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 |
|---|---|---|---|---|
351,325 | <p>It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is <em>perfectly normal</em> if for any two disjoint nonempty closed subsets, there is a continuous function <span class="math-container">$f$</span> to <span class="math-container">$[0,1]$</span> that "precisely separates" the two sets, meaning that the two closed sets are <span class="math-container">$f^{-1}(0)$</span> and <span class="math-container">$f^{-1}(1)$</span>.) And how do we prove it?</p>
| Will Brian | 70,618 | <p>As Nate Eldredge points out in the comments, the book <em>Counterexamples in Topology</em> states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.</p>
<p>A <em>perfectly normal</em> space is a normal space in which every closed set is a <span class="math-container">$G_\delta$</span>. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a <span class="math-container">$G_\delta$</span>. I claim that the subspace <span class="math-container">$C = [0,1] \times \{0,1\}$</span> works. (<span class="math-container">$C$</span> is the two horizontal lines on the top and bottom.) The reason is that if <span class="math-container">$U \subseteq [0,1] \times [0,1]$</span> is an open set containing <span class="math-container">$C$</span>, then
<span class="math-container">$$A_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$</span>
(i.e., the set of all vertical lines contained in <span class="math-container">$U$</span>) is a co-countable subset of <span class="math-container">$[0,1]$</span>. (This is because for every <span class="math-container">$p \in (0,1]$</span>, <span class="math-container">$U$</span> must contain a neighborhood of <span class="math-container">$(p,1)$</span>, and therefore <span class="math-container">$A_U$</span> must contain an interval of the form <span class="math-container">$(p-\varepsilon,p)$</span>. Likewise, <span class="math-container">$U$</span> must contain a neighborhood of <span class="math-container">$(p,0)$</span>, and therefore <span class="math-container">$A_U$</span> must contain an interval of the form <span class="math-container">$(p,p+\varepsilon)$</span>. So every point of <span class="math-container">$[0,1] \setminus A_U$</span> is isolated.) Therefore any countable intersection of open neighborhoods of <span class="math-container">$C$</span> contains a horizontal line.</p>
<p><strong>EDIT:</strong> Thanks to Nate Eldredge for helping me to simplify my original argument.</p>
|
351,325 | <p>It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is <em>perfectly normal</em> if for any two disjoint nonempty closed subsets, there is a continuous function <span class="math-container">$f$</span> to <span class="math-container">$[0,1]$</span> that "precisely separates" the two sets, meaning that the two closed sets are <span class="math-container">$f^{-1}(0)$</span> and <span class="math-container">$f^{-1}(1)$</span>.) And how do we prove it?</p>
| Ramiro de la Vega | 17,836 | <p>It is a known fact that any perfectly normal (countably) compact space <span class="math-container">$X$</span> is ccc. The proof is what you would try: start with an uncountable cellular family <span class="math-container">$\mathcal{U}$</span>, choose a point <span class="math-container">$p_U \in U$</span> for each <span class="math-container">$U \in \mathcal{U}$</span> and consider the closed set <span class="math-container">$C=X \setminus \bigcup \mathcal{U}$</span>. Since the space is perfectly normal, <span class="math-container">$C$</span> is a <span class="math-container">$G_\delta$</span> (say <span class="math-container">$C= \bigcap_nV_n$</span>) and since <span class="math-container">$\mathcal{U}$</span> is uncountable you can find <span class="math-container">$n$</span> such that infinitely many of the <span class="math-container">$p_U$</span>'s are outside of <span class="math-container">$V_n$</span>. Now this infinite set of <span class="math-container">$p_U$</span>´s has no limit points, contradicting the compactness of <span class="math-container">$X$</span>.</p>
<p>Since <span class="math-container">$I^2_{lex}$</span> is compact and not ccc (both standard facts), it cannot be perfectly normal.</p>
|
351,325 | <p>It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is <em>perfectly normal</em> if for any two disjoint nonempty closed subsets, there is a continuous function <span class="math-container">$f$</span> to <span class="math-container">$[0,1]$</span> that "precisely separates" the two sets, meaning that the two closed sets are <span class="math-container">$f^{-1}(0)$</span> and <span class="math-container">$f^{-1}(1)$</span>.) And how do we prove it?</p>
| Henno Brandsma | 2,060 | <p>For a compact Hausdorff <span class="math-container">$X$</span> it is equivalent that <span class="math-container">$X$</span> is hereditarily Lindelöf or that <span class="math-container">$X$</span> is perfectly normal. Sketch: <span class="math-container">$X$</span> is hereditarily Lindelöf implies that every open set is an <span class="math-container">$F_\sigma$</span> (as <span class="math-container">$X$</span> is regular) so dually every closed set is a <span class="math-container">$G_\delta$</span>. OTOH if <span class="math-container">$X$</span> is perfectly normal, every open set is an <span class="math-container">$F_\sigma$</span> so <span class="math-container">$\sigma$</span>-compact and Lindelöf, making all open sets Lindelöf and <span class="math-container">$X$</span> hereditarily Lindelöf again. </p>
<p><span class="math-container">$[0,1]^2$</span> in the lexicographic order topology has a discrete subset <span class="math-container">$[0,1]\times \{\frac12\}$</span> so is definitely not hereditarily Lindelöf, so not perfectly normal either.</p>
|
1,506,805 | <p>$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi }{2}\right)}{x}\right)\:$</p>
| Angelo Mark | 280,637 | <p>$\cos(x+\frac{\pi}{2})=-\sin(x)$</p>
<p>So $\displaystyle\lim_{x \to 0} \frac{-\sin(x)}{x}=-1$</p>
|
1,506,805 | <p>$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi }{2}\right)}{x}\right)\:$</p>
| Alekos Robotis | 252,284 | <p>Another suggestion which may work more generally (although the above solutions are better in this case) is to apply L'Hôpital's Rule for an indeterminate form $\frac{0}{0}$. That is, in this case, the limit of the derivatives of the numerator and denominator is equal to the limit of the original expression:</p>
<p>$$ \lim_{x\to 0}\frac{\cos(x+\frac{\pi}{2})}{x} = \lim_{x\to 0}\frac{-\sin(x+\frac{\pi}{2})}{1}=\lim_{x\to0}-\sin(x+\frac{\pi}{2})=-\sin(\frac{\pi}{2})=-1.$$</p>
<p>The same result as in the other answers. Use this technique any time you have an indeterminate form.</p>
|
2,759,063 | <p>$A=\left\{n+\frac{1}{2n}|n\in\mathbb{N}\right\}$<br/>
Is this set closed?
<br/>
I was reading answer form <a href="https://math.stackexchange.com/questions/125709/example-to-show-the-distance-between-two-closed-sets-can-be-0-even-if-the-two-se">Example to show the distance between two closed sets can be 0 even if the two sets are disjoint</a>
where Nishrito mentioned above set is closed .
<br/>
But On that account then
$B=\left\{\frac{1}{2n}|n\in\mathbb{N}\right\}$<br/> This set also become closed but I know which is not as $0$ is limit point of it which is not belong to that set.
Any Help in this regard will be appreciated .
<br/>
Thanks a lot</p>
| user | 293,846 | <p>By <a href="https://en.wikipedia.org/wiki/Law_of_sines" rel="nofollow noreferrer">law of sines</a>:
$$
\frac{\sin CAM}{\sin BAM}=\frac{\sin C}{\sin B}=:\lambda.
$$</p>
<p>Setting $\sin BAM=x$, and solving the resulting equation for cosine of the sum:
$$
\sqrt{1-x^2}\cdot\sqrt{1-\lambda^2x^2}-x\cdot\lambda x=\cos A\\
$$
one obtains:
$$
\sin BAM=\frac{\sin A}{\sqrt{1+\lambda^2+2\lambda\cos A}}
=\frac{\sin A\sin B}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}
$$
and similarly
$$
\sin CAM=\frac{\sin A\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}.
$$
Correspondingly:
$$
\cos BAM=\frac{\sin B\cos A+\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}},\\
\cos CAM=\frac{\sin B+\sin C\cos A}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}.
$$</p>
<p>Writing:
$$\small{
\cos(BAM-CAM)=\frac{(\sin B\cos A+\sin C)(\sin B+\sin C\cos A)+\sin B\sin C\sin^2A}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}\\
=\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}
}
$$
and noticing that $|\widehat{BAM}-\widehat{CAM}|$ is the doubled value of the angle in question one finally obtains:
$$
\theta=\frac{1}{2}\arccos\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}.
$$</p>
|
2,759,063 | <p>$A=\left\{n+\frac{1}{2n}|n\in\mathbb{N}\right\}$<br/>
Is this set closed?
