-\frac{1}{x}y'=2x\cdot cos(x)$$</span></p>
<p>For the homogeneous part I multiplied through with <span class="math-container">$x^2$</span> and got a second order Cauchy Euler equation with the general solution: <span class="math-container">$$y_h (x)=A x^2 +B$$</span></p>
<p>Then for the particular solution I tried using the method of undetermined coefficients but the whole thing became too entangled to solve and I couldn't get anywhere!</p>
<p>Maple tells me the solution is: <span class="math-container">$$2sin(x) - 2x\cdot cos(x)+ \frac{C_1\cdot x^2 }{2}+C_2$$</span>
but I can't figure it out...
Any help would be appreciated!</p>
| user577215664 | 475,762 | <p><span class="math-container">$$y''-\frac{1}{x}y'=2x\cos(x)$$</span>
<span class="math-container">$$\dfrac {xy''-y'}{x^2}=2\cos(x)$$</span>
<span class="math-container">$$\left (\dfrac {y'}{x} \right)'=2\cos(x)$$</span>
Integrate.</p>
|
95,964 | <p>On the page 43 of <em>Real Analysis</em> by H.L. Royden (1st Edition) we read: "(Ideally) we should like $m$ (the measure function) to have the following properties:</p>
<ol>
<li>$m(E)$ is defined for each subset $E$ of real numbers.</li>
<li>For an interval $I$, $m(I) = l(I)$ (the length of $I$).</li>
<li>If $\{E_n\}$ is a sequence of disjoint sets (for which $m$ is defined),
$m(\bigcup E_n)= \sum m (E_n)$."</li>
</ol>
<p>Then at the end of page 44 we read : "If we assume the Continuum Hypothesis (that every non countable set of real numbers can be put in one to one correspondence with the set of all real numbers) then such a measure is impossible," and no more explanation was given.</p>
<p>Now assuming the Continuum Hypothesis I am not able to see why such a measure is not possible. Would you be kind enough to help me?</p>
| Michael Greinecker | 21,674 | <p>This is a consequence of Ulam's theorem: If a finite countable additive measure $\mu$ is defined on all subsets of a set of cardinality $\aleph_1$ (the first uncountable cardinal) and vanishes on all singletons, then it is identically zero.</p>
<p>This is the version of the theorem from the book of <a href="http://rads.stackoverflow.com/amzn/click/3540345132">Bogachev</a> (1.12.40), the <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm16/fm16114.pdf">original result</a> is a bit stronger. Anyways, under CH, $\mathbb{R}$ has cardinality $\aleph_1$, so the theorem applies to $\mathbb{R}$. I'm pretty sure this is what Royden refers to.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Michael Hardy | 11,667 | <p>$3\times8$ means $8+8+8$.</p>
<p>$8\times3$ means $3+3+3+3+3+3+3+3$.</p>
<p>Why should those both be the same number?</p>
<p>Why should $a\times b$ always be the same as $b\times a$?</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Jyrki Lahtonen | 11,619 | <p>I am bit undecided about the following riddles, whether they are too difficult for 9-year-olds. Here comes anyway. </p>
<ol>
<li>A boy was selling eggs (replace eggs with whatever works locally) to people in a building with four floors. The person living in the fourth floor bought half the eggs plus half an egg. The person living in the third floor bought half of the remaining eggs plus half an egg. The same with the persons living the second floor and in the ground floor. After his tour the boy had sold all his eggs. How many did he have in the beginning? (ok, the fraction may be a dealbreaker here - anyway, the idea is to work it out backwards)</li>
<li>The combined ages (in full years) of a girl and her mother are 43, those of the girl and her grandmother 69, and the ages of the mother and the grandmother add up to 94. How old are they? (Don't use a system of equations, just add them all up, divide by two, and... The method by trial and error is not to be sneered at either.)</li>
<li>Have them draw a 4x4 checkerboard, and cut out two opposite corners. Can they cover the remaining 14 squares with 7 domino pieces? Each domino piece is the size of exactly two squares. How to do it, or why cannot we do it all? (a classic - possibly tough to figure out, but easy to grasp).</li>
</ol>
<p>Hmm. 11-13 years might be closer to the mark with the above. Rewinding my tape further...</p>
<ol>
<li>They are probably working their way throught multiplication tables. Ask them to calculate 3x3 and 2x4, then 5x5 and 4x6, then 8x8 and 7x9? Is somebody seeing a pattern? Prove it using a square! If somebody has a sparkle in his/her eye, also try 7x7 and 9x5.</li>
</ol>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Naffi | 82,231 | <p>Show them a pizza. Ask how many pieces they can make by cutting it once.
