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 |
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3,491,028 | <p>Problem:<br>
Suppose that <span class="math-container">$x_1$</span>, <span class="math-container">$x_2$</span> and <span class="math-container">$x_3$</span> are independent uniformly distributed on the interval <span class="math-container">$[1,3]$</span>. What is the probability that
<span class="math-container">$x_1 + x_2 + x_3 < 8$</span>.<br>
Answer:<br>
Let <span class="math-container">$p$</span> be the probability we seek. The density for these three random variables is:
<span class="math-container">$$ f(x) = \begin{cases}
\frac{1}{2} & \text{for } 1 \leq x \leq 3 \\
0, & \text{otherwise }
\end{cases} $$</span>
<span class="math-container">\begin{align*}
p &= \int_{1}^{3} \int_{1}^{5-x_1} \int_{1}^{8-x_1-x_2} \left( \frac{1}{2}\right)^3 \, dx_3 \, dx_2 \, dx_1 \\
p &= \int_{1}^{3} \int_{1}^{5-x_1} \frac{x_3}{8} \, \Big|_{x_3 = 1}^{x_3 = 8-x_1-x_2} \, dx_2 \, dx_1 \\
p &= \int_{1}^{3} \int_{1}^{5-x_1} \frac{8 - x_1 - x_2}{8} - \frac{1}{8} \, dx_2 dx_1 \\
p &= \int_{1}^{3} \int_{1}^{5-x_1} \frac{7 - x_1 - x_2}{8} \, dx_2 \, dx_1 \\
p &= \int_{1}^{3} \frac{7x_2 - x_1 x_2 - \frac{x_2^2}{2}}{8} \Big|_{1}^{5-x_1} \, dx_1 \\
p &= \int_{1}^{3} \frac{7(5-x_1) - x_1(5-x_1) - \frac{(5-x_1)^2}{2} }{8} - \frac{1}{8} \, dx_1 \\
p &= \int_{1}^{3} \frac{14(5-x_1) - 2x_1(5-x_1) - (5-x_1)^2 - 2 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ 70 - 14x_1 - 2x_1(5-x_1) - ( 25 - 10x_1 + x_1^2 ) - 2 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ 70 - 14x_1 - 2x_1(5-x_1) - 25 + 10x_1 - x_1^2 - 2 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ - 14x_1 - 2x_1(5-x_1) + 10x_1 - x_1^2 + 43 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ -4x_1 - 2x_1(5-x_1) - x_1^2 + 43 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ -4x_1 - 10x_1 + 2x_1^2 - x_1^2 + 43 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ x_1^2 - 14x_1 + 43 }{16} \, dx_1 \\
p &= \int_{1}^{3} \frac{ x_1^2 - 14x_1 }{16} \, dx_1 + (3-1)\left( \frac{43}{16} \right) \\
p &= \int_{1}^{3} \frac{ x_1^2 - 14x_1 }{16} \, dx_1 + \frac{43}{8} \\
p &= \left( \frac{1}{16 }\right) \int_{1}^{3} ( x_1^2 - 14x_1 ) \, dx_1 + \frac{43}{8} \\
\end{align*}</span>
Using an online integral calculator, I find:
<span class="math-container">$$ \int_{1}^{3} ( x_1^2 - 14x_1 ) \, dx_1 = - \frac{142}{3} $$</span>
<span class="math-container">\begin{align*}
p &= \left( \frac{1}{16 }\right) \left( - \frac{142}{3} \right) \,+ \frac{43}{8} \\
p &= -\frac{71}{3(8)} + \frac{43}{8} = \frac{139 - 71}{24}
\end{align*}</span>
Since <span class="math-container">$p$</span> is greater than <span class="math-container">$1$</span>, my answer cannot be right. Were did I go wrong?</p>
<p>I would also like to know if I setup the integral correctly.</p>
<p>I ran the following R script:</p>
<pre><code>count = 0
limit = 10*1000*1000
for ( i in 1:limit ) {
num = sum( runif( 3, 1, 3 ) )
if ( num <= 8 )
count = count + 1
}
</code></pre>
<p>The result was around 0.979. Therefore, I question the answer of <span class="math-container">$\frac{7}{8}$</span>. </p>
| V.J. | 736,680 | <p>As @Leander point out, you missed in the limits that the information of <span class="math-container">$x_1$</span>, <span class="math-container">$x_2$</span> and <span class="math-container">$x_3$</span> must lie on <span class="math-container">$[1,3]$</span>. Thus your correct integral should be</p>
<p><span class="math-container">$$p = \int_{1}^{3} \int_{1}^{\min\{5-x_1,\,3\}} \int_{1}^{\min\{8-x_1-x_2,\,3\}} \left( \frac{1}{2}\right)^3 \, dx_3 \, dx_2 \, dx_1 = \frac{7}{8}\lt 1.$$</span></p>
|
823,055 | <p>This may be a naive question. I am reading the definition of differetiablity of a function $f:\mathbb{R^n}\rightarrow \mathbb{R^m}$ in the book Calculus Manifolds. I already know that all norms on $\mathbb{R}^n$ induce the same metric topology. If we change the norms in the definition (for example we can use the manhattan norm), does the set of differentiable functions change ?</p>
<p>I already know that all norms on $\mathbb{R}^n$ induce the same metric topology but that doesn't seem to imply a negative answer to my question.</p>
<p>Another Question: If the set of differentiable functions changes, is there any reason why we are defining differentiablity using the Pythagorean norm ? </p>
<p>Thank you</p>
| Ted Shifrin | 71,348 | <p>Unless you mean a Möbius band that's modeled on an open interval $(-1,1)$ rather than the usual closed interval $[-1,1]$, this will be (in the smooth world) a manifold with corners, not a manifold with boundary; in the topological world, no problem. </p>
<p>In general, $\partial(M\times N) = \partial M\times N \cup M\times\partial N$. So $$\partial (M\times I) \approx S^1\times I \cup M\times \partial I \approx S^1\times I \cup M\times\{0,1\}.$$ The latter factor is certainly not orientable. </p>
|
3,068,197 | <p>I'm working through this problem and I haven't been able to make any progress. The textbook provides the answer of <span class="math-container">$ {9 \choose 4}$</span> but I'm not sure as to how they got this result. </p>
| Yuval Filmus | 1,277 | <p>Here is a more sophisticated way to solve this, using DFAs. We can construct a state machine accepting the language of all strings without two consecutive zeroes as follows:</p>
<ul>
<li>There are two states, <span class="math-container">$q_0$</span> and <span class="math-container">$q_1$</span>.</li>
<li>The initial state is <span class="math-container">$q_1$</span>.</li>
<li>At state <span class="math-container">$q_0$</span>, there is a <span class="math-container">$1$</span>-transition leading to <span class="math-container">$q_1$</span>.</li>
<li>At state <span class="math-container">$q_1$</span>, there is a <span class="math-container">$0$</span>-transition leading to <span class="math-container">$q_0$</span>, and a <span class="math-container">$1$</span>-transition leading to <span class="math-container">$q_1$</span>.</li>
</ul>
<p>The transition matrix ("transfer matrix") corresponding to this DFA is as follows, where <span class="math-container">$0$</span>-transitions are indicated by the variable <span class="math-container">$x$</span> and <span class="math-container">$1$</span>-transitions are indicated by the variable <span class="math-container">$y$</span>:
<span class="math-container">$$
\begin{pmatrix}
0 & x \\
y & y
\end{pmatrix}
$$</span>
The number of words of length <span class="math-container">$n$</span>, counted according to the number of <span class="math-container">$0$</span>s and <span class="math-container">$1$</span>s, is
<span class="math-container">$$
\begin{pmatrix} 1 & 1 \end{pmatrix}
\begin{pmatrix}
0 & x \\
y & y
\end{pmatrix}^n
\begin{pmatrix} 0 \\ 1 \end{pmatrix}
$$</span>
For example, when <span class="math-container">$n = 12$</span> we get
<span class="math-container">$$
7x^6y^6 + 56x^5y^7 + 126x^4y^8 + 120x^3y^9 + 55x^2y^{10} + 12xy^{11} + y^{12}.
$$</span>
The coefficient of <span class="math-container">$x^4y^8$</span> is what you wanted.</p>
<hr>
<p>Why do we need this very elaborate method? Since it works whenever your constraint can be described by a succinct DFA. For example, you can answer questions of the form "how many strings of length 100 have exactly 50 zeroes and 50 ones, and no copies of 000 or 1101?"</p>
<p>The calculation will amount to a dynamic programming algorithm. Using DFAs is just a principled way of designing the dynamic programming recurrence.</p>
|
965,851 | <p>I have a computer problem that I was able to reduce to an equation in quadratic form, and thus I can solve the problem, but it's a little messy. I was just wondering if anybody sees any tricks to simplify it?</p>
<p>$$\sin^2\beta ⋅ d^4 + c^2\left(\cos^2\beta⋅\cos^2\alpha-\frac{\cos^2\beta}{2}-\frac12\right)d^2 + sin^2\beta⋅\frac{c^4}{16} = 0$$</p>
<p>Obviously I am using the quadratic formula to solve for $d^2$.</p>
<p>$\beta$, $\alpha$, and $c$ are known. </p>
<p>That middle term is so ugly. Perhaps this could even be simplified and solved without the quadratic formula? </p>
<p>By dividing through with $\sin^2\beta$ I came up with (assuming I did it correctly, I didn't double check it):</p>
<p>$$d^4 + \left[\cot^2\beta\cos^2\alpha - \frac{\csc^2\beta}{2}-\frac{\cot^2\beta}{2}\right]c^2d^2 + \frac{c^4}{16} = 0$$</p>
<p>Which is a little less ugly. Am I missing a cool trick to simplify this?</p>
| egreg | 62,967 | <p>The equation can be written $x\log2=2\log|x|$. Let's consider the function
$$
f(x)=x\log2-2\log|x|
$$
defined for $x\ne0$.
We have easily
$$
\lim_{x\to-\infty}f(x)=-\infty,
\qquad
\lim_{x\to\infty}f(x)=\infty
$$
and
$$
\lim_{x\to0}f(x)=\infty.
$$
Moreover
$$
f'(x)=\log2-\frac{2}{x}=\frac{x\log2-2}{x}
$$
Set $\alpha=2/\log2$; then $f'(x)$ is positive for $x<0$ and for $x>\alpha$, while it's negative for $0<x<\alpha$.</p>
<p>Thus the function is increasing in $(-\infty,0)$, which accounts for a solution in this interval. In the interval $(0,\infty)$ the function has a minimum at $\alpha$ and
$$
f(\alpha)=\frac{2}{\log2}\log2-2\log\frac{2}{\log2}
=2(1-\log2+\log\log2)\approx-0.85
$$
Since the minimum is negative, this accounts for two solutions in $(0,\infty)$, which clearly are $x=2$ and $x=4$.</p>
|
1,522,929 | <p>For every fixed $t\ge 0$ I need to prove that the sequence $\big\{n\big(t^{\frac{1}{n}}-1\big) \big\}_{n\in \Bbb N}$ is non-increasing, i.e.
$$n\big(t^{\frac{1}{n}}-1\big)\ge (n+1)\big(t^{\frac{1}{n+1}}-1\big)\;\ \forall n\in \Bbb N$$
I'm trying by induction over $n$, but got stuck in the proof for $n+1$:
<br/>
For n=2 its clear that follows since
$$t-1\ge 2(t^{1/2}-1)\Leftrightarrow t-1\ge 2t^{1/2}-2\Leftrightarrow t+1\ge 2t^{1/2}\Leftrightarrow t^2+2t+1\ge 4t\Leftrightarrow t^2-2t+1\ge 0\Leftrightarrow (t-1)^2\ge 0$$</p>
<p>So, we suppose that $\;\ n\big(t^{\frac{1}{n}}-1\big)\ge (n+1)\big(t^{\frac{1}{n+1}}-1\big)\;\ $ is valid. (I.H.)
<br/>
So I need to prove that:
$$(n+1)\big(t^{\frac{1}{n+1}}-1\big)\ge (n+2)\big(t^{\frac{1}{n+2}}-1\big)\ $$</p>
<p>But I have not reached anywhere helpful expanding all. Any ideas or different approaches to porve this will be appreciated.</p>
| TravorLZH | 748,964 | <p>This answer is dedicated to give an alternative proof of</p>
<p><span class="math-container">$$
\lim_{R\to+\infty}\int_\Gamma e^{iz^2}\mathrm dz=0
$$</span></p>
<p>where <span class="math-container">$\Gamma$</span> denotes a circular arc connecting <span class="math-container">$R$</span> and <span class="math-container">$e^{i\pi/4}R$</span>. Using the parametrization that <span class="math-container">$z=Re^{i\theta}$</span>, we have</p>
<p><span class="math-container">\begin{aligned}
\left|\int_\Gamma e^{iz^2}\mathrm dz\right|
&\le R\int_0^{\pi/4}e^{-R^2\sin(2\theta)}\mathrm d\theta \\
&=R\left(\int_0^\delta+\int_\delta^{\pi/4}\right)e^{-R^2\sin(2\theta)}\mathrm d\theta \\
&<R\delta+{R\pi\over4}e^{-R^2\sin(2\delta)}
\end{aligned}</span></p>
<p>Since <span class="math-container">$\sin u\sim u$</span> when <span class="math-container">$u\to0$</span>, we see that for small <span class="math-container">$\delta>0$</span> there is <span class="math-container">$\sin(2\delta)>\delta$</span>, which means when <span class="math-container">$\delta=R^{-3/2}$</span> and <span class="math-container">$R$</span> is large there is</p>
<p><span class="math-container">$$
\left|\int_\Gamma e^{iz^2}\mathrm dz\right|<{1\over\sqrt R}+{\pi R\over2}e^{-\sqrt R}=O\left(1\over\sqrt R\right)
$$</span></p>
|
2,691,266 | <p>The quotient ring $\mathcal{O}/\mathfrak{a}$ of a Dedekind domain by an ideal $\mathfrak{a}\ne 0$ is a principal ideal domain. </p>
<p>I am trying to show $\mathcal{O}/\mathfrak{p}^n$ is principal ring. Let $\mathfrak{p}^i/\mathfrak{p}^n$ be an ideal and choose $\pi\in\mathfrak{p}\setminus\mathfrak{p}^2$. Then how I show that $\mathfrak{p}^i = \mathcal{O}\pi^i + \mathfrak{p}^n$ ? Any help/hint in this regards would be highly appreciated. Thanks in advance!</p>
| Notone | 408,724 | <p>You have
$(\pi)^i \not\subset \mathfrak{p}^{i+1}$, but $(\pi)^i\subset \mathfrak{p}^i$. </p>
<p>Thus $(\pi)^i=J\cdot \mathfrak{p}^i$ for an ideal $J$ with $\mathfrak{p}\nmid J$ . </p>
<p>Now we have $(\pi)^i+\mathfrak{p}^{n}=\mathfrak{p}^{i}(J+\mathfrak{p}^{n-i})$. But $J$ and $\mathfrak{p}^{n-i}$ have no common prime ideal factors, thus $J+\mathfrak{p}^{n-i}=\mathcal{O}$ and we get the desired result. </p>
|
825,703 | <p>I have been working with vector spaces for a while and I now take for granted what the vector space does. I feel like I dont really understand why multiplication and addition must be defined on a vector space. For example, it feels like adding two vectors and having their sum contained within the space is just a name for a vector space and I dont get what necessarily happens IF the two vector's sum arent in the space. In other words, I dont know why must addition and multiplication must be defined on a vector space, is it to take advantage of nice properties? Thanks!</p>
| Tom Collinge | 98,230 | <p>A vector space is a generalisation of the 3D space we live in and the behaviour of vector addition and scalar multiplication is modelled on that.</p>
<p>Another consideration is that mathematical structures are generally <em>closed</em> - groups, rings, fields, vector spaces are all closed. This means that the operations that can be performed on elements of the structure result in another element of the structure. I can't think of a mathematical structure that is not closed in this way. The reason for defining structures in this way is clear, if a structure were not closed then the results of some operations being outside the structure would be undefined in terms of the rules of the structure.</p>
|
825,703 | <p>I have been working with vector spaces for a while and I now take for granted what the vector space does. I feel like I dont really understand why multiplication and addition must be defined on a vector space. For example, it feels like adding two vectors and having their sum contained within the space is just a name for a vector space and I dont get what necessarily happens IF the two vector's sum arent in the space. In other words, I dont know why must addition and multiplication must be defined on a vector space, is it to take advantage of nice properties? Thanks!</p>
| Lee Mosher | 26,501 | <p>To see why this property is important, consider an example where the property fails. Let $X$ be the union of the $x$ and $y$ axes in the Cartesian plane, so
$$X = \{(x,0) \, | \, x \in \mathbb{R}\} \cup \{(0,y) \, | \, y \in \mathbb{R}\}
$$
Is $X$ a vector space? If not, what axiom fails? And (perhaps of more importance to extending your understanding of vector spaces) what part of your intuition for vector spaces does this example effect?</p>
|
3,027,286 | <p>I am a little confused as to proving that <span class="math-container">$(C^*)^{-1} = (C^{-1})^*$</span> where <span class="math-container">$C$</span> is an invertible matrix which is complex. </p>
<p>Initially, I thought that it would have something to do with the identity matrix where <span class="math-container">$CC^{-1}=C^{-1}$</span>. <span class="math-container">$C = I$</span> but don't seem to be getting anywhere with that. </p>
<p>Thank you! </p>
| Acccumulation | 476,070 | <p>Another counterexample can be derived from the <a href="https://en.wikipedia.org/wiki/St._Petersburg_paradox" rel="nofollow noreferrer">St. Petersburg paradox</a>. Suppose that <span class="math-container">$p_j=\frac1j$</span> if <span class="math-container">$j$</span> is a power of <span class="math-container">$2$</span>, and <span class="math-container">$0$</span> otherwise. Then <span class="math-container">$\sum p_j = \sum 2^{-k}=1$</span>, but <span class="math-container">$jp_j=1$</span> whenever <span class="math-container">$j$</span> is a power of <span class="math-container">$2$</span>, and thus <span class="math-container">$\sum jp_j$</span> diverges.</p>
|
2,245,631 | <blockquote>
<p>$x+x\sqrt{(2x+2)}=3$</p>
</blockquote>
<p>I must solve this, but I always get to a point where I don't know what to do. The answer is 1.</p>
<p>Here is what I did: </p>
<p>$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$</p>
<p>Then I got:
$-2x^{3}-x^{2}-6x+9=0$ </p>
| fleablood | 280,126 | <p>I wouldn't have gotten there is such a convoluted way (and I wouldn't have divided by $x$ without checking that $x \ne 0$ first) but that would really make a difference.</p>
<p>And I'd watch out for extraneous roots.</p>
<p>$x +x\sqrt{2x+2} = 3$</p>
<p>$x\sqrt{2x+2} = 3-x$ $2x+2 > 0$ so $x > -1$</p>
<p>$x^2(2x+2) = 9 -6x + x^2; x > -1$</p>
<p>$2x^3 + x^2 + 6x -9=0; x > -1$ which is basically what you have.</p>
<p>Rational root theoreom says that if there are any rational roots the will be of the form $\frac {\pm 1,3,9}{\pm 1,2}$. </p>
<p>It's pretty clear that $x = 1$ is solution so $x-1$ is a factor and so </p>
<p>$(2x^3 + x^2 + 6x - 9)=(x-1)(2x^2 + bx + 9)$ where $bx^2 -2x^2 = x^2$ and $9x-bx=6x$ so $b=3$ so $(x-1)(2x^2 + 3x + 9)=0$.</p>
<p>$x =1$ is a solution and $2x^2 + 3x + 9 = 0$ is another. But that'd imply $2x^2 = -3x - 9$. But that's imply $3x < -9$ and $x<-3$ which is impossible as $(2x + 2) > 0$.</p>
<p>So $x=1$ is the only real solution.</p>
<p>====</p>
<p>Another way to factor. If you <em>KNOW</em> $x-1$ is a factor, which we do we have:</p>
<p>$2x^3 + x^2 + 6x - 9=$</p>
<p>$2x^2(x-1) + 2x^2 + x^2 + 6x - 9 =$</p>
<p>$2x^2(x-1) + 3x^2 + 6x - 9=$</p>
<p>$2x^2(x-1) + 3x(x-1) + 3x + 6x - 9=$</p>
<p>$2x^2(x-1) + 3x(x-1) + 9x - 9 = $</p>
<p>$2x^2(x-1) + 3x(x-1) + 9(x-1) + 9 - 9=$</p>
<p>$2x^2(x-1) + 3x(x-1) + 9(x-1)=$</p>
<p>$(x-1)(2x^2 + 3x + 9) = 0$.</p>
<p>We can try to solve $2x^2 + 3x + 9 =0$ by the quadratic equation. But we don't actually have to. We know as $\sqrt{2x+2}$ must be real that $x \ge -1$ So $2x^2 + 3x + 9 \ge 2x^2 + 3(-1) + 9 = 2x^2 + 9 > 9 > 0$ so there is no solution that will satisfy <em>both</em> $2x^2 + 3x + 9 = 0$ and $\sqrt{2x+2}$ is real.</p>
<p>(There's actually no real solution to just $2x^2 + 3x + 9=0$ at all as the quadratic equation or completeing the square would show us.)</p>
|
4,271,909 | <p>I'm struggling with improper integrals (Calc I). I've calculated the following:</p>
<p>If <span class="math-container">$a = 0$</span>:
<span class="math-container">$$\int_{0}^{\infty}\cos(x)dx =\lim\limits_{R\to\infty} \int_{0}^{R}\cos(x)dx =\lim\limits_{R\to\infty} \sin(R) $$</span>
Which diverges?</p>
<p>If <span class="math-container">$a > 0$</span>:</p>
<p><span class="math-container">$$\int_{0}^{\infty}e^{-ax}\cos(x)dx = \lim\limits_{R\to\infty}\int_{0}^{R}e^{-ax}\cos(x)dx = \lim\limits_{R\to\infty}\left[\frac{a}{a+1}+\left(\frac{\sin(R)-a\cos(R)}{a^{2}+1}\right)e^{-aR}\right] $$</span>
But what happens to this expression? Does it equal <span class="math-container">$\frac{a}{a+1}$</span>, and how can I decide the greatest area from this? (That is if the expression I got is correct...)</p>
| Giorgos Giapitzakis | 907,711 | <p>For <span class="math-container">$a > 0$</span> the integral evaluates to <span class="math-container">$\frac{a}{a^2+1}$</span> (you forgot a square in the denominator of the first fraction). Now set <span class="math-container">$g(a) = \frac{a}{a^2+1}$</span>, <span class="math-container">$a \in (0, \infty)$</span> and find its derivative. Using that you can find that the maximum value is <span class="math-container">$\frac{1}{2}$</span>, achieved when <span class="math-container">$a=1$</span>.</p>
|
4,271,909 | <p>I'm struggling with improper integrals (Calc I). I've calculated the following:</p>
<p>If <span class="math-container">$a = 0$</span>:
<span class="math-container">$$\int_{0}^{\infty}\cos(x)dx =\lim\limits_{R\to\infty} \int_{0}^{R}\cos(x)dx =\lim\limits_{R\to\infty} \sin(R) $$</span>
Which diverges?</p>
<p>If <span class="math-container">$a > 0$</span>:</p>
<p><span class="math-container">$$\int_{0}^{\infty}e^{-ax}\cos(x)dx = \lim\limits_{R\to\infty}\int_{0}^{R}e^{-ax}\cos(x)dx = \lim\limits_{R\to\infty}\left[\frac{a}{a+1}+\left(\frac{\sin(R)-a\cos(R)}{a^{2}+1}\right)e^{-aR}\right] $$</span>
But what happens to this expression? Does it equal <span class="math-container">$\frac{a}{a+1}$</span>, and how can I decide the greatest area from this? (That is if the expression I got is correct...)</p>
| Robert Lee | 695,196 | <p>You can actually continue your method to get the correct solution! You want to evaluate the limit
<span class="math-container">$$
\lim_{R\to\infty} \left[\frac{a}{a^{\color{red}{2}}+1} + \left(\frac{\sin(R)-a\cos(R)}{a^{2}+1}\right)e^{-aR}\right]
$$</span></p>
<p>By the <a href="https://mathworld.wolfram.com/HarmonicAdditionTheorem.html" rel="nofollow noreferrer">harmonic addition theorem</a>, <span class="math-container">$\sin(R)-a\cos(R)$</span> can be written as <span class="math-container">$$\sin(R)-a\cos(R)=-\sqrt{a^{2}+1}\cos\left(R+\arctan\left(\frac{1}{a}\right)\right) $$</span>
But since <span class="math-container">$-1\le \cos(x) \le 1$</span> for all <span class="math-container">$x$</span>, by the <a href="https://en.wikipedia.org/wiki/Squeeze_theorem" rel="nofollow noreferrer">squeeze theorem</a> you get:
<span class="math-container">$$
\begin{align*}
0= \lim_{R\to\infty}- \sqrt{a^2+1}e^{-aR} \le \lim_{R\to\infty} \left(\sin(R)-a\cos(R)\right)e^{-aR}\le \lim_{R\to\infty}\sqrt{a^2+1}e^{-aR}=0
\end{align*}
$$</span>
since <span class="math-container">$e^{-x}\to 0$</span> as <span class="math-container">$x \to \infty$</span>. This means that your integral evaluates to
<span class="math-container">$$
\int_{0}^{\infty} e^{-ax}\cos(x) \, dx =\frac{a}{a^2+1} + \frac{1}{a^2+1} \underbrace{\lim_{R\to\infty} \left(\sin(R)-a\cos(R)\right)e^{-aR}}_{0} =\frac{a}{a^2+1}
$$</span>
