qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,736,341 | <p>As stated in the title: Find the eigenvalues and eigenfunctions for $y''+\lambda y=0$, where $y'(1)=0$ and $y'(2)=0$.</p>
<p>So I have already eliminated the cases for $\lambda=0$ and $\lambda<0$ and I'm focused on the case for $\lambda>0$ now.
The characteristic equation for this case is $r^2+\lambda=0$, so $r=\pm i \sqrt{\lambda}$. Then the possible solution is
$$y(x)=k_1\cos(\sqrt{\lambda}t)+k_2\sin(\sqrt{\lambda}t)$$
and the derivative is
$$y'(x)=-k_1\sqrt{\lambda}\sin(\sqrt{\lambda}t) +k_2\sqrt{\lambda}\cos(\sqrt{\lambda}t)$$
Substituting the boundary values into the above equation gives:
$$y'(1)=-k_1\sqrt{\lambda}\sin(\sqrt{\lambda}) +k_2\sqrt{\lambda}\cos(\sqrt{\lambda})=0$$
and
$$y'(2)=-k_1\sqrt{\lambda}\sin(2\sqrt{\lambda}) +k_2\sqrt{\lambda}\cos(2\sqrt{\lambda})=0$$
I have no idea how to go forward from here. All the other BVP i've done had boundary values (usually one of the BVs IS zero) that led to one of the constants equaling zero or a substitution that was obvious. I'm having trouble seeing an obvious path forward with this problem. </p>
| user5713492 | 316,404 | <p>If $y_n(x)=\sin(n\pi x)$, then $y_n^{\prime}(1)=(-1)^n\cdot n\pi$, so that's a problem. Let's go back and eliminate cases. If $\lambda<0$, then the solutions are
$$y(x)=k_1\cosh(\sqrt{-\lambda}(x-1))+k_2\sinh(\sqrt{-\lambda}(x-1))$$
$$y^{\prime}(x)=\sqrt{-\lambda}k_1\sinh(\sqrt{-\lambda}(x-1))+\sqrt{-\lambda}k_2\cosh(\sqrt{-\lambda}(x-1))$$
Then $y^{\prime}(1)=\sqrt{-\lambda}k_2=0$ and
$$y^{\prime}(2)=\sqrt{-\lambda}k_1\sinh(\sqrt{-\lambda})=0$$
Which has no solution for $\lambda<0$. If $\lambda=0$, then the solutions are
$$y(x)=k_1+k_2(x-1)$$
$$y^{\prime}(x)=k_2$$
So applying boundary conditions $y^{\prime}(1)=y^{\prime}(2)=k_2=0$ so we get the solution $y_0(x)=k_1$. Normalizing to unity, this would be $y_0(x)=1$.
Now if $\lambda>0$, then the solutions are
$$y(x)=k_1\cos(\sqrt{\lambda}(x-1))+k_2\sin(\sqrt{\lambda}(x-1))$$
$$y^{\prime}(x)=-\sqrt{\lambda}k_1\sin(\sqrt{\lambda}(x-1))+\sqrt{\lambda}k_2\cos(\sqrt{\lambda}(x-1))$$
Then $y^{\prime}(1)=\sqrt{\lambda}k_2=0$ and now
$$y^{\prime}(2)=-\sqrt{\lambda}k_1\sin(\sqrt{\lambda})=0$$
so $\lambda_n=n^2\pi^2$ and $y_n(x)=k_1\cos(n\pi(x-1))=(-1)^nk_1\cos(n\pi x)$. Normalizing to unity, we have $y_n(x)=\sqrt2\cos(n\pi x)$. </p>
<p>In summary, it would have been easier to have taken functions of $(x-1)$ instead of $x$, and the solutions were cosines, not sines, and there was a constant solution as well.</p>
|
233,075 | <p>I am trying to solve the following equation in the Natural Numbers, with the condition <span class="math-container">$a\ge1$</span>, <span class="math-container">$b\ge1$</span>, and <span class="math-container">$r\ge3$</span>:</p>
<p><span class="math-container">$$\frac{a(a + 3)(a(r - 5) + (12 - r))}{9}=\frac{b (9 + b (-14 + r) - r)}{3}\tag1$$</span></p>
<p>The method I know use is, that I solve the equation for <span class="math-container">$b$</span> and I got:</p>
<p><span class="math-container">$$b=\displaystyle\frac{1}{6} \left(\sqrt{3\cdot\frac{4 a (a+3) (r-14) (a (r-5)-r+12)+3 (r-9)^2}{(r-14)^2}}+\frac{15}{r-14}+3\right)\tag2$$</span></p>
<p>Now, I used Mathematica to check when the function under the square root is a perfect square, with the following code:</p>
<pre><code>ParallelTable[
If[IntegerQ[
FullSimplify[
Sqrt[3*((
4 a (3 + a) (12 + a (-5 + r) - r) (-14 + r) +
3 (-9 + r)^2)/(-14 + r)^2)]]], {a, r}, Nothing], {a, 1,
10^5}, {r, 3, 10^5}] //. {} -> Nothing
</code></pre>
<p>And the solutions I got, I put in equation <span class="math-container">$(1)$</span> to check if I can find a solution to the original problem.</p>
<blockquote>
<p>This method takes a very very long time, but I am not knowing if there is a faster and smarter way to program this. Can you help me with this. Thanks a lot in advance.</p>
</blockquote>
| Roman | 26,598 | <p>Much faster to solve for <span class="math-container">$r$</span> and check that it's an integer: (it's always rational; no square-roots involved)</p>
<pre><code>Solve[(a(a+3)(a(r-5)+(12-r)))/9 == (b(9+b(-14+r)-r))/3, r] // FullSimplify
(* {{r -> (a(3+a)(-12+5a)+3(9-14b)b)/((-1+a)a(3+a)-3(-1+b)b)}} *)
R = (a(3+a)(-12+5a)+3(9-14b)b)/((-1+a)a(3+a)-3(-1+b)b);
With[{s = 10^3},
Do[If[IntegerQ[R] && R >= 3, Sow[{a, b, R}]], {a, s}, {b, s}] // Reap // Last // First]
(* {{3, 6, 24}, {5, 8, 244}, {5, 10, 31}, {5, 14, 19}, {9, 18, 177},
{9, 20, 46}, {12, 30, 45}, {32, 112, 139}, {33, 114, 573},
{35, 126, 220}, {45, 180, 553}, {47, 450, 16}, {48, 204, 129},
{63, 294, 3750}, {77, 396, 3889}, {116, 728, 46750}} *)
</code></pre>
<p>Even faster: instead of calculating <span class="math-container">$r$</span> and checking if it's an integer, we can just check if its numerator is divisible by its denominator:</p>
<pre><code>R = (a(3+a)(-12+5a)+3(9-14b)b)/((-1+a)a(3+a)-3(-1+b)b);
With[{s = 10^3},
Do[If[Divisible[a(3+a)(-12+5a)+3(9-14b)b, (-1+a)a(3+a)-3(-1+b)b] && R>=3,
Sow[{a, b, R}]], {a, s}, {b, s}] // Reap // Last // First]
(* {{3, 6, 24}, {5, 8, 244}, {5, 10, 31}, {5, 14, 19}, {9, 18, 177},
{9, 20, 46}, {12, 30, 45}, {32, 112, 139}, {33, 114, 573},
{35, 126, 220}, {45, 180, 553}, {47, 450, 16}, {48, 204, 129},
{63, 294, 3750}, {77, 396, 3889}, {116, 728, 46750}} *)
</code></pre>
|
377,925 | <p>This post comes from the suggestion of <a href="https://mathoverflow.net/users/18698/joel-moreira">Joel Moreira</a> in a <a href="https://mathoverflow.net/questions/377706/an-alternative-to-continued-fraction-and-applications#comment958452_377706">comment</a> on <a href="https://mathoverflow.net/q/377706/34538">An alternative to continued fraction and applications</a> (itself inspired by the Numberphile video <a href="https://youtu.be/_gCKX6VMvmU" rel="nofollow noreferrer">2.920050977316</a> and <a href="https://doi.org/10.1080/00029890.2019.1530554" rel="nofollow noreferrer">Fridman, Garbulsky, Glecer, Grime, and Tron Florentin - A prime-representing constant</a>).</p>
<p>Let <span class="math-container">$u_0 \ge 2$</span> be a rational, and <span class="math-container">$u_{n+1}=⌊u_n⌋(u_n - ⌊u_n⌋ + 1)$</span>.<br />
<strong>Question</strong>: Does the sequence <span class="math-container">$(u_n)$</span> reach an integer?</p>
<p><span class="math-container">$\to$</span> see below the application to irrational number theory.</p>
<p><em>Remark</em>: It is true for <span class="math-container">$u_0=\frac{p}{q}$</span> with <span class="math-container">$p \le 40000$</span> (see Appendix).</p>
<p><strong>Proposition</strong>: It is always true for <span class="math-container">$u_0 = \frac{p}{2}$</span>.<br />
<em>Proof by contradiction</em>: Assume that the sequence never reach an integer, then <span class="math-container">$u_n = k_n + \frac{1}{2}$</span> for all <span class="math-container">$n$</span>. Next note that <span class="math-container">$u_{n+1} = k_n + \frac{k_n}{2}$</span>, so <span class="math-container">$k_n$</span> must be odd for all <span class="math-container">$n$</span>. Let write <span class="math-container">$k_n = 2 h_n +1$</span>, then <span class="math-container">$u_n = 2h_n+1+\frac{1}{2}$</span> (with <span class="math-container">$h_n \ge 1$</span>) and <span class="math-container">$u_{n+1} = 3h_n+1+\frac{1}{2}$</span>. It follows that <span class="math-container">$2h_{n+1} = 3h_n$</span>, and so <span class="math-container">$h_n = (\frac{3}{2})^nh_0$</span>, which implies that <span class="math-container">$2^n$</span> divides <span class="math-container">$h_0$</span> for all <span class="math-container">$n$</span>, contradiction. <span class="math-container">$\square$</span></p>
<p>For <span class="math-container">$u_0=\frac{11}{5}$</span>, then <span class="math-container">$$(u_n)= (\frac{11}{5}, \frac{12}{5}, \frac{14}{5}, \frac{18}{5}, \frac{24}{5}, \frac{36}{5}, \frac{42}{5}, \frac{56}{5}, \frac{66}{5}, \frac{78}{5}, 24, \dots).$$</span> Here is a picture of the dynamic:<br />
<a href="https://i.stack.imgur.com/viHif.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/viHif.png" alt="enter image description here" /></a></p>
<p>By regarding (for example) when <span class="math-container">$u_0=\frac{15}{7}$</span> below, we guess that the general proof should be hard. <span class="math-container">$(u_n) = (\frac{15}{7}, \frac{16}{7}, \frac{18}{7}, \frac{22}{7}, \frac{24}{7}, \frac{30}{7}, \frac{36}{7}, \frac{40}{7}, \frac{60}{7}, \frac{88}{7}, \frac{132}{7}, \frac{234}{7}, \frac{330}{7}, \frac{376}{7}, \frac{636}{7}, \frac{1170}{7}, \frac{1336}{7}, \frac{2470}{7}, \frac{4576}{7}, \frac{7836}{7}, \frac{11190}{7}, \frac{17578}{7}, \frac{20088}{7}, \frac{34428}{7}, \frac{44262}{7}, \frac{50584}{7}, \frac{65034}{7}, \frac{102190}{7}, \frac{160578}{7}, 39324, \dots)$</span></p>
<p>For <span class="math-container">$u_0=\frac{10307}{4513}=\frac{k_0}{q}$</span>, the sequence <span class="math-container">$(\frac{k_n}{q})=(u_n)$</span> reaches an integer at <span class="math-container">$n=58254$</span>. The sequence <span class="math-container">$(u_n)$</span> reaches an integer once <span class="math-container">$k_n \text{ mod } q=0$</span>. Below is the picture for <span class="math-container">$(n,k_n \text{ mod } q)$</span>; it looks completely random. The probability for <span class="math-container">$s$</span> random integers between <span class="math-container">$0$</span> and <span class="math-container">$q-1$</span> to never be zero is about <span class="math-container">$e^{-s/q}$</span> when <span class="math-container">$q$</span> is large enough.</p>
<p><a href="https://i.stack.imgur.com/xdKmq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xdKmq.png" alt="enter image description here" /></a></p>
<blockquote>
<p><strong>Application to irrational number theory</strong></p>
<p>According to <a href="https://doi.org/10.1080/00029890.2019.1530554" rel="nofollow noreferrer">the paper</a> mentioned above, there is a bijection between the set of numbers <span class="math-container">$u_0 \ge 2$</span>, and the set of
sequences <span class="math-container">$(a_n)$</span> such that for all <span class="math-container">$n$</span>:</p>
<ul>
<li><span class="math-container">$a_n \in \mathbb{N}_{\ge 2}$</span>,</li>
<li><span class="math-container">$a_n \le a_{n+1} < 2a_n$</span>.</li>
</ul>
<p>The bijection is given by: <span class="math-container">$$u_0 \mapsto (a_n) \text{ with } a_n =
⌊u_n⌋ \text{ and } u_{n+1}=⌊u_n⌋(u_n - ⌊u_n⌋ + 1),$$</span> <span class="math-container">$$(a_n) \mapsto
u_0 = \sum_{n=0}^{\infty}\frac{a_n-1}{\prod_{i=0}^{n-1}a_i}.$$</span> A
positive answer to the question would provide a kind of alternative to
<a href="https://en.m.wikipedia.org/wiki/Continued_fraction" rel="nofollow noreferrer">continued fraction</a>, in the sense of a natural way to represent
the numbers, with a complete characterization of the irrational ones,
which here would be that <span class="math-container">$\lim_{n \to \infty} (a_n)=\infty$</span>.</p>
</blockquote>
<hr />
<p><strong>Appendix</strong></p>
<p>In the following list the datum <span class="math-container">$[r,(p,q)]$</span> means that the sequence <span class="math-container">$(u_n)$</span>, with <span class="math-container">$u_0=\frac{p}{q}$</span>, reaches an integer at <span class="math-container">$n=r$</span>. The list provides the ones with the longest <span class="math-container">$r$</span> according the lexicographic order of <span class="math-container">$(p,q)$</span>.</p>
<p><em>Computation</em></p>
<pre><code>sage: search(40120)
[1, (2, 1)]
[2, (5, 2)]
[3, (7, 2)]
[4, (7, 3)]
[11, (11, 5)]
[30, (15, 7)]
[31, (29, 14)]
[45, (37, 17)]
[53, (39, 17)]
[124, (41, 19)]
[167, (59, 29)]
[168, (117, 58)]
[358, (123, 53)]
[380, (183, 89)]
[381, (201, 89)]
[530, (209, 97)]
[532, (221, 97)]
[622, (285, 131)]
[624, (295, 131)]
[921, (359, 167)]
[1233, (383, 181)]
[1365, (517, 251)]
[1482, (541, 269)]
[2532, (583, 263)]
[3121, (805, 389)]
[3586, (1197, 587)]
[3608, (1237, 607)]
[3860, (1263, 617)]
[4160, (1425, 643)]
[6056, (1487, 743)]
[9658, (1875, 859)]
[9662, (1933, 859)]
[10467, (2519, 1213)]
[10534, (2805, 1289)]
[11843, (2927, 1423)]
[12563, (3169, 1583)]
[13523, (3535, 1637)]
[14004, (3771, 1871)]
[14461, (4147, 2011)]
[17485, (4227, 1709)]
[18193, (4641, 1987)]
[18978, (4711, 2347)]
[22680, (5193, 2377)]
[23742, (5415, 2707)]
[24582, (5711, 2663)]
[27786, (5789, 2837)]
[27869, (6275, 2969)]
[29168, (6523, 3229)]
[32485, (6753, 2917)]
[33819, (7203, 3361)]
[41710, (7801, 3719)]
[49402, (8357, 3863)]
[58254, (10307, 4513)]
[58700, (10957, 4943)]
[81773, (12159, 5659)]
[85815, (16335, 7963)]
[91298, (16543, 7517)]
[91300, (17179, 7517)]
[98102, (19133, 9437)]
[100315, (19587, 8893)]
[100319, (20037, 8893)]
[102230, (20091, 9749)]
[102707, (21289, 10267)]
[103894, (21511, 10151)]
[105508, (22439, 11149)]
[107715, (22565, 10729)]
[142580, (23049, 11257)]
[154265, (24915, 12007)]
[177616, (27461, 13421)]
[178421, (32063, 15377)]
[190758, (34141, 16547)]
[228068, (34783, 15473)]
[228876, (35515, 17477)]
[277844, (40119, 19391)]
</code></pre>
<p><em>Code</em></p>
<pre><code>def Seq(p,q):
x=Rational(p/q)
A=[floor(x)]
while not floor(x)==x:
n=floor(x)
x=Rational(n*(x-n+1))
m=floor(x)
A.append(m)
return A
def search(r):
m=0
for p in range(2,r):
for q in range(1,floor(p/2)+1):
A=Seq(p,q)
l=len(A)
if l>m:
m=l
print([m,(p,q)])
</code></pre>
| Ben Burns | 51,771 | <p>I want to leave a few elementary comments, maybe they will be helpful.</p>
<p>The question asks about recurrence relation</p>
<p><span class="math-container">$$
u_{n+1}= \lfloor u_n \rfloor (u_n − \lfloor u_n \rfloor + 1)
$$</span></p>
<p>Suppose you write rational <span class="math-container">$u$</span> in terms of natural numbers <span class="math-container">$\frac{pq+r}{p}$</span>. Then <span class="math-container">$\lfloor u \rfloor = q$</span>. The recurrence relation is now</p>
<p><span class="math-container">$$
u_{n+1} = q_n (\frac{pq_n+r_n}{p} − q_n + 1)
\\ = q_n (\frac{pq_n+r_n− pq_n + p}{p} )
\\ = \frac{q_n(r_n + p)}{p}
$$</span></p>
<p>Therefore, we can investigate the same series in natural numbers <span class="math-container">$u_{n+1}=q_n(r_n + p)$</span> asking when the series converges to a multiple of <span class="math-container">$p$</span>. Now we have to deal with <span class="math-container">$r_{n+1} = q_n * r_n \mod p$</span> and <span class="math-container">$q_{n+1} = \lfloor u_{n+1} / p\rfloor$</span> but let me show the benefit. First compare to the original <span class="math-container">$u_0 = 11/5$</span>.</p>
<p>Now <span class="math-container">$p=5, q_0=2,r_0=1$</span>.</p>
<p><span class="math-container">$$
u_0 = 2*5 + 1 \\
u_{n+1} = 2*(5+1) = 12 \\
u_{n+2} = 2*(5+2) = 14\\
u_{n+3} = 2*(5+4) = 18\\
u_{n+4} = 3*(5+3) = 24\\
u_{n+5} = 4*(5+4) = 36\\
u_{n+5} = 7*(5+1) = 42\\
u_{n+6} = 8*(5+2) = 56\\
u_{n+7} = 11*(5+1) = 66\\
u_{n+8} = 13*(5+1) = 78\\
u_{n+9} = 15*(5+3) = 120\\
u_{n+10} = 24*(5+0) = 120\\
$$</span></p>
<p>The sequence continues indefinitely once <span class="math-container">$r_n=0$</span>.</p>
<hr />
<p>Minor remark: <span class="math-container">$q_{n+1} \ge q_n$</span>. By definition, <span class="math-container">$q_{n+1} = \lfloor u_{n+1} / p\rfloor \rightarrow \lfloor q_n(r_n + p) / p\rfloor$</span>.</p>
<hr />
<p>Minor remark: There are no cycles of length 1. A cycle would require <span class="math-container">$q_{n+1}=q_n$</span> and <span class="math-container">$r_{n+1}=r_n$</span>. The same <span class="math-container">$q_{n+1}$</span> implies <span class="math-container">$q_n * r_n \lt p$</span>. And therefore the only way for <span class="math-container">$r_{n+1} =r_n$</span> is if <span class="math-container">$q_n=1$</span>, but it is required <span class="math-container">$q_n>2$</span> (by definition on <span class="math-container">$u_0$</span>).</p>
<hr />
<p>Now a rough sketch showing that "many" <span class="math-container">$u_0$</span> converge. Consider when <span class="math-container">$p=10$</span>. The same logic will apply to other <span class="math-container">$p$</span> (prime and composite) but the analogy here is easiest. Note that <span class="math-container">$10=5*2$</span>. Therefore, when <span class="math-container">$u_n$</span> "steps" into values <span class="math-container">$50-59$</span>, half the time (even numbers) the following step will terminate:</p>
<p><span class="math-container">$$
50 => 5*(10 + 0) => 50\\
51 => 5*(10 + 1) => 55\\
52 => 5*(10 + 2) => 60\\
53 => 5*(10 + 3) => 65\\
54 => 5*(10 + 4) => 70\\
55 => 5*(10 + 5) => 75\\
56 => 5*(10 + 6) => 80\\
57 => 5*(10 + 7) => 85\\
58 => 5*(10 + 8) => 90\\
59 => 5*(10 + 9) => 95
$$</span></p>
<p>This will also be the case for other values that factor to <span class="math-container">$5*m$</span> (for <span class="math-container">$p=10$</span>) such as <span class="math-container">$150-159(=3*5), 250-259(=5*5)$</span> etc. Other composite <span class="math-container">$2*p$</span> behave similar, e.g., <span class="math-container">$p=14$</span> then half of <span class="math-container">$70-79$</span> terminate in one step. This can be applied to factors other than <span class="math-container">$2$</span>. Additionally, this idea of finding "terminating zones" also applies to a number of <span class="math-container">$u_n$</span> for prime <span class="math-container">$p$</span>, which can be easily decided as terminating in the area of <span class="math-container">$p^2-p$</span>. For instance, with <span class="math-container">$p=5$</span>, then there is a value <span class="math-container">$q_n=p-1$</span> and some <span class="math-container">$r$</span> such that <span class="math-container">$p^2<u_n<p^2+p$</span>; in this example, <span class="math-container">$q_n=4,r_n=2\rightarrow 4*(5+2)=28$</span> and the sequence terminates after <span class="math-container">$5*(5+3)$</span>.</p>
<p>There are still many other cases yet to prove, but maybe this post was helpful.</p>
|
541,890 | <p>Prove or disprove:
For all integers $m$ and $n$, if $m+n$ is even then so is $m-n$.
