qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,981,745 | <p>Wolfram Alpha shows that <span class="math-container">$$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$</span>
I tried to use the Fourier series
<span class="math-container">$$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$</span>
I am not sure how to continue from this point. I need some help.</p>
| Larry | 587,091 | <p>Thanks mrtaurho, I am able to finish it.
<span class="math-container">$$\begin{align}
\int_{0}^{2\pi}\pi^2\ln(1-\cos x)dx
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\left[-x^2\frac{\sin(2nx)}{2n}+\frac{2x\cos(2nx)}{4n^2}-\frac{2\sin(2nx)}{8n^3}\right]_{0}^{\pi}\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}{n}\frac{2\pi}{4n^2}\\
&= -\frac{8\pi^3}3\ln(2)-8\pi\sum_{n=1}^{\infty}\frac{1}{n^3}\\
&=-\frac{8}{3}\pi(\pi^2\ln2+3\zeta(3))
\end{align}$$</span></p>
|
666,461 | <p>The function $f(x)=x+\log x$ has only one root on $(0,\infty)$ which is in $(0,1)$.</p>
<p>Using the Intermediate value theorem: $f$ is continuous on $(0,\infty)$ and $f(0)=0+\log(0)=-\infty<0$ and $f(1)=1+\log(1)=1>0$. So there exists an $x$ such $f(x)=0$.</p>
<p>But how to show that this $x$ is the only root? </p>
| Mhenni Benghorbal | 35,472 | <p><strong>Hints:</strong></p>
<p>1) $\lim_{x\to 0^{+}}f(x)=-\infty .$</p>
<p>2) $f(1)=1$</p>
<p>3) $f(x)$ is increasing.</p>
<p>Can you conclude?</p>
|
2,930,413 | <p>The problem is as shown. I tried using gradient and Hessian but can not make any conclusions from them. Any ideas?</p>
<p><span class="math-container">$$\max x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$$</span></p>
<p>subject to</p>
<p><span class="math-container">$$\sum_{i=1}^nx_i=1,\quad x_i\geq 0,\quad i=1,2,\ldots,n,$$</span></p>
<p>where <span class="math-container">$a_i$</span> are given positive scalars. Find a global maximum and show that it is unique.</p>
| David G. Stork | 210,401 | <p><span class="math-container">$$x(t) = A \cos (2 \pi \omega t + \phi)$$</span></p>
<p>where <span class="math-container">$\omega = 1$</span>.</p>
|
2,930,413 | <p>The problem is as shown. I tried using gradient and Hessian but can not make any conclusions from them. Any ideas?</p>
<p><span class="math-container">$$\max x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$$</span></p>
<p>subject to</p>
<p><span class="math-container">$$\sum_{i=1}^nx_i=1,\quad x_i\geq 0,\quad i=1,2,\ldots,n,$$</span></p>
<p>where <span class="math-container">$a_i$</span> are given positive scalars. Find a global maximum and show that it is unique.</p>
| user0102 | 322,814 | <p><strong>HINT</strong></p>
<p>Suppose <span class="math-container">$x(t) = e^{kt}$</span>, where <span class="math-container">$k\in\mathbb{C}$</span>:
<span class="math-container">\begin{align*}
x^{\prime\prime} + x = 0 \Longleftrightarrow k^{2}e^{kt} + e^{kt} = 0 \Longleftrightarrow k^{2} + 1 = 0 \Longleftrightarrow k = \pm i
\end{align*}</span>
Therefore <span class="math-container">$x(t) = Ae^{it} + Be^{-it}$</span>. Can you proceed from here?</p>
|
743,473 | <p>A long Weierstrass equation is an equation of the form
$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$
Why are the coefficients named $a_1, a_2, a_3, a_4$ and $a_6$ in this manner, corresponding to $xy, x^2, y, x$ and $1$ respectively? Why is $a_5$ absent?</p>
| drhab | 75,923 | <p>1) $$P\left(M\right)=P\left(M|L\right)P\left(L\right)+P\left(M|L^{c}\right)P\left(L^{c}\right)$$ </p>
<p>2) $$P\left(L\mid M^{c}\right)=\frac{P\left(L\cap M^{c}\right)}{P\left(M^{c}\right)}=\frac{P\left(L\cap M^{c}\right)}{P\left(L\right)}\frac{P\left(L\right)}{P\left(M^{c}\right)}=P\left(M^{c}\mid L\right)\frac{P\left(L\right)}{1-P\left(M\right)}=\left(1-P\left(M\mid L\right)\right)\frac{P\left(L\right)}{1-P\left(M\right)}$$ </p>
|
283,824 | <p>Let $x$ be a random vector uniformly distributed on the unit sphere $\mathbb{S}^{n-1}$. Let $V$ be a linear subspace of dimension $k$ and let $P_V(x)$ be the orthogonal projection of $x$ onto $V$.
I have seen quoted in the literature that
\begin{align}
\mathbb{P}[|\left\| P_V(x)\right\|_2 - \sqrt{k/n} | \le \epsilon] \ge 1 -2\exp(-n\epsilon^2/2). \, \, \, \, \, \, \, (1)
\end{align} However, i can still not find a concrete proof. What i do understand is that for a $1$-Lipschitz function $f:\mathbb{S}^{n-1} \rightarrow \mathbb{R}$ such as $x \mapsto |\left\| P_V(x)\right\|_2$, we have that
\begin{align}
\mathbb{P}[|f - M_f | \le \epsilon] \ge 1 -2\exp(-n\epsilon^2/2), \, \, \, \, \, \, \, (2)
\end{align} where $M_f$ is the median of $f$. (2) mostly follows from the isoperimetric inequality on the sphere. The issue though with (1) is that $\sqrt{k/n}$ does not seem to be the median of $x \mapsto |\left\| P_V(x)\right\|_2$. Is anyone able to provide a clean argument for (1) or a self-contained reference in the literature? Many thanks. </p>
| Iosif Pinelis | 36,721 | <p>In general, under regularity conditions such as compactness, one can use standard minimax duality argument ($\min_y\max_x F(x,y)=\max_x\min_y F(x,y)$ -- for bilinear or, more generally, concave-convex functions $F$) to quickly show that
\begin{multline}
W(\mu_1, \mu_2, \dots, \mu_m):=\inf_\gamma\int c d\gamma \\
=\inf_\gamma^*\sup_{f_1,\dots,f_m}^*\Big(
\int c(x) \gamma(dx)
+\sum_{j=1}^m \int f_i d\mu_i-\sum_{j=1}^m \int f_i(x_i) \gamma(dx)
\Big) \\
=\sup_{f_1,\dots,f_m}^*\inf_\gamma^*\Big(
\int c(x) \gamma(dx)
+\sum_{j=1}^m \int f_i d\mu_i-\sum_{j=1}^m \int f_i(x_i) \gamma(dx)
\Big) \\
=\sup
\sum_{j=1}^m \int f_i d\mu_i,
\end{multline}
where $\inf_\gamma$ is taken over all probability measures $\gamma$ with marginals $\mu_1, \mu_2, \dots, \mu_m$, $\inf\limits_\gamma^*$ is taken over all (nonnegative) measures $\gamma$ on $\mathcal{X}^m$,
$\sup\limits_{f_1,\dots,f_m}^*$ is taken over all appropriately integrable real-valued measurable functions $f_1,\dots,f_m$ on $\mathcal{X}$,
$\sup$ is taken over all such functions $f_1,\dots,f_m$ on $\mathcal{X}$ that satisfy the condition
\begin{equation}
c(x)\ge \sum_{j=1}^m f_i(x_i)\text{ for all }x\in\mathcal{X}^m,
\end{equation}
$x=(x_1,\dots,x_m)$, $dx=dx_1\times\dots\times dx_m$.<br>
The second equality above follows because
\begin{multline}
\sup_{f_1,\dots,f_m}^*\Big(
\sum_{j=1}^m \int f_i d\mu_i-\sum_{j=1}^m \int f_i(x_i) \gamma(dx)
\Big) \\
=\begin{cases}
0&\text{ if the marginals of $\gamma$ are $\mu_1, \mu_2, \dots, \mu_m$}, \\
\infty&\text{ otherwise.}
\end{cases}
\end{multline}
The third equality above is the minimax duality.
The last equality there
follows because
\begin{multline}
\inf_\gamma^*\Big(
\int c(x) \gamma(dx)
-\sum_{j=1}^m \int f_i(x_i) \gamma(dx)
\Big) \\
=\begin{cases}
0&\text{ if $c(x)\ge \sum_{j=1}^m f_i(x_i)\text{ for all }x\in\mathcal{X}^m$}, \\
-\infty&\text{ otherwise.}
\end{cases}
\end{multline}</p>
<blockquote>
<p>So, in general $\sum_{j=1}^m \int f_i d\mu_i$ and the condition $c(x)\ge \sum_{j=1}^m f_i(x_i)$ for all $x\in\mathcal{X}^m$ replace, respectively,
$\mathbb{E}_{\mu}f(X)-\mathbb{E}_\nu f(X)=\int f d\mu-\int f d\nu$
and the condition $|f(x)-f(y)|\leq d(x, y)$ for all $x, y\in \mathcal{X}$
in the definition of the Wasserstein distance. </p>
</blockquote>
<p>Even in the case $m=2$, the reduction from two functions $f_1$ and $f_2$ in the condition $c(x)\ge \sum_{j=1}^2 f_i(x_i)$ for all $x\in\mathcal{X}^2$ to one function $f$ in the condition $|f(x)-f(y)|\leq d(x, y)$ for all $x, y\in \mathcal{X}$ is based on the special properties of a metric, available only if $c=d$, a metric. This is how it is done: Suppose $d(x_1,x_2)\ge \sum_{j=1}^2 f_i(x_i)$ for all $x\in\mathcal{X}^2$, that is, $f(s):=\inf\limits_t[d(s,t)-f_2(t)]\ge f_1(s)$ for all $s$. Also, $f(s)\le d(s,s)-f_2(s)=-f_2(s)$ for all $s$, so that $f_1\le f$ and $f_2\le-f$ and hence
\begin{equation}
\sum_{j=1}^2 \int f_i\, d\mu_i\le \int f\, d\mu_1-\int f\, d\mu_2.
\end{equation}
Also,<br>
$$d(s,t)+f(t)=\inf\limits_u[d(s,t)+d(t,u)-f_2(u)]
\ge\inf\limits_u[d(s,u)-f_2(u)]=f(s)$$
and hence $d(s,t)\ge f(s)-f(t)$ for all $s,t$ or, equivalently,
$|f(t)-f(s)|\le d(s,t)$ for all $s,t$. </p>
|
4,138,292 | <p>Can anybody help me understand why these terms are always reciprocals? (theta <= 45°)</p>
<p><span class="math-container">$$ x = \frac{1}{\cos \theta} + \tan{\theta} $$</span>
<span class="math-container">$$ \frac{1}{x} = \frac{1}{\cos \theta} - \tan{\theta} $$</span></p>
<p>I understand that if we multiply them, they equal <span class="math-container">$1$</span> (because of the equation for a circle).</p>
<p><span class="math-container">$$\begin{align}
1 &= (\frac{1}{\cos \theta} + \tan{θ})(\frac{1}{\cos \theta} - \tan{θ}) \\[4pt]
1 &= \frac{1}{(\cos{\theta})^{2}} - \frac{\tan{\theta}}{\cos{\theta}} + \frac{\tan{\theta}}{\cos{\theta}} - (\tan{\theta})^2 \\[4pt]
1 &= \frac{1}{(\cos{\theta})^2} - (\tan{\theta})^2 \\[4pt]
(\cos{\theta})^2 &= 1 - (\cos{\theta})^2(\tan{\theta})^2 \\[4pt]
(\cos{\theta})^2 &= 1 - (\sin{\theta})^2
\end{align}$$</span></p>
<p>But I am looking for a deeper understanding? Regards</p>
| bjcolby15 | 122,251 | <p>Proof:</p>
<p>If <span class="math-container">$$x = \sec \theta + \tan \theta \ (\text {here} \sec \theta = \dfrac {1}{\cos \theta})$$</span></p>
<p>Then <span class="math-container">$$ \dfrac {1}{x} = \dfrac {1}{\sec \theta+\tan \theta}$$</span></p>
<p>Rationalizing the denominator by multiplying by <span class="math-container">$\sec \theta - \tan \theta$</span> we have <span class="math-container">$$ \dfrac {1}{x} = \dfrac {\sec \theta-\tan \theta}{\sec^2 \theta - \tan^2 \theta}$$</span></p>
<p>As <span class="math-container">$\sec^2 \theta - \tan^2 \theta = 1$</span> via the Pythagorean theorem, we then have <span class="math-container">$$ \dfrac {1}{x} = {\sec \theta-\tan \theta}.$$</span></p>
|
881,141 | <p>Let $A$ and $B$ be two covariance matrices such that $AB=BA$. Is $AB$ a covariance matrix?</p>
<p>A covariance matrix must be symmetric and positive semi definite. The symmetry of $AB$ can be proved as follows:
$$(AB)^T = B^TA^T = BA = AB$$</p>
<p>The question is, how to prove or disprove the positive semi definitive character of $AB$?</p>
| Quang Hoang | 91,708 | <p>Two commuting matrices can be diagonalized by the same matrix. The positive semi definite follows immediately.</p>
|
881,141 | <p>Let $A$ and $B$ be two covariance matrices such that $AB=BA$. Is $AB$ a covariance matrix?</p>
<p>A covariance matrix must be symmetric and positive semi definite. The symmetry of $AB$ can be proved as follows:
$$(AB)^T = B^TA^T = BA = AB$$</p>
<p>The question is, how to prove or disprove the positive semi definitive character of $AB$?</p>
| Qaswed | 333,427 | <p>Your question about "the positive semi definitive character of $AB$" can be answered as follows: $A\cdot B$ is positive semi definite if and only if $A \cdot B$ is normal (i.e. $(A\cdot B)^T\cdot (A\cdot B) = (A\cdot B)\cdot (A \cdot B)^T$). Reference: Meenakshi, A. and C. Rajian (1999). <em>On a product of positive semidefinite matrices</em>. Linear Algebra and its Applications 295 (1-3), 3–6. </p>
<p>Using ist symmetry $(A \cdot B)^T = (A\cdot B)$ we can in fact show that $A \cdot B$ is normal: $(A \cdot B)^T \cdot (A \cdot B) = (A \cdot B) \cdot (A \cdot B) = (A \cdot B) \cdot (A \cdot B)^T$. So $A B$ is symmetric and positive semi definite and thus a covariance Matrix.</p>
|
228,036 | <p>I quote from the <a href="http://en.wikipedia.org/wiki/Von_Neumann_cardinal_assignment" rel="nofollow">Wikipedia article</a>:</p>
<p>"So (assuming the axiom of choice) we identify $\omega_\alpha$ with $\aleph_\alpha$, except that the notation $\aleph_\alpha$ is used for writing cardinals, and $\omega_\alpha$ for writing ordinals. "</p>
<p>Let's remind ourselves of the definitions: </p>
<p><strong>(Def.1)</strong> The $\aleph_\alpha$ numbers are defined recursively, with $\aleph_0 = |\mathbb N| = \omega = \omega_0$ and $\aleph_{\alpha + 1} =$ the least ordinal such that it has strictly greater cardinality than $\aleph_\alpha$ for $\alpha \in \mathbf{ON}$.</p>
<p><strong>(Def.2)</strong> The initial ordinal of a cardinal $\kappa$ is defined to be the least ordinal of cardinality $\kappa$. Then $\omega_\alpha$ are the <a href="http://mathworld.wolfram.com/InitialOrdinal.html" rel="nofollow">transfinite initial ordinals</a>.</p>
<p>Then by definition, every $\aleph$ is an ordinal and also by definition, $\aleph_\alpha = \omega_\alpha$, with or without AC. What am I missing? Thanks for clarification. </p>
| Asaf Karagila | 622 | <p>The $\aleph$ numbers are <em>defined</em> as the initial ordinals, without any appeal to the axiom of choice. This is done by a transfinite recursion over the ordinals. $\aleph_\alpha$ is therefore the ordinal $\omega_\alpha$, which is the unique initial ordinal such that the set of all initial ordinals strictly smaller than itself has order type $\alpha$.</p>
<p>In some places define cardinals <em>only</em> as $\aleph$ numbers, and then you need AC to assert that every set has a cardinal number. Of course you can do it without AC but then there are non-$\aleph$ cardinals as well.</p>
<hr>
<p>One should note that before Dana Scott suggested the definition of a cardinal without the axiom of choice much later than von Neumann suggested using the $\aleph$ numbers. Historically, if so, it was "obvious" how to assign cardinals when the axiom of choice was present, but not without it. </p>
|
2,441,630 | <p>The operator given is the right-shift operator $T$ on $l^2$. We show that $\lambda=1$ is in the residual spectrum. Therefore we show that $(I-T)$ is injective but fails to have a dense range. While injectivity is clear, I fail to understand why the following shows that the range is not dense:</p>
<p>Let $y=(I-T)x$. Then $y(k)+ \cdots + y(1) = x(k)$ where $x(k)$ is the $k$-th entry of the sequence. Now for every $x\in l^2$ we have $\lim_{k\rightarrow \infty} x(k) = 0$ which then forces $|y(k) + \cdots y(1)| \rightarrow 0$ as well. Now this means that $(\alpha,0,0,...)$ cannot be in the range of $(I-T)$ which implies that the range of $(I-T)$ lies in the orthogonal complement of $(1,0,0,...)$.</p>
<p>Now first of all: <strong>why does it imply that it lies in the complement?</strong> and second of all <strong>why does that imply that the range isn't dense?</strong></p>
| John Hughes | 114,036 | <p>Not a complete answer, but at least some direction: </p>
<ol>
<li>I don't have an answer, although the fact that $T$ preserves orthogonality probably comes into it somewhere. </li>
</ol>
<p>But assuming this part to have been addressed, we have:</p>
<ol start="2">
<li>It's not dense because, for instance, the open ball of radius $1/2$ around $(1, 0, 0, \ldots)$ is entirely outside the orthogonal complement of $(1, 0, \ldots)$ (which consists entirely of things of the form $(0, *, *, \ldots)$, of course). </li>
</ol>
|
1,946,438 | <p>I solved the equation $e^{e^z}=1$ and it seemed to easy so I suspect I must be missing something.</p>
<blockquote>
<p>Would someone please check my answer?</p>
</blockquote>
<p>My original answer:</p>
<p>$e^{e^z}=1$ if and only if $e^z = 2\pi i k$ for $k\in \mathbb Z$ if and only if $z=\ln(2\pi i k)$ for $k\in \mathbb Z$.</p>
<p><strong>Edit</strong></p>
<p>After reading the comments and answers I tried to do it again. Unfortunately, I still do not get the same result as in the answers.</p>
<p><strong>My second attempt:</strong></p>
<p>We have </p>
<p>$$ e^x = 1 \iff x = 2 \pi i k$$</p>
<p>hence </p>
<p>$$ e^z = 2 \pi i k$$</p>
<p>for some $k$ in $\mathbb Z$. </p>
<p>Letting $e^z = e^x (\cos y + i \sin y)$ we get </p>
<p>$$ e^x \cos y + i e^x \sin y = 2 \pi k i$$</p>
<p>which implies that $\cos y = 0$ which happens if and only if $y_j = {\pi \over 2} + \pi j$ where $j\in \mathbb Z$. At $y_j$ we have
$\sin y = \pm 1$ hence if $j$ is even</p>
<p>$$ e^x = 2 \pi i k$$</p>
<p>and if $j$ is odd </p>
<p>$$ e^x = -2 \pi i k$$</p>
<p>Hence if $j$ is even,</p>
<p>$$ x = {\pi \over 2} + \ln(2 \pi k)$$</p>
<p>and if $j$ is odd,</p>
<p>$$ x = {3\pi \over 2} + \ln(2 \pi k)$$</p>
<p>So we see that the solutions are</p>
<p>$$
z_{t,k}=\begin{cases}
{\pi \over 2} + \ln(2 \pi k) + i ({\pi \over 2} + 2t \pi )\\
{3\pi \over 2} + \ln(2 \pi k) + i({\pi \over 2} + (2 +1)t \pi )
\end{cases}
$$
for $k,t \in \mathbb Z$.</p>
<blockquote>
<p>What am I doing wrong?</p>
</blockquote>
| Community | -1 | <p>Since $1$ can be written $1=e^0$ it follows that a first solution is
$$e^z=i2\pi k,\qquad (k \in \mathbb{Z})$$</p>
<p>If $k=0$ there are no solutions since $e^z$ is never zero.</p>
<p>If $k>0$ write $i2 \pi k$ in exponential polar form and you should find that
$$|i2 \pi k|=2 \pi |k| = 2 \pi k, \qquad (\text{since $k>0$})$$</p>
<p>So
\begin{align}
i 2 \pi k &= e^{i \pi /2}e^{\ln(2 \pi k)} \\
&= e^{i \pi /2 + \ln(2 \pi k)}
\end{align}
Hence solution for $k>0$ is of the form
$$z=\ln(2 \pi k)+i\left(\frac{\pi}{2}+2 \pi n \right), \qquad \text{for $n \in \mathbb{Z}, k \in \mathbb{Z} \cap (0, + \infty)$}$$</p>
<p>I leave the case $k < 0$ as an exercise for the OP.</p>
|
4,151 | <p>Since it's currently summer break, and I've a bit more time than normal, I've been organizing my old notes. I seem to have an almost unwieldy amount of old handouts and tests from classes previously taught. I'm hesitant to get rid of these, as they may provide useful for some future course. Because I adjunct at a few different colleges, with slightly different course content, I find that a lot of these materials are useful for multiple classes. I've begun organizing them according to specific content, but I'm curious, does anyone have an organizational style that they have found useful?</p>
| JPBurke | 759 | <p>While my data organizing problems are usually more related to research, they still often involve notes, papers, and sheets of data collection interview items that are much like handouts.</p>
<p>One solution that I have used (although I am still perfecting it) is to scan everything and get rid of the paper copies.</p>
<p>Once I have electronic copies of things, a number of organizational methods are possible. The most obvious is a hierarchical organization, which can be accomplished in any computer's file system folder structure. As a bonus, if you have some sort of cloud storage, store or mirror your file structure there so you can browse your scans any time. </p>
<p>The limitation of a hierarchical system is that you can only put a scan into one folder. That handout may have been useful for multiple projects or classes. So, a system that allows tagging is better if you need to be able to find one handout but you would like to remember that you used it in your most recent class (because you don't remember what class you first used it in). </p>
<p>Tags allow you to create a new tag for every use (or class). Then you can (and should) tag the handout for every use you have considered. </p>
<p>Example: lets say you decide to use Evernote to store all your scanned notes. It allows tagging. You have a handout you particularly like. You put it into Evernote and tag it with a subject matter tag. But then you also create a tag for the class you used it in. Finally, you add the "handout" tag that you give all your handouts. Now you can find the thing by scanning the list of handouts, or by remembering one of the classes you used it in, or by remembering the subject matter.</p>
<p>You can see that my main concern is finding stuff later. Information can pile up, and it's nice to have things like this easily retrievable. </p>
<p><a href="http://evernote.com/evernote/guide/" rel="nofollow">Here's the getting started guide for Evernote.</a> If you're going to scan your notes, that can be tedious. But you did say you had some time.</p>
|
57,769 | <p>Consider a finitely axiomatized theory $T$ with axioms $\phi_1,...,\phi_n$ over a first-order language with relation symbols $R_1,...,R_k$ of arities $\alpha_1,...,\alpha_k$. Consider the atomic formulas written in the form $(x_1,...,x_{\alpha_j})\ \varepsilon R_j$.</p>
<p>Translate this theory into a (finite) set-theoretic definition</p>
<p>$T(X) :\equiv (\exists R_1)...(\exists R_k) R_i \subseteq X^{\alpha_i} \wedge \phi'_1 \wedge ... \wedge \phi'_n$</p>
<p>where $\phi'_i$ is $\phi_i$ with $(\forall x)$ replaced by $(\forall x \in X)$ and $(x_1,...,x_{\alpha_j})\ \varepsilon R_j$ replaced by $(x_1,...,x_{\alpha_j})\ \in R_j$ with $(x_1,...,x_{\alpha_j})$ an abbreviation for ordered tuples.</p>
<p>To show that $T$ has a model — i.e. to show that $T$ is consistent — is to prove the statement $(\exists x) T(x)$ from the axioms of set theory. </p>
<p>It is essential that the relations fulfill the conditions $\phi_i$ simultaneously. Thus it is not clear at first sight, how the existence of a model of a theory can be proved (or even be stated set-theoretically) that is <em>not</em> finitely axiomatizable, since it cannot be translated into a finite sentence.</p>
<p>Some other things are not clear (to me):</p>
<ol>
<li><p>In this setting, doesn't the consistency of every theory dependend on the consistency of the choosen set theory? (If the set theory isn't consistent, every theory has a model.)</p></li>
<li><p>Furthermore, doesn't the consistency of a theory depend on the choice of the set theory in which $(\exists x) T(x)$ is proved? (In some set theories $(\exists x) T(x)$ can be proved, in others maybe not.)</p></li>
<li><p>What conditions has a theory to fulfill to be able to play the role of set theory in this setting? [It doesn't have to be the element relation $\in$ which $\varepsilon$ is mapped on. But one needs to be able to build ordered tuples of arbitrary length. What else? Something like powersets (since $R_i \subseteq X^{\alpha_i}$ is $R_i \in \mathcal{P}(X^{\alpha_i})$)? Is extensionality necessary? What is the general framework to discuss such questions?]</p></li>
</ol>
| Robert Israel | 8,508 | <p>In dealing with gravitational waves, yes. A quick Google search came up with lots of hits, e.g. <a href="http://arxiv.org/ftp/gr-qc/papers/0701/0701008.pdf" rel="nofollow">http://arxiv.org/ftp/gr-qc/papers/0701/0701008.pdf</a></p>
|
118,232 | <p>For example I have </p>
<pre><code>square = Graphics[Polygon[{{0, 0} ,{0, 1}, {1, 1}, {1, 0}}]]
</code></pre>
<p>What functions can I apply to <code>sqaure</code> to extract the coordinates of the polygon? It is necessary to do this kind of extraction when I have a graphics object as an argument of a function, and want to use coordinates taken from primitives within it to do some calculation. </p>
| bill s | 1,783 | <p>I'm afraid that you have calculated the stability incorrectly. Here is the Jacobian of your system:</p>
<pre><code>a = 2; b = 1; c = 2.5; d = 1.2; k = 2.8;
jac[x_, y_] := {{D[a x*(1 - x/k) - b x*y, x], D[a x*(1 - x/k) - b x*y, y]},
{D[-c y + d x*y, x], D[-c y + d x*y, y]}};
</code></pre>
<p>At the equilibrium, this is:</p>
<pre><code>jacEq = jac[x, y] //. {x -> 2.08333, y -> 0.511905}
</code></pre>
<p>The eigenvalues of this are:</p>
<pre><code>Re[Eigenvalues[jacEq]]
{-0.744047, -0.744047}
</code></pre>
<p>so the system is stable about this equilibrium.</p>
|
1,487,966 | <p>I have been looking at stereographic projections in books, online but they all seem...I don't know how else to put this, but very pedantic yet skipping the details of calculations.</p>
<p>Say, I have a problem here which asks;</p>
<blockquote>
<p>Let <span class="math-container">$n \geq 1$</span> and put <span class="math-container">$S^n=\{(x_0,x_1...,x_n) \in \mathbb{R}^{n+1}|{x_0}^2+...+{x_n}^2=1\}$</span> (So I understand this is a unit sphere in <span class="math-container">$n+1$</span> dimensions). Let <span class="math-container">$P=\{(1,0,...,0)\}$</span> and consider <span class="math-container">$S^n$</span>\ <span class="math-container">$P$</span> and <span class="math-container">$Y=\{(y_0...y_n)\in \mathbb{R}^{n+1}|y_0=0\}$</span> both with Euclidean metric. Thus <span class="math-container">$X$</span> is an <span class="math-container">$n$</span>-sphere with a point removed and <span class="math-container">$Y \cong \mathbb{R}^n$</span>.</p>
</blockquote>
<p>That is the set up. The problem I don't know how to approach is,</p>
<blockquote>
<p><span class="math-container">$i$</span>) For <span class="math-container">$x=(x_0,...,x_n) \in X$</span> and let <span class="math-container">$f(x)$</span> be a unique point of <span class="math-container">$Y$</span> such that <span class="math-container">$P=(1,0...0)$</span> and <span class="math-container">$x$</span>, <span class="math-container">$f(x)$</span> are collinear. So, find <span class="math-container">$ \lambda(x)$</span> such that <span class="math-container">$f(x)=\lambda(x)P+(1-\lambda(x))x$</span> fo some <span class="math-container">$\lambda(x) \in \mathbb{R}$</span>.</p>
<p><span class="math-container">$ii$</span>) For <span class="math-container">$y=(y_0,...y_n) \in Y$</span> let <span class="math-container">$g(y)$</span> be a unique point of <span class="math-container">$X$</span> such that <span class="math-container">$(1,0,...,0)$</span>, <span class="math-container">$y$</span> and <span class="math-container">$g(y)$</span> are collinear, by similar method to <span class="math-container">$i$</span>), find a formula for <span class="math-container">$g(y)$</span>.</p>
<p><span class="math-container">$iii$</span>) Prove that <span class="math-container">$f : X \rightarrow Y$</span> and <span class="math-container">$g : Y \rightarrow X$</span> are inverse to each other and deduce that <span class="math-container">$X$</span> is homeomorphic to <span class="math-container">$Y$</span></p>
</blockquote>
<p>Leaving aside part iii, I don't get how to do i.
I have seen some examples on removing the North Pole(which is NOT my case) but then the equation for <span class="math-container">$f(x)$</span> appears out of nowhere in those cases.
I don't know where and how they were obtained, no explanations and steps were given.</p>
<p>So, <span class="math-container">$\lambda(x)$</span> is a real number or, at least a scalar I understand. Given my conditions, I tried substituting the points to the given collinear form but found only that <span class="math-container">$\frac{y_0-x_0}{1-x_0} = \lambda(x)$</span> and also <span class="math-container">$\frac{x_i-y_i}{x_i} = \lambda(x)$</span> unless <span class="math-container">$i=0$</span>. Which doesn't make sense to me, as I found well, 2 different <span class="math-container">$\lambda(x)$</span>s (haven't I??)</p>
<p>I just don't get this stereographic projection thing "analytically". I have seen pictures and diagrams which visualises it and that's all very nice but algebraically/analytically, I cannot make sense of it.</p>
<p>Would anyone care helping me out at all?? Thank you so much....</p>
| mr_e_man | 472,818 | <p>For the function $g$, we need to find a point $x=g(y)$ on the unit sphere $x^2=1$, which is also on the line between $y$ and the pole $p$. (Also $p^2=1$.) So there should be some scalar $k$ such that</p>
<p>$$x = p + k(y-p)$$</p>
<p>$$x^2 = p^2 + 2k\,p\cdot(y-p) + k^2(y-p)^2$$</p>
<p>Cancel $x^2=p^2$, and $k\neq0$, so</p>
<p>$$0 = 2p\cdot(y-p) + k(y-p)^2$$</p>
<p>$$k = -2\frac{p\cdot(y-p)}{(y-p)^2}$$</p>
<p>Substitute this $k$ in the first equation:</p>
<p>$$x = p - 2\Big(\frac{p\cdot(y-p)}{(y-p)\cdot(y-p)}\Big)(y-p)$$</p>
<p>[The second term can be recognized as the projection of $p$ parallel to $(y-p)$ :</p>
<p>$$x = p - 2p_\parallel = (p_\parallel + p_\perp) - 2p_\parallel = -p_\parallel + p_\perp$$</p>
<p>So $x=g(y)$ is the reflection of $p$ along the vector $(y-p)$. This makes sense, because a reflection of a unit vector is also a unit vector.]</p>
<p>We also have $p\cdot y=0$, because $y$ is in the plane normal to $p$, so</p>
<p>$$x = p - 2\Big(\frac{0-1}{y^2+1}\Big)(y-p)$$</p>
<p>$$= p + 2\frac{y-p}{y^2+1}$$</p>
<p>Now you can find the coordinates (though I think the vector expression is neater):</p>
<p>$$x_0 = x\cdot p = 1 + 2\frac{0-1}{y^2+1} = \frac{(y^2+1)+2(-1)}{y^2+1} = \frac{y^2-1}{y^2+1}$$</p>
<p>$$x_i = 0 + 2\frac{y_i-0}{y^2+1} = \frac{2y_i}{y^2+1},\qquad 1 \leq i \leq n$$</p>
<hr>
<p>For the function $f$, we need to find a point $y=f(x)$ in the plane $y\cdot p=0$, which is also on the line between $x$ and $p$. So, again,</p>
<p>$$y = p + k(x-p)$$</p>
<p>$$y\cdot p = p^2 + k(x-p)\cdot p$$</p>
<p>$$0 = 1 + k(x-p)\cdot p$$</p>
<p>$$k = \frac{-1}{(x-p)\cdot p}$$</p>
<p>Substitute $k$ :</p>
<p>$$y = p - \frac{x-p}{(x-p)\cdot p}$$</p>
<p>$$= \frac{\big((x-p)\cdot p\big)p - (x-p)}{(x-p)\cdot p} = \frac{\big(x\cdot p\big)p - x}{(x-p)\cdot p} = \frac{x - \big(x\cdot p\big)p}{-(x-p)\cdot p}$$</p>
<p>[The second term can be recognized as the projection of $x$ parallel to $p$ :</p>
<p>$$y = \frac{x - x_\parallel}{-(x-p)\cdot p} = \frac{x_\perp}{-(x-p)\cdot p}$$]</p>
<p>Using that $x\cdot p=x_0$, the coordinates are</p>
<p>$$y_0 = \frac{x_0 - (x_0)1}{1-x_0} = 0$$</p>
<p>$$y_i = \frac{x_i - (x_0)0}{1-x_0} = \frac{x_i}{1-x_0},\qquad 1 \leq i \leq n$$</p>
<hr>
<p>It can be seen that $f$ and $g$ are inverses:</p>
<p>$$g(f(x)) = p + 2\frac{f(x) - p}{f(x)^2+1}$$</p>
<p>$$= p + 2\frac{\Big(\frac{x - (x\cdot p)p}{1-x\cdot p}\Big) - p}{\Big(\frac{x - (x\cdot p)p}{1-x\cdot p}\Big)^2+1}$$</p>
<p>$$= p + 2\big(1-x\cdot p\big)\frac{\big(x - (x\cdot p)p\big) - p\big(1-x\cdot p\big)}{\big(x - (x\cdot p)p\big)^2+\big(1-x\cdot p\big)^2}$$</p>
<p>$$= p + 2\big(1-x\cdot p\big)\frac{x - p}{\big(1 - 2(x\cdot p)^2 + (x\cdot p)^2\big)+\big(1 - 2x\cdot p + (x\cdot p)^2\big)}$$</p>
<p>$$= p + 2\big(1-x\cdot p\big)\frac{x - p}{2 - 2x\cdot p}$$</p>
<p>$$= p + (x - p)$$</p>
<p>$$= x$$</p>
<p>and</p>
<p>$$f(g(y)) = \frac{g(y) - \big(g(y)\cdot p\big)p}{1-g(y)\cdot p}$$</p>
<p>$$= \frac{\big(p + 2\frac{y-p}{y^2+1}\big) - \Big(\big(p + 2\frac{y-p}{y^2+1}\big)\cdot p\Big)p}{1-\big(p + 2\frac{y-p}{y^2+1}\big)\cdot p}$$</p>
<p>$$= \frac{0 + 2\frac{y}{y^2+1} - 2\frac{y\cdot p}{y^2+1}p}{0 - 2\frac{y\cdot p-1}{y^2+1}}$$</p>
<p>$$= \frac{y - (y\cdot p)p}{1 - y\cdot p}$$</p>
<p>$$= \frac{y - 0}{1 - 0}$$</p>
<p>$$= y$$</p>
<hr>
<p>Note that the projection functions $f$ and $g$ can easily have their domains extended to all of space (except the plane $x\cdot p=1$), but their ranges are still the sphere and the plane, so the extensions are not invertible.</p>
|
8,023 | <p>I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for $n=1$, all Riemannian manifolds are modeled on $\mathbb R$. When $n=2$, I believe that it suffices for the scalar curvature to vanish everywhere (this is certainly necessary). But my intuition is poor for higher-dimensional structures.</p>
<p>Put another way: given a Riemannian structure $g$ on a smooth manifold, when can I find coordinates $x^1,\dots,x^n$ so that $g_{ij}(x) = \delta_{ij}$?</p>
| Deane Yang | 613 | <p>If the Riemannian metric is twice differentiable in some co-ordinate system, then this holds in any dimension if and only if the Riemann curvature tensor vanishes identically.</p>
<p>In dimension 2, it suffices for the scalar curvature to vanish.