<br/>
I was reading answer form <a href="https://math.stackexchange.com/questions/125709/example-to-show-the-distance-between-two-closed-sets-can-be-0-even-if-the-two-se">Example to show the distance between two closed sets can be 0 even if the two sets are disjoint</a>
where Nishrito mentioned above set is closed .
<br/>
But On that account then
$B=\left\{\frac{1}{2n}|n\in\mathbb{N}\right\}$<br/> This set also become closed but I know which is not as $0$ is limit point of it which is not belong to that set.
Any Help in this regard will be appreciated .
<br/>
Thanks a lot</p>
| Blue | 409 | <p>Let $A = 2\alpha$, $B = 2\beta$, $C = 2\gamma$. We can coordinatize, with
$$A = (0,0) \qquad B = c\left(\cos\alpha,-\sin\alpha\right) \qquad C = b\left(\cos\alpha,\sin\alpha\right) \qquad M = \frac12(B+C)$$ </p>
<p>Since bisector $\overline{AD}$ coincides with the $x$-axis, $\angle DAM$ is simply the direction angle of $M$. Perhaps the simplest representation is</p>
<blockquote>
<p>$$\tan DAM = \frac{M_y}{M_x} = \frac{B_y+C_y}{B_x+C_x} = \frac{(b-c)\sin\alpha}{(b+c)\cos\alpha} \tag{$\star$}$$</p>
</blockquote>
<p>The <a href="https://en.wikipedia.org/wiki/Law_of_sines" rel="nofollow noreferrer">Law of Sines</a> tells us that $b = d \sin B$ and $c = d \sin C$, where $d$ is the circumdiameter of $\triangle ABC$. With trig's <a href="http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html" rel="nofollow noreferrer">prosthaphaeresis formulas</a>, and recalling that $\alpha + \beta + \gamma = 90^\circ$, we can rewrite $(\star)$ as</p>
<blockquote>
<p>$$\tan DAM = \frac{\sin^2\alpha \sin(\beta-\gamma)}{\cos^2\alpha\cos(\beta-\gamma)} = \tan^2\alpha \tan(\beta-\gamma) \tag{$\star\star$}$$</p>
</blockquote>
<p>Alternatively, using the <a href="https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Half-angle_formulae" rel="nofollow noreferrer">half-angle formulas</a>, we have
$$\tan DAM = \frac{(1-\cos A)\sqrt{1-\cos(B-C)}}{(1+\cos A)\sqrt{1+\cos(B-C)}}$$
which, with the help of the <a href="https://en.wikipedia.org/wiki/Law_of_cosines" rel="nofollow noreferrer">Law of Cosines</a>, is manipulatable into this lengths-only form</p>
<blockquote>
<p>$$\tan DAM = \frac{b-c}{b+c}\;\sqrt{\frac{\phantom{-}(a - b + c)(a + b - c)}{(-a + b + c) (a + b + c)}} \tag{$\star\star\star$}$$</p>
</blockquote>
|
1,321,826 | <p>I need to find residue of function $f(z) = \frac{z}{1-\cos(z)}$ at $z=2\pi k$, where $k\in \Bbb Z$.
I know residue at $z=0$ from <a href="https://math.stackexchange.com/questions/168355/residue-of-fz-fracz1-cosz-at-z-0">here</a>.
I got a hint that need to substitute $z=\hat z+2\pi k$, so $\hat z=z-2\pi k$, so I have to find residue at $\hat z=0$. So $f(\hat z)=\frac{\hat z+2\pi k}{1-\cos(\hat z)}$.
One more hint - I have to split function in two parts, when k is odd and k is even. </p>
| Nitin Uniyal | 246,221 | <p>Hint: As the denominator 1-cosz has a simple pole at z=2kπ,
Res(f;2kπ)= lim┬(z→2kπ)〖z/(d/dz(1-cosz))〗</p>
|
1,321,826 | <p>I need to find residue of function $f(z) = \frac{z}{1-\cos(z)}$ at $z=2\pi k$, where $k\in \Bbb Z$.
I know residue at $z=0$ from <a href="https://math.stackexchange.com/questions/168355/residue-of-fz-fracz1-cosz-at-z-0">here</a>.
I got a hint that need to substitute $z=\hat z+2\pi k$, so $\hat z=z-2\pi k$, so I have to find residue at $\hat z=0$. So $f(\hat z)=\frac{\hat z+2\pi k}{1-\cos(\hat z)}$.
One more hint - I have to split function in two parts, when k is odd and k is even. </p>
| klackna | 472,041 | <p>no it has pole of order $2$ for $2k\pi$.see image.</p>
<p><a href="https://i.stack.imgur.com/LW1Zc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LW1Zc.png" alt="enter image description here"></a></p>
|
3,557,398 | <p>Here are two second-order differential equations.</p>
<p><span class="math-container">$$ y''+9y=\sin(2t) \tag 1 $$</span></p>
<p><span class="math-container">$$ y'' +4y =\sin(2t) \tag 2 $$</span></p>
<p>I am told to use undetermine coefficients method to solve.</p>
<p>For 1), I use <span class="math-container">$y_p=A \cos(2t)+B \sin(2t)$</span> to get <span class="math-container">$A=0$</span> and B=<span class="math-container">$\frac{1}{5}$</span> and get <span class="math-container">$y_p=\frac{1}{5} \sin(2t)$</span></p>
<p>For 2), I realize that that method doesn't work and told to do <span class="math-container">$y_p=t(A \cos(2t)+B \sin(2t)$</span> Why does it work then?</p>
| Community | -1 | <p>Yes, if <span class="math-container">$x+y<z$</span> then <span class="math-container">$x<z$</span> and <span class="math-container">$y<z$</span>.