Would you just set them even to each other because you are given $m+n$ is even?</p>
| George Tomlinson | 92,587 | <p>The base case is correct. </p>
<p>Inductive step:</p>
<p>Assume that the result is true for $n = k$, which is to say that everybody in a group of $k$ people has the same hair colour. </p>
<p>For the proof to work, you now have to prove that $P(k)$ implies $P(k+1)$, but it doesn't. Just because it's true that for any group of $k$ people, everyone in the group has the same hair colour, it doesn't follow that for any group of $k+1$ people, everyone in the group has the same hair colour. One counterexample suffices to disprove the claim: take a group with one person in, who has red hair. Now add someone with blonde hair.</p>
|
2,692,341 | <blockquote>
<p>Is it possible for a commutative ring with 1 to be isomorphic to the
power series ring over itself? $$R \cong R[[x]].$$ Bonus marks if you
can answer for $n$ variables more generally, $R \cong R[[x_1, \dots, x_n]]$.</p>
</blockquote>
<p>Obviously, $R$ would have to be huge (infinite Krull dimension). It's not clear to me that taking a colimit of iterates of the functor $S \mapsto S[[x_1, \dots, x_n]]$ produces a ring of the desired kind, essentially because
$$R[[x]][[y]] \not \cong R[[x,y]].$$
So you can't make a construction analogous to what's done for $R \cong R[X]$.</p>
<p>EDIT: Looks like I was under a misapprehension about the above-stated non-isomorphism. Thank you to several people for correcting this important mistake!</p>
| Pierre-Yves Gaillard | 660 | <p>As I said in the comments, the zero ring is a first example. Another example is $A[[X_1,X_2,...]]$ where $A$ is a commutative ring with one and the $X_i$ are indeterminates.</p>
<p><strong>Edit.</strong> Here is a brief reminder.</p>
<p>If $A$ is a commutative ring with one, if $(X_i)_{i\in I}$ is a family of indeterminates, and if $M(I)\subset A[(X_i)_{i\in I}]$ is the set of monomials, then $A[[(X_i)_{i\in I}]]$ is the set of formal expressions of the form
$$
\sum_{m\in M(I)}a_mm,
$$
with $a_m$ in $A$, equipped with its natural ring structure.</p>
<p>If $I=J\sqcup K$ is a partition of $I$, then the natural morphism
$$
\left(A[[(X_j)_{j\in J}]]\right)[[(X_k)_{k\in K}]]\to A[[(X_i)_{i\in I}]]
$$
is easily seen to be an isomorphism.</p>
|
2,692,341 | <blockquote>
<p>Is it possible for a commutative ring with 1 to be isomorphic to the
power series ring over itself? $$R \cong R[[x]].$$ Bonus marks if you
can answer for $n$ variables more generally, $R \cong R[[x_1, \dots, x_n]]$.</p>
</blockquote>
<p>Obviously, $R$ would have to be huge (infinite Krull dimension). It's not clear to me that taking a colimit of iterates of the functor $S \mapsto S[[x_1, \dots, x_n]]$ produces a ring of the desired kind, essentially because
$$R[[x]][[y]] \not \cong R[[x,y]].$$
So you can't make a construction analogous to what's done for $R \cong R[X]$.</p>
<p>EDIT: Looks like I was under a misapprehension about the above-stated non-isomorphism. Thank you to several people for correcting this important mistake!</p>
| jgon | 90,543 | <p>Let $M$ be a monoid which is "locally finite" by which we mean, for all $m\in M$, $\{(a,b)\in M\times M : ab =m \}$ is finite.</p>
<p>For such a monoid, for a ring $R$, we can make $R^M$ into a ring with the convolution product instead of $R^{\oplus M}$. Lets denote $R^M$ with the convolution product by $R[[M]]$. Now $R[[x]]=R[[\Bbb{N}]]$. Now observe that as sets, by the usual product-hom adjunction, $(R^M)^N = R^{M\times N}$. Moreover, if $M$ and $N$ are locally finite, then $M\times N$ is locally finite. Indeed, one can check that the preserves sums and products, so $R[[M]][[N]]\cong R[[M\times N]]$. Thus we just need to find a locally finite monoid $M$ such that $M\times \Bbb{N}\cong M$. An easy choice is
$$M=\{u\in \Bbb{N}^\Bbb{N}:u_n\ne 0\text{ for only finitely many $n$}\}.$$
This is intuitively probably the monoid being referred to when people talk about $R[[x_1,x_2,\ldots]]$.</p>
<p>Thus this gives a proof outline that when $A=R[[x_1,x_2,\ldots]]$, then $A[[y]]\cong A$.</p>
|
31,100 | <p>I have just begun to learn about the fundamental group.
An exercise asks me to prove that $$X=\{(x,y,z): z \ge 0\}-\{(x,y,z): y=0,0\leq z \leq 1\}$$ has trivial fundamental group.</p>
<p>What I know is:</p>
<p>1) the definition of the fundamental group.</p>
<p>2) X has trivial fundamental group iff any loop in X can be shrunk into a constant loop at the base point.</p>
<p>3) Homeomorphic (path-connected) spaces have isomorphic fundamental groups.</p>
<p>4) Any convex subset of $\mathbb{E}^n$ and $S^m,m\ge 2$ has trivial fundamental group.</p>
<p>I tried to construct a homeomorphism from X to a convex subset of $\mathbb{E}^3$ such as an area like this:
$$\{(x,y,z): -1\leq y \leq 1,z>0\}$$
But I failed.</p>
<p>Can you please help me? Thank you!</p>
| user8268 | 8,268 | <p>You know that convex subsets of $\mathbb{R}^n$ have trivial fundamental group. Check the proof to see that it proves also this: if $X\subset \mathbb{R}^n$ and $a\in X$ are such that for every $b\in X$ the segment connecting $a$ and $b$ is in $X$ (i.e. if $X$ is "star-shaped") then $\pi_1(X,a)$ is trivial. (Since $X$ is then contractible, along the segments going to $a$.)</p>
<p>In your case $a=(0,0,2)$. </p>
|
31,100 | <p>I have just begun to learn about the fundamental group.
An exercise asks me to prove that $$X=\{(x,y,z): z \ge 0\}-\{(x,y,z): y=0,0\leq z \leq 1\}$$ has trivial fundamental group.</p>
<p>What I know is:</p>
<p>1) the definition of the fundamental group.</p>
<p>2) X has trivial fundamental group iff any loop in X can be shrunk into a constant loop at the base point.</p>
<p>3) Homeomorphic (path-connected) spaces have isomorphic fundamental groups.</p>
<p>4) Any convex subset of $\mathbb{E}^n$ and $S^m,m\ge 2$ has trivial fundamental group.</p>
<p>I tried to construct a homeomorphism from X to a convex subset of $\mathbb{E}^3$ such as an area like this:
$$\{(x,y,z): -1\leq y \leq 1,z>0\}$$
But I failed.</p>
<p>Can you please help me? Thank you!</p>
| Thomas Andrews | 7,933 | <p>If $T$ is your space, then $T$ has the property that if $(x,y,z)\in T$ and $t >0$ then $(x,y,z+t)\in T$.</p>
<p>This lets you find a homotopy between any loop in $T$ to a loop in $T_0=\{(x,y,z): z>=1\}$ But $T_0$ is convex, so it is simply connected.</p>
|
2,331,191 | <p>Use either direct proof, proof by contrapositive, or proof by contradiction.</p>
<p>Using proof by contradiction method</p>
<blockquote>
<p>Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)</p>
<p>There exists integers x and y such that <span class="math-container">$n = x^2$</span> and <span class="math-container">$n+3 = y^2$</span></p>
<p>Then <span class="math-container">$x^2 + 3 = y^2$</span></p>
<p>Then <span class="math-container">$3 = y^2 - x^2$</span></p>
<p>Then <span class="math-container">$3 = (y-x)(y+x)$</span></p>
<p>Then <span class="math-container">$y+x = 3$</span> and <span class="math-container">$y-x=1$</span></p>
<p>Then <span class="math-container">$x = 1, y = 2$</span></p>
<p>Since <span class="math-container">$x = 1$</span>, that implies <span class="math-container">$n = 1$</span></p>
<p><em><strong>this is how far I got</strong></em></p>
</blockquote>
<p>Anyone know what I should do now?</p>
| Ross Millikan | 1,827 | <p>You have proved that if $n$ and $n+3$ are perfect squares, $n=1=1^2, n+3=4=2^2$ The statement you were trying to prove is false. Maybe you were also given $n \gt 1$?</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Dave Rusin | 463 | <p>You can understand the squares purely algebraically: let $R = F_2[x_1, x_2, \ldots, x_n]$ be a polynomial ring over the field $F_2$ of $2$ elements.
A ring homomorphism $f$ on $R$ is completely determined by the values of $f(x_i)$, and those values are unrestricted, right? So define $f \colon R \to R$ by $f(x) = x + x^2$ for each $x = x_i$. (Since $\operatorname{char}(F)=2$, we actually have $f(x) = x + x^2$ for each $F$-linear combination of the $x_i$, too, i.e. this is a basis-free -- "gentlemanly"? -- definition.) Then for any homogeneous polynomial $p$ in $R$, we have $f(p) = p + p^2 + \text{other stuff}$; separating the components of $f(p)$ by degree, we can write $f(p) = \sum Sq^i ( p )$.
E.g. $f(x_1 x_2) = (x_1 + x_1^2) \cdot (x_2 + x_2^2) = (x_1 x_2) + (x_1^2 x_1 + x_2^2 x_1) + (x_1 x_2)^2$, so $Sq^1( x_1 x_2 ) = x_1 x_2 (x_1 + x_2)$.</p>
<p>You can extend the definition of the $Sq^i$ to non-homogeneous polynomials by additivity if you like. This definition of the $Sq^i$ is consistent for polynomials in arbitrary numbers of variables (i.e. there are commutative diagrams involving the inclusions $F_2[x_1] \to F_2[x_1,x_2]$ etc.) So you can visualize the whole Steenrod algebra as an algebra of endomorphisms on an infinite polynomial ring.</p>
<p>This definition is sufficient to check the Adem relations.</p>
<p>Topologically, this mechanism defines $Sq^i$ on $H^\ast( (\mathbb{RP}^\infty)^n, Z/2Z)$. Then by naturality it defines the $Sq^i$ on cohomology classes that are pullbacks from those rings, i.e. you have a description of how $Sq^i$ acts on cohomology classes defined by vector bundles. This really is how I think of them when I play with classifying spaces $BG$.</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Andrew Ranicki | 732 | <p>Section 4L of Hatcher's <a href="http://www.math.cornell.edu/~hatcher/AT/ATpage.html">Algebraic Topology</a> is a downloadable reference.</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Robert Bruner | 6,872 | <p>I give an intuitive account of the origin of Steenrod operations in cohomology, Dyer-Lashof operations in homology of infinite loop spaces, and Steenrod operations in the cohomology of cocommutative Hopf algebras in Chapter 3 of my (unfinished)
<a href="http://www.rrb.wayne.edu/papers/adams.pdf" rel="nofollow noreferrer"> Adams Spectral Sequence Primer</a>. Then in Chapter 4, I define the precise conditions needed and the resulting theorems, following Peter May's <a href="http://www.math.uchicago.edu/~may/PAPERS/10.pdf" rel="nofollow noreferrer"> A General Algebraic Approach to Steenrod Operations</a>. The role of symmetries in producing the operations in the first place, and then in producing their properties, is the focus. A key point that May makes is that all these operations have the same source, a category I call <span class="math-container">$Sym_\infty$</span> in my Primer.
It is a category of DGAs with some additional structure encoding the symmetries.</p>
<p>One other beautiful way to see the Steenrod operations is through the dual of the Steenrod algebra. This is the Hopf algebra of coordinate functions on the automorphisms of the additive group. Precisely, let <span class="math-container">$F_2[[x]]$</span> be the power series ring in one variable <span class="math-container">$x$</span> with primitive coproduct <span class="math-container">$\psi(x) = 1 \otimes x + x \otimes 1$</span>, corepresenting the additive group.</p>
<p>The automorphism which sends <span class="math-container">$x$</span> to a power series <span class="math-container">$f = \sum a_i x^i$</span> commutes with <span class="math-container">$\psi$</span> iff the only non-zero coefficients are those with <span class="math-container">$i$</span> a power of two, so we may write an automorphism <span class="math-container">$f$</span> as <span class="math-container">$\sum c_i x^{2^i}$</span> (with <span class="math-container">$c_0=1$</span> of course). Let <span class="math-container">$\xi_i(f) = c_i$</span>. There are no relations between these coordinate functions <span class="math-container">$\xi_i$</span>, so we get a polynomial ring <span class="math-container">$F_2[\xi_1, \xi_2,\ldots]$</span> of functions on the group of automorphisms.</p>
<p>Now consider <span class="math-container">$\xi_n(f\circ g)$</span>. We easily calculate that <span class="math-container">$\xi_n(f \circ g) = \sum_{i+j=n} \xi_i(f) \xi_j(g)^{2^i}$</span>, so that the coproduct on these coordinate functions is <span class="math-container">$\psi(\xi_n) = \sum \xi_i \otimes \xi_j^{2^i}$</span>, which is the dual Steenrod algebra, as shown by Milnor in 1956. (Note that these <span class="math-container">$\xi_i%$</span> are the conjugates of Milnor's generators, so the coproduct is reversed.) Jack Morava showed me this derivation of the dual Steenrod algebra.</p>
|
2,153,421 | <p>I want to find curvatures and torsions for the following curves but get stuck with their natural parametrizations ($s$ is natural if $|\dot{\gamma}(s)| = 1$). Can anyone help me?</p>
<p>(a) $e^t(\cos t,\sin t,1)$</p>
<p>(b) $(t^3+t,t^3-t,\sqrt{3}t^2)$</p>
<p>(c) $3x^2+15y^2=1, z=xy$</p>
<p><strong>Update</strong>:</p>
<p>Here are my attempts on solving (a):</p>
<p>$\dot{\gamma}(t) = (\dot{t}e^t \cos t - \dot{t} e^t \sin t, \dot{t} e^t \sin t + \dot{t}e^t \cos t, \dot{t} e^t)$ which gives $|\dot{\gamma}(t)| = \sqrt{2}\dot{t}e^t=1$ and the solution for this ODE is $t = \ln\frac{\tau}{2}$.</p>
<p>But if I substitute $t$ with $t=\ln\frac{\tau}{2}$ the result will be $\dot{\gamma}(\tau) = (\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2} - \frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2},\frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2}+\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2},\frac{1}{2})$ and $|\dot{\gamma}(\tau)| = \sqrt{\frac{3}{2}}$. So for $|\dot{\gamma}(\tau)| = 1$ we should take $t = \ln\frac{\tau}{3}$. Where is my mistake?</p>
<p>Any help and hints will be very appreciative.</p>
| Ted Shifrin | 71,348 | <p>You do not <em>need</em> an arclength or "natural" parametrization to do these computations. It's just a matter of using the chain rule. I've posted about this numerous times on here. Rather than write out another lecture, I'll just refer you to my (free) <a href="https://math.franklin.uga.edu/sites/default/files/inline-files/ShifrinDiffGeo.pdf" rel="nofollow noreferrer">differential geometry text</a>. Of course, if you have further questions, feel free to ask :)</p>
|
4,069,053 | <blockquote>
<p>If a finite field has characteristic 2 why is every element a square?</p>
</blockquote>
<p>I've attempted this problem by calculating the square root as follows: <span class="math-container">$\mathbb{F}$</span> has <span class="math-container">$q = 2^m$</span> elements, so if <span class="math-container">$a \in \mathbb{F}$</span> then <span class="math-container">$a = a^q = (a^{2^{(n-1)}})^2$</span>. Since, <span class="math-container">$a \in \mathbb{F}$</span> is a square elements if there exist <span class="math-container">$b \in \mathbb{F}$</span> s.t <span class="math-container">$b^2 = a$</span>, I believe the next step would be to show that the square root <span class="math-container">$b = a^{2^{(n-1)}} \in \mathbb{F}$</span>. Is this reasonable or should I take a different approach to the problem (and if so what approach?)?</p>
| Aryaman Maithani | 427,810 | <p>Consider the function <span class="math-container">$f : \Bbb F \to \Bbb F$</span> given by <span class="math-container">$f(x) = x^2$</span>.<br />
Then, <span class="math-container">$$f(x) = f(y) \iff x^2 = y^2 \iff x^2 - y^2 = 0 \color{red}{\iff} (x - y)^2 = 0 \iff x - y = 0.$$</span>
The <span class="math-container">$\color{red}{\iff}$</span> is true since the field has characteristic <span class="math-container">$2$</span>. Thus, <span class="math-container">$f$</span> is actually injective. An injective map from a finite set to itself is always surjective and thus, every element of <span class="math-container">$\Bbb F$</span> is a square.</p>
<hr />
<p>In general, the above argument shows that if <span class="math-container">$\Bbb F$</span> is a finite field with characteristic <span class="math-container">$p$</span>, then <span class="math-container">$x \mapsto x^p$</span> is a bijection. In fact, this is actually an isomorphism since <span class="math-container">$(x + y)^p = x^p + y^p$</span>.</p>
<hr />
<p>Note that finiteness is crucial. For example, consider the field <span class="math-container">$\Bbb F = \Bbb F_{2}(T)$</span> of rational functions over <span class="math-container">$\Bbb F_{2}$</span> in the indeterminate <span class="math-container">$T$</span>. In this case, the map above is an injection but the image is strictly smaller than <span class="math-container">$\Bbb F$</span>. For example, <span class="math-container">$T$</span> is not in the image.</p>
|
2,232,390 | <p>$X$ is a subset of $\mathbb{R}$. It contains $\sqrt{3}$ and $\sqrt{2}$ and is closed under addition, subtraction and multiplication.</p>
<ol>
<li>Prove that $X$ contains $\sqrt{8}$.</li>
<li>Prove that $X$ contains $1$.</li>
<li>Prove that $X$ contains $\frac{1}{\sqrt{2}+1}$.</li>
<li>Is it true that $X$ is necessarily a field?</li>
</ol>
<p>So far I have only been successful with 1 and 2. I've just used the elements that are already in the group and performed operations on the them to try and get $\sqrt{8}$ and $1$. </p>
| Ian | 83,396 | <p>Heuristically, if $X_1,X_2$ are independent and have one sign (as in this case) you should expect $f_{X_1/X_2}(r)=\int_{-\infty}^\infty f_{X_1}(rx) f_{X_2}(x) dx$, because if $X_1/X_2=r$ and $X_2=x$ then $X_1=rx$. </p>
<p>Plugging in the relevant densities in this case gives $\int_0^\infty 9 e^{-3(rx)} e^{-3x} dx$. This integral is of course quite easy to calculate, but can you prove that this formula is correct? (Probably the best way to carry out this proof is to pass through the CDF $F_{X_1/X_2}$.)</p>
|
3,355,544 | <p><a href="https://i.stack.imgur.com/oYf7f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oYf7f.jpg" alt="enter image description here"></a></p>
<p>Given the following equations:</p>
<p><span class="math-container">$$x=t^2$$</span></p>
<p><span class="math-container">$$y=t^2$$</span></p>
<p><span class="math-container">$$z=2t+3$$</span></p>
<p>Show that the graph of the curve <span class="math-container">$(t^2,t^2,2t+3)$</span> lives in a plane and that, within that plane, the graph is a parabola.</p>
<p>My attempt, so far:</p>
<p>From the first two equations <span class="math-container">$x=y$</span>, meaning the curve belongs to the <span class="math-container">$x-y=0$</span> plane.</p>
<p>Plugging the first or second equation into the third one:</p>
<p><span class="math-container">$$((z-3)/2)^2=x,\>\>\> ((z-3)/2)^2=y$$</span></p>
<p>This tells me it is parabola in the zx and zy planes respectively , I guess ??</p>
<p>A matlab plot shows it is , in fact, a parabola in the <span class="math-container">$x-y=0$</span> plane. What I am doing wrong and how do I show it?</p>
| Intelligenti pauca | 255,730 | <p>If a curve <span class="math-container">$u(t)$</span> lies on a plane, then the plane defined by any three points <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span> on the curve must be always the same, i.e. the vector product
<span class="math-container">$\vec n=(B-A)\times(C-A)$</span> must have the same direction for any choice of the points.</p>
<p>In practice we can keep two points fixed and choose, for instance: <span class="math-container">$A=u(0)=(0,0,3)$</span>, <span class="math-container">$B=u(1)=(1,1,5)$</span>, <span class="math-container">$C=u(t)=(t^2,t^2,2t+3)$</span>, to find:
<span class="math-container">$$
\vec n=(1,1,2)\times(t^2,t^2,2t)=(2t-t^2,2t-t^2,0)=(2t-t^2)(1,1,0).