In dimension 3, it suffices for the Ricci curvature to vanish.
In higher dimensions, you need to have the full Riemann curvature vanish.</p>
|
8,023 | <p>I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for $n=1$, all Riemannian manifolds are modeled on $\mathbb R$. When $n=2$, I believe that it suffices for the scalar curvature to vanish everywhere (this is certainly necessary). But my intuition is poor for higher-dimensional structures.</p>
<p>Put another way: given a Riemannian structure $g$ on a smooth manifold, when can I find coordinates $x^1,\dots,x^n$ so that $g_{ij}(x) = \delta_{ij}$?</p>
| José Figueroa-O'Farrill | 394 | <p>Deane already answered the question. I just want to add that knowing the existence of local flat coordinates (by the vanishing of the curvature) and actually <em>finding</em> the flat coordinates are two very different things. I've had "fun" in the past finding explicit flat coordinates for flat metrics and it can be nontrivial, albeit highly satisfying when you work them out!</p>
|
3,869,237 | <p>I know this is quite weird or it does not make much sense, but I was wondering, does <span class="math-container">$\int e^{dx}$</span> has any meaning or whether it makes sense at all? If it does means something, can it be integrated and what is the result?</p>
| Community | -1 | <p>The differential in an integral is essentially a symbolic way to show on which variable you integrate and is not to be taken as a factor. With this in mind,</p>
<p><span class="math-container">$$\int e^{dx}$$</span> is just syntactically unparsable, in the same way as</p>
<p><span class="math-container">$$\sin(x=(+$$</span></p>
|
3,421,858 | <p><span class="math-container">$\sqrt{2}$</span> is irrational using proof by contradiction.</p>
<p>say <span class="math-container">$\sqrt{2}$</span> = <span class="math-container">$\frac{a}{b}$</span> where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers. </p>
<p><span class="math-container">$b\sqrt{2}$</span> is an integer. ----[Understood]</p>
<p>Let <span class="math-container">$b$</span> denote the smallest such positive integer.----[My understanding of this is </p>
<p>that were are going to assume b is the smallest possible integer such that <span class="math-container">$\sqrt{2}$</span> = <span class="math-container">$\frac{a}{b}$</span>, ... Understood]</p>
<p>Then <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span>----[I'm not sure I understand the point that's being made here,</p>
<p>from creating a variable <span class="math-container">$b^{*} = a - b$</span> ]</p>
<p>Next, <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span> is a positive integer such that <span class="math-container">$b^{*}\sqrt{2}$</span> is an integer.----[ I get that (<span class="math-container">$a - b$</span>) has to</p>
<p>be a positive integer, why does it follow that then <span class="math-container">$b^{*}\sqrt{2}$</span> is an integer?]</p>
<p>Lastly, <span class="math-container">$b^{*}<b$</span>, which is a contradiction.----[I can see that given <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span>, we then have </p>
<p><span class="math-container">$b^{*}<b$</span>, I don't get how that creates a contradiction]</p>
<p>Any help is appreciated, thank you. </p>
| Noah Schweber | 28,111 | <p>This is a bit oddly phrased. I think it would be better to write it as follows:</p>
<ul>
<li><p>Suppose <span class="math-container">$\sqrt{2}$</span> is rational. Let <span class="math-container">$b$</span> be the smallest positive integer such that <span class="math-container">$b\sqrt{2}$</span> is a positive integer. </p>
<ul>
<li>Note the difference between this and your second [comment].</li>
</ul></li>
<li><p>Let <span class="math-container">$a:=b\sqrt{2}$</span>, and let <span class="math-container">$b^*=b(\sqrt{2}-1)$</span>. A priori this doesn't look like an integer, but it is: we have <span class="math-container">$$b^*=b\sqrt{2}-b=a-b.$$</span></p></li>
<li><p>OK, now multiply both sides of the definition of <span class="math-container">$b^*$</span> by <span class="math-container">$\sqrt{2}$</span>; this yields <span class="math-container">$$b^*\sqrt{2}=b(\sqrt{2}-1)\sqrt{2}=2b-b\sqrt{2}=2b-a.$$</span> Clearly <span class="math-container">$2b-a$</span> is an integer, so <span class="math-container">$b^*\sqrt{2}$</span> is an integer. </p>
<ul>
<li>This addresses your fourth [comment]: the point is just that by expanding things out appropriately we can re-apply the assumption that <span class="math-container">$b\sqrt{2}$</span> is an integer.</li>
</ul></li>
<li><p>But <span class="math-container">$0<b^*<b$</span> (since <span class="math-container">$0<\sqrt{2}-1<1$</span>); so <span class="math-container">$b^*$</span> is a positive integer smaller than <span class="math-container">$b$</span> which when multiplied by <span class="math-container">$\sqrt{2}$</span> yields an integer. <strong>This contradicts the definition of <span class="math-container">$b$</span>.</strong> </p>
<ul>
<li>... and addresses your third and fifth [comments] - the number <span class="math-container">$b^*$</span> has a property it can't possibly have.</li>
</ul></li>
</ul>
|
3,730,083 | <p>If <span class="math-container">$a,b>0$</span> and <span class="math-container">$Q=\{x_1, x_2, x_3,..., x_a\}$</span> a subset of the natural numbers <span class="math-container">$1, 2, 3,..., b$</span> such that, for <span class="math-container">$x_i+x_j<b+1$</span> with <span class="math-container">$1 ≤ i ≤ j ≤ a$</span>, then <span class="math-container">$x_i+x_j$</span> is also an element of Q. Prove that:<br> <br>
<span class="math-container">$ \frac{x_1+x_2+x_3+...+x_a}{a} ≥ \frac{b+1}{2}$</span> <br>so basically, you have to prove that the arithmetic mean of Q satisfying the condition is greater than equal to the arithmetic mean of the set the natural numbers <span class="math-container">$1, 2, 3,..., b$</span>.<br><br>
Any help would be appreciated. Thanks in advance!</p>
| user | 293,846 | <p>Yes there is. The number of vertices is <span class="math-container">$2^n$</span> and the number of edges is <span class="math-container">$n2^{n-1}$</span>. Generally the number of <span class="math-container">$d$</span>- dimensional elements in <span class="math-container">$n$</span>- dimensional cube is
<span class="math-container">$$\binom nd 2^{n-d}.$$</span></p>
|
178,823 | <p>How would I prove the following trig identity? </p>
<blockquote>
<p><span class="math-container">$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$</span></p>
</blockquote>
<p>My work thus far has been:
<span class="math-container">$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\dfrac{A-B}{2}} =-\cot\dfrac{A+B}{2} \cot\dfrac{A-B}{2} \ .$$</span></p>
| Lance Helsten | 26,902 | <p>A couple of hints: use the $\cot$ identity on the RHS to put in terms of $\sin$ and $\cos$, use a product to sum identity, and pay attention to odd and even functions. This should simplify it to an easier expression.</p>
|
428,530 | <p>Let $\Omega := [0, 1] \times [0,\pi]$. We are searching for a function $u$ on $\Omega$ s.t.
$$
\Delta u =0
$$
$$
u(x,0) = f_0(x), \quad u(x,1) = f_1(x), \quad u(0,y) = u(\pi,y) = 0
$$ with
$$
f_0(x) = \sum_{k=1}^\infty A_k \sin kx \quad, f_1(x) = \sum_{k=1}^\infty B_k \sin kx
$$
If I use seperation of variables, say $u(x,y) = f(x)g(y)$ I get
$$
f''(x)+\lambda f(x) = 0 , \quad g''(y)-\lambda g(y) = 0
$$ with $f(0) = f(\pi) = 0$ where I use that $f,g \neq 0$. $\lambda$ is some constant. How can I proceed ?</p>
<p>Thanks in advance.</p>
| Community | -1 | <p>After some time, work and help by Avitus I finally got it :</p>
<p>Assume <span class="math-container">$u(x,y) = A(x)B(y)$</span> with <span class="math-container">$A,B \neq 0$</span> on <span class="math-container">$\Omega$</span>. This yields to
<span class="math-container">$$
A''(x)+\lambda A(x) = 0 , \quad B''(y)-\lambda B(y) =0
$$</span> General solutions are
<span class="math-container">$$
A(x) = \gamma_1 e^{\sqrt \lambda i x} + \gamma_2 e^{-\sqrt \lambda ix}
$$</span> which gives with the conditions <span class="math-container">$u(0,y)=u(0,\pi) = 0$</span> that <span class="math-container">$A_k(x) =\gamma_k \sin |k| x$</span> for <span class="math-container">$k \in \mathbb Z$</span>. For <span class="math-container">$B(y)$</span> we find then
<span class="math-container">$$
B_k(y) = c_k \sinh |k|y + d_k \cosh |k|y
$$</span> Then we have that
<span class="math-container">$$
u(x,y) = \sum_{k \in \mathbb Z} A_k(x)B_k(y) = \sum_{k=1}^\infty \sin kx \left ( \lambda_k \sinh ky + \mu_k \cosh ky \right )
$$</span> where <span class="math-container">$\lambda_k = \gamma_kc_k +\gamma_{-k}c_{-k}$</span> and similar for <span class="math-container">$\mu_k$</span>.
By using <span class="math-container">$u(x,0)= f_0(x)$</span> and <span class="math-container">$u(x,1)=f_1(x)$</span> we find by comparing coefficients that
<span class="math-container">$$
\mu_k = A_k, \quad \lambda_k = \frac{B_k -A_k\cosh k}{\sinh k}
$$</span> Writing out <span class="math-container">$\lambda_k \sinh ky + \mu_k \cosh ky$</span> and using that <span class="math-container">$\sinh(a-b) = \sinh a\cosh b - \sinh b \cosh a$</span> we find that
<span class="math-container">$$
u(x,y) = \sum_{k=1}^\infty \sin kx \left ( A_k \frac{\sinh k(1-y)}{\sinh k} + B_k \frac{ \sinh ky }{\sinh k} \right )
$$</span></p>
|
237,142 | <p>I am having a problem with the final question of this exercise.</p>
<p>Show that $e$ is irrational (I did that). Then find the first $5$ digits in a decimal expansion of $e$ ($2.71828$).</p>
<p>Can you approximate $e$ by a rational number with error $< 10^{-1000}$ ? </p>
<p>Thank you in advance</p>
| lhf | 589 | <p>In the standard proof that $e$ is irrational, one first proves that
$$
0 < e -s_n < \frac1{n!n}
\qquad\mbox{where}\qquad
s_n = \sum_{k=0}^n \frac1{k!}
$$
So you only need to find $n$ such that $\frac1{n!n}< 10^{-1000}$ or $n!>10^{1000}$. You can use Stirling's approximation for that I guess. <a href="http://www.wolframalpha.com/input/?i=solve+n%21%3E+10%5E1000" rel="nofollow">Wolfram Alpha</a> says $n=450$ suffices.</p>
|
2,130,658 | <p>How would I go about proving this mathematically? Having looked at a proof for a similar question I think it requires proof by induction. </p>
<p>It seems obvious that it would be even by thinking about the first few cases. As for $n=0$ there will be no horizontal dominoes which is even, and for $n=1$ there can only be one vertical domino so there are $0$ horizontal dominoes, which is again even. Then for $n=2$ you can have either two horizontal or two vertical dominoes which again gives $0$ or $2$ horizontal dominoes which is again an even number. And so on for n greater than $2$.</p>
<p>I would like to prove that the number of way of dividing a 2-high-by-n-wide rectangle into dominoes so that $2j$ dominoes are horizontal is ${n-j\choose j}$ and deduce that $U_n$ (where $U_n$ is the number of ways to divide a 2-high-by-n-wide rectangle into 2-wide-by-1-high dominoes) = $$\sum_j {n-j\choose j}$$ where this sum is over all the integers $j$ with $0\le j\le \frac{n}{2}$.</p>
<p>I understand that trivially for a 2-high-by-n-wide rectangle you can divide it by exactly $2j=n$ horizontal dominoes or by $n$ vertical dominoes or some combination of vertical and horizontal dominoes, but how can I use this knowledge to construct the proof?</p>
| fleablood | 280,126 | <p>Ah, you mean the number of horizontal dominos in <em>BOTH</em> rows combined is even!</p>
<p>Consider each square in the grid. Either it is occupied by a vertical or horizontal domino. If it is occupied by a vertical domino, than so is the square in the grid directly below or directly above. If it is occupied by a horizontal domino then it is not occupied by a vertical domino and neither is the square directly above or below.</p>
<p>So each of the two rows have the exact same composition. So the number of horizontal dominos in the first row are the same number (and int the exact same positions) as the number of horizontal dominoes in the second row. </p>
<p>Thus there is an even number of them.</p>
|
1,635,136 | <p>In Kelley's "General topology" (in the "Appendix") the <em>full classes</em> $X$ are defined as those with the property
$$
\forall A\in X\quad A\subseteq X.
$$
In the Russian translation it is added that this is equivalent to the property that the relation $\in$ is transitive on $X$:
$$
\forall A,B,C\in X\quad A\in B\in C \ \Rightarrow\ A\in C.
$$
I think there must be an easy trick for proving this equivalence, but I don't see it. Can anybody help me?</p>
| hmakholm left over Monica | 14,366 | <p>The first property (the one you call "full class") is more commonly known as a <strong>transitive</strong> class/set in set theory.</p>
<p>It is not entirely uncommon to be confused by this name and get the misconception that this is the same as "a class on which the $\in$ relation is transitive" -- and apparently the translator thought this -- but that is not true, as Pedro's example $\{\{0\}\}$ shows.</p>
<p>Being transitive/full <em>is</em>, of course equivalent to
$$ \forall A, B[ A\in B\in X \Rightarrow A\in X \bigl] $$</p>
|
1,635,136 | <p>In Kelley's "General topology" (in the "Appendix") the <em>full classes</em> $X$ are defined as those with the property
$$
\forall A\in X\quad A\subseteq X.
$$
In the Russian translation it is added that this is equivalent to the property that the relation $\in$ is transitive on $X$:
$$
\forall A,B,C\in X\quad A\in B\in C \ \Rightarrow\ A\in C.
$$
I think there must be an easy trick for proving this equivalence, but I don't see it. Can anybody help me?</p>
| Pedro Sánchez Terraf | 212,120 | <p>I already posted one counterexample in the comment section. The one for the other direction is the following. </p>
<p>Intuitively, the class $V$ of all sets is transitive but it contains non transitive sets. So it is enough to get a “smaller” counterexample. For instance, $$
V_3=\mathcal{P}^3(\emptyset)=
\{0,\{0\},\{\{0\}\},\{0,\{0\}\}\}.
$$
The first three would do the job.</p>
|
2,965,082 | <blockquote>
<p>Suppose that <span class="math-container">$(X,\ d)$</span> and <span class="math-container">$(Y,\ \rho)$</span> are metric spaces, that
<span class="math-container">$f_n:X\to Y$</span> is continuous for each <span class="math-container">$n$</span>, and that <span class="math-container">$(f_n)$</span> convergence
pointwise to <span class="math-container">$f$</span> on <span class="math-container">$X$</span>. If there exists a sequence <span class="math-container">$(x_n)$</span> in <span class="math-container">$X$</span>
such that <span class="math-container">$x_n\to x$</span> in <span class="math-container">$X$</span> but <span class="math-container">$f_n(x_n)\not\to f(x)$</span>, show that
<span class="math-container">$(f_n)$</span> does not converge uniformly to <span class="math-container">$f$</span> on <span class="math-container">$X$</span>.</p>
</blockquote>
<p>I've managed to "prove" that the question is self-contradictory, so please find my error.</p>
<p><span class="math-container">$\forall n\in\mathbb{N}$</span>, <span class="math-container">$f_n$</span> is continuous at <span class="math-container">$x$</span>. Let <span class="math-container">$\epsilon>0$</span>. <span class="math-container">$\forall n\in\mathbb{N}$</span> <span class="math-container">$\exists\delta>0$</span> s.t. <span class="math-container">$\rho(f_n(y),\ f_n(x))<\epsilon/2$</span> for all <span class="math-container">$y\in X$</span> s.t. <span class="math-container">$d(y,\ x)<\delta$</span>. ------------ (1)</p>
<p>Since <span class="math-container">$x_m\to x$</span>, <span class="math-container">$\exists N_1\in\mathbb{N}$</span> s.t. <span class="math-container">$d(x_m,\ x)<\delta$</span> <span class="math-container">$\forall m\ge N_1$</span>. ------------- (2)</p>
<p>From (1) and (2),</p>
<p><span class="math-container">$\forall n$</span>, <span class="math-container">$\forall m\ge N_1\implies d(x_m,\ x)<\delta\implies \rho(f_n(x_m),\ f_n(x))<\epsilon/2$</span> ------------- (3)</p>
<p>Also, <span class="math-container">$f_n(y)\to f(y)$</span> for all <span class="math-container">$y\in X$</span> due to pointwise convergence. So, <span class="math-container">$\exists N_2\in\mathbb{N}$</span> s.t. <span class="math-container">$\rho(f_n(y),\ f(y))<\epsilon/2$</span> <span class="math-container">$\forall n\ge N_2$</span> and <span class="math-container">$\forall y\in X$</span>. ----------- (4)</p>
<p>Let <span class="math-container">$N_3=\max(N_1,\ N_2)$</span>. Suppose <span class="math-container">$n\ge N_3$</span>. Then <span class="math-container">$n\ge N_1$</span> and <span class="math-container">$n\ge N_2$</span>.</p>
<p><span class="math-container">$\begin{aligned}
\implies\rho(f_n(x_n), f(x))&\le\rho(f_n(x_n),\ f_n(x))+\rho(f_n(x),\ f(x)) \\
&<\epsilon/2+\epsilon/2\text{ [Using (3) and (4)]} \\
&=\epsilon
\end{aligned}$</span></p>
<p>I have thus "proved" that <span class="math-container">$f_n(x_n)\to f(x)$</span>, contradicting the question. Where have I gone wrong?</p>
| Alex Ortiz | 305,215 | <p>If we are more precise, we may see where the error is. I will use <span class="math-container">$|y-x|$</span> to mean <span class="math-container">$d(y,x)$</span> since I think it makes it clearer than working with <span class="math-container">$d$</span> and <span class="math-container">$\rho$</span>. It won't change any of the important details. </p>
<p>For each <span class="math-container">$n$</span>, <span class="math-container">$f_n$</span> is continuous at <span class="math-container">$x$</span>. Let <span class="math-container">$\epsilon>0$</span>. Then, for each <span class="math-container">$n\in\Bbb N$</span>, there is a <span class="math-container">$\delta = \delta(n)$</span>, which depends on <span class="math-container">$n$</span>, such that <span class="math-container">$|f_n(y)-f_n(x)|<\epsilon/2$</span> whenever <span class="math-container">$|y-x|<\delta(n)$</span>.</p>
<p>Since <span class="math-container">$x_m\to x$</span>, there is an <span class="math-container">$N_1 = N_1(n)\in\Bbb N$</span> (note that <span class="math-container">$N_1$</span> also depends on <span class="math-container">$n$</span> here!) such that <span class="math-container">$|x_m - x| < \delta(n)$</span> for each <span class="math-container">$m\ge N_1$</span>.</p>
<p>Now you claim that for each <span class="math-container">$k,m\ge N_1(n)$</span>, whenever <span class="math-container">$|x_m-x|<\delta(n)$</span>, you have <span class="math-container">$|f_k(x_m)-f_k(x)| < \epsilon/2$</span>.</p>
<p>This is the first serious point you erred. The reason this is an incorrect deduction is that for different <span class="math-container">$n$</span>, you don't know that <span class="math-container">$N_1 = N_1(n)$</span> is large enough to guarantee that <span class="math-container">$|f_k(x_m)-f_k(x)|<\epsilon/2$</span>. Only for the specific <span class="math-container">$n$</span> for which <span class="math-container">$\delta = \delta(n)$</span> is this true.</p>
<hr>
<p><strong>Edit:</strong> RRL summed it up nicely in the comment. You are assuming <em>equicontinuity</em> of the sequence <span class="math-container">$\{f_n\}$</span> when you are neglecting the <span class="math-container">$n$</span> that the <span class="math-container">$\delta = \delta(n)$</span> depends on.</p>
|
1,707,132 | <blockquote>
<p>Let $X$ be a contractible space (i.e., the identity map is homotopic to the constant map). Show that $X$ is simply connected.</p>
</blockquote>
<p>Let $F$ be the homotopy between $\mathrm{id}_X$ and $x_0$, that is $F:X\times [0,1]\to X$ is a continuous map such that
$$ F(x,0)=x,\quad F(x,1)=x_0$$
for all $x\in X$.
Next let $f:[0,1]\to X$ be a loop based $x_0$, I need to show that $f$ is path homotopic to the constant path $e_{x_0}$, The continuous map $G:[0,1]\times [0,1]\to X$ defined by $G(s,t)=F(f(s),t))$ is a homotopy between $f$ and $e_{x_0}$, but it may not be a path homotopy (i.e., $G(0,t)=G(1,t)=x_0$ for all $t\in [0,1]$), how to get around this?</p>
| Faraad Armwood | 317,914 | <p>Since $X$ is contractible there exists a homotopy $h=h_t: id_X \to x_0$ where $x_0 \in X$. Therefore, if we take any loop $\alpha: I \to X$ then $h|_{\alpha}$ takes $\alpha$ to $x_0$. Can you show $X$ is path-connected? You know $h_t$ takes every point in $X$ to $x_0$ therefore given any $x,y \in X$ then $h_{2t}(x) * h^{-1}_{2t-1}(y)$ is a path from $x$ to $y$ where $h^{-1}$ represents to reverse of the homotopy i.e $h^{-1}_0(y) = x_0$ and $h^{-1}_1(y) = y$. </p>
|
1,707,132 | <blockquote>
<p>Let $X$ be a contractible space (i.e., the identity map is homotopic to the constant map). Show that $X$ is simply connected.</p>
</blockquote>
<p>Let $F$ be the homotopy between $\mathrm{id}_X$ and $x_0$, that is $F:X\times [0,1]\to X$ is a continuous map such that
$$ F(x,0)=x,\quad F(x,1)=x_0$$
for all $x\in X$.
Next let $f:[0,1]\to X$ be a loop based $x_0$, I need to show that $f$ is path homotopic to the constant path $e_{x_0}$, The continuous map $G:[0,1]\times [0,1]\to X$ defined by $G(s,t)=F(f(s),t))$ is a homotopy between $f$ and $e_{x_0}$, but it may not be a path homotopy (i.e., $G(0,t)=G(1,t)=x_0$ for all $t\in [0,1]$), how to get around this?</p>
| Ivin Babu | 704,464 | <p>Given that X is contractible.<br />
Hence there exists a function <span class="math-container">$α:X \to X$</span>, <span class="math-container">$α(x) =x_0 $</span> which is homotopic to the identity map on <span class="math-container">$X$</span>.<br />
Hence the subspace {<span class="math-container">$x_0$</span>} is a deformation retract of X and hence X is simply connected.</p>
|
2,666,568 | <p>I have a dynamical system: $\dot{\mathbf x}$= A$\mathbf x$ with $\mathbf x$=
$\bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ and A =
$\bigl( \begin{smallmatrix} 3 & 0 \\ \beta & 3 \end{smallmatrix} \bigr). \beta$ real, time-independent.</p>
<p>I calculated the eigenvalue $\lambda$ = 3 with the algebraic multiplicity of 2.</p>
<p>The <strong>first question</strong> is about eigenvectors when $\beta = 0$ and when $\beta \neq$ 0:</p>
<p>1) when $\beta$ = 0, I have $\bigl( \begin{smallmatrix} 0 & 0 \\ \beta & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 \\ 0\end{smallmatrix} \bigr)$.
Does this allow for eigenvectors calculation? Does it tell me anything at all?</p>
<p>2) when $\beta \neq$ 0, I have $\bigl( \begin{smallmatrix} 0 & 0 \\ \beta & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 \\ 0\end{smallmatrix} \bigr)$, so my eigenvector is $\bigl( \begin{smallmatrix} 0 \\ 1\end{smallmatrix} \bigr)$ and A is defective? Is there any other eigenvector?</p>
<p>The <strong>other question</strong> is further on case 2) when $\beta$ = 3. I am to find any fixed points + their stability, but first I am wondering whether I did the above correctly. I am not sure how to approach it, thought about a trajectory expression, but I am confused by the defective A.</p>
<p>Thanks!</p>
| Will Jagy | 10,400 | <p>Although it does not really matter, it is traditional (well, in the U.S.) to put the $1$ in the Jordan form above the diagonal. I have been noticing students lately getting to the Jordan form but failing to write things in the reverse( and actually useful) order.
$$
\left(
\begin{array}{rr}
0 & \frac{1}{\beta} \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
3 & 0 \\
\beta & 3
\end{array}
\right)
\left(
\begin{array}{rr}
0 & 1 \\
\beta & 0
\end{array}
\right) =
\left(
\begin{array}{rr}
3 & 1 \\
0 & 3
\end{array}
\right)
$$</p>
<p>$$
\left(
\begin{array}{rr}
0 & 1 \\
\beta & 0
\end{array}
\right)
\left(
\begin{array}{rr}
3 & 1 \\
0 & 3
\end{array}
\right)
\left(
\begin{array}{rr}
0 & \frac{1}{\beta} \\
1 & 0
\end{array}
\right) =
\left(
\begin{array}{rr}
3 & 0 \\
\beta & 3
\end{array}
\right)
$$</p>
<p>Then the exponential of $Jt$ with
$$
J =
\left(
\begin{array}{rr}
3 & 1 \\
0 & 3
\end{array}
\right)
$$
is
$$
e^{3t} \; \;
\left(
\begin{array}{rr}
1 & t \\
0 & 1
\end{array}
\right)
$$
and you use the second matrix identity above to finish</p>
<p>$$
e^{At} =
e^{3t}
\left(
\begin{array}{rr}
0 & 1 \\
\beta & 0
\end{array}
\right)
\left(
\begin{array}{rr}
1 & t \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
0 & \frac{1}{\beta} \\
1 & 0
\end{array}
\right) =
e^{3t}
\left(
\begin{array}{rr}
1 & 0 \\
\beta t & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
e^{3t} & 0 \\
\beta t e^{3t}& e^{3t}
\end{array}
\right)
$$</p>
|
394,321 | <p>I'm studying the asymptotic behavior <span class="math-container">$(n \rightarrow \infty)$</span> of the following formula, where <span class="math-container">$k$</span> is a given constant.
<span class="math-container">$$ \frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$</span></p>
<p>I'm trying to do a series expansion on this equation to give the denominator a simpler form so that it is easier to make an asymptotic analysis.</p>
<p>I used mathematica/wolframalpha to expand the formula (the documents say Taylor expansion is used).
<a href="http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29</a></p>
<p>However in series expansion at <span class="math-container">$n \rightarrow \infty$</span>, the result still has <span class="math-container">$\ln n$</span>. This is actually a form I prefer, compared to the form
<span class="math-container">$$a_0 + a_1x + a_2 x^2+...$$</span>
Does anyone see how the result may be produced? Any help is much appreciated. Thanks.</p>
| mrf | 19,440 | <p>It depends on whether you view the expression as a function of $x$ or as a function of $y$. If you view it as a function of $x$, the decomposition is already done.</p>
<p>If you view it as a function of $y$, make the usual ansatz:</p>
<p>$$\frac{y}{(x-y)(y-1)} = \frac{A}{x-y} + \frac{B}{y-1}.$$</p>
<p>$A$ and $B$ will depend on $x$ but not on $y$.</p>
|
394,321 | <p>I'm studying the asymptotic behavior <span class="math-container">$(n \rightarrow \infty)$</span> of the following formula, where <span class="math-container">$k$</span> is a given constant.
<span class="math-container">$$ \frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$</span></p>
<p>I'm trying to do a series expansion on this equation to give the denominator a simpler form so that it is easier to make an asymptotic analysis.</p>
<p>I used mathematica/wolframalpha to expand the formula (the documents say Taylor expansion is used).
<a href="http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29</a></p>
<p>However in series expansion at <span class="math-container">$n \rightarrow \infty$</span>, the result still has <span class="math-container">$\ln n$</span>. This is actually a form I prefer, compared to the form
<span class="math-container">$$a_0 + a_1x + a_2 x^2+...$$</span>
Does anyone see how the result may be produced? Any help is much appreciated. Thanks.</p>
| Community | -1 | <p>$$\text{Plot3D}\left[\left\{\frac{y^4}{(x-1) (y-1)}-\frac{y \left(x^2 y^2+x^2 y+x^2+x y^2+x y+y^2\right)}{x^3},\frac{y}{(y-1) (x-y)},x \left(y+\frac{1}{y}+1\right)+y+1\right\},\{x,-2,2\},\{y,-2,2\},\text{PlotLegends}\to \text{Automatic}\right]$$
use this
$$\frac{y^4}{(x-1) (y-1)}-\frac{y \left(x^2 y^2+x^2 y+x^2+x y^2+x y+y^2\right)}{x^3}$$
<a href="https://i.stack.imgur.com/GmHXO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GmHXO.jpg" alt="enter image description here"></a>
as you can see the orange (above ) and the blue (your fraction ) are very near is the generalitated laurent series</p>
|
2,965,459 | <p>Some curves defined by polynomial equations are disconnected over reals but not over complexes, e.g., <span class="math-container">$x y - 1 = 0$</span>. How can we convince someone with background only on equations over reals that the curve drawn by above equation is connected over complexes? Is a plot or something possible, for example? It will be a 4d plot if x and y are expanded to real and imaginary parts.
Any other plot, or algebraic way to show connectedness?</p>
| Community | -1 | <p>You can connect any two complex points <span class="math-container">$u, v$</span> with a spiral</p>
<p><span class="math-container">$$u^{1-t}v^t$$</span> where <span class="math-container">$t$</span> runs from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>.</p>
<p>More precisely, in parametric polar coordinates,</p>
<p><span class="math-container">$$\begin{cases}\theta=\theta_u(1-t)+\theta_vt,\\\rho=\rho_u^{1-t}\rho_v^t.\end{cases}$$</span></p>
<p>(You can very well eliminate <span class="math-container">$t$</span>.)</p>
<p>Then the curve <span class="math-container">$(-\theta,\rho^{-1})$</span> corresponds to the inverse points and is a similar continuous spiral.</p>
<p><a href="https://i.stack.imgur.com/14UJ0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/14UJ0.png" alt="enter image description here"></a></p>
|
1,722,287 | <p>So far I know that when matrices A and B are multiplied, with B on the right, the result, AB, is a linear combination of the columns of A, but I'm not sure what to do with this. </p>
| Bernard | 202,857 | <p>Consider the linear maps associated to $A$ and $B$ in canonical bases, respectively ($K$ denotes the base field):
\begin{align*}
f\colon K^p\to K^q\\
g\colon K^n\to K^p
\end{align*}
$\DeclareMathOperator\rk{rank}\DeclareMathOperator\img{Im}$The rank of a matrix is the dimension of the image of the associated linear map. Thus
$$\rk(AB)=\rk(fg),\quad \rk A=\rk f,\quad \rk B=\rk g,$$
and $\;\rk(fg)=\dim f(g(K^n))=\dim(\img(f\,\vert_{\img g})\le \dim(\img g)=\rk B$.</p>
|
1,007,399 | <p>I came across following problem</p>
<blockquote>
<p>Evaluate $$\int\frac{1}{1+x^6} \,dx$$</p>
</blockquote>
<p>When I asked my teacher for hint he said first evaluate</p>
<blockquote>
<p>$$\int\frac{1}{1+x^4} \,dx$$</p>
</blockquote>
<p>I've tried to factorize $1+x^6$ as</p>
<p>$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing</p>
<p>$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$</p>
<p>However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps</p>
<p>I've also tried to reverse engineer the <a href="http://www.wolframalpha.com/input/?i=integrate+%281%29%2F%281%2Bx%5E6%29" rel="noreferrer">solution given by Wolfram Alpha</a></p>
<p>And I need to have terms similar to<br>
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?</p>
<p>Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?</p>
| MathArt | 319,307 | <p>By partial fraction expansion, <span class="math-container">$$I=\int{1\over x^6+1}dx=\int{1\over f(x)}={1\over f'(x_1)}\int{dx\over x-x_1}+{1\over f'(x_2)}\int{dx\over x-x_2}\ldots+{1\over f'(x_6)}\int{dx\over x-x_6}=\sum_{k=0}^6\ln(x-x_k)^{1\over f'(x_k)}$$</span> where <span class="math-container">$x_k$</span> are the roots of the denominator polynomial. The six complex roots of −1 are <span class="math-container">$(-1)^{1/6}=e^{(2k+1)\pi i/6},\;k=0\ldots5$</span> So the integral is</p>
<p>[Correction/simplification: <span class="math-container">${1\over f'(x_k)}={1\over6x_k^5}=-{x_k\over6}$</span>]
<span class="math-container">$$I={1\over6}\ln\left(\Pi_{k=0}^5\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)^{e^\frac{-5(2k+1)\pi i}{6}}\right)={1\over6}\ln\left(\Pi_{k=0}^5\left(x-e^{\frac{(2k+1)\pi}{6}i}\right)^{-e^\frac{(2k+1)\pi i}{6}}\right)$$</span></p>
|
1,007,399 | <p>I came across following problem</p>
<blockquote>
<p>Evaluate $$\int\frac{1}{1+x^6} \,dx$$</p>
</blockquote>
<p>When I asked my teacher for hint he said first evaluate</p>
<blockquote>
<p>$$\int\frac{1}{1+x^4} \,dx$$</p>
</blockquote>
<p>I've tried to factorize $1+x^6$ as</p>
<p>$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing</p>
<p>$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$</p>
<p>However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps</p>
<p>I've also tried to reverse engineer the <a href="http://www.wolframalpha.com/input/?i=integrate+%281%29%2F%281%2Bx%5E6%29" rel="noreferrer">solution given by Wolfram Alpha</a></p>
<p>And I need to have terms similar to<br>
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?</p>
<p>Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p><span class="math-container">$$1+x^6$$</span> factors with the sixth roots of minus one, <span class="math-container">$\pm i$</span> and <span class="math-container">$\dfrac{\pm\sqrt3\pm i}2$</span> and by grouping the conjugate roots, we obtain a real factorization:</p>
<p><span class="math-container">$$(1+x^2)(1+\sqrt3 x+x^2)(1-\sqrt3 x+x^2).$$</span></p>
<p>From this we deduce a decomposition in simple fractions,</p>
<p><span class="math-container">$$\frac1{1+x^6}=a\frac{2x+b}{1+x^2}+c\frac{(2x+\sqrt3)+d}{1+\sqrt3x+x^2}+e\frac{(2x-\sqrt3)+f}{1-\sqrt3x+x^2}.$$</span></p>
<p>Then by shitfing the variables,</p>
<p><span class="math-container">$$a\frac{2x+b}{1+x^2}+c\frac{2x'+d}{1+x'^2}+e\frac{2x''+f}{1+x''^2}$$</span> are easy to integrate.</p>
|
3,616,969 | <p>How to prove that if <span class="math-container">$n$</span> a natural number then <span class="math-container">$n^2$</span> never ends with the digits 2,3,7,8</p>
| Daniel | 530,757 | <p>Every natural number can be represented as :<span class="math-container">$$n=10k+r$$</span>where <span class="math-container">$k$</span> is an integer and <span class="math-container">$r$</span> is the remainder <span class="math-container">$(0,1,...,9)$</span> , now look at <span class="math-container">$$(10k+r)^{2}=100k^{2}+20kr+r^{2}$$</span> so the last digit of this number must be the same as the last digit of <span class="math-container">$r^{2}$</span> and a simple check for all the possible <span class="math-container">$r$</span> will give you all the possibilities for the last digit of a square number.</p>
|
3,616,969 | <p>How to prove that if <span class="math-container">$n$</span> a natural number then <span class="math-container">$n^2$</span> never ends with the digits 2,3,7,8</p>
| Mostafa Ayaz | 518,023 | <p>They do not end with those numbers because they end with <span class="math-container">$$0^2\equiv0\\1^2\equiv1\\2^2\equiv4\\3^2\equiv9\\4^2\equiv6\\5^2\equiv5\\6^2\equiv6\\7^2\equiv9\\8^2\equiv4\\9^2\equiv1$$</span>hence <span class="math-container">$\{0,1,4,5,6,9\}$</span></p>
|
3,616,969 | <p>How to prove that if <span class="math-container">$n$</span> a natural number then <span class="math-container">$n^2$</span> never ends with the digits 2,3,7,8</p>
| Community | -1 | <p>Assume that <span class="math-container">$n$</span> end with <span class="math-container">$0$</span> then <span class="math-container">$n^2$</span> end with <span class="math-container">$0$</span> too.
