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 |
|---|---|---|---|---|
3,314,561 | <p>Consider the triangle <span class="math-container">$PAT$</span>, with angle <span class="math-container">$P = 36$</span> degres, angle <span class="math-container">$A = 56$</span> degrees and <span class="math-container">$PA=10$</span>. The points <span class="math-container">$U$</span> and <span class="math-container">$G$</span> lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?</p>
<p>It would be very helpful if anyone had a solution using complex numbers to this problem.</p>
| Community | -1 | <p>People can maybe talk more generally but I have a really simple example (but helpful in my opinion):</p>
<blockquote>
<p>Not all waves are differentiable. We want all waves to satisfy the wave equation (in some sense). That sense is weak. </p>
</blockquote>
|
3,314,561 | <p>Consider the triangle <span class="math-container">$PAT$</span>, with angle <span class="math-container">$P = 36$</span> degres, angle <span class="math-container">$A = 56$</span> degrees and <span class="math-container">$PA=10$</span>. The points <span class="math-container">$U$</span> and <span class="math-container">$G$</span> lie on sides TP and TA respectively, such that PU = AG = 1. Let M and N be the midpoints of segments PA and UG. What is the degree measure of the acute angle formed by the lines MN and PA?</p>
<p>It would be very helpful if anyone had a solution using complex numbers to this problem.</p>
| Jonas Lenz | 450,140 | <p>Let's have a look at the Dirichlet problem on some (say smoothly) bounded domain <span class="math-container">$\Omega$</span>, i.e.
<span class="math-container">$$
-\Delta u=f \text{ in } \Omega\\
u=0~ \text{ on } \partial \Omega
$$</span>
for <span class="math-container">$f \in \text{C}^0(\overline{\Omega})$</span>.
Then, <a href="https://en.wikipedia.org/wiki/Dirichlet%27s_principle" rel="noreferrer">Dirichlet's principle</a> states a classical solution is a minimizer of an energy functional, namely <span class="math-container">$E(u):=\dfrac{1}{2}\int_\Omega \left|\nabla u\right|^2 \mathrm{d}x-\int_\Omega f u ~\mathrm{d}x$</span>. (Here we need some boundary condition on <span class="math-container">$\Omega$</span> for the first integral to be finite).</p>
<p>So the question one may ask is, if I have some PDE why not just take corresponding the energy functional, minimize it in the right function space and obtain a solution of the PDE.
So far so good. But the problem that may occur is finding this minimizer.
It can be shown that such functionals are bounded by below, so we have some infimum.
As also stated in the Wikipedia article, it was just assumed (e.g. by Riemann) that this infimum will always be attained, which shown by Weierstrass unfortunately not always is the case (see also <a href="https://mathoverflow.net/a/42195">this answer</a> on MO).</p>
<p>Hence, we find differentiable functions which are "close" (in some sense) to a "solution" of the PDE, but no actual differentiable solution. I feel that this is quite unsatisfactory.</p>
<p>So have could we save this? We can multiply the PDE (take the Laplace equation for simplicity) with some test function and integrate by parts to obtain
<span class="math-container">$$
\int_\Omega \nabla u \cdot \nabla v~\mathrm{d}x= \int_\Omega fv~\mathrm{d}x
$$</span>
for all test functions <span class="math-container">$v$</span>.
But from what space should <span class="math-container">$u$</span> come from? What do we need to make sense to the integral?</p>
<p>Well, <span class="math-container">$\nabla u \in \text{L}^2(\Omega)$</span> would be nice, because then the first integral is well-defined via Cauchy-Schwarz.
But as shown by Weierstrass, classical derivatives are not enough, so we need some weaker sense. And here we got to Sobolev Spaces and looking again at the last formula, we see the weak formulation.</p>
<p>I am aware that this does not give a full explanation to why one should "believe" in weak solutions, Sobolev spaces and so on.
What I stated above is a quick run through how in my course on PDE the step from classical to weak theory was motivated and at least I was quite happy about it.</p>
|
3,854,785 | <p>Considering <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>
to show <span class="math-container">$$8\sin^4(x) = 3 - 4\cos(2x) +\cos(4x)$$</span>
Assuming I did not how to initially do this proof properly, how would I be able to set up a proof that is still valid to show that <span class="math-container">$$1 = 1$$</span></p>
<p>The steps were square both sides of <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>, then you get <span class="math-container">$$2(1 - \cos(2x))^2 = 3 - 4\cos(2x) +\cos(4x)$$</span>
which can lead to <span class="math-container">$$\cos^2(2x) + \sin^2(x) = 1$$</span> which is true
Apparently this is not properly valid. My feedback was to put "RTP" before <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>. How can you make a proof that meets in the middle valid?</p>
| Michael Rozenberg | 190,319 | <p><span class="math-container">$$8\sin^4x=2(1-\cos2x)^2=2-4\cos2x+2\cos^22x=$$</span>
<span class="math-container">$$=2-4\cos2x+1+\cos4x=3-4\cos2x+\cos4x.$$</span>
If really want to use <span class="math-container">$2\sin^2x=1-\cos2x$</span> only, we obtain:
<span class="math-container">$$2-4\cos2x+2\cos^22x=2-4\cos2x+2-2\sin^22x=$$</span>
<span class="math-container">$$=4-4\cos2x-1+\cos4x=3-4\cos2x+\cos4x.$$</span></p>
|
3,854,785 | <p>Considering <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>
to show <span class="math-container">$$8\sin^4(x) = 3 - 4\cos(2x) +\cos(4x)$$</span>
Assuming I did not how to initially do this proof properly, how would I be able to set up a proof that is still valid to show that <span class="math-container">$$1 = 1$$</span></p>
<p>The steps were square both sides of <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>, then you get <span class="math-container">$$2(1 - \cos(2x))^2 = 3 - 4\cos(2x) +\cos(4x)$$</span>
which can lead to <span class="math-container">$$\cos^2(2x) + \sin^2(x) = 1$$</span> which is true
Apparently this is not properly valid. My feedback was to put "RTP" before <span class="math-container">$$2\sin^2(x) = 1 - \cos(2x)$$</span>. How can you make a proof that meets in the middle valid?</p>
| robjohn | 13,854 | <p><span class="math-container">$$
\begin{align}
4\sin^4(x)
&=1-2\cos(2x)+\cos^2(2x)\tag1\\[6pt]
&=1-2\cos(2x)+1-\sin^2(2x)\tag2\\
&=1-2\cos(2x)+1-\frac12+\frac12\cos(4x)\tag3\\
8\sin^4(x)
&=3-4\cos(2x)+\cos(4x)\tag4
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$(1)$</span>: square <span class="math-container">$2\sin^2(x)=1-\cos(2x)$</span><br />
<span class="math-container">$(2)$</span>: <span class="math-container">$\cos^2(x)=1-\sin^2(x)$</span><br />
<span class="math-container">$(3)$</span>: apply <span class="math-container">$\frac12$</span> of <span class="math-container">$2\sin^2(x)=1-\cos(2x)$</span><br />
<span class="math-container">$(4)$</span>: collect terms on the right and multiply by <span class="math-container">$2$</span></p>
|
255,827 | <p>I've had trouble coming up with one.</p>
<p>I know that if I could find </p>
<p>an irreducible poly $p(x)$ over $\mathbb{Q}$
which has roots $\alpha, \beta, \gamma\in Q(\alpha)$,</p>
<p>then $|\mathbb{Q}(\alpha) : \mathbb{Q}| $ = 3 and would be a normal extension,
as $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\beta,\gamma)$ would be a splitting field of $f$ over $\mathbb{Q}$.</p>
<p>However, this is a lot of conditions to find by luck...</p>
<p>Any help appreciated!</p>
| Gerry Myerson | 8,269 | <p>You may know that the Galois group of $x^n-1$ over the rationals is cyclic of order $\phi(n)$ (that's the Euler phi-function). If $\phi(n)$ is a multiple of $3$ (and it's not hard to find such $n$), then you can find a normal extension of the rationals of degree $3$ as a subfield of the splitting field of $x^n-1$. </p>
<p>This is easiest to do if you don't choose $n$ any bigger than necessary. </p>
|
1,026,506 | <p>If $I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta $, show that
$I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $</p>
<p>Hence calculate $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$</p>
<p>I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$ ,, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.</p>
| Idris Addou | 192,045 | <p>Here is the last part of the problem you post.
You can verify the value of the definite integral in the wolfram, at </p>
<p><a href="http://www.wolframalpha.com/input/?i=integral_0" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=integral_0</a>^%28pi%29+sin^4%28x%29cos^6%28x%29dx</p>
<p><img src="https://i.stack.imgur.com/lJwWO.jpg" alt="Last part"></p>
|
2,524,890 | <p>I know that if matrix $a$ is similar to matrix $b$ then $\operatorname{trace} a=\operatorname{trace} b$.</p>
<p>Does it go to the other side?</p>
<p>Thanks.</p>
| Erik T. | 501,005 | <p>You could just use the common second degree equation formula.</p>
<p>So:</p>
<p>$$ b^2-8b+16a=0\implies b = \frac{8 \pm \sqrt{64-64a}}{2}$$</p>
<p>From this we can conclude that there will only exist a solution over $\mathbb{R}$ when $\sqrt{64-64a}$ exist. Is easy to check that the root only exists when the inside part is greater than 0, so $64-64a\geq 0 \implies 64\geq 64a\implies 1 \geq a$. So a can take any value from $(-\infty,1]$.</p>
|
2,600,776 | <blockquote>
<p>A continuos random variable $X$ has the density
$$
f(x) = 2\phi(x)\Phi(x), ~x\in\mathbb{R}
$$
then</p>
<p>(<em>A</em>) $E(X) > 0$</p>
<p>(<em>B</em>) $E(X) < 0$</p>
<p>(<em>C</em>) $P(X\leq 0) > 0.5$</p>
<p>(<em>D</em>) $P(X\ge0) < 0.25$</p>
<p>\begin{eqnarray}
\Phi(x) &=& \text{Cumulative distribution function of } N(0,1)\\
\phi(x) &=& \text{Density function of } N(0, 1)
\end{eqnarray}</p>
</blockquote>
<p>I don't have a slightest clue where to start with. Can someone give me a little push. I saw some answers on same question like this but I didn't understand how should I integrate it when calculating expectation. </p>
| Matt | 9,666 | <p>This question requires no calculations.
You should not integrate anything to answer it.</p>
<p><strong>The Key</strong><br>
If
$\phi(x)$ is the density function of a distribution $D$,<br>
and
$\Phi(x)$ is the cumulative distribution function of $D$,<br>
then
the density $f(x) = 2\phi(x)\Phi(x)$
corresponds to the distribution of a random variable $X$ defined
as the greater of two random variables (say $D_1,D_2$) picked according to $D$.</p>
<p><strong>Why you would see this</strong><br>
To see this from the formula for $f$, see that $\phi(x)$ corresponds to the chance that $D_1$ has a certain value, and $\Phi(x)$ is the chance that another sample $D_2$ has a lesser value. Finally, the factor of 2 is for the alternative case that actually $D_2$ had this value and was greater, so together, $f(x)$ is the chance that $x$ is the greater of two randomly picked values.</p>
<p>Once you see this, answering the question is easy and requires no calculations.</p>
<p>A: Yes, you expect the higher of two samples to be greater on average than what you expect for just the first sample, since the second sample can only make the maximum go up. (Remember $0$ is the expected value of a single sample of $N(0,1)$.)</p>
<p>B: No, you don't expect it to be lower.</p>
<p>C: Will the greater of two samples of $N(0,1)$ be negative more than half the time? Of course not! (It will be negative 1/4 of the time -- when both samples are negative.)</p>
<p>D: Will the greater sample be positive less than 1/4 of the time? Of course not! (It will be positive 3/4 of the time -- when either sample is positive.)</p>
|
1,921,101 | <p>$∃x.P(x) \Rightarrow ∀x.P(x) $</p>
<p>How can I read this in simple English? I translated it as: There exists an element x for which P(x) implies that for all elements x, P(x) is true - but I feel like this doesn't make much sense. What am I doing wrong here?</p>
| Hayden | 27,496 | <p>$\exists x P(x)$ is read as </p>
<blockquote>
<p>there exists $x$ such that $P(x)$ holds</p>
</blockquote>
<p>Likewise $\forall x P(x)$ is read as </p>
<blockquote>
<p>for every $x$, $P(x)$ holds.</p>
</blockquote>
<p>$A \implies B$ is read as</p>
<blockquote>
<p>If $A$ then $B$.</p>
</blockquote>
<p>Thus, bringing all of this together, $\exists x P(x) \implies \forall x P(x)$ is read as </p>
<blockquote>
<p>If there exists $x$ such that $P(x)$, then $P(x)$ holds for every $x$.</p>
</blockquote>
<p>Which is the same as what you have. </p>
|
19,596 | <p>I am trying to rearrange and manipulate some vector differential equations in <em>Mathematica</em>. As far as I understand you have to tell <em>Mathematica</em> that a variable is a vector by specifying the components of the vector. For example</p>
<pre><code>r = {x, y, z};
</code></pre>
<p>If I want to define vector fields I have to do it in the following way</p>
<pre><code>v = {vx[r, t], vy[r, t], vz[r, t]};
n = {nx[r, t], ny[r, t], nz[r, t]};
</code></pre>
<p>Now I can use the <code>Div</code> operator to express a differential equation, in this case the continuity equation in hydrodynamics:</p>
<pre><code>D[n, t] + Div[n*v, r] == 0
</code></pre>
<p>My problem is the output I get from this. It looks absolutely horrible and I can't do anything with it.</p>
<blockquote>
<p>({0, 0, 0},1) ({0, 0, 1},0) {nx
[{x, y, z}, t] + vz[{x, y, z}, t] nz [{x, y, z}, t] +
({0, 0, 1},0) ({0, 1, 0},0) nz[{x, y, z}, t] vz [{x, y, z}, t] + vy[{x,
y, z}, t] ny [{x, y, z}, t] +
({0, 1, 0},0) ({1, 0, 0},0) ny[{x, y, z}, t] vy [{x, y, z}, t] + vx[{x,
y, z}, t] nx [{x, y, z}, t] +
({1, 0, 0},0) nx[{x, y, z}, t] vx [{x, y, z}, t], ({0, 0, 0},1)<br>
({0, 0, 1},0) ny [{x, y, z}, t] + vz[{x, y, z}, t] nz<br>
[{x, y, z}, t] +
({0, 0, 1},0) ({0, 1, 0},0) nz[{x, y, z}, t] vz [{x, y, z}, t] + vy[{x,
y, z}, t] ny [{x, y, z}, t] +
({0, 1, 0},0) ({1, 0, 0},0) ny[{x, y, z}, t] vy [{x, y, z}, t] + vx[{x,
y, z}, t] nx [{x, y, z}, t] +
({1, 0, 0},0) nx[{x, y, z}, t] vx [{x, y, z}, t], ({0, 0, 0},1)<br>
({0, 0, 1},0) nz [{x, y, z}, t] + vz[{x, y, z}, t] nz<br>
[{x, y, z}, t] +
({0, 0, 1},0) ({0, 1, 0},0) nz[{x, y, z}, t] vz [{x, y, z}, t] + vy[{x,
y, z}, t] ny [{x, y, z}, t] +
({0, 1, 0},0) ({1, 0, 0},0) ny[{x, y, z}, t] vy [{x, y, z}, t] + vx[{x,
y, z}, t] nx [{x, y, z}, t] +
({1, 0, 0},0) nx[{x, y, z}, t] vx [{x, y, z}, t]} == 0</p>
</blockquote>
<p>I would like <em>Mathematica</em> to write the equation in vectorial form so that I can rearrange it and use vector identities to manipulate it.</p>
<p>Is there a way to do this?</p>
| Jens | 245 | <p>The answer depends a lot on what you mean by "doing" vector calculus. You want results to be displayed without using component notation, and that's in general a difficult thing to achieve.</p>
<p>A prerequisite about doing <strong>completely symbolic</strong> vector calculus is to define the simplification rules. But even in "non-vector" algebra it's often hard to get <code>Simplify</code> to reach an expression that you would like to obtain. This certainly won't get any easier when going to vector calculus. Trying to go the completely symbolic route may then not be worth your time, especially since I'm guessing you're doing everything in three dimensions, not "n dimensions."</p>
<p>So <strong>instead</strong> I usually follow a different approach in which <em>Mathematica</em> doesn't <em>derive</em> theorems but instead <em>verifies</em> them. That looks less impressive, but it's a <em>lot</em> easier to do. In your case, it means taking a vector calculus identity <code>lhs == rhs</code> and feeding it to <code>Simplify</code> in hopes of getting the result <code>True</code>. By contrast, starting only with <code>Simplify[lhs]</code>, your chances of arriving exactly at <code>rhs</code> will in general by slim to nil.</p>
<p>To verify vector calculus identities, it's typically <em>necessary</em> to define your fields and coordinates in component form, but if you're lucky you won't have to display those components in the end result.</p>
<p>To show some examples, I wasn't able to make up my mind if I should use the <code>VectorAnalysis</code> package or the new version 9 functions. So I'll include my own versions of these functions here, just for the cartesian coordinate case.</p>
<p>Then I'll define something analogous to your vector <code>r</code>, and a general vector field notation. My choice for the coordinate symbols is to define them in a <code>Array</code> using a string <code>"x"</code> as the name for all components, so that there's no potential confusion with existing symbols.</p>
<pre><code>Clear[grad, curl, div, cross, r];
grad[f_] := D[f, {Array["x", Length[f]]}]
div[f_] := Inner[D, f, Array["x", Length[f]]]
curl[f_?VectorQ] :=
Inner[D, f.Normal[LeviCivitaTensor[Length[f]]], Array["x", Length[f]]]
cross[f_?VectorQ, g_?VectorQ] :=
Dot[g.Normal[LeviCivitaTensor[Length[g]]], f]
r = Array["x", 3];
field[name_, dimension_: 3] :=
Through[Array[name, dimension] @@ Array["x", dimension]]
</code></pre>
<p>Now check the vector identity that you mentioned in your comment:</p>
<pre><code>Simplify[
div[cross[field[A], field[B]]] ==
field[B].curl[field[A]] - field[A].curl[field[B]]]
</code></pre>
<blockquote>
<p><code>True</code></p>
</blockquote>
<p>Next, check the <strong>continuity equation</strong> that motivated the question, by defining things in component form but displaying only the truth value of the identity:</p>
<pre><code>Clear[jVec, ρ];
jVec[rVec_?VectorQ, t_] :=
E^(-(rVec.rVec/(1 + t)^2))/(Pi^(3/2) (1 + t)^4) rVec
ρ[rVec_?VectorQ, t_] :=
1/(Sqrt[Pi] Pi (1 + t)^3) Exp[-(rVec.rVec)/(1 + t)^2]
Simplify[
div[jVec[r, t]] + D[ρ[r, t], t] == 0
]
</code></pre>
<blockquote>
<p><code>True</code></p>
</blockquote>
<p>When you do look at the coordinate form of the expressions, they will aways look somewhat uglier of course. </p>
|
66,671 | <p>$$\text{ABC- triangle:} A(4,2); B(-2,1);C(3,-2)$$<br>
Find a D point so this equality is true:</p>
<p>$$5\vec{AD}=2\vec{AB}-3\vec{AC}$$</p>
| Peđa | 15,660 | <p>So,let's observe picture below.first of all you will need to find point $E$...use that $E$ lies on $p(A,B)$ and that $\left\vert AB \right\vert = \left\vert BE \right\vert $. Since $ p(A,C)\left\vert \right\vert p(F,E)$ we may write next equation: $\frac{y_C-y_A}{x_C-x_A}=\frac{y_E-y_F}{x_E-x_F}$ and $\left\vert EF \right\vert=3 \left\vert AC \right\vert$ so we may find point F.Since $\left\vert AF \right\vert=5 \left\vert AD \right\vert$ we may write next equations: $x_D=\frac{x_F+4x_A}{5}$ and $y_D=\frac{y_F+4y_A}{5}$</p>
<p><img src="https://i.stack.imgur.com/rkNFi.jpg" alt="enter image description here"></p>
|
66,671 | <p>$$\text{ABC- triangle:} A(4,2); B(-2,1);C(3,-2)$$<br>
Find a D point so this equality is true:</p>
<p>$$5\vec{AD}=2\vec{AB}-3\vec{AC}$$</p>
| robjohn | 13,854 | <p>Recall that the vector $\overrightarrow{PQ}$ is the difference of two points $Q{-}P$. In this way,
$$
5\overrightarrow{AD}=2\overrightarrow{AB}-3\overrightarrow{AC}
$$
becomes
$$
5(D-A)=2(B-A)-3(C-A)
$$
All that is left is to solve for $D$.</p>
|
3,786,654 | <blockquote>
<p>Let <span class="math-container">$x_1, x_2, x_3 \in \Bbb R$</span>, satisfy <span class="math-container">$0 \leq x_1 \leq x_2 \leq x_3 \leq 4$</span>. If their squares form an arithmetic progression with common difference <span class="math-container">$2$</span>, determine the minimum possible value of <span class="math-container">$$|x_1-x_2|+|x_2-x_3|$$</span></p>
</blockquote>
<hr />
<p>So far, I've started with the fact <span class="math-container">$x_2^2 - x_1^2 = x_3^2 - x_2^2 = 2$</span> since we know their squares form an arithmetic progression with common difference <span class="math-container">$2.$</span> We can solve this to obtain</p>
<p><span class="math-container">$$x_2 = \pm\sqrt{x_1^2+2}, \qquad x_3 = \pm\sqrt{x_1^2+4}$$</span></p>
<p>I'm not sure how to continue. Thanks in advance for the help.</p>
| dezdichado | 152,744 | <p>Notice that:</p>
<p><span class="math-container">$$x_3 - x_1 = \dfrac{x_3^2-x_1^2}{x_3+x_1} = \dfrac{4}{\sqrt{x_1^2+4}+x_1}$$</span>
and this is obviously minimized at the largest possible value of <span class="math-container">$x_1.$</span> That value is obtained by observing:
<span class="math-container">$$16\geq x_3^2 = x_1^2+4\iff x_1\leq\sqrt{12}$$</span>
and so the minimum is <span class="math-container">$\dfrac{4}{4+\sqrt{12}} = 4-\sqrt{12}.$</span></p>
|
4,068,314 | <p>I do know that double negation and LEM are equivalent, but can we prove
<span class="math-container">$$\vdash \neg \neg (p \vee \neg p)$$</span>
without using either of them, in a Fitch-style proof?</p>
| Marc van Leeuwen | 18,880 | <p>What you may use is the fact that in intuitionistic logic one can derive from <span class="math-container">$\lnot(p\lor q)$</span> that <span class="math-container">$\lnot p\land\lnot q$</span> (and vice versa: the two ways to interpret <span class="math-container">$p$</span> NOR <span class="math-container">$q$</span> are equivalent)*. And substituting <span class="math-container">$q:=\lnot p$</span> into those expressions, the second form easily leads to a contradiction. With this, you should be able to prove what you want to without much difficulty.</p>
<p>*The same is <strong>not true</strong> for NAND: while from <span class="math-container">$\lnot p\lor\lnot q$</span> one easily deduces <span class="math-container">$\lnot(p\land q)$</span>, the opposite deduction is not possible in intuitionistic logic.</p>
|
804,283 | <p>I have the equation $ t\sin (t^2) = 0.22984$. I solved this with a graphing calculator, but is there any way to solve for $ t$ without graphing? </p>
<p>Thanks!</p>
| Community | -1 | <p>As is said, there is no closed form solution to this equation. No formula if you prefer.</p>
<p>In such cases, numerical methods are used, which means that different values for $t$ are tried, using specific strategies to get closer and closer to the solution.</p>
<p>It is useful to carry out the study of the function to get a rough idea where the solutions can be.</p>
<p>In this case, if you rewrite as $sin(p)=\frac{0.22984}{\sqrt p}$, to make more familiar functions appear, you see that you intersect a sinusoid with a kind of hyperbola, having an horizontal asymptote.</p>
<p>The first root can also be estimated from $sin(p)\approx p$ (for small $p$), so that $t\sin(t^2)\approx t^3$, and $t\approx\sqrt[3]{0.22984}=0.61255$. (Check: $0.61255\sin(0.61255^2)=0.22448$).</p>
<p>For large $p$, the equation becomes $sin(p)\approx 0$, meaning that you will find infinitely many solutions close to $p=k\pi$, i.e. $t=\sqrt{k\pi}$.</p>
<p>A yet better approximation is obtained by setting $p=q+k\pi$ (for small $q$), so that $sin(q+k\pi)\approx \pm q\approx\pm\frac{0.22984}{\sqrt{k\pi}}$ ($+$ for even $k$, $-$ for odd), i.e. $p\approx(-1)^k\frac{0.22984}{\sqrt{k\pi}}+k\pi$, and $t\approx\sqrt{(-1)^k\frac{0.22984}{\sqrt{k\pi}}+k\pi}$. </p>
<p>There could be additional solutions in between.</p>
|
804,283 | <p>I have the equation $ t\sin (t^2) = 0.22984$. I solved this with a graphing calculator, but is there any way to solve for $ t$ without graphing? </p>
<p>Thanks!</p>
| Richard | 14,493 | <p>Using a <a href="https://en.wikipedia.org/wiki/Taylor_series" rel="nofollow">Taylor series</a>, $\sin(x)$ can be written as</p>
<p>$\sin(x)\approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots$</p>
<p>Replacing $x$ for $t^2$ gives:</p>
<p>$\sin(t^2)\approx t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} - \frac{t^{14}}{7!} + \ldots$</p>
<p>Plugging this into your original equation gives:</p>
<p>$t\sin(t^2)=0.22984$</p>
<p>$=t^3 - \frac{t^7}{3!} + \frac{t^{11}}{5!} - \frac{t^{15}}{7!} + \ldots=0.22984$</p>
<p>So you can see why solving this in a closed-form sense might be difficult.</p>
<p>That said, it's reasonable to think that there might be a value of $t$ less than one, in which case you can try neglecting the higher level terms (this is the <a href="https://en.wikipedia.org/wiki/Small-angle_approximation" rel="nofollow">small angle approximation</a>).</p>
<p>This gives</p>
<p>$t^3=0.22984$</p>
<p>$t=0.61255046092664577035$</p>
<p>Graphically, you find a root at $\sim0.617544$. The difference is 0.8%.</p>
|
3,128,862 | <p>I'm really stuck in this fairly simple example of conditional probability, I don't understand the book reasoning:</p>
<blockquote>
<p>An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each.
