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 |
|---|---|---|---|---|
1,489,078 | <p>Let $P(S)$ denotes the power set of set $S$. Which of the following is always true$?$</p>
<ol>
<li>$P(P(S)) = P(S)$</li>
<li>$P(S) ∩ P(P(S)) = \{ Ø \}$</li>
<li>$P(S) ∩ S = P(S)$</li>
<li>$S ∉ P(S)$</li>
</ol>
<hr>
<p>I try to explain $:$
If $S$ is the set $\{x, y, z \},$ then the subsets of S are:</p>
<p>${}$ (also denoted $\phi,$ the empty set)
$\{x\},
\{y\},
\{z\},
\{x, y \},
\{x, z \},
\{y, z \},
\{x, y, z \}$
and hence the power set of S is i.e.,</p>
<p>$P(S) =$ $\{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\}$.</p>
<p>Similarly , </p>
<p>$P(P(S))=\{\{\{\}\}, \{\{x\}\}, \{\{y\}\}, \{\{z\}\}, \{\{x, y\}\}, \{\{x, z\}\}, \{\{y, z\}\}, \{\{x, y, z\}\}\}.....\}$</p>
<p>therefore , </p>
<p>$P(S) ∩ P(P(S)) = \{ Ø \}$</p>
<p>Note that $\{ Ø \}$ is always element of powerset of a set , and also $\{ Ø \}$ is the subset of a set , in other words all subset of a set is a powerset .</p>
<hr>
<blockquote>
<p>My question is $:$ both $\{\}$ and $\{\{\}\}$ same element ?</p>
</blockquote>
| hmakholm left over Monica | 14,366 | <p>No, $\{\}$ and $\{\{\}\}$ are two different sets. You can see that by noting that there's something that is an element of one but not of the other:</p>
<p>$$ \{\}\notin \{\} \qquad\text{but}\qquad \{\}\in\{\{\}\} $$</p>
<p>$\{\}$ is a set that <em>has no elements</em>. It is the same set that is also notated $\varnothing$.</p>
<p>$\{\{\}\}$ is a set which has exactly one element, namely $\{\}$. Since $\{\}$ is the same as $\varnothing$, a different name for $\{\{\}\}$ would be $\{\varnothing\}$.</p>
|
3,624,953 | <p>Let <span class="math-container">$(X_t)_{t \in [0, T]}$</span> be a continuous stochastic process with paths which are a.s. continuous, the underlying space of which is irrelevant but is well defined. Let <span class="math-container">$a$</span> be a constant
Define two stopping times
<span class="math-container">$$\tau_1 = \inf\{t \geq 0: X_t > a\}$$</span>
<span class="math-container">$$\tau_2 = \inf\{t \geq 0: X_t = a\}$$</span>
Evidently, <span class="math-container">$X_{\tau_2} = a$</span>. However, can we claim <span class="math-container">$X_{\tau_1} = a$</span> ? This "feels like" having something to do with continuity/topology but I cannot figure it out. Any help would be greatly appreciated.</p>
| Surb | 154,545 | <p>Yes. Indeed, let <span class="math-container">$\omega \in \{X_{\tau_1}>a\}$</span>. In particular, <span class="math-container">$X_{\tau_1(\omega )-h}\leq a$</span>. Therefore, <span class="math-container">$$\lim_{h\to 0}X_{\tau_1(\omega )-h}=X_{\tau_1^-(\omega )}\leq a<X_{\tau_1(\omega )}.$$</span></p>
<p>Since <span class="math-container">$(X_{t})$</span> is continuous, <span class="math-container">$\mathbb P\{X_{\tau_1}>a\}=0$</span>.</p>
|
92,613 | <p>This is a question in Pinter's <em>A Book of Abstract Algebra</em>.</p>
<blockquote>
<p>Let $S=\{g\in G\mid \operatorname{ord}(g)=p\}$. Prove the order of $S$ is a multiple of $p-1$.</p>
</blockquote>
<p>In his solution Pinter says $a \in S$ implies that $a$ generates a subgroup with $p-1$ elements. Shouldn't there be $p$ elements $\{1,a^1,\dots,a^{p-1}\}$? Or is it typical to only count the non-trivial elements in a subgroup?</p>
| Dan Petersen | 677 | <p>No, you are right, every element $a \in S$ generates a subgroup with $p$ elements. However, only $p-1$ of those will lie in the set $S$, which I guess is what Pinter means.</p>
|
1,123,247 | <p>let $a_n$ be a sequence of real numbers such that the series $\sum |a_n|^2$ is convergent.Find range of $p$ such that the series $\sum |a_n|^p$ is convergent.</p>
<p>My try:</p>
<p>To show the series it is necessary to show that sequence of partial sums of $\sum |a_n|^p$ is bounded.</p>
<p>Consider $S_n=|a_1|^p+|a_2|^p+...+|a_n|^p$. How to proceed next? Any inequality which can be used?</p>
| Cm7F7Bb | 23,249 | <p>Suppose that $\ell_q$ is the space of real sequences $x=\{x_n\}$ such that $\sum|x_n|^q<\infty$ with the norm
$$
\|x\|_q=\biggl(\sum|x_n|^q\biggr)^{1/q}
$$
for $q\ge1$. We have that $\|x\|_p\le\|x\|_q$ for $1\le q\le p<\infty$ (see <a href="https://math.stackexchange.com/questions/69125/inequality-between-ellp-norms?lq=1">here</a>). Hence, the series $\sum|a_n|^p$ converges for all $p>2$.</p>
<p>To see that the series might not necessarily converge for $p<2$, let's take, for example, $a_n=n^{-1/p}$ with $p<2$.</p>
|
102,932 | <p>This is a naive question but I hope that the answers will be educational. When is it the case that a finitely presented group $G$ admits a faithful $2$-dimensional complex representation, e.g. an embedding into $\text{GL}_2(\mathbb{C})$? (I am mostly interested in sufficient conditions.) </p>
<p>I think I can figure out the finite groups with this property (they can be conjugated into $\text{U}(2)$ and taking determinants reduces to the classification of finite subgroups of $\text{SU}(2)$ and an extension problem) as well as the f.g. abelian groups with this property (there can't be too much torsion). But already I don't know what finitely presented groups appear as, say, congruence subgroups of $\text{GL}_2(\mathcal{O}_K)$ for $K$ a number field. </p>
<p>What can be said if you are given, say, a nice space $X$ with fundamental group $G$? I hear that in this case linear representations of $G$ are related to vector bundles on $X$ with flat connection. </p>
| Geoff Robinson | 14,450 | <p>I'm not sure whether it is worth remarking on, but if $G$ is any irreducible subgroup of ${\rm GL}(2,\mathbb{C}),$ its non-central Abelian subgroups have some very restrictive properties:
If $A$ is a non-central Abelian subgroup of $G$, then $C_{G}(A)$ is also Abelian and $[N_{G}(A):C_{G}(A)] \leq 2.$</p>
|
3,029,519 | <p>I have a variable, <span class="math-container">$X$</span>, which is normally distributed</p>
<p><span class="math-container">$$X \sim \mathcal{N}(\mu, \sigma^2)$$</span></p>
<p>I also have an event, <span class="math-container">$A$</span>, which measures the probability that <span class="math-container">$X$</span> is greater than or equal to 0.
<span class="math-container">$$P[A] = P[X \geq 0] = \lim_{b \to +\infty} \int_{x=0}^{x=b} f(x)\,dx \quad \text{f(x) is PDF of X}$$</span></p>
<p>I'd like to know what the derivative of <span class="math-container">$P[A]$</span> is w.r.t. <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma^2$</span>. </p>
<p><span class="math-container">$$\frac{\partial P[A]}{\partial \mu} $$</span>
<span class="math-container">$$\frac{\partial P[A]}{\partial \sigma^2}$$</span></p>
<p>I think this is really simple because I'm looking for the derivative of an integral, but would like to make sure I'm doing it right:</p>
<p><span class="math-container">$$\frac{\partial P[A]}{\partial \mu} = \frac{\partial}{\partial \mu} \lim_{b->+\infty} f(b) - f(0) = \frac{\partial}{\partial \mu} f(0) $$</span>
<span class="math-container">$$\frac{\partial P[A]}{\partial \sigma^2} = \frac{\partial}{\partial \sigma^2} \lim_{b->+\infty} f(b) - f(0) = \frac{\partial}{\partial \sigma^2} f(0)$$</span></p>
<p>where</p>
<p><span class="math-container">$$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$</span></p>
<p>Are the above equations correct?</p>
| lab bhattacharjee | 33,337 | <p>I think it is simpler to utilize binomial identities:</p>
<p><span class="math-container">$$\dfrac{\binom nk}{\binom{k+3}k}=\dfrac{n!3!}{(n-k)!(k+3)!}=\dfrac6{(n+1)(n+2)(n+3)}\binom{n+3}{k+3}$$</span></p>
<p><span class="math-container">$$\sum_{k=0}^n\dfrac{\binom nk}{\binom{k+3}k}=\dfrac6{(-1)^3(n+1)(n+2)(n+3)}\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$</span></p>
<p><span class="math-container">$$\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$</span></p>
<p><span class="math-container">$$=\sum_{r=0}^{n+3}\binom{n+3}r(-1)^r-\sum_{r=0}^2\binom{n+3}r(-1)^r$$</span></p>
<p><span class="math-container">$$=(1-1)^n-\sum_{k=0}^2\binom{n+3}k(-1)^k=?$$</span></p>
|
2,263,520 | <p>I'm trying to understand how to simplify summations. My text says that: $$\sum_{i=1}^{\frac{n}{2}} \sum_{j=i}^{n-i} \sum_{k=1}^{j} 1 = \frac{n^3}{8}$$</p>
<p>But does not explain how to get to the right-hand side.</p>
<p>I think the above nested summation evaluates to $\sum_{i=1}^{n/2} \left[\sum_{j=i}^{n-i} j = i + (i + 1) + (i + 2) + .. + (n-i-1) + (n-i)\right]$, but I don't know how to proceed from here.</p>
| Ng Chung Tak | 299,599 | <p>\begin{align}
\sum_{k=1}^{j} 1 &= j \\
\sum_{j=i}^{n-i} j
&= \frac{n(n-2i+1)}{2} \\
S &= \sum_{i=1}^{n/2} \frac{n(n-2i+1)}{2} \\
&= \frac{n(n+1)}{2} \sum_{i=1}^{n/2} 1-
n \sum_{i=1}^{n/2} i
\end{align}</p>
<ul>
<li><p>If $n=2m$,
\begin{align}
S &= \frac{2m(2m+1)}{2} \times m-2m\times \frac{m(m+1)}{2} \\
&= m^3 \\
&= \frac{n^3}{8}
\end{align}</p></li>
<li><p>If $n=2m-1$,
\begin{align}
S &= \frac{(2m-1)(2m)}{2} \times (m-1)-(2m-1)\times \frac{(m-1)m}{2} \\
&= \left( \frac{2m-1}{2} \right)^3-\frac{2m-1}{8} \\
&= \frac{n^3-n}{8}
\end{align}</p></li>
</ul>
<blockquote>
<p>Your equality holds for even $n$ only.</p>
</blockquote>
|
2,263,520 | <p>I'm trying to understand how to simplify summations. My text says that: $$\sum_{i=1}^{\frac{n}{2}} \sum_{j=i}^{n-i} \sum_{k=1}^{j} 1 = \frac{n^3}{8}$$</p>
<p>But does not explain how to get to the right-hand side.</p>
<p>I think the above nested summation evaluates to $\sum_{i=1}^{n/2} \left[\sum_{j=i}^{n-i} j = i + (i + 1) + (i + 2) + .. + (n-i-1) + (n-i)\right]$, but I don't know how to proceed from here.</p>
| Hypergeometricx | 168,053 | <p>Exploit the symmetry.</p>
<p>Put $n=2m$ (assuming $n$ is even):</p>
<p>$$\begin{align}
S
=\sum_{i=1}^{\frac n2}\sum_{j=i}^{n-i}\sum_{k=1}^j 1
=\sum_{i=1}^{\frac n2}\sum_{j=i}^{n-i}j
&=\sum_{i=1}^m\sum_{j=i}^{2m-i}j&&(n=2m)\tag{1}\\
&=\sum_{i=1}^m\sum_{j=i}^{2m-i}(2m-j)&&(j\leftarrow 2m-j)\tag{2}\\
&=m\sum_{i=1}^m\sum_{j=i}^{2m-i}1
&&\dfrac{(1)+(2)}2\\
&=m\sum_{i=1}^m\left(\sum_{j=i}^m1+\sum_{j=m+1}^{2m-i}1\right)\\
&=m\sum_{i=1}^m\left(\sum_{j=i}^m1+\sum_{j=1}^{m-i}1\right)\\
&=m\sum_{i=1}^m\left(\sum_{j=i}^m1+\sum_{j=1}^{i-1}1\right)
&&\big(i\leftarrow m-i+1\atop\scriptsize\text{for second summation}\big)\\
&=m\sum_{i=1}^m\sum_{j=1}^m 1\\
&=m^3\\
&=\color{red}{\frac {n^3}{8\;}}
\end{align}$$</p>
<p>The solution is derived only by manipulating summation limits and indices without requiring expansion or evaluation of sum of squares or integers. </p>
|
1,818,557 | <p>I will be teaching some "topology" to high school students. I was wondering how to explain to such a school student that on a sphere the shortest path between 2 points is given by a great circle?</p>
<p>Also, how to explain that if they lived on a sphere they would have no notion of "above" or "below"? I cannot find a nice way to convince them since they see the sphere embedded on 3d? </p>
| Narasimham | 95,860 | <p>There are two ways I tried with students. </p>
<p>Case 1. Equator of ball</p>
<p>On a plastic ball toy carefully tie a string around any great circle, ( use a smal cellulose tape/tab if needed, to prevent side slippage ,) for exactly one rotation. Make the string taut by pulling in opposite directions. The ball will be compressed, tension in taut string increases.</p>
<p>Case 2. Parallel circles of ball</p>
<p>Next repeat the same by marking a smaller or latitude circle, stick the string on it. When pulled the string easily slips out of its place.</p>
<p>Case 3. Winding and unwinding a cone</p>
<p>Next roll a rectangular sheet of paper into a cone. Look at the edges in two situations when rolled and when flattened out. The sideways straightness is preserved in either case... as zero curvature.</p>
<p>In Case 1 a sidewise straightness existed, in Case 2 no sidewise straightness existed, in fact the edge became base of a cone.</p>
<p>These concepts of geodesy and geodesic curvature in differential geometry can be thus demonsrated.</p>
<p>Next, <em>up-down</em> feeling is <em>conditioned</em> by 1) Gravity or Force vector 2) Fluid sensation in Cochlea semi-circular coils inside human ear as a biological response to such forces..</p>
|
1,818,557 | <p>I will be teaching some "topology" to high school students. I was wondering how to explain to such a school student that on a sphere the shortest path between 2 points is given by a great circle?</p>
<p>Also, how to explain that if they lived on a sphere they would have no notion of "above" or "below"? I cannot find a nice way to convince them since they see the sphere embedded on 3d? </p>
| zyx | 14,120 | <p>Set up lines of latitude and longitude on the sphere so that the two points, call them $A$ and $B$, are on the same meridian.</p>
<p>Then any motion from $A$ to $B$ can be "tracked" along the meridian by taking a moving point on the meridian that has the same latitude as the point moving from $A$ to $B$. The motion along the meridian (an arc of a great circle) is shorter, because at every instant it sweeps out length at a rate less than or equal to that of any other motion at the same latitudes.</p>
<p>In the extreme case, when the points are antipodal, this is all still true but the inequality becomes an equality if the other motion is along a different great circle.</p>
<hr>
<blockquote>
<p>Also, how to explain that if they lived on a sphere they would have no notion of "above" or "below"? I cannot find a nice way to convince them since they see the sphere embedded on 3d? </p>
</blockquote>
<p>They do live on a sort of sphere. Do two people on opposite sides of the Earth have a way of deciding which one is upside down? Assigning directions North/South or East/West relies on extra information such as a magnetic pole or the motion of the sun relative to the Earth.</p>
|
1,812,525 | <p>Let $T:H \to H$ be defined as $Tx=\sum_{n=1}^{\infty} \lambda_n \langle x,\varphi _n \rangle \varphi _n$, given that $\{\varphi _n\}_{n=1}^\infty$ is an orthonormal sequence (not necessarily a basis) and $\{\lambda_n\}_{n=1}^\infty$ is a sequence of numbers (which may be complex if the Hilbert space is complex).</p>
<p>Show that $\ker (T)=\{\varphi _n\mid\lambda_n\neq 0\}^\perp $.</p>
<p>What does this $\{\}^\perp $ notation mean? Do I need to show that $\varphi _n$ are perpendicular to each other? If so how?</p>
| walkar | 98,077 | <p>HINT: You need to use the fact that $\phi_n$ is orthonormal. That is, $\left\langle \phi_n,\phi_m \right\rangle = \left\lbrace \begin{array}{cc}
0 & n \neq m \\
1 & n = m
\end{array} \right.$. Start by supposing that $f \in \ker(T)$ so that $T(f) = \sum_{n=1}^\infty \lambda_n \left\langle f,\phi_n \right\rangle \phi_n=0$. What can you say if $f$ is one of the $\phi_m$s? Try working with the definition of orthogonal.</p>
|
702,506 | <p>We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals.</p>
<p>How to solve</p>
<p>$$\sin x - \cos x = -1$$</p>
<p>for the interval $(0, 2\pi)$.</p>
| Community | -1 | <p>$$\sin x-\cos x=-1\\
\implies (\sin x-\cos x)^2=\sin^2x+\cos^2x-2\sin x\cos x=1+\sin(2x)=1$$
So,
$$\sin(2x)=0$$
Therefore,
$$x=3\pi/2,\mbox{ as }x\in(0,2\pi)$$</p>
|
702,506 | <p>We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals.</p>
<p>How to solve</p>
<p>$$\sin x - \cos x = -1$$</p>
<p>for the interval $(0, 2\pi)$.</p>
| lab bhattacharjee | 33,337 | <p><strong>Method</strong> $\#1$ </p>
<p>Avoid squaring which immediately introduces <a href="http://en.wikipedia.org/wiki/Extraneous_and_missing_solutions" rel="nofollow">extraneous roots</a> which demand exclusion </p>
<p>We have $\displaystyle\sin x-\cos x=-1$</p>
<p>$$\iff\sin x=-(1-\cos x)\iff2\sin\frac x2\cos\frac x2=-2\sin^2\frac x2$$</p>
<p>$$\iff2\sin\frac x2\left(\cos\frac x2+\sin\frac x2\right)=0$$</p>
<p>If $\displaystyle \sin\frac x2=0,\frac x2=n\pi\iff x=2n\pi$ where $n$ is any integer</p>
<p>If $\displaystyle\cos\frac x2+\sin\frac x2=0\iff\sin\frac x2=-\cos\frac x2$</p>
<p>$\displaystyle\iff\tan\frac x2=-1=-\tan\frac\pi4=\tan\left(-\frac\pi4\right)$</p>
<p>$\displaystyle\iff\frac x2=m\pi-\frac\pi4\iff x=2m\pi-\frac\pi2$ where $m$ is any integer</p>
<p><strong>Method</strong> $\#2$</p>
<p>Let $\displaystyle1=r\cos\phi,-1=r\sin\phi\ \ \ \ (1)$ where $r>0$</p>
<p>$\displaystyle\cos\phi=\frac1r>0$ and $\displaystyle\sin\phi=-\frac1r<0$</p>
<p>$\displaystyle\implies\phi$ lies in the fourth Quadrant</p>
<p>On division, $\displaystyle\frac{r\sin\phi}{r\cos\phi}=-1\iff\tan\phi=-1$
$\displaystyle\implies\phi=-\frac\pi4$</p>
<p>$\displaystyle\sin x-\cos x=-1\implies r\cos\phi\sin x+r\sin\phi\cos x=r\sin\phi$</p>
<p>$\displaystyle\implies\sin(x+\phi)=\sin\phi$</p>
<p>$\displaystyle\implies x+\phi=r\pi+(-1)^r\phi$ where $r$ is any integer</p>
<p>If $r$ is even $=2a$(say) $\displaystyle\implies x=2a\pi$</p>
<p>If $r$ is odd $=2a+1$(say) $\displaystyle\implies x+\phi=(2a+1)\pi-\phi\iff x=2a\pi+\pi-2\phi$</p>
<p><strong>Method</strong> $\#3$</p>
<p>Using <a href="http://www.cut-the-knot.org/arithmetic/algebra/WeierstrassSubstitution.shtml" rel="nofollow">Weierstrass Substitution</a>,</p>
<p>$\displaystyle\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}=-1\ \ \ \ (2)$ where $u=\tan\frac x2$</p>
<p>$\displaystyle\iff2u^2+2u=0\iff u(u+1)=0$</p>
<p>If $\displaystyle u=0,\tan\frac x2=0\iff\frac x2=b\pi$ where $b$ is any integer</p>
<p>If $\displaystyle u=-1,\tan\frac x2=-1$ which has been addressed in <strong>Method</strong> $\#1$ </p>
|
2,380,880 | <p>The following question was asked in an exam:<br>
Consider the problem:<br>
Maximize $2y_1+3y_2+5y_3+4y_4$<br>
subject to<br>
$y_1+y_2\leq 1,$ $y_2+y_3\leq 1,$<br>
$y_4+y_1\leq 1,$ $y_3+y_4\leq 1$ and $y_i\geq 0$ for i=1,2,3,4.<br>
Then the optimum value is<br>
1. equal to 8<br>
2. between 8 and 9<br>
3. greater than or equal to 7<br>
4. less than or equal to 7<br>