Now assume that end with <span class="math-container">$1$</span> then <span class="math-container">$n^2$</span> end with <span class="math-container">$1$</span> too.
You can repeat this for each digit and newer meet <span class="math-container">$2,3,7,8$</span>.</p>
|
85,470 | <p>We decided to do secret Santa in our office. And this brought up a whole heap of problems that nobody could think of solutions for - bear with me here.. this is an important problem.</p>
<p>We have 4 people in our office - each with a partner that will be at our Christmas meal.</p>
<p>Steve,
Christine,
Mark,
Mary,
Ken,
Ann,
Paul(me),
Vicki</p>
<p>Desired outcome</p>
<blockquote>
<p>Nobody can know who is buying a present for anybody else. But we each
want to know who we are buying our present for before going to the
Christmas party. And we don't want to be buying presents for our partners.
Partners are not in the office.</p>
</blockquote>
<p>Obvious solution is to put all the names in the hat - go around the office and draw two cards.</p>
<p>And yes - sure enough I drew myself and Mark drew his partner. (we swapped)</p>
<p>With that information I could work out that Steve had a 1/3 chance of having Vicki(he didn't have himself or Christine - nor the two cards I had acquired Ann or Mary) and I knew that Mark was buying my present. Unacceptable result.</p>
<p>Ken asked the question: "What are the chances that we will pick ourselves or our partner?"</p>
<p>So I had a stab at working that out.</p>
<p>First card drawn -> 2/8
Second card drawn -> 12/56</p>
<p>Adding them together makes 28/56 i.e. 1/2.</p>
<p>i.e. This method won't ever work... half chances of drawing somebody you know means we'll be drawing all year before we get a solution that works.</p>
<p>My first thought was that we attach two cards to our backs... put on blindfolds and stumble around in the dark grabbing the first cards we came across... However this is a little unpractical and I'm pretty certain we'd end up knowing who grabbed what anyway.</p>
<p>Does anybody have a solution for distributing cards that results in our desired outcome?</p>
<hr>
<p><em><strong>I'd prefer a solution without a third party..</em></strong></p>
| Dimitar Slavchev | 20,048 | <p>1) You could have a third party party handle the distribution of cards in the hat so that every draw will be valid.</p>
<p>And after each draw the third party will remove invalid cards and put valid ones for the next draw. That should happen without any of the participants knowledge of how much cards are placed and removed.</p>
<p>The downside is that you will have that person know (or at least have enough information to deduce it) about who draws who.</p>
<p>2) Another solution would be if you take 8 cards with each persons name on one side and then every pair makes a unique mark on the other side and puts them in envelopes.</p>
<p>Then you shuffle the envelops, put the cards on the table unique mark up and everyone draws in sequence. In this way you could distinguish the invalid choice without knowing who's who.</p>
<p>3) A modification of the above idea. Use cards with names, then every couple puts the in unique marked box (or envelopes) and then in a bigger box identical to the other ones.</p>
<p>Then someone puts the boxes in one of the offices, opens the big boxes and choses one card from one of the small boxes. Then everyone goes in one by one.</p>
<p>If no one looks at his own box no one knows if he/she was drawn or not.</p>
<p><strong>UPDATE</strong> There is a chance that this scheme will fail if the last pair have less then 2 valid choices. </p>
<p><strong>UPDATE</strong> If you have people enter the room randomly and without knowing the order of entrance that will solve the problem. That could be done (without third party) if you have one of the participants pick people at random to go there and then taking the last one.</p>
<p>This I hope works just as needed. :)</p>
<p>On a side Note : I would go for the script - it's easier, but organizing all of the above might be a lot of fun to all.</p>
<p><strong>UPDATE</strong> Well the stared look of someone looking <strong>at you</strong> will be the hint no algorithm can deal with.</p>
|
842,365 | <blockquote>
<p>Show that a field <span class="math-container">$\mathbb{F}$</span> is finite if and only if its multiplicative group <span class="math-container">$\mathbb{F}^{\times}$</span> is finitely generated.</p>
</blockquote>
<p>The "<span class="math-container">$\Rightarrow$</span>" implication is obvious, but how to prove the otherwise?</p>
| Martin Argerami | 22,857 | <p>Even if you had only two elements per "axis", the number of points in your array is uncountable. That's precisely what Cantor's diagonal argument shows.</p>
<p>Technically, what I'm saying is that the Cartesian product of countably many copies of $\{0,1\} $ is uncountable. </p>
|
2,952,556 | <p>I tried to to solve this but what I found is </p>
<pre><code>A is not necessary for B
</code></pre>
<blockquote>
<p>I could be wrong</p>
</blockquote>
<pre><code>= not(A is necessary for B)
= not(not(B) or A)
= not(A) and B
</code></pre>
<p>but it doesn't make sense.
Let's take an example:</p>
<pre><code>A = understand things
B = argue about things
A is not necessary to B = not(A) and B
</code></pre>
<p>so</p>
<blockquote>
<p>Understand things is not necessary to argue about things = Not understand things and argue about things</p>
</blockquote>
<p>should be the same thing.
I really appreciate any kind of help.</p>
| Vera | 169,789 | <p>"<span class="math-container">$A$</span> necessary for <span class="math-container">$B$</span>" is actually "<span class="math-container">$B$</span> implies <span class="math-container">$A$</span>"</p>
<p>So "<span class="math-container">$A$</span> not necessary for <span class="math-container">$B$</span>" is actually "<span class="math-container">$B$</span> does not imply <span class="math-container">$A$</span>"</p>
<p>In mathematical notation:<span class="math-container">$$\neg[B\implies A]\text{ or equivalently }\neg A\wedge B$$</span></p>
<p>This states that <span class="math-container">$B$</span> can be true while at the same time <span class="math-container">$A$</span> is not true (so no necessity for <span class="math-container">$A$</span> to be true in order to achieve that <span class="math-container">$A$</span> is true).</p>
|
2,952,556 | <p>I tried to to solve this but what I found is </p>
<pre><code>A is not necessary for B
</code></pre>
<blockquote>
<p>I could be wrong</p>
</blockquote>
<pre><code>= not(A is necessary for B)
= not(not(B) or A)
= not(A) and B
</code></pre>
<p>but it doesn't make sense.
Let's take an example:</p>
<pre><code>A = understand things
B = argue about things
A is not necessary to B = not(A) and B
</code></pre>
<p>so</p>
<blockquote>
<p>Understand things is not necessary to argue about things = Not understand things and argue about things</p>
</blockquote>
<p>should be the same thing.
I really appreciate any kind of help.</p>
| Graham Kemp | 135,106 | <p>When you interpret "is necessary" as "is implied by" then indeed we have "<span class="math-container">$A$</span> is not neccessary for <span class="math-container">$B$</span>" exactly where "not <span class="math-container">$A$</span> yet <span class="math-container">$B$</span>".</p>
<p>Check the truth table: <span class="math-container">$$\begin{array}{l:l|cc}A & B & A\gets B\\\hline \top & \top & \top \\ \top & \bot & \top \\ \bot & \top & \bot &\star \\\bot& \bot & \top \end{array}$$</span> </p>
<hr>
<p>The issue is you are interpreting "is not necessary" as "is <em>maybe</em> not implied by"; which requires modal logic quantifiers.</p>
<p><span class="math-container">$$\neg\Box(B\to A)\equiv \Diamond(B\land\lnot A)$$</span></p>
|
3,510,233 | <blockquote>
<p>If <span class="math-container">$\sin\left(\operatorname{cot^{-1}}(x + 1)\right) = \cos\left(\tan^{-1}x\right)$</span>, then find the value of <span class="math-container">$x$</span>.</p>
</blockquote>
<p>Please solve this question by using <span class="math-container">$\cos\left(\dfrac\pi2 - \theta\right) = \sin\theta$</span> by changing <span class="math-container">$\cos\left(\tan^{-1}x\right) = \sin\left(\dfrac\pi2 - \tan^{-1}x\right)$</span> and then equate both LHS and RHS. If not then why? How does the contradiction below occur?</p>
<p><a href="https://i.stack.imgur.com/WDXoM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WDXoM.jpg" alt="enter image description here"></a></p>
| Math1000 | 38,584 | <p>The rate at which the wasp population is growing is given by
<span class="math-container">$$
P'(t) = \frac{25000000 e^{-\frac{t}{2}}}{\left(1000 e^{-\frac{t}{2}}+1\right)^2}.
$$</span>
We want to find the value of <span class="math-container">$t$</span> that maximizes this function. Since <span class="math-container">$P'$</span> is continuous on <span class="math-container">$[0,25]$</span> and differentiable on <span class="math-container">$(0,25)$</span>, this must occur at a boundary point (<span class="math-container">$t=0$</span> or <span class="math-container">$t=25$</span>) or at a point such that <span class="math-container">$P''(t)=0$</span>. We evaluate <span class="math-container">$P$</span> at the boundary points:
<span class="math-container">\begin{align}
P'(0) &= \frac{25000000}{1002001}\approx 24.95007\\
\quad P'(25) &= \frac{25000000}{\left(1+\frac{1000}{e^{25/2}}\right)^2 e^{25/2}}\approx 92.47579,
\end{align}</span>
then compute the derivative:
<span class="math-container">$$
P''(t) = \frac{12500000 e^{t/2} \left(1000-e^{t/2}\right)}{\left(e^{t/2}+1000\right)^3}.
$$</span>
We see that <span class="math-container">$P''(t)= 0$</span> if and only if <span class="math-container">$12500000 e^{t/2} \left(1000-e^{t/2}\right)=0$</span>, and since the first factor is always positive, this is the case if and only if <span class="math-container">$e^{t/2}=1000$</span>. Taking logs and solving for <span class="math-container">$t$</span> yields <span class="math-container">$t= 2\log1000\approx 13.81551$</span>. Since <span class="math-container">$P''$</span> is positive on <span class="math-container">$[0,2\log1000)$</span> and negative on <span class="math-container">$(2\log1000,25]$</span>, it follows that <span class="math-container">$t=2\log1000$</span> is a local (and hence global) maximum for <span class="math-container">$P'$</span>.</p>
|
592,560 | <p>Let G be an abelian group. Show that, if G is not cyclic, then for all $x\in G$, there is a divisor $d$ of $n = |G|$ which is strictly smaller than n satisfying $x^d=1$. </p>
<p>I'm guessing that this is a consequence of Lagrange's Theorem. We can have that G is a disjoint union of left cosets that all have the same cardinality. So $|H| < |G|$ since G is composed with more than just one left coset. By Lagrange's Theorem, we have that $|H|=d$ and then $d$ divides $n$. However the "if G is not cyclic" part is bothering me. Does the fact that G is not cyclic put a restriction?</p>
| lhf | 589 | <ul>
<li><p>Consider $E=\{ e \in \mathbb Z : x^e =1 \mbox{ for all } x\in G \} $, the set of <em>exponents</em> of $G$.<br>
Then $E$ is a subgroup of $\mathbb Z$ and so $E=m\mathbb Z$.</p></li>
<li><p>By Lagrange's Theorem, $n\in E$ and so $m$ divides $n$.</p></li>
<li><p>$m$ is the lcm of the orders of all elements of $G$.</p></li>
<li><p>If $a,b\in G$ with $r=ord(a)$ and $s=ord(b)$ and $\gcd(r,s)=1$, then $ord(ab)=rs=lcm(r,s)$.</p></li>
</ul>
<p>The last two facts imply that there is an element of $G$ having order $m$.</p>
<p>Thus, either $G$ is cyclic or has exponent a proper divisor of $n$.</p>
|
1,557,688 | <p>I want to show that the two metrics are equivalent. </p>
<p>Suppose we have a metric space $X \times Y$. Two metrics are defined as:</p>
<p>$d_{X \times Y}((x, y), (x', y')) := \max\{d_X(x, x'), d_Y(y, y')\}$</p>
<p>$d'_{X \times Y}((x, y), (x', y')) := d_X(x, x')+d_Y(y, y')$</p>
<p>Here is my attempt at proof:</p>
<p>Define $\tau$ to be the collection of open sets with respect to $d_{X \times Y}((x, y), (x', y'))$ . Also define $\tau*$ to be the collection of open sets with respect to $d'_{X \times Y}((x, y), (x', y'))$.</p>
<p>First, assume an open set $U \in \tau$. We want to show that $U \in \tau*$.</p>
<p>Since $U \in \tau$, $\forall (x, y) \in U$, $\exists p > 0$ such that $B_{p}^{d} (x, y) \subset U$.</p>
<p>Then $\forall (x, y) \in U$, $B_{p/2}^{d} (x, y) \subset U.$
Since $d_X(x,x') + d_Y(y, y') \leqslant 2 \cdot \max\{d_X(x, x'), d_Y(y, y')\}$,</p>
<p>If we set $r := 2 \cdot \max\{d_X(x, x'), d_Y(y, y')\}$ then
$B_{r}^{d'}(x,y) \subset B_{p/2}^{d} \subset U$.</p>
<p>As you can see, this reasoning is confusing, in fact I got confused writing this. I can kind of see the direction I should head to but this is not it. Can someone enlighten me?</p>
<p>Thanks.</p>
<p>If I get this part correct I think I can handle the converse.</p>
| Asker | 201,024 | <p>According to your book, if you have $N$ different things that can be picked, $K$ is the number of things from those $N$ things which would be considered "successes". $X$ is then the number of successes from those $K$ successes <em>that are actually picked.</em> </p>
<p>The number of trials is not $K$. The number of trials is $n$.
So $X$ is the number of successes in $n$ trials, and $K$ is the number of successes "waiting to be picked" at the beginning.</p>
|
3,756,436 | <p>Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:</p>
<p><span class="math-container">$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$</span></p>
<p>I want to find <span class="math-container">$0 < \theta < \frac{\pi}2$</span> for which I can later take the largest <span class="math-container">$X$</span> value that solves this equation, i.e. optimize the implicit curve to maximize <span class="math-container">$X$</span>.</p>
<p>I tried solving this by implicit differentiation (assuming <span class="math-container">$X$</span> can be written as a function of <span class="math-container">$\theta$</span>) with respect to <span class="math-container">$\theta$</span> and then by setting <span class="math-container">$\frac{dX}{d\theta} = 0$</span>:</p>
<p><span class="math-container">\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}</span></p>
<p>This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of <span class="math-container">$X$</span>, and solve for <span class="math-container">$\theta$</span> such that <span class="math-container">$D=0$</span>.</p>
<p>Taking discriminant and equating to 0, I get</p>
<p><span class="math-container">$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$</span></p>
<p>and, the angle from it is, 24.45 degrees</p>
<p>I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of <span class="math-container">$X$</span> but different angles: <span class="math-container">$\theta =24.45^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from discriminant method), and
<span class="math-container">$\theta = 47^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from implicit differentiation).</p>
<p>I later realized the original quadratic can only have solutions if <span class="math-container">$D(\theta) > 0$</span>, where <span class="math-container">$D$</span> is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that <span class="math-container">$X$</span> decreases monotonically as a function of <span class="math-container">$\theta$</span>, then I can use the lower bound for further calculations of <span class="math-container">$\theta$</span>.</p>
<p>So then I used the implicit function theorem and got</p>
<p><span class="math-container">$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$</span></p>
<p>Now the problem here is that I can't prove this function is in monotonic in terms of <span class="math-container">$\theta$</span> as the implicit derivative is a function of both <span class="math-container">$\theta$</span> and <span class="math-container">$X$</span>.</p>
| Community | -1 | <p>This is a similar but simple approach which gives the same result. Seeing that <span class="math-container">$y=\tan(\theta)$</span> can take any positive value, we go for maximizing <span class="math-container">$x$</span> and get <span class="math-container">$\max(x) = 1123$</span>.</p>
<p>We have: <span class="math-container">$105 = x\tan\theta-\frac{g}{2} \frac{x^2\sec^2\theta}{(110)^2}$</span></p>
<p>Let <span class="math-container">$\ a=105,\ y=\tan\theta,\ b = \frac{g}{2} \frac{1}{(110)^2}.\quad$</span> Then, <span class="math-container">$\ a=xy-bx^2(y^2+1)$</span></p>
<p>Taking <span class="math-container">$xy=c,\; x^2=\frac{c-bc^2-a}{b}$</span> (Note that since <span class="math-container">$y$</span> can take any positive value, so can <span class="math-container">$c$</span>)</p>
<p>So, for <span class="math-container">$\,x_{\max},\quad \frac{d}{dc}\ (\frac{c-bc^2-a}{b}) = 0$</span> which gives <span class="math-container">$c=\frac{1}{2b}$</span></p>
<p>Now, <span class="math-container">$x_{\max} = \sqrt{\frac{c-bc^2-a}{b}}\quad at\, c=\frac{1}{2b},$</span> i.e. <span class="math-container">$\,x_{\max} \approx 1123$</span> by plugging in values of <span class="math-container">$a$</span> and <span class="math-container">$b.$</span></p>
|
3,756,436 | <p>Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:</p>
<p><span class="math-container">$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$</span></p>
<p>I want to find <span class="math-container">$0 < \theta < \frac{\pi}2$</span> for which I can later take the largest <span class="math-container">$X$</span> value that solves this equation, i.e. optimize the implicit curve to maximize <span class="math-container">$X$</span>.</p>
<p>I tried solving this by implicit differentiation (assuming <span class="math-container">$X$</span> can be written as a function of <span class="math-container">$\theta$</span>) with respect to <span class="math-container">$\theta$</span> and then by setting <span class="math-container">$\frac{dX}{d\theta} = 0$</span>:</p>
<p><span class="math-container">\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}</span></p>
<p>This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of <span class="math-container">$X$</span>, and solve for <span class="math-container">$\theta$</span> such that <span class="math-container">$D=0$</span>.</p>
<p>Taking discriminant and equating to 0, I get</p>
<p><span class="math-container">$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$</span></p>
<p>and, the angle from it is, 24.45 degrees</p>
<p>I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of <span class="math-container">$X$</span> but different angles: <span class="math-container">$\theta =24.45^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from discriminant method), and
<span class="math-container">$\theta = 47^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from implicit differentiation).</p>
<p>I later realized the original quadratic can only have solutions if <span class="math-container">$D(\theta) > 0$</span>, where <span class="math-container">$D$</span> is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that <span class="math-container">$X$</span> decreases monotonically as a function of <span class="math-container">$\theta$</span>, then I can use the lower bound for further calculations of <span class="math-container">$\theta$</span>.</p>
<p>So then I used the implicit function theorem and got</p>
<p><span class="math-container">$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$</span></p>
<p>Now the problem here is that I can't prove this function is in monotonic in terms of <span class="math-container">$\theta$</span> as the implicit derivative is a function of both <span class="math-container">$\theta$</span> and <span class="math-container">$X$</span>.</p>
| River Li | 584,414 | <p>By letting <span class="math-container">$u = \tan \theta \in (0, \infty)$</span>, the equation is written as
<span class="math-container">$$\frac{g}{2} \frac{ X^2 (1+u^2) }{ (110)^2 } - X u + 105 = 0.$$</span>
The equation has real roots if and only if its discriminant is non-negative, that is,
<span class="math-container">$D = \frac{(110^2 - 210 g) u^2 - 210 g}{110^2} \ge 0$</span> (equivalently, <span class="math-container">$u \ge \sqrt{\frac{210g}{110^2 - 210g}}$</span>).
Also, if <span class="math-container">$D \ge 0$</span>, the equation has two positive real roots (by Vieta's formulas), given by
<span class="math-container">\begin{align}
X_1 &= \frac{110^2 u + 110^2\sqrt{D}}{g(u^2+1)}, \\
X_2 &= \frac{110^2 u - 110^2\sqrt{D}}{g(u^2+1)}.
\end{align}</span>
Clearly, we should choose the larger root, that is,
<span class="math-container">$$X = X_1 = \frac{110^2 (u + \sqrt{D})}{g(u^2+1)}.$$</span>
Clearly, if <span class="math-container">$D > 0$</span> (equivalently, <span class="math-container">$u > \sqrt{\frac{210g}{110^2 - 210g}}$</span>), then <span class="math-container">$X$</span> is differentiable with
<span class="math-container">\begin{align}
\frac{\mathrm{d} X}{\mathrm{d} u} &= \frac{110^2}{g}
\cdot \frac{(1 + \frac{1}{2\sqrt{D}}\frac{\mathrm{d} D}{\mathrm{d} u})(u^2+1) -
(u+\sqrt{D})\cdot 2u}{(u^2+1)^2}\\
&= \frac{110^2}{g(u^2+1)^2}
\left(1 - u^2 + \left(\frac{(110^2-210g)u(u^2+1)}{110^2 D} - 2u\right)\sqrt{D}\right).
\end{align}</span>
We have
<span class="math-container">\begin{align}
&\frac{\mathrm{d} X}{\mathrm{d} u} = 0 \\
\Longrightarrow \quad & u^2 - 1 = \left(\frac{(110^2-210g)u(u^2+1)}{110^2 D} - 2u\right)\sqrt{D}\\
\Longrightarrow \quad & (u^2 - 1)^2 = \left(\frac{(110^2-210g)u(u^2+1)}{110^2 D} - 2u\right)^2D \\
\Longrightarrow \quad & \frac{210g}{110^4 D}(u^2+1)^2(110^2-(110^2-210g)u^2) = 0\\
\Longrightarrow \quad & u = \frac{110}{\sqrt{110^2 - 210g}}.
\end{align}</span>
Comparing <span class="math-container">$X(\sqrt{\frac{210g}{110^2 - 210g}})$</span>, <span class="math-container">$X(\frac{110}{\sqrt{110^2 - 210g}})$</span> and <span class="math-container">$X(\infty)$</span>, we know that
<span class="math-container">$X$</span> achieves its maximum <span class="math-container">$\frac{110}{g}\sqrt{110^2-210g}$</span> at <span class="math-container">$u = \frac{110}{\sqrt{110^2 - 210g}}$</span>.</p>
<p>For <span class="math-container">$g = 9.81$</span>, <span class="math-container">$X$</span> achieves its maximum <span class="math-container">$1123.539567$</span> at <span class="math-container">$\theta = 47.66952494^\circ$</span>.</p>
|
1,276,957 | <p>These are the provided notes:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/NesWm.png" alt="Blockquote"></p>
</blockquote>
<p>These are the provided questions:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/t0Ta7.png" alt="Blockquote"></p>
</blockquote>
<p>I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)</p>
| danimal | 202,026 | <p>The point is, I think here, that the sine, cosine and tangent functions are all many-to-one, that is there are an infinite number of different values of $\theta$ that will give the <em>same</em> value of $\sin\theta$ (and the same for $\cos\theta$ and $\tan\theta$). If you consider the graph of, for example, $y=\sin\theta$, you can see it goes up and down (oscillates) indefinitely, so that if say $\sin\theta=1/2$, the horizontal line $y=1/2$ will intersect the sine curve at many (infinite) places, at $30^o, 150^o, 390^o$ etc. </p>
<p>For answering questions, I would recommend doing a sketch of the relevant trig function and the horizontal line corresponding to the value of $\sin\theta$ or $\cos\theta$ etc you are given, and then looking at the intersection points and using the symmetry of the curve to determine the values you don't know. </p>
|
1,276,957 | <p>These are the provided notes:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/NesWm.png" alt="Blockquote"></p>
</blockquote>
<p>These are the provided questions:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/t0Ta7.png" alt="Blockquote"></p>
</blockquote>
<p>I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)</p>
| Stephen P | 233,086 | <p>If we are just looking at acute and obtuse angles</p>
<p>$\sin (180 - \theta ) = \sin \theta $ tells you that sin is positive for both acute and obtuse angles. Try it, type in to your calculator any angle from 0 to 180 and you get a positive numbers.</p>
<p>The cos and tan expressions tell you that for obtuse angles that is bigger than 90 and less than 180 you get negative answers. Again try it on your calculator.</p>
<p>So looking at the questions provided</p>
<p>a) -0.2566 is negative for cos so it is an obtuse angle. So if your calculator gives you an acute angled answer you subtract is from 180 to get an obtuse angle.</p>
<p>b) same applies swapping tan for cas in a) as when tan is negative it is an obtuse angle.</p>
<p>C) sin is positive for both acute and obtuse angles so you get 2 answers. One is acute and than subtracting the acute angle from 180 gives you the second answer which is obtuse.</p>
|
1,276,957 | <p>These are the provided notes:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/NesWm.png" alt="Blockquote"></p>
</blockquote>
<p>These are the provided questions:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/t0Ta7.png" alt="Blockquote"></p>
</blockquote>
<p>I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)</p>
| Fax | 239,640 | <p>The question asks what the angles are if they are obtuse, i.e. between 90 and 180 degrees. The notes provide the conversions, e.g. if you get sin(30) = 0.5 then you convert it to an obtuse angle through sin(30) = sin(180 - 30) from the first note.</p>
|
1,276,957 | <p>These are the provided notes:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/NesWm.png" alt="Blockquote"></p>
</blockquote>
<p>These are the provided questions:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/t0Ta7.png" alt="Blockquote"></p>
</blockquote>
<p>I do not understand when I should subtract, add or leave the answer as is (The notes do not make sense to me very much). I do not, clearly, have an intuitive understanding of this. I hope someone can please please please show me thank you :)</p>
| John Joy | 140,156 | <p>Try Googling the following terms.</p>
<ul>
<li>unit circle</li>
<li>standard position</li>
<li>terminal size</li>
<li>initial side</li>
<li>reference angle</li>
</ul>
<p>Draw a diagram, and label the point where the terminal side intersects the unit circle as $A(\sin\theta, \cos\theta)$. Plot also point $B(\sin(180-\theta), \cos(180-\theta))$.</p>
<p>Look at diagram and compare points $A$ and $B$.</p>
|
754,888 | <p>The letters that can be used are A, I, L, S, T. </p>
<p>The word must start and end with a consonant. Exactly two vowels must be used. The vowels can't be adjacent.</p>
| fgp | 42,986 | <p>If $a$ has prime factorization $$
a = \prod_{k=1}^n p_k^{e_k} \text{ where } e_k \in \mathbb{N}, p_k \text{ prime}
$$
then $$
\sqrt[n]{a} \in \mathbb{N} \text{ exactly if $n \mid e_k$ for all $k$, meaning if } n \mid \textrm{gcd}(e_1,\ldots,d_k).
$$
The largest such $n$ is thus $\textrm{gcd}(e_1,\ldots,d_k)$.</p>
|
167,446 | <p>Let $p$ be a prime number, $C_p$-cyclic group of order $p$, and $G$ an elementary p-group of order $p^n$. Let us denote by Cext$(G,C_p)$ the group of all central extensions of $C_p$ by $G$. Is the number of non isomorphic groups in Cext$(G,C_p)$ known as a function of $n$? </p>
| Lenny Krop | 50,922 | <p>I'd like to thank everyone who has responded to my query.
The question was whether $I(n,p)$= # of non isomorphic groups in $\mathrm{Cext}(G,C_p)$ is known. My impression so far is that it is not.
Regarding discussion, first $\mathrm{Cext}(G,A)$, with $A$ abelian stands for $\mathrm{Opext}(G,A,\text{triv})$, hence by the standard convention $A$ is a normal subgroup of extension.
The number in question $I(n,p)=\lfloor\frac{3n+2}{2}\rfloor$ for odd $p$. The reason for exclusion of $p=2$ lies in the fact that $H^2(G,C_p)$ fits into the exact sequence
$$0\to\widehat G\to H^2(G,C_p)\to\mathrm{Alt}(G)\to 0,$$
where $\widehat G$ is the dual group, splits up as $\mathrm{Aut}(G)$-module iff $p$ is odd. I also show, as a consequence of a general theorem on abelian extensions of Hopf algebras, that the so-called weak isoclasses of some previous posts coincide with isoclasses of noncommutative extensions. Commutative extensions labeled by $\mathrm{Ext}(G,C_p)$ are also in bijection with the orbits, this is trivial. It follows that one has to classify the orbits of $\mathrm{Aut}(G)\times\mathrm{Aut}(C_p)$ in $H^2(G,C_p)$. Roughly speaking, for a pair $(f,\alpha)$ its invariant is $(\ker f,\text{rad}(\alpha)$. From this one gets the number.
George Glauberman seems to have a proof that the formula for $I(n,p)$ holds for $p=2$.