Compute the probability that each pile has exactly 1 ace. </p>
<p><strong>Solution.</strong> Define events <span class="math-container">$E_i, i = 1, 2, 3, 4$</span>, as follows:</p>
<p><span class="math-container">$E_1$</span> = {the ace of spades is in any one of the piles}</p>
<p><span class="math-container">$E_2$</span> = {the ace of spades and the ace of hearts are in different piles}</p>
<p><span class="math-container">$E_3$</span> = {the aces of spades, hearts, and diamonds are all in different piles}</p>
<p><span class="math-container">$E_4$</span> = {all 4 aces are in different piles}</p>
<p>The desired probability is <span class="math-container">$P(E_1E_2E_3E_4)$</span>, and by the multiplication rule, </p>
<p><span class="math-container">$P(E_1E_2E_3E_4) = P(E_1)P(E_2|E_1)P(E_3|E_1E_2)P(E_4|E_1E_2E_3)$</span></p>
<p>Now, <span class="math-container">$P(E_1) = 1$</span> since <span class="math-container">$E_1$</span> is the sample space S. Also, <span class="math-container">$P(E_2|E_1) = \frac{39}{51}$</span> since the pile containing the ace of spades will receive 12 of the remaining 51 cards (...)</p>
</blockquote>
<p>I was lost with <span class="math-container">$P(E_2|E_1)$</span>, I don't understand why it's <span class="math-container">$\frac{39}{51}$</span>. I tried to think like this: </p>
<p>by definition, <span class="math-container">$P(E_2|E_1) = P(E_1E_2)/P(E_1)$</span> and, since <span class="math-container">$P(E_1) = 1$</span>, <span class="math-container">$P(E_2|E_1) = P(E_1E_2)$</span>. But it's obvious that <span class="math-container">$E_2 \subset E_1 \Rightarrow E_1 \cap E_2 = E_2 \Rightarrow (E_2|E_1) = P(E_2)$</span>. </p>
<p>So I tried to calculate <span class="math-container">$P(E_2)$</span> to see if it matched the answer of the book. By definition, <span class="math-container">$E_2$</span> is the event where the ace of spades and the ace of hearts are in different piles. So the sample space is <span class="math-container">$52\choose13,13,13,13$</span>. Now, suppose you take out the ace of hearts and the ace of spades of your deck, now you have 50 cards and there are <span class="math-container">$50\choose12,12,13,13$</span> ways of dividing the deck and <span class="math-container">$4\choose2$</span> ways of deciding which piles receive 12 cards and which receives 13. After making this division, there are <span class="math-container">$2!$</span> ways, for each result, to put back the 2 aces you took off (each one in one of the piles with 12 cards). So</p>
<p><span class="math-container">$P(E_2) =$</span> <span class="math-container">${50}\choose{12,12,13,13}$$ {4}\choose{2}$$2!$$/$${52}\choose{13,13,13,13}$$= \frac{39}{51}$</span></p>
<p>In the end, I found the right answer for <span class="math-container">$P(E_2)$</span>, but I went through a whole line of reasoning that was not trivial. In the other hand, the book apparently deduces it in a trivial way:</p>
<blockquote>
<p>since the pile containing the ace of spades will receive 12 of the remaining 51 cards</p>
</blockquote>
<p>What I'm not getting?</p>
| Servaes | 30,382 | <p>The parallel lines <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> are contained in a plane <span class="math-container">$V$</span>. That <span class="math-container">$L^{\phi}$</span> and <span class="math-container">$L'^{\phi}$</span> are contained in <span class="math-container">$V^{\phi}$</span> is immediate; all points of <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> are points of <span class="math-container">$V$</span>, so because <span class="math-container">$\phi$</span> is a function, all points of <span class="math-container">$L^{\phi}$</span> and <span class="math-container">$L'^{\phi}$</span> are points of <span class="math-container">$V^{\phi}$</span>. </p>
<p>To see that <span class="math-container">$L^{\phi}$</span> and <span class="math-container">$L'^{\phi}$</span> are parallel, suppose towards a contradiction that <span class="math-container">$L^{\phi}$</span> and <span class="math-container">$L'^{\phi}$</span> are <em>not</em> parallel. Then they meet in a point, say <span class="math-container">$p$</span>. Because <span class="math-container">$\phi$</span> is a permutation of the set of points, there exists a point <span class="math-container">$q$</span> such that <span class="math-container">$p=q^{\phi}$</span>. Because <span class="math-container">$q^{\phi}$</span> is contained in both <span class="math-container">$L^{\phi}$</span> and <span class="math-container">$L'^{\phi}$</span> and <span class="math-container">$\phi$</span> is a permutation, applying <span class="math-container">$\phi^{-1}$</span> shows that <span class="math-container">$q$</span> is contained in both <span class="math-container">$L$</span> and <span class="math-container">$L'$</span>. This contradicts the fact that <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> are parallel.</p>
|
14,612 | <p>For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ? </p>
| Adam | 4,791 | <p>I think that you may be selling short the value of a counterexample! They are quite useful for making sure that you have not <em>proven</em> <em>too</em> <em>much</em>. i.e. When you have a plausible but only semi-formal argument, how do you tell if it is worth the effort in making it rigorous? Checking against counterexamples is often a useful step.</p>
<p>Still, there are relations to unbounded paths in an infinite binary tree without leaves. That is, suppose that you start at the root and write <span class="math-container">$0.$</span>, thought of as a ternary number. As we travel to the left or right, write a <span class="math-container">$0$</span> or <span class="math-container">$2$</span>, respectively, for the following digit. Continuing on we get a ternary expansion defining a real number in the Cantor set. Correspondingly, the ternary expansion of an element of the Cantor set gives rise to a path from the root in the tree.</p>
|
2,396,073 | <p>Let $\omega_1$ be the first uncountable ordinal. In some book, the set $\Omega_0:=[1,\omega_1)=[1,\omega_1]\backslash\{\omega_1\}$ is called the set of countable ordinals. Why? It is obvious that it is an uncountable set, because $[1,\omega_1]$ is uncountable. The most possible reason I think is that for any $x\prec \omega_1$, the set $[1,x)$ is countable. </p>
| bof | 111,012 | <p>The set $\Omega=[0,\omega_1)$ is a set of countable ordinals because every element of $\Omega$ is a countable ordinal. To see this, suppose that $x\in\Omega$ is <em>not</em> a countable ordinal. Since $x$ is an ordinal, it follows that $x$ is an <em>uncountable</em> ordinal; but $x\lt\omega_1,$ contradicting the fact that $\omega_1$ is the <em>first</em> uncountable ordinal.</p>
|
295,773 | <p>What would be a good Riemannian Geometry (or Differential Geometry) book that would go well with a General Relativity class (offered by a physics department)? I'm in one right now, but I'd like a pure math perspective on the math that's introduced as I can imagine, inevitably some things would be swept under the rug and I'd like a fuller picture. I'm looking at John M. Lee's "Riemannian Manifolds" and Jeffrey Lee's "Manifolds and Differential Geometry".</p>
<p>Are these books suitable? What parts should I study?</p>
| Will Jagy | 10,400 | <p><a href="http://www.math.harvard.edu/~shlomo/docs/semi_riemannian_geometry.pdf" rel="nofollow">STERNBERG_PDF</a> and <a href="http://rads.stackoverflow.com/amzn/click/0125267401" rel="nofollow">O'NEILL</a> ${}{}{}{}{}{}{}$</p>
|
295,773 | <p>What would be a good Riemannian Geometry (or Differential Geometry) book that would go well with a General Relativity class (offered by a physics department)? I'm in one right now, but I'd like a pure math perspective on the math that's introduced as I can imagine, inevitably some things would be swept under the rug and I'd like a fuller picture. I'm looking at John M. Lee's "Riemannian Manifolds" and Jeffrey Lee's "Manifolds and Differential Geometry".</p>
<p>Are these books suitable? What parts should I study?</p>
| Javier Álvarez-Vizoso | 4,058 | <p>The new book by Sternberg (a freely available version is linked in the answer by Will Jagy) is very affordable and focused on just what you may need:</p>
<ul>
<li><strong>Sternberg</strong> - <a href="http://amzn.com/0486478556" rel="nofollow noreferrer"><em>Curvature in Mathematics and Physics</em></a>; Dover 2012.</li>
</ul>
<p>If you want something more detailed and explanatory than Nakahara's, as a good bridge to the more purely mathematical references, you should take a look at these excellent titles:</p>
<ul>
<li><strong>Eschrig</strong> - <em><a href="http://books.google.es/books?id=-iMyoHDdk1wC&lpg=PP1&pg=PP1#v=onepage&q&f=false" rel="nofollow noreferrer">Topology and Geometry for Physics</a></em>; Springer 2011.</li>
<li><strong>Frankel</strong> - <em><a href="http://books.google.es/books?id=gXvHCiUlCgUC&lpg=PP1&dq=1107602602&pg=PP1#v=onepage&q&f=false" rel="nofollow noreferrer">The Geometry of Physics, An Introduction</a></em>; Cambridge University Press 2008.</li>
</ul>
<p>In particular I would recommend the new book by Eschrig to complement any general relativity or gauge theory courses, with the formal background; it is filled with geometric and visual motivation alongside the formal concepts and arguments. Nevertheless, these books do not focus on (pseudo)-Riemannian geometry per se, but on general differential geometry, trying to introduce as many concepts as possible for the needs of modern theoretical physics.</p>
<p>Most purely mathematical books on Riemannian geometry do not treat the pseudo-Riemannian case (although many results are exactly the same). That is why the books on "geometry for physicists", from easy to very formal level, are the best first approach to the mathematical background. Once you get into advanced general relativity, the most mathematical treatments are given by books like Wald, Hawking/Ellis, de Felice, Fré et al... where the physical study of the mathematics of pseudo-Riemannian geometry is explained. Check out the references mentioned in <a href="https://math.stackexchange.com/a/242898/4058">this other answer</a>. You may also find useful <a href="https://math.stackexchange.com/a/22583/4058">this other answer</a> on self-learning differential topology and differential geometry.</p>
|
259,308 | <p>The output of <code>ListPointPlot3D</code> is shown below:
<a href="https://i.stack.imgur.com/ypt73.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ypt73.png" alt="enter image description here" /></a>
I only want to connect the dots in such a way that it forms a ring-like mesh. However, when I use <code>ListPlot3D</code>, the output is like this:
<a href="https://i.stack.imgur.com/zmaI1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zmaI1.png" alt="enter image description here" /></a>
Which is not what I want.</p>
<p>May I know how I should modify the code to achieve the intended result? Thank you.</p>
<p>Here is the data I used (with fewer data points):</p>
<pre><code>{{15.,-4.33154,0.015},{15.4114,-3.89839,0.0154114},{15.7792,-3.46523,0.0157792},{16.1037,-3.03208,0.0161037},{16.3847,-2.59893,0.0163847},{16.6224,-2.16577,0.0166224},{16.8169,-1.73262,0.0168169},{16.968,-1.29946,0.0169681},{17.076,-0.866309,0.017076},{17.1408,-0.433154,0.0171408},{17.1624,-6.01122*10^-16,0.0171624},{17.1408,0.433154,0.0171408},{17.076,0.866309,0.017076},{16.968,1.29946,0.0169681},{16.8169,1.73262,0.0168169},{16.6224,2.16577,0.0166224},{16.3847,2.59893,0.0163847},{16.1037,3.03208,0.0161037},{15.7792,3.46523,0.0157792},{15.4114,3.89839,0.0154114},{14.698,-4.79395,2.99474},{15.1194,-4.31455,3.08061},{15.4564,-3.83516,3.14927},{15.7246,-3.35576,3.2039},{15.9361,-2.87637,3.24701},{16.1009,-2.39697,3.28058},{16.2264,-1.91758,3.30616},{16.3185,-1.43818,3.32492},{16.3815,-0.95879,3.33775},{16.4181,-0.479395,3.34521},{16.4301,-6.65294*10^-16,3.34766},{16.4181,0.479395,3.34521},{16.3815,0.95879,3.33775},{16.3185,1.43818,3.32492},{16.2264,1.91758,3.30616},{16.1009,2.39697,3.28058},{15.9361,2.87637,3.24701},{15.7246,3.35576,3.2039},{15.4564,3.83516,3.14927},{15.1194,4.31455,3.08061},{13.8101,-5.25915,5.85509},{14.2225,-4.73323,6.02996},{14.5189,-4.20732,6.15563},{14.7316,-3.6814,6.24579},{14.8837,-3.15549,6.31026},{14.9917,-2.62957,6.35607},{15.0675,-2.10366,6.3882},{15.1193,-1.57774,6.41015},{15.1527,-1.05183,6.42432},{15.1714,-0.525915,6.43225},{15.1774,-7.29853*10^-16,6.4348},{15.1714,0.525915,6.43225},{15.1527,1.05183,6.42432},{15.1193,1.57774,6.41015},{15.0675,2.10366,6.3882},{14.9917,2.62957,6.35607},{14.8837,3.15549,6.31026},{14.7316,3.6814,6.24579},{14.5189,4.20732,6.15563},{14.2225,4.73323,6.02996},{12.3716,-5.68073,8.48201},{12.7528,-5.11265,8.74341},{13.0014,-4.54458,8.91387},{13.1635,-3.97651,9.02496},{13.2689,-3.40844,9.09726},{13.3374,-2.84036,9.14419},{13.3815,-2.27229,9.17442},{13.4094,-1.70422,9.19355},{13.4263,-1.13615,9.20515},{13.4353,-0.568073,9.21134},{13.4382,-7.88359*10^-16,9.21329},{13.4353,0.568073,9.21134},{13.4263,1.13615,9.20515},{13.4094,1.70422,9.19355},{13.3815,2.27229,9.17442},{13.3374,2.84036,9.14419},{13.2689,3.40844,9.09726},{13.1635,3.97651,9.02496},{13.0014,4.54458,8.91387},{12.7528,5.11265,8.74341},{10.4398,-6.03187,10.7708},{10.7685,-5.42868,11.1099},{10.9656,-4.8255,11.3133},{11.0838,-4.22231,11.4352},{11.1547,-3.61912,11.5083},{11.1971,-3.01594,11.552},{11.2223,-2.41275,11.5781},{11.2372,-1.80956,11.5935},{11.2457,-1.20637,11.6022},{11.25,-0.603187,11.6066},{11.2513,-8.3709*10^-16,11.608},{11.25,0.603187,11.6066},{11.2457,1.20637,11.6022},{11.2372,1.80956,11.5935},{11.2223,2.41275,11.5781},{11.1971,3.01594,11.552},{11.1547,3.61912,11.5083},{11.0838,4.22231,11.4352},{10.9656,4.8255,11.3133},{10.7685,5.42868,11.1099},{8.09191,-6.3005,12.6302},{8.35003,-5.67045,13.033},{8.49481,-5.0404,13.259},{8.57602,-4.41035,13.3858},{8.62155,-3.7803,13.4569},{8.64707,-3.15025,13.4967},{8.66132,-2.5202,13.5189},{8.66922,-1.89015,13.5313},{8.67348,-1.2601,13.5379},{8.67557,-0.63005,13.5412},{8.67619,-8.74371*10^-16,13.5421},{8.67557,0.63005,13.5412},{8.67348,1.2601,13.5379},{8.66922,1.89015,13.5313},{8.66132,2.5202,13.5189},{8.64707,3.15025,13.4967},{8.62155,3.7803,13.4569},{8.57602,4.41035,13.3858},{8.49481,5.0404,13.259},{8.35003,5.67045,13.033},{5.42138,-6.48148,13.986},{5.5955,-5.83333,14.4352},{5.68862,-5.18519,14.6754},{5.73842,-4.53704,14.8039},{5.76505,-3.88889,14.8726},{5.77928,-3.24074,14.9093},{5.78687,-2.59259,14.9289},{5.79089,-1.94444,14.9393},{5.79297,-1.2963,14.9446},{5.79396,-0.648148,14.9472},{5.79425,-8.99486*10^-16,14.9479},{5.79396,0.648148,14.9472},{5.79297,1.2963,14.9446},{5.79089,1.94444,14.9393},{5.78687,2.59259,14.9289},{5.77928,3.24074,14.9093},{5.76505,3.88889,14.8726},{5.73842,4.53704,14.8039},{5.68862,5.18519,14.6754},{5.5955,5.83333,14.4352},{2.53472,-6.57254,14.7843},{2.6163,-5.91529,15.2601},{2.65878,-5.25803,15.5079},{2.6809,-4.60078,15.6369},{2.69241,-3.94352,15.704},{2.69841,-3.28627,15.739},{2.70152,-2.62902,15.7572},{2.70313,-1.97176,15.7666},{2.70395,-1.31451,15.7713},{2.70433,-0.657254,15.7735},{2.70444,-9.12123*10^-16,15.7742},{2.70433,0.657254,15.7735},{2.70395,1.31451,15.7713},{2.70313,1.97176,15.7666},{2.70152,2.62902,15.7572},{2.69841,3.28627,15.739},{2.69241,3.94352,15.704},{2.6809,4.60078,15.6369},{2.65878,5.25803,15.5079},{2.6163,5.91529,15.2601},{-0.452986,-6.5727,14.9932},{-0.467535,-5.91543,15.4747},{-0.475076,-5.25816,15.7243},{-0.478985,-4.60089,15.8537},{-0.481011,-3.94362,15.9207},{-0.48206,-3.28635,15.9555},{-0.482603,-2.62908,15.9734},{-0.482883,-1.97181,15.9827},{-0.483024,-1.31454,15.9873},{-0.483089,-0.65727,15.9895},{-0.483108,-9.12146*10^-16,15.9901},{-0.483089,0.65727,15.9895},{-0.483024,1.31454,15.9873},{-0.482883,1.97181,15.9827},{-0.482603,2.62908,15.9734},{-0.48206,3.28635,15.9555},{-0.481011,3.94362,15.9207},{-0.478985,4.60089,15.8537},{-0.475076,5.25816,15.7243},{-0.467535,5.91543,15.4747},{-3.42264,-6.48177,14.6043},{-3.5319,-5.8336,15.0705},{-3.58953,-5.18542,15.3164},{-3.61993,-4.53724,15.4462},{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</code></pre>
| Daniel Huber | 46,318 | <p>A solution with a hole:</p>
<pre><code>d1 = Select[pts, #[[3]] > -0.1 &];
d2 = Select[pts, #[[3]] < 0 &];
ListPlot3D[{d1, d2}, BoxRatios -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/uUUBr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uUUBr.png" alt="enter image description here" /></a></p>
<p>You could try to patch the hole with a third surface:</p>
<pre><code>d1 = Select[pts, #[[3]] > -0.1 &];
d2 = Select[pts, #[[3]] < 0 &];
d3 = Select[pts, #[[1]] > 0 && -3 < #[[3]] < 3 &];
ListPlot3D[{d1, d2, d3}, BoxRatios -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/dyf2y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dyf2y.png" alt="enter image description here" /></a></p>
|
1,854,823 | <p>How do I express the hyperplane $x+y=1$ as the span of two vectors or more?</p>
<p>P. S. We have a 3D space.</p>
| Chee Han | 242,589 | <p>Note that it cannot be expressed as span of some vectors since the set of vectors satisfying the hyplerplane equation does not form a subspace. Nevertheless, let $u=\begin{bmatrix} x \\ y \end{bmatrix}$ be any vector satisfying the equation $x+y=1$. Rearranging gives $y=1-x$ and substituting this into $u$ gives
\begin{align*}
u = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ 1-x \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} + x\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\end{align*}</p>
|
138,520 | <p>I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$</p>
<p>So i have:
$$y=\sum_{n=0}^{\infty} a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$
$$y''=\sum_{n=0}^{\infty}a_{n}n(n-1)x^{n-2}$$</p>
<p>then substituting these into the differential equation I get:</p>
<p>$$(1-2x+x^2)\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>$$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>relabeling the indexes: </p>
<p>$$\sum_{n=-2}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=-1}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>and then cancelling the $n=-2$ and $n=-1$ terms:</p>
<p>$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>but this doesn't give me what I want (I don't think) as I have $n^2$ terms as I would need</p>
<p>$(n^2+3n+2)a_{n+2}-(2n^2+n)a_{n+1}+(n^2-n-2)a_{n}=0$</p>
<p>I'm not sure where I have gone wrong?</p>
<p>Thanks very much for any help</p>
| Alex R. | 22,064 | <p>Everything seems correct. Before you expand out powers of $n$, notice that equating like powers of $x^n$ in your last sum gives:</p>
<p>$(n+2)(n+1)a_{n+2}-2n(n+1)a_{n+1}+[n(n-1)-2]a_n=0$</p>
<p>Notice that $[n(n-1)-2]=n^2-n-2=(n+1)(n-2)$, so since $n\geq 0$, you can divide the recurrence equation by $(n+1)$ to get the desired result.</p>