$$</span>
The direction <span class="math-container">$\vec n/|\vec n|$</span> is then <span class="math-container">${1\over\sqrt2}(1,1,0)$</span> and does not depend on <span class="math-container">$t$</span>: hence the curve lies on a plane whose normal unit vector is <span class="math-container">${1\over\sqrt2}(1,1,0)$</span>.</p>
|
2,269,917 | <p>I've come across this question in a university exam paper. It's causing me a huge headache due to the fact that it goes from one vector space to another ($\def\R{\Bbb R}\R^3 \to \R^2$), otherwise it would be fairly standard. If anyone could shed some light on what I'm missing it would be much appreciated.</p>
<p>A4. Let $T : \R^3 → \R^2$ be given by
$T((x, y, z)) = (2x + 3y − 5z, x − 2y + z)$.</p>
<p>(i) Find a basis for the kernel $\ker T$ of $T$.</p>
<p>(ii) Find a basis for the image $\operatorname{im}T$ of $T$.</p>
| egreg | 62,967 | <p>A standard method is to begin with the matrix representation of $T$: denoting by $e_1$, $e_2$ and $e_3$ the vectors in the standard basis of $\mathbb{R}^3$, we have
$$
T(e_1)=\begin{bmatrix}2\\1\end{bmatrix}
\qquad
T(e_2)=\begin{bmatrix}3\\-2\end{bmatrix}
\qquad
T(e_3)=\begin{bmatrix}-5\\1\end{bmatrix}
$$
so we have
$$
T\begin{bmatrix}x\\y\\z\end{bmatrix}=
\begin{bmatrix}
2 & 3 & -5 \\
1 & -2 & 1
\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
The kernel and image of $T$ are the same as the null space and the column space of the representing matrix. With Gaussian elimination
$$
\begin{bmatrix}
2 & 3 & -5 \\
1 & -2 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 3/2 & -5/2 \\
1 & -2 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 3/2 & -5/2 \\
0 & -7/2 & 7/2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 3/2 & -5/2 \\
0 & 1 & -1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & -1\\
0 & 1 & -1
\end{bmatrix}
$$
This means that the matrix has rank $2$, so the column space is $\mathbb{R}^2$. Hence the map $T$ is surjective: any basis of $\mathbb{R}^2$ will do, in particular
$$
\left\{
\begin{bmatrix}2\\1\end{bmatrix},
\begin{bmatrix}3\\-2\end{bmatrix}
\right\}
$$
(the pivot columns). However also the standard basis would do.</p>
<p>The last matrix also shows that the null space is generated by the vector
$$
\begin{bmatrix}1\\1\\1\end{bmatrix}
$$
which forms a basis for the kernel of $T$.</p>
|
569,927 | <p>$p^{\; \left\lfloor \sqrt{p} \right\rfloor}\; -\; q^{\; \left\lfloor \sqrt{q} \right\rfloor}\; =\; 999$</p>
<p>How do you find positive integer solutions to this equation?</p>
| hmakholm left over Monica | 14,366 | <p>Calculate $x^{\lfloor x^{1/2}\rfloor}$ for the first dozen or so integers. Soon the function starts to increase so quickly that there's no hope of getting 999 as the difference between its values. Check whether any of the values you get until that happens happen to be an earlier value plus 999.</p>
<p>(There are two solutions).</p>
|
699,872 | <p>Compute the double integral of f(x, y) = 8 x^3 y over the given shaded domain in the following Figure :</p>
<p><img src="https://i.stack.imgur.com/RHimo.png" alt="enter image description here"></p>
<p>I am confused about the limits,
my work so far for the limits is x from 0 to 4 and y from 0 to 2 </p>
| MPW | 113,214 | <p>Note that $f$ doesn't enter into it.</p>
<p>It's easiest in this case to let $y$ range between $0$ and $2$. Then for a fixed $y$, $x$ ranges between $y$ and $4$.</p>
<p>You could do it the other way around, but you would have to break it into two cases. First let $x$ range between $0$ and $2$; for a fixed $x$ in this range, $y$ ranges between $0$ and $x$. Then let $x$ range between $2$ and $4$; for a fixed $x$ in this range, $y$ ranges between $0$ and $2$.</p>
<p>The first method would give you
$$\int_{y=0}^2 \int_{x=y}^4 8x^3y \,dx\, dy$$</p>
<p>The second method would give you
$$\int_{x=0}^2 \int_{y=0}^x 8x^3y \,dy\, dx + \int_{x=2}^4 \int_{y=0}^2 8x^3y \,dy\, dx$$</p>
<p>They should be equal, but the first looks like less work.</p>
|
699,872 | <p>Compute the double integral of f(x, y) = 8 x^3 y over the given shaded domain in the following Figure :</p>
<p><img src="https://i.stack.imgur.com/RHimo.png" alt="enter image description here"></p>
<p>I am confused about the limits,
my work so far for the limits is x from 0 to 4 and y from 0 to 2 </p>
| David | 119,775 | <p>The important thing to realise about an iterated integral is that there are implied brackets:
$$\int\!\!\int f(x,y)\,dx\,dy=\int\left(\int f(x,y)\,dx\right)dy\ .$$
To evaluate the integral we have to do the part in brackets first. This means integrating with respect to $x$, taking $y$ to be a fixed value. Therefore the limits for $x$ on the inner integral might not be constants, but might be expressions in $y$. On the other hand, the integral with respect to $y$ is performed after $x$ has been eliminated from the calculation, so the limits for $y$ will be constants.</p>
<p>Some people differ on this, but I recommend establishing the limits from the outside in. So, start with $y$. As pointed out already, by the time we come to deal with $y$ we will have removed $x$ from the expression. Therefore we need the minimum and maximum values of $y$ over the whole region. I hope it is clear in your problem that this means $y=0$ and $y=2$. So we have
$$\int_0^2\left(\int_?^? f(x,y)\,dx\right)dy\ .$$
Now the situation regarding $x$ is different. Remember that we will integrate with respect to $x$, taking $y$ as a constant. To find the relevant $x$-values, draw on your diagram a line $y=\hbox{constant}$. (Please do it yourself: I am no good at posting diagrams online.) That is, a horizontal straight line passing through the shaded region. The limits for $x$ are the minimum and maximum $x$-values within the shaded region <strong>on this line</strong>, that is, $x=y$ and $x=4$. So the integral is
$$\int_0^2\left(\int_y^4 f(x,y)\,dx\right)dy\ .$$</p>
|
13,030 | <p>At work, we were discussing when is it the best time to change to winter tires for bikes and/or cars.</p>
<p>Using <code>WeatherData[]</code> and <code>DateListPlot[]</code>, it was fairly straightforward for me to create the diagram below:</p>
<p><img src="https://i.stack.imgur.com/Y5wNT.png" alt="Mean temperature per day in Stockholm"> </p>
<p><em>Fig 1 Mean temperature per day in Stockholm, red is negative and a risk without winter tires.</em></p>
<p>The code for this is</p>
<pre><code>cityTemp = WeatherData["Stockholm", "MeanTemperature", {{1977, 1, 1}, {2011, 12, 31}, "Day"}];
iceRiskDays = Select[cityTemp, Last[#] < 0 &];
yearStrip[ dataItem_] := {{0, Part[First[dataItem], 2], Last[First[dataItem]]}, Last[dataItem]}
DateListPlot[{yearStrip[#] & /@ cityTemp, yearStrip[#] & /@ iceRiskDays} ]
</code></pre>
<p>My question is: <strong>How do I calculate for each day the proportion of values for that day that are below 0 Celsius?</strong> (e.g. for a date at the end of November, the proportion is likely to be bigger than 0.5)?</p>
<p>My attempts to do this ended with trying to create separate lists for each date, but I felt that this was a less elegant way and also creates less "fit" with DateListPlot.</p>
| chris | 1,089 | <p>Following @J.M. if you are after the likelihood of icy days as a function of month</p>
<pre><code>ff[x_] := Select[cityTemp, #[[1, 2]] == x &] // (Count[#, {d_List, t_?Negative}]/Length[#]) &
{Range[12] - 1/2, ff /@ Range[12]} // Transpose //
ListLinePlot[#, InterpolationOrder -> 0,AxesLabel -> {"Month", "likelihood"}] &
</code></pre>
<p>which suggests in January its about 15% (from where I live) and 60 % in Stockholm (Brrr...)</p>
<p><img src="https://i.stack.imgur.com/yB3ET.png" alt="Mathematica graphics"></p>
<p>Comparing Paris to Stockholm (using the nice package <a href="https://mathematica.stackexchange.com/a/4028/1089">autoLegend</a>)</p>
<pre><code>ff[x_, temp_] := Select[temp, #[[1, 2]] == x &] // (Count[#, {d_List, t_?Negative}]/
Length[#]) &
{{Range[12] - 1/2, ff[#, cityTempS] & /@ Range[12]} // Transpose,
{Range[12] - 1/2, ff[#, cityTempP] & /@ Range[12]} // Transpose} //
ListLinePlot[#, InterpolationOrder -> 0,
AxesLabel -> {"Month", "likelihood"}] &//autoLegend[#, {"Stockholm","Paris"}] &
</code></pre>
<p><img src="https://i.stack.imgur.com/f5FNd.png" alt="Mathematica graphics"></p>
<p>Stealing yet again another suggestion of @J.M. we can look at global warming ;-)
This time we look at the faction of days with ice as a function of year:</p>
<pre><code> dat = Map[Count[#, {d_List, t_?Negative}]/Length[#] &,
SplitBy[cityTempP, {#[[1, 1]] &, #[[1, 2]] &}], {2}] // Rest;
fP = LinearModelFit[Mean /@ dat, {1, x}, x];
dat = Map[Count[#, {d_List, t_?Negative}]/Length[#] &,
SplitBy[cityTempS, {#[[1, 1]] &, #[[1, 2]] &}], {2}] // Rest;
fS = LinearModelFit[Mean /@ dat, {1, x}, x];
Show[Plot[{fS[x], fP[x]}, {x, 1, 31}, AxesLabel -> {"year", "Likelihood"}],
ListPlot[{fS["Data"], fP["Data"]}]]// autoLegend[#, {"Stockholm", "Paris"}] &
</code></pre>
<p><img src="https://i.stack.imgur.com/slAz2.png" alt="Mathematica graphics"></p>
<p>And finally for the sake of contributing something original (the code for <a href="http://pastebin.com/08KWB9aT" rel="nofollow noreferrer">FilledErrorPlot</a>), let us launch the climate-skeptic versus climato-believer debate while illustrating the famous <a href="http://en.wikipedia.org/wiki/Lies,_damned_lies,_and_statistics" rel="nofollow noreferrer">Lies, damned lies, and statistics</a> methodology on two different ways of plotting the same data; guess which is which ;-)</p>
<pre><code>Show[dat // Transpose // FilledErrorPlot[#] &,
Plot[{fS[x]}, {x, 1, 33},
PlotStyle -> Directive[AbsoluteThickness[2], ColorData[10][1]],
AxesLabel -> {"year", "Likelihood"}], AspectRatio -> 1.5,
PlotRange -> {0.035, 0.45}, AxesOrigin -> {0., 0.035},
AxesLabel -> {"year", "Fraction of \n icy days"}]
</code></pre>
<p><img src="https://i.stack.imgur.com/kawk4.png" alt="Mathematica graphics"></p>
<p>versus</p>
<pre><code>Show[dat // Transpose // FilledErrorPlot[#, ErrorOnMean -> False] &,
Plot[{fS[x]}, {x, 1, 33},
PlotStyle -> Directive[AbsoluteThickness[2], ColorData[10][1]],
AxesLabel -> {"year", "Likelihood"}], AspectRatio -> 0.5, PlotRange -> {-0.5, 1},
AxesLabel -> {"year", "Fraction of \n icy days"}]
</code></pre>
<p><img src="https://i.stack.imgur.com/aRrxn.png" alt="Mathematica graphics"></p>
|
1,449,776 | <p>I have always known that $a^n=a*a*a*.....$(n times)</p>
<p>Then what exactly is the meaning if $a^0$ and why will it be equal to $1$?</p>
<p>I have checked it in the internet but everywhere the solution is based on the principle that $a^m*a^n=a^{m+n}$ and when $n=0$ it will be $a^m$ and clearly $a^0$ is equal to $1$. </p>
<p>But what exactly does $a^0$ mean does it mean $a*a*a*...$(zero times)?</p>
<p>Any help is highly appreciated.</p>
| egreg | 62,967 | <p>That's the intuition; but the powers are properly defined by recursion from
$$
a^0=1,\quad
a^{n+1}=a\cdot a^n
$$
so $a^0=1$ <em>by definition</em>. It is a sound definition, because it agrees with the property $a^{m+n}=a^m\cdot a^n$ for any natural $m$ and $n$.</p>
<hr>
<p>Think to what you do when you have a heap of candies to count. You start from zero and take one candy at a time, uttering the corresponding number: one, two, three, and so on.</p>
<p>Similarly, if you have to know how many candies are in a bunch of heaps, you can count each heap and write down the number. Then you start from zero, add the first number, then the second and so on (at this stage you already know how to perform symbolic sums).</p>
<p>For multiplication it's the same, but you start from one! So $a^0=1$, then $a^1=a\cdot 1$, $a^2=a\cdot a^1$, and so on, each time multiplying by $a$ until you arrive at $n$ and you have your $a^n$.</p>
|
126,074 | <p>I would like to know some applications of Anick's resolution in non-commutative algebras.</p>
| Noah Schweber | 8,133 | <p>The first paragraph of David Anick's paper, "On the Homology of Associative Algebras" (<a href="http://www.jstor.org/stable/2000383" rel="nofollow">http://www.jstor.org/stable/2000383</a>): </p>
<blockquote>
<p>Let $k$ be a field and let $G$ be an associative augmented $k$-algebra. For many purposes one wishes to have a projective resolution of $k$ as a $G$-module. The bar resolution is always easy to define, but it is often too large to use in practice. At the other extreme, minimal resolutions may exist, but they are often hard to write down in a way that is amenable to calculations. The main theorem of this paper presents a compromise resolution. Though rarely minimal, it is small enough to offer some bounds but explicit enough to facilitate calculations. As it relies heavily upon combinatorial constructions, it is best suited for analyzing otherwise tricky algebras given via generators and relations.</p>
</blockquote>
|
1,407,700 | <p>I am stuck on solving the following systems of equations with 3 variables. The textbook asks to use the addition method so can we please stick to that.</p>
<p>${5x -y = 3}$</p>
<p>${3x + z = 11}$</p>
<p>${y - 2z = -3}$</p>
<p>I am used to systems of equations where each equation has at least one instance of the variable e.g. ${x + y + z = 1}$ but in each of the above, one of the variables is omitted in each equation.</p>
<p>Could somebody explain what to do in this situation? Should I multiply both sides by one of the variables to balance it up? </p>
| pisco | 257,943 | <p>It is not uniformly continuous on $(0,+\infty)$. We apply the following simple theorem, which immediately says the function not uniform continuous on $[1,+\infty)$</p>
<hr>
<p><strong>Theorem</strong>: Suppose $f(x)$ is uniformly continuous on $[1,+\infty)$, then $f(x)\over x$ is bounded on $[1,+\infty)$.</p>
<p><strong>Proof</strong>:
By uniform coninuity, $\exists \delta > 0$, such that when $\left| {x - y} \right| < \delta ,{\rm{ }}\left| {f(x) - f(y)} \right| < 1$.</p>
<p>Hence $$\begin{array}{l}
- 1 < f(1 + \delta ) - f(1) < 1\\
- 1 < f(1 + 2\delta ) - f(1 + \delta ) < 1\\
...\\
- 1 < f(1 + n\delta ) - f(1 + (n - 1)\delta ) < 1
\end{array}$$
Therefore, we can prove that $\forall n \in \mathbb{N}, - n + f(1) < f(1 + n\delta ) < n + f(1)$. Hence for $n \in \mathbb{N},\frac{{f(1 + n\delta )}}{{1 + n\delta }}$ is bounded.</p>
<p>Now, $\forall x > 1$, there is an integer $N$ such that $1 + N\delta \le x < 1 + (N + 1)\delta $, which implies $\left| {f(x) - f(1+N\delta )} \right| < 1$. This concludes the proof.</p>
|
1,217,557 | <p>I was tasked with drawing the contour lines of $ z = \sqrt{xy} $, which I find a bit problematic since I can see no way in which one can plot (by hand, and not with wolfram and others....) the $ z = \sqrt{xy} $ graph in $R^2( x-, y- $ projection} to begin with for this surface...</p>
<p>How can one draw this contour graph manually? </p>
| Jonathan Hebert | 120,932 | <p>There are ultimately a lot of different ways we use the term "linear". In the context of linear transformations, $L(x) = ax+b$ is linear if and only if $b=0$. Perhaps you could explore a proof for that instead.</p>
|
1,402,953 | <ol>
<li>How can one intuitively understand the definition of a bilinear map? Is there some way of looking at it geometrically? I found the following definition:</li>
</ol>
<p>Let $\mathit{A}$,$\mathit{B}$,$\mathit{C}$ be vector spaces. A map $f:\mathit{A}\times \mathit{B}\to C$ is said to be bilinear if for each fixed element $b\in \mathit{B}$, $f(.,b):\mathit{A}\to\mathit{C}$ is a linear map. Similarly, for each fixed element of $\mathit{A}$.</p>
<p>Matrix multiplication is an example of a bilinear map.Following my definition, I can prove that it is a bilinear map, but I don't understand the intuitive idea behind it. In my opinion, it is simply a linear map with one element fixed.</p>
<ol start="2">
<li>Is there some formal definition of a bilinear algorithm? I could find an explanation for it only in the context of matrix multiplication: <a href="http://www.issac-symposium.org/2014/tutorials/ISSAC2014_tutorial_handout_LeGall.pdf" rel="nofollow">http://www.issac-symposium.org/2014/tutorials/ISSAC2014_tutorial_handout_LeGall.pdf</a></li>
</ol>
<p>Kindly help me with these questions.
Thanks!</p>
| Luckyluck63 | 206,839 | <p>For a definition of a bilinear algorithm, please look at:</p>
<p>Michael Kaminski, David G Kirkpatrick, Nader H Bshouty, Addition requirements for matrix and transposed matrix products, Journal of Algorithms, Volume 9, Issue 3, 1988, Pages 354-364.</p>
<p>(<a href="http://www.sciencedirect.com/science/article/pii/0196677488900260" rel="nofollow noreferrer">http://www.sciencedirect.com/science/article/pii/0196677488900260</a>)</p>
|
2,679,821 | <p>$K=\{(x,y,z)\in \mathbb{R}^{3}|1\leq x^{2}+y^{2}+z^{2}\leq 2,x+y\geq0,\sqrt3x-y\leq0,z\geq0\} \rightarrow \\
\rightarrow\{spherical\quad coordinates\}\rightarrow K:1\leq r\leq \sqrt2,\quad0\leq \theta\leq \pi/2,\quad\pi/3\leq\phi\leq3\pi/4.$</p>
<p>My question is how we arrive at these inequalities/intervals for the new variables? I can understand the $r$ part, just plug all new expressions into the first inequality and we get the answer. As for the others, I don't see it. Some kind combinations of the inequalities?</p>
<p>Edit:</p>
<p><strong>Spherical coordinates</strong> $ :\\ x=r\sin{\theta}\cos{\phi}\\
y=r\sin{\theta}\sin{\phi}\\
z=r\cos{\theta}$</p>
| Noah Schweber | 28,111 | <p>"$\vee$" by itself is <strong>inclusive or</strong>. To express <strong>exclusive or</strong>, we need to write a more complicated expression.</p>
<p>Remember that "$A$ xor $B$" just means "$A$ or $B$, but not $A$ and $B$." With that in mind, "$A$ xor $B$" can be expressed as $$(A\vee B)\wedge\neg(A\wedge B).$$ Another useful way to express it is $$(\neg A\wedge B)\vee(A\wedge\neg B).$$ It's a good exercise to check that these two expressions are in fact equivalent.</p>
|
19,996 | <p>In 1556, Tartaglia claimed that the sums<br>
1 + 2 + 4<br>
1 + 2 + 4 + 8<br>
1 + 2 + 4 + 8 + 16<br>
are alternative prime and composite. Show that his conjecture is false. </p>
<p>With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:<br>
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime.
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?</p>
| N. S. | 9,176 | <p>It is easy to see that $2^{2n}-1 \equiv 0 \mod 3$. This shows that for $n \geq 2$, $2^{2n}-1$ is never prime.</p>
<p>Also, since $2^3 \equiv 1 \mod 7$ it follows that $2^{3n} -1$ is always divisible by $7$.</p>
<p>Hence $2^{6n+2}-1, 2^{6n+3}-1, 2^{6n+4}-1$ are always composite. </p>
|
1,641,076 | <p>Let's say I have the following decomposition: </p>
<p>$$\{100,10011,00110\}^*$$</p>
<p>How would I determine if the decomposition is ambiguous or unambiguous?</p>
| Maverick | 171,392 | <p>Putting $x=0$ and solving the two equations we obtain a point which satisfies both equations i.e.$(0,-1,3)$.</p>
<p>Therefore this point would lie on the line of intersection of the two planes. Let equation of line be
$$ \frac{x-0}{l}=\frac{y+1}{m}=\frac{z-3}{n}$$.</p>
<p>Angle between this line and both the planes will be $0$.</p>
<p>Therefore,</p>
<p>$3l-m+n=0$</p>
<p>$m+n=0$</p>
<p>Solving we get$\displaystyle \frac{l}{2}=\frac{m}{3}\frac{n}{-3}$.</p>
<p>Thus equation of line will be $\displaystyle \frac{x-0}{2}=\frac{y+1}{3}=\frac{z-3}{-3}=t(say)$.</p>
<p>Thus $x=2t,y=3t-1,3-3t$ </p>
|
103,960 | <pre><code>tt = Flatten[Table[{x, y, z, btot[x, y, z]}, {x, -1, 1, 0.1}, {y, -1, 1,0.1}, {z, -1, 1, 0.1}], 2];
ff = Interpolation[tt]
</code></pre>
<p>Till here it is working fine as it is returning the values of the interpolated function at various <code>{x,y,z}</code> points.</p>
<p>Then I want to find the gradient of this interpolated function. But when I am using</p>
<pre><code>ffd[x_,y_,z_]:= D[ff[x,y,z],{{x,y,z}}]
</code></pre>
<p>I am not getting the gradient.</p>
| Kuba | 5,478 | <p>When you define your function like this, then e.g. <code>ffd[.5, .5, .5]</code> is really <code>D[ff[.5, .5, .5], {{.5, .5, .5}}]</code>.</p>
<p>to avoid scoping issues with </p>
<pre><code>x=5;
ffd[x_, y_, z_] = D[ff[x, y, z], {{x, y, z}}]
</code></pre>
<p>you can use</p>
<pre><code>ffd = Evaluate[D[ff[#, #2, #3], {{#, #2, #3}}]] &
</code></pre>
<p><code>&</code> means there is a held <code>Function</code> so we have to use <code>Evaluate</code> to force computation of gradien so it is not repeated each time.</p>
|
191,307 | <pre><code>roots = NSolve[Sin[2z] == Cos[Sin[z]] + 1 && -2π <= Re[z] <= 2π && -2π <= Im[z] <= 2π, z];
w = z\.roots;
ListPlot[{Re[w], Im[w]}, Ticks -> {{-2π, -π, 0, π, 2π}, {-2π, -π, 0, π, 2π}}]
</code></pre>
<p>In this code, <code>Ticks</code> command is not working properly, that is, real axis range is not showing between <span class="math-container">$-2\pi$</span> to <span class="math-container">$2\pi$</span>, where as imaginary axis is showing properly.</p>
<p>Also, is there any command for complex list plot and complex plot? I mean a command such that the following command works </p>
<pre><code>ComplexListPlot[w, Ticks -> {{-2π, -π, 0, π, 2π}, {-2π, -π, 0, π, 2π}}]
</code></pre>
<p>instead of </p>
<pre><code>ListPlot[{Re[w], Im[w]}, Ticks -> {{-2π, -π, 0, π, 2π}, {-2π, -π, 0, π, 2π}}]
</code></pre>
| Greg Hurst | 4,346 | <p>Hard to know exactly what you want without the explicit data, but maybe you'd like to bin the data?</p>
<pre><code>SeedRandom[1234];
pts = RandomReal[{0, 1}, {1000, 3}];
centers = Tuples[Range[.1, .9, .2], 3];
nf = Nearest[centers];
ListDensityPlot3D[KeyValueMap[Append, CountsBy[pts, First@*nf]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/V9Nsa.png" width="512"></p>
|
275,310 | <p>I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated</p>
| mdp | 25,159 | <p>A linear function fixes the origin, whereas an affine function need not do so. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else.</p>
<p>Linear functions between vector spaces preserve the vector space structure (so in particular they must fix the origin). While affine functions don't preserve the origin, they do preserve some of the other geometry of the space, such as the collection of straight lines.</p>
<p>If you choose bases for vector spaces $V$ and $W$ of dimensions $m$ and $n$ respectively, and consider functions $f\colon V\to W$, then $f$ is linear if $f(v)=Av$ for some $n\times m$ matrix $A$ and $f$ is affine if $f(v)=Av+b$ for some matrix $A$ and vector $b$, where coordinate representations are used with respect to the bases chosen.</p>
|
275,310 | <p>I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated</p>
| Francis Zerbib | 853,307 | <p>Answer for any confused French reader.</p>
<p>In France/French, the distinction between linear and affine appears to be different from other countries.</p>
<p>An affine function in the French sense is a any function that may be written</p>
<p><span class="math-container">$f(x) = a + bx$</span></p>
<p>where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are independent of <span class="math-container">$x$</span> (not necessarily Real). For example, <span class="math-container">$f(n) = a + bn$</span>, where <span class="math-container">$n \in N$</span> is an affine function over the Natural number set <span class="math-container">$N$</span>.</p>
<p>A linear function in the French sense is an affine function that passes through the origin, that is <span class="math-container">$a=0$</span> and <span class="math-container">$f(x)=bx$</span> for some number <span class="math-container">$b$</span> independent of <span class="math-container">$x$</span>.</p>
<p>Reference: <a href="https://fr.wikipedia.org/wiki/Fonction_affine" rel="nofollow noreferrer">wiki/Fonction_affine</a></p>
|
3,580,293 | <blockquote>
<p>The value of <span class="math-container">$$\lim\limits_{x \rightarrow \infty} \left(5^x + 5^{3x}\right)^{\frac{1}{x}}$$</span> is...</p>
</blockquote>
<p>My approach :</p>
<blockquote>
<p><span class="math-container">$$\lim\limits_{x \rightarrow \infty} \left(5^x + 5^{3x}\right)^{\frac{1}{x}}$$</span>
<span class="math-container">$$=\lim\limits_{x\rightarrow\infty} \left(5^x\left(1 + 5^{2x}\right)\right)^\frac{1}{x}$$</span>
<span class="math-container">$$=\lim\limits_{x\rightarrow\infty} \left(5^x\right)^\frac{1}{x}\left(1 + 5^{2x}\right)^\frac{1}{x}$$</span>
<span class="math-container">$$=\lim\limits_{x\rightarrow\infty} 5\left(1 + 5^{2x}\right)^\frac{1}{x}$$</span></p>
</blockquote>
<p>But I don't know how to proceed. Can anyone help? Thanks!</p>
| bjorn93 | 570,684 | <p>Hint:
<span class="math-container">$$\left(5^x+5^{3x}\right)^{1/x}=5^3\left(1+5^{-2x}\right)^{1/x}$$</span>
Can you finish now? The limit of the last expression in parentheses is not an indeterminate form now.</p>
|
1,234,661 | <p>I lost my baby Rudin book on real analysis book but I recall a pair of results in homework exercises that he seemed to indicate that there is no "boundary" between convergent and divergent series of positive decreasing terms. One result was that if $a_n$ is positive decreasing, and $\sum_n a_n$ is divergent, then $\sum_n a_n/s_n$ is also divergent where $s_n$ is the $n$th partial sum of the $a_i$. So, since the original series diverges, we can keep repeating this construction to get series that diverge slower and slower, since $\lim_{n \to \infty} s_n = \infty$.</p>
<p>Rudin paired this with another homework exercise result about convergent series, something showing that given any convergent series (possibly of positive decreasing terms) you could construct a new series that converged "slower" in some obvious sense. Can someone recall that result?</p>
| Will Jagy | 10,400 | <p>Never seen it, but I can repeat the standard collection of examples that exhibit this behavior. The series with terms</p>
<p>$$ \frac{1}{n}, $$
$$ \frac{1}{n \log n}, $$
$$ \frac{1}{n \log n \log \log n}, $$
$$ \frac{1}{n \log n \log \log n \log \log \log n}, $$
all diverge (we need to start at $n = n_0$ large enough to get all terms defined)</p>
<p>The series with terms
$$ \frac{1}{n^2}, $$
$$ \frac{1}{n \log^2 n}, $$
$$ \frac{1}{n \log n \left( \log \log n \right)^2 }, $$
$$ \frac{1}{n \log n \log \log n \left( \log \log \log n \right)^2}, $$
all converge.</p>
<p>In both families we are comparing with easy explicit integrals. For the first bunch, divergent, what is the derivative of $\log \log \log \log x?$
Trying to remember second bunch.</p>
<p>There we go, what is the derivative of
$$ \frac{-1}{\log \log \log x}? $$</p>
|
3,684,048 | <p>I understand how to prove that algebraically using implicit differentiation:
<a href="https://i.stack.imgur.com/wvANe.png" rel="nofollow noreferrer">1</a></p>
<p>However, when I hope to gain an understanding through the graphs, I had a hard time wrapping my head around why the transformed lnx function still has the same slope at every point of x as the parent function.</p>
<p><a href="https://i.stack.imgur.com/2Ob5v.png" rel="nofollow noreferrer">Graphs of the 3 functions that I talked about</a></p>
| symplectomorphic | 23,611 | <p>Here's a more general way of asking the question. Let <span class="math-container">$c\neq0$</span>.</p>
<p>In order for <span class="math-container">$f(x)$</span> and <span class="math-container">$g(x):=f(cx)$</span> to have the same slope for all <span class="math-container">$x$</span>, we need <span class="math-container">$f'(x)\equiv g'(x)\equiv c\cdot f'(cx)$</span>. (The <span class="math-container">$\equiv$</span> sign means that these equations need to hold for all values of <span class="math-container">$x$</span>: in order words, these <em>functions</em> must be equal, not just coincide at a particular point <span class="math-container">$x$</span>.)</p>
<p>Which functions <span class="math-container">$h$</span> have the property that <span class="math-container">$h(x)\equiv c\cdot h(cx)$</span>? Or, in other words, <span class="math-container">$h(cx)\equiv c^{-1}\cdot h(x)$</span>?</p>
<p>Geometrically, this last identity says that compressing the graph of <span class="math-container">$h$</span> horizontally by a factor of <span class="math-container">$c$</span> is the same as stretching the graph of <span class="math-container">$h$</span> vertically by a factor of <span class="math-container">$c^{-1}$</span>. </p>
<p>Algebraically, you can see that any function of the form <span class="math-container">$ax^{-1}$</span> solves this functional equation. In fact, any sum of such functions will work, too: we can have <span class="math-container">$h(x)=\sum a_ix^{-1}$</span>.</p>
<p>Note that you can see clearly that no polynomial will work, though: substituting <span class="math-container">$cx$</span> for <span class="math-container">$x$</span> in <span class="math-container">$\sum a_ix^i$</span> for <span class="math-container">$i\geq0$</span> will not allow you to factor <span class="math-container">$c^{-1}$</span> out of each term.</p>
|
102,357 | <p>Let $G=PSL(2,q)$ where $q$ is prime power. What is Aut$(G\times G)$ and Aut$(G\times G\times G)$? Also if $G=A_{n}$ where $A_{n}$ is the alternating group of degree $n$, then what is Aut$(G\times G)$? </p>
<p>Thanks in advance</p>
| Geoff Robinson | 14,450 | <p>In general, if $S$ is a finite non-Abelian simple group, and $E$ is a direct product of $n$ copies of $S$, then ${\rm Out}(E) = {\rm Aut}(E)/E$ is isomorphic to ${\rm Out}(S) \wr S_{n}.$ This is because every minimal normal subgroup of $E$ is isomorphic to $S$ (in fact, is one of the obvious simple direct factors of $E$) and the automorphism group of $S$ permutes the minimal normal subgroups of $E$.