<img src="https://i.stack.imgur.com/Penoh.jpg" alt="enter image description here"></p>
<p>The answer should be 2. Now ask them for 2 cutting. The answer should be 4. Then ask them to try for 3 cuttings. I hope some of them will the correct answer, which is 7. Then explain them what is happening in each cutting, how the number of pieces are increasing and why. </p>
<p>Then help them to formulate the problem for n cutting. The formula is</p>
<p>$C(n) = \frac{n(n+1)}{2} + 1$</p>
<p>If they understand everything so far, you can then extend the problem from pizza to cake :-)</p>
<p>Hope they will have fun! :-)</p>
<p><img src="https://i.stack.imgur.com/opB2z.jpg" alt="enter image description here"></p>
|
4,206,286 | <p><span class="math-container">$$\int_0^1\int_0^\infty ye^{-xy}\sin x\,dx\,dy$$</span></p>
<p>How can I calculate out the value of this integral?</p>
<p>P.S. One easy way is to calculate this integral over <span class="math-container">$dy$</span> first, to get an integral form <span class="math-container">$\frac{1-e^{-x}(x+1)}{x^2}\sin x$</span>, if I calculated correctly, but I don't know any way to calculate out this value other than a hard work with contour integral. - so I wonder if there be a way other than this (=integrate over <span class="math-container">$dy$</span> and do contour integral)</p>
<p>(I tried some integration by parts and substitutions but it seems it does not work well, probably the <span class="math-container">$\infty$</span> at the second integral is crucial. I think this requires capturing a term in the integrand and converting to a further integral, and reverting the order of integral by Fubini's theorem.. but I'm not sure)</p>
| sirous | 346,566 | <p><a href="https://i.stack.imgur.com/t4Q8T.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t4Q8T.jpg" alt="" /></a></p>
<p>Hints:</p>
<p>For equilateral triangle shown in figure a it is obvious:</p>
<p><span class="math-container">$$(d_c=r)^2=(OB=R)^2-\big(\frac{AB=c}2\big)$$</span></p>
<p><span class="math-container">$$\frac c2=R\cos 30^o$$</span></p>
<p><span class="math-container">$\Rightarrow d_c=r=\sqrt{R^2-\frac 34 R^2}=\frac R2$</span></p>
<p><span class="math-container">$\Rightarrow d_a+d_b+d_c=3d_c=3r=r+2r=R+r$</span></p>
<p>For other type I was thinking to use Euler,s nine point circle shown in figure b and following formula:</p>
<p><span class="math-container">$\frac{d_a}{h_a} +\frac{d_b}{h_b}+\frac{d_c}{h_c}=1$</span></p>
<p>Should I find anything I will include in my answer.</p>
|
3,779,589 | <p>Let the metric <span class="math-container">$d$</span> be defined as
<span class="math-container">$$
d(f,g) =\sup_{x\in[0,1]}|f(x)-g(x)|,
$$</span>
and let<br />
<span class="math-container">$$
H(x) = \begin{cases} 0 \text{ if } x \leq \frac{1}{2}\\ 1 \text { if } x > \frac{1}{2} \end{cases}.
$$</span>
Is <span class="math-container">$f(x) = x$</span> in <span class="math-container">$B_\frac{1}{2}(H)$</span> ?</p>
<p><strong>My answer</strong>. No, because
<span class="math-container">$$
d(H(x),f(x)) = \sup_{x\in[0,1]}|f(x)-H(x)| = \frac{1}{2}.
$$</span>
Therefore, <span class="math-container">$f(x) \not \in B_\frac{1}{2}(H)$</span></p>
<p>I am not sure if my answer is correct, and I found that it is hard to visualize this metric. Can someone helps me on this?</p>
| user2661923 | 464,411 | <p>I agree with the other responses but offer an <strong>informal</strong> [i.e. intuitive rather than proven] alternative approach.</p>
<p>Consider the numbers <span class="math-container">$1, 2, \cdots, 10.$</span> Of these 10 numbers, exactly 4 are not divisible by either 2 or 5. Further, <span class="math-container">$10$</span> is a common multiple of <span class="math-container">$2$</span> and <span class="math-container">$5$</span>. Therefore, I would intuitively (that is <strong>informally</strong>) expect the pattern to repeat for the numbers 11 through 20, 21 through 30, ...</p>
<p>Since there are 4 such numbers in the set <span class="math-container">$\{1,2, \cdots, 10\}$</span> and since
<span class="math-container">$\frac{200}{10} = 20$</span>, I would expect the computation to be <br>
<span class="math-container">$4 \times 20 = 80.$</span></p>
<p><strong>Edit</strong><br>
In hindsight, it occurs to me that since 10 is a common multiple of both 2 and 5, then given that <span class="math-container">$k$</span> is not divisible by either 2 or 5, it seems immediate that any number of the form <span class="math-container">$k + 10r$</span> [where <span class="math-container">$r$</span> is a positive <strong>or negative</strong> integer] must also not be divisible by either 2 or 5.</p>