|
4,089,114 | <p>I'm a newbie for mathematics and now I'm learning PDE and stuck on that. Could anyone help me out to understand this elimination from PDE. The equation is similar to solve <span class="math-container">$$(D^2 -6DD'+9D'^2)u = y\cos x$$</span></p>
| jacopoburelli | 530,398 | <p>By <a href="https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem" rel="nofollow noreferrer">Rank-nullity theorem</a> it's sufficient (finite-dimensional case) to prove that the sequence <span class="math-container">$b_k := \dim(B_k) - \dim(B_{k-1})$</span> are non decreasing.</p>
<p>Let's denote with <span class="math-container">$B_i=\ker (A^i)$</span>. Since we have <span class="math-container">$\{0\}=B_0 \subset B_1 \subset B_2 \subset \dots \subset B_n \subset \cdots $</span> we have the following : <span class="math-container">$\forall 1 \le i \le n-1, \dim(B_i) - \dim(B_{i-1}) \geq \dim(B_{i+1}) - \dim(B_i)$</span>, i.e <span class="math-container">$b_i \geq b_{i-1}$</span>.</p>
<p>Consider the following homomorphism :</p>
<p><span class="math-container">$$\begin{array}{ccccc}
& B_{i+1} & \xrightarrow{f} & B_i & \xrightarrow{\pi_i} & B_i /B_{i-1}
\end{array}$$</span></p>
<p><span class="math-container">$\ker(\pi_i \circ f)=f^{-1}(\pi_i^{-1}(\{0\}))=f^{-1}(B_{i-1})=B_i$</span>, so by <a href="https://en.wikipedia.org/wiki/Isomorphism_theorems" rel="nofollow noreferrer">1-st isomorphism theorem</a> we have <span class="math-container">$B_{i+1} /B_{i} \simeq \text{Im}(\pi_i \circ f) < B_i /B_{i-1}$</span>. So <span class="math-container">$\dim(B_{i+1} /B_{i}) \le \dim(B_i /B_{i-1})$</span> which was what we wanted to prove.</p>
|
2,792,751 | <p>Prove that $p_2(n) = \left \lfloor{\frac{n}{2}}\right \rfloor+1$ using the identity $$\frac{1}{(1-x)(1-x^2)}=\frac{1}{2}\left(\frac{1}{(1-x)^2}\right)+\frac{1}{2}\left(\frac{1}{1-x^2}\right)$$</p>
<p>where $p_k(n)$ is the number of partitions of an integer $n$ into a most $k$ parts. The generating function $P_k(x)$ of $\{p_k(n)\}$ is $$P_k(x) = \frac{1}{\prod_{r=0}^{k}(1-x^k)}$$
Therefore, the generating function $P_2(x)$ of ${p_2(n)}$ is
$$P_2(x) = \frac{1}{(1-x)(1-x^2)}$$
Now I see here that we can use the identity given above, but I am confused of how to apply it to prove the desired statement. </p>
| Václav Mordvinov | 499,176 | <p>Note that this is a <a href="https://en.wikipedia.org/wiki/Bipartite_graph" rel="nofollow noreferrer">bipartite graph</a> $G=(V,E)$. You can find a partition of $V$ into two subsets $V_1,V_2\subset V$, such that $u,w\in V_1\implies \{u,w\}\not\in E$ and similarly $u,w\in V_2\implies \{u,w\}\not\in E$. Now we see that $\sum_{v_i\in V_j}d(v_i)=11$ for $j=1,2$. This is impossible if $|V_1|=1,2$.</p>
<p>Therefore, without loss of generality, we may assume $|V_1|=3$. Let $V_1=\{v_1,v_2,v_3\}$ and $V_2=\{v_4,v_5,v_6,v_7\}$. Then let $G'=(V,E')=K_{3,4}$ and let $E=E'\setminus\{v_3,v_7\}$. Here $K_{n,m}$ denotes a <a href="https://en.wikipedia.org/wiki/Complete_bipartite_graph" rel="nofollow noreferrer">complete bipartite graph</a> with $V=V_1\cup V_2$, $V_1\cap V_2=\varnothing$, and $|V_1|=n$, $|V_2|=m$. </p>
|
3,715,715 | <p>Let’s say I have a set <span class="math-container">$X$</span> and a set <span class="math-container">$Y$</span>, and <span class="math-container">$X \subseteq Y$</span>. Is it possible to state that <span class="math-container">$|X| \leq |Y|$</span> (<span class="math-container">$|X|$</span> cardinality of <span class="math-container">$X$</span>)? How can I demonstrate that?</p>
| Community | -1 | <p>From Wikipedia: "<span class="math-container">$A$</span> has cardinality less than or equal to the cardinality of <span class="math-container">$B$</span> if there exists an injective function from <span class="math-container">$A$</span> into <span class="math-container">$B$</span>." (That is a definition.)</p>
<p>An injective function is obvious: the identity.</p>
|
2,539,693 | <p>A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$. In fact, these are rather close:</p>
<p>\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &=& 0.46364 \\ \\
\frac{1}{\sqrt{5}} &=& 0.44721
\end{eqnarray*}</p>
<p>So at least numerically I think we have the answer that the first one is bigger. Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different. So we are left with:</p>
<p>$$ \tan^{-1} \frac{1}{2} > \frac{1}{\sqrt{5}} > 0 $$</p>
<p>It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that. And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.</p>
<p>For now I just want proof of the inequality as stated above.</p>
| Amos Quito | 290,715 | <p>Suppose $\tan(x)=\frac{1}{2}$. Then in particular you know $\frac{\sin(x)}{\cos(x)}=\frac{1}{2}$. You also know $\sin^2(x)+\cos^2(x)=1$. Using the last two equalities you find that $\sin(x)=\frac{1}{\sqrt{5}}$. Since $x\geq\sin(x)$, you're done.</p>
|
2,539,693 | <p>A number theory textbook asked us to compare $\tan^{-1}(\frac{1}{2})$ and $\sqrt{5}$. In fact, these are rather close:</p>
<p>\begin{eqnarray*}
\tan^{-1} \frac{1}{2} &=& 0.46364 \\ \\
\frac{1}{\sqrt{5}} &=& 0.44721
\end{eqnarray*}</p>
<p>So at least numerically I think we have the answer that the first one is bigger. Momentarily, I thought we had an exact answer: $\tan^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} $, but that's totally different. So we are left with:</p>
<p>$$ \tan^{-1} \frac{1}{2} > \frac{1}{\sqrt{5}} > 0 $$</p>
<p>It's impressive that we could have so many decimal places, and I wonder if I should take the computer on faith for that. And I noticed these two answers are close, so I also wonder if we estimate the difference... I don't have any conjecture in either case yet.</p>
<p>For now I just want proof of the inequality as stated above.</p>
| Jack D'Aurizio | 44,121 | <p>By <a href="https://arxiv.org/abs/1304.0753" rel="nofollow noreferrer">the Shafer-Fink inequality</a>$^{(*)}$:
$$ \arctan\left(\tfrac{1}{2}\right)> \frac{3\cdot\frac{1}{2}}{1+2\sqrt{1+\left(\frac{1}{2}\right)^2}}=\tfrac{3}{8}\left(\sqrt{5}-1\right) $$
and $\frac{3}{8}\left(\sqrt{5}-1\right)\geq \frac{1}{\sqrt{5}}$ is equivalent to $5-\sqrt{5}\geq\frac{8}{3}$, or to $7\geq 3\sqrt{5}$, or to $49\geq 45$, which is trivial.</p>
<hr>
<p>$(*)$ It can be proved by noticing that all the derivatives at the origin of $\tan(x)$ <a href="http://mathworld.wolfram.com/TangentNumber.html" rel="nofollow noreferrer">are natural numbers</a> and by playing a bit with the tangent duplication formulas and polynomial interpolation.</p>
|
26,152 | <p>In my textbook, they said:</p>
<p>$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$</p>
<p>The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$</p>
<p>And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:</p>
<p>Let $y = 2x^{3} + 7x - 4$, we have:<br>
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$</p>
<p>What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this? </p>
<p>Thanks, </p>
| Community | -1 | <p>What your textbook means is $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \iff \left(x \equiv 1 \pmod{5} \right)$$</p>
<p>This is what you checked by plugging in the different cases for $x$.</p>
<p>By taking $x \equiv 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \equiv 0 \pmod{5}$ and hence $$ \left(x \equiv 1 \pmod{5} \right) \Rightarrow \left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right)$$</p>
<p>Now by taking $x \neq 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \neq 0 \pmod{5}$ and hence $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \Rightarrow \left(x \equiv 1 \pmod{5} \right)$$</p>
|
758,274 | <p>reference: <a href="https://math.stackexchange.com/questions/428313/what-is-the-orbit-of-a-permutation">What is the orbit of a permutation?</a></p>
<p>To be honest, i don't understand the answer in the link.</p>
<p>The orbit of a group action is defined as follows:</p>
<blockquote>
<p>Let <span class="math-container">$G$</span> be a group acting on a set <span class="math-container">$X$</span>.</p>
<p>Define <span class="math-container">$G.x=\{g.x\in X: g\in G\}$</span> where <span class="math-container">$x\in X$</span>.</p>
<p>Then <span class="math-container">$G.x$</span> is called the orbit of <span class="math-container">$x$</span>.</p>
</blockquote>
<p>Below is the definition in Fraleigh:</p>
<blockquote>
<p>Let <span class="math-container">$\sigma\in S_A$</span></p>
<p>Give an equivalence relation on <span class="math-container">$A$</span> as <span class="math-container">$a\sim b$</span> iff <span class="math-container">$\exists n\in \mathbb{Z}$</span> such that <span class="math-container">$b=\sigma^n(a)$</span>.</p>
<p>Those equivalence classes are called th orbits of <span class="math-container">$\sigma$</span></p>
</blockquote>
<p>I don't understand why these two definitions are consistent.</p>
<p>What would be the group action makes these consistent? <span class="math-container">$S_A\times S_A \rightarrow S_A$</span> or <span class="math-container">$\mathbb{Z}\times S_A \rightarrow S_A$</span> or what..?</p>
| Hagen von Eitzen | 39,174 | <p>If $S_A$ is the group of permutations of $A$, then the very definition of $S_A$ gives you an action of the group on the set $A$. But that is not meant here! Instead, each (fixed) $\sigma\in S_A$ also gives us an action of the group $\mathbb Z$ on the set $A$, namely $\mathbb Z\times A\to A$, $(n,a)\mapsto \sigma^n(a)$. </p>
|
758,274 | <p>reference: <a href="https://math.stackexchange.com/questions/428313/what-is-the-orbit-of-a-permutation">What is the orbit of a permutation?</a></p>
<p>To be honest, i don't understand the answer in the link.</p>
<p>The orbit of a group action is defined as follows:</p>
<blockquote>
<p>Let <span class="math-container">$G$</span> be a group acting on a set <span class="math-container">$X$</span>.</p>
<p>Define <span class="math-container">$G.x=\{g.x\in X: g\in G\}$</span> where <span class="math-container">$x\in X$</span>.</p>
<p>Then <span class="math-container">$G.x$</span> is called the orbit of <span class="math-container">$x$</span>.</p>
</blockquote>
<p>Below is the definition in Fraleigh:</p>
<blockquote>
<p>Let <span class="math-container">$\sigma\in S_A$</span></p>
<p>Give an equivalence relation on <span class="math-container">$A$</span> as <span class="math-container">$a\sim b$</span> iff <span class="math-container">$\exists n\in \mathbb{Z}$</span> such that <span class="math-container">$b=\sigma^n(a)$</span>.</p>
<p>Those equivalence classes are called th orbits of <span class="math-container">$\sigma$</span></p>
</blockquote>
<p>I don't understand why these two definitions are consistent.</p>
<p>What would be the group action makes these consistent? <span class="math-container">$S_A\times S_A \rightarrow S_A$</span> or <span class="math-container">$\mathbb{Z}\times S_A \rightarrow S_A$</span> or what..?</p>
| blue | 34,139 | <p>If $G$ is a group acting on a set $X$ and $H\le G$ is a subgroup, then in particular $H$ also acts on $X$.</p>
<p>If $g\in G$ and $x\in X$ then the orbit of $x$ under $g$ means the orbit of $x$ under the action of the cyclic group $\langle g\rangle$, whose action on $X$ is determined by $G$ since $\langle g\rangle\le G$ is a subgroup.</p>
|
4,336,659 | <p>For a beta distribution with parameters <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, we can interpret it as the distribution of the probability of heads for a coin we tossed <span class="math-container">$a+b$</span> times and saw <span class="math-container">$a$</span> heads and <span class="math-container">$b$</span> tails. At the same time, if we draw <span class="math-container">$n$</span> uniform random numbers and sort them, the <span class="math-container">$k$</span>-th order statistic is also Beta distributed with parameters <span class="math-container">$a=k$</span> and <span class="math-container">$b=n+1-k$</span>. So, its like we tossed <span class="math-container">$n+1$</span> coins and got <span class="math-container">$k$</span> heads. Is there an intuitive explanation for this? I can see the derivations mechanically but any logical reason the two distributions should be the same?</p>
| Rohit Pandey | 155,881 | <p>I'm putting in a partial answer in the hope that either I or someone else will be able to fill in the missing pieces. At a high level, we can condition on the <span class="math-container">$k$</span>-th order statistic. Then, all of the other samples of the uniform can lie either before it or after it. This becomes like tossing a coin and observing <span class="math-container">$n-k$</span> tosses are heads and <span class="math-container">$k-1$</span> are tails. This is off by one per the notation in the question and there might be a bug somewhere (like @LeeDavidChingLin suggested).</p>
|
3,335,081 | <p>Is a temperature change in Celsius larger than a temperature change in Fahrenheit?</p>
<p><strong>The teacher offers this second way of thinking about the question.</strong></p>
<blockquote>
<p>If the temperature increases by 1 degree Celsius, does it also increase by 1 degree Fahrenheit? Or is one temperature change larger than the other?</p>
</blockquote>
<p><strong>My Attempt:</strong>
<span class="math-container">$$T_F = \frac95T_C + 32 \implies (T_F + T_{\triangle F}) = \frac95(T_C+T_{\triangle C}) + 32$$</span>
Where <span class="math-container">$T_{\triangle F}$</span> is some change on the Fahranheit scale and <span class="math-container">$T_{\triangle C}$</span> is the respective change on the Celsius scale. (I am confident this is true as long as we assume the changes in temperature are respective to each other.)</p>
<p>Thus, <span class="math-container">$$T_{\triangle F} = \frac95T_C + 32 - T_F + \frac95T_{\triangle C} \implies T_{\triangle F} = T_F - T_F + \frac95T_{\triangle C}$$</span>
<span class="math-container">$$\implies T_{\triangle C} = \frac59T_{\triangle F} \checkmark$$</span></p>
<p>For some degree change on the Fahrenheit scale, <span class="math-container">$T_{\triangle F}$</span>, the resulting change on the Celsius scale, <span class="math-container">$T_{\triangle C}$</span>, is <span class="math-container">$\frac59$</span> that change. Therefore, a temperature change on the Celsius scale is larger than a temperature change on the Fahrenheit scale.</p>
<p><span class="math-container">$$$$</span>
<span class="math-container">$$$$</span></p>
<p><strong>Question:</strong> Does this show what I want? And does it answer the original question? Thanks in advance! This one is making me over think a bit I think.</p>
| Dilip Sarwate | 15,941 | <blockquote>
<p>Is a temperature change in Celsius larger than a temperature change in Fahrenheit?</p>
</blockquote>
<p>is a very poorly-phrased question for which there can be two diametrically-different interpretations. </p>
<ol>
<li><p>A temperature change of <em>one degree Celsius</em> is a larger temperature change than a temperature change of <em>one degree Fahrenheit</em>. This is the interpretation that the OP has come up with.</p></li>
<li><p>A fixed temperature difference (say the difference at sea level between the boiling point of water and the freezing point of water) measures 100 degrees on the Celsius scale but 180 degrees on the Fahrenheit scale; that is, a fixed temperature difference is more when measured on the Fahrenheit scale than when it is measured on the Celsius scale.</p></li>
</ol>
|
316,866 | <p>Suppose $(a_n)$ is a real sequence and $A:=\{a_n \mid n\in \Bbb N \}$ has an infinite linearly independent subset (with respect to field $\Bbb Q$). Is $A$ dense in $\Bbb R?$</p>
| Jonas Meyer | 1,424 | <p>If $A$ is a linearly independent subset of $\mathbb R$, for each $a\in A$ there is a positive integer $n(a)$ such that $n(a)>|a|$. The set $\left\{\dfrac{a}{n(a)}:a\in A\right\}$ is a linearly independent set with the same cardinality and span as $A$, but it is a subset of $(-1,1)$.</p>
|
118,540 | <p>Let $X$ be a projective surface defined over a field $k$ of characteristic $0$, and let $G$ be a finite group acting biregularly on $X$.</p>
<p>Assuming that $X$ is rational over $k$, is the quotient $X/G$ always rational?</p>
<p>If $k=\mathbb{C}$, we can use Castelnuovo's theorem and see that $X/G$ is unirational and hence rational. If $k=\mathbb{R}$, then $X/G$ is geometrically rational and also connected for the transcendental topology, and is thus rational.</p>
<p>But what happens for a general $k$, in particular when $k=\mathbb{Q}$?</p>
| Christian Liedtke | 16,751 | <p>Just to round out the picture: if the characteristic of $k$ is positive and $G$ is a finite, but non-reduced group scheme (for example, the infinitesimal group scheme $\mu_p$ of $p$.the roots of unity), then the quotient $X/G$ need not even have Kodaira dimension $-\infty$ after desingularization. Moreover, if $G$ is not linearly reductive (for example, the infinitesmial group scheme $\alpha_p$), then a resolution of singularities $f:Y\to X/G$ need not satisfy $R^if_\ast{\mathcal O}_Y=0$ for $i\geq1$. Both phenomenons occur already if $\dim X=2$ - for example, there are many examples of ``unirational surfaces of general type'' in positive characteristic.</p>
|
800,363 | <p>What is </p>
<blockquote>
<p>$$\lim_{x\to 0}\left(\frac{x}{e^{-x}+x-1}\right)^x$$</p>
</blockquote>
<p>Using the expansion of <a href="http://en.wikipedia.org/wiki/Exponential_function" rel="nofollow">$e^x$</a>, I get that the function</p>
<blockquote>
<p>$$y=\left(\frac{x}{e^{-x}+x-1}\right)^x$$</p>
</blockquote>
<p>is not defined for negative numbers.</p>
<p>Hence the limit at $0^{-}$ must not exist.$\implies$The limit at $0$ does not exist.</p>
<p>However <a href="http://www.wolframalpha.com/input/?i=lim%28x%2F%28%28e%5E%28-x%29%29%2Bx-1%29%29%5Ex+as+x+tends+to+0" rel="nofollow">WA</a> says that it should be $1$. :(</p>
<p>Am I wrong?</p>
| Community | -1 | <p>We have using the Taylor series</p>
<p>$$e^{-x}+x-1\sim_0\frac{x^2}{2}$$
hence
$$\frac{x}{e^{-x}+x-1}\sim_0\frac2x$$
and then
$$\left(\frac{x}{e^{-x}+x-1}\right)^x=\exp\left(x\log \left(\frac{x}{e^{-x}+x-1}\right)\right)\sim_0\exp\left(x\log\left(\frac2x\right)\right)\xrightarrow{x\to0}e^0=1$$</p>
|
119,636 | <p>I want to know the general formula for $\sum_{n=0}^{m}nr^n$ for some constant r and how it is derived.</p>
<p>For example, when r = 2, the formula is given by:
$\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)$
according to <a href="http://www.wolframalpha.com/input/?i=partial+sum+of+n+2%5En" rel="noreferrer">http://www.wolframalpha.com/input/?i=partial+sum+of+n+2%5En</a></p>
<p>Thanks!</p>
| Julián Aguirre | 4,791 | <p>I suppose you are familiar with the sum of an geometric progression:
<span class="math-container">$$
1+x+x^2+\dots+x^m=\frac{x^{m+1}-1}{x-1}.