Can you see how to approach the other cases?</p>
|
1,799,559 | <blockquote>
<p>Let $f:[a,b]\to\mathbb{R}$ be a continious function. Show that if $$\int_a^b f(x)g(x)dx=0$$
for all continious functions $g:[a,b]\to\mathbb{R}$ with $g(a)=g(b)=0$, then $f(x)=0$ $\forall x\in[a,b]$</p>
</blockquote>
<p>I have difficulties proving this question. Consider for example $g(x)=0$ $\forall x\in[a,b]$, then assumptions still hold but $f(x)$ can be anything (f.i. $f(x)=1\neq0$). Could anyone tell me if this reasoning is correct? </p>
| mariob6 | 326,028 | <p>The proof starts like this:</p>
<p>Suppose, by contraddiction that </p>
<p>$ \exists x_0 \ \in \ (a,b) \ :\ f(x_0)>0 $ then $ \exists \delta>0 \ :\ f(x)>0 \ \ \forall x \in(x_0-\delta,x_0+\delta) $</p>
<p>Can you reach the contraddiction?</p>
|
3,810,856 | <p>I'm having trouble understanding how this form of the principle (<a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">on wiki</a>) results in the form below.</p>
<p>Wiki form:
<a href="https://i.stack.imgur.com/6IXQZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6IXQZ.png" alt="wiki form" /></a></p>
<p>Using three sets:
<a href="https://i.stack.imgur.com/6maIp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6maIp.png" alt="example" /></a></p>
<p>My confusion is the last <span class="math-container">$(-1)^{n-1} |A_1 \cap \cdots \cap A_n|$</span>. Where does this last form appear in the three set example?</p>
<p>Also, if using two sets vs three, is the <span class="math-container">$\sum_{1 \le i \lt j \lt k \le n}$</span> unsatisfiable because there are no three <span class="math-container">$i, j, k$</span> terms to satisfy it? In that case this summation term disappears?</p>
| mjw | 655,367 | <p>We want <span class="math-container">$$\frac{\pi}{\sin^2 \frac{\pi}{x}} \gt \frac{x}{\tan \frac{\pi}{x}}$$</span></p>
<p>When the denominators are non-zero, we can rewrite this as:</p>
<p><span class="math-container">$$\frac{\pi}{x} > \sin \frac{\pi}{x} \cos \frac{\pi}{x}$$</span></p>
<p>Well, for all <span class="math-container">$p\ne 0$</span>, <span class="math-container">$$|p|>|\sin p| \ge |\sin p \cos p|.$$</span></p>
<p>The expression is true for all <span class="math-container">$x \ne 0 \text{ (and where defined).}$</span></p>
|
501,993 | <p>I have never encounter "proof" questions before in my career, but this question in textbook troubled me and I have totally no clue where to start.</p>
<p>Prove that the limit
$$\lim_{x\rightarrow0} \frac{1-\sqrt{1-x^2}}{x^2} = \frac{1}{2}$$</p>
| Pedro | 23,350 | <p><strong>Hint</strong> Multiply by the conjugate of the numerator.</p>
|
2,803,956 | <p>Let $\mathfrak{g}$ be a Lie algebra. Then its automorphism group $Aut(\mathfrak{g})$ is a Lie group, and hence we may take its Lie algebra $Lie(Aut(\mathfrak{g}))$.</p>
<p>I'd like to say that this is equal to the Lie algebra of derivations $Der(\mathfrak{g})$. Is this true? Where can I find a reference?</p>
| Dietrich Burde | 83,966 | <p>This is a standard fact from the theory of Lie algebras. There are several references, say over $\Bbb{R}$ and $\Bbb{C}$, e.g., Proposition 1.25 in the book <a href="https://books.google.at/books?id=RXbLBQAAQBAJ&pg=PA35&lpg=PA35&dq=show+that+Der(L)+is+the+Lie+algebra+of+the+automorphism+group+of+L&source=bl&ots=9CAeAIz5dr&sig=9UdtZGsc4eakMDaI833Suqwpv50&hl=de&sa=X&ved=0ahUKEwiPnYiXhrLbAhWSZ1AKHVvpCsQ4FBDoAQhjMAg#v=onepage&q=show%20that%20Der(L)%20is%20the%20Lie%20algebra%20of%20the%20automorphism%20group%20of%20L&f=false" rel="nofollow noreferrer">The Structure of Complex Lie Groups</a>.</p>
|
4,632,869 | <p>Suppose <span class="math-container">$\Omega \subset \mathbb{R}$</span> is a bounded domain. Then using
Poincare-Wirtinger, one can prove that for functions <span class="math-container">$f \in W^{1,2}(\Omega)$</span> there exists <span class="math-container">$C>0$</span> such that</p>
<p><span class="math-container">$$\|u\|_{L^{2}}^{2} \le C\|u'\|_{L^{2}}^{2} + C\left( \int u(x)~dx
\right)^{2}. $$</span></p>
<p><strong>Question:</strong> Suppose we have a function <span class="math-container">$u$</span> which satisfies</p>
<p><span class="math-container">$$ \|u'\|_{L^{2}}^{2} + \left( \int u(x)~dx \right)^{2} \le C .
$$</span></p>
<p>Can we conclude that <span class="math-container">$u \in W^{1,2}(\Omega)$</span>? The reason I am not sure is because in order for us to use the first inequality we need that <span class="math-container">$u \in W^{1,2}(\Omega)$</span> which a-priori we don't have. I think the statement could be true and we can prove it by some density argument but I am not sure how to proceed.</p>
| LL 3.14 | 731,946 | <p>If <span class="math-container">$\Omega$</span> has regular boundary (say Lipshitz) you can use the fact that since <span class="math-container">$u' \in L^2(\Omega)$</span> then by Sobolev-Morrey inequality, <span class="math-container">$u\in C^{0,\alpha}(\Omega)$</span> for some <span class="math-container">$\alpha>0$</span>, and so is bounded on <span class="math-container">$\Omega$</span>. This implies that <span class="math-container">$u\in L^2$</span> since <span class="math-container">$\Omega$</span> is bounded.</p>
|
101,256 | <p>Can you please help me finding an exact description of the set:</p>
<p>$$ E_{R}=\{\cos{z} | z \in \mathbb{C}, |z|>R\} $$</p>
<p>For any $0<R \in \mathbb{R}$.</p>
<p>My feeling is the $E_R = \mathbb{C}$, for any $R$, but I don't know how to show it, if it's true.</p>
| Fabian | 7,266 | <p><em>Hint:</em>
Make use of <a href="http://en.wikipedia.org/wiki/Picard_theorem" rel="nofollow">Picard theorem</a> to show that it the image is all of $\mathbb{C}$ with at most a single point missing.</p>
|
1,951,964 | <p>I think the answer is <span class="math-container">$\mathbb{Z}_n$</span> where <span class="math-container">$n$</span> is all the factors of <span class="math-container">$26$</span>. Is this right?</p>
| E.H.E | 187,799 | <p>$${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^nx^{n}$$
let $x\rightarrow x^2$
$${\frac {1}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n}$$
then multiply by $x$
$${\frac {x}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n+1}$$</p>
|
4,023,342 | <p>Blue: <span class="math-container">$x \ln x-x$</span></p>
<p>Brown: <span class="math-container">$\ln \frac{\Gamma(x+1/2)}{\sqrt{\pi}}$</span></p>
<p><a href="https://i.stack.imgur.com/CbVYi.png" rel="noreferrer"><img src="https://i.stack.imgur.com/CbVYi.png" alt="enter image description here" /></a></p>
| Anixx | 2,513 | <p>Well, I post my own answer. The real reason is that (up to some constant terms and factors), the first function is <span class="math-container">$D^{-1}D^{-1}[\frac1x]$</span> and the second function is <span class="math-container">$D^{-1}\Delta^{-1}[\frac1x]$</span>. So, in one case we replace one normal integration operator with discrete integration. There is no surprise, the functions should be similar.</p>
|
3,652,087 | <p>I need to find an example of a sequence that on the one hand is divergent and on the other hand has two subsequences that <span class="math-container">$$\lim_{n\to \infty} \mathit{a}_{3n} = \lim_{n\to \infty} \mathit{a}_{5n}.$$</span></p>
<p>Thanks a lot.</p>
| J. W. Tanner | 615,567 | <p>What about <span class="math-container">$1,2,0,4,0,0,7,8,0,0,11,0,13...$</span>?</p>
|
3,652,087 | <p>I need to find an example of a sequence that on the one hand is divergent and on the other hand has two subsequences that <span class="math-container">$$\lim_{n\to \infty} \mathit{a}_{3n} = \lim_{n\to \infty} \mathit{a}_{5n}.$$</span></p>
<p>Thanks a lot.</p>
| Sahiba Arora | 266,110 | <p>Let <span class="math-container">$$a_n=\begin{cases}1, \text{if } 3\mid n \text{ or }5\mid n \\0, \text{otherwise }\end{cases},$$</span>
then <span class="math-container">$a_{3n}=a_{5n}=1$</span> for all <span class="math-container">$n$</span> and hence <span class="math-container">$$\lim_{n\to \infty} \mathit{a}_{3n} = \lim_{n\to \infty} \mathit{a}_{5n}$$</span> but <span class="math-container">$a_n$</span> doesn't converge.</p>
|
1,794,025 | <p>Show that <span class="math-container">$Z(G) = \cap_{a \in G} C(a)$</span>.</p>
<p>Let <span class="math-container">$a \in Z(G)$</span>. Then <span class="math-container">$ax=xa$</span> for all <span class="math-container">$x$</span> in <span class="math-container">$G$</span>. In particular we can say that <span class="math-container">$ax_1=x_1a$</span> and <span class="math-container">$ax_2=x_2a$</span> and <span class="math-container">$ax_3=x_3a$</span> and so on (<span class="math-container">$x_i$</span> are elements of <span class="math-container">$G$</span>). This is nothing but intersection of all subgroups of form <span class="math-container">$C(a)$</span>. However I doubt my way of doing this question. Please guide me</p>
<p>Thanks</p>
| Bernard | 202,857 | <p>Maybe this is the shortest:
<span class="math-container">$$\begin{align}
x\in Z(G)&\iff \forall a\in G,\;ax=xa\\
&\iff\forall a\in G,\;x\in C(a)\\
&\iff x\in\bigcap_{a\in G}C(a).