and from here, as others have pointed out, you just need to differentiate to get that the maximum is at <span class="math-container">$a=1$</span> which gives the maximum value of the integral <span class="math-container">$\frac{1}{1^2+1} = \frac{1}{2}$</span>.</p>
|
4,271,909 | <p>I'm struggling with improper integrals (Calc I). I've calculated the following:</p>
<p>If <span class="math-container">$a = 0$</span>:
<span class="math-container">$$\int_{0}^{\infty}\cos(x)dx =\lim\limits_{R\to\infty} \int_{0}^{R}\cos(x)dx =\lim\limits_{R\to\infty} \sin(R) $$</span>
Which diverges?</p>
<p>If <span class="math-container">$a > 0$</span>:</p>
<p><span class="math-container">$$\int_{0}^{\infty}e^{-ax}\cos(x)dx = \lim\limits_{R\to\infty}\int_{0}^{R}e^{-ax}\cos(x)dx = \lim\limits_{R\to\infty}\left[\frac{a}{a+1}+\left(\frac{\sin(R)-a\cos(R)}{a^{2}+1}\right)e^{-aR}\right] $$</span>
But what happens to this expression? Does it equal <span class="math-container">$\frac{a}{a+1}$</span>, and how can I decide the greatest area from this? (That is if the expression I got is correct...)</p>
| Claude Leibovici | 82,404 | <p><span class="math-container">$$I=\int_{0}^{\infty}e^{-ax}\cos(x)dx=\Re \int_{0}^{\infty}e^{-ax}e^{ix}dx=\Re \int_{0}^{\infty}e^{-(a-i)x}dx=\Re\left(\frac 1{a-i}\right)$$</span>
<span class="math-container">$$I=\Re\left(\frac{a}{a^2+1}+\frac{i}{a^2+1}\right)=\frac{a}{a^2+1}$$</span></p>
|
1,685,895 | <blockquote>
<blockquote>
<p>Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.</p>
</blockquote>
</blockquote>
<hr>
<p>My attempt:</p>
<p>$\displaystyle \binom{n}{7}=\binom{n}{8} $</p>
<p>$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$</p>
<p>$$ \frac{6}{7!} = \frac{n-7}{40320} $$</p>
<p>$$ n-7 = 48 $$</p>
<p>$$ n=55 $$</p>
| Community | -1 | <p>$\displaystyle \left(2+\frac{x}{3}\right)^{n}=\sum_{i=0} ^n {n \choose k}({\frac{x}{3}})^i 2^{n-i}$<br>
Now for $i =7\implies{n \choose 7}({\frac{x}{3}})^7 2^{n-7}$ </p>
<p>and for $i =8\implies{n \choose 8}({\frac{x}{3}})^8 2^{n-8}$ </p>
<p>now to get the coefficients of $x^8$ & $x^7$ equal. $\implies$ </p>
<p>${n \choose 7}({\frac{1}{3}})^7 2^{n-7}={n \choose 8}({\frac{1}{3}})^8 2^{n-8} \implies \frac{1}{n-7}=\frac{1}{8}* \frac{1}{3}*\frac{1}{2} \implies n=55$</p>
|
2,554,153 | <p>I have some problem with writing character table of a group. For instance, a group $S_4$. When we write character table, we write irreducible representations of group. So, how can I quickly find them? Then how to Fill the table? Can someone explain me upon this example?</p>
| Somos | 438,089 | <p>You want to find the character table of a small finite group. The quickest way is to look them up in a book or <a href="http://brauer.maths.qmul.ac.uk/Atlas/v3/" rel="nofollow noreferrer">website</a>, or use a program like <a href="https://www.gap-system.org/" rel="nofollow noreferrer">GAP</a>. Failing that, you can find it for yourself with a general method. There are a lot of shortcuts depending on the particular group. We could use orthogonality properties and other properties of characters. I suggest first finding the conjugacy classes and then number them. For example $C_1$ is the conjugacy class of the identity. For $S_4$ there are four others.</p>
<p>We now identify a representative element of each conjugacy class. $C_1$ has the identity permutation $[1,2,3,4]$ as its only element. Let $C_2$ be the class of
permutation $d:=abab=[2,1,4,3]$ with $3$ elments of order $2$. Let $C_3$ be the class of permutation $a:=[1,3,4,2]$ with $8$ elements of order $3$. Let $C_4$ be the class of permutation $b:=[4,2,3,1]$ with $6$ elements of order $2$. Let $C_5$ be the class of permutation $c:=ab=[4,3,1,2]$ with $6$ elements of order $4$. A total of $24$ group elements.</p>
<p>Now the fun part. We define a multiplication of conjugacy classes as multisets of group elements. That is, given two multisets of group elments, the product is the multiset of all pairwise products of elments of the two multisets. Thus, $C_1$ is also the identity for conjugacy class multiplication. Now $C_2C_2=3C_1+2C_2$ because the $3$ elements of order $2$ gives the $3C_1$ where we multiply an element by itself, and for the product of two distinct elements of $C_2$ we get an element of $C_2$ in two ways.</p>
<p>Now we get a system of ten equations in the conjugacy classes and also require $C_1=1$. We got $C_2C_2=3C_1+2C_2$. Next, $C_3C_2=3C_3, C_3C_3=8C_1+8C_2+4C_3$. There are eight more equations. The last is $C_5C_5=6C_1+2C_2+3C_3$. Solving the system of equations gives five solutions, namely, $[1,3,8,6,6], [1,3,8,-6,-6], [1,3,-4,0,0], [1,-1,0,2,-2], [1,-1,0,-2,2].$ These solutions are almost the irreducible characters of our group, but we make two adjustments.</p>
<p>First, divide each solution value of $C_n$ by $c_n$, the number of elements of class $C_n$. This gives us $[1,1,1,1,1], [1,1,1,-1,-1], [1,1,-1/2,0,0], [1,-1/3,0,1/3,-1/3], [1,-1/3,0,-1/3,1/3].$ Now, we scale each vector so that the sum of $c_n$ multiplied by the squares of each value is $24$, the number of elements of the group. This scale factor turns out to be the dimension of each character representation. In our case they are $\{1,1,2,3,3\}$ respectively. Thus the characters are: $[1,1,1,1,1],[1,1,1,-1,-1],[2,2,-1,0,0],[3,-1,0,1,-1],[3,-1,0,-1,1].$</p>
<p>This method is completely computational and can be quickly and easily done (for small groups) by computer algorithms. You don't need to know any theorems in representation theory. The hardest part is solving the class multiplication equations, and fortunately, this can be easily done by using computer algebra systems in one command. There are other methods that require knowing special properties of group characters and special knowledge about the group in question.</p>
|
2,264,614 | <p>Is there a way to evaluate, </p>
<p>$$
\large \cos x \cdot \cos \frac{x}{2} \cdot \cos \frac{x}{4} ... \cdot \cos \frac{x}{2^{n-1}} \tag*{(1)}
$$</p>
<p>I asked this to one of my teachers and what he told is something like this, </p>
<p>Multiply and divide the last term of $(1)$ with $\boxed{\sin \frac{x}{2^{n-1}}}$</p>
<p>So,
$$
\large \frac{\cos \frac{x}{2^{n-1}} \cdot \sin \frac{x}{2^{n-1}}}{\sin \frac{x}{2^{n-1}}} \\ \tag*{(2)}
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \sin \frac{x}{2^{n-1}}} \\
$$
$$
\large \implies \frac{\sin (2x)}{2^n \cdot \frac{\sin \frac{x}{2^{n-1}}}{\frac{x}{2^{n-1}}} \cdot \frac{x}{2^{n-1}}} \tag*{(3)}
$$</p>
<p>Now, as $n \to \infty$ , we have $x \to 0$,
Using this, $\lim$ we have, </p>
<p>$$
\boxed{ \lim_{x \to 0} \frac{\sin x}{x} = 1}
$$</p>
<p>Using this in $(3)$, we have,
$$
\large \boxed{\frac{\sin (2x)}{2x}} \tag*{(4)}
$$</p>
<p>All the steps <strong>sort of</strong> make sense. My doubts are, </p>
<ol>
<li>How do I do this for other trigonometric ratios?</li>
<li>How does the step 2 happen? </li>
</ol>
<p>I need help looking into it more intuitionally. </p>
<p>Please provide necessary reading suggestions. </p>
<p>Regards.</p>
| Jack D'Aurizio | 44,121 | <p>By the sine duplication formula $\sin(2z)=2\sin(z)\cos(z)$ we have $\cos(z)=\frac{1}{2}\cdot\frac{\sin(2z)}{\sin(z)}$.<br>
In particular
$$ \prod_{k=0}^{n-1}\cos\left(\frac{x}{2^k}\right)=\frac{1}{2^n}\prod_{k=0}^{n-1}\frac{\sin\frac{x}{2^{k-1}}}{\sin\frac{x}{2^k}}=\frac{\sin(2x)}{2^n\sin\frac{x}{2^{n-1}}} \tag{1} $$
is a simple telescopic product, that can be written in the form
$$ \frac{\sin(2x)}{2x}\cdot\frac{x}{2^{n-1}\sin\frac{x}{2^{n-1}}}\tag{2}$$
and since $\lim_{w\to 0}\frac{\sin w}{w}=\lim_{w\to 0}\frac{w}{\sin w}=1$ it follows that
$$ \prod_{k\geq 0}\cos\left(\frac{x}{2^k}\right)=\frac{\sin(2x)}{2x}.\tag{3}$$
In general, as soon as we have a differentiable function $f(x)$ such that $f(2x)=2 g(x)\,f(x)$ and $f(0)=0, f'(0)\neq 0$ we have the identity
$$ \prod_{k\geq 0}g\left(\frac{x}{2^k}\right) = \frac{f(2x)}{2x}\cdot \frac{1}{f'(0)}\tag{4} $$
by de l'Hospital theorem.</p>
|
108,060 | <p>Suppose:
$$\sum_{n=2}^{\infty} \left( \frac{1}{n(\ln(n))^{k}} \right) =\frac{1}{ 2(\ln(2))^{k} } +\frac{1}{ 3(\ln(3))^{k} }+...,
$$
by which $k$ does it converge?</p>
<p>When I use comparison test I get inconclusive result:</p>
<p>$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_{n}}=\frac{n\ln(n)^{k}}{(n+1)\ln(n+1)^{k}}
=\lim_{n\rightarrow\infty} =\frac{n\ln(n)^{k}+\ln(n+1)^{k}-\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}\approx
1- \frac{\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}=\\1-\lim_{n\rightarrow \infty}\frac{1}{n+1}=1$</p>
<p>Now my conclusion would be when $k\in\mathbb R$ but I feel I am doing something wrong because I am pretty sure I have done this kind of problems earlier where I used some well-known series comparison. WA nothing <a href="http://www.wolframalpha.com/input/?i=Integrate%28x%5E%7Bk%7Dln%28x%29,x,0,%5Cinfty%29" rel="nofollow">here</a>.</p>
<p><strong>[Update] Trying to use Cauchy condensation test</strong></p>
<p>$$\sum_{n=2}^{\infty}\left( n^{1/k} ln(n)\right)^{-k}$$</p>
<p>and now C-test:</p>
<p>$$\sum_{n=2}^{\infty}\left( \left(2^{n} \right) 2^{n/k} ln(2^{n})\right)^{-k}=$$
$$\sum_{n=2}^{\infty} e^{-k \left( n(1+\frac{1}{k})ln(2)+ln(n)+ln(ln(2) \right) }$$</p>
<p>so now as a geometric series, can I conclude something in terms of $k$? Look $k$ is still in one denominator not the just first factor in the exponent. </p>
| Selim Ghazouani | 24,713 | <p>You can use integral/series comparison.</p>
|
327,860 | <p>Let <span class="math-container">$A$</span> be a symmetric <span class="math-container">$d\times d$</span> matrix with integer entries such that the quadratic form <span class="math-container">$Q(x)=\langle Ax,x\rangle, x\in \mathbb{R}^d$</span>, is non-negative definite. For which <span class="math-container">$d$</span> does it imply that <span class="math-container">$Q$</span> is a sum of finitely many squares of linear forms with integer coefficients
<span class="math-container">$$
Q(x)=\sum_{i=1}^N (\ell_i(x))^2\quad \text{for some}\, N?
$$</span>
For <span class="math-container">$d=2$</span> this is true, I know it from the <a href="https://artofproblemsolving.com/community/c6h219889p1219430" rel="noreferrer">problem</a> proposed by Sweden to IMO in 1995, but probably all this stuff is known for a longer time. </p>
<p>I think, I may prove it for some other small dimensions, although not so elementary (using Minkowski theorem on lattice points in convex bodies: if <span class="math-container">$Q$</span> is positive definite, we may find a linear form <span class="math-container">$\ell(x)$</span> such that <span class="math-container">$Q-\ell^2$</span> is still non-negative definite, this is equivalent to finding an integer point in an ellipsoid), but for large <span class="math-container">$d$</span> this argument fails. </p>
| WKC | 29,241 | <p>I should add that there have been many results on this problem since the 1990s. The existence of the number <span class="math-container">$g(n)$</span> was established by Maria Icaza in her Ohio State 1992 thesis. You should check out her papers. A bit later Myung Kwan Kim and Byeong Kweon Oh published a few papers. They showed that <span class="math-container">$g(6) = 10$</span> (not 9 as Mordell and Ko thought in the 1930s) and this is the last known exact value of the function <span class="math-container">$g(n)$</span>. </p>
|
2,778,422 | <p>The generalized $\lambda-\text{eigenspace}$ is defined by:
$V^f_{(\lambda)}=\bigl\lbrace v\in V\mid\exists j\,\text{ such that }\,(f-\lambda)^jv=0 \bigr\rbrace$.
Suppose that $V$ is a vector space over the field $k$ and $f,g\in \operatorname{End}_k(V)$ satisfy $f\circ g=g\circ f$. Show that $g(V^f_{(\lambda)})\subseteq V^f_{(\lambda)}$.</p>
<p>Own work: Well I tried choosing an element of $v\in V^f_{(\lambda)}$ and I have to see if $(f-\lambda)^j(g(v))=0$. But I don't know how to link the commutativity of f and g with the application of $(f-\lambda)^j(g(v)$.</p>
| mechanodroid | 144,766 | <p>For $x \in V^f_{(\lambda)}$ take $i \in \mathbb{N}$ such that $(f-\lambda I)^ix = 0$. We have</p>
<p>$$(f-\lambda I)^i(g(x)) = g((f-\lambda I)^ix) = g(0) = 0$$</p>
<p>because $g$ commutes with $f$. Therefore $g(x) \in V^f_{(\lambda)}$. </p>
|
1,423,252 | <p>The proposition is:</p>
<blockquote>
<p>If $\lim S_n = L$ and for every $n$, $S_n$ is in the interval $[a,b]$, then $L$ is also in $[a,b]$.</p>
</blockquote>
<p>I have proved this effectively, but now the question is to provide a counterexample to the stronger assumption, for the interval $(a,b)$. </p>
<p>Basically, where this proposition does not guarantee that $L$ is also in $(a,b)$.</p>
| Dontknowanything | 240,480 | <p>What about $1/(n+1)$ in $(0,1)$?</p>
|
3,096,115 | <p>By using a sieve created by Prime Number Tables set up by the formula PN+(PNx6) for numbers generated by 6n+or-1, takes 182 calculations to identify 170 composite numbers. Using the Sieve of Eratosthenes would take around 1600 calculations. The Prime Number Tables identify all the composite numbers on the the list of 332 possible prime numbers: </p>
<p>5,11,17,23,29,35,41,47,53,59,65,71,77,83,89,95,101,107,113,119,125,131,137,143,149,155,161...</p>
<p>7,13,19,25,31,37,49,55,61,67,73,79,85,91,97,103,109,115,121,127,133,139,145,151,157,163,169...</p>
<p>Prime Number Table for:</p>
<p>5 identifies the multiples of 5 with 67 calculations (the Sieve of E: 249)</p>
<p>7: 23 calculations (the Sieve of E: 133)</p>
<p>11: 29 calculations (the Sieve of E: 98)</p>
<p>13: 11 calculations (the Sieve of E: 75)</p>
<p>17: 17 calculations (the Sieve of E: 57)</p>
<p>19: 4 calculations (the Sieve of E: 51)</p>
<p>23: 12 calculations (the Sieve of E: 42)</p>
<p>29: 5 calculations (the Sieve of E: 34)</p>
<p>31: 1 calculations (the Sieve of E: 31)</p>
<p>41: 3 47: 3 53: 2 59: 2 calculations (the Sieve of E:0)</p>
<p>71: 2 89: 1 101: 1 calculations (the Sieve of E: 0)</p>
<p>107: 1 113: 1 131: 1 137: 1 calculations (the Sieve of E: 0)</p>
<p>2: 0 calculations (the Sieve of E:499)</p>
<p>3: 0 calculations (the Sieve of E: 332)</p>
<p>Total Calculations:</p>
<p>A Sieve using PN Tables: 187 calculations to find 166 Prime Numbers by identifying 166 composite numbers (10 of 11 duplicate of multiples of 5)</p>
<p>Sieve of Eratosthenes: 1601 calculations to find 168 Prime Numbers by identifying 832 composite numbers (769 duplication of calculations)</p>
<p><em>Note: What I am really hoping for is some help. I have tested this up to 1411. There is no reason to believe it wouldn't go to whatever number. It seems since it deals with less numbers and less calculations, it would use less memory. If you look at the tables and what I have been able to research it makes Primes numbers even more interesting for children who might then take up more interest in math. Hey, I am a guy who works in a grocery store who just likes to think about things. I need help. People keep telling me about the Sieve of Eratosthenes. I have given a comparison between the 2 sieves. Would you rather make 1600 calculations or 187?</em></p>
<p>You can check on my website: <a href="https://mrspudgetty.wixsite.com/mr-spudgetty/prime-numbers" rel="nofollow noreferrer">https://mrspudgetty.wixsite.com/mr-spudgetty/prime-numbers</a></p>
| D Left Adjoint to U | 26,327 | <p>What you are looking for is a computational function that returns the list of primes less than 1000, and works as efficiently as possible, on just that problem.</p>
<p>There are <a href="https://www.wolframalpha.com/input/?i=pi(1000)" rel="nofollow noreferrer">168 primes</a> between 1 and 1000, so any method will take at least 168 operations, obviously.</p>
<p>Suppose you could find approximately a list of 13 numbers and a disjoint other list of 13 numbers, such that precisely every prime occurs when you sum the two sets together elementwise, and no more than that (no composites; otherwise more testing would be required to filter them out).</p>
<p>What I mean is:</p>
<p><span class="math-container">$$
A := \{0, 2, 6\} \\
B := \{5, 11, 17\} \\
A + B = \{ 5, 11, 17, 7, 13, 19, 23\}
$$</span></p>
<p>You can most likely find two sets of around 13-20 elements each such that when you sum the sets you get your list. The sets would then be constants in your program, and the program that generates your output in under 200 operations, actually in 169 arithmetic operations (less if you're allowed to include <span class="math-container">$B$</span> which could be made to be all prime numbers if there's a <span class="math-container">$0$</span> included in <span class="math-container">$A$</span>).</p>
<p>See:</p>
<p><a href="https://math.stackexchange.com/questions/2962971/compressing-the-primes-using-simple-addition">Compressing the primes using addition</a></p>
|
2,041,839 | <p>$$\int_{1}^{x}\frac{dt}{\sqrt{t^3-1}}$$ does this have a closed form involving jacobi elliptic functions of parameter $k$?</p>
<p><strong>N.B</strong> I tried with the change of variables $t=1+k\frac{1-u}{1+u}$. But this leads no where. <a href="http://mathworld.wolfram.com/JacobiEllipticFunctions.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/JacobiEllipticFunctions.html</a></p>
<p><strong>update</strong> the above integral is equivalent to $$\int\limits_{0}^{\sec^{-1}x^{\frac{3}{2}}}\sec^{\frac{2}{3}}tdt$$.</p>
| B. Goddard | 362,009 | <p>We don't have good intuition about what time-squared means, so maybe you should use length, since length squared is area. If you had a square which was $x$-by-$x$ feet and then increased both sides by $dx$, then the area of your new square is $x^2 +2x \; dx+ (dx)^2$. The change in the area is two skinny $x$-by-$dx$ strips and then a little corner piece of area $(dx)^2.$ So it's not really the change in $x^2$, but it's a measure of how much the change in $x^2$ is not linear.</p>
<p>Or you could just consider the skinny strip. Then $(dx)^2$ is (sort of) the change from $x$-by-$dx$ to $x+dx$-by-$dx$.</p>
<p>As others have said, it takes a decent sized pile of mathematical machinery to really define and deal with differentials rigorously.</p>
|
617,163 | <p>I need to find a proper definition of a quantile.
It says:
a p-th quantile $x_p$ is a number, that satisfies the following conditions:
$$
0<p<1
$$
and
$$
P(X \le x_{p}) \ge p
$$
and
$$
P(X \ge x_{p}) \ge 1-p
$$
is this definition right?</p>
| Eric Stucky | 31,888 | <p>The total amount of money he brought in was the selling price of $250$ chairs. But $50$ of those chairs were pure gain, so when <em>he</em> bought the chairs, he only spent the equivalent of the selling price of $200$ chairs.</p>
<p>The profit percent is the amount of gain divided by the amount he spent. This is different from what you have calculated (gain divided by total money brought in). So that is
$$\frac{\text{selling price of } 50 \text{ chairs gained}}{\text{selling price of } 200 \text{ chairs spent}}\times 100 = 25\%\frac{\text{gained}}{\text{spent}}.$$</p>
|
1,038,198 | <p>How do you prove that
$8 \cos{(x)}\cos{(2x)}\cos{(3x)} - 1 = \dfrac{\sin{(7x)}}{\sin{(x)}}$?</p>
| Petite Etincelle | 100,564 | <p>Multiply both sides by $\sin x$, then</p>
<p>$$2\cos x \sin x = \sin 2x$$</p>
<p>$$2\cos 2x \sin 2x = \sin 4x$$</p>
<p>$$2\sin 4x \cos 3x = \sin 7x + \sin x$$</p>
|
2,275,604 | <p>We can break up the circle into an infinite amount of rings with perimeter $2\pi r$. For a given circle $r$, the outside ring has perimeter of $2\pi r$ and the smallest one has of course perimeter $0$. We can add up all the area of the infinite rings using the arithmetic series concept; we can get the average of the first term and the last term and multiply by the number of terms. Therefore, we can find the area of the circle as $\frac{2\pi r + 0}{2} \times r$ = $\pi r^2$.</p>
<p>I thought of this proof and was wondering if it was valid. I know the phrase "arithmetic series" to add up/integrate the rings together is not really the right wording, but is it the right idea?</p>
| Paramanand Singh | 72,031 | <p>You started off with the right idea. Area of any plane region is found by splitting into smaller regions and adding their areas. The assumption here is that the smaller regions can be chosen with familiar shapes like squares, rectangles or even triangles whose areas can be calculated using well known formulas. Naturally any shape can not be broken into an exact number of smaller regions of such shapes and hence the technique works via approximation. There are some parts of region left out when we try to divide it into a number of square shaped regions. Our hope is that the approximation has less and less error as we make finer and finer divisions.</p>
<p>And all this procedure can be formalized into the language of limits and techniques of integral calculus can be used to get exact values of the area of the plane region.</p>
<p>But your approach has a fundamental problem. You divide circle into a number of rings and you don't have the formula for area of a ring. Some people try to approximate the ring as a rectangle with circumereference as length and width of ring as width of rectangle. This approximation works well and gives area of circle correctly. But note that this is not exactly an arithmetic progression and there is no $r$ number of terms. So that kind of sum of an arithmetic progression does not make sense here.</p>
<hr>
<p>There is another aspect here which is worth mentioning and that is definition of area of plane regions. Most ancient mathematicians never bothered to formulate an exact definition and instead regarded it as obvious that any plane region has an area which can be found by the technique of subdivision into smaller regions. However it is important to distinguish definition of area with methods of its calculation. I have provided one definition of area in <a href="https://math.stackexchange.com/a/2271494/72031">this answer</a> which you may find interesting.</p>
|
316,965 | <p>How do I interpret following types of matrices as special types of transformations?