$$</span>
Take derivatives an multiply by <span class="math-container">$x$</span>.</p>
|
621,742 | <p>How do you get from$$\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$$to
$$\frac{1}{2}\int^\infty_0\int^u_{-u}e^{-u^2} dv\ du?$$ I have tried using a change of variables formula but to no avail.<br>
Edit: Ok as suggested I set $u=x+y$ and $v=x-y$, so I can see this gives $dx dy=\frac{1}{2}dudv$ but I still can't see how to get the new integration limits. Sorry if I'm being slow.</p>
| Robert Israel | 8,508 | <p>Hint: Try $u = x+y$, $v = x-y$</p>
|
3,180,914 | <p>Let <span class="math-container">$G$</span> be a cyclic group of order <span class="math-container">$n$</span>. Let <span class="math-container">$G_k$</span> the subgroup
<span class="math-container">$$G_k=\left\{x^k: x\in G\right\}.$$</span>
Is it true that <span class="math-container">$[G:G_k]\in\{1,k\}$</span>?</p>
<p>If <span class="math-container">$n=p-1$</span> and <span class="math-container">$k=2$</span> this is true and I used many times in some number theory exercises. How much can I generalize this thing?</p>
<p>What if <span class="math-container">$G$</span> is any (maybe abelian) group?</p>
| lhf | 589 | <p>No: For <span class="math-container">$G=C_{12}$</span> we have <span class="math-container">$[G:G_8]=3$</span>.</p>
<p>In general, for <span class="math-container">$G=C_{n}$</span> we have <span class="math-container">$[G:G_k]=\dfrac{n}{\gcd(n,k)}$</span>.</p>
|
3,547,529 | <p>I did the following: I set <span class="math-container">$3^m+3^n+1=x^2$</span> where <span class="math-container">$x\in\Bbb{N}$</span> and assumed it was true for positive integer exponents and for all whole numbers x so that I can later on prove it's invalidity with contradiction. Since <span class="math-container">$3^m+3^n+1$</span> is odd we can write <span class="math-container">$3^m+3^n+1 = (2k+1)^2$</span> for <span class="math-container">$k\in\Bbb{N}$</span>. After a while I can't seem to prove it and I don't have any ideas on how else to approach this problem. </p>
<p>It would be very helpful if someone could give me a brief explanation on a possible proof since I'm currently practicing for a math competition. </p>
<p>Thanks in advance </p>
| Piquito | 219,998 | <p>COMMENT.-Let <span class="math-container">$F_n(x)$</span> defined as in your problem
<span class="math-container">$$F_n(x)=ax^{2n}+bx^{2n-1}+cx^{2n-2}+dx^{2n-3}\ldots+px+q$$</span> Assuming <span class="math-container">$c\ne0$</span> one has
<span class="math-container">$$F_n(x)=ax^{2n}+bx^{2n-1}\pm F_{n-1}(x)\hspace{10mm}(*)$$</span> (the sign <span class="math-container">$+$</span> when <span class="math-container">$c$</span> is positive and the sign <span class="math-container">$-$</span> when <span class="math-container">$c$</span> is negative).</p>
<p>Besides the property is easily verified for <span class="math-container">$n=1$</span>. In fact <span class="math-container">$$F_1(x)=ax^2+bx+c\Rightarrow a(x_1^2-x_2^2)+b(x_1-x_2)=0\Rightarrow x_1+x_2=\frac{-b}{a}$$</span></p>
<p>Can you now apply induction in any way using the equation <span class="math-container">$(*)$</span>?</p>
|
3,449,589 | <p>In Example 1.4 of <em>Lee's Introduction to Smooth Manifolds</em>, which is showing that the <span class="math-container">$n$</span>-sphere, <span class="math-container">$\mathbb{S}^n$</span> is a topological <span class="math-container">$n$</span>-manifold, the following is stated.</p>
<p>In the part where the author shows that <span class="math-container">$\mathbb{S}^n$</span> is locally Euclidean, he does the following. For <span class="math-container">$1\leq i \leq n+1$</span> he let's <span class="math-container">$$U_i^+ = \{(x^1, \cdots, x^{n+1}) \in \mathbb{R}^{n+1} \ | \ x^i > 0 \}$$</span> and <span class="math-container">$$U_i^{-} = \{(x^1, \cdots, x^{n+1}) \in \mathbb{R}^{n+1} \ | \ x^i < 0 \}.$$</span> Then he defines <span class="math-container">$f : \mathbb{B}^n \to \mathbb{R}$</span> by <span class="math-container">$$f(u) = \sqrt{1-|u|^2}$$</span> and claims that <span class="math-container">$U_i^+ \cap \mathbb{S}^n$</span> is the graph of the function <span class="math-container">$$x^i = f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$$</span> and <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is the graph of the function <span class="math-container">$$x^i = -f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$$</span> and where the hat indicates omission of the <span class="math-container">$x^i$</span>.</p>
<p>Now here is where my confusion comes about. Let me call the first function that <span class="math-container">$U_i^+ \cap \mathbb{S}^n$</span> is supposedly the graph of to be <span class="math-container">$g_i$</span> and let me label <span class="math-container">$h_i$</span> as the function which <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is the graph of. So now <span class="math-container">$$g_i = f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$$</span> and <span class="math-container">$$h_i = -f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1}).$$</span> The problem that I'm having is that if I want to write down the domain and codomain of <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> set-theoretically, they are both going to be maps from <span class="math-container">$\mathbb{B}^{n+1} \to \mathbb{R}$</span>, because surely one would need both <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> to take inputs of the form <span class="math-container">$(x^1, \dots, x^i, \dots, x^{n+1})$</span> to even begin talking about removing the <span class="math-container">$x^i$</span>, so we must have <span class="math-container">$g_i, h_i : \mathbb{B}^{n+1} \to \mathbb{R}$</span>, but the graph of these functions would then be a subset of <span class="math-container">$\mathbb{R}^{n+2}$</span> whereas <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is a subset of <span class="math-container">$\mathbb{R}^{n+1}$</span>.</p>
<p>So my question is how can we rigorously (in terms of set theory) define these functions <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> so that <span class="math-container">$U_i^+ \cap \mathbb{S}^n$</span> is the graph of <span class="math-container">$g_i$</span> and <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is the graph of <span class="math-container">$h_i$</span>?</p>
| Lee Mosher | 26,501 | <p>You might be misunderstanding the meaning of the notation for the input parameters of <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span>. For example, consider
<span class="math-container">$$g_i = f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})
$$</span>
What this notation means is that from the list of <span class="math-container">$n+1$</span> input parameters, you <em>remove</em> the <span class="math-container">$i^{\text{th}}$</span> parameter <span class="math-container">$x^i$</span>, leaving you with only <span class="math-container">$n$</span> input parameters.</p>
<p>For example, if <span class="math-container">$n=4$</span> and <span class="math-container">$i=2$</span> then
<span class="math-container">$$g_2 = f(x^1,\widehat{x^2},x^3,x^4,x^5) = f(x^1,x^3,x^4,x^5)
$$</span></p>
<p>You also asked for something more rigorous from set theory, so how about this. For each <span class="math-container">$i \in \{1,...,n+1\}$</span> define a function <span class="math-container">$r_i : \{1,...,n\} \to \{1,...,n+1\}$</span> by the formula
<span class="math-container">$$r_i(j) = \begin{cases} j &\quad\text{if $j < i$} \\ j+1 &\quad\text{if $j \ge i$}
\end{cases}
$$</span>
Define <span class="math-container">$g_i = f \circ r_i$</span>.</p>
|
1,789,373 | <p>I'm trying to figure out why the following is true:</p>
<p>Let $ \kappa $ be an uncountable, regular cardinal. Suppose we turn it into a group (i.e. there are operations $ (\cdot, ^{-1}, e) $ with which $ \kappa $ is a group. My aim is to prove that the set</p>
<p>$$ \{ \alpha \in \kappa : \alpha \text{ is a subgroup of } \kappa\}$$</p>
<p>is a club. It's rather obvious that it is closed, however I'm not quite sure why it would be unbounded. Suppose we have an $ \alpha \in \kappa $ which is a subgroup. Somehow we need to enlarge $ \alpha $ so that it will still be both a group and an ordinal.</p>
<p>We can close $ \alpha + 1 $ under the group operations, but we have no warranty that it will be an ordinal (it wont be equal to $ \kappa $ since $ \kappa $ is regular).</p>
<p>I don't see any way to approach this. I would appreciate a hints</p>
| Batominovski | 72,152 | <p><strong>Physics (Classical Mechanics) Solution:</strong></p>
<p>Consider a $1$-dimensional elastic collision of a particle $X$ of mass $4$ moving at velocity $2$ into a particle $E$ of mass $1$, initially at rest. Due to this collision, $X$ breaks into $4$ smaller particles $A$, $B$, $C$, and $D$ (of course, we are ignoring their binding energy, which would lead to an inelastic collision) with identical mass $1$, with velocities $a$, $b$, $c$, and $d$, and $E$ attains a velocity $e$. Let $T_e$ denote the total energy of the particles $A$, $B$, $C$, and $D$ as a function of $e$. Since $T_e$ is at least the kinetic energy of the center-of-mass frame of the particles $A$, $B$, $C$, and $D$, $$T_e\geq \frac{1}{2}\cdot (1+1+1+1)\cdot \left(\frac{a+b+c+d}{4}\right)^2=\frac{(a+b+c+d)^2}{8}\,.$$
The equality case of the inequality above is when $a=b=c=d$ (i.e., when the particles $A$, $B$, $C$, and $D$ are at rest in their center-of-mass frame). By the Conservation Law of Momentum, $$a+b+c+d=4\cdot 2-1\cdot e=8-e\,.$$ By the Conservation Law of Energy, $$T_e=\frac{1}{2}\cdot4\cdot 2^2-\frac{1}{2}\cdot 1\cdot e^2=\frac{16-e^2}{2}\,.$$
Hence,
$$\frac{16-e^2}{2}=T_e\geq \frac{(a+b+c+d)^2}{8}=\frac{(8-e)^2}{8}\,,$$
whence $0\leq e\leq \frac{16}{5}$. The maximum $e=\frac{16}{5}$ is attained iff $a=b=c=d=\frac{6}{5}$. The minimum $e=0$ is attained iff $a=b=c=d=2$.</p>
|
446,326 | <blockquote>
<p>Let $Q$ be a $3\times3$ special orthogonal matrix. Show that $Q(u\times v)=Q(u)\times Q(v)$ for any vectors $u, v\in\mathbb R^3$.</p>
</blockquote>
<p>I have no idea how to start. I'm not sure if $Q(u)\cdot Q(V)=Q(u\cdot v)$ would helps. Please give me some help. Thanks.</p>
| user81327 | 81,327 | <p>Maybe it will be easiest to show this explicitly for the basis vectors $ \{ (1,0,0) \cdots \} $ , and then the general case follows from linearity of all things involved. It will be useful to note that if $ \vec{Q_1}, \vec{Q_2}, \vec{Q_3} $ are the column vectors of $ Q $, then the fact that $ \det(Q) = 1 = \vec{Q1} \cdot (\vec{Q_2} \times \vec{Q_3}) $ gives the "right hand rule" that $\vec{Q_1} \times \vec{Q_2} = \vec{Q_3} $. </p>
<p>Please note, I have been very sloppy and did not check the signs and orders of things. You should check that all formulae are indeed right.</p>
|
4,000,576 | <blockquote>
<p>What is the value of the following integral:
<span class="math-container">$$\int_0^{2\pi}\frac{1}{4\cos^2(t)+9\sin^2(t)}\mathrm{d}t$$</span>
<span class="math-container">$\frac\pi9$</span> ; <span class="math-container">$\frac\pi6$</span> ; <span class="math-container">$\frac\pi3$</span> ; <span class="math-container">$\frac\pi2$</span> or <span class="math-container">$\frac\pi4$</span>?</p>
</blockquote>
<p>a full solution for this problem would be much appreciated</p>
| Community | -1 | <p>Here is a solution using differentiation under integral sign. Consider a more general integral, namely:</p>
<p><span class="math-container">\begin{align*}
I_1(\alpha,\beta) & = \int_{0}^{2\pi}\frac{dx}{\alpha \cos^2x+\beta \sin^2x} \\
& = \int_{0}^{2\pi} \frac{\sec^2x}{\alpha + \beta \tan^2x} \,\mathrm{d}x\\
& = \frac{1}{\beta} \int_0^{2\pi} \frac{1}{\left(\sqrt{\frac{\alpha}{\beta}}\right)^2 + \tan^2 (x)}\; \mathrm{d}(\tan x)\,\\
& = \frac{1}{\sqrt{\alpha \beta}} \left[ \left(\tan^{-1}\left(\sqrt{\frac{\beta}{\alpha}}\tan (x)\right)\right) \right]_0^{2\pi} = \frac{2\pi}{\sqrt{\alpha\beta}}
\end{align*}</span></p>
|
35,688 | <p>I'm looking for a fun (not too many tedious calculations) calculus one problem that uses the concept that, after subsitution, you have two integrals of diffent functions with different limits, but equal area. For example:</p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%28%28pi%5E2%29/x%29%29/%28x%5E2%29%20from%20pi%20to%20pi/2" rel="nofollow">Function one</a></p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%20x%29%2a%28-1/pi%5E2%29%20from%20pi%20to%202%20pi" rel="nofollow">Function two</a></p>
<p>Those two regions are the same. I'm thinking of a problem that asks the student to prove two regions are of equal area, and once they set up the integrals they can see it is a case of subsitution. </p>
<p>Do you know of such a problem?</p>
| Michael Lugo | 173 | <p>In general there are two ways to compute $E(X^2)$ where $X$ is a random variable with density $f(x)$; for simplicity I'll say $X$ always takes values between $0$ and $a$, so $f(x) = 0$ when $x < 0$ or $x > a$. One is to simply take $\int_0^a x^2 f(x) \: dx$. The other is to find the density $g$ of the random variable $Y = X^2$; this is $g(y) = f(\sqrt{y})/(2\sqrt{y})$. Then </p>
<p>$$ E(Y) = \int_0^{a^2} y g(y) \: dy = \int_0^{a^2} {\sqrt{y} \over 2} f(\sqrt{y}) \: dy. $$</p>
<p>These two integrals are related by the change of variables $y = x^2$. You can generate many examples from this. For example take $f(x) = 6x(1-x)$ on $0 \le x \le 1$; then this gives </p>
<p>$$ \int_0^1 6x^3 - 6x^4 \: dx = \int_0^1 3y - 3y^{3/2} \: dy. $$</p>
<p>Although the inspiration for this comes from probability (which I'm teaching currently, which probably explains why this came to mind) the result is true even if $f$ takes on negative values or doesn't integrate to one.</p>
|
35,688 | <p>I'm looking for a fun (not too many tedious calculations) calculus one problem that uses the concept that, after subsitution, you have two integrals of diffent functions with different limits, but equal area. For example:</p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%28%28pi%5E2%29/x%29%29/%28x%5E2%29%20from%20pi%20to%20pi/2" rel="nofollow">Function one</a></p>
<p><a href="http://www.wolframalpha.com/input/?i=int%20%28sin%20x%29%2a%28-1/pi%5E2%29%20from%20pi%20to%202%20pi" rel="nofollow">Function two</a></p>
<p>Those two regions are the same. I'm thinking of a problem that asks the student to prove two regions are of equal area, and once they set up the integrals they can see it is a case of subsitution. </p>
<p>Do you know of such a problem?</p>
| Américo Tavares | 752 | <p>If we make the substitution $t=\frac{u}{1+u}$ in the beta function defined by the following first integral, we get the second:</p>
<p>$$B(p,q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}\;\mathrm{d}t=\int_{0}^{\infty }\frac{%
u^{p-1}}{(1+u)^{p+q}}\;\mathrm{d}u.$$</p>
<p>Since the relation $B(p,q)=\frac{\Gamma (p)\Gamma (q)}{\Gamma (p+q)}$, for $p=q=\frac{1}{2}$ yields</p>
<p>$$B(1/2,1/2)=\frac{\left( \Gamma (1/2)\right) ^{2}}{\Gamma (1)%
}=\pi,$$</p>
<p>we obtain</p>
<p>$$\int_{0}^{1}\frac{1}{\sqrt{t\left( 1-t\right) }}\;\mathrm{d}%
t=\int_{0}^{\infty }\frac{1}{\left( 1+u\right) \sqrt{u}}\;\mathrm{d}u=\pi.$$</p>
|
3,613,235 | <p>I know such integral: <span class="math-container">$\int_0^{\infty}\frac{\ln x}{e^x}\,dx=-\gamma$</span>. What about the integral <span class="math-container">$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx$</span>? </p>
<p>The answer seems very nice: <span class="math-container">$-\frac{1}{2}{\ln}^22$</span> but how it could be calculated? I tried integration by parts but the limit <span class="math-container">$\displaystyle{\lim_{x\to 0}\ln x\ln(1+e^{-x})}$</span> doesn't exist. Or I can also write the following equality
<span class="math-container">$$\int_0^{\infty}\frac{\ln x}{e^x+1}\,dx=\lim\limits_{t\to 0}\frac{d}{dt}\left(\int_0^{\infty}\frac{x^t}{e^x+1}\, dx\right)$$</span>
but I don't know what to do next. </p>
| CHAMSI | 758,100 | <p>Let <span class="math-container">$ m \in\mathbb{N}^{*} : $</span></p>
<p><span class="math-container">\begin{aligned}\sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n}\right)}\\ &=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n+1}\right)}+\sum_{n=1}^{m}{\left(\sqrt{n+1}-\sqrt{n}\right)}\\ \sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sqrt{m+2}-\sqrt{2}+\sqrt{m+1}-1\end{aligned}</span></p>
|
323,665 | <p>Given the base case <span class="math-container">$a_0 = 1$</span>, does <span class="math-container">$a_n = a_{n-1} + \frac{1}{\left\lfloor{a_{n-1}}\right \rfloor}$</span> have a closed form solution? The sequence itself is divergent and simply goes {<span class="math-container">$1, 2, 2+\frac{1}{2}, 3, 3+\frac{1}{3}, 3+\frac{2}{3}, 4, 4+\frac{1}{4}, 4+\frac{2}{4}, 4+\frac{3}{4}, . . .$</span>} and so forth. It seems like it should be easy but I can't seem to find a solution. Any suggestions?</p>
| Carlo Beenakker | 11,260 | <p>The sequence <span class="math-container">$a_n$</span> for <span class="math-container">$n\geq 1$</span> has the following formula:
<span class="math-container">$$a_n=\left\lfloor \sqrt{2n}+\tfrac{1}{2}\right\rfloor +\frac{\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor-\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor ^2 +2 n}{2 \left\lfloor \sqrt{2n}+\frac{1}{2}\right\rfloor }.$$</span>
Here is the <A HREF="https://www.wolframalpha.com/input/?i=Table%5BFloor%5B1%2F2+%2B+Sqrt%5B2%5D+Sqrt%5Bn%5D%5D+%2B+(++2+n+%2B+Floor%5B1%2F2+(1+%2B+Sqrt%5B-7+%2B+8+n%5D)%5D+-++++Floor%5B1%2F2+(1+%2B+Sqrt%5B-7+%2B+8+n%5D)%5D%5E2)%2F(2+Floor%5B1%2F2+%2B+Sqrt%5B2%5D+Sqrt%5Bn%5D%5D),%7Bn,1,15%7D%5D" rel="noreferrer">Wolfram Alpha</A> link to check it.</p>
<p>It is related to <A HREF="https://oeis.org/A002024" rel="noreferrer">OEIS A002024</A> and <A HREF="https://oeis.org/A002262" rel="noreferrer">OEIS A002262.</A></p>
|
323,665 | <p>Given the base case <span class="math-container">$a_0 = 1$</span>, does <span class="math-container">$a_n = a_{n-1} + \frac{1}{\left\lfloor{a_{n-1}}\right \rfloor}$</span> have a closed form solution? The sequence itself is divergent and simply goes {<span class="math-container">$1, 2, 2+\frac{1}{2}, 3, 3+\frac{1}{3}, 3+\frac{2}{3}, 4, 4+\frac{1}{4}, 4+\frac{2}{4}, 4+\frac{3}{4}, . . .$</span>} and so forth. It seems like it should be easy but I can't seem to find a solution. Any suggestions?</p>
| Stuart LaForge | 23,508 | <p>I have decided to call this sequence <span class="math-container">$\Theta_n$</span> for the triangular-harmonic number sequence because it clearly has properties related to both triangular and harmonic numbers.</p>
<p>I had been studying it by using the recursive definition <span class="math-container">$$\Theta_0 = 1\mid\Theta_n = \Theta_{n-1} + \frac{1}{\lfloor\Theta_{n-1}\rfloor}$$</span>
The simplest closed form of the sequence is<br>
<span class="math-container">$$\Theta_n = \frac{T_n^{-1}}{2} + \frac{n}{T_n^{-1}} + \frac{1}{2} \mid n\geq1$$</span> with <span class="math-container">$$T_n^{-1}=\lfloor\sqrt{2n}+\frac{1}{2}\rfloor$$</span> being the inverse triangular number function. I have been researching its many fascinating properties.</p>
|
1,649,907 | <p>Please kindly forgive me if my question is too naive, i'm just a <em>prospective</em> undergraduate who is simply and deeply fascinated by the world of numbers.</p>
<p>My question is: Suppose we want to prove that $f(x) > \frac{1}{a}$, and we <em>know</em> that $g(x) > a$, where $f,g$ and $a$ are all positive and $a$ is a nonzero real number.
