qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,934,974 | <p>I'm wondering how to prove the associativity and identity to prove that the Möbius transformations forms a group.</p>
<p>A Möbius transformation is a complex function of the form <span class="math-container">$M(z)=\dfrac{az+b}{cz+d}$</span>.</p>
<p>Thank you in advance :)</p>
| John Clever | 783,246 | <p>I don't know how much you've done on your own, but to help subsequent users, I'll put the full answer. Stop after each hint and try and do it yourself.</p>
<p><strong>Hint 1:</strong>
Prove that the Möbius transformations are bijective from the extended complex numbers to themselves.</p>
<p><strong>Hint 2:</strong>
The Möbius transformations are therefore a subset of the set of bijections on the extended complex numbers.</p>
<p><strong>Hint 3:</strong>
The set of all bijections of the complex numbers is a group.</p>
<p><strong>Hint 4:</strong>
Composition of two Möbius transformations and inversion of one Möbius transformation is another Möbius transformation.</p>
<p><strong>Hint 5:</strong>
A closed subset of a group is a group.</p>
<p><strong>Hint 6:</strong>
The set of Möbius transformations is closed.</p>
|
1,176,098 | <p>Here are some of my ideas:</p>
<p><strong>1. Addition Formula:</strong> <span class="math-container">$\sin{x}$</span> and <span class="math-container">$\cos{x}$</span> are the unique functions satisfying:</p>
<ul>
<li><p><span class="math-container">$\sin(x + y) = \sin x \cos y + \cos x \sin y $</span></p>
</li>
<li><p><span class="math-container">$\cos(x + y) = \cos x \cos y - \sin x \sin y$</span></p>
</li>
<li><p><span class="math-container">$\sin 0 = 0\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{\sin x }{x} = 1}$</span></p>
</li>
<li><p><span class="math-container">$\cos 0 = 1\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos x}{x} = 0}$</span></p>
</li>
</ul>
<p><strong>2. Taylor Series:</strong></p>
<ul>
<li><p><span class="math-container">$\displaystyle{\sin x = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!}\;x^{2n+1}}$</span></p>
</li>
<li><p><span class="math-container">$\displaystyle{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\;x^{2n}}$</span></p>
</li>
</ul>
<p><strong>3. Differential Equations:</strong> <span class="math-container">$\sin(x)$</span> and <span class="math-container">$\cos(x)$</span> are the unique solutions to <span class="math-container">$y'' = -y$</span>, where <span class="math-container">$\sin(0) = \cos^\prime(0) = 0$</span> and <span class="math-container">$\sin^\prime(0) = \cos(0) = 1$</span>.</p>
<p><strong>4. Inverse Formula:</strong> We have:</p>
<p><span class="math-container">$$\begin{align}
\arcsin x &= \phantom{\frac{\pi}{2} + } \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt \\[6pt]
\arccos x &= \frac{\pi}{2} - \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt
\end{align}$$</span></p>
<p>Then <span class="math-container">$\sin x$</span> is the inverse of <span class="math-container">$\arcsin x$</span>, extended appropriately to the real line, and <span class="math-container">$\cos x$</span> is similar.</p>
<p><strong>Question:</strong> Are there any others that you like? In particular, are there any good rigorous ones coming from the original geometric definition?</p>
| Emilio Novati | 187,568 | <p>My preferred definition is:</p>
<blockquote>
<p>$\cos x$ and $\sin x$ are the real and imaginary parts of the
exponential function $\exp(ix)$.</p>
</blockquote>
<p>Since we have:
$$
\begin{split}
e^{ix}= \sum_{k=0}^\infty\dfrac{(ix)^k}{k!}&=1+ix+\dfrac{(ix)^2}{2!}+\dfrac{(ix)^3}{3!}+\cdots+\dfrac{(ix)^n}{n!}+\cdots\\
&=1-\dfrac{(x)^2}{2!}+\dfrac{(x)^4}{4!}-\dfrac{(x)^6}{6!}+ \cdots +i\left[ x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+\cdots \right]\\
&=
\sum_{k=0}^\infty\dfrac{(-1)^k\,x^{2k}}{(2k)!}+i \sum_{k=0}^\infty\dfrac{(-1)^k\,x^{2k+1}}{(2k+1)!}\end{split}
$$
we find the usual series definitions.</p>
|
1,176,098 | <p>Here are some of my ideas:</p>
<p><strong>1. Addition Formula:</strong> <span class="math-container">$\sin{x}$</span> and <span class="math-container">$\cos{x}$</span> are the unique functions satisfying:</p>
<ul>
<li><p><span class="math-container">$\sin(x + y) = \sin x \cos y + \cos x \sin y $</span></p>
</li>
<li><p><span class="math-container">$\cos(x + y) = \cos x \cos y - \sin x \sin y$</span></p>
</li>
<li><p><span class="math-container">$\sin 0 = 0\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{\sin x }{x} = 1}$</span></p>
</li>
<li><p><span class="math-container">$\cos 0 = 1\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos x}{x} = 0}$</span></p>
</li>
</ul>
<p><strong>2. Taylor Series:</strong></p>
<ul>
<li><p><span class="math-container">$\displaystyle{\sin x = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!}\;x^{2n+1}}$</span></p>
</li>
<li><p><span class="math-container">$\displaystyle{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\;x^{2n}}$</span></p>
</li>
</ul>
<p><strong>3. Differential Equations:</strong> <span class="math-container">$\sin(x)$</span> and <span class="math-container">$\cos(x)$</span> are the unique solutions to <span class="math-container">$y'' = -y$</span>, where <span class="math-container">$\sin(0) = \cos^\prime(0) = 0$</span> and <span class="math-container">$\sin^\prime(0) = \cos(0) = 1$</span>.</p>
<p><strong>4. Inverse Formula:</strong> We have:</p>
<p><span class="math-container">$$\begin{align}
\arcsin x &= \phantom{\frac{\pi}{2} + } \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt \\[6pt]
\arccos x &= \frac{\pi}{2} - \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt
\end{align}$$</span></p>
<p>Then <span class="math-container">$\sin x$</span> is the inverse of <span class="math-container">$\arcsin x$</span>, extended appropriately to the real line, and <span class="math-container">$\cos x$</span> is similar.</p>
<p><strong>Question:</strong> Are there any others that you like? In particular, are there any good rigorous ones coming from the original geometric definition?</p>
| Math2718 | 685,688 | <p>One definition that I think has a very clear geometric meaning is this: On the unit circle in the <span class="math-container">$xy$</span>-plane, draw a line segment from the origin to a point on the circle. Call the angle that the line segment makes with the <span class="math-container">$x$</span>-axis <span class="math-container">$t$</span> as shown in the picture. We define the <span class="math-container">$x$</span>-component of the endpoint of the line segment as <span class="math-container">$\cos(t)$</span> and the <span class="math-container">$y$</span>-component as <span class="math-container">$\sin(t)$</span>.</p>
<img src="https://i.stack.imgur.com/lAfA8.png" width="300" height="300">
<p>The trig class "opposite, ajacent, hypotenuse" definition follows from this one. It is also not too difficult to derive inverse function integral definition from this. The addition formula definition is a bit more work to obtain, but there are geometric proofs of all of the facts listed.</p>
|
1,461,254 | <p>I am about to take an undergraduate course in Mathematical logic any textbooks to recommend.Want it to be rigorous and not missing things. I am an Math undergraduate. Got still 2 years for my degree. I have taken courses mostly in Algebra (ring theory etc.). Also taken Real analysis courses. So my mathematical maturity is mediocre I would say. (I have also done calculus courses etc on my first year.)</p>
| David | 297,532 | <p>I would recommend Cori and Lascar's <a href="http://rads.stackoverflow.com/amzn/click/0198500483" rel="nofollow"><em>Mathematical Logic: A Course with Exercises</em>.</a> It has a lot of examples from algebra, so is appropriate for a math major who has taken abstract algebra.</p>
|
16,733 | <p>This is a variant on the question <a href="https://matheducators.stackexchange.com/questions/14492/small-real-numbers">small real numbers</a>.</p>
<p>I have a disagreement with someone about the meaning of "bigger" real numbers.</p>
<p>Say we have the real number <span class="math-container">$-1.$</span> Is <span class="math-container">$0$</span> "bigger" or "smaller" than <span class="math-container">$-1$</span>? In other words, should we interpret "bigger" to be a synonym of "greater than"?</p>
<p>This may be considered an "opinion-based" question, but is there a consensus about how to assign meaning to "bigger" or "smaller" (real) numbers?</p>
<p>My interpretation is that "bigger" real numbers have larger magnitude, irrespective of sign. For example, I consider that <span class="math-container">$-2$</span> is "bigger" than <span class="math-container">$-1.$</span> To avoid ambiguity (we are mathematicians!), I think it better to say/write "<span class="math-container">$-2$</span> has a larger magnitude/size than <span class="math-container">$1$</span>".</p>
<p>Further, I try to avoid using "bigger" or "smaller" with respect to comparing numbers unless it is clear we are comparing their magnitude (e.g. vectors or complex numbers), where we are only comparing non-negative values. I would stick to "greater / less than" or "closer to / further from zero".</p>
<p>Regards,</p>
| amWhy | 238 | <p>The words "big" and "small" are relative. Asking a student: "Is <span class="math-container">$1$</span> small?" is confusing, and meaningless. Likewise, "bigger" and "smaller" are relative. If you consider six numbers, <span class="math-container">$1, 2, 4, 5, 7, 8$</span>, then each of <span class="math-container">$2, 4, 5, 7, 8$</span> is bigger <strong><em>than</em></strong> <span class="math-container">$1,$</span> <em>and</em> each of them is "smaller" <em>than</em> <span class="math-container">$8$</span>. </p>
<p>What <em>you</em> seem to want to convey, when comparing two real numbers and calling one "bigger than" the other, is that one is "bigger in magnitude" or one is "smaller in magnitude", where <em>magnitude</em> can be thought of as the <em>distance</em> of a number from zero. So <span class="math-container">$-2$</span> is bigger (greater) in <em>magnitude</em> than is <span class="math-container">$1$</span>, where the magnitude of a real number <span class="math-container">$x$</span> is given by <span class="math-container">$|x|$</span>. But without adding the qualifier <em>in magnitude,</em> I'm afraid you will not be well understood. </p>
<p>So, indeed, <span class="math-container">$|-2| > |1|$</span>.</p>
<p>The more common and well understood terms used to compare two real numbers, 'greater than' and 'less than', (or, 'equal to'), are used to refer to the stricter interpretation in which the one further to the left on the real number line is less than the other. (And the number further to the right on the real number is greater than the other.) </p>
<p>Hence we write, e.g., <span class="math-container">$-83\lt -3 \lt 0 \lt 1$</span>.</p>
<p>I'd suggest that the use of "greater than" or "less than," when comparing magnitudes of two numbers, that, e.g., "greater in magnitude" ("lesser in magnitude") is more widely used than is "bigger than, smaller than."</p>
|
2,091,589 | <p>The remainder when a polynomial $f(x)$ is divided by $(x-2)(x+3)$ is $ax+b$. When $f(x)$ is divided by $(x-2)$, then remainder is $5$. $(x+3)$ is a factor of $f(x)$. Find the values of $a$ and $b$. I am thinking of using the remainder and factor theorem to solve this however their quotients are different. Can anyone please show me how? Thanks</p>
| Ian S | 405,681 | <p>Yes you are on the right track.</p>
<p>Let <em>f</em>(<em>x</em>) = (<em>x</em>-2)(<em>x</em>+3) x <em>m</em> + (<em>ax</em> + <em>b</em>).</p>
<p>We know <em>f</em>(2) = 5, so (2-2)(2+3) x <em>m</em> + (2 <em>a</em> + <em>b</em>) = 5
And similarly <em>f</em>(-3) = 0, so (-3-2)(-3+3) x <em>m</em> + (-3 <em>a</em> + <em>b</em>) = 0</p>
<p>This gives 2 equations </p>
<p>(2 <em>a</em> + <em>b</em>) = 5, and</p>
<p>(-3 <em>a</em> + <em>b</em>) = 0</p>
<p>Which you can solve to give <em>a</em>=1 and <em>b</em>=3</p>
|
2,091,589 | <p>The remainder when a polynomial $f(x)$ is divided by $(x-2)(x+3)$ is $ax+b$. When $f(x)$ is divided by $(x-2)$, then remainder is $5$. $(x+3)$ is a factor of $f(x)$. Find the values of $a$ and $b$. I am thinking of using the remainder and factor theorem to solve this however their quotients are different. Can anyone please show me how? Thanks</p>
| Roman83 | 309,360 | <h3>Hint:</h3>
<p>If <span class="math-container">$$f(x)=(x-\alpha)g(x)+r$$</span>
then <span class="math-container">$r=f(\alpha)$</span></p>
<p>Really, <span class="math-container">$x=\alpha$</span> <span class="math-container">$f(\alpha)=0+r$</span></p>
<p>Then <span class="math-container">$$f(x)=(x-2)(x+3)+ax+b$$</span></p>
<p><span class="math-container">$f(2)=5$</span> and <span class="math-container">$f(-3)=0$</span></p>
<p><span class="math-container">$$2a+b=5$$</span>
<span class="math-container">$$-3a+b=0$$</span></p>
<p><span class="math-container">$$a=1, b=3$$</span></p>
|
1,509,007 | <blockquote>
<p>Consider an unweighted and undirected graph $G=(V,E)$, where the vertices $V$ of $G$ lie on the unit n-sphere. If we choose a normal vector uniformly at random on this $n$-sphere, then the corresponding hyperplane (shifted to go through the origin) splits the vertices into two disjoint sets $A$ and $B$, and the expected number of edges cut by this hyperplane is
$$
\mathbb{E}[Cut(A,B)] =
\sum_{(i,j)\in E}\frac{\arccos(v_i\cdot v_j)}{\pi}\; .
$$</p>
</blockquote>
<p>So I understand that $\frac{1}{\pi}\arccos(v_i\cdot v_j)$ is basically the probability that a randomly chosen hyperplane which goes through the origin will separate $v_i$ and $v_j$. But I can't grasp how this sum is equal to the expectation described above.</p>
<p>Also, can this sum be extended to work for weighted graphs?</p>
| davidlowryduda | 9,754 | <p>Roughly speaking, your result is that
$$ \sigma(n) \ll n \log n,$$
where I use $\ll$ as a sort of <a href="https://en.wikipedia.org/wiki/Big_O_notation" rel="nofollow">big Oh</a> notation. It's actually known that
$$ \sigma(n) \ll n \log \log n.$$
The difference between what you've shown and what is known is the difference between $\log n$ and $\log \log n$. In short, the second is much, much smaller. For instance, $\log 10^6 \approx 13.81$ and $\log \log 10^6 \approx 2.62.$ And this difference becomes more pronounced as $n$ gets large, even though it happens slowly.</p>
<p>As $n$ gets large, the ratio between your error and the known error is unboundedly large. But to be fair, the rate of divergence is quite slow.</p>
|
215,898 | <p>This is a quick follow up to my other <a href="https://mathematica.stackexchange.com/q/215884/44420">question</a> which I thought was different enough to warrant a separate post.</p>
<p><strong>My Question</strong></p>
<blockquote>
<p>How do you plot a region (such as <span class="math-container">$f(x,y)>z$</span>) on a 2D surface <span class="math-container">$g(x,y)=z$</span> and change the color of the region?</p>
</blockquote>
<p><strong>A Simple Example</strong></p>
<pre><code>f[x_, y_] := Sin[x] + Sin[y];
g[x_, y_] := y x Sin[x y];
img = RegionPlot[3/2 > f[x, y] > 1, {x, 0, 2}, {y, 0, 2},
Frame -> False, PlotRangePadding -> None, PlotStyle -> Orange,
BoundaryStyle -> Black]
Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2}, PlotStyle -> {Texture[img]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/yQF6w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yQF6w.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/W3V5K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W3V5K.png" alt="enter image description here"></a></p>
<p>So I have figured out how to overlay the region plot on the surface, but the only color the region appears is black. This is in spite that I specified the <code>RegionPlot</code> to be <code>Orange</code>. How can I change the region plot on the surface to anything but black.</p>
<p><strong>Notes</strong></p>
<ul>
<li>If you have any questions or need clarification please ask.</li>
<li>Again this is a quick follow up on my previous <a href="https://mathematica.stackexchange.com/q/215884/44420">question</a>.</li>
</ul>
| Rupesh | 63,381 | <p>You may also try Piecewise:
Something similar to this for defining Colorfunction of separate regions.</p>
<pre><code> colfn = Piecewise[{{Red, 3/2 > f[#, #2] > 1}, {White, True}}] &
Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2},
PlotPoints -> 200, PlotRange -> All, ColorFunction -> colfn,
ColorFunctionScaling -> False, Mesh -> None]
</code></pre>
<p>Also, check Exclusions in the plot options.</p>
|
907,850 | <p>We have tiles of size <code>2 * 1</code>. We need to arrange the tiles to get the floor size of <code>m * n</code>.</p>
<p>For <code>m = 3</code>, we get arrangement like this:</p>
<p><img src="https://i.stack.imgur.com/SicS8.png" alt="enter image description here">
<br></p>
<p>My question is: In how many different ways can we arrange the tiles?</p>
<p>For <code>m = 3</code>, we get recurrence relation as:</p>
<pre><code>T(n) = 4*T(n-2) - T(n-4)
</code></pre>
<p>For <code>m = 4</code>, we get</p>
<pre><code>T(n) = T(n−1) + 5*T(n−2) + T(n−3) − T(n−4)
</code></pre>
<p>How to solve it for <code>n=5</code> and <code>m = n</code>? Is there any generalized solution?</p>
<p>Problem Definitions :</p>
<p>1) <a href="http://www.spoj.com/problems/M5TILE/" rel="nofollow noreferrer">http://www.spoj.com/problems/M5TILE/</a><br>
2) <a href="http://www.spoj.com/problems/MNTILE/" rel="nofollow noreferrer">http://www.spoj.com/problems/MNTILE/</a></p>
| MHS | 158,061 | <p>There is a general formula for the number of valid domino tilings in an (m x n) rectangle, which was first found by Kastelyn (around 1960).</p>
<p>$
\prod_{j=1}^{\lceil\frac{m}{2}\rceil} \prod_{k=1}^{\lceil\frac{n}{2}\rceil} \left ( 4\cos^2 \frac{\pi j}{m + 1} + 4\cos^2 \frac{\pi k}{n + 1} \right )
$</p>
<p>This formula was then used to compute the topological entropy of the domino tilings (the corresponding two-dimensional subshift of finite type) - one of the few results where an explicit formula for the entropy value of a two dimensional (non-degenerate) shift of finite type is know.</p>
<p>Also if you already know some values for the first sizes (small n, m) a good way to find general formulas like this one is to take a look at Sloane's database of integer sequences (see <a href="https://oeis.org/" rel="nofollow">https://oeis.org/</a> ). There you can put in the first few terms of a sequence and you will probably find the correct one in the database, together with additional terms (for larger values of n,m), a detailed description of where those sequences occur and in many cases a closed formula or a recursion to produce the terms in the corresponding sequence.</p>
|
907,850 | <p>We have tiles of size <code>2 * 1</code>. We need to arrange the tiles to get the floor size of <code>m * n</code>.</p>
<p>For <code>m = 3</code>, we get arrangement like this:</p>
<p><img src="https://i.stack.imgur.com/SicS8.png" alt="enter image description here">
<br></p>
<p>My question is: In how many different ways can we arrange the tiles?</p>
<p>For <code>m = 3</code>, we get recurrence relation as:</p>
<pre><code>T(n) = 4*T(n-2) - T(n-4)
</code></pre>
<p>For <code>m = 4</code>, we get</p>
<pre><code>T(n) = T(n−1) + 5*T(n−2) + T(n−3) − T(n−4)
</code></pre>
<p>How to solve it for <code>n=5</code> and <code>m = n</code>? Is there any generalized solution?</p>
<p>Problem Definitions :</p>
<p>1) <a href="http://www.spoj.com/problems/M5TILE/" rel="nofollow noreferrer">http://www.spoj.com/problems/M5TILE/</a><br>
2) <a href="http://www.spoj.com/problems/MNTILE/" rel="nofollow noreferrer">http://www.spoj.com/problems/MNTILE/</a></p>
| MHS | 158,061 | <p>Here is a link to a presentation (by some undergraduate student) about the computation of those numbers:</p>
<p><a href="http://math.cmu.edu/~bwsulliv/domino-tilings.pdf" rel="nofollow">http://math.cmu.edu/~bwsulliv/domino-tilings.pdf</a></p>
<p>It also contains references to Kastelyn's original paper and other interesting material.</p>
|
1,833,854 | <p>We have $M$ Binomial random variables, where $X_0 \sim $ Bin$(n,p)$ and $X_i \sim $ Bin$(n,1/2)$. </p>
<p>Suppose $p > 1/2$. I'm interested in the probability that $\mathbb{P}(\max \{X_1,\dots,X_M\} \geq X_0)$. Is this tractable? </p>
<p>If not, is it tightly boundable/approximable? If this is a very difficult question, I'd accept a reference providing some insight on the problem as well.</p>
| Graham Kemp | 135,106 | <p>The maximum of $\{X_i\}_{i\in\{1;M\}}$ will be at least as great as $X_0$ when it is not that all values are less than $X_0$. (Employ the rule of complements.) Assuming that $X_0$ is <em>also</em> independent of the iid variables then:
$$\begin{align}
\mathsf P\left(\max_{i=1}^M\{X_i\} \geqslant X_0\right) =&~ 1-\sum_{k=1}^n\mathsf P(X_0=k)\mathsf P\Big(\bigcap_{i=1}^M X_i<k\Big)
\\ =&~ 1- 2^{-nM} \sum_{k=1}^n\binom{n}{k}p^k(1-p)^{n-k}\left(\sum_{h=0}^{k-1}\binom{n}{h}\right)^M
\end{align}$$</p>
|
2,748,495 | <p>I have a numeric table for artillery operations (Royal Italian Army, year 1940), in the instructions it refers to a measure of a planar angle as $32.00^{\circ\circ}$ and it seems to me that this angle is equivalent to $\pi$ radian. I had a look at <a href="https://en.wikipedia.org/wiki/Gradian" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gradian</a> but it states that $\pi$ radian is equal to 200 gradian.</p>
<p>Referring to the following figure, given the angles $\alpha$, $\beta$ and the distance $b$, the numeric table will give the distance $X$.</p>
<p><a href="https://i.stack.imgur.com/mkwD7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mkwD7.jpg" alt="enter image description here"></a></p>
<p>My rough translation of the interesting part of the instruction is this:</p>
<blockquote>
<p>The angles $\alpha$ and $\beta$ will be measured by means of
goniometers. So in each end-point of the baseline you will set up a
goniometer, the direction of the goniometer is the direction of the
baseline $SD$ and the orientation of the goniometer will be from $D$
to $S$. So the angle of the line-of-sight from $D$ to $S$ will be set
up, in the goniometer in $D$, equal to $0.00^{\circ\circ}$ and the
angle of the line-of-sight from $S$ to $D$ will be set up, in the
goniometer in $S$, equal to $32.00^{\circ\circ}$.</p>
</blockquote>
<p><strong>Edit</strong></p>
<p>After the <a href="https://math.stackexchange.com/a/2748607/10799">answer</a> by <a href="https://math.stackexchange.com/users/1303/christian-blatter">Christian Blatter</a> I have found some nice pictures of the Swiss Army Compass with the "Art. ‰" scale:</p>
<p><a href="https://image.jimcdn.com/app/cms/image/transf/none/path/s904213d06cc35930/image/ifba6fe2fc60a4024/version/1424088065/image.jpg" rel="nofollow noreferrer"><img src="https://image.jimcdn.com/app/cms/image/transf/none/path/s904213d06cc35930/image/ifba6fe2fc60a4024/version/1424088065/image.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/QulF9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QulF9.jpg" alt="enter image description here"></a></p>
| David K | 139,123 | <p>Working from context, in order for the angles to be measured as shown in the figures, the direction from $S$ to $D,$ measured by the goniometer at $S,$ must be $\pi$ radians.