Details can be found at arXiv:1211.5621, look up the latest version (should appear shortly) </p>
|
269,178 | <p>i would like to know where i could find a plot of</p>
<p>$$ J_{ia}(2\pi i)$$ (1)</p>
<p>using Quantum mechanics i have conjectured that if $ a= \frac{x}{2} $ and $ i= \sqrt{-1} $ then </p>
<p>$$ J_{it}(2\pi i)\approx0=\zeta (1/2+2it)$$ at least for big $ t \to \infty $ (2)</p>
<p>however i do not know how to check or disprove this fact.</p>
<p>the idea is that the Operator $$ -D^{2}y(x)+4\pi ^{2}e^{4x}y(x)=E_{n}y(x) $$ (3)</p>
<p>has a Weyl term for the Eigenvalues as $ N(T)= \frac{\sqrt{T}}{2\pi}log( \frac{\sqrt{T}}{2\pi e}) $</p>
<p>inthe same fashion as the Riemann zeta function</p>
<p>the condition (2) is stablished by imposing that the eigenvalue problem satisfy $ y(0)=0=y(\infty) $</p>
| GEdgar | 442 | <p>Here is what I get in Maple (first real part, then imaginary part)</p>
<p><img src="https://i.stack.imgur.com/SwjTu.jpg" alt="Bessel"></p>
<p>But here is your zeta</p>
<p><img src="https://i.stack.imgur.com/FfXZ2.jpg" alt="zeta"></p>
|
3,541,910 | <p>Let </p>
<p><span class="math-container">$$a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1} + \cdots +\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$$</span></p>
<p>Does this sequence converge to a simple number?</p>
<p>My thought was to compute each denominator:</p>
<p><span class="math-container">$$1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1=(n+1)\sum_{k=1}^n k^3-\sum_{k=1}^nk^4$$</span></p>
<p>and this are known, I find <span class="math-container">$\frac{1}{60}n(n+1)(2n+1)(3n^2+6n+1)$</span>. But can we find maybe a closed formula for <span class="math-container">$a_n$</span> after this?</p>
| Community | -1 | <p>Your sum is:
<span class="math-container">$$\sum_{k=1}^{n}\frac{1}{\left(1^{3}\cdot k\right)+\left(2^{3}\cdot\left(k-1\right)\right)+...+\left(k^{3}\cdot\left(1\right)\right)}=\sum_{k=1}^{n}\frac{1}{\color{red}{\sum_{m=1}^{k}m^{3}\cdot\left(k+1-m\right)}}$$</span>
For the red part we have:
<span class="math-container">$$\sum_{m=1}^{k}m^{3}\cdot\left(k+1-m\right)=\color{blue}{\left(k+1\right)\sum_{m=1}^{k}m^{3}}-\color{green}{\sum_{m=1}^{k}m^{4}}$$</span>
Using <a href="https://en.wikipedia.org/wiki/Faulhaber%27s_formula" rel="nofollow noreferrer">Faulhaber's formula</a>
follows:</p>
<p><span class="math-container">$$\color{blue}{\left(k+1\right)\cdot\frac{k^{4}+2k^{3}+k^{2}}{4}}-\color{green}{\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^{2}+3k-1\right)}{30}}$$</span>
Replace this relation in the main sum:</p>
<p><span class="math-container">$$\sum_{k=1}^{n}\frac{1}{\left(k+1\right)\cdot\frac{k^{4}+2k^{3}+k^{2}}{4}-\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^{2}+3k-1\right)}{30}}$$</span><span class="math-container">$$=60\sum_{k=1}^{n}\frac{1}{15\left(k+1\right)\cdot\left(k^{4}+2k^{3}+k^{2}\right)-2k\left(k+1\right)\left(2k+1\right)\left(3k^{2}+3k-1\right)}$$</span><span class="math-container">$$=60\sum_{k=1}^{n}\frac{1}{3k^{5}+15k^{4}+25k^{3}+15k^{2}+2k}$$</span></p>
<p>Notice that :
<span class="math-container">$$3k^{5}+15k^{4}+25k^{3}+15k^{2}+2k$$</span><span class="math-container">$$=k\left(3k^{4}+15k^{3}+25k^{2}+15k+2\right)$$</span></p>
<p>Clearly one of the roots is <span class="math-container">$k=0$</span>.</p>
<p>Assume rational roots of the other part are in the form <span class="math-container">$\frac{p}{q}$</span> where <span class="math-container">$p,q∈ℤ$</span> and <span class="math-container">$q≠0$</span>, also assume this fraction is in the simplest form ,<a href="https://en.wikipedia.org/wiki/Rational_root_theorem" rel="nofollow noreferrer">using rational root theorem</a> implies <span class="math-container">$p$</span> must divide <span class="math-container">$2$</span> and <span class="math-container">$q$</span> must divide <span class="math-container">$3$</span>, so the whole fractions with these assumptions are:</p>
<p><span class="math-container">$$\pm1 , \pm2 ,\pm\frac{1}{3} , \pm\frac{2}{3}$$</span></p>
<p>Checking them implies <span class="math-container">$-1,-2$</span> are two integer roots of the equation.</p>
<p>So we apply what we derived:
<span class="math-container">$$=60\sum_{k=1}^{n}\frac{1}
{k\left(k+1\right)\left(k+2\right)\left(3k^{2}+6k+1\right)}$$</span>
Using <a href="https://en.wikipedia.org/wiki/Partial_fraction_decomposition" rel="nofollow noreferrer">partial fraction decomposition</a> we have:</p>
<p><span class="math-container">$$=60\left[\sum_{k=1}^{n}\frac{1}{2k}+\sum_{k=1}^{n}\frac{1}{2\left(k+2\right)}+\sum_{k=1}^{n}\frac{1}{2\left(k+1\right)}+\sum_{k=1}^{n}-\frac{9}{2}\cdot\frac{k+1}{3k^{2}+6k+1}\right]$$</span></p>
<p><span class="math-container">$$=30\left[\color{blue}{\sum_{k=1}^{n}\frac{1}{k}}+\color{red}{\sum_{k=1}^{n}\frac{1}{k+2}}+\color{green}{\sum_{k=1}^{n}\frac{1}{k+1}}-\color{orange}{9\sum_{k=1}^{n}\frac{k+1}{3k^{2}+6k+1}}\right]$$</span></p>
<p>For calculating the orange part we have:</p>
<p><span class="math-container">$$9\sum_{k=1}^{n}\frac{k+1}{3k^{2}+6k+1}$$</span><span class="math-container">$$=\frac{9}{2}\left[\sum_{k=1}^{n}\frac{1}{3k+3+\sqrt{6}}+\sum_{k=1}^{n}\frac{1}{3k+3-\sqrt{6}}\right]$$</span></p>
<p><span class="math-container">$$=\frac{3}{2}\left[\sum_{k=1}^{n}\frac{1}{k+1+\frac{\sqrt{6}}{3}}+\sum_{k=1}^{n}\frac{1}{k+1-\frac{\sqrt{6}}{3}}\right]$$</span></p>
<p>Setting <span class="math-container">$k+1+\frac{\sqrt{6}}{3} \mapsto k$</span> and <span class="math-container">$k+1-\frac{\sqrt{6}}{3} \mapsto k'$</span> yields:</p>
<p><span class="math-container">$$=\frac{3}{2}\left[\sum_{k=2+\sqrt{\frac{2}{3}}}^{n+1+\sqrt{\frac{2}{3}}}\frac{1}{k}+\sum_{k'=2-\sqrt{\frac{2}{3}}}^{n+1-\sqrt{\frac{2}{3}}}\frac{1}{k'}\right]$$</span><span class="math-container">$$=\frac{3}{2}\left[\sum_{k=1}^{n+1+\sqrt{\frac{2}{3}}}\frac{1}{k}-\sum_{k=1}^{1+\sqrt{\frac{2}{3}}}\frac{1}{k}+\sum_{k'=1}^{n+1-\sqrt{\frac{2}{3}}}\frac{1}{k'}-\sum_{k'=1}^{1-\sqrt{\frac{2}{3}}}\frac{1}{k'}\right]=\frac{3}{2}\left[H_{n+1+\sqrt{\frac{2}{3}}}-H_{1+\sqrt{\frac{2}{3}}}+H_{n+1-\sqrt{\frac{2}{3}}}-H_{1-\sqrt{\frac{2}{3}}}\right]$$</span>
<span class="math-container">$$=30\left(\color{blue}{H_{n}}+\color{red}{H_{n+2}-\frac{3}{2}}+\color{green}{H_{n+1}-1}-\color{orange}{\frac{3}{2}\left[H_{n-\sqrt{\frac{2}{3}}+1}+H_{n+\sqrt{\frac{2}{3}}+1}-H_{1+\sqrt{\frac{2}{3}}}-H_{1-\sqrt{\frac{2}{3}}}\right]}\right)$$</span></p>
<p>Where <span class="math-container">$H_n$</span> is the n-th <a href="https://en.wikipedia.org/wiki/Harmonic_number" rel="nofollow noreferrer">harmonic number</a>. </p>
<p>And that is the <strong>closed form</strong> you where looking for.</p>
<p>Now you need some simple addition subtraction tricks and using the following fact:
<span class="math-container">$$\lim_{n\to\infty}\left(H_{n}-\ln\left(n\right)\right)$$</span><span class="math-container">$$=\lim_{n\to\infty}\left(H_{n+1}-\ln\left(n+1\right)\right)$$</span><span class="math-container">$$=\lim_{n\to\infty}\left(H_{n+2}-\ln\left(n+2\right)\right)$$</span><span class="math-container">$$=\lim_{n\to\infty}\left(H_{n-\sqrt{\frac{2}{3}}+1}-\ln\left(n-\sqrt{\frac{2}{3}}+1\right)\right)$$</span><span class="math-container">$$=\lim_{n\to\infty}\left(H_{n+\sqrt{\frac{2}{3}}+1}-\ln\left({n+\sqrt{\frac{2}{3}}+1}\right)\right)$$</span><span class="math-container">$$=\gamma$$</span></p>
<p><span class="math-container">$\gamma$</span> is <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant" rel="nofollow noreferrer">Euler–Mascheroni constant</a>.</p>
<p>Finally taking the limit of the relation we get:</p>
<p><span class="math-container">$$=30\lim_{n\to\infty}\left(3\gamma-\frac{3}{2}(2\gamma)-\frac{5}{2}\right)$$</span>
<span class="math-container">$$+30\lim_{n\to\infty}\ln\left(\frac{n\left(n+2\right)\left(n+1\right)}{\sqrt{\left(\left(n-\sqrt{\frac{2}{3}}+1\right)\left(n+\sqrt{\frac{2}{3}}+1\right)\right)^{3}}}\right)$$</span></p>
<p><span class="math-container">$$-45\lim_{n\to\infty}\left(-H_{1+\sqrt{\frac{2}{3}}}-H_{1-\sqrt{\frac{2}{3}}}\right)$$</span></p>
<p><span class="math-container">$$\simeq\bbox[5px,border:2px solid #C0A000]{1.134103506}$$</span></p>
|
3,541,910 | <p>Let </p>
<p><span class="math-container">$$a_n = \frac{1}{1^3\cdot 1}+\frac{1}{1^3\cdot 2+2^3\cdot 1} + \cdots +\frac{1}{1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1}$$</span></p>
<p>Does this sequence converge to a simple number?</p>
<p>My thought was to compute each denominator:</p>
<p><span class="math-container">$$1^3\cdot n+2^3\cdot (n-1)+\cdots+n^3\cdot 1=(n+1)\sum_{k=1}^n k^3-\sum_{k=1}^nk^4$$</span></p>
<p>and this are known, I find <span class="math-container">$\frac{1}{60}n(n+1)(2n+1)(3n^2+6n+1)$</span>. But can we find maybe a closed formula for <span class="math-container">$a_n$</span> after this?</p>
| Gary | 83,800 | <p>Continuing in a different way, you can express the limit in terms of the digamma function as follows:
<span class="math-container">$$
\mathop {\lim }\limits_{n \to + \infty } 30\left[ {\sum\limits_{k = 1}^n {\frac{1}{k}} + \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} + \sum\limits_{k = 1}^n {\frac{1}{{k + 2}}} - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 + \sqrt {2/3} }}} - \frac{3}{2}\sum\limits_{k = 1}^n {\frac{1}{{k + 1 - \sqrt {2/3} }}} } \right]
\\
= \mathop {\lim }\limits_{n \to + \infty } 30\left[ {\frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]} + \frac{3}{2}\sum\limits_{k = 1}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right]
\\
= \mathop {\lim }\limits_{n \to + \infty } 30\left[ {6 + \frac{1}{2} - \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 + \sqrt {2/3} }}} \right]} + \frac{3}{2}\sum\limits_{k = 0}^n {\left[ {\frac{1}{{k + 1}} - \frac{1}{{k + 1 - \sqrt {2/3} }}} \right]} } \right]
\\
= 30\left[ {6 + \frac{1}{2} + \frac{3}{2}\left( {\psi \left( {\sqrt {2/3} } \right) + \gamma } \right) + \frac{3}{2}\left( {\psi \left( { - \sqrt {2/3} } \right) + \gamma } \right)} \right]
\\
= 195 + 90\gamma + 45\left( {\psi \left( {\sqrt {2/3} } \right) + \psi \left( { - \sqrt {2/3} } \right)} \right)
\\
= 195 + 90\gamma + 45\pi \cot (\pi \sqrt {2/3} ) + 45\sqrt {\frac{3}{2}} + 90\psi (\sqrt {2/3} ).
$$</span></p>
|
201,576 | <p>Struggling with something basic. Suppose when defining f[x_] the outcome depends on Sign[x]. When calling this function, how do I tell Mathematica the sign of the argument?
My attempt:</p>
<pre><code>f[x_]=x Sign[x];
Assuming[a>0, f[a]]
</code></pre>
<p>The output I get is </p>
<pre><code>a Sign[a]
</code></pre>
<p>Obviously, I want to get simply <code>a</code> because I am trying to tell Mathematica <code>a</code> is positive</p>
<p><strong>Additional details:</strong></p>
<p>Thanks for the answers below. What I am really trying to do is as follows:</p>
<pre><code>p = x /. Solve[(a - x)/y^2 == 1, x];
q = ((3 a - 2 p)/(4 y^2));
f[x_] = x Sign[x] ;
Assuming[q > 0, Simplify[f[q]]]
</code></pre>
<p>If I delete the first line in which <code>p</code> is defined via Solve, everything works fine. If, however, the first line is present, the output is: </p>
<pre><code>{((a + 2 y^2) Sign[a + 2 y^2])/(4 y^2 Sign[y]^2)}
</code></pre>
<p>So, the Sign operator is still there.</p>
<p>What's wrong? </p>
| Bill | 18,890 | <pre><code>f[x_]=x Sign[x];
Assuming[a>0,FunctionExpand[f[a]]]
</code></pre>
<p>returns a</p>
<pre><code>Assuming[a>0,Simplify[f[a]]]
</code></pre>
<p>returns a</p>
|
19,253 | <p><strong>Bug introduced in 7.0.1 or earlier and fixed in 10.0.2 or earlier</strong></p>
<hr>
<p>I have access to two versions of Mathematica, version 7.0.1 on Linux and version 8 on Windows. When I try the following two lines on version 7, the kernel quits when it tries to plot. In version 8 it plots just fine. You can import the data using:</p>
<pre><code>temp = Import["http://pastie.org/pastes/6098244/download", "Table"];
</code></pre>
<p>Any idea why this crashes in version 7? </p>
<pre><code>ListContourPlot[temp]
</code></pre>
<p>It may also crash in Mathematica 7 for Windows too, but I don't personally have a copy of that on my machine.</p>
<p>ETA: Other users have reported this also crashes in some versions 8 and 9 across different platforms, so I changed the name of the post accordingly.</p>
<hr>
<p>If the pastebin link fails, please copy the data from here:</p>
<pre><code>temp={{-0.1,11000,1.00214},{-0.1,11050,0.998176},
{-0.1,11100,0.994088},{-0.1,11150,0.990016},{-0.1,11200,0.986111},
{-0.1,11250,0.982526},{-0.1,11300,0.979411},{-0.1,11350,0.976904},
{-0.1,11400,0.97513},{-0.1,11450,0.974195},{-0.1,11500,0.974179},
{-0.1,11550,0.975137},{-0.1,11600,0.977087},{-0.1,11650,0.98002},
{-0.1,11700,0.983885},{-0.1,11750,0.988602},{-0.1,11800,0.994051},
{-0.1,11850,1.00008},{-0.1,11900,1.00652},{-0.1,11950,1.01315},
{-0.1,12000,1.01975},{-0.1,12050,1.02608},{-0.1,12100,1.03189},
{-0.1,12150,1.03694},{-0.1,12200,1.04098},{-0.1,12250,1.04378},
{-0.1,12300,1.04513},{-0.1,12350,1.04486},{-0.1,12400,1.04282},
{-0.1,12450,1.03889},{-0.1,12500,1.03303},{-0.1,12550,1.02522},
{-0.1,12600,1.01549},{-0.1,12650,1.00392},{-0.1,12700,0.990664},
{-0.1,12750,0.975894},{-0.1,12800,0.959842},{-0.1,12850,0.942777},
{-0.1,12900,0.924999},{-0.1,12950,0.906838},{-0.1,13000,0.888643},
{-0.1,13050,0.870773},{-0.1,13100,0.853588},{-0.1,13150,0.837442},
{-0.1,13200,0.822673},{-0.1,13250,0.809592},{-0.1,13300,0.798481},
{-0.1,13350,0.789577},{-0.1,13400,0.783075},{-0.1,13450,0.779116},
{-0.1,13500,0.777787},{-0.098,11000,0.990164},{-0.098,11050,0.986921},
{-0.098,11100,0.983944},{-0.098,11150,0.981344},{-0.098,11200,0.979226},
{-0.098,11250,0.977683},{-0.098,11300,0.976797},{-0.098,11350,0.976633},
{-0.098,11400,0.977237},{-0.098,11450,0.978633},{-0.098,11500,0.980822},
{-0.098,11550,0.983781},{-0.098,11600,0.98746},{-0.098,11650,0.991787},
{-0.098,11700,0.996664},{-0.098,11750,1.00197},{-0.098,11800,1.00756},
{-0.098,11850,1.01329},{-0.098,11900,1.01897},{-0.098,11950,1.02443},
{-0.098,12000,1.02947},{-0.098,12050,1.03391},{-0.098,12100,1.03756},
{-0.098,12150,1.04024},{-0.098,12200,1.0418},{-0.098,12250,1.04207},
{-0.098,12300,1.04095},{-0.098,12350,1.03834},{-0.098,12400,1.03416},
{-0.098,12450,1.0284},{-0.098,12500,1.02104},{-0.098,12550,1.01214},
{-0.098,12600,1.00175},{-0.098,12650,0.989997},{-0.098,12700,0.977021},
{-0.098,12750,0.962997},{-0.098,12800,0.948131},{-0.098,12850,0.932651},
{-0.098,12900,0.916808},{-0.098,12950,0.900866},{-0.098,13000,0.885099},
{-0.098,13050,0.869785},{-0.098,13100,0.855197},{-0.098,13150,0.841603},
{-0.098,13200,0.829253},{-0.098,13250,0.818377},{-0.098,13300,0.809181},
{-0.098,13350,0.80184},{-0.098,13400,0.796493},{-0.098,13450,0.793244},
{-0.098,13500,0.792153},{-0.096,11000,0.98162},{-0.096,11050,0.980727},
{-0.096,11100,0.980311},{-0.096,11150,0.980411},{-0.096,11200,0.981055},
{-0.096,11250,0.982257},{-0.096,11300,0.984019},{-0.096,11350,0.986327},
{-0.096,11400,0.989153},{-0.096,11450,0.992454},{-0.096,11500,0.99617},
{-0.096,11550,1.00023},{-0.096,11600,1.00455},{-0.096,11650,1.00903},
{-0.096,11700,1.01357},{-0.096,11750,1.01804},{-0.096,11800,1.02233},
{-0.096,11850,1.02632},{-0.096,11900,1.02987},{-0.096,11950,1.03285},
{-0.096,12000,1.03516},{-0.096,12050,1.03668},{-0.096,12100,1.03729},
{-0.096,12150,1.03692},{-0.096,12200,1.03549},{-0.096,12250,1.03294},
{-0.096,12300,1.02922},{-0.096,12350,1.02432},{-0.096,12400,1.01825},
{-0.096,12450,1.01102},{-0.096,12500,1.00268},{-0.096,12550,0.993304},
{-0.096,12600,0.98298},{-0.096,12650,0.971818},{-0.096,12700,0.959949},
{-0.096,12750,0.94752},{-0.096,12800,0.934692},{-0.096,12850,0.921638},
{-0.096,12900,0.90854},{-0.096,12950,0.895584},{-0.096,13000,0.882961},
{-0.096,13050,0.870857},{-0.096,13100,0.859456},{-0.096,13150,0.848933},
{-0.096,13200,0.839451},{-0.096,13250,0.831159},{-0.096,13300,0.824186},
{-0.096,13350,0.818643},{-0.096,13400,0.81462},{-0.096,13450,0.81218},
{-0.096,13500,0.811362},{-0.094,11000,0.98692},{-0.094,11050,0.988506},
{-0.094,11100,0.990436},{-0.094,11150,0.99269},{-0.094,11200,0.995243},
{-0.094,11250,0.998058},{-0.094,11300,1.00109},{-0.094,11350,1.0043},
{-0.094,11400,1.00762},{-0.094,11450,1.011},{-0.094,11500,1.01436},
{-0.094,11550,1.01764},{-0.094,11600,1.02076},{-0.094,11650,1.02365},
{-0.094,11700,1.02624},{-0.094,11750,1.02844},{-0.094,11800,1.03019},
{-0.094,11850,1.03142},{-0.094,11900,1.03206},{-0.094,11950,1.03205},
{-0.094,12000,1.03135},{-0.094,12050,1.02991},{-0.094,12100,1.02769},
{-0.094,12150,1.02468},{-0.094,12200,1.02086},{-0.094,12250,1.01624},
{-0.094,12300,1.01081},{-0.094,12350,1.00462},{-0.094,12400,0.997686},
{-0.094,12450,0.990066},{-0.094,12500,0.981817},{-0.094,12550,0.973011},
{-0.094,12600,0.963727},{-0.094,12650,0.954056},{-0.094,12700,0.944096},
{-0.094,12750,0.933951},{-0.094,12800,0.923733},{-0.094,12850,0.913554},
{-0.094,12900,0.903532},{-0.094,12950,0.893781},{-0.094,13000,0.884419},
{-0.094,13050,0.875557},{-0.094,13100,0.867302},{-0.094,13150,0.859756},
{-0.094,13200,0.853013},{-0.094,13250,0.847157},{-0.094,13300,0.84226},
{-0.094,13350,0.838386},{-0.094,13400,0.835583},{-0.094,13450,0.833887},
{-0.094,13500,0.833319},{-0.092,11000,1.00347},{-0.092,11050,1.00579},
{-0.092,11100,1.00815},{-0.092,11150,1.01053},{-0.092,11200,1.01289},
{-0.092,11250,1.0152},{-0.092,11300,1.0174},{-0.092,11350,1.01948},
{-0.092,11400,1.02137},{-0.092,11450,1.02304},{-0.092,11500,1.02446},
{-0.092,11550,1.02558},{-0.092,11600,1.02636},{-0.092,11650,1.02677},
{-0.092,11700,1.02678},{-0.092,11750,1.02636},{-0.092,11800,1.02547},
{-0.092,11850,1.02411},{-0.092,11900,1.02226},{-0.092,11950,1.01989},
{-0.092,12000,1.01702},{-0.092,12050,1.01362},{-0.092,12100,1.00972},
{-0.092,12150,1.00532},{-0.092,12200,1.00044},{-0.092,12250,0.995097},
{-0.092,12300,0.989324},{-0.092,12350,0.983155},{-0.092,12400,0.976629},
{-0.092,12450,0.969789},{-0.092,12500,0.962686},{-0.092,12550,0.955372},
{-0.092,12600,0.947903},{-0.092,12650,0.940341},{-0.092,12700,0.932747},
{-0.092,12750,0.925184},{-0.092,12800,0.917719},{-0.092,12850,0.910417},
{-0.092,12900,0.903341},{-0.092,12950,0.896558},{-0.092,13000,0.890127},
{-0.092,13050,0.884109},{-0.092,13100,0.87856},{-0.092,13150,0.873532},
{-0.092,13200,0.869073},{-0.092,13250,0.865224},{-0.092,13300,0.862024},
{-0.092,13350,0.859502},{-0.092,13400,0.857683},{-0.092,13450,0.856584},
{-0.092,13500,0.856217},{-0.09,11000,1.01694},{-0.09,11050,1.01807},
{-0.09,11100,1.01906},{-0.09,11150,1.0199},{-0.09,11200,1.02057},
{-0.09,11250,1.02104},{-0.09,11300,1.0213},{-0.09,11350,1.02134},
{-0.09,11400,1.02113},{-0.09,11450,1.02066},{-0.09,11500,1.01993},
{-0.09,11550,1.01891},{-0.09,11600,1.01761},{-0.09,11650,1.01602},
{-0.09,11700,1.01412},{-0.09,11750,1.01193},{-0.09,11800,1.00943},
{-0.09,11850,1.00664},{-0.09,11900,1.00356},{-0.09,11950,1.00019},
{-0.09,12000,0.996544},{-0.09,12050,0.992639},{-0.09,12100,0.988487},
{-0.09,12150,0.984107},{-0.09,12200,0.979518},{-0.09,12250,0.974743},
{-0.09,12300,0.969804},{-0.09,12350,0.964729},{-0.09,12400,0.959544},
{-0.09,12450,0.954279},{-0.09,12500,0.948964},{-0.09,12550,0.943631},
{-0.09,12600,0.938312},{-0.09,12650,0.933039},{-0.09,12700,0.927846},
{-0.09,12750,0.922765},{-0.09,12800,0.917829},{-0.09,12850,0.913071},
{-0.09,12900,0.908522},{-0.09,12950,0.904211},{-0.09,13000,0.900169},
{-0.09,13050,0.896423},{-0.09,13100,0.892997},{-0.09,13150,0.889917},
{-0.09,13200,0.887202},{-0.09,13250,0.884872},{-0.09,13300,0.882943},
{-0.09,13350,0.881428},{-0.09,13400,0.880339},{-0.09,13450,0.879682},
{-0.09,13500,0.879463},{-0.088,11000,1.01491},{-0.088,11050,1.01441},
{-0.088,11100,1.01377},{-0.088,11150,1.01298},{-0.088,11200,1.01205},
{-0.088,11250,1.01096},{-0.088,11300,1.00972},{-0.088,11350,1.00832},
{-0.088,11400,1.00677},{-0.088,11450,1.00506},{-0.088,11500,1.00319},
{-0.088,11550,1.00118},{-0.088,11600,0.999012},{-0.088,11650,0.996702},
{-0.088,11700,0.994251},{-0.088,11750,0.991665},{-0.088,11800,0.988952},
{-0.088,11850,0.986118},{-0.088,11900,0.983171},{-0.088,11950,0.980122},
{-0.088,12000,0.976979},{-0.088,12050,0.973754},{-0.088,12100,0.970457},
{-0.088,12150,0.967101},{-0.088,12200,0.963698},{-0.088,12250,0.960262},
{-0.088,12300,0.956805},{-0.088,12350,0.953341},{-0.088,12400,0.949885},
{-0.088,12450,0.946452},{-0.088,12500,0.943054},{-0.088,12550,0.939708},
{-0.088,12600,0.936428},{-0.088,12650,0.933227},{-0.088,12700,0.930121},
{-0.088,12750,0.927123},{-0.088,12800,0.924246},{-0.088,12850,0.921505},
{-0.088,12900,0.918911},{-0.088,12950,0.916477},{-0.088,13000,0.914214},
{-0.088,13050,0.912132},{-0.088,13100,0.910242},{-0.088,13150,0.908553},
{-0.088,13200,0.907072},{-0.088,13250,0.905806},{-0.088,13300,0.904762},
{-0.088,13350,0.903945},{-0.088,13400,0.903359},{-0.088,13450,0.903006},
{-0.088,13500,0.902888},{-0.086,11000,0.99908},{-0.086,11050,0.997876},
{-0.086,11100,0.996606},{-0.086,11150,0.995273},{-0.086,11200,0.993877},
{-0.086,11250,0.99242},{-0.086,11300,0.990905},{-0.086,11350,0.989333},
{-0.086,11400,0.987707},{-0.086,11450,0.98603},{-0.086,11500,0.984305},
{-0.086,11550,0.982535},{-0.086,11600,0.980724},{-0.086,11650,0.978874},
{-0.086,11700,0.976991},{-0.086,11750,0.975078},{-0.086,11800,0.973139},
{-0.086,11850,0.97118},{-0.086,11900,0.969203},{-0.086,11950,0.967215},
{-0.086,12000,0.96522},{-0.086,12050,0.963223},{-0.086,12100,0.96123},
{-0.086,12150,0.959244},{-0.086,12200,0.957272},{-0.086,12250,0.955319},
{-0.086,12300,0.953389},{-0.086,12350,0.951489},{-0.086,12400,0.949624},
{-0.086,12450,0.947798},{-0.086,12500,0.946017},{-0.086,12550,0.944286},
{-0.086,12600,0.94261},{-0.086,12650,0.940994},{-0.086,12700,0.939443},
{-0.086,12750,0.93796},{-0.086,12800,0.936551},{-0.086,12850,0.93522},
{-0.086,12900,0.93397},{-0.086,12950,0.932806},{-0.086,13000,0.93173},
{-0.086,13050,0.930747},{-0.086,13100,0.929859},{-0.086,13150,0.929069},
{-0.086,13200,0.928379},{-0.086,13250,0.927792},{-0.086,13300,0.927309},
{-0.086,13350,0.926932},{-0.086,13400,0.926662},{-0.086,13450,0.926499},
{-0.086,13500,0.926445},{-0.084,11000,0.983926},{-0.084,11050,0.983048},
{-0.084,11100,0.982159},{-0.084,11150,0.981259},{-0.084,11200,0.980349},
{-0.084,11250,0.979431},{-0.084,11300,0.978505},{-0.084,11350,0.977573},
{-0.084,11400,0.976636},{-0.084,11450,0.975695},{-0.084,11500,0.974751},
{-0.084,11550,0.973806},{-0.084,11600,0.972862},{-0.084,11650,0.971919},
{-0.084,11700,0.970979},{-0.084,11750,0.970042},{-0.084,11800,0.969112},
{-0.084,11850,0.968189},{-0.084,11900,0.967274},{-0.084,11950,0.966368},
{-0.084,12000,0.965474},{-0.084,12050,0.964593},{-0.084,12100,0.963726},
{-0.084,12150,0.962875},{-0.084,12200,0.96204},{-0.084,12250,0.961224},
{-0.084,12300,0.960427},{-0.084,12350,0.959651},{-0.084,12400,0.958898},
{-0.084,12450,0.958168},{-0.084,12500,0.957463},{-0.084,12550,0.956784},
{-0.084,12600,0.956132},{-0.084,12650,0.955508},{-0.084,12700,0.954914},
{-0.084,12750,0.95435},{-0.084,12800,0.953818},{-0.084,12850,0.953318},
{-0.084,12900,0.952852},{-0.084,12950,0.95242},{-0.084,13000,0.952022},
{-0.084,13050,0.951661},{-0.084,13100,0.951335},{-0.084,13150,0.951047},
{-0.084,13200,0.950796},{-0.084,13250,0.950582},{-0.084,13300,0.950407},
{-0.084,13350,0.950271},{-0.084,13400,0.950173},{-0.084,13450,0.950114},
{-0.084,13500,0.950095},{-0.082,11000,0.983051},{-0.082,11050,0.982741},
{-0.082,11100,0.982433},{-0.082,11150,0.982125},{-0.082,11200,0.981819},
{-0.082,11250,0.981516},{-0.082,11300,0.981214},{-0.082,11350,0.980915},
{-0.082,11400,0.980618},{-0.082,11450,0.980324},{-0.082,11500,0.980034},
{-0.082,11550,0.979746},{-0.082,11600,0.979463},{-0.082,11650,0.979183},
{-0.082,11700,0.978907},{-0.082,11750,0.978635},{-0.082,11800,0.978368},
{-0.082,11850,0.978106},{-0.082,11900,0.977849},{-0.082,11950,0.977597},
{-0.082,12000,0.977351},{-0.082,12050,0.97711},{-0.082,12100,0.976875},
{-0.082,12150,0.976647},{-0.082,12200,0.976424},{-0.082,12250,0.976209},
{-0.082,12300,0.975999},{-0.082,12350,0.975797},{-0.082,12400,0.975602},
{-0.082,12450,0.975415},{-0.082,12500,0.975234},{-0.082,12550,0.975062},
{-0.082,12600,0.974897},{-0.082,12650,0.97474},{-0.082,12700,0.974591},
{-0.082,12750,0.974451},{-0.082,12800,0.974319},{-0.082,12850,0.974196},
{-0.082,12900,0.974081},{-0.082,12950,0.973975},{-0.082,13000,0.973878},
{-0.082,13050,0.973789},{-0.082,13100,0.97371},{-0.082,13150,0.97364},
{-0.082,13200,0.973579},{-0.082,13250,0.973528},{-0.082,13300,0.973486},
{-0.082,13350,0.973453},{-0.082,13400,0.973429},{-0.082,13450,0.973415},
{-0.082,13500,0.97341},{-0.08,11000,0.99337},{-0.08,11050,0.993324},
{-0.08,11100,0.993278},{-0.08,11150,0.993232},{-0.08,11200,0.993188},
{-0.08,11250,0.993143},{-0.08,11300,0.9931},{-0.08,11350,0.993057},
{-0.08,11400,0.993015},{-0.08,11450,0.992973},{-0.08,11500,0.992932},
{-0.08,11550,0.992892},{-0.08,11600,0.992853},{-0.08,11650,0.992814},
{-0.08,11700,0.992777},{-0.08,11750,0.99274},{-0.08,11800,0.992704},
{-0.08,11850,0.992668},{-0.08,11900,0.992634},{-0.08,11950,0.9926},
{-0.08,12000,0.992568},{-0.08,12050,0.992536},{-0.08,12100,0.992505},
{-0.08,12150,0.992475},{-0.08,12200,0.992446},{-0.08,12250,0.992418},
{-0.08,12300,0.992391},{-0.08,12350,0.992365},{-0.08,12400,0.99234},
{-0.08,12450,0.992316},{-0.08,12500,0.992294},{-0.08,12550,0.992272},
{-0.08,12600,0.992251},{-0.08,12650,0.992231},{-0.08,12700,0.992212},
{-0.08,12750,0.992195},{-0.08,12800,0.992178},{-0.08,12850,0.992163},
{-0.08,12900,0.992149},{-0.08,12950,0.992135},{-0.08,13000,0.992123},
{-0.08,13050,0.992112},{-0.08,13100,0.992103},{-0.08,13150,0.992094},
{-0.08,13200,0.992086},{-0.08,13250,0.99208},{-0.08,13300,0.992075},
{-0.08,13350,0.992071},{-0.08,13400,0.992068},{-0.08,13450,0.992066},
{-0.08,13500,0.992066},{-0.078,11000,0.999253},{-0.078,11050,0.999251},
{-0.078,11100,0.999248},{-0.078,11150,0.999246},{-0.078,11200,0.999243},
{-0.078,11250,0.999241},{-0.078,11300,0.999239},{-0.078,11350,0.999236},
{-0.078,11400,0.999234},{-0.078,11450,0.999232},{-0.078,11500,0.999229},
{-0.078,11550,0.999227},{-0.078,11600,0.999225},{-0.078,11650,0.999223},
{-0.078,11700,0.999221},{-0.078,11750,0.999219},{-0.078,11800,0.999217},
{-0.078,11850,0.999215},{-0.078,11900,0.999213},{-0.078,11950,0.999211},
{-0.078,12000,0.99921},{-0.078,12050,0.999208},{-0.078,12100,0.999206},
{-0.078,12150,0.999205},{-0.078,12200,0.999203},{-0.078,12250,0.999202},
{-0.078,12300,0.9992},{-0.078,12350,0.999199},{-0.078,12400,0.999198},
{-0.078,12450,0.999196},{-0.078,12500,0.999195},{-0.078,12550,0.999194},
{-0.078,12600,0.999193},{-0.078,12650,0.999192},{-0.078,12700,0.999191},
{-0.078,12750,0.99919},{-0.078,12800,0.999189},{-0.078,12850,0.999188},
{-0.078,12900,0.999187},{-0.078,12950,0.999187},{-0.078,13000,0.999186},
{-0.078,13050,0.999186},{-0.078,13100,0.999185},{-0.078,13150,0.999185},
{-0.078,13200,0.999184},{-0.078,13250,0.999184},{-0.078,13300,0.999184},
{-0.078,13350,0.999183},{-0.078,13400,0.999183},{-0.078,13450,0.999183},
{-0.078,13500,0.999183},{-0.076,11000,0.999963},{-0.076,11050,0.999963},
{-0.076,11100,0.999963},{-0.076,11150,0.999962},{-0.076,11200,0.999962},
{-0.076,11250,0.999962},{-0.076,11300,0.999961},{-0.076,11350,0.999961},
{-0.076,11400,0.999961},{-0.076,11450,0.99996},{-0.076,11500,0.99996},
{-0.076,11550,0.99996},{-0.076,11600,0.99996},{-0.076,11650,0.999959},
{-0.076,11700,0.999959},{-0.076,11750,0.999959},{-0.076,11800,0.999958},