<em>If we can show</em> that $f(x)g(x) > 1$, would that imply our required proof ?</p>
<p>EDIT: As demonstrated by various users in the solutions below, the answer is definitely <em>no</em>.
What about if we now want to prove the <em>reverse</em> inequality $f(x) \leq \frac{1}{a}$ given that $g(x) < a$, if we can show that $f(x)g(x)<1$, i guess our required result would follow ?</p>
| Win Vineeth | 311,216 | <p>No, it wouldn't imply that. Because, if $g(x)>a$ , Let's take $g(x) = na ; n>1$
$f(x)g(x)=naf(x)$ <br> If you prove $f(x)g(x)>1$ , You get $naf(x) > 1$ $\implies$ $f(x) > $$1\over na$ ; $n>1$ Which is not what you wanted. <br> $f(x)$ can lie between $1\over na$ and $1\over a$ !!</p>
|
3,394,277 | <p>I am new to this site and not familiar with how to type out math notation so I will do my best. I have a problem I am working on regarding the volume of a circle wrapped around a cylinder of variable radius. For the first part of the problem I had no issue creating a function to represent the cross sectional area. Using this function and I am not trying to integrate to find the volume for different values of cylindrical radius r. The first half of the integral was painless but I have been stuck on the second half for a while now and am looking for some help as I cannot find a solution anywhere. Here is the integral;</p>
<p><span class="math-container">$$\int_{-4}^4\sin(\sqrt{16-x^2})dx$$</span></p>
<p>This integral does include a couple other variable terms but they are treated as constants so there is little point in including them here as they will just complicate the problem. I am basically just looking for the technique used to integrate something like this because I am clueless. Thanks in advance.</p>
| Karthik Kannan | 245,965 | <p>As pointed out in the comments it is sufficient to prove that <span class="math-container">$\sigma(\{(-\infty, F(x)]\cap(0, 1):x\in\mathbb{R}\})\subseteq \mathcal{B}(0, 1)$</span>. If <span class="math-container">$F(x)<1$</span> then clearly <span class="math-container">$(-\infty, F(x)]\cap(0, 1) = (0, F(x)]\in\mathcal{B}(0, 1)$</span> and if <span class="math-container">$F(x) = 1$</span> then again <span class="math-container">$(-\infty, F(x)]\cap(0, 1) = (0, 1)\in\mathcal{B}(0, 1)$</span>. So <span class="math-container">$\forall x\in\mathbb{R}$</span> we have <span class="math-container">$(-\infty, F(x)]\cap(0, 1)\in\mathcal{B}(0, 1)$</span> and therefore the result follows.</p>
|
2,087,235 | <p>I have a question about this question. Find all complex numbers $z$ such that the equation
$$t^2 + [(z+\overline z)-i(z-\overline z)]t + 2z\overline z\ =\ 0$$
has a real solution $t$.</p>
<p><strong>Attempt at a solution</strong></p>
<p>The discriminant is</p>
<p>$[(z+\overline z) - i(z-\overline z)]^2 - 4(2z\overline z)$<br>
$=\ (z+\overline z)^2 - 2i(z+\overline z)(z-\overline z) + [i(z-\overline z)]^2 -8z\overline z$<br>
$=\ (z^2+2z\overline z+\overline z^2) -2i(z^2-\overline z^2) - (z^2-2z\overline z+\overline z^2)-8z\overline z$<br>
$=\ -4z\overline z - 2iz^2 + 2i\overline z^2$</p>
<p>For real solutions, the discriminant must be non-negative. But $z$ is a complex number; how can complex numbers be positive or negative? This is what I don't understand.</p>
<p>Would appreciate any help. Thanks.</p>
| George Law | 141,584 | <p>Hint: Letting $z=a+ib$ reduces the quadratic equation to one with only real coefficients.</p>
|
1,519,952 | <p>Show that
$$S(n,k) = \sum_{m = k-1}^{n-1} {n-1 \choose m} S(m,k-1) $$</p>
<p>-I was having trouble with this proof in class and my professor suggested to look at it as another proof of the following theorem which states:</p>
<p>-For all $n\ge1$
$$B(n) = \sum_{k=0}^{n-1} {n-1 \choose k} B(k) $$
-Unfortunately I still do not understand how to solve this proof, I do see the similarities in structure although I am brand new to Stirling numbers and am unsure of how this would affect the proof. Any help is appreciated.</p>
| Marko Riedel | 44,883 | <p>It may interest the reader to see how this can be done using generating functions.</p>
<p>Fixing the parameter $k$ we seek to show that
$${n\brace k} = \sum_{m=0}^{n-1} {n-1\choose m} {m\brace k-1}.$$</p>
<p>Here we have extended the summation back to zero because the second Stirling number produces zero for those extra values.</p>
<p>Recall the species for set partitions
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields the generating function
$$G(z, u) = \exp(u(\exp(z)-1)).$$</p>
<p>It follows that the EGF of the LHS is
$$\frac{(\exp(z)-1)^k}{k!}.$$</p>
<p>Observe that when we multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the exponential
generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$</p>
<p>In the present case we have $A(z)=\exp(z)$ and
$$B(z) = \frac{(\exp(z)-1)^{k-1}}{(k-1)!}.$$</p>
<p>Therefore on the RHS we are extracting the coefficient
$$(n-1)! [z^{n-1}] \exp(z) \frac{(\exp(z)-1)^{k-1}}{(k-1)!}.$$</p>
<p>Integrating we see that this is
$$n! [z^n] \frac{(\exp(z)-1)^{k}}{k!},$$</p>
<p>the same as the LHS as claimed.</p>
|
2,163,067 | <p>Prove that $\mathbb{Z}_5[x]$ is a unique factorization domain.</p>
<p>My approach is to prove that $\mathbb{Z}_5[x]$ is a PID, which implies that it is a UFD.</p>
<p>Proof:</p>
<p>Suppose there exists an ideal $I$ in $\mathbb{Z}_5[x]$ such that it is generated by two or more elements of $\mathbb{Z}_5[x]$. That is, $I = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Then $I=\{a_1(x)g_1(x)+a_2(x)g_2(x)+...+a_n(x)g_n(x):a_i(x)\in \mathbb{Z}_5[x] \}$. Consider $\max\{a_i(x)g_i(x)\}=\deg_{max} (I)$. Then, since $\mathbb{Z}_5$ is a PID, $\langle a_i(x)g_i(x)\rangle = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Hence, $\mathbb{Z}_5[x]$ is a PID. This implies that $\mathbb{Z}_5[x]$ is a UFD.</p>
<p>It would be interesting to know one's opinion on my proof.</p>
| Bernard | 202,857 | <p><strong>Hint:</strong></p>
<p>You have to use that, in a polynomial ring over a field, you can perform Euclidean divisions, and consider a non-zero polynomial of least degree in the ideal.</p>
|
8,878 | <p>We can restrict the <strong>movement</strong> of locators in a <code>LocatorPane</code> as follows:</p>
<p><img src="https://i.stack.imgur.com/Is904.png" alt="locator movement"></p>
<p>In the following example, the first locator's movement is confined to the x-axis and the second locator's movement is confined to the y-axis.</p>
<pre><code>Manipulate[LocatorPane[Dynamic[pts],
Graphics[{}, PlotRange -> max, Axes -> True, ImageSize -> 350],
{{{-max, 0}, {max, 0}},
{{0, -max}, {0, max}}}],
{{pts, {{9, 0}, {0, 11}}}, ControlType -> None}, Initialization :> {max = 25}]
</code></pre>
<p><img src="https://i.stack.imgur.com/NvUL8.png" alt="enter image description here"></p>
<p>However, a locator can be dragged without necessarily being selected.
You can begin dragging a locator by clicking down on just about any point in the four quadrants and dragging from there.</p>
<p>I would like to have the locator respond only to drags that are initiated by "mousedown"ing on the locator itself.</p>
<p>Any ideas as to how I might achieve this? </p>
<hr>
<p><strong>Addendum</strong>
In my the more complicated example I've been working with, I've made use of jump parameters to confine locators to integer values:</p>
<pre><code>{{{-maxX, 0}, {maxX, 0}, {1, 0}},
{{0, -maxY}, {0, maxY}, {0, 1}}}
</code></pre>
<p>I'm not sure whether such constraints (to integers) can be imposed on locators outside of <code>LocatorPane</code>s.</p>
| rm -rf | 5 | <p>That is the default (and expected) behaviour of <code>LocatorPane</code>. This is useful in implementing things like colour pickers, for example, where it is convenient to simply click on any point to select that colour and have the locator move there automatically to indicate selection. </p>
<p>To create locators that move only when explicitly clicked and dragged, use a <code>Locator</code>. Here's an example for the scenario in the question:</p>
<pre><code>DynamicModule[{pt1 = {-5, 0}, pt2 = {0, 5}},
With[{max = 25},
Graphics[{
Locator[Dynamic[pt1, (pt1 = {First@#, 0}) &]],
Locator[Dynamic[pt2, (pt2 = {0, Last@#}) &]]
}, PlotRange -> max, Axes -> True, ImageSize -> 350
]
]
]
</code></pre>
|
8,878 | <p>We can restrict the <strong>movement</strong> of locators in a <code>LocatorPane</code> as follows:</p>
<p><img src="https://i.stack.imgur.com/Is904.png" alt="locator movement"></p>
<p>In the following example, the first locator's movement is confined to the x-axis and the second locator's movement is confined to the y-axis.</p>
<pre><code>Manipulate[LocatorPane[Dynamic[pts],
Graphics[{}, PlotRange -> max, Axes -> True, ImageSize -> 350],
{{{-max, 0}, {max, 0}},
{{0, -max}, {0, max}}}],
{{pts, {{9, 0}, {0, 11}}}, ControlType -> None}, Initialization :> {max = 25}]
</code></pre>
<p><img src="https://i.stack.imgur.com/NvUL8.png" alt="enter image description here"></p>
<p>However, a locator can be dragged without necessarily being selected.
You can begin dragging a locator by clicking down on just about any point in the four quadrants and dragging from there.</p>
<p>I would like to have the locator respond only to drags that are initiated by "mousedown"ing on the locator itself.</p>
<p>Any ideas as to how I might achieve this? </p>
<hr>
<p><strong>Addendum</strong>
In my the more complicated example I've been working with, I've made use of jump parameters to confine locators to integer values:</p>
<pre><code>{{{-maxX, 0}, {maxX, 0}, {1, 0}},
{{0, -maxY}, {0, maxY}, {0, 1}}}
</code></pre>
<p>I'm not sure whether such constraints (to integers) can be imposed on locators outside of <code>LocatorPane</code>s.</p>
| Vitaliy Kaurov | 13 | <p>This is the bare bone implementation with direct control of mouse events:</p>
<pre><code>DynamicModule[{p = {0, 0}}, EventHandler[Framed@Dynamic[Style[
Graphics[{Red, Disk[p, 0.2]}, PlotRange -> 2], Selectable -> False]],
{"MouseDragged" :> (p = MousePosition["Graphics"])}]]
</code></pre>
<p>And this is more or less what you need:</p>
<pre><code>DynamicModule[{p1 = {0, 2}, p2 = {2, 0}}, EventHandler[
Dynamic[Style[Graphics[{{Red, Disk[{0, Round@p1[[2]]}, .2]}, {Blue,
Disk[{Round@p2[[1]], 0}, .2]}}, PlotRange -> 5, Axes -> True],
Selectable -> False]],{"MouseDragged" :> (If[
EuclideanDistance[p1, MousePosition["Graphics"]] <
EuclideanDistance[p2, MousePosition["Graphics"]],
p1 = MousePosition["Graphics"], p2 = MousePosition["Graphics"]];)}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/lxQrg.png" alt="enter image description here" /></p>
<p>It is a bit hacky, but because of integer settings it works.</p>
|
3,784,471 | <p>To solve this exercise,</p>
<p><span class="math-container">$$|\arccos(\cos(x))|<\pi/4$$</span></p>
<p>I have thought to apply this condition,
<span class="math-container">$$|f(x)|<k, \quad k\in \Bbb R^+, \iff -k<f(x)<k$$</span></p>
<p>Hence,</p>
<p><span class="math-container">$$-\frac \pi4<\arccos(\cos(x))<\frac \pi4$$</span>
Being <span class="math-container">$\arccos\colon [-1,1]\to [0,\pi]$</span>, I can have</p>
<p><span class="math-container">$$\cos\left(-\frac \pi4\right)<\cos(\arccos(\cos(x)))<\cos\left(\frac \pi4\right) \iff \frac{\sqrt2}2<\cos(x)<\frac{\sqrt2}2$$</span></p>
<p>false for all <span class="math-container">$x\in \mathbb{R}$</span>. Are they correct my steps?</p>
| Michael Kinyon | 444,012 | <p>Nonassociative division rings with unity are known as <em>semifields</em>. They come up in the coordinatization of projective planes. Their study in the finite case started with</p>
<p>Donald Knuth, Finite semifields and projective planes. <em>J. Algebra</em> <strong>2</strong> (1965), 182-217.</p>
<p>This published version was based on Knuth's 1963 PhD dissertation.</p>
<p>There is a considerable literature on semifields, but note that the term "semifield" is also used in a conflicting sense in other parts of mathematics as an associative semiring with unity in which every nonzero element has a multiplicative inverse. Papers on semifields in the nonassociative sense are generally tagged with the MSC classification 17A35 (Nonassociative division algebras).</p>
|
479,697 | <p>I have some quetion on essential singularity.</p>
<p>Let $f(z)$, $g(z)$ have the same essential singularity at $z=z_0$.</p>
<p>Then, if $\frac{f(z)}{g(z)}$ is not a constant function on some neighborhood of $z_0$, then $ \frac{f(z)}{g(z)}$ also has essential singularity at $z=z_0$?</p>
<p>If not, could you give me some counter example?</p>
| Dennis Fleming | 92,318 | <p>I'm going to assume that by essential singularity you mean is 0 at the given point and that the singularity occurs when the function is in the denominator. The classic example for this is the sinc function, sin(x)/x. This is the Fourier transform for a rectangular window. Clearly both functions approach 0 at x = 0, but if you take the derivative of each you will see that x approaches 0 at a constant rate and sin(x) approaches 0 as a function of cos(x). The limit as x ->0 is 1, so that the value is non-singular and the function itself is analytic at all points of the curve.</p>
|
479,697 | <p>I have some quetion on essential singularity.</p>
<p>Let $f(z)$, $g(z)$ have the same essential singularity at $z=z_0$.</p>
<p>Then, if $\frac{f(z)}{g(z)}$ is not a constant function on some neighborhood of $z_0$, then $ \frac{f(z)}{g(z)}$ also has essential singularity at $z=z_0$?</p>
<p>If not, could you give me some counter example?</p>
| Silvia Ghinassi | 258,310 | <p>Dominic Michaelis provided the following counterexample in the comments section: if $f$ has an essential singularity at $z=z_0$, then also $g(z)=zf(z)$ has an essential singularity at $z=z_0$ but $\frac{g(z)}{f(z)}=\frac{zf(z)}{f(z)}=z$ which does not have an essential singularity at $z=z_0$.</p>
|
775,265 | <p>Please help me get the answer to this question.</p>
<p>Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)</p>
| re0 | 144,190 | <p>We have $|x-4|< \delta$.</p>
<p>$|f(x)-f(4)|=|\sqrt {2x-6}-\sqrt {2}|=|\sqrt{2}(\sqrt{x-3}-1)|$</p>
<p>Now, multiplying the numerator and denominator by $(\sqrt{x-3}+1)$,</p>
<p>$=\sqrt{2} |\frac{x-4}{\sqrt{x-3}+1}$|</p>
<p>Now, ${\sqrt{x-3}+1}$ is always greater than or equal to $1$. Thus, $\frac{x-4}{\sqrt{x-3}+1}\leqslant (x-4)$.</p>
<p>Therefore, $|f(x)-f(4)|\leqslant \sqrt{2}|(x-4)| < \sqrt{2} \delta=\epsilon$</p>
<p>Thus, $f(x)$ is continuous at $x=4$.</p>
|
1,662,876 | <p>Now we have some examples of what I mean $$\int_0^{2\pi} \sin x~dx=0$$
$$\int_0^{8\pi} \cos 4x~dx=0$$</p>
<p>$$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$</p>
<p>Looking at the graph of $f(x)=\sin (x)$ for example it makes some sense to me that $$\int_0^{2\pi} \sin x~dx=0$$ because the region below the $x$ axis will "cancel" with the part above but I don't understand how I can make claims about more general integrals like $$\int_{\pi}^{2\pi} \sin^3 10x~dx=0$$ which I just got from testing with an online integral tool.</p>
<p>Obviously there is something more deeper going on and I would like to understand when and why I can just claim that an integral will be zero as it will save me a lot of time.</p>
<p>I have had a thought but can't seem to conclude how it would be obvious that any of the less obvious integrals would be zero (like the third one I stated), I imagine that there is some kind of symmetry but I'm just not seeing it. </p>
<p>So if anyone could explain when (in a more general sense not just for the integrals above) and why this works I would appreciate it. </p>
| Mathstudent | 307,000 | <p>Hint : for the first one, use the change of variable $y = x-\pi$ on $[\pi,2\pi]$, the fact that the function $sin$ is an odd function and $$\int_0^{2\pi} \sin x~dx=\int_0^{\pi} \sin x~dx+\int_{\pi}^{2\pi} \sin x~dx$$</p>
|
2,003,916 | <p>Probably a very simple question:</p>
<p>Suppose a hospital orders defibrillators from a manufacturer. It is well known that defibrillations are often not effective, even when the defibrillators themselves are working properly. Suppose research shows that only 15% percent of defibrillations are effective. Over the next few months the hospital performs about 3000 defibrillations. What is the probability that none of these are successful?</p>
| Thomas Andrews | 7,933 | <p>$ab^p$ and $ba^p$ are both odd if $a,b$ are both odd, and both even otherwise, so $ab^p-ba^p$ is divisible by $2.$</p>
<p>$b^p-b$ is divisible by $p$ for any $b$, so $ab^p-ab$ is divisible by $p$. Similarly, $ba^p-ba$ is divisible by $p$. So $ab^p-ba^p=(ab^p-ab)-(ba^p-ba)$ is divisible by $p$.</p>
<p>The last thing you need to show is that $3\mid ab^p-ba^p$. For this case, you need $p$ odd. I'll leave that to you. </p>
<p>(You need $p>3$ only because once you have $p\mid x$, $2\mid x$ and $3\mid x$, you need $p\neq 2,3$ to get that $6p\mid x$.)</p>
|
1,033,383 | <p>$ABCD$ is a rectangle and the lines ending at $E$, $F$ and $G$ are all parallel to $AB$ as shown. </p>
<p>If $AD = 12$, then calculate the length of $AG$.<img src="https://i.stack.imgur.com/OUQZ8.png" alt="enter image description here"></p>
<p>Ok, I started by setting up a system of axes where $A$ is the origin and the $x$-axis along $AB$ and the $y$-axis along $AD$. So $D(0;-12)$ AND $C(x;-12)$ and $B(x;0)$. I am really stuck now on how to proceed so can someone please help me?</p>
| Anatoly | 90,997 | <p>I would suggest you to set the origin in $B$. Setting $AB=DC=d$, line $AC$ has equation $y=-\frac{12}{d}x-12$, and point $E$ has coordinates $(-d,-6)$.</p>
<p>Since line $EB$ has equation $y=\frac{6}{d}x$, the $x$-coordinate of its intersection with $AC$ is given by the solution of</p>
<p>$$\frac{6}{d}x=-\frac{12}{d}x-12$$</p>
<p>which is $x=-\frac{2}{3}d$. The $y$-coordinate of the intersection is then $y=4$. So we get that the $y$-coordinate of point $F$ is equal to $4$ as well.</p>
<p>Repeating the procedure, we can note that because line $FB$ has equation $y=\frac{4}{d}x$, the $x$-coordinate of its intersection with $AC$ is given by the solution of</p>
<p>$$\frac{4}{d}x=-\frac{12}{d}x-12$$</p>
<p>which is $x=-\frac{3}{4}d$. The $y$-coordinate of the intersection is then $y=3$. So we get that the $y$-coordinate of point $G$ is equal to $3$ as well, and then $AG= 3$.</p>
|
2,747,753 | <p>Let $x\in\mathbb{R}$. Demonstrate that if the numbers $a = x^3–x$ and $b = x^2 +1$ are rational, then $x$ is rational.</p>
| José Carlos Santos | 446,262 | <p><strong>Hint:</strong> $x^3-x=x\bigl((x^2+1)-2\bigr)$</p>
|
8,382 | <h3>Context</h3>
<p>I'm writing a function that look something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[#2[[1]]-#2[[2]]] <= 1 &, mat, {2}] //
Flatten // And @@ # &
</code></pre>
<p>Now, things like <code>#2[[1]]</code> and <code>#2[[2]]</code> are somewhat hard to read. I'd prefer to do something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[i-j] <= 1 &, mat, {2}] //
Flatten // And @@ # & (* with a {i, j} <- #2 somewhere *)
</code></pre>
<h3>Question</h3>
<p>Is there someway to do something like "destructuring" in <em>Mathematica</em>?</p>
<hr>
<p>The following links convey what I mean by "destructuring":</p>
<ul>