Otherwise the angles $\alpha$ and $\beta$ would not be measured in the
indicated orientations from the same base line.</p>
<p>Therefore I would conclude that
$$32.00^{\circ\circ} = \pi \text{ radians}.$$</p>
<p>As Christian Blatter points out, a division of the circle into
$6400$ parts <a href="https://en.wikipedia.org/wiki/Milliradian#Definitions_for_maps_and_artillery" rel="nofollow noreferrer">is used by the military in various countries</a>.
If one full circle ($2\pi$ radians) is equal to $6400$ NATO angular mils,
then we can infer that this Italian manual defines $1^{\circ\circ}$
so that
$$ 1^{\circ\circ} = 100 \text{ NATO angular mils}.$$</p>
|
1,330,858 | <p>Suppose I'm at $(x=0,y=0)$ and I want to get to $(x=1,y=1)$. The shortest path is the diagonal and it has length $\sqrt{2}$. But what if I'm only allowed to make moves in coordinate directions---e.g., $1/2$ along $x$, $1/2$ along $y$, another $1/2$ along $x$, and a final $1/2$ along $y$. Then the length of my path is $2$. In fact, any coordinate-constrained path has length $2$. Let the path $p_n$ be $1/n$ along $x$, followed by $1/n$ along $y$, followed by $1/n$ along $x$, etc., until I get to $(1,1)$. Presumably, the limit of $p_n$ as $n\rightarrow\infty$ is the diagonal line. But the path length of each $p_n$ is $2$, while the path length of the limit is $\sqrt{2}$.</p>
<p>Weird, right? Is this just an example that shows that you can't exchange limit and path length?</p>
| Zach L | 249,196 | <p>This is indeed because you cannot exchange the limit that you have described and the path length. No matter how large your n gets, if we look at one change in x and one change in y, the total distance from start to finish will always be √2 * the change in x (or y, because they are both the same). The error does not disappear as n gets larger. So, yes this example just shows that you cannot exchange the limit and path length. </p>
|
52,802 | <p>This is partly a programming and partly a combinatorics question.</p>
<p>I'm working in a language that unfortunately doesn't support array structures. I've run into a problem where I need to sort my variables in increasing order.</p>
<p>Since the language has functions for the minimum and maximum of two inputs (but the language does not allow me to nest them, e.g. <code>min(a, min(b, c))</code> is disallowed), I thought this might be one way towards my problem.</p>
<p>If, for instance, I have two variables $a$ and $b$, I only need one temporary variable so that $a$ ends up being less than or equal to $b$:</p>
<pre><code>t = min(a, b);
b = max(a, b);
a = t;
</code></pre>
<p>for three variables $a,b,c$, the situation is a little more complicated, but only one temporary variable still suffices so that $a \leq b \leq c$:</p>
<pre><code>a = min(a, b);
t = max(a, b);
c = max(t, c);
t = min(t, c);
b = max(a, t);
a = min(a, t);
</code></pre>
<p>Not having a strong combinatorics background, however, I don't know how to generalize the above constructions if I have $n$ variables in general. In particular, is there a way to figure out how many temporary variables I would need to sort out $n$ variables, and to figure out what is the minimum number of assignment statements needed for sorting?</p>
<p>Thanks in advance!</p>
| Patrick Da Silva | 10,704 | <p>$e-A = A$ is clearly impossible... $e-A$ contains no element of the set $A$ by construction. The only thing that could define an identity is clearly the empty set, because $(e-A)$ doesn't contain any element of $A$, hence must not contain any element at all, for if $e$ contains an element, there exists a set $A$ which does not contain this element and therefore $(e-A)$ contains an element not in $A$, contradicting the fact that $(A-e) \cup (e-A) = A$. Now $A-\varnothing = A$ so that $\varnothing$ is your identity.</p>
<p>Hope that helps,</p>
|
2,439,744 | <p>How to prove that $\lim_{x\to 0,y\to 0}{\frac {\sqrt {a+x^2y^2} -1} {x^2+y^2}} (a>0)$ doesn't exist while a $\ne 1$?</p>
<p>I already calculated that when a = 1 by multiplying $\sqrt {a+x^2y^2} + 1$ on both denominator and numerator and using the fact $x^2y^2<(x^2+y^2)^2/4$.</p>
<p>Any help will be appreciated.</p>
| Community | -1 | <p>If your function <em>can</em> have limit at $(0,0)$ for some $a_0$ then it can be defined at $a_0$ to be continuous at $a_0$. So suppose that we have some $a_0$ for which your function can be continuous at $a_0$.</p>
<p>If I recall corectly if $f$ is continuous then the limit is equal to iterated limits so we have $\lim_{(x,y) \to (0,0)}{\dfrac {\sqrt {a+x^2y^2} -1} {x^2+y^2}} =\lim_{x \to 0}(\lim_{y \to 0}\dfrac {\sqrt {a+x^2y^2} -1} {x^2+y^2})=\lim_{x \to 0}(\dfrac {\sqrt {a}-1}{x^2})$ and this limit can only exist if $\sqrt {a} =1$, that is, only if $a=1$.</p>
<p>Someone should correct me if I am wrong somewhere.</p>
|
2,662,605 | <p>Problem: If ${F_n}$ is a sequence of bounded functions from a set $D \subset \mathbb R^p$ into $ \mathbb R^q$ and if ${F_n}$ converges uniformly to $F$ on $D$, then $F$ is also bounded. </p>
<p>Proof(Attempt): Let $\epsilon >0$. Since ${F_n}$ converges uniformly to $F$ on $D$, then there is an $N \in\mathbb R$ such that $||F(x)-F_n(x)||< \epsilon $ whenever $x \in D$ and $n\ge N$. </p>
<p>I'm using Joseph Taylor's Foundation of Analysis textbook.
Since ${F_n}$ is bounded we know that $||F_n|| \le M $ for every $x\in D$.<br>
At this point, I'm thinking that I need to use a trick that gets $||F|| \le M$ from the inequality $||F(x)-F_n(x)||< \epsilon $. </p>
<p>I would appreciate advise and hints that will help guide me. </p>
| user284331 | 284,331 | <p>The assumption that each $F_{n}$ is bounded does not necessarily mean that there is an $M>0$ such that $|F_{n}(x)|\leq M$ for all $x\in D$ and $n=1,2,...$</p>
<p>Rather, we need to establish the uniform bound for all $F_{n}$. So the sequence is uniformly Cauchy, so $|F_{n}(x)-F_{m}(x)|<1$ for all $x\in D$ and $n,m\geq N$.</p>
<p>So $|F_{n}(x)|\leq|F_{1}(x)|+\cdots+|F_{N}(x)|+1$ for all $x\in D$ and $n=1,2,...$, now we are to take $M=M_{1}+\cdots+M_{N}+1$, where $|F_{n}(x)|\leq M_{n}$ for all $x\in D$ and $n=1,2,...,N$. </p>
|
4,416,001 | <p>Given <span class="math-container">$f(xy) = f(x+y)$</span> and <span class="math-container">$f(11) = 11$</span>, what is <span class="math-container">$f(49)$</span>?</p>
| Maksim | 617,634 | <p><span class="math-container">$f(49)=f(0+49)=f(0\times49)=f(0)=f(11 \times 0)=f(11+0)=f(11)=11$</span></p>
|
2,709,273 | <blockquote>
<p>Determine the kernel of the following group homomorphism:
$$
\phi\colon\mathbb Z/270\mathbb Z\to\mathbb Z/270\mathbb Z\colon\overline x\mapsto\overline{6x}.
$$
Then find the solutions of the following system of equations in $\mathbb Z/270\mathbb Z$:
\begin{align}
6x=3\mod 27\\
6x=2\mod 10
\end{align}</p>
</blockquote>
<p>Since $6*45=270$, the kernel is $\{\overline{45}, \overline{90}, \overline{135}, \overline{180}, \overline{225}, \overline{270}\}$. Instead of working with $6x$, I solved the following equations:
\begin{align}
a=3\mod 27\\
a=2\mod 10,
\end{align}
using the Chinese remainder theorem. I found: $a=192=32*6$. So I would guess they are looking for this answer:
$$
\left\{\overline{32+k45}:k\in\{1,2,3,4,5,6\}\right\}.
$$
But I'm a bit confused by their phrasing, because we initially solved for $x\in\{0,1,\dots,269\}$, and not for $\overline a\in\mathbb Z/270\mathbb Z$... So I could only sort of guess what I had to do, but could someone clarify their wording? Why can it be interpreted as: find $\overline a\in\mathbb Z/270\mathbb Z$, such that for
$$
\phi'=\theta\circ\phi,
$$
where
$$
\theta\colon\mathbb Z/270\mathbb Z\to\mathbb Z/27\mathbb Z\times\mathbb Z/10\mathbb Z\colon a\mapsto (a,a),
$$
we have $\phi'(\overline a)=(3,2)$.</p>
<p><em>edit</em></p>
<p>I think I got it: to find solutions in $\mathbb Z/n\mathbb Z$ just means to find solutions modulo $n$.</p>
| Steven Alexis Gregory | 75,410 | <p><span class="math-container">\begin{align}
6x &\equiv 3 \pmod {27}\\
2x &\equiv 1 \pmod{9} \\
\color{red}{x} &\color{red}{\equiv} \color{red}{5 \pmod{9}} \\
\hline
6x &\equiv 2 \pmod{10} \\
3x &\equiv 1 \pmod{5} \\
\color{red}x &\color{red}{\equiv} \color{red}{2 \pmod{5}} \\
\end{align}</span></p>
<p><span class="math-container">\begin{array}{c|cc}
&\mod 5 &\mod 9 \\
\hline
9 & 4 & 0 \\
5 & 0 & 5 \\
\hline
\color{brown}{-9} & 1 & 0 \\
\color{brown}{10} & 0 & 1 \\
\hline
\end{array}</span></p>
<p><span class="math-container">$$
\left. \begin{array}{c}
x \equiv 5 \pmod 9 \\
x \equiv 2 \pmod 5
\end{array} \right\}
\implies x \equiv \color{brown}{-9}(\color{red}2)
+\color{brown}{10}(\color{red}5)
\equiv 32 \pmod{45}
$$</span></p>
|
2,861,293 | <p>I found this statement with the proof:</p>
<p><a href="https://i.stack.imgur.com/bGRiZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bGRiZ.jpg" alt="enter image description here" /></a></p>
<p>But I don't understand the proof. Where is the contradiction? We have a nonempty interval <span class="math-container">$J$</span> contained in the nonempty interval <span class="math-container">$(f(a),f(b))$</span>. Where is the problem?</p>
| Paul Frost | 349,785 | <p>In the above form the proof isn't done well. You know that $y_0^- < f(x_0)$ or $f(x_0) < y_0^+$. Hence at least one of the intervals $J^- = (y_0^-, f(x_0))$ and $J^+ = (f(x_0),y_0^+)$ is non-empty and does not contain any point of the form $f(x)$. This comes from the fact that $f(x_0)$ is contained in none of these intervals and $f(x) \le y_0^-$ for $x < x_0$ and $f(x) \ge y_0^+$ for $x > x_0$. W.l.o.g. assume $J^- \ne \emptyset$ and $J^- \cap f(I) = \emptyset$. </p>
<p>This is a contradiction: By assumption $f(I)$ is an interval which must contain $J^-$ because $f(a) \le y_0^-$ and $y_0^+ \le f(b)$.</p>
|
4,419,565 | <p>Consider the function <span class="math-container">$f(x)$</span> and let <span class="math-container">$g(x)=f(cx)$</span>.</p>
<p>By the definition of derivative</p>
<p><span class="math-container">$$f'(x)=\frac{df(x)}{dx}=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}\tag{1}$$</span></p>
<p>and so the definition of <span class="math-container">$f'(cx)$</span> is</p>
<p><span class="math-container">$$f'(cx)=\frac{df(cx)}{d(cx)}=\lim\limits_{h \to 0} \frac{f(cx+h)-f(cx)}{h}\tag{2}$$</span></p>
<p>Also</p>
<p><span class="math-container">$$g'(x) = \lim\limits_{h \to 0} \frac{g(x+h)-g(x)}{h}$$</span>
<span class="math-container">$$=\lim\limits_{h \to 0} \frac{f(c(x+h))-f(cx)}{h}$$</span>
<span class="math-container">$$=\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{h}$$</span>
<span class="math-container">$$=c\lim\limits_{h \to 0} \frac{f(cx+ch)-f(cx)}{ch}$$</span>
<span class="math-container">$$=cf'(cx)$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$g'(x)=cf'(cx)$$</span></p>
<p>I am not sure if I am seeing an inexistent ambiguity, but <span class="math-container">$f'(cx)$</span> seems like ambiguous notation.</p>
<p>As defined above in <span class="math-container">$(2)$</span>, <span class="math-container">$f'(cx)$</span> means the derivative of <span class="math-container">$f$</span> relative to <span class="math-container">$cx$</span>, evaluated at a point we call <span class="math-container">$cx$</span>, .</p>
<p>Note that this is different than the derivative of <span class="math-container">$f$</span> relative to <span class="math-container">$x$</span> evaluated at a point <span class="math-container">$cx$</span>: this derivative is <span class="math-container">$g'(x)=\frac{df(cx)}{dx}$</span>. In "prime" notation, how do we denote this latter derivative? It would seem to be <span class="math-container">$f'(cx)$</span>, but I think either this is incorrect, or the definition given in <span class="math-container">$(2)$</span> is somehow non-standard.</p>
<p>For example, let <span class="math-container">$$f(x)=3x^3$$</span> <span class="math-container">$$g(x)=f(cx)=3c^3x^3$$</span></p>
<p>Then</p>
<p><span class="math-container">$$f'(x)=\frac{df(x)}{dx}=9x^2$$</span>
<span class="math-container">$$\left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2\ \ (=f'(cx)???)$$</span>
<span class="math-container">$$g'(x)=\frac{df(cx)}{dx}=c\frac{df(cx)}{d(cx)}=c\cdot f'(cx)=c\cdot 9c^2x^2= 9c^3x^2\tag{3}$$</span></p>
<p>In this example, <span class="math-container">$f'(cx)=\frac{df(cx)}{d(cx)}=9c^2x^2$</span> according to definition I gave in <span class="math-container">$(2)$</span>.</p>
<p>But perhaps more intuitively, it could also be <span class="math-container">$f'(cx)= \left.\frac{df(x)}{dx}\right \vert_{x=cx}=9c^2x^2$</span></p>
<p>Note that both uses of <span class="math-container">$f'(cx)$</span> lead to the same result in this example.</p>
<p>Which use of <span class="math-container">$f'(cx)$</span> is the "correct" or "standard" one?</p>
| epi163sqrt | 132,007 | <p>This is just a supplementary part to the nice answer of @KarlSchildkraut. I think it is helpful to consider the <em>complete</em> definition of functions and look at their relationship somewhat more detailed.</p>
<p><strong>The setting more detailed:</strong></p>
<p>We start with a differentiable function
<span class="math-container">\begin{align*}
&f:\mathbb{R}\to\mathbb{R}\\
&x\mapsto f(x)
\end{align*}</span></p>
<p>When we now consider
<span class="math-container">\begin{align*}
g(x)=f(cx)
\end{align*}</span>
we have a function <span class="math-container">$h$</span>
<span class="math-container">\begin{align*}
&h:\mathbb{R}\to\mathbb{R}\\
&x\mapsto h(x)=cx\\
\end{align*}</span>
and a function <span class="math-container">$g$</span>
<span class="math-container">\begin{align*}
&g:\mathbb{R}\to\mathbb{R}\\
&x\mapsto g(x)=\left(f\circ h\right)(x)\\
\end{align*}</span>
which is the <em>composition</em> of <span class="math-container">$f$</span> with <span class="math-container">$h$</span>. It follows according to the <em>definition</em> of <span class="math-container">$h$</span> and <span class="math-container">$g$</span> :
<span class="math-container">\begin{align*}
g(x)=\left(f\circ h\right)(x)=f(h(x))=f(cx)
\end{align*}</span></p>
<p><strong>Differentiation: Two different situations</strong></p>
<p>As @KarlSchildkraut said there is no ambiguity, but we have different situations instead. On the one hand we have</p>
<p><span class="math-container">\begin{align*}
\color{blue}{f^{\prime}(cx)=f^{\prime}(u)\big|_{u=cx}}
\end{align*}</span>
which is the differentiated function <span class="math-container">$f$</span> <em>evaluated</em> at a point <span class="math-container">$u=cx$</span>. On the other hand we have
<span class="math-container">\begin{align*}
\color{blue}{g^{\prime}(x)}&\color{blue}{=\left(f\circ h\right)^{\prime}(x)}=\left(f(h(x)\right)^{\prime}=\left(f(cx)\right)^{\prime}\\
&=f^{\prime}(h(x))h^{\prime}(x)\\
&\,\,\color{blue}{=f^{\prime}(cx)c}
\end{align*}</span>
which is the differentiation of a composition of functions. So, we have two different situations.</p>
|
2,482,868 | <p>I am trying to find</p>
<p>$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$</p>
<p>where $p>0$. I have tried to factor out as</p>
<p>$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
but still was not able to make progress. Any other approach to this is welcome.</p>
| user70925 | 479,902 | <p>Let try to use a general method to solve this kind of limits by
looking at the first order Taylor expansion of your expression:</p>
<p>\begin{aligned}\left( 1+x^{p+1}\right)^{\frac{1}{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} &=
x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}} \\
&= x\left[ \left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- \left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}}\right] \\
&= x\left[1+\frac{1}{x^{p+1}{p+1}}+o(x^{-p-1}) - 1 -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&= x\left[\frac{1}{x^{p+1}{p+1}} -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&=\frac{1}{x^{p}{p+1}} -\frac{1}{x^{p-1}{p}}+o(x^{-p+1})
\end{aligned}
So that the whole expression tends to 0 when $x\rightarrow \infty$.</p>
|
815,739 | <p>Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?</p>
| fosco | 685 | <p>My two cents: to a certain extent your question is not silly, and can be partially answered.</p>
<p>Let's start unbiased: you have this big box where you put all vector spaces $\mathbb R^n$ and linear maps between them. If you think up to isomorphism it's a category, the category $\bf Vect$ of finite dimensional real vector spaces (in functorial salons people say that it is the skeleton of the category of real vector spaces, but don't worry about this). </p>
<p>Now, notice that $\mathbb R^n\oplus \mathbb R^m = \mathbb R^{n+m}$ and $\mathbb R^n\otimes \mathbb R^m = \mathbb R^{nm}$, and suchlike relations, so direct sum and multiplication "behave like" addition and multiplication of natural numbers.</p>
<p>In fact this is not so silly: (iso classes of) vector spaces <a href="http://ncatlab.org/nlab/show/vertical+categorification" rel="nofollow">categorify</a> natural numbers.</p>
<p>Now, there is a machinery you feed with a commutative, cancellative monoid (like $\mathbb N$) and which returns you an abelian group (like $\mathbb Z$): it is the <a href="http://ncatlab.org/nlab/show/Grothendieck+group" rel="nofollow">Grothendieck group</a> of the monoid. Basically you formally add inverses to those elements which do not have one, and define a compatible group operation extending the old monoid operation.</p>
<p>Now.</p>
<p>What if I was able to do "the same thing" in the upper floor, "formally inverting" vector spaces? In the end, "grouping" the monoid $\mathbb N$ you used some sort of adjunction $G\colon {\bf Mon\leftrightarrows Grp}\colon U$: it is reasonable to expect that this extends (internalizes?) to a 2-adjunction
$$
\underline G\colon {\bf Mon(Cat) \leftrightarrows Grp(Cat)}\colon \underline U
$$
sending a monoidal category (like, for example, the categoy of vector spaces) into its <em>2-Grothendieck group</em>. Google in fact gives a reference about something like this: an old Joyal paper called "Traced Monoidal Categories".</p>
<p>In principle, you should be able to build a new category (which can have really nothing to share with the old $\bf Vect$ you started with, but which we call $\overline{\bf Vect}$), which is a <a href="http://en.wikipedia.org/wiki/2-group" rel="nofollow">categorical group</a>, i.e. a monoidal category "where every object $M$ has a multiplicative inverse $\bar M$", such that $M\otimes \bar M\cong \bar M\otimes M\cong I$, if $I$ is the identity object ($\mathbb R^1$, or $\mathbb R^0$, in the case of $\bf Vect$, according to the case you consider $\oplus$- or $\otimes$-monoidal structure).</p>
<p>Now, I've absolutely no clue about the shape of your $\overline{\bf Vect}$, or how does it behave under "natural" constructions, or whether it has some "nice" properties.</p>
|
184,940 | <p>Gauss has proven in his famous Theorema Egregium, that it is possible, to calculate the gaussian curvature from measuring angles and distances on the surface, irrespective of how the surface is embedded into space.</p>
<blockquote>
<p><strong>Question:</strong></p>
<p>is it also possible, to calculate the euclidean distance of two points on the surface, also from distance- and angle-measurements on the surface, alone?</p>
</blockquote>
| Sebastian | 4,572 | <p>My answer is contained in the one of Thomas which was posted when I wrote my answer.</p>
<p>No, consider the plane $\{(x,y,0)\mid x,y\in\mathbb,0<y<2\pi\}$ and the subset
$\{(x,\cos y, \sin y)\mid x,y\in\mathbb,0<y<2\pi\}$ of the round cylinder in euclidean 3-space. Both are globally isometric, so all distances and angles are the same. Clearly, their euclidean distances are not the same.</p>
|
162,520 | <p>This is a cross-post from <a href="https://math.stackexchange.com/questions/738094/good-book-on-analytic-continuation">MSE</a>.</p>
<p>For my Bachelor's thesis, I am investigating divergent series summation methods. One of those is analytic continuation. There are quite a few books on complex analysis that include a chapter or two on analytic continuation, but I would like to study this phenomenon in more detail. Could you please suggest a book on analytic continuation? Or a book on complex analysis with a heavy focus on analytic continuation?</p>
| user74550 | 74,550 | <p>There is a book on Complex Variables with Physical Applications by Arthur A.Hauser, Jr.
is theory and step-by step solutions to 760 problems.Chapter 10 of this book deals at grant extend on Analytic Continuation.</p>
|
1,412,862 | <p>The author of my textbook has an unsatisfactory proof when it is describing the properties of the closure of a set.