\end{align}$$</span></p>
|
459,067 | <p>In set theory, numbers are often constructed, e.g. from nestings of sets which eventually contain the empty set. The operations are defined in term of taking unions etc.etc. </p>
<p>The extended real number line hat plus and minus infinity $\infty,-\infty$ in addition to the reals, and they fulfill <a href="https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations" rel="noreferrer">certain axioms</a>.</p>
<blockquote>
<p>How are the two objects "$\infty$" and "$-\infty$" of the <a href="https://en.wikipedia.org/wiki/Extended_real_number_line" rel="noreferrer">extended real number line</a> $\mathbb R\cup\{-\infty,\infty\}$ modeled in rigorous set theory? </p>
</blockquote>
<p>I figure it might be related to ordinal numbers. But these are <em>two</em> things and they behave somewhat similarly. So I wonder what they are made of, set theoretically.</p>
| Hagen von Eitzen | 39,174 | <p>All you need is two objects that are not in $\mathbb R$, so you may let $\infty=\{\mathbb R\}$ and $-\infty=\{\{\mathbb R\}\}$, for example.</p>
|
1,185,846 | <p>Hey all I am looking for help on a problem. I will post it, and than I will add what I have tried and my ideas etc. The question has been up now for a few days, I'm sure someone out there can help! I even put a bounty, I have spent a lot of time on this question!</p>
<p>I am interested in calculating the torsion ($\tau$) and curvature ($\kappa$) of the curve $$X(t)=(at,bt^2,ct^3), \quad t \ge 0 $$ and $a$, $b$, and $c$ are all positive constants.</p>
<p>So here is what I am having problems with. It seems like there are so many different formulas for curvature, and there are also the Frenet–Serret formulas so I am having issues deciding how to do it. I was thinking maybe I could reparametrize with respect to arc length, which would give me it in terms of unit length so I could use some of Frenet–Serret formulas, but I am not confident in that.</p>
<p>What I did so far was I calculated $X'(t)=(a,2bt,3ct^2)$ and $|X'(t)|=\sqrt{a^2+4b^2 t^2 +9c^2 t^4}$.</p>
<p>Then I calculated $$X''(t)=(0, 2b, 6ct)$$ and $$|X''(t)|=2\sqrt{b^{2}+9c^{2}t^{2}}.$$</p>
<p>I also know that the unit tangent, $$T(t)={X'(t)\over|X'(t)|}$$ and the unit normal is $$N(t)={T'(t) \over |T'(t)|}$$ and that the binormal is $B= T \times N$.</p>
<p>But I am really not sure how to take it further.</p>
<p>I know that in arc length parametrzation we would have $dT/ds= \kappa N$ and $dN/ds=-\kappa T + \tau B$ and $dB/ds= -\tau N$.</p>
<p>Should I keep it in the form it is and use the equation $$\kappa={|X'(t) \times X''(t)|\over |X'(t)|^3}?$$</p>
<p>A few of the other things I am thinking of is that maybe I could solve for torsion by the following:</p>
<p>I know that $T$, $B$ and $N$ form an orthonormal basis of $\mathbb R^{3}$ so can and when we write $N'=\alpha T + \tau B$ , is the coefficient, $\tau$ the definition of torsion?</p>
<p>Moreover, noting $B= T \times N$, we have $$B'= T'\times N + T \times N' = T \times N' =T \times (\alpha T+\tau B) = \tau T \times B= -\tau N $$ (because $N=B \times T$ so $T \times B=-N)$. I think this is how I understand the derivation for that equation.</p>
<p>Is it only in arc length parametrization that I can use the Frenet equations for example?</p>
<p>However, I think I should be able to do it just using the regular formulas, and not an arc length parametrization as the integral would be tough. I did conform this as well, so in terms of the basics of this question I'd like to do it without arc length parametrization.</p>
<p>I am also interested in seeing a intuitive derivation of the torsion formula (the one with the triple product). </p>
<p>I apologize if I didn't show enough work, this is all I could do but I am very happy to learn it. Thanks a lot to any help!</p>
<p>UPDATE:</p>
<p>I am wondering about the validity of what I now have.</p>
<p>In addition to above I computed $X'''(t)=(0,0,6c)$</p>
<p>$X' \times X''= (6bct^2,-6act,2ab)$</p>
<p>$(X' \times X'') \cdot X''' = (12abc)$</p>
<p>$|X' \times X''| = 2 \sqrt {9b^2c^2t^4+9a^2c^2t^2+a^2b^2}$.</p>
<p>Now can I just apply the formulas</p>
<p>$$\tau = \frac{ (X' \times X'') \cdot X'''}{|X' \times X''|^2}.$$</p>
| Frieder | 221,616 | <p>First trace (red)
$x-2 y=6$
has parametrization like this
$(2 t+6,t,0) $
for $-5<t<-1$
Straight-Line is complete in x-y- plane</p>
<p>Second trace (green)
$x+3 z=6$
has parametrization like this
$(6 - 3 t, 0, t) $
for $-1<t<1$
Straight-Line is complete in x-z- plane</p>
<p>Third trace (magenta)
$-2 y + 3 z = 6$
has parametrization like this
$(0,t,\frac{2 t}{3}+2) $
for $-1<t<1$
Straight-Line is complete in y-z- plane.</p>
<p>Here's the picture:</p>
<p><img src="https://i.stack.imgur.com/UV0Pi.png" alt="enter image description here"></p>
<p>Here is another one:</p>
<p><img src="https://i.stack.imgur.com/CDt8H.png" alt="enter image description here"></p>
|
2,702,347 | <blockquote>
<p>Prove that $$\left|\dfrac{1}{z^4-4z^2+3}\right|\leq \dfrac{1}{3},\quad\text{ if}\quad |z|=2$$</p>
</blockquote>
<p>I know that I can complete squares on the denominator ($(z^2-2)^2-1$). I know the triangular inequality and some other few things... but I can't work so well with absolute value on $\mathbb{C}$, if anyone could help me...</p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Recall that $$\left|\frac1w \right|=|w|^{-1}$$</p>
<p>and then refer to Martin R answer.</p>
|
949,320 | <p>how can I solve this:</p>
<blockquote>
<p>How many different ways can 8 children be divided into two groups of 3
and one group of 2?</p>
</blockquote>
<p>My method was:</p>
<p>8C3*5C3*2C2+8C3*5C2*3C3+8C2*6C3*3C3=560+560+560=1680</p>
<p>But the answer is:</p>
<p>8C3*5C3*2C2=560</p>
<p>I feel close to getting the logic but I always seem to miss it... Can anyone point it out to me? I'm bad in perms and combs. Thanks.</p>
| Arthur | 15,500 | <p>It doesn't matter what order you pick out the groups. So first picking $3$ children, then $3$ children then $2$ (the first term in your sum) yields a result that in the end is indistinguishable from any of the two other orders you can do it.</p>
<p>Which group you pick out children for <em>first</em> is irrelevant to what groups you end up with at the end, and as such having one term for each way it can be done leads to overcounting.</p>
|
1,143,832 | <p>Let $\zeta$ be a 12th primitive root over $\mathbb Q$. Determine all intermediate field of $\mathbb Q(\zeta)/\mathbb Q$.</p>
<p>My problem is that this is a task from an old exam where you were not allowed to use some sheets which you prepared at home.</p>
<p>So I would like to know if there are some possibilities in terms of determining the intermediate field <strong>without</strong> knowing the 12th cyclotomic polynomial (I guess nobody have to memorize all of them..)</p>
<p>So I am looking for something like a "trick" or way how to solve this task. Maybe something which I dont know yes.</p>
<p>Thanks in advance!</p>
| MooS | 211,913 | <p>The Galois group is $(\mathbb Z/12\mathbb Z)^* \cong V_4$, hence there are $3$ proper intermediate fields of degree $2$, which are made up of square roots.</p>
<p>Obviously the fourth and the third root of unity give rise to the intermediate fields $\mathbb Q(i)$ and $\mathbb Q(i\sqrt{3})$. So the third one is given by $\mathbb Q(\sqrt{3})$.</p>