<li><a href="http://clojure.org/special_forms">http://clojure.org/special_forms</a></li>
<li><a href="http://java.dzone.com/articles/clojure-destructuring">http://java.dzone.com/articles/clojure-destructuring</a></li>
<li><a href="http://blog.jayfields.com/2010/07/clojure-destructuring.html">http://blog.jayfields.com/2010/07/clojure-destructuring.html</a></li>
</ul>
<p>(These have nothing to do with <em>Mathematica</em>; they're posted mainly to demonstrate what is meant by "destructuring")</p>
| M.R. | 403 | <p>Perhaps you could just use With in a Function?</p>
<pre><code>lst = ConstantArray[0, {3, 3}];
MapIndexed[
Function[{value, pos},
With[{i = pos[[1]], j = pos[[2]]},
{value, i, j}
]
],
lst, {2}]
</code></pre>
|
8,382 | <h3>Context</h3>
<p>I'm writing a function that look something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[#2[[1]]-#2[[2]]] <= 1 &, mat, {2}] //
Flatten // And @@ # &
</code></pre>
<p>Now, things like <code>#2[[1]]</code> and <code>#2[[2]]</code> are somewhat hard to read. I'd prefer to do something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[i-j] <= 1 &, mat, {2}] //
Flatten // And @@ # & (* with a {i, j} <- #2 somewhere *)
</code></pre>
<h3>Question</h3>
<p>Is there someway to do something like "destructuring" in <em>Mathematica</em>?</p>
<hr>
<p>The following links convey what I mean by "destructuring":</p>
<ul>
<li><a href="http://clojure.org/special_forms">http://clojure.org/special_forms</a></li>
<li><a href="http://java.dzone.com/articles/clojure-destructuring">http://java.dzone.com/articles/clojure-destructuring</a></li>
<li><a href="http://blog.jayfields.com/2010/07/clojure-destructuring.html">http://blog.jayfields.com/2010/07/clojure-destructuring.html</a></li>
</ul>
<p>(These have nothing to do with <em>Mathematica</em>; they're posted mainly to demonstrate what is meant by "destructuring")</p>
| Mr.Wizard | 121 | <p>Leonid provides a nice method for doing this within "pure functions" but I think it should be pointed out that the common method for doing this is pattern matching.</p>
<p><strong>I argue that destructuring is the <em>foundational use</em> of pattern matching in <em>Mathematica</em>.</strong></p>
<p>Every replacement pattern, be it an explicit rule (<code>:></code>, <code>-></code>) or part of a definition (<code>:=</code>, <code>=</code>), that uses a named pattern on the left-hand side that does not match the entire expression or argument is doing destructuring.</p>
<p>Applied to your specific example:</p>
<pre><code>f[a_, {i_, j_}] := a == 0 || Abs[i - j] <= 1
triDiagonalQ[mat_] := And @@ Flatten @ MapIndexed[f, mat, {2}]
</code></pre>
<p>Or:</p>
<pre><code>triDiagonalQ[mat_] := And @@ Flatten @
MapIndexed[#2 /. {i_, j_} :> # == 0 || Abs[i - j] <= 1 &, mat, {2}]
</code></pre>
<p>The second example is almost exactly what you asked for: "with a <code>{i, j} <- #2</code> somewhere"<br>
It's just turned around: <code>#2 /. {i_, j_}</code>.</p>
<p>This destructuring is common in <em>Mathematica</em> programming for experienced users.</p>
<p>Among many examples:</p>
<p><a href="https://mathematica.stackexchange.com/a/2450/121">Here</a> I use it to separate <code>a + b + c</code>:</p>
<pre><code>(a + b + c) /. head_[body___] :> {head, body} (* Out= {Plus, a, b, c} *)
</code></pre>
<p><a href="https://mathematica.stackexchange.com/a/5756/121">Here</a> Leonid uses it in a recursive function. (<code>{x_, y_List}</code>)</p>
<p>Szabolcs uses it in <a href="https://mathematica.stackexchange.com/a/1328/121"><code>iter</code></a>, also recursive.</p>
<p>Heike uses it with <code>/.</code> in <a href="https://mathematica.stackexchange.com/a/5891/121"><code>PerforatePolygons</code></a> and with <code>:=</code> in <a href="https://mathematica.stackexchange.com/a/4155/121"><code>torn</code></a>.</p>
<p><a href="https://mathematica.stackexchange.com/a/7882/121">Here</a> I used it simply in <code>formula</code> but also in <code>MakeBoxes[defer[args__], fmt_] :=</code> where the parameter pattern <code>defer[args__]</code> serves to match the literal head <code>defer</code> while also destructuring.</p>
<p>In <a href="https://mathematica.stackexchange.com/a/3259/121"><code>withOptions</code></a> it is used both in the function definition and in the replacement rule.</p>
<p>The <a href="https://mathematica.stackexchange.com/a/1937/121">"injector pattern"</a> is a form of destructing. </p>
<p>I also used it in <a href="https://mathematica.stackexchange.com/a/9410/121"><code>inside</code></a>, <a href="https://mathematica.stackexchange.com/a/8219/121"><code>withTaggedMsg</code></a>, <a href="https://mathematica.stackexchange.com/a/1129/121"><code>pwSplit</code></a>, <a href="https://mathematica.stackexchange.com/a/7512/121"><code>dPcore</code></a> etc.</p>
<hr>
<p>Another, simpler form of destructuring exists in the form of <code>Set</code> and <code>List</code> (<code>{}</code>). A matching <code>List</code> structure on the left and right sides of <code>=</code> will assign values part-wise.</p>
<pre><code>{{a, b}, c, {d}} = {{1, 2}, 3, {4}};
{a, b, c, d}
</code></pre>
<blockquote>
<pre><code> {1, 2, 3, 4}
</code></pre>
</blockquote>
<p>This is used e.g. in the first <a href="http://reference.wolfram.com/mathematica/ref/LUDecomposition.html" rel="nofollow noreferrer"><code>LUDecomposition</code></a> example, and R.M uses it <a href="https://mathematica.stackexchange.com/a/4027/121">here</a>.</p>
|
8,382 | <h3>Context</h3>
<p>I'm writing a function that look something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[#2[[1]]-#2[[2]]] <= 1 &, mat, {2}] //
Flatten // And @@ # &
</code></pre>
<p>Now, things like <code>#2[[1]]</code> and <code>#2[[2]]</code> are somewhat hard to read. I'd prefer to do something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[i-j] <= 1 &, mat, {2}] //
Flatten // And @@ # & (* with a {i, j} <- #2 somewhere *)
</code></pre>
<h3>Question</h3>
<p>Is there someway to do something like "destructuring" in <em>Mathematica</em>?</p>
<hr>
<p>The following links convey what I mean by "destructuring":</p>
<ul>
<li><a href="http://clojure.org/special_forms">http://clojure.org/special_forms</a></li>
<li><a href="http://java.dzone.com/articles/clojure-destructuring">http://java.dzone.com/articles/clojure-destructuring</a></li>
<li><a href="http://blog.jayfields.com/2010/07/clojure-destructuring.html">http://blog.jayfields.com/2010/07/clojure-destructuring.html</a></li>
</ul>
<p>(These have nothing to do with <em>Mathematica</em>; they're posted mainly to demonstrate what is meant by "destructuring")</p>
| jVincent | 1,194 | <p>I think Mr. Wizard provided a very thorough answer to the question. I would however like to add a slight example of wrapping this up nicely in a format similar to <code>Function[]</code> but using destructuring:</p>
<pre><code>SetAttributes[dFunction, HoldAll]
dFunction[pattern_, body_][arg___] /;MatchQ[{arg}, pattern] := {arg} /. pattern :> body
</code></pre>
<p>This then allows you to have nice syntax for an anonymous destructuring function using pattern matching:</p>
<pre><code>triDiagonalQ3[mat_] := And @@ Flatten@ MapIndexed[
dFunction[{v_, {i_, j_}}, v == 0 || Abs[i - j] <= 1], mat, {2}]
</code></pre>
<p><strong>Syntax</strong></p>
<p>As to the "clumsiness" of this syntax, I have to agree with Mr. Wizard that it's not as nicely looking as replacement rules, but then again I don't particularly like <code>Function[x,x^2]</code> either, and much prefer the aesthetics of the built-in shorthand which looks like <code>x ↦ x^2</code>(entered via <kbd>Esc</kbd><kbd>f</kbd><kbd>n</kbd><kbd>Esc</kbd>). The above code can be made to use a similar shorthand rather simply by setting <code>RightArrowBar = dFunction;</code> which allows the definition to look like : </p>
<pre><code>{v_, {i_, j_}} ⇥ (v == 0 || Abs[i - j] <= 1)
</code></pre>
<p>And if one is afraid of using the build-in undefined infix operators, you could of course define a custom symbol for it which doesn't lay claim on <code>RightArrowBar</code>. Personally I find this appearance of an anonymous replacement rule based function to be quite nice.</p>
<p><strong>The value of pattern matching</strong></p>
<p>As was discussed in comments, one of the interesting behaviors of this construct is that it closely mirrors the behavior of a function defined through pattern matching. if <code>f[x_Integer]:=...</code> is defined and called with arguments not matching the pattern the function remains unevaluated. For instance <code>f[2.3]</code> will return itself. The same behavior can be observed with <code>dFunction</code> where for instance <code>dFunction[{x_Integer}, x][3.2]</code> will return itself unevaluated. </p>
<p>Sadly due to the nature of <code>SubValues</code>, it's not possible to fully simulate pattern matching functions in an anonymous form using this method, since attributes such as <code>HoldAll</code> cannot be implemented. I do belive that this could be implimented using devious hacks similar to the method employed here <a href="https://mathematica.stackexchange.com/a/5458/1194">https://mathematica.stackexchange.com/a/5458/1194</a> by Leonid Shifrin. However I fear that even if I did mange to implement it I would never use the result for fear of unexpected behavior, while I quite like the simplicity of my current <code>dFunction</code>. </p>
|
1,794,072 | <p>My attempt :</p>
<p>If $n$ is odd, then the square must be 2 (mod 3), which is not possible.</p>
<p>Hence $n =2m$</p>
<p>$2^{2m}+3^{2m}=(2^m+a)^2$</p>
<p>$a^2+2^{m+1}a=3^{2m}$</p>
<p>$a (a+2^{m+1})=3^{2m} $</p>
<p>By fundamental theorem of arithmetic, </p>
<p>$a=3^x $</p>
<p>$3^x +2^{m+1}=3^y $</p>
<p>$2^{m+1}=3^x (3^{y-x}-1) $</p>
<p>Which is not possible by Fundamental theorem of Arithmetic </p>
<p>Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 $(n) $are used.</p>
| Alijah Ahmed | 124,032 | <p>Yet another approach. </p>
<p>For the odd case of $n$, as you mentioned the result is $2\mod 3$ which cannot be a square number.</p>
<p>As for the even case $n=2m$, we have $2^{2m}+3^{2m}\mod 10$ being equal to $2$ (for when $m=0$) or either $3$ (for $n=2+4k,k\geq0$) or $7$ (for $n=4k,k\geq1$), which clearly cannot be square (as a square number modulus $10$ takes values $0,1,4,9,6,5$). </p>
|
105,750 | <p>Given a <code>ContourPlot</code> with a set of contours, say, this:</p>
<p><a href="https://i.stack.imgur.com/cKoyo.jpg"><img src="https://i.stack.imgur.com/cKoyo.jpg" alt="enter image description here"></a></p>
<p>is it possible to get the contours separating domains with the different colors in the form of lists? </p>
<p>For example, how to extract the boundaries of the blue domain in the image above?
Or just for the sake of trial, from such a simple example:</p>
<pre><code> ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 3}, {y, -3, 3},
PlotRange -> {0, 0.5}, ColorFunction -> "Rainbow"]
</code></pre>
<p><a href="https://i.stack.imgur.com/Beuzu.jpg"><img src="https://i.stack.imgur.com/Beuzu.jpg" alt="enter image description here"></a></p>
<p>The same task, let us find the lists corresponding to the blue domain boundaries.</p>
<p>To make it clear, I am not asking of how to get the lines from the function behind. This I understand. I ask of how to extract the contour lines that are generated by Mma.</p>
<p>Let us put this question another way around. Is it possible to define the areas with the same color as separate geometric regions in the sense of the computation geometry, and then work with these domains separately?</p>
| Basheer Algohi | 13,548 | <pre><code>p = ContourPlot[x*Exp[-x^2 - y^2], {x, 0, 3}, {y, -3, 3},
PlotRange -> {0, 0.5}, ColorFunction -> "Rainbow"];
colors = Cases[p, _RGBColor, -1];
poly = Cases[Cases[Normal@p, {__, colors[[2]], __}, -1],Polygon[__], -1];
r = RegionUnion[poly];
lines = Cases[Normal@RegionPlot[r], Line[__], -1];
Graphics[lines]
</code></pre>
<p><a href="https://i.stack.imgur.com/xOxSJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xOxSJ.jpg" alt="enter image description here"></a></p>
|
881,282 | <p>Same as above, how to simplify it. I am to calculate its $n$th derivative w.r.t x where t is const, but I can't simplify it. Any help would be appreciated. Thank you.</p>
| DeepSea | 101,504 | <p>Let $u = \sqrt{3}x$, and $v = 4 - \sqrt{3}x$, then:</p>
<p>$u + v = 4$, and $\dfrac{3}{u^2} + \dfrac{1}{v^2} = 1 \to 3v^2 + u^2 = u^2v^2 \to 3v^2 = u^2(v^2 - 1) \to 3v^2 = (4-v)^2(v^2 - 1)$. Observe that $v=2$ is a root to the above equation. From this we can use synthetic division to factor the polynomial and finish it.</p>
|
2,872,807 | <p>I was browsing through facebook and came across this image: <a href="https://i.stack.imgur.com/jozSg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jozSg.png" alt="enter image description here"></a></p>
<p>I was wondering if we can find more examples where this happens?</p>
<p>I guess this reduces to finding integer solutions for the equation </p>
<p>$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for integers a,b,c</p>
<p>Or can we even further extend to when they are all distinct that is finding solutions to </p>
<p>$$ \frac{a^3+b^3}{c^3+d^3} = \frac{a+b}{c+d} $$ for integers a,b,c</p>
<p>I don't really have that much knowledge in the number theory area so I have come here</p>
| mengdie1982 | 560,634 | <p>$$\frac{a^3+b^3}{a^3+c^3}=\frac{(a+b)(a^2-ab+b^2)}{(a+c)(a^2-ac+c^2)}=\frac{a+b}{a+c}$$</p>
<p>If $a+c \neq 0$ and $a+b \neq 0,$ then $$a^2-ab+b^2=a^2-ac+c^2,$$namely $$(b+c-a)(b-c)=0.$$</p>
<p>If $b=c$, the case is trivial. If $b \neq c$, then $$b+c=a.$$</p>
|
2,783,423 | <p>If I have a line formed by points A and B, how can I find the distance of another point from that line. Also, whether that line is clockwise or CCW from point A.
<a href="https://i.stack.imgur.com/dpazD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dpazD.png" alt="enter image description here"></a></p>
<p>I'm not really even sure how to get started with this. I know the midpoint of A and B, but after that I am out of my depth. </p>
<p>Would appreciate advice on how to proceed.</p>
| G Cab | 317,234 | <p><a href="https://i.stack.imgur.com/Ou2nG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ou2nG.png" alt="punto_linea_1"></a></p>
<p>Take the vectors
$$
{\bf v} = \mathop {BA}\limits^ \to \quad \;{\bf z} = \mathop {BZ}\limits^ \to
$$</p>
<p>Compute the unitary vector $\bf t$ parallel to $\bf v$
$$
{\bf t} = {{\bf v} \over {\left| {\bf v} \right|}}
$$
and the unitary vector $\bf n$ normal to it
$$
{\bf n} = \left( {t_y , - t_x } \right)
$$</p>
<p>You can see that with the signs chosen for $\bf n$, the sign of
$\bf n \times \bf t$ is positive, according to the right hand rule,
that is the axes $(n,t)$ are oriented same as $(x,y)$.</p>
<p>Now the dot products
$$
\left( {{\bf z} \cdot {\bf n},\;{\bf z} \cdot {\bf t}} \right)
$$
will give you the coordinates of $\bf z$ in that system, and their signs the
position of $\bf z$ in the relative quadrants.<br>
${\bf z} \cdot {\bf n}$ is the distance of $Z$ from the line $AB$, with the sign
corresponding to the direction of $\bf n$.</p>
|
2,783,423 | <p>If I have a line formed by points A and B, how can I find the distance of another point from that line. Also, whether that line is clockwise or CCW from point A.
<a href="https://i.stack.imgur.com/dpazD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dpazD.png" alt="enter image description here"></a></p>
<p>I'm not really even sure how to get started with this. I know the midpoint of A and B, but after that I am out of my depth. </p>
<p>Would appreciate advice on how to proceed.</p>
| Toby Mak | 285,313 | <p>This might be a better method:</p>
<p><img src="https://i.stack.imgur.com/sPY41.png" alt="enter link description here"></p>
<p>Let $A$ be at $(0,2)$, $B$ be at $(3, -2)$ , and $C$ be at $(7, -3)$. </p>
<p>The distance from $A$ to $B$ is $\sqrt{(-2-2)^2+(3-0)^2} = 5$. Let $AD$ be $x$ and $BD$ be $5-x$. You can then use find $AC$ and $BC$ and use the Pythagorean theorem to find $x$, and after that, find how long the perpendicular is.</p>
|
3,596,566 | <p>I've been going through <a href="https://math.stackexchange.com/a/27177/288450">this proof</a>.</p>
<p>And I'm wondering what allows me to change the order of the integral and the infinite sum.</p>
<p><span class="math-container">$$\int_{-\infty}^{\infty} \left( \sum_{n \ge 0} \frac{2^n t^n x^n}{n!} \right) e^{-x^2} dx = \sum_{n \ge 0} \frac{2^n t^n}{n!} \int_{-\infty}^{\infty} x^n e^{-x^2} \, dx.$$</span></p>
<p>I know the result that for a function series that converges uniformly on <span class="math-container">$[a, b]$</span> the following holds:</p>
<p><span class="math-container">$$ \int^{b}_{a} \sum_{n=0}^{\infty} f_n(x) \ dx = \sum_{n=0}^{\infty} \int_a^b f_n(x) \ dx$$</span></p>
<p>But here the limits of integration are <span class="math-container">$-\infty$</span> and <span class="math-container">$+\infty$</span>, and I'm not sure what to make of it.</p>
| zhw. | 228,045 | <p>Let <span class="math-container">$\mu$</span> be a positive measure on <span class="math-container">$X.$</span> If <span class="math-container">$S_N(x)\to S(x)$</span> pointwise <span class="math-container">$\mu$</span> a.e. on <span class="math-container">$X,$</span> and there exists <span class="math-container">$f\in L^1(X,\mu)$</span>
such that <span class="math-container">$|S_N(x)|\le f(x)$</span> on <span class="math-container">$X$</span> for each <span class="math-container">$N,$</span> then</p>
<p><span class="math-container">$$\int_X S_N \,d\mu \to \int_X S \,d\mu.$$</span></p>
<p>This is just straight up DCT. All that's left to do is think of our problem, where <span class="math-container">$t$</span> is fixed, identify <span class="math-container">$S_N$</span> as</p>
<p><span class="math-container">$$\sum_{n=0}^{N}\frac{2^nt^nx^n}{n!}e^{-x^2}$$</span></p>
<p>and identify <span class="math-container">$f(x)= \exp (2|t||x|-x^2)\dots$</span></p>
|
3,970,641 | <p>I have 927 unique sequences of the numbers 1, 2 and 3, all of which sum to 12 and represent every possible one-octave scale on the piano, with the numbers representing the intervals between notes in half-steps (i.e., adjacent keys). For example, the <a href="https://en.wikipedia.org/wiki/Major_scale" rel="nofollow noreferrer">Major Scale</a> is { 2,2,1,2,2,2,1 }, while the hexatonic (6-note) <a href="https://en.wikipedia.org/wiki/Blues_scale" rel="nofollow noreferrer">Blues Scale</a> is { 3, 2, 1, 1, 3, 2 }. It can help mentally to add a zero at the beginning to represent the root note.</p>
<p>I'm trying to decide how to categorize them, and believe Shannon entropy is a natural place to begin, with one problem: The default Shannon Equation returns the same figure agnostic of the order of the sequence (as we would expect), whereas this is a case where the order is supremely relevant. (Music theory and information theory have a lot in common, though this appears to be largely unexplored.)</p>
<p>Let's take a simple example: Both of these are valid scales, though neither has a name, to my knowledge:</p>
<pre><code>{ 1,2,1,2,1,2,1,2 }
{ 2,2,1,2,1,1,1,2 }
</code></pre>