To provide more detail in order to make up for the lack of a reference: The $n$ "obvious" simple direct factors of $E$ are called the components of $E.$ The direct product of $n$ copies of ${\rm Aut}(S)$ obviously sits inside ${\rm Aut}(E).$ Furthermore, the assumed isomorphisms between the $n$ components may be included to show that ${\rm Aut}(S) \wr S_{n}$ embeds in ${\rm Aut}(E).$ On the other hand, the permutation action of ${\rm Aut}(E)$ on the components of $E$ gives a homomorphism from ${\rm Aut}(E)$ to $S_{n}.$
The kernel of this homomorphism is the intersection $K$ of the normalizers of the individual components. Since $E$ contains its centralizer in ${\rm Aut}(E)$, the group $K/E$ is isomorphic to a subgroup of a direct product of $n$ copies of ${\rm Out}(S).$
Hence this establishes that $|{\rm Aut}(E)| \leq | {\rm Aut}(S) \wr S_{n}|.$ But we have the inequality the other way round, so ${\rm Aut}(E) \cong {\rm Aut}(S) \wr S_{n}.$</p>
|
1,742,409 | <blockquote>
<p>Let <span class="math-container">$(Y, \langle .,. \rangle )$</span> be an inner product space with <span class="math-container">$Y=C[0,1]$</span> and <span class="math-container">$$\langle f,g \rangle = \int_0^1 f(x)g(x)dx$$</span></p>
<p>In <span class="math-container">$C[0,1]$</span>, let <span class="math-container">$f(t) =t$</span>, and find <span class="math-container">$g \in f^{\perp}$</span> such that <span class="math-container">$g(0) = 5$</span>.</p>
</blockquote>
<hr />
<p>Am I right in saying <span class="math-container">$f^{\perp}= \{ g \in Y:g \perp f \}$</span>?</p>
<p>So we need to find <span class="math-container">$g$</span> so that <span class="math-container">$ \langle f,g \rangle =0$</span>. If we have <span class="math-container">$g(t)=0$</span> when <span class="math-container">$t=1$</span> and <span class="math-container">$g(t)=5$</span> when <span class="math-container">$0 \le t <1$</span>, and <span class="math-container">$f(t)=1$</span> then would that work?</p>
<p>Because then <span class="math-container">$f(t)g(t)=0$</span> when <span class="math-container">$t=1$</span> and <span class="math-container">$f(t)g(t)=5t$</span> (this is the part I am unsure about) when <span class="math-container">$0 \le t <1$</span>.</p>
<p>Then let the integral of <span class="math-container">$f(t)g(t)$</span> be <span class="math-container">$F(t)$</span>. Then <span class="math-container">$F(t)=0$</span> when <span class="math-container">$t=1$</span> and <span class="math-container">$F(t)=5t^2/2$</span> when <span class="math-container">$0 \le t <1$</span>. So then find <span class="math-container">$F(1)-F(0)=0- 5(0)^2/2=0$</span>?</p>
| Hagen von Eitzen | 39,174 | <p>Your $g$ is not continuous. Also, you will have $\langle f,g\rangle = 5\int_0^1f(x)\,\mathrm dx\ne 0$ in general.</p>
<hr>
<p>If $f=0$, any $g$ will do, for example $g(t)=5$.</p>
<p>Otherwise, pick $a\in(0,1)$ with $f(a)\ne 0$. Wlog. $f(a)>0$. Then for some small $r>0$ we have $f(x)>0$ for all $x\in(a-r,a+r)$. Let $h(x)=\max\{0,1-\frac{|x-a|}r\}$. Then $\langle f,h\rangle >0$. Let $g_0(t)=5$. Then $g = g_0-\frac{\langle f,g_0\rangle}{\langle f,h\rangle}h$ has the desired properties. - The key point was that $h(0)=0$ and $\langle f,h\rangle \ne 0$.</p>
|
1,742,409 | <blockquote>
<p>Let <span class="math-container">$(Y, \langle .,. \rangle )$</span> be an inner product space with <span class="math-container">$Y=C[0,1]$</span> and <span class="math-container">$$\langle f,g \rangle = \int_0^1 f(x)g(x)dx$$</span></p>
<p>In <span class="math-container">$C[0,1]$</span>, let <span class="math-container">$f(t) =t$</span>, and find <span class="math-container">$g \in f^{\perp}$</span> such that <span class="math-container">$g(0) = 5$</span>.</p>
</blockquote>
<hr />
<p>Am I right in saying <span class="math-container">$f^{\perp}= \{ g \in Y:g \perp f \}$</span>?</p>
<p>So we need to find <span class="math-container">$g$</span> so that <span class="math-container">$ \langle f,g \rangle =0$</span>. If we have <span class="math-container">$g(t)=0$</span> when <span class="math-container">$t=1$</span> and <span class="math-container">$g(t)=5$</span> when <span class="math-container">$0 \le t <1$</span>, and <span class="math-container">$f(t)=1$</span> then would that work?</p>
<p>Because then <span class="math-container">$f(t)g(t)=0$</span> when <span class="math-container">$t=1$</span> and <span class="math-container">$f(t)g(t)=5t$</span> (this is the part I am unsure about) when <span class="math-container">$0 \le t <1$</span>.</p>
<p>Then let the integral of <span class="math-container">$f(t)g(t)$</span> be <span class="math-container">$F(t)$</span>. Then <span class="math-container">$F(t)=0$</span> when <span class="math-container">$t=1$</span> and <span class="math-container">$F(t)=5t^2/2$</span> when <span class="math-container">$0 \le t <1$</span>. So then find <span class="math-container">$F(1)-F(0)=0- 5(0)^2/2=0$</span>?</p>
| Travis Willse | 155,629 | <p>The candidate function $g$ you mention is not continuous, i.e., not in $C[0, 1]$.</p>
<p>One simple approach is the following: Choose a simple $1$-parameter family of functions $g$ satisfying the given condition $g(0) = 5$, compute $\langle f, g \rangle$, and then find a parameter value (if one exists) such that the inner product is $0$.</p>
<blockquote class="spoiler">
<p>A natural choice for the family is the affine functions with $y$-intercept $5$, namely, $$g_a(t) := a t + 5 .$$ Now, $$\langle f , g \rangle = \int_0^1 f g \,dt = \int_0^1 t (a t + 5) \,dt .$$ Can you find a value of the parameter $a$ for which this is zero?</p>
</blockquote>
|
360,063 | <p>Let <span class="math-container">$\mathbb{N}$</span> denote the set of positive integers. For any prime <span class="math-container">$p\in\mathbb{N}$</span> let <span class="math-container">$p\mathbb{N} = \{np: n\in \mathbb{N}\}$</span>. Is there a partition <span class="math-container">${\cal P}$</span> of <span class="math-container">$\mathbb{N}\setminus\{1\}$</span> such that for all <span class="math-container">$B \in {\cal P}$</span> and every prime <span class="math-container">$p\in\mathbb{N}$</span> we have <span class="math-container">$|B \cap p\mathbb{N}|=1$</span>?</p>
| Jan Kyncl | 24,076 | <p>Place each <span class="math-container">$n$</span> that is not a prime power into its own <span class="math-container">$B=B(n)$</span>. Then fill the rest of <span class="math-container">$B(n)$</span> with powers of primes that do not divide <span class="math-container">$n$</span>.</p>
|
1,580,270 | <p>Consider the groups $G = \{0,1,2\} = \mathbb Z_3$ and $H = \{a,b,c\}$
given by the following multiplication tables:</p>
<p><a href="https://i.stack.imgur.com/hXgBb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hXgBb.jpg" alt="enter image description here"></a></p>
<p>The first one isn't really multiplication but in my notes it said it doesn't really matter.</p>
<p>So how do I show an isomorphism? The groups have the same size so they can be bijective right? But it just seems so abstract to show if there's an isomorphism... Exactly what do we have to check.</p>
| copper.hat | 27,978 | <p>Suppose that there is an isomorphism $\phi$.</p>
<p>Then you must have $\phi(0 + 0) = \phi(0) * \phi(0)$, so
from the table we must have $\phi(0) = b$.</p>
<p>Now there are only two possibilities for $\phi(1),\phi(2)$.</p>
<p>If we try $\phi(1) = a,\phi(2) = c$, it is easy to check that
$\phi$ is an isomorphism (that is, $\phi(x+y) = \phi(x) * \phi(y)$ and $\phi$ is a bijection).</p>
<p>In fact, if we let $\phi(1) = c,\phi(2) = a$, it is easy to check that
$\phi$ is also an isomorphism.</p>
|
659,254 | <p>Say $X_1, X_2, \ldots, X_n$ are independent and identically distributed uniform random variables on the interval $(0,1)$.</p>
<p>What is the product distribution of two of such random variables, e.g.,
$Z_2 = X_1 \cdot X_2$?</p>
<p>What if there are 3; $Z_3 = X_1 \cdot X_2 \cdot X_3$?</p>
<p>What if there are $n$ of such uniform variables?
$Z_n = X_1 \cdot X_2 \cdot \ldots \cdot X_n$?</p>
| Julian Wergieluk | 731 | <p>If $X_1$ is uniform, then $-\log X_1 \sim \textrm{Exp}(1)$. Therefore,
$$- \log X_1 \dots X_n = -\log X_1 + \dots -\log X_n$$
is a sum of independent exponential random variables and has Gamma distribution with parameters $(n,1)$ and density $g(y) = \frac{1}{(n-1)!} y^{n-1}e^{-y}$ for $y\geq 0$. Let $f$ be the density of the product $X_1 \dots X_n$, then the Jacobi's transformation formula yields
$$ f( h^{-1}(y) ) | \partial h^{-1}(y) | = g(y), $$
with $h(x) = -\log x$ and $h^{-1}(y) = \exp(-y)$. The substitution $y=h(x)$ in the above equation gives
$$ f(x) = \frac{1}{(n-1)!}(-\log x)^{n-1} \, 1_{ (0,1]}(x).$$</p>
|
3,742,517 | <p>Let <span class="math-container">$f(x)=x^3-1$</span>. To approximate the root <span class="math-container">$x^{\star}=1$</span>, we consider the sequence <span class="math-container">$(x_n)$</span> that we get if we apply Newton's method with <span class="math-container">$x_0>0$</span>. Show that the sequence converges to <span class="math-container">$1$</span>.</p>
<p>I have done the following:</p>
<p>I used <span class="math-container">$x_0=0,5$</span> and applied the method and in that way we see that the sequence converges to <span class="math-container">$1$</span>.</p>
<p>Is that correct?</p>
<p>Now I think that we could also do the following:</p>
<p>From Newton's method we get <span class="math-container">$x_{n+1}=x_n-\frac{x_n^3-1}{3x_n^2}=\frac{2x_n^3+1}{3x_n^2}$</span> and we have to show that this sequence converges to <span class="math-container">$1$</span>, or not?</p>
| Robert Israel | 8,508 | <p>You can try doing it in two stages. First solve the problem with the original objective to minimize <span class="math-container">$\sum x_i$</span> and original constraints. Suppose the optimal solution has <span class="math-container">$\sum x_i = s$</span>. Then solve the problem with additional constraint <span class="math-container">$\sum x_i = s$</span> and objective to minimize <span class="math-container">$\sum y_i$</span>.</p>
|
942,651 | <p>Let $n,r,k$ be non negative integers such that $r,k\leq n$.</p>
<p>If $r^{n-r}=k^{n-k}$, when is this true other than $r=k$ ?</p>
<p>For example it is holds for $n=6,r=2,k=4$.</p>
| SPK.z | 171,119 | <p>This equation has many solutions. I have no general form to find all these solutions, but if you're just interested in finding some other values for which it holds, you can do the following:
$$r^{n-r} = k^{n-k} \quad\Longrightarrow\quad 1=\frac{k^{n-k}}{r^{n-r}}=\left(\frac{k}{r}\right)^{n-r}k^{r-k} \quad\Longrightarrow\quad k^{k-r}=\left(\frac{k}{r}\right)^{n-r}.$$
This leads to an expression for $n$, depending on $r$ and $k$:
$$n=r+\frac{\ln{k}}{\ln{k}-\ln{r}}(k-r).$$
You can choose any $r$ and $k$ and the equation will give you $n$. You can make a loop going over many $r$'s and $k$'s and check if the corresponding $n$ is an integer.</p>
<p>For example, in Matlab you could do:</p>
<pre><code>for r = 1:r_end
for k = 1:k_end
n = r + log(k^(k-r)) / (log(k)-log(r));
if mod(n,1) == 0
disp(['r = ', num2str(r), ' , k = ', num2str(k), ' , n = ', num2str(n)])
end
end
end
</code></pre>
<p>This gives many solutions, such as $[r,k,n] = [27,3,39]$, $[49,7,91]$ or $[9,81,153]$.</p>
|
1,134,510 | <p>Regarding My Background I have covered stuff like </p>
<p>1.Single Variable Calculus</p>
<p>2.Multivariable Calculus (Multiple Integration,Vector Calculus etc) (Thomas Finney)</p>
<p>3.Basic Linear Algebra Course (Containing Vector spaces,Linear Transformation)</p>
<p>4.Ordinary Differential Equation</p>
<p>5.Real Analysis (Sequences And series)</p>
<p>I am interested In Number theory and i am big fan of Ramanujan .I have not been through rigorous proofs before . But i want to dig deep into number theory especially area where Ramanujan was working .Anyone researcher,Professor can advice me about which preparations are needed for going into number theory and which books should i need .I will be highly obliged .I am asking this for self study or you can say pursuing research at home .</p>
<p>Note - All stuff i have covered is with help of youtube videos and self study .Thanks </p>
| JMP | 210,189 | <p>There are $\binom{7}{3}$ ways to choose 3 from 7, which is 35. The first letter can be one of 26, the second one of 25 (unless both letters are allowed to be the same!). So the answer is $35*26*25=22750$.</p>
|
152,582 | <p>I have tried to use <code>OpenRead</code> for my application as it appears to be less of a burden on the memory.</p>
<p>I <code>OpenRead</code> a .csv file to extract coordinates in the form {x,y} in the following way</p>
<pre><code>ClearAll[f];
f = OpenRead["mathfile.csv"];
g = ReadList[f, String, RecordLists -> True];
Close[f];
</code></pre>
<p>I can access each record seperately by just calling <code>g[[1]]</code>. However the format is not as numbers and so a function like <code>ListPlot[g[[1]]]</code> doesn't work.</p>
<p>If I use <code>"Number"</code> or <code>"Real"</code> as an option within <code>ReadList</code> I get an error:</p>
<p><strong>"Read:readn: Invalid real number found when reading from 'f'."</strong></p>
<p>Using <code>"String"</code> or <code>"Record"</code> within <code>ReadList</code> doesn't cause a problem. However I would have to convert the <code>"String"</code> or <code>"Record"</code> into a number format.</p>
<p>What am I doing wrong ? I would like to call <code>g[[1]]</code> and have the co-ordinate in a form that can be read by <code>ListPlot</code></p>
<p>Thanks</p>
| ZachB | 1,043 | <p>Mathematica doesn't have configurable separators for <code>Read</code> and <code>ReadList</code>, except when reading <code>Word</code>s and <code>Record</code>s, so importing a CSV of numbers is trickier. Some workarounds:</p>
<ol>
<li><p>Use TSV format, which will "just work."</p></li>
<li><p>Read a <code>Character</code> to get rid of the comma separator:</p>
<pre><code>ReadList[stream, {Number, Character, Number}, RecordLists -> True][[All, {1,3}]]
</code></pre></li>
<li><p>Set <code>WordSeparators -> {","}</code>, the read <code>Word</code>s and apply <code>ToExpression</code>. This is maybe the slowest.</p></li>
</ol>
<hr>
<p>BTW, there are a few nice (albeit maybe hefty) answers in <a href="https://mathematica.stackexchange.com/questions/3890/stream-csv-or-tsv-files?rq=1">this related question</a> that you might want to look at.</p>
|
3,753,839 | <p>Upon inspection, I found out that: <span class="math-container">$3!\cdot5! =6! $</span> and <span class="math-container">$6!\cdot7! =10!.$</span> (couldn't dare to search for more such triplets)</p>
<p>How to check whether there exist more such triplets or not? Please show me a proper way to handle such problems.</p>
| Henry | 6,460 | <p>Your <span class="math-container">$3!\cdot 5! =6!$</span> is an example of the case for general <span class="math-container">$n$</span>:</p>
<p><span class="math-container">$$n! \cdot (n!-1)! = (n!)!$$</span></p>
<p>and when <span class="math-container">$n=4$</span> this gives <span class="math-container">$4! \cdot 23! = 24!$</span></p>
|
3,753,839 | <p>Upon inspection, I found out that: <span class="math-container">$3!\cdot5! =6! $</span> and <span class="math-container">$6!\cdot7! =10!.$</span> (couldn't dare to search for more such triplets)</p>
<p>How to check whether there exist more such triplets or not? Please show me a proper way to handle such problems.</p>
| hamam_Abdallah | 369,188 | <p>For example, you can observe that</p>
<p><span class="math-container">$$(n!-1)!.(n!)=(n!)!$$</span></p>
<p>If <span class="math-container">$ n=3$</span>, it gives</p>
<p><span class="math-container">$$5!.3!=6!$$</span>
for <span class="math-container">$ n=4 $</span>, we get</p>
<p><span class="math-container">$$23!.4!=24!$$</span>
and so on.</p>
|
4,621,390 | <p>I'm studying Linear Algebra and have come to think of a column vector as an ordered bunch of objects, where each object is the product of a scalar and a basis vector, vis:</p>
<p><span class="math-container">$\begin{bmatrix}
a \\
b \\
\vdots \\
\end{bmatrix} = a \hat{i} + b \hat{j} + \dots$</span></p>
<p>Where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are scalars and <span class="math-container">$\hat{i}$</span> and <span class="math-container">$\hat{j}$</span> are <em>unique</em> basis vectors. By that I mean <span class="math-container">$\hat{i}$</span> must not equal <span class="math-container">$\hat{j}$</span>.</p>
<p>Is that correct?</p>
| Seeker | 1,050,393 | <p>That’s not exactly correct. The entries in a column vector come from a linear combination of the basis vectors.</p>
<p>This would be the case if you use the basis vectors <span class="math-container">$$\begin{bmatrix}1 \\ 0 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \end{bmatrix} $$</span> The entries of a vector written as a linear combination of these these vectors would be a scalar multiple of the basis vectors.</p>
<p>Whereas in a basis like <span class="math-container">$$\begin{bmatrix}1 \\ 2 \end{bmatrix}, \begin{bmatrix}3 \\ 4 \end{bmatrix} $$</span> There are vectors whose entries come from a the sum of the scalar multiples of both vectors. Such as <span class="math-container">$$2\begin{bmatrix}1 \\ 2 \end{bmatrix}+ 3\begin{bmatrix}3 \\ 4 \end{bmatrix} = \begin{bmatrix}2 \\ 4 \end{bmatrix}, \begin{bmatrix}9 \\ 12 \end{bmatrix}=\begin{bmatrix}11 \\ 16 \end{bmatrix}$$</span></p>
|
4,651,364 | <p>I am trying to evaluate the following contour integral by evaluating the residue at <span class="math-container">$z=0$</span> of the integrand.</p>
<p><span class="math-container">$$I=\oint_{|z|=1} \frac{z^n+z^{-n}}{2iz(1-rz)(1-rz^{-1})}dz.$$</span></p>
<p>We can manipulate the integrand (which we denote by <span class="math-container">$f(z)$</span>) and ignore the term with positive <span class="math-container">$z$</span> exponent to get</p>
<p><span class="math-container">$$\mathop{\rm Res}_{z=0}f(z) = \mathop{\rm Res}_{z=0} \frac{z^{-n}}{2i(1-rz)(z-r)}.$$</span></p>
<p>I am not quite sure what to do here. I'm not sure how to deal with a pole that isn't just a simple pole. How do I deal with this?</p>
| ecrin | 629,202 | <p>Let <span class="math-container">$x,y\in X$</span>. Since <span class="math-container">$f(X_0)$</span> is dense, there exists two sequence <span class="math-container">$z_n,w_n$</span> in <span class="math-container">$f(X_0)$</span> converging to <span class="math-container">$f(x),f(y)$</span>, respectively.</p>
<p><span class="math-container">$f\colon X_0\to f(X_0)$</span> is an isometry by assumption, hence we can take the preimages of <span class="math-container">$x_n:=f^{-1}(z_n),y_n:=f^{-1}(w_n)$</span> to produce two sequences in <span class="math-container">$X$</span>, which are Cauchy sequences, since <span class="math-container">$z_n,w_n$</span> converges and <span class="math-container">$f$</span> is an isometry on <span class="math-container">$X_0$</span>.</p>
<p>It follows that <span class="math-container">$x_n$</span> and <span class="math-container">$y_n$</span> converges, and in particular they converges to <span class="math-container">$x,y$</span> by continuity.</p>
<p>One concludes remarking that
<span class="math-container">\begin{align}
d_X(x_n,y_n)=d_Y(f(x_n),f(y_n))\implies d_X(x,y)=d_Y(f(x),f(y)).