|
1,143,832 | <p>Let $\zeta$ be a 12th primitive root over $\mathbb Q$. Determine all intermediate field of $\mathbb Q(\zeta)/\mathbb Q$.</p>
<p>My problem is that this is a task from an old exam where you were not allowed to use some sheets which you prepared at home.</p>
<p>So I would like to know if there are some possibilities in terms of determining the intermediate field <strong>without</strong> knowing the 12th cyclotomic polynomial (I guess nobody have to memorize all of them..)</p>
<p>So I am looking for something like a "trick" or way how to solve this task. Maybe something which I dont know yes.</p>
<p>Thanks in advance!</p>
| Lubin | 17,760 | <p>It’s worth while recognizing what the twelfth roots of unity <em>are</em>. You know that the cube roots of unity are, they’re $1$ and $(-1\pm i\sqrt3)/2$. You also know what the fourth roots of unity are, they’re $\pm1$ and $\pm i$. Combine them in an appropriate way and you’ll get the twelfth roots; but looking at the irrationalities I’ve written out, you can see immediately what the intermediate quadratic fields are.</p>
<p>And although you certainly didn’t need the twelfth cyclotomic polynomial, it’s easy to calculate, as $X^4-X^2+1$. You do this by looking first for the sixth polynomial, which is $X^2-X+1$, and noticing that the primitive twelfth roots of unity are the square roots of the primitive sixth roots.</p>
|
41,940 | <p>For example, the square can be described with the equation $|x| + |y| = 1$. So is there a general equation that can describe a regular polygon (in the 2D Cartesian plane?), given the number of sides required?</p>
<p>Using the Wolfram Alpha site, this input gave an almost-square:
<code>PolarPlot(0.75 + ArcSin(Sin(2x+Pi/2))/(Sin(2x+Pi/2)*(Pi/4))) (x from 0 to 2Pi)</code></p>
<p>This input gave an almost-octagon:
<code>PolarPlot(0.75 + ArcSin(Sin(4x+Pi/2))/(Sin(4x+Pi/2)*Pi^2)) (x from 0 to 2Pi)</code></p>
<p>The idea is that as the number of sides in a regular polygon goes to infinity, the regular polygon approaches a circle. Since a circle can be described by an equation, can a regular polygon be described by one too? For our purposes, this is a regular convex polygon (triangle, square, pentagon, hexagon and so on).</p>
<p>It can be assumed that the centre of the regular polygon is at the origin $(0,0)$, and the radius is $1$ unit.</p>
<p>If there's no such equation, can the non-existence be proven? If there <em>are</em> equations, but only for certain polygons (for example, only for $n < 7$ or something), can those equations be provided?</p>
| André Nicolas | 6,312 | <p>The following is probably not in the spirit of the game, but what about a <em>parametric equation</em>? If we are willing to use complex numbers to represent points in the plane, we could use
$$z=t\exp(2\pi ik/n) +(1-t)\exp(2\pi i(k+1)/n),\qquad 0 \le t<1,\quad k=0, 1, \dots, n-1$$ </p>
|
41,940 | <p>For example, the square can be described with the equation $|x| + |y| = 1$. So is there a general equation that can describe a regular polygon (in the 2D Cartesian plane?), given the number of sides required?</p>
<p>Using the Wolfram Alpha site, this input gave an almost-square:
<code>PolarPlot(0.75 + ArcSin(Sin(2x+Pi/2))/(Sin(2x+Pi/2)*(Pi/4))) (x from 0 to 2Pi)</code></p>
<p>This input gave an almost-octagon:
<code>PolarPlot(0.75 + ArcSin(Sin(4x+Pi/2))/(Sin(4x+Pi/2)*Pi^2)) (x from 0 to 2Pi)</code></p>
<p>The idea is that as the number of sides in a regular polygon goes to infinity, the regular polygon approaches a circle. Since a circle can be described by an equation, can a regular polygon be described by one too? For our purposes, this is a regular convex polygon (triangle, square, pentagon, hexagon and so on).</p>
<p>It can be assumed that the centre of the regular polygon is at the origin $(0,0)$, and the radius is $1$ unit.</p>
<p>If there's no such equation, can the non-existence be proven? If there <em>are</em> equations, but only for certain polygons (for example, only for $n < 7$ or something), can those equations be provided?</p>
| NJS | 129,614 | <p>How's this (Mathematica code)? It gives the n-gons inscribed in the unit circle, with vertices at the nth roots of unity.</p>
<p>Manipulate[PolarPlot[Cos[Pi/n] Sec[(2/n) ArcTan[Cot[(n t/2)]]], {t, 0, 2 Pi}], {n, 3, 40, 1}]</p>
|
41,940 | <p>For example, the square can be described with the equation $|x| + |y| = 1$. So is there a general equation that can describe a regular polygon (in the 2D Cartesian plane?), given the number of sides required?</p>
<p>Using the Wolfram Alpha site, this input gave an almost-square:
<code>PolarPlot(0.75 + ArcSin(Sin(2x+Pi/2))/(Sin(2x+Pi/2)*(Pi/4))) (x from 0 to 2Pi)</code></p>
<p>This input gave an almost-octagon:
<code>PolarPlot(0.75 + ArcSin(Sin(4x+Pi/2))/(Sin(4x+Pi/2)*Pi^2)) (x from 0 to 2Pi)</code></p>
<p>The idea is that as the number of sides in a regular polygon goes to infinity, the regular polygon approaches a circle. Since a circle can be described by an equation, can a regular polygon be described by one too? For our purposes, this is a regular convex polygon (triangle, square, pentagon, hexagon and so on).</p>
<p>It can be assumed that the centre of the regular polygon is at the origin $(0,0)$, and the radius is $1$ unit.</p>
<p>If there's no such equation, can the non-existence be proven? If there <em>are</em> equations, but only for certain polygons (for example, only for $n < 7$ or something), can those equations be provided?</p>
| JoeB | 243,738 | <p>$\arcsin \left(\sin \left(\left[\sin \left(60\cdot \frac{\pi }{180}\right)\left(\frac{\sqrt{3}}{3}\cdot x+y\right),x\right]\right)\right)+\left(\frac{1}{3}\right)\cdot \arcsin \left(\sin \left(\left[\cos \left(30\cdot \frac{\pi }{180}\right)\left(x-\frac{\sqrt{3}}{3}y\right),y\right]\cdot \sqrt{3}\right)\right)=0$</p>
<p><a href="https://www.desmos.com/calculator/yptmqucuwl" rel="nofollow">https://www.desmos.com/calculator/yptmqucuwl</a></p>
<p>I came up with this to plot a hexagon tessellation in desmos....the [list] funct "hides" the overlay ...so it isn't really a single EQ.....actually there was very little info on this.....took me weeks....enjoy</p>
|
3,753,205 | <p>In my general topology textbook there is the following exercise:</p>
<blockquote>
<ul>
<li>(i) - Let <span class="math-container">$\mathcal B$</span> be a basis for a topology <span class="math-container">$\tau$</span> on a non-empty set <span class="math-container">$X$</span>. If <span class="math-container">$\mathcal B_1$</span> is a collection of subsets of <span class="math-container">$X$</span> such that <span class="math-container">$\mathcal B \subseteq \mathcal B_1 \subseteq \tau$</span>, prove that <span class="math-container">$\mathcal B_1$</span> is also a basis of <span class="math-container">$\tau$</span></li>
<li>(ii) - Deduce from (i) that there exists an uncountable number of distinct basis of the euclidean topology on <span class="math-container">$\mathbb R$</span></li>
</ul>
</blockquote>
<p>I already proved statement number (i), but I'm having some trouble to think of a proof for the second statement.</p>