I solved the above problem by the usual simplex method and got the optimal value to be 7. Now, I am just curious to know if there's any other simpler and faster method to solve such problems in competitive exams where time is a constraint.<br>
Thank you in advance!</p>
| Michael Hartley | 96,763 | <p>You could let $z_1=y_1+y_2$, $z_2=y_2+y_3$ and $z_3=y_3+y_4$. Then the problem becomes:</p>
<p>Minimise $2z_1+z_2+4z_3$ subject to $z_1\leq 1$, $z_2\leq 1$, $z_3\leq 1$, $z_1-z_2+z_3\leq 1$, </p>
<p>Then, by inspection $z_1=z_2=z_3=1$ satisfies the constraints and attains the maximum. </p>
|
198,298 | <p>Is there a tool for editing the spelling dictionary? Now that <em>Mathematica</em> 12 does spellchecking on the fly, I've been using it. And of course, as I was wondering how I might delete a word accidentally added to the dictionary, I accidentally added a misspelled word to the dictionary. I'm sure I can hunt it down and edit it by hand, but it seems safer to use a proper tool if one exists. Is there one? </p>
| Alexey Popkov | 280 | <p><a href="https://mathematica.stackexchange.com/a/198308/280">Answer by ciao</a> shows a way to edit user-supplied list of correct words via GUI. The same can be done via setting <code>$FrontEnd</code> option:</p>
<pre><code>Options[$FrontEnd, SpellingDictionaries]
</code></pre>
<blockquote>
<pre><code>{SpellingDictionaries -> {"CorrectWords" -> {}, "IncorrectWords" -> {}, "Suggestions" -> {}}}
</code></pre>
</blockquote>
<p>The meaning of the sub-options is <a href="http://reference.wolfram.com/language/ref/SpellingDictionaries.html" rel="nofollow noreferrer">documented</a>:</p>
<blockquote>
<ul>
<li><p><code>"CorrectWords"->{"word1",""word2"",...}</code> means that if any word in the list is found during a spell check, it is skipped even if it does
not exist in the main dictionary.</p></li>
<li><p><code>"IncorrectWords"->{"word1",""word2"",...}</code> means that if any word in the list is found, the spell check stops, even if the word exists
in the main dictionary.</p></li>
<li><p><code>"Suggestions"->{{"word1"->"newword1"},...}</code> means that if any word in the list is found, the spell check stops and the corresponding new
word is offered as an alternative in the <strong>Check Spelling</strong> dialog
box.</p></li>
</ul>
</blockquote>
<p>Note that there is also the <code>"IncorrectWords"</code> option with rather strange <em>documented</em> meaning: "... if any word in the list is found, the spell check <strong>stops</strong>...". I would assume that the spell checking doesn't <em>stop</em> at such word but rather marks it as misspelled.</p>
<hr>
<blockquote>
<p>Can there be other hidden locations for accidental additions? Or does Mma actually recognize Euclidian (vs Euclidean) as a correct word, which just happened in my v.12.</p>
</blockquote>
<p>Searching through all textual files in the directory</p>
<pre><code>FileNameJoin[{$InstallationDirectory, "SystemFiles", "Components", "SpellingData", "SpellingDictionaries"}]
</code></pre>
<p>reveals that the whole word "Euclidian" is indeed present in the files "en_Supplemental.dic" and "ro_RO.dic" located in that directory. So it looks like the word "Euclidian" is indeed considered as a correct word by <em>Mathematica</em>'s spellchecker because it is included in one of the basic dictionaries for English. We can try to add this word to the list of incorrect words via the <code>"IncorrectWords"</code> option:</p>
<pre><code>SetOptions[$FrontEnd, SpellingDictionaries -> {"IncorrectWords" -> {"Euclidian"}}]
</code></pre>
<p>Unfortunately it doesn't have any effect: even after restarting <em>Mathematica</em> this word isn't marked as misspelled. I would consider this as a bug.</p>
<p>P.S. It is also interesting that this word is present in the file </p>
<pre><code>FileNameJoin[{$InstallationDirectory, "SystemFiles", "Components", "OpenPHACTS", "Kernel", "OpenPHACTS.m"}]
</code></pre>
<hr>
<p><strong>UPDATE1</strong>. I've moved the file "en_Supplemental.dic" outside of the <code>$InstallationDirectory</code> and then restarted <em>Mathematica</em>. After the restart the word "euclidian" is still considered as correctly spelled. So there seems to be some other place from where the spellchecker takes it...</p>
<p><strong>UPDATE2</strong>. The word "Euclidian" is so well-protected that even <a href="https://mathematica.stackexchange.com/a/123510/280">setting the language of spell checking</a> to Russian doesn't affect it: now all usual English words are marked as misspelled including the word "Euclidean" but <em>excepting</em> the word "Euclidian":</p>
<p><a href="https://i.stack.imgur.com/LqEfR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LqEfR.png" alt="screenshot"></a></p>
<p><strong>UPDATE3</strong>. Searching for the word "euclidian" in the all files in the <code>$InstallationDirectory</code> (including binary files) reveals that it is present in the following files:</p>
<pre><code>FileNameJoin[{<span class="math-container">$InstallationDirectory,"Documentation","English","Index","_e.cfs"}]
FileNameJoin[{$InstallationDirectory,"Documentation","English","SearchIndex","2","_1.cfs"}]
FileNameJoin[{$</span>InstallationDirectory,"Documentation","English","SpellIndex","_19s.cfs"}]
</code></pre>
<p>Especially suspicious is the file in the directory "SpellIndex". But even moving this directory outside of <code>$InstallationDirectory</code> and restarting <em>Mathematica</em> doesn't change the described behavior... </p>
<p><strong>UPDATE4</strong>. The <code>Entity</code> framework considers the words "euclidean" and "euclidian" as synonyms:</p>
<pre><code>Entity["Word", "euclidian"]@EntityProperty["Word", "SynonymsList"]
Entity["Word", "euclidean"]@EntityProperty["Word", "SynonymsList"]
</code></pre>
<blockquote>
<pre><code>{"euclidean"}
{"euclidian"}
</code></pre>
</blockquote>
|
387,365 | <p>I have a physics project, and I have to develop an argument, but am not allowed to use phrases like "From the graph you can tell..."</p>
<p>How can the nature of the graph be determined manually, e.g. finding that it is expontential rather than quadratic?</p>
| Dale M | 55,635 | <p>I assume by disjoint squares you mean squares made with 2 cuts, each parallel to an axis and that the leftover rectangles are thrown away.</p>
<p>A PDF uniform on $x$ when $y=y_0$ and 0 otherwise would give Alice nothing.</p>
<p>In light of @RossMillikan comment</p>
<p>$$P(x,y)=\begin{cases}
\frac{1}{5}&[0,0],[1,0],[0,1],[1,1],[\frac{1}{4},\frac{1}{4}]\\
0&\text{otherwise}
\end{cases}$$</p>
|
387,365 | <p>I have a physics project, and I have to develop an argument, but am not allowed to use phrases like "From the graph you can tell..."</p>
<p>How can the nature of the graph be determined manually, e.g. finding that it is expontential rather than quadratic?</p>
| Erel Segal-Halevi | 29,780 | <p>I think Alice can always assure herself at least $1 \over 4$ cdf, in the following way.</p>
<p>First, in each of the 4 corners, mark a square that contains $1 \over 4$ cdf. Since the pdf is finite, it is always possible to construct such a square, by starting from the corner and increasing the square gradually, until it contains exactly $1 \over 4$ cdf.</p>
<p>There is at least one corner, in which the side length of such a square will be at most $1 \over 2$ . Suppose this is the lower-left corner, and the side length is a, so square #1 is [0,a]x[0,a], with a <= $1 \over 2$.</p>
<p>Now, consider the following 3 squares:</p>
<ul>
<li>To the right of square #1: [a,1]x[0,1-a]</li>
<li>On top of square #1: [0,1-a]x[a,1]</li>
<li>On the top-right of square #1: [a,1]x[a,1]</li>
</ul>
<p>The union of these squares covers the entire remainder after we remove square #1. This remainder contains $3 \over 4$ cdf. So, the sum of cdf in all 3 squares is <em>at least</em> $3 \over 4$ (probably more, because the squares overlap).</p>
<p>Among those 3, select the one with the greatest cdf. It must contain at least $1 \over 3$ of $3 \over 4$, i.e., at least $1 \over 4$. This is square #2.</p>
<p>So, Alice can always cut two squares that contain at least $1 \over 4$ cdf.</p>
<p>Note that this procedure relies on the fact (that I mentioned in the original question) that the pdf is finite. Otherwise, it may not always be possible to construct a square with $1 \over 4$ cdf.</p>
|
2,783,129 | <p>How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?</p>
<p>The solution shows that one of these is true:</p>
<p>$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$ </p>
<p>$\cos x = -1$ and thus $x = 180^\circ$</p>
<p><strong>Question:</strong> Inserting the $\arcsin$ of $1/2$ will yield to $30°$, how do I get $120^\circ$? and what is that $120^\circ$, why is there $2$ value but when you substitute $\frac12$ as $x$, you'll only get $1$ value which is the $30^\circ$?</p>
<p>Also, when I do it inversely: $\sin(30^\circ)$ will result to 1/2 which is true as $\arcsin$ of $1/2$ is $30^\circ$. But when you do $\sin(120^\circ)$, it will be $\frac{\sqrt{3}}{2}$, and when you calculate the $\arcsin$ of $\frac{\sqrt{3}}{2}$, it will result to $60^\circ$ and not $120^\circ$. Why?</p>
| Christopher Marley | 510,133 | <p>If $ab = 0$, then either $a=0$ or $b=0$.<br>
Given $(2\sin x - 1)(\cos x + 1) = 0$, we can separate these and solve separately.</p>
<p>So... $\ \ \ \ \ 2\sin x -1 = 0$<br>
$\Rightarrow \sin x = \frac12 \Rightarrow x = 30^\circ, 150^\circ$<br>
We get those two values because $\sin x$ is symmetrical across the $y$-axis.</p>
<p>Also... $\ \ \ \ \ \cos x + 1 = 0$<br>
$\Rightarrow \cos x = -1 \Rightarrow x = 180^\circ$</p>
|
1,217,771 | <p>Let $x,y \in R$.
If $0 \leq y < x$ for all $x > 0$, then $y=0$.</p>
<p>Proof by contradiction: </p>
<p>Assume the opposite that is; "If $0 \leq y < x$ for all $x > 0$, then $y\neq0$".
Subtract $x$ from each part of the inequality to get,
$0-x \leq y-x < 0$
Then multiply through by -1 to get,
$x \geq y+ x > 0$
Since $x>0$, this implies a contradiction of the original statement, therefore we conclude that if $0 < y < x$ for all $x > 0$, then $y=0$.
Is my reasoning correct or is there something I can improve upon?</p>
| Olivier Oloa | 118,798 | <p>You may just observe that, as $x\to \infty$:
$$
\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\sim \frac{\sqrt{x}\left(1+\frac1{\sqrt{x}}\right)}{\sqrt{x}\left(\sqrt{1+\frac{1}{\sqrt{x}}}\right)+\sqrt{x}\left(\sqrt{1-\frac{1}{x}}\right)}\sim \frac12.
$$</p>
|
1,217,771 | <p>Let $x,y \in R$.
If $0 \leq y < x$ for all $x > 0$, then $y=0$.</p>
<p>Proof by contradiction: </p>
<p>Assume the opposite that is; "If $0 \leq y < x$ for all $x > 0$, then $y\neq0$".
Subtract $x$ from each part of the inequality to get,
$0-x \leq y-x < 0$
Then multiply through by -1 to get,
$x \geq y+ x > 0$
Since $x>0$, this implies a contradiction of the original statement, therefore we conclude that if $0 < y < x$ for all $x > 0$, then $y=0$.
Is my reasoning correct or is there something I can improve upon?</p>
| davidlowryduda | 9,754 | <p>Let's start from your last line:
$$\begin{align}
\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} &= \lim \frac{\sqrt x}{\sqrt x} \frac{\frac{1}{\sqrt x} + 1}{\sqrt{1 + \frac{1}{\sqrt x}} + \sqrt{1 - \frac{1}{ x}}} \\
&= \frac{1}{1 + 1} = \frac{1}{2}
\end{align}$$
where we note that everywhere we have $\frac{1}{\sqrt x}$, those terms go to $0$ as $x \to \infty$. The method of factoring out the largest element in the numerator and denominator very often works. $\diamondsuit$</p>
|
138,866 | <p>I have data in a csv file. The first row has labels, and the first column, too.</p>
<pre><code>Datos = Import["C:\\Users\\jodom\\Desktop\\Data.csv"]
</code></pre>
<p>Tha data in the csv file is that:</p>
<pre><code>{{"No", "Vol", "Vel"}, {1, 500, 45}, {2, 700, 67}, {3, 350, 87}, {4,
123, 23}, {5, 587, 45}, {6, 435, 89}, {7, 896, 65}, {8, 125,
45}, {9, 476, 27}, {10, 987, 80}}
</code></pre>
<p>I put those csv data into a dataset:</p>
<pre><code>B = Dataset[Datos]
</code></pre>
<p>You can check it out as an image here,on how it has seen on wolfram after the import:
<a href="https://drive.google.com/file/d/0B56r_V66BiodQUhUMWNHcHZFOWc/view?usp=sharing" rel="noreferrer">https://drive.google.com/file/d/0B56r_V66BiodQUhUMWNHcHZFOWc/view?usp=sharing</a></p>
<p>Now I want to convert the first row that has the labels, into a head or label of the dataset, and the first column into a label column, so I can get data from this dataset, like </p>
<pre><code>Dataset[labelrow, labelcolumn]
</code></pre>
| Ali Hashmi | 27,331 | <pre><code>data = {{"No", "Vol", "Vel"}, {1, 500, 45}, {2, 700, 67}, {3, 350,
87}, {4, 123, 23}, {5, 587, 45}, {6, 435, 89}, {7, 896, 65}, {8, 125, 45},
{9, 476, 27}, {10, 987, 80}}
dataset = Dataset[Association@MapThread[(#1 -> <|"Vol" -> #2, "Vel" -> #3 |> &),
Transpose@Rest@data]]
dataset[1, "Vol"]
(* 500 *)
dataset[1,"Vel"]
(* 45 *)
</code></pre>
|
4,231,580 | <p>I have been working on th integral
<span class="math-container">$$\int_0^\infty \frac{\sin x}{1+x^2} dx$$</span>
for a short while now trying substitutions and even Laplace transforms and other stuff but gave up. I looked to Wolfram Alpha but how did it get the series expansion of the constant that it did (see <a href="https://www.wolframalpha.com/input/?i=series+integral+of+sin%28x%29%2F%28x%5E2%2B1%29+from+0+to+inf" rel="nofollow noreferrer">here</a>)? I tried doing a little algebra with the expression from Wolfram Alpha it looks like an integral and there are more things that could be done but I don't know. Thanks.</p>
| robjohn | 13,854 | <p><span class="math-container">$$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}\newcommand{\Ei}{\operatorname{Ei}}\newcommand{\PV}{\operatorname{PV}}
\begin{align}
\int_0^\infty\frac{\sin(x)}{1+x^2}\,\mathrm{d}x
&=\Im\left(\int_0^\infty\frac{e^{ix}}{1+x^2}\,\mathrm{d}x\right)\tag1\\
&=\frac12\Im\left(\int_0^\infty e^{ix}\left(\frac1{1-ix}+\frac1{1+ix}\right)\mathrm{d}x\right)\tag2\\
&=\frac12\Im\left(\int_0^{i\infty}e^{ix}\left(\frac1{1-ix}+\frac1{1+ix}\right)\mathrm{d}x\right)\tag3\\
&=\frac12\Re\left(\int_0^\infty e^{-x}\left(\frac1{x+1}-\frac1{x-1}\right)\mathrm{d}x\right)\tag4\\
&=\frac e2\int_1^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\frac1{2e}\PV\!\!\int_{-1}^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\tag5\\
&=\frac1{2e}\left(\Ei(1)-e^2\Ei(-1)\right)\tag6
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$(1)$</span>: <span class="math-container">$\sin(x)=\Im\left(e^{ix}\right)$</span><br />
<span class="math-container">$(2)$</span>: partial fractions<br />
<span class="math-container">$(3)$</span>: move contour to the positive imaginary axis<br />
<span class="math-container">$\phantom{\text{(3):}}$</span> with a small counterclockwise bump around <span class="math-container">$i$</span><br />
<span class="math-container">$(4)$</span>: substitute <span class="math-container">$x\mapsto ix$</span> using <span class="math-container">$\Im(iz)=\Re(z)$</span><br />
<span class="math-container">$(5)$</span>: the small counterclockwise bump around <span class="math-container">$1$</span><br />
<span class="math-container">$\phantom{\text{(5):}}$</span> gets turned into <span class="math-container">$\PV$</span> by <span class="math-container">$\Re$</span><br />
<span class="math-container">$(6)$</span>: apply <span class="math-container">$\Ei(z)=-\PV\!\int_{-z}^\infty\frac{e^{-x}}{x}\,\mathrm{d}x$</span></p>
|
196,002 | <p>The paper <a href="http://www.sciencedirect.com/science/article/pii/S002212369690110X" rel="nofollow noreferrer">Lattices of Intermediate Subfactors</a> of Y. Watatani, received on December 1994, finishes by: </p>
<blockquote>
<p><strong>Prop. 6.2.</strong> $ \ $ Any finite lattice with at most five elements can be
realized as an intermediate subfactor lattice.</p>
</blockquote>
<p>In fact he has investigated all the lattices with at most six elements, and they can be realized as an intermediate subfactor lattice, except the following two lattices for which he didn't know:</p>
<p><img src="https://i.stack.imgur.com/RpIjp.png" alt="enter image description here"> </p>
<p><strong>Question</strong>: Can any finite lattice with at most six elements be realized as an intermediate subfactor lattice?<br>
In others words: Can $L_{19}$ and $L_{20}$ be realized as intermediate subfactor lattices?</p>
<p>Today is 20 years after this paper of Y. Watatani, and perhaps subfactors realizing these lattices has been found or perhaps we now know how to prove they don't exist.<br>
Of course, if they exist we should ask the same question for seven elements, eight elements... and finally:<br>
Can any finite lattice be realized as an intermediate subfactor lattice?<br>
We've sketched a planar algebraic approach for this question in the optional part <a href="https://mathoverflow.net/q/195806/34538">here</a>, but we don't know if the skein theory is practicable or not in these cases. </p>
| Sergei | 49,208 | <p>Some results for differences $A_n-G_n, G_n-H_n$ and so after summing up for $A_n-H_n$ you may find in the book: Classical and new inequalities in analysis by D. S. Mitrinovic; J. E. Pecaric; A. M. Fink on pages 25,39. </p>
|
163,153 | <p>How do i find the absolute maximum and absolute minimum values of f on this given interval.</p>
<p>$f(x) = 6x^3 − 9x^2 − 36x + 7, \
[−2, 3]$</p>
| Roman Chokler | 38,328 | <p>I am not sure on how to prove this.
I myself have heard about other properties on periodicity.
A particularly interisting one is as follows:
Label a each primary bulb with a fraction $\frac pq$ where the bulb has period $q$ and the $p$th spoke the going counterclockwise from the main spoke is the biggest spoke of the antenna. Note, that $(p,q)=1$ for every primary bulb.