I'm using <span class="math-container">$E^*$</span> for E closure. Also, <span class="math-container">$E'$</span> indicates the set of limit points of <span class="math-container">$E$</span>.</p>
<blockquote>
<p><strong>Theorem:</strong> <span class="math-container">$E^*\subset F$</span> for every closed set <span class="math-container">$F\subset X$</span> such that <span class="math-container">$E\subset F$</span></p>
<p><em>Proof</em>: If <span class="math-container">$F$</span> is closed and <span class="math-container">$E\subset F$</span>, then <span class="math-container">$F'\subset F$</span>, hence <span class="math-container">$E'\subset F$</span>. Thus <span class="math-container">$E^*\subset F$</span>.</p>
</blockquote>
<p>My question is, why is he able to conclude that <span class="math-container">$F$</span> contains <span class="math-container">$E$</span>'? Why is <span class="math-container">$E'$</span> a subset of <span class="math-container">$F'$</span>?</p>
| Rupsa | 264,612 | <p>for your 2nd ques prove if F contains E then F' contains E' if E'=null set then there nothing to prove . Let E' is not a null set then x<em>e</em>E' implies x is a limit point of the set E in X implies E meet [B(x,r){x}] is not equal to null set for all +ve real r implies F meet[B(x,r){x}] is not equal to null set as F contains E implies x is a limit point of F in X implies x belongs to F' therefore F' contains E'</p>
|
1,992,789 | <blockquote>
<p>Let $\{x_n\}$ be a sequence that does not converge and let L be a real
number. Prove that there exist $\epsilon >0$ and a sub-sequence
$\{x_{p_n}\}$ of $\{x_n\}$ such that $|x_{p_n}-L|>\epsilon$ for all n.</p>
</blockquote>
<p>I don't have any idea on how to prove this. Any advice and suggestions will be greatly appreciated.</p>
| barak manos | 131,263 | <p>Use <em>inclusion/exclusion</em> principle:</p>
<ul>
<li>Include the total number of sequences, which is $\frac{8!}{2!2!2!2!}$</li>
<li>Exclude the number of sequences containing $11$, which is $\frac{7!}{1!2!2!2!}$</li>
<li>Exclude the number of sequences containing $22$, which is $\frac{7!}{1!2!2!2!}$</li>
<li>Exclude the number of sequences containing $33$, which is $\frac{7!}{1!2!2!2!}$</li>
<li>Exclude the number of sequences containing $44$, which is $\frac{7!}{1!2!2!2!}$</li>
<li>Include the number of sequences containing $11$ and $22$, which is $\frac{6!}{1!1!2!2!}$</li>
<li>Include the number of sequences containing $11$ and $33$, which is $\frac{6!}{1!1!2!2!}$</li>
<li>Include the number of sequences containing $11$ and $44$, which is $\frac{6!}{1!1!2!2!}$</li>
<li>Include the number of sequences containing $22$ and $33$, which is $\frac{6!}{1!1!2!2!}$</li>
<li>Include the number of sequences containing $22$ and $44$, which is $\frac{6!}{1!1!2!2!}$</li>
<li>Include the number of sequences containing $33$ and $44$, which is $\frac{6!}{1!1!2!2!}$</li>
<li>Exclude the number of sequences containing $11$ and $22$ and $33$, which is $\frac{5!}{1!1!1!2!}$</li>
<li>Exclude the number of sequences containing $11$ and $22$ and $44$, which is $\frac{5!}{1!1!1!2!}$</li>
<li>Exclude the number of sequences containing $11$ and $33$ and $44$, which is $\frac{5!}{1!1!1!2!}$</li>
<li>Exclude the number of sequences containing $22$ and $33$ and $44$, which is $\frac{5!}{1!1!1!2!}$</li>
<li>Include the number of sequences containing $11$ and $22$ and $33$ and $44$, which is $\frac{4!}{1!1!1!1!}$</li>
</ul>
<hr>
<p>Hence the number of such sequences is $\sum\limits_{n=0}^{4}(-1)^n\cdot\binom4n\cdot\frac{(8-n)!}{(1!)^n(2!)^{4-n}}=864$.</p>
|
1,027,235 | <p>Let N = 12345678910111213141516171819. How can I use modular arithmetic to show that N is (or isn't) divisible by 11? In general, how can I apply modular arithmetic to determine the divisibility of an integer by a smaller integer? I am finding modular arithmetic very confusing and unintuitive. I can understand "simple" modular arithmetic like the 24-hour day etc. but when it comes down to finding the modulo of high raised powers, or checking divisibility of large integers, I am totally lost. </p>
| Silvio Moura Velho | 74,553 | <p>$N = a,bcd$; $a' \equiv (− cd \mod 11 + a) \mod 11 \rightarrow a'b$</p>
<p>$N = bcd;$ $a' \equiv (−cd \mod 11 + 0) \mod 11 \rightarrow a'b$</p>
<p>If $11 |a 'b$ then $11|N$</p>
<p>Apply the algorithm repetitively from right to left, always eliminating the last two digits.If the result is a multiple of $11$ then the tested number is also a multiple of $11$.
This algorithm works quickly for divisibility by $7$, $11$ and $13$.</p>
<p>Regarding divisibility by $7$, Google: Youtube, "divisibility by $7$", "large number"</p>
|
454,426 | <blockquote>
<p>In set theory and combinatorics, the cardinal number $n^m$ is the size of the set of functions from a set of size m into a set of size $n$.</p>
</blockquote>
<p>I read this from this <a href="http://en.wikipedia.org/wiki/Empty_product#0_raised_to_the_0th_power" rel="nofollow noreferrer">Wikipedia page</a>.</p>
<p>I don't understand, however, why this is true. I reason with this example in which $M$ is a set of size $5$, and $N$ is a set of size $3$. For each element in set $M$, there are three functions to map the element from the set of size $5$ to an element in the set of size $3$. </p>
<p>By my reasoning, that means the total number of functions is just $3*5$, i.e. $3$ functions for each of the $5$ elements in the set. Why is it actually $3^5$? I saw on <a href="https://math.stackexchange.com/questions/209361/size-of-the-set-of-functions-from-x-to-y">this thread</a> that the number of functions from a set of size $n$ to a set of size $m$ is equivalent to "How many $m$-digit numbers can I form using the digits $1,2,...,n$ and allowing repetition?" I know how to answer that question, but I don't know why it's the same thing as finding the number of functions from the size $n$ set to the size $m$ set. </p>
| Stefan Hamcke | 41,672 | <p>An $m$-digit number $x_1 x_2 ... x_m$ where each $x_i$ can be a digit from the set $\{1,...,n\}$ is basically the same as a function from $M:=\{1,2,...,m\}$ to $N:=\{1,2,...,n\}$. Let $x:M\to N$ be such a map. Then you can construct the number $x(1)x(2)...x(m)$. If on the other hand you have a number $x_1 x_2 ... x_m$, then $x(i)=x_i$ yields a function from $M$ to $N$.</p>
|
501,250 | <p>I want to say that $|\textbf{x}-\textbf y|<\delta$ implies $|x_1- y_1|<\delta$ and $|x_2- y_2|<\delta$ for a proof I am working on. This is assuming that $\textbf{x}=(x_1,x_2) \in \text R^2$ and $\textbf{y}=(y_1,y_2) \in \text R^2$. If true, I'd also like to extend this to $\textbf{x} \in \text R^{n_1+n_2}$ and $\textbf{x} \in \text R^{n_1+n_2}$ where $n_1 , n_2 \in \text N$.</p>
<p>I am wondering if this is obvious enough to state or would it be better to prove it? If I should prove it then what would be the best way? My guess would be to use the distance formula. </p>
| Arash | 92,185 | <p>The set $C =\{ \frac{1}{n}|n \in \mathbb{N} \}$ is not open in $\mathbb R$. The reason is that for each $x\in C$ you cannot find a neighborhood of $x$ such that $N(x,r)\subset C$. For each neighborhood of $x=\frac{1}{n}$, it is possible to find $y$ such that $\frac{1}{n+1}<y<\frac{1}{n}$ which means that $y\notin C$. </p>
<p><strong>Some remarks:</strong> The concept of open and closed is meaningful only with respect to a given topological space. If your space is $\mathbb R$ with Euclidean metric, then the metric is $|x-y|$ for $x,y\in\mathbb R$. It is also not pertinent to represent a set $C$ of $\mathbb R$ in two dimensional space.</p>
|
1,105,126 | <p><img src="https://i.stack.imgur.com/XtrB7.png" alt="enter image description here"></p>
<p>My attempt at the solution is to let P(n) be $10^{3n} + 13^{n+1}$</p>
<p>P(1)= $10^3 + 13^2 = 1169$</p>
<p>Thus P(1) is true.</p>
<p>Suppose P(k) is true for all $k \in N$
$\Rightarrow P(k) = 10^{3k} + 13^{k+1} = 10^{3k} + 13\cdot13^{k}$</p>
<p>$P(k+1) = 10^{3k+3} + 13^{k+2} \\ P(k+1) = 1000\cdot10^{3k} + 169\cdot13^k \\ P(k+1) = (10^{3k} + 13\cdot13^k) + 999\cdot10^{3k} + 12\cdot13^{k+1}$</p>
<p>My solution is not divisible by 7. I've always use this method to solve these types of questions. Can someone point out my error?</p>
| orangeskid | 168,051 | <p>Consider $P$ the amount of papers. In one minute the first person delivers $\frac{P}{40}$, the second one $\frac{P}{50}$. Together in one minute they deliver $\frac{P}{40}+\frac{P}{50}$. So they need </p>
<p>$$\frac{P}{ \frac{P}{40}+\frac{P}{50}}$$ minutes to deliver the whole thing. </p>
<p>The quantity $P$ magically disappears...</p>
|
1,521,649 | <p>I need to show that at most finitely many terms of this sequence are greater than or equal to $c$. </p>
<p>I don't know if it is the wording of the problem but I don't know what this is asking me to do. Help on this would be amazing! And thank you in advance.</p>
| egreg | 62,967 | <p>I guess you have already proved the main property
$$
\ln(xy)=\ln x+\ln y
$$
(for positive reals $x$ and $y$). By an easy induction you get that
$$
\ln(a^m)=m\ln a
$$
for a positive integer $m$. If $m<0$, you have
$$
\ln(a^m)=\ln\frac{1}{a^{-m}}=-\ln(a^{-m})=-(-m\ln a)=m\ln a
$$</p>
<p>Now suppose $b=m/n$, where, without loss of generality, $n>0$. Then
$$
\ln(a^b)=\ln((a^{1/n})^m)=m\ln(a^{1/n})
$$
by the above argument, so we're left to prove that
$$
\ln(a^{1/n})=\frac{1}{n}\ln a
$$
Since
$$
n\left(\frac{1}{n}\ln a\right)=\ln a
$$
and
$$
n\ln(a^{1/n})=\ln((a^{1/n})^n)=\ln a
$$
we get the desired result.</p>
|
4,292,815 | <p>Compute line integral <span class="math-container">$\int_a^b (y^2z^3dx + 2xyz^3dy + 3xy^2z^2dz)$</span> where <span class="math-container">$a = (1,1,1)$</span> and <span class="math-container">$b = (2,2,2)$</span></p>
<p>What I have done:</p>
<p>To find <span class="math-container">$t$</span> I used the calculation for slope:</p>
<p><span class="math-container">$\frac{x-1}{2-1}=t, \frac{y-1}{2-1}=t, \frac{z-1}{2-1}=t$</span> and then re-arranged for x,y,z to calculate the derivate to find that <span class="math-container">$dx =1, dy=1, dz=1$</span> and <span class="math-container">$t+1 = (x, y, z)$</span></p>
<p>Plugging this back into the integral:
<span class="math-container">$$6\int_a^b (t+1)^5 dt$$</span></p>
<p>However, what do I do with the integral bounds? Do I set them from <span class="math-container">$t \in [1,2]$</span>?</p>
| José Carlos Santos | 446,262 | <p>Since <span class="math-container">$(x,y,z)=(t+1,t+1,t+1)$</span> and you want to go from <span class="math-container">$(1,1,1)$</span> to <span class="math-container">$(2,2,2)$</span>, you should compute that integral with <span class="math-container">$a=0$</span> and <span class="math-container">$b=1$</span>.</p>
<p>Note that talking about <span class="math-container">$\int_a^b(y^2z^3\,\mathrm dx+2xyz^3\,\mathrm dy+3xy^2z^2\,\mathrm dz)$</span> only makes sense because your vector field is conservative. Otherwise, the integral would depend upon the path going from <span class="math-container">$(1,1,1)$</span> to <span class="math-container">$(2,2,2)$</span>.</p>
|
858,353 | <p>$$x^2(y')^2+3xyy'+2y^2=0$$
I have no idea how to start, I probably need to do some tricky substitution but as of now I cant see any options.</p>
| johannesvalks | 155,865 | <p>Hint:</p>
<p>$$
\Big( x y' + \frac{3}{2} y \Big)^2 = x^2 (y')^2 + 3xy y' + \frac{9}{4} y^2
$$</p>
|
3,489,280 | <p>If <span class="math-container">$f_n → f$</span> and <span class="math-container">$g_n → g$</span>, does <span class="math-container">$f_n g_n → fg$</span> in the space <span class="math-container">$C[0, 1]$</span> for the norms <span class="math-container">$||.||_1$</span> and <span class="math-container">$||.||_∞$</span></p>
<p>Give a proof or counterexample for each.</p>
<p>I know that <span class="math-container">$||.||_1$</span> is the sum of the magnitudes of the values, and <span class="math-container">$||.||_∞$</span> is the biggest magnitude of the values. From this I assume <span class="math-container">$||.||_1$</span> DOES work whilst <span class="math-container">$||.||_∞$</span> doesn't, but I am unsure on if this is true, and if so, how to show it.</p>
<p>Thanks in advance</p>
| zhw. | 228,045 | <p>Unfortunately, both of your guesses are incorrect.</p>
<p>To see the <span class="math-container">$\|\,\|_1$</span> result fails, let</p>
<p><span class="math-container">$$f_n(x)= \frac{1}{[\ln (n+1)(x+1/n)]^{1/2}}.$$</span></p>
<p>Verify that <span class="math-container">$f_n\to 0$</span> in the <span class="math-container">$\|\,\|_1$</span>-norm. However <span class="math-container">$f_n\cdot f_n$</span> does not converge to <span class="math-container">$0\cdot 0=0$</span> in the <span class="math-container">$\|\,\|_1$</span>-norm, as you can check.</p>
<p>The <span class="math-container">$\|\,\|_\infty$</span> result is true. This is classic and is easier to show. We have <span class="math-container">$f_n\to f,g_n\to g$</span> uniformly on <span class="math-container">$[0,1].$</span> This implies there is a uniform bound on all of these functions. Now use</p>
<p><span class="math-container">$$f_ng_n-fg= f_ng_n-fg_n + fg_n-fg$$</span></p>
<p>to see <span class="math-container">$f_ng_n\to fg$</span> uniformly.</p>
|
3,489,280 | <p>If <span class="math-container">$f_n → f$</span> and <span class="math-container">$g_n → g$</span>, does <span class="math-container">$f_n g_n → fg$</span> in the space <span class="math-container">$C[0, 1]$</span> for the norms <span class="math-container">$||.||_1$</span> and <span class="math-container">$||.||_∞$</span></p>
<p>Give a proof or counterexample for each.</p>
<p>I know that <span class="math-container">$||.||_1$</span> is the sum of the magnitudes of the values, and <span class="math-container">$||.||_∞$</span> is the biggest magnitude of the values. From this I assume <span class="math-container">$||.||_1$</span> DOES work whilst <span class="math-container">$||.||_∞$</span> doesn't, but I am unsure on if this is true, and if so, how to show it.</p>
<p>Thanks in advance</p>
| mechanodroid | 144,766 | <p>For a counterexample to the <span class="math-container">$\|\cdot\|_1$</span> norm, consider
<span class="math-container">$$f_n(t) = \begin{cases} n-n^3t, &\text{ if } t \in \left[0,\frac1{n^2}\right]\\
0, &\text{ if } t \in \left[\frac1{n^2},1\right]\\
\end{cases}$$</span></p>
<p>Then <span class="math-container">$\|f_n\|_1 = \frac1{2n} \to 0$</span> but <span class="math-container">$\|f_n^2\|_1 = \frac13 \not\to 0$</span>.</p>
|
1,914,752 | <p>dividing by a whole number i can describe by simply saying split this "cookie" into two pieces, then you now have half a cookie. </p>
<p>does anyone have an easy way to describe dividing by a fraction? 1/2 divided by 1/2 is 1</p>
| John Joy | 140,156 | <p>Dividing a whole number (lets say one) into two pieces can be restated as "If one (lets say gallon) fills two containers, then how much does one container hold?". Of course the answer is $\frac{1}{2}$ gal.</p>
<p>Using that same language consider "If $\frac{5}{16}$ Gal. fills $\frac{3}{7}$ of a container, then how much does one container hold?". If we scale both numbers by a factor of seven, we see that $3$ containers will hold $\frac{5\times 7}{16}$ gal.. Finally, we scale by a factor of $\frac{1}{3}$, giving $3\times\frac{1}{3} = 1$ container contains $\frac{5\times 7}{16\times 3}$ gal..</p>
|
1,479,822 | <p>$f\cdot g$ is Lebesgue integrable, g is Lebesgue integrable, can we deduce that f is Lebesgue integrable? </p>
<p>$f\cdot g$ is integrable, g is integrable, can we deduce that f is finite a.e.? </p>
| Chappers | 221,811 | <p>If the support of $g$ and the support of $f$ have intersection with measure zero, you can't conclude anything.</p>
|
203,827 | <p>Suppose I have the following lists: </p>
<pre><code>prod = {{"x1", {"a", "b", "c", "d"}}, {"x2", {"e", "f",
"g"}}, {"x3", {"h", "i", "j", "k", "l"}}, {"x4", {"m",
"n"}}, {"x5", {"o", "p", "q", "r"}}}
</code></pre>
<p>and </p>
<pre><code>sub = {{"m", "n"}, {"o", "p", "r", "q"}, {"g", "f", "e"}};
</code></pre>
<p>for each element in <code>sub</code> I want to go through <code>prod</code> and select if the element exist such that I get the following output, </p>
<pre><code> {{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q","r"}}}
</code></pre>
<p>I tried doing: </p>
<pre><code>Table[Select[
prod[[All, 2]][[i]], # == ContainsAny[Map[Sort, sub]][[i]] &], {i,
Length[sub]}]
</code></pre>
<p>yet it doesn't work, am I missing something? </p>
| Chris Degnen | 363 | <pre><code>getMatches[prod_, sub_] := Module[{test},
Scan[(test[Sort[#]] = True) &, sub];
Cases[prod, {_, y_?test}]]
getMatches[prod, sub]
</code></pre>
<blockquote>
<p>{{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q", "r"}}}</p>
</blockquote>
<p>Also</p>
<pre><code>getMatches[prod_, sub_] := Module[{test},
Scan[(test[Sort[#]] = True) &, sub];
Select[prod, test@*Last]]
</code></pre>
|
1,997,513 | <p>I spend so much time for proving this triangle and i still don't know. </p>
<p>Question :</p>
<p>Given Triangle ABC, AD and BE are altitudes of the triangle. Prove that Triangle DEC similarity with triangle ABC</p>
| Vidyanshu Mishra | 363,566 | <p>Construct a figure from the data you have given, now follow these steps(you just have to prove equality of any two angles of respective triangles):</p>
<p>Since BE and AD are perpendiculars so you get ∠BEA = ∠ADB =90 Degrees.Now you can see that ABDE is a cyclic quadrilateral {as ∠BEA = ∠ADB(you can call them angles in the same segment)}</p>
<p>Now as you know sum of opposite angles in a cyclic quadrilateral is 180 degrees so ∠EDB= 180-∠A = ∠EDC. Therefore:</p>
<p>For triangles CDE and CAB:
∠CDE=∠A
and ∠C = ∠C </p>
<p>Hence,triangles CDE and CAB are similar.</p>
|
4,263,631 | <p>so i have question about existence of function <span class="math-container">$f:\mathbb{R} \to \mathbb{R}$</span> such <span class="math-container">$f$</span> is not the pointwise limit of a sequence of continuous functions <span class="math-container">$\mathbb{R} \to \mathbb{R}$</span>.</p>
<p>i'm created a family of continuous functions <span class="math-container">$k_i:\mathbb{R} \to \mathbb{R}$</span> for any function <span class="math-container">$f:\mathbb{R} \to \mathbb{R}$</span> such that <span class="math-container">$\lim_{n \to \infty} k_n(x)=f(x)$</span> , how ever i am pretty sure my construction have a flaw in it but i could not understand why it's wrong can some one tell me what am i doing wrong , because it seems very unrealistic to be able to do it for any <span class="math-container">$f$</span>.</p>
<p>here is my construction steps:</p>
<p>1.pick any arbitrary <span class="math-container">$x,y \in \mathbb{R}$</span> such <span class="math-container">$x<y$</span> , and connect <span class="math-container">$f(x)$</span> to <span class="math-container">$f(y)$</span>.</p>
<p>2.pick any arbitrary <span class="math-container">$z \in \mathbb{R}-\{x,y\}$</span> , if <span class="math-container">$f(y)<f(z)$</span> connect <span class="math-container">$f(z)$</span> to <span class="math-container">$f(y)$</span> and if <span class="math-container">$f(z)<f(x)$</span> connect <span class="math-container">$f(z)$</span> to <span class="math-container">$f(x)$</span>, if <span class="math-container">$f(x)<f(z)<f(y)$</span> , connect <span class="math-container">$f(x)$</span> to <span class="math-container">$f(z)$</span> and <span class="math-container">$f(z)$</span> to <span class="math-container">$f(y)$</span>.</p>
<p>now we are again doing this uncountable times with other elemnts in <span class="math-container">$\mathbb{R}-\{x,y,z\}$</span> and do the step 2 for that element over all choosed elements by uncountable number of comparisions (after uncountable number of iteration we will have uncountable number of picked elements of <span class="math-container">$\mathbb{R}$</span>).</p>
<p>call set of picked elements <span class="math-container">$S$</span>, when we pick an other element from <span class="math-container">$\mathbb{R}-S$</span> , like <span class="math-container">$t$</span> we will consider <span class="math-container">$ x_1= \max \{ x ; x \in S \land x<t \}$</span> and <span class="math-container">$x_2 = \min \{ x ; x \in S \land x>t \}$</span> and we connect <span class="math-container">$f(x_1)$</span> to <span class="math-container">$f(t)$</span> and then <span class="math-container">$f(t)$</span> to <span class="math-container">$f(x_2)$</span> and call all the connected line with <span class="math-container">$S \cup {t}$</span> function <span class="math-container">$k_t$</span>. we will do this uncountable number of times.</p>
<p>i'm pretty sure there is a flaw in my argument , but can you please help me to find it.i really appreciate your kindness and support.</p>
| Mark | 470,733 | <p>You are close. Measure of a set is equal to the Lebesgue integral of the constant function <span class="math-container">$1$</span> on that set. So the measure of your set is equal to the double integral <span class="math-container">$\iint\limits_A 1 dxdy$</span>. Since the function has an absolutely convergent improper Riemann integral, the Lebesgue integral is equal to the improper Riemann integral here. And by Fubini's theorem the Riemann integral is indeed equal to <span class="math-container">$\int_1^\infty\int_0^{\frac{1}{x^2}}1dydx=\int_1^\infty\frac{1}{x^2}dx$</span>, which is a finite number.</p>
|
13,989 | <p>Suppose $E_1$ and $E_2$ are elliptic curves defined over $\mathbb{Q}$.