I mean what are the transformative properties of following types of matrices, from $\mathbb{R}^n $ to $ \mathbb{R}^n$, or $\mathbb{C^n}$ to $\mathbb{C^n}$?</p>
<p><strong>Normal and Anti Hermitian Matrices</strong>?</p>
<p><strong>ADDED</strong></p>
<p>I expect answer something like this for orthogonal matrices <a href="http://en.wikipedia.org/wiki/Orthogonal_matrix">Quoting Wikipedia:</a>
As a linear transformation, an orthogonal matrix preserves the dot product of vectors, and therefore acts as an isometry of Euclidean space, such as a rotation or reflection. In other words, it is a unitary transformation.</p>
<p><strong>ADDED</strong></p>
<p>I managed to collect some more information:
Unitary matrices as Linear Maps <strong>preserve inner products</strong>, may be not necessarily for reals only, as orthogonal ones(???).</p>
| Branimir Ćaćić | 49,610 | <p>Let $V$ be a finite-dimensional complex inner product space. </p>
<ol>
<li><p>By the finite-dimensional spectral theorem, <a href="http://en.wikipedia.org/wiki/Normal_matrix" rel="noreferrer">one has that</a> $T \in L(V) := L(V,V)$ is normal if and only if there exists an orthonormal basis for $V$ consisting of eigenvectors of $T$. Geometrically, this means that $T$ acts on individual each coordinate axis (with respect to this orthonormal basis) by rescaling by a complex scaling factor (i.e., the eigenvalue for the eigenvector spanning that axis). In other words, $T$ is normal if and only if there exists some orthonormal coordinate system for $V$ such that $T$ fixes each coordinate axis, i.e., for each coordinate axis $\mathbb{C} v_k$, $T(\mathbb{C} v_k) \subseteq \mathbb{C} v_k$.</p></li>
<li><p>A <em>$1$-parameter groups of unitaries</em> is a map $U : \mathbb{R} \to U(V)$, the group of unitaries in $L(V)$, such that $U(0) = I$ and $U(t_1+t_2) = U(t_1)U(t_2)$ for any $t_1$, $t_2$; geometrically, given a fixed orthonormal basis defining your initial orthonormal coordinate system, you can view $U$ as defining a a time-dependent orthonormal coordinate system on $V$. </p>
<p>So, let $U : \mathbb{R} \to U(V)$ be a $1$-parameter group of unitaries. Observe that for $t \in \mathbb{R}$ and $h \neq 0$,
$$
\frac{1}{h}(U(t+h)-U(t)) = \left(\frac{1}{h}\left(U(h)-I\right)\right)U(t),
$$
so that $U$ is differentiable at $t$ if and only if it is differentiable at $0$. So, suppose that $U$ is differentiable at $0$. Then
$$
(U^\prime(0))^\ast = \left(\lim_{h \to 0}\frac{1}{h}\left(U(h)-I)\right)\right)^\ast = \lim_{h\to 0}\frac{1}{h}\left(U(-h)-I\right) = -\lim_{k\to 0} \frac{1}{k} \left(U(k)-I\right) = -U^\prime(0),
$$
so that $U(t)$ satisfies the ODE
$$
U^\prime(t) = A U(t),
$$
where $A := U^\prime(0)$ is skew-adjoint. Thus, from a geometric standpoint, skew-adjoint operators are precisely the <em>infinitesimal generators</em> of differentiable time-dependent orthonormal coordinate systems. </p>
<p>In fact, bearing in mind that $T$ is skew-adjoint if and only if $iT$ is self-adjoint (Hermitian), and that in mathematical physics and theoretical physics one generally uses $e^{iA}$ instead of $e^A$, this is essentially why self-adjoint operators (on infinite-dimensional Hilbert space!) become essential in quantum mechanics. This is because in quantum mechanics, the time evolution (in the so-called Heisenberg picture) is effected by a $1$-parameter group of unitaries whose infinitesimal generator is the all-important <em>Hamiltonian</em> of your quantum mechanical system.</p></li>
</ol>
<p>Now, suppose that $V$ is a finite-dimensional <em>real</em> inner product space.</p>
<ol>
<li><p>By the finite-dimensional spectral theorem one has that $T \in L(V)$ is symmetric (i.e., self-adjoint) if and only if there exists an orthonormal basis for $V$ consisting of eigenvectors. In other words, $T$ is symmetric if and only if there exists some orthonormal coordinate system for $V$ such that $T$ fixes each coordinate axis, i.e., for each coordinate axis $\mathbb{R}v_k$, $T(\mathbb{R}v_k) \subseteq \mathbb{R} v_k$. Observe that the notion of normal operator is no longer nearly as useful as in the complex case, since the finite-dimensional spectral theorem now characterizes precisely the symmetric operators. In particular, even though orthogonal operators are normal, examples of orthogonal operators with non-real complex eigenvalues are plentiful.</p></li>
<li><p>When it comes to skew-symmetric (i.e., skew-adjoint) operators, everything carries over exactly as before, except that we now find that they are precisely the infinitesimal generators of $1$-parameter groups of orthogonal operators, viz, time-dependent orthonormal coordinate systems. Note, however, that skew-symmetric and symmetric operators are now fundamentally different now, since we can no longer multiply by $i$ to go between the skew-adjoints and self-adjoint, as we could in the complex case. In particular, symmetric operators necessarily have real eigenvalues, whilst examples of skew-symmetric operators with non-real complex eigenvalues are absolutely plentiful.</p></li>
</ol>
|
1,936,260 | <p>We have a binary sequence of 1s and 0s, and the length is 10.
I wonder how many binary sequence of length 10 with four 1's can be created such that the 1's do not appear consecutively?</p>
| carmichael561 | 314,708 | <p>Since $\log_{10}(xy)=\log_{10}(x)+\log_{10}(y)$, it's enough to prove that $\lfloor a+b\rfloor \geq \lfloor a\rfloor +\lfloor b\rfloor$ for all real numbers $a,b$.</p>
|
4,037,697 | <p><span class="math-container">$a_n $</span> is a sequence defined this way:
<a href="https://i.stack.imgur.com/nHdiD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nHdiD.png" alt="enter image description here" /></a></p>
<p>and we define: <a href="https://i.stack.imgur.com/b01hX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b01hX.png" alt="enter image description here" /></a></p>
<p>need to prove that this happens:
<a href="https://i.stack.imgur.com/YMuE6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YMuE6.png" alt="enter image description here" /></a></p>
<p>i got exam soon and a lot of simmilar questions but i have no idea where to start</p>
| Beyond Infinity | 683,570 | <p>Let <span class="math-container">$\mathbf {x, y} \in \mathbb{R}^n$</span> be unit vectors.<br />
Then, the angle <span class="math-container">$\theta_1$</span> between them is given by <span class="math-container">$ \cos \theta_1 = \mathbf{x^\top y}$</span>.</p>
<p>Also, angle between <span class="math-container">$\mathbf{Ax}$</span> and <span class="math-container">$\mathbf{Ay}$</span> is given by :</p>
<p><span class="math-container">$\cos \theta_2 = \dfrac{(\mathbf{Ax})^\top (\mathbf{Ay})}{\|\mathbf{Ax}\|\|\mathbf{Ay}\|}$</span>.<br />
Now, <span class="math-container">$\|\mathbf{Ax}\|_2^2 = (\mathbf{Ax})^\top(\mathbf{Ax}) = \mathbf{x^\top A^\top Ax} = \mathbf{\lambda ^2 x^\top x}$</span>.<br />
Thus, <span class="math-container">$\mathbf{\|Ax\|} = \mathbf{|\lambda|}$</span>.<br />
<span class="math-container">$\implies \cos \theta_2 = \mathbf{\dfrac{\lambda^2 x^ \top y}{\lambda^2}} = \mathbf{x^ \top y} \implies \theta _1 = \theta _2$</span>.<br />
Thus, <span class="math-container">$\mathbf{A}$</span> preserves angles and is a similarity transformation.</p>
|
2,525,573 | <p>Find the domain and range of $y=\sqrt {x-2}$</p>
<p>My Attempt:
$$y=\sqrt {x-2}$$
For $y$ to be defined,
$$(x-2)\geq 0$$
$$x\geq 2$$
So $dom(f)=[2,\infty)$.</p>
| randomgirl | 209,647 | <p>Write $5e^{2x}$ as $e^{ \ln(5)}e^{2x}=e^{\ln(5)+2x}$</p>
|
2,525,573 | <p>Find the domain and range of $y=\sqrt {x-2}$</p>
<p>My Attempt:
$$y=\sqrt {x-2}$$
For $y$ to be defined,
$$(x-2)\geq 0$$
$$x\geq 2$$
So $dom(f)=[2,\infty)$.</p>
| Sarvesh Ravichandran Iyer | 316,409 | <blockquote>
<p>$\ln ab = \ln a + \ln b$, $\ln b^a = a \ln b$.</p>
</blockquote>
<p>You seem to have confused $\ln 5 \times e^{2x}$ with $\ln (5e)^{2x}$. The first would simplify according to the first rule, as $\ln 5 + 2x$, while the second would simplify as $2x \ln (5e)$. You should have obtained the first expression, and not the second, which lead to the incorrect answer.</p>
<p>After taking $\ln$ then, you should get $3x - 1 = 2x + \ln 5$, which would give you $x = \ln 5 + 1$, which should be correct.</p>
|
121,403 | <blockquote>
<p>A manifold $M$ of dimension n is a topological space with the following properties:<br>
a) $M$ is Hausdorff<br>
b)$M$ is locally Euclidean of dimension n<br>
c) $M$ has a countable basis of open sets. </p>
</blockquote>
<p>Why is the first property necessary? I do not have much experience with Hausdorff spaces, hence I am not able to see the importance of that condition, probably it is something obvious. </p>
<p>Since I come from a physics background, I can understand the importance of the second property. But what is the strong mathematical motivation to study such special subset of a topological space? Also the second property can be alternatively stated as: each point $p\in M$ has a neighborhood $U$ which is homeomorphic to an open subset $U$ of $\mathbb R^n$. Why should it be homeomorphic to an open set and not closed?</p>
<p>Similarly in the third property,why a countable basis of open sets? </p>
| Samuel Reid | 19,723 | <p>Let us first review our topological notions to ensure there isn't any confusion there.</p>
<p>A topological space is a set $X$ together with a collection $U$ of subsets of $X$, called a topology on $X$, so that the following hold,</p>
<ul>
<li>The empty set $\varnothing$ and the set $X$ are both open.</li>
<li>$U$ is closed under arbitrary union: $$\text{If } U_{1},...,U_{k} \text{ are open, then } \bigcup_{i =1}^k U_{i} \text{ is open.}$$</li>
<li>$U$ is closed under finite intersections: $$\text{If } U_{1},...,U_{k} \text{ are open, then } \bigcap_{i=1}^k U_{i} \text{ is open.}$$</li>
</ul>
<p>We can define maps between topological spaces $X$ and $Y$ where,
A map $f: X \rightarrow Y$ is continuous if $f^{-1}(U)$ is open for every open set $U \subset Y$. A continuous map that has a continuous inverse is a homeomorphism and if there exists a homeomorphism between $X$ and $Y$, then $X$ and $Y$ are considered topologically equivalent.</p>
<p>Generally, a topological manifold can be thought of as some space which locally looks like $\mathbb{R}^n$ endowed with a metric by which calculus can be performed. More precisely, a topological space $\mathcal{M}$ is locally Euclidean of dimension $n$ if and only if every point of $\mathcal{M}$ has a neighbourhood that is homeomorphic to an open subset of $\mathbb{R}^n$. We then have our initial definition of a topological manifold as follows,</p>
<p>A topological space $\mathcal{M}$ is called a topological $n$-manifold when,
$\mathcal{M}$ is locally Euclidean of dimension $n$; that is, for each $p \in \mathcal{M}$ there exists, </p>
<ul>
<li>An open set $U \subset \mathcal{M}$ with $p \in U$, </li>
<li>An open set $\widetilde{U} \subset \mathbb{R}^d$ such that $\varphi : U \rightarrow \widetilde{U}$ is a homeomorphism.</li>
</ul>
<p>Furthermore, $\mathcal{M}$ is a Hausdorff space: For every two distinct points $p,q \in \mathcal{M}$, there are disjoint open subsets $U,V \subset \mathcal{M}$ such that $p \in U$ and $q \in V$. An additional property that $\mathcal{M}$ is second countable (there is a countable basis for the topology of $\mathcal{M}$) gives a more general definition</p>
<p>Examples of topological manifolds, are $\mathbb{R}^n$, the unit sphere $\mathbb{S}^n$, graphs of continuous functions $f: U \rightarrow \mathbb{R}^d=n$, and the real projective plane $\mathbb{RP}^n$.</p>
<hr>
<p>In general the reason for defining a topological $n$-manifold with the additional characteristics to locally Euclidean is to avoid certain pathological examples that we do not want to count as a manifold. For example, to explain why we would want a space to be Hausdorff, I'll present the best illustration I have found so far: we want all points to be, in a sense, separable. Consider the real line $\mathbb{R}$ glued to another real line $\mathbb{R}'$ along the open sets $(-\infty,0) \cup (0,\infty)$ away from the origin. Then we cannot embed this space in Euclidean space because there are multiple ways to fill in a jump discontinuity of a function, and compact sets are not necessarily closed. </p>
<p>For a space to be second-countable we need to avoid the uncountable well-ordered sets guaranteed to exist by Zorn's Lemma. Firstly recall that if $(X,<)$ is a partial order, then $C \subseteq X$ is a chain in $X$ if $C$ is linearly order by $<$.</p>
<p>That is, Zorn's Lemma states,</p>
<blockquote>
<p>If $(X,<)$ is a partial order such that for every chain $C \subseteq X$ there is $x \in X$ such that $c \leq x$ for all $c \in C$, then there is $y \in X$ such that the is no $z \in X$ with $z > x$. In other words, if every chain has an upper bound, then there is a maximal element of $X$.</p>
</blockquote>
<p>From this we can give a proof of the well-ordering principle which states any there is always a well-ordering of $A$ (for any nonempty $C \subseteq A$, there is an $a \in C$ such that $a \leq b$ for all $b \in C$). Now, the actual reason we need to have our manifold be second countable, is that Zorn's lemma guarantees the existence on an uncountable well-ordered set. So, given the order topology we can take half-open intervals indexed by our uncountable well-ordered set and have that the space cannot be embedded in $\mathbb{R}^n$ and that there is no countable dense subset (not separable). So, we require our space to be second countable because we want to avoid such pathological examples to do with bizzare uncountable orderings that lose analytic separability.</p>
<p>I will stop rambling now, but hopefully this is helpful; the general idea that I would take away is there there are good reasons for defining a topological $n$-manifold the way it is commonly defined because there are certain "counterexamples" or pathological examples that do not have the nice properties. If we are to classify such a vast collection of spaces, as "all topological manifolds of any dimension", we certainly what the power of generality to not have to deal with pathological examples that cause general theorems to not hold because of some bizzare uncountable ordering or the like.</p>
|
1,742,768 | <p>How to examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$ for $a, b, c> 0$ using Taylor's theorem?</p>
| Clement C. | 75,808 | <p>You have, from Taylor expansion of $e^u$ around $0$ to second order,
$$\begin{align}
\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2} &=
e^{\frac{1}{n}\ln a} - \frac{1}{2}\left(e^{\frac{1}{n}\ln c}+e^{\frac{1}{n}\ln b}\right)\\ &=
1+\frac{1}{n}\ln a + \frac{1}{2n^2}\ln ^2 a \\&\qquad- \frac{1}{2}\left(2+\frac{1}{n}\ln b + \frac{1}{2n^2}\ln ^2 b+\frac{1}{n}\ln c + \frac{1}{2n^2}\ln ^2 c\right) + o\left(\frac{1}{n^2}\right) \\
&= \frac{1}{2n}(2\ln a - (\ln b + \ln c)) + \frac{1}{4n^2}(2\ln ^2 a - ( \ln ^2 b+\ln ^2 c)) + o\left(\frac{1}{n^2}\right).
\end{align}$$</p>
<p>If $(2\ln a - (\ln b + \ln c)) \neq 0$, can you conclude (by theorems of comparisons)? And if $(2\ln a - (\ln b + \ln c)) = 0$?</p>
|
2,516,023 | <blockquote>
<p>Why does taking logarithms on both sides of $0<r<s$ reverse the inequality for logarithms with base $a$, $0<a<1$?</p>
</blockquote>
<p>I would like some intuition on why this works. I tried graphing $\log_{0.5}(x)$ on Desmos, for example, and if the graph were true this would be evident from the graph, but the graph seems wrong because I don't get why as $x\to0^+$, $y\to \infty$ since $0.5^x$ should be $\le 0.5$ where $0<x<1$.</p>
| Ross Millikan | 1,827 | <p>Because logs with bases less than $1$ are decreasing, not increasing functions. As an example, we have $8 \lt 16$, but $\log_{0.5} 8=-3 \gt \log_{0.5}16=-4$. It is like multiplying by a negative number, which is a decreasing function and reverses the inequality.</p>
|
148,313 | <p>Someone has claimed that he has constructed a quaternion representation of the one dimensional (along the x axis) Lorentz Boost.</p>
<p>His quaternion Lorentz Boost is $v'=hvh^*+ 1/2( [hhv]^*-[h^*h^*v^*]^*)$ where h is (sinh(x),cosh(x),0,0). He derived this odd transform by substituting the hyperbolic sine and cosine for the sine and cosine in the usual unit quaternion rotation $v'=hvh^*$ and then subtracting out unwanted factors. You can see the short "proof" here: </p>
<p><a href="http://visualphysics.org/preprints/qmn10091026" rel="nofollow">http://visualphysics.org/preprints/qmn10091026</a></p>
<p>I have argued that, whereas the quaternion rotations form a group, his newly devised transform probably does not. Two transformations of the form $v'=hvh^*+ 1/2( [hhv]^*-[h^*h^*v^*]^*)$ almost certainly do not make another transformation of the form $v'=fvf^*+ 1/2( [ffv]^*-[f^*f^*v^*]^*)$ He has not attempted to prove that and he won't because he thinks it is unnecessary.</p>
<p>He has responded that, even if I'm correct, he still has created a Lorentz Boost despite the fact that he has not created a group. I've argued that that it is essential that two Lorentz Boosts make another Lorentz Boost. Without that group structure there is no Boost.
Is this correct? </p>
<p>Can you have a Lorentz Boost along the x axis without having the group structure Boost+Boost=Boost? (for the case of Boosts along the x axis. I know that, for boosts in different directions, Boost+Boost=Rotation)</p>
| Jeremy Stein | 31,954 | <p>Not mathematically rigorous, but perhaps more practical:</p>
<p>Let's assume we can narrow down the numbers to one of ten values. Perhaps you're willing to round your salaries to the nearest 5K and you think they're between 50K and 95K, or you were both willing to announce that you had less than 10 speeding tickets.</p>
<p>Each of you should take half a deck of cards. Pick one card to be your marker (perhaps an Ace - something easy to remember) and show the other person. Now, as you together count up through the possible values, you take turns placing a card face down onto a common pile. When you get to your number, you put your marker card down. When you're done counting, you have 20 cards in the pile and whichever marker card is on top would indicate the person who has the higher number, but you can't look yet!</p>
<p>For example, if the first person has 2 speeding tickets (marker card in green), and the second has 4 (marker card in red), the pile would grow like this:</p>
<p><img src="https://i.stack.imgur.com/NsseA.png" alt="card piles"></p>
<p>Now you each take turns holding the pile under the table while you put some of your remaining cards on the top of the pile and some underneath the pile. After you've each done this, you don't know how deep in the deck your original pile of 20 cards is. But it's somewhere, and the top marker card still indicates whose number is higher.</p>
<p>For the above example, the deck would now look something like this:</p>
<p><img src="https://i.stack.imgur.com/HCDh3.png" alt="sideways card pile"></p>
<p>Then you can flip the cards off the top of the pile until you hit one of your marker cards. If it's the marker of the person who was placing his card second on each number announcement, you should check the next card also, to see if you have a tie. Immediately shuffle the deck.</p>
|
1,979,226 | <p>Use Bayes' theorem or a tree diagram to calculate the indicated probability. Round your answer to four decimal places.