{-0.076,11850,0.999958},{-0.076,11900,0.999958},{-0.076,11950,0.999958},
{-0.076,12000,0.999957},{-0.076,12050,0.999957},{-0.076,12100,0.999957},
{-0.076,12150,0.999956},{-0.076,12200,0.999956},{-0.076,12250,0.999956},
{-0.076,12300,0.999956},{-0.076,12350,0.999956},{-0.076,12400,0.999955},
{-0.076,12450,0.999955},{-0.076,12500,0.999955},{-0.076,12550,0.999955},
{-0.076,12600,0.999955},{-0.076,12650,0.999954},{-0.076,12700,0.999954},
{-0.076,12750,0.999954},{-0.076,12800,0.999954},{-0.076,12850,0.999954},
{-0.076,12900,0.999954},{-0.076,12950,0.999954},{-0.076,13000,0.999954},
{-0.076,13050,0.999953},{-0.076,13100,0.999953},{-0.076,13150,0.999953},
{-0.076,13200,0.999953},{-0.076,13250,0.999953},{-0.076,13300,0.999953},
{-0.076,13350,0.999953},{-0.076,13400,0.999953},{-0.076,13450,0.999953},
{-0.076,13500,0.999953},{-0.074,11000,0.999983},{-0.074,11050,0.999983},
{-0.074,11100,0.999982},{-0.074,11150,0.999982},{-0.074,11200,0.999982},
{-0.074,11250,0.999982},{-0.074,11300,0.999981},{-0.074,11350,0.999981},
{-0.074,11400,0.999981},{-0.074,11450,0.999981},{-0.074,11500,0.99998},
{-0.074,11550,0.99998},{-0.074,11600,0.99998},{-0.074,11650,0.99998},
{-0.074,11700,0.999979},{-0.074,11750,0.999979},{-0.074,11800,0.999979},
{-0.074,11850,0.999979},{-0.074,11900,0.999979},{-0.074,11950,0.999978},
{-0.074,12000,0.999978},{-0.074,12050,0.999978},{-0.074,12100,0.999978},
{-0.074,12150,0.999978},{-0.074,12200,0.999977},{-0.074,12250,0.999977},
{-0.074,12300,0.999977},{-0.074,12350,0.999977},{-0.074,12400,0.999977},
{-0.074,12450,0.999977},{-0.074,12500,0.999976},{-0.074,12550,0.999976},
{-0.074,12600,0.999976},{-0.074,12650,0.999976},{-0.074,12700,0.999976},
{-0.074,12750,0.999976},{-0.074,12800,0.999976},{-0.074,12850,0.999976},
{-0.074,12900,0.999976},{-0.074,12950,0.999975},{-0.074,13000,0.999975},
{-0.074,13050,0.999975},{-0.074,13100,0.999975},{-0.074,13150,0.999975},
{-0.074,13200,0.999975},{-0.074,13250,0.999975},{-0.074,13300,0.999975},
{-0.074,13350,0.999975},{-0.074,13400,0.999975},{-0.074,13450,0.999975},
{-0.074,13500,0.999975},{-0.072,11000,0.999986},{-0.072,11050,0.999986},
{-0.072,11100,0.999986},{-0.072,11150,0.999985},{-0.072,11200,0.999985},
{-0.072,11250,0.999985},{-0.072,11300,0.999985},{-0.072,11350,0.999985},
{-0.072,11400,0.999984},{-0.072,11450,0.999984},{-0.072,11500,0.999984},
{-0.072,11550,0.999984},{-0.072,11600,0.999984},{-0.072,11650,0.999984},
{-0.072,11700,0.999983},{-0.072,11750,0.999983},{-0.072,11800,0.999983},
{-0.072,11850,0.999983},{-0.072,11900,0.999983},{-0.072,11950,0.999983},
{-0.072,12000,0.999983},{-0.072,12050,0.999982},{-0.072,12100,0.999982},
{-0.072,12150,0.999982},{-0.072,12200,0.999982},{-0.072,12250,0.999982},
{-0.072,12300,0.999982},{-0.072,12350,0.999982},{-0.072,12400,0.999981},
{-0.072,12450,0.999981},{-0.072,12500,0.999981},{-0.072,12550,0.999981},
{-0.072,12600,0.999981},{-0.072,12650,0.999981},{-0.072,12700,0.999981},
{-0.072,12750,0.999981},{-0.072,12800,0.999981},{-0.072,12850,0.999981},
{-0.072,12900,0.999981},{-0.072,12950,0.999981},{-0.072,13000,0.99998},
{-0.072,13050,0.99998},{-0.072,13100,0.99998},{-0.072,13150,0.99998},
{-0.072,13200,0.99998},{-0.072,13250,0.99998},{-0.072,13300,0.99998},
{-0.072,13350,0.99998},{-0.072,13400,0.99998},{-0.072,13450,0.99998},
{-0.072,13500,0.99998},{-0.07,11000,0.999989},{-0.07,11050,0.999989},
{-0.07,11100,0.999989},{-0.07,11150,0.999989},{-0.07,11200,0.999989},
{-0.07,11250,0.999988},{-0.07,11300,0.999988},{-0.07,11350,0.999988},
{-0.07,11400,0.999988},{-0.07,11450,0.999988},{-0.07,11500,0.999988},
{-0.07,11550,0.999988},{-0.07,11600,0.999988},{-0.07,11650,0.999987},
{-0.07,11700,0.999987},{-0.07,11750,0.999987},{-0.07,11800,0.999987},
{-0.07,11850,0.999987},{-0.07,11900,0.999987},{-0.07,11950,0.999987},
{-0.07,12000,0.999987},{-0.07,12050,0.999987},{-0.07,12100,0.999987},
{-0.07,12150,0.999986},{-0.07,12200,0.999986},{-0.07,12250,0.999986},
{-0.07,12300,0.999986},{-0.07,12350,0.999986},{-0.07,12400,0.999986},
{-0.07,12450,0.999986},{-0.07,12500,0.999986},{-0.07,12550,0.999986},
{-0.07,12600,0.999986},{-0.07,12650,0.999986},{-0.07,12700,0.999986},
{-0.07,12750,0.999986},{-0.07,12800,0.999986},{-0.07,12850,0.999985},
{-0.07,12900,0.999985},{-0.07,12950,0.999985},{-0.07,13000,0.999985},
{-0.07,13050,0.999985},{-0.07,13100,0.999985},{-0.07,13150,0.999985},
{-0.07,13200,0.999985},{-0.07,13250,0.999985},{-0.07,13300,0.999985},
{-0.07,13350,0.999985},{-0.07,13400,0.999985},{-0.07,13450,0.999985},
{-0.07,13500,0.999985},{-0.068,11000,0.999992},{-0.068,11050,0.999992},
{-0.068,11100,0.999992},{-0.068,11150,0.999992},{-0.068,11200,0.999992},
{-0.068,11250,0.999992},{-0.068,11300,0.999992},{-0.068,11350,0.999992},
{-0.068,11400,0.999992},{-0.068,11450,0.999992},{-0.068,11500,0.999991},
{-0.068,11550,0.999991},{-0.068,11600,0.999991},{-0.068,11650,0.999991},
{-0.068,11700,0.999991},{-0.068,11750,0.999991},{-0.068,11800,0.999991},
{-0.068,11850,0.999991},{-0.068,11900,0.999991},{-0.068,11950,0.999991},
{-0.068,12000,0.999991},{-0.068,12050,0.999991},{-0.068,12100,0.999991},
{-0.068,12150,0.999991},{-0.068,12200,0.999991},{-0.068,12250,0.999991},
{-0.068,12300,0.999991},{-0.068,12350,0.99999},{-0.068,12400,0.99999},
{-0.068,12450,0.99999},{-0.068,12500,0.99999},{-0.068,12550,0.99999},
{-0.068,12600,0.99999},{-0.068,12650,0.99999},{-0.068,12700,0.99999},
{-0.068,12750,0.99999},{-0.068,12800,0.99999},{-0.068,12850,0.99999},
{-0.068,12900,0.99999},{-0.068,12950,0.99999},{-0.068,13000,0.99999},
{-0.068,13050,0.99999},{-0.068,13100,0.99999},{-0.068,13150,0.99999},
{-0.068,13200,0.99999},{-0.068,13250,0.99999},{-0.068,13300,0.99999},
{-0.068,13350,0.99999},{-0.068,13400,0.99999},{-0.068,13450,0.99999},
{-0.068,13500,0.99999},{-0.066,11000,0.999995},{-0.066,11050,0.999995},
{-0.066,11100,0.999995},{-0.066,11150,0.999995},{-0.066,11200,0.999995},
{-0.066,11250,0.999995},{-0.066,11300,0.999995},{-0.066,11350,0.999995},
{-0.066,11400,0.999995},{-0.066,11450,0.999995},{-0.066,11500,0.999995},
{-0.066,11550,0.999995},{-0.066,11600,0.999995},{-0.066,11650,0.999995},
{-0.066,11700,0.999995},{-0.066,11750,0.999995},{-0.066,11800,0.999995},
{-0.066,11850,0.999995},{-0.066,11900,0.999995},{-0.066,11950,0.999995},
{-0.066,12000,0.999995},{-0.066,12050,0.999995},{-0.066,12100,0.999995},
{-0.066,12150,0.999995},{-0.066,12200,0.999995},{-0.066,12250,0.999995},
{-0.066,12300,0.999994},{-0.066,12350,0.999994},{-0.066,12400,0.999994},
{-0.066,12450,0.999994},{-0.066,12500,0.999994},{-0.066,12550,0.999994},
{-0.066,12600,0.999994},{-0.066,12650,0.999994},{-0.066,12700,0.999994},
{-0.066,12750,0.999994},{-0.066,12800,0.999994},{-0.066,12850,0.999994},
{-0.066,12900,0.999994},{-0.066,12950,0.999994},{-0.066,13000,0.999994},
{-0.066,13050,0.999994},{-0.066,13100,0.999994},{-0.066,13150,0.999994},
{-0.066,13200,0.999994},{-0.066,13250,0.999994},{-0.066,13300,0.999994},
{-0.066,13350,0.999994},{-0.066,13400,0.999994},{-0.066,13450,0.999994},
{-0.066,13500,0.999994},{-0.064,11000,0.999998},{-0.064,11050,0.999998},
{-0.064,11100,0.999998},{-0.064,11150,0.999998},{-0.064,11200,0.999998},
{-0.064,11250,0.999998},{-0.064,11300,0.999998},{-0.064,11350,0.999998},
{-0.064,11400,0.999998},{-0.064,11450,0.999998},{-0.064,11500,0.999998},
{-0.064,11550,0.999998},{-0.064,11600,0.999998},{-0.064,11650,0.999998},
{-0.064,11700,0.999998},{-0.064,11750,0.999998},{-0.064,11800,0.999998},
{-0.064,11850,0.999998},{-0.064,11900,0.999998},{-0.064,11950,0.999998},
{-0.064,12000,0.999998},{-0.064,12050,0.999998},{-0.064,12100,0.999998},
{-0.064,12150,0.999998},{-0.064,12200,0.999998},{-0.064,12250,0.999998},
{-0.064,12300,0.999998},{-0.064,12350,0.999998},{-0.064,12400,0.999998},
{-0.064,12450,0.999998},{-0.064,12500,0.999998},{-0.064,12550,0.999998},
{-0.064,12600,0.999998},{-0.064,12650,0.999998},{-0.064,12700,0.999998},
{-0.064,12750,0.999998},{-0.064,12800,0.999998},{-0.064,12850,0.999998},
{-0.064,12900,0.999998},{-0.064,12950,0.999998},{-0.064,13000,0.999998},
{-0.064,13050,0.999998},{-0.064,13100,0.999998},{-0.064,13150,0.999998},
{-0.064,13200,0.999998},{-0.064,13250,0.999998},{-0.064,13300,0.999998},
{-0.064,13350,0.999998},{-0.064,13400,0.999998},{-0.064,13450,0.999998},
{-0.064,13500,0.999998},{-0.062,11000,0.999999},{-0.062,11050,0.999999},
{-0.062,11100,0.999999},{-0.062,11150,0.999999},{-0.062,11200,0.999999},
{-0.062,11250,0.999999},{-0.062,11300,0.999999},{-0.062,11350,0.999999},
{-0.062,11400,0.999999},{-0.062,11450,0.999999},{-0.062,11500,0.999999},
{-0.062,11550,0.999999},{-0.062,11600,0.999999},{-0.062,11650,0.999999},
{-0.062,11700,0.999999},{-0.062,11750,0.999999},{-0.062,11800,0.999999},
{-0.062,11850,0.999999},{-0.062,11900,0.999999},{-0.062,11950,0.999999},
{-0.062,12000,0.999999},{-0.062,12050,0.999999},{-0.062,12100,0.999999},
{-0.062,12150,0.999999},{-0.062,12200,0.999999},{-0.062,12250,0.999999},
{-0.062,12300,0.999999},{-0.062,12350,0.999999},{-0.062,12400,0.999999},
{-0.062,12450,0.999999},{-0.062,12500,0.999999},{-0.062,12550,0.999999},
{-0.062,12600,0.999999},{-0.062,12650,0.999999},{-0.062,12700,0.999999},
{-0.062,12750,0.999999},{-0.062,12800,0.999999},{-0.062,12850,0.999999},
{-0.062,12900,0.999999},{-0.062,12950,0.999999},{-0.062,13000,0.999999},
{-0.062,13050,0.999999},{-0.062,13100,0.999999},{-0.062,13150,0.999999},
{-0.062,13200,0.999999},{-0.062,13250,0.999999},{-0.062,13300,0.999999},
{-0.062,13350,0.999999},{-0.062,13400,0.999999},{-0.062,13450,0.999999},
{-0.062,13500,0.999999},{-0.06,11000,1.},{-0.06,11050,1.},{-0.06,11100,1.},
{-0.06,11150,1.},{-0.06,11200,1.},{-0.06,11250,1.},{-0.06,11300,1.},
{-0.06,11350,1.},{-0.06,11400,1.},{-0.06,11450,1.},{-0.06,11500,1.},
{-0.06,11550,1.},{-0.06,11600,1.},{-0.06,11650,1.},{-0.06,11700,1.},
{-0.06,11750,1.},{-0.06,11800,1.},{-0.06,11850,1.},{-0.06,11900,1.},
{-0.06,11950,1.},{-0.06,12000,1.},{-0.06,12050,1.},{-0.06,12100,1.},
{-0.06,12150,1.},{-0.06,12200,1.},{-0.06,12250,1.},{-0.06,12300,1.},
{-0.06,12350,1.},{-0.06,12400,1.},{-0.06,12450,1.},{-0.06,12500,1.},
{-0.06,12550,1.},{-0.06,12600,1.},{-0.06,12650,1.},{-0.06,12700,1.},
{-0.06,12750,1.},{-0.06,12800,1.},{-0.06,12850,1.},{-0.06,12900,1.},
{-0.06,12950,1.},{-0.06,13000,1.},{-0.06,13050,1.},{-0.06,13100,1.},
{-0.06,13150,1.},{-0.06,13200,1.},{-0.06,13250,1.},{-0.06,13300,1.},
{-0.06,13350,1.},{-0.06,13400,1.},{-0.06,13450,1.},{-0.06,13500,1.}};
</code></pre>
| Alexey Popkov | 280 | <p>I also had such problem in version 7 when the data grid is too regular. I think it is a limitation of the triangulation algorithm used in v.7. To avoid this you could try to perturbate original grid a bit by adding a small random noise to the {x ,y} values: <a href="https://groups.google.com/d/msg/comp.soft-sys.math.mathematica/Zvz2_wZmQyI/nGJRukiBWJ8J" rel="nofollow">Patrick Scheibe's method</a>, <a href="https://groups.google.com/d/msg/comp.soft-sys.math.mathematica/Zvz2_wZmQyI/lOXvIxPiLHkJ" rel="nofollow">Peter Pein's method</a>. Another possibility is to use <code>Interpolation</code> with <code>InterpolationOrder->1</code> and <code>ContourPlot</code> instead of <code>ListContourPlot</code>. </p>
|
19,253 | <p><strong>Bug introduced in 7.0.1 or earlier and fixed in 10.0.2 or earlier</strong></p>
<hr>
<p>I have access to two versions of Mathematica, version 7.0.1 on Linux and version 8 on Windows. When I try the following two lines on version 7, the kernel quits when it tries to plot. In version 8 it plots just fine. You can import the data using:</p>
<pre><code>temp = Import["http://pastie.org/pastes/6098244/download", "Table"];
</code></pre>
<p>Any idea why this crashes in version 7? </p>
<pre><code>ListContourPlot[temp]
</code></pre>
<p>It may also crash in Mathematica 7 for Windows too, but I don't personally have a copy of that on my machine.</p>
<p>ETA: Other users have reported this also crashes in some versions 8 and 9 across different platforms, so I changed the name of the post accordingly.</p>
<hr>
<p>If the pastebin link fails, please copy the data from here:</p>
<pre><code>temp={{-0.1,11000,1.00214},{-0.1,11050,0.998176},
{-0.1,11100,0.994088},{-0.1,11150,0.990016},{-0.1,11200,0.986111},
{-0.1,11250,0.982526},{-0.1,11300,0.979411},{-0.1,11350,0.976904},
{-0.1,11400,0.97513},{-0.1,11450,0.974195},{-0.1,11500,0.974179},
{-0.1,11550,0.975137},{-0.1,11600,0.977087},{-0.1,11650,0.98002},
{-0.1,11700,0.983885},{-0.1,11750,0.988602},{-0.1,11800,0.994051},
{-0.1,11850,1.00008},{-0.1,11900,1.00652},{-0.1,11950,1.01315},
{-0.1,12000,1.01975},{-0.1,12050,1.02608},{-0.1,12100,1.03189},
{-0.1,12150,1.03694},{-0.1,12200,1.04098},{-0.1,12250,1.04378},
{-0.1,12300,1.04513},{-0.1,12350,1.04486},{-0.1,12400,1.04282},
{-0.1,12450,1.03889},{-0.1,12500,1.03303},{-0.1,12550,1.02522},
{-0.1,12600,1.01549},{-0.1,12650,1.00392},{-0.1,12700,0.990664},
{-0.1,12750,0.975894},{-0.1,12800,0.959842},{-0.1,12850,0.942777},
{-0.1,12900,0.924999},{-0.1,12950,0.906838},{-0.1,13000,0.888643},
{-0.1,13050,0.870773},{-0.1,13100,0.853588},{-0.1,13150,0.837442},
{-0.1,13200,0.822673},{-0.1,13250,0.809592},{-0.1,13300,0.798481},
{-0.1,13350,0.789577},{-0.1,13400,0.783075},{-0.1,13450,0.779116},
{-0.1,13500,0.777787},{-0.098,11000,0.990164},{-0.098,11050,0.986921},
{-0.098,11100,0.983944},{-0.098,11150,0.981344},{-0.098,11200,0.979226},
{-0.098,11250,0.977683},{-0.098,11300,0.976797},{-0.098,11350,0.976633},
{-0.098,11400,0.977237},{-0.098,11450,0.978633},{-0.098,11500,0.980822},
{-0.098,11550,0.983781},{-0.098,11600,0.98746},{-0.098,11650,0.991787},
{-0.098,11700,0.996664},{-0.098,11750,1.00197},{-0.098,11800,1.00756},
{-0.098,11850,1.01329},{-0.098,11900,1.01897},{-0.098,11950,1.02443},
{-0.098,12000,1.02947},{-0.098,12050,1.03391},{-0.098,12100,1.03756},
{-0.098,12150,1.04024},{-0.098,12200,1.0418},{-0.098,12250,1.04207},
{-0.098,12300,1.04095},{-0.098,12350,1.03834},{-0.098,12400,1.03416},
{-0.098,12450,1.0284},{-0.098,12500,1.02104},{-0.098,12550,1.01214},
{-0.098,12600,1.00175},{-0.098,12650,0.989997},{-0.098,12700,0.977021},
{-0.098,12750,0.962997},{-0.098,12800,0.948131},{-0.098,12850,0.932651},
{-0.098,12900,0.916808},{-0.098,12950,0.900866},{-0.098,13000,0.885099},
{-0.098,13050,0.869785},{-0.098,13100,0.855197},{-0.098,13150,0.841603},
{-0.098,13200,0.829253},{-0.098,13250,0.818377},{-0.098,13300,0.809181},
{-0.098,13350,0.80184},{-0.098,13400,0.796493},{-0.098,13450,0.793244},
{-0.098,13500,0.792153},{-0.096,11000,0.98162},{-0.096,11050,0.980727},
{-0.096,11100,0.980311},{-0.096,11150,0.980411},{-0.096,11200,0.981055},
{-0.096,11250,0.982257},{-0.096,11300,0.984019},{-0.096,11350,0.986327},
{-0.096,11400,0.989153},{-0.096,11450,0.992454},{-0.096,11500,0.99617},
{-0.096,11550,1.00023},{-0.096,11600,1.00455},{-0.096,11650,1.00903},
{-0.096,11700,1.01357},{-0.096,11750,1.01804},{-0.096,11800,1.02233},
{-0.096,11850,1.02632},{-0.096,11900,1.02987},{-0.096,11950,1.03285},
{-0.096,12000,1.03516},{-0.096,12050,1.03668},{-0.096,12100,1.03729},
{-0.096,12150,1.03692},{-0.096,12200,1.03549},{-0.096,12250,1.03294},
{-0.096,12300,1.02922},{-0.096,12350,1.02432},{-0.096,12400,1.01825},
{-0.096,12450,1.01102},{-0.096,12500,1.00268},{-0.096,12550,0.993304},
{-0.096,12600,0.98298},{-0.096,12650,0.971818},{-0.096,12700,0.959949},
{-0.096,12750,0.94752},{-0.096,12800,0.934692},{-0.096,12850,0.921638},
{-0.096,12900,0.90854},{-0.096,12950,0.895584},{-0.096,13000,0.882961},
{-0.096,13050,0.870857},{-0.096,13100,0.859456},{-0.096,13150,0.848933},
{-0.096,13200,0.839451},{-0.096,13250,0.831159},{-0.096,13300,0.824186},
{-0.096,13350,0.818643},{-0.096,13400,0.81462},{-0.096,13450,0.81218},
{-0.096,13500,0.811362},{-0.094,11000,0.98692},{-0.094,11050,0.988506},
{-0.094,11100,0.990436},{-0.094,11150,0.99269},{-0.094,11200,0.995243},
{-0.094,11250,0.998058},{-0.094,11300,1.00109},{-0.094,11350,1.0043},
{-0.094,11400,1.00762},{-0.094,11450,1.011},{-0.094,11500,1.01436},
{-0.094,11550,1.01764},{-0.094,11600,1.02076},{-0.094,11650,1.02365},
{-0.094,11700,1.02624},{-0.094,11750,1.02844},{-0.094,11800,1.03019},
{-0.094,11850,1.03142},{-0.094,11900,1.03206},{-0.094,11950,1.03205},
{-0.094,12000,1.03135},{-0.094,12050,1.02991},{-0.094,12100,1.02769},
{-0.094,12150,1.02468},{-0.094,12200,1.02086},{-0.094,12250,1.01624},
{-0.094,12300,1.01081},{-0.094,12350,1.00462},{-0.094,12400,0.997686},
{-0.094,12450,0.990066},{-0.094,12500,0.981817},{-0.094,12550,0.973011},
{-0.094,12600,0.963727},{-0.094,12650,0.954056},{-0.094,12700,0.944096},
{-0.094,12750,0.933951},{-0.094,12800,0.923733},{-0.094,12850,0.913554},
{-0.094,12900,0.903532},{-0.094,12950,0.893781},{-0.094,13000,0.884419},
{-0.094,13050,0.875557},{-0.094,13100,0.867302},{-0.094,13150,0.859756},
{-0.094,13200,0.853013},{-0.094,13250,0.847157},{-0.094,13300,0.84226},
{-0.094,13350,0.838386},{-0.094,13400,0.835583},{-0.094,13450,0.833887},
{-0.094,13500,0.833319},{-0.092,11000,1.00347},{-0.092,11050,1.00579},
{-0.092,11100,1.00815},{-0.092,11150,1.01053},{-0.092,11200,1.01289},
{-0.092,11250,1.0152},{-0.092,11300,1.0174},{-0.092,11350,1.01948},
{-0.092,11400,1.02137},{-0.092,11450,1.02304},{-0.092,11500,1.02446},
{-0.092,11550,1.02558},{-0.092,11600,1.02636},{-0.092,11650,1.02677},
{-0.092,11700,1.02678},{-0.092,11750,1.02636},{-0.092,11800,1.02547},
{-0.092,11850,1.02411},{-0.092,11900,1.02226},{-0.092,11950,1.01989},
{-0.092,12000,1.01702},{-0.092,12050,1.01362},{-0.092,12100,1.00972},
{-0.092,12150,1.00532},{-0.092,12200,1.00044},{-0.092,12250,0.995097},
{-0.092,12300,0.989324},{-0.092,12350,0.983155},{-0.092,12400,0.976629},
{-0.092,12450,0.969789},{-0.092,12500,0.962686},{-0.092,12550,0.955372},
{-0.092,12600,0.947903},{-0.092,12650,0.940341},{-0.092,12700,0.932747},
{-0.092,12750,0.925184},{-0.092,12800,0.917719},{-0.092,12850,0.910417},
{-0.092,12900,0.903341},{-0.092,12950,0.896558},{-0.092,13000,0.890127},
{-0.092,13050,0.884109},{-0.092,13100,0.87856},{-0.092,13150,0.873532},
{-0.092,13200,0.869073},{-0.092,13250,0.865224},{-0.092,13300,0.862024},
{-0.092,13350,0.859502},{-0.092,13400,0.857683},{-0.092,13450,0.856584},
{-0.092,13500,0.856217},{-0.09,11000,1.01694},{-0.09,11050,1.01807},
{-0.09,11100,1.01906},{-0.09,11150,1.0199},{-0.09,11200,1.02057},
{-0.09,11250,1.02104},{-0.09,11300,1.0213},{-0.09,11350,1.02134},
{-0.09,11400,1.02113},{-0.09,11450,1.02066},{-0.09,11500,1.01993},
{-0.09,11550,1.01891},{-0.09,11600,1.01761},{-0.09,11650,1.01602},
{-0.09,11700,1.01412},{-0.09,11750,1.01193},{-0.09,11800,1.00943},
{-0.09,11850,1.00664},{-0.09,11900,1.00356},{-0.09,11950,1.00019},
{-0.09,12000,0.996544},{-0.09,12050,0.992639},{-0.09,12100,0.988487},
{-0.09,12150,0.984107},{-0.09,12200,0.979518},{-0.09,12250,0.974743},
{-0.09,12300,0.969804},{-0.09,12350,0.964729},{-0.09,12400,0.959544},
{-0.09,12450,0.954279},{-0.09,12500,0.948964},{-0.09,12550,0.943631},
{-0.09,12600,0.938312},{-0.09,12650,0.933039},{-0.09,12700,0.927846},
{-0.09,12750,0.922765},{-0.09,12800,0.917829},{-0.09,12850,0.913071},
{-0.09,12900,0.908522},{-0.09,12950,0.904211},{-0.09,13000,0.900169},
{-0.09,13050,0.896423},{-0.09,13100,0.892997},{-0.09,13150,0.889917},
{-0.09,13200,0.887202},{-0.09,13250,0.884872},{-0.09,13300,0.882943},
{-0.09,13350,0.881428},{-0.09,13400,0.880339},{-0.09,13450,0.879682},
{-0.09,13500,0.879463},{-0.088,11000,1.01491},{-0.088,11050,1.01441},
{-0.088,11100,1.01377},{-0.088,11150,1.01298},{-0.088,11200,1.01205},
{-0.088,11250,1.01096},{-0.088,11300,1.00972},{-0.088,11350,1.00832},
{-0.088,11400,1.00677},{-0.088,11450,1.00506},{-0.088,11500,1.00319},
{-0.088,11550,1.00118},{-0.088,11600,0.999012},{-0.088,11650,0.996702},
{-0.088,11700,0.994251},{-0.088,11750,0.991665},{-0.088,11800,0.988952},
{-0.088,11850,0.986118},{-0.088,11900,0.983171},{-0.088,11950,0.980122},
{-0.088,12000,0.976979},{-0.088,12050,0.973754},{-0.088,12100,0.970457},
{-0.088,12150,0.967101},{-0.088,12200,0.963698},{-0.088,12250,0.960262},
{-0.088,12300,0.956805},{-0.088,12350,0.953341},{-0.088,12400,0.949885},
{-0.088,12450,0.946452},{-0.088,12500,0.943054},{-0.088,12550,0.939708},
{-0.088,12600,0.936428},{-0.088,12650,0.933227},{-0.088,12700,0.930121},
{-0.088,12750,0.927123},{-0.088,12800,0.924246},{-0.088,12850,0.921505},
{-0.088,12900,0.918911},{-0.088,12950,0.916477},{-0.088,13000,0.914214},
{-0.088,13050,0.912132},{-0.088,13100,0.910242},{-0.088,13150,0.908553},
{-0.088,13200,0.907072},{-0.088,13250,0.905806},{-0.088,13300,0.904762},
{-0.088,13350,0.903945},{-0.088,13400,0.903359},{-0.088,13450,0.903006},
{-0.088,13500,0.902888},{-0.086,11000,0.99908},{-0.086,11050,0.997876},
{-0.086,11100,0.996606},{-0.086,11150,0.995273},{-0.086,11200,0.993877},
{-0.086,11250,0.99242},{-0.086,11300,0.990905},{-0.086,11350,0.989333},
{-0.086,11400,0.987707},{-0.086,11450,0.98603},{-0.086,11500,0.984305},
{-0.086,11550,0.982535},{-0.086,11600,0.980724},{-0.086,11650,0.978874},
{-0.086,11700,0.976991},{-0.086,11750,0.975078},{-0.086,11800,0.973139},
{-0.086,11850,0.97118},{-0.086,11900,0.969203},{-0.086,11950,0.967215},
{-0.086,12000,0.96522},{-0.086,12050,0.963223},{-0.086,12100,0.96123},
{-0.086,12150,0.959244},{-0.086,12200,0.957272},{-0.086,12250,0.955319},
{-0.086,12300,0.953389},{-0.086,12350,0.951489},{-0.086,12400,0.949624},
{-0.086,12450,0.947798},{-0.086,12500,0.946017},{-0.086,12550,0.944286},
{-0.086,12600,0.94261},{-0.086,12650,0.940994},{-0.086,12700,0.939443},
{-0.086,12750,0.93796},{-0.086,12800,0.936551},{-0.086,12850,0.93522},
{-0.086,12900,0.93397},{-0.086,12950,0.932806},{-0.086,13000,0.93173},
{-0.086,13050,0.930747},{-0.086,13100,0.929859},{-0.086,13150,0.929069},
{-0.086,13200,0.928379},{-0.086,13250,0.927792},{-0.086,13300,0.927309},
{-0.086,13350,0.926932},{-0.086,13400,0.926662},{-0.086,13450,0.926499},
{-0.086,13500,0.926445},{-0.084,11000,0.983926},{-0.084,11050,0.983048},
{-0.084,11100,0.982159},{-0.084,11150,0.981259},{-0.084,11200,0.980349},
{-0.084,11250,0.979431},{-0.084,11300,0.978505},{-0.084,11350,0.977573},
{-0.084,11400,0.976636},{-0.084,11450,0.975695},{-0.084,11500,0.974751},
{-0.084,11550,0.973806},{-0.084,11600,0.972862},{-0.084,11650,0.971919},
{-0.084,11700,0.970979},{-0.084,11750,0.970042},{-0.084,11800,0.969112},
{-0.084,11850,0.968189},{-0.084,11900,0.967274},{-0.084,11950,0.966368},
{-0.084,12000,0.965474},{-0.084,12050,0.964593},{-0.084,12100,0.963726},
{-0.084,12150,0.962875},{-0.084,12200,0.96204},{-0.084,12250,0.961224},
{-0.084,12300,0.960427},{-0.084,12350,0.959651},{-0.084,12400,0.958898},
{-0.084,12450,0.958168},{-0.084,12500,0.957463},{-0.084,12550,0.956784},
{-0.084,12600,0.956132},{-0.084,12650,0.955508},{-0.084,12700,0.954914},
{-0.084,12750,0.95435},{-0.084,12800,0.953818},{-0.084,12850,0.953318},
{-0.084,12900,0.952852},{-0.084,12950,0.95242},{-0.084,13000,0.952022},
{-0.084,13050,0.951661},{-0.084,13100,0.951335},{-0.084,13150,0.951047},
{-0.084,13200,0.950796},{-0.084,13250,0.950582},{-0.084,13300,0.950407},
{-0.084,13350,0.950271},{-0.084,13400,0.950173},{-0.084,13450,0.950114},
{-0.084,13500,0.950095},{-0.082,11000,0.983051},{-0.082,11050,0.982741},
{-0.082,11100,0.982433},{-0.082,11150,0.982125},{-0.082,11200,0.981819},
{-0.082,11250,0.981516},{-0.082,11300,0.981214},{-0.082,11350,0.980915},
{-0.082,11400,0.980618},{-0.082,11450,0.980324},{-0.082,11500,0.980034},
{-0.082,11550,0.979746},{-0.082,11600,0.979463},{-0.082,11650,0.979183},
{-0.082,11700,0.978907},{-0.082,11750,0.978635},{-0.082,11800,0.978368},
{-0.082,11850,0.978106},{-0.082,11900,0.977849},{-0.082,11950,0.977597},
{-0.082,12000,0.977351},{-0.082,12050,0.97711},{-0.082,12100,0.976875},
{-0.082,12150,0.976647},{-0.082,12200,0.976424},{-0.082,12250,0.976209},
{-0.082,12300,0.975999},{-0.082,12350,0.975797},{-0.082,12400,0.975602},
{-0.082,12450,0.975415},{-0.082,12500,0.975234},{-0.082,12550,0.975062},
{-0.082,12600,0.974897},{-0.082,12650,0.97474},{-0.082,12700,0.974591},
{-0.082,12750,0.974451},{-0.082,12800,0.974319},{-0.082,12850,0.974196},
{-0.082,12900,0.974081},{-0.082,12950,0.973975},{-0.082,13000,0.973878},
{-0.082,13050,0.973789},{-0.082,13100,0.97371},{-0.082,13150,0.97364},
{-0.082,13200,0.973579},{-0.082,13250,0.973528},{-0.082,13300,0.973486},
{-0.082,13350,0.973453},{-0.082,13400,0.973429},{-0.082,13450,0.973415},
{-0.082,13500,0.97341},{-0.08,11000,0.99337},{-0.08,11050,0.993324},
{-0.08,11100,0.993278},{-0.08,11150,0.993232},{-0.08,11200,0.993188},
{-0.08,11250,0.993143},{-0.08,11300,0.9931},{-0.08,11350,0.993057},
{-0.08,11400,0.993015},{-0.08,11450,0.992973},{-0.08,11500,0.992932},
{-0.08,11550,0.992892},{-0.08,11600,0.992853},{-0.08,11650,0.992814},
{-0.08,11700,0.992777},{-0.08,11750,0.99274},{-0.08,11800,0.992704},
{-0.08,11850,0.992668},{-0.08,11900,0.992634},{-0.08,11950,0.9926},
{-0.08,12000,0.992568},{-0.08,12050,0.992536},{-0.08,12100,0.992505},
{-0.08,12150,0.992475},{-0.08,12200,0.992446},{-0.08,12250,0.992418},
{-0.08,12300,0.992391},{-0.08,12350,0.992365},{-0.08,12400,0.99234},
{-0.08,12450,0.992316},{-0.08,12500,0.992294},{-0.08,12550,0.992272},
{-0.08,12600,0.992251},{-0.08,12650,0.992231},{-0.08,12700,0.992212},
{-0.08,12750,0.992195},{-0.08,12800,0.992178},{-0.08,12850,0.992163},
{-0.08,12900,0.992149},{-0.08,12950,0.992135},{-0.08,13000,0.992123},
{-0.08,13050,0.992112},{-0.08,13100,0.992103},{-0.08,13150,0.992094},
{-0.08,13200,0.992086},{-0.08,13250,0.99208},{-0.08,13300,0.992075},
{-0.08,13350,0.992071},{-0.08,13400,0.992068},{-0.08,13450,0.992066},
{-0.08,13500,0.992066},{-0.078,11000,0.999253},{-0.078,11050,0.999251},
{-0.078,11100,0.999248},{-0.078,11150,0.999246},{-0.078,11200,0.999243},
{-0.078,11250,0.999241},{-0.078,11300,0.999239},{-0.078,11350,0.999236},
{-0.078,11400,0.999234},{-0.078,11450,0.999232},{-0.078,11500,0.999229},
{-0.078,11550,0.999227},{-0.078,11600,0.999225},{-0.078,11650,0.999223},
{-0.078,11700,0.999221},{-0.078,11750,0.999219},{-0.078,11800,0.999217},
{-0.078,11850,0.999215},{-0.078,11900,0.999213},{-0.078,11950,0.999211},
{-0.078,12000,0.99921},{-0.078,12050,0.999208},{-0.078,12100,0.999206},
{-0.078,12150,0.999205},{-0.078,12200,0.999203},{-0.078,12250,0.999202},
{-0.078,12300,0.9992},{-0.078,12350,0.999199},{-0.078,12400,0.999198},
{-0.078,12450,0.999196},{-0.078,12500,0.999195},{-0.078,12550,0.999194},
{-0.078,12600,0.999193},{-0.078,12650,0.999192},{-0.078,12700,0.999191},
{-0.078,12750,0.99919},{-0.078,12800,0.999189},{-0.078,12850,0.999188},
{-0.078,12900,0.999187},{-0.078,12950,0.999187},{-0.078,13000,0.999186},
{-0.078,13050,0.999186},{-0.078,13100,0.999185},{-0.078,13150,0.999185},
{-0.078,13200,0.999184},{-0.078,13250,0.999184},{-0.078,13300,0.999184},
{-0.078,13350,0.999183},{-0.078,13400,0.999183},{-0.078,13450,0.999183},
{-0.078,13500,0.999183},{-0.076,11000,0.999963},{-0.076,11050,0.999963},
{-0.076,11100,0.999963},{-0.076,11150,0.999962},{-0.076,11200,0.999962},
{-0.076,11250,0.999962},{-0.076,11300,0.999961},{-0.076,11350,0.999961},
{-0.076,11400,0.999961},{-0.076,11450,0.99996},{-0.076,11500,0.99996},
{-0.076,11550,0.99996},{-0.076,11600,0.99996},{-0.076,11650,0.999959},
{-0.076,11700,0.999959},{-0.076,11750,0.999959},{-0.076,11800,0.999958},
{-0.076,11850,0.999958},{-0.076,11900,0.999958},{-0.076,11950,0.999958},
{-0.076,12000,0.999957},{-0.076,12050,0.999957},{-0.076,12100,0.999957},
{-0.076,12150,0.999956},{-0.076,12200,0.999956},{-0.076,12250,0.999956},
{-0.076,12300,0.999956},{-0.076,12350,0.999956},{-0.076,12400,0.999955},
{-0.076,12450,0.999955},{-0.076,12500,0.999955},{-0.076,12550,0.999955},