Then the largest primary bulb between primary bulbs $\frac ab$ and $\frac cd$ is labeled with $\frac {a+c}{b+d}$ reduced to lowest terms.</p>
|
1,990,638 | <p>This is a purely affine variety. Define $F_1=y-x^2, F_2=z-xy, F_3=xz-y^2$. </p>
<p>It is clear that $(x,x^2,x^3)$ is a solution to $F_1=F_2=F_3=0$. Since projective twisted cubic is not intersection of two quadractics which contains an extra line, I would not expect this to be the case in affine as well. However, $F_1\cap F_2$ seems to be exactly twisted cubic curve and I could not see any extra information on the other line. The other two intersections(i.e. $F_2\cap F_3$ and $F_1\cap F_3$ do yield two extra lines.</p>
<p>What have I done wrong here?</p>
| Ted | 15,012 | <blockquote>
<p>Since projective twisted cubic is not intersection of two quadratics
which contains an extra line, I would not expect this to be the case in
affine as well.</p>
</blockquote>
<p>This is not correct. When you go from projective to the affine space, you remove some points. Claims like "projective twisted cubic is not intersection of two quadratics" only hold in projective space, which is complete.</p>
<p>If this is not clear, consider even more basic statements like "in the projective plane, two distinct lines intersect in exactly one point." If you go to the affine plane, this isn't true.</p>
|
313,254 | <p>I know this question has been asked before on MO and MSE (<a href="https://mathoverflow.net/questions/59605/reference-in-riemann-surfaces">here</a>, <a href="https://math.stackexchange.com/questions/407004/good-book-for-riemann-surfaces">here</a>, <a href="https://math.stackexchange.com/questions/1839673/books-on-riemann-surfaces">here</a>, <a href="https://math.stackexchange.com/questions/200537/complex-analysis-book-with-a-view-toward-riemann-surfaces">here</a>) but the answers that were given were only partially helpful to me, and I suspect that I am not the only one.</p>
<p>I am about to teach a first course on Riemann surfaces, and I am trying to get a fairly comprehensive view of the main references, as a support for both myself and students.</p>
<p>I compiled a list, here goes in alphabetical order. Of course, it is necessarily subjective. For more detailed entries, I made a bibliography using the bibtex entries from MathSciNet: <a href="https://www.brice.loustau.eu/teaching/RiemannSurfaces2018/References.pdf" rel="noreferrer">click here</a>.</p>
<ol>
<li>Bobenko. Introduction to compact Riemann surfaces. </li>
<li>Bost. Introduction to compact Riemann surfaces, Jacobians, and abelian varieties.</li>
<li>de Saint-Gervais. Uniformisation des surfaces de Riemann: retour sur un théorème centenaire.</li>
<li>Donaldson. Riemann surfaces.</li>
<li>Farkas and Kra. Riemann surfaces.</li>
<li>Forster. Lectures on Riemann surfaces.</li>
<li>Griffiths. Introduction to algebraic curves.</li>
<li>Gunning. Lectures on Riemann surfaces.</li>
<li>Jost. Compact Riemann surfaces.</li>
<li>Kirwan. Complex algebraic curves.</li>
<li>McMullen. Complex analysis on Riemann surfaces.</li>
<li>McMullen. Riemann surfaces, dynamics and geometry.</li>
<li>Miranda. Algebraic curves and Riemann surfaces.</li>
<li>Narasimhan. Compact Riemann surfaces.</li>
<li>Narasimhan and Nievergelt. Complex analysis in one variable.</li>
<li>Reyssat. Quelques aspects des surfaces de Riemann.</li>
<li>Springer. Introduction to Riemann surfaces.</li>
<li>Varolin. Riemann surfaces by way of complex analytic geometry.</li>
<li>Weyl. The concept of a Riemann surface.</li>
</ol>
<p>Having a good sense of what each of these books does, beyond a superficial first impression, is quite a colossal task (at least for me).</p>
<p>What I'm hoping is that if you know very well such or such reference in the list, you can give a short description of it: where it stands in the existing literature, what approach/viewpoint is adopted, what are its benefits and pitfalls. Of course, I am also happy to update the list with new references, especially if I missed some major ones.</p>
<p>As an example, for Forster's book (5.) I can just use the accepted answer <a href="https://math.stackexchange.com/questions/407004/good-book-for-riemann-surfaces">there</a>: According to <a href="https://math.stackexchange.com/users/71348/ted-shifrin">Ted Shifrin</a>:</p>
<blockquote>
<p>It is extremely well-written, but definitely more analytic in flavor.
In particular, it includes pretty much all the analysis to prove
finite-dimensionality of sheaf cohomology on a compact Riemann
surface. It also deals quite a bit with non-compact Riemann surfaces,
but does include standard material on Abel's Theorem, the Abel-Jacobi
map, etc.</p>
</blockquote>
| Chris Judge | 7,120 | <p>For a hyperbolic perspective: Buser's "geometry and spectrum of compact riemann surfaces" is a nice book. </p>
|
234,668 | <p>I have an equation in $x$ and I would like to determine if it has any solutions modulo a large prime $p$. Suppose $p$ is large enough that I can factor numbers up to $p$, but I cannot test all values up to $p$. (Actually, so far, I have been doing just that -- but I'd like to avoid this as it takes a long time. If you can avoid factoring, all the better.)</p>
<p>The particular equation I have is
$$
x^4-x^2\equiv4\pmod p
$$
but I would be interested in</p>
<ol>
<li>Solutions to this particular problem, or more generally</li>
<li>Solutions to other quadratics$\pmod p$ in $x^2$, or more generally</li>
<li>Solutions to quartics$\pmod p$.</li>
</ol>
<p>I'm familiar with quadratic reciprocity but not with cubic or biquadratic. (It's not clear to me if this can be transformed so they can be used; if so, demonstrating the transformation and giving a pointer to a good source on higher reciprocity would suffice as an answer.)</p>
| Cameron Buie | 28,900 | <p>This is false, as written. If $k$ is even, then so is $kn$, regardless of whether $n$ is odd or not.</p>
<p>Now, suppose that $k,n$ are <em>both</em> odd. Then $k=2j+1$ and $n=2m+1$ for some integers $j,m$. Therefore $kn=2(\text{stuff})+1$. (I leave the determination of "stuff" to you, though I will tell you it's an integer.)</p>
|
854,079 | <blockquote>
<p>Let $|A|=\mathcal c, \ |B|=\aleph_0, \ A\cap B=\emptyset,$ Prove that $ |A\cup B|=\mathcal c$</p>
</blockquote>
<p>So $|A\cup B|=|A|+|B|$ but this just leads to cardinal arithmetic which I don't think is the right approach. </p>
<p>Maybe embedding could work ? There has to be a 1-1 function from $A$ to $(0,1)$ and from $B$ to the naturals, so $A\cup B = \{x\in [0,1], y | x\in \mathbb R, y\in \mathbb N\}$ and now it's enough to know that for all $x$ the cardinality of this set is $\mathcal c$.</p>
| martini | 15,379 | <p>I will suppose $a_n \ge 0$, as otherwise the statement is wrong.</p>
<p>By induction, we have $a_n < k^na_0$, hence $0 \le a_n \le k^n a_0$. As $0 \to 0$ and $k^n a_0 \to 0$ (we have $k \in (0,1)$), by the squeeze theorem $a_n \to 0$.</p>
|
854,079 | <blockquote>
<p>Let $|A|=\mathcal c, \ |B|=\aleph_0, \ A\cap B=\emptyset,$ Prove that $ |A\cup B|=\mathcal c$</p>
</blockquote>
<p>So $|A\cup B|=|A|+|B|$ but this just leads to cardinal arithmetic which I don't think is the right approach. </p>
<p>Maybe embedding could work ? There has to be a 1-1 function from $A$ to $(0,1)$ and from $B$ to the naturals, so $A\cup B = \{x\in [0,1], y | x\in \mathbb R, y\in \mathbb N\}$ and now it's enough to know that for all $x$ the cardinality of this set is $\mathcal c$.</p>
| user157745 | 157,745 | <p>we have a theorem if a sequence is bounded and descend this sequence ins convergence and this condition is true for your question .and i think we have $a_n\geq 0 $</p>
|
4,116,732 | <p>If we have a topological category and the underlying category forgetting the topological structure, are the nerves same. They should be, is what my guess is from the definition of nerve of a category. Then, I have some facts which leads to contradiction (which should not be, thus I am missing something). The facts are below:</p>
<p>(For a category C, classifying space of C is the realization of the nerve of C.)</p>
<ol>
<li><p>If G is a group consider it a category with the only object is * and set of morphisms is G. Then the nerve of this category has realization K(G,1).</p>
</li>
<li><p>If G is a topological group, then the classifying space (for Principal G-bundle) is not always (weakly-)equivalent to K(G,1). For example, take unit circle. They are same (upto weak-equivalence) when G has discrete topology.</p>
</li>
<li><p>In the paper of Graeme Segal (<a href="https://www.maths.ed.ac.uk/%7Ev1ranick/papers/segalclass.pdf" rel="nofollow noreferrer">https://www.maths.ed.ac.uk/~v1ranick/papers/segalclass.pdf</a>) his model of classifying space (in the sense of principal G-bundles) is in terms of realization of the nerve if the category in 1 (assuming say G is a locally finite CW-complex).</p>
<p>Now, any two classifying spaces (in the sense of Principal G-bundle) are weakly equivalent, then 2 gives contradiction to 1 and 3 . I know I am mistaken at some stage. Any help is welcome.</p>
</li>
</ol>
| Connor Malin | 574,354 | <p>No, these two nerves will not be the same. For example, let <span class="math-container">$G$</span> be a topological group. We can form its classifying category <span class="math-container">$BG$</span>, and this is a topological category. If <span class="math-container">$G^\partial$</span> denotes the discrete group <span class="math-container">$G$</span>, because we have a map <span class="math-container">$G^\partial \rightarrow G$</span> we get a map of classifying categories <span class="math-container">$BG^\partial \rightarrow BG$</span>. This is a specific example of your comparison. This will almost never induce an equivalence on nerves.</p>
<p>Concretely, let <span class="math-container">$G=\mathbb{R}$</span>. Then since the nerve of <span class="math-container">$BG$</span> is a model for the classifying space of <span class="math-container">$\mathbb{R}$</span>, we see that it is contractible since all principal <span class="math-container">$\mathbb{R}$</span> bundles are trivial (alternatively you can see it has trivial homotopy groups).</p>
<p>However, <span class="math-container">$G^\partial = \mathbb{R}^\partial$</span> has <span class="math-container">$BG^\partial$</span> very far from trivial. It's fundamental group is uncountable!</p>
|
2,682,093 | <blockquote>
<p>Let $\circ$ be an inequality.</p>
<p>Prove $|x| \circ a \equiv -a \circ x \circ a$.</p>
</blockquote>
<p>If $x$ is positive, then $|x| \circ a = x \circ a$.</p>
<p>If $x$ is negative, then $|x| \circ a = x \circ a$ ?</p>
| farruhota | 425,072 | <p>Note $\circ$ can be either $<$ or $>$.</p>
<p>If $x>0$, then $|x|\circ a \equiv x\circ a$.</p>
<p>If $x<0$, then $ |x|\circ a \equiv -x\circ a$.</p>
|
2,741,511 | <p><a href="https://i.stack.imgur.com/eh6nF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eh6nF.png" alt="enter image description here"></a></p>
<ul>
<li>How long does it take this object to arrive at the point $M$? Assume that $|KL| =|LM|$. </li>
</ul>
<p><strong>Options:</strong> </p>
<p>a) $\frac{3}{2}\sqrt \frac{h}{g\sin^2\theta}$ b) $\frac{3}{2}\sqrt \frac{2h}{g\sin^2\theta}$ c) $\frac{3}{2}\sqrt \frac{2h}{g\cos^2\theta}$ d) $\frac{3}{2}\sqrt {2h} \sin \theta$ e) $\frac{3}{2}\sqrt {2h} \cos^2 \theta$ </p>
<p>I'm out of my mind right now. It seems very complex to me, doesn't it? In fact, I've to be familiar with free-body diagram to solve this question easily. Can I take your professional helps? </p>
<p>Regards!</p>
| Takahiro Waki | 268,226 | <p>Use <a href="http://www.mathsmutt.co.uk/files/newt.htm" rel="nofollow noreferrer">Newton's equations of motion</a></p>
<p>$mg\sinθ=ma$ </p>
<p>$⇒a=g\sinθ$</p>
<p>$v^2=2g\sinθ*h/\sinθ$ (velocity on point L).</p>
<p>$v=g\sinθt_1$</p>
<p>$t_1=\dfrac{\sqrt{2gh}}{g\sinθ}$</p>
<p>$t=t_1+t_2=\dfrac{\sqrt{2gh}}{g\sinθ}+\dfrac{h}{\sqrt{2gh}\sinθ}
=\dfrac{3\sqrt h}{\sqrt{2g}\sinθ}$</p>
<p>this solution holds on difficult situation which square-like object continue to slide with vertex from point L.</p>
|
2,741,511 | <p><a href="https://i.stack.imgur.com/eh6nF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eh6nF.png" alt="enter image description here"></a></p>
<ul>
<li>How long does it take this object to arrive at the point $M$? Assume that $|KL| =|LM|$. </li>
</ul>
<p><strong>Options:</strong> </p>
<p>a) $\frac{3}{2}\sqrt \frac{h}{g\sin^2\theta}$ b) $\frac{3}{2}\sqrt \frac{2h}{g\sin^2\theta}$ c) $\frac{3}{2}\sqrt \frac{2h}{g\cos^2\theta}$ d) $\frac{3}{2}\sqrt {2h} \sin \theta$ e) $\frac{3}{2}\sqrt {2h} \cos^2 \theta$ </p>
<p>I'm out of my mind right now. It seems very complex to me, doesn't it? In fact, I've to be familiar with free-body diagram to solve this question easily. Can I take your professional helps? </p>
<p>Regards!</p>
| Peter Szilas | 408,605 | <p>1)$ |KL|:=s.$</p>
<p>Note : $h/s= \sin \theta$; or </p>
<p>$s= h/(\sin \theta)$.</p>
<p>Constant $a=g \sin\theta$ down the slope.</p>
<p>A) Final speed at $L$:</p>
<p>$v_f = at$ , where $t$ is the time taken down the slope, and</p>
<p>$s= (1/2)at^2$, or</p>
<p>$t_1:= t =((2s)/a)^{1/2}$; and</p>
<p>$v_f = a((2s)/a)^{1/2}$.</p>
<p>2) Constant speed $v_f$ along $|LM|=s$ ;</p>
<p>$t_2:= t= s/v_f.$</p>
<p>3) Total time taken from $K$ to $L$:</p>
<p>$T= t_1+t_2$.</p>
|
701,176 | <p>A function $f$ is differentiable over its domain and has the following properties:</p>
<ol>
<li><p>$\displaystyle f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$</p></li>
<li><p>$\lim_{h \to 0} f(h) = 0$</p></li>
<li><p>$\lim_{h \to 0} f(h)/h = 1$</p></li>
</ol>
<p>i) Show that $f(0)=0$</p>
<p>ii) show that $f'(x)=1+[f(x)]^2$ by using the def of derivatives Show how the above properties are involved.</p>
<p>iii) find $f(x)$ by finding the antiderivative. Use the boundary condition from part (i).</p>
<hr>
<p>So basically I think I found out how to do part 1 because if $x+y=0$ then the top part of the fraction will always have to be zero.</p>
<p>part 2 and 3 are giving me trouble. The definition is the limit $(f(x+h)-f(x))/h$</p>
<p>So I can set $x+y=h$ and make the numerator equal to $f(h)$?</p>
<p>Thanks for all who help</p>
| Barry Cipra | 86,747 | <p>All you need to do is show that the equation $a^3-6a^2-2=0$ has no positive integer solution. If it did, then you'd have $a^2(a-6)=2$. But $a^2\mid2$ implies $a=1$, in which case $a^2(a-6)=-5\not=2$.</p>
|
235,319 | <p>The problem states: Suppose $f'(b) = M$ and $M <0$. Find $\delta>0$ so that if $x\in (b-\delta, b)$, then $f(x) > f(b).$</p>
<p>This intuitively makes sense, but I am not exactly sure how to find $\delta$. I greatly appreciate any help I can receive. </p>
| Robert Israel | 8,508 | <p>You could use dynamical programming. Let $a(m,n)$ be the number of $m$-digit nonnegative integers with each digit 0 to 7 and the sum of digits is $n$. Then $a(1,n) = 1$ for $0 \le n \le 7$, $0$ otherwise, and
$$a(m+1,n) = \sum_{d=0}^{\min(n,7)} a(m,n-d)$$</p>
<p>EDIT: That solution allows leading zeros. If leading zeros are not allowed, take $a(1,0) = 0$ as well.</p>
<p>EDIT: Let $a(m,n)$ be as above in the version with leading zeros not allowed, and $b(m,n)$ the version with leading zeros allowed. Then $a(6,20) = \sum_{k=0}^{20} a(3,k) b(3,20-k)$.