Y1, Y2, Y3 form a partition of S.</p>
<p>P(X | Y1) = .8, P(X | Y2) = .1, P(X | Y3) = .9, P(Y1) = .1, P(Y2) = .4. </p>
<p>Find P(Y1 | X).</p>
<p>P(Y1 | X) =</p>
<p>For this one I thought that all I had to do was P(X | Y1)*P(Y1)/P(X | Y1)*P(Y1)+P(X | Y2)*P(Y2)+P(X | Y3)*P(Y3)</p>
<p>But when I do that I am not getting the correct answer, is it possible that the value for P(Y3) is not .1 and if it is not, what is it? </p>
| mercio | 17,445 | <p>Since every $u_n$ is positive, we have $u_n \ge \sqrt {u_0}$ for $n \ge 1$, and then $u_n \ge \sqrt {(n-1)\sqrt {u_0}}$ for $n \ge 2$, which shows that $(u_n)$ diverges.</p>
|
4,492,566 | <blockquote>
<p>To which degree must I rotate a parabola for it to be no longer the graph of a function?</p>
</blockquote>
<p>I have no problem with narrowing the question down by only concerning the standard parabola: <span class="math-container">$$f(x)=x^2.$$</span></p>
<p>I am looking for a specific angle measure. One such measure must exist as the reflection of <span class="math-container">$f$</span> over the line <span class="math-container">$y=x$</span> is certainly no longer well-defined. I realize that preferentially I should ask the question on this site with a bit of work put into it but, alas, I have no intuition for where to start. I suppose I know immediately that it must be less than <span class="math-container">$45^\circ$</span> as such a rotation will cross the y-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,\sqrt{2})$</span>.</p>
<p>Any insight on how to proceed?</p>
| PDE | 300,754 | <p>My attempt to phrase Brian Drake's answer even more simply:</p>
<p>The original problem can be phrased geometrically as, "What is the smallest angle <span class="math-container">$\theta$</span> by which we can rotate the standard parabola <span class="math-container">$P$</span> around the origin in order to make its rotated version <span class="math-container">$R$</span> intersect a given vertical test line <span class="math-container">$v$</span> at two points?"</p>
<p>There are two keys to solving the problem.</p>
<p><strong>Key #1: In this problem, position is relative!</strong> The absolute positions of the rotated parabola <span class="math-container">$R$</span> and vertical line <span class="math-container">$v$</span> on the <span class="math-container">$xy$</span>-plane do not change whether or not <span class="math-container">$R$</span> describes a function.</p>
<p><strong>Key #2: Rotations are also always relative!</strong></p>
<p>Let's now bring these two ideas together.</p>
<p>According to Key #1, we can choose the <span class="math-container">$y$</span>-axis for our vertical line <span class="math-container">$v$</span>. Why this choice? Some reasons:</p>
<ul>
<li>Lines passing through the origin have an easily written function</li>
<li>The <span class="math-container">$y$</span>-axis passes through the origin and so has an easy function</li>
<li>Standard parabola <span class="math-container">$P$</span> also passes through the origin</li>
<li>Rotated parabola <span class="math-container">$R$</span> was generated from rotating <span class="math-container">$P$</span> around the origin and so also passes through the origin</li>
<li><strong>So, the <span class="math-container">$y$</span>-axis passes through both <span class="math-container">$R$</span> and <span class="math-container">$P$</span></strong></li>
</ul>
<p>Then, the relativity of rotation from Key #2 mean it's completely OK to think of <span class="math-container">$\theta$</span> as a <strong>rotating <span class="math-container">$v$</span></strong> around the origin while keeping the standard parabola <span class="math-container">$P$</span> fixed.</p>
<p>And I think that makes it obvious that any non-zero <span class="math-container">$\theta$</span> will make the point that <strong>any</strong> diagonal line passing through the origin will intersect the standard parabola at two points.</p>
<p>Of course, if you're not a visual/intuitive person, then you can formally justify this result in many ways.</p>
<ul>
<li>5xum does it above using rotations. He finds that <span class="math-container">$x_{intersection} = tan(\theta)$</span>.</li>
<li>Brian Drake does it above by noting that a diagonal line passing through the origin will have an equation <span class="math-container">$y = mx$</span> for some slope <span class="math-container">$m$</span>. He then finds the (other) intersection of this line with the standard parabola by setting <span class="math-container">$y_{intersection} = mx_{intersection} = x_{intersection}^2 \implies x_{intersection} = m$</span>.</li>
<li>These two solutions are related by the fact that for all lines, <span class="math-container">$tan(\theta) = \frac{sin(\theta)}{cos(\theta)} = \frac{\Delta y}{\Delta x} = slope = m$</span>.</li>
</ul>
|
4,492,566 | <blockquote>
<p>To which degree must I rotate a parabola for it to be no longer the graph of a function?</p>
</blockquote>
<p>I have no problem with narrowing the question down by only concerning the standard parabola: <span class="math-container">$$f(x)=x^2.$$</span></p>
<p>I am looking for a specific angle measure. One such measure must exist as the reflection of <span class="math-container">$f$</span> over the line <span class="math-container">$y=x$</span> is certainly no longer well-defined. I realize that preferentially I should ask the question on this site with a bit of work put into it but, alas, I have no intuition for where to start. I suppose I know immediately that it must be less than <span class="math-container">$45^\circ$</span> as such a rotation will cross the y-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,\sqrt{2})$</span>.</p>
<p>Any insight on how to proceed?</p>
| Narasimham | 95,860 | <p>A rigid parabolic arc <em>remains a parabola</em> <strong>for any arbitrary displacement or rotation</strong> imparted to the parabolic arc.</p>
<p>That is why we can describe any parabola in the plane by an intrinsic or natural equation</p>
<p><span class="math-container">$$ \kappa = \dfrac{\cos^3\phi}{2f} $$</span></p>
<p>where <span class="math-container">$\kappa$</span> is curvature and <span class="math-container">$ \phi $</span> the rotation of tangent reckoned from its vertex V and <span class="math-container">$f= VF, $</span> its focal length.</p>
<p><a href="https://i.stack.imgur.com/znyOj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/znyOj.png" alt="enter image description here" /></a></p>
<p>In rectangular coordinates with any constants <span class="math-container">$(a,b,c,d)$</span> we have two connected arcs of the same parabola separated by a tangent parallel to x-axis
(not shown):</p>
<p><span class="math-container">$$ y = ax+b \pm \sqrt{cx+d}.$$</span></p>
|
4,492,566 | <blockquote>
<p>To which degree must I rotate a parabola for it to be no longer the graph of a function?</p>
</blockquote>
<p>I have no problem with narrowing the question down by only concerning the standard parabola: <span class="math-container">$$f(x)=x^2.$$</span></p>
<p>I am looking for a specific angle measure. One such measure must exist as the reflection of <span class="math-container">$f$</span> over the line <span class="math-container">$y=x$</span> is certainly no longer well-defined. I realize that preferentially I should ask the question on this site with a bit of work put into it but, alas, I have no intuition for where to start. I suppose I know immediately that it must be less than <span class="math-container">$45^\circ$</span> as such a rotation will cross the y-axis at <span class="math-container">$(0,0)$</span> and <span class="math-container">$(0,\sqrt{2})$</span>.</p>
<p>Any insight on how to proceed?</p>
| Blue | 409 | <p>A little more generally ... Consider a <em>non-degenerate</em> (and, for the sake of this discussion, non-circular) conic with eccentricity <span class="math-container">$e>0$</span>, focus <span class="math-container">$(f_x,f_y)$</span>, focus-to-directrix length <span class="math-container">$d>0$</span>, and major/transverse axis direction vector <span class="math-container">$(\cos\theta,\sin\theta)$</span>. By the focus-directrix definition of a conic, the curve has this equation:
<span class="math-container">$$\sqrt{(x-f_x)^2+(y-f_y)^2}=e \left(\; (x-f_x)\cos\theta+(y-f_x)\sin\theta-d \;\right) \tag1$$</span>
Squaring and gathering terms gives us
<span class="math-container">$$x^2 (1-e^2\cos^2\theta) - 2 x y e^2\cos\theta\sin\theta + y^2 (1-e^2\sin^2\theta) + \cdots = 0 \tag2$$</span>
As a quadratic in <span class="math-container">$y$</span>, the equation cannot represent a function of <span class="math-container">$x$</span>. (Conceivably, the equation can give double-roots for every <span class="math-container">$x$</span>, but this is clearly a degenerate case, which we're ignoring. (But see the Note below.)) Therefore, to have <span class="math-container">$y$</span> as a function of <span class="math-container">$x$</span>, we need to eliminate the equation's quadratic-ness, which we do by making the coefficient of <span class="math-container">$y^2$</span> vanish.
<span class="math-container">$$\sin\theta=\pm\frac1e \tag{$\star$}$$</span>
We can conclude ...</p>
<ul>
<li>For ellipses (<span class="math-container">$e<1$</span>), equation <span class="math-container">$(\star)$</span> has no solutions, confirming the fact that no (non-degenerate) ellipse is the graph of a function.</li>
<li>For parabolas (<span class="math-container">$e=1$</span>), equation <span class="math-container">$(\star)$</span> has the solutions <span class="math-container">$\theta=\pm90^\circ$</span>. Thus, a (non-degenerate) parabola must have a vertical axis to be the graph of a function. (This answers OP's question.)</li>
<li>For hyperbolas (<span class="math-container">$e>1$</span>), equation <span class="math-container">$(\star)$</span> has exactly two solutions. We can interpret this as saying that any (non-degenerate) hyperbola has exactly two orientations that make it the graph of a function; the reader can check that these orientations correspond to making one or the other asymptote vertical.</li>
</ul>
<hr />
<p><strong>Note.</strong> Equation <span class="math-container">$(2)$</span> gives double-root <span class="math-container">$y$</span> value when its <span class="math-container">$y$</span>-discriminant vanishes; that is,
<span class="math-container">$$x^2 (1 - e^2)
- 2 x ( f_x (1 - e^2) - e^2 d \cos\theta)
+ f_x^2 (1 - e^2)
- 2 \cos\theta d e^2 f_x
- d^2 e^2
= 0$$</span>
For a polynomial in <span class="math-container">$x$</span> to vanish <em>across a range</em> of <span class="math-container">$x$</span>-values, each coefficient must vanish. Thus, <span class="math-container">$1-e^2=0$</span>, which implies <span class="math-container">$d\cos\theta=0$</span>, and then also <span class="math-container">$d=0$</span>. So, this situation applies only to a parabola with <span class="math-container">$d=0$</span>, for which equation <span class="math-container">$(2)$</span> factors as
<span class="math-container">$$\left(\;(x-f_x)\sin\theta - (y-f_y)\cos\theta\;\right)^2 = 0$$</span>
making the parabola (a portion of) a "double-line" coinciding with its axis. (The "portion" consists of the focus and the points on the appropriate side of the focus to satisfy <span class="math-container">$(1)$</span>.) In this case, the graph is a function <em>except</em> when that axis is vertical.</p>
|
2,880,566 | <p>In an optimization problem I finally get to the point where I have to solve</p>
<p>$$x +\sec(x)(\tan(x)\cos(2x)+\tan(x)-2\sin(2x)) =0$$</p>
<p>which obviously leads to</p>
<p>$$x=-\sec(x)(\tan(x)\cos(2x)+\tan(x)-2\sin(2x))$$</p>
<p>Nevertheless, this couldn't in any case help knowing the optimal size of the angle $x$. Wolfram Alpha can give me the solutions, but is not able to give me a step-by-step explanation of how to reach an approximate value for $x$. </p>
<p>Then, how do we solve manually such an equation?</p>
| mfl | 148,513 | <p>We have that $\cos 2x=\cos^2x-\sin^2x=2\cos^2x-1,\sin 2x=2\sin x\cos x.$ Now</p>
<p>$$\tan(x)\cos(2x)+\tan(x)= \tan x(\cos 2x+1)=2\tan x\cos^2x=2\sin x\cos x.$$</p>
<p>Thus</p>
<p>$$\tan(x)\cos(2x)+\tan(x)-2\sin(2x)=-2\sin x\cos x.$$</p>
<p>So</p>
<p>$$\sec(x)(\tan(x)\cos(2x)+\tan(x)-2\sin(2x))=-2\sin x.$$</p>
<p>So, you equation is just $x=2\sin x.$</p>
<p>It is easy to see that $x=0$ is a solution. The other solutions are not so easy to get manually.</p>
|
48,864 | <p>I can't resist asking this companion question to the <a href="https://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking"> one of Gowers</a>. There, Tim Dokchitser suggested the idea of Grothendieck topologies as a fundamentally new insight. But Gowers' original motivation is to probe the boundary between a human's way of thinking and that of a computer. I argued, therefore, that Grothendieck topologies might be more natural to computers, in some sense, than to humans. It seems Grothendieck always encouraged people to think of an object in terms of the category that surrounds it, rather than its internal structure. That is, even the most lovable mathematical structure might be represented simply as a symbol $A$, and its special properties encoded in arrows $A\rightarrow B$ and $C\rightarrow A$, that is, a grand combinatorial network. I'm tempted to say that the idea of a Grothendieck topology is something of an obvious corollary of this framework. It's not something I've devoted much thought to, but it seems this is exactly the kind of reasoning more agreeable to a computer than to a woolly, touchy-feelly thinker like me.</p>
<p>So the actual question is, what other mathematical insights do you know that might come more naturally to a computer than to a human? I won't try here to define computers and humans, for lack of competence. I don't think having a deep knowledge of computers is really a prerequisite for the question or for an answer. But it would be nice if your examples were connected to substantial mathematics. </p>
<p>I see that this question is subjective (but not argumentative in intent), so if you wish to close it on those grounds, that's fine.</p>
<p>Added, 11 December: Being a faulty human, I had an inexplicable attachment to the past tense. But, being weak-willed on top of it all, I am bowing to peer pressure and changing the title.</p>
| Michael Renardy | 12,120 | <p>Grothendieck seems to be still alive. So should not the question be: Is Grothendieck a computer? (Ask him, good luck!) Or perhaps: How did he morph from a computer to ... whatever it may be?</p>
|
349,309 | <p>I seem to be short on examples for $I$-adic completions of rings.</p>
<p>I know that a ring is $I$-adically complete if the canonical homomorphism into the inverse limit is an isomorphism. My thinking and searching on the internet has been surprisingly fruitless, though, for examples where the map is either surjective and not injective, or injective and not surjective. (Am I mistaken that both are possible?)</p>
<blockquote>
<p>So, the main question here is for one or more useful examples of both of these types of $I$-adically incomplete rings.</p>
</blockquote>
<p>If possible, it would be nice to have the surjective-not-injective example be with respect to an ideal $I$ which contains a nonzero idempotent $e$.</p>
| Andreas Caranti | 58,401 | <p>For injective but not surjective, take the ring $R = \Bbb{Z}$ and the ideal $I = \langle p \rangle$, for a prime $p$, to obtain the $p$-adic integers $\Bbb{Z}_{p}$. </p>
<p>For surjective but not injective, I think you might take the ring $R = \Bbb{Z}_{p} \oplus B$, where $B$ is any Boolean ring, and $I = \langle (p, 0), \{0\} \oplus B \rangle$, the completion here being once more $\Bbb{Z}_{p}$, as $I^{n} = \langle (p^{n}, 0), \{0\} \oplus B \rangle$.</p>
<p><strong>PS</strong> In a previous version I had $\lvert B \rvert = 1$. I expanded $B$ to any Boolean ring after <strong>rschwieb</strong>'s comment below.</p>
|
3,903,774 | <p><span class="math-container">$30$</span> red balls and <span class="math-container">$20$</span> black balls are being distributed to <span class="math-container">$5$</span> kids, so that each kid gets at least one red ball. In how many ways can we distribute balls?</p>
<p>Circle the correct answers:</p>
<p>a) <span class="math-container">$\binom{29}{4}$</span> <span class="math-container">$\binom{24}{20}$</span></p>
<p>b) <span class="math-container">$\binom{29}{5}$</span> <span class="math-container">$\binom{24}{5}$</span></p>
<p>c)<span class="math-container">$|Sur(N_{30},N_{5})|S(20,5)$</span> Note: <span class="math-container">$|Sur(N_{30},N_{5})|$</span> is the number of surjections, and <span class="math-container">$S(20,5)$</span> is a Stirling number of the second kind</p>
<p>d) None of <span class="math-container">$3$</span> previous answers are correct</p>
<p>My approach:</p>
<p>First I gave each of <span class="math-container">$5$</span> kids one red ball, which leaves me with <span class="math-container">$25$</span> red balls. Now I used the stars and bars method to distribute the balls I am left with.</p>
<p>Red balls: <span class="math-container">$x_1+x_2+x_3+x_4+x_5=25,x_i\geq 0, i=1,..,5$</span>. This equation has <span class="math-container">$\binom{5+25-1}{25}=\binom{29}{25}=\binom{29}{4}$</span>.</p>
<p>Black balls: <span class="math-container">$x_1+x_2+x_3+x_4+x_5=20,x_i\geq 0, i=1,..,5$</span>. This equation has <span class="math-container">$\binom{5+20-1}{20}=\binom{24}{20}=\binom{24}{4}$</span>.</p>
<p>So I would say a) is the correct answer.. am I right?</p>
| PierreCarre | 639,238 | <p>The subscript is relative to the norm being used. The "2" stands for the euclidean norm. The superscript is a power.
<span class="math-container">$$
\|X\|_2^2 = \left(\left(\sum_{i=1}^n X_i^2 \right)^{1/2}\right)^2 = \sum_{i=1}^n X_i^2
$$</span></p>
<p>Other examples of norms could be</p>
<p><span class="math-container">$$
\|X\|_1 = \sum_{i=1}^n |X_i|, \quad \|X\|_{\infty}= \max_{i=1,\cdots, n}|X_i|.
$$</span></p>
|
735,015 | <p>I don't have to write a proof I just have to show that $(^{n}_{k})(^k_m)=(^n_m)(^{n-m}_{k-m})$
But I am struggling to expand this.
$$\frac{n!}{k!(n-k)!}*\frac{k!}{m!(k-m)!}$$</p>
<p>Once I get it to this point I am having no luck multiplying it out to make it look like the other side should look</p>
| Thanos Darkadakis | 105,049 | <p>Just use the expansion of $(^{n}_{k})$ and simplify the fractions..</p>
<p>$(^{n}_{k})(^k_m)=(^n_m)(^{n-m}_{k-m})$</p>
<p>$\frac{n!}{k!(n-k)!}\cdot \frac{k!}{m!(k-m)!}=\frac{(n-m)!}{(k-m)!(n-k)!}$</p>
<p>$\frac{n!}{m!}=(n-m)!$, which is true.</p>
|
735,015 | <p>I don't have to write a proof I just have to show that $(^{n}_{k})(^k_m)=(^n_m)(^{n-m}_{k-m})$
But I am struggling to expand this.
$$\frac{n!}{k!(n-k)!}*\frac{k!}{m!(k-m)!}$$</p>
<p>Once I get it to this point I am having no luck multiplying it out to make it look like the other side should look</p>
| Nigel Overmars | 96,700 | <p>LHS: You're forming a committee of $k$ people that can be chosen out of $n$ people, $n \choose k$ combinations. Then, from those $k$ people you choose a board of $m$ people, $k \choose m$ combinations. So in total there are ${n \choose k} {k \choose m}$ combinations.</p>
<p>RHS: Reversed order of the LHS, first you choose a board of $m$ persons out of all ($n$) people. Then, from the $n-m$ people left, you choose the committee of $k-m$ persons.</p>
<p>Since they both count the same things, they must be equal.</p>
|
1,211,978 | <p>I cannot find the roots of the characteristic equation to get a solution. I only know the basic way to solve these equations. I factored out an $r^2$.</p>
<p>$2r^5-7r^4+12r^3-8r^2 = 0$</p>
<p>$r^2(2r^3-7r^2+12r-8) = 0$</p>
| parsiad | 64,601 | <p>Assuming $y^{\left(n\right)}\equiv\frac{\partial^{n}y}{\partial t^{n}}$.
Let $w=y^{\prime\prime}$ so that the equation becomes
$$
2w^{\prime\prime\prime}-7w^{\prime\prime}+12w^{\prime}-8w=0.
$$
The characteristic polynomial, as you pointed out, is
$$
2r^{3}-7r^{2}+12r-8
$$
with imaginary and real roots (solvable in $\mathbb{Q}$). As @Pieter21 points out, you can use Wolfram to get the roots; they are not "nice" numbers. Perhaps you can just use the approximate forms $1.2581$ and $1.210\pm 1.3867i$.</p>
<p>With a solution for $w$, you can derive a solution for $y$ since $y=\int\int w$.</p>
|
2,532,280 | <p>If a N×N (N≥3) Hermitian matrix <strong>A</strong> meets the following conditions: </p>
<ol>
<li><strong>A</strong> is positive semi-definite (not positive definite, i.e. <strong>A</strong> has at least M zero eigenvalue, where M is a given paremeter with 1≤M≤N-1).</li>
<li>The sum of each off diagonal results in 0, and the main diagonal elements are non-negative, which is shown in the figure (set N=4 as example).</li>
</ol>
<p>Then what the general solution of <strong>A</strong> is?</p>
<p>For example, a particular solution of <strong>A</strong> can be $$
\begin{matrix}
I_{M'} & 0 &\\
0 & 0 \\
\end{matrix}
$$ where M≤N-M'≤N-1. It is just a particular solution, I wonder what is the general solution under these two conditions.</p>
<p><a href="https://i.stack.imgur.com/biXfS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/biXfS.png" alt="enter image description here"></a></p>
| Michael Rozenberg | 190,319 | <p>For $$ax^2+bx+c=0,$$ where $a\neq0$ we need
$b=0$ and $\frac{c}{a}<0$.</p>
<p>Thus, $k=-\frac{3}{5}$ and check that $2k^2-1<0$.</p>
|
1,764,106 | <p>In my book I have the following definition for subgroups of a group $G$ generated by $A$, a subset of G:</p>
<p>$$\langle A\rangle=\{x_1^{\epsilon_1}x_2^{\epsilon_2}...x_n^{\epsilon_n}\mid x_i\in A,~\epsilon_i\in\mathbb Z,~ x_i \neq x_{i+1} ,~ n=1,2,3...\}$$</p>
<p>I have no trouble understanding this. But then we define the subgroup of an abelian group $G$ generated by a subset $A$:</p>
<p>$$\langle A\rangle=\{x_1^{\epsilon_1}x_2^{\epsilon_2}...x_k^{\epsilon_k}\mid x_i\in A,~\epsilon_i\in\mathbb Z \text{ for each } i\}$$</p>
<p>I fail to understand how this implies that we can commute the $x_i$'s. Could somebody please explain to me how the commutative property follows from this?</p>
| Christian Gaetz | 75,296 | <p>The commutative property doesn't follow from that definition per se. However since $G$ is abelian and the elements of $A$ multiply according to the multiplication rules for $G$, we get commutativity for free.</p>
|
4,482,707 | <p><a href="https://math.stackexchange.com/questions/942263/really-advanced-techniques-of-integration-definite-or-indefinite/1885401#1885401">Here</a>, I saw the following formula:</p>
<p><span class="math-container">$$\int_{0}^{\infty }\frac{f(t)}{t}dt=\int_{0}^{\infty }\mathcal{L}\left \{ f(t) \right \}ds$$</span></p>
<hr />
<p>Say we have the integrand only <span class="math-container">$f(t)$</span>, not <span class="math-container">$\frac{f(t)}{t}$</span>, then I think we can multiply and divide this integrand by <span class="math-container">$t$</span>, then we apply the formula. For instance, if the integrand is <span class="math-container">$te^{-t}$</span>, then we should rewrite it as</p>
<p><span class="math-container">$$\frac{t^2e^{-t}}{t}$$</span></p>
<p>So <span class="math-container">$f(t)$</span> is actually <span class="math-container">$t^2e^{-t}$</span>, not <span class="math-container">$te^{-t}$</span>.</p>
<p>I know that if the integrand is <span class="math-container">$te^{-t}$</span>, then it can be done by integrating by part. However, I think it is a good looking function to illustrate my question.</p>
<p>Am I right about multiplying and dividing by <span class="math-container">$t$</span> just to make a use of the formula?</p>
<p>In other words, is the following true?</p>
<p><span class="math-container">$$\int_{0}^{\infty }f(t)dt=\int_{0}^{\infty }\mathcal{L}\left \{ tf(t) \right \}ds$$</span></p>
<hr />
<p>Second Question:</p>
<p>Can we, somehow, generalize the formula to any interval <span class="math-container">$(a,b)$</span> instead of <span class="math-container">$(0,\infty)$</span>?</p>
<p>I know that the Laplace will have the same limits of integration, <span class="math-container">$(0,\infty)$</span> but I am asking about the original integral, not the integral from the <span class="math-container">$\mathcal{L}$</span>.</p>
<hr />
<p>Sorry for my bad English, hope my questions are clear. Your help would be appreciated. THANKS!</p>
<hr />
<h2>Edit:</h2>
<p>For the second question, what I mean is</p>
<p>Given <span class="math-container">$a,b,$</span> and <span class="math-container">$h(x)$</span>, and that <span class="math-container">$\int_{a}^{b}h(x)dx = \int_{0}^{\infty}r(t)dt$</span>. How to find <span class="math-container">$r(t)$</span>?</p>
<p>This will make use of the first formula in this post.</p>
<p>For instance, <span class="math-container">$\int_{1}^{2}\sqrt{4x(2-x)}dx$</span> can be replaced by <span class="math-container">$\int_{0}^{\infty} \frac{dt}{1+t^2}$</span>. And now we can use the above formula.</p>
| j4nd3r53n | 446,918 | <p>As Theo Bendt says in his comment, a relation between sets <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is simply a subset of <span class="math-container">$X \times Y$</span>, that is, if you choose <span class="math-container">$R \subseteq X \times Y$</span>, then you can define a relation, let's call it <span class="math-container">$\lt_R$</span>, by stating that <span class="math-container">$x \lt_R y$</span> for <span class="math-container">$x \in X, y \in Y$</span> if and only if <span class="math-container">$(x,y) \in X \times Y$</span>.</p>
<p>Thus, <span class="math-container">$R_1$</span> does not define a relation, since <span class="math-container">$R_1=\{(1,3),(2,4),(3,5),(4,6),(5,7)\}$</span>, which is not a subset of <span class="math-container">$X \times Y$</span>, as <span class="math-container">$4 \notin Y$</span>.</p>
<p>Applying the same analysis shows that <span class="math-container">$R_2$</span> also doesn't define a relation, but <span class="math-container">$R_3$</span> and <span class="math-container">$R_4$</span> both do.</p>
|
2,414,011 | <p>In my recent works in PDEs, I'm interested in finding a family of cut-off functions satisfying following properties:</p>
<p>For each $\varepsilon >0$, find a function ${\psi _\varepsilon } \in {C^\infty }\left( \mathbb{R} \right)$ which is a non-decreasing function on $\mathbb{R}$ such that:</p>
<ol>
<li>${\psi _\varepsilon }\left( x \right) = \left\{ {\begin{array}{*{20}{l}}
{0 \mbox{ if } x \le \varepsilon ,}\\
{1\mbox{ if } x \ge 2\varepsilon ,}
\end{array}} \right.$ and</li>
<li>The function $x \mapsto x{\psi _\varepsilon }'\left( x \right)$ is bounded uniformly with respect to $\varepsilon$ as $\varepsilon \to 0$.</li>
</ol>
<p>The main problem here is ${\psi _\varepsilon }'\left( x \right) \to \infty $ for some $x \in \left( {\varepsilon ,2\varepsilon } \right)$ as $\varepsilon \to 0$. I also start with <a href="https://en.wikipedia.org/wiki/Non-analytic_smooth_function" rel="nofollow noreferrer">this function</a> to define explicitly ${\psi _\varepsilon }$ in the interval $\left( {\varepsilon ,2\varepsilon } \right)$ but my attempts to adjust the referenced function failed. </p>
<p>Can you find an example of these cut-off functions?</p>
<p>Thanks in advanced.</p>
| Rigel | 11,776 | <p>You can start with the piecewise affine Lipschitz function
$$
h_\varepsilon(x) :=
\begin{cases}
0, & \text{if}\ |x| \leq 5\varepsilon/4,\\
1, & \text{if}\ |x| \geq 7\varepsilon/4
\end{cases}
$$
(and affine for the remaining values of $x$).