Now we know that both curves are isomorphic over $\mathbb{C}$ iff
they have the same $j$-invariant.</p>
<p>But $E_1$ and $E_2$ could also be isomorphic over a subfield of $\mathbb{C}$.
As is the case for $E$ and its quadratic twist $E_d$. Now the question general is.</p>
<blockquote>
<p>$E_1$ and $E_2$ defined over $\mathbb{Q}$ and isomorphic over $\mathbb{C}$. Let $K$
the smallest subfield of $\mathbb{C}$ such that $E_1$ and $E_2$ become isomorphic over $K$.
What can be said about $K$. Is it always a finite extension of $\mathbb{Q}$. If so, what can be
said about the extension $K|\mathbb{Q}$.</p>
</blockquote>
<p>My second question is something goes something like in the opposite direction. I start again with
quadratic twists. Let $E$ be an elliptic curve over $\mathbb{Q}$ and consider the quadratic extension
$\mathbb{Q}|\mathbb{Q}(\sqrt{d})$. Describe the curves over $\mathbb{Q}$(or isomorphism classes over $\mathbb{Q}$)
which become isomorphic to $E$ over $\mathbb{Q}(\sqrt{d})$. I think the answer is $E$ and $E_d$.
Again I would like to know what happens if we take a larger extension.</p>
<blockquote>
<p>Let $E$ be an elliptic curve over $\mathbb{Q}$ and $K|\mathbb{Q}$ a finite extension.
Describe the isomorphism classes of elliptic curves over $\mathbb{Q}$ which become isomorphic
to $E$ over K.</p>
</blockquote>
<p>I have no idea what is the right context to answer such questions.</p>
| Andrea Mori | 3,602 | <p>As for your first question: if you think of your elliptic curves as plane cubics (Weierstrass'model) the isomorphism between them is a polynomial function. Polynomials include only finitely many coefficients and the isomorphism is defined over the field generated by them, which is finite over $\Bbb Q$.</p>
|
497,546 | <p>Let $A$ be an infinite set.</p>
<p>Then, we can construct an injective function $f:\omega \rightarrow A$. </p>
<p>But how do i construct this via orginal statement of $AC_\omega$? (i.e. $\forall countable X, [\emptyset \notin X \Rightarrow \exists f:X\rightarrow \bigcup X \forall A\in X, f(A)\in A$)</p>
<p>So my question is:</p>
<p>(1) how do i prove that every infinite set has a countable subset?</p>
<p>(2) how do i write down a situation such as: "after choosing $x_1,...,x_k$, choosing $x_k$ satisfying a given condition. Continue this and form a sequence"</p>
| Herman Jaramillo | 62,024 | <p>In general, a generating function for a sequence of functions $P_n(x)$, is a function
$G(x,t)$, such that</p>
<p>\begin{eqnarray*}
G(x,t)= \sum_{n=0}^{\infty} P_n(x) t^n,
\end{eqnarray*}
where, by matching equal powers of $t$, the Taylor series expansion of $G(x,t)$ provides
the functions $P_n(x)$. In particular we find $G(x,t)$ when the $P_n(x)$ are Legendre
polynomials. Historically,
<a href="https://en.wikipedia.org/wiki/Legendre_polynomials" rel="nofollow noreferrer">Legendre</a>
defined the polyomials $P_n(x)$ knowing that $G(x,t)$ was the Newtonian potential
(gravitational potential) but here we assumed that $P_n$ were found from solving
the Laplace equation with azimuthal symmetry and some data given on the surface of
a sphere. That is, we set up the problem of solving </p>
<p>\begin{eqnarray*}
\nabla^2 u(\mathbf{r}) =0
\end{eqnarray*}
subjected to some boundary conditions. Let us assume the Dirichlet boundary condition
at a sphere $S$, $|\mathbf{r}|=r_1$,</p>
<p>\begin{eqnarray*}
u(\mathbf{r} \in S)=\frac{1}{|\mathbf{r}_1-\mathbf{r}_0|}
\label{bcdiri}
\end{eqnarray*}
with $\mathbf{r}_0$ some fixed vector in $\mathbb{R}^3$.
Then the solution of the Laplace's equation</p>
<p>\begin{eqnarray}
u(\mathbf{r}) = \frac{1}{|\mathbf{r} - \mathbf{r}_0|}
\label{sollap}
\end{eqnarray}
satisfies the Dirichlet boundary condition above. Since
the solution of the Laplace's equation is unique with Dirichlet boundary conditions
we claim that this equation is THE solution of the Laplace's equation
with the given Dirichlet boundary conditions. Let us further
assume that $\mathbf{r}_0$ is aligned with the positive $z$ axis.</p>
<p>The solution of the Laplace's equation using separation of variables
provides </p>
<p>\begin{eqnarray*}
u_{\mathbf{r}_0}(r, \theta, \phi) = \sum_{n=0}^{\infty} A_n r^n P_n( \cos \theta).
\end{eqnarray*}
Here $r$ is $|\mathbf{r}|$, $\theta$ is
the polar angle that $\mathbf{r}$ makes with the positive axis $z$, and $\phi$
is the azimutal angle. The angle of the projection of $\mathbf{r}$ on the plane
$z=0$ and the $x$ axis. It is then implicit here that we are assuming azimuthal
symmetry. If there is no azimuthal symmetry we are forced to use spherical
harmonic functions $Y_{nm}(\theta,\phi)$ instead of Legendre polynomials
$P_n(\cos \theta)$ for this representation. Still, we show at the end how to get around this assumption. When solving the Laplace's equation using
separation of variables the radial functions $r^n$ are solutions of the resulting Euler ODE
and the polar functions $P_n(\cos \theta)$ are solutions of the Legendre differential
equation. We have
\begin{eqnarray*}
|\mathbf{r}-\mathbf{r}_0|^2 = \mathbf{r} \cdot \mathbf{r} - 2 \mathbf{r} \cdot \mathbf{r}_0 + \mathbf{r}_0
\mathbf{r}_0 = r^2 - 2 r \cos \alpha + 1,
\end{eqnarray*}
where $\alpha$ is the angle between $\mathbf{r}$ and $\mathbf{r}_0$. However, since
we assume that $\mathbf{r}_0$ is aligned with the positive $z$ axis $\alpha=\theta$ and
the function</p>
<p>\begin{eqnarray*}
\frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}}.
\end{eqnarray*}
satisfies the Laplace's equation with Dirichlet boundary conditions, so we can write</p>
<p>\begin{eqnarray*}
\frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}} = \sum_{n=0}^{\infty} A_n r^n P_n(\cos
\theta).
\end{eqnarray*}
We should now find $A_n$ to finish the problem. We show $A_n=1$.
Since the identity should be valid for all angles, let us assume $\theta=0$,
so $\cos \theta=1$ and we know already that $P_n(1)=1$, so</p>
<p>\begin{eqnarray*}
\frac{1}{|r-1|} = \frac{1}{1-r} = \sum_{n=0}^{\infty} A_n r^n \quad,
\quad \mathrm{please \; observe \; that \; we \; are \; assuming} \quad r<1
\end{eqnarray*}
but</p>
<p>\begin{eqnarray*}
\frac{1}{1-r}=1 + r + \cdots +r^n + \cdots,
\end{eqnarray*}
so $A_n=1$ and </p>
<p>\begin{eqnarray}
\frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}} = \sum_{n=0}^{\infty} r^n P_n(\cos
\theta).
\label{generatingLeg}
\end{eqnarray}
Hence</p>
<p>\begin{eqnarray}
G(x,r) = \frac{1}{\sqrt{r^2 - 2 r \cos \theta + 1}} =
\frac{1}{\sqrt{r^2 - 2 r x + 1}}
\label{torecur}
\end{eqnarray}
with $x=\cos \theta$ is the generating function for the Legendre polynomials.</p>
<p>At this point $x$ could represent any value between -1 and 1. In particular
if $\mathbf{r}$ and $\mathbf{r}_0$ are any two vectors with $|\mathbf{r}_0|=1$,
then $\mathbf{r} \cdot \mathbf{r}_0 = r \cos \alpha$, and we can write </p>
<p>\begin{eqnarray*}
\frac{1}{\sqrt{r^2 - 2 r \cos \alpha + 1}} = \sum_{n=0}^{\infty} r^n
P_n(\cos \alpha ),
\end{eqnarray*}
where now $\alpha$ is the angle between $\mathbf{r}$ and $\mathbf{r}_0$.</p>
<p>It is common in the literature to find the expansion of the generating function
$G(x,r)$ in Taylor series in $r$ and from their coefficients verify that they
correspond to $P_n(x)$. However this method, even though valid, implies more cumbersome algebra (infinity binomial expansions with fractional coefficients, Gamma functions with fractional coefficients...) and less physical insight.</p>
<hr>
|
4,258,226 | <p>Suppose a group <span class="math-container">$G$</span> splits as a semidirect product <span class="math-container">$N\rtimes\mathbb{Z}_2$</span>, and let <span class="math-container">$\phi:G\to\mathbb{Z}_2$</span> the the associated quotient map. If I have a subset of elements <span class="math-container">$\{g_1,\dots,g_n,h\}$</span> of <span class="math-container">$G$</span> such that <span class="math-container">$g_i\in N$</span> for <span class="math-container">$1\le i\le n$</span>, and <span class="math-container">$\phi(h)=[1]_2$</span> which I know <em>a priori</em> generates <span class="math-container">$G$</span>, <strong>can I conclude that <span class="math-container">$\{g_1,\dots,g_n\}$</span> generates <span class="math-container">$N$</span> as a subgroup of <span class="math-container">$G$</span>?</strong></p>
<p>I am particularly interested in the case the <span class="math-container">$G$</span> is a Coxeter group, so <span class="math-container">$N$</span> can be thought of (morally at any rate) as the subgroup of orientation preserving transformations. I can't see how I might prove this, but I also haven't been able to think of a counter example.</p>
| Adayah | 149,178 | <p>No, you can't.</p>
<p>Consider <span class="math-container">$G = \mathbb{Z}^2 \rtimes \mathbb{Z}_2$</span> where <span class="math-container">$1 \in \mathbb{Z}_2$</span> corresponds to the automorphism <span class="math-container">$\varphi \in \operatorname{Aut}(\mathbb{Z}^2, +)$</span> such that <span class="math-container">$\varphi(x, y) = (y, x)$</span>. Then it's easy to see that <span class="math-container">$\{ g, h \} = \{ ((1, 0), 0), ((0, 0), 1) \}$</span> generate <span class="math-container">$G$</span> but <span class="math-container">$(1, 0)$</span> does not generate <span class="math-container">$\mathbb{Z}^2$</span>.</p>
|
893,822 | <p>If $p(x)$ has integer coefficients and $p(100)$ equals $100$ what is the maximum number of integer solutions $k$ to the equation $p(k)=k^3$.</p>
<p>I have tried hard to solve this problem but I could not figure it out. I tried some particular cases but got nowhere, could someone please show me how to get the answer.</p>
| ShakesBeer | 168,631 | <p>The easy way:
<br>
Let $f(x)=RHS-LHS$.
<br>
$f'(x)=2n^2(1+n(x-1))^{n-1}-n(1+\frac{x}{\sqrt{x^2-1}})(x+\sqrt{x^2-1})^{n-1}-n(1-\frac{x}{\sqrt{x^2-1}})(x-\sqrt{x^2-1})^{n-1}$
<br><br>
$f'(x)=2n^2(1+n(x-1))^{n-1}-\frac{n}{\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}+\frac{n}{\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$
<br>
$f'(x) > 2n^2(1+n(x-1))^{n-1}-\frac{n^2}{x+\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}-\frac{n^2}{x-\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$.
<br>
That, by induction, gives $f'(x) \geq 0$, so we are done.
<br>
Verifying that $\frac{n}{\sqrt{x^2-1}}<\frac{n^2}{x+\sqrt{x^2-1}}$, for $n \geq 2, x>1$, and formalising the inductive argument, are exercises left to the reader.</p>
|
3,878,380 | <p>Do the columns of a matrix always represent different vectors? If so, I don't understand how if I have a <span class="math-container">$3\times3$</span> matrix where the rows represent the dimensions and I multiply it by a <span class="math-container">$3\times1$</span> column vector with the same dimensions, it will give me a vector. Some sources say the result comes from the dot product of each line - is this correct? If so, <span class="math-container">$a_{12}\cdot b_{21}$</span> would give zero right?</p>
| Steven Alexis Gregory | 75,410 | <p><span class="math-container">$$\left(\begin{array}{c}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right)
\cdot
\left( \begin{array}{c}
x \\ y \\ z
\end{array}
\right)
=
x\left( \begin{array}{c}
a \\ d \\ g
\end{array}
\right) +
y\left( \begin{array}{c}
b \\ e \\ h
\end{array}
\right) +
z\left( \begin{array}{c}
c \\ f \\ i
\end{array}
\right)
$$</span></p>
<p>and a linear combination of three vectors is a vector.</p>
|
175,971 | <p>Let's $F$ be a field. What is $\operatorname{Spec}(F)$? I know that $\operatorname{Spec}(R)$ for ring $R$ is the set of prime ideals of $R$. But field doesn't have any non-trivial ideals.</p>
<p>Thanks a lot!</p>
| Axel Boldt | 133,539 | <p>When people in modern algebraic geometry talk about <span class="math-container">$\operatorname{Spec}(k)$</span>, the prime spectrum of a field <span class="math-container">$k$</span>, they mean it in the sense of schemes.</p>
<p>Every scheme has an underlying topological space; in the case of <span class="math-container">$\operatorname{Spec}(k)$</span> this is a space with a single point, namely the zero ideal in <span class="math-container">$k$</span> (the only prime ideal in <span class="math-container">$k$</span>).</p>
<p>But a scheme is more than just a topological space: it is a "ringed space" satisfying certain properties. In particular, there's a commutative ring assigned to every open set of the space. In our case of <span class="math-container">$\operatorname{Spec}(k)$</span>, there are only two open sets: the empty set and the entire space. The ring assigned to the empty set is the trivial ring <span class="math-container">$\{0\}$</span>, and the ring assigned to the entire space is our field <span class="math-container">$k$</span>.</p>
<p>So in a sense, <span class="math-container">$\operatorname{Spec}(k)$</span> is a "single point that remembers the field it's defined over, namely <span class="math-container">$k$</span>". This is useful in order to define more general schemes defined over <span class="math-container">$k$</span>.</p>
|
1,132,599 | <p>For which prime numbers p does the congruence $x^2+x+1\equiv0$ mod p have solutions? </p>
<p>I am new to the topic of quadratic reciprocity and I know how to answer this question had it been for which prime numbers p does the congruence $x^2\equiv-6$ mod p have solutions? </p>
<p>Can I perhaps split the congruence into two parts, solve them individually and then combine solutions? </p>
<p>Thanks in advance. </p>
| David | 119,775 | <p><strong>Hint</strong>. For $p\ne2$ we have
$$x^2+x+1\equiv0\pmod p\quad\Leftrightarrow\quad
(2x+1)^2\equiv-3\pmod p\ .$$</p>
|
2,041,484 | <p>Solve the system of equations for all real values of $x$ and $y$
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$</p>
<p>I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.<br>
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5}}\le y \le {\frac 45}$}</p>
<p>I don't know what to do next.</p>
| Stefan4024 | 67,746 | <p>Let $z=x + iy \in \mathbb{C}$, now for $|z|^2 = x^2+y^2 \not = 0$ have that:</p>
<p>$$\left(5x + \frac{5x}{x^2+y^2}\right) + i\left(5y - \frac{5y}{x^2+y^2}\right) = 12 + 4i$$</p>
<p>$$5(x+iy) + \frac{5(x-iy)}{x^2+y^2} = 12 + 4i$$</p>
<p>$$5(x+iy) + \frac{5}{x+iy} = 12 + 4i$$</p>
<p>$$5z + \frac{5}{z} = 12 + 4i$$</p>
<p>$$5z^2 - (12+4i)z + 5 = 0$$</p>
<p>Solving this complex equation we get $z_1=\frac 25 - \frac 15i$ and $z_2 =2+i$, corresponding to the solution $(x,y) = \left\{\left(\frac 25, - \frac 15 \right), (2,1)\right\}$</p>
|
739,516 | <p>Is there a closed form for $$\int_{0}^{\infty}e^{ax^3+bx^2}\,\mathrm{d}x $$?</p>
| Sasha | 11,069 | <p>Let $a<0$, and changing variable $y = (-a x^3)$:
$$
\int_0^\infty \exp(a x^3 + b x^2) \mathrm{d}x = \alpha\int_0^\infty \exp\left(-y + \beta y^{2/3} \right) y^{-2/3} \mathrm{d}y
$$
where $\alpha = 1/\left(3 (-a)^{1/3}\right)$ and $\beta = b (-a)^{-2/3}$.</p>
<p>Now, using $y^{-2/3} \exp\left(\beta y^{2/3}\right) = \sum_{m=0}^\infty \frac{1}{m!} \beta^{m} y^{2 (m-1)/3}$, and interchanging the order summation and integration as warranted by Tonelli theorem:
$$
\int_0^\infty \exp\left(-y + \beta y^{2/3} \right) y^{-2/3} \mathrm{d}y = \sum_{m=0}^\infty \frac{\beta^m}{m!} \int_0^\infty \mathrm{e}^{-y} y^{2 (m-1)/3} \mathrm{d}y = \sum_{m=0}^\infty \frac{\beta^m}{m!} \Gamma\left(\frac{2m}{3} + \frac{1}{3}\right)
$$
Using duplication and triplication formulae for $\Gamma$-function
$$
\Gamma(2x) = \frac{2^{2x}}{2 \sqrt{\pi}} \Gamma(x) \Gamma\left(x+\tfrac{1}{2}\right) \quad \Gamma(3x) = \frac{3^{3x}}{2 \sqrt{3} \pi} \Gamma(x) \Gamma\left(x+\tfrac{1}{3}\right) \Gamma\left(x+\tfrac{2}{3}\right)
$$
we can rewrite the summand:
$$
\frac{\Gamma\left(\frac{2m}{3} + \frac{1}{3}\right)}{m!} = \frac{\Gamma\left(\frac{2m}{3} + \frac{1}{3}\right)}{\Gamma(m+1)} = \frac{2^{1/3} \sqrt{\pi}}{\sqrt{3}} \left(\frac{2^{2/3}}{3}\right)^m \frac{\Gamma\left(\frac{m}{3} + \frac{1}{6}\right)}{\Gamma\left(\frac{m}{3} + 1\right) \Gamma\left(\frac{m}{3} + \frac{1}{3}\right) }
$$
It now suffices to split the sum
$$\sum_{m=0}^\infty g(m) = \sum_{k=0}^\infty g(3k) + \sum_{k=0}^\infty g(3k+1) + \sum_{k=0}^\infty g(3k+2)
$$
and write the answer in terms of hypergeometric functions:
$$
\begin{eqnarray}
\int_0^\infty \exp(a x^3 + b x^2) \mathrm{d}x &=& \tilde{\alpha} \frac{\Gamma(1/6)}{\Gamma(1/3)} \cdot {}_1F_1\left(\tfrac{1}{6}; \tfrac{1}{3}; \left(2^{2/3} \frac{\beta}{3}\right)^3 \right) \\
&+& \tilde{\alpha} \cdot \frac{2^{2/3} \beta}{3} \cdot \frac{\Gamma(1/2)}{\Gamma(2/3)\Gamma(4/3)} \cdot {}_2F_2\left(1, \tfrac{1}{2}; \tfrac{2}{3},\tfrac{4}{3}; \left(2^{2/3} \frac{\beta}{3}\right)^3 \right) \\
&+& \tilde{\alpha} \cdot \left(\frac{2^{2/3} \beta}{3}\right)^2 \cdot \frac{\Gamma(5/6)}{\Gamma(1)\Gamma(5/3)} \cdot {}_1F_1\left(\tfrac{5}{6}; \tfrac{5}{3}; \left(2^{2/3} \frac{\beta}{3}\right)^3 \right)
\end{eqnarray}
$$
where $\tilde{\alpha} = \frac{2^{1/3} \sqrt{\pi}}{\sqrt{3}}\alpha $</p>
|
81,588 | <p>A certain function contains points $(-3,5)$ and $(5,2)$. We are asked to find this function,of course this will be simplest if we consider slope form equation </p>
<p>$$y-y_1=m(x-x_1)$$</p>
<p>but could we find for general form of equation? for example quadratic? cubic?</p>
| Ayman Hourieh | 4,583 | <p>$f(x)$ is separable since its derivative is $f'(x) = -1 \ne 0$.</p>
<p>Suppose $\theta$ is a root of $f(x) = x^p - x + a$. Using the Frobenius automorphism, we have:
\begin{align}
f(\theta + 1) &= (\theta + 1)^p - (\theta + 1) + a\\
&= \theta^p + 1^p - \theta - 1 + a\\
&= \theta^p - \theta + a\\
&= f(\theta) = 0
\end{align}</p>
<p>Thus, by induction, if $\theta$ is a root of $f(x)$, then $\theta + j$ is also a root for all $j \in \mathbb F_p$.</p>
<p>By above, if $f(x)$ were to have a root in $\mathbb F_p$, then $0$ would a be a root too, but this contradicts $a \ne 0$. Thus, $f(x)$ has no roots in $\mathbb F_p$. (This can also be shown using Fermat's little theorem.)</p>
<p>Suppose $\theta$ is a root of $f(x)$ in some extension of $\mathbb F_p$. We know that $\theta + j$ is also a root for all $j \in \mathbb F_p$. Since $f(x)$ is of degree $p$, these are all of the roots of $f(x)$.</p>
<p>Clearly, $\mathbb F_p(\theta) = \mathbb F_p(\theta + j)$ for all $j \in \mathbb F_p$. Thus, all $\{\theta + j\}$ have the same degree over $\mathbb F_p$. Since $f(x)$ is separable, it follows that $f(x)$ must be the product of all minimal polynomials of $\{\theta + j\}$. Suppose the minimal polynomials have degree $m$. We have $p = km$ for some $k$. Since $p$ is prime, either $m = 1$; hence $\theta \in \mathbb F_p$, a contradiction. Or $k = 1$; hence $f(x)$ is irreducible because it's the minimal polynomial.</p>
|
81,588 | <p>A certain function contains points $(-3,5)$ and $(5,2)$. We are asked to find this function,of course this will be simplest if we consider slope form equation </p>
<p>$$y-y_1=m(x-x_1)$$</p>
<p>but could we find for general form of equation? for example quadratic? cubic?</p>
| Or Kedar | 460,215 | <p>A little bit late, but I have no doubt people will still come back to this question.