For $m = 1,2,3$ we make the following tables:</p>
<pre><code> n a(m,n) b(m,n)
m=1 2 3 1 2 3
0 0 0 0 1 1 1
1 1 1 1 1 2 3
2 1 2 3 1 3 6
3 1 3 6 1 4 10
4 1 4 10 1 5 15
5 1 5 15 1 6 21
6 1 6 21 1 7 28
7 1 7 28 1 8 36
8 0 7 35 0 7 42
9 0 6 40 0 6 46
10 0 5 43 0 5 48
11 0 4 44 0 4 48
12 0 3 43 0 3 46
13 0 2 40 0 2 42
14 0 1 35 0 1 36
15 0 0 28 0 0 28
16 0 0 21 0 0 21
17 0 0 15 0 0 15
18 0 0 10 0 0 10
19 0 0 6 0 0 6
20 0 0 3 0 0 3
</code></pre>
<p>Then (going down the $a(3,n)$ column and up the $b(3,n)$ column) your answer is
$$\eqalign{&0 \times 3 + 1 \times 6 + 3 \times 10 + 6 \times 15 + 10 \times 21 + 15 \times 28 + 21 \times 36 + \cr &28 \times 42 + 35 \times 46 + 40 \times 48 + 43 \times 48 + 44 \times 46 + 43 \times 42 + 40 \times 36 + \cr &35 \times 28 + 28 \times 21 + 21 \times 15 + 15 \times 10 + 10 \times 6 + 6 \times 3 + 3 \times 1 \cr}$$</p>
|
183,039 | <p>I'm pretty weak in the field of mathematics, but a strong programmer. I am looking for a mathematical solution that, given two points on a line will give me a curve between them, including those two points within the curve itself.</p>
<p>For instance, if I have a set of points { (0, 3) (1,10) } I'd like a mathematical way to generate points between the two (I believe this is called interpolate) to create a curve that will contain { (0,3) (1,10) }</p>
<p>Will Linear Interpolation give me this? </p>
<p>Thank you</p>
| RougeSegwayUser | 15,870 | <p>You also might want to check out Lagrange interpolation. The idea is that you can create a polynomial that goes through the given points, provided that the points arn't lined up vertically. Here's the link to the wikipedia article.</p>
<p><a href="http://en.wikipedia.org/wiki/Lagrange_polynomial" rel="nofollow">http://en.wikipedia.org/wiki/Lagrange_polynomial</a></p>
|
1,335,446 | <p>Ordinary sphere in $\mathbb{R}^3$ is two-dimensional object (2-sphere), i.e. it requires at least two coordinates to define point on a surface. As I notice, however, there is a catch. <br/>
If we use spherical coordinates,$(\phi,\theta)$, then there are two points where $\phi$ is not defined (polar regions). If we use stereographic coordinates, $j=x+iy$, then there is one point where both $x$ and $y$ are infinite, hence not defined. But if we use three coordinates (e.g. cartesian), then we have no such problem. <br/>
In that sense, is 2-sphere a three-dimensional object, i.e. there is some dimension function that is equal three for 2-sphere (but equal two for 2-sphere with two points excluded)? Or is there a way to use two coordinates and not to encounter such 'problematic' points? </p>
| Andrew D. Hwang | 86,418 | <p>First, there is no way to cover the sphere with a single coordinate system. If there were, the sphere would be diffeomorphic to an open set in the plane, which it isn't. (Generally, a manifold of positive dimension can be covered by a single "coordinate chart" only if it is diffeomorphic to some open subset of a Euclidean space. Particularly, no compact manifold has this property. Some non-compact manifolds, such as the complement of a point in a torus, also cannot be covered by one coordinate system.)</p>
<p>Second, "ambient" Cartesian coordinates aren't suitable as "local coordinates" on a sphere because "they don't vary independently". Generally, I know of no intrinsic sense in which the unit sphere in $\mathbf{R}^{3}$ is anything but "$2$-dimensional".</p>
|
18,224 | <p>I need to generate a very large sparse block matrix, with blocks consisting only of ones along the diagonal. I have tried several ways of doing this, but I seem to always run out of memory.</p>
<p>The fastest way of doing this that I've come up with so far is as follows:</p>
<p>(typically, I will need <code>n</code> to be at least 2500 and m of the order 50).</p>
<pre><code>tmp= SparseArray[{}, {n,n}, 1];
SparseArray@
ArrayFlatten@
Table[If[i == j, tmp, 0], {i, m}, {j, m}]
</code></pre>
<p>Example when n=2, m=4:</p>
<p><img src="https://i.stack.imgur.com/PIpvo.png" alt="Matrix"></p>
<p>The problem with this construction is that <code>ArrayFlatten</code> for some reason converts the result to a normal matrix, and I run out of memory. That is, when it works, this code computes the end result very quickly and the result does not take up much memory. At some specific number however, it suddenly crashes as the intermediate <code>ArrayFlatten</code> step clogs up the memory.</p>
<p>Any help will be greatly appreciated!</p>
| b.gates.you.know.what | 134 | <p>Maybe this ?</p>
<pre><code>SparseArray[{Band[{1, 1}, Dimensions[tmp] {m, m}] -> {tmp}}, Dimensions[tmp] {m, m}]
</code></pre>
|
18,224 | <p>I need to generate a very large sparse block matrix, with blocks consisting only of ones along the diagonal. I have tried several ways of doing this, but I seem to always run out of memory.</p>
<p>The fastest way of doing this that I've come up with so far is as follows:</p>
<p>(typically, I will need <code>n</code> to be at least 2500 and m of the order 50).</p>
<pre><code>tmp= SparseArray[{}, {n,n}, 1];
SparseArray@
ArrayFlatten@
Table[If[i == j, tmp, 0], {i, m}, {j, m}]
</code></pre>
<p>Example when n=2, m=4:</p>
<p><img src="https://i.stack.imgur.com/PIpvo.png" alt="Matrix"></p>
<p>The problem with this construction is that <code>ArrayFlatten</code> for some reason converts the result to a normal matrix, and I run out of memory. That is, when it works, this code computes the end result very quickly and the result does not take up much memory. At some specific number however, it suddenly crashes as the intermediate <code>ArrayFlatten</code> step clogs up the memory.</p>
<p>Any help will be greatly appreciated!</p>
| Mr.Wizard | 121 | <p>I find that this performs equivalently to b.gatessucks' code while being significantly cleaner.</p>
<pre><code>n = 3;
m = 2;
SparseArray[Band[{1, 1}] -> SparseArray[{}, {m, n, n}, 1]]
</code></pre>
<p>$\left(
\begin{array}{cccccc}
1 & 1 & 1 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 \\
\end{array}
\right)$</p>
<p>However <a href="http://reference.wolfram.com/language/ref/ArrayFlatten.html" rel="nofollow noreferrer"><code>ArrayFlatten</code></a> is still worth your consideration. Using a modified form of a method I posted for this slightly more general Q&A <a href="https://mathematica.stackexchange.com/q/19778/121">How to form a block-diagonal Matrix from a list of matrices?</a>:</p>
<pre><code>IdentityMatrix[m] /. {1 -> SparseArray[{}, {n, n}, 1]} // ArrayFlatten
(* same output as above, but not a SparseArray *)
</code></pre>
<p>In cases were n >> m (as described in the question) the peak memory consumption is actually not much more, it is easily an order of magnitude faster, and conversion into a sparse array is very fast:</p>
<pre><code>{n, m} = {300, 20};
MaxMemoryUsed[
r1 = SparseArray[Band[{1, 1}] -> SparseArray[{}, {m, n, n}, 1]];] // AbsoluteTiming
MaxMemoryUsed[
r2 = IdentityMatrix[m] /. {1 -> SparseArray[{}, {n, n}, 1]} //
ArrayFlatten;] // AbsoluteTiming
r1 == r2
SparseArray[r2]; // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{3.37077, 459439920}
{0.13651, 576105096}
True
{0.091, Null}
</code></pre>
</blockquote>
<p>For optimal memory performance a variation of ybeltukov's <code>blockArray</code> from <a href="https://mathematica.stackexchange.com/q/99467/121#99468">Speeding up generation of block diagonal matrix</a> surpasses both methods:</p>
<pre><code>fn[n_, m_] :=
SparseArray[Tuples[Range@# - {1, 0, 0}].{Rest@#, {1, 0}, {0, 1}} &@{m, n, n} -> 1]
MaxMemoryUsed[r3 = fn[300, 20];] // RepeatedTiming
r1 == r3
</code></pre>
<blockquote>
<pre><code>{0.769, 379069184}
True
</code></pre>
</blockquote>
|
2,002,672 | <p>I am ok with finding a power series for $\tanh^{-1}{z}$ since I can just see that since $\tanh iz = i\tan z$, then $\tanh^{-1}z = \frac1i \tan^{-1}{iz}$ and use the power series for $\tan$ to get:</p>
<p>$$\tanh^{-1}{z} = \sum^\infty_{k=0} \frac{z^{2k+1}}{2k+1}$$</p>
<p>But then I am asked to deduce that:</p>
<p>$$1- \frac15 + \frac19 - \frac1{13} + \ldots = \frac{\pi + 2 \ln(1+\sqrt{2})}{4 \sqrt 2}$$</p>
<p>Looking at this sum, it looks like it's the power series for something involving $\tanh^{-1} 1$ and $\tan^{-1} 1$. But surely $\tanh^{-1} 1$ won't be defined?</p>
<p>I'm stumped, there's probably something I'm not seeing here.</p>
| Martin Argerami | 22,857 | <p>For the functions you mention you can estimate the error in the Taylor polynomial. For cases like $\sin x$ and $\cos x$, it is easy to show that the error term goes to zero for each fixed $x\in\mathbb R$. This makes the function equal to its Taylor series everywhere, and thus analytic. </p>
<p>If you can bound the error term within an interval, then you would have proven that the function is analytic on an interval. </p>
|
2,388,613 | <p>1) a right cone has a surface area 12 m${^2}$ and radius 1.3 m</p>
<p><strong>here is my answers:</strong></p>
<p>1) $s$ = 1.28</p>
<p><a href="https://i.stack.imgur.com/a8ovb.jpg" rel="nofollow noreferrer">(photo of how i got my answer)</a></p>
<p><strong>textbook answer:</strong> 1.64 m</p>
<p>how did they get that??</p>
| Cye Waldman | 424,641 | <p>Your error is in thinking that they want the vertical height of the cone. For a cone, $s$ usually represents the slant height. So the first thing you want to is to express the total surface are as</p>
<p>$$S=\pi r s+\pi r^2$$</p>
<p>where $\pi r s$ is the lateral area. Then</p>
<p>$$
s=\frac{S-\pi r^2}{\pi r}\approx1.638
$$</p>
<p>as in your text!</p>
|
87,271 | <p>For a category $C$, let $C-Set$ denote the category of set-valued functors $\delta\colon C\to Set$. Given categories $C$ and $D$, and a functor $F\colon C\to D$, composition with $F$ yields a functor that I'll denote by $$\Delta_F\colon D-Set\longrightarrow C-Set.$$ The functor $\Delta_F$ has both a left adjoint, which I'll denote by $\Sigma_F\colon C-Set\longrightarrow D-Set$, and a right adjoint $\Pi_F\colon C-Set\longrightarrow D-Set$. One then has a monad $M = \Delta_F\Sigma_F$ on $C-Set$ and a monad $N=\Pi_F\Delta_F$ on $D-Set$. Let $M-alg=(C-Set)^M$ denote the category of $M$-algebras on $C-Set$, and similarly, let $N-alg=(D-Set)^N$ denote the category of $N$-algebras on $D-Set$.</p>
<p>I'll say that $\Delta_F$ is monadic if the obvious functor $D-Set\longrightarrow M-alg$ is an equivalence of categories, and I'll say that $\Pi_F$ is monadic if the obvious functor $C-Set\longrightarrow N-alg$ is an equivalence of categories.</p>
<p>Questions: </p>
<ol>
<li><p>Under what conditions on $F$ is $\Delta_F$ monadic?</p></li>
<li><p>Under what conditions on $F$ is $\Pi_F$ monadic?</p></li>
</ol>
<p>Thanks!</p>
| Karol Szumiło | 12,547 | <p>I've been wondering about the same thing recently and here's my best guess. Let $\widetilde C$ denote the Cauchy completion of $C$. if the associated functor $\widetilde F : \widetilde C \to \widetilde D$ induces a bijection of the sets isomorphism classes of objects in $\widetilde C$ and $\widetilde D$, then $\Delta_F$ is monadic.</p>
<p>Here's a rough idea why this is true. The monad $M$ is given by a coend formula $M X = \int^c X c \times D(F c, F-)$. For simplicity let's assume that $C$ and $D$ are Cauchy complete (we can do this since $C$ and $\widetilde C$ are Morita equivalent) and $F$ is actually bijective on objects (replace $C$ and $D$ by their skeleta). Now a map $M X \to X$ is a transformation $X c \times D(F c, F c') \to X c'$ dinatural in $c$ and natural in $c'$ i.e. a transformation $D(F c, F c') \to \mathrm{Set}(X c, X c')$ natural in both in $c$ and $c'$. Since $F$ is bijective on objects we can rewrite it as $D(d, d') \to \mathrm{Set}(X d, X d')$ where $X (F c) = X c$. At this point this is just a family of maps and I do not claim any compatibility with composition in $D$. However, if $M X \to X$ is an $M$-algebra structure on $X$, then this family actually obtains a structure of a functor $D \to \mathrm{Set}$. It shouldn't be to difficult to verify (I admit I haven't done it in full detail myself) that this is an inverse to the canonical functor $\mathrm{Set}^D \to M$-$\mathrm{Alg}$.</p>
<p>I also have reasons to believe that the essential surjectivity is necessary. Here's a <em>very rough</em> idea why I suspect this. If $F$ is not essentially surjective, then we seem not to be able to control the behavior of functors $D \to \mathrm{Set}$ by means of $M$-algebra structure. If I wanted to try to actually verify this, I would try to show that in that case $\Delta_F$ fails to create limits. (This is what fails when $C$ is the terminal category and $D$ is a two-point discrete category.)</p>
<p>On the other hand I don't know whether the injectivity on the isomorphism classes is necessary.</p>
<p>I also don't know any answer to your second question and in fact I don't see any non-trivial example of $F$ such that $\Pi_F$ is monadic (if you have such an example, please share it).</p>
|
109,037 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/107336/why-doesnt-dx-n-x-n1-rightarrow-0-as-n-rightarrow-infty-imply-x-n">Why doesn't $d(x_n,x_{n+1})\rightarrow 0$ as $n\rightarrow\infty$ imply ${x_n}$ is Cauchy?</a> </p>
</blockquote>
<p>my question is this:</p>
<hr>
<p><em>The following definition is weaker than the definition of Cauchy sequences:</em> </p>
<p>$\forall \; \epsilon > 0, \;\exists N \in \mathbb{N} \;s.t.\; \forall\; n \geq N, \; |a_{n+1}-a_n | < \epsilon.$</p>
<p><em>Show that this is not equivalent to $(a_n)$ being a Cauchy sequence.</em></p>
<hr>
<p>The definition of Cauchy sequence is: </p>
<p>A sequence $(s_n)$ is Cauchy if (and only if) for each $\epsilon > 0$ there exists an integer $N$ with the property that $|s_n-s_m| < \epsilon$ whenever $n\geq N$ and $m \geq N$.</p>
<p>Note that a sequence (of real numbers) is convergent if and only if it is Cauchy.</p>
<hr>
<p>So I see an (the?) obvious difference between these two in that the Cauchy criteria demands that all values in a sequence above a certain index ($N$) are within a prescribed tolerance of each other, whether adjacent or not. This is where the question is weaker, in that it only requires the immediately adjacent values of the sequence to be within a tolerance of $\epsilon$. This would then allow, by taking successive differences of adjacent values, to accumulate a difference greater than $\epsilon$. This is seen as,</p>
<p>$$\left| \sum_{i=1}^{n+1}\,a_i - \sum_{i=1}^{n}\,a_i\right| < \epsilon,\quad
\left|\sum_{i=1}^{n+2}\,a_i - \sum_{i=1}^{n+1}\,a_i\right| < \epsilon,\quad
\left| \sum_{i=1}^{n+3}\,a_i - \sum_{i=1}^{n+2}\,a_i\right| < \epsilon,$$
and summing each side of the inequalities gives (after reverting to sequence-notation and employing the triangle inequality),
$$
\left(|a_{n+1}-a_n| + |a_{n+2}-a_{n+1}| + |a_{n+3}-a_{n+2}| + \cdots + |a_{K} - a_{K-1}| \right) \leq \left( \epsilon_{1,2} + \epsilon_{2,3} + \epsilon_{3,4} + \cdots + \epsilon_{K-1,K} \right)
$$
which implies
$$\left( \epsilon_{1,2} + \epsilon_{2,3} + \epsilon_{3,4} + \cdots + \epsilon_{K-1,K} \right)_{\textrm{ by weaker criteria }} \geq |a_n - a_{n+K}|_{\textrm{ by Cauchy criteria }}$$</p>
<p>If I understand these differences correctly, then my main problem is putting these into a formal mathematical proof. Unless this would qualify?</p>
<p>Thanks much for the help and the site!</p>
| Dave L. Renfro | 13,130 | <p>Your intuition and argument are correct as far as they go, but note that you've only shown that the Cauchy condition is at least as strong, not strictly stronger. You can show "strictly stronger" by finding a counterexample, or by finding a property that all Cauchy sequences have and the weaker formulation doesn't have. A counterexample is probably easier to come by. Think about how this could happen. Can you think of a function $f$ defined for $x>0$ (hence, it is defined for each $n = 1, \; 2, \; 3,\;...$) that approaches $\infty$ as $x \rightarrow \infty$ (hence, $f(n) \rightarrow \infty$ as $n \rightarrow \infty$), but it does so in such a way that its everywhere positive slope approaches $0$ as $x \rightarrow \infty$ (thereby ensuring that $f(n+1) - f(n) \rightarrow 0$)? More specifically, think about how $f(x) = \log x$ or $f(x) = \sqrt x$ could help in light of what I've said.</p>
|
31,992 | <p>I am attempting to generate partially transparent images for PNG <code>Export</code>, but seem to run to the following issue. If I <code>Rasterize</code> a simple piece of <code>Graphics</code> with <code>Background -> None</code> (transparent) it looks worse:</p>
<pre><code>rings = Image[
Rasterize[
Graphics[{Black, Disk[], White, Disk[{1, 1}/64, 1 - Sqrt[2]/64]}],
Background -> #, RasterSize -> 400, ImageSize -> 400]] & /@ {None, White}
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/ho1BD.png" alt="enter image description here"></p>
</blockquote>
<p>First version appears jagged, as a circle drawn using a too simple polygon would do. I can verify this with <code>ImageDifference</code>:</p>
<pre><code>ImageAdjust[ImageDifference @@ (RemoveAlphaChannel[#, White] & /@ rings)]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/JZsO6.png" alt="enter image description here"></p>
</blockquote>
<p>Jagged pattern is clearly visible.</p>
<p>How to use <code>Rasterize</code> with <code>Background -> None</code> (or anything similar generating an alpha-channel image) and achieve good output quality without resorting to generating primitives such as <code>Disk</code> using hand-crafted code?</p>
<p><strong>Clarification:</strong></p>
<p>I want alpha channel on the output to behave as it does on <code>Rasterize[..., Background -> None]</code>. That is, the image having conceptually three regions: transparent background, black outer disk and white (non-transparent) inner disk.</p>
<p>(These screenshots were taken on Mathematica 9.0.1.0 running on OS X 10.8.4.)</p>
| Michael Hale | 5,568 | <p>Oops. I was too tired to fully verify the expected output before I wrote my comment. Here is a simple hack to make the image you want though.</p>
<pre><code>disk[c_, r_] :=
Polygon@Table[c + r {Cos@a, Sin@a}, {a, 0, 2 Pi, Pi/50}]
Image[Rasterize[
Graphics[{Black, disk[{0, 0}, 1], White,
disk[{1, 1}/78, 1 - Sqrt[2]/64]}], Background -> None,
RasterSize -> 400, ImageSize -> 400]]
</code></pre>
|
417,131 | <p>Does limit $\frac{4xy}{\sqrt{x^2+y^2}}$ as $(x,y) \to (0,0)$ exist or not?</p>
<p>To remove the root, I squared the whole equation and I get
limit $\frac{16x^2y^2}{\sqrt{x^2+y^2}}$. Then I dont know how to continue working it. Can anyone help to do this questions for me please.</p>
<p>Thank you for your effort in advance!</p>
| A.S | 24,829 | <p>Remember that $|\sin(z)| < z$ for $z \ne 0$. So $0 \le \sin^2(z) < z^2$. Set $z = x-y$. Your denominator satisfies $|z| = |x-y| \le |x| + |y|$ by the triangle inequality.</p>
<p>$$0 \le \lim_{(x,y) \to (0,0)} \frac{\sin^2(x-y)}{|x|+|y|} \le \lim_{(x,y) \to (0,0)} \frac{z^2}{|z|} = \lim_{(x,y) \to (0,0)} |z| = 0$$</p>
<p>You can easily convert this proof, which uses the squeeze theorem, into an $\epsilon-\delta$ proof by noting the inequalities and setting $\delta = \epsilon$. For $\Vert ( x,y ) \Vert < \epsilon$, we have $|z| < \epsilon$ since $|x-y| \le \sqrt{x^2 + y^2}$.</p>
|
3,435,286 | <p>Suppose that <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are random variables with joint density function</p>
<p><span class="math-container">$$f_{X,Y}(x,y)=\begin{cases} 8xy \space \text{ for } 0<y<x<1\\ 0 \space\space\space\space\space\text{ otherwise.}\end{cases}$$</span> </p>
<p>Determine the conditional probability <span class="math-container">$P\big( X>\frac{3}{4} \big| Y = \frac{1}{2}\big)$</span>.</p>
<p>Is the answer <span class="math-container">$\frac{5}{12}$</span>?</p>
| J.G. | 56,861 | <p>With <span class="math-container">$c:=\cos x$</span>, <span class="math-container">$f=2c^2+c-1$</span>, so find the extrema of this function on <span class="math-container">$[-1, 1]$</span>. <strong>Do not</strong> assume each extremum is at an endpoint of this range. The quadratic has its global minimum at <span class="math-container">$c=-\frac14$</span> in this range, so its maximum is at whichever endpoint is furthe from this value of <span class="math-container">$c$</span>. So the extrema differ by<span class="math-container">$$[2c^2+c-1]_{-1/4}^1=[2x^2]_0^{5/4}=\frac{25}{8}.$$</span></p>
|
1,132,922 | <p>Let $f$ and $g$ be two differentiable functions s.t $ f '(x) \le g '(x) $ for all $ x\lt 1$ and $ f '(x) \ge g'(x) $ for all $ x\gt 1$ then</p>
<ol>
<li><p>If $f(1) \ge g(1)$, then $f(x)\ge g(x)$ for all $x$</p></li>
<li><p>If $f(1) \le g(1)$, then $f(x)\le g(x)$ for all $x$</p></li>
<li><p>$f(1) \le g(1)$</p></li>
<li><p>$f(1) \ge g(1)$</p></li>
</ol>
<p>In this I am having difficulty in guessing fxn to counter options.</p>
| Subhra Mazumdar | 329,168 | <p>Let $f(x)=(x-1)^{2}$ and $g(x)=(x-1)$. Then $f'(x)=2(x-1)$ and $g'(x)=1$. This clearly satisfies the constraint $f '(x) \ge g'(x)$ for all $x>1$ and $f '(x) \le g'(x)$ for all $x<1$. $f(1)=0$ and $g(1)=0$, but for any value of $x , f (x) > g(x)$. So option (1)</p>
|
706,514 | <p>I know that the fundamental group of homeomorphic spaces are isomorphic. Is the converse true? I mean, can we say the two spaces with isomorphic fundamental groups are homeomorphic? </p>
| Brian Fitzpatrick | 56,960 | <p>No. In fact, there are spaces that are not <em>homotopy equivalent</em> (isomorphic in the homotopy category of topological spaces) but have the same fundamental group. Take $X=S^2$ and $Y=\{\operatorname{pt}\}$.</p>
|
706,514 | <p>I know that the fundamental group of homeomorphic spaces are isomorphic. Is the converse true? I mean, can we say the two spaces with isomorphic fundamental groups are homeomorphic? </p>
| Moishe Kohan | 84,907 | <p>This is an addendum to Jason DeVito's answer; it concerns the current status of Borel conjecture: </p>
<p><em>If two closed aspherical (i.e., with contractible universal covers) manifolds $M, N$ have isomorphic fundamental groups then they are homeomorphic.</em> </p>
<p>The strongest to date result towards Borel conjecture in dimensions $\ge 5$ is in the paper </p>
<p>"The Borel Conjecture for hyperbolic and CAT(0)-groups" by Bartels and Lueck which can be found <a href="http://arxiv.org/pdf/0901.0442.pdf">here</a>. </p>
<p>Their theorem states that Borel conjecture (in dimensions $\ge 5$) holds for CAT(0) and hyperbolic groups; this result includes the earlier results by Farrel and Jones on Borel conjecture for nonpositively curved manifolds. </p>
<p>Furthermore, Borel conjecture holds in dimension 3 thanks to Perelman's solution of Thurston's Geometrization Conjecture. People usually are aware of the fact that Perelman proved Poincare conjecture in dimension 3, but his theorem proves much more than this. Here is how Borel conjecture in 3d follows from Perelman's work:</p>
<p>Perelman's result immediately implies that every closed 3-manifold $M$ with contractible universal cover is one of the following:</p>
<ol>
<li><p>Hyperbolic. </p></li>
<li><p>Seifert. </p></li>
<li><p>Haken.</p></li>
</ol>
<p>Borel conjecture for hyperbolic manifolds follows from <a href="http://en.wikipedia.org/wiki/Mostow_rigidity_theorem">Mostow rigidity theorem</a>. For Haken manifolds Borel conjecture was proven by Waldhausen (see <a href="http://en.wikipedia.org/wiki/Haken_manifold">here</a>). For Seifert manifolds with contractible universal cover Seifert invariants (which completely determine topology of the manifolds) can be read off the fundamental group (my guess is that this was known to Seifert). </p>
<p>Borel conjecture in dimension 4 is wide-open as far as I know. </p>
|
209,044 | <p>Please help me deal with this kind of question about ODEs.