<p>Since there are equal numbers of 1s and 2s in each set, we don't need a calculator to see that the Shannon Entropy for each is 4:</p>
<p><span class="math-container">$$-\sum_{i=1}^8 \frac{1}{2} \log_{2}\frac{1}{2}$$</span></p>
<p>However, I believe--I may be very wrong here--that there is significantly more information contained in the second sequence, given that the first can be communicated as "{1,2}, four times" while the second is considerably more arbitrary. (I believe this is debate in bioinformatics, where one can locate long spans of simple repeating sequences of nucleotides that some consider "junk DNA," though others dispute that adjective.)</p>
<p><strong>How does one account for order when measuring surprisal?</strong> If I were to take, say, <em>The Great Gatsby</em> and sort the ~261,000 characters (as in letters, etc., not fictional people!) in ASCII order, I don't think it would contain the same amount of information as the original text.</p>
<p>One strategy I tried was a rolling measure of probability, where the algorithm is initially unaware of the total frequency of each number but accumulates the odds as it goes. (I suppose this is like measuring the accumulating surprisal of a die you cannot see and don't know how many sides it has.) This did not produce very intuitive or useful results.</p>
<p>Another thought I had was to divide each sequence into "words" that are repeating sub-sequences of at least two digits, then measure the entropy of each word and ... I'm not sure what to do with them. A sum of <span class="math-container">$H[{1,2}]$</span> four times still adds to 4. I'm a bit out of my depth here both in reducing sequences to the most efficient subsequences and knowing what to do with them.</p>
<p>I've read Shannon's 1948 paper twice, including the discussion of the relative odds of letters in a given language, which seems relevant but is, in fact, not useful in this case, where every possible ordered combination is present and the intervals are added independently of those preceding them, with the edge case that they cannot exceed a sum of 12.</p>
<p><strong>Which is all to say, how does one quantify the entropy of an <em>ordered</em> sequence, when the order is of extreme relevance but each item is independent of the others?</strong> Below I'll include how I generated the 927 sequences for anyone interested, but it's ancillary to the question, I think.</p>
<p>***</p>
<p>First, I wrote a simple recursive algorithm that begins with an empty set and calls itself three times after adding 1, 2 or 3, terminating when the sum is 12 and tossing any that go over. Happy to provide code or pseudo-code, but I think it's intuitive.</p>
<p>To check myself, I generated the all the <em>combinations</em> of [1,2,3] that add to 12--there are 19--using a <a href="https://algorithmist.com/wiki/Coin_change" rel="nofollow noreferrer">Coin-Change algorithm</a>, and then calculated the unique, distinct <em>permutations</em> of each of them (treating identical digits as indistinguishable). The sum of the permutations is also 927, which range from 1 for the sequence of twelve 1s (the Chromatic Scale) or four 3s (a sparse diminished scale) to 140 possible scales for the intervals { 1, 1, 1, 2, 2, 2, 3 }, which <em>almost</em> subsumes the Harmonic Minor Scale.</p>
<h3>One more thought</h3>
<p>Regarding the above example of four 1s and four 2s: If I was flipping an unbiased coin 12 times and coding Heads as "1" and Tails as "2", I would be more surprised, in a general sense of the word, to get alternating results than to get the second sequence with small clusters. So it's possible that "surprisal" is the wrong frame of mind here versus information.</p>
| obscurans | 619,038 | <p>You're starting to think of <a href="https://en.wikipedia.org/wiki/Kolmogorov_complexity" rel="nofollow noreferrer">Kolmogorov complexity</a>, which is a (almost uncomputable) measure of "how hard it is to describe" the sequence. It is completely dependent on "what is allowed to be used to describe" sequences (as computer programs, actually).</p>
<p>Shannon entropy fundamentally describes how much information <em>per character</em> there is when a <em>stream</em> of the specified probability distribution arrives. Serial correlations and so on are specifically not accounted for (you may of course consider characters to be multi-interval, and then your second sequence does have higher entropy).</p>
<p>Another point: surprisal in terms of "the intervals being in an unexpected mathematical sequence" is drastically different from surprisal per "this is a dissonant interval relative to the previous ones", but you're into atonality so that may not matter.</p>
<p>This is also heavy math, but in the direction of the lattice of rationals (abstract algebra), not information theory.</p>
|
3,294,082 | <p>The exercise is to prove that the minimum value between <span class="math-container">$a^{1/b}$</span> and <span class="math-container">$b^{1/a}$</span> is no greater than <span class="math-container">$3^{1/3}$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers. As it was presented in an introductory calculus class, I tried using brute Mathematics, ploting graphs, but was unable to develop and was looking for some elegant ideas.</p>
| Hw Chu | 507,264 | <p>Suppose <span class="math-container">$a \leq b$</span>. Then <span class="math-container">$1/a \geq 1/b$</span> and <span class="math-container">$a^{1/b} \leq b^{1/a}$</span>. Seeing that <span class="math-container">$a^{1/b} \leq a^{1/a}$</span>, It suffices to prove <span class="math-container">$a^{1/a} \leq 3^{1/3}$</span>. Raising both sides to the <span class="math-container">$3a$</span>-th power, it is <span class="math-container">$a^3 \leq 3^a$</span>, or
<span class="math-container">$$
a \cdots a \leq 3\cdots3,
$$</span>
where <span class="math-container">$a$</span> and <span class="math-container">$3$</span> appeared <span class="math-container">$3$</span> and <span class="math-container">$a$</span> times, respectively.</p>
<p>This is related to the following question: Let <span class="math-container">$N = 3a$</span> be an integer, we want to find a partition of <span class="math-container">$N$</span>, which is a sequence <span class="math-container">$(a_1, a_2, \cdots, a_l)$</span>, where <span class="math-container">$l$</span> is any integer, <span class="math-container">$a_i$</span> are all integers, and <span class="math-container">$\sum_{i=1}^l a_i = N$</span>. We want to find <span class="math-container">$\max\{\prod_{i=1}^l{a_i}\}$</span>.</p>
<p>The maximum is attained when the partition only contains <span class="math-container">$3$</span>. We can prove this by improving the sequence to get a greater product:</p>
<ul>
<li>If the partition contains a 1, clearly we can merge the 1 into any term and increase the product.</li>
<li>If the partition contains a <span class="math-container">$n \geq 4$</span>, we can split <span class="math-container">$n$</span> into <span class="math-container">$2$</span> and <span class="math-container">$n-2$</span> and increase the product.</li>
<li>Up to this step, the candidate sequence contains only 2 and 3. And 3 is more efficient than 2.</li>
</ul>
|
66,463 | <p>Hi,</p>
<p>Let $\Gamma$ be a free subgroup of rank 2 in $\mathbb{G}_m^2(\mathbb{Q})$. For all but finitely many primes p we can reduce $\Gamma$ modulo p. Let $S$ be the of primes for which $\Gamma$ does not reduce modulo p, and for any $p$ not in $S$, let $\gamma_p$ be the size of $\Gamma \mod p$. My question is what is known about the function</p>
<blockquote>
<p>$f(x)= \sum_{p\not\in S,\ p\leq x}\frac{\log p }{\gamma_p}$</p>
</blockquote>
<p>In particular what is the asymptotic behavior of $f$? Is the corresponding infinite series convergent whenever $\Gamma$ is <em>not</em> contained in an algebraic subgroup of $\mathbb{G}_m^2$? Do you know of any references that might be relevant to those questions?</p>
<p>Thanks in advance,</p>
| Joe Silverman | 11,926 | <p>Presumably "exceptional" means primes where either one of the generators of $\Gamma$ is 0 or $\infty$ mod p, or where $\Gamma$ mod $p$ has rank smaller than $2$. The following reference is possibly relevant to your question, although we consider a somewhat different sum. We give an upper bound (that should be fairly sharp) for the sum
$$\sum_{p} \frac{\log p}{p\cdot\gamma_p^\epsilon}.$$
In particular, we prove that
$$\limsup_{\epsilon\to0} ~~\epsilon \cdot \sum_{p} \frac{\log p}{p\cdot\gamma_p^\epsilon}
\le 1+\frac{1}{\text{rank}~\Gamma}.$$
The article is</p>
<p>Murty, M. Ram and Rosen, Michael and Silverman, Joseph H., Variations on a theme of Romanoff, <em>Internat. J. Math.</em> <strong>7</strong> (1996), 373-391 (MR1395936).</p>
|
352,983 | <p>How to find this expression $(1000!\mod 3^{300})$?</p>
| Clive Newstead | 19,542 | <p>$3$ goes into $1000!$ at least $300$ times, since it divides $3, 6, 9, \dots, 900$, and hence $3^{300} \mid 1000!$.</p>
|
99,237 | <p>If we have a directed graph $G = (V,E)$ and we want to find if there is such node $s \in V$ that we can reach all other nodes of $G$</p>
<p>What is a good algorithm to solve this problem and what is its execution time?</p>
| hmakholm left over Monica | 14,366 | <p>Run Tarjan's linear-time <a href="http://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm" rel="nofollow">algorithm for finding strongly connected components</a>.</p>
<p>If there is more than one component with <em>no incoming edges</em>, then there can be no node that can reach everywhere.</p>
<p>On the other hand, if there is exactly one component with no incoming edges, each node in that component (and no other nodes) can reach everywhere in the graph.</p>
<p>(If you're looking for nodes that are reachable <em>from</em> everywhere -- as in the original the title of the question said -- look instead for components with no <em>outgoing</em> edges).</p>
|
3,753,060 | <blockquote>
<p>If <span class="math-container">$\int f(x)dx =g(x)$</span> then <span class="math-container">$\int f^{-1}(x)dx $</span> is equal to</p>
<p>(1) <span class="math-container">$g^{-1}(x)$</span></p>
<p>(2) <span class="math-container">$xf^{-1}(x)-g(f^{-1}(x))$</span></p>
<p>(3) <span class="math-container">$xf^{-1}(x)-g^{-1}(x)$</span></p>
<p>(4) <span class="math-container">$f^{-1}(x)$</span></p>
</blockquote>
<p>My approach is as follows:
Let <span class="math-container">$f(x)=y$</span>, therefore <span class="math-container">$f^{-1}(y)=x$</span>, <span class="math-container">$\int f^{-1}(f(x))dx =g(f(x))$</span></p>
<p>On differentiating we get <span class="math-container">$x=g'(f(x))f'(x)$</span></p>
<p>After this step, I am not able to proceed.</p>
| Rivers McForge | 774,222 | <p>There's a nice visual computation of the antiderivative of an inverse function: <span class="math-container">$$F(x) := \int_0^x f^{-1}(t) dt$$</span> is an antiderivative for <span class="math-container">$f^{-1}(x)$</span>, and for <span class="math-container">$x = a$</span>, <span class="math-container">$F(a)$</span> is equal to the green area in the picture below<sup>†</sup>:
<a href="https://i.stack.imgur.com/gmfaQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gmfaQ.png" alt="area under the curve of the inverse function" /></a></p>
<p>If we could figure out the blue area, we would be set, because</p>
<p><span class="math-container">\begin{align*}
\text{ (green area) } &= \text{ (rectangle area) } - \text{ (blue area) } \\
F(a) &= af^{-1}(a) - \text{ (blue area) }.\
\end{align*}</span></p>
<p>But if we reflect this picture across the line <span class="math-container">$y = x$</span>, we see the blue area is just the antiderivative of <span class="math-container">$f$</span>, namely <span class="math-container">$g(x) := \int_0^x f(t) dt$</span>, evaluated at <span class="math-container">$f^{-1}(a)$</span>:</p>
<p><a href="https://i.stack.imgur.com/AGWSc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AGWSc.png" alt="antiderivative of the original function" /></a></p>
<p>So we get <span class="math-container">$$\text{ (blue area) } = g(f^{-1}(a)),$$</span></p>
<p>and plugging this in, we get that <span class="math-container">$F(a)$</span> is equal to the second of the four choices:</p>
<p><span class="math-container">\begin{align*}
F(a) &= af^{-1}(a) - \text{ (blue area) } \\
&= af^{-1}(a) - g(f^{-1}(a)). \
\end{align*}</span></p>
<hr />
<p><sup>† To make the picture look nice, we assumed <span class="math-container">$f(0) = f^{-1}(0) = 0$</span> without loss of generality. </sup></p>
|
265,189 | <p>Integrate, $$\int_{0}^{\frac{\pi}{2}}\sin (\tan\theta) \mathrm{d\theta}$$</p>
| Sasha | 11,069 | <p>Making a change of variables $u=\tan(\theta)$:
$$
\int_0^{\pi/2} \sin\left(\tan \theta\right) \mathrm{d} \theta = \int_0^\infty \frac{\sin(u)}{1+u^2} \mathrm{d} u \tag{1}
$$
In order to evaluate this we use the technique of <a href="http://en.wikipedia.org/wiki/Mellin_transform">Mellin transform</a>.</p>
<ol>
<li><p>Evaluate the Mellin transforms of $\sin(u)$ and $\left(1+u^2\right)^{-1}$:
$$
\mathcal{M}_s(\sin(u))= \int_0^\infty u^{s-1} \sin(u) \mathrm{d} u = \frac{\sqrt{\pi}}{2} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)}
$$
defined in the strip $-1<\Re(s)<1$. Applying the inverse transform:
$$
\sin(u) = \frac{1}{2 \pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} \frac{\sqrt{\pi}}{2} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} u^{-s} \mathrm{d} s \tag{2}
$$
where $\gamma$ is an arbitrary real constant such that $-1<\gamma<1$.
The Mellin transform of the rational function $(1+u^2)^{-1}$ reads
$$
\mathcal{M}_s\left(\frac{1}{1+u^2}\right)= \int_0^\infty \frac{u^{s-1}}{1+u^2}\mathrm{d} u = \frac{1}{2} \Gamma\left(\frac{s}{2}\right) \Gamma\left(1-\frac{s}{2}\right) \tag{3}
$$
and is defined in the strip $0<\Re(s)<2$.</p></li>
<li><p>Now substituting $(2)$ into $(1)$, choosing $\gamma$ such that $-1<\gamma<0$, and interchanging the order of integration:
$$\begin{eqnarray}
\int_0^\infty \frac{\sin(u)}{1+u^2} \mathrm{d} u &=& \frac{1}{2\pi i} \frac{\sqrt{\pi}}{2} \int_{\gamma-i \infty}^{\gamma+i \infty} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} \left( \int_0^\infty \frac{u^{-s}}{1+u^2} \mathrm{d}u \right) \mathrm{d} s \\ &\stackrel{(2)}{=}& \frac{1}{2\pi i} \frac{\sqrt{\pi}}{2} \int_{\gamma-i \infty}^{\gamma+i \infty} 2^{s} \frac{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} \left(\frac{1}{2} \Gamma\left(\frac{1-s}{2}\right) \Gamma\left(1-\frac{1-s}{2}\right) \right)\mathrm{d} s \\
&=& \frac{1}{2\pi i} \frac{\sqrt{\pi}}{4} \int_{\gamma-i \infty}^{\gamma+i \infty} \Gamma\left(\frac{1}{2}-\frac{s}{2}\right) \frac{\Gamma\left(\frac{1}{2}+\frac{s}{2}\right)^2}{\Gamma\left(1-\frac{s}{2}\right)} \left(\frac{1}{2}\right)^{-s} \mathrm{d}s \\
&\stackrel{\text{reflection}}{=}& \frac{1}{2\pi i} \frac{\pi}{2} \int_{\gamma - i \infty}^{\gamma+i \infty} \Gamma(s) \tan\left(\frac{\pi}{2} s \right) \mathrm{d}s
\end{eqnarray}
$$
The latter integral can be evaluated as a sum over residues at poles to the left of the integration contour, situated at odd negative integers. The poles of $\Gamma(s)$ are even non-positive integers are canceled by zeros of the tangent function. The poles at the odd negative integers are double poles:
$$
\operatorname{Res}_{s=-2k-1} \left(\frac{\pi}{2} \Gamma\left(s\right) \tan\left(\frac{\pi}{2} s\right) \right) = \frac{1}{\Gamma\left(2k+2\right)} \psi\left(2k+2\right)
$$
where $\psi(x)$ is the <a href="http://en.wikipedia.org/wiki/Digamma_function">digamma function</a>. </p></li>
</ol>
<p>Thus we have the result:
$$\begin{eqnarray}
\int_0^\infty \frac{\sin u}{1+u^2} \mathrm{d} u &=& \sum_{k=0}^\infty \frac{\psi(2k+2)}{\Gamma(2k+2)} = \frac{1}{2} \left( \sum_{k=0}^\infty \frac{\psi(k+1)}{\Gamma(k+1)} - \sum_{k=0}^\infty (-1)^k \frac{\psi(k+1)}{\Gamma(k+1)}\right) \\ &=&
\frac{1}{2} \left(f(1) - f(-1)\right) = \frac{\exp(-1)}{2} \operatorname{Ei}(1) - \frac{\exp(1)}{2} \operatorname{Ei}(-1)
\end{eqnarray}
$$
where $\operatorname{Ei}(x)$ denotes the <a href="http://en.wikipedia.org/wiki/Exponential_integral">exponential integral</a> special function.
We now proceed to prove that, for real non-zero $x$:
$$
f(x) = \sum_{k=0}^\infty x^k \frac{\gamma(k+1)}{\Gamma(k+1)} = \mathrm{e}^{x} \left( \frac{1}{2} \log(x^2) - \operatorname{Ei}(-x) \right)$$
Indeed, denoting the <a href="http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant">Euler-Mascheroni constant</a> as $C$, and using $H_n = -\sum_{k=1}^n \frac{(-1)^k}{k} \binom{n}{k}$:
$$\begin{eqnarray}
\sum_{k=0}^\infty x^k \frac{\gamma(k+1)}{\Gamma(k+1)} &=& \sum_{k=0}^\infty x^k \frac{-C +H_{k}}{k!} = - C \mathrm{e}^{x} - \sum_{k=1}^\infty \frac{x^k}{k!} \sum_{m=1}^k \frac{(-1)^m}{m} \binom{k}{m} \\ &=&
-C \mathrm{e}^{x} - \sum_{m=0}^\infty \frac{x^m}{m!} \sum_{k=1}^\infty \frac{(-x)^k}{k \cdot k!} = -C \mathrm{e}^{x} - \mathrm{e}^{x} \left( \operatorname{Ei}(-x) - \log |x| -C \right) \\ &=& \mathrm{e}^{x} \left( \log |x| - \operatorname{Ei}(-x) \right)
\end{eqnarray}$$</p>
|
2,820,796 | <p>In How many ways can a 25 Identical books can be placed in 5 identical boxes. </p>
<p>I know the process by counting but that is too lengthy .
I want different approach by which I can easily calculate required number in Exam hall in few minutes. </p>
<p>Process of Counting :
This problem can be taken partitions of 25 into 5 parts.</p>
<p>25 = 25+0+0+0+0</p>
<p>25 = 24 +1 + 0 + 0 +0</p>
<p>25 = 23+ 1 +1 +0 + 0
... ....
Like this way many combinations are made.: about 377 </p>
<p>How can we calculate it without this process of manual counting. </p>
| Jaroslaw Matlak | 389,592 | <p>You could use a recurrence.</p>
<p>For example: put $a=0,1,2,...,n$ books into the first box and calculate, in how many ways you can put $n-a$ books in $m-1$ boxes. To prevent repetitions, assume, that in the next box you will put no less books, than to the previous one.</p>
<p>Of course:</p>
<ul>
<li>if $n<a(m-1)$, then there are no options</li>
<li>if $n\geq a$ and $m-1=1$, then there is one option</li>
</ul>
<p>This function would look like this:</p>
<p>$$F_m(n,a)=\begin{cases}1, & m=1 \wedge n\geq a\\
0, & n<am\\
\sum_{k=0}^n F_{m-1}(n-k, k),&\text{in other cases}\end{cases} $$ </p>
<p>What you need to calculate is $F_5(25,0)$</p>
|
2,555,399 | <p>The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$</p>
<p>I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>The coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$</p>
<p>$=$</p>
<p>the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3+\cdots)^{1/2}$</p>
<p>Using <a href="https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot">Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $</a>,</p>
<p>$$1-2x+3x^2-4x^3+\cdots=(1+x)^{-2}$$</p>
<p>$$(1-2x+3x^2-4x^3+\cdots)^{1/2}=(1+x)^{-1}=1-x+x^2-x^3+\cdots$$</p>
|
3,864,729 | <p>I am studying for an exam and I am almost grasping compactness. However, some examples are still unclear. E.g. for <span class="math-container">$A = (a, b)$</span> with <span class="math-container">$a < b$</span> and <span class="math-container">$a, b \in \mathbb{R}$</span>.