For this function you have $|x h_\varepsilon'(x)| \leq 4$.</p>
<p>Let $\varphi_\rho$ be the standard family of mollifiers.</p>
<p>For every $\rho \in (0, \varepsilon/4)$ the mollified function
$$
\psi_\varepsilon := \varphi_\rho \ast h_\varepsilon
$$
is a $C^\infty$ function satisfying your requirements.</p>
<p>Namely, since $\psi_\varepsilon' = \varphi_\rho \ast h_\varepsilon'$, by the definition of convolution and the fact that $\varphi_\rho\geq 0$ you get
$$
\left| \psi_\varepsilon'(x) \right|
= \left| \int \varphi_\rho(x-y) h_\varepsilon'(y) dy \right|
\leq \int \varphi_\rho(x-y) | h_\varepsilon'(y) | dy
\leq 2 \int \varphi_\rho(x-y) dy = 4.
$$</p>
|
69,948 | <p>Has anyone ever created a "pairing function" (possibly non-injective)
with the property to be nondecreasing wrt to product of arguments, integers n>=2, m>=2. (We can also assume that n and m are bounded by an integer K, if useful) :</p>
<p>n m > n' m' => p(n,m) > p(n',m') </p>
<p>If yes what does it look like, does it have a name ? </p>
<p>-Luna</p>
| Gerhard Paseman | 3,206 | <p>To pull this observation out of the comments, suppose we had a pairing function which was monotonic increasing with respect to products of 2 or more integers each larger than 1, and which had nice inverses, say one of them was F(p) and had a nice formula for it which was quickly computable and returned an integer greater than or equal to 2. If the pairing function did not grow too fast, I could take a large odd number 2n+1, feed 2 and n to the pairing function, and feed 2 and n+ 1 to the pairing function again, and get lower and upper bounds on a range of values to invert with F. If F returns a value, I can test it as a nontrivial factor of my odd number. With some assumptions on how nice the pairing function and its inverses are, I could make a poly-time factoring algorithm.</p>
<p>Because of this result, I would instead believe that such a pairing function and its inverse are not so nice, either being trivial and not giving useful information, or not being quick or easy to compute the inverse.</p>
<p>Gerhard "Email Me About System Design" Paseman, 2011.07.11</p>
|
1,611,730 | <p>I am a linguist, not a mathematician, so I apologize if there's something wrong with my terminology and/or notation.</p>
<p>I have two structures that I want to merge (partially or completely). To generate a list of all possible combinations, I compute the Cartesian product of the two sets of objects, which gives me a set of pairs, and then I compute the [1, ..., <em>n</em>]-fold Cartesian product of my set of pairs with itself where <em>n</em> is the highest cardinality out of the two structures (here, 5).</p>
<p><a href="https://i.stack.imgur.com/WO79P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WO79P.png" alt="Two structures with 5 objects each"></a></p>
<pre><code>A = (a1, a2, a3, a4, a5) and B = (b1, b2, b3, b4, b5)
</code></pre>
<p>Basically, I'm generating 1-tuples like ((a1, b1)), ((a1, b2)), ..., ((a5, b5)) that merge one pair of objects, 2-tuples that merge two pairs of objects, etc. up to <em>n</em>-tuples. I end up with $\sum\limits_{i=1}^{n} (\bar{\bar{A}}\times \bar{\bar{B}})^i$ tuples, where $\bar{\bar{A}}$ and $\bar{\bar{B}}$ represent the cardinalities of the two structures.</p>
<p>In this case, I get 10172525 tuples, which is way too high for my needs. I filter my list to only keep tuples if the pairs they contain are all different and cannot be found in another tuple with the same length. This removes up to 99% of the original tuples. For example:</p>
<p>((a1, b1), (a2, b2), (a3, b3))<br>
((a1, b1), (a3, b3), (a2, b2)) has the same pairs as the preceding tuple, but in a different order<br>
((a1, b1), (a1, b1), (a3, b3)) has the pair (a1, b1) more than once</p>
<p>I'm looking for an equation that will help me predict the number of unique tuples. For 1-tuples, there's $\bar{\bar{P}}$ unique tuples where $\bar{\bar{P}}$ is the number of pairs. For 2-tuples, I do $\frac{\bar{\bar{P}}\times (\bar{\bar{P}}-1)}{2}$. I'm sure there's an equation that works for any tuple length, but I can't figure it out.</p>
<p>Here are the numbers of tuples generated vs. unique tuples for two structures with 4 objects each:</p>
<pre><code>+--------+-----------+--------+
| Length | generated | unique |
+--------+-----------+--------+
| 1 | 16 | 16 |
| 2 | 256 | 120 |
| 3 | 4096 | 560 |
| 4 | 65536 | 1820 |
| Total | 69904 | 2516 |
+--------+-----------+--------+
</code></pre>
| JasonRobinson | 333,310 | <p>If you want to fully enumerate all of those combinations (and save it to csv format along with the sets that generated that Cartesian product) and you need to account for variable size sets (set A has 3 distinct members, set B has 5 distinct members, C has 2 distinct members, etc) then I have written an algorithm that enumerates all the resulting combinations. It makes sure no two results contain the same elements (Combinations not Permutations).</p>
<p><a href="https://github.com/jjrobinson/TestingUtilities" rel="nofollow">https://github.com/jjrobinson/TestingUtilities</a></p>
<p>I wrapped my algorithm in a cmd line interface with some text banner prompts to help keep the input organized. The next release of my program will allow ingesting a csv file that describes each set (one set per column, set members in rows) to fully automate the process end to end.</p>
|
209,116 | <p>Well, I just want to know if is there any significance of the term "linear" in the of name "General Linear Group" - for example, $\text{GL}_ n(\mathbb{R})$?</p>
| Alexander Gruber | 12,952 | <p>$GL(V)$ is the group of <em>linear</em> transformations over a vector space $V$. You can also, as you have, write it $GL_n(K)$ if $V$ is an $n$-dimensional vector space over a field $K$, and thus isomorphic to $K^n$.</p>
<p>So, the "linear" part refers to the linearity property of the transformations: given vectors $v,w\in V$, scalars $\alpha,\beta\in K$, and a transformation $T\in GL(V)$, $$T(\alpha v+\beta w)=\alpha T(v) + \beta T(w).$$</p>
|
209,116 | <p>Well, I just want to know if is there any significance of the term "linear" in the of name "General Linear Group" - for example, $\text{GL}_ n(\mathbb{R})$?</p>
| Tobias Kildetoft | 2,538 | <p>The term linear here refers to the fact that it is a group consisting of linear transformations of some vector space.
In some sense, all groups are "linear" like this, but usually if one refers to something as a linear group, then a specific realization as a group of linear transformations is usually (at least implicitly) meant.</p>
|
324,385 | <p>I'm going through Wallace Clarke Boyden's <a href="http://books.google.com/books?id=OhMAAAAAYAAJ&pg=PA71#v=onepage&q&f=false" rel="noreferrer">A First Book in Algebra</a>, and there's a section on finding the square root of a perfect square polynomial, eg. <span class="math-container">$4x^2-12xy+9y^2=(2x-3y)^2$</span>. He describes an algorithm for finding the square root of such a polynomial when it's not immediately apparent, but despite my best efforts, I find the language indecipherable. Can anyone clarify the process he's describing? The example I'm currently wrestling with is <span class="math-container">$x^6-2x^5+5x^4-6x^3+6x^2-4x+1$</span>.</p>
<p>It's a lot of language to parse, but if anyone wants to take a stab at it, here's the original text:</p>
<blockquote>
<p><em>To find the square root of a polynomial, arrange the terms with reference to the powers of some number; take the square root of the first term of the polynomial for the first term of the root, and subtract its square from the polynomial; divide the first term of the remainder by twice the root found for the next term of the root, and add the quotient to the trial divisor; multiply the complete divisor by the second term of the root, and subtract the product from the remainder. If there is still a remainder, consider the root already found as one term, and proceed as before.</em></p>
</blockquote>
<p>I did some hunting online but didn't turn up anything useful. Is it possible this is an outdated method that's been abandoned for something cleaner?</p>
| Will Jagy | 10,400 | <p>It's just saying start with the highest degree and work down. This may not be the fastest but will work or tell you that the thing is not really a square. So, begin with $x^3,$ since the square must be $x^6$ and we get one free choice, $\pm x^3.$ Next, $(x^3 + A x^2)^2 = x^6 + 2 A x^5 + \mbox{more}.$ So $2A = -2, A = -1.$</p>
<p>Alright, $(x^3 -x^2 + B x)^2 = x^6 - 2 x^5 + (2B+1)x^4 + \mbox{more}.$ So $2B+1 = 5$ and $B=2.$ </p>
<p>Finally $(x^3 - x^2 + 2 x + C)^2 = x^6 - 2 x^5 + 5 x^4 +(2C -4)x^3 + \mbox{more}.$ So $2C-4 = -6$ and $C=-1.$</p>
<p>Then check
$$ (x^3 - x^2 + 2 x -1)^2 = x^6 - 2 x^5 +5 x^4 - 6 x^3 + 6 x^2 - 4 x + 1. $$
So it worked.</p>
|
4,565,728 | <p>Given a collection of topological spaces <span class="math-container">$X_i$</span> indexed by the elements <span class="math-container">$i$</span> of a set <span class="math-container">$I$</span>, we consider the set product <span class="math-container">$P = \prod_{i \in I} X_i$</span> with projections <span class="math-container">$p_i : P \to X_i$</span>. There are two methods of topologizing <span class="math-container">$P$</span>.</p>
<ol>
<li><p>Take the <em>product topology</em> having as a subbase all sets of the form <span class="math-container">$p_i^{-1}(U_i)$</span> with <span class="math-container">$i \in I$</span> and <span class="math-container">$U_i \subset X_i$</span> open. Therefore a base for the product topology consists of all products of form <span class="math-container">$\prod_{i \in I} U_i$</span> with <span class="math-container">$U_i \subset X_i$</span> open and <span class="math-container">$U_i \ne X_i$</span> only for finitely many <span class="math-container">$i$</span>.</p>
</li>
<li><p>Take the <em>box topology</em> having as a base all products of form <span class="math-container">$\prod_{i \in I} U_i$</span> with <span class="math-container">$U_i \subset X_i$</span> open.</p>
</li>
</ol>
<p>By definition the product topology is the coarsest topology on <span class="math-container">$P$</span> such that all <span class="math-container">$p_i^{-1}(U_i)$</span>, with open <span class="math-container">$U_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are open. This is equivalent to defining the product topology as the coarsest topology on <span class="math-container">$P$</span> such that all projections <span class="math-container">$p_i$</span> become continuous.</p>
<p>It is standard to endow <span class="math-container">$P$</span> with the product topology. This makes <span class="math-container">$(P,(p_i)_{i \in I})$</span> the product of the objects <span class="math-container">$X_i$</span> in the category of topological spaces which is characterized by the following universal property:</p>
<blockquote>
<p>A function <span class="math-container">$f : Y \to P$</span> defined on a topological space <span class="math-container">$Y$</span> is continuous if and only if all <span class="math-container">$p_i \circ f : Y \to X_i$</span> are continuous.</p>
</blockquote>
<p>This seems to be the optimum what can be expected from a topology on <span class="math-container">$P$</span>.</p>
<p>On infinite products the box topology is in general strictly finer than the product topology, thus it does in general not have this universal property. As far as I know, the box topology does not have <em>any really nice universal property</em>; see <a href="https://math.stackexchange.com/q/3095816">Does the box topology have a universal property?</a></p>
<p>On the other hand, the box topology is the coarsest topology such that all products of the form <span class="math-container">$\prod_{i \in I} U_i$</span> with open <span class="math-container">$U_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are open. On the level of sets this seems to be a very natural requirement, even more natural than requiring that all special products of the form <span class="math-container">$\prod_{i \in I} U_i$</span> with open <span class="math-container">$U_i \subset X_i$</span> open and <span class="math-container">$U_i \ne X_i$</span> only for finitely many <span class="math-container">$i$</span>, are open.</p>
<p><strong>Question:</strong> Are there other characterizations of product or box topology than those described above? If so, do they occur somewhere in the literature?</p>
| Paul Frost | 349,785 | <p>Here is a fairly obvious alternative characterization of the product topology:</p>
<blockquote>
<p>The product topology is the coarsest topology on <span class="math-container">$P$</span> such that all products of the form <span class="math-container">$\prod_{i \in I} A_i$</span> with closed <span class="math-container">$A_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are closed.</p>
</blockquote>
<p>This resembles the characterization of the box topology as the coarsest topology on <span class="math-container">$P$</span> such that all products of the form <span class="math-container">$\prod_{i \in I} U_i$</span> with open <span class="math-container">$U_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are open; it is also a very natural requirement.</p>
<p>Note that the products of the form <span class="math-container">$\prod_{i \in I} A_i$</span> with closed <span class="math-container">$A_i \subset X_i$</span> are also closed in the box topology since it is finer than the product topology.</p>
<p>Let us prove the above characterization.</p>
<p>We know that the product topology is the coarsest topology on <span class="math-container">$P$</span> such that all <span class="math-container">$p_i$</span> become continuous. The continuity of all <span class="math-container">$p_i$</span> is equivalent to</p>
<ul>
<li>All <span class="math-container">$p_i^{-1}(A_i)$</span> with closed <span class="math-container">$A_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are closed.</li>
</ul>
<p>But this is equivelent to</p>
<ul>
<li>All products of the form <span class="math-container">$\prod_{i \in I} A_i$</span> with closed <span class="math-container">$A_i \subset X_i$</span>, <span class="math-container">$i \in I$</span>, are closed.</li>
</ul>
<p>In fact, <span class="math-container">$p_i^{-1}(A_i)$</span> is a special case of such a product (take <span class="math-container">$A_j = X_j$</span> for <span class="math-container">$j \ne i$</span>). Conversely, <span class="math-container">$\prod_{i \in I} A_i = \bigcap_{i \in I} p_i^{-1}(A_i)$</span> which is closed if all <span class="math-container">$p_i^{-1}(A_i)$</span> are closed.</p>
|
1,040,136 | <p>Just a quick question:</p>
<p>Is the size of the set of real numbers from 1 to 2 greater, or equal in size to the number of real numbers between 1 and 10?</p>
<p>I'm a Physicist so I'm not totally clued up on Mathematical jargon pertaining to set theory...</p>
| Adriano | 76,987 | <p>They're equal. Indeed, it's not too hard to show that the function $f\colon [1, 2] \to [1, 10]$ defined by:
$$
f(x) = 9x - 8
$$
is a bijection (which shows that the two sets have equal cardinality).</p>
|
1,555,548 | <p>There are $8$ people and they want to sit in a bus which has $2$ single front seats and $4$ sets of $3$ seats with $1$ person that is always the designated driver. How many ways are there for the people to sit in the bus?</p>
<p>I solved it by using:</p>
<p>$6!*(\binom{9}{3}) - 4((6*5*4*3)*2(\binom{4}{2})+(6*5*4*3*2)(\binom{3}{2}) + 6!(\binom{2}{2})) + 7!*(\binom{10}{3})-4((7*6*5*4)*3!*(\binom{5}{2})+(7*6*5*4*3)*2!*(\binom{4}{2})+(7*6*5*4*3*2)(\binom{3}{2}) + 7!(\binom{2}{2}) = \boxed{233280}$</p>
<p>I did complementary counting and took out the cases where there were more than $3$ people in a set of rows. Can anyone tell me if my answer is right?</p>
| fleablood | 280,126 | <p>If it's the same person driving then the answer is $14!/6!$ or 14*13*... 7. There are 14 choices for the first person, 13 for the second, and so on. </p>
<p>The arrangements of the seats and the driver are utterly irrelevant.</p>
<p>====</p>
<p>Why isn't it just $8*14!/7!$? One of the eight people has to drive and the remaining 7 have to sit in the fourteen seats. </p>
<p>You can't move the seats around so specifying that there are 2 front seats and 4 sets or 3 is of no importance any more than indicating what color they are or whether they have gum stuck under them.</p>
|
2,774,923 | <blockquote>
<p>$ABC$ is a triangle where $AE$ and $EB$ are angle bisectors, $|EC| = 5$, $|DE| = 3$, $|AB| = 9$. Find the perimeter of the triangle $ABC$.
<a href="https://i.stack.imgur.com/4nsiM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4nsiM.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>I realized that the length of the side $|DC| = 8$. In the $\triangle BEC$, we have special triangle $3-4-5$. This is where I'm stuck. Can I take your thinkings? </p>
| g.kov | 122,782 | <p><a href="https://i.stack.imgur.com/NtslB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NtslB.png" alt="enter image description here"></a></p>
<p>$|DE|=3=u$,
$|EC|=5=v$,
$|AB|=9=c$.</p>
<p>The question is not that innocent, as it looks at first glance.</p>
<p>Yes, following
the angle bisector theorem
we can calculate that
$|BC|+|AC|=15$ and
$|AB|+|BC|+|AC|=24$,
as it was shown in the other answers.</p>
<p>But the question is,
is it possible to construct
a triangle with claimed properties?</p>
<p>Let $r,R$ be the radii
of inscribed and circumscribed circles
of $\triangle ABC$, respectively,
and let $\rho=\tfrac12(|AB|+|BC|+|AC|)=12$.</p>
<p>\begin{align}
\triangle CEG:\quad
r&=v\sin\tfrac\gamma2
,\\
\text{on the other hand, in }\triangle ABC\quad
r&=(\rho-c)\tan\tfrac\gamma2
,\\
\text{hence }\quad
\cos\tfrac\gamma2&=
\frac{\rho-c}{v}=\frac35
,\\
\sin\tfrac\gamma2&=\frac45
,\\
r&=4
,\\
\sin\gamma&=2\cdot\frac35\cdot\frac45=\frac{24}{25}
,\\
R&=\frac{c}{2\sin\gamma}
=\frac{9}{2\cdot\tfrac{24}{25}}
=\frac{75}{16}
.
\end{align}</p>
<p>So far, so good. But, the validity test:
\begin{align}
\frac{r}{R}
&=\frac{4\cdot16}{75}
=\frac{64}{75}\approx 0.85
\end{align}</p>
<p>shows that this ratio
is bigger than the known maximum for triangles,
$\max\frac{r}{R}=0.5$.</p>
<p>We can actually calculate
the sided of this "triangle",
given that the known $\rho,r$ and $R$
uniquely define the sides
of the triangle as the roots of cubic polynomial</p>
<p>\begin{align}
x^3-2\rho x^2+(\rho^2+r^2+4r\,R)x-4r\,\rho\,R&=0
,\\
x^3-24x^2+235x-900 &=0
,\\
x_1&=9
,\\
x_{2,3}&=
\tfrac{15}2\pm\tfrac{5\sqrt7}2\cdot i
.
\end{align}</p>
<p>Indeed, the sum of the two other "sides" is $15$,
but they are not real.</p>
<p>Thus, the answer is: <strong>a triangle with claimed properties does not exist.</strong></p>
<p><strong>Edit</strong></p>
<p>In the following illustration, $|AB|=9$,
and the locus of points $C$ is the ellipse with
foci $A,B$ and $|CA|+|CB|=15$.
The green line is the locus of points $D\in AB:\ |CD|=8$,
the red curve is the locus of points
$E\in CD:\ |CE|=5,\,|ED|=3$,
and the blue curve is the locus of
the centers $I$ of inscribed circles of
corresponding triangles $ABC$.</p>
<p>The blue and red curves does not intersect,
and $E=I$ never holds.</p>
<p><a href="https://i.stack.imgur.com/MbHXQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MbHXQ.png" alt="enter image description here"></a></p>
|
1,548,771 | <p>I have come up with the following constrained minimization problem:
\begin{eqnarray}
\min\ \sum_{i=1}^\infty x_i^2\\
\sum_{i=1}^\infty a_ix_i=1
\end{eqnarray}
If it were a finite-dimensional case it would be easily solved via Lagrange multipliers; in this case I ask your help since I don't know where to begin.</p>
| Kay K. | 292,333 | <p>Assuming the Cauchy's inequality still holds for infinite series,
$$\left(\sum_{i=1}^{\infty}x_i^2\right)\left(\sum_{i=1}^{\infty}a_i^2\right)\geq \sum_{i=1}^{\infty}x_ia_i=1$$
$$\sum_{i=1}^{\infty}x_i^2\geq \frac{1}{\sum_{i=1}^{\infty}a_i^2}$$</p>
|
3,193,696 | <p>Could someone explain what are (at least the four first) moments ? (normalized moment to be more precise) Let <span class="math-container">$X$</span> a r.v. </p>
<ul>
<li><p>So the first moment is the expectation. This will correspond to <span class="math-container">$\mathbb E[X]$</span> and is going to be the "barycenter" (in the sense that if <span class="math-container">$X$</span> is supported on <span class="math-container">$[a,b]$</span> and <span class="math-container">$\rho(t)$</span> is the density of the line <span class="math-container">$[a,b]$</span> (i.e. <span class="math-container">$\rho(dt)=\mathbb P_X(dt)$</span>), then the barycenter of the line is <span class="math-container">$\mathbb E[X]$</span> (I like to see this in physic term).</p></li>
<li><p>So the second moment is the variance. So, it's going to represent the dispersion of the mass around <span class="math-container">$\mathbb E[X]$</span>, i.e. <span class="math-container">$66\%$</span> of the total mass is going to be concentrate on <span class="math-container">$[\mathbb E[X]-\sqrt{Var(X)},\mathbb E[X]+\sqrt{Var(X)}]$</span>, but I don't really understand why... </p></li>
</ul>
<p>Is my intuition for <span class="math-container">$\mathbb E[X]$</span> and <span class="math-container">$\text{Var}(X)$</span> correct ? What about the the third and the fourth moments ? (I think of higher dimension it's much more abstract).</p>
| BigBendRegion | 472,987 | <p>Actually, the geometric interpretation of kurtosis is quite simple: Following J.G.'s notation, let <span class="math-container">$V = Z^4$</span>, and let <span class="math-container">$p_V(v)$</span> denote the pdf of <span class="math-container">$V$</span>. Then the kurtosis of <span class="math-container">$X$</span> is the center of mass (point of balance) of <span class="math-container">$p_V(v)$</span>. </p>
<p>Now, place a fulcrum on the horizontal axis of the graph of <span class="math-container">$p_V(v)$</span> at 3.0: If <span class="math-container">$p_V(v)$</span> falls to the right, then kurtosis of <span class="math-container">$X$</span> is >3, and the distribution of <span class="math-container">$X$</span> is heavier-tailed than the normal distribution. If it falls to the left, then the kurtosis is <3, and the distribution of <span class="math-container">$X$</span> is lighter-tailed than the normal distribution. </p>
<p>Yes, "flatness" as a descriptor of kurtosis is quite silly - kurtosis has nothing to do with that. Just look at the beta(.5,1) distribution: It has kurtosis less than 3.0, and hence is classified as "platykurtic." But it is nowhere near flat-topped. In fact, it is infinitely peaked. </p>
<p>Nor does large kurtosis have anything to do with "peakedness," or "concentration toward the mean." Think about it: With kurtosis > 3, why does the pdf of <span class="math-container">$Z^4$</span> (<span class="math-container">$p_V(v)$</span>) fall to the <em>right</em> when the fulcrum is placed at 3.0: Is it because of the "peakedness," or concentration near the mean (which translates to large mass where <span class="math-container">$X$</span> is near to <span class="math-container">$\mu$</span>, and hence <span class="math-container">$Z^4$</span> is near <span class="math-container">$0$</span>), or is it due to tail extremity (where <span class="math-container">$X$</span> is far from <span class="math-container">$\mu$</span>, and hence <span class="math-container">$Z^4$</span> is large)? The answer is obviously that <span class="math-container">$p_V(v)$</span> falls to the right due to tail extremity.</p>
<p>You can make a similar geometric argument for skewness: Consider the pdf of <span class="math-container">$Z^3$</span>: Its center of mass is the skewness. Place a fulcrum at <span class="math-container">$0.0$</span>: If the pdf of <span class="math-container">$Z^3$</span> falls to the right, then the distribution of <span class="math-container">$X$</span> is positively skewed, meaning that the right tail of the distribution of <span class="math-container">$X$</span> is heavier (as measured by combined leverage of cubed deviations from the mean) than the left tail of the distribution of <span class="math-container">$X$</span>. The converse description applies for negative skewness.</p>
|
52,841 | <p>In classical Mechanics, momentum and position can be paired together to form a symplectic manifold. If you have the simple harmonic oscillator with energy $H = (k/2)x^2 + (m/2)\dot{x}^2$. In this case, the orbits are ellipses. How is the vector field determined by the (symplectic) gradient, then? </p>
<p>Also, does anyone know an interpretation for the area inside a closed curve in phase space?</p>
| Yakov Shlapentokh-Rothman | 4,345 | <p>For one interpretation of the area inside a curve in phase space, see Arnold's <em>Mathematical Methods of Classical Mechanics</em> page 20. In case you do not have a copy of the book, he defines a function $S: (E_0 - \epsilon, E_0 + \epsilon) \to \mathbb{R}$ which gives the area inside the curve associated to an energy level $E$ (assuming this is well defined). The problem on page 20 asserts that $T = \frac{dS}{dE}$ where $T$ is the period of motion along the curve.</p>
<p>Note: The relevant page is available on google books.</p>
|
4,172,964 | <p>EDIT: Agreed, this isn't a well formed question. But responses below have at least given me a different way to think about it.</p>
<p>EDIT2: Thanks the answers I discovered the Google S2 library (<a href="http://s2geometry.io/devguide/s2cell_hierarchy" rel="nofollow noreferrer">http://s2geometry.io/devguide/s2cell_hierarchy</a>) which provides methods for tracing a space filling curve on a sphere. I'm working this into my camera position algorithm instead of just gimballing through altitude/azimuth. Thanks!</p>
<p>First post here because I'm not a math guy, but I have a feeling the confusion I have is due to something mathy. Perhaps this isn't even the right forum; if not, I'll delete.</p>
<p>Here's the problem:</p>
<p>I built an apparatus to rotate a sensor through a 360 circle and measure incident light at each point. It's pretty cool, I can automatically orient a sensor on the end of arm to maximize incident light (looks like a radar antenna) using a stepper.</p>
<p>I tried extending this to a sphere, but what I found was: I need to discretize the the second angle to some minimum delta-phi. It seems odd to me that I can sweep through an infinite number of degrees in a circle, continuously as my theta angle, but then I need to choose a delta-phi to rotate that circle through to scan the sphere surface.</p>
<p>How is it that I can rotate through an infinite number of points on a circle in one rotation in finite time, but cannot cover the infinite number of circles that comprise the surface of a sphere in finite time because I can always increment a smaller and smaller phi angle...</p>
<p>Or maybe the question is: how is it that an rotating circle traces out a sphere surface: aren't there always going to be more points "covered" by the circle at the poles, compared to the equator, since the distance covered by an infinitesimal delta-phi is short at the poles than the equator?</p>
<p>Obviously I ultimately have to discretize everything since I'm using a digital circuit, but this sudden appearance if finite/infinite stopped me in my tracks...</p>
<p>I'm not even sure what I'm asking, just puzzled. Anyone know what I'm up against here?</p>
| Harsh joshi | 940,974 | <p>This problem can be solved by simple geometry.</p>
<p>Let <span class="math-container">$z^4=w$</span></p>
<p>Using <span class="math-container">$|z^4|=|z|^4$</span>, as <span class="math-container">$|z|=1$</span> then <span class="math-container">$|w|=1$</span>.</p>
<p>Now <span class="math-container">$w$</span> satisfies two properties</p>
<ul>
<li><p><span class="math-container">$|w+4|=\sqrt{5}$</span> and <span class="math-container">$|w|=1$</span>, plotting in Argand plane we get that former is a circle with centre at <span class="math-container">$(-4,0)$</span> and radius <span class="math-container">$\sqrt{5}$</span></p>
</li>
<li><p>while the latter is a circle with centre at <span class="math-container">$(0,0)$</span> and radius <span class="math-container">$1$</span>.</p>
</li>
</ul>
<p>Now the solutions exist on the Argand plane where these two circles intersect and it can be clearly observed that they do not intersect at all .