My codes are as follows</p>
<pre><code>m = 100;
a = D[x[t], {t, 2}];
t1up = 2 x''[t] + 1/2 (490 + 34 x''[t] + 2 (490 + 50 x''[t]));
t1down = 490 + 53 x''[t];
t1 = Piecewise[{{t1up, x'[t] >= 0}, {t1down, x'[t] < 0}}]
equa00 = t1 == m*a
t0 = 50;
s1 = NDSolve[{equa00, x[0] == 1, x'[0] == 1}, x, {t, 0, 50}]
</code></pre>
<p>However, I get an error:</p>
<blockquote>
<p>NDSolve::ntdvdae: Cannot solve to find an explicit formula for the derivatives. NDSolve will try solving the system as differential-algebraic equations. >></p>
</blockquote>
<p>So is it a differential-algebraic equation? How to solve it?</p>
<p>I have another question, too: How to plot the <code>t1-t</code> figure after we get the <code>s1</code>?
I have tried the following codes:</p>
<pre><code>t1upvalue = (t1up /. {x'[t] -> (x'[t] /. s1), x''[t] -> (x''[t] /. s1)})
t1downvalue = (t1down /. {x'[t] -> (x'[t] /. s1), x''[t] -> (x''[t] /. s1)})
t1value = Piecewise[{{t1upvalue, (x'[t] /. s1) >= 0}, {t1downvalue, (x'[t] /. s1) < 0}}],
Plot[t1value[[1]], {t, 0, t0},PlotRange -> All]
</code></pre>
<p>However it doesn't work.</p>
| xzczd | 1,871 | <p>Another solution is to use <a href="https://mathematica.stackexchange.com/a/133530/1871"><code>Simplify`PWToUnitStep</code></a>:</p>
<pre><code>s1 = NDSolve[{equa00 // Simplify`PWToUnitStep, x[0] == 1, x'[0] == 1}, x, {t, 0, 50}]
</code></pre>
|
2,936,236 | <p>I have two 2D disks, <span class="math-container">$(C_1, r_1)$</span> (blue) and <span class="math-container">$(C_2, r_2)$</span> (red), where the blue disk somehow overlaps the red disk (the figure below shows one example). I’m interested in finding the positive distance along the <span class="math-container">$y$</span> axis the blue disk needs to be moved so that it touches but does not overlap the red disk. If I move the blue disk radially outward from the red disk then it’s trivial to figure out how far to move in order to avoid overlap; but in my case, where I only want to move the blue disk in the positive <span class="math-container">$y$</span> direction, the general formula (for any configuration of the two disks, assuming there is any overlap) seems more difficult. Could someone provide some insight on what a general formula for this would be, if it's feasible?</p>
<p><a href="https://i.stack.imgur.com/CMCrI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CMCrI.png" alt="enter image description here"></a></p>
| amd | 265,466 | <p>Find the point(s) on the vertical line through the center of the blue disk that are at a distance equal to the sum of the two disks’ radii from the red disk’s center. That’s where the center of the blue disk should be for it to be tangent to the red. Looks like a job for the Pythagorean theorem.</p>
|
2,936,236 | <p>I have two 2D disks, <span class="math-container">$(C_1, r_1)$</span> (blue) and <span class="math-container">$(C_2, r_2)$</span> (red), where the blue disk somehow overlaps the red disk (the figure below shows one example). I’m interested in finding the positive distance along the <span class="math-container">$y$</span> axis the blue disk needs to be moved so that it touches but does not overlap the red disk. If I move the blue disk radially outward from the red disk then it’s trivial to figure out how far to move in order to avoid overlap; but in my case, where I only want to move the blue disk in the positive <span class="math-container">$y$</span> direction, the general formula (for any configuration of the two disks, assuming there is any overlap) seems more difficult. Could someone provide some insight on what a general formula for this would be, if it's feasible?</p>
<p><a href="https://i.stack.imgur.com/CMCrI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CMCrI.png" alt="enter image description here"></a></p>
| Narasimham | 95,860 | <p>Shift the green circle upwards until increasing distance between circle centers just equals</p>
<p><span class="math-container">$$(C_1 C_2) =(r_1+r_2).$$</span></p>
|
2,279,660 | <p>On $C^1([0,1],\mathbb{R})$ I have two norms
\begin{align*}
N_1(f)&=|f(0)|+\|f'\|_{\infty}\\
N_2(f)&=\|f\|_{\infty}+\|f'\|_{\infty}.
\end{align*}</p>
<p>We have always that $N_1(f)\leq N_2(f)$ but what about the other inequality please?</p>
<p>And please how to see If $N_1$ and $N_2$ are equivalent to $\left\|\cdot\right\|_{\infty}$. </p>
<p>Thank you. </p>
| zhw. | 228,045 | <p>Hint for the first question: By the mean value theorem, $f(x) = f(0) + f(x)-f(0) = f(0) + f'(c_x)x.$</p>
<p>Hint for the second question: Consider the functions $x^n,n=1,2,\dots $</p>
|
249,597 | <p>I am suppose to find all the solutions to this problem, I think some theorem states that there can only be as many solutions to the problem as the highest degree. I know that calculus reinforces this so I know that</p>
<p>$2x^2 + 4x + 1 = 0$</p>
<p>Can have at most two solutions. In calculus this is proven by the derivative being zero at only somewhere. I can't remember and it isn't important yet.</p>
<p>Anyways I have no idea what to do with this problem. I don't think I can factor it conventionally because of the 2 coefficient so what is the method at this point? I tried guessing and it didn't work at all for -2 - 3.</p>
| Tengu | 37,026 | <p>If you don't want to use the formula, you can do the following.(Actually this is how the formula is derived though. I used to use this method when I didn't memorize the formula.)</p>
<p>$2x^2+4x+1=0$</p>
<p>$x^2+2x+1/2=0$</p>
<p>$(x+1)^2-1/2=0$</p>
<p>$(x+1)^2=1/2$</p>
<p>$x+1=+1/\sqrt 2$ or $-1/\sqrt 2$</p>
|
249,597 | <p>I am suppose to find all the solutions to this problem, I think some theorem states that there can only be as many solutions to the problem as the highest degree. I know that calculus reinforces this so I know that</p>
<p>$2x^2 + 4x + 1 = 0$</p>
<p>Can have at most two solutions. In calculus this is proven by the derivative being zero at only somewhere. I can't remember and it isn't important yet.</p>
<p>Anyways I have no idea what to do with this problem. I don't think I can factor it conventionally because of the 2 coefficient so what is the method at this point? I tried guessing and it didn't work at all for -2 - 3.</p>
| P.K. | 34,397 | <p>The <a href="http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula" rel="nofollow">quadratic formula</a> works for all quadratic equations. For any equation [in the form] $ax^2 + bx + c = 0$, this is true:$$x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$$</p>
|
996,103 | <p>Suppose that $A$, $B$, and $C$ are sets. Prove that $(A\cap B)\times C =(A\times C)\cap(B\times C)$. Prove the statement both ways or use only if and only if statements.</p>
| marty cohen | 13,079 | <p>Here is a way
to show convergence
without finding the sum.</p>
<p>If $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
$.
Therefore
(the $1+x$ in the denominator
of the next integral
is intentional)
$\int_1^{1+x} \frac{dt}{1+x}
< \ln(1+x)
< \int_1^{1+x} \frac{dt}{1}
$
or
$\frac{x}{1+x}
< \ln(1+x)
< x
$.</p>
<p>Since
$\frac{n^2}{n^2-1}
=\frac{n^2-1+1}{n^2-1}
=1+\frac{1}{n^2-1}
$,
$\ln \frac{n^2}{n^2-1}
<\frac{1}{n^2-1}
$.</p>
<p>Since
$\sum^\infty_{n=2} \frac{1}{n^2-1}
$ converges,
so does
$\sum^\infty_{n=2} \ln(n^2/(n^2-1))$.</p>
<p>For a lower bound,
$\ln \frac{n^2}{n^2-1}
>\frac{\frac{1}{n^2-1}}{1+\frac{1}{n^2-1}}
=\frac{1}{n^2}
$.</p>
<p>Note that,
in general,
if $m > 1$,
$\frac{1}{m}
<\ln \frac{m}{m-1}
<\frac{1}{m-1}
$.</p>
|
475,221 | <p>I am working on problem #25 of <em>Linear Algebra and its Applications</em> and the question asks:</p>
<blockquote>
<p>Find an equation involving $g$, $h$, and $k$ that makes this augmented matrix correspond to a consistent system:
$$\left(\begin{array}{ccc|c}
1& -4& 7& g \\
0& 3& -5& h \\
-2& 5& -9& k
\end{array}\right).$$
After I do $R_3 \gets 2R_1 + R_3$ and $R_3 \gets R_2 + R_3$
I end up with
$$\left(\begin{array}{ccc|c} 1& -4& 7& g \\ 0& 3& -5& h \\ 0& 0& 0& 2g+k+h
\end{array}\right).$$</p>
</blockquote>
<p>For this to be a consistent system the third row should be $\begin{pmatrix}0& 0& 0& 0\end{pmatrix}$,
so in order for this augmented matrix to be a consistent system then $2g + k + h =0$</p>
<p>The answer in the back of the book is $k - 2g + k = 0$.</p>
<p>Where am I going wrong with my calculation? Or is the book wrong?</p>
| Community | -1 | <p><strong>Hint</strong></p>
<ul>
<li>$$\frac{y^2}{y^2+d^2}=\frac{y^2+d^2-d^2}{y^2+d^2}=1-\frac{d^2}{y^2+d^2}$$</li>
<li>$$\frac{d^2}{y^2+d^2}=\frac{1}{(y/d)^2+1}=\frac{1}{t^2+1}$$</li>
<li>$$\int\frac{dt}{t^2+1}=\arctan t+C$$</li>
</ul>
|
4,095,361 | <p>Given <span class="math-container">$f : [0, 1] \rightarrow \mathbb{R}$</span> be a bounded function which is continuous on <span class="math-container">$[0,1]$</span> except at <span class="math-container">$1/2$</span>. Let <span class="math-container">$\alpha(x) = x^2$</span>. No further description of function <span class="math-container">$f$</span> is given. How do I show that <span class="math-container">$f$</span> Riemann Stieltjes integrable with respect to <span class="math-container">$\alpha$</span>?</p>
| tommik | 791,458 | <p>your distribution is not absolutely continuous thus your expectation is wrong. Your random variable has a positive probability mass in <span class="math-container">$x=5/6$</span></p>
<p>An easy way to calculate its expectation is the following purple area <span class="math-container">$=35/72$</span></p>
<p><a href="https://i.stack.imgur.com/C38tz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C38tz.jpg" alt="enter image description here" /></a></p>
<p>...you calculated only the area of the triangle...<span class="math-container">$=25/72$</span></p>
<p>Try yourself to reason about my hint and calculate the variance...</p>
|
820,490 | <p>I am studying pre-calculus mathematics at the moment, and I need help in verifying if $\sin (\theta)$ and $\cos (\theta)$ are functions? I want to demonstrate that for any angle $\theta$ that there is only one associated value of $\sin (\theta)$ and $\cos (\theta)$. How do I go about showing this?</p>
| Christian Blatter | 1,303 | <p>Usually one begins studying the $\sin$ function in connection with right triangles. Here the notion of angle is not questioned, so I shall stick with this intuitive concept. It follows that we have $\sin \theta$ defined for $0\leq\theta\leq{\pi\over2}$.</p>
<p>As a next step one considers polar angles in the $(x,y)$-plane, measured counterclockwise from the positive $x$-axis, and one identifies these angles with the length of the corresponding arc on the unit circle $S^1$. When $0\leq\theta\leq{\pi\over2}$ then $\sin\theta$ is the $y$-coordinate of the point where the second leg of the angle in question intersects $S^1$. It is then <em>natural to define</em> $\sin \theta$ for all $\theta\in[0,2\pi]$ in this way, and a further step leads to the definition of $\sin\theta$ for all $\theta\in{\mathbb R}$: For arbitrary $\theta>0$ spool a thread of length $\theta$ counterclockwise around $S^1$ with initial point at $(1,0)$. The $y$-coordinate of the endpoint is then <em>defined</em> to be $\sin\theta$. It is then intuitively obvious that the function $\sin$ is periodic.</p>
<p>This sounds all very simple and natural. The difficulties begin when we try to make the above thought process mathematically precise. In the first place we need a precise notion of angle: axiomatic euclidean geometry allows comparison and addition of angles, but does not provide an identification of angles with real numbers, which is inherent in the concept of the $\sin$ function.</p>
|
549,254 | <blockquote>
<p>$$\frac{1}{3}=.33\bar{3}$$ </p>
</blockquote>
<p>is a rational number, but the $3$ keeps on repeating indefinitely (infinitely?). How is this a ratio if it shows this continuous pattern instead of being a finite ratio? </p>
<p>I understand that $\pi$ is irrational because it extends infinitely <em>without repetition</em>, but I am confused about what makes $1/3=.3333\bar{3}$ rational. It is clearly repeating, but when you apply it to a number, the answers are different: $.33$ and $.3333$ are part of the same concept, $1/3$, yet:</p>
<p>$.33$ and $.3333$ are different numbers: </p>
<p>$.33/2=.165$ and $.3333/2=.16665$, yet they are both part of $1/3$. </p>
<p>How is $1/3=.33\bar{3}$ rational?</p>
| Khosrotash | 104,171 | <p>Let <span class="math-container">$$a=0.333333333333333...$$</span></p>
<p>Then multiply by <span class="math-container">$10$</span></p>
<p><span class="math-container">$$10a=3.333333333333333...$$</span></p>
<p>Substract <span class="math-container">$a$</span> from both sides</p>
<p><span class="math-container">$$10a-a=3.3333333....- 0.3333333...$$</span></p>
<p>So we get</p>
<p><span class="math-container">$$9a=3$$</span></p>
<p>And therefore</p>
<p><span class="math-container">$$a=1/3$$</span></p>
<p>and it’s rational</p>
|
1,336,209 | <p>I need to understand very good how the properties of this formula</p>
<p>$\frac{4}{\pi} = \frac{5}{4} + \sum_{N \geq 1} \left[ 2^{-12N + 1} \times(42N + 5)\times {\binom {2N-1} {N}}^3 \right] $</p>
<p>Taken from the paper "Radian Reduction for Trigonometric Function" (Hanek Payne Algorithm)</p>
<p>Some remarkable properties are stated, specifically these four ones</p>
<ol>
<li>The $k^{th}$ term of the formula is exactly representable in $6k$ bits;</li>
<li>The first $n$ terms of the sum can be represented exactly in $12n$ bits;</li>
<li>The most significant bit of the $k^{th}$ term has weight at most $2^{1-6k}$ and hence each successive term increases the number of valid bits in the sum by at least $6$;</li>
<li>If $12k < m + 1 \leq 12(k+1)$, then the $m^{th}$ bit of $\frac{4}{\pi}$ may be computed using only terms beyond the $k^{th}$.</li>
</ol>
<p>My questions are:
1. How to prove the formula?
2. How to prove the properties stated above?</p>
<p>PS. I guess with terms the paper means the generic term $a_N$ of the sum... </p>
| Ant | 66,711 | <p>The Stolz Cesaro theorem yields $$\lim_{n \to \infty} \frac{(n+1)^4}{(n+1)^k-n^k}$$</p>
<p>The degree of the numerator is $4$ while the degree of the denominator is $k-1$. So if $k \le 4$, the limit is $\infty$.
If $k \ge 6$, the limit is $0$. </p>
<p>If $k = 5$, you only need to compute the coefficient of $n^4$ in the denominator, and the binomial theorem easily yields $\binom 51 = 5$. So the limit if $k=5$ is $\frac 15$</p>
|
1,336,209 | <p>I need to understand very good how the properties of this formula</p>
<p>$\frac{4}{\pi} = \frac{5}{4} + \sum_{N \geq 1} \left[ 2^{-12N + 1} \times(42N + 5)\times {\binom {2N-1} {N}}^3 \right] $</p>
<p>Taken from the paper "Radian Reduction for Trigonometric Function" (Hanek Payne Algorithm)</p>
<p>Some remarkable properties are stated, specifically these four ones</p>
<ol>
<li>The $k^{th}$ term of the formula is exactly representable in $6k$ bits;</li>
<li>The first $n$ terms of the sum can be represented exactly in $12n$ bits;</li>
<li>The most significant bit of the $k^{th}$ term has weight at most $2^{1-6k}$ and hence each successive term increases the number of valid bits in the sum by at least $6$;</li>
<li>If $12k < m + 1 \leq 12(k+1)$, then the $m^{th}$ bit of $\frac{4}{\pi}$ may be computed using only terms beyond the $k^{th}$.</li>
</ol>
<p>My questions are:
1. How to prove the formula?
2. How to prove the properties stated above?</p>
<p>PS. I guess with terms the paper means the generic term $a_N$ of the sum... </p>
| user26486 | 107,671 | <p>You can prove $1^4+2^4+3^4+\cdots+n^4=\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n$ using simple induction.</p>
<p>$$\lim_{n\to \infty} \frac{1^4+2^4+3^4+\ldots+n^4}{n^k}$$</p>
<p>$$=\lim_{n\to \infty} \frac{\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n}{n^k}$$</p>
<p>If $k<5$, the limit is $\infty$, if $k>5$, the limit is $0$. </p>
<p>For $k=5$, the limit is $1/5$.</p>
<p>$$\lim_{n\to \infty} \frac{\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n}{n^5}$$</p>
<p>$$=\lim_{n\to\infty} \left(\frac{1}{5}+\frac{1}{2n}+\frac{1}{3n^2}-\frac{1}{30n^4}\right)$$</p>
|
3,542,374 | <p>I have the matrix of stacked constraints</p>
<p><span class="math-container">$$\begin{bmatrix} x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_1^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{bmatrix} \mathbf{c} = \mathbf{0},$$</span></p>
<p>where <span class="math-container">$\mathbf{c} = (a, b, c, d, e, f)^T$</span> is a conic.</p>
<p>So <span class="math-container">$\mathbf{c}$</span> is the null vector of this <span class="math-container">$5 \times 6$</span> matrix. Apparently, this shows that <span class="math-container">$\mathbf{c}$</span> is determined uniquely (up to scale) by five points in general position. What is the concept from linear algebra that tells us that this shows that <span class="math-container">$\mathbf{c}$</span> is determined uniquely? And what is meant by "up to scale"?</p>
<p>Thank you.</p>
| user8675309 | 735,806 | <p><strong>a linear algebra (+ Kronecker products) proof of the rank of your interpolation matrix</strong></p>
<p><span class="math-container">$\begin{bmatrix} x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{bmatrix} \mathbf{c} = \mathbf{0}$</span><br />
and you want to prove that the nullspace has dimension 1 -- so up to rescaling, there is one and only one nonzero vector in the nullspace of that matrix. By rank-nullity this is equivalent to proving the above matrix has rank 5.</p>
<p>Permuting columns doesn't change rank. Also appending columns that are copies of existing columns doesn't change rank,so it becomes convenient to consider instead the rank of</p>
<p><span class="math-container">$\begin{bmatrix}
x_1^2 & x_1y_1 & x_1 & x_1 y_1& y_1^2 & y_1 & x_1 & y_1 & 1
\\\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots
\\x_5^2 & x_5y_5 & x_5 & x_5 y_5& y_5^2 & y_5 & x_5 & y_5 & 1 \end{bmatrix} = \begin{bmatrix} \mathbf x_1^T\otimes \mathbf x_1^T \\ \vdots \\ \mathbf x_5^T\otimes \mathbf x_5^T\\ \end{bmatrix}$</span></p>
<p>where<br />
<span class="math-container">$\mathbf x_k := \begin{bmatrix} x_k \\ y_k \\ 1\end{bmatrix}$</span><br />
and <span class="math-container">$\otimes$</span> denotes Kronecker Product</p>
<p>again it must be the case that<br />
<span class="math-container">$\text{rank}\left(\begin{bmatrix} x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{bmatrix}\right) = \text{rank}\left(\begin{bmatrix} \mathbf x_1^T\otimes \mathbf x_1^T \\ \vdots \\ \mathbf x_5^T\otimes \mathbf x_5^T\\ \end{bmatrix}\right)$</span></p>
<p>so we want to prove that<br />
<span class="math-container">$\text{rank}\left(\begin{bmatrix} \mathbf x_1^T\otimes \mathbf x_1^T \\ \vdots \\ \mathbf x_5^T\otimes \mathbf x_5^T\\ \end{bmatrix}\right) = 5$</span><br />
or using the equivalence of row and column rank, it's equivalent to to prove that</p>
<p><span class="math-container">$\Big\{\mathbf x_1\otimes \mathbf x_1, \mathbf x_2\otimes \mathbf x_2, \mathbf x_3\otimes \mathbf x_3. \mathbf x_4\otimes \mathbf x_4, \mathbf x_5\otimes \mathbf x_5\Big\}$</span><br />
is a linearly independent set (of 5 vectors)</p>
<p><em>now using the fact that (no subset of 3) of the 5 points selected for interpolation are colinear</em> we choose 3 (WLOG assume the first 3) and form a basis to write the others in terms of. Since the original points are not colinear <em>this implies</em> many things, including<br />
(i) <span class="math-container">$\det\big(A\big) \neq 0$</span>, (ii) <span class="math-container">$\mathbf z_4$</span> and <span class="math-container">$\mathbf z_5$</span> have no components equal to zero and (iii) <span class="math-container">$\mathbf z_4 \not\propto \mathbf z_5$</span></p>
<p>So<br />
<span class="math-container">$A :=\bigg[\begin{array}{c|c|c}
\mathbf x_1 & \mathbf x_2 & \mathbf x_3
\end{array}\bigg]$</span><br />
and<br />
<span class="math-container">$\mathbf x_1 = A\mathbf e_1$</span><br />
<span class="math-container">$\mathbf x_2 = A\mathbf e_2$</span><br />
<span class="math-container">$\mathbf x_3 = A\mathbf e_3$</span><br />
<span class="math-container">$\mathbf x_4 = A\mathbf z_4$</span><br />
<span class="math-container">$\mathbf x_5 = A\mathbf z_5$</span><br />
where <span class="math-container">$\mathbf e_k$</span> is the kth standard basis vector in <span class="math-container">$\mathbb R^3$</span>.</p>
<p>applying the Kronecker product<br />
<span class="math-container">$\mathbf x_1\otimes \mathbf x_1 = \big(A\mathbf e_1\big)\otimes \big(A\mathbf e_1\big) = \big(A\otimes A\big)\big(\mathbf e_1 \otimes \mathbf e_1\big)$</span><br />
<span class="math-container">$\mathbf x_2\otimes \mathbf x_2 =\big(A\otimes A\big)\big(\mathbf e_2 \otimes \mathbf e_2\big)$</span><br />
<span class="math-container">$\mathbf x_3\otimes \mathbf x_3 = \big(A\otimes A\big)\big(\mathbf e_3 \otimes \mathbf e_3\big)$</span><br />
<span class="math-container">$\mathbf x_4\otimes \mathbf x_4 = \big(A\otimes A\big)\big(\mathbf z_4 \otimes \mathbf z_4\big)$</span><br />
<span class="math-container">$\mathbf x_5\otimes \mathbf x_5 = \big(A\otimes A\big)\big(\mathbf z_5 \otimes \mathbf z_5\big)$</span></p>
<p>so our linearly independent set at least includes<br />
<span class="math-container">$\Big\{\mathbf e_1\otimes \mathbf e_1,\mathbf e_2\otimes \mathbf e_2, \mathbf e_3\otimes \mathbf e_3\Big\}$</span><br />
i.e. 3 vectors that are all zero except they have a single one in the 1st, 5th, and 9th components respectively (i.e. they are <span class="math-container">$\mathbf e_1, \mathbf e_5, \mathbf e_9 \in \mathbb R^9$</span>)<br />
Now <span class="math-container">$\mathbf z_4$</span> has every component non-zero so it can't possibly be a linear combination of those three vectors. Thus we have a linearly independent set including at least<br />
<span class="math-container">$\Big\{\mathbf e_1\otimes \mathbf e_1,\mathbf e_2\otimes \mathbf e_2, \mathbf e_3\otimes \mathbf e_3, \mathbf z_4 \otimes \mathbf z_4\Big\}$</span></p>
<p>it remains to prove <span class="math-container">$\mathbf z_5 \otimes \mathbf z_5$</span> cannot be written as a linear combination of vectors in that set. In particular we'll prove that<br />
<span class="math-container">$\alpha \mathbf z_4 \otimes \mathbf z_4 + \mathbf z_5 \otimes \mathbf z_5\neq \sum_{k=1}^3 \gamma_k\mathbf e_k\otimes \mathbf e_k$</span></p>
<p>the problem is easy to finish by using a simple isomorphism. I.e. consider<br />
<span class="math-container">$\text{vec}\big(\mathbf z_j \mathbf z_j^T \big) =\big(\mathbf z_j \otimes \mathbf z_j \big)$</span><br />
where the vec operator just takes a matrix and converts it into a 'big vector' by stacking one column on top of the other.</p>
<p><em>so to finish, it's sufficient to prove that it's impossible to have</em><br />
<span class="math-container">$\alpha \mathbf z_4 \mathbf z_4^T + \mathbf z_5 \mathbf z_5^T=D$</span><br />
for some diagonal matrix <span class="math-container">$D \in \mathbb R^\text{3 x 3}$</span></p>
<p>note: if <span class="math-container">$D$</span> exists, then <span class="math-container">$3 =\text{rank}\big(D\big)$</span>. If this wasn't the case then there is (at least one) diagonal component <span class="math-container">$d_{k,k} = 0$</span>, which implies<br />
<span class="math-container">$\alpha \mathbf z_4 \mathbf z_4^T\mathbf e_k + \mathbf z_5 \mathbf z_5^T\mathbf e_k = \alpha z_4^{(k)}\mathbf z_4 +z_5^{(k)} \mathbf z_5 =\mathbf 0 = D\mathbf e_k$</span> or<br />
<span class="math-container">$\mathbf z_4 \propto \mathbf z_5 $</span><br />
since all components of <span class="math-container">$\mathbf z_4$</span> and <span class="math-container">$\mathbf z_5$</span> are non-zero. But the above is impossible since no points are colinear-- i.e. recall (ii) and (iii). Note: the trivial case of setting <span class="math-container">$\alpha:=0$</span> is also covered because that would imply <span class="math-container">$\mathbf z_5=\mathbf 0 $</span> but that is impossible as well -- (ii) or (iii) will do it.</p>
<p>Thus if <span class="math-container">$D$</span> exists it must be the case that<br />
<span class="math-container">$3 =\text{rank}\big(D\big) = \text{rank}\big(\alpha \mathbf z_4 \mathbf z_4^T +\mathbf z_5 \mathbf z_5^T\big) \leq 2$</span><br />
where the right inequality follows because the sum of 2 rank one matrices is at most rank 2. Thus<br />
<span class="math-container">$\alpha \mathbf z_4 \mathbf z_4^T + \mathbf z_5 \mathbf z_5^T \neq D$</span></p>
<p>which proves<br />
<span class="math-container">$\Big\{\mathbf e_1\otimes \mathbf e_1,\mathbf e_2\otimes \mathbf e_2, \mathbf e_3\otimes \mathbf e_3, \mathbf z_4 \otimes \mathbf z_4, \mathbf z_5 \otimes \mathbf z_5 \Big\}$</span><br />
is a linearly independent set and by the invertibility of <span class="math-container">$\big(A\otimes A\big)$</span> we know</p>
<p><span class="math-container">$\Big\{\mathbf x_1\otimes \mathbf x_1, \mathbf x_2\otimes \mathbf x_2, \mathbf x_3\otimes \mathbf x_3. \mathbf x_4\otimes \mathbf x_4, \mathbf x_5\otimes \mathbf x_5\Big\}$</span><br />
is a linearly independent set as well, which proves</p>
<p><span class="math-container">$5 =\text{rank}\left(\begin{bmatrix} \mathbf x_1^T\otimes \mathbf x_1^T \\ \vdots \\ \mathbf x_5^T\otimes \mathbf x_5^T\\ \end{bmatrix}\right)= \text{rank}\left(\begin{bmatrix} x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{bmatrix} \right)$</span></p>
<p>and completes the proof</p>
<p><strong>post script</strong><br />
a convenient property of the Kronecker product is<br />
<span class="math-container">$\text{vec}\big(\mathbf {XYZ}\big) = \big(\mathbf Z^T \otimes \mathbf X\big)\text{vec}\big(\mathbf {Y}\big)$</span></p>
<p>In context of the interpolation problem here, the problem is to collect, with (non-colinear) <span class="math-container">$\mathbf x_k$</span>, the values of</p>
<p><span class="math-container">$\mathbf x_k^T C \mathbf x_k = 0$</span><br />
for <span class="math-container">$k\in\{1,2,3,4,5\}$</span>, where <span class="math-container">$C := \begin{bmatrix} a & b/2 & d/2 \\ b/2 & c & e/2 \\ d/2 & e/2 & f \end{bmatrix}$</span></p>
<p>so using the Kronecker product we can organize the quadratic form into a convenient a system of equations</p>
<p><span class="math-container">$0 = \mathbf x_k^T C \mathbf x_k \longrightarrow 0 = \text{vec}\big(0\big) = \text{vec}\big(\mathbf x_k^T C \mathbf x_k\big) =\big(\mathbf x_k^T \otimes \mathbf x_k^T\big) \text{vec}\big( C\big)$</span><br />
for <span class="math-container">$k\in\{1,2,3,4,5\}$</span>. And we can collect this system of equations as</p>
<p><span class="math-container">$\begin{bmatrix} \mathbf x_1^T\otimes \mathbf x_1^T \\ \vdots \\ \mathbf x_5^T\otimes \mathbf x_5^T\\ \end{bmatrix}\text{vec}\big( C\big) = \mathbf 0$</span></p>
<p>after deleting redundant columns (and associated components in <span class="math-container">$\text{vec}\big( C\big)$</span>), we recover the original problem of</p>
<p><span class="math-container">$\begin{bmatrix} x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1 \\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1 \\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1 \\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1 \\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{bmatrix} \mathbf{c} = \mathbf{0}$</span></p>
|
544,673 | <p>let $\alpha$ be a cycle of length $s$, say $\alpha = (a_1, a_2, \ldots, a_s)$</p>
<p>Prove $\alpha^2$ is a cycle if and only if $s$ is odd.</p>
<p>Let me start off by saying I am in my 5th week of Group Theory. I often have trouble getting these problems started. This is my first proof based course. </p>
<p>I believe $\alpha^2 = (a_2, a_3, \ldots, a_1)$</p>
<p>Any tips on where to go from here would be great. Also...if there are any tips for starting proofs like these in general, I could really use them! My teacher teaches as if a proofing class was a pre-req, which it was not. </p>
| Dan Shved | 47,560 | <p>Here is my somewhat crude illustration of what happens when you multiply a cycle with itself.</p>
<p>If the cycle has odd length, it just changes the order in which vertices are visited. The pentagon on the left is the graph of a $5$-cycle. The star on the right is the graph of that cycle multiplied with itself.</p>
<p><img src="https://i.stack.imgur.com/wDrg5.png" alt="enter image description here"></p>
<p>If the cycle has even length, things are different: it splits in two.</p>
<p><img src="https://i.stack.imgur.com/J35sO.png" alt="enter image description here"></p>
<p>When you see this clearly in your mind, it should not be hard to write a formal proof.</p>
|
66,480 | <p>I have the following problem:</p>
<blockquote>
<p>Given $P(A)=0.2$, $P(B)=0.4$, $P(C)=0.8$, $P(D)=0.5$, find $P(A\cup B\cup C\cup D)$</p>
</blockquote>
<p>And the final answer should be 0.952</p>
<p>I know how to find the union of two and three elements (for 2, its: $A+B-AB$), but the formula becomes clumsy after 3. The best things I've found says that to find the union for n elements, I add as follows $$0.2-(0.2\times0.4)+(0.2\times0.4\times0.8)-(0.2\times0.4\times0.8\times0.5) = 0.152$$ which is wrong.</p>
<p>What is a good general rule for n events?</p>
| Dilip Sarwate | 15,941 | <p>The question you pose cannot be solved; all that can be said is that
$$
\max\{P(A), P(B), P(C), P(D)\} = 0.8 \leq P(A \cup B \cup C \cup D) \leq 1
$$
However, the answer $0.952$ that you give corresponds to the case when $A$, $B$, $C$, and $D$ are <em>independent</em> events. Did you leave out this important piece of information when you typed in your question?</p>
<p>Generally, if $A_i, 1 \leq i \leq n$ are <em>independent</em> events, then
using DeMorgan's laws
$$
P(A_1 \cup \cdots \cup A_n) = 1 - P((A_1 \cup \cdots \cup A_n)^c) = 1 - P(A_1^c\cap\cdots \cap A_n^c) = 1 - P(A_1^c)\cdots P(A_n^c)
$$
In order to avoid unnecessary arithmetic calculations, it is of the utmost importance that the expression on the right <em>not</em>
be expressed as
$$
1 - [1-P(A_1)]\cdots[1-P(A_n)]
$$
and multiplied out to get
$$P(A_1) + \cdots + P(A_n) - [P(A_1)P(A_2) + P(A_1)P(A_3) + \cdots P(A_{n-1})P(A_n)] + \cdots
$$
The latter is an expression based on the <em>principle of inclusion and exclusion</em> and is a lot more work to evaluate.</p>
|
4,270,857 | <p>Solve by separation of variables
<span class="math-container">$$\frac{\partial u}{\partial t}=k\frac{\partial^2 u}{\partial x^2}$$</span>
given intitial conditions:</p>
<p><span class="math-container">$$\frac{\partial u}{\partial x}=0 \text{ at } x=a \text{ and } x=-a, \forall t\geq 0;$$</span>
<span class="math-container">$$u \text{ is bounded for }-a\leq x\leq a\text{ as }t\rightarrow\infty;$$</span>
<span class="math-container">$$u=|x|\text{ for }-a\leq x\leq a\text{ at }t=0.$$</span></p>
<p>So I've been stuck on this problem for a few days now. I've tried a bunch of different things but I'm unsure of how to properly approach the problem. I have the answer, but it's unclear on how to get there. I originally thought that I could start with the general solution of the heat equation which is already known.
<span class="math-container">$$u(x,t)=X(x)T(t)=(A\cos{wx}+B\sin{wx})Ce^{-w^2kt}$$</span>
I started to play with initial conditions using this solution and ultimately I got with <span class="math-container">$B^*=BC, A^*=AC$</span> that <span class="math-container">$B^*=A^*\tan{wa}$</span> which lead to a weird <span class="math-container">$u$</span> that was defined on an interval using the boundary conditions. I couldn't gather much information.</p>
<p>After reviewing notes and the text, "Geoff Stephenson - PDE's for Scientists & Engineers, Ch. 4.3". I thought it would be wise to assume a solution of the form <span class="math-container">$u(x,t)=v(x)+w(x,t)$</span> where <span class="math-container">$\frac{\partial^2 v}{\partial x^2}=0$</span> and <span class="math-container">$w(x,t)=X(x)T(t)$</span>. Using boundary conditions, this lead me to
<span class="math-container">$$u(x,0)=|x|\implies |x|=v(x)+w(x,0)\implies v(x)=|x|-w(x,0)$$</span>
From this can I then say <span class="math-container">$u(x,t)=|x|-w(x,0)+w(x,t)$</span>?</p>
<p>I can't figure this stuff out for the life of me.</p>
| Community | -1 | <p>First, for <span class="math-container">$1\le i\le n+1$</span>,
<span class="math-container">$$
\mathsf{P}(x_i=a)=\frac{n!}{(n+1)!}=\frac{1}{n+1},
$$</span>
and for <span class="math-container">$i\ne j$</span> and <span class="math-container">$a\ne b$</span>,
<span class="math-container">$$
\mathsf{P}(x_i=a,x_j=b)=\frac{(n-1)!}{(n+1)!}=\frac{1}{n(n+1)}.
$$</span>
Thus,
<span class="math-container">$$
\mathsf{E}x_i^2=\sum_{k=0}^n\frac{k^2}{n+1}=\frac{n^2}{3}+\frac{n}{6},
$$</span>
and for <span class="math-container">$i\ne j$</span>,
<span class="math-container">$$
\mathsf{E}x_ix_j=\sum_{k=0}^n\sum_{l\ne k}\frac{kl}{n(n+1)}=\frac{n^2}{4}-\frac{n}{12}-\frac{1}{6}.
$$</span>
Finally, since <span class="math-container">$x_ix_j\overset{d}{=}y_iy_j$</span>,
<span class="math-container">$$
\mathsf{E}Z^2=(n+1)\left(\frac{n^2}{3}+\frac{n}{6}\right)^2+n(n+1)\left(\frac{n^2}{4}-\frac{n}{12}-\frac{1}{6}\right)^2.
$$</span></p>
|
1,557,686 | <p>I have a definition of a <a href="https://en.wikipedia.org/wiki/Hypergeometric_distribution" rel="nofollow">Hypergeometric distribution</a> as follows:</p>
<blockquote>
<p>Definition: the Hypergeometric distribution is a discrete probability distribution that describes the probability of $k$ successes in $n$ draws, without replacement, from a finite population of size $N$ that contains exactly $K$ successes, wherein each draw is either a success or a failure. In contrast, the binomial distribution describes the probability of $k$ successes in $n$ draws with replacement.</p>
</blockquote>
<p>A random variable $X$: no. of successes in <strong>$K$ successes</strong>. The pdf is</p>
<p>$$P(X=k)=\frac{(\text{#ways for $k$ successes})\times (\text{# ways for $n-k$ failures})}{(\text{total number of way to select})}=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$</p>
<p>My question is what is definition of r.v $X$? In my word, I write as <em>"A random variable $X$: no. of successes in <strong>$K$ successes</strong>. "</em>, Is it correct? I am confusing about "no. of successes in <strong>$K$ successes</strong>" or "no. of successes in <strong>$K$ trails</strong>." Thanks in advance</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>
|
4,249,191 | <p>I need to find chromatic number of graph <span class="math-container">$G$</span> obtained from <span class="math-container">$K_{2019}$</span> , by removing one Hamiltonian cycle from it. I already did this example for even <span class="math-container">$n$</span>. Now if <span class="math-container">$V(G)=\{1,2,...,n\}$</span>, and if <span class="math-container">$n$</span> is even, "odd" vertices are forming graph <span class="math-container">$K_{\frac{n}{2}}$</span>, so chromatic number was <span class="math-container">$\ge \frac{n}{2}$</span>. Then it was easy to find one proper coloring with <span class="math-container">$\frac{n}{2}$</span> colors. But now, for <span class="math-container">$2019$</span>, "even" vertices are forming graph <span class="math-container">$K_{1009}$</span> so chromatic number is <span class="math-container">$\ge 1009$</span>. Now, I am not sure that there exists proper coloring with <span class="math-container">$1009$</span> colors, so chromatic number should be <span class="math-container">$1010$</span>. But how to prove that there is not proper coloring with <span class="math-container">$1009$</span> colors?</p>
| Empy2 | 81,790 | <p>Suppose a colouring with 1009 colours is possible.<br />
Take 1008 consecutive even points. They have 1008 different colours.<br />
There is a gap of 5 between the ends. Either of the middle two can complete a set of 1009, so they are the same colour.<br />
Likewise, all adjacent points are the same colour, which gives a contradiction.</p>
|
4,249,191 | <p>I need to find chromatic number of graph <span class="math-container">$G$</span> obtained from <span class="math-container">$K_{2019}$</span> , by removing one Hamiltonian cycle from it. I already did this example for even <span class="math-container">$n$</span>. Now if <span class="math-container">$V(G)=\{1,2,...,n\}$</span>, and if <span class="math-container">$n$</span> is even, "odd" vertices are forming graph <span class="math-container">$K_{\frac{n}{2}}$</span>, so chromatic number was <span class="math-container">$\ge \frac{n}{2}$</span>. Then it was easy to find one proper coloring with <span class="math-container">$\frac{n}{2}$</span> colors. But now, for <span class="math-container">$2019$</span>, "even" vertices are forming graph <span class="math-container">$K_{1009}$</span> so chromatic number is <span class="math-container">$\ge 1009$</span>. Now, I am not sure that there exists proper coloring with <span class="math-container">$1009$</span> colors, so chromatic number should be <span class="math-container">$1010$</span>. But how to prove that there is not proper coloring with <span class="math-container">$1009$</span> colors?</p>
| kabenyuk | 528,593 | <p>Let the vertices of the complete graph <span class="math-container">$K_n$</span> be located at the vertices of a regular <span class="math-container">$n$</span>-gon. Then the edges of <span class="math-container">$K_n$</span> are all possible diagonals of <span class="math-container">$n$</span>-gon and all its sides. Now remove all sides of the <span class="math-container">$n$</span>-gon. We obtain our graph <span class="math-container">$G$</span>. Let <span class="math-container">$n=2019$</span>. Suppose the vertices of <span class="math-container">$G$</span> are painted in <span class="math-container">$1009$</span> colors. Then we find three vertices colored in the same color. But then this coloring is not proper.</p>
|
1,704,707 | <p>The function $f: ℝ → ℝ$ defined by $f(x) = x^{3}$ is onto because for any real number $r$, we have that $\sqrt[3]r$ is a real number and $f(\sqrt[3]r)=r$. Consider the same function defined on the integers $g: ℤ → ℤ$ by $g(n) = n^3.$ Explain why $g$ is not onto $ℤ$ and give one integer that $g$ cannot output. </p>
<p>I can't think of any integer that cannot be cubed, so this problem has me confused.</p>
| KGD | 322,022 | <p>Note that each $f_k$ is 1-lipschitz. In particular, this implies that, for each $\varepsilon > 0$, you can write $(|x-y| < \varepsilon) \Rightarrow (|f_k(x)-f_k(y)| < \varepsilon)$ for each $ k \ge 1$, therefore the family is equicontinuous.</p>
|
1,704,707 | <p>The function $f: ℝ → ℝ$ defined by $f(x) = x^{3}$ is onto because for any real number $r$, we have that $\sqrt[3]r$ is a real number and $f(\sqrt[3]r)=r$. Consider the same function defined on the integers $g: ℤ → ℤ$ by $g(n) = n^3.$ Explain why $g$ is not onto $ℤ$ and give one integer that $g$ cannot output. </p>
<p>I can't think of any integer that cannot be cubed, so this problem has me confused.</p>
| Lutz Lehmann | 115,115 | <p>Note that with $k=0$ this sequence contains its own limit. This is the second most trivial situation for a compact set, the most trivial being a set of finitely many points.</p>
|
3,271,917 | <p><span class="math-container">$$ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $$</span></p>
<p>I have done the<span class="math-container">$ \sum_{n=1}^\infty \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $</span> part , and showed it divergent using Gauss test .</p>
<p>But i am not able to do this part <span class="math-container">$ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $</span> ,tried leibniz test to do , but could not do that.</p>
<p>I have no idea how to do this please help.</p>
| Jack D'Aurizio | 44,121 | <p>It is enough to prove that <span class="math-container">$\frac{\Gamma(n+1/3)}{\Gamma(n+2/3)}$</span> is decreasing to zero, then invoke Leibiz' criterion. On the other hand</p>
<p><span class="math-container">$$\frac{\Gamma(n+1/3)}{\Gamma(n+2/3)}=\frac{1}{\Gamma(1/3)}B(n+1/3,1/3)=\frac{3}{\Gamma(1/3)}\int_{0}^{1}\color{red}{x^{3n}}(1-x^3)^{-2/3}\,dx$$</span>
is <em>obviously</em> decreasing to zero: for any fixed <span class="math-container">$x\in(0,1)$</span>, <span class="math-container">$x^n\searrow 0$</span> as <span class="math-container">$n\to +\infty$</span>.</p>
|
71,354 | <p>I've been reading Moroianu's Kahler geometry notes and found a unattributed quote that says that if the Kahler properties hold, then
"a long list of miracles occur"</p>
<p>I am guessing that this quote belongs to Kahler himself, but I can't back this up. Does anyone know?</p>
| Igor Rivin | 11,142 | <p>I believe the answer is yes, see:</p>
<p><a href="http://books.google.com/books?id=u6WFVmoHxFkC&pg=PA740&dq=Kahler+%2B+%22a+long+list+of+miracles%22" rel="nofollow">http://books.google.com/books?id=u6WFVmoHxFkC&pg=PA740&dq=Kahler+%2B+%22a+long+list+of+miracles%22</a></p>
|
1,443,166 | <blockquote>
<p>If the proposition ¬p→v is true, then the truth value of the
proposition ¬p∨(p→q), where ¬ is negation, ∨ is inclusive OR and → is
implication, is</p>
<ol>
<li>True</li>
<li>False</li>
<li>Multiple Values</li>
<li>Cannot be determined</li>
</ol>
</blockquote>
<hr>
<p><strong>I try to explain</strong></p>
<hr>
<p>Given ,¬p→v = p+v is valid
now , proposition ¬p∨(p→q) = ¬p+q ,</p>
<p>hence using rule of inference ,
= (p+v)→(¬p+q) = ¬p.¬v + ¬p + q = ¬p+q ,</p>
<p>so it cannot be determined.</p>
<p><strong>Edit :</strong> Is my method correct ?</p>
| Eric Auld | 76,333 | <p>I don't see what $v$ has to do with the question. It seems that the statement is true if $\neg p$, and false if $p \land \neg q$.</p>
|
76,457 | <p>I have an ellipse centered at $(h,k)$, with semi-major axis $r_x$, semi-minor axis $r_y$, both aligned with the Cartesian plane.</p>
<p>How do I determine if a point $(x,y)$ is within the area bounded by the ellipse? </p>
| user3731622 | 341,117 | <p>1) Consider the point as a vector</p>
<p>$$
p=\begin{bmatrix}
x \\
y \\
\end{bmatrix}
$$</p>
<p>2) Consider the center of the ellipse as</p>
<p>$$
c=\begin{bmatrix}
h \\
k \\
\end{bmatrix}
$$</p>
<p>3) Subtract the center of the ellipse
$$
p_{centered}= p - c
$$
4) Create a whitening matrix
$$
W = \Lambda^{-1/2}E^T
$$
where
$$
\Lambda = \begin{bmatrix}
r_x & 0 \\
0 & r_y \\
\end{bmatrix}
$$
and
$$
E = \begin{bmatrix}
e_{major} & e_{minor}\\
\end{bmatrix}
$$
where $e_{major}$ and $e_{major}$ are the unit vectors in the direction of the ellipse's major and minor axes. Since you're example is for a non-rotated matrix with major axis along x-axis and minor axis along y-axis</p>
<p>$e_{major}=\begin{bmatrix}
1\\
0\\
\end{bmatrix}$ and $e_{minor}=\begin{bmatrix}
0\\
1\\
\end{bmatrix}$</p>
<p>5) Whiten the point
$$
p_{white} = Wp_{centered}
$$
6) Check if the length of the vector is less than 1.