To the proof:</p>
<p>If <span class="math-container">$r$</span> is a root of <span class="math-container">$f$</span> then <span class="math-container">$r+j$</span> is a root of <span class="math-container">$f$</span>, for <span class="math-container">$j\in \mathbb F_p$</span>. If <span class="math-container">$r\in \mathbb F_p$</span>, then <span class="math-container">$0$</span> is a root of <span class="math-container">$f$</span>, but <span class="math-container">$f(0)=a\neq 0$</span>. Therefore, <span class="math-container">$f$</span> has no roots in <span class="math-container">$\mathbb F_p$</span>. Let <span class="math-container">$r$</span> be a root of <span class="math-container">$f$</span>. Then <span class="math-container">$\mathbb F_p(r)$</span> is Galois over <span class="math-container">$\mathbb F_p$</span>, since all roots are of the form <span class="math-container">$r+j$</span>, and <span class="math-container">$f'(x)=-1\neq 0$</span> so <span class="math-container">$f$</span> is separable.</p>
<p>There exists an automorphism of <span class="math-container">$\mathbb F_p(r)$</span> given by <span class="math-container">$\sigma(r)=r+1$</span>. Therefore, <span class="math-container">$\sigma^j(r)=r+j$</span>, and therefore <span class="math-container">$r+j$</span> is a root of the minimal polynomial of <span class="math-container">$f$</span>, for all <span class="math-container">$j\in \mathbb F_p$</span>. Therefore, <span class="math-container">$f$</span> is irreducible.</p>
|
1,056,041 | <p>Look at problem 8 :</p>
<blockquote>
<p>Let $n\geq 1$ be a fixed integer. Calculate the distance:
$$\inf_{p,f}\max_{x\in[0,1]}|f(x)-p(x)|$$ where $p$ runs over
polynomials with degree less than $n$ with real coefficients and $f$
runs over functions $$ f(x)=\sum_{k=n}^{+\infty}c_k\, x^k$$ defined on
the closed interval $[0,1]$, where $c_k\geq 0$ and
$\sum_{k=n}^{+\infty}c_k = 1.$</p>
</blockquote>
<p>This is what I have so far.</p>
<p>Clearly for $n=1$, we have $1/2$.
I am conjecturing for $n>1$, we have $(n-1)^{(n-1)} / n^n$ or something similar to that? (just put $x^{(n-1)}$ and $x^n$ then use AM-GM). it's just weird that the pattern does not fit, so it's probably wrong. Any ideas?</p>
| Jimmy R. | 128,037 | <p>Your approach (although nice) has a flaw in the second bullet. The problem is that there you count two different things: on the one hand ways to choose a box and on the other hand ways to choose a ball and this results to a confusion. In detail</p>
<ol>
<li>Your denominator is correct,</li>
<li>Your numerator is missing one term that should express the number of ways in which you can choose the $2$ balls out of $n$ that you will put in the choosen box with the $2$ balls. This can be done in $\dbinom{n}{2}$ ways.</li>
<li>The other terms in your numerator are correct. Note that your numerator can be written more simple as $$\dbinom{n}{1}\dbinom{n-1}{1} (n-2)!=n\cdot(n-1)\cdot(n-2)!=n!$$</li>
</ol>
<p>Adding the ommitted term, gives the correct result which differs from yours only in this term (the highlighted one)
$$\frac{\dbinom{n}{1}\color{blue}{\dbinom{n}{2}} \dbinom{n-1}{1} (n-2)!}{n^n}=\frac{\dbinom{n}{2}n!}{n^n}$$</p>
|
2,663,130 | <p>Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be function $f(x,y)=(\frac{1}{2}x+y,x-2y)$. Find a image of set $A\subset\mathbb{R}^2$ bounded with lines $x-2y=0, x-2y+2=0, x+2y-2=0, x+2y-3=0.$</p>
<p>Set $A$ is parallelogram with vertices $(1,\frac{1}{2}), (\frac{3}{2},\frac{3}{4}), (\frac{1}{2},\frac{5}{4}), (0,1)$.</p>
<p>What is $f(A)$?</p>
<p>Any help is welcome. Thanks in advance. </p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Note that</p>
<p>$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{n(n+2)}=\frac{n^2+2n+1}{n^2+2n}=1+\frac{1}{n^2+2n}>1$$</p>
<p>thus $a_n$ is <strong>strictly increasing</strong>.</p>
|
2,953,757 | <p>My try to solve this question goes as follows:</p>
<p><span class="math-container">$g=gcd(n^2+1, (n+1^2)+1) = gcd(n^2+1, 2n+1) = gcd(n^2-2n, 2n+1)$</span>.</p>
<p>By long division: </p>
<p><span class="math-container">$$n^2-2n = -2n(2n+1) + 5n^2$$</span></p>
<p>Since <span class="math-container">$g$</span> divides <span class="math-container">$n^2-2n$</span> and <span class="math-container">$g$</span> divides <span class="math-container">$(2n+1)$</span>, <span class="math-container">$g$</span> divides <span class="math-container">$5n^2$</span>. However, I need to show that <span class="math-container">$g$</span> is prime in order to show that it divides <span class="math-container">$5$</span> or <span class="math-container">$n^2$</span>. I am stuck at this point.</p>
<p>Any Ideas,,</p>
| Bill Dubuque | 242 | <p>mod <span class="math-container">$\, (f(n),\!\!\overbrace{n^2\!+\!1}^{\color{#c00}{\Large n^2\ \equiv\ -1}}\!\!\!)\!:\ \ 0\equiv \overbrace{\color{#c00}{n^2}\!+\!2n\!+\!2}^{\large f(n)}\equiv 2n\!+\!1\ \Rightarrow\ 0\equiv \overbrace{(1\!+\!2n}^{\Large \color{#0a0}w})(\overbrace{1\!-\!2n}^{\Large\color{#0a0}{ \bar w}})\equiv 1\!-\!4\color{#c00}{n^2}\equiv 5$</span></p>
<p><strong>Or</strong> in Gaussian integers <span class="math-container">$\, \Bbb Z[i] \cong \Bbb Z[n]/(n^2\!+\!1)\!:\,$</span> <span class="math-container">$ w\! =\! f(i)\! =\! 1\!+\!2i\,\Rightarrow\, \bmod w\!:\ 0 \equiv \color{#0a0}{w\,\bar w} = 5$</span></p>
<p><strong>Corollary</strong> <span class="math-container">$\ \gcd(f(n),n^2\!+\!1)\mid f(i)\overline{f(i)}\ $</span> for any polynomial <span class="math-container">$f\in\Bbb Z[x]$</span></p>
|
3,066,913 | <p>Given a linear operator <span class="math-container">$T$</span> on a finite-dimensional vector space <span class="math-container">$V$</span>, satisfying <span class="math-container">$T^2 = T$</span> answer the following:</p>
<p>(a) Using the dimension theorem, show that <span class="math-container">$N(T) \bigoplus R(T) = V$</span>;</p>
<p>(b) Identify all eigenvalues of <span class="math-container">$T$</span> and the corresponding eigenspaces and show that <span class="math-container">$T$</span> is diagonalizable.</p>
<p>I have already proven part (a). Basically define <span class="math-container">$Tv = w$</span> and the rest follows.</p>
<p>Part (b) however I am honestly confused on where to start. </p>
| Community | -1 | <p>Let <span class="math-container">$ST(X)=X$</span> for any <span class="math-container">$X \in \mathbb{R^n}$</span>. Since <span class="math-container">$S, T$</span> are linear and these have inverses, then we have <span class="math-container">$T^{-1}S^{-1}$</span> is linear an exist so</p>
<p><span class="math-container">$$T^{-1}S^{-1} S T (X)= X = T^{-1}S^{-1} (X)$$</span>,</p>
<p>hence <span class="math-container">$T^{-1}S^{-1}$</span> is an inverse of <span class="math-container">$ST$</span>.</p>
<p><strong>Can you check my answer?</strong></p>
|
3,596,514 | <p>Let <span class="math-container">$0\le x \le 1$</span>
find the maximum value of <span class="math-container">$x(9\sqrt{1+x^2}+13\sqrt{1-x^2})$</span></p>
<p>I try to use am-gm inequality to solve this because it's similar to CMIMC2020 team prob.12 but i don't know how to do next. </p>
| Michael Rozenberg | 190,319 | <p>By C-S and AM-GM we obtain:
<span class="math-container">$$x\left(9\sqrt{1+x^2}+13\sqrt{1-x^2}\right)\leq x\sqrt{(27+13)(3(1+x^2)+13(1-x^2))}=$$</span>
<span class="math-container">$$=\sqrt{16\cdot5x^2(8-5x^2)}\leq\sqrt{16\left(\frac{5x^2+8-5x^2}{2}\right)^2}=16.$$</span>
The equality occurs for <span class="math-container">$x=\frac{2}{\sqrt5},$</span> which says that <span class="math-container">$16$</span> is a maximal value.</p>
|
352,305 | <p>For <span class="math-container">$\mathcal{S}$</span> the <span class="math-container">$(\infty,1)$</span>-category of spaces <a href="https://mathoverflow.net/questions/239383/the-homotopy-category-is-not-complete-nor-cocomplete">its homotopy category <span class="math-container">$h\mathcal{S}$</span> does not have pushouts or pullbacks</a>. Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in <span class="math-container">$\mathcal{S}$</span>.</p>
<p>Generally, filtered colimits are much better behaved than general ones. For example <a href="https://mathoverflow.net/questions/235526/when-do-colimits-agree-with-homotopy-colimits">filtered colimits in the (Quillen) model category <span class="math-container">$\mathit{sSet}$</span> compute homotopy colimits in <span class="math-container">$\mathcal{S}$</span>.</a>
In light of this I was wondering the following:</p>
<blockquote>
<p>Does <span class="math-container">$h\mathcal{S}$</span> have filtered colimits and, if so, does the functor <span class="math-container">$\mathcal{S} \to h\mathcal{S}$</span> preserve them?</p>
</blockquote>
<p>More generally one could ask the same for any presentable <span class="math-container">$(\infty,1)$</span>-category <span class="math-container">$\mathcal{C}$</span>; I particularly care about the case of the derived category <span class="math-container">$\mathcal{D}(R)$</span> of a ring <span class="math-container">$R$</span>.</p>
| Kevin Arlin | 43,000 | <p>As has already been said, the homotopy category does not admit filtered colimits in general, but it’s much worse than that. Even colimits in an <span class="math-container">$\infty$</span>-category which don’t give rise to colimits in the homotopy category sometimes do give rise to <em>weak</em> colimits. (A weak colimit cocone gives the existence, but not the uniqueness, of the factorizations a colimit cocone gives.) This is the case, for instance, with sequential colimits, as well as those along any free category, at least in spaces. </p>
<p>So one might ask whether at least every filtered colimit in <span class="math-container">$\mathcal S$</span> gives rise to a <em>weak</em> filtered colimit in <span class="math-container">$h\mathcal S$</span>. Alas, this is still not true. In our paper kindly referenced by Tim, Christensen and I give an <span class="math-container">$\aleph_1$</span>-indexed sequence of spaces whose homotopy colimit is not a weak colimit in the homotopy category, namely the sequence mapping a countable ordinal <span class="math-container">$\alpha$</span> to the wedge of <span class="math-container">$\alpha$</span> circles. The homotopy colimit is a wedge of <span class="math-container">$\aleph_1$</span> circles, and the problem is that a map out of <em>that</em> just requires too much coherence to be constructed out of a cocone over countable wedges in <span class="math-container">$h\mathcal S$</span>. So there is not much hope for filtered colimits in <span class="math-container">$h\mathcal S$</span>. I expect the same counterexample would work, though I have no idea how to make the argument, in higher homotopy categories <span class="math-container">$h_n\mathcal S$</span>. </p>
<p>Regarding “minimal” or “distinguished” weak colimits, the general idea is that you want some weak colimits which are distinguished up to at least non-unique isomorphism, as occurs for cones in triangulated categories. Since “homotopy colimits” of sequences in triangulated categories with countable coproduct a are constructed out of those coproduct together with cones, they are also distinguished in this sense. </p>
<p>It is possible to get at the idea of minimal weak colimit of at least a filtered diagram in a category which may not be triangulated, but which has some set of objects detecting isomorphisms, by asking that <span class="math-container">$Hom(S,\mathrm{wcolim} D_i)\cong \mathrm{colim} Hom(S, D_i)$</span> for every <span class="math-container">$S$</span> in your isomorphism-detecting set. Such weak colimits are then indeed determined up to isomorphism, and they’re also nice because they see the objects <span class="math-container">$S$</span> as compact. </p>
<p>However, this is not to say that such distinguished weak colimits are common! Our diagram from above actually admits no weak colimit which views even <span class="math-container">$S^1$</span> as compact in this way. (Though note that some weak colimit always exists-homotopy pushouts give weak pushouts, coproducts exist, and then the usual construction applies.) </p>
<p>If your category actually has a set of compact generators in a model, as for <span class="math-container">$D(R)$</span>, then a distinguished weak colimit must come from a homotopy colimit. Franke gives an argument, cited in our paper, that on these grounds distinguished weak colimits of uncountable chains should essentially never exist in <span class="math-container">$D(R)$</span>. The problem is that there’s a spectral sequence converging to homs out of a homotopy colimit indexed by <span class="math-container">$J$</span> whose <span class="math-container">$E_2$</span> page involves the derived functors <span class="math-container">$R^n\mathrm{lim}^J$</span> for all <span class="math-container">$n$</span>. These derived functors were shown by Osofsky to be non vanishing up through <span class="math-container">$n$</span> when <span class="math-container">$J=\aleph_n$</span>, and the homotopy colimit is a weak colimit only if the spectral sequence collapses, so this probably shouldn’t happen. However Franke doesn’t give an argument that there couldn’t in principle be enough unlikely differentials to produce the collapse.</p>
<p>Christensen and I tried for a while to work with the analogous spectral sequence for spaces, but it seemed to require a proficiency with calculating higher derived limits unsupported by the literature-Osofsky gives a special example, and for all I can tell nobody else has ever calculated a derived limit over <span class="math-container">$\aleph_n$</span>. So our approach turns out to be entirely different and doesn’t immediately apply to the stable case. Thus I think it’s unknown, though highly doubtful, whether <span class="math-container">$D(R)$</span> admits minimal filtered colimits in general.</p>
|
744,787 | <p>I need the equation of the line passing through a given point $A(2,3,1)$ perpendicular to the given line
$$
\frac{x+1}{2}= \frac{y}{-1} = \frac{z-2}{3}.
$$</p>
<p>I think there must bee some kind of rule to do this, but I can't find it anyway.</p>
| MarkisaB | 139,345 | <p>Imagine you put point A in a plane that is perpendicular to given line, then the line you're looking for be in this plane, and could be found using the plane line intersection point and point A.
<img src="https://i.stack.imgur.com/S1uzQ.png" alt="enter image description here"></p>
<p>something like this. </p>
<ol>
<li>So normal vector of the plane is the direction vector of the given line. </li>
<li>Point A is located in this plane</li>
<li>I can construct equation of the plane</li>
<li>I can find intersection point B, where line intersects the plane</li>
<li>Line you're looking for is the line passing through the points A and B.
<img src="https://i.stack.imgur.com/2HOJ2.png" alt="enter image description here"></li>
</ol>
<p>Calculus would look something like this:</p>
<p>$\vec c=(2,-1,3)=\vec{n_{\pi}}$</p>
<p>$2(x-x_A)-1(y-y_A)+3(z-z_A)=0$</p>
<p>$2(x-2)-1(y-3)+3(z-1)=0$</p>
<p>$2x-y+3z-4=0$</p>
<p>Line parametric: </p>
<p>$$x(t)=2t-1$$
$$y(t)=-t$$
$$z(t)=3t+2$$</p>
<p>Throw it in the plane equ: </p>
<p>$4t-2+t+9t+6-4=0\rightarrow t_B=0$</p>
<p>So coordinates of point B are:
$$x(t_B)=2t-1=-1$$
$$y(t_B)=-t=0$$
$$z(t_B)=3t+2=2$$</p>
<p>We have now point B(-1,0,-2), we can construct direction vector of the line as $\vec{AB}$ we have a point on a line and that's it. </p>
|
285,719 | <p>Why is $(-3)^4 =81$ and $-3^4 =-81 $?This might be the most stupidest question that you might have encountered,but unfortunately i'am unable to understand this.</p>
| P.K. | 34,397 | <p>$$(-3)^4 = -3 \cdot -3\cdot -3\cdot -3 = 9\cdot9=81 $$Now, when we have to "simplify" the expression $-3^4$, first we do the exponent and then the rest. So this simplifies to $-(81) = -81$</p>
<hr>
<p>This question had confused me a lot too, and then I got my enlightenment (now waiting for the badge)...</p>
|
3,456,351 | <pre><code>Apples: 1
Apple Value: 2500
Pears: lowest = 1, highest = 10
</code></pre>
<p>If I have 1 apple and my apple is worth <strong>2500</strong> if I have 1 pear, how can I calculate the value of my apple if I have X pears, at a maximum of 10 pears and a minimum of 1 pear, where 1 pear represents 100% value and 10 pears represents an increase of that value by 150%?</p>
<pre><code>Apple Worth = 2500 * (1+((pears*15)/100))
</code></pre>
<p>This formula sort of works. The problem here is that if I have 1 pear, the value becomes <code>2500 * 1.15 = 2875</code> which is incorrect, as the value of my apple always is 100% of it's original value if I have at least 1 pear (pear count can never go below 1).</p>
<p>What am I missing here?</p>
<p>Edit:</p>
<p>Some clarification.</p>
<pre><code>2500 = 1p = 100% (do not add 0.15 until p > 1)
2875 = 2p = 115%
3250 = 3p = 130%
3625 = 4p = 145%
4000 = 5p = 160%
4375 = 6p = 175%
4750 = 7p = 190%
5125 = 8p = 205%
5500 = 9p = 220%
5875 = 10p = 235%
6250 = 11p = 250% (this is what 10p should be)
</code></pre>
<p>At 1 pear, my apple is worth 100% of it's original value, so this means that unless I have more than 1 pear, I can't add 0.15% per pear. The first pear is, in other words, not worth 0.15%, but rather 0%. This also means that the 10th pear should be 250% and not 235%.</p>
| John Omielan | 602,049 | <p>The identity is true for <span class="math-container">$n = 0$</span>, so consider <span class="math-container">$n \ge 1$</span>. Also, let</p>
<p><span class="math-container">$$m = \sqrt{n} + \sqrt{n + 1} \tag{1}\label{eq1A}$$</span></p>
<p>With <span class="math-container">$m \gt 0 \; \to \; m = \sqrt{m^2}$</span>, we get</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
m & = \sqrt{\left(\sqrt{n} + \sqrt{n + 1}\right)^2} \\
& = \sqrt{n + 2\sqrt{n(n+1)} + n + 1} \\
& = \sqrt{2n + 1 + 2\sqrt{n(n+1)}}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$</span></p>
<p>We also have</p>
<p><span class="math-container">$$\color{blue}{n} \lt \color{red}{\sqrt{n(n+1)}} \lt \color{green}{n + 1} \tag{3}\label{eq3A}$$</span></p>
<p>Thus, using this with \eqref{eq2A} results in</p>
<p><span class="math-container">$$\begin{equation}\begin{aligned}
\sqrt{2n + 1 + 2(\color{blue}{n})} & \lt \sqrt{2n + 1 + 2\color{red}{\sqrt{n(n+1)}}} \lt \sqrt{2n + 1 + 2(\color{green}{n + 1})} \\
\sqrt{4n + 1} & \lt m \lt \sqrt{4n + 3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$</span></p>
<p>Since natural numbers squared are congruent to <span class="math-container">$0$</span> modulo <span class="math-container">$4$</span> for even values and to <span class="math-container">$1$</span> modulo <span class="math-container">$4$</span> for odd values, neither <span class="math-container">$4n + 2$</span> or <span class="math-container">$4n + 3$</span> can be a perfect square. Thus, the largest perfect square less than or equal to values in \eqref{eq4A}, say it's <span class="math-container">$k^2$</span>, must be less than or equal to <span class="math-container">$4n + 1$</span>. This therefore gives</p>
<p><span class="math-container">$$k \le \sqrt{4n + 1} \lt m \lt \sqrt{4n + 3} \lt k + 1 \tag{5}\label{eq5A}$$</span></p>
<p>Finally, we get that</p>
<p><span class="math-container">$$k = \lfloor m \rfloor = \lfloor \sqrt{n} + \sqrt{n + 1} \rfloor = \lfloor \sqrt{4n + 2} \rfloor \tag{6}\label{eq6A}$$</span></p>
|
182,101 | <p>With respect to assignments/definitions, when is it appropriate to use $\equiv$ as in </p>
<blockquote>
<p>$$M \equiv \max\{b_1, b_2, \dots, b_n\}$$</p>
</blockquote>
<p>which I encountered in my analysis textbook as opposed to the "colon equals" sign, where this example is taken from Terence Tao's <a href="http://terrytao.wordpress.com/">blog</a> :</p>
<blockquote>
<p>$$S(x, \alpha):= \sum_{p\le x} e(\alpha p) $$</p>
</blockquote>
<p>Is it user-background dependent, or are there certain circumstances in which one is more appropriate than the other?</p>
| Christian Blatter | 1,303 | <p>An "equality by definition" is a directed mental operation, so it is <em>nonsymmetric</em> to begin with. It's only natural to express such an equality by a nonsymmetric symbol such as $:=\, .\ $ Seeing a formula like $e=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$ for the first time a naive reader would look for an $e$ on the foregoing pages in the hope that it would then become immediately clear why such a formula should be true.</p>
<p>On the other hand, symbols like $=$, $\equiv$, $\sim$ and the like stand for <em>symmetric</em> relations between predeclared mathematical objects or variables. The symbol $\equiv$ is used , e.g., in elementary number theory for a "weakened" equality (equality modulo some given $m$), and in analysis for a "universally valid" equality: An "identity" like $\cos^2 x+\sin^2 x\equiv1$ is not meant to define a solution set (like $x^2-5x+6=0$); instead, it is expressing the idea of "equal for all $x$ under discussion".</p>
|
2,696,579 | <p>So I understand that in order to have a particular solution you have to have a non homogenous second order differential equation. However I have a slightly difficult time comprehending how to pick the particular solution given $g(t)$. </p>
<p>The reason I ask is our teacher skimmed over it and hardly covered it in class and I would like some more information if possible. The question in short is given the forms and the function $g(t)$ in a second order non homogenous differential equation what is the best way to pick your potential particular solution?</p>
| Martin Argerami | 22,857 | <p>Since $f$ is monotonic, it is continuous almost everywhere. Let $x_0$ be a point such that $f$ is continuous at $x_0$. Then, noting that $f(0)=0$ and $-f(x_0)=f(-x_0)$,
$$
\lim_{h\to0}f(h)=\lim_{h\to0}f(x_0+h-x_0)=\lim_{h\to0} f(x_0+h)-f(x_0)=f(x_0)-f(x_0)=0.