<p>I think that if we consider <span class="math-container">$\mathcal B= \{]a,b[:a,b\in \mathbb R\}$</span> as a basis for the euclidean topology <span class="math-container">$\tau$</span>, then we just need to show that there exists an uncountable number of sets <span class="math-container">$\mathcal B_1$</span> such that <span class="math-container">$\mathcal B \subseteq \mathcal B_1 \subseteq \tau$</span>. But how can I show that?</p>
| freakish | 340,986 | <p>You are on the right track. So you have</p>
<p><span class="math-container">$$\mathcal B= \{(a,b):a,b\in \mathbb R\}$$</span></p>
<p>Now for any <span class="math-container">$x\in\mathbb{R}$</span> add a single open set to <span class="math-container">$\mathcal B$</span>:</p>
<p><span class="math-container">$$\mathcal B_x= \mathcal B\ \cup\{(x,\infty)\}$$</span></p>
<p>What can you say about <span class="math-container">$\mathcal B_x$</span>? Is it a basis? How many different <span class="math-container">$\mathcal B_x$</span> are there?</p>
|
343,993 | <p>Let $I$ be an ideal of $\mathbb{C}[x,y]$ such that its zero set in $\mathbb{C}^2$ has cardinality $n$. Is it true that $\mathbb{C}[x,y]/I$ is an $n$-dimensional $\mathbb{C}$-vector space (and why)?</p>
| Jim | 56,747 | <p>No, the zero set of $I = (x, y^k)$ has cardinality $1$, it's just the point $(0, 0)$. But $\mathbb C[x, y]/(x, y^k)$ has dimension $k$.</p>
|
343,993 | <p>Let $I$ be an ideal of $\mathbb{C}[x,y]$ such that its zero set in $\mathbb{C}^2$ has cardinality $n$. Is it true that $\mathbb{C}[x,y]/I$ is an $n$-dimensional $\mathbb{C}$-vector space (and why)?</p>
| Alex Becker | 8,173 | <p>As Jim points out, this is false in general. The correct statement is that $\mathbb C[x,y]/\sqrt{I}$ is an $n$-dimensional vector space.</p>
<p><strong>Proof:</strong> Write $\sqrt{I} = P_1\cap \cdots \cap P_m$, where $P_i$ are the primes minimal over $I$. Since $V(P_i)\subseteq V(I)$, we see that each $V(P_i)$ is a single point. Thus since each $V(P_i)$ is distinct, we get $m=n$. Furthermore, each $P_i$ is maximal, so the $P_i$ are comaximal hence by the Chinese Remainder Theorem we get
$$\frac{\mathbb C[x,y]}{P_1\cap\cdots\cap P_n}\cong\frac{\mathbb C[x,y]}{P_1}\times\cdots\times\frac{\mathbb C[x,y]}{P_n}\cong \mathbb C\times \cdots \times \mathbb C$$
with $n$ copies of $\mathbb C$.</p>
|
281,387 | <p>I've been reading Girard et al's 'Proofs and Types', which in Chapter 6 presents a proof of strong normalisation for the simply typed lambda calculus with products and base types. The proof is based on Tait's method of defining a set of 'reducible terms for type $T$' by induction on the type $T$. For example a term $t$ is reducible for type $A\to B$ if for all reducible $u$ for type $A$, $t\,u$ is reducible for type $B$.</p>
<p>However I don't see how to extend this proof to (binary) sums, which are covered in a later chapter, because the elimination for sums involves a 'parasitic' type which prevents one from defining reducible terms of type $A+B$ via the elimination by induction on $A+B$.</p>
<p>How (if at all) does this proof style extend to sums? I can't find a good reference that presents this.</p>
| gallais | 14,972 | <p>The typical strategy when you're building a model to perform normalisation by evaluation (i.e. the computational part of Tait's method, see [1]) with sum types is to have as a semantics for <code>A + B</code>:</p>
<ul>
<li>either the semantics of <code>A</code></li>
<li>or the semantics of <code>B</code></li>
<li>or a neutral term of type <code>A + B</code> (i.e. a head variable with a spine of eliminators)</li>
</ul>
<p>This makes it possible for the proof to go through. I'd wager that the same strategy works for the proof of SN. In other words, your reducibility candidate $R_{A+B}(t)$ would be:</p>
<ul>
<li>either $t \leadsto^* \texttt{inl}(a)$ and $R_A(a)$</li>
<li>or $t \leadsto^* \texttt{inr}(b)$ and $R_B(b)$</li>
<li>or $t \leadsto^* \textit{ne} = x(t_1)\dots(t_n)$ and $SN(\textit{ne})$</li>
</ul>
<p>[1] <a href="https://www.cambridge.org/core/journals/mathematical-structures-in-computer-science/article/div-classtitleintuitionistic-model-constructions-and-normalization-proofsdiv/15AE4B790FF9E4B1998CE92054DBD3CF" rel="noreferrer">COQUAND, T., & DYBJER, P. (1997). Intuitionistic model constructions and normalization proofs. Mathematical Structures in Computer Science, 7(1), 75-94.</a></p>
|
281,387 | <p>I've been reading Girard et al's 'Proofs and Types', which in Chapter 6 presents a proof of strong normalisation for the simply typed lambda calculus with products and base types. The proof is based on Tait's method of defining a set of 'reducible terms for type $T$' by induction on the type $T$. For example a term $t$ is reducible for type $A\to B$ if for all reducible $u$ for type $A$, $t\,u$ is reducible for type $B$.</p>
<p>However I don't see how to extend this proof to (binary) sums, which are covered in a later chapter, because the elimination for sums involves a 'parasitic' type which prevents one from defining reducible terms of type $A+B$ via the elimination by induction on $A+B$.</p>
<p>How (if at all) does this proof style extend to sums? I can't find a good reference that presents this.</p>
| cody | 36,103 | <p>Guillaume's answer is correct, that is you may define the candidate for $A+B$ as the smallest set containing
$$ \mathrm{inl}(a)\ a\in R_A(a)$$
$$ \mathrm{inr}(b)\ b\in R_B(b)$$</p>
<p>and which is a candidate, i.e. closed under anti-reductions and inclusion of neutral terms. But it is possible to define an "elim" candidate as well which states that $t$ is "well behaved in elimination contexts". This would look something like:</p>
<p>$$R_{A+B}(t)\Longleftrightarrow \forall C\ l\ r, R_{A\rightarrow C}(l)\wedge R_{B\rightarrow C}(r)\Longrightarrow R_C(\mathrm{case}\ t\ \mathrm{of}\ \mathrm{inl} \Rightarrow l,\mathrm{inr}\Rightarrow r)$$</p>
<p>This is a little impredicative, since nothing forces $C$ to be simpler than $A+B$, but that's ok: impredicative definitions can be done by taking the intersection of all "candidate" relations $R_T$! This is exactly the trick for defining interpretations for system $F$.</p>
|
98 | <p>I am looking for a reference (book or article) that poses a problem that seems to be a classic, in that I've heard it posed many times, but that I've never seen written anywhere: that of the possibility of a man in a circular pen with a lion, each with some maximum speed, avoiding capture by that lion. </p>
<p>References to pursuit problems in general would also be appreciated, and the original source of this problem.</p>
| Jan Gorzny | 50 | <p>Since you're asking for a reference, perhaps this will do?</p>
<p>Wolfram Mathworld says the problem was listed by Rado in 1925. The reference is on the problem description page, <a href="http://mathworld.wolfram.com/LionandManProblem.html">here</a>.</p>
|
334,541 | <p>The standard frame for <span class="math-container">$S^3$</span> consists of <span class="math-container">$X_i,X_j,X_k$</span> with <span class="math-container">$X_i(a)=ia, X_j(a)=ja, X_k(a)=ka$</span> where <span class="math-container">$i,j,k$</span> are standard quaternion numbers, <span class="math-container">$a\in S^3$</span>, and the multiplication is the quaternion multiplication.