Hence no solution exists for the complex equations.</p>
|
85,841 | <p><a href="http://reference.wolfram.com/language/ref/Binomial.html" rel="nofollow"><code>Binomial[n, k]</code></a> is converted to a polynomial only for <code>k</code> less than 6.</p>
<pre><code>Table[Binomial[n, k], {k, 1, 8}]
(* {n,
1/2 (-1 + n) n,
1/6 (-2 + n) (-1 + n) n,
1/24 (-3 + n) (-2 + n) (-1 + n) n,
1/120 (-4 + n) (-3 + n) (-2 + n) (-1 + n) n,
Binomial[n, 6],
Binomial[n, 7],
Binomial[n, 8]} *)
</code></pre>
<p>For <code>k</code> greater or equal 6 is an output only copy of <a href="http://reference.wolfram.com/language/ref/Binomial.html" rel="nofollow"><code>Binomial[n, k]</code></a>. Why?
It's possible change this boundary? (but I not accept a solution like <a href="http://reference.wolfram.com/language/ref/Product.html" rel="nofollow"><code>Product[...]</code></a>)</p>
| Vaclav Kotesovec | 29,655 | <p>Solution of my problem is</p>
<pre><code>FunctionExpand[Table[Binomial[n, k], {k, 1, 8}]]
</code></pre>
|
8,699 | <p>I love your site.... but the your question does not meet our quality standards thing is really annoying... I have wasted lots of time trying to figure out what this message means.....maybe someone could explain it to me.....whats wrong with this question:</p>
<p>Find numbers a and b such that: </p>
<p>$ lim =((sqrt(ax+b)-2)/(x))=1$<br>
$x->0$ </p>
<p>I don't quite understand what I a supposed to do with this?</p>
<p>Any help would be appreciated......</p>
| zyx | 14,120 | <p>As this is turning into a kind of case study of how the meta operates, I did a tiny bit of research that merits separate posting.</p>
<p>The source of this question shows that the OP did use TeX/MathJax dollar signs for the formulas, and tried to render the $x \to 0$ below the limit by placing it on the next line. The syntax is imperfect, but reveals an obvious effort to use MathJax, probably based on the documentation in the FAQ or the meta.</p>
<p>Whether any effort was made to comply with MSE practices has no effect on the actions of the filter, but the attempt to use correct math formatting indicates that even the more innocuous suggestions made in the answers, about using LaTeX, were not dealing with the real question and the real person. The posted answers were derived from assumptions that lead to soapbox speeches of "helpful hints". In reality, the OP was aware that mathematical typesetting exists and tried to use it in accordance with the recommendations on the site.</p>
<p>In addition, there was a supportive answer (from MJD?) that was immediately downvoted below $0$ and deleted. I interpret the voting as herd pressure and (as argued before) part of the general pattern of humans rationalizing the actions of Stackexchange computers. I do not think it healthy for MSE if users over-identify with the SE network or policies. One can participate and/or support the policies, without identifying.</p>
|
1,879,129 | <p>If $0 < y < 1$ and $-1 < x<1$, then prove that $$\left|\frac{x(1-y)}{1+yx}\right| < 1$$</p>
| lab bhattacharjee | 33,337 | <p>This will hold true iff $$x^2(1-y)^2<(1+xy)^2$$</p>
<p>$$\iff0>x^2-2x^2y-2xy-1=(x+1)(x-1-2xy)$$</p>
<p>$$\iff0>x-1-2xy\iff1>x(1-2y)$$</p>
<p>WLOG let $y=\sin^2A,x=\cos B$</p>
<p>$x(1-2y)=\cos B\cos2A$ which is $<1$</p>
<p>OR</p>
<p>as $0<y<1\iff0>-2y>-2\iff1>1-2y>-1$ and we have $-1<x<1$</p>
|
2,410,517 | <p>I feel like I'm missing something very simple here, but I'm confused at how Rudin proved Theorem 2.27 c:</p>
<p>If <span class="math-container">$X$</span> is a metric space and <span class="math-container">$E\subset X$</span>, then <span class="math-container">$\overline{E}\subset F$</span> for every closed set <span class="math-container">$F\subset X$</span> such that <span class="math-container">$E\subset F$</span>. Note: <span class="math-container">$\overline{E}$</span> denotes the closure of <span class="math-container">$E$</span>; in other words, <span class="math-container">$\overline{E} = E \cup E'$</span>, where <span class="math-container">$E'$</span> is the set of limit points of <span class="math-container">$E$</span>.</p>
<p>Proof: If <span class="math-container">$F$</span> is closed and <span class="math-container">$F \supset E$</span>, then <span class="math-container">$F\supset F'$</span>, hence <span class="math-container">$F\supset E'$</span>. Thus <span class="math-container">$F \supset \overline{E}$</span>.</p>
<p>What I'm confused about is how we know <span class="math-container">$F \supset E'$</span> from the previous facts?</p>
| fleablood | 280,126 | <p>If <span class="math-container">$A \subset B$</span> then <span class="math-container">$A' \subset B'$</span>.</p>
<p>Pf: If <span class="math-container">$a \in A'$</span> then <span class="math-container">$a$</span> is a limit point of <span class="math-container">$A$</span>. So every neighborhood of <span class="math-container">$a$</span> contains a point <span class="math-container">$b \in A$</span> with <span class="math-container">$b \ne a$</span>. But if <span class="math-container">$b \in A$</span> then <span class="math-container">$b \in B$</span> as <span class="math-container">$A \subset B$</span>. So every neighborhood of <span class="math-container">$a$</span> a conains a point <span class="math-container">$b \in B$</span> with <span class="math-container">$b \ne a$</span>. So <span class="math-container">$a$</span> is a limit point of <span class="math-container">$B$</span>. And <span class="math-container">$a \in B'$</span>. </p>
<p>And <span class="math-container">$A' \subset B'$</span>.</p>
<p>====</p>
<p>So if <span class="math-container">$F \supset F'$</span>, and <span class="math-container">$F\supset E$</span>. Then <span class="math-container">$F' \supset E'$</span> and <span class="math-container">$F \supset F' \supset E'$</span>.</p>
|
2,410,517 | <p>I feel like I'm missing something very simple here, but I'm confused at how Rudin proved Theorem 2.27 c:</p>
<p>If <span class="math-container">$X$</span> is a metric space and <span class="math-container">$E\subset X$</span>, then <span class="math-container">$\overline{E}\subset F$</span> for every closed set <span class="math-container">$F\subset X$</span> such that <span class="math-container">$E\subset F$</span>. Note: <span class="math-container">$\overline{E}$</span> denotes the closure of <span class="math-container">$E$</span>; in other words, <span class="math-container">$\overline{E} = E \cup E'$</span>, where <span class="math-container">$E'$</span> is the set of limit points of <span class="math-container">$E$</span>.</p>
<p>Proof: If <span class="math-container">$F$</span> is closed and <span class="math-container">$F \supset E$</span>, then <span class="math-container">$F\supset F'$</span>, hence <span class="math-container">$F\supset E'$</span>. Thus <span class="math-container">$F \supset \overline{E}$</span>.</p>
<p>What I'm confused about is how we know <span class="math-container">$F \supset E'$</span> from the previous facts?</p>
| Community | -1 | <p>$F\supset F'$ because $F $ is closed. $F'\supset E'$ because $F\supset E $, by assumption. Therefore $F\supset E' $.</p>
|
1,611,390 | <p>How to show that the following function is an injective function?</p>
<p>$ \varphi : \mathbb{N}\times \mathbb{N} \rightarrow \mathbb{N} \\
\varphi(\langle n, k\rangle) = \frac{1}{2}(n+k+1)(n+k)+n$</p>
<p>I'm starting with $ \frac{1}{2}(a+b+1)(a+b)+a = \frac{1}{2}(c+d+1)(c+d)+c$, but how am I supposed to show from this equality that $\langle a, b\rangle = \langle c, d\rangle$, where $\langle a, b\rangle \in \mathbb{N}\times \mathbb{N}$ ?</p>
| Sam Birns | 291,254 | <p>It might be easier to start by letting $(n, k)$ be such that $\varphi( \langle n, k\rangle ) = 0$. Then, $(n + k + 1)(n+k) + 2n = 0$, and since $n, k \in \mathbb{N}$, $(n + k + 1)(n + k), 2n \ge 0$, so $2n = n = 0$, so $k = 0$, and since $(n, k) = (0, 0)$, $\varphi$ is injective. </p>
|
1,611,390 | <p>How to show that the following function is an injective function?</p>
<p>$ \varphi : \mathbb{N}\times \mathbb{N} \rightarrow \mathbb{N} \\
\varphi(\langle n, k\rangle) = \frac{1}{2}(n+k+1)(n+k)+n$</p>
<p>I'm starting with $ \frac{1}{2}(a+b+1)(a+b)+a = \frac{1}{2}(c+d+1)(c+d)+c$, but how am I supposed to show from this equality that $\langle a, b\rangle = \langle c, d\rangle$, where $\langle a, b\rangle \in \mathbb{N}\times \mathbb{N}$ ?</p>
| BrianO | 277,043 | <p>Note that
$$\begin{align}
\varphi(0, n+k) &= \frac 1 2 (n+k)(n+k+1) \\
&\le \varphi(n,k) \\
&= \frac 1 2 (n+k)(n+k+1) + n \\
&= n + \sum_{i \le (n+k)} i \\
&< \sum_{i \le (n+k+1)} i \\
&= \varphi(0, n+k+1).
\end{align}$$</p>
<p>Suppose $\varphi(a,b) = \varphi(c,d)$. If $a+b < c+d$, then
$$\begin{align}
\varphi(a,b) &< \varphi(0,a+b+1) \\
&\le \varphi(0,c+d) \\
&\le \varphi(c,d), \\
\end{align}$$
so $a+b \ge c+d$. Similarly, $a+b \le c+d$, so $a+b = c+d$. By definition of $\varphi$, we have $a=c$. It follows that
$$\begin{align}
(a+b)(a+b+1) &= (c+d)(c+d+1) \\
&= c^2 + 2cd + d^2 + c + d \tag{i}\\
&= (c+b)(c+b+1) \\
&= c^2 + 2cb + b^2 + c + b.\tag{ii}
\end{align}$$</p>
<p>Subtracting (ii) from (i),
$$\begin{align}
0 &= (c^2 + 2cd + d^2 + c + d) - (c^2 + 2cb + b^2 + c + b) \\
&= 2c(d-b) + (d^2-b^2) + (d - b) \\
&= 2c(d-b) + (b+d)(d-b) + (d - b) \\
&= (2c+b+d+1)(d-b),\\
\end{align}$$
so $(2c+b+d+1) = 0$ or $(d-b) = 0$. But $(2c+b+d+1) > 0$ as $b,c,d\in \Bbb N$, so we must have $d-b = 0$, which is to say, $b = d$.</p>
|
894,159 | <p>I was assigned the following problem: find the value of $$\sum_{k=1}^{n} k \binom {n} {k}$$ by using the derivative of $(1+x)^n$, but I'm basically clueless. Can anyone give me a hint?</p>
| Caddyshack | 168,872 | <p>Notice that $\displaystyle S = \sum_{k=0}^{n} k \binom {n} {k} = \sum_{k=0}^{n}(n-k)\binom {n} {n-k} = n\sum_{k=0}^{n}\binom {n} {n-k}-S$</p>
<p>so, as $\displaystyle \binom {n} {n-k}=\binom {n} {k}$ we have $\displaystyle 2S = n\sum_{k=0}^{n}\binom {n} {k} = n2^n$ and so $\displaystyle S = n2^{n-1}.$</p>
|
512,590 | <p>According to the definition my professor gave us its okay for a matrix in echelon form to have a zero row, but a system of equations in echelon form cannot have an equation with no leading variable.</p>
<p>Why is this? Aren't they supposed to represent the same thing?</p>
| egreg | 62,967 | <p>Your hypothesis means that
$$
z^2 - 2z\cos\alpha + 1 = 0
$$
Solve the quadratic equation:
$$
z=\cos\alpha\pm\sqrt{\cos^2\alpha-1}
$$
so
$$
z=\cos\alpha+i\sin\alpha
\quad\text{or}\quad
z=\cos\alpha-i\sin\alpha=\cos(-\alpha)+i\sin(-\alpha).
$$
Apply de Moivre's formula.</p>
|
325,765 | <p>Is there any method which allows us to describe all continuous functions (maps to $\mathbb{R}$) on the quotient space?</p>
<p>For examle, how could I classify all continuous functions on $\mathbb{R}/[x\sim2x]$?</p>
| Community | -1 | <p>Let $Y=X/\sim$ be a quotient space equipped with the quotient topology. $f:Y\to \mathbb{R}$ is continuous if and only if $f \circ \pi$ is continuous, where $\pi:X\to Y$ is the natural quotient map. The reason is $U\subseteq Y$ is open in the quotient topology if and only if $\pi^{-1}(U)$ is open in $X$.</p>
|
1,826,964 | <blockquote>
<p>A fair die is tossed n times (for large n). Assume tosses are independent. What is the probability that the sum of the face showing is $6n-3$?</p>
</blockquote>
<p>Is there a way to do this without random variables explicitly? This is in a basic probability theory reviewer, and random variables was not yet discussed. It's Larsen and Marx before Chapter 3 (where random variables starts)</p>
<p>One thing I tried:</p>
<p>$$P(X_1 = 3) = 1/6$$</p>
<p>$$P(\sum_{i=1}^{2} X_i = 9) = (1/6)^2 4$$</p>
<p>$$P(\sum_{i=1}^{3} X_i = 15) = (1/6)^3 (10)$$</p>
<p>$$P(\sum_{i=1}^{4} X_i = 21) = (1/6)^4 (20)$$</p>
<p>I'm not seeing any pattern for the multiplicand of $(1/6)^n$</p>
<p>$$P(\sum_{i=1}^{n} X_i = 6n-3) = (1/6)^n (?)$$</p>
<p>Another thing I tried:</p>
<p>$$P(\sum_{i=1}^{n} X_i = 6n-3)$$</p>
<p>$$= P(X_i = 6 \ \text{except for 3 which are 5's}) \tag{1}$$</p>
<p>$$+ P(X_i = 6 \ \text{except for 2 which are a 4 and a 5}) \tag{2}$$</p>
<p>$$+ P(X_i = 6 \ \text{except for 1 which is 3}) \tag{3}$$</p>
<p>where I think</p>
<p>$$(1) = (1/6)^{n-3}(1/6)^{3}\binom{n}{3}$$</p>
<p>$$(2) = (1/6)^{n-3}(1/6)^{2}\binom{n}{2}(1/6)^{1}\binom{n-1}{1}$$</p>
<p>$$(3) = (1/6)^{n-2}(1/6)^{1}\binom{n}{1}$$</p>
<p>Is any approach going somewhere? If not, please suggest</p>
| Emre | 321,157 | <p>Your way is correct, but your calculations have some flaws. For (2), you choose $1$ place for $4$ and $1$ place for $5$. So, the binomial coefficients should be ${n\choose 1},{n-1\choose 1}$. So, $(2)={n\choose 1}(1/6)^1{n-1\choose 1}(1/6)^1(1/6)^{n-2}$</p>
<p>For three you need to have $(1/6)^{n-1}$, instead of $(1/6)^{n-2}$.</p>
<p>An alternative way would be defining $a_n=P(X=6n),b_n=P(X=6n-1),c_n=(X=6n-2)$, which are easier to calculate, and $d_n=P(X=6n-3)$. Then, using the recurrence relation
$$d_{n+1}=1/6(d_n+c_n+b_n+a_n)$$</p>
|
2,909,480 | <p>Please notice the following before reading: the following text is translated from Swedish and it may contain wrong wording. Also note that I am a first year student at an university - in the sense that my knowledge in mathematics is limited.</p>
<p>Translated text:</p>
<p><strong>Example 4.4</strong> Show that it for all integers $n$ is true that $n^3 - n$ is evenly divisible by $3$.<sup>1</sup> </p>
<p>Here we are put in front of a situation with a statement for <em>all integers</em> and not all positive. But it is enough that we treat the cases when $n$ is non-negative, for if $n$ is negative, put $m = -n$. Then $m$ is positive, $n^3 - n = -(m^3 - m)$ and if $3$ divides $a$, then $3$ also divides $-a$.</p>
<p>Now here also exists a statement for $n = 0$ so that we have a sequence $p_0, p_1, p_2, \; \ldots$ of statements, but that the first statement has the number $0$ and not $1$ is of course not of any higher meaning. Statement number $0$ says that $0^3 - 0$, which equals $0$, is evenly divisible by $3$, which obviously is true. If the statement number $n$ now is true, id est $n^3 - n = 3b$ for some integer $b$, then the statement number $n+1$ also must be true for</p>
<p>$
\begin{split}
(n + 1)^3 - (n + 1) & = n^3 - n + 3n^2 + 3n \\
& = 3b + 3n^2 + 3n \\
& =3(b + n^2 + n)
\end{split}
$</p>
<p>and $b + n^2 + n$ is an intege. What we was supposed to show now follows from the induction principle. $\square$</p>
<p><p>1. That an integer $a$ is "evenly divisible by 3" are everyday language rather than mathematical. The precise meaning is that it exists another integer $b$ such that $a = 3b$.</p></p>
<p><strong>In the above written text</strong>, I understanding everything (or I at least think so) except </p>
<p>$
\begin{split}
(n + 1)^3 - (n + 1) & = n^3 - n + 3n^2 + 3n \\
& = 3b + 3n^2 + 3n \\
& =3(b + n^2 + n).
\end{split}
$</p>
<p>Could someone please explain what happened, because I am totally lost?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Since we have assumed the $$n^3-n$$ is divisible by $3$ we can write $$n^3-n=3b$$ with $b$ is an integer number.
The proof becomes very easy if we write $$n^3-n=(n-1)n(n+1)$$</p>
|
409,220 | <p>$$f(x,y)=6x^3y^2-x^4y^2-x^3y^3$$
$$\frac{\delta f}{\delta x}=18x^2y^2-4x^3y^2-3x^2y^3$$
$$\frac{\delta f}{\delta y}=12x^3y-2x^4y-3x^3y^2$$
Points, in which partial derivatives ar equal to 0 are: (3,2), (x,0), (0,y), x,y are any real numbers. Now I find second derivatives
$$\Delta_1=\frac{\delta f}{\delta x^2}=36xy^2-12x^2y^2-6xy^3$$
$$\frac{\delta f}{\delta y^2}=12x^3-2x^4-6x^3y$$
$$\frac{\delta f}{\delta x \delta y}=\frac{\delta f}{\delta y \delta x} = 36x^2y-8x^3y-9x^2y^2$$
$$\Delta_2=\begin{vmatrix}\frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial y\partial x}\\\frac{\partial^2 f}{\partial x\partial y}& \frac{\partial^2 f}{\partial y^2} \end{vmatrix}$$
After plugging in the point (3,2) we get $\Delta_1<0$ and $\Delta_2>0$, so (3,2) is maxima. Now then I try to plug in (x,0) and (0,y) I obviously get $\Delta_1=0$ and $\Delta_2=0$ and I can't tell, using Sylvester's criterion, if those points are minima or maxima or neither. What should I do?</p>
| PierreCarre | 639,238 | <p>You just need to use the definition of maximum/minimum, together with the knowledge of the sign of <span class="math-container">$f$</span>.</p>
<p>For instance, if you take a point of the form <span class="math-container">$(x,0), x<0$</span>, you see that <span class="math-container">$f(x,0)=0$</span> and <span class="math-container">$f(x,y)<0$</span> for all neighbouring points. This means that points of the form <span class="math-container">$(x,0), x<0$</span> lead to local maxima.</p>
<p>As another example, you see that <span class="math-container">$f(0,0)=0$</span> and around that point <span class="math-container">$f$</span> always takes both positive and negative values. The origin is a saddle point.</p>
<p>Now you can use similar arguments for all other cases.</p>
<p><a href="https://i.stack.imgur.com/Hacwf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hacwf.png" alt="enter image description here" /></a></p>
|
1,416,972 | <p>Recently I started reading about <em>graph embeddings</em>, but I am unable to grasp its definition from <a href="https://en.wikipedia.org/wiki/Graph_embedding" rel="nofollow">Wikipedia</a>. Can anyone explain this term with an example.</p>
| Mike Pierce | 167,197 | <p>A <em>graph</em> is just an abstract idea, whereas a <em>graph embedding</em> is an actual physical instance of a graph that has to be drawn (or embedded) onto some surface.</p>
<hr />
<p>In order to understand what a <em>graph embedding</em> is, it helps to first define what a <em>graph</em> is.</p>
<blockquote>
<p><strong>Definition:</strong> A <em>graph</em> is a set <em>vertices</em> and a set of <em>edges</em> connecting those vertices. In a way we can think of a graph as a set of pairs of values, where each pair is an edge and each value is a vertex.