{-0.076,12600,0.999955},{-0.076,12650,0.999954},{-0.076,12700,0.999954},
{-0.076,12750,0.999954},{-0.076,12800,0.999954},{-0.076,12850,0.999954},
{-0.076,12900,0.999954},{-0.076,12950,0.999954},{-0.076,13000,0.999954},
{-0.076,13050,0.999953},{-0.076,13100,0.999953},{-0.076,13150,0.999953},
{-0.076,13200,0.999953},{-0.076,13250,0.999953},{-0.076,13300,0.999953},
{-0.076,13350,0.999953},{-0.076,13400,0.999953},{-0.076,13450,0.999953},
{-0.076,13500,0.999953},{-0.074,11000,0.999983},{-0.074,11050,0.999983},
{-0.074,11100,0.999982},{-0.074,11150,0.999982},{-0.074,11200,0.999982},
{-0.074,11250,0.999982},{-0.074,11300,0.999981},{-0.074,11350,0.999981},
{-0.074,11400,0.999981},{-0.074,11450,0.999981},{-0.074,11500,0.99998},
{-0.074,11550,0.99998},{-0.074,11600,0.99998},{-0.074,11650,0.99998},
{-0.074,11700,0.999979},{-0.074,11750,0.999979},{-0.074,11800,0.999979},
{-0.074,11850,0.999979},{-0.074,11900,0.999979},{-0.074,11950,0.999978},
{-0.074,12000,0.999978},{-0.074,12050,0.999978},{-0.074,12100,0.999978},
{-0.074,12150,0.999978},{-0.074,12200,0.999977},{-0.074,12250,0.999977},
{-0.074,12300,0.999977},{-0.074,12350,0.999977},{-0.074,12400,0.999977},
{-0.074,12450,0.999977},{-0.074,12500,0.999976},{-0.074,12550,0.999976},
{-0.074,12600,0.999976},{-0.074,12650,0.999976},{-0.074,12700,0.999976},
{-0.074,12750,0.999976},{-0.074,12800,0.999976},{-0.074,12850,0.999976},
{-0.074,12900,0.999976},{-0.074,12950,0.999975},{-0.074,13000,0.999975},
{-0.074,13050,0.999975},{-0.074,13100,0.999975},{-0.074,13150,0.999975},
{-0.074,13200,0.999975},{-0.074,13250,0.999975},{-0.074,13300,0.999975},
{-0.074,13350,0.999975},{-0.074,13400,0.999975},{-0.074,13450,0.999975},
{-0.074,13500,0.999975},{-0.072,11000,0.999986},{-0.072,11050,0.999986},
{-0.072,11100,0.999986},{-0.072,11150,0.999985},{-0.072,11200,0.999985},
{-0.072,11250,0.999985},{-0.072,11300,0.999985},{-0.072,11350,0.999985},
{-0.072,11400,0.999984},{-0.072,11450,0.999984},{-0.072,11500,0.999984},
{-0.072,11550,0.999984},{-0.072,11600,0.999984},{-0.072,11650,0.999984},
{-0.072,11700,0.999983},{-0.072,11750,0.999983},{-0.072,11800,0.999983},
{-0.072,11850,0.999983},{-0.072,11900,0.999983},{-0.072,11950,0.999983},
{-0.072,12000,0.999983},{-0.072,12050,0.999982},{-0.072,12100,0.999982},
{-0.072,12150,0.999982},{-0.072,12200,0.999982},{-0.072,12250,0.999982},
{-0.072,12300,0.999982},{-0.072,12350,0.999982},{-0.072,12400,0.999981},
{-0.072,12450,0.999981},{-0.072,12500,0.999981},{-0.072,12550,0.999981},
{-0.072,12600,0.999981},{-0.072,12650,0.999981},{-0.072,12700,0.999981},
{-0.072,12750,0.999981},{-0.072,12800,0.999981},{-0.072,12850,0.999981},
{-0.072,12900,0.999981},{-0.072,12950,0.999981},{-0.072,13000,0.99998},
{-0.072,13050,0.99998},{-0.072,13100,0.99998},{-0.072,13150,0.99998},
{-0.072,13200,0.99998},{-0.072,13250,0.99998},{-0.072,13300,0.99998},
{-0.072,13350,0.99998},{-0.072,13400,0.99998},{-0.072,13450,0.99998},
{-0.072,13500,0.99998},{-0.07,11000,0.999989},{-0.07,11050,0.999989},
{-0.07,11100,0.999989},{-0.07,11150,0.999989},{-0.07,11200,0.999989},
{-0.07,11250,0.999988},{-0.07,11300,0.999988},{-0.07,11350,0.999988},
{-0.07,11400,0.999988},{-0.07,11450,0.999988},{-0.07,11500,0.999988},
{-0.07,11550,0.999988},{-0.07,11600,0.999988},{-0.07,11650,0.999987},
{-0.07,11700,0.999987},{-0.07,11750,0.999987},{-0.07,11800,0.999987},
{-0.07,11850,0.999987},{-0.07,11900,0.999987},{-0.07,11950,0.999987},
{-0.07,12000,0.999987},{-0.07,12050,0.999987},{-0.07,12100,0.999987},
{-0.07,12150,0.999986},{-0.07,12200,0.999986},{-0.07,12250,0.999986},
{-0.07,12300,0.999986},{-0.07,12350,0.999986},{-0.07,12400,0.999986},
{-0.07,12450,0.999986},{-0.07,12500,0.999986},{-0.07,12550,0.999986},
{-0.07,12600,0.999986},{-0.07,12650,0.999986},{-0.07,12700,0.999986},
{-0.07,12750,0.999986},{-0.07,12800,0.999986},{-0.07,12850,0.999985},
{-0.07,12900,0.999985},{-0.07,12950,0.999985},{-0.07,13000,0.999985},
{-0.07,13050,0.999985},{-0.07,13100,0.999985},{-0.07,13150,0.999985},
{-0.07,13200,0.999985},{-0.07,13250,0.999985},{-0.07,13300,0.999985},
{-0.07,13350,0.999985},{-0.07,13400,0.999985},{-0.07,13450,0.999985},
{-0.07,13500,0.999985},{-0.068,11000,0.999992},{-0.068,11050,0.999992},
{-0.068,11100,0.999992},{-0.068,11150,0.999992},{-0.068,11200,0.999992},
{-0.068,11250,0.999992},{-0.068,11300,0.999992},{-0.068,11350,0.999992},
{-0.068,11400,0.999992},{-0.068,11450,0.999992},{-0.068,11500,0.999991},
{-0.068,11550,0.999991},{-0.068,11600,0.999991},{-0.068,11650,0.999991},
{-0.068,11700,0.999991},{-0.068,11750,0.999991},{-0.068,11800,0.999991},
{-0.068,11850,0.999991},{-0.068,11900,0.999991},{-0.068,11950,0.999991},
{-0.068,12000,0.999991},{-0.068,12050,0.999991},{-0.068,12100,0.999991},
{-0.068,12150,0.999991},{-0.068,12200,0.999991},{-0.068,12250,0.999991},
{-0.068,12300,0.999991},{-0.068,12350,0.99999},{-0.068,12400,0.99999},
{-0.068,12450,0.99999},{-0.068,12500,0.99999},{-0.068,12550,0.99999},
{-0.068,12600,0.99999},{-0.068,12650,0.99999},{-0.068,12700,0.99999},
{-0.068,12750,0.99999},{-0.068,12800,0.99999},{-0.068,12850,0.99999},
{-0.068,12900,0.99999},{-0.068,12950,0.99999},{-0.068,13000,0.99999},
{-0.068,13050,0.99999},{-0.068,13100,0.99999},{-0.068,13150,0.99999},
{-0.068,13200,0.99999},{-0.068,13250,0.99999},{-0.068,13300,0.99999},
{-0.068,13350,0.99999},{-0.068,13400,0.99999},{-0.068,13450,0.99999},
{-0.068,13500,0.99999},{-0.066,11000,0.999995},{-0.066,11050,0.999995},
{-0.066,11100,0.999995},{-0.066,11150,0.999995},{-0.066,11200,0.999995},
{-0.066,11250,0.999995},{-0.066,11300,0.999995},{-0.066,11350,0.999995},
{-0.066,11400,0.999995},{-0.066,11450,0.999995},{-0.066,11500,0.999995},
{-0.066,11550,0.999995},{-0.066,11600,0.999995},{-0.066,11650,0.999995},
{-0.066,11700,0.999995},{-0.066,11750,0.999995},{-0.066,11800,0.999995},
{-0.066,11850,0.999995},{-0.066,11900,0.999995},{-0.066,11950,0.999995},
{-0.066,12000,0.999995},{-0.066,12050,0.999995},{-0.066,12100,0.999995},
{-0.066,12150,0.999995},{-0.066,12200,0.999995},{-0.066,12250,0.999995},
{-0.066,12300,0.999994},{-0.066,12350,0.999994},{-0.066,12400,0.999994},
{-0.066,12450,0.999994},{-0.066,12500,0.999994},{-0.066,12550,0.999994},
{-0.066,12600,0.999994},{-0.066,12650,0.999994},{-0.066,12700,0.999994},
{-0.066,12750,0.999994},{-0.066,12800,0.999994},{-0.066,12850,0.999994},
{-0.066,12900,0.999994},{-0.066,12950,0.999994},{-0.066,13000,0.999994},
{-0.066,13050,0.999994},{-0.066,13100,0.999994},{-0.066,13150,0.999994},
{-0.066,13200,0.999994},{-0.066,13250,0.999994},{-0.066,13300,0.999994},
{-0.066,13350,0.999994},{-0.066,13400,0.999994},{-0.066,13450,0.999994},
{-0.066,13500,0.999994},{-0.064,11000,0.999998},{-0.064,11050,0.999998},
{-0.064,11100,0.999998},{-0.064,11150,0.999998},{-0.064,11200,0.999998},
{-0.064,11250,0.999998},{-0.064,11300,0.999998},{-0.064,11350,0.999998},
{-0.064,11400,0.999998},{-0.064,11450,0.999998},{-0.064,11500,0.999998},
{-0.064,11550,0.999998},{-0.064,11600,0.999998},{-0.064,11650,0.999998},
{-0.064,11700,0.999998},{-0.064,11750,0.999998},{-0.064,11800,0.999998},
{-0.064,11850,0.999998},{-0.064,11900,0.999998},{-0.064,11950,0.999998},
{-0.064,12000,0.999998},{-0.064,12050,0.999998},{-0.064,12100,0.999998},
{-0.064,12150,0.999998},{-0.064,12200,0.999998},{-0.064,12250,0.999998},
{-0.064,12300,0.999998},{-0.064,12350,0.999998},{-0.064,12400,0.999998},
{-0.064,12450,0.999998},{-0.064,12500,0.999998},{-0.064,12550,0.999998},
{-0.064,12600,0.999998},{-0.064,12650,0.999998},{-0.064,12700,0.999998},
{-0.064,12750,0.999998},{-0.064,12800,0.999998},{-0.064,12850,0.999998},
{-0.064,12900,0.999998},{-0.064,12950,0.999998},{-0.064,13000,0.999998},
{-0.064,13050,0.999998},{-0.064,13100,0.999998},{-0.064,13150,0.999998},
{-0.064,13200,0.999998},{-0.064,13250,0.999998},{-0.064,13300,0.999998},
{-0.064,13350,0.999998},{-0.064,13400,0.999998},{-0.064,13450,0.999998},
{-0.064,13500,0.999998},{-0.062,11000,0.999999},{-0.062,11050,0.999999},
{-0.062,11100,0.999999},{-0.062,11150,0.999999},{-0.062,11200,0.999999},
{-0.062,11250,0.999999},{-0.062,11300,0.999999},{-0.062,11350,0.999999},
{-0.062,11400,0.999999},{-0.062,11450,0.999999},{-0.062,11500,0.999999},
{-0.062,11550,0.999999},{-0.062,11600,0.999999},{-0.062,11650,0.999999},
{-0.062,11700,0.999999},{-0.062,11750,0.999999},{-0.062,11800,0.999999},
{-0.062,11850,0.999999},{-0.062,11900,0.999999},{-0.062,11950,0.999999},
{-0.062,12000,0.999999},{-0.062,12050,0.999999},{-0.062,12100,0.999999},
{-0.062,12150,0.999999},{-0.062,12200,0.999999},{-0.062,12250,0.999999},
{-0.062,12300,0.999999},{-0.062,12350,0.999999},{-0.062,12400,0.999999},
{-0.062,12450,0.999999},{-0.062,12500,0.999999},{-0.062,12550,0.999999},
{-0.062,12600,0.999999},{-0.062,12650,0.999999},{-0.062,12700,0.999999},
{-0.062,12750,0.999999},{-0.062,12800,0.999999},{-0.062,12850,0.999999},
{-0.062,12900,0.999999},{-0.062,12950,0.999999},{-0.062,13000,0.999999},
{-0.062,13050,0.999999},{-0.062,13100,0.999999},{-0.062,13150,0.999999},
{-0.062,13200,0.999999},{-0.062,13250,0.999999},{-0.062,13300,0.999999},
{-0.062,13350,0.999999},{-0.062,13400,0.999999},{-0.062,13450,0.999999},
{-0.062,13500,0.999999},{-0.06,11000,1.},{-0.06,11050,1.},{-0.06,11100,1.},
{-0.06,11150,1.},{-0.06,11200,1.},{-0.06,11250,1.},{-0.06,11300,1.},
{-0.06,11350,1.},{-0.06,11400,1.},{-0.06,11450,1.},{-0.06,11500,1.},
{-0.06,11550,1.},{-0.06,11600,1.},{-0.06,11650,1.},{-0.06,11700,1.},
{-0.06,11750,1.},{-0.06,11800,1.},{-0.06,11850,1.},{-0.06,11900,1.},
{-0.06,11950,1.},{-0.06,12000,1.},{-0.06,12050,1.},{-0.06,12100,1.},
{-0.06,12150,1.},{-0.06,12200,1.},{-0.06,12250,1.},{-0.06,12300,1.},
{-0.06,12350,1.},{-0.06,12400,1.},{-0.06,12450,1.},{-0.06,12500,1.},
{-0.06,12550,1.},{-0.06,12600,1.},{-0.06,12650,1.},{-0.06,12700,1.},
{-0.06,12750,1.},{-0.06,12800,1.},{-0.06,12850,1.},{-0.06,12900,1.},
{-0.06,12950,1.},{-0.06,13000,1.},{-0.06,13050,1.},{-0.06,13100,1.},
{-0.06,13150,1.},{-0.06,13200,1.},{-0.06,13250,1.},{-0.06,13300,1.},
{-0.06,13350,1.},{-0.06,13400,1.},{-0.06,13450,1.},{-0.06,13500,1.}};
</code></pre>
| Simon Woods | 862 | <p>You can use the undocumented <code>Method</code> option <code>"DelaunayDomainScaling"</code> to deal with the problem. I don't know any details but I assume it works by rescaling the data inside the triangulation algorithm:</p>
<pre><code>ListContourPlot[temp, Method -> {"DelaunayDomainScaling" -> True}, PlotRange -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/cHzuS.png" alt="enter image description here"></p>
|
114,754 | <p>I have several questions concerning the proof. I don't think I quite understand the details and motivation of the proof. Here is the proof given by our professor.</p>
<p>The space of polynomials $F[x]$ is not finite-dimensional.</p>
<p><em>Proof</em>. Suppose
$$F[x] = \operatorname{span}\{f_1,f_2,\dots,f_n\}$$</p>
<p>Let us choose a positive integer $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n$. As $\{f_1,f_2,\dots,f_n\}$
spans $F[x]$ we can find scalars $a_1, a_2, \dots,a_n$ such that $x^N = a_1f_1 + a_2f_2 + \cdots + a_nf_n$.
Then the polynomial </p>
<p>$$G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$$</p>
<p>is a polynomial of degree $N$ which is identically zero. This is a contradiction since $G(x)$ cannot
have more than $N$ roots. </p>
<p><strong>Questions</strong></p>
<ul>
<li>Why is $G(x)$ <strong>identically</strong> zero and what does it mean for it to be <strong>identically</strong> zero? </li>
<li>After obtaining that $G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$, why must we have more than $N$ roots?</li>
<li><p>What is the <strong>motivation</strong> behind choosing $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n?$ </p></li>
<li><p>What are the <strong>implications</strong> if we chose $N$ less than or equal to $f_i$ such that $f_i$ has the greatest degree? How will the proof fail in this case? </p></li>
<li><p>If there are any other important <strong>details</strong> and key <strong>insights</strong> that are worthy to be pointed out please let me know so that I can better my understanding of the argument presented.</p></li>
</ul>
| Brian M. Scott | 12,042 | <p>To say that a polynomial $p(x)$ is <em>identically zero</em> is to say that $p(x)=0$ for <strong>all</strong> values of $x$. Here the coefficients $a_1,\dots,a_n$ were chosen so that $$x^N=a_1f_1(x)+a_2f_2(x)+\dots+a_nf_n(x)$$ for all values of $x$, and $G(x)$ was defined by $$G(x)=x^N-a_1f_1(x)-a_2f_2(x)-\dots-a_nf_n(x)\;,$$ so for each $x$ we have $$G(x)=x^N-\Big(a_1f_1(x)+a_2f_2(x)+\dots+a_nf_n(x)\Big)=0\;.$$ This is exactly what’s meant when we say that $G(x)$ is identically zero and write $G(x)\equiv 0$.</p>
<p>In particular, $G(x)=0$ for <strong>every</strong> value of $x$, so every possible value of $x$ is a root of $G(x)$. Assuming that your field of scalars is infinite, that means that $G(x)$ has infinitely many roots and therefore certainly has more than $N$ roots.</p>
<p>The reason for choosing $N$ to be greater than the degree of any of the $f_i$ is to ensure that the polynomial $$a_1f_1(x)+a_2f_2(x)+\dots+a_nf_n(x)$$ has degree less than $N$, so that when it is subtracted from $x^N$, the resulting polynomial $G(x)$ still has an $x^N$ term, i.e., is still of degree $N$. Otherwise $G(x)$ might actually be the zero polynomial: all of the terms might cancel out. The whole point is to get a polynomial that on the one hand cannot be the zero polynomial, because its leading term is $x^N$, but on the other hand must be the zero polynomial, because it’s the difference of two equal polynomials.</p>
|
3,345,063 | <p><a href="https://i.stack.imgur.com/T3Ue9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T3Ue9.png" alt="Given a triangle whose apex angle is \theta" /></a></p>
<p><strong>Given a triangle with two circles and apex angle equals <span class="math-container">$\theta$</span>.</strong></p>
<p><em><strong>Find the ratio of radius of the two circles in terms of <span class="math-container">$\theta$</span></strong></em>.</p>
<p>My approach: treat the circles as incircle and excircle by drawing a line parallel to base.<br />
We know that <span class="math-container">$$ \frac{r}{R}=s-\frac{a}{s}$$</span> where <span class="math-container">$s=$</span>semiperimeter and <span class="math-container">$$\sin \frac\theta 2= \sqrt{(s-b)(s-c)}/bc$$</span>
and proceeding for equilateral triangle and right angled triangle,
I get <span class="math-container">$$ \left(1+\sin \frac\theta 2\right)/\left(1-\sin \frac\theta 2\right).$$</span></p>
<p>I have proved it for the following angles: <span class="math-container">$\frac{\pi}{4}$</span> and <span class="math-container">$\frac{\pi}{3}$</span>.</p>
<p>But I am yet to prove it for a scalene triangle.</p>
<p>Thanks.</p>
| MachineLearner | 647,466 | <p>Use the lower picture and note that the red lines are perpendicular to the outer edges.</p>
<p>Using this knowledge we can set up the following system.</p>
<p><span class="math-container">$$\sin\dfrac{\theta}{2} = \dfrac{r}{r+s}$$</span>
<span class="math-container">$$\sin\dfrac{\theta}{2} = \dfrac{R}{R+2r+s}$$</span></p>
<p>Equate both equations and solve for <span class="math-container">$s=s(r,R)$</span>. Then take the first equation. Plug in <span class="math-container">$s=s(r,R)$</span> divide the denominator and numerator by <span class="math-container">$R$</span> and you should only get ratios <span class="math-container">$r/R$</span>. Then solve for this ratio.</p>
<p><a href="https://i.stack.imgur.com/cZkgX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cZkgX.png" alt="enter image description here" /></a></p>
|
1,855,650 | <p>Need to solve:</p>
<p>$$2^x+2^{-x} = 2$$</p>
<p>I can't use substitution in this case. Which is the best approach?</p>
<p>Event in this form I do not have any clue:</p>
<p>$$2^x+\frac{1}{2^x} = 2$$</p>
| Aakash Kumar | 346,279 | <p>$$2^x =y $$
$$y+ \frac{1}{y} =2$$
Using AM-GM INEQUALITY
$$\frac{({y+ \frac{1}{y}})}{2} \ge \sqrt {y.\frac{1}{y}}$$</p>
<p>$$y+ \frac{1}{y}\ge 2$$</p>
|
1,855,650 | <p>Need to solve:</p>
<p>$$2^x+2^{-x} = 2$$</p>
<p>I can't use substitution in this case. Which is the best approach?</p>
<p>Event in this form I do not have any clue:</p>
<p>$$2^x+\frac{1}{2^x} = 2$$</p>
| Community | -1 | <p><strong>A contrived solution</strong>:</p>
<p>Write</p>
<p>$$\frac{e^{x\ln2}+e^{-x\ln2}}2=1=\cosh(x\ln2),$$</p>
<p>then</p>
<p>$$x\ln2=\cosh^{-1}1=0.$$</p>
<hr>
<p><strong>Another one</strong>:</p>
<p>By inspection, $x=0$ is a solution.</p>
<p>The derivative of the LHS is</p>
<p>$$2^x\ln2-2^{-x}\ln2,$$ which is positive for $x>0$ and negative for $x<0$, so the function is monotonic on both sides and there are no other roots.</p>
<hr>
<p><strong>Yet another</strong>:</p>
<p>Observe</p>
<p>$$2^x-2+2^{-x}=\sqrt2^{2x}-2\sqrt2^x\sqrt2^{-x}+\sqrt2^{-2x}=(\sqrt2^x-\sqrt2^{-x})^2.$$</p>
<p>The only root is when</p>
<p>$$\sqrt2^x=\sqrt2^{-x},$$ i.e. $$x=-x.$$</p>
<hr>
<p><strong>A last one</strong>:</p>
<p>The equation can be rewritten</p>
<p>$$2^x-1=1-2^{-x},$$ i.e. unless the RHS is zero</p>
<p>$$\frac{2^x-1}{1-2^{-x}}=2^x=1.$$</p>
<p>There is no other solution than $x=0$.</p>
|
161,029 | <p>I have not seen a problem like this so I have no idea what to do.</p>
<p>Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.</p>
<p>$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$</p>
<p>I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$</p>
<p>But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.</p>
| Américo Tavares | 752 | <ol>
<li><p><strong>Eliminating the parameter</strong> $t$. The given system of two parametric equations $$\begin{eqnarray*}\left\{
\begin{array}{c}
x=1+\ln t \\
y=t^{2}+2
\end{array}
\right. \end{eqnarray*} \tag{A}$$ is equivalent successively to
$$\begin{eqnarray*}
\left\{
\begin{array}{c}
x-1=\ln t \\
y=t^{2}+2
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
t=e^{x-1} \\
y=t^{2}+2
\end{array}
\right.
\Leftrightarrow \left\{
\begin{array}{c}
t=e^{x-1} \\
y=\left( e^{x-1}\right) ^{2}+2
\end{array}
\right. \end{eqnarray*}$$ and finally to $$\left\{
\begin{array}{c}
t=e^{x-1} \\
y=e^{2\left( x-1\right) }+2.
\end{array}
\right. \tag{B}
$$
We have thus eliminated the parameter $t$.
Here is a plot of $y=e^{2\left( x-1\right) }+2$
<img src="https://i.stack.imgur.com/hFWNM.jpg" alt="enter image description here">
Differentiating the equation of the curve $$y=e^{2\left( x-1\right) }+2\tag{C}$$ gives by the chain rule applied to $e^u$, with $u=2(x-1)$
$$\begin{eqnarray*}
\frac{dy}{dx} &=&\frac{d}{dx}\left( e^{2\left( x-1\right) }+2\right) =\frac{d
}{dx}e^{2\left( x-1\right) }=\left( \frac{
d}{du}e^{u}\right) \frac{du}{dx}
=e^{u}\times 2=2e^{2\left( x-1\right) }.
\end{eqnarray*}$$
At $x=1$ this derivative has the value
$$\begin{equation*}
\left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2e^{2\left( x-1\right)
}\right\vert _{x=1}=2e^{2(1-1)}=2.\tag{1}
\end{equation*}$$
Since at $x=1$, $$y=e^{2(1-1)}+2=3,$$ we have confirmed that the point $(x,y)=(1,3)$ belongs to the graph of the curve given by equation $(C)$. Hence the equation of the tangent line to the graph of the curve at $(1,3)$ is
$$\begin{equation*}
y-3=2(x-1)\Leftrightarrow y=2x+1\tag{2}
\end{equation*}$$</p></li>
<li><p><strong>Without eliminating the parameter</strong> $t$. <em>(Reformulated in view of OP's comment.)</em> To compute the derivative we use now the parametric equations $(A)$ and the formula $$\frac{dy}{dx} =\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}/\frac{dx}{dt}.\tag{D}$$ We have
$$\begin{eqnarray*}
\frac{dy}{dx} &=&\frac{dy}{dt}/\frac{dx}{dt}=
\left( \frac{d}{dt}\left( t^{2}+2\right) \right) /\frac{d}{dt}\left( 1+\ln
t\right) \\
&=&2t/\frac{1}{t}=2t^2,\tag{3}
\end{eqnarray*}$$
<em>which confirms your result</em>. Since for $x=1$ the equation $$x=1+\ln t $$ gives $$1=1+\ln t\Leftrightarrow 0=\ln t \Leftrightarrow t=1,$$ we get the same value as in $(1)$ for the derivative $$\begin{equation*}
\left. \frac{dy}{dx}\right\vert _{x=1}=\left. 2t^2\right\vert
_{t=1}=2\cdot 1^2=2.\tag{3a}
\end{equation*}$$
The equation of the tangent line is as above
$$\begin{equation*}
y=2x+1.\tag{4}
\end{equation*}$$
In terms of the parameter $t$ the tangent line at $t=1$, i.e. at $(x,y)=(1,3)$ is given by the parametric equations
$$\begin{equation*}
\left\{
\begin{array}{c}
x=t \\
y=2t+1,
\end{array}
\right.\tag{4a}
\end{equation*}$$
because
$$\begin{equation*}
\left. \frac{dx}{dt}\right\vert _{t=1}=\left. \frac{1}{t}\right\vert _{t=1}=1
\end{equation*}$$
and
$$\begin{equation*}
\left. \frac{dy}{dt}\right\vert _{t=1}=\left. 2t\right\vert _{t=1}=2.
\end{equation*}$$</p></li>
</ol>
|
192,883 | <p>Can anyone please give an example of why the following definition of $\displaystyle{\lim_{x \to a} f(x) =L}$ is NOT correct?:</p>
<p>$\forall$ $\delta >0$ $\exists$ $\epsilon>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$</p>
<p>I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work.</p>
| Fly by Night | 38,495 | <p>Let's consider a counter example. Let's use your definition to prove that the limit of $x$, as $x$ tends towards 1, is 2. For all $\delta > 0$, we claim that there exists $\varepsilon_{\delta} > 0$ such that: </p>
<p>If $|x - 1| < \delta$ then $|x-2| <\varepsilon_{\delta}$. </p>
<p>We could define $\varepsilon_{\delta} := \delta + 2$. That seems to satisfy your definition.</p>
|
3,900,962 | <p>In Valter Moretti's "<a href="https://rads.stackoverflow.com/amzn/click/com/8847028345" rel="nofollow noreferrer" rel="nofollow noreferrer">Spectral Theory and Quantum Mechanics</a>," Remark 4.2 (1), he claims:</p>
<blockquote>
<p>Compactness is hereditary, in the sense that it is passed on to induced topologies.</p>
</blockquote>
<p>I am not sure exactly what he means by this. In particular, one might interpret this in the most strong/general sense possible:</p>
<blockquote>
<p>Let <span class="math-container">$(X,\mathcal T_X)$</span> be a topological space, and <span class="math-container">$(A,\mathcal T_A$</span>) be a subspace, i.e., <span class="math-container">$A \subseteq X$</span> and <span class="math-container">$\mathcal T_A$</span> is the induced topology from <span class="math-container">$\mathcal T_X$</span>. Let <span class="math-container">$K⊆ X$</span> be compact. Then <span class="math-container">$K\cap A$</span> is compact in <span class="math-container">$(A,\mathcal T_A)$</span>.</p>
</blockquote>
<p>Is this true? Please explain why or why not.</p>
| QuantumSpace | 661,543 | <p>False. <span class="math-container">$X=K=[0,1]$</span> is compact but <span class="math-container">$K \cap A=A=(0,1)$</span> is non-compact.</p>
<p>It is true that compactness is inherited by closed subsets. Indeed, let <span class="math-container">$A\subseteq X$</span> be a closed set. If <span class="math-container">$\{U_i\}$</span> is an open cover of <span class="math-container">$A$</span>, then <span class="math-container">$\{U_i\}\cup\{A^c\}$</span> is an open cover of <span class="math-container">$X$</span> and by compactness we may extract a finite subcover for <span class="math-container">$X$</span>, which also gives a finite subcover for <span class="math-container">$A$</span>.</p>
|
3,900,962 | <p>In Valter Moretti's "<a href="https://rads.stackoverflow.com/amzn/click/com/8847028345" rel="nofollow noreferrer" rel="nofollow noreferrer">Spectral Theory and Quantum Mechanics</a>," Remark 4.2 (1), he claims:</p>
<blockquote>
<p>Compactness is hereditary, in the sense that it is passed on to induced topologies.</p>
</blockquote>
<p>I am not sure exactly what he means by this. In particular, one might interpret this in the most strong/general sense possible:</p>
<blockquote>
<p>Let <span class="math-container">$(X,\mathcal T_X)$</span> be a topological space, and <span class="math-container">$(A,\mathcal T_A$</span>) be a subspace, i.e., <span class="math-container">$A \subseteq X$</span> and <span class="math-container">$\mathcal T_A$</span> is the induced topology from <span class="math-container">$\mathcal T_X$</span>. Let <span class="math-container">$K⊆ X$</span> be compact. Then <span class="math-container">$K\cap A$</span> is compact in <span class="math-container">$(A,\mathcal T_A)$</span>.</p>
</blockquote>
<p>Is this true? Please explain why or why not.</p>
| Anatoliy Lotkov | 422,463 | <p>The author probably means that compactness is inherent property of subspace: you don't need the space in which it is contained to deduce its compactness.</p>
<p>In more rigorous terms. Let <span class="math-container">$ (X, \, \tau_X)$</span> be a topological space. Consider its compact subspace <span class="math-container">$ K \subset X $</span>. Then for any <span class="math-container">$ Y $</span> such that <span class="math-container">$ K \subset Y \subset X $</span> with topology <span class="math-container">$ \tau_Y $</span> induced from <span class="math-container">$ X $</span>, <span class="math-container">$K$</span> as subspace of <span class="math-container">$ Y $</span> is also compact.</p>
<p>In particularly, when <span class="math-container">$ Y = K $</span> space <span class="math-container">$K$</span> is compact in <span class="math-container">$K$</span> itself. What does it mean? Consider a collection of open sets <span class="math-container">$\{U_{\alpha}\}$</span> with <span class="math-container">$\forall \alpha:\: U_{\alpha} \subset K $</span> that covers <span class="math-container">$K \subset \bigcup_{\alpha}U_{\alpha} $</span>, or as everything is contained in <span class="math-container">$K$</span> we have <span class="math-container">$K = \bigcup_{\alpha} U_{\alpha}$</span>. Then we have a finite subcover: <span class="math-container">$K = \bigcup_{i=1}^{N} U_i $</span>.</p>
<p>It is easily proved by using the definition of subspace topology. For <span class="math-container">$ Y \subset X $</span> <span class="math-container">$\tau_Y$</span> is defined by the map <span class="math-container">$ i\ :\ Y \to X $</span>. The set <span class="math-container">$U \subset Y $</span> is open, iff <span class="math-container">$\exists V \subset X $</span> such that <span class="math-container">$ U = i^{-1}(V)$</span>. In other words <span class="math-container">$i$</span> is an open map (it maps open sets to open sets). Thus we can push any open cover <span class="math-container">$\{U_{\alpha}\}$</span> in <span class="math-container">$Y $</span> to open cover <span class="math-container">$\{f(U_{\alpha})\}$</span> in <span class="math-container">$X$</span>. Therefore, because we consider <span class="math-container">$K$</span> compact in <span class="math-container">$X$</span>, <span class="math-container">$\{f(U_{\alpha})\}$</span> has a finite subcover <span class="math-container">$\{f(U_i)\}$</span>, which we pull back to <span class="math-container">$Y$</span> and get a finite subcover <span class="math-container">$\{U_i\}$</span> in <span class="math-container">$Y$</span>. The statement is proved.</p>
|
137,434 | <p>I would like to draw the solution of the following equation in a log-log plot of $\lambda$ against $M$.</p>
<p>$$10^{-10} = \frac{\lambda^{2}}{(4\pi)^{2}}\int_{0}^{1}dz\frac{(1-x)^{3}}{(1-x)^{2}+zM^{2}}$$</p>
<p>Here is my code:</p>
<pre><code>table = Table[{Exp[M], λ /.