This global frame on <span class="math-container">$S^3$</span> gives us a trivialization of <span class="math-container">$TS^3\simeq S^3\times \mathbb{R}^3$</span>. So the Hopf map <span class="math-container">$P:S^3\to S^2\subset \mathbb{R}^3$</span> is counted as a unit vector field on <span class="math-container">$S^3$</span>. This vector field is called "Hopf vector field".</p>
<p>Does the Hopf vector field have a periodic orbit? Is this vector field discussed in various attempts to find an analytic counter example to <a href="https://en.m.wikipedia.org/wiki/Seifert_conjecture" rel="nofollow noreferrer">Seifert conjecture</a>?</p>
| Gael Meigniez | 105,095 | <p>Seifert actually proved that every vector field close enough to X_i has a periodic orbit. (In other words, if you perturbate slightly X_i whose all orbits are compact, then most orbits will of course in general become noncompact, but there will remain at least one compact orbit).
K. Kuperberg built on S^3 a smooth (C^infty) nonsingular vector field without periodic orbit; W. Thurston noticed that one can even make Kuperberg's example real analytic. So, it is not "attempts" any more.</p>
|
2,658,368 | <p>I know that one can use Category Theory to formulate polynomial equations by modeling solutions as limits. For example, the sphere is the equalizer of the functions
\begin{equation}
s,t:\mathbb{R}^3\rightarrow\mathbb{R},\qquad s(x,y,z):=x^2+y^2+z^2,~t(x,y,z)=1.
\label{equalizer}
\end{equation}
One could then find out more about the solution set by mapping the equalizer diagram into other categories. More generally, solution sets of polynomial equations (and more generally, <a href="https://en.wikipedia.org/wiki/Algebraic_variety" rel="noreferrer">algebraic varieties</a>) are a central study object of algebraic geometry.</p>
<p>As differential equations are central to all areas of physics, I assume that there have been made a lot of attempts to generalise these ideas to solution sets of these. However, I do not yet have a lot of knowledge about algebraic geometry, topos theory or synthetic differential geometry. Thus I would be grateful if someone could explain roughly where and how Category Theory is used to study differential equations. </p>
<p>Can Category Theory really <strong>help to solve</strong> differential equations (for example by mapping diagrams of equations to other categories, similarly to how problems of topology are often solved by mapping topological spaces to algebraic ones in algebraic topology) or can it "only" provide schemes for generalising differential equations to other spaces/categories?</p>
<p>I am particularly interested in names of areas I have to look into if I want to understand this better. Also literature recommendation would be very welcome.
<br><br><br>
<strong>EDIT</strong>: I found a book by Vinogradov called <a href="http://books.google.dk/books?id=XIve9AEZgZIC" rel="noreferrer">Cohomological Analysis of Partial Differential Equations and Secondary Calculus</a> where "the main result [...] is Secondary Calculus on diffieties". </p>
<p>However, the material is very deep and thus I am still not completely able to say whether these "new geometrical objects which are analogs of algebraic varieties" can be used to help solving PDEs or if they serve to structure the theory of PDEs or result in other applications I am not aware of. Thus further information would be very appreciated!</p>
| Community | -1 | <p>There is this observation of Marvan <a href="https://ncatlab.org/nlab/files/MarvanJetComonadPDE.pdf" rel="nofollow noreferrer">A Note on the Category of PDEs</a> that the jet bundle construction in ordinary differential geometry has the structure of a <a href="https://en.wikipedia.org/wiki/Monad_(category_theory)#Comonads_and_their_importance" rel="nofollow noreferrer">Comonad</a>, whose <a href="https://en.wikipedia.org/wiki/Monad_(category_theory)#Algebras_for_a_monad" rel="nofollow noreferrer">Eilenberg-Moore</a> category of coalgebras is equivalent
to <a href="https://www.researchgate.net/publication/225690652_Category_of_nonlinear_differential_equations" rel="nofollow noreferrer">Vinogradov’s category of PDEs</a>.</p>
<p>This 'synthetic' generalization of the jet bundle construction exhibits as the base change a comonad along the unit of the “infinitesimal shape” functor, which can then be shown to be the differential geometric analog of Simpson’s “de Rham shape” operation in algebraic geometry.</p>
<p><strong>Edit</strong>
In response to @Exchange's request for more recent developments, the work by Khavkine and Schreiber comes to mind, in a similar vein to the work undertaken by the authors above.</p>
<p>They were able to expand on the work of Marvan,and present a formal theory of PDEs in Synthetic Differential Geometry. Using <a href="https://en.wikipedia.org/wiki/Topos#Geometric_morphisms" rel="nofollow noreferrer">Topos</a> theory Khavkine and Screiber exhibit a synthetic generalization of the jet bundle construction. Under this generalisation the authors show that this is always equivalent to the Eilenberg-Moore category over the synthetic jet comonad.</p>
<p>They expand on this result by showing that whenever the unit of the “infinitesimal shape” <span class="math-container">$\scr{S}$</span> operation is epimorphic the category of formally integrable PDEs with independent variables ranging in some <span class="math-container">$\Sigma$</span> is also equivalent simply to the slice category over <span class="math-container">$\scr{S} \Sigma$</span>. This yields in particular a convenient presentation of the categories of PDEs in general contexts.</p>
<p>For a more comprehensive account, see <a href="https://arxiv.org/pdf/1701.06238.pdf" rel="nofollow noreferrer">the paper from 2017</a></p>
<p>Let me know if you need any more info,</p>
<p>Best</p>
<p>Kevin</p>
|
391,572 | <p>I never really understood what $e$ means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that?</p>
| Hagen von Eitzen | 39,174 | <p>On can show that there are functions $f(x)$ with the property
$$\tag1f'(x)=f(x)$$ for all $x$. If one additionally requires $f(0)=1$, the solution is <em>unique</em> and called $\exp(x)$. From this property alone, it follows that the general solution to $f'(x)=f(x)$ is $f(x)=c\exp(x)$ with $c=f(0)$. And also that $$\tag2\exp(x+y)=\exp(y)\exp(x)$$ for all $x,y$
because $x\mapsto \exp(x+y)$ is also a solution of $(1)$. Especially, $\exp(n)=\exp(1)\exp(1)\ldots\exp(1)=(\exp(1))^n$. In effect, because of $(2)$, the function $x\mapsto \exp(x)$ suspiciously behaves like exponentiation and is therefore usually suggestively written like exponentiation, i.e. we write $e^x$ instead of $\exp(x)$. Of course for this to make sense, we must define $e=\exp(1)$. It turns out that $\exp(1)\approx 2.71828$.</p>
|