$$
So $f$ is continuous at $0$. But then $f$ is continuous at any point:
$$
f(x+h)=f(x)+f(h)\xrightarrow[h\to0]{}f(x).
$$
Now do the usual $f(n)=nf(1)$, $f(1/n)=f(1)/n$, $f(m/n)=mf(1)/n$. By continuity,
$$
f(x)=f(1)x.
$$</p>
|
214,475 | <p>Function:
<a href="https://i.stack.imgur.com/sH7mh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sH7mh.png" alt="enter image description here"></a></p>
<p>I am to solve for <span class="math-container">$T_{12}(4.8), T_{24}(1.2)$</span>, using <strong>If</strong> and <strong>Which</strong> functions.</p>
<p>I started with this function and keep getting a recursion limit error:</p>
<pre><code>t[n_] := (7/2 x) t[n - 1] - (7/2) t[n + 1]
</code></pre>
| Vasily Mitch | 53,855 | <p>Try</p>
<pre><code>T[0] = 1;
T[1] = x;
T[n_] := (1/x) T[n - 2] - (2/7) T[n - 1];
</code></pre>
|
2,909,626 | <p>I'm having trouble remembering how to solve when you have an equation such as </p>
<p>$$0.3=(1-0.63x)^{5.26}$$</p>
<p>any help would be appreciated.</p>
| Joseph Adams | 583,129 | <p>Try thinking about it this way: your functional is a linear continuous function from the step functions to $\mathbb{R}$. Because of the linearity the continuity of the functional is equivalent to its uniform continuity. Now it is well known that a uniformly continuous function can be uniquely extended (as a uniformly continuous function) to the completion of its domain (here the completion is $L^p$, since the step functions are dense). That is what is meant by that phrase.</p>
|
413,882 | <p>Let $\mathbb F$ be a field and $\mathbb F[x]$ the ring of polynomials with coefficients in $\mathbb F$. Let $p(x)$ be an irreducible polynomial in $\mathbb F[x]$. Let $k$ be a positive integer and consider the vector space $V$, over the field $\frac{\mathbb F[x]}{(p(x))}$with basis </p>
<p>$$1, p(x), p(x)^2, \ldots, p(x)^{k-1}.$$</p>
<p>That is, $$V=\left\{q_0+q_1p(x)+\cdots +q_{k-1}p(x)^{k-1}\;\;:\;\;q_i\in\frac{\mathbb F[x]}{(p(x))}\right\}.$$ </p>
<p>Define in $V$ a product modulo $p(x)^k$. That is, if </p>
<p>$$v=v_0+v_1p(x)+\cdots +v_{k-1}p(x)^{k-1}\;\;\text{and}\;\;u=u_0+u_1p(x)+\cdots +u_{k-1}p(x)^{k-1},$$
then we multiply $v$ by $u$ the obvious way, using the distributivity and assuming that $p(x)^k=0$. This makes $V$ an algebra. My question is:</p>
<blockquote>
<p>Is this algebra $V$ isomorphic to $\frac{\mathbb F[x]}{(p(x)^k)}$? </p>
</blockquote>
| Konstantin Ardakov | 119,170 | <p>See <a href="https://math.stackexchange.com/questions/416688/isomorphic-rings-or-not/628040#628040">this follow-up question</a> for a discussion of <em>coefficient fields</em>.</p>
<p>By the proof of Cohen's Structure Theorem, there is a coefficient field inside $A = \mathbb{F}[X]/(p^k)$, hence an injective map $L \to A$, whose image is a complement to the maximal ideal $\bar{p}A$ of $A$. Extend it to an $L$-algebra homomorphism $\varphi : B = L[Y]/(Y^k) \to A$ by sending $\bar{Y}$ to $\bar{p}$. We can now view both $A$ and $B$ as $L$-vector spaces. Since $(p^i)/(p^{i+1})$ is isomorphic to $L$ for all $i \geq 0$, $\dim_L A = \dim_LB = k$. Therefore $\varphi$ is an isomorphism.</p>
|
10,622 | <p>Given an open set $U \subset \mathbb R ^n $, there exists an exhaustion by compact sets, i.e. a sequence of compact sets $K_i$, s.t.</p>
<p>$\cup _{i=0}^{\infty} K_i = U$ and $\forall i \in \mathbb N : K_i \subset K_{i+1} ^{\circ}$
</p>
<p>We can imagine that different exhaustions by compact sets 'propagate' at different speeds for the various parts $U$.</p>
<p>Let us call an exhaustion $(K_i)_i$ 'stronger' than another exhaustions $(L_i)_i$, whenever we have</p>
<p>$\forall i \in \mathbb N \exists j \in \mathbb N : L_i \subset K_j$.</p>
<p>We call two exhaustions equivalent, if each one is stronger than the other - i.e. for each compact set in the first, we have a compact superset in the second, and vice versa.</p>
<p><strong>Question</strong>: Are all exhaustions of $U$ equivalent?</p>
<p>Purpose: You encounter various settings, where you define a certain structure by such an exhaustion. If the above question has a positive answer, this would spare from wondering whether your construction is independent of such an exhaustion.</p>
| Nate Eldredge | 822 | <p>Yes. Let $\{L_i\}_{i=1}^\infty$, $\{K_i\}_{i=1}^\infty$ be two exhaustions. It follows that $\bigcup_{i=1}^\infty K_i^\circ = U$, so for each $n$, $\{K_i^\circ\}_{i=1}^\infty$ is an open cover of $L_n$. By compactness, it has a finite subcover $\{K_{i_1}^\circ, \dots, K_{i_l}^\circ\}$. But if $m = \operatorname{max}(i_1, \dots, i_l)$, we have $K_{i_j}^\circ \subset K_{i_j + 1} \subset K_{m+1}$ for all $j=1,\dots, l$. Thus $L_n \subset K_{m+1}$, and so $K_i$ is stronger than $L_i$. By symmetry they are equivalent.</p>
|
4,058,319 | <p>My background in mathematical logic, model theory, etc. is patchy, so I'm looking for a clearer way to think about this. (Edit: I'm not asking what an isomorphism is, I'm asking how to formalize the idea of "preserving all logical properties" in order to state the principle described below in its full power.)</p>
<p>Motivating example: we know that <span class="math-container">$\Bbb N$</span> and <span class="math-container">$\Bbb Z$</span> are not order-isomorphic, because <span class="math-container">$\Bbb N$</span> has a least element whereas <span class="math-container">$\Bbb Z$</span> does not. The principle underlying this argument seems to be that an order isomorphism preserves the truth values of (first-order) sentences: if <span class="math-container">$(A,<_A)\cong(B,<_B)$</span> then <span class="math-container">$\operatorname{Th}(A,<_A)=\operatorname{Th}(B,<_B)$</span>. Since the property "there is a least element" corresponds to the sentence <span class="math-container">$\exists x\forall y. x\le y$</span>, the principle applies. The principle also seems to apply to higher-order properties. For example, we can conclude that <span class="math-container">$(\Bbb R\setminus\{0\},<)\not\cong(\Bbb R,<)$</span> by checking the least-upper-bound property, which is second-order.</p>
<p>More generally (if we think of set theory as modeling higher-order logic in first-order logic), I'm tempted to claim that the whole "set theory of <" (I'm not sure what this is exactly - a two-sorted first-order logic with relation symbols <span class="math-container">$<,\in$</span>, maybe?) is preserved by isomorphisms. I think I can convince myself of this fact by observing that when <span class="math-container">$f:A\to B$</span> is an isomorphism, certain substitution rules involving <span class="math-container">$f$</span> in various syntactic contexts are valid, allowing us to prove equivalences between propositions about <span class="math-container">$<_A$</span> and the corresponding propositions about <span class="math-container">$<_B$</span>.</p>
<p>I'm calling this a "principle" instead of a "theorem" because it seems easier to apply in specific cases (by reasoning directly about the elements and the isomorphism, without formalizing the notion of sentences) than to express in its full generality. My claim that "the set theory of <span class="math-container">$<$</span> is preserved" seems closely related to the fact that set theory is my metatheory (if that's the right word), so some kind of circularity or self-reference is appearing when I attempt to state the principle, and I wonder whether any formal version of the principle can fully eliminate the need to apply the principle directly.</p>
<p>So I guess my questions are:</p>
<ul>
<li>In what sense (if any) is there a most general version of this principle, applying to all properties preserved by order isomorphisms?</li>
<li>Can it be expressed as a theorem in the deductive system where we want to apply it, or is there a theoretical obstacle to this?</li>
<li>Am I getting anything wrong? Or missing a clarifying perspective?</li>
</ul>
<p>Thanks!</p>
| Rob Arthan | 23,171 | <p>If <span class="math-container">$\mathbf{M_1}$</span> and <span class="math-container">$\mathbf{M_2}$</span> are structures for some signature <span class="math-container">$\Sigma$</span> (so <span class="math-container">$\Sigma = (<)$</span>) in your example. Then, by definition, <span class="math-container">$\mathbf{M_1}$</span> and <span class="math-container">$\mathbf{M_2}$</span> are isomorphic iff there are functions <span class="math-container">$f : M_1 \to M_2$</span> and <span class="math-container">$g : M_2 \to M_1$</span> (where <span class="math-container">$M_i$</span> is the universe of discourse of <span class="math-container">$\mathbf{M_i}$</span>), such that <span class="math-container">$g \circ f = 1_{M_1}$</span> and <span class="math-container">$g \circ f = 1_{M_2}$</span> and such that <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are <strong>homomorphisms</strong> for the relations and functions in <span class="math-container">$\Sigma$</span> (so, in your example, <span class="math-container">$x < y$</span> iff <span class="math-container">$f(x) < g(y)$</span> and <span class="math-container">$u < v$</span> iff <span class="math-container">$g(u) < g(v)$</span>). So if <span class="math-container">$\mathbf{M_1}$</span> and <span class="math-container">$\mathbf{M_2}$</span> are isomorphic, a logical language <span class="math-container">$\cal L$</span> (first-order or higher-order) over the signature <span class="math-container">$\Sigma$</span> cannot distinguish between <span class="math-container">$\mathbf{M_1}$</span> and <span class="math-container">$\mathbf{M_2}$</span>, since <span class="math-container">$f$</span> induces a truth-preserving mapping from interpretations of <span class="math-container">$\cal L$</span> in <span class="math-container">$\mathbf{M_1}$</span> and interpretations of <span class="math-container">$\cal L$</span> in <span class="math-container">$\mathbf{M_2}$</span> (and vice versa for <span class="math-container">$g$</span>).</p>
|
3,166,419 | <blockquote>
<p>A fair coin is tossed three times in succession. If at least one of
the tosses has resulted in Heads, what is the probability that at
least one of the tosses resulted in Tails?</p>
</blockquote>
<p>My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is <span class="math-container">$ \frac 12 + \frac 12 - \frac 14 = \frac 34 $</span>. </p>
<p>The given answer: <span class="math-container">$ \frac 67 $</span>. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is <span class="math-container">$ \frac 67 $</span>.</p>
<p>Why is my answer wrong? What am I missing?</p>
| Graham Kemp | 135,106 | <blockquote>
<p>The coin flips are all independent of each other, </p>
</blockquote>
<p>Here's the error in your reasoning.</p>
<p>The coin flips are not <em>conditionally</em> independent under constraint that <em>at least one</em> is a head. </p>
<hr>
<p>Let <span class="math-container">$H_k$</span> be the event that trial <span class="math-container">$k$</span> shows a head, <span class="math-container">$T_k$</span> be the event that it is a tail, and the evidence that at least one among them shows a head is: <span class="math-container">$E = H_1\cup H_2\cup H_3$</span></p>
<p>Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail. </p>
<p><span class="math-container">$$\mathsf P(T_1\mid E)~\mathsf P(T_2\mid E)~\mathsf P(T_3\mid E) >0$$</span></p>
<p>However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.</p>
<p><span class="math-container">$$\mathsf P(T_1\cap T_2\cap T_3\mid E)=0$$</span></p>
|
2,650,634 | <p>EDIT: I know how to integrate the last part. I'm just try to find mistake in converting Sum to integral</p>
<p>Question: </p>
<blockquote>
<p>$$a_n=\left(\left(1+\left(\frac1n\right)^2\right)\left(1+\left(\frac2n\right)^2\right)\cdots\left(1+\left(\frac{n}n\right)^2\right)\right)^n$$ find<br>
$$\lim_{n\to\infty}a_n^{-1/n^2}$$</p>
</blockquote>
<p>My Approach:
Let $$y=\lim_{n\to\infty}a_n^{-1/n^2}$$ I Converted this to
$$\ln y=\lim_{n\to\infty}{-1\over n}\sum^n_{k=1}\left(\ln\left(1+{k^2\over n^2}\right) \right)$$</p>
<p>and then to (here's where i think the mistake is in conversion, not integration):</p>
<p>$$\ln y=-\int_0^1\ln(1+x^2)dx$$ </p>
<p>However, the answer is $\ln y=1/2-\ln2$</p>
<p>Please the help me find mistake</p>
| hamam_Abdallah | 369,188 | <p>You are correct but..</p>
<p>By parts,</p>
<p>$$\int_0^1\ln (1+x^2)dx=$$
$$[x\ln (1+x^2)]_0^1-2\int_0^1\frac {x^2+1-1}{1+x^2}dx =$$</p>
<p>$$\ln (2)-2+2\arctan (1) =$$
$$\ln (2)-2+\frac {\pi}{2} =-\ln y$$</p>
|
3,148,094 | <p>In class, my professor computed the density of a gamma random variable by taking the derivative of its CDF, but he skipped many steps. I am trying to go through the derivation carefully but cannot reproduce his final result.</p>
<p>Let <span class="math-container">$k$</span> be the shape and <span class="math-container">$\mu$</span> be the scale. Then the CDF for a gamma random variable <span class="math-container">$T$</span> is</p>
<p><span class="math-container">$$
F(t) = 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!}
$$</span></p>
<p>Using the product rule, I get</p>
<p><span class="math-container">$$
\begin{align}
f(t)
&= \frac{\partial}{\partial t} F(t)
\\
&= \frac{\partial}{\partial t} \Big( 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big)
\\
&= - \sum_{i=0}^{k-1} \frac{\partial}{\partial t} \Big( \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big)
\\
&= - \sum_{i=0}^{k-1} \frac{1}{i!} \frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big)
\end{align}
$$</span></p>
<p>where</p>
<p><span class="math-container">$$
\frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big) = e^{-\mu t} (-\mu)(\mu t)^i + e^{-\mu t} i (\mu t)^{i-1}
$$</span></p>
<p>Putting everything together, we get</p>
<p><span class="math-container">$$
f(t)
=
\sum_{i=0}^{k-1} \frac{\mu e^{-\mu t} (\mu t)^i}{i!}
-
\sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i-1}}{(i-1)!}
$$</span></p>
<p>But here I am stuck. I know that</p>
<p><span class="math-container">$$
f(t) = \frac{\mu e^{-\mu t} (\mu t)^{k-1}}{(k-1)!}
$$</span></p>
<p>but am not sure how to get there.</p>
| Parcly Taxel | 357,390 | <p>Simple manipulations will give us inequalities like <span class="math-container">$A$</span>:</p>
<ul>
<li>Since <span class="math-container">$2<x$</span>, <span class="math-container">$x^2>4$</span> and <span class="math-container">$x^2-4>0$</span>.</li>
<li>Since <span class="math-container">$x\le 3$</span> <em>and</em> <span class="math-container">$x$</span> is constrained to be positive by <span class="math-container">$2<x$</span>, <span class="math-container">$x^2\le9$</span> and <span class="math-container">$x^2-9\le0$</span>.</li>
</ul>
<p>Thus <span class="math-container">$x\in B\implies x\in A$</span> and <span class="math-container">$B\subseteq A$</span>. To show properness of this inclusion (<span class="math-container">$A\subset B$</span>), see that <span class="math-container">$x=-3$</span> is in <span class="math-container">$A$</span> but not <span class="math-container">$B$</span>.</p>
|
241,210 | <p>I am confused with the concept of topology base. Which are the properties a base has to have?</p>
<p>Having the next two examples for $X=\{a,b,c\}$:</p>
<p>1) $(X,\mathcal{T})$ is a topological space where $\mathcal{T}=\{\emptyset,X,\{a\},\{b\},\{a,b\}\}$. Which is the general procedure to follow in order to get a base for $(X,\mathcal{T})$? Can $\mathcal{B}=\{\{a\},\{b\}\}$ be a base?</p>
<p>2) Having just $X$ and no topology $\mathcal{T}$ defined for $X$, is $\mathcal{A}=\{X,\{a\},\{c\}\}$ a base? What is the topology that it generates? </p>
<p>Thank you very much.</p>
| Brian M. Scott | 12,042 | <p>Let $X$ be a non-empty set. A collection $\mathscr{B}$ of subsets of $X$ is a <em>base</em> for some topology on $X$ if it satisfies two conditions:</p>
<ol>
<li>$\mathscr{B}$ covers $X$. That is, every point of $X$ belongs to at least one member of $\mathscr{B}$. </li>
<li>If $B_1,B_2\in\mathscr{B}$ and $x\in B_1\cap B_2$, then there is a $B_3\in\mathscr{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.</li>
</ol>
<p>These two conditions are exactly what is needed to ensure that</p>
<p>$$\mathscr{T}=\Big\{\bigcup\mathscr{A}:\mathscr{A}\subseteq\mathscr{B}\Big\}$$</p>
<p>is a topology on $X$. In words, the collection of all unions of members of $\mathscr{B}$ is a topology on $X$, the <em>topology generated by the base</em>.</p>
<p>Note that a topology may have many different bases. The topology $\big\{\varnothing,\{a\},\{b\},\{a,b\}\big\}$ on the set $\{a,b\}$ has the following bases:</p>
<ol>
<li>$\big\{\varnothing,\{a\},\{b\},\{a,b\}\big\}$ </li>
<li>$\big\{\{a\},\{b\},\{a,b\}\big\}$ </li>
<li>$\big\{\varnothing,\{a\},\{b\}\big\}$</li>
<li>$\big\{\{a\},\{b\}\big\}$</li>
</ol>
<p>Conditions (1) and (2) above are the easiest way to characterize the families of sets that are bases for <strong>some</strong> topology on $X$. If you already have a topology $\mathscr{T}$ on $X$, you can say simply that a subset $\mathscr{B}$ of $\mathscr{T}$ is a base for $\mathscr{T}$ if and only if every member of $\mathscr{T}$ (i.e., every open set in the space $\langle X,\mathscr{T}\rangle$) is a union of members of $\mathscr{B}$.</p>
<hr>
<p>Yes, $\big\{\varnothing,\{a\},\{c\},X\big\}$ is a base for a topology on $X$: it satisfies the two conditions given at the beginning of this answer. The topology that it generates is</p>
<p>$$\big\{\varnothing,\{a\},\{c\},\{a,c\},\{a,b,c\}\big\}\;.$$</p>
|
141,522 | <p>First a summary of the general problem I'm trying to solve:
I want to get <strong>a</strong> set of inequalities for a very complex function (If you are interested is the no-arbitrage conditions Black-Scholes equation with a volatility given by an SVI function)</p>
<p>So basically I'm trying to find the parameters that best fit a model given some conditions.</p>
<p>Long story short these are my problems:</p>
<p>1) I want only ONE set of inequalities, Reduce and Solve give me many solutions but in doing so take a little to much time. (This shouldn't be difficult)</p>
<p>2) Only get "general" inequalities: (this is the hard one). I have parameters and variables: I can set the parameters but the variables are independant. There, I need solutions that don't depend on the variables, but only on the parameters.</p>
<p>As an easy example, let's say my model is: <code>a*b*(1+x^2)</code>, where x is a Real variable and <em>a</em> and <em>b</em> are my variables. Note, if my condition is something like: <code>f(a,b,x)>2</code> I want this result:</p>
<p><code>a*b>2</code></p>
<p>Instead of this result which is what I got:</p>
<p><code>a*b>2/(1+x^2)</code></p>
<p>The second term depends on x^2 so it doesn't give me defined boundaries for my conditions which I can use to fit the model to the data (As I need general terms of <em>a</em> and <em>b</em> that should fit any x)</p>
<p>EDIT: I found of the function <code>ForAll</code> which solves my simple example, but doesn't work for me on the actual problem as I have also conditions on x (Is there any similar command with no such conditions?)</p>
<p>Thanks in advance for your time, and sorry if this is simple, I couldn't find a solution within this site.</p>
| mikado | 36,788 | <p>One approach is to eliminate x using <code>Minimize</code>. Additional constraints can be applied </p>
<pre><code>f = a*b*(1 + x^2);
min = Minimize[f, x];
</code></pre>
<p>The minimum can be compared with the threshold and solved</p>
<pre><code>Reduce[First[min] > 2, {a, b}]
(* (a < 0 && b < 2/a) || (a > 0 && b > 2/a) *)
</code></pre>
|
141,522 | <p>First a summary of the general problem I'm trying to solve:
I want to get <strong>a</strong> set of inequalities for a very complex function (If you are interested is the no-arbitrage conditions Black-Scholes equation with a volatility given by an SVI function)</p>
<p>So basically I'm trying to find the parameters that best fit a model given some conditions.</p>
<p>Long story short these are my problems:</p>
<p>1) I want only ONE set of inequalities, Reduce and Solve give me many solutions but in doing so take a little to much time. (This shouldn't be difficult)</p>
<p>2) Only get "general" inequalities: (this is the hard one). I have parameters and variables: I can set the parameters but the variables are independant. There, I need solutions that don't depend on the variables, but only on the parameters.</p>
<p>As an easy example, let's say my model is: <code>a*b*(1+x^2)</code>, where x is a Real variable and <em>a</em> and <em>b</em> are my variables. Note, if my condition is something like: <code>f(a,b,x)>2</code> I want this result:</p>
<p><code>a*b>2</code></p>
<p>Instead of this result which is what I got:</p>
<p><code>a*b>2/(1+x^2)</code></p>
<p>The second term depends on x^2 so it doesn't give me defined boundaries for my conditions which I can use to fit the model to the data (As I need general terms of <em>a</em> and <em>b</em> that should fit any x)</p>
<p>EDIT: I found of the function <code>ForAll</code> which solves my simple example, but doesn't work for me on the actual problem as I have also conditions on x (Is there any similar command with no such conditions?)</p>
<p>Thanks in advance for your time, and sorry if this is simple, I couldn't find a solution within this site.</p>
| user64494 | 7,152 | <p>You wrote "I found of the function ForAll which solves my simple example, but doesn't work for me on the actual problem as I have also conditions on x (Is there any similar command with no such conditions?) ". </p>
<p>The <code>ForAll</code> function works with restricted variables too. How about the following?</p>
<pre><code>ForAll[x, x >= -1 && x^2 < 2, a*b*(1 + x^2) > 2]
</code></pre>
<blockquote>
<p>$$ \forall _{x,x\geq -1\land x^2<2}a b \left(x^2+1\right)>2$$</p>
</blockquote>
<pre><code>Resolve[%, {a, b}]
</code></pre>
<blockquote>
<p>(a < 0 && b < 2/a) || (a > 0 && b > 2/a) </p>
</blockquote>
<p>Addition. This makes the difference:</p>
<pre><code>ForAll[x, x >= 1/2 && x^2 < 2, a*b*(1 + x^2) > 2];
Resolve[%]
</code></pre>
<blockquote>
<p>$$ (a|b)\in \mathbb{R}\land a b>\frac{8}{5} $$ </p>
</blockquote>
|
3,424,720 | <p>I want to calculate the above limit. Using sage math, I already know that the solution is going to be <span class="math-container">$-\sin(\alpha)$</span>, however, I fail to see how to get to this conclusion.</p>
<h2>My ideas</h2>
<p>I've tried transforming the term in such a way that the limit is easier to find:
<span class="math-container">\begin{align}
\frac{\cos(\alpha + x) - \cos(\alpha)}{x}
&= \frac{\cos(x)\cos(\alpha)-\sin(x)\sin(\alpha)-\cos(\alpha)}{x} & (1) \\
&= \frac{\cos(x)\cos(\alpha)-\cos(x)}{x} - \frac{\sin(x)\sin(\alpha)}{x} & (2) \\
&= \frac{\cos(\alpha)(\cos(x)-1)}{x} - \sin(\alpha) & (3) \\
\end{align}</span></p>
<p>However, I'm not sure whether term <span class="math-container">$(3)$</span> is enough to solve the problem. Surely, for <span class="math-container">$x \to 0$</span>, this evaluates to <span class="math-container">$\frac{0}{0} - \sin(\alpha)$</span>, which is not enough to determine the limit.</p>
<p>Another try was to transform the term in such a way that I can use <span class="math-container">$\lim_{x \to 0} \frac{\sin x}{x} = 1$</span>. For instance, I found that
<span class="math-container">$$
\frac{\cos(\alpha + x) - \cos(\alpha)}{x} = \frac{\sin(\alpha - 90^° + x) - \sin(\alpha-90^°)}{x} \qquad (4)
$$</span></p>
<p>However, this only seems to lead to more complicated term manipulations that I did not manage to bring to a useful point.</p>
<p>What can I do to solve this problem?</p>
| Bernard | 202,857 | <p>Don't make things more complicated than they are. This quotient is the rate of variation of the function of <span class="math-container">$x$</span>: <span class="math-container">$\cos(\alpha +x)$</span>, hence its limit is the derivative of the function at <span class="math-container">$x=0$</span>.</p>
|
3,999,325 | <p>Let me start with some objects. Consider the <span class="math-container">$\mathrm{C}^*$</span>-algebra <span class="math-container">$A$</span> defined by:
<span class="math-container">$$A=M_1(\mathbb{C})\oplus M_2(\mathbb{C})\subset B(\mathbb{C}^3).$$</span>
Let <span class="math-container">$x=\mathbb{C}^3$</span> be given by <span class="math-container">$(e_1+e_2)/\sqrt{2}\,$</span> (the one dimensional factor acts on <span class="math-container">$e_1$</span>).</p>
<p>We have a vector state given by <span class="math-container">$\rho_x(f)=\langle x,f(x)\rangle$</span>, and</p>
<p><span class="math-container">$$\rho_x\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1+\frac{1}{2}c_{11}.$$</span></p>
<p>If I read something about quantum mechanics for such a state, as it is is given by a unit vector, it is called a pure state.</p>
<p>However this contradicts <a href="https://math.stackexchange.com/a/3273820/19352">the answer</a> to this question... and when I pick up Murphy he says that a pure state is such that if <span class="math-container">$\rho_0\leq \rho_x$</span> is a positive linear functional, then <span class="math-container">$\rho_0=t\rho_x$</span> for <span class="math-container">$t\in[0,1]$</span>.</p>
<p>However for <span class="math-container">$\rho_x$</span> we have the linear functional:</p>
<p><span class="math-container">$$\rho_0\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1$$</span></p>
<p>is a bounded linear functional such that <span class="math-container">$\rho_x-\rho_{0}$</span> is a positive linear functional --- half the state given by the vector <span class="math-container">$e_2$</span>, not <span class="math-container">$t\rho_x$</span>... and so Murphy would say <span class="math-container">$\rho_x$</span> is <em>not</em> pure.</p>
<p><strong>Can you help me with my confusion?</strong></p>
| Ruy | 728,080 | <p>If <span class="math-container">$\pi :A\to B(H)$</span> is any non-degenerate representation of the C*-algebra <span class="math-container">$A$</span>, and if <span class="math-container">$x$</span> is a unit vector in <span class="math-container">$H$</span>, then the state
<span class="math-container">$$
\varphi (a) = \langle x,\pi(a)x\rangle
$$</span>
is pure iff the cyclic space
<span class="math-container">$$
[\pi (A)x] = \overline{\text{span}}\{\pi (a) x: a\in A\}
$$</span>
is irreducible for the action of <span class="math-container">$A$</span> there.</p>
<p>This is because the representation of <span class="math-container">$A$</span> on <span class="math-container">$[\pi (A)x]$</span> is equivalent to the GNS representation associated to <span class="math-container">$\varphi $</span>, and
it is well known that a GNS representation is irreducible iff the state is pure.</p>
<p>In the example given, one can show that <span class="math-container">$[Ax]=\mathbb C^3$</span>, which is not irreducible, so <span class="math-container">$\rho _x$</span> is not pure.</p>
<p>On the other hand,
as noted by @MaoWao,
if <span class="math-container">$A$</span> is an irreducible algebra of operators on <span class="math-container">$H$</span>, such as <span class="math-container">$B(H)$</span> itself, <span class="math-container">$[\pi (A)x] = H$</span>, for every unit vector
<span class="math-container">$x$</span>, whence every vector state is pure.</p>
|
109,961 | <p>Suppose $x^2\equiv x\pmod p$ where $p$ is a prime, then is it generally true that $x^2\equiv x\pmod {p^n}$ for any natural number $n$? And are they the only solutions?</p>
| Lubin | 17,760 | <p>The English word “any” is slippery, but I think that the question asks whether if $x\equiv x^2 \pmod{p}$, then $x\equiv x^2 \pmod{p^n}$ is true for all $n$. One possible way to look at a congruence that’s true modulo all powers of a prime $p$ is that it’s really a statement about equality in the field ${\mathbb Q}_p$ of $p$-adic numbers. In particular, if $x\equiv x^2$ modulo all powers of $p$, then $x=x^2$ in ${\mathbb Q}_p$. But since the $p$-adic numbers are a field, high-school algebra tells you that either $x=0$ or $x=1$.</p>
|
4,011,864 | <p><span class="math-container">$$\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}$$</span></p>
<p>I'm fairly sure I can't bring the limit inside the 1/n and I don't think I can use l'Hôpital's rule. I'm pretty sure I'm meant to use the sandwich theorem but I'm not quite sure how to do that in this circumstance.</p>
| DatBoi | 734,160 | <p><span class="math-container">$$\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}=\lim_{n \to \infty}3(1+\frac{1}{3^n})^{\frac{1}{n}}=\lim_{n \to \infty}3e^{\frac{1}{3^nn}}=3$$</span></p>
|
4,011,864 | <p><span class="math-container">$$\lim_{n \to \infty}(3^n+1)^{\frac{1}{n}}$$</span></p>
<p>I'm fairly sure I can't bring the limit inside the 1/n and I don't think I can use l'Hôpital's rule. I'm pretty sure I'm meant to use the sandwich theorem but I'm not quite sure how to do that in this circumstance.</p>
| QC_QAOA | 364,346 | <p>We have</p>
<p><span class="math-container">$$3= (3^n)^{1/n}\leq (3^n+1)^{1/n}\leq (3^n+3^n)^{1/n}=(2\cdot 3^n)^{1/n}=3\sqrt[n]{2}$$</span></p>
<p>Thus</p>
<p><span class="math-container">$$3\leq \lim_{n\to\infty}(3^n+1)^{1/n}\leq \lim_{n\to\infty}3\sqrt[n]{2}=3$$</span></p>
<p>We conclude the limit is <span class="math-container">$3$</span>.</p>
|
621,109 | <p>I need to find all the numbers that are coprime to a given $N$ and less than $N$.
Note that $N$ can be as large as $10^9.$ For example, numbers coprime to $5$ are $1,2,3,4$.</p>
<p>I want an efficient algorithm to do it. Can anyone help? </p>
| mathlove | 78,967 | <p>What you want is <a href="http://en.wikipedia.org/wiki/Euler%27s_totient_function" rel="noreferrer">Euler's totient function</a>.
You'll find a formula there.</p>
|
621,109 | <p>I need to find all the numbers that are coprime to a given $N$ and less than $N$.
Note that $N$ can be as large as $10^9.$ For example, numbers coprime to $5$ are $1,2,3,4$.</p>
<p>I want an efficient algorithm to do it. Can anyone help? </p>
| Sabareesh Muralidharan | 241,483 | <p>First find the factors of it by factorisation
second get the prime numbers involved in that say $(a,b,\ldots)$.
E.g., $18=(2^1)(3^2)$. The prime factors are $2$ and $3$.
Third formula:
$$\text{No. of coprimes to $N$}= N(1-1/a)(1-1/b)\cdots$$</p>
|
102,624 | <p>I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one. </p>
<p>Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder
1) for which lattices this conjecture had been conjectured?
2) what is the motivation for this conjecture?
3) to whom this conjecture is due?
4) is it published somewhere?
5) is it proven in some cases?</p>
| Marc Palm | 10,400 | <p>Multiplicity one refers to something else, related but much weaker.</p>
<p>For the analogue question for lattices, there are trivial counter examples: Induction by steps for example suggests on the level of Lie groups
$$ Ind_{\Gamma(N)} ^{PSL_2(\mathbb{R})} 1 \cong Ind_{PSL_2(\mathbb{Z})} ^{PSL_2(\mathbb{R})} Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$
and e.g. by the Peter-Weyl theorem, we know that
$$ Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$
the multiplicity of an irreducible representation equals its dimension.</p>
<p>Note that GH's example is less trivial, since
$$ Ind_{\Gamma_0(N)}^{PSL_2(\mathbb{Z})} 1$$
decomposes with multiplicity one.</p>
|
102,624 | <p>I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one. </p>
<p>Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder
1) for which lattices this conjecture had been conjectured?
2) what is the motivation for this conjecture?
3) to whom this conjecture is due?
4) is it published somewhere?
5) is it proven in some cases?</p>
| Marc Palm | 10,400 | <p>I rethought your question and have discovered a partial answer for 1) and 2). I add this as a disjoint answer, since my other answer adresses a totally different (negative) issue.</p>
<p>In Sarnak's article <a href="http://web.math.princeton.edu/sarnak/baltimore.pdf" rel="nofollow">http://web.math.princeton.edu/sarnak/baltimore.pdf</a>, he recalls one famous conjecture (Conjecture I, due to himself) that $\Gamma \backslash \mathbb{H}$ should have very few Maass forms for "most" Fuchsian lattices $\Gamma$.</p>
<p>Btw, he attributes the simplicity conjecture (Conjecture 3) for $SL_2(\mathbb{Z})$ to Cartier.</p>
<p>Wolpert (Theorem I) has shown that Sarnak's conjecture would follow if the simplicity conjecture holds for the congruence subgroup $\Gamma(2)$.</p>
<p>Also GH last conjecture that the multiplicity is uniformly bounded in $N$ would suffice for Sarnak's conjecture, but current knowledge is that the multiplicity of an eigenvalue of magnitude $T$ is at most $\ll_N \sqrt{T}/ \log T$ and not $\ll_N 1$. </p>
<p>In fact, Sarnak conjectures that there exist $\Gamma$ with only finitely many Maass wave forms, but this does not follow from the simplicity conjecture for $\Gamma(2)$.</p>
|
4,000,089 | <p>Let <span class="math-container">$x$</span> be an element of a Banach Algebra. Let <span class="math-container">$\lambda \in \rho(x)$</span>, where <span class="math-container">$\rho(x)$</span> is the resolvent of <span class="math-container">$x$</span>.</p>
<p>Let <span class="math-container">${d(\lambda, \sigma(x))}$</span> be the distance between <span class="math-container">$\lambda$</span> and <span class="math-container">$\sigma(x)$</span>.</p>
<p>Show that <span class="math-container">$\| (\lambda-x)^{-1} \| \geq \frac{1}{d(\lambda, \sigma(x))}$</span>.</p>
<p>Information I know:</p>
<p><span class="math-container">$\rho(x)$</span> is an open set.</p>
<p><span class="math-container">$\sigma(x)$</span> is compact.</p>
<p>I've seen many places using this inequality, but I cannot find a proof of it. And I'm not sure how to prove this by breaking down the definitions I know.</p>
<p>Thanks in advance!</p>
| Intelligenti pauca | 255,730 | <p><a href="https://math.stackexchange.com/questions/3455731/curvature-calculation-of-the-ellipse-at-the-end-of-its-axes/3457101#3457101">It is well known that</a> (see figure below, <span class="math-container">$F$</span> and <span class="math-container">$G$</span> are the foci):</p>
<p><span class="math-container">$$
\kappa={\cos\alpha\over2}\left({1\over p}+{1\over q}\right)
={a\cos\alpha\over pq}.
$$</span>
On the other hand, from the formula for the length of a bisector we get:
<span class="math-container">$$
d^2={b^2\over a^2}pq
$$</span>
while from the cosine rule applied to triangle <span class="math-container">$FPG$</span> we obtain
<span class="math-container">$$
\cos^2\alpha={b^2\over pq}.
$$</span>
Inserting these results into the formula for <span class="math-container">$\kappa$</span> we finally get:
<span class="math-container">$$
\kappa={b^4\over a^2d^3}.
$$</span></p>
<p><a href="https://i.stack.imgur.com/MhVl9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MhVl9.png" alt="enter image description here" /></a></p>
|
1,569,936 | <p>Given that $a_0, a_1,...,a_{n-1} \in \mathbb{C}$ I am trying to understand how the following calculation for the determinant of the following matrix follows:
$$
\text{det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_0 \\
-1 & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ =
(x) \text{ det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_1 \\
-1 & x & 0 & ... & 0 & a_2 \\
0 & -1 & x & ... & 0 & a_3 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
+
\text{det}
\begin{bmatrix}
0 & 0 & ... & 0 & a_0 \\
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ = x(x^{n-1} + a_{n-1}x^{n-2}+...+a_1) + (-1)^{n-1}\text{det}
\begin{bmatrix}
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
0 & 0 & ... & 0 & a_0 \\
\end{bmatrix}
$$</p>
<p>I do not understand:
(1) how the determinant can be broken up into the sum of the determinants of the 2 smaller matrices and (2) how are the determinants of the 2 smaller matrices what they are?</p>
| Thomas | 26,188 | <p>The <a href="https://en.wikipedia.org/wiki/Laplace_expansion" rel="nofollow">determinant was expanded</a> along the first column, so you get
$$
\begin{bmatrix}
\color{red}{x} & 0 & 0 & ... & 0 & a_0 \\
\color{blue}{-1} & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ =
(\color{red}{x}) \text{ det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_1 \\
-1 & x & 0 & ... & 0 & a_2 \\
0 & -1 & x & ... & 0 & a_3 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
- (\color{blue}{-1})^{1+2}
\text{det}
\begin{bmatrix}
0 & 0 & ... & 0 & a_0 \\
-1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
$$
After this just keep expanding along the first column.</p>
|
4,222,110 | <p>I am following course on topology that is kind of lack luster (not made for mathematicians). The course starts off with predicate logic and axiomatic set theory (ZFC). Now, I reached a point where the author defined the partition of unity and used the set of all continuous functions between 2 sets. But at the starts of the course, we learned about the principle of restricted comperhension, which requires us to state a set, say <span class="math-container">$D$</span>, in order for <span class="math-container">$\{\phi\in D|\phi:M\rightarrow N\,continuous\}$</span> to be a set. So, my question is: in what set are the maps from <span class="math-container">$M$</span> to <span class="math-container">$N$</span> found?</p>
| OmG | 356,329 | <p>First, reformulate the problem as the following:</p>
<p><span class="math-container">$$
y^2 - 9x^2 = p \Rightarrow(y-3x)(y+3x)= p
$$</span></p>
<p>Now, for any given <span class="math-container">$p$</span>, find its prime factors. Then, for any 2-partitions of them, solve a simple equation system.</p>
<p>To simplify some cases, suppose <span class="math-container">$p$</span> is factorized to <span class="math-container">$p_1 \times p_2$</span>. Now, solve the following system:</p>
<p><span class="math-container">$$
y-3x = p_1
$$</span>
<span class="math-container">$$
y+3x = p_2
$$</span></p>
<p>So, <span class="math-container">$y = \frac{p_1 + p_2}{2}$</span>, and <span class="math-container">$x = \frac{p_2 - p_1}{6}$</span>. It gives us a heuristic to find potential answers more quickly. <span class="math-container">$p_2$</span> is in the form of <span class="math-container">$6k + p_1$</span>.</p>
|
1,042,375 | <p><strong>Question:</strong></p>
<blockquote>
<p>Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Prove that $\{ x \in \mathbb{R} \mid f(x) > 0\}$ is an open subset of $\mathbb{R}$.</p>
</blockquote>
<p>At first I thought this was quite obvious, but then I came up with a counterexample. What if $f(x)=1$ for all $x$? Then the set becomes $\{ x \in \mathbb{R} \mid f(x) = 1\}$, a proper subset of the set in question, which is a closed subset.</p>
| Balbichi | 24,690 | <p><img src="https://i.stack.imgur.com/n6Mu0.png" alt="sign preserving property"> </p>
<p>Suppose $S$ be the set which you want to prove open.</p>
<p>$x\in S$ then $f(x)>0$ by Sign Preserving property of continuous function there is a $\delta>0$ such that $f(x)>0\forall x\in (x-\delta,x+\delta)$</p>
<p>so $x\in S\Rightarrow \exists \delta>0\ni (x-\delta,x+\delta)\subsetneq S$ </p>
<p>so $x$ is an interior point of $S$. $x$ was chosen arbitrary from $S$</p>
<p>So every point of $S$ is an interior point of $S$. So $S$ is open set.</p>
|
41,302 | <p>This is a two part question, and for that I apologize.. but they're related!</p>
<p>Here's what I'm working with:</p>
<pre><code>d1 = Import["file.CSV", "List"]
size = Length[d1]
dis1 = RandomChoice[{d1}, {100, size}]
</code></pre>
<ul>
<li><p><strong>Q1</strong>: Length views <code>d1</code> as $300,000$ individual elements but <code>RandomChoice</code> views it as a single element. Thus when I execute I get $100$ repetitions of <code>d1</code> in the exact order <code>d1</code> is presented in. </p>
<p>Have I made an annotation error in assigning the <code>RandomChoice</code> number pool as <code>{d1}</code>? If I replace <code>{d1}</code> with a hand written list of values it executes perfectly, so I assume it is a presentation issue.. but I can't tell what I've done wrong from the documentation of the function.</p></li>
<li><p><strong>Q2</strong>: Asking to make $100$ random lists of <code>size</code> ~ $300,000$ elements requires a lot of memory. The thing is I don't need to save each <code>RandomChoice</code> output, I just need the mean/median/SD/SEM for each of the $100$ sets I've tasked <code>RandomChoice</code> with. </p>
<p>Is there a way to tell the program to spit out those end-point values and dump the accumulated list after each one is generated?</p></li>
</ul>
| ciao | 11,467 | <p>Q1: You don't need the curly-braces around <code>d1</code> in the <code>RandomChoice</code>: That turns it into a list with one element - <code>d1</code>.</p>
<p>Q2: If memory utilization is more important than speed (because it is usually far better to generate samples/variates/etc. <em>en masse</em>), you can do something like this:</p>
<pre><code>myBigList = RandomInteger[100, 20000];
listLen = Length[myBigList];
results =
Reap[Do[picks = RandomChoice[myBigList, listLen];
Sow[{Mean[#], Median[#], StandardDeviation[#]} &@picks], {100}]][[2,
1]]
</code></pre>
<p>(Replace my <code>RandomInteger</code> with your import).</p>
|
59,965 | <p>If I have a function $f(x,y)$, is it always true that I can write it as a product of single-variable functions, $u(x)v(y)$?</p>
<p>Thanks.</p>
| ShawnD | 14,467 | <p>If $f(x,y)=x^2+y^2$ could be written as $u(x)v(y)$, then since $f(0,0)=0$, this means either $u(0)=0$ or $v(0)=0$. If this were the case, $f(x,y)$ would be equal to zero on either the entire $y$-axis or the entire $x$-axis. This clearly is not the case, so $f(x,y)$ can not be written as $u(x)v(y)$.</p>
<p>You might try graphing a few functions that are of the form $u(x)v(y)$ and see what the graph looks like. This might give some insight to why the statement is not true.</p>
|
908,083 | <p>I'd like to know what methods can I apply to simplify the fraction $\frac{4x + 2}{12 x ^2}$ </p>
<p>Is it valid to divide above and below by 2? (I didn't know it but Geogebra's Simplify aparantly does this)</p>
<p>Thanks in advance</p>
| Ross Millikan | 1,827 | <p>Yes, you are multiplying by $1$ in the form $\dfrac {\frac 12}{\frac 12}$, which takes you to $\frac {2x+1}{6x^2}$ That is all you can do</p>
|
2,311,848 | <p>$X$ and $Y$ are independent r.v.'s and we know $F_X(x)$ and $F_Y(y)$. Let $Z=max(X,Y)$. Find $F_Z(z)$.</p>
<p>Here's my reasoning: </p>
<p>$F_Z(z)=P(Z\leq z)=P(max(X,Y)\leq z)$. </p>
<p>I claim that we have 2 cases here: </p>
<p>1) $max(X,Y)=X$. If $X<z$, we are guaranteed that $Y<z$, so $F_Z(z)=P(Z\leq z)=P(X<z)=F_X(z)$</p>
<p>2) $max(X,Y)=Y$. Similarly, $F_Z(z)=P(Z\leq z)=P(Y<z)=F_Y(z)$</p>
<p>Since we're interested in either case #1 or #2, </p>
<p>$F_Z(z)=F_X(z)+F_Y(z)-F_X(z)*F_Y(z)$</p>
<p>However, it's wrong and I know it. But I would like to know where the flaw in my reasoning is. I <strong><em>know the answer</em></strong> to this problem, I just want to know at what moment my reasoning fails.</p>
| Graham Kemp | 135,106 | <p>Your reasoning, of using the Principle of Inclusion and Exclusion, would be fine if you were dealing with minimum rather than maximum.</p>
<p>$\bullet \quad F_{\min \{X,Y\}}(z) ~{=~ \mathsf P(X\leq z~\cup~Y\leq z) \\=~ F_X(x)+F_Y(z)-F_X(z)\cdot F_Y(z)}\\\bullet\quad F_{\max\{X,Y\}}(z)~{=~\mathsf P(X\leq z~\cap~Y\leq z)\\=~F_X(z)\cdot F_Y(z)}$</p>
|
1,442,344 | <p>I would need a little help in finding the inverse Laplace transform of the function:
$$f(s)=\frac{s}{(s+1)^2(s+2)}.$$</p>
<p>Thanks in advance.</p>
| Jack D'Aurizio | 44,121 | <p><strong>Hint:</strong> recall that:
$$ \mathcal{L}^{-1}\left(\frac{1}{s+a}\right)=e^{-ax},\qquad \mathcal{L}^{-1}\left(\frac{1}{(s+a)^2}\right) = xe^{-ax} $$
and apply a partial fraction decomposition.</p>
|
1,442,344 | <p>I would need a little help in finding the inverse Laplace transform of the function:
$$f(s)=\frac{s}{(s+1)^2(s+2)}.$$</p>
<p>Thanks in advance.</p>
| Mark Watson | 272,109 | <p><strong>Hint:</strong> the partial fraction decomposition could begin with something like:</p>
<p>$$ \frac{s}{(s+1)^2(s+2)} = \frac{2s+1}{(s+1)^2} + \frac{-2}{s+2} $$</p>
<p>Now you continue from here.</p>
|
61,316 | <p>Hi all,</p>
<p>I heard a claim that if I have a matrix $A\in\mathbb R^{n\times n}$ such that $A^n \to 0 \ (\text{for }n\to\infty )$
(that is, every entry of $A^n$ converges to $0$ where $n\to \infty$)
then $I-A$ is invertible.</p>
<p>anyone knows if there is a name for such a matrix or how (for general knowledge) to prove this ?</p>
| Robert Israel | 13,650 | <p>No need for an infinite series. If $I - A$ was not invertible, there would be a nonzero vector $v$ with $A v = v$, and then $A^n v = v$ for all $n$, implying $A^n$ can't go to 0 as $n \to \infty$. </p>
|
3,436,891 | <p>I have the following stupid question in my mind while i am studying for exams.