NSolve[((λ^2)/(4 *Pi)^(2)) *
NIntegrate[(((1-x)^3)/((1-x)^2 + Exp[2 M] x)), {x, 0, 1}] ==
10^(-10), λ][[2]]},
{M, -15, 15, 0.03}];
ListLogLogPlot[table, Joined \[RightArrow] True,
AxesOrigin \[RightArrow] {1, 1}]
</code></pre>
<p>With this code, I get a list of error messages and a list of the output values, but not the plot. How do I get the plot?</p>
| David G. Stork | 9,735 | <p>Here is your graph, once you delete the non-converged values:</p>
<pre><code>ListLogLogPlot[table, AxesOrigin -> {1,1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/LXLwL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LXLwL.png" alt="enter image description here"></a></p>
|
137,434 | <p>I would like to draw the solution of the following equation in a log-log plot of $\lambda$ against $M$.</p>
<p>$$10^{-10} = \frac{\lambda^{2}}{(4\pi)^{2}}\int_{0}^{1}dz\frac{(1-x)^{3}}{(1-x)^{2}+zM^{2}}$$</p>
<p>Here is my code:</p>
<pre><code>table = Table[{Exp[M], λ /.
NSolve[((λ^2)/(4 *Pi)^(2)) *
NIntegrate[(((1-x)^3)/((1-x)^2 + Exp[2 M] x)), {x, 0, 1}] ==
10^(-10), λ][[2]]},
{M, -15, 15, 0.03}];
ListLogLogPlot[table, Joined \[RightArrow] True,
AxesOrigin \[RightArrow] {1, 1}]
</code></pre>
<p>With this code, I get a list of error messages and a list of the output values, but not the plot. How do I get the plot?</p>
| Dr. Wolfgang Hintze | 16,361 | <p>Why not make things easier and solve the integral exactly?</p>
<p>After making the integration variable consistently $\text{x}$ and changing it to $\text{x}\to 1-\text{x}$, the integral can be calculated explicitly. (In order to assist Mathematica in solving the integral we also assume temporarily that $\text{M}<-2$ such that the roots of the denominator are real.)</p>
<p>We get</p>
<pre><code>f[M_] = Integrate[x^3/(x^2 + (1 - x) M^2), {x, 0, 1}, Assumptions -> M < -2] //
FullSimplify
(* Out[111]= (Sqrt[-4 + M^2] (1 + 2 M^2) - M^2 Sqrt[-4 + M^2] (-1 + M^2) Log[M^2] -
M^3 (-3 + M^2) Log[1/2 (-2 + M^2 - M Sqrt[-4 + M^2])])/(2 Sqrt[-4 + M^2]) *)
</code></pre>
<p>$$f(\text{M})=\int_0^1 \frac{x^3}{x^2+(1-x)M^2 } \, dx\\= \frac{1}{2} \left(\left(2 M^2+1\right)-M^2 \left(M^2-1\right) \log \left(M^2\right)\right)\\-\frac{\left(M^3 \left(M^2-3\right)\right) \log \left(\frac{1}{2} \left(M^2- M\sqrt{M^2-4}-2\right)\right)}{2 \sqrt{M^2-4}}$$</p>
<p>The result now holds for all real $\text{M}$.</p>
<p>At the special points it is</p>
<pre><code>Limit [f[M], M -> 0]
(* Out[113]= 1/2 *)
{Limit [f[M], M -> 2], Limit [f[M], M -> -2]}
(* Out[115]= {17/2 - 12 Log[2], 17/2 - 6 Log[4]} *)
Series[f[M], {M, \[Infinity], 2}] // Normal
(* Out[148]= (-(11/6) - 2 Log[1/M])/M^2 *)
</code></pre>
<p>And here's the graph</p>
<pre><code>Plot[f[M], {M, -10, 10},
PlotLabel -> "The integral as a function of M"]
</code></pre>
<p><a href="https://i.stack.imgur.com/DbCE5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DbCE5.jpg" alt="enter image description here"></a></p>
<p>Now the function $\lambda $ is defined as</p>
<pre><code>ld[M_] := 4 \[Pi] 10^-5/Sqrt[f[M]]
</code></pre>
<p>A log-log-plot (which is naturally confined to positive quantities) is then</p>
<pre><code>LogLogPlot[ld[M], {M, .001, 15}, PlotLabel -> "Log-log-plot",
AxesLabel -> {"Log[M]", "log[\[Lambda]]"}]
</code></pre>
<p><a href="https://i.stack.imgur.com/Jr6xH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jr6xH.jpg" alt="enter image description here"></a></p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Hollis Williams | 119,114 | <p>The only generic advice that can be offered here is that deciding what to do in this situation is quite a difficult skill, but one that you are expected to develop as a PhD student. You ultimately might have to think of the “cost-benefit” analysis.</p>
<p>Is the time and energy which you would expend continuing to work on this problem likely to be worthwhile? Or could you say “I learned a lot by working on this” but move on as it will probably not be worthwhile to stay with it further?</p>
<p>As you know the area, the problem and the techniques, you are probably the only one who can answer that question in a definitive way, but if in doubt make sure to discuss it with your supervisor.</p>
<p>Finally, I think this question would be much better suited to <a href="https://academia.stackexchange.com">Academia Stack Exchange</a> (where I think there a few mathematicians anyway who might be better positioned to answer such a question).</p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Geordie Williamson | 919 | <p>One stochastic algorithm that sometimes works for me:</p>
<ol>
<li><p>Take Pólya's "How to solve it" from bookshelf.</p>
</li>
<li><p>Open at random page, and read for a few pages.</p>
</li>
<li><p>Now attempt problem again.</p>
</li>
</ol>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Gabe K | 125,275 | <p>In my experience, one key aspect to avoid getting stuck is the choice of problem in the first place. With a good project, there are many smaller questions or computations to be done when you are stuck at a particularly difficult point with the main project. Doing these tasks might not solve your particular issue, but are still productive research hours.</p>
<p>Of course, choosing good problems is a difficult skill and requires a lot of practice. Here are a few guidelines which I have found helpful.</p>
<ol>
<li><p>Choose a problem which you believe you can make some progress with, rather than one you would really likely to solve. In particular, it is generally a good idea to minimize how "clever" you need to be in order to make progress. Trying to come up with a big new idea is a high-risk, high-reward gamble and generally not advisable unless you have a good reason to believe you will be successful.
(1a. Personally, learning this skill forced me to put aside my ego and take a more professional approach to math, in which producing research is my job. Too often, mathematical culture ties people's worth to their research ability, and consciously decoupling from this was really helpful for me.)</p>
</li>
<li><p>Building on the previous advice, once you have an established set of tools, it is often possible to find questions where these techniques may be applied. This can be the source of relatively low-hanging fruit, because it is possible to apply familiar ideas in a new setting.
(2a. One particularly good source of problems come from papers which relate several different areas of mathematics. Oftentimes, having expertise in one area provides the insights to solve problems which arise in these works. So these papers are worth reading.)</p>
</li>
<li><p>With a good project, there are often many opportunities to prove smaller results. For example, it is often possible to sharpen existing results or establish quantitative estimates using existing techniques, and this can be a fruitful activity when you are stuck. Personally, I tend to favor projects with a quantitative aspect rather than trying to prove novel qualitative results. The reason is that even if I fail at my main task, it still might be possible to improve some epsilon from <span class="math-container">$\frac{1}{100}$</span> to <span class="math-container">$\frac{1}{10}$</span> and make some small progress.</p>
</li>
<li><p>It can be tempting to attack a problem endlessly (and I am can certainly be guilty of this), but it is also a mathematical skill to understand when the methods you are using are unable to resolve the issue at hand. As such, when you get to a sticking point, it is worthwhile to really understand the strengths and limitations of your techniques to determine whether it is merely a technical issue that you can resolve or a fundamental difficulty. As such, learning when to quit a problem, at least temporarily, is an invaluable skill.</p>
</li>
<li><p>(Addition, as requested) Consider what advantage you have over other mathematicians in working on this problem. It could be that you think you have the right combination of knowledge or have done some calculations you think no one else has. It could simply be that no one else has spent much time on the problem, or no one else has the right combination of time to spend and access to your advisor. It could be that no one else cares about this problem. (Hopefully, that is because you think there is some reason to care about this problem that others haven't realized.) But if you think Person X is better at solving this problem than you from every angle, and you think Person X has worked on this problem but not solved it, then there is probably a reason you don't know why Person X hasn't solved it, and you will probably end up running into it also.</p>
</li>
</ol>
<p>Feel free for anyone else to add advice on how they choose research problems to minimize the probability of getting stuck.</p>
|
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? </p>
| Milo Brandt | 174,927 | <p>Well, if you are considering that $y=\sqrt{x}$ is the <em>relation</em> $y^2=x$, then, yes, $\pm i$ are both solutions to $\sqrt{-1}$. However, this is not usually how square roots are defined. Typically we say:
$$\sqrt{1}=1$$
Not plus or minus $1$ - just $1$. This means that $\sqrt{x}$ is a "right inverse" of $x^2$ - that is, we have that
$$\left(\sqrt{x}\right)^2=x$$
but not necessarily that
$$\sqrt{(x^2)}=x.$$
The difference here is that, to make $\sqrt{x}$ a function, we need it to yield unique output values - so we choose a principal root. We generally define that whenever we take the square root of a positive number, we get a positive number. And the definition $\sqrt{-1}=i$ is equally innocuous, since if we defined it as $\sqrt{-1}=-i$, we could just relabel every number by its complex conjugate, and get back to our typical definition (that is to say, $\sqrt{-1}=i$ more defines $i$ than it defines $\sqrt{-1}$). So, though the equation $x^2+1$ has two solutions (as does any equation $x^2-c$ for complex $c\neq 0$), for the sake of making square roots act like a function, we must choose one of them to be the square root of $-1$, and we call this number $i$.</p>
|
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? </p>
| Andrew D. Hwang | 86,418 | <p>If $x$ is a non-negative real number, there's an unambiguous interpretation for the expression $\sqrt{x}$, namely, the <em>non-negative</em> square root of $x$. (The $\pm$ signs aren't "part of the square root function", which is why they have to be included explicitly when "solving an equation by taking square roots", e.g., passing from $x^{2} = 1$ to $x = \pm 1$.) </p>
<p>When one tries to extend the square root function to the complex numbers, there are tricky domain issues. Rather than write "$\sqrt{-1} = \pm i$" (which can't be true if the radical symbol denotes a function), it's safer to stick with $(\pm i)^{2} = -1$ (which is unambiguously true) until one is invested in understanding the fine points of functions of a complex variable.</p>
<p>In case you or your student are curious: Each non-zero complex number has two square roots, but there is no <em>continuous</em> choice of square root on the complex plane.</p>
<p>To get a continuous "branch of square root" it's necessary to remove enough of the plane that "the domain of the square root doesn't encircle the origin". The customary choice is to remove the <em>non-positive reals</em>. (Ironically, this explicitly excludes $-1$ from the domain of the square root.)</p>
<p>A common alternative choice is to remove the non-positive imaginary axis. In this event, $\sqrt{-1} = i$ for the continuous branch of square root that satisfies $\sqrt{1} = 1$.</p>
<p>The take-away points are:</p>
<ul>
<li><p>If $\sqrt{\ }$ denotes a function, then it must be <em>single-valued</em> (no $\pm$).</p></li>
<li><p>When allowing complex numbers (other than non-negative reals) under a radical sign, You Really Need To Be Careful.</p></li>
</ul>
|
1,680,269 | <p>Here $\mathbb{Z}_{n}^{*}$ means $\mathbb{Z}_{n}-{[0]_{n}}$</p>
<p>My attempt:</p>
<p>$(\leftarrow )$</p>
<p>$p$ is a prime, then, for every $[x]_{n},[y]_{n},[z]_{n}$ $\in (\mathbb{Z}_{n}^{*},.)$ are verified the following:</p>
<p>1) $[x]_{n}.([y]_{n}.[z]_{n}) = ([x]_{n}.[y]_{n}).[z]_{n}$, since from the operation . we have $[a]_{n}.[b]_{n}=[a.b]_{n}$ and . is associative in $\mathbb{Z}$.</p>
<p>2) There is an element $e$ such that $[x]_{n}.e = e.[x]_{n} = [x]_{n}$, since $ [x]_{n}.[1]_{n} = [x.1]_{n} = [x]_{n} = [x.1]_{n} = [x]_{n}[1]_{n}$</p>
<p>But I don't know how to check the inverse property, neither how to do the $(\rightarrow)$ part. </p>
<p>Thanks!</p>
| Community | -1 | <p>By definition of infimum there are sequences $(x_n)_{n\in\mathbb{N}}\subseteq K, (y_n)_{n\in\mathbb{N}}\subseteq L$ with $\lim_{n\to\infty}|x_n-y_n|=d$. From compactness we have convergent subsequences $x_{n_k}, y_{n_k}$ (it's possible to take the same indices $n_k$!), so there are $x_0\in K, y_0\in L$ with $\lim_{k\to\infty}x_{n_k}=x_0$ and $\lim_{k\to\infty}y_{n_k}=y_0$.</p>
<p>So form continuity of $|\cdot|$ we get
$$|x_0-y_0|=|\lim_{k\to\infty} (x_{n_k}-y_{n_k})|=\lim_{k\to\infty}| (x_{n_k}-y_{n_k})|=d.$$</p>
<p>For $K, L$ disjoint, $d=0$ we get the contradiction $x_0=y_0\in K\cap L=\emptyset$.</p>
|
1,298,971 | <p>Any help on this problem is greatly appreciated! I'm completely stuck</p>
<p>School board officials are debating whether to require all high school seniors to take a proficiency exam before graduating. A student passing all three parts (mathematics, language skills, and general knowledge) would be awarded a diploma; otherwise, he or she would receive only a certificate of attendance. A practice test given to this year’s ninety-five hundred seniors resulted in the
following numbers of failures:
$$\begin{array}{|l|l|}
\hline \\
\textbf{Subject Area} & \textbf{Number Failing} \\\hline \\
\text{Mathematics} & 3325 \\\hline \\
\text{Language Skills} & 1900 \\\hline \\
\text{General Knowledge} & 1425\\ \hline
\end{array}$$</p>
<p>If “Student fails mathematics,” “Student fails language skills,” and “Student fails general knowledge” are independent events, what proportion of next year’s seniors can be expected to fail to qualify for a diploma?</p>
| André Nicolas | 6,312 | <p>Consider a square such that the distance from the centre to any side is $r$. Then the area of the square is $4r^2$, and the perimeter of the square is $8r$, which is the derivative of $4r^2$. </p>
<p>So your circle rule works for the square, if we use the right parameter to describe its size.</p>
<p><strong>Exploration:</strong> Let us play the same game with an equilateral triangle $T$. Again use as parameter $r$ the distance from the centre of $T$ to a side. Drop a perpendicular from the centre to a side, and join the centre to a vertex on that side. We get a $30$-$60$-$90$ triangle. Using this triangle we find that the perimeter of $T$ is $6\sqrt{3}r$, and its area is $3\sqrt{3}r^2$. Again the derivative of area is the perimeter.</p>
<p>The same pattern holds for all regular polygons. I will leave it to you to write out the argument, either by scaling (better) or by computation.</p>
|
1,869,119 | <p>Show that the Monotone Convergence Theorem may not hold for decreasing sequences of functions.</p>
<p>Suppose $\left\{f_{n}\right\}$ is a sequence of nonnegative decreasing functions converging to $f$ pointwise. I know that if $f_{1}$ is finite,we can construct the sequence say $\left\{f_{1}-f_{n}\right\}$ which is increasing to prove that MCT still hold for decreasing sequence. So the only way in which the MCT is not valid is when $f_{1}$ is not finite. I don't know how to construct such sequence.</p>
| Aweygan | 234,668 | <p>Define the sequence $\{f_n\}$ on $[0,\infty)$ by letting $f_n(0)=0$ for all $n$, and
$$ f_n(x)=\frac{1}{nx}\qquad (x\in(0,\infty). $$
Then $f_n$ is monotonically decreasing, and converges pointwise to $0$ on $[0,\infty)$. However,
$$\int_{[0,\infty)}f_n\ dm=\infty,$$
while
$$\int_{[0,\infty)}\lim_{n\to\infty}f_n\ dm=0. $$</p>
|
2,231,949 | <p>To find the minimal polynomial of $i\sqrt{-1+2\sqrt{3}}$, I need to prove that
$x^4-2x^2-11$ is irreducible over $\Bbb Q$. And I am stuck. Could someone please help? Thanks so much!</p>
| Alex Macedo | 400,433 | <p>If you have a candidate $f$ for the minimal polynomial of an algebraic element $\alpha$ over $\mathbf Q$ but you don't know if $f$ is really irreducible and want to avoid trying possible factorizations, you might want to compute the degree $$[\mathbf Q(\alpha): \mathbf Q]$$ and check if it agrees with the degree of $f$.</p>
<p>For instance, in your question, let $$\alpha = i\sqrt{-1 + 2\sqrt 3}$$.</p>
<blockquote>
<p><strong>Exercise</strong>: show that $[\mathbf Q(\alpha): \mathbf Q] = 4$ using the inclusions $$\mathbf Q \subset \mathbf Q(\sqrt{3}) \subset \mathbf Q(\alpha).$$</p>
</blockquote>
<hr>
<p>See also <a href="https://math.stackexchange.com/a/2231196/400433">this answer</a> for a similar application of this philosophy.</p>
|
720,969 | <p>While studying real analysis, I got confused on the following issue.</p>
<p>Suppose we construct real numbers as equivalence classes of cauchy sequences. Let $x = (a_n)$ and $y= (b_n)$ be two cauchy sequences, representing real numbers $x$ and $y$.</p>
<p>Addition operation $x+y$ is defined as $x+y = (a_n + b_n)$.</p>
<p>To check if this operation is well defined, we substitute $x = (a_n)$ with some real number $x' = (c_n)$ and verify that $x+y = x'+y$. We also repeat it for $y$. i.e. we verify that $x+y = x+y'$.</p>
<p><strong>Question:</strong></p>
<blockquote>
<p>Instead of checking that $x+y = x+y'$ and $x'+y = x+y$ seperately, would it suffice to check that $x+y = x' + y'$ in a single operation in order to show that addition is well defined for real numbers. Would it hurt to checking well definedness? Can any one explain me the logic behind ?</p>
</blockquote>
| Christian Blatter | 1,303 | <p>You have to differentiate typographically between (a) sequences and (b) equivalence classes of sequences, i.e., real numbers.</p>
<p>Write $x$ for the Cauchy sequence $(x_n)_{n\geq1}$ and $[x]$ for the equivalence class represented by $x$.</p>
<p>Since addition of real numbers is described in terms of representants:
$$[x]+[y]:=[x+y]\ ,$$
we have to check whether this actually defines a binary operation on ${\mathbb R}$. It is sufficient to prove that
$$x\sim x'\qquad\Rightarrow\qquad x+y\quad \sim\quad x'+y\ ,$$
for then we can argue as follows: When $x'\sim x$ and $y'\sim y$ then using commutativity we have the following chain:
$$x+y\ \sim x'+y\ =\ y+x'\ \sim \ y'+x'\ =x'+y'\ .$$</p>
|
124,662 | <p>Is topology on $\mathbb{R}/\mathbb{Z}$ compact? If it is, how to prove it? </p>
<p>$\mathbb{R}/\mathbb{Z}$ denotes the set of equivalence classes of the set of real numbers, two real numbers being equivalent if and only if their difference is an integer.</p>
| Rudy the Reindeer | 5,798 | <p>Or use that $S^1$ is compact and try to write down a homeomorphism between $\mathbb R/ \mathbb Z$ and $S^1$.</p>
|
92,660 | <p>Let $X$ be a nonsingular projective variety over $\mathbb{C}$, and let $\widetilde{X}$ be the blow-up of X at a point $p\in X$.
What relationships exist between the degrees of the Chern classes of $X$ (i.e. of the tangent bundle of $X$) and the degrees of the Chern classes of $\widetilde{X}$?</p>
<p>Thanks.</p>
| Georges Elencwajg | 450 | <p>For the first Chern class you get the simple formula<br>
$$c_1(\tilde X)=p^*c_1(X)- (n-1)E$$
where $p:\tilde X \to X$ is the projection and $E$ the exceptional divisor. </p>
<p>In general the formula is more complicated and I'll refer you to Fulton's <em>Intersection Theory</em>, where the formula you require is given in Theorem 15.4. </p>
<p>In particular cases the relation may be quite simple: for example if $X$ is of dimension 3, it is just $c_2(\tilde X)=p^*(c_2(X)$ for the second Chern class, as proved in Griffiths-Harris's <em>Principles of Algebraic Geometry</em>, page 609. </p>
|
1,550,841 | <p>$\int e^{2\theta}\ \sin 3\theta\ d\theta$</p>
<p>After Integrating by parts a second time, It seems that the problem will repeat for ever. Am I doing something wrong. I would love for someone to show me using the method I am using in a clean and clear fashion. Thanks. <a href="https://i.stack.imgur.com/FipiE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FipiE.jpg" alt="enter image description here"></a></p>
| Ian | 83,396 | <p>When you do it with integration by parts, you have to go in the "same direction" both times. For instance, if you initially differentiate $e^{2 \theta}$, then you need to differentiate $e^{2 \theta}$ again; if you integrate it, you will wind up back where you started. If you do this, you should find something of the form</p>
<p>$$\int e^{2 \theta} \cos(3 \theta) d \theta = f(\theta) + C \int e^{2 \theta} \cos(3 \theta) d \theta$$</p>
<p>where $C$ is not $1$. Therefore you can solve the equation for the desired quantity:</p>
<p>$$\int e^{2 \theta} \cos(3 \theta) d \theta = \frac{f(\theta)}{1-C}.$$</p>
<p>There is also a nice approach with complex numbers: $\cos(3 \theta)=\frac{e^{3 i \theta}+e^{-3 i \theta}}{2}$, so your integral is</p>
<p>$$\frac{1}{2} \int e^{(2+3i) \theta} + e^{(2-3i) \theta} d \theta$$</p>
<p>which are pretty easy integrals. You do some complex number arithmetic and it works out.</p>
|
1,550,841 | <p>$\int e^{2\theta}\ \sin 3\theta\ d\theta$</p>
<p>After Integrating by parts a second time, It seems that the problem will repeat for ever. Am I doing something wrong. I would love for someone to show me using the method I am using in a clean and clear fashion. Thanks. <a href="https://i.stack.imgur.com/FipiE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FipiE.jpg" alt="enter image description here"></a></p>
| Yes | 155,328 | <p>Here is a plain answer for your reference:</p>
<p>Let $\simeq$ denote the equality sign up to a constant.
We have
$$
\int e^{2x}\sin 3x dx = \int \sin 3x de^{2x}\frac{1}{2} \simeq \frac{1}{2}[ e^{2x}\sin 3x - 3 \int e^{2x} \cos 3x dx ]\\ = \frac{1}{2}e^{2x}\sin 3x - \frac{3}{2} \int e^{2x}\cos 3x dx;
$$
we have
$$
\int e^{2x}\cos 3x dx = \frac{1}{2}\int \cos 3x de^{2x} \simeq \frac{1}{2}e^{2x}\cos 3x + \frac{3}{2}\int e^{2x}\sin 3x dx;
$$
hence
$$
\frac{13}{4}\int e^{2x}\sin 3x dx \simeq \frac{1}{2}e^{2x}\sin 3x - \frac{3}{4}e^{2x}\cos 3x.
$$</p>
|
143,655 | <p>According to <a href="http://en.wikipedia.org/wiki/Lipschitz_continuity#Properties" rel="nofollow noreferrer">wikipedia</a> a function <span class="math-container">$f\colon \mathbb{R}^n\to\mathbb{R}^n$</span> that is continuously differentiable, is also locally Lipschitz.</p>
<p>I there someone who knows a good reference which contains a proof of this statement?</p>
| copper.hat | 27,978 | <p>The proof on $\mathbb{R}^n$ is fairly straightforward.</p>
<p>Choose some ball $B(\hat{x},\epsilon)$. The closure is compact, so the derivative $\frac{\partial f}{\partial x}$ is bounded by some $L$ on the ball. Now suppose $x,y \in B(\hat{x},\epsilon)$, then using Taylor's formula, we have:
$$f(x)-f(y) = \int_o^1 \frac{\partial f (y+t(x-y))}{\partial x}(x-y)\;dt.$$
Hence we can get the bound:
$$\|f(x)-f(y) \| \leq \int_o^1 \|\frac{\partial f (y+t(x-y))}{\partial x}\|\; \|(x-y)\| \;dt \leq L \|x-y\|.$$ </p>
|
4,549,300 | <p>Matrix C of size n<span class="math-container">$\times$</span>n is symmetric . Zero is a simple eigenvalue of C. The associated eigenvector is q. For <span class="math-container">$\epsilon$</span>>0, the equation <span class="math-container">$Cx+\epsilon x=d$</span> in x, where x and d are n-dimensional Column vectors and d is known, has a solution that depends on <span class="math-container">$\epsilon$</span>. Call this solution <span class="math-container">$x(\epsilon)$</span>. Express <span class="math-container">$ \lim\limits_{\epsilon \to 0^+} \epsilon x(\epsilon)$</span> in terms of vectors q and d.</p>
<p>I have some inspirations:</p>
<ol>
<li>When <span class="math-container">$\epsilon \to 0$</span>, <span class="math-container">$\epsilon q$</span> also comes to <span class="math-container">$0$</span>. So we have: <span class="math-container">$C(x+q)+\epsilon (x+q)=d$</span></li>
<li>Since C is a symmetrix matrix, we use diagonalization like: <span class="math-container">$U\Lambda U^Hx+\epsilon x=d$</span></li>
</ol>
<p>But to be honest, I have no clue on this and it bothers me a lot. I can't figure out how to use these conditions we have.</p>
<p>Thank you for reading my question!</p>
<p>Could you give me some clues on this problem?</p>
| Ted Shifrin | 71,348 | <p>I assume here that <span class="math-container">$C$</span> is a <em>real</em> symmetric matrix.</p>
<p><strong>HINT</strong>: We assume <span class="math-container">$q$</span> is a unit vector. Apply the Spectral Theorem to write
<span class="math-container">$$C+\epsilon I = \epsilon qq^\top + \sum_{i=1}^{n-1} (\lambda_i + \epsilon) v_iv_i^\top,$$</span>
where <span class="math-container">$q,v_1,\dots,v_{n-1}$</span> form an orthonormal basis. Then
<span class="math-container">$$(C+\epsilon I)^{-1} = \frac1{\epsilon} qq^\top + \sum_{i=1}^{n-1} \frac1{\lambda_i + \epsilon} v_iv_i^\top.$$</span>
Of course, we assume <span class="math-container">$\epsilon>0$</span> small enough so that no <span class="math-container">$\lambda_i+\epsilon$</span> becomes <span class="math-container">$0$</span>.