To give an abstract example consider the graph <span class="math-container">$G = \{\{1,a\},\{\text{blue},a\},\{1,\text{blue}\}\}$</span>.</p>
</blockquote>
<p>Note that this definition doesn't at all mention a drawing or a picture of the graph; it's just a list of pairs. A <em>graph embedding</em> is where we have to take a graph and actually draw a picture of it on some surface. For example, consider these three drawings of <span class="math-container">$K_4$</span> in <span class="math-container">$\mathbb{R}^2$</span> (a flat surface):</p>
<p><img src="https://i.stack.imgur.com/yoZht.png" alt="drawings of K4" /></p>
<p>Typically we require that no edges cross in a drawing of a graph to call it an embedding. This because we want to be consistent with the <a href="https://en.wikipedia.org/wiki/Embedding#Topology_and_geometry" rel="nofollow noreferrer">topological definition of an embedding</a>. So really only the second and third drawings of <span class="math-container">$K_4$</span> above are embeddings into <span class="math-container">$\mathbb{R}^2$</span>. Because we can embed <span class="math-container">$K_4$</span> in the plane, we say it is a <em>planar graph</em>. And of course you can start talking about embedding graphs on other surfaces. For example here is a graph embedded on a <a href="https://en.wikipedia.org/wiki/Torus" rel="nofollow noreferrer">torus</a>:</p>
<p><img src="https://i.stack.imgur.com/8orTf.png" alt="embedding of graph on a torus" />
<br><sub>Image originally from <a href="https://www.ics.uci.edu/%7Eeppstein/0xDE/F24/3dtorus.png" rel="nofollow noreferrer">www.ics.uci.edu/~eppstein/0xDE/F24/3dtorus.png</a> </sub></p>
<p>And then an obvious question arises:</p>
<blockquote>
<p><strong>Question</strong>: Given a surface, what graphs can be embedded on that surface?</p>
</blockquote>
<p>We've seen above that <span class="math-container">$K_4$</span> can be embedded on the plane, but it turns out that <span class="math-container">$K_5$</span> cannot! In fact we have a <a href="https://en.wikipedia.org/wiki/Kuratowski%27s_theorem" rel="nofollow noreferrer">complete characterization of all graphs that can and cannot be embedded on the plane</a>. The question of which graphs can and cannot be embedded on a torus, though, is still unanswered.</p>
|
1,416,972 | <p>Recently I started reading about <em>graph embeddings</em>, but I am unable to grasp its definition from <a href="https://en.wikipedia.org/wiki/Graph_embedding" rel="nofollow">Wikipedia</a>. Can anyone explain this term with an example.</p>
| elmodeer | 523,196 | <p>I think the best example to understand the nuanced difference between Topological vs Embedded Graphs is the example Steven Skiena gave in his <a href="https://www.google.com/search?q=The%20Algorithm%20Design%20Manual&rlz=1C5CHFA_enDE999DE999&sxsrf=ALiCzsbGZ7iG-u3J7kVPWVSjEg_SKrvJFQ%3A1654439483011&ei=O76cYtEnqYfFzw-UjoS4Dw&ved=0ahUKEwiRsb-tw5b4AhWpQ_EDHRQHAfcQ4dUDCA4&uact=5&oq=The%20Algorithm%20Design%20Manual&gs_lcp=Cgdnd3Mtd2l6EAMyBAguEEMyBAgAEEMyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDoHCAAQRxCwAzoHCAAQsAMQQzoMCC4QyAMQsAMQQxgBOg8ILhDUAhDIAxCwAxBDGAFKBAhBGABKBAhGGABQuANYuANgvwZoAnABeACAAUOIAUOSAQExmAEAoAECoAEByAENwAEB2gEECAEYCA&sclient=gws-wiz" rel="nofollow noreferrer">book</a>.</p>
<p>If we assumed a social graph where Vertices are people and Edges are the connections between them, we can assume a friendship relation between two people would be modeled as an undirected edge between two people/points.</p>
<p>and if we asked the question "Do my friends live near me?", we would find that most of our friends live near us or has lived near us (old college roommates).
One could argue that they are only friends because they live/have lived near us. This geographic info strongly influences our understanding of the Social Graph, and it may not be explicitly encoded, but the fact that
the graph is inherently <em><strong>embedded</strong></em> in the plane shapes our interpretation of any analysis.</p>
<p>such info would not be possible from a topological Graph for example.</p>
|
3,006,511 | <p>Let <span class="math-container">$c = \{ x = \{x_k\}_{k=1}^{\infty} \in l^\infty \vert \exists \lim_{k \to \infty} x_k \in \mathbb{C} \}$</span>. </p>
<p>Let <span class="math-container">$x_n \in c$</span>, with <span class="math-container">$x_n \to x = \{x_k\}$</span> with the sup norm. </p>
<p>I want to prove that <span class="math-container">$ x \in c$</span>.</p>
<p>So I am a bit stuck here, I want to use the fact that <span class="math-container">$x_n$</span> belongs to c, so it has a limit that I call <span class="math-container">$l_n$</span>. But from now on I do not know how to keep going. Any help?</p>
| DanielWainfleet | 254,665 | <p>With the notation of the Answer given by JustDroppedIn.</p>
<p>Since you want to show that <span class="math-container">$c$</span> is closed in <span class="math-container">$l^{\infty},$</span> you can prove that <span class="math-container">$l^{\infty}\setminus c$</span> is open, as follows: </p>
<p>Let <span class="math-container">$x \in l^{\infty}\setminus c.$</span> Then <span class="math-container">$x$</span> is not a Cauchy sequence, so there exists <span class="math-container">$r>0,$</span> and functions <span class="math-container">$f:\Bbb N\to \Bbb N,\, g:\Bbb N\to \Bbb N,$</span> both strictly increasing, such that <span class="math-container">$$\forall k\in \Bbb N\,(\,|x^{f(k)}-x^{g(k)}|>r\,).$$</span> Consider the open ball <span class="math-container">$B(x,r/3)=\{y\in l^{\infty}: \|y-x\|<r/3\}.$</span> If <span class="math-container">$y\in B(x,r/3)$</span> then for all <span class="math-container">$k\in \Bbb N$</span> we have <span class="math-container">$$|y^{f(k)}-y^{g(k)}|=$$</span> <span class="math-container">$$=|(y^{f(k)}-x^{f(k)})+(x^{f(k)}-x^{g(k)})+(x^{g(k)}-y^{g(k)})|\geq$$</span> <span class="math-container">$$\geq -|y^{f(k)}-x^{f(k)}|+|x^{f(k)}-x^{g(k)}|-|x^{g(k)}-y^{g(k)}|\geq$$</span> <span class="math-container">$$\geq -\|y-x\|+|x^{f(k)}-x^{g(k)}|-\|x-y\|>$$</span> <span class="math-container">$$>-r/3+r-r/3=r/3.$$</span> Since <span class="math-container">$\{f(k):j\in \Bbb N\}$</span> and <span class="math-container">$\{g(k):k\in \Bbb N\}$</span> are infinite sets, this implies that any <span class="math-container">$y\in B(x,r/3)$</span> is not a Cauchy sequence. So <span class="math-container">$B(x,r/3) \cap c =\emptyset.$</span></p>
<p>The idea is that you cannot uniformly approximate a non-convergent sequence <span class="math-container">$x$</span> to an arbitrary degree by a convergent sequence because of the "<span class="math-container">$r$</span>". For a bounded non-convergent sequence <span class="math-container">$x$</span> we can take <span class="math-container">$0<r<(\lim \sup_{j\to \infty} x^j)-(\lim \inf_{j\to \infty} x^j).$</span></p>
|
677,241 | <p>Let $A$ be a list of $n$ numbers in range $[1,100]$ (numbers can repeat).
I'm looking for the number of permutations of $A$ which start with a non-decreasing part, where this part ends with the first instance of the highest number, call this "index $i$" (1 based)from the left. After $i$, the remaining permutation is an arbitrary arrangement.</p>
<p><strong>Examples</strong> </p>
<ul>
<li><p>For $n=1$ and $A=\{9\}$, we have $1$ way only.</p></li>
<li><p>For $n=2$ and $A=\{2,5\}$, we have $2!$ ways in total.</p>
<ul>
<li>For $i=1$ we have $1$ way $(5,2)$ only.</li>
<li>For $i=2$ we have $1$ way $(2,5)$ only.</li>
</ul></li>
<li><p>If we were given , $A=\{5,5\}$, we have 2 ways totally:</p>
<ul>
<li>For $i=1$ we have one way, namely $(5,5)$, where the first $5$ coincides with first occurrence of $5$ in $A$, and</li>
<li>For $i=2$ we have one way, namely $(5,5)$, where the second $5$ coincides with first occurrence of $5$ in $A$</li>
</ul></li>
<li><p>For $n=3$ and $A=\{1, 4, 3\}$, we have $3!$ ways in total.</p>
<ul>
<li>For $i=1$, we have $2!$ permutations starting with $4$, namely $(4,3,1)$ and $(4,1,3)$.</li>
<li>For $i=2$ , $2!$ ways $(1,4,3)$ and $(3,4,1)$ in which $4$ is at $2$nd place.</li>
<li>For $i=3$ , $2!$ ways $(1,3,4)$ and $(3,1,4)$ ending with $4$.</li>
</ul></li>
</ul>
<p>To be exact : the number of permutations, without replacement, of given $n$ numbers such that the numbers up to $i$th place from left (say index) are in non-decreasing order, $i$-th place is occupied by first instance of largest number and rest can be random...</p>
<p>p.s. I cannot be more specific...</p>
| Marc Romaní | 179,483 | <p>Let's say you have $t_i$ times integer $i$, with $i$ from $0$ to $9$. The number of permutations of the tuple $(a_1, \ldots, a_{t_1 + \cdots + t_9})$ with $0 \leq a_i \leq 9$, such that $a_i \leq a_j$ if $i < j$, is $$t_1!\cdots t_9!$$</p>
|
1,274,816 | <p>It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: <a href="https://math.stackexchange.com/questions/1192338/number-theory-taxicab-number">Number Theory Taxicab Number</a>).</p>
<p>We know that</p>
<p>$$1729 = 10^3+9^3 = 12^3 + 1^3,$$</p>
<p>and I am wondering if there are infinitely many numbers like this that can be expressed as the sum of two positive cubes in <em>exactly</em> two ways?</p>
<p>In fact, are there even any other such numbers?</p>
<p>EDIT:
As provided by MJD in the comments section, here are other examples:
$$4104 = 2^3+16^3 = 9^3+15^3,$$
$$13832 = 20^3+18^3=24^3+2^3,$$
$$20683 = 10^3 +27^3 = 19^3 +24^3.$$ </p>
| Jibran | 515,854 | <p>As for the question regarding whether or not there are infinitely many numbers that can be expressed as the sum of two cubes in two different ways or not, there's a very quick and simple way to prove this. Since you have
<span class="math-container">$$ 1729 = 10^3 + 9^3 = 12^3 + 1^3 $$</span>
Multiply both sides by <span class="math-container">$ n^3 $</span> where <span class="math-container">$ n $</span> is a positive integer to get
<span class="math-container">$$ 1729 n^3 = (10n)^3 + (9n)^3 = (12n)^3 + n^3 $$</span>
and there you have it, plug in any value for <span class="math-container">$ n $</span> and and you have your proof for the infinitude of such numbers. Obviously this doesn't include all the solutions but it proves that there certainly are an infinite number of them </p>
|
2,498,628 | <p>This was a question in our exam and I did not know which change of variables or trick to apply</p>
<p><strong>How to show by inspection ( change of variables or whatever trick ) that</strong></p>
<p><span class="math-container">$$ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx \tag{I} $$</span></p>
<p>Computing the values of these integrals are known as routine. Further from their values, the equality holds. But can we show equality beforehand?</p>
<blockquote>
<p><strong>Note</strong>: I am not asking for computation since it can be found <a href="https://math.stackexchange.com/questions/187729/evaluating-int-0-infty-sin-x2-dx-with-real-methods?noredirect=1&lq=1.">here</a>
and we have as well that,
<span class="math-container">$$ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx =\sqrt{\frac{\pi}{8}}$$</span>
and the result can be recover here, <a href="https://math.stackexchange.com/questions/187729/evaluating-int-0-infty-sin-x2-dx-with-real-methods?noredirect=1&lq=1">Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?</a>.</p>
</blockquote>
<p>Is there any trick to prove the equality in (I) without computing the exact values of these integrals beforehand?</p>
| Giuseppe Negro | 8,157 | <blockquote>
<p><strong>This is an extended comment not a proper answer.</strong></p>
</blockquote>
<p>The question can be rewritten as follows: to show that
$$\tag{1}
\Re \int_{-\infty}^\infty e^{-i|\xi|^2}\, d\xi + \Im \int_{-\infty}^\infty e^{-i|\xi|^2}\, d\xi=0,$$
where the integral is in the principal-value sense. This integral arises in PDEs as evaluation at the spatial origin of the fundamental solution to the Schrödinger equation. More precisely, if $E=E(t, \mathbf x)$ solves
$$\tag{2}
\begin{cases} (i\partial_t + \Delta) E(t, \mathbf x)=0, & t\in \mathbb R, \mathbf x\in\mathbb R^n\\ E(0, \mathbf x)=\delta(\mathbf x)\end{cases}$$
(where $\delta$ is <a href="https://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow noreferrer">the Dirac distribution</a>) then the Fourier transform of $E(t, \cdot)$ is
$$
\hat{E}(t,\boldsymbol \xi)=\int_{\mathbb R^n} e^{-i\mathbf x\cdot \boldsymbol\xi}E(t, \mathbf x)\, d\mathbf x= e^{-it|\boldsymbol\xi|^2},\quad \boldsymbol\xi\in\mathbb R^n.$$
Now we observe that, for all suitable function $u$, we have that $\int_{\mathbb R^n} \hat{u}(t,\boldsymbol \xi)\, d\boldsymbol\xi=u(t, \mathbf 0)$. This gives a reformulation of problem (1) that generalizes to arbitrary dimension: </p>
<blockquote>
<p>Is it true that $$\tag{3}\Re E(1,\mathbf 0)+ \Im E(1, \mathbf 0) =0, $$ where $E$ is the solution to (2)?</p>
</blockquote>
<p>I found it surprising that the answer is affirmative if and only if $n=1\mod 4$: this follows from <a href="https://math.stackexchange.com/a/292552/8157">the explicit formula</a>
$$E(1,\mathbf 0)=\lim_{R\to \infty}\int_{[-R, R]^n}e^{-i|\boldsymbol \xi|^2}\, d\boldsymbol\xi = \frac{\pi^\frac{n}{2}}{i^\frac{n}{2}}=\pi^{\frac n 2}e^{-i \frac{n}{4}\pi}.$$</p>
<blockquote>
<p><strong>Conclusion</strong>. The OP asks for a solution that relies purely on change of variable in the integrals. In view of the reformulation (3), such changes of variable correspond to the symmetries of the PDE (2). These symmetries are dimension-independent, but the solution to the problem (3) is dependent on the dimension. Therefore, it seems to me that this approach is unlikely to work.</p>
</blockquote>
|
3,792,954 | <p>For vector space <span class="math-container">$V$</span> and <span class="math-container">$v \in V$</span>, there is a natural identification <span class="math-container">$T_vV \cong V$</span> where <span class="math-container">$T_vV$</span> is the tangent space of <span class="math-container">$V$</span> at <span class="math-container">$v$</span>. Can somebody explain this identification to me? Thank you</p>
| LSpice | 87,579 | <p>Note that <span class="math-container">$1 + x y$</span> is positive for all <span class="math-container">$x, y \in (-1, 1)$</span>; so, since our function is odd in <span class="math-container">$(x, y)$</span> and hence takes values in <span class="math-container">$(-1, 1)$</span> if it takes values in <span class="math-container">$(-\infty, 1)$</span>, it suffices to show that <span class="math-container">$x + y$</span> is strictly less than <span class="math-container">$1 + x y$</span>. This is true because <span class="math-container">$1 + x y - x - y = (1 - x)(1 - y)$</span> is strictly positive when <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are both strictly less than <span class="math-container">$1$</span>.</p>
|
959,201 | <p>I am confused about the following.</p>
<p>Could you explain me why if $A=\varnothing$,then $\cap A$ is the set of all sets?</p>
<p>Definition of $\cap A$:</p>
<p>For $A \neq \varnothing$:</p>
<p>$$x \in \cap A \leftrightarrow (\forall b \in A )x \in b$$</p>
<p><strong>EDIT</strong>:</p>
<p>I want to prove that $\cap \varnothing$ is not a set.</p>
<p>To do that, do I have to begin, supposing that it is a set?</p>
| Mauro ALLEGRANZA | 108,274 | <p>See Herbert Enderton, <a href="http://rads.stackoverflow.com/amzn/click/0122384407" rel="noreferrer">Elements of Set Theory</a> (1977), page 24 :</p>
<blockquote>
<p>Suppose we want to take the intersection of infinitely many sets $b_0, b_1, \ldots$. Then where $A = \{ b_0, b_1,\ldots \}$ the desired intersection can be informally characterized as $\cap A = \cap_i b_i = \{ x | x$ belongs to every $b_i \in A \}$.</p>
<p>In general, we define for every non empty set $A$, the intersection $\cap A$ of $A$
by the condition </p>
<blockquote>
<p>$x \in \cap A \Leftrightarrow x$ belongs to every member of $A$.</p>
</blockquote>
<p>What happens if $A = \emptyset$ ? For any $x$ at all, it is vacuously true that $x$ belongs to every member of $\emptyset$. (There can be no member of $\emptyset$
to which $x$ fails to belong.) Thus it looks as if $\cap \emptyset$ should be the class $V$
of all sets. By <strong>Theorem 2A</strong> [page 22 : There is no set to which every set belongs], there is no set $C$ such that for all $x$, </p>
<blockquote>
<p>$x \in C \Leftrightarrow x$ belongs to every member of $\emptyset$</p>
</blockquote>
<p>since the right side is true of every $x$. This presents a mild notational problem: How do we define $\cap \emptyset$ ? The situation is analogous to division by zero in arithmetic. How does one define $a/0$ ? One option is to leave $\cap \emptyset$ undefined, since there is no very satisfactory way of defining it. This option works perfectly well, but some logicians dislike it. It leaves $\cap \emptyset$ as an untidy loose end, which they may later trip over. The other option is to select some arbitrary scapegoat (the set $\emptyset$ is always used for this) and define $\cap \emptyset$ to equal that object. </p>
<p>Either way, whenever one forms $\cap A$ one must beware the possibility that perhaps $A = 0$. Since it makes no difference which of the two options one follows, we will not bother to make a choice between them at all.</p>
</blockquote>
|
784,258 | <p>I understand that this is an induction question. </p>
<p>I start with the base case (n=1):</p>
<p>$$1 < 2 \tag{That works!}$$</p>
<p>Induction step: Assume the statement works for all $n = k$, Prove for all $n = k+1$</p>
<p>Assume $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}+ ... +\frac{1}{\sqrt{k+1}}\le 2\sqrt{k+1}$</p>
<p>I'm a bit confused as to where to go next, may I please have some assistance?</p>
| Srinivasa Ramanujan | 147,778 | <p>Compare the area below the red curve ($y=1/\sqrt{x}$) and the blue curve from $x=0$ to $x=\sqrt{n}$.</p>
<p><img src="https://i.stack.imgur.com/elri5.png" alt="enter image description here"></p>
|
87,948 | <p>Let $\mu_t, t \geq 0,$ be a family of probability measures on the real line. One can assume whatever one wishes about them, although typically they will be continuous in some topology (usually at least the topology of weak convergence of measures), and they will be absolutely continuous with respect to Lebesgue measure. The basic question is as follows:</p>
<p>Is there a Markov process $X_t$ such that its marginal distribution at each time is $\mu_t$?</p>
<p>An obvious example is when $$d \mu_t = \frac{e^{-x^2/2t}}{\sqrt{2 \pi t}} dx$$ and $\mu_0 = \delta_0$, in which case we know that Brownian motion is such a Markov process. I am curious to know if there is any general theory along these lines.</p>
<h2>Edit</h2>
<p>As per Byron's comment below, I would like the Markov process to be continuous. Ideally I would like to have an SDE description of the process. </p>
<p>The SDE description actually suggests one possible answer: simply compute and play with the time and space derivatives of the density function to see if they satisfy some sort of parabolic equation (like the heat equation), use this to get the adjoint of the generator, and then compute the generator itself. This is a very plausible option, but I was hoping that there might be something more systematic.</p>
| The Bridge | 2,642 | <p>Hi, </p>
<p>This is not the general answer you are looking for but it might be sufficient for your needs, here are my two cents. </p>
<p>If you are given a Semimartingale then there exists conditions that ensures that there exists a Markov process such that this markov process has the same marginal distributions that the marginal distributions of your semimartingale. </p>
<p>Take a look <a href="http://arxiv.org/abs/0910.3992" rel="nofollow">here</a>, where it is motivated by application to mathematical finance and where the first version of this theorem is known as "Gyongy's Lemma" I believe. </p>
<p>Best Regards</p>
|
2,353,272 | <p>Suppose that we are given the function $f(x)$ in the following product form:
$$f(x) = \prod_{k = -K}^K (1-a^k x)\,,$$
Where $a$ is some real number. </p>
<p>I would like to find the expansion coefficients $c_n$, such that:
$$f(x) = \sum_{n = 0}^{2K+1} c_n x^n\,.$$</p>
<p>A closed form solution for $c_n$, or at least a relation between the coefficients $c_n$ (e.g. between $c_n$ and $c_{n+1}$) would be great! </p>
| Vassilis Markos | 460,287 | <p>Well, we have that - we need to suppose that $a\neq0$:
$$f(x)=\prod_{k=-K}^K(1-a^kx)=(1-a^{-K}x)(1-a^{-K+1}x)\dots(1-a^Kx)$$
We can expand this product to a sum of powers of $x$, by considering in how many ways can we get $x^n$, where $n=0,1,2,\dots,2K+1$.</p>
<p>While expanding this product, we take each time exactly one of either $1$ or $-a^kx$ from each parenthesis and we multiply them. So, to take $x^n$, we need to take exactly $n$ times $-a^kx$ and $2K+1-n$ time $1$.</p>
<p>Now, we have $b_n=\binom{2K+1}{n}$ choices that give us exactly $n$ out of the $2K+1$ $(-a^kx)$'s, so, we can see that:
$$f(x)=\sum_{n=0}^{2K+1}(-1)^n\left(\sum_{m=1}^{b_n}a^{s_{m,n}}\right)x^n$$
where $s_{m,n}$ is a - different for every $m$ - sum of $n$ distinct numbers from $\{-K,-K+1,\dots,K\}$.</p>
<p>From now own one can elaborate on the latter relation.</p>
|
2,353,272 | <p>Suppose that we are given the function $f(x)$ in the following product form:
$$f(x) = \prod_{k = -K}^K (1-a^k x)\,,$$
Where $a$ is some real number. </p>
<p>I would like to find the expansion coefficients $c_n$, such that:
$$f(x) = \sum_{n = 0}^{2K+1} c_n x^n\,.$$</p>
<p>A closed form solution for $c_n$, or at least a relation between the coefficients $c_n$ (e.g. between $c_n$ and $c_{n+1}$) would be great! </p>
| epi163sqrt | 132,007 | <p>We derive a representation of $c_j$ in
\begin{align*}
f_K(x)=\prod_{k=-K}^K(1-a^kx)=\sum_{j=0}^{2K+1}\color{blue}{c_j}x^j\qquad\qquad \qquad K\geq 0
\end{align*}
based upon the <em><a href="https://en.wikipedia.org/wiki/Q-Pochhammer_symbol" rel="nofollow noreferrer">q-Pochhammer symbol</a></em> $(x;a)_K$.</p>
<blockquote>
<p>According to the referred Wiki page the following is valid.
\begin{align*}
(x;a)_{K}=\prod_{k=0}^{K-1}(1-a^kx)=\sum_{j=0}^K\binom{K}{j}_qa^{\binom{j}{2}}(-x)^j\tag{1}
\end{align*}
Here we denote with $\binom{K}{j}_q$ the <em><a href="https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient" rel="nofollow noreferrer">q-binomial coefficient</a></em>, a generalisation of the binomial coefficients.</p>
</blockquote>
<p>We obtain from the product representation in (1)
\begin{align*}
(xa^{-K};a)_K&=\prod_{k=0}^{K-1}(1-a^{k-K}x)
=\prod_{k=0}^{K-1}(1-a^{-k-1}x)\tag{2}\\
&=\prod_{k=1}^{K}(1-a^{-k}x)=\prod_{k=-K}^{-1}(1-a^kx)\tag{3}
\end{align*}</p>
<p><em>Comment:</em></p>
<ul>
<li><p>In (2) we revert the order of the terms of the product: $k\to K-1-k$.</p></li>
<li><p>In (3) we shift the index $k$ by one and replace the index $k$ with $-k$.</p></li>
</ul>
<blockquote>
<p>Putting (1) and (3) together we derive the following representation of $f_K(x)$
\begin{align*}
\color{blue}{f_K(x)}&=\prod_{k=-K}^{-1}(1-a^kx)\cdot\prod_{k=0}^{K}(1-a^kx)\\
&=(xa^{-K};a)_K\cdot(x;a)_{K+1}\\
&=\left(\sum_{j=0}^K\binom{K}{j}_qa^{\binom{j}{2}}(-xa^{-K})^j\right)\cdot\left(\sum_{l=0}^{K+1}\binom{K+1}{l}_qa^{\binom{l}{2}}(-x)^l\right)\\
&\color{blue}{
=\sum_{n=0}^{2K+1}\left(\sum_{{{j+l=n}\atop{0\leq j\leq K}}\atop{\ \ \,0\leq l\leq K+1}}\binom{K}{j}_q\binom{K+1}{l}_q(-1)^{j+l}a^{-Kj+\binom{j}{2}+\binom{l}{2}}\right)x^n}
\end{align*}</p>
</blockquote>
|
4,013,559 | <p>In the video <a href="https://youtu.be/eI4an8aSsgw?t=16354" rel="nofollow noreferrer">https://youtu.be/eI4an8aSsgw?t=16354</a> the professor says that the following equation
<span class="math-container">$$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a$$</span>
simplifies to
<span class="math-container">$$
(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)
$$</span>
I don't get it. I need someone who can explain it step by step</p>
| Z Ahmed | 671,540 | <p>Let
<span class="math-container">$$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a=\sqrt{P}+\sqrt{Q}~~~(1)$$</span>
<span class="math-container">$$\implies P-Q=4cx \implies \sqrt{P}-\sqrt{Q}=2cx/a~~~(2)$$</span>
Adding (1) and (2) get
<span class="math-container">$$\sqrt{P}=xc/a+a \implies (x+c)^2+y^2=(cx/a+a)^2 \implies (1-c^2/a^2)x^2+y^2=a^2-c^2$$</span>
<span class="math-container">$$\implies \frac{x^2}{a^2}+\frac{y^2}{(1-c^2/a^2)}=1$$</span>
Let <span class="math-container">$c=ae$</span>, then we get ellips as
<span class="math-container">$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$</span></p>
|
681,737 | <p>What is the simplest way we can find which one of $\cos(\cos(1))$ and $\cos(\cos(\cos(1)))$ [in radians] is greater without using a calculator [pen and paper approach]? I thought of using some inequality relating $\cos(x)$ and $x$, but do not know anything helpful.