\end{align}</span></p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| E.H.E | 187,799 | <p>The inequality equivalent to
$$\tan^{-1}(x)< x$$</p>
<p>take the derivative for both sides to get
$$\frac{1}{1+x^2}< 1$$ </p>
|
2,623,924 | <p>My textbook explains the proof, which I don't understand:</p>
<p>"Consider two bases $v_1...v_p$ and $w_1...w_q$ of V. Since the vectors $v$ are linearly independent and the vectors $w$ span V..."</p>
<p>How exactly does $w$ span V? </p>
<p>The book then says the same for vectors $v$, that $v$ spans V and hence $p=q$, but I don't really understand how you can assume that given two sets of vectors that are basis, one set must span V.</p>
| dxiv | 291,201 | <p>Take for example $\,f(x) = g(x) = \dfrac{1}{2}\,$, $\,c=2\,$, then:</p>
<ul>
<li><p>$f(x) \le c g(x)\,$ holds because $\,\dfrac{1}{2} \le 2 \cdot \dfrac{1}{2}$</p></li>
<li><p>$\log\big(f(x)\big) \gt c \cdot \log\big(g(x)\big)\,$ holds because $\log\big(\dfrac{1}{2}\big) \gt 2 \cdot \log\big(\dfrac{1}{2}\big) = \log\big(\dfrac{1}{4}\big)$</p></li>
</ul>
|
1,023,463 | <p>I know $x \sin(1/x)$ is not Lipschitz on $[0,1]$, but some experimentation makes me conjecture that it is $1/2$-Holder. What is a good way to prove this?</p>
| Gaultier | 193,240 | <p>I will sketch the proof that $f(x)=x \sin(x^{-1})$ is 1/2-Holder on $[0,1/2\pi]$.</p>
<p>There are two cases.</p>
<ol>
<li>If $x,y\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$.</li>
</ol>
<p>Write $x= \frac{1}{2\pi n+ \xi}$ and $y=\frac{1}{2\pi n+\zeta}$ where $0 \le \xi,\zeta \le 2\pi$.
There exists a constant $0<c_1$ such that
$$| f(x)- f(y) | \le c_1 \frac{|\xi-\zeta|}{n} \le c_2 \frac{\sqrt{|\xi-\zeta|}}{n} $$
On the other hand
$$ \left| x-y \right| \ge \frac{|\xi-\zeta|}{\left( 2\pi (n+1) \right)^2} $$
So that
$$| f(x)- f(y) | \le c_3 \sqrt{|x-y|} $$</p>
<ol start="2">
<li>Without loss of generality, we can assume $y\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$ and that $x<\frac{1}{2\pi (n+1)}$. </li>
</ol>
<p>Then, by periodicity of sinus, there exists $x'\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$ such that $f(x')=f(x)$ and by 1. we get
\begin{align*}| f(x)- f(y) |=| f(x')- f(y) | &\le c_3 \sqrt{|x'-y|} \\
&\le c_3 \sqrt{|x-y|}
\end{align*}</p>
<p>The exponent $1/2$ is sharp and I guess the best constant is $\sqrt{2}$. </p>
|
300,944 | <p>Show that there are no intergers $x$ and $y$ such that</p>
<p>$P(x,y)=x^2-5y^2=2$</p>
<p>Hint from professor:</p>
<p>Consider the equation in a convenient $\mod (n)$ so that you end up with a polynomial in a single variable. Then proceed as solving number of congruence.</p>
<hr>
<p>Im not sure how to approach this question</p>
<p>Since $P(x,y)=x^2-5y^2=2$</p>
<p>then $x^2-5y^2=0$ $\to$ $x^2=5y^2$</p>
<p>we have $5y^2\equiv0\mod(x)$</p>
<p>then how do I continue..?</p>
<p>Thank you!!</p>
| Math Gems | 75,092 | <p>Following the hint: $\rm\ mod\ 5\!:\ x^2\! - 5y^2 \equiv x^2\in \{0, \pm1, \pm 2\}^2\! \equiv \{0, 1, 4\},\:$ so is not $\equiv 2.$</p>
|
3,159,884 | <p>Prove that if <span class="math-container">$|z+w|=|z-w|$</span> then <span class="math-container">$z\overline{w}$</span> is purely imaginary.</p>
<p>To start off, I said let <span class="math-container">$z=a+bi$</span> and let <span class="math-container">$w=p+qi$</span>. Not sure where to go from here after subbing in those for <span class="math-container">$z$</span> and <span class="math-container">$w$</span>.</p>
| 1123581321 | 444,046 | <p><span class="math-container">$z + w = (a + p) + (b + q)i$</span> </p>
<p><span class="math-container">$|z + w|^2 = (a + p)^2 + (b + q)^2$</span> </p>
<p><span class="math-container">$z - w = (a - p) + (b - q)i$</span> </p>
<p><span class="math-container">$|z - w|^2 = (a - p)^2 + (b - q)^2$</span> </p>
<p><span class="math-container">$a^2 + 2ap + p^2 + b^2 + 2bq + q^2 = a^2 - 2ap + p^2 + b^2 - 2bq + q^2$</span></p>
<p><span class="math-container">$4ap + 4bq = 0$</span></p>
<p><span class="math-container">$ap + bq = 0$</span> [1]</p>
<p><span class="math-container">$z\bar{w} = (a + bi)(p - qi) = (ap + bq) + i(bp - aq) = (bp - aq)i$</span>, which is purely imaginary (using [1])</p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Mozibur Ullah | 26,254 | <p>Seeing how software has infiltrated our lives in so many ways it seems that category theoretic applications to programming is pretty much 'real world' stuff. </p>
<p>Category theory is more geared up to clarifying conceptual structures, so I imagine that there isn't likely to be real world applications in a very direct way soon, and I say this as some-one who likes the general theory. </p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Dan Piponi | 80 | <p>I recently wrote some software to simulate a real world physical system to show that a hardware technology has a chance of doing what we want it to do. That project has grown to a pretty substantial piece of engineering with sizable budget. I wrote the simulation in a programming language whose syntax could be described as the "internal language" of a Cartesian closed category with a bunch of extensions, many of which were categorically motivated.</p>
<p>Sadly I eventually had to switch to Python because I couldn't find libraries for everything I wanted.</p>
<p>The language was Haskell.</p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| Harley Eades | 208,820 | <p>This answer is also very late. There is a recent book by David Spivak titled "Category Theory for the Sciences." Its sole purpose is to connect the "real world" with category theory. You can find an older draft <a href="http://math.mit.edu/~dspivak/CT4S.pdf">here</a> from the authors homepage.</p>
|
148,809 | <p>Let $M$ be a complete Riemannian manifold, does there exists a positive non-constant harmonic function $f \in L^1(M)$? Who can answer me or give me a counter example? Thank you very much!</p>
| Willie Wong | 1,543 | <p>Consider the surface of revolution (so topologically we are dealing with $\mathbb{R}\times\mathbb{S}$) with standard coordinates $(z,\theta)$. Let the metric be
$$ \mathrm{d}s^2 = \mathrm{d}z^2 + h^2(z) \mathrm{d}\theta^2$$
This manifold is clearly complete (it is a warped product of two geodesically complete manifolds). </p>
<p>The Laplace-Beltrami operator associated to it is
$$ \triangle = \frac{1}{h} \partial_z h \partial_z + \frac{1}{h^2} \partial^2_\theta $$
and the volume/area form is $h \mathrm{d}z \mathrm{d}\theta$. </p>
<p>Let $f = f(z)$ be a function. It being $L^1(M)$ is equivalent to
$$ \int_{-\infty}^\infty |f(z)| h(z) \mathrm{d}z < \infty $$
It being harmonic is the same as
$$ h \partial_z f \equiv c $$
for some constant $c$ (which we can assume, WLOG, to be 1). So this implies that we need to find a monotonic function $f$ such that $f / f'$ is absolutely integrable. This requires that $\frac{d}{dz} \log f$ to grow superlinearly in $z$. </p>
<p>So we can consider the following: let $f(z) = \exp (z + z^3)$. Define $h(z) = \frac{1}{(1 + 3z^2) \exp (z + z^3)}$. Then $h f' = 1$ so $f$ is harmonic. On the other hand, $hf = \frac{1}{1+3z^2}$ is integrable in $z$, and hence $f\in L^1(M)$. </p>
<p>Note that the scalar curvature can be computed to be $$ - \frac{2}{h} h''$$ which has fourth order growth in $z$ and so violates the hypotheses of Li's theorem. (In fact, the order of growth of the scalar curvature will be roughly twice that of the growth of $\frac{d}{dz} \log f$. So in this sense the quadratic growth assumption in Li's theorem is sharp.)</p>
|
2,164,823 | <p>Could someone please explain mathematical explanation behind this?</p>
<p>You can win three kinds of basketball points, 1 point, 2 points, and 3 points. Given a total score $n$, print out all the combination to compose $n$.</p>
<p>Examples:<br>
For n = 1, the program should print following:<br>
1 </p>
<p>For n = 2, the program should print following:<br>
1 1<br>
2 </p>
<p>For n = 3, the program should print following:<br>
1 1 1<br>
1 2<br>
2 1<br>
3 </p>
<p>For n = 4, the program should print following:<br>
1 1 1 1<br>
1 1 2<br>
1 2 1<br>
1 3<br>
2 1 1<br>
2 2<br>
3 1 </p>
<p>Algorithm:</p>
<ul>
<li>At first position we can have three numbers 1 or 2 or 3. First put 1 at first position and recursively call for n-1. </li>
<li>Then put 2 at first position and recursively call for n-2. </li>
<li>Then put 3 at first position and recursively call for n-3. </li>
<li>If n becomes 0 then we have formed a combination that compose n, so print the current combination. </li>
</ul>
<p>I've solved in using JS as per below. But I don't quite understand the mathematical reasoning behind it.</p>
<p><a href="https://jsfiddle.net/d2Lft7d1/" rel="nofollow noreferrer">https://jsfiddle.net/d2Lft7d1/</a></p>
| benguin | 121,903 | <p>Consider the set of sequences of points that add up to $n$ and let's call it $S(n)$. There are three different cases to consider regarding the type of sequence we might have.</p>
<ol>
<li><p>The first point value is 1. This can only happen if the sum of the remaining points is $n-1$ so the rest of the subsequence must be an element of $S(n-1)$.</p></li>
<li><p>The first point value is 2. This can only happen if the sum of the remaining points is $n-1$ so the rest of the subsequence must be an element of $S(n-2)$.</p></li>
<li><p>The first point value is 3. This can only happen if the sum of the remaining points is $n-3$ so the rest of the subsequence must be an element of $S(n-3)$.</p></li>
</ol>
<p>These are the only possible cases, each of which is mutually exclusive so the algorithm will print out every possible sequence exactly once.</p>
<p>EDIT: As an example, look at your $n=4$ case, in other words, sequences in $S(4)$. Notice that every sequence is one of the three forms:</p>
<ol>
<li>The number $1$ followed by some sequence in the $n=3$ case since $S(n-1) = S(3).$</li>
<li>The number $2$ followed by some sequence in the $n=2$ case since $S(n-1) = S(2).$</li>
<li>The number $3$ followed by some sequence in the $n=1$ case since $S(n-1) = S(1).$</li>
</ol>
|
4,575,165 | <p>Let <span class="math-container">$\alpha$</span> be an arbitrary positive real number in:
<span class="math-container">$$
F_1 = \int_0^1 x^2 \left[ \int_{-1}^{+1} \frac{e^{-\alpha\sqrt{1+x^2+2xy}}(xy+1)}{(1+x^2+2xy)^{3/2}}dy\right]dx $$</span> <span class="math-container">$$
F_2 = \int_1^\infty x^2 \left[ \int_{-1}^{+1} \frac{e^{-\alpha\sqrt{1+x^2-2xy}}(xy-1)}{(1+x^2-2xy)^{3/2}}dy\right]dx
$$</span>
<b>Prove or disprove that <span class="math-container">$F_1 = F_2$</span></b>.
<br>Source of the problem are the equations (3) and (4) in
<a href="https://www.ias.ac.in/article/fulltext/reso/021/05/0447-0452" rel="nofollow noreferrer"><i>A Paradox of Newtonian Gravitation and Laplace’s Solution</i></a>
by Amitabha Ghosh and Ujjal Dey.
They have done already numerical experiments that seem to confirm equality.
<br>Quote: <i>an analytical proof showing that F1 and F2 are exactly equal will be an interesting mathematical exercise.</i>
And that's it. I have no idea how to proceed.</p>
<h3>Progress so far</h3>
I promised myself not to do numerical experiments.
But the outcome of the inner integral - the one between square brackets - is indeed terrible.
So what else would be an option?
With <a href="https://en.wikipedia.org/wiki/Maple_(software)" rel="nofollow noreferrer">MAPLE</a> 8 some values in the publication can be reproduced.
<pre>
for k from 0 to 10 do
alpha := k*0.1;
g(x,alpha) := int(exp(-alpha*sqrt(1+x^2+2*x*y))*(x*y+1)/(1+x^2+2*x*y)^(3/2),y=-1..1);
F1 := evalf(int(g(x,alpha)*x^2,x=0..1));
F2 := evalf(int(-g(x,alpha)*x^2,x=1..10^3));
end do;
</pre>
The special case <span class="math-container">$\alpha = 0$</span> gives <span class="math-container">$\,F_1=\frac{2}{3}\,$</span> and <span class="math-container">$\,F_2=0\,$</span> exactly.
So it appears that nearby <span class="math-container">$\alpha = 0$</span> the integrals must be unequal.
But MAPLE keeps calculating endlessly for those low values and I had to manually stop the program.
<p><p>
<b><a href="https://www.cantorsparadise.com/richard-feynmans-integral-trick-e7afae85e25c" rel="nofollow noreferrer">Feynman trick</a>.</b>
<span class="math-container">$F_1$</span> and <span class="math-container">$F_2$</span> are both a function of <span class="math-container">$\alpha$</span> . We can take derivatives under the integral sign and see what happens near <span class="math-container">$\alpha=0$</span>.
<span class="math-container">$$
\left.\frac{dF_1}{d\alpha}\right|_{\alpha=0} =
- \int_0^1 x^2 \left[ \int_{-1}^{+1} \frac{xy+1}{1+x^2+2xy}dy\right]dx = -\frac{1}{2} \\
\Longrightarrow \quad dF_1 = -\frac{1}{2}d\alpha
$$</span> <span class="math-container">$$
\left.\frac{dF_2}{d\alpha}\right|_{\alpha=0} =
- \int_1^\infty x^2 \left[ \int_{-1}^{+1} \frac{xy-1}{1+x^2-2xy}dy\right]dx = \infty \\
\Longrightarrow \quad dF_2 = \infty\,d\alpha
$$</span>
Leading to the following heuristics.
<span class="math-container">$F_1(\alpha)$</span> is somewhat decreasing from <span class="math-container">$F_1(0)=2/3$</span> to lower values, but the increase in <span class="math-container">$F_2(0)=0$</span> is infinitely large at that place.
Based upon this, it's impossible to <a href="https://en.wikipedia.org/wiki/Keeping_Up_Appearances" rel="nofollow noreferrer">keep up appearances</a> )-:
we can actually say nothing yet whether the outcome is <span class="math-container">$\,F_1(\alpha) = F_2(\alpha)\,$</span> for all <span class="math-container">$\,\alpha \gt 0\,$</span>.
| Gary | 83,800 | <p>Proof sketch:
<span class="math-container">\begin{align*}
\frac{{{\rm d}^3 F_1 }}{{{\rm d}\alpha ^3 }} & = -\int_0^1 {x^2 \int_{ - 1}^1 {{\rm e}^{ - \alpha \sqrt {1 + x^2 + 2xy} } (xy + 1){\rm d}y} \,{\rm d}x} \\ & = -\frac{{{\rm e}^{ - 2\alpha } (4\alpha ^4 + 14\alpha ^3 + 27\alpha ^2 + 30\alpha + 15)}}{{\alpha ^6 }} - \frac{{3(\alpha ^2 - 5)}}{{\alpha ^6 }}
\end{align*}</span>
and
<span class="math-container">\begin{align*}
\frac{{{\rm d}^3 F_2 }}{{{\rm d}\alpha ^3 }} & = -\int_1^{ + \infty } {x^2 \int_{ - 1}^1 {{\rm e}^{ - \alpha \sqrt {1 + x^2 - 2xy} } (xy - 1){\rm d}y} \,{\rm d}x} \\ & = -\frac{{{\rm e}^{ - 2\alpha } (4\alpha ^4 + 14\alpha ^3 + 27\alpha ^2 + 30\alpha + 15)}}{{\alpha ^6 }} - \frac{{3(\alpha ^2 - 5)}}{{\alpha ^6 }}.
\end{align*}</span>
Thus,
<span class="math-container">$$
\frac{{{\rm d}^3 (F_1 - F_2 )}}{{{\rm d}\alpha ^3 }} = 0,
$$</span>
i.e., <span class="math-container">$F_1-F_2$</span> is a polynomial of degree at most <span class="math-container">$2$</span>. By changing the variable <span class="math-container">$y\to -y$</span> in the definition of <span class="math-container">$F_2$</span>, we see that
<span class="math-container">$$
F_1 - F_2 = \int_0^{ + \infty } {x^2 \int_{ - 1}^1 {\frac{{{\rm e}^{ - \alpha \sqrt {1 + x^2 + 2xy} } ( xy+1)}}{{(1 + x^2 + 2xy)^{3/2} }}{\rm d}y}\, {\rm d}x} .
$$</span>
The right-hand side tends to <span class="math-container">$0$</span> as <span class="math-container">$\alpha\to +\infty$</span>, i.e., the polynomial <span class="math-container">$F_1-F_2$</span> must be the zero polynomial.</p>
|
179,659 | <p>I want to assign each cell in my current notebook a tag so that I can rerun cells with specific tags later in the notebook <a href="https://mathematica.stackexchange.com/a/71333/44420">(Using a method similar to this answer)</a>.</p>
<p>My questions are</p>
<ul>
<li>How do I assign cells in my current notebook tags?</li>
<li>How do I view the tags that I have assigned?</li>
</ul>
| AzJ | 44,420 | <p>For thoose looking at this question in the future if you want to label the first 3 input cells cell tags use the following</p>
<pre><code>{cell1, cell2, cell3} = Cells[CellStyle -> {"Input"}][[1 ;; 3]]
SetOptions[cell1, CellTags -> {"FOO"}]
SetOptions[cell2, CellTags -> {"FOO2"}]
SetOptions[cell3, CellTags -> {"FOO3"}]
(*Check cell tags*)
CurrentValue[{cell1, cell2, cell3}, CellTags]
</code></pre>
|
28,892 | <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p>
<p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p>
<p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p>
<p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p>
<p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p>
<hr>
<p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p>
| user23201 | 23,201 | <p>How about Ravi Agarwal and Donal O'Regan with 445 joint journal publications on MathSciNet?</p>
|
4,272,755 | <p>Calculate the triple integral using spherical coordinates: <span class="math-container">$\int_C z^2dxdydz$</span> where C is the region in <span class="math-container">$R^3$</span> described by <span class="math-container">$1 \le x^2+y^2+z^2 \le 4$</span></p>
<p>Here's what I have tried:</p>
<p>My computation for <span class="math-container">$z$</span> is: <span class="math-container">$\sqrt{1-x^2-y^2} \le z \le \sqrt{4-x^2-y^2}$</span>, as for y I get: <span class="math-container">$-\sqrt{1-x^2}\le y \le \sqrt{4-x^2}$</span> and for x I get: <span class="math-container">$1 \le x \le 2$</span></p>
<p>The triple integral becomes:</p>
<p><span class="math-container">$$\int_{1}^2 dx \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dy \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}z^2dz$$</span></p>
<p>The way I have pictured theta is as so:</p>
<p><a href="https://i.stack.imgur.com/Y1fvn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y1fvn.png" alt="enter image description here" /></a></p>
<p>Where The Red + Green is equal to <span class="math-container">$\frac{\pi}{2}$</span> But because we're only interested in the region from <span class="math-container">$1 \le x \le 2$</span> this covers only <span class="math-container">$\frac{\pi}{4}$</span></p>
<p>The integral becomes:</p>
<p>I presumed <span class="math-container">$\phi$</span> only goes from <span class="math-container">$\frac{\pi}{4}$</span> also, and we know that <span class="math-container">$r$</span> goes from <span class="math-container">$1\le r \le 2$</span></p>
<p>So the final integral becomes:</p>
<p><span class="math-container">$$\int_0^{\frac{\pi}{4}} d\theta \int_0^{\frac{\pi}{4}} \cos^2(\phi)\sin(\phi) \space d\phi \int_1^2 r^4 dr$$</span></p>
<p>Because <span class="math-container">$z^2 = r^2 \cos^2(\phi)$</span></p>
<p>However my answer that I get is <span class="math-container">$2\pi(\sqrt{2}-4)$</span> but the answer should be <span class="math-container">$\frac{124\pi}{15}$</span>. I would greatly appreciate the communities assistance</p>
| Gon | 969,491 | <p>This is the kind of exercise that is perfect to use the spherical coordinates.