If it is, then the point is within the ellipse.</p>
<p>Note: this is inspired by my experience with Covariance matrices.
I'll try to update this answer with an intuitive relation b/w ellipses and covariance matrices. For now you can take a peak at <a href="http://www.visiondummy.com/2014/04/draw-error-ellipse-representing-covariance-matrix/" rel="noreferrer">http://www.visiondummy.com/2014/04/draw-error-ellipse-representing-covariance-matrix/</a></p>
|
898,082 | <p>I have the answer for this, but my teacher hadn't taught the whole "when cosine is an even, the value of $-\arccos (-0.7)$ is a solution too." </p>
<p>Please: </p>
<p>-tell me when a $\cos$/$\sin$ function is even/odd </p>
<p>-what happens if its odd? </p>
<p>-how to use the "$±\arccos(-0.7) + 2kπ$" ( don't understand why you add 2kπ) </p>
<p>-and how to find the solutions! </p>
<p>Another example is $$\sec( x )= -3, -π ≤ x < π $$</p>
<p>I really don't understand what happens if the function is "even" or "odd." And how to determine if it is.</p>
| chaffdog | 168,595 | <p>As others have said, a function $f(x)$ is called even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.</p>
<p>We add $2k\pi$ to the answer because both sine and cosine are periodic of period $2\pi$. To see this, you can think of sine and cosine as the $x$ and $y$ values respectively of a point on the circumference of the unit circle:
<img src="https://i.stack.imgur.com/54ssa.png" alt="Sine and Cosine on the unit circle."></p>
<p>Notice that once we go around the circle by $2\pi$, all the $x$ and $y$ values are repeated. Hence, for any equation like $\cos(x) = -0.7$, there are an infinite number of answers. However the "primary" answers all occur within a $2\pi$ window, all other answers are in some sense duplicates of these answers (they correspond to the same points on the unit circle, but after having gone around the circle more than once).</p>
<p>The value of $\arccos(-0.7)$ is defined to be <strong>unique</strong> value of $x$ between $0$ and $\pi$ such that $\cos(x) = -0.7$. There is another solution to this equation on the bottom half of the circle ($\pi$ to $2\pi$ or $0$ to $-\pi$). This is where odd and even functions come in. Because cosine is even, $cos(-x) = cos(x)$, so we know that if $\cos(x) = =0.7$, $\cos(-x) = -0.7$ as well.</p>
<p>To get all the other infinitely many answers to the problem, we can add multiples of $2\pi$ to our original two answers. This is why the general solution is of the problem has $2k\pi$ added.</p>
|
944,965 | <p>Question: whats the order of the element a=33 in Z60 (under modular addition)?</p>
<p>Answer: $\langle 33 \rangle= \{33,6,39,12,45,18,51,24,57,30,3,36,9,42,15,48,21,54,27,0,\}$. Therefore, $\text{order}(33)=20$</p>
<p>Whats the inverse of $33$ in $\mathbb{Z}_{60}$ (under modular addition)?</p>
<p>Im struggling with finding the inverse. Can anyone show me how to find it? </p>
<p>I would deeply appreciate your work and efforts</p>
<p>Thanks</p>
| Adriano | 76,987 | <p><strong>Hint:</strong> From your previous work, we know that:
$$
0 = \underbrace{33 + 33 + \cdots + 33}_{20 \text{ times}} = 33 + \underbrace{33 + \cdots + 33}_{19 \text{ times}}
$$
Thus, we know that the inverse of $33$ is:
$$
\underbrace{33 + \cdots + 33}_{19 \text{ times}}
$$
Incidentally, this is just $60 - 33$.</p>
|
3,552,779 | <p>In a 3D world, given a box <code>B</code>, a pivot <code>P</code>, and a direction vector <code>V</code>, how can I find out how much to rotate at <code>P</code> such that <code>V</code> points towards an arbitrary point <code>A</code>?</p>
<p>Problem source:</p>
<p>I am a software developer that come across the need to rotate an object in the above manner, where a 3d model need to be rotated in this way for the user to interact with.</p>
<p>Current Attempts:</p>
<p>I tried using an offset between the direction vector and the pivot, and calculate the rotation required between the offseted target and the pivot.</p>
<p>However all my current attempts is done in code, and I left the mathematical calculation to the libraries due to my limited knowledge - which means to be honest I am not very clear what they actually do.</p>
<p>Note:</p>
<ol>
<li><code>B</code> can be of any arbitrary size, </li>
<li><code>P</code> can be anywhere within the box</li>
<li><code>V</code> can be anywhere within the box</li>
<li><code>A</code> can be anywhere in the world</li>
</ol>
<p><a href="https://i.stack.imgur.com/sEGP9.png" rel="nofollow noreferrer">An illustration of what I am aiming for in 2D</a></p>
| user170231 | 170,231 | <p>Without too much simplification, the substitution you cite yields</p>
<p><span class="math-container">$$\int_2^3\frac9{(x-2)^{1/4}}\,\mathrm dx=36\int_0^1\frac{u^3}u\,\mathrm du$$</span></p>
<p>which you certainly welcome to treat as an improper integral,</p>
<p><span class="math-container">$$36\left(\frac13-\lim_{u\to0^+}\frac{u^3}3\right)$$</span></p>
<p>but since <span class="math-container">$u=0$</span> is a removable discontinuity and the limand reduces to <span class="math-container">$u^2$</span>, you may as skip this treatment altogether.</p>
|
2,012,080 | <p>Consider the PDE for $v(x,y)$:</p>
<blockquote>
<p>$v_{xx} − v_{xy} − 6v_{yy} − 5v_{x} − 10v_{y} + 25 = 0$</p>
</blockquote>
<p>Taking $\xi = y-2x$ and $\eta=y+3x$ that I found from $b^2-4ac$ I have transformed the equation to the form</p>
<p>$v_{\eta\xi} + v_{\eta}=1$</p>
<p>I have been posed the question:</p>
<blockquote>
<p>Using this result obtain the general solution, casting your answer in terms of
the x and y variables.</p>
</blockquote>
<p>I am not sure how to solve this problem. Could anyone point me in the right direction?</p>
| Aleksas Domarkas | 562,074 | <p>Solve PDE
$$v_{xx} − v_{xy} − 6v_{yy} − 5v_{x} − 10v_{y} + 25 = 0 $$</p>
<ul>
<li>$D_x^2-D_xD_y-6D_y^2-5D_x-10D_y=
\left( {D_x}-3 {D_y}-5\right) \, \left( {D_x}+2 {D_y}\right)$ </li>
<li>General solution of $\quad v_x+2v_y=0\quad$ is $\quad v_1=f(y-2x)$</li>
<li>General solution of $\quad v_x-3v_y-5v= 0\quad$ is $\quad v_2=e^{5x}g(3x+y)$</li>
<li>Partial solution of PDE is $\quad v_p=5x$</li>
<li>General solution of PDE is $v=v_1+v_2+v_p=f(y-2x)+e^{5x}g(3x+y)+5x$</li>
</ul>
|
165,385 | <p>I am generating a list of 1's, 2's, and 3's with different probabilities for each number. I then convert this list into three binarized lists, giving a list of the locations of each digit in the original list. There must be a more efficient way to do this? Perhaps changing the way I make the original list?</p>
<pre><code>list = RandomChoice[{0.5, 0.3, 0.2} -> {1, 2, 3}, 20]
list1 = ReplacePart[ReplacePart[ReplacePart[list, Position[list, 2] -> 0],
Position[list, 3] -> 0], Position[list, 1] -> 1]
list2 = ReplacePart[ReplacePart[ReplacePart[list, Position[list, 1] -> 0],
Position[list, 3] -> 0], Position[list, 2] -> 1]
list3 = ReplacePart[ReplacePart[ReplacePart[list, Position[list, 1] -> 0],
Position[list, 2] -> 0], Position[list, 3] -> 1]
</code></pre>
| Carl Woll | 45,431 | <p>There are many possibilities, but I like using <a href="http://reference.wolfram.com/language/ref/Clip" rel="nofollow noreferrer"><code>Clip</code></a>:</p>
<pre><code>Unitize@Clip[list,{1,1},{0,0}]
Unitize@Clip[list,{2,2},{0,0}]
Unitize@Clip[list,{3,3},{0,0}]
</code></pre>
<blockquote>
<p>{0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0}</p>
<p>{0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1}</p>
<p>{1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0}</p>
</blockquote>
<p><strong>Addendum</strong></p>
<p>If your lists consist of just integers, than a compiled version is possible, e.g.:</p>
<pre><code>fc = Last @ Compile[{{d, _Integer, 1}, {t,_Integer}},
Table[Boole[Compile`GetElement[d, i]==2], {i, Length@d}],
CompilationTarget->"C",
RuntimeOptions->"Speed"
];
</code></pre>
<p>Comparison:</p>
<pre><code>lst = Developer`ToPackedArray @ RandomChoice[{.5,.2,.3}->{1,2,3}, 10^7];
r1 = fc[lst, 2]; //RepeatedTiming
r2 = BitXor[1, Unitize @ BitXor[2, lst]]; //RepeatedTiming
r1 === r2
</code></pre>
<blockquote>
<p>{0.042, Null}</p>
<p>{0.0614, Null}</p>
<p>True</p>
</blockquote>
|
165,385 | <p>I am generating a list of 1's, 2's, and 3's with different probabilities for each number. I then convert this list into three binarized lists, giving a list of the locations of each digit in the original list. There must be a more efficient way to do this? Perhaps changing the way I make the original list?</p>
<pre><code>list = RandomChoice[{0.5, 0.3, 0.2} -> {1, 2, 3}, 20]
list1 = ReplacePart[ReplacePart[ReplacePart[list, Position[list, 2] -> 0],
Position[list, 3] -> 0], Position[list, 1] -> 1]
list2 = ReplacePart[ReplacePart[ReplacePart[list, Position[list, 1] -> 0],
Position[list, 3] -> 0], Position[list, 2] -> 1]
list3 = ReplacePart[ReplacePart[ReplacePart[list, Position[list, 1] -> 0],
Position[list, 2] -> 0], Position[list, 3] -> 1]
</code></pre>
| Alucard | 18,859 | <pre><code>list1e = Boole[# == 1] & /@ list;
list2e = Boole[# == 2] & /@ list;
list3e = Boole[# == 3] & /@ list;
</code></pre>
<p>another option is using Piecewise. For example:</p>
<pre><code>fun[x_] := Piecewise[{{1, x == 1}}, 0]
SetAttributes[fun, Listable]
list1c = fun[list];
</code></pre>
<p>or the combination Reap/Scan:</p>
<pre><code>list1d = Reap[Scan[If[# == 1, Sow[1], Sow[0]] &, list]][[2, 1]];
</code></pre>
|
141,128 | <p>I have a complicated function for <code>StreamPlot</code> to call that includes some error checking and the use of <code>Message</code>. Somehow the messages are not displayed when the function is called from <code>StreamPlot</code>. Here's a toy example:</p>
<p>Define the function & message:</p>
<pre><code>f[x_] := Module[{},
If[x > 1, Print["print: ", x]; Message[f::msg, x]];
x];
f::msg = "`1` is greater than 1";
</code></pre>
<p>Test it by itself:</p>
<pre><code>f[1.2]
(* print: 1.2 *)
(* f::msg f:1.2` is greater than 1 *)
(* 1.2 *)
</code></pre>
<p>Works as expected, but put it in a <code>StreamPlot</code> and only the <code>Print</code>ed messages are shown:</p>
<pre><code>StreamPlot[{f[x], y}, {x, 0, 1.2}, {y, 0, 1}]
(* txt: 1.22403 *)
(* txt: 1.22403 *)
(* ... *)
</code></pre>
<p>How can I get those <code>Messages</code> to show?</p>
<p><strong>Update:</strong></p>
<p><code>Plot</code> eats <code>Messages</code> too.</p>
| Michael E2 | 4,999 | <p>You need to reset <code>$Messages</code>.</p>
<pre><code>f[x_] := Module[{},
If[x > 1, Print["print: ", x];
Block[{$Messages = Streams["stdout"]},
If[! ListQ@$MessageList, $MessageList = {}];
Message[f::msg, x]]];
x];
</code></pre>
<p><code>$MessageList</code> should be a list, which if is not when <code>f</code> is called. Just what to do about it, whether to <code>Block</code> it too or reset it globally as above, I am not completely sure. It worked both ways with the following plot:</p>
<pre><code>Plot[f[x], {x, 0, 1.01}]
</code></pre>
|
289,757 | <p>I am writing this, as I am a currently an intern at an aircraft manufactur. I am studying a mixture of engineering and applied math. During the semester I focussed on numerical courses and my applied field is CFD. Even though every mathematician would say I have not heard a lot of math, for myself I would say that I get the "most amount of math" you can get while not studying math.</p>
<p>In my courses I have done deep theoretical analysis for numerical concepts and application in CFD. But currently I am starting to wonder, how much the e.g. Calculus of Variation course really helps me in my future career. The theory you learn at university seems to get only a little application in the <em>real word</em>. </p>
<p>Example: In my numerics for PDE class I have spent (wasted?) so many hours on trying to figure out the CFL number of certain schemes, but what I am doing right now has nothing to do with that. <em>Oh your simulation does diverge? Well let's take 2 instead of 4 as our CFL number.</em>
Furthermore, I am not really programming stuff as I hoped I could, but I am rather scripting. Fact is, 99 out of 100 people are not going to program a CFD solver. You rather use the code and apply it to your needs.</p>
<p>I am aware that university always follows a way more theoretical path than industry, but I am actually disappointed how little math I am really doing. Okay you might, say that's due to the fact that I am an intern and of course you are right. But I am in the lucky situation, that my team comes really close to research. Most of the members hold a PhD and studied engineering or math, and the focus is definetely on research ( in this departure of the company). But if the amount of math is that small in such an environment, where are you really able to make use of what you have learned at university.</p>
<p>So here comes my question</p>
<blockquote>
<p>How much math are you actually doing at your job?