Does <span class="math-container">$X<\infty \ a.s$</span>, implies that <span class="math-container">$\mathbb E(X)<\infty$</span>? </p>
<p>Further on this, is the converse of the above statement true? Do give me a bit summary on this. Thanks very much.</p>
<p>I thought this was true until I realize the following example: Let's consider a simple symmetric random walk, we know that each state is null-recurrent. Let <span class="math-container">$\tau_L$</span> be a stopping time when the walk first hits <span class="math-container">$L$</span> started from <span class="math-container">$0$</span>. then
<span class="math-container">$$\mathbb P(\tau_L<\infty)=1$$</span>
so <span class="math-container">$\tau_L<\infty \ a.s.$</span>
but we also know that
<span class="math-container">$$\mathbb E(\tau_L)=\infty$$</span>
Is this a counter-example? Thanks, i am a bit weak on measure theory. </p>
| roundsquare | 706,295 | <p>The simplest counter example I can think of a random variable <span class="math-container">$X$</span> which can take values <span class="math-container">$\{1, 2, 4, ...\}$</span> where <span class="math-container">$P(X=n)=\frac{1}{2^n}$</span> if <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> and <span class="math-container">$0$</span> otherwise.. Then</p>
<p><span class="math-container">$E(X)=\sum\frac{2^{n-1}}{2^{n}}=\infty $</span></p>
|
115,269 | <p>I'm studying the remainder class $\mathbb{Z}_n$, I've grabbed something, but something else is unfocused. Let
$$20x \equiv 4\pmod{34}$$
then GCD(20,34)=2 so I rewrite as:
$$10x \equiv 2\pmod{17}$$
and successively:
$$10x \equiv 1\pmod{17}$$
Now I know $\gcd(10, 17)=1$</p>
<blockquote>
<p>Question 1: Why? Is this cause I've divided both $20$ and $34$ for the
same $\gcd=2$? If $d$ is a $\gcd$ for $a$ and $b$, then $\gcd(a/d, b/d)=1$?</p>
</blockquote>
<p>At this point I can:
$$1=10\alpha + 17\beta$$
that will be:
$$1=10(-5)+17\cdot 3$$</p>
<blockquote>
<p>Question 2: I know that $-5$ is the solution of equation, but why I have to choose $-5$ more than $3$?</p>
</blockquote>
<p>Now $-5$ is a solution for:
$$10x \equiv 1\pmod{17}$$
and $-5\cdot2$ is a solution for:
$$10x \equiv 2\pmod{17}$$
$[-10]_{17}$ is class, the entire set is $\{-10+17k: k\in\mathbb{Z}\}$.</p>
<blockquote>
<p>Question 3: This set will contain all invertible elements. Right? Now
If I took $[8]_{17}$ and $[7]_{17}$ why the class $[56]_{17}$ is not
in the same class of $[1]_{17}$ since it is invertible, matter of
fact, $\gcd(56,17)=1$? Is $\gcd(x,y)=1$ the only way to test if $[x]_y$ is
invertible? I know that element $a$ was invertible if $ba=ab=1$, but this
contradicts my knowledge.</p>
</blockquote>
| Gigili | 181,853 | <p>Prove <em>by induction</em> that:</p>
<p>$$\sum\limits_{i=1}^k i= 1 + 2+ 3 +4+...+k=\frac{k(k+1)}{2}$$</p>
<p>If it holds for $k$, it should be true for $k+1$:</p>
<p>$$\sum\limits_{i=1}^{k+1} i= \color{red}{1 + 2+ 3 +4+...+k}+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{(k+1)(k+2)}{2}$$</p>
<p>The same applies for your second question:
$$\sum\limits_{i=1}^k i^2 + 3i$$</p>
<p>$$=\sum\limits_{i=1}^k i^2+ 3\sum\limits_{i=1}^k i$$</p>
<p>$$=\frac{k(k+1)(k+2)}{6}+3(\frac{k(k+1)}{2})$$</p>
|
550,441 | <p>Say I roll a 6-sided die until its sum exceeds $X$. What is E(rolls)?</p>
| Did | 6,179 | <p>One looks for $n(x+1)$ where, for every integer $k$, $n(k)$ denotes the expected number of rolls needed to exceed $x$ starting from $x+1-k$. Thus, $n(k)=1+\frac16\sum\limits_{i=1}^6n(k-i)$ for every $k\geqslant1$ and $n(k)=0$ for every $k\leqslant0$.</p>
<p>For every $|s|\lt1$, let $N(s)=\sum\limits_{k}n(k)s^k$, then the one-step recursion above yields
$$
N(s)=\sum\limits_{k\geqslant1}s^k+\frac16\sum\limits_{i=1}^6s^i\sum\limits_{k\geqslant1}n(k)s^k,
$$
that is,
$$
N(s)=\frac{\frac{s}{1-s}}{1-\frac16\sum\limits_{i=1}^6s^i}=\frac{6s}{(1-s)^2Q(s)},
$$
where
$$
Q(s)=6+5s+4s^2+3s^3+2s^4+s^5.
$$
This shows that $n(x+1)\sim\nu x$ when $x\to\infty$, with
$$
\nu=\lim_{s\to1}(1-s)^2N(s)=\frac6{Q(1)}=\frac27.
$$
To go further, consider $\bar n(k)=n(k)-\nu k$ and $\bar N(s)=\sum\limits_{k\geqslant1}\bar n(k)s^k$, then
$$
(1-s)\bar N(s)=\frac{6s}{Q(s)Q(1)}\,\frac{Q(1)-Q(s)}{1-s}\to\mu,\qquad \mu=\frac{6Q'(1)}{Q(1)^2},
$$
hence $\bar n(k)\to\mu$ with $\mu=10/21$. Finally,
$$
n(x+1)=\frac27(x+1)+\frac{10}{21}+o(1),
$$
where $o(1)$ is geometrically small when $x\to\infty$.</p>
|
3,931,246 | <p>I want to find the range of <span class="math-container">$x$</span> on which <span class="math-container">$f$</span> is decreasing, where
<span class="math-container">$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$</span></p>
<p>Let <span class="math-container">$u=x^2-x$</span>, then <span class="math-container">$\frac{du}{dx}=2x-1$</span>, then <span class="math-container">$$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$</span></p>
<p>Since <span class="math-container">$e^{x^4-2x^3+x^2-1}>0$</span> for all <span class="math-container">$x\in \Bbb R$</span> and <span class="math-container">$2x-1<0\iff x<\frac{1}{2}$</span>. <span class="math-container">$f$</span> is decreasing on <span class="math-container">$(-\infty,\frac{1}{2})$</span>.</p>
<p>Furthermore, <span class="math-container">$f$</span> is increasing on <span class="math-container">$(\frac{1}{2},\infty)$</span>, <span class="math-container">$f$</span> is differentiable at <span class="math-container">$x=\frac{1}{2}$</span>, and <span class="math-container">$f'(\frac{1}{2})=0$</span>, <span class="math-container">$f$</span> attains its minimum value at <span class="math-container">$x=\frac{1}{2}$</span>.</p>
<p>Am I right?</p>
| Fred | 380,717 | <p>Everything is fine ! A little bit more can be said:</p>
<p><span class="math-container">$f$</span> is strictly decreasing on <span class="math-container">$(-\infty,\frac{1}{2}]$</span></p>
<p>and</p>
<p><span class="math-container">$f$</span> is strictly increasing on <span class="math-container">$[\frac{1}{2}, \infty).$</span></p>
|
4,463,559 | <p>Let <span class="math-container">$K \subset \mathbb{R}^n\times [a,b]$</span> a compact subset. For each <span class="math-container">$t \in [a,b]$</span>, let <span class="math-container">$K_t= \{x \in \mathbb{R}^n : (x,t) \in K\}$</span>. Suppose that, for all <span class="math-container">$t \in [a,b]$</span>, <span class="math-container">$K_t$</span> has measure zero in <span class="math-container">$\mathbb{R}^n$</span>. Thus, <span class="math-container">$K$</span> has measure zero in <span class="math-container">$\mathbb{R}^{n+1}$</span>.</p>
<p>The definition of measure zero in <span class="math-container">$\mathbb{R}^n$</span> here is:</p>
<p><em><span class="math-container">$A$</span> has measure zero if given <span class="math-container">$\epsilon > 0$</span>, there is <span class="math-container">$\{Q_i\}_{i \in \mathbb{N}}$</span> rectangles in <span class="math-container">$\mathbb{R}^n$</span> such that
<span class="math-container">$$A \subset \bigcup_{i \in \mathbb{N}} Q_i,$$</span>
and,
<span class="math-container">$$\sum_{i \in \mathbb{N}}v(Q_i) < \epsilon,$$</span>
where <span class="math-container">$v(Q) = (b_1 - a_1) \cdots (b_n - a_n)$</span>, if <span class="math-container">$Q = [a_1,b_1]\times \cdots \times [a_n, b_n]$</span>. The previous summation is called total volume of the cover.</em></p>
<p>There are some interesting points here:</p>
<ol>
<li>If <span class="math-container">$K_t$</span> has measure zero in <span class="math-container">$\mathbb{R}^n$</span>, then <span class="math-container">$K_t \times \{t\}$</span> has measure zero in <span class="math-container">$\mathbb{R}^{n+1}$</span>.</li>
<li>For every <span class="math-container">$t \in [a,b]$</span>, <span class="math-container">$K_t$</span> is compact. Then, <span class="math-container">$K_t \times \{t\}$</span> is compact.</li>
<li><span class="math-container">$K = \bigcup_{t \in [a,b]}K_t \times \{t\}$</span>.</li>
</ol>
<p>Then, I tried to take an finite open cover (of rectangles) of <span class="math-container">$K_t \times \{t\}$</span> which total volume is less then <span class="math-container">$\epsilon '$</span>. Then, use it to construct an open cover of <span class="math-container">$K$</span>, and using Lindelöf theorem to get a countable cover of <span class="math-container">$K$</span>. But I'm stuck because I can't prove that total volume is less then <span class="math-container">$\epsilon$</span>.</p>
| André Rasera | 577,754 | <p>The use of the Lebesgue measure would simplify things a lot. But it's possible to use only the definition of the total volume of countable unions of rectangles (not necessarily open). So, let <span class="math-container">$v_p$</span> denote the total volume of a countable union of rectangles <span class="math-container">$V \subseteq \mathbb{R}^p$</span>.</p>
<p>The solution below is based on <a href="https://www.jstor.org/stable/2324722" rel="nofollow noreferrer">this article</a>.</p>
<p>It relies on the following lemma, which says that a cover of a <span class="math-container">$K_t$</span> is actually a cover of the sets <span class="math-container">$K_u$</span> that are "close enough".</p>
<p><strong>Lemma.</strong> Let <span class="math-container">$K \subseteq \mathbb{R}^n \times [a,b]$</span> be compact, let <span class="math-container">$t \in \mathbb{R}$</span> and let <span class="math-container">$T$</span> be an open set in <span class="math-container">$\mathbb{R}^n$</span> with <span class="math-container">$K_t \subseteq T$</span>. Then there is an interval <span class="math-container">$(c,d) \ni t$</span> such that <span class="math-container">$K_u \subseteq T$</span> for every <span class="math-container">$u \in (c,d)$</span>.</p>
<p>The proof goes as follows:</p>
<p>Let <span class="math-container">$(a_k)$</span> be a strictly increasing sequence converging to <span class="math-container">$t$</span> and <span class="math-container">$(b_k)$</span> a strictly decreasing sequence converging to <span class="math-container">$t$</span>. Define the sequence of sets <span class="math-container">$(F_k)$</span> in the following manner:
<span class="math-container">$$F_k = K \cap ((\mathbb{R}^n \setminus T) \times [a_k,b_k]).$$</span>
It is clear that this sequence is decreasing and all the <span class="math-container">$F_k$</span> are compact sets. It follows that
<span class="math-container">$$\bigcap_{k = 1}^{\infty} F_k = K \cap \left( \bigcap_{k = 1}^{\infty} (\mathbb{R}^n \setminus T) \times [a_k, b_k] \right) = K \cap ((\mathbb{R}^n \setminus T) \times \{ t \}) = \varnothing,$$</span>
because <span class="math-container">$K_t \cap (\mathbb{R}^n \setminus T) = \varnothing$</span>. From compactness, it follows that <span class="math-container">$F_{k_0} = \varnothing$</span> for some <span class="math-container">$k_0 \in \mathbb{N}$</span>. Let <span class="math-container">$c = a_{k_0}$</span> and <span class="math-container">$d = b_{k_0}$</span>. Then,
<span class="math-container">$$K \cap ((\mathbb{R}^n \setminus T) \times [c,d]) = \varnothing \Rightarrow K_u \cap (\mathbb{R}^n \setminus T) = \varnothing \ \ \ \forall u \in [c,d],$$</span>
and then <span class="math-container">$K_u \subseteq T \ \ \ \forall u \in (c,d)$</span>, as desired.</p>
<p>This result allows to prove the initial proposition.</p>
<p>Consider an arbitrary <span class="math-container">$\delta > 0$</span>. As the projection <span class="math-container">$P \, \colon \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$</span>, <span class="math-container">$(x,t) \mapsto t$</span>, is a continuous function, <span class="math-container">$K$</span> is compact implies that so is the set
<span class="math-container">$$L = P(K) = \{ t \in \mathbb{R} : (x,t) \in K \ \text{ for some } \ x \in \mathbb{R}^n \}.$$</span>
Therefore, there is a bounded open interval <span class="math-container">$Q$</span> such that <span class="math-container">$L \subseteq Q$</span>. Choose <span class="math-container">$\varepsilon > 0$</span> such that <span class="math-container">$\varepsilon < \delta/v_1(Q)$</span>.</p>
<p>Call a set <span class="math-container">$U \subseteq \mathbb{R}$</span> <span class="math-container">$\varepsilon$</span>-good if there is a countable union of open rectangles <span class="math-container">$V \subseteq \mathbb{R}^n$</span> with total volume less than <span class="math-container">$\varepsilon$</span> and <span class="math-container">$K_u \subseteq V$</span> for every <span class="math-container">$u \in U$</span>. Let <span class="math-container">$\mathcal{F}$</span> be the set of open <span class="math-container">$\varepsilon$</span>-good sets contained in <span class="math-container">$Q$</span>. It follows that
<span class="math-container">$$L \subseteq \bigcup_{U \in \mathcal{F}} U \subseteq Q.$$</span>
Indeed, if <span class="math-container">$t \in L$</span>, then <span class="math-container">$(y,t) \in K$</span> for some <span class="math-container">$y \in \mathbb{R}^n$</span>. As <span class="math-container">$K_t$</span> has measure zero, there is an open cover of rectangles <span class="math-container">$\bigcup_{j = 1}^{\infty} I_{j,t}$</span> of <span class="math-container">$K_t$</span> such that <span class="math-container">$\sum_{j = 1}^{\infty} v_n(I_{j,t}) < \varepsilon$</span>. Call this cover <span class="math-container">$V$</span> and see that, according to the previous lemma, there exists an interval <span class="math-container">$(c,d)$</span> that contains <span class="math-container">$t$</span> and satisfies <span class="math-container">$K_u \subseteq V$</span> for every <span class="math-container">$u \in (c,d)$</span>. As <span class="math-container">$Q$</span> is an open interval, <span class="math-container">$Q \cap (c,d) \ni t$</span> also satisfies this property and is, therefore, <span class="math-container">$\varepsilon$</span>-good. This implies that <span class="math-container">$t \in \bigcup_{U \in \mathcal{F}} U$</span>.</p>
<p>As each open set in <span class="math-container">$\mathbb{R}$</span> is a countable union of open intervals, it follows that <span class="math-container">$\bigcup_{U \in \mathcal{F}} U$</span> can be written as a countable union of open intervals <span class="math-container">$\bigcup_{i = 1}^{\infty} G_i$</span> that covers <span class="math-container">$L$</span>. Using the fact that <span class="math-container">$L$</span> is compact, there is a finite subcover <span class="math-container">$\bigcup_{i = 1}^{m} G_i$</span>. As each <span class="math-container">$G_i$</span> is a subset of an <span class="math-container">$\varepsilon$</span>-good set, the <span class="math-container">$G_i$</span> themselves are <span class="math-container">$\varepsilon$</span>-good.</p>
<p>To make this union disjoint, let
<span class="math-container">$$S_j = G_j \setminus \bigcup_{i < j} G_i$$</span>
for every <span class="math-container">$1 \leq j \leq m$</span>. It is evident that the <span class="math-container">$S_j$</span> are finite unions of not necessarily open intervals (here, I consider a point an interval of length zero, to avoid any posssible complications - this is one of the reasons this proof is simpler when using the Lebesgue measure). As <span class="math-container">$S_j \subseteq G_j$</span> for every <span class="math-container">$j$</span>, i.e., every <span class="math-container">$S_j$</span> is a subset of an <span class="math-container">$\varepsilon$</span>-good set, they are <span class="math-container">$\varepsilon$</span>-good sets themselves. Therefore, for every <span class="math-container">$j$</span>, there is a countable union of open rectangles <span class="math-container">$V_j = \bigcup_{l = 1}^{\infty} T_{j,l}$</span> with total volume less than <span class="math-container">$\varepsilon$</span> and satisfying <span class="math-container">$K_s \subseteq V_j$</span> for every <span class="math-container">$s \in S_j$</span>.</p>
<p>It follows that
<span class="math-container">$$K \subseteq \bigcup_{j = 1}^{m} V_j \times S_j = \bigcup_{j = 1}^{m} \bigcup_{l = 1}^{\infty} T_{j,l} \times S_j.$$</span>
Indeed, if <span class="math-container">$(x,t) \in K$</span>, then <span class="math-container">$t \in S_k$</span> for some <span class="math-container">$1 \leq k \leq m$</span>, as the union of the <span class="math-container">$S_j$</span> covers <span class="math-container">$L$</span>. Then, <span class="math-container">$x \in K_t \subseteq V_k$</span>, by construction of the <span class="math-container">$V_j$</span>, and <span class="math-container">$(x,t) \in \bigcup_{j = 1}^{m} V_j \times S_j$</span>. The expression on the right indicates that this cover is a countable cover of rectangles of total volume equal to
<span class="math-container">$$\sum_{j = 1}^{m} \sum_{l = 1}^{\infty} v_n(T_{j,l}) \cdot v_1(S_j) = \sum_{j = 1}^{m} v_1(S_j) \sum_{l = 1}^{\infty} v_n(T_{j,l}) < \varepsilon \sum_{j = 1}^{m} v_1(S_j) = \varepsilon \cdot v_1 \left( \bigcup_{j = 1}^{m} S_j \right).$$</span>
However, <span class="math-container">$\bigcup_{j = 1}^{m} S_j \subseteq Q$</span>, hence
<span class="math-container">$$\varepsilon \cdot v_1 \left( \bigcup_{j = 1}^{m} S_j \right) \leq \varepsilon \cdot v_1(Q) < \delta.$$</span>
As the choice of <span class="math-container">$\delta$</span> is arbitrary, it follows by definition that <span class="math-container">$K$</span> has measure zero.</p>
|
65,166 | <p>For a graph $G$, let its Laplacian be $\Delta = I - D^{-1/2}AD^{-1/2}$, where $A$ is the adjacency matrix, $I$ is the identity matrix and $D$ is the diagonal matrix with vertex degrees. I'm interested in the spectral gap of $G$, i.e. the first nonzero eigenvalue of $\Delta$, denoted by $\lambda_{1}(G)$.</p>
<p>Is it true that a randomly chosen (with uniform distribution) $d$-regular bipartite graph on $(n, n)$ vertices (with multiple edges allowed) has, with probability approaching $1$ as $n \to \infty$, $\lambda_1$ arbitrarily close to $1$ (i.e. we can make arbitrarily close by taking $d$ large enough)?</p>
<p>If yes, is there a reference for this fact?</p>
<p>Proofs of expanding properties for random regular graphs which I have found in the literature usually give the probability only bounded from below by a constant, i. e. $1/2$, although I imagine that actually almost all random graphs have good spectral gap.</p>
<p>Note: by $d$-regular bipartite graph I mean a graph in which each vertex (on the left and on the right) has degree $d$.</p>
| Aaron Meyerowitz | 8,008 | <p>I assume you mean to fix $d$ and let $n$ grow. When $d \gt 3$ the graph (at least in the case that all vertices have degree $r$) will likely be connected and even $d$-connected. But in case $d=2$ one has a disjoint union of cycles. Then it seems likely that for fixed $m$ there is, with probability approaching 1, a cycle of size over $m$ as $n$ grows. That would give the result in that case. </p>
|
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