Lets start with</p>
<p><span class="math-container">$f(x,y,z)=z^2 \rightarrow f(r,\theta,\phi)=(rcos(\phi))^2=r^2cos^2(\phi)$</span></p>
<p><span class="math-container">$C=\{(x,y,z)\in\mathbb{R^3} | 1\leq x^2+y^2+z^2\leq4\} \rightarrow C=\{(r,\theta,\phi)\in\mathbb{R^3} | 1\leq r^2\leq4\}$</span></p>
<p>In fact <span class="math-container">$\theta\in[0,2\pi]$</span> and <span class="math-container">$\phi\in[0,\pi]$</span> and <span class="math-container">$r\in[1,2]$</span>.
With this I mean that <span class="math-container">$\phi$</span> is the angle between the vector <span class="math-container">$(x,y,z)$</span> and positive semi axis <span class="math-container">$Oz$</span> and <span class="math-container">$\theta$</span> is the angle between the projection of <span class="math-container">$(x,y,z)$</span> in <span class="math-container">$xOy$</span> and positive semi axis <span class="math-container">$Ox$</span>.</p>
<p>Therefore the integral you want to calculate is</p>
<p><span class="math-container">$\int_0^\pi\int_0^{2\pi}\int_1^2r^2cos^2(\phi)r^2sin(\phi)drd\theta d\phi$</span></p>
<p>Ps: What does the red and green areas mean? In fact that area is <span class="math-container">$\frac{\pi2^2}{4}=\pi$</span>. How did you get the red area is half of the red+green area?
The domain of the integral is the space between the spheres. I post it so it is a bit more clear what the angles are.
<a href="https://i.stack.imgur.com/peqMZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/peqMZ.png" alt="enter image description here" /></a></p>
|
606,843 | <p><strong>Definitions:</strong></p>
<ol>
<li><p>$[x]$ is the integer value of $x$. For example: $[4.3]=4$.</p></li>
<li><p>$\{x\}=x-[x]$.</p></li>
</ol>
<p>Can someone help me calculate the derivative of the following functions, and determine where the derivative are not defined?</p>
<ol>
<li><p>$[x^2]\sin^2(\pi x)$</p></li>
<li><p>$\{x^2\}\sin^2(\pi x)$</p></li>
</ol>
<p>Thank you!!</p>
| Parth Thakkar | 70,311 | <p>For the first one:</p>
<p>Consider 2 intervals $ A = [ \sqrt n, \sqrt{n+1} )$ and $ B = [ \sqrt {n+1}, \sqrt{n+2} )$. Of course, $n$ is an integer.</p>
<p>Now, in the first interval, your function becomes $ n\sin^2(\pi x) $. It's derivative is $ n\pi\sin(2\pi x) $. Call it $ f'_1(x) $.</p>
<p>Consider the second interval. Your function becomes $ (n+1)\sin^2(\pi x) $. It's derivative is $ (n+1)\pi\sin(2\pi x) $. Call it $ f'_2(x) $</p>
<p>Clearly it is differentiable <em>within</em> each interval. We now have to check if it is differentiable even at the point $ x = \sqrt {n+1} $. </p>
<p>For that, we must have $ \lim _{x\to(\sqrt{n+1})^-} f'_1(x) = \lim _{x\to(\sqrt{n+1})^+} f'_2(x) $.</p>
<p>That is, we must have:
$ n\pi\sin(2\pi \sqrt{n+1}) = (n+1)\pi\sin(2\pi \sqrt{n+1}) $</p>
<p>Which will be true when $ \sqrt{n+1} $ is an integer. In other words, it is differentiable when $n+1$ is a perfect square.</p>
<hr>
<p>So, your function is differentiable everywhere, except at those integers which are not perfect squares. Note, here our intervals consisted only of positive integers, but a simple sign change will give the same result for negative integers.</p>
<p>I think the same logic can be applied to the second function.</p>
|
1,808,881 | <p>I am struggling with finding all roots of unity in $\mathbb{Q}(i)$. I know that if $a+bi$ is a root of unity in $\mathbb{Q}(i)$, then $a^2+b^2=1$, and I know how to find all $a, b \in \mathbb{Q}$ that satisfy that equation. However, I do not know how to filter out the roots of unity. I think there should be a somewhat easier way, for example looking at cyclotomic polynomials. Any help would be very nice.</p>
| Eric Towers | 123,905 | <p>This might help:</p>
<p>Dresden, G., "<a href="http://home.wlu.edu/~dresdeng/papers/rattan.pdf" rel="nofollow">A New Approach to Rational Values of Trigonometric Functions</a>", (preprint).</p>
<p>This paper attacks: "for $c,d$ relatively prime integers (with $d > 0$), the primitive $d^\text{th}$ root of unity $\mathrm{e}^{2\pi \mathrm{i}c/d}$ has degree $\leq 2$ over $\Bbb{Q}$ iff $d \in \{1,2,3,4,6\}$."</p>
|
4,516,356 | <p>In Ch. 20 of Spivak's <em>Calculus</em>, he shows that the remainder terms for <span class="math-container">$\arctan$</span> and <span class="math-container">$\log{(1+x)}$</span> become large with the order of the Taylor polynomial used to approximate these functions. Thus these approximations</p>
<blockquote>
<p>are of no use whatsoever in computing <span class="math-container">$\arctan{x}$</span> and <span class="math-container">$\log{(1+x)}$</span>.
This is no tragedy, <strong>because the values of these functions can be
found for any <span class="math-container">$x$</span> once they are known for all <span class="math-container">$x$</span> with <span class="math-container">$|x|<1$</span></strong>.</p>
</blockquote>
<p>How do we find the values of these functions for any <span class="math-container">$x$</span> if we know the functions for <span class="math-container">$|x|<1$</span>?</p>
| Suzu Hirose | 190,784 | <p>With base ten logarithms, we have <span class="math-container">$\log_{10}(1234)=4+\log_{10}(0.1234)$</span>, and in general,
<span class="math-container">$$\log_{10} x=n+\log_{10} (x/10^n),$$</span>
where <span class="math-container">$n$</span> is the number of digits of <span class="math-container">$x$</span> before the decimal point. Natural logarithms are just this multiplied by a constant: <span class="math-container">$\log_e(x)=\log_{10}(x) / \log_{10}(e)$</span>.</p>
<p>For tangents,
<span class="math-container">$$
\begin{align}
\tan(90^\circ-x)&={\sin (90^\circ-x)\over\cos (90^\circ-x)}\\
&={\cos x\over \sin x}\\
&={1\over \tan x}
\end{align}
$$</span>
All the values of <span class="math-container">$\tan$</span> from <span class="math-container">$0$</span> to <span class="math-container">$90^\circ$</span> are known once the values are known up to <span class="math-container">$45^\circ$</span>, so <span class="math-container">$$\arctan (1/x)=90^\circ-\arctan(x).$$</span> Thus you only need to compute values from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>.</p>
|
685,918 | <p>I'm doing exercises in Real Analysis of Folland and got stuck on this problem. I don't know how to calculate limit with the variable on the upper bound of the integral. Hope some one can help me solve this. I really appreciate.</p>
<blockquote>
<blockquote>
<p>Show that $\lim\limits_{k\rightarrow\infty}\int_0^kx^n(1-k^{-1}x)^k~dx=n!$</p>
</blockquote>
</blockquote>
<p>Thanks so much for your consideration.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\down}{\downarrow}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Hereafter we'll use the <a href="http://en.wikipedia.org/wiki/Beta_function" rel="nofollow">Beta Function</a>
$\ds{{\rm B}\pars{x,y} = \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$
with $\ds{\pars{~\Re\pars{x} > 0\,,\Re\pars{y} > 0~}}$ and its property
$\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$.
$\ds{\Gamma\pars{z}}$ is the <a href="http://en.wikipedia.org/wiki/Gamma_function" rel="nofollow">Gamma Function</a>. For $n\ \in\ {\mathbb N}$, it
satisfies $\ds{\Gamma\pars{n + 1} = n!}$.</p>
<blockquote>
<p>\begin{align}
&\lim_{k \to \infty}\int_{0}^{k}x^{n}\pars{1 -k^{- 1}x}^{k}\,\dd x=
\lim_{k \to \infty}\bracks{k^{n + 1}\int_{0}^{1}x^{n}\pars{1 - x}^{k}\,\dd x}
\\[3mm]&=\lim_{k \to \infty}\bracks{k^{n + 1}{\rm B}\pars{n + 1,k + 1}}
=\lim_{k \to \infty}\bracks{%
k^{n + 1}\,{\Gamma\pars{n + 1}\Gamma\pars{k + 1}\over \Gamma\pars{n + k + 2}}}
\\[3mm]&=n!\lim_{k \to \infty}\bracks{{k^{n + 1}k! \over \pars{n + k + 1}!}}
\tag{1}
\end{align}</p>
</blockquote>
<p>However,
\begin{align}
&\lim_{k \to \infty}\bracks{k^{n + 1}k! \over \pars{n + k + 1}!}
=\lim_{k \to \infty}
\bracks{k^{n + 1} \over \pars{k + n + 1}\pars{k + n}\ldots\pars{k + 1}}
\\[3mm]&=
\lim_{k \to \infty}\bracks{%
\pars{1 + {n + 1 \over k}}\pars{1 + {n \over k}}\ldots\pars{1 + {1 \over k}}}^{-1}
= 1
\end{align}</p>
<blockquote>
<p>By replacing this result in $\pars{1}$, we get:
$$\color{#00f}{\large%
\lim_{k \to \infty}\int_{0}^{k}x^{n}\pars{1 -k^{- 1}x}^{k}\,\dd x
=n!}
$$</p>
</blockquote>
|
226,449 | <p>Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1 $; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:</p>
<blockquote>
<p>If $A$ is an associative and commutative ring, then we can define an
unary operation on the set of all the finite subsets of our ring,
denoted by $+ \left(A\right) $ and $\times \left(A\right)$. While it
is intuitive to define $+ \left( \emptyset \right) =0$, why should
the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?</p>
</blockquote>
| GEdgar | 442 | <p>In general, we want the "associative" law to hold in the form:
$$
\times(A \cup B) = (\times A) (\times B)
$$
whenever $A$ and $B$ are disjoint. What would this mean when $B = \emptyset$ ??</p>
|
226,449 | <p>Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1 $; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:</p>
<blockquote>
<p>If $A$ is an associative and commutative ring, then we can define an
unary operation on the set of all the finite subsets of our ring,
denoted by $+ \left(A\right) $ and $\times \left(A\right)$. While it
is intuitive to define $+ \left( \emptyset \right) =0$, why should
the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?</p>
</blockquote>
| Michael Hardy | 11,667 | <p>Multiplying by $1$ is the same as not multiplying by anything at all.</p>
|
2,464,641 | <p>Let we have an equation $x^2+4x+2$ and we want to divide it with $x-4$. By remainder theorem it gives the value of $34$. But when we put the value of $x=6$ in equations it gives zero remainder. And quotient is $31$. How's that? Where I am mistaking?</p>
| Sagar Chand | 477,029 | <p>You are getting remainder 34. This is because you haven't placed value of x yet. If you place value of x, you will see that divisor is (6-4) = 2. Now obviously you can't get remainder of 34 when divisor is 2.(remainder can't be greater than divisor) You get remainder 0.</p>
<p>The equation you solved earlier was general case. So if x had been any other number greater than 38(so that divisor is greater than 34), then remainder would have been 34. Otherwise you would have to divide remaining term with divisor</p>
|
2,464,641 | <p>Let we have an equation $x^2+4x+2$ and we want to divide it with $x-4$. By remainder theorem it gives the value of $34$. But when we put the value of $x=6$ in equations it gives zero remainder. And quotient is $31$. How's that? Where I am mistaking?</p>
| Francisco José Letterio | 482,896 | <p>$x^2 + 4x +2=(x+8)(x-4)+34$. I think you're mistaking what dividing a polynomial means</p>
|
1,851,084 | <p>I have to solve the following problem:
find the matrix $A \in M_{n \times n}(\mathbb{R})$ such that:
$$A^2+A=I$$ and $\det(A)=1$.
How many of these matrices can be found when $n$ is given?
Thanks in advance.</p>
| Mark Fischler | 150,362 | <p>Let the eigenvalues of $A$ be $\{e_i\}$ and let $P$ be a similarity transformation that diagonalizes $A$. Then $P$ also diagonalizes $A^2$ and the equation becomes
$$
\forall i: e_i^2 + e_i = 1
$$</p>
<p>Since there are two real roots to that equation, there are $2^n$ candidates for $A$, before imposing that the determinant is one. </p>
<p>When we impose that the diagonal is one, we see that the eigenvalues have to come in pairs
$$
e_i = \frac{-1-\sqrt{5}}{2}. e_{i'} = \frac{-1+\sqrt{5}}{2}.
$$
and that there must be an even number of such pairs. So no such $A$ exists, unless $N=4k$ for$k\in\Bbb{Z}^+$. For a given such $N$ there are
$$
\binom{N}{N/2}
$$
possible diagonal determinant one $N\times N$ matrices satisfying the equation.</p>
<p>However, if $A$ satisfies the equation, then so does any similar matrix
$$
p^{-1}AP
$$
so as long as $N=4k$ there are a continuum of solutions for $A$.</p>
|
377,266 | <p>My question is very direct:</p>
<blockquote>
<p>What are the motivations for the name "jet"(subjet, superjet) in the context of viscosity solutions for second order fully nonlinear elliptic PDE?</p>
</blockquote>
<p>The definition of which can be seen in Crandall, Ishii, Lions:</p>
<p><em>Crandall, Michael G.; Ishii, Hitoshi; Lions, Pierre-Louis</em>, <a href="http://dx.doi.org/10.1090/S0273-0979-1992-00266-5" rel="nofollow noreferrer"><strong>User’s guide to viscosity solutions of second order partial differential equations</strong></a>, Bull. Am. Math. Soc., New Ser. 27, No. 1, 1-67 (1992). <a href="https://zbmath.org/?q=an:0755.35015" rel="nofollow noreferrer">ZBL0755.35015</a>.</p>
<p>see also <a href="https://arxiv.org/pdf/math/9207212.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/math/9207212.pdf</a>.</p>
| Mozibur Ullah | 35,706 | <p>The machinery of Jets were introduced by Ehresmann as an geometrically invariant way of describing PDEs and their solutions in analogy to the way that vector fields and integral curves invariantly describe ODEs and their solutions.</p>
<p>For example, there is a way of geometrically describing symmetries here and so Noether's theorem.</p>
<p>It's worth noting that sprays describe 2nd order ODEs and are a sort of midway house. It comes from iterating tangent bundles - but this doesn't appear to generalise well - at least I haven't seen any generalisations of such, except in passing. Whereas the notion of a jet generalises the equivalence class of curves of a certain tangent order.</p>
|
1,356,932 | <blockquote>
<p><strong>Problem.</strong> Let $h\in C(\mathbb{R})$ be a continuous function, and let
$\Phi:\Omega:=[0,1]^{2}\rightarrow\mathbb{R}^{2}$ be the map defined
by \begin{align*}
\Phi(x_{1},x_{2}):=\left(x_{1}+h(x_{1}+x_{2}),x_{2}-h(x_{1}+x_{2})\right) \tag{1}
\end{align*}
What is the (Lebesgue) measure (denoted by $\left|\cdot\right|$) of the set $\Phi(\Omega)$?</p>
</blockquote>
<p>Implicit in the problem statement is that $\Phi(\Omega)$ is Lebesgue measurable, but this is obvious as $\Phi$ is the composition of continuous functions, so $\Phi(\Omega)$ is a compact subset of $\mathbb{R}^{2}$.</p>
<p>If we knew that $h$ was $C^{1}$, then the Jacobian of $\Phi$ is
\begin{align*}
J\Phi(x_{1},x_{2})=\begin{bmatrix} {1+h'(x_{1}+x_{2})} & {h'(x_{1}+x_{2})} \\ {-h'(x_{1}+x_{2})} & {1-h'(x_{1}+x_{2})} \end{bmatrix}
\end{align*}
which has determinant $1$ everywhere. Whence,
\begin{align*}
\left|\Phi(\Omega)\right|=\int_{\mathbb{R}^{2}}\chi_{\Phi(\Omega)}(x_{1},x_{2})\mathrm{d}x_{1}\mathrm{d}x_{2}=\int_{\mathbb{R}^{2}}\chi_{\Omega}(x_{1},x_{2})\left|J\Phi(x_{1},x_{2})\right|\mathrm{d}x_{1}\mathrm{d}x_{2}=1
\end{align*}</p>
<p>My thought was to approximate $h$ uniformly in a neighborhood of $[-2,2]$ by a smooth function $g$, so that $\left|h(x_{1}+x_{2)}-g(x_{1}+x_{2})\right|<\epsilon$ for all $(x_{1},x_{2})\in\Omega$, given $\epsilon>0$. Denote the analogue of $\Phi$ with $g$ instead of $h$ by $\Phi_{\epsilon}$. One can verify that</p>
<p>$$\forall x=(x_{1},x_{2})\in\Omega,\quad \left|\Phi(x)-\Phi_{\epsilon}(x)\right|=\sqrt{2}\left|h(x_{1}+x_{2})-g(x_{1}+x_{2})\right|<\sqrt{2}\epsilon$$</p>
<p>So $\Phi(\Omega)\subset U_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ and $\Phi_{\epsilon}(\Omega)\subset V_{\epsilon}:=\left\{y\in\mathbb{R}^{2} : d(y,\Phi(\Omega))<\sqrt{2}\epsilon\right\}$.</p>
<p>We obtain that
\begin{align*}
1=\left|\Phi_{\epsilon}(\Omega)\right|\leq\int_{\mathbb{R}^{2}}\chi_{V_{\epsilon}}(x)\mathrm{d}x
\end{align*}
Since $\Phi(\Omega)$ is compact, in particular closed, $\chi_{V_{\epsilon}}\rightarrow\chi_{\Phi(\Omega)}$ a.e. as $\epsilon\downarrow 0$. From monotone convergence, we obtain that
\begin{align*}
1\leq\left|\Phi(\Omega)\right| \tag{2}
\end{align*}</p>
<p>My issue is with establishing the reverse inequality. I know that
\begin{align*}
\left|\Phi(\Omega)\right|\leq\left|U_{\epsilon}\right|=1+\left|\left\{x\in\mathbb{R}^{2}:0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}\right|
\end{align*}
But I do not know how to control the measure of set $\left\{x\in\mathbb{R}^{2}: 0<d(x,\Phi_{\epsilon}(\Omega))<\sqrt{2}\epsilon\right\}$ as $\epsilon\downarrow 0$, as $\Phi(\Omega)\setminus\Phi_{\epsilon}(\Omega)$ is contained in this set for all $\epsilon>0$.</p>
| Disintegrating By Parts | 112,478 | <p>Perform the Gram-Schmidt process on $\{1,x,x^{2},\cdots\}$ to obtain an orthonormal sequence of polynomials $\{p_0,p_1,p_2,\cdots\}$, where $p_n$ is a polynomial of order $n$. Note that $p_n$ is orthogonal to $1,x,\cdots,x^{n-1}$. This sequence must be a complete orthonormal sequence because the polynomials are dense in $L^{2}[0,1]$.</p>
<p>If $g$ is a unit vector in $L^{2}$ for which $(g,x^{2})$ is maximized and $(g,x)=(g,1)=0$, then
$$
g = \sum_{k=0}^{\infty}(g,p_k)p_k=\sum_{k=2}^{\infty}(g,p_k)p_k, \\
1=\|g\|^{2}=\sum_{k=2}^{\infty}|(g,p_k)|^{2} \\
(g,x^{2})=\sum_{n=2}^{\infty}(g,p_k)(p_k,x^{2})=(g,p_2)(p_2,x^{2}).
$$
It follows that $(g,x^{2})$ is maximized when $g=p_2$ or $g=-p_2$, and that maximum value is $|(p_2,x^{2})|$. This is because $|(g,p_2)| \le \|g\|\|p_2\|=1$ and $(p_2,p_2)=1$.
Hence, $p_2$ is determined by Gram-Schmidt:
\begin{align}
p_0 &= 1,\\
p_1 &= \frac{x-(x,1)1}{\|x-(x,1)1\|} =\frac{x-1/2}{\|x-1/2\|}=2\sqrt{3}(x-1/2)\\
\pm g=p_2 &= \frac{x^{2}-(x^{2},p_1)p_1-(x^{2},p_0)p_0}{\|x^{2}-(x^{2},p_1)p_1-(x^{2},p_0)p_0\|} \\
& = \frac{x^{2}-12\int_{0}^{1}y^{2}(y-1/2)dy(x-1/2)-\int_{0}^{1}y^{2}dy}{\|x^{2}-12\int_{0}^{1}y^{2}(y-1/2)dy(x-1/2)-\int_{0}^{1}y^{2}dy\|} \\
% & = \frac{x^{2}-12(1/4-1/6)(x-1/2)-1/3}{\|x^{2}-12(1/4-1/6)(x-1/2)-1/3\|} \\
% & = \frac{x^{2}-(x-1/2)-1/3}{\|x^{2}-(x-1/2)-1/3\|}
& =\frac{x^{2}-x+1/6}{\|x^{2}-x+1/6\|}.