<p><strong>Edit-2</strong><br>
I'm not sure if what I am about to write has been covered by what others are referring to as the Euler function. Anyway...<br>
With <span class="math-container">$2$</span> and <span class="math-container">$5$</span> relatively prime, you can <strong>intuitively regard</strong> divisibility by 2 (chance = 1/2) and divisibility by 5 (chance = 1/5) as <strong>independent events.</strong></p>
<p>This means that the desired computation can be <strong>informally regarded as</strong><br>
<span class="math-container">$200 \times [1 - (1/2)] \times [1 - (1/5)].$</span><br>
This approach obviously requires that if the range contains <span class="math-container">$n$</span> consecutive numbers, then <span class="math-container">$n$</span> must be a common multiple of <span class="math-container">$2$</span> and <span class="math-container">$5.$</span></p>
|
1,005,193 | <p>My problem is as follows:</p>
<p>I have a point $A$ and a circle with center $B$ and radius $R$. Points $A$ and $B$ are fixed, also $A$ is outside of the circle. A random point $C$ is picked with uniform distribution in the area of disk $B$. My question is how to calculate the expected value of $AC^{-4}$. I am working with the path loss in Wireless Communication so $AC^{-4}$ measures how much energy is dissipated along the distance $AC$</p>
<p>My approach is to first denote $\theta$ as the angle between AB and BC then $\theta$ is uniformly distributed between $[0,2\pi]$. Denote $r$ as the distance of BC then distribution of $r$ in $[0,R]$ is $\frac{2r}{R^2}$. Using the formula $AC^2 = AB^2 + BC^2 - 2AB\times BC \times \cos\theta$ , I have</p>
<p>\begin{align}
E[AC^{-4}] & = \int_0^{2\pi}\int_0^R (AB^2 + BC^2 - 2AB\times BC \times
\cos\theta)^{-2} f_\theta f_r \, dr \, d\theta \\
& = \int_0^{2\pi}\int_0^R (AB^2 + r^2 - 2AB\times r \times \cos\theta)^{-2} \frac{1}{2\pi} \frac{2r}{R^2} \, dr \, d\theta
\end{align}</p>
<p>However, I am unable to solve this integration. I want to ask if anyone know any method that can give me the closed-form of the above expected value. If not, then maybe an approximation method that can give a closed-form is also good. Thanks in advance.</p>
| Vladimir Vargas | 187,578 | <p>Notice that:</p>
<p>$$f_{WVU}(w,v,u)=|\boldsymbol{J(h)}|f_{XYZ}(h(x,t,z))=|\boldsymbol{J(h)}|f_X\left(\dfrac{w}{u}\right)\chi_{[0,1]}(w)f_Y(u)\chi_{[w,1]}(u)f_Z(\sqrt{v})\chi_{[0,1]}(v).$$</p>
|
34,215 | <p>How do professional mathematicians learn new things? How do they expand their comfort zone? By talking to colleagues? </p>
| JSE | 431 | <p>Slowly and with difficulty, just like amateur mathematicians.</p>
|
240,461 | <p>What's the mathematica command to get the <strong>numerical value</strong> of :</p>
<p><span class="math-container">$$PV\int_0^\infty \frac{\tan x}{x}\text{d}x?$$</span></p>
<p>where <span class="math-container">$PV$</span> is the principal value.</p>
| Dominic | 47,466 | <p>You can evaluate principal-valued integrals using NIntegrate using Method->"PrincipalValue" but you must specify the singular point(s). In your case, there are an infinite number of them. For starters, integrating over just the first one at <span class="math-container">$\pi/2$</span>, we can write:</p>
<pre><code>NIntegrate[Tan[x]/x, {x, 0, Pi/2, 3},
Method -> "PrincipalValue"]
Out[42]= 1.36501
</code></pre>
<p>Notice how I inserted the first singular point <span class="math-container">$\pi/2$</span> between the limits of integration. If we wanted to integrate over two of them, then:</p>
<pre><code>In[44]:= NIntegrate[Tan[x]/x, {x, 0, Pi/2, 3 Pi/2, 6},
Method -> "PrincipalValue"]
Out[44]= 1.46884
</code></pre>
<p>And if I wanted to integrate over 25 singular points, I would sequence-in a table of 25 singular points into the limits of integration and notice in the plot below it's convergence to <span class="math-container">$\pi/2$</span>:</p>
<pre><code>totalN = 25;
myTable = Table[
{index,
NIntegrate[
Tan[x]/x, {x, 0,
Sequence @@ Table[Pi/2 + n Pi, {n, 0, index}], (index + 1) Pi},
Method -> "PrincipalValue"]},
{index, 0, totalN}];
ListPlot[myTable, Joined -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/EtM2D.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/EtM2D.jpg" alt="plot of principal-valued integrals" /></a></p>
|
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