How can <span class="math-container">$(a, b) \subseteq \cup _{k \in \mathbb{N}} \in (a, b − \frac{1}{k})$</span>, making this an open covering. Even though <span class="math-container">$\frac{1}{k} \ge 0$</span> for <span class="math-container">$k\ge\infty$</span>, how is this set still covering <span class="math-container">$(a,b)$</span> when <span class="math-container">$b -\frac{1}{k}$</span> is smaller than <span class="math-container">$b$</span>? I hope you can make this clear for me, thanks!</p>
| Darsen | 637,522 | <p>If <span class="math-container">$x\in(a,b)$</span> then <span class="math-container">$a<x<b$</span>. Consider <span class="math-container">$b-x>0$</span>. There is some <span class="math-container">$n_0\in\Bbb N$</span> such that <span class="math-container">$\frac{1}{n_0}<b-x$</span>, so <span class="math-container">$x<b-\frac{1}{n_0}$</span>. Then <span class="math-container">$x\in(a,b-\frac{1}{n_0})$</span>, and since <span class="math-container">$(a,b-\frac{1}{n_0})\subset\cup_{k\in\Bbb N}(a,b-\frac{1}{k})$</span>, we have <span class="math-container">$x\in\cup_{k\in\Bbb N}(a,b-\frac{1}{k})$</span>. Therefore <span class="math-container">$(a,b)\subseteq\cup_{k\in\Bbb N}(a,b-\frac{1}{k})$</span>.</p>
<p>In fact it is an equality, since <span class="math-container">$(a,b-\frac{1}{k})\subset(a,b)\quad \forall k\in\Bbb N\Rightarrow\cup_{k\in\Bbb N}(a,b-\frac{1}{k})\subseteq (a,b)$</span></p>
|
1,748,719 | <p>So there are a few basic formulas I'd like to start with, $W=\int_0^bFdx$, $F=ma$, and $a=\frac{d^2}{dt^2}x$.</p>
<p>In words, Work $(W)$ is defined as the area under a Force versus Displacement $(F/x)$ graph, Force is defined mass times acceleration $(m\cdot a)$, and acceleration is defined as the second derivative of displacement with respect to time $(t)$.</p>
<p>While this may sound very simple, I am just confused on all of this.</p>
<p>Putting all of this together, I got</p>
<p>$$W=\int_0^bFdx=\int_0^bmadx$$</p>
<p>Assuming mass $(m)$ doesn't change,</p>
<p>$$W=m\int_0^badx$$</p>
<blockquote>
<p>$$W=m\int_0^b\left(\frac{d^2}{dt^2}x\right)dx$$</p>
</blockquote>
<p>At this point, I am slightly confused. How can I take the integral with respect to $x$ when $x$ is part of the definition of acceleration? An example situation and how to tackle it would be nice (assume acceleration is not constant).</p>
<p>I was wondering if it were possible to simplify the integral, given acceleration as a function of time.</p>
| Brian | 315,119 | <p>When work is done by a variable force, one would calculate the area under the Force $(F(x))$ vs. Position $(x)$ graph.</p>
<p>So, the work integral would be</p>
<p>$$W=\int_{a}^b F(x) dx=\int_a^b m\cdot a(x) dx$$</p>
<p>Since $a(x)=\frac{dv(x)}{dt}=\frac{dv(x)}{dt}\frac{dx}{dt}=\frac{dv(x)}{dx}v(x)$, then $a(x)dx=v(x)dv(x)$</p>
<p>You can work with this integral, but you would still need to know the relevant information with respect to the position.</p>
|
1,748,719 | <p>So there are a few basic formulas I'd like to start with, $W=\int_0^bFdx$, $F=ma$, and $a=\frac{d^2}{dt^2}x$.</p>
<p>In words, Work $(W)$ is defined as the area under a Force versus Displacement $(F/x)$ graph, Force is defined mass times acceleration $(m\cdot a)$, and acceleration is defined as the second derivative of displacement with respect to time $(t)$.</p>
<p>While this may sound very simple, I am just confused on all of this.</p>
<p>Putting all of this together, I got</p>
<p>$$W=\int_0^bFdx=\int_0^bmadx$$</p>
<p>Assuming mass $(m)$ doesn't change,</p>
<p>$$W=m\int_0^badx$$</p>
<blockquote>
<p>$$W=m\int_0^b\left(\frac{d^2}{dt^2}x\right)dx$$</p>
</blockquote>
<p>At this point, I am slightly confused. How can I take the integral with respect to $x$ when $x$ is part of the definition of acceleration? An example situation and how to tackle it would be nice (assume acceleration is not constant).</p>
<p>I was wondering if it were possible to simplify the integral, given acceleration as a function of time.</p>
| amd | 265,466 | <p>Even though you've accepted a previous answer, I wanted to expand a bit on vnd's comments to the question because they lead to a useful physical result. </p>
<p>Following his and Mackenzie's suggestions, let $v=\frac{da}{dt}$. The integral then becomes, per vnd's comments, $$W = m\int_{x_1}^{x_2}\left(\frac{d^2}{dt^2}x\right)\,dx = m\int_{v_1}^{v_2} v\,dv = \frac12mv_2^2-\frac12mv_1^2.$$ In other words, the net work is equal to the change in kinetic energy between the start and end points. You still need to find the velocity as a function of either time or position, but since you say that you have acceleration as a function of time, you can compute $$v(t) = v_0 + \int_0^t a(t)\,dt,$$ where $v_0$ is a known initial velocity.</p>
|
1,435,269 | <p>Let a sequence $x_n$ be defined inductively by $x_{n+1}=F(x_n)$. Suppose that $x_n\to x$ as $n\to \infty$ and $F'(x)=0$. Show that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.</p>
<p>I'm not sure how to do this. Any solutions are greatly appreciated. I think The Mean-Value Theorem will be useful and we can assume that $F$ is continuously differentiable.</p>
| user402543 | 402,543 | <p>Assuming that $F'\in C^1$, we have:</p>
<p>$\lim_{n\to \infty}\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}=\lim_{n\to \infty}\frac{F(x_{n+1})-F(x_n)}{x_{n+1}-x_n}=F'(x)=0$</p>
<p>Where the last equality has ben obtained by the fact that the sequence is a cauchy sequence (since converges to $x$), and thus:</p>
<p>$\forall m \lim_{n \to \infty} x_{m+n} \to x_n$</p>
|
45,911 | <p>I've been wondering for some time now about the difference between a point and a vector. In high school, it was very important to distinguish them from each other, and we used the notation $(x,y,z)$ for points and $[x,y,z]$ for vectors. We always had to translate the point $P=(a,b,c)$ to the vector $\overrightarrow{OP} =[a,b,c]$ before we started calculating with them. </p>
<p>Now, after I started at the university, people don't seem to care anymore. My professors either say that they're the same, or that they're almost the same, and the books I have seem to share that view. The book I use for my calculus course (Colley's Vector Calculus) says, among other things, the following:</p>
<blockquote>
<p>[...] we adopt the point of view that a vector field assigns to each <em>point</em> $\textbf{x}$ in X a <em>vector</em> $\textbf{F}(\textbf{x})$ in $\mathbb{R}^n$, represented by an arrow whose tail is at the point $\textbf{x}$.</p>
</blockquote>
<p>So it seems like a point is also a vector.</p>
<p>My question is this: Do mathematicians distinguish between points and vectors, and if they do, in what circumstances?</p>
| Hans Lundmark | 1,242 | <p>I would say it's a good habit to distinguish points from vectors (in the context that I think you're referring to), even at university!</p>
<p>Geometrically, any point looks just the same as any other point, whereas not all vectors are equal; two vectors can have different lengths, for example, and there is one very special vector which has length zero. And to talk about the coordinates of a vector, what you need is only a basis, but to talk about the coordinates of a point you need a basis <em>and an origin</em> (an arbitrarily selected reference point).</p>
<p>However, converting points $P$ to vectors $\overrightarrow{OP}$ is strictly speaking not necessary (and in my opinion a bit artificial actually). You can instead use the geometrically natural operations "point + vector = point" and "point – point = vector" (but, as Qiaochu already said, <em>not</em> "point + point", which is geometrically meaningless). The textbooks insist on using the vector $\overrightarrow{OP}$ just so that they can express things in terms of the operation "vector + vector = vector" and don't have to introduce those other operations.</p>
|
4,539,167 | <p><span class="math-container">$$
g(x) =\min_y f(x, y) =\min_y x^TAx + 2x^TBy + y^TCy
$$</span>
where <span class="math-container">$x\in \mathbb R^{n\times 1}$</span>, <span class="math-container">$y\in \mathbb R^{m\times 1}$</span>, <span class="math-container">$A\in \mathbb R^{n\times n}$</span>, <span class="math-container">$B\in \mathbb R^{n\times m}$</span>, <span class="math-container">$C\in \mathbb R^{m\times m}$</span>.
How to compute the derivative: <span class="math-container">$\frac{d g}{d x}$</span>?</p>
| wxydx00 | 410,047 | <p>The <a href="https://en.wikipedia.org/wiki/Envelope_theorem" rel="nofollow noreferrer">envelope theorem</a> should be helpful.</p>
|
1,986,249 | <blockquote>
<p>Let q be a positive integer such that $q \geq 2$ and such that for any
integers a and b, if $q|ab$, then $q|a$ or $q|b$. Show that $\sqrt{q}$
is irrational.</p>
</blockquote>
<p>Proof;</p>
<p>Let assume $\sqrt{q}$ is a rational number, where $n \neq 0$ and $\gcd (m,n)=1$, meaning $\sqrt{q} = \frac{m}{n} \Rightarrow q=\frac{m^2}{n^2} $</p>
<p>Since $n^2 \nmid m^2$, $q|m^2 \Rightarrow q|m$, so $m=qt$ where $t\in \mathbb{Z}$</p>
<p>By substitute $m=qt$ in the equation $qn^2 = m^2$, we get $n^2=qt^2$.</p>
<p>Since tells us that $q|n^2$ and $t^2|n^2$, it contradicts with the assumption $\gcd (m,n)=1$; therefore, $\sqrt{q}$ is irrational.</p>
<p>I get this proof with the assistant of the course, but is there any flaw or mistake? What are the other methods for proving this statement, can you at least give one different method? And how can I improve this proof?</p>
| Jack D'Aurizio | 44,121 | <p>We just have to show that for any prime number $p$, $\sqrt{p}\not\in\mathbb{Q}$.<br>
If we assume $\sqrt{p}=\frac{a}{b}$ with $a,b\in\mathbb{Z}^+$ we get the identity $pb^2=a^2$.<br>
For any $n\in\mathbb{Z}^+$, let $\nu_p(n)=\max\{m\in\mathbb{N}: p^m\mid n\}$. The identity $pb^2=a^2$ implies
$$ \nu_p(pb^2) = \nu_p(a^2) $$
but that is impossible, since $\nu_p(a^2)$ is an even number and $\nu_p(pb^2)=1+\nu_p(b^2)$ is an odd number. $\sqrt{p}\not\in\mathbb{Q}$ trivially follows.</p>
|
867,209 | <p>I tried to do the implication part. Please, see what I need to do to fix it.</p>
<p>claim: $n|a – b → n|a^2 – b^2$.</p>
<p>claim: $nk = a – b$ for some $k \in \mathbb Z \to nk' = a^2 – b^2$ for some $k' \in Z$.</p>
<p>$(a + 1)^2 – (b +1)^2$</p>
<p>$= a^2 + 2a + 1 -(b^2 +2b + 1)$</p>
<p>$= a^2 + 2a -b^2 -2b$</p>
<p>$= a^2 – b^2 + 2(a – b)$</p>
<p>$= (a + b)(a – b) + 2(a – b)$</p>
<p>$= (a – b)[(a + b) + 2]$</p>
<p>$= nk[(a + b) + 2]$</p>
<p>$= n[(a + b) + 2k]$</p>
| hexaflexagonal | 162,141 | <p>A relation on $A$ is simply a subset of the Cartesian product $A \times A$. For $A=\{a,b,c,d\}$, both $(b,c)$ and $(b,d)$ are contained in $A \times A$; therefore, $\{(b,c),(b,d)\}$ is a relation on $A$. </p>
|
3,598,476 | <p>I have proven that if <span class="math-container">$|x|<\varepsilon,\forall\varepsilon>0$</span>, then <span class="math-container">$x=0$</span>. Further I have proven that ,<span class="math-container">$L=\displaystyle\lim_{n\to\infty}\frac{1}{n} = 0$</span> so that by definition <span class="math-container">$(\forall\varepsilon >0)(\exists N>0)(\forall n>N)\left|\frac{1}{n}-L\right|<\varepsilon$</span>, but <span class="math-container">$\frac{1}{n}\ne 0, \forall n\in\mathbb N$</span>.</p>
<p>Is there a contradiction or have I made an error in analysis?</p>
| Julio Ignacio Quijas Aceves | 764,383 | <p>You are not wrong nor you are arriving to contradiction. The deal with the first statement, is that <span class="math-container">$x$</span> is fixed, not matter which epsilon you take <span class="math-container">$|x|<\epsilon$</span>. But in the other case, you are taking an <span class="math-container">$\epsilon$</span> and then finding that for some <span class="math-container">$N$</span> if <span class="math-container">$n>N$</span>, <span class="math-container">$\frac{1}{n}<\epsilon$</span>, you are not proving that for every <span class="math-container">$\epsilon>0$</span>, <span class="math-container">$\frac{1}{n} < \epsilon$</span>, is subtle, but they are drastically different statements.</p>
|
2,054,676 | <p>I know that m is even and m/2 is odd, but I don't know where/how I can use this. Also, 3y^2 is odd and the sum is odd when x^2 is even. I'm trying to prove that its always odd, but I'm stuck.
Can someone please help?
Thanks</p>
| Peter | 82,961 | <p>Suppose : $m=x^2+3y^2$ is even. </p>
<p>Case $1$ : $x$ is even. Then $y$ must be even as well. So, $m$ is divisble by $4$.</p>
<p>Case $2$ : $x$ is odd. Then, $\ x^2\equiv 1 \mod 4\ $. Since $y$ must be odd as well, we also have $\ y^2\equiv 1\mod 4\ $. Hence, $x^2+3y^2\equiv 0\mod 4$. </p>
<p>So, again, $m$ is divisble by $4$</p>
|
1,379,283 | <p>In my notes it shows how to calculate by using the unit circle. But I do not know why the value of sin is the y coordinate and the value of cos is the x coordinate.</p>
| 3SAT | 203,577 | <p>It is very easy to remember the graph of $\cos(x)$ </p>
<p>[<img src="https://i.stack.imgur.com/iTr0H.png" alt="enter image description here[1"></p>
|
151,425 | <p>I've considered the following spectral problems for a long time, I did not kow how to tackle them. Maybe they needs some skills with inequalities.</p>
<p>For the first, suppose $T:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by
$$Tf(x)=\int_{0}^{x} \! f(t) \, dt$$</p>
<p>How can I calculate:</p>
<ul>
<li>the radius of the spectrum of $T$?</li>
<li>$T^{*}T$?</li>
<li>the norm of $T$?</li>
</ul>
<p>I guess $r(T)$ should be $0$. but I did know how to prove it. My idea is use Fourier to tackle it, however it does not seem to work.</p>
<p>The other problem may be very similar to this one. Let $T:C[0,1]\rightarrow C[0,1]$ be defined by
$$Tf(x)=\int_{0}^{1-x}f(t)dt$$</p>
<p>It is obvious that $T$ is compact and I guess its radius of spectrum is zero, but I do not know how to prove it.</p>
<p>Any references and advice will be much appreciated.</p>
| Community | -1 | <p>I'm not sure if this can be done with Fourier series, although that's interesting to think about. Anyway, the usual way to see these things is by applying the spectral theorems for compact operators as is done in what follows.</p>
<p>For $f \in L^2[0,1]$ with $\|f\|_2 \le 1$, the Cauchy-Schwarz Inequality shows that</p>
<p>$$|(Tf)(x) - (Tf)(y)| \le \int_y^x \! |1 \cdot f(t)| \, dt \le \left( \int_y^x \! \, dt \right)^{1/2}\left( \int_y^x \! |f(t)|^2 \, dt \right)^{1/2}$$
$$ \le \|f\|_2|x-y|^{1/2} = |x-y|^{1/2}$$</p>
<p>so $Tf$ is Hölder continuous of exponent $1/2$. It now follows from the <a href="http://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem">Arzela-Ascoli Theorem</a> that $T: L^2[0,1] \to C[0,1]$ is compact and since the identity mapping from $C[0,1]$ to $L^2[0,1]$ is bounded, this shows that $T: L^2[0,1] \to L^2[0,1]$ is compact as well. Thus, by the <a href="http://en.wikipedia.org/wiki/Spectral_theory_of_compact_operators">Spectral Theorem for Compact Operators</a>, all non-zero elements of $\sigma(T)$ are eigenvalues of $T$. Now suppose that $\lambda \neq 0, f \neq 0$ are an eigenvalue and eigenfunction of $T$ respectively. Then</p>
<p>$$\int_0^x \! f(t) \, dt = \lambda f(x)$$</p>
<p>Since the left hand side is continuous, the right hand side must be also so that $f$ is continuous. But since now $f$ is continuous, this makes the left hand side continuously differentiable, so $f$ must be continuously differentiable. We can now use the fundamental theorem of calculus to differentiate both sides thereby getting
$f(x) = \lambda f'(x)$ which has only solutions of the form $f(x) = Ce^{t/\lambda}$. But plugging in $x = 0$ on both sides of the equation $Tf = \lambda f$ above, we see that we must also have $f(0) = 0$. Thus $C = 0$ and therefore $f = 0$. This contradiction shows that $T$ has no non-zero eigenvalues and hence we have shown $\sigma(T) = \{0\}$. Now it follows immediately from the definition that the spectral radius $r(T)$ is $0$. This argument does not depend on the fact that $T$ is defined on a Hilbert space, so slight modifications also show that the spectral radius of your second operator is also $0$.</p>
<p>Consider now the function $k:[0,1]^2 \to \mathbb{R}$ defined by
$$k(x,y) = \begin{cases}
\ 1 & \text{if $x \geq y$},\\
\ 0 & \text{otherwise}
\end{cases}$$</p>
<p>Then we have
$$Tf = \int_0^x \! f(y) \, dy = \int_0^1 \! k(x,y)f(y) \, dy$$</p>
<p>Thus, it follows by <a href="http://en.wikipedia.org/wiki/Fubini%27s_theorem">Fubini's Theorem</a> that
$$\langle g, Tf \rangle = \int_0^1\! \overline{g(x)}\int_0^1 \! k(x,y)f(y) \, dy \, dx$$
$$ = \int_0^1 \! f(y) \int_0^1 \! \overline{g(x)}k(x,y) \, dx \, dy$$
$$ = \int_0^1\! \overline{\int_0^1 \! g(x) \overline{k(y,x)} \, dx} \, f(y)\, dy$$
$$ = \langle \int_0^1 \! g(x) \overline{k(y,x)} \, dx, f \rangle$$
which shows that we must have</p>
<p>$$(T^*f)(y) = \int_0^1\int_0^1\! \overline{k(y,x)} f(x) \, dx = \int_y^1 \! f(t) \, dt$$</p>
<p>so you can now compute $T^*T$ as</p>
<p>$$T^*Tf(x) = \int_x^1 \int_0^y \! f(t) \, dt dy$$</p>
<p>Finally, to get the norm of $T$ we will use the "$B^*$-Identity" $\|T\| = \sqrt{\|T^*T\|}$. Since $S = T^*T$ is compact and self-adjoint, it's set of eigenvalues has an element of largest absolute value, and this absolute value is equal to the norm of $S$ (<a href="http://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Compact_self_adjoint_operator">Spectral Theorem for Compact Self-Adjoint Operators</a>). Using an argument very similar to the one above showing that $T$ has no non-zero eigenvalues, we see that if $\lambda \neq 0, f \neq 0$ are an eigenvalue and an eigenfunction of $S$ respectively then $f$ is twice continuously differentiable and</p>
<p>$$\lambda f''(x) = - f(x)$$</p>
<p>so that we must have $f(x) = \alpha e^{i \omega x} + \beta e^{-i\omega x}$ for some $\alpha, \beta \in \mathbb{C}$ and $\omega^2 = 1/\lambda$. A direct but somewhat tedious computation yields</p>
<p>$$S(\alpha e^{i \omega x} + \beta e^{-i\omega x}) = \frac{1}{\omega^2}(\alpha e^{i \omega x} + \beta e^{-i\omega x}) + \left(\frac{\alpha}{i\omega} - \frac{\beta}{i\omega} \right)x - \left(\frac{\alpha e^{i \omega}}{\omega^2} - \frac{\beta e^{-i\omega}}{\omega^2}\right) - \left(\frac{\alpha}{i\omega} - \frac{\beta}{i\omega} \right) $$</p>
<p>This formula shows that if $\alpha e^{i \omega x} + \beta e^{-i\omega x}$ is to be an eigenfunction then we must have $\alpha = \beta$ and thus an eigenfunction must be of the form
$$f(x) = \alpha(e^{i \omega x} + e^{-i\omega x}) = 2\alpha \cos (\omega x)$$
with eigenvalue $1/\omega^2$. Now we can again just as we did before read from the equation $\omega^2Sf = f$ that $f(1) = 0$, which since $f \neq 0$ (and therefore $\alpha \neq 0$) implies</p>
<p>$$\omega = \frac{(2n + 1)\pi}{2}$$</p>
<p>and thus the eigenvalues are given by</p>
<p>$$\lambda_n = \frac{4}{(2n + 1)^2\pi^2}$$</p>
<p>Maximizing this over $n$ yields a largest eigenvalue of $4/\pi^2$ at $n = 0$. This shows that the norm of $S$ is $4/\pi^2$ and hence the norm of $T$ is $\sqrt{4/\pi^2} = 2/\pi$.</p>
<p>If you want to search for some references about this stuff, the keywords here are "Volterra Operator" and "Hilbert-Schmidt Kernel."</p>
|
4,187,238 | <p>How many points are common to the graphs of the two equations <span class="math-container">$(x-y+2)(3x+y-4)=0$</span> and <span class="math-container">$(x+y-2)(2x-5y+7)=0$</span>?</p>
<p><span class="math-container">\begin{align*}
(x-y+2)(3x+y-4) &= 0\tag{1}\\
(x+y-2)(2x-5y+7) &= 0\tag{2}
\end{align*}</span></p>
<p>In equation <span class="math-container">$1$</span>, it is not possible for both <span class="math-container">$x-y+2$</span> and <span class="math-container">$3x+y-4$</span> to be nonzero. Like-wise in equation <span class="math-container">$2$</span>, it is not possible for both <span class="math-container">$x+y-2$</span> and <span class="math-container">$2x-5y+7$</span> to be nonzero.
Therefore the LHS of either equation must involve a product of <span class="math-container">$0$</span> and some number.
This means there are infinitely many solutions to both equations.</p>
<p>However this doesn't shead any light on which points are common between the two equations. I'm confused. It seems I've made an error in my judgement. What should I reconsider?</p>
| Rhys Hughes | 487,658 | <p>Compare equations indivually, and work case-by-case.</p>
<p><span class="math-container">$$x-y+2=0\implies y=x+2, x+y-2=0\implies y=2-x$$</span>
<span class="math-container">$$x+2=2-x\implies x=0\to y=2$$</span></p>
<p>So <span class="math-container">$(0,2)$</span> is a solution.</p>
<p><span class="math-container">$$3x+y-4=0\implies y=4-3x,\\ x+2=4-3x\implies x=\frac12\to y=\frac52$$</span></p>
<p>So <span class="math-container">$(\frac12, \frac52)$</span> is a solution.</p>
<p>You can do the other two.</p>
|
392,442 | <p>What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer, however, I would like to know some of these implications - like the breaking of many security systems based on the primes. Moreover, there are examples of practical Math's implication, not just theoretical, of a possible prime number formula discovery? There is for example in Physics, Chemistry, Geography or Astronomy any field which would be very improved with a so great and dreamed <em>Eureka</em>?</p>
| Charles | 1,778 | <p>There are dozens, probably hundreds, of formulas for prime numbers. It's a very well-studied problem. Guy's <em>Unsolved Problems in Number Theory</em> has a section devoted to this, and Ribenboim's books cover this in some depth. Many formulas have been published in mathematical journals, and I've seen at least one survey paper in a journal dedicated to giving an overview of the various types of formulas for primes.</p>
<p>Finding a new formula for the primes might be interesting enough to be published, but not in a first-rate journal unless there's something special about it.</p>
<p>As for security concerns, what would be much more important would be a way to solve integer factorization quickly -- say, in polynomial time. (At the risk of verbosity, this means the time needed to factor $n$ is less than $(\log n)^k$ for some fixed $k$.) Proving that a number is prime can already be done in polynomial time, see the famous AKS algorithm (or the more practical ECPP).</p>
|
2,802,959 | <p>If I write
$$
x\in [0,1] \tag 1
$$
does it mean $x$ could be ANY number between $0$ and $1$?</p>
<p>Is it correct to call $[0,1]$ a set? Or should I instead write $\{[0,1]\}$? </p>
<p>Q2:</p>
<p>If I instead have
$$
x\in \{0,1\} \tag 2
$$
does it mean $x$ could be only $0$ OR $1$?</p>
| poyea | 498,637 | <blockquote>
<p>Is it correct to call $[0,1]$ a set?</p>
</blockquote>
<p>Yes, although it doesn't sound natural to me if you <em>"call $[0,1]$ a set"</em>. I'd rather call it the closed interval. You can also write as $\{x|x\in[0,1]\}$ (trivial). </p>
<p>$x\in \{0,1\}$ means $x=1$ or $x=0$.</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Charles Matthews | 6,153 | <p>The standard conjectures (<a href="http://en.wikipedia.org/wiki/Standard_conjectures_on_algebraic_cycles">http://en.wikipedia.org/wiki/Standard_conjectures_on_algebraic_cycles</a>) were pretty much designed to be used in this way (and then proved); but proofs are lacking, and some of the results now have non-conditional proofs. There are many related results in the theory of motives.</p>
<p>In number theory, Vandiver's conjecture (<a href="http://en.wikipedia.org/wiki/Vandiver%27s_conjecture">http://en.wikipedia.org/wiki/Vandiver%27s_conjecture</a>) has begun to stand out, because of its connection with K-theory (which is another area in which there are large scale conjectures used in this way).</p>
|
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