\end{align}
Because
$$
(p_2,x^{2})=\frac{(x^{2}-x+1/6,x^{2})}{\|x^{2}-x+1/6\|}=\frac{1/5-1/4+1/18}{\|x^{2}-x+1/6\|} > 0,
$$
then $g=\|p_2\|^{-1}p_2$ <em>maximizes</em> $(g,x^{2})$ subject to $(g,1)=(g,x)=0$.</p>
|
4,035,080 | <p>Let <span class="math-container">$E$</span> be a Polish space, and let <span class="math-container">$\mu$</span> be a measure on <span class="math-container">$E$</span>. Define the following properties:</p>
<ul>
<li><span class="math-container">$E$</span> is <span class="math-container">$\sigma$</span>-<em>compact</em> if <span class="math-container">$E$</span> is the countable union of compact sets.</li>
<li><span class="math-container">$E$</span> is <em>locally compact</em> if every <span class="math-container">$x \in E$</span> has an open neighborhood <span class="math-container">$U$</span> whose closure is compact.</li>
<li><span class="math-container">$\mu$</span> is <em>locally finite</em> if for every <span class="math-container">$x \in E$</span>, there is an open set <span class="math-container">$U \subset E$</span> containing <span class="math-container">$x$</span> with <span class="math-container">$\mu(U) < \infty$</span>.</li>
<li><span class="math-container">$\mu$</span> is a <em>Borel measure</em> if for every compact <span class="math-container">$K \subset E$</span>, we have <span class="math-container">$\mu(K) < \infty$</span>.</li>
</ul>
<p>Clearly a locally finite measure is Borel. And if <span class="math-container">$E$</span> is locally compact (not even necessarily separable or complete), Borel measures are necessarily locally finite. But are there more general conditions for which Borel measures are locally finite?</p>
<p><strong>Question:</strong> If <span class="math-container">$E$</span> is a <span class="math-container">$\sigma$</span>-compact Polish space and <span class="math-container">$\mu$</span> is a Borel measure, is <span class="math-container">$\mu$</span> locally finite?</p>
<p>I can’t think of a counter example to this, but I’m having trouble proving it. My original strategy was to prove that a <span class="math-container">$\sigma$</span>-compact Polish space is locally compact. However, as the comments demonstrate, <span class="math-container">$\sigma$</span>-compact Polish spaces are not necessarily locally compact, so that strategy doesn’t work. A counter example is the subset <span class="math-container">$X \subset \ell^2$</span> given by the union of the lines <span class="math-container">$X_k = \{\lambda e_k : \lambda \in \mathbb R\}$</span>, where <span class="math-container">$\{e_k\}_{k \geq 1}$</span> is the standard orthonormal basis of <span class="math-container">$\ell^2$</span>. Then <span class="math-container">$X$</span> is <span class="math-container">$\sigma$</span>-compact, but not locally compact (at the origin specifically).</p>
<p>But I’m not sure of a measure <span class="math-container">$\mu$</span> on this space that is Borel, but not locally finite; the problem is there are compact subsets of <span class="math-container">$X$</span> containing <span class="math-container">$0$</span> and intersecting infinitely many of the <span class="math-container">$X_k$</span>. Can anyone think of a counter example?</p>
| Keen-ameteur | 421,273 | <p>Following the answer I wrote in <a href="https://math.stackexchange.com/questions/4633639/when-is-a-borel-measure-locally-finite/4635045?noredirect=1#comment9780717_4635045">your follow up thread</a>, I would like to try and propose a counter example for this more restrictive question. I think this is in the same spirit as @Henno Brandsma's answer.</p>
<p>Consider the space of <a href="https://topology.pi-base.org/spaces/S000112/properties" rel="nofollow noreferrer">nested rectangles in the real plane</a>. This is a subspace of <span class="math-container">$\mathbb{R}^2$</span> with the induced topology, consisting of <span class="math-container">$L_1=\{x=1\}$</span>, <span class="math-container">$L_2=\{ x=-1 \}$</span> and a sequence of boundary of rectangles <span class="math-container">$R_n=\partial B_n$</span>, given by <span class="math-container">$B_n=\{ -\frac{n}{n+1} \leq x\leq \frac{n}{n+1} \}\cap \{ -n \leq y\leq n \}$</span>. This is a Polish space which is <span class="math-container">$\sigma$</span>-compact but not locally compact.</p>
<p>Define the measure <span class="math-container">$\mu=\sum_{n=1}^\infty \delta_{ R_n}$</span>. Any neighborhood of a point on <span class="math-container">$L_1$</span> has to intersect infinitely many <span class="math-container">$\partial R_n$</span>, and so such a neighborhood would have infinite measure.</p>
<p>I think that any set intersecting <span class="math-container">$L_1$</span> or <span class="math-container">$L_2$</span> nontrivially would not be compact. So any compact subset in the space would have to intersect finitely many of the <span class="math-container">$R_n$</span>, because of the topology induced from <span class="math-container">$\mathbb{R}^2$</span>.</p>
|
3,104,051 | <p>I have the work of the proof done, but at the end after showing
<span class="math-container">$3^{2(n+1)} - 1=9(3^{2n} - 1)+8$</span>
I make the statement that since <span class="math-container">$9(3^{2n} - 1)$</span> is a multiple of 8 and 8 is a multiple of 8 then <span class="math-container">$3^{2n} - 1$</span> is a multiple of 8. I need to quote the facts about divisibility I use in this statement and I'm not sure exactly what to put for that, I have already included the definition of divisibility in the proof</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Note that <span class="math-container">$$3^{2n}-1=9^n-1$$</span> and <span class="math-container">$$9\equiv 1 \mod 8$$</span></p>
|
231,360 | <p>Let $A,B\subseteq\mathbb R^d$ with $A$ closed and $B$ open and such that $A\cap B\neq\emptyset$. Assume that there exists a sequence $(x_k)_{k\in\mathbb N}\subseteq\mathbb R^d$ conveging in Euclidean norm to some $x\in A\cap B$. Do we have $x_k\in A\cap B$ for some $k\in\mathbb N$? If $x$ is an interior point of $A$ then of course this is true, since $B$ is an open set and $A$ then contains a ball around $x$ of size $\varepsilon>0$. So the question becomes problematic if $x\in\partial A$ (boundary of $A$).</p>
| P.. | 39,722 | <p>For a counterexample take $A=[-1,0], \ B=(-1,1) \subseteq \mathbb{R}$ and $x_n=\frac{1}{n}$. </p>
<p>This sequence $(x_n)\to 0 \in A \cap B$ but $x_n \not \in (-1,0]=A \cap B, \ \forall n \in \mathbb{N} $ </p>
|
231,360 | <p>Let $A,B\subseteq\mathbb R^d$ with $A$ closed and $B$ open and such that $A\cap B\neq\emptyset$. Assume that there exists a sequence $(x_k)_{k\in\mathbb N}\subseteq\mathbb R^d$ conveging in Euclidean norm to some $x\in A\cap B$. Do we have $x_k\in A\cap B$ for some $k\in\mathbb N$? If $x$ is an interior point of $A$ then of course this is true, since $B$ is an open set and $A$ then contains a ball around $x$ of size $\varepsilon>0$. So the question becomes problematic if $x\in\partial A$ (boundary of $A$).</p>
| fgp | 42,986 | <p>Let $x_i = (1 + \frac{1}{i},0)$, let $A$ be the closed unit ball and $B$ some small open ball around $(1,0)$. $(x_i)_{i\in\mathbb{N}}$ the converges to $(1,0) \in A \cap B$, yet none of the $x_i$ lie in $A$ and hence not in $A \cap B$. The answer is therefore no.</p>
|
3,390,979 | <blockquote>
<p>Let <span class="math-container">$\sum_i^na_i=n$</span>, <span class="math-container">$a_i>0$</span>. Then prove that <span class="math-container">$$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$</span></p>
</blockquote>
<p>I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The fourth power in the LHS really evades me, and I struggle to see what can be done.</p>
<p>My attempts didn’t lead me to any result ... Simply cauchy , where <span class="math-container">$a_i=x,$</span> <span class="math-container">$b_i=1$</span> to find an inequality involving <span class="math-container">$\sum x^2$</span> . I also tried finding an inequality involving <span class="math-container">$\sum x^3$</span> using <span class="math-container">$a_i=\frac{x^3}{2}$</span> and <span class="math-container">$b_i=x^{\frac{1}{2}}$</span></p>
| Batominovski | 72,152 | <p><strong>A Hint for an Alternative Solution.</strong></p>
<p>We want to show that <span class="math-container">$$\left(\frac{x^3+1}{x^2+1}\right)^4\geq 2x-1$$</span> for every <span class="math-container">$x\in\mathbb{R}$</span>. By the AM-GM Inequality,
<span class="math-container">$$\left(\frac{x^3+1}{x^2+1}\right)^4+1\geq 2\,\left(\frac{x^3+1}{x^2+1}\right)^2\,.$$</span>
Hence, it suffices to verify that
<span class="math-container">$$\left(\frac{x^3+1}{x^2+1}\right)^2\geq x$$</span>
for all <span class="math-container">$x\in\mathbb{R}$</span>. This part is left for the OP. </p>
<p><strong>Remark.</strong> From this solution, we need not require that <span class="math-container">$a_1,a_2,\ldots,a_n$</span> be positive. That is, for any real numbers <span class="math-container">$a_1,a_2,\ldots,a_n$</span> such that <span class="math-container">$\sum\limits_{i=1}^n\,a_i=n$</span>, we always have
<span class="math-container">$$\sum_{i=1}^n\,\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n\,.$$</span>
However, the sole equality case is when <span class="math-container">$a_1=a_2=\ldots=a_n=1$</span>.</p>
|
2,329,600 | <p>I haven't studied any maths since I was at university 20 years ago. Yesterday, however, I came across a pair of equations in an online article about gaming and I couldn't understand how they'd been derived. </p>
<p>Here's the scenario. If we make a single trial of generating a number between 1 and 20, there's an even 5% chance of getting any given number. Therefore to get "at least", say, a 11, you can just add up the percentages and subtract them from 100: 100 - (5% x 10) (because 1 is the minimum, not zero) = 50%. </p>
<p>What happens, though, if you make two trials and take either the highest or the lowest number? How then do you calculate the chance of getting "at least" a certain number? For this, I was given the following equations:</p>
<ul>
<li>If taking the highest, it's <em>1-(1-P)^2</em></li>
<li>If taking the lowest it's <em>P^2</em> </li>
</ul>
<p>It's clear these are correct. Rolling "at least" an 11 when taking the lowest values is therefore 0.5 * 0.5 ... 0,25. But what I want to understand is how someone arrived at these equations without using trial and error.</p>
<p>From the limited maths I can recall, P^2 looks not unlike the binomial distribution formula of μ = np - but of course it's raising the probability to the power of two rather than multiplying by two for two trials.</p>
<p>Can someone please explain to me where these come from?</p>
| praveen kr | 453,426 | <p>Given $P$ is the probability of getting a number $N$ or greater with one trial.</p>
<p>1)For two trials, if you are taking the lower number, then both results should he greater than $N$. Since the two trials can be assumed to be independent, we have to multiply the two probabilities. Thus we get $P\times P=P^2$</p>
<p>2)Now consider the case of taking higher of the two.
In this case we have to find (1-probability(getting number N or lower in both trials)). This is seen by similar arguments to give $(1-(1-P)^2)$</p>
|
2,942,879 | <p>Suppose that <span class="math-container">$lim_{n\rightarrow \infty} a_n = L$</span> and <span class="math-container">$L \neq 0$</span>. Prove there is some <span class="math-container">$N$</span> such that <span class="math-container">$a_n \neq 0$</span> for all <span class="math-container">$n \geq N$</span>.</p>
<p>We know by the definition of convergence of a sequence, <span class="math-container">$\forall \epsilon > 0, \exists\ N \in \mathbb{N}$</span> such that <span class="math-container">$\forall n \geq N$</span>, <span class="math-container">$|x_n - L| < \epsilon$</span>.</p>
<p>So take such an <span class="math-container">$N$</span> which we know exists since we're given the limit <span class="math-container">$L$</span>.
So for the condition <span class="math-container">$|a_n - L| < \epsilon$</span> to hold <span class="math-container">$\forall \epsilon > 0$</span>, <span class="math-container">$a_n \neq 0$</span>, as otherwise if <span class="math-container">$a_n = 0$</span>, since <span class="math-container">$L \neq 0$</span>, we could find an <span class="math-container">$\epsilon$</span> such that <span class="math-container">$\epsilon < |-L| = L$</span>.</p>
<p>Proof seems rather short and maybe not rigorous enough. Wanted thoughts or if there's a better way to do this.</p>
| Peter Szilas | 408,605 | <p>For fun.</p>
<p>Given :</p>
<p><span class="math-container">$\lim_{n \rightarrow \infty}a_n=L \not = 0$</span>.</p>
<p>Statement :</p>
<p>There is a <span class="math-container">$N \in \mathbb{Z^+}$</span> s.t. for</p>
<p><span class="math-container">$ n \ge N$</span> <span class="math-container">$a_n \not =0$</span>.</p>
<p>Assume the statement is not true:</p>
<p>For all <span class="math-container">$N \in \mathbb{Z^+}$</span> exists a <span class="math-container">$n \ge N$</span> s.t. <span class="math-container">$a_n=0$</span>.</p>
<p>Let <span class="math-container">$a_{n_k}$</span> , <span class="math-container">$k=1,2,3...,$</span> be a subsequence with <span class="math-container">$a_{n_k}=0$</span>.</p>
<p><span class="math-container">$\lim_{k \rightarrow \infty} a_{n _k}=0$</span>, a contradiction.</p>
<p>.</p>
|
307,545 | <p>If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. </p>
| Community | -1 | <p>$$\begin{align}
\gcd(a+b, a^2 + b^2)
&= \gcd(a+b, a^2 + b^2 - a(a+b))
\\&= \gcd(a+b, b^2 - ab)
\\&= \gcd(a+b, b^2 - ab + b(a+b))
\\&= \gcd(a+b, 2b^2)
\end{align}
$$</p>
<p>Now, $\gcd(a+b,b) = \gcd(a,b) = 1$, so we can get rid of the factors of $b$ and have</p>
<p>$$ \gcd(a+b, a^2 + b^2) = \gcd(a+b, 2) $$</p>
<p>The strategy I used was still the basic idea of the Euclidean algorithm; since I couldn't compare numeric values, I instead simplified by working to eliminate the variable $a$, starting with the largest power of $a$.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| Eric Wofsey | 86,856 | <p>To a first approximation, there is a spoonful of wine in the cup and a spoonful of tea in the barrel. How much are each of these approximations off by? Well, there is a bit less than a spoonful of wine in the cup, since a bit of the wine was removed in the second step. And, there is a bit less than a spoonful of tea in the barrel, since there was a little wine mixed into the spoonful that was put into it. But these errors are exactly the same: both are the amount of wine that was in the second spoonful. So the two quantities are the same: both are one spoonful minus the amount of wine that was in the second spoonful.</p>
<p>Or, here's an even slicker way. Notice that the total volumes of liquid of the cup and barrel have not changed, since the two spoonfuls they exchanged cancelled out. So, the overall change must be that the barrel exchanged some volume of wine for the same volume of tea from the cup.</p>
<p>Note that your solution is actually wrong--when you compute the amounts of wine and tea in the second spoonful, you are assuming the cup was mixed uniformly after the first spoonful, which the problem tells you not to assume (that's what the "(nonuniform!)" is all about).</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| Kinro | 850,504 | <p>After step 1, there is always one spoonful of wine in your cup.</p>
<p>If, on step two:</p>
<p>You place one spoonful of wine back into your barrel, then there is 0 tea in your wine and 0 wine in your barrel.</p>
<p>You place one spoonful of tea back into in your barrel, then there is 1 tea in your wine and 1 tea in your barrel.</p>
<p>You place half a spoonful of tea and half of wine back into in your barrel, then there is 0.5 tea in your wine and 0.5 wine in your tea.</p>
<p>It seems that the quantity (not ratio) of tea in your wine and wine in your tea remains equal regardless.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| Dr. Michael W. Ecker | 438,184 | <p>Imagine the tea in the wine in the form of a tiny ball within the wine. Then that ball must be exactly the amount of wine that is missing from the wine. Ergo, it is the amount of wine that is in the tea. So, the two amounts are the same, with exactly as much tea in the wine as wine in the tea.</p>
|
3,910,623 | <p>There is a problem that appears in an interview<span class="math-container">$^\color{red}{\star}$</span> with <a href="https://en.wikipedia.org/wiki/Vladimir_Arnold" rel="nofollow noreferrer">Vladimir Arnol'd</a>.</p>
<blockquote>
<p>You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?</p>
</blockquote>
<p>This problem is also quoted <a href="https://math.stackexchange.com/q/895627">here</a>.</p>
<hr />
<p>Here's my solution:</p>
<p>The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let <span class="math-container">$s$</span> be the volume of a spoonful and <span class="math-container">$c$</span> be the volume of a cup. The quantity of wine in this second spoonful is <span class="math-container">$\frac{s}{s+c}\cdot s$</span> and the quantity of tea in this spoonful is <span class="math-container">$\frac{c}{s+c}\cdot s$</span>. Then the quantity of wine left in the cup is <span class="math-container">$$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$</span> and the quantity of tea in the barrel now is also <span class="math-container">$\frac{cs}{s+c}.$</span> So the quantities that we are asked to compare are the same.</p>
<p>However, Arnol'd also says</p>
<blockquote>
<p>Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.</p>
</blockquote>
<p>Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.</p>
<hr />
<p><span class="math-container">$\color{red}{\star}\quad$</span> S. H. Lui, <a href="https://www.ams.org/notices/199704/arnold.pdf" rel="nofollow noreferrer">An interview with Vladimir Arnol′d</a>, Notices of the AMS, April 1997.</p>
| Bram28 | 256,001 | <p>Huh. I must be a 5 or 6 year old since I thought this was completely trivial. I have noticed though that I tend to reason visually far more often than algebraically.</p>
<p>Before:</p>
<p><a href="https://i.stack.imgur.com/IQgGB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IQgGB.png" alt="enter image description here" /></a></p>
<p>After:</p>
<p><a href="https://i.stack.imgur.com/m6Hf6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6Hf6.png" alt="enter image description here" /></a></p>
<p>I mean, it <em>has</em> to be! No matter whether it's a spoonful, or a pinch, or whether you moved stuff back and forth 3 or 4 times, or what have you ... in the end, whatever amount of wine gets replaced by tea must have ended up in the cup.</p>
|
3,446,663 | <blockquote>
<p>Let <span class="math-container">$\sigma\in S_{14}$</span> which is an even permutation of the order of <span class="math-container">$28$</span>.<br> Prove that exist <span class="math-container">$x\in \left\{ 1,...,14 \right\}$</span> such that <span class="math-container">$\sigma(x)=x$</span>.</p>
</blockquote>
<p><strong>My try:</strong>
<br>We know that the permutation order is equal to the least common multiple of cycles that make up a given permutation and <span class="math-container">$28=2\cdot2\cdot7$</span>.</p>
<p>So <span class="math-container">$\sigma$</span> must be character <span class="math-container">$(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$</span> - composition of row cycle <span class="math-container">$4$</span> and row cycle <span class="math-container">$7$</span>
because if <span class="math-container">$\sigma$</span> would be a character <span class="math-container">$(a_1
,a_2)(b_1b_2)(c_1...c_7)$</span> then <span class="math-container">$|\sigma|=2\cdot7=14$</span> which is contrary to the assumption.</p>
<p>That's why <span class="math-container">$4+7=11$</span> elements elements undergo nontrivial permutations and <span class="math-container">$14-11=3$</span> elements pass on to each other.</p>
<p>So <span class="math-container">$\sigma$</span> has a character:</p>
<p><span class="math-container">$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_1 & c_2 & c_3\end{pmatrix}$$</span></p>
<p>Moreover we have information that <span class="math-container">$\sigma=(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$</span> is composition an even number of transpositions.</p>
<p>However these are my only thoughts and I don't know what to do next to come to the thesis.</p>
<p><strong>EDIT:</strong></p>
<p>According to the remark from @EricTowers <span class="math-container">$\sigma$</span> can still have a character <span class="math-container">$(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)(c_1c_2)$</span> then <span class="math-container">$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_2 & c_1 & c_3\end{pmatrix}$$</span></p>
| lhf | 589 | <p>Let <span class="math-container">$n_k$</span> be the number of <span class="math-container">$k$</span>-cycles in the disjoint cycle decomposition of <span class="math-container">$\sigma$</span>. Then</p>
<ul>
<li><p><span class="math-container">$k \in \{1,2,4,7,14,28\}$</span></p></li>
<li><p><span class="math-container">$n_{28}=0$</span> because <span class="math-container">$28>14$</span></p></li>
<li><p><span class="math-container">$n_{14}=0$</span> because a <span class="math-container">$14$</span>-cycle is not even</p></li>
<li><p><span class="math-container">$n_4 \ge 1$</span></p></li>
<li><p><span class="math-container">$n_7 \ge 1$</span></p></li>
<li><p><span class="math-container">$n_1 + 2n_2 + 4n_4 +7n_7 = 14$</span></p></li>
</ul>
<p>The last equation has no solutions if <span class="math-container">$n_1=0$</span>. Thus <span class="math-container">$n_1\ge1$</span>, as required.</p>
<p>Actually, the only solutions are
<span class="math-container">$(n_1,n_2,n_4,n_7)=(1,1,1,1)$</span> and <span class="math-container">$(3,0,1,1)$</span>.