Can you finish now?</p>
|
34,557 | <p>I just came across a spam answer which is extremely vulgar (sexual). I flagged it for moderator attention, and then it occurred to me that I could edit it and erase its contents by blanking it till a moderator gets to look at it. Is this an acceptable thing to do?</p>
| KReiser | 21,412 | <p>From the comments: No. Instead, flag it as <em>rude or abusive</em> and move on.</p>
|
107,399 | <p>Let's say we have a set a\of associations:</p>
<pre><code>dataset = {
<|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "a", "subtype" -> "II", "value" -> 2|>,
<|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "b", "subtype" -> "II", "value" -> 2|>
}
</code></pre>
<p>where every entry is unique in terms of <code>{#type, #subtype}</code>, </p>
<h3>I'd like to build a nested association for more handy querying, e.g. I would like to have:</h3>
<pre><code>nested["a", "II", "value"]
</code></pre>
<blockquote>
<pre><code>2
</code></pre>
</blockquote>
<h3>I can start with</h3>
<pre><code>GroupBy[dataset, {#type &, #subtype &}]
</code></pre>
<blockquote>
<pre><code><|
"a" -> <|
"I" -> {<|"type" -> "a", "subtype" -> "I", "value" -> 1|>},
"II" -> {<|"type" -> "a", "subtype" -> "II", "value" -> 2|>}|>,
"b" -> <|
"I" -> {<|"type" -> "b", "subtype" -> "I", "value" -> 3|>},
"II" -> {<|"type" -> "b", "subtype" -> "II", "value" -> 4|>}
|>|>
</code></pre>
</blockquote>
<p>But <code>nested["a", "I"]</code> points to a <strong>list with one association</strong>, what is expected but I would like to drop that list. </p>
<p>It seems that the third argument of <code>GroupBy</code> isn't generalized to handle nested grouping... </p>
<p>So basically I would like to have <code>... "I" -> <|"type" -> "a", ...</code>.</p>
<h3>What is a generic way to go?</h3>
<p>I can do:</p>
<ul>
<li><p>nested <code>GroupBy</code>: </p>
<pre><code>GroupBy[dataset, #type &, GroupBy[#, #subtype &, First] &]
</code></pre></li>
<li><p><code>Map</code> later:</p>
<pre><code>GroupBy[dataset, {#type &, #subtype &}] // Map[First, #, {-3}] &
</code></pre></li>
</ul>
<p>But the first is not handy in general while the second is ugly (and not general either). </p>
<hr>
<p>Acceptable outputs are:</p>
<pre><code><|
"a" -> <|
"I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> <|"value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> 1 ,
...|>
</code></pre>
<p>but the first is the most desired one because we may have more that one ("value") key left.</p>
| Leonid Shifrin | 81 | <p>I have posted code doing a very similar thing <a href="https://mathematica.stackexchange.com/a/54493/81">here</a> - the functions <code>pushUp</code> and <code>pushUpNested</code>. That code was more general, since there I provided a declarative interface to group by values <em>or their parts</em>. To do what you need, I'll redefine slightly (assuming you run that code):</p>
<pre><code>ClearAll[pushUpNested];
pushUpNested[{}, elemF_: Identity] := elemF;
pushUpNested[specs : {_List ..}, elemF_: Identity ] :=
Composition[
Map[pushUpNested[Rest[specs], elemF]],
pushUp@First[specs]
];
</code></pre>
<p>Now we create a transform:</p>
<pre><code>transform = pushUpNested[{{"type"}, {"subtype"}}, First]
(*
Map[Map[First]@*GroupBy[#1[[Sequence[Key["subtype"]]]] &]]@*
GroupBy[#1[[Sequence[Key["type"]]]] &]
*)
</code></pre>
<p>which we can now apply to get the nested structure:</p>
<pre><code>nested = transform@dataset
(*
<|
"a" -> <|
"I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
"II" -> <|"type" -> "a", "subtype" -> "II", "value" -> 2|>
|>,
"b" -> <|
"I" -> <|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
"II" -> <|"type" -> "b", "subtype" -> "II", "value" -> 2|>
|>
|>
*)
</code></pre>
<p>The advantage of using <code>pushUpNested</code> is that it makes it very easy and declarative to construct such transforms, and the transform is available for inspection as a stand-alone fully-prepared function.</p>
|
2,152,914 | <p>Say I have a random function $f: \mathbb{Z}_N \rightarrow \mathbb{Z}_N$. (by random I mean for each possible input we choose an output in an uniform distribution from $\mathbb{Z}_N$) <br/>
So I know that for all $x,y \in \mathbb{Z}_N$ I have $Pr[f(x)=y]={1 \over |\mathbb{Z}_N|}$. <br/>
Now I look at $f'(x):=f(f(x))$. Is $f'$ a random function like $f$? <br/></p>
| Peter | 82,961 | <p>Yes, because the value of $f(x)$ is irrelevant for the determination of $f(f(x))$</p>
|
2,152,914 | <p>Say I have a random function $f: \mathbb{Z}_N \rightarrow \mathbb{Z}_N$. (by random I mean for each possible input we choose an output in an uniform distribution from $\mathbb{Z}_N$) <br/>
So I know that for all $x,y \in \mathbb{Z}_N$ I have $Pr[f(x)=y]={1 \over |\mathbb{Z}_N|}$. <br/>
Now I look at $f'(x):=f(f(x))$. Is $f'$ a random function like $f$? <br/></p>
| Empy2 | 81,790 | <p>Here is one difference:
$f(x)$ probably has around $N(1-1/e)$ different values. The chance a value never appears is $(1-1/N)^N\approx 1/e$</p>
<p>$f(f(x))$ probably has around $N(1-1/e)^2$ different values.</p>
<p>EDIT:
Let $M=N/e$ be the number of 'orphans' that are not a value of $f(x)$.<br>
Let $P=N-M$ be the number with parents. </p>
<p>The chance that a number has a parent but not a grandparent is the chance that at least one orphan, but none with parents, chose it, which would be </p>
<p>$$(1-1/N)^P(1-(1-1/N)^M)\approx e^{1/e-1}(1-e^{-1/e})$$</p>
<p>So I think $f(f(x))$ has about $N(1-e^{1/e}/e) \approx 0.4685 N$ different values.</p>
|
97,393 | <p>The polynomial</p>
<p>$F(x) = x^5-9x^4+24x^3-24x^2+23x-15$</p>
<p>has roots $x=1$ and $x=j$. Calculate all the roots of the polynomial.</p>
<p>I was told I had to use radicals or similar to solve this but after reading up on it I'm still confused about how to solve it.</p>
| John R Ramsden | 18,548 | <p>Knowing that one root is $x = 1$ means $F(x)$ has a factor $x - 1$. So you can either obtain the complementary factor by long division, or note that:</p>
<p>$\begin{align}F(x) &= x^4(x - 1) + 24x^2(x - 1) - 8x^4 + 23x - 15
\\ &= x^4(x - 1) + 24x^2(x - 1) - 8x(x^3 - 1) + 15(x - 1)\end{align}$</p>
<p>so using $x^3 - 1 = (x - 1)(x^2 + x + 1)$, the complementary factor can be read off as:</p>
<p>$x^4 - 8x^3 + 16x^2 - 8x + 15$</p>
<p>Now if a polynomial with real coefficients has a complex root then it must have the conjugate of that value as another root, and thus having a root $j$ (which I presume denotes $-1^\frac{1}{2}$) means it must have a root $-j$, and thus a factor $(x - j)(x + j)$ which is $x^2 + 1$.</p>
<p>Dividing this out leaves a quadratic factor $x^2 - 8 x + 15$, which by inspection (or solving in the usual way) factors as $(x - 3)(x - 5)$.</p>
<p>edit: DOH! I didn't see yoyo's reply until I had posted mine! May as well leave it now.</p>
|
3,172,485 | <p>Consider the familiar trigonometric identity: <span class="math-container">$\cos^3(x) = \frac{3}{4} \cos(x) + \frac{1}{4} \cos(3x)$</span></p>
<p>Show that the identity above can be interpreted as Fourier series expansion.</p>
<p>so we know that cos is periodic between <span class="math-container">$\pi$</span> and <span class="math-container">$-\pi$</span> and <span class="math-container">$\cos$</span> is an even function, therefore, <span class="math-container">$\cos^3$</span> is even. </p>
<p>so we need to compute <span class="math-container">$a_0$</span> ( the integral of <span class="math-container">$f(x)$</span> and it will equal <span class="math-container">$0$</span>) and <span class="math-container">$a_n$</span> ( the integral from <span class="math-container">$\pi$</span> to <span class="math-container">$-\pi$</span> of <span class="math-container">$\cos^3(x) \cos(nx)$</span> )</p>
<p>how to compute <span class="math-container">$a_0$</span></p>
<p>thanks </p>
| PrincessEev | 597,568 | <p>Recall: from a familiar trigonometric identity,</p>
<p><span class="math-container">$$\cos^3(x) = \cos(x)\color{blue}{\cos^2(x)} = \cos(x)\color{blue}{(1 - \sin^2(x))}$$</span></p>
<p>Thus,</p>
<p><span class="math-container">$$\int \cos^3(x)dx = \int (1-\sin^2(x))\cos(x)dx$$</span></p>
<p>Make the <span class="math-container">$u$</span>-substitution <span class="math-container">$u = \sin(x), du = \cos(x)dx$</span> and you should be able to find the result easily. You should get</p>
<p><span class="math-container">$$\int \cos^3(x)dx = \sin(x) - \frac 1 3 \sin^3(x) + C$$</span></p>
<p>Apply the fundamental theorem of calculus on the bounds <span class="math-container">$0,\pi$</span>. Since <span class="math-container">$\sin(n\pi) \equiv 0 \; \forall n \in \Bbb Z$</span>,</p>
<p><span class="math-container">$$\int_0^\pi \cos^3(x)dx = \sin(0) - \frac 1 3 \sin^3(0) - \left( \sin(\pi) - \frac 1 3 \sin^3(\pi) \right) = 0$$</span></p>
<p>and thus, <span class="math-container">$a_0 = 0$</span>.</p>
|
255,164 | <p>$\newcommand{\al}{\alpha}$
$\newcommand{\euc}{\mathcal{e}}$
$\newcommand{\Cof}{\operatorname{Cof}}$
$\newcommand{\Det}{\operatorname{Det}}$</p>
<p>Let $M,N$ be smooth $n$-dimensional Riemannian manifolds (perhaps with smooth boundary), and let $\, f:M \to N$ be a smooth <strong>immersion</strong>. Let $\Omega^k(M,f^*TN)$ be the space of $f^*TN$-valued $k$-forms. </p>
<p>Let $d:\Omega^k(M,f^*TN) \to \Omega^{k+1}(M,f^*TN)$ be the covariant exterior derivative associated with the pullback connection of the Levi-Civita connection on $N$ (via $f$), and let $\delta$ be its adjoint.</p>
<p>Denote $\delta_{k}:\Omega^k(M,f^*TN) \to \Omega^{k-1}(M,f^*TN)$.</p>
<p><strong>Question:</strong> Is $\ker \delta_1$ is infinite dimensional?</p>
<p>Since $\delta=\star d \star$ (up to sign), $\dim(\ker \delta_1)=\dim(\ker d_{n-1})$. So, this is a question about $d$.</p>
<hr>
<p><strong>Edit:</strong></p>
<p>In the special case where $N$ is flat, $\delta_{k} \circ \delta_{k+1}=0$, so $\operatorname{Image}(\delta_{k+1}) \subseteq \ker(\delta_k)$.</p>
<p>In particulr, $\operatorname{Image}(\delta_{2}) \subseteq \ker(\delta_1)$.</p>
<p>Is $\operatorname{Image}(\delta_{2})$ is infinite-dimensional? Can we at least something when $N=\mathbb{R}^n$? or even when $M=N=\mathbb{R}^n$?</p>
<p>Let's see what happens when $M=N=\mathbb{R}^n,f=\operatorname{Id}$:</p>
<p>$$ d_{n-1}(\sigma)(e_1,...,e_n)=\sum_{j=1}^{d} (-1)^{j+1} \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)$$</p>
<p>Write $\sigma=(-1)^{j+1}a_j^idx^1 \wedge \dots \wedge \hat{dx^j} \wedge \dots \wedge dx^d \otimes e_i, a_j^i \in C^{\infty}(\mathbb{R}^n)$. </p>
<p>Then $ \sigma(e_1,...,\hat{e_j},...,e_d)=(-1)^{j+1}a_j^i e_i $, so </p>
<p>$$ \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)= \nabla^{T\mathbb{R}^n}_{e_j}((-1)^{j+1}a_j^i e_i)=(-1)^{j+1}\frac{\partial a_j^i}{\partial x_j}e_i.$$</p>
<p>Thus, $$\sigma \in \ker d_{n-1} \iff \frac{\partial a_j^i}{\partial x_j}e_i=0 $$ (in the last term there is a double summation, on $i,j$).</p>
<p>Since $e_i$ are independent, this is equivalent to
$$\sum_{j=1}^d \frac{\partial a_j^i}{\partial x_j}=0 \, \text{ for all } \, i=1,...,d $$</p>
<p>Denoting $\bar a^i=(a_1^i,...,a_d^i)$, we get that $\operatorname{div} (\bar a^i)=0$.</p>
<p>I guess it shouldn't be too hard to see now that $\dim(\ker d_{n-1}) = \infty$. (By taking the $a_j^i$ to be constants, one immediately gets $\dim(\ker d_{n-1}) \ge n^2$. Since the condition on the divergence do not touch $\frac{\partial a_j^i}{\partial x_k}$ for $k \neq j$, it seems plausible that the dimension is indeed infinite. Perhaps someone can come with a slick argument to show this.)</p>
<p>The next case we should try is $M=N=\mathbb{R}^d$, and $f$ an arbitrary mapping....</p>
<hr>
<p><strong>Comment:</strong></p>
<p>I know that $\operatorname{Cof}(df) \in \ker \delta_1$, where $\operatorname{Cof}(df)$ is the corresponding <em>cofactor map</em> of $df$:
$$ \Cof df= (-1)^{d-1} \star_{f^*TN}^{n-1} (\wedge^{n-1} df) \star_{TM}^1. $$ </p>
<p>Since $f$ is an immersion, $\operatorname{Cof}(df) \neq 0$, so $\dim (\ker \delta_1 ) \ge 1$.</p>
<p>For my purposes, It would suffice to know that $\ker \delta_1$ always contains elements which are linearly independent of $\Cof df$.</p>
| Igor Khavkine | 2,622 | <p>Suppose that the vector field $v$ is a non-vanishing covariantly constant on $N$, then its pullback $f^*v$ is a constant section of $f^*TN \to M$ with respect to the pulled back connection. Then, for any $(n-1)$-form $\alpha$ and $(n-2)$-form $\beta$ on $M$, we have $d_{n-2} (\beta \otimes f^*v) = (d\beta) \otimes f^*v$ and $d_{n-1} (\alpha \otimes f^*v) = (d\alpha) \otimes f^*v$, where I used $d$ to denote the usual de Rham exterior derivative. Then clearly, $\mathrm{im}(d) \otimes f^*v \subseteq \ker d_{n-1}$ is an infinite dimensional subspace.</p>
<p>On the other hand, suppose that there are no non-vanishing constant sections of $f^*TN \to M$. We might still be able to find some differential operator $q \colon \Gamma(E) \to \Gamma(\Lambda^{n-1}T^*M \otimes_M f^*TN)$ whose image lands inside $\ker d_{n-1}$, that is $d_{n-1} \circ q = 0$. The trick is to first dualize. As you noted, $d_{n-1}^* = d_0$, up to conjugation by invertible operators. Then any "compatibility operator" $q^*$, such that $q^* \circ d_0 = 0$, gives us a candidate $q = (q^*)^*$, because $d_{n-1} \circ q = d_0^* \circ (q^*)^* = (q^* \circ d_0)^*$.</p>
<p>The non-existence of non-trivial constant sections of $f^*TN \to M$ means that there exist integrability conditions for the equation $d_0 v = 0$. In particular, after some massaging, we can come up with an operator $C\colon \Gamma(T^*M \otimes_M f^*TN) \to \Gamma(f^*TN)$ such that $C\circ d_0 = \mathrm{id}$. As a consequence, we have the identity $(\mathrm{id} - d_0 \circ C) \circ d_0 = d_0 - d_0 \circ (C\circ d_0) = 0$. Hence we have found a compatibility operator $q^* = \mathrm{id} - d_0 \circ C$, from which $q = \mathrm{id} - C^* \circ d_{n-1}$. In conclusion, we have $\mathrm{im}(q) \subseteq \ker d_{n-1}$.</p>
<p>It remains to check that $\mathrm{im}(q)$ is indeed infinite dimensional. I don't have a slick argument for that, at the moment. But I'm sure that in individual cases it should not be hard to establish, at the very least by looking at the effect of $q$ on jets of higher and higher order at some fixed point $x\in M$.</p>
|
2,484 | <p>It's been quite a while since I was tutoring a high school student and even longer since not a gifted one.</p>
<p>However, this time, something was amiss. I have asked him to show me how he does some exercise, and then another and the only thing I wanted to do was to shout:</p>
<blockquote>
<p><strong>You are doing it <em>wrong</em>!</strong></p>
</blockquote>
<p>I hope I didn't let it show and tried to work it out in small steps. However, it was like he didn't wanted to learn, he just wanted to get the problem set done.</p>
<p>One mistake he would frequently do, is to mess the signs in trigonometric reductions (e.g. $\cos(90^\circ + x)$ would be $\sin x$ instead of $-\sin x$). In fact, he memorized all the formulas, but couldn't recall them properly. I've tried to teach him how to recover formulas from graphs of $\sin$ and $\cos$, but the response was along the lines of "I don't need to understand it" and "I would like to do it without thinking".</p>
<p>For example, there was a pair of tasks 1) prove that $X(\alpha) = Y(\alpha)$ and 2) calculate $X(30^\circ)$, where $Y$ was simple and $X$ was not. However, in the class they covered such calculations, so he evaluated the $X$ directly. Yet, pointing out the simpler way didn't changed his approach.</p>
<p>Another example could be an expression with non-round numbers, where I suggested to hide them behind $\alpha$ and its variations (esp. since such substitution made the necessary formulas and reductions easily visible). Nevertheless, he tirelessly rewrote them all the way through calculations and clung to their concreteness like to some kind of lifeline.</p>
<p>I am aware that an experienced tutor would do much better than I did, but that's not what I'm after. What shook me most strongly was this strange quality of his approach to mathematics that would produce a strong feeling of dissatisfaction (similar to the one you have when seeing something ugly). What I would like to ask about is:</p>
<p><strong>How can I change such a student's approach to math?</strong></p>
<p>It seems like a Catch-22. He sees math as horrendous, but won't learn otherwise, because he doesn't want to deal with it. On the other hand, when he does, it's so wrong that no wonder he sees math this way.</p>
<p>For a change I wanted to show him something engaging, e.g. how math allows us to build nice things using 3D animation software, but he would not make a connection. It was just a few cool pics useful only as long as they can be used to raise his status (e.g. his friends think them awesome).</p>
<p>I have no illusions on whether I could teach him to enjoy math. But, is it possible to make him not hurt mathematics? Right now, I admit defeat.</p>
| JvR | 1,193 | <p><em>(This is a very, very long answer because I want to highlight both the likely mindset of such a pupil, and possible approaches to win them over. Also, please check <a href="https://matheducators.stackexchange.com/questions/895/student-poisoned-experience-with-math">Student Poisoned Experience with Math</a> in case it may help you.)</em></p>
<h1>Understanding the Mindset</h1>
<blockquote>
<p>Another example could be an expression with non-round numbers, where I suggested to hide them behind α and its variations (esp. since such substitution made the necessary formulas and reductions easily visible). Nevertheless, he tirelessly rewrote them all the way through calculations and <strong>clung to their concreteness like to some kind of lifeline</strong>.</p>
</blockquote>
<p>You sound like you're teaching a younger me. I feel sorry for you.</p>
<p>This was exactly what I did. I would reduce the problem, not to its simplest abstract form from which a solution becomes more apparent, but <em>to the form conceptually most recognisable to me</em>, which would lead me around and often down a blind alley. And when you would say that there is an easier way to do it, and you'd even show it, I'd just ignore it because I didn't understand what you were saying, and I wanted to get there on my own steam, dammit.</p>
<p>In short, you have a pupil who is disenchanted, depressed, and close to belligerent.</p>
<p>I feel that, as tutors, we tend to make a mistake. We see someone take the long way around and we offer a short-cut. We approach the subject from the perspective of a master, with insights of simplicity and foresight. By imparting our knowledge, we're hoping something will stick, and that their performance will improve.</p>
<p>The perspective of our pupils, however, is that of a frog trying to cross a busy highway: at some point, they just make the plunge, race screaming and flailing for the other side, and hope not to get hit by a truck. They can't see the short-cut because they don't understand the short-cut, even when it's pointed out with neon signs.</p>
<p>And that's the really hardest part here, I think: debugging the misconceptions op your pupil. Stepping out of our frame of reference, and getting into theirs.</p>
<blockquote>
<p>"I don't need to understand it" and "I would like to do it without thinking". <sup>(*)</sup></p>
</blockquote>
<p><a href="http://www.youtube.com/watch?feature=player_detailpage&v=pIwToj3p3vM#t=62" rel="noreferrer">How's that working out for you, partner?</a></p>
<p>Help him out with the rules by putting them on display. For trigonometry, show the unit circle, and put an example in each quadrant. It'll help him visualize and internalize the relations between the trigonometric functions. At some point, remove the unit circle--but let him draw it himself, if he so chooses.</p>
<p>It will also help him realise that these 'rules' are not axiomatic, independent laws, but properties and relations; tools rather than obstacles.</p>
<p><sup>(*)</sup> <sub>Objects follow the path of least physical resistance. People follow the path of least intellectual resistance.</sub></p>
<hr />
<h1>Changing the Approach</h1>
<blockquote>
<p>How can I change such a student's approach to math?</p>
</blockquote>
<p>Sneak the math onto him. Really. Seriously.</p>
<p>My brother had the same problem tutoring me in physics. I was horrible. I didn't get anything right. I hated it. I wanted to burn my course book after smothering it in the blood of a sacrificial goat.</p>
<p>We had been covering potential and kinetic energy, and then he decided we needed a better exercise. So we were going to calculate the horsepower of <em>throwing a horse off the Niagara falls.</em> Because the exercise was so far removed from what I personally experienced as the boring, horrible physics, I did not have the instant mindblock I would otherwise have <sup>(*)</sup>. Suddenly, <strong>physics was a means to an end</strong>, and I had the (admittedly morbid) drive to find the answer.</p>
<p>A logic and philosophy professor at our university had this approach to teaching where he would start with anecdotes and slowly, sneakily slide into his courses. By the time students 'caught on' that he was back to teaching, he brought up another anecdote to reel them in.</p>
<p>Our probability and statistics professor would take topics from recent news for exercises and show either why the news report was highly misleading, or what the implications of something would be: error margins on polling, probabilities on game shows, casino games, poker, stuff like that.</p>
<p><sup>(*)</sup> <sub>We didn't actually throw a horse off a cliff. My brother wasn't too big on empirical physics.</sub></p>
<blockquote>
<p>For a change I wanted to show him something engaging, e.g. how math allows us to build nice things using 3D animation software, but he would not make a connection. It was just a few cool pics useful only as long as they can be used to raise his status (e.g. his friends think them awesome).</p>
</blockquote>
<p>3D animations can be cool and flashy, but they don't immediately, intuitively link back to math for a layperson.</p>
<p>Stick closer to home. Does he like video games? Maybe he played Call of Duty, and heard of the <a href="http://www.youtube.com/watch?v=9cf6wfVyb1Q" rel="noreferrer" title="Call of Duty 4 - One Shot, One Kill">Coriolis effect</a>. Or maybe he played Realm of the Mad God; that uses Voronoi diagrams for world generation. Maybe he's addicted to his cell- or smartphone; introduce satellite orbits or talk about GPS systems. Math itself is difficult to connect directly, so you may have to go through physics.</p>
<p>You're right: you can't make him go from loathe to love in an instant. Even now, I don't love mathematics, but I am no longer terrified of it, and I regularly use it as a powerful tool of abstraction.</p>
|
405,953 | <p>Let</p>
<ul>
<li><span class="math-container">$E$</span> be the usual sobolev space <span class="math-container">$H^{1}_{0}(\Omega)$</span> on a smoothly bounded domain <span class="math-container">$\Omega$</span>,</li>
<li><span class="math-container">$E_{k}$</span> be its subspace spanned by the first <span class="math-container">$k$</span> eigenfunctions of the Laplace operator, i.e.
<span class="math-container">$$E_{k}:=\text{span}\{\varphi_{j}\in E: -\Delta\varphi_{j}=\lambda_{j}\varphi_{j},~j=1,2\dots,k \},$$</span></li>
<li><span class="math-container">$P$</span> be the positive cone in <span class="math-container">$E$</span>, i.e,
<span class="math-container">$$P:=\{u \in E~; u \geq 0 ~a.e.\},$$</span></li>
</ul>
<p>Now set <span class="math-container">$P_{k}=P\cap E_{k}~~u^{-}=min\{u,0\}$</span>.</p>
<p><strong>My question</strong>: does it exist <span class="math-container">$C_{k}>0$</span>, such that
<span class="math-container">$$\text{dist}_{L^{2}}(u,P_{k}) \leq C_{k} \text{dist}_{L^{2}}(u,P)~\text{ holds }~\forall u\in E_{k}\; ?$$</span>
or
<span class="math-container">$$\text{dist}_{E}(u,P_{k}) \leq C_{k} \text{dist}_{E}(u,P)~\text{ holds }~\forall u\in E_{k}\; ?$$</span>
or <span class="math-container">$$\text{dist}_{E}(u,P_{k}) \leq C_{k} \|u^{-}\|_{E}~\text{ holds }~\forall u\in E_{k}\; ?$$</span>
If not, could you please show me a counterexample? Thanks.</p>
<p>Moreover, this problem is raised from the proof of lemma 3.6 in "Infinitely many solutions to perturbed elliptic equations",<a href="https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123605X05468/1-s2.0-S0022123605002582/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEIf%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJIMEYCIQCA3DmdTW7ToN5U0cdmubTTGzQJCXX38jpSYRu%2FaTQqOgIhANqGdebL%2FSXPmzKTqmmN0Xr%2F8itx2hkp7WGf6i72BrqTKvoDCBAQBBoMMDU5MDAzNTQ2ODY1Igz5FlE6QYbbcb7YH8kq1wMEn%2BjofTAgOGQ%2B7YALRJ5jJO286FtV2mHjiOBbpm1uIdXmNcC7GQJyghCxd4oRxfBUUIvPrbOwfdDH09ESAULHmDMXJNKlFyTF%2F33Jn2DSngmTTTq%2F8JFpuR2Bqe7p4k%2FDYnmnBm%2BMIsCoevLmtmJQ%2BNbL0bx%2FnvwCnvW%2BI45ClJddGgxdHCYBLrLxg0AOtNgy%2FIEagUE8CxkIXDro7QeNKau9lwQfEWrVa9lakzcMfXC70%2BnlVk2VzXAIdX2cKYDsf%2Fl8wl7Dkcs1%2FlqbbXNxz95n8CuAg7ifFqc0dfnetoIf02pLxKaw4hMx40lKKaWRdMLyKnT9C5fFToz5n%2Ff1liqy%2BDPvphnNV18lN7fZ%2Bbw5hUYj8q8rCxCIc8DnSeExbF78GjTXwR%2FsHHuHUZoXjIah2vK8SJUeVspjWfFK3CDxHokr%2BibxlkNhX2pjXIrZeAHqwMavPCT0bkvyU90XbZuI1vOc3h0FxnHfFXhyReh%2Bz6NjMl7OUv2YiaQgoxQEFRp4g2nNLau0RaPfxLTG77edF5nlL6b3bmSyu6KV%2FzWE2dOGEu9vk9cG%2F4ID3R1jUl8sSwEDiCMHZAaAMxjAld0l%2BtLwvKGq0046wxg5Uy9klhYCFbUwit2UiwY6pAHeFguTRVuilMp%2BddCPI10eVsY1ugttXA0kflCcFX73U8SfGzOTjmoheDKj%2FCNTQrzgGp8p4yhBjLvODQ8croHzoNyV6PoOl92rAARM5%2Fa9NOlTDvT2RTMpQzOI0sfSCvXEuUClp%2BrdylgeZJSLSMIT%2FQ6lyj12iGeJWZpoygEIvc3oTNdx8YjZSwbV4cv8smABlGArjB4p3eRWS0R5gFzpL8SZgA%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20211012T080454Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTY4Y4CPKBJ%2F20211012%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=e9f461d9634a663aa1a87583e2578ad501f687454a69dff0e960c653c345a3f2&hash=ebaa2e2d930797073bf18eae43724b9df1cc260ef6f19e04102732392edbc536&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123605002582&tid=spdf-ca8bb609-0d0a-4a63-8c85-58d7098f6e09&sid=e27151265f16754ed3986a79985943117767gxrqa&type=client" rel="nofollow noreferrer">doi:10.1016/j.jfa.2005.06.014</a>
<a href="https://i.stack.imgur.com/ruowq.png" rel="nofollow noreferrer">enter image description here</a></p>
<p>also see lemma 5.4 in "On finding sign-changing solutions",<a href="https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123606X05591/1-s2.0-S0022123605003319/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEJ7%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJIMEYCIQDyfP9hsP1tHkBhbzcqwuQOpM51AhMSmUw8lQjARVyomQIhAL%2BkqkTXKKLwXuf08PPLGidPv2ZzyX2LYEMUTEyKhGbgKvoDCCcQBBoMMDU5MDAzNTQ2ODY1Igy%2F9ma67gic8nnsmhIq1wMLrIT%2FDxI9G6FQavBW3P21mCaVAw3de%2B4W15bgV7bTepA4sSaYvFGgm84PGZscT2JoxYpbJXPpHh%2FHncPeCKdSwMB6KisZ3qj3TUi5UVctygkh7PsrPp48LPms%2B98r%2FmEHdrrGD3PejQlZKrHT6eZ7S48OpXV7qtNt3FBqPGj12kZU3npDDdO4y3eOdo4vTyBkab7tyPAovwCr4frZiAooONzBayGkUEmKZ0bAnYgkElIXAgUbkGawBjFzPsfoiXmPsgX0GwiSIIV%2BBj4BEW4VglGabDqRtAO7xjGPegxoVM7chmabrwHwcOIuzT6pZhD4rqOeTjTtq88dBlRgUWVm3Ko6j%2BWDcfNVQxsbpgZZ3FjDKlJCt2Ki%2FKhBoET%2BC2CNNWMww8XJvT789JwPTI5MR7icHX5g54pxcbZV7iSgTrFHCUjE%2B%2BYyGg3cRNz73Wv65FvL1YnebwAeAGinMYhKSGPA5Vujv4f%2B3UrZM7TWOEK76t8ASgklluYS2B4hNdfNxMuhElgannyqfO8Mao2JVUcag5mBM1dJHcPc6xGbRgJQJkKG9AZ01CQ6trBMe9gCDy3ZSwRZAsgSu9xf0gr6%2BjjH8WzmMqoyE%2B7KLyqOS5i2ecsyQcgw%2F%2BaZiwY6pAEBkjdSZRAyQgflJe%2FR0CGWu1dHufRVHsGeRCeSEwzyfaBkii6sCmJv5IjB7QLnMiHrZYogDf4FdczVdD5ij%2Bp%2BL5Kd7q7FWkUAhLTx9Gm32EiH%2B4Nn8UO7Iczh0aMSuqPXdyec3P%2BRiF50K5bXrOKwo2fWF9nfUbGZMTFAh4UjeuIU1WD9OKgaIcMDNlMNzFMWJCUN4bS5AVZ1pV8UsBE5IpoaVQ%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20211013T062711Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYY3F5SIHU%2F20211013%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=74d463a3fdfe47d9fe039c0ce0d0c8d8647cae0dbe5669c7c1dd2f6fe8e6ddac&hash=4ab29c5d1d48cc6b4b9a75d0b843958d134747e1605579583d6db363c5a6e3b2&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123605003319&tid=spdf-9e5fb13a-6657-456c-a6e4-cfaa6bcc0bf1&sid=8b90046339b6054a78486f6-8787b72cf34agxrqa&type=client" rel="nofollow noreferrer">doi:10.1016/j.jfa.2005.09.004</a>
<a href="https://i.stack.imgur.com/CLUMD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CLUMD.png" alt="enter image description here" /></a></p>
| user378654 | 378,654 | <p>This does not answer the question asked (see the other answer for a good counterexample) and I don't know if it's relevant to the paper. However, if <span class="math-container">$u$</span> is sufficiently far from <span class="math-container">$\varphi_1$</span>, the first eigenfucntion, you do get a positive answer. Below I assume <span class="math-container">$\Omega$</span> is connected (you'd have to work out what happens more carefully if it's not).</p>
<p>First, we have that
<span class="math-container">$$
\|u\|_{L^\infty} \leq C(k) \|u\|_{L^2}
$$</span>
for any <span class="math-container">$u \in E_k$</span>; this can be checked for each <span class="math-container">$\phi_j$</span> and is standard (there is a clean argument with the heat kernel, or you can just apply the local maximum principle repeatedly on balls). Also, <span class="math-container">$\varphi_1 > 0$</span> on <span class="math-container">$\Omega$</span>.</p>
<p>Now consider <span class="math-container">$u \in E_k$</span> with <span class="math-container">$\int u^2 = 1$</span> and
<span class="math-container">$$
\int u \varphi_1 = a.
$$</span>
We have that
<span class="math-container">$$
\int |u|\varphi_1 \geq c(k)
$$</span>
using only that <span class="math-container">$|u|\leq C(k)$</span> and <span class="math-container">$\int u^2 = 1$</span>. Indeed, this integral is minimized by a piecewise constant function <span class="math-container">$u$</span> which is <span class="math-container">$C(k)$</span> on <span class="math-container">$F = \{\varphi_1 < t\}$</span> and <span class="math-container">$0$</span> outside of <span class="math-container">$F$</span>, where <span class="math-container">$t$</span> is chosen so that <span class="math-container">$|F| = 1/C(k)^2$</span>. Setting <span class="math-container">$c(k) = C(k) \int_F \varphi_1 > 0$</span> gives the inequality.</p>
<p>Combining, we have that
<span class="math-container">$$
\|u_-\|_{L^2} \geq \int u_- \varphi_1 \geq \frac{c(k) - a}{2}.
$$</span>
Undoing the normalization, we have shown that for <span class="math-container">$u \in E_k$</span>, if <span class="math-container">$\int u \varphi_1 \leq \frac{c(k)}{2} \|u\|_{L^2}$</span>, then
<span class="math-container">$$
\|u\|_{L^2} \leq \frac{4}{c(k)} \|u_-\|_{L^2}.
$$</span>
This implies both inequalities in the question in this case (the <span class="math-container">$L^2$</span> and <span class="math-container">$H^1_0$</span> norms of each side are comparable). In particular, it applies to any <span class="math-container">$u \in E_k$</span> orthogonal to <span class="math-container">$\varphi_1$</span>, so to any eigenfunction <span class="math-container">$\varphi_j$</span>.</p>
|
8,816 | <p>What is the result of multiplying several (or perhaps an infinite number) of binomial distributions together?</p>
<p>To clarify, an example.</p>
<p>Suppose that a bunch of people are playing a game with k (to start) weighted coins, such that heads appears with probability p < 1. When the players play a round, they flip all their coins. For each heads, they get a coin to flip in the next round. This is repeated every round until they have a round with no heads.</p>
<p>How would I calculate the probability distribution of the number or coins a player will have after n rounds? Especially if n is extremely high and p extremely close to 1?</p>
| Mikhail Bondarko | 2,191 | <p>If A is an elliptic curve then G (in your notation) is finite. Yet it seems that for a square of an elliptic curve you get something infinite and very far from being algebraic. If $A=B\times B$, $B$ is a 'general' elliptic curve, then it seems that $G(A)=GL_2(\Bbb Z)$; this is a 'large' discrete group. Another example is an abelian variety with complex multplication: you get the group of units in some order in some CM-field over $\Bbb Q$, and this is infinite if this field is not quadratic. I think, in these examples it is not difficult to prove that Aut(A) is not an algebraic group.</p>
|
3,632,576 | <p>Considering that input <span class="math-container">$x$</span> is a scalar, the data generation process works as follows:</p>
<ul>
<li>First, a target t is sampled from {0, 1} with equal probability.</li>
<li>If t = 0, x is sampled from a uniform distribution over the interval
[0, 1]. </li>
<li>If t = 1, x is sampled from a uniform distribution over the
interval [0, 2].</li>
</ul>
<p>I'm trying to find the formula for <span class="math-container">$P(t=1)$</span>, <span class="math-container">$P(t=0)$</span>, <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span> and then find the posterior probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>.</p>
<p>So far I have that <span class="math-container">$P(t=1)$</span> and <span class="math-container">$P(t=0)$</span> <span class="math-container">$=$</span> <span class="math-container">$\frac12$</span> but I wasn't sure how to find <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span>.</p>
<p>I know from there we can just use <span class="math-container">$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$</span> to compute the probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>. Is that correct?</p>
| Sam | 584,704 | <p>It's simpler if you keep the absolute values and use the same idea: <span class="math-container">$$\left|\frac{1}{x}\int_0^x f(t)dt - a\right| $$</span> <span class="math-container">$$= \left|\frac{1}{x}\int_0^N f(t)dt + \frac{1}{x}\int_N^xf(t)dt - a\right|$$</span>
<span class="math-container">$$\le\left|\frac{1}{x}\int_0^N f(t)dt\right| + \left|\frac{1}{x}\int_N^xf(t)dt - a\right|$$</span>
<span class="math-container">$$ \le \left|\frac{1}{x}\int_0^N f(t)dt\right| + \frac{1}{x}\int_N^x|f(t) - a|dt + \left |a\left (\frac{x-N}{x}-1\right)\right|$$</span>
<span class="math-container">$$\le \left|\frac{1}{x}\int_0^N f(t)dt\right| +\frac{(x-N)}{x}\epsilon + \left |a\left (\frac{x-N}{x}-1\right)\right|$$</span></p>
<p>Now choose <span class="math-container">$M \gt N$</span> such that for <span class="math-container">$x \gt M$</span> the first term is <span class="math-container">$\lt \epsilon$</span> and <span class="math-container">$\left|\frac{(x-N)}{x} - 1\right|\lt \min(1,\epsilon/|a|)$</span>. Then the sum above is <span class="math-container">$\lt \epsilon + 2\epsilon+\epsilon$</span> and the result follows.</p>
|
186,878 | <p>Is there a way to find the geoposition of a given distance from start in a <code>GeoPath</code>? I want to mark equidistant positions along a track, for example, a mark every 500 km along the path given by</p>
<pre><code>path=GeoGraphics[
GeoPath[{
Entity["City", {"Boston", "Massachusetts", "UnitedStates"}],
Entity["City", {"Rochester", "NewYork", "UnitedStates"}],
Entity["City", {"Chicago", "Illinois", "UnitedStates"}]
}, "Rhumb"]
]
</code></pre>
<p>Is there a way to find the pos that gives <code>GeoDistance[Entity["City", {"Boston", "Massachusetts", "UnitedStates"}],pos]==500 km</code> along the path? </p>
| Greg Hurst | 4,346 | <p>Here's my attempt to parametrize the path by arclength, where here arclength is <code>GeoLength</code>.</p>
<p>First I build up a function that can be used on many values:</p>
<pre><code>ParametrizeGeoPath[g_GeoPath, t_] := ParametrizeGeoPath[g][t]
ParametrizeGeoPath[GeoPath[locs_, args___]] :=
Block[{line, nodes, lens, acc, nf, n1, n2, solver},
line = GeoGraphics`GeoEvaluate[GeoPath[locs, args]];
nodes = GeoPosition /@ Reverse[line[[1]], {2}];
lens = QuantityMagnitude[UnitConvert[GeoLength[GeoPath[#, args]]& /@ Partition[nodes, 2, 1], "Kilometers"]];
acc = Accumulate[lens];
nf = Nearest[acc -> {"Index", "Element"}];
GeoPathParametricFunction[acc, nodes, nf, args]
]
</code></pre>
<p>Given a target distance, we can invert <code>GeoLength</code> with <code>FindRoot</code>:</p>
<pre><code>GeoPathParametricFunction[_, nodes_, __][d_] /; d == 0 := First[nodes]
GeoPathParametricFunction[acc_, nodes_, __][d_] /; d == Last[acc] := Last[nodes]
GeoPathParametricFunction[acc_, nodes_, nf_, args___][d_] /; 0 < d < Last[acc] :=
Block[{i, v, s, n1, n2, dist, root, t = Unique["t"]},
{i, v} = First[nf[d]];
If[v > d, i--];
s = If[i == 0, 0, acc[[i]]];
n1 = nodes[[i+1, 1]];
n2 = nodes[[i+2, 1]];
dist[t_?NumericQ] := s + QuantityMagnitude[UnitConvert[GeoLength[GeoPath[{GeoPosition[n1], GeoPosition[(1-t)n1 + t n2]}, args]], "Kilometers"]];
root = Quiet@FindRoot[dist[t] == d, {t, .5, 0, 1}];
(
GeoPosition[(1-t)n1 + t n2] /. root
) /; MatchQ[root, {t -> _Real}]
]
GeoPathParametricFunction[___][d_?NumericQ] = Indeterminate;
GeoPathParametricFunction /: MakeBoxes[expr:GeoPathParametricFunction[__], _] := InterpretationBox[RowBox[{"GeoPathParametricFunction", "[", "\"\[Ellipsis]\"", "]"}], expr]
</code></pre>
<p>Your example:</p>
<pre><code>path = GeoPath[{
Entity["City", {"Boston", "Massachusetts", "UnitedStates"}],
Entity["City", {"Rochester", "NewYork", "UnitedStates"}],
Entity["City", {"Chicago", "Illinois", "UnitedStates"}]
}, "Rhumb"];
gpf = ParametrizeGeoPath[path];
gpf[500]
</code></pre>
<blockquote>
<pre><code> GeoPosition[{43.0932, -77.0359}]
</code></pre>
</blockquote>
<pre><code>Manipulate[GeoGraphics[{
path,
GeoMarker[gpf[d]]
},
PlotLabel -> Row[{"Distance: ", Quantity[d, "Kilometers"]}]],
{d, 0, gpf[[1, -1]]}
]
</code></pre>
<p><a href="https://i.stack.imgur.com/ClnQx.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ClnQx.gif" alt="enter image description here"></a></p>
<p>The points returned are very close to the initial path:</p>
<pre><code>ListLinePlot[
GeoDistance[path, g /@ Range[0, 1300, 100]],
TargetUnits -> "Meters",
AxesLabel -> Automatic,
DataRange -> {0, 1300}
]
</code></pre>
<p><a href="https://i.stack.imgur.com/oeac0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oeac0.png" alt="enter image description here"></a></p>
|
359,212 | <p>I mean, $\Bbb Z_p$ is an instance of $\Bbb F_p$, I wonder if there are other ways to construct a field with characteristic $p$?
Thanks a lot!</p>
| Ittay Weiss | 30,953 | <p>Just to supplement the other answers: As stated in the other answers, for every prime power $p^r$, $r>0$, there is a unique (up to isomorphism) field with $p^r$ elements. There are also infinite fields of characteristic $p$, for instance if $F$ is any field of characteristic $p$ (e.g., $\mathbb Z_p$), the field $F(t)$ (the field of fractions of the polynomial ring $F[t]$, with $t$ an indeterminate), is an infinite field of characteristic $p$. </p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.