We can use basic calculus. Please help. </p>
| Yiorgos S. Smyrlis | 57,021 | <p>This statement is correct is "bounded" is replaced by "compact".</p>
<p>Also, it is correct if $\mathbb R$ is replaced by "a compact metric space".</p>
|
3,852,952 | <p>Given a projective space <span class="math-container">$\mathbb{P}^n(\mathbb{C})$</span>, I can consider the Grasmannian of lines <span class="math-container">$G(2,n+1)$</span>, which has a structure of projective variety inside <span class="math-container">$\mathbb{P}^N$</span>, where <span class="math-container">$N=\binom{2+n+1}{2}-1$</span> thanks to the Pl<span class="math-container">$\ddot{u}$</span>cker embedding.</p>
<p>It has been told me -literally, I have no references, it was a speech- that while a point <span class="math-container">$p\in G(2,n+1)$</span> obviously represents a line in <span class="math-container">$\mathbb{P}^n$</span> by definition, a <em>line</em> inside <span class="math-container">$G(2,n+1)$</span> -viewd as a projective variety in <span class="math-container">$\mathbb{P}^N$</span>-corresponds to a pencil of planes.</p>
<p>Unfortunately I'm still trying to properly understand why this work -because unfortunately it is not crystal for me -, but I was also wondering if it makes sense -that is, if there is a geometrical meaning-, also for quadrics contained in <span class="math-container">$G(2,n+1)$</span>.</p>
<p>Therefore I'd like to have an idea of what happens for quadrics, and moreover even a little help for lines in <span class="math-container">$G(2,n+1)$</span>.</p>
<p>Since this question comes primarily from my curiosity, I apologize for the vagueness: any comment, reference or answer would be much appreciate!</p>
| Gustavo Labegalini | 356,961 | <p>End up here with the same question, but I could not understand the answers very well. I'll let a proposal of an answer out here, and see what's up.</p>
<p><strong>Plucker relations for <span class="math-container">$k=2$</span>:</strong> An element <span class="math-container">$\omega \in \bigwedge^2 V$</span> is decomposable, meaning of type <span class="math-container">$v \wedge w$</span> (in the image of the Plucker embedding <span class="math-container">$G(2,V) \hookrightarrow \mathbb{P}(\bigwedge^2 V)$</span>, <span class="math-container">$\langle v,w \rangle \mapsto [v \wedge w]$</span>) iff <span class="math-container">$\omega \wedge \omega = 0$</span>.</p>
<p>Now, a line in <span class="math-container">$\mathbb{P}(\bigwedge^2 V)$</span> is a 2-plane through the origin in <span class="math-container">$\bigwedge^2 V$</span>. Take the image of two non-trivial Grassmannians <span class="math-container">$\alpha \doteq a_{1} \wedge a_{2}$</span> and <span class="math-container">$\beta \doteq b_{1} \wedge b_{2}$</span>, not a scalar multiple one of each other. The projective line <span class="math-container">$\mathfrak{L}$</span>, given by <span class="math-container">$[\lambda \alpha + (1-\lambda) \beta]\subset \mathbb{P}(\bigwedge^2 V)$</span> is inside the image of the Plucker embedding iff</p>
<p><span class="math-container">$$ (\lambda \alpha + (1-\lambda) \beta) \wedge (\lambda \alpha + (1-\lambda) \beta) = 0 $$</span></p>
<p>Notice that <span class="math-container">$\alpha \wedge \alpha = \beta \wedge \beta = 0$</span> and you need an even number of permutations to transform <span class="math-container">$\beta \wedge \alpha$</span> into <span class="math-container">$\alpha \wedge \beta$</span>. We conclude <span class="math-container">$\alpha \wedge \beta = 0$</span>. But then, we have one of the following:</p>
<ul>
<li>some <span class="math-container">$a_i \in \mathrm{span}(b_1,b_2)$</span> or</li>
<li>some <span class="math-container">$b_i \in \mathrm{span}(a_1,a_2)$</span>,</li>
</ul>
<p>but then this vector represents a common vector <span class="math-container">$v \in \mathrm{span}(a_1,a_2) \cap \mathrm{span}(b_1,b_2)$</span>.</p>
<p>Now the point <span class="math-container">$[v] \in \mathbb{P}V$</span> is common to both projective lines <span class="math-container">$\ell_1 = [\mathrm{span}(a_1,a_2)]$</span> and <span class="math-container">$\ell_2 = [\mathrm{span}(b_1,b_2)]$</span>, and all of the lines in <span class="math-container">$\mathfrak{L}$</span> can be considered to be, w.l.o.g., inside the projective plane <span class="math-container">$\pi \doteq [\mathrm{span}(a_1,a_2,b_1)]$</span> (when <span class="math-container">$b_2$</span> is the L.D. one).</p>
|
926,804 | <p>Is there a word for the quality of a number to be either positive or negative? Consider this question:</p>
<p><em>What's the ... (sign/positivity/negativity, but a word that could describe either) of number <strong>x</strong>?</em></p>
<p>Also, is there an all-encompassing word for the sign put in front of a number (-5 or +5)? Word that describes both a plus and a minus sign?</p>
| k170 | 161,538 | <p>The word that you are looking for is the very word that you are using and that word is <strong>sign</strong>.</p>
<p>If you are looking for something more "academic", then you can use its Latin variant, <strong>signum</strong>.</p>
<p>Context can also effect what word you might choose. For instance, in physics it's common to use the word <strong>direction</strong> in place of the word "sign."</p>
<p>But don't take my word for it; have a look at page <span class="math-container">$229$</span> of "<a href="http://www.abstractmath.org/Handbook/handbook.pdf" rel="nofollow noreferrer">A Handbook of Mathematical Discourse</a>".</p>
|
926,804 | <p>Is there a word for the quality of a number to be either positive or negative? Consider this question:</p>
<p><em>What's the ... (sign/positivity/negativity, but a word that could describe either) of number <strong>x</strong>?</em></p>
<p>Also, is there an all-encompassing word for the sign put in front of a number (-5 or +5)? Word that describes both a plus and a minus sign?</p>
| Neville | 493,521 | <p>The search is for a PHILOSOPHICAL term to denote positive / negative. I agree that "sign" lacks something - hence, consider "polarity" and "charge".</p>
|
4,346,338 | <p>In Loring Tu Introduction to Manifolds, pg. 328:</p>
<blockquote>
<p>The projection <span class="math-container">$\pi:X \times Y \rightarrow X$</span>, <span class="math-container">$\pi(x,y)=x$</span>, is
continuous.</p>
<p>Proof: Let <span class="math-container">$U$</span> be open in <span class="math-container">$X$</span>. Then <span class="math-container">$\pi^{-1}(U)=U\times Y$</span>, which is open in the product topology on <span class="math-container">$X \times Y$</span>.</p>
</blockquote>
<p>Let's say we have <span class="math-container">$W$</span> which is an open set of <span class="math-container">$X$</span> and <span class="math-container">$V$</span> which is a closed set of <span class="math-container">$Y$</span>. Then the projection map will map <span class="math-container">$(W, V) \rightarrow W$</span>. The inverse map will map <span class="math-container">$W\rightarrow (W, V).$</span> Since <span class="math-container">$W$</span> is an open set in <span class="math-container">$X$</span> and <span class="math-container">$W\times V$</span> is not an open set in the product topology, we can say that the projection map is not continuous.</p>
<p>What is wrong with my argument? Perhaps the projection map only maps <span class="math-container">$(W,Y) \rightarrow W$</span> instead of <span class="math-container">$(W,Z)\rightarrow W$</span> where <span class="math-container">$Z$</span> is a subset of <span class="math-container">$Y$</span> and is not <span class="math-container">$Y$</span>?</p>
| Intelligenti pauca | 255,730 | <p>Relabelling <span class="math-container">$z=x-x_0$</span> and <span class="math-container">$y=f(z)$</span> your first equation reads
<span class="math-container">$$
y={a\over2}\big(z-\sqrt{z^2+b^2}\big).
$$</span>
Rearranging and squaring yields
<span class="math-container">$$
y^2-ayz={a^2b^2\over4},
$$</span>
that is:
<span class="math-container">$$
y(y-az)={a^2b^2\over4}.
$$</span>
This is the equation of a hyperbola with asymptotes <span class="math-container">$y=0$</span> and <span class="math-container">$y=az$</span>, confirming what you found. But the axes are the angle bisectors of the asymptotes, hence their equations you cited in the question cannot be correct. The right result for the equation of the axes is
<span class="math-container">$$
y={\pm\sqrt{1+a^2}-1\over a}z.
$$</span></p>
|
4,346,338 | <p>In Loring Tu Introduction to Manifolds, pg. 328:</p>
<blockquote>
<p>The projection <span class="math-container">$\pi:X \times Y \rightarrow X$</span>, <span class="math-container">$\pi(x,y)=x$</span>, is
continuous.</p>
<p>Proof: Let <span class="math-container">$U$</span> be open in <span class="math-container">$X$</span>. Then <span class="math-container">$\pi^{-1}(U)=U\times Y$</span>, which is open in the product topology on <span class="math-container">$X \times Y$</span>.</p>
</blockquote>
<p>Let's say we have <span class="math-container">$W$</span> which is an open set of <span class="math-container">$X$</span> and <span class="math-container">$V$</span> which is a closed set of <span class="math-container">$Y$</span>. Then the projection map will map <span class="math-container">$(W, V) \rightarrow W$</span>. The inverse map will map <span class="math-container">$W\rightarrow (W, V).$</span> Since <span class="math-container">$W$</span> is an open set in <span class="math-container">$X$</span> and <span class="math-container">$W\times V$</span> is not an open set in the product topology, we can say that the projection map is not continuous.</p>
<p>What is wrong with my argument? Perhaps the projection map only maps <span class="math-container">$(W,Y) \rightarrow W$</span> instead of <span class="math-container">$(W,Z)\rightarrow W$</span> where <span class="math-container">$Z$</span> is a subset of <span class="math-container">$Y$</span> and is not <span class="math-container">$Y$</span>?</p>
| user2603428 | 1,010,688 | <p>Thanks to @Intelligenti pauca for pointing to my error and how to get to the answer that I find satisfactory.</p>
<p>First I redefine f(x) such that it the independent and dependent variables are unitless. Analytic geometry requires that all dimensions have the same unit for anything to be sensible. What is the length of the major and minor axes in a rotated frame supposed to be? What does <span class="math-container">$\sqrt{(5\text{GeV})^2+3^2}$</span> mean? It doesn't mean anything. So best to keep things consistent.
<span class="math-container">$$z:=a(x-x_0)$$</span>
<span class="math-container">$$f(z):=\frac{1}{2}(z-\sqrt{a^2b^2+z^2})$$</span>
This gives me an asymptote of <span class="math-container">$y=z$</span> and <span class="math-container">$y=0$</span>. The major and minor axes bisect the angles, not what would be the perpendicular bisector of a triangle. That gives axes of <span class="math-container">$y=z\tan(\frac{\pi}{8})$</span> and <span class="math-container">$y=-z\cot(\frac{\pi}{8})$</span> or <span class="math-container">$y=a(x-x_0)\tan(\frac{\pi}{8})$</span> and <span class="math-container">$y=-a(x-x_0)\cot(\frac{\pi}{8})$</span> in the physical space, which don't appear to be perpendicular. But what is perpendicular in the physical space? Two lines that are <span class="math-container">$\pi/4$</span> radians apart. What is a radian? The angle where the radius is equal to the arclength. What is the radius of a circle where the length is 1 GeV in one direction and 1 in the other direction? That clearly isn't a sensible question. This will lead to funny results in the physical space. The function in the physical space is also clearly a hyperbola with it's own perpendicular axes. Using those axes will be consistent with the answer given by @Intelligenti pauca, but the answer will have unit conflicts.</p>
<p>Now that I have axes, I can run the major axis to the intersection with the hyperbola. That gives me a vertex at <span class="math-container">$\left(-\frac{2 \sqrt[4]{2} a b \sin \left(\frac{\pi }{8}\right)}{\sqrt{2-\sqrt{2}}-\sqrt{1+\cot \left(\frac{\pi }{8}\right)}},-\frac{2 \sqrt[4]{2} a b \sin \left(\frac{\pi }{8}\right)}{\sqrt{2-\sqrt{2}}-\sqrt{1+\cot \left(\frac{\pi }{8}\right)}}\right)$</span>. I would have been happy with this answer, as I could take it and run with it to the actual answer that I needed. Having found the answer final answer that I need, I can't keep it to myself, so here's the plot of everything in the unitless space.</p>
<p><a href="https://i.stack.imgur.com/M6zIo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M6zIo.png" alt="A plot of both branches of a hyperbola with its asymptotes, axes, and latera recta." /></a></p>
<p>The z coordinates of the primary latus rectum are <span class="math-container">$$z_1=-\frac{a b \csc \left(\frac{\pi }{8}\right) \left(8-4 \cot \left(\frac{\pi }{8}\right)+\sqrt[4]{2} \sqrt{3 \sqrt{2}+4} \csc \left(\frac{\pi }{8}\right)\right)}{4\ 2^{3/4} \left(\cot \left(\frac{\pi }{8}\right)-1\right)}\approx-1.99546ab$$</span> <span class="math-container">$$z_2=\frac{2 \sqrt[4]{2} ab \sin \left(\frac{\pi }{8}\right)}{\sqrt{2-\sqrt{2}}-\sqrt{1+\cot \left(\frac{\pi }{8}\right)}}\approx2.90564ab$$</span></p>
<p>Now back in the physical space, the vertex is located at <span class="math-container">$\left(\frac{1}{2} \sqrt{\frac{3}{\sqrt{2}}-2} b+x_0,-\frac{1}{8} a \left(\sqrt{6 \sqrt{2}+8} b-\sqrt{6 \sqrt{2}-8} b\right)\right)$</span>. For <span class="math-container">$a=5\text{GeV}^{-1}$</span>, <span class="math-container">$b=.75\text{GeV}$</span>, and <span class="math-container">$x_0=2.5\text{GeV}$</span>, that works out to <span class="math-container">$\left(\left(\frac{3}{8} \sqrt{\frac{3}{\sqrt{2}}-2}+\frac{5}{2}\right)\text{GeV},-\frac{15}{8 \sqrt[4]{2}}\right)\approx(2.63062\text{GeV},-1.57668)$</span>. The end points for that latus rectum are approximately <span class="math-container">$(1.0034\text{GeV},-7.92651)$</span> and <span class="math-container">$(4.67923\text{GeV},-.313621)$</span>. How long is that latus rectum? No idea, that isn't a sensible question. But how long is the x interval is a sensible question, and the complete answer I was actually after. I thought it would be discourteous to ask this forum for the complete answer. That interval is<span class="math-container">$\frac{\sqrt{17 \sqrt{2}+24}}{\cot \left(\frac{\pi }{8}\right)-1}b\approx4.9011b$</span>.</p>
<p><a href="https://i.stack.imgur.com/phJLh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/phJLh.png" alt="The above hyperbola with the axes, vertex, and latus rectum. The axes of the hyperbola are not perpendicular. The problem is the plot's x-axis is in GeV and y-axis is unitless. In that case, what does do angles even mean much less what is perpendicular in any orientiation?" /></a></p>
|
643,560 | <p>Let $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ be sequences of real numbers. Does the following hold:</p>
<p>$$
\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n).
$$ </p>
<p>This is what I have tried but I am not quite sure if it is correct.
$\text{Fix } K>1. \text{ Let }L=\inf_{1\le i \le k}y_i$. Now
$$
\sup_{1 \le i \le k}(x_i+y_i)\ge \sup_{1\le i \le k}(x_i+L)=L+\sup_{1\le i \le k}(x_i) =
\inf_{1\le i \le k}(y_i)+\sup_{1\le i \le k}(x_i).
$$
Now we take the limit as $k \rightarrow \infty$ and we have the desired result. Does this look OK?</p>
| taritari | 122,374 | <p>$$\limsup (x_n+y_n) = \lim_{n\to\infty} \left(\sup_{k>n}(x_k+y_k)\right) \leq \lim_{n\to\infty}\left(\sup_{k>n}x_k+\sup_{k>n}y_k\right)=\\=\limsup(x_n)+\limsup (y_n)$$</p>
<p>The inequality is because a property of supreme: $\sup (a+b) \leq \sup (a) + \sup (b)$.</p>
|
268,152 | <p>I often see proofs, that claim to be <em>by induction</em>, but where the variable we induct on <em>doesn't</em> take value is $\mathbb{N}$ but only in some set $\{1,\ldots,m\}$.</p>
<p>Imagine for example that we have to prove an equality that encompasses $n$ variables on each side, where $n$ can <em>only</em> range through $n\in\{1,\ldots,m\}$ (imagine that for $n>m$ the function relating the variables on one of the sides isn't welldefined): For $n=1$ imagine that the equality is easy to prove by manipulating the variables algebraically and that for <em>some</em> $n\in\{1,\ldots,m\}$ we can also show the equation holds provided it holds for $n\in\{1,\ldots,n-1\}$. Then we have proved the equation for every $n\in\{1,\ldots,m\}$, <em>but</em> does this qualify as a <strong>proof by induction</strong> ?</p>
<p>(Correct me if I'm wrong: If the equation were indeed definable and true for every $n\in\mathbb{N}$ we could - although we are only interested in the case $n\in\{1,\ldots,m\}$ - "extend" it to $\mathbb{N}$ and then use "normal" induction to prove it holds for every $n\in \mathbb{N}$, since then it would also hold for $n\in\{1,\ldots,m\}$ .)</p>
| Hagen von Eitzen | 39,174 | <p>If the statement in question really does not "work" if $n>m$, then necessarily the induction step $n\to n+1$ at least <em>somewhere</em> uses that $n<m$. You may view this as actually proving by induction
$$\tag1\forall n\in \mathbb N\colon (n>m\lor \phi(m,n))$$
That is, you first show
$$\tag2\phi(m,1)$$
(which of course implies $1>m\lor \phi(m,1)$) and then show
$$\tag3(n>m\lor \phi(m,n))\Rightarrow (n+1>m\lor \phi(m,n+1)),$$
where you can more or less ignore the trivial case $n\ge m$.
Of course $(2)$ and $(3)$ establish a perfectly valid proof of $(1)$ by induction on $n$.</p>
|
1,530,848 | <p>Let $F(\mathbb{R})$ be the set of all functions $f : \mathbb{R} → \mathbb{R}$. Define pointwise addition and multiplication as follows. For any $f$ and $g$ in $F(\mathbb{R})$ let:</p>
<p>(i) $(f + g)(s) = f(x) + g(x)$ for all $x \in \mathbb{R}$</p>
<p>(ii) $(f · g)(s) = f(x) · g(x)$ for all $x \in \mathbb{R}$</p>
<p>Prove that $F(\mathbb{R})$ forms a ring under these two operations.</p>
<p>I know the axioms of rings that I need to show hold for this to be true. My question is, when I am showing closure for instance,</p>
<p>$(f + g)(s) +(f + g)(t) = f(x) + g(x)+ f(y) + g(y) \in \mathbb{R}$ and</p>
<p>$(f · g)(s)·(f · g)(t) = f(x) · g(x) · f(y) · g(y) \in \mathbb{R}$</p>
<p>is this the correct way to apply the axioms? My book was very general about the definition of a ring. The whole pointwise notation versus the definition is confusing me. Any help would be appreciated.</p>
| beroal | 7,011 | <p>“This is not really an answer, but it was getting too long to be a comment.”</p>
<blockquote>
<p>I see in each new chapter that I study in calculus or linear algebra,
there is an inherent geometrical intuition that can be easily
visualized</p>
</blockquote>
<p>…And you expect it to generalize to the whole mathematics. Unfortunately, it does not. For example, what you saw in linear algebra does not generalize to algebra. Everything can be proved by symbolic manipulation, but different visual tools are needed for different topics in mathematics, and sometimes there are no visual tool at all, and we have no other choice than to do without it. Hence visual tools are supplements, not a foundation. Finding them is hit-or-miss, I would say it is like collector’s job. ☺</p>
<p>As <a href="https://math.stackexchange.com/a/1530911/7011">this answer</a> points out, we can gain intuition about manipulating symbols by practice, I mean, what rule will likely lead to a useful result. Then we can develop <strong>visual tools for manipulating symbols</strong>. It is the approach which I use. Diagrams in category theory are a famous example, and value-relation diagrams are similar and suitable for more elementary thinking about points, sets, and relations.</p>
|
2,214,137 | <p>How many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?</p>
<p>I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.</p>
<p>How would I solve this?</p>
| Arthur | 15,500 | <p>Hint: For each possible value of $a$, count the number of possible values of $b$ and $c$. It's a quite regular pattern, so you don't have to brute force much to find it.</p>
|
2,214,137 | <p>How many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?</p>
<p>I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.</p>
<p>How would I solve this?</p>
| Christian Blatter | 1,303 | <p>By stars and bars there are ${99\choose2}$ nonnegative solutions to the equation $x_1+x_2+x_3=97$. Among these there are $3\cdot49$ having two equal $x_i\in[0\ ..\ 48]$, so that there are $${1\over6}\left({99\choose2}-3\cdot49\right)=784$$
solutions satisfying $x_1<x_2<x_3$. For each of these the numbers $a:=1+x_1$, $b:=1+x_2$, $c:=1+x_3$ form an admissible triple.</p>
|
186,146 | <p>On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?</p>
<p>How about $T:V\rightarrow W$ linear surjection in general?</p>
| Robert Israel | 8,508 | <p>Let $T$ be any map of $V$ to $W$ that is onto but not one-to-one. A right inverse for $T$ is any $R: W \to V$ such that $T(R(x)) = x$ for every $x \in W$. In particular, for any $x$ such that there are $y_1 \ne y_2$ with $T(y_1) = T(y_2) = x$, you are free to make $R(x) = y_1$ or $R(x) = y_2$. If $x = u + v$ (and neither $u$ nor $v$ is $0$, so neither is $x$), at least one of these choices will not be $R(u) + R(v)$. So there will always be a nonlinear right inverse.</p>
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