However, since <span class="math-container">$\sigma$</span> is even, we must have <span class="math-container">$n_2+n_4$</span> even and so the only solution is <span class="math-container">$(n_1,n_2,n_4,n_7)=(1,1,1,1)$</span>.</p>
|
35,393 | <p>I have some data that represents small-scale surface topography. The data is planar but not flat because it's impractical to level this microscope to a tolerance of microns. It looks like this:</p>
<p><img src="https://i.stack.imgur.com/nQcLB.jpg" alt="An example of 3D profilometric data with a nonzero planar slope">
I'd like to postprocess that slope out, such that the major axis is colinear with the long side of the blue rectangle and the vertical axis can then be rescaled to show meaningful topographic variation instead of the (really large) planar slope.</p>
<p>One approach here might be to do an orthogonal transformation via (basically) principal component analysis, eg. find the three eigenvalue/eigenvector pairs, and then treat the eigenvectors corresponding to the largest two eigenvalues as the new X and Y axes. I'd be very interested to see how this (or some better way) might be done efficiently in Mathematica (this data is generally in the millions of rows).</p>
<p>An example of my data is <a href="https://dl.dropboxusercontent.com/u/28347262/6000s.txt" rel="nofollow noreferrer">here.</a></p>
<p>and this is the code I've written so far (which depends on version 9+ for <code>Downsample[]</code>):</p>
<pre><code> pathToDatafile = "path\\to\\file\\6000s.xyz";
data = Import[pathToDatafile,"Table","HeaderLines"->15];
data = DeleteCases[data,{__,_String,_String}]; (*remove the "no data" lines that the instrument creates*)
data=Delete[data,-1]; (*throw away the instrument's pound sign at the end of the file*)
data = Downsample[data,{100,1}];(*downsample by a lot because we have way too many datapoints*)
ListPlot3D[data,
Mesh->None,
BoxRatios->Automatic,
PlotStyle->LightBlue,
Boxed->False,
Axes->{True,True,True},
AxesLabel->{"\[Mu]m","","\[Mu]m"},
PlotRange->{Automatic,Automatic,{0,250}},
ImageSize->Large
]
</code></pre>
| bobthechemist | 7,167 | <p>This solution is giving some artifacts, I believe, due to the downsampling, but the problem can be solved by using <a href="http://reference.wolfram.com/mathematica/ref/RotationTransform.html" rel="nofollow noreferrer"><code>RotationTransform</code></a> similar to <a href="https://mathematica.stackexchange.com/questions/28752/rotating-a-a-set-of-points-in-a-square-boxed-region-by-90-degree-increments">this</a></p>
<pre><code>ListPlot3D[data]
ListPlot3D[RotationTransform[-2.2 Degree, {1, 0, 0}, data[[1]]]@data]
</code></pre>
<p><img src="https://i.stack.imgur.com/7MPd2.png" alt="Mathematica graphics"></p>
<p><img src="https://i.stack.imgur.com/RKewy.png" alt="Mathematica graphics"></p>
<p>You can make an interactive leveler. This requires a severely downsampled dataset in order to work smoothly (at least on my computer)</p>
<pre><code>subdata = Downsample[data, {100, 1}];
Manipulate[Module[{d},
d = RotationTransform[x Degree, {1, 0, 0}, subdata[[1]]]@subdata;
d = RotationTransform[y Degree, {0, 1, 0}, d[[1]]]@d;
ListPlot3D[d]
], {{x, 0}, -5, 5}, {{y, 0}, -5, 5}]
</code></pre>
<p><img src="https://i.stack.imgur.com/c4Uht.png" alt="Mathematica graphics"></p>
<p>And then use those results on the full set of data</p>
<pre><code>Module[{d},
d = RotationTransform[-2.16 Degree, {1, 0, 0}, data[[1]]]@data;
d = RotationTransform[0.12 Degree, {0, 1, 0}, d[[1]]]@d;
ListPlot3D[d]
]
</code></pre>
<p>Which gets you a slightly tweaked version of the image I have above, so I won't repost it.</p>
|
4,181,442 | <p>I know this is a dumb question but...</p>
<p>Is <span class="math-container">$x ≠ 1,3,5$</span> the same as <span class="math-container">$x$</span> does not belong to {<span class="math-container">$1,3,5$</span>}, for example ?</p>
<hr />
<p>Sorry for the formatting.</p>
<p>Btw, anyone has a link with mathjax's commands I guess?</p>
<hr />
<p>Thanks in advance.</p>
| Soham | 242,402 | <p>Using the set notation has benefits of its own. For eg. once you define how to want an arbitrary set <span class="math-container">$S$</span> to look like, the name <span class="math-container">$S$</span> becomes an alias for that definition forever. So, you are saved from a ton of writing by referring to <span class="math-container">$S$</span> only and not individually to its elements.</p>
<p>So, while writing a proof, instead of writing and rewriting <span class="math-container">$x\ne1$</span> and <span class="math-container">$x\ne2$</span> and <span class="math-container">$x\ne3$</span>, you can just write <span class="math-container">$x\notin S$</span> where <span class="math-container">$S=\{1,2,3\}$</span> and you are done. More over, you can't just give commas in between as in the question. One good reason for that is because the operations "logical-and" & "logical-or" are not conveyed very properly by the comma notation.</p>
<p>Note: Refer to the comment by @Erick Wong in the comments below. This provides another valid perspective.</p>
|
1,484,736 | <p>[<img src="https://i.stack.imgur.com/i5iPj.jpg" alt="enter image description here">]</p>
<p>$$\lim_{x\to 0}f\big(f(x)\big)$$</p>
<p>(Original image <a href="https://imgur.com/a/Uogqk" rel="nofollow noreferrer">here</a>.)</p>
<p>I don't need an answer, I just want to know how I can calculate the limit based on the given information.</p>
| Stefan | 280,501 | <p>The limit at a point exists in general if the right hand and left hand limit at that point equal the actual value of the function at that point. For example at $x=2$, we see that the function maps to $-1$. So if the left hand and right hand limit as $x$ approaches $x=2$ both equal $-1$, then the function is continuous at $x=2$ and the limit exists in general. From the left, the function approaches $-2$ as $x$ approaches $x=2$. From the right the function approaches $2$ as $x$ approaches $x=2$. Therefore since the left hand and right hand limits are not equal to the function value at the point $x=2$ the limit does not exist and the function is discontinuous at $x=2$.</p>
<p>I meant above to be a general case for limits not a direct answer to the question. For composite functions we check the limit of the inside function as $x$ approaches a given value. If this limit exists and equals $L$, all we need to do is check the limit of the outside function as it approaches $L$. We check this limit from the left if the inside function strictly approaches $L$ from underneath, and we check from the right if the inside function approaches $L$ from above. We check the right hand and left hand limit if the inside function approaches $L$ from above and below.</p>
|
3,518,719 | <p>Evaluate <span class="math-container">$$\lim_{n \to \infty} \sqrt[n^2]{2^n+4^{n^2}}$$</span></p>
<p>We know that as <span class="math-container">$n\to \infty$</span> we have <span class="math-container">$2^n<<2^{2n^2}$</span> and therefore the limit is <span class="math-container">$4$</span></p>
<p>In a more formal way I started with:</p>
<p><span class="math-container">$$\log(L)=\lim_{n \to \infty} \log(2^n+4^{n^2})^{\frac{1}{n^2}}=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n+2^{2n^2})$$</span></p>
<p>Continuing to <span class="math-container">$$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\log\left[2^n(1+2^{2n})\right]$$</span></p>
<p>Did not help much</p>
<p>As I arrived to <span class="math-container">$$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\left[\log(2^n)+\log(1+2^{2n})\right]=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n)+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})=0+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})$$</span></p>
| TeM | 247,735 | <p>Given a <em>regular curve</em> <span class="math-container">$\boldsymbol{\alpha}(t)$</span> (therefore of class <span class="math-container">$C^1$</span> and with <span class="math-container">$\boldsymbol{\alpha}'(t) \ne \boldsymbol{0}$</span>), <em>tangent</em>, <em>normal</em> and <em>binormal versors</em> are defined as:</p>
<p><span class="math-container">$$
\mathbf{T}(t) = \frac{\boldsymbol{\alpha}'(t)}{||\boldsymbol{\alpha}'(t)||}\,,
\; \; \; \; \; \;
\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \, (\text{when} \; \mathbf{T}'(t) \ne \boldsymbol{0})\,,
\; \; \; \; \; \;
\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)\,.
$$</span></p>
<p>So, starting from these definitions, in turn we define the <em>curvature</em> and the <em>torsion</em>:</p>
<p><span class="math-container">$$
\kappa(t) = \frac{||\mathbf{T}'(t)||}{||\boldsymbol{\alpha}'(t)||}\,,
\; \; \; \; \; \;
\tau(t) = -\frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{||\boldsymbol{\alpha}'(t)||}\,,
$$</span> </p>
<p>which respectively measure how far a curve moves away from being a straight line and from being contained in a plane.</p>
<hr>
<p>In particular, given a <em>regular curve</em> parameterized as:</p>
<p><span class="math-container">$$
\boldsymbol{\alpha}(t) := \left(r\,\cos t, \; r\,\sin t, \; c\,t\right) \; \; \; \text{with} \; t \in [0,\,2\pi)
$$</span></p>
<p>we get:</p>
<p><span class="math-container">$$
\begin{aligned}
& \boldsymbol{\alpha}'(t) = \left(-r\,\sin t, \; r\,\cos t, \; c\right); \\
& ||\boldsymbol{\alpha}'(t)|| = \sqrt{r^2+c^2}\;; \\
& \mathbf{T}(t) = \left(-\frac{r}{\sqrt{r^2+c^2}}\,\sin t, \; \frac{r}{\sqrt{r^2+c^2}}\,\cos t, \; \frac{c}{\sqrt{r^2+c^2}}\right); \\
& \mathbf{T}'(t) = \left(-\frac{r}{\sqrt{r^2+c^2}}\,\cos t, \; -\frac{r}{\sqrt{r^2+c^2}}\,\sin t, \; 0\right); \\
& ||\mathbf{T}'(t)|| = \frac{r}{\sqrt{r^2+c^2}}\;; \\
& \mathbf{N}(t) = \left(-\cos t, \; -\sin t, \; 0\right); \\
& \mathbf{B}(t) = \left(\frac{c}{\sqrt{r^2+c^2}}\,\sin t, \; -\frac{c}{\sqrt{r^2+c^2}}\,\cos t, \; \frac{r}{\sqrt{r^2+c^2}}\right); \\
& \mathbf{B}'(t) = \left(\frac{c}{\sqrt{r^2+c^2}}\,\cos t, \; \frac{c}{\sqrt{r^2+c^2}}\,\sin t, \; 0\right);
\end{aligned}
$$</span></p>
<p>from which:</p>
<p><span class="math-container">$$
\kappa(t) = \frac{r}{r^2+c^2}\,,
\; \; \; \; \; \;
\tau(t) = \frac{c}{r^2+c^2}\,,
$$</span></p>
<p>i.e.</p>
<ul>
<li>if <span class="math-container">$c = 0$</span> (<em>circle</em>) we have <span class="math-container">$\tau(t) = 0$</span>, then we deduce that <strong>it is</strong> a <em>plane curve</em>;</li>
<li>if <span class="math-container">$c \ne 0$</span> (<em>helix</em>) we have <span class="math-container">$\tau(t) \ne 0$</span>, then we deduce that <strong>it isn't</strong> a <em>plane curve</em>.</li>
</ul>
|
76,420 | <p>The following code segment shows what I'd like to do. I'm a procedural programmer trying to learn the Mathematica functional style. Any help on this would be appreciated.</p>
<pre><code>B = {{1, 3}, {1, 5}, {4, 2}, {5, 2}, {5, 5}}
u = SparseArray[{{1, 2} -> 5, {1, 3} -> 9, {1, 4} -> 6,
{1, 5} -> Infinity, {3, 2} -> 2, {3, 4} -> 4,
{4, 2} -> 9, {5, 2} -> Infinity, {5, 3} -> Infinity,
{5, 4} -> Infinity, {5, 5} -> Infinity}];
r = {3, 0, -2, 10, 19}
(* Want to convert this python-looking code to functional Mathematica code
for {i, j} in B :
r[i] = r[i] - u[[i, j]]
r[j] = r[j] + u[[i, j]]
*)
</code></pre>
| Simon Woods | 862 | <p>You could also use</p>
<pre><code>Function[{i, j},
r[[i]] -= u[[i, j]];
r[[j]] += u[[i, j]]
] @@@ B;
r
</code></pre>
<p>or</p>
<pre><code>With[{e = Extract[u, B]},
r[[B[[All, 1]]]] -= e;
r[[B[[All, 2]]]] += e];
r
</code></pre>
|
119,696 | <p>Let us suppose i have two graphs for sequences A and B as follows</p>
<pre><code>a1 := {{0.9, 0.086133}, {0.086133, 0.0082432}, {0.0082432,
0.0007889}, {0.0007889, 0.0000755}, {0.0000755,
7.2256*10^-6}, {7.2256*10^-6, 6.9151*10^-7}, {6.9151*10^-7,
6.618*10^-8}, {6.618*10^-8, 6.3336*10^-9}, {6.3336*10^-9,
6.0615*10^-10}, {6.0615*10^-10, 5.801*10^-11}};
plot1 = ListPlot[a1, Joined -> True,
PlotStyle -> {Red, Thick, Dashing[{0.01}]}]
</code></pre>
<p>and </p>
<pre><code>a2 := {{0.9, 0.21797}, {0.21797, 0.052789}, {0.052789,
0.012785}, {0.012785, 0.0030963}, {0.0030963,
0.0007499}, {0.0007499, 0.00018162}, {0.00018162,
0.000043985}, {0.000043985, 0.000010653}, {0.000010653,
2.5799*10^-6}, {2.5799*10^-6, 6.24831*10^-7}};
plot2 = ListPlot[a2, Joined -> True, PlotStyle -> {Yellow, Thick}]
</code></pre>
<p>I combine both the graphs by </p>
<pre><code>Show[plot1, plot2, PlotRange -> All]
</code></pre>
<p>But cant do any things with usings Legends.
I want to use Legends to show that "Red colour box"=A and "yellow colour box"=B</p>
<p>Thanks in Advance</p>
| gwr | 764 | <p>Actually the answer is already given in the documentation for <a href="http://reference.wolfram.com/language/ref/SwatchLegend.html" rel="nofollow noreferrer"><code>SwatchLegend</code></a>. All you need to do is to wrap <a href="http://reference.wolfram.com/language/ref/Legended.html" rel="nofollow noreferrer"><code>Legended</code></a> around the <code>Show</code>:</p>
<pre><code>Legended[
Show[ plot1, plot2, PlotRange -> All ],
SwatchLegend[ {Red, Yellow}, {"A", "B"} ]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/1RzIn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1RzIn.png" alt="SwatchLegend"></a></p>
<p>Note: Should the legend become larger and is entered by hand, using <code>Transpose</code> may make this easier, e.g.</p>
<pre><code>SwatchLegend @@ Transpose @ { {Red, "A"}, {Yellow, "B"} }
</code></pre>
<p><strong>Update:</strong> By wrapping <a href="http://reference.wolfram.com/language/ref/Placed.html" rel="nofollow noreferrer"><code>Placed</code></a> around <code>SwatchLegend[ ... ]</code> you can control placement.</p>
|
619,370 | <p>Let $C,Q$ is complex numbers Field and Rational number
Field,respectively,if $f(x),g(x)\in Q[x]$,</p>
<p>if $g(x)|f(x)$ on $C[x]$,show that
$$g(x)|f(x)$$ on $Q[x]$</p>
<p>My try: since
$g(x)|f(x)$,then we have
$$f(x)=g(x)h(x)$$
where $h(x)\in C[x]$.
Then I can't prove also have
$$g(x)|f(x)$$ on $Q[x]$.</p>
<p>Thank you</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ It is an immediate consequence of the <a href="https://math.stackexchange.com/a/168853/242"><em>uniqueness</em> of the quotient (and remainder)</a> in the Division Algorithm (which is the same in $\rm\:\Bbb Q[x]\:$ and $\rm\:\Bbb C[x]).$ Since dividing $\rm\:f\:$ by $\rm\:g\:$ in $\rm\:\Bbb C[x]\:$ leaves remainder $\,0$, by uniqueness, the remainder is $\,0\,$ in $\rm\:\Bbb Q[x],\:$ i.e. $\rm\:g\ |\ f\ $ in $\rm\:\Bbb C[x]\,$ $\, \Rightarrow\, $ $\rm\, g\ |\ f\ $ in $\rm\:\Bbb Q[x]$. </p>
<p>This is but one of <a href="http://www.google.com/search?q=site:http%3A%2F%2Fmath.stackexchange.com+dubuque++%22uniqueness+theorems%22+proving+equalities" rel="nofollow noreferrer">many examples</a> of the <em>power of uniqueness theorems for proving equalities</em>.</p>
|
474,807 | <p>I am studying a text on permutation groups, which has the following example in a section on regular normal subgroups: </p>
<blockquote>
<p>If $Z(N)=1$, then $N \cong \mathrm{Inn}(N)$, the group of inner automorphisms of $G$, and the semidirect product $N \rtimes \mathrm{Inn}(N)$ is the diagonal group $N^*=N \times N$ described earlier.</p>
</blockquote>
<p>My question is on why the group $N \rtimes \mathrm{Inn}(N)$ "is" the group $N \times N$? What would be the group isomorphism? I presume it is not necessary that $N$ be regular and normal since this is not stated explicitly in this example, and I'm not sure why a $G$ enters the picture (is this a typo - should this be an $N$ instead?). </p>
<p>Let me recall a couple facts described earlier in the text: </p>
<p>(a) If $G$ is a group, then $G^* := G \times G$ acts on $G$ by the rule $\mu(x,(g,h)):=g^{-1} xh$. Thus, $G^*$ is the product action that contains the left and right regular actions as normal subgroups. </p>
<p>(b) If $N$ is a regular normal subgroup of $G$, then $G=N \rtimes G_1$; and conversely, if $N$ is any group and $H \le \mathrm{Aut}(N)$, then $N \rtimes H$ acts as a permutation group on $N$, with $N$ as a regular normal subgroup and $H$ as the stabilizer of the identity; the element $hn$ of $N \rtimes H$ acts on $N$ by the rule $\mu(x,hn)=x^h n$. [The group operation in $N \rtimes H$ is not specified in the text, but we do get an action if the binary operation is assumed to be $(n_1,h_1)(n_2,h_2)=(n_1^{h_2} n_2, h_1 h_2)$.]</p>
<p>Since $N/Z(N) \cong \mathrm{Inn}(N)$, the assumption $Z(N)=1$ implies $N \cong \mathrm{Inn}(N)$. To get an isomorphism from $N^*:= N \times N$ to $N \rtimes \mathrm{Inn}(N)$, I tried the map $(g,h) \mapsto (g, c(h))$, where $c(h)$ denotes conjugation by $h$. But this map does not seem to be a homomorphism: $\phi((g_1,h_1)(g_2,h_2))= \phi((g_1 g_2, h_1 h_2)) = (g_1 g_2, c(h_1 h_2))$, but $\phi((g_1,h_1)) \phi((g_2,h_2)) = (g_1, c(h_1))(g_2,c(h_2)) = (g_1 ^{c(h_2)} g_2, c(h_1 h_2))$. </p>
| spin | 12,623 | <p>If $N \trianglelefteq H$, then $H$ acts on $N$ by conjugation. If this action induces only inner automorphisms (ie. for all $h \in H$, the map $n \mapsto n^h$ is an inner automorphism of $N$), it follows that $H = N C_H(N)$. If also $Z(N) = 1$, then $H = N \times C_H(N)$.</p>
<p>Thus in your case of $H = N \rtimes \operatorname{Inn}(N)$, it follows that $H = N \times C_H(N)$. Here $C_H(N) \cong H/N \cong \operatorname{Inn}(N) \cong N$, so $H \cong N \times N$.</p>
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23,192 | <p>The question if there is an upper bound known for Brun's constant was discussed briefly here: <a href="http://gowers.wordpress.com/2009/05/22/what-is-wolfram-alpha-good-for/" rel="nofollow">http://gowers.wordpress.com/2009/05/22/what-is-wolfram-alpha-good-for/</a> but no sure answer was given. </p>
<p>So I thought I'd ask the question here. Can one get any upper bound for the sum of the reciprocals of the twin primes?</p>
| David Hansen | 1,464 | <p>This seems to me like a simple matter of enumerating all the small twin primes, and then estimating the resulting error using a sieve bound. In particular, if $\pi_2(x)$ is the number of twin primes $\leq x$ then we know $\pi_2(x) \ll x (\log x)^{-2}$. By a simple summation-by-parts exercise, this gives</p>
<p>$B=\sum_{p\,twin,p < X}\frac{1}{p}+\frac{1}{p+2}+O(\log{X}^{-1})$.</p>
<p>I'm not sure what the numerical constant in the O-term is, but presumably it can be computed.</p>
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133,794 | <p>Prove the existence of infinite number of infinite cardinals </p>
<p>1) $\alpha$, such that $\alpha<\alpha^\aleph$ </p>
<p>2) $\beta$, such that $\beta=\beta^\aleph$</p>
| André Nicolas | 6,312 | <p>The first problem is quite a bit harder than the second. </p>
<p>For $1$), use <a href="http://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_%28set_theory%29" rel="nofollow">König's theorem.</a> We still then need to show that, for example, there are infinitely many cardinals of cofinality say $\omega$, but that part is not hard. For suppose we start at the infinite cardinal $\kappa_0$. Let $\kappa_1=2^{\kappa_0}$, $\kappa_2=2^{\kappa_1}$, and so on, and let $\kappa_\omega=\bigcup \kappa_n$. Then $\kappa_\omega$ has a right cofinality, and König's theorem applies. For the next one, start at $\kappa_\omega$.</p>
<p>For $2$), we can use for $\beta$ anything of shape say $\kappa^\aleph$, where $\kappa$ is an infinite cardinal.</p>
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2,333,857 | <p>I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!</p>
<p>Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid. </p>
<p>$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$</p>
<p><a href="https://i.stack.imgur.com/FtgoC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FtgoC.jpg" alt="Surface area problem worked out"></a></p>
| DMcMor | 155,622 | <p>Here's one way to look at it:</p>
<p>When you write $x = \lbrace -2, -1, \dotsc ,3\rbrace$ you are saying "$x$ is equal to the set consisting of the integers $-2$ through $3$". If $x$ were really a set then you'd be fine, but if you want to say that the set consists of the possible values for $x$, that is it's the domain of $f$, then saying $x = \lbrace -2, -1, \dotsc ,3\rbrace$ isn't true. When you write $x \in \lbrace -2, -1, \dotsc, 3\rbrace$ you are saying "$x$ is an element of the set consisting of the integers $-2$ through $3$" which is correct.</p>
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3,499,352 | <p>Define a function <span class="math-container">$f:X\to Y$</span> to be a strict contraction, if there exists a <span class="math-container">$c\in [0,1[$</span> such that
<span class="math-container">$$\forall x,y \in X: d_Y\left( f(x),f(y)\right)\le cd_X(x,y)$$</span>
Now, consider a metric space <span class="math-container">$(X,d)$</span> and a sequence <span class="math-container">$(f_n)_n$</span> of strict contractions <span class="math-container">$f_n:X \to X$</span>, which is converging to a function <span class="math-container">$f$</span>. My question is under what conditions will the limit function be a strict contraction again?</p>
<p>I made the following estimate
<span class="math-container">$$d(f(x),f(y))\le d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(x),f(y))$$</span>
Now each <span class="math-container">$f_n$</span> has its own contraction factor <span class="math-container">$c_n$</span> such that this can be rewritten as
<span class="math-container">$$d(f(x),f(y))\le d(f(x),f_n(x))+c_nd(x,y)+d(f_n(x),f(y))$$</span>
for all <span class="math-container">$n\in \mathbb{N}$</span>. Now by taking the limit <span class="math-container">$n\to\infty$</span> we obtain
<span class="math-container">$$d(f(x),f(y))\le c d(x,y)$$</span>
where <span class="math-container">$c = \sup\{c_n|n\in\mathbb{N}\}$</span>. This <span class="math-container">$c$</span> is not necessarily strictly smaller than <span class="math-container">$1$</span>.</p>
<p>So I was looking for additional conditions such that we are guaranteed the limit function to be a strict contraction (i.e. <span class="math-container">$c<1$</span>). The easiest way, would be to demand that all <span class="math-container">$f_n$</span> have the same contraction factor. But that's quite a strong constrain, so my suggestions would be to demand the convergence <span class="math-container">$f_n\to f$</span> to be uniformly and <span class="math-container">$(X,d)$</span> to be a complete metric space. </p>
<p><strong>My sketch of the proof</strong>:
Each <span class="math-container">$f_n$</span> has fixed point <span class="math-container">$x_n^*$</span>. In addition we have uniform convergence, so there exists some <span class="math-container">$n_0$</span> such that <span class="math-container">$$d(f_n(x_n^*),f_m(x_m^*))=d(x^*_n,x^*_m)< \epsilon$$</span>
for all <span class="math-container">$n,m\ge n_0$</span>. So <span class="math-container">$(x^*_n)_n$</span> is Cauchy sequence in complete X, so it converges to some <span class="math-container">$x^*$</span>. Using the fact that uniform convergence, we have
<span class="math-container">$$x^*=\lim_{n\to\infty}x^*_n = \lim_{n\to \infty}f_n(x_n^*) = f(x^*)$$</span>
This shows that <span class="math-container">$f$</span> has a unique fixed point in a complete metric space.</p>
<p><strong>Edit:</strong> But the existence of a fixed point does not imply that <span class="math-container">$f$</span> will be a strict contraction. So how can I make sure <span class="math-container">$f$</span> to be a strict contraction? </p>
| bign | 735,454 | <p>It seems to me you don't need the completeness of <span class="math-container">$X$</span>. As you pointed out, all you need is uniform convergence of the sequence of contractions mappings <span class="math-container">$(f_{n})$</span>.</p>
<h1>Here's my argument, using much of what you've already written</h1>
<p>Since each <span class="math-container">$c_{i} \in [0,1]$</span>, you may write <span class="math-container">$c = sup \{ c_{i} | i \in \mathbb{N} \}$</span>; however, <span class="math-container">$c$</span> is not necessarily the limit of the sequence <span class="math-container">$(c_{n})$</span>. I think your initial estimate needs a tweak in the third term on the right hand side as follows:</p>
<p><span class="math-container">$d(f(x),f(y)) \leq d(f(x),f_{n}(x)) + d(f_{n}(x),f_{n}(y)) + d(f_{n}(y),f(y))$</span></p>
<p>Your arguments still holds though. <span class="math-container">$d(f_{n}(x),f_{n}(y)) \leq cd(x,y)$</span> and there exists <span class="math-container">$N \in \mathbb{N}$</span> sufficiently large as to guarantee that <span class="math-container">$n \geq N$</span> implies <span class="math-container">$d(f(z),f_{n}(z)) \leq \frac{\epsilon}{2}$</span> for all <span class="math-container">$z \in X$</span>. So for any <span class="math-container">$\epsilon > 0$</span> we have,</p>
<p><span class="math-container">$d(f(x),f(y)) \leq cd(x,y) + \epsilon$</span></p>
<p>This means <span class="math-container">$f$</span> is a contraction mapping.</p>
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