And I don't mean, how much math is helping you to understand things, but how often does it happen, that you sit down and really <strong>do math</strong> in your non-academic job?</p>
</blockquote>
<p>Personally I get the impression that I could do the exact work without having heard most of my courses. Don't get me wrong, I really enjoy the theory, but currently I am rather frustrated.</p>
<p>Note: As this is my first Question, I hope I did not screw up completely. I did not found similar questions on this side. And feel free to edit or ask questions if thinks are not clear.</p>
| TakeS | 38,349 | <p>In my job as a (thus far) applied computer science researcher, I can sympathize with you in that I am not utilizing as much math directly from my courses into my job. Sure, I have to know some basic probability, but much of my job consists of reading research papers where most math consists of the standard operations along with some tricky, but not difficult, summations and integrals. I am a math major, but I could easily perform my research if I had never taken abstract algebra or real analysis. If I ever did need knowledge from abstract algebra, the fact that I took the class makes it much easier for me to "re-understand" concepts again.</p>
<p>But one thing that may be misleading is that, no matter what you major in, you're unlikely to use a significant fraction of whatever you study in your everyday job. In my opinion, it's far more likely that you've encountered a small topic in one course that you end up working on in depth, as deep knowledge of one subject trumps general breadth.</p>
<p>Regardless, I still take math courses as they are among my favorites. And the point of university isn't solely to provide preparation for <em>specific jobs</em>; that's why people do internships. Experience can also trump coursework.</p>
|
2,466,022 | <p>In Real and Complex Analysis, 3rd Edition, Walter Rudin advances the following:</p>
<p><a href="https://i.stack.imgur.com/XEVwK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XEVwK.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/dTwbg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dTwbg.png" alt="enter image description here"></a></p>
<p>How does $e^z \cdot e^{-z} = 1$ entail $(a)$?</p>
| Sunyam | 463,614 | <p>Another way to see is a proof by contradiction. Assume there exists a complex number z for which $e^z=0$.Then $|e^z|=0$. Which implies for this z, $e^{\Re{z}}=0$. Which is not true for any real number $\Re{z}$. Hence proved.</p>
|
2,447,850 | <p>So I have to prove 2 things:</p>
<ol>
<li><p>That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R, x>0$. </p></li>
<li><p>That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R$. </p></li>
</ol>
<p>For #1, I know that $\frac{x^n}{n!} >0$, which means that I can find an upper bound and use squeeze theorem. For #2, I have no idea where to start.</p>
| Zestylemonzi | 270,448 | <p>Hint: Given $x$, take $N \in \mathbb{N}$ such that $N>x$. Then, for $n >N$,
$$\frac{x^n}{n!} = \frac{x^N}{N!} \cdot \frac{x^{n-N}}{(n!/N!)} \le \frac{x^N}{N!} \cdot \left(\frac{x}{N+1}\right)^{n-N} .$$</p>
|
2,447,850 | <p>So I have to prove 2 things:</p>
<ol>
<li><p>That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R, x>0$. </p></li>
<li><p>That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R$. </p></li>
</ol>
<p>For #1, I know that $\frac{x^n}{n!} >0$, which means that I can find an upper bound and use squeeze theorem. For #2, I have no idea where to start.</p>
| Michael Hardy | 11,667 | <p>$$
\frac {x^n}{x!} = \frac{\overbrace{x\cdot x\cdot x\cdots\cdots x}^{\Large n \text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots\cdots n}_{\Large n \text{ factors}}}
$$
When $n$ gets to be more than twice as big as $x,$ then multiplying by $\dfrac x n$ reduces the thing to less than half what it was. Then $n$ keeps growing (while $x$ does not) and at some point $n$ is more than three times as big as $x$ and then you're multiplying by $\dfrac x n < \dfrac 1 3$ and making the thing at each step less than $1/3$ what it had been. And so on$\,\ldots$ So it approaches $0.$</p>
<p>(That assumes $x$ is positive. If $x$ is negative, then think about $\dfrac{|x|^n}{n!}.$ It that approaches $0,$ then without the absolute value that still approaches $0.$</p>
|
128,651 | <p>Let $M$ be a smooth manifold (maybe compact, if that helps). Denote by $\operatorname{Diff}(M)$ the group of diffeomorphisms $M\to M$ and by $R(M)$ the space of Riemannian metrics on $M$. We obtain a canonical group action
$$ R(M) \times \operatorname{Diff}(M) \to R(M), (g,F) \mapsto F^*g, $$
where $F^*g$ denotes the pullback of $g$ along $F$. Is this action transitive? In other words, is it possible for any two Riemannian metrics $g,h$ on $M$ to find a diffeomorphism $F$ such that $F^*g=h$? Do you know any references for this type of questions?</p>
| Holonomia | 73,953 | <p>The group of C^1 diffeomorphism does not act transitively on the space of Riemannian metrics on a compact manifold. For example, two circles of different radius have different diameters and by pullback we can "copy" the metric of one of them on the other. Doing so we get two metrics on a circle with different diameters, hence they can not be C^1 related. </p>
<p>Thus, the point is that "the diameter" is a C^1-invariant hence an homotety of the metric change the diameter and the "new metric" have different diameter. </p>
<p>For non compact manifolds, by using a theorem of Nomizu, there exist both complete and non-complete riemannian metrics. See <a href="http://www.oberlin.edu/faculty/jcalcut/Nomizu_Ozeki_1961.pdf" rel="nofollow">http://www.oberlin.edu/faculty/jcalcut/Nomizu_Ozeki_1961.pdf</a> </p>
<p>Since "completeness" is a C^1 invariant we get that the group of C^1 diffeomorphism does not act transitively on the space of riemannian metrics.</p>
<p>h. </p>
|
1,515,900 | <p>In many books and papers on analysis I met this equality without proof:</p>
<p>$$\sup \limits_{t\in[a,b]}f(t)-\inf \limits_{t\in[a,b]}f(t)=\sup \limits_{t,s\in[a,b]}|f(t)-f(s)|$$</p>
<p>Can anyone show strict and nice proof of that equality?</p>
<p>I would really grateful foe your help!</p>
| robjohn | 13,854 | <p>$$
\begin{align}
\sup_{t\in[a,b]}f(t)-\inf_{s\in[a,b]}f(s)
&=\sup_{t\in[a,b]}f(t)+\sup_{s\in[a,b]}(-f(s))\\
&=\sup_{s,t\in[a,b]}\big(f(t)+(-f(s))\big)\\
&=\sup_{s,t\in[a,b]}\big(f(t)-f(s)\big)\\
&=\sup_{s,t\in[a,b]}\left|f(t)-f(s)\right|
\end{align}
$$
The last equality follows since $\big\{f(t)-f(s):s,t\in[a,b]\big\}$ is symmetric about $0$.</p>
|
1,051,357 | <p>$$
\begin{bmatrix}
-1&&2&&-3&&4\\
5&&0&&2&&-2\\
2&&1&&1&&2\\
0&&0&&3&&-2\\
\end{bmatrix}
$$</p>
<p>I wanted to confirm that if I use $(w, x,$$ y, z)$ would the image of the vector be:</p>
<pre><code>-w + 2x - 3y + 4z
5w + 2y - 2z
2w + x + y + 2z
3y - 2z
</code></pre>
<p>Could anybody provide some help?</p>
<p>If i wanted to check if this matrix is invertible how would i go about doing that?</p>
| NicNic8 | 24,205 | <p>Your statement is correct, however, it may be able to be simplified.</p>
<p>One thing you could do is determine the reduced row echelon form (rref) of your matrix. Then the image (or column space) of your matrix is the column space of the rref of your matrix.</p>
|
830,599 | <p>The function $f$ is defined as follows:
$$f(x):=\sum_{j=1}^{\infty} \frac{x^j}{j!} e^{-x}$$</p>
<p>It's easy to see that $f(0)=0$. But I am interested in the value
$$\lim_{x \rightarrow 0^+} f(x).$$</p>
<p>Even <a href="https://www.wolframalpha.com/input/?i=lim_%28x-%3E0%29+%28sum_%28i%3D1%29%5Einfinity+x%5Ej%2F%28j%21%29+e%5E%28-x%29%29+" rel="nofollow">Wolfram Alpha</a> does not help here. I tried to plot this function, but this doesn't work neither. And my calculator doesn't give a solution for concrete values of $x$, so I have no idea how to get on here. </p>
| zuriel | 147,911 | <p>Notice that $e^x=\displaystyle\sum_{j=0}^{\infty}\displaystyle\frac{x^j}{j!}$. </p>
<p>So $\displaystyle\sum_{j=1}^{\infty} \frac{x^j}{j!} e^{-x}=\displaystyle\sum_{j=0}^{\infty} \frac{x^j}{j!} e^{-x}-e^{-x}=1-e^{-x}$</p>
|
1,371,549 | <p>Differentiate the Function : $y=\log_2(e^{-x} \cos(\pi x))$</p>
<p>Here is my work. What I have I done wrong?
<a href="https://i.stack.imgur.com/w9RSN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w9RSN.jpg" alt="enter image description here"></a></p>
| mathlove | 78,967 | <p>$$\frac{-\sin(\pi x)\cdot \pi+\cos(\pi x)\cdot (-1)}{e^x(e^{-x}\cos(\pi x))\cdot \ln 2}$$
is correct, <strong>but the last expression you wrote is not correct</strong>. You have a mistake when you simplify the expression (you miss minus signs. Also, note that $e^x\cdot e^{-x}=1$). The correct answer will be
$$\frac{-\pi\sin(\pi x)-\cos(\pi x)}{\cos(\pi x)\cdot\ln 2}$$</p>
|
2,135,811 | <p>$ \frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 $</p>
<p>I've tried putting $t= \frac{6x}{x-2} $ and play algebraically, using square of sum, but still no luck, any help?</p>
| Michael Rozenberg | 190,319 | <p>Let $\sqrt[4]{\frac{12x}{x-2}}=t$. Hence, $t\geq0$ and
$$\frac{t^4}{2}-t^2-2t>0$$ or
$$t(t^3-2t-4)>0$$ or
$$t(t^3-2t^2+2t^2-4t+2t-2)>0$$ or
$$t(t-2)(t^2+2t+2)>0$$ or
$$t(t-2)((t+1)^2+1)>0$$ or
$$t(t-2)>0$$ and since $t\geq0$, we obtain $t>2$ or
$$\frac{12x}{x-2}>16$$ or
$$\frac{3x}{x-2}-4>0$$ or
$$\frac{8-x}{x-2}>0$$ or$$2<x<8$$
Done!</p>
|
4,279,409 | <p>Consider the ring <span class="math-container">$\mathbb{R}[x]$</span>, determine the ideal <span class="math-container">$$(x^3-5x^2+6x-2,x^4-4x^3+3x^2-4x+2)$$</span></p>
<p>Is it principal? Is it prime? Is it maximal?</p>
<p>Attempt:</p>
<p>We have that <span class="math-container">$x^3-5x^2+6x-2=0$</span> in this ideal. Thus, we have that <span class="math-container">$(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))=0$</span>.</p>
<p>If <span class="math-container">$x=1$</span> the ideal reduces to <span class="math-container">$$(x^3-5x^2+6x-2,-2)$$</span></p>
<p>which contains a unit. Similarly, if <span class="math-container">$x=2+\sqrt{2},2-\sqrt{2}$</span>, the ideal contains a unit. So the ideal is the whole ring. Therefore, it is principal. The ideal is neither prime nor maximal, since such ideals are proper by definition.</p>
<p>Is any of this right? I am wondering if <span class="math-container">$(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))=0$</span> in the ideal means we can have <span class="math-container">$x$</span> being equal to any one of these roots. Or is just equal to one of these, and we do not know which one?</p>
| user7427029 | 513,618 | <p>I would try something named »coefficient comparison«. For example, for <span class="math-container">$jh + 1 = 0$</span>, we get <span class="math-container">$1 = \dots x^{jh + 1}$</span></p>
|
4,279,409 | <p>Consider the ring <span class="math-container">$\mathbb{R}[x]$</span>, determine the ideal <span class="math-container">$$(x^3-5x^2+6x-2,x^4-4x^3+3x^2-4x+2)$$</span></p>
<p>Is it principal? Is it prime? Is it maximal?</p>
<p>Attempt:</p>
<p>We have that <span class="math-container">$x^3-5x^2+6x-2=0$</span> in this ideal. Thus, we have that <span class="math-container">$(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))=0$</span>.</p>
<p>If <span class="math-container">$x=1$</span> the ideal reduces to <span class="math-container">$$(x^3-5x^2+6x-2,-2)$$</span></p>
<p>which contains a unit. Similarly, if <span class="math-container">$x=2+\sqrt{2},2-\sqrt{2}$</span>, the ideal contains a unit. So the ideal is the whole ring. Therefore, it is principal. The ideal is neither prime nor maximal, since such ideals are proper by definition.</p>
<p>Is any of this right? I am wondering if <span class="math-container">$(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))=0$</span> in the ideal means we can have <span class="math-container">$x$</span> being equal to any one of these roots. Or is just equal to one of these, and we do not know which one?</p>
| B. Goddard | 362,009 | <p>If you expand <span class="math-container">$(a+bx^h)^4$</span> you get</p>
<p><span class="math-container">$$a^4 + 4a^3bx^h + 6 a^2b^2x^{2h} + 4ab^3x^{3h} + b^4x^{4h}.$$</span></p>
<p>So your equation is</p>
<p><span class="math-container">$$-1 + x+ ax(a+bx^h)^4 = 0.$$</span></p>
|
516,244 | <p>My professor gave us this example on her notes:</p>
<p>$$\sum_{n = 1}^\infty \left(\frac{3}{n(n+3)}+\frac{1}{2^n}\right)$$</p>
<p>So I know we're supposed to find the partial fraction, which ends up being</p>
<p>$$\left(\frac{3}{n(n+3)}=\frac{A}{n}+\frac{B}{n+3}=
\frac{1}{n}-\frac{1}{n+3}\right)$$</p>
<p>So based on how she did the other examples, I would expect her to do:</p>
<p>$$\sum_{n = 1}^\infty \left(\frac{3}{n(n+3)}=\frac{1}{1}-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}\right.....)$$, because I'd be plugging in numbers for n starting with n=1. However, she instead did the following:</p>
<p>$$\sum_{n = 1}^\infty \left(\frac{3}{n(n+3)}=\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}\right)$$,</p>
<p>which would definitely be a lot more helpful in helping cancel out terms like you're supposed to when doing telescoping series, BUT I don't know why she's doing this. I thought we were supposed to plug in values from n and that's what should be increasing each time, but instead the number being added to n is the one going up and I have no clue why. I don't think I'm asking this question in the best way possible, but I'm kinda confusing myself because she did other examples and they feel nothing like this and I'm just starting to learn all this, so can somebody please give me some insight as to what is going on?</p>
<p>(and I know I'm supposed to also deal with the sum of the $$\frac{1}{2^n}$$ term but I'm kinda ignoring it for now since I don't even know what's going on with the first one</p>
| Shobhit | 79,894 | <p>if $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the vertices of a triangle then its area is given by:-</p>
<p>$$\frac{
1}{2}|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|$$</p>
|
516,244 | <p>My professor gave us this example on her notes:</p>
<p>$$\sum_{n = 1}^\infty \left(\frac{3}{n(n+3)}+\frac{1}{2^n}\right)$$</p>
<p>So I know we're supposed to find the partial fraction, which ends up being</p>
<p>$$\left(\frac{3}{n(n+3)}=\frac{A}{n}+\frac{B}{n+3}=
\frac{1}{n}-\frac{1}{n+3}\right)$$</p>
<p>So based on how she did the other examples, I would expect her to do:</p>
<p>$$\sum_{n = 1}^\infty \left(\frac{3}{n(n+3)}=\frac{1}{1}-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}\right.....)$$, because I'd be plugging in numbers for n starting with n=1. However, she instead did the following:</p>
<p>$$\sum_{n = 1}^\infty \left(\frac{3}{n(n+3)}=\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}\right)$$,</p>
<p>which would definitely be a lot more helpful in helping cancel out terms like you're supposed to when doing telescoping series, BUT I don't know why she's doing this. I thought we were supposed to plug in values from n and that's what should be increasing each time, but instead the number being added to n is the one going up and I have no clue why. I don't think I'm asking this question in the best way possible, but I'm kinda confusing myself because she did other examples and they feel nothing like this and I'm just starting to learn all this, so can somebody please give me some insight as to what is going on?</p>
<p>(and I know I'm supposed to also deal with the sum of the $$\frac{1}{2^n}$$ term but I'm kinda ignoring it for now since I don't even know what's going on with the first one</p>
| Beedassy Lekraj | 243,296 | <p>The area $A$ of the triangle two of whose vertices lie on the axes, with coordinates $(a, 0)$, $(0, b)$, and a third vertex $(c, d)$ is obtained from previous formula by a mere horizontal axis shift of -a units as
$$A = \frac{|-ad + b(a - c)|}{2}$$</p>
|
801,680 | <p>Is it true that every point in a space equipped with the Hausdorff topology is closed?</p>
<p>I am assuming this is true but am having a difficult time proving it. Fix $a \in X$ where $X$ is equipped with the Hausdorff topology. If I can show that $X \setminus \{a\}$ is open then I am finished.</p>
<p>I know that $\forall y \in X \setminus \{a\}$ there exists $U$ and $V$ such that $a \in U$ and $y \in V$ and $U \cap V = \emptyset$. </p>
<p>In general, I have been having a difficult time with these types of proofs. The fact that $U$ and $V$ change based on my choice of $y$ causes some confusion. </p>
| SomeOne | 87,286 | <p><strong>Hint:</strong> we could use the fact: Intersections of closed sets is a closed set, in order to construct singleton <span class="math-container">$\{x\}$</span>,if we can show that in Hausdorff then automatically it will be a closed set.</p>
<ol>
<li><p>For any <span class="math-container">$y \in X \ \& \ y \neq x $</span> there exist two open set such that</p>
<p><span class="math-container">$x \in U_y \ \& \ y \in V_y: U_y \bigcap V_y = \phi \Rightarrow y \notin \overline {U_{y}}$</span> (since it's Hausdorff)</p>
</li>
</ol>
<p>2.for any <span class="math-container">$y \neq x,\Rightarrow y \notin \overline U_y \Rightarrow $</span> <span class="math-container">$y \notin \bigcap_{y \in X -\{x\} } \overline {U_{y}} \Rightarrow \bigcap_{y \in X -\{x\} } \overline {U_{y}} =\{x\}$</span></p>
<p>but <span class="math-container">$\overline {U_{y}} $</span> is close set for each <span class="math-container">$y$</span> therefore <span class="math-container">$ \bigcap_{y \in X -\{x\} } \overline {U_{y}} =\{x\} $</span> is close set</p>
<p><em>Q.E.D</em></p>
|
1,743,465 | <p>If $a=(1,2,3,4,5)$ is an example of a vector in $\mathbb R^5$, what could be an example of a vector in $\mathbb C^5$ ? is it $a=(1,2,3,4,i5)$ ?</p>
<p>Also, $x=a+ib$ is $2$ dimensional, can a complex number be one dimensional? like when $a=0$ or $b=0$, but if $b=0$ then it is a real number, so can we say that all real numbers (scalars) are one dimensional complex numbers?</p>
| EHH | 133,303 | <p>A vector in $\mathbb{P}^n$ is anything of the form $(x_1,\ldots,x_n)$ for $x_i \in \mathbb{P}$, so you $(1,2,3,4,5i)$ is indeed in $\mathbb{C}^5$ and so is any $(x_1,\ldots,x_n)$ with all $x_i \in \mathbb{C}$ (remember $\mathbb{R}$ is a subset of $\mathbb{C}$).</p>
<p>Dimensionality then depends on your field of scalars, i.e. the set of number you are allowed to multiply the vectors by. Over $\mathbb{R}$ $\{x = a + ib\}$ is 2 dimensional, but over $\mathbb{C}$ it is just 1 dimensional since we can hit every complex number by taking the number 1 as our basis vector and then multiplying it by any complex number. </p>
|
1,022,485 | <p>I'm working on a homework problem for my discrete math class, and I'm stuck. (Note: I made a post about this earlier, but I read the problem incorrectly, thus the work was wrong, so I deleted the post.)</p>
<p>Prove by mathematical induction that for every integer n $\ge$ 0, $12\mid8^{2n+1}+2^{4n+2}$</p>
<p>I start out by proving the base case, $F(0)$, to be true:</p>
<p>$$
F(0)=8^1+2^2=12\quad \text{Obviously, 12 is divisible by 12}
$$</p>
<p>I then move on to the induction step to prove the $F(n+1)$ is true:</p>
<p>I assume that $8^{2n+1}+2^{4n+2}$ is divisible by 12, and then plug in $(n+1)$:</p>
<p>$$
F(n+1)=8^{2(n+1)+1}+2^{4(n+1)+2}=8^{2n+3}+2^{4n+6}
$$</p>
<p>I then do $F(n+1)-F(n)$:</p>
<p>$$
F(n+1)-F(n)=(8^{2n+3}+2^{4n+6})-(8^{2n+1}+2^{4n+2})
$$</p>
<p>$$
=8^{2n+3}-8^{2n+1}+2^{4n+6}-2^{4n+2}
$$</p>
<p>I then factor out the terms used in $F(n)$:</p>
<p>$$
8^{2n+1}(8^2-1)+2^{4n+2}(2^4-1)=8^{2n+1}(63)+2^{4n+2}(15)
$$</p>
<p>I can re-write the result as:</p>
<p>$$
8^{2n+1}(63)+2^{4n+2}(12+3)
$$</p>
<p>This is where I'm stuck. I broke up the $15$ into $12+3$ since I need to prove that there is a multiple of 12, but I don't know what to do with the 63, since (I think) you're supposed to have the terms in $F(n)$ multiplied by 3 after you distribute so that you can factor out the 3 and have $F(n)$ in the equation, which is proven to be divisible by 12. </p>
<p>I tried splitting the $63$ into $(21*3)$</p>
<p>$$
8^{2n+1}(21*3)+2^{4n+2}(12+3)
$$</p>
<p>But I'm not sure what to do next. Any ideas?</p>
| Przemysław Scherwentke | 72,361 | <p>If you already have $F(n)=12k$ for some $k\in\mathbb{N}$ and
$$
F(n+1)=8^{2(n+1)+1}+2^{4(n+1)+2}=8^{2n+3}+2^{4n+6},
$$
then
\begin{align}
8^{2n+3}+2^{4n+6}
&=64\cdot8^{2n+1}+16\cdot2^{4n+2}\\
&=12(5\cdot8^{2n+1}+2^{4n+2})+4(8^{2n+1}+2^{4n+2})\\
&=12(5\cdot8^{2n+1}+2^{4n+2}+4k),
\end{align}
which is divisible by 12.</p>
|
1,573,947 | <p>I'm trying to see the relationship between the sample variance equation </p>
<p>$\sum(X_i- \bar X)^2/(n-1)$ and the variance estimate, $\bar X(1-\bar X),$ in case of binary samples. </p>
<p>I wonder if the outputs are the same, or if not, what is the relationship between the two??</p>
<p>I'm trying to prove their relationship but it's quite challenging to me.. </p>
<p>Please help!</p>
<p><a href="https://i.stack.imgur.com/xYZAH.jpg" rel="nofollow noreferrer">Sigma(Xi-Xbar)/(n-1)</a>
<a href="https://i.stack.imgur.com/EMi0G.jpg" rel="nofollow noreferrer">Xbar(1-Xbar)</a></p>
| zyx | 14,120 | <p>The first quantity is the standard variance estimator that is unbiased for i.i.d samples from any distribution.</p>
<p>The second quantity is a simplified formula (the simplification being valid only for 0-1 binary data) for calculating exactly, not estimating, the variance of the sample.</p>
<p>Using the second instead of the first to estimate the distribution variance will, on average, lead to slight underestimates. This is equivalent to the use of $n$ instead of $n-1$ in the denominator of the estimator.</p>
|
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