1,609,854 | <p>If a polyhedron is made only of pentagons and hexagons, how many pentagons can it contain? With the assumption of three polygons per vertex, one can prove there are 12 pentagons.</p>
<p>Let's not make that assumption, and only use pentagons. </p>
<p>12 pentagons: dodecahedron and <a href="https://en.wikipedia.org/wiki/Dodecahedron#Tetartoid" rel="nofollow noreferrer">tetartoid</a>.<br>
24 pentagons: <a href="https://en.wikipedia.org/wiki/Pentagonal_icositetrahedron" rel="nofollow noreferrer">pentagonal icositetrahedron</a>.<br>
60 pentagons: <a href="https://en.wikipedia.org/wiki/Pentagonal_hexecontahedron" rel="nofollow noreferrer">pentagonal hexecontahedron</a>.<br>
72 pentagons: <a href="http://dmccooey.com/polyhedra/DualSnubTruncatedOctahedron.html" rel="nofollow noreferrer">dual snub truncated octahedron</a>.<br>
132 pentagons: <a href="http://dmccooey.com/polyhedra/132Pentagons.html" rel="nofollow noreferrer">132-pentagon polyhedron</a>.<br>
180 pentagons: <a href="http://dmccooey.com/polyhedra/DualSnubTruncatedIcosahedron.html" rel="nofollow noreferrer">dual snub truncated icosahedron</a>. </p>
<p>Here's what the 132 looks like. </p>
<p><a href="https://i.stack.imgur.com/TW9h8.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TW9h8.gif" alt="132 pentagon polyhedron"></a></p>
<p>In that range of 12 to 180, what values are missing? For values missing here where an all-pentagon polyhedron exists, what is the most symmetrical polyhedron for that value?</p>
<p>Edit: According to <a href="http://www.emis.de/journals/JGAA/accepted/2011/HasheminezhadMcKayReeves2011.15.3.pdf" rel="nofollow noreferrer">Hasheminezhad, McKay, and Reeves</a>, there are planar graphs that lead to 16, 18, 20, and 22 pentagonal faces, but I've never seen these polyhedra.</p>
<p>16 would be the dual of the <a href="https://en.wikipedia.org/wiki/Snub_square_antiprism" rel="nofollow noreferrer">snub square antiprism</a>.<br>
20 would be the dual of this graph:<br>
<a href="https://i.stack.imgur.com/GDnez.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GDnez.gif" alt="quintic 20,1"></a></p>
| Vincenzo Zaccaro | 269,380 | <p>Another way is: </p>
<p>$\dfrac {1}{z^2}=\dfrac {\left(1-\dfrac{1}{z}\right)^2}{(z-1)^2}=
\dfrac {(1-z)^2+2(1-z)^3+3(1-z)^4+4(1-z)^5+ \ldots}{(1-z)^2}=1+2(1-z)+3(1-z)^2+4(1-z)^3+ \ldots$ </p>
<p>using the product of convergent power series.</p>
|
1,869,981 | <p>I have a question in a book which says in how many ways can 6 different things be divided between 2 boys and (my understanding of) the explanation goes something along the lines of:</p>
<p>Items: 1 1 1 1 1 1</p>
<p>The first item can be distributed to either boy 1 or boy 2, i.e. 2 choices the second item could be distributed to either boy 1 or boy 2 therefore you get 2 choices again.
Therefore the number of choices are $2\times2\times2\times2\times2\times2$.</p>
<p>However if I thought of it in another way such as you have 6 items therefore boy 1 can get all 6 and boy 2 could get 0, you could organise it into pairs such that:</p>
<p>\begin{array}{c c}
boy 1 & boy 2 \\
0 & 6 \\
1 & 5 \\
2 & 4 \\
3 & 3 \\
4 & 2 \\
5 & 1 \\
6 & 0
\end{array}</p>
<p>Therefore there are 7 ways of distributing the items. Where am I wrong with my thoughts? I cannot see why my way of thinking is wrong.</p>
| smcc | 354,034 | <p>Your first method correctly counts the number of ways of distributing six different things to the two boys.</p>
<p>Your second method is wrong because it does not take account of the fact that the items are different. </p>
<p>Your second method (if you do it correctly) would give the number of ways of dividing six identical things between two people. (I think with the second method you meant to count the 7 possible cases: Boy 1 gets $0,1,2,3,4,5$ or $6$ things.)</p>
<p>When the objects are distinguishable from each other, there are more distinguishable ways to distribute them. Suppose the objects are balls. If the balls are identical then there is only one way to give the first boy one ball (the balls cannot be distinguished from each other). If the balls are all different colours, then there are multiple ways of giving the first boy one ball (the ball can be six different colours).</p>
|
1,057,172 | <p>How many 3digit numbers can be written with $2,4,4,6,6$ ? </p>
<p>I tried $\frac{5.4.3}{2!.2!} = 15$ but it's wrong. </p>
<p>when I solved the question "how many 3digit numbers can be written with $1,1,2$"
the solution $\frac{3.2.1}{2!}$ was correct but why this way doesn't work for above question?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$</p>
<blockquote>
<p>$\ds{\, P_{\sigma} = \half}$ is the probability of taking a jump of length $\ds{\sigma\ =\ \pm 1}$. So, the desired answer is given by:</p>
</blockquote>
<p>\begin{align}&\color{#66f}{\large%
\sum_{\sigma_{1}\ =\ \pm}\sum_{\sigma_{2}\ =\ \pm}\ldots\sum_{\sigma_{N}\ =\ \pm} \, P_{\sigma_{1}}\, P_{\sigma_{2}}\ldots \, P_{\sigma_{N}}
\delta_{\sigma_{1} + \sigma_{2}+\cdots+\sigma_{N},1}}
\\[5mm]&=\sum_{\sigma_{1}\ =\ \pm}\sum_{\sigma_{2}\ =\ \pm}\ldots\sum_{\sigma_{N}\ =\ \pm} \, P_{\sigma_{1}}\, P_{\sigma_{2}}\ldots \, P_{\sigma_{N}}
\oint_{\verts{z}\ =\ 1}
{1 \over z^{\sigma_{1} + \sigma_{2} + \cdots + \sigma_{N}}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}
\pars{\sum_{\sigma\ =\ \pm}P_{\sigma}z^{-\sigma}}^{N}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}
\pars{\half\, z + \half\,{1 \over z}}^{N}\,{\dd z \over 2\pi\ic}
\\[5mm]&={1 \over 2^{N}}\oint_{\verts{z}\ =\ 1}{\pars{1 + z^{2}}^{N} \over z^{N}}
\,{\dd z \over 2\pi\ic}
={1 \over 2^{N}}\sum_{\ell\ =\ 0}^{N}{N \choose \ell}\oint_{\verts{z}\ =\ 1}
{1 \over z^{N - 2\ell}}\,{\dd z \over 2\pi\ic}
\\[5mm]&={1 \over 2^{N}}\sum_{\ell\ =\ 0}^{N}{N \choose \ell}\delta_{2\ell,N - 1}
=\color{#66f}{\large{1 \over 2^{N}}\,{N \choose \bracks{N - 1}/2}}
\quad\dsc{{\tt\mbox{if}}\ N\ {\tt\mbox{is odd and}}\ N \geq 1}.
\\&\dsc{\mbox{Otherwise, it vanishes out}}.
\end{align}</p>
<p><img src="https://i.stack.imgur.com/TLdld.png" alt="enter image description here"></p>
|
112,889 | <p>I am new in Wolfram Math and I need a help in making a simple program. Let's suppose we have a list:</p>
<pre><code>list = {{1, 11}, {2, 7}, {4, 2}, {7, 9}, {-2, 3}, {-1, 10}};
</code></pre>
<p>Now, I need to collect the first elements of sublists, but not all of them, only those whose second elements are larger than 8. </p>
<p>Thanks in advance. </p>
| kglr | 125 | <pre><code>Pick[list[[All, 1]], # > 8 & /@ list[[All, 2]]]
</code></pre>
<p>or</p>
<pre><code>Pick[#[[1]], # > 8 & /@ #[[2]]] &@Transpose[list]
</code></pre>
<blockquote>
<p>{1, 7, -1}</p>
</blockquote>
|
427,663 | <p>If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $</p>
<p>i found one link that had a similar problem.
<a href="https://math.stackexchange.com/questions/343982/prove-if-z-1-and-w-1-then-1-zw-neq-0-and-z-w-over-1">Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$</a></p>
| user1337 | 62,839 | <p>We have $|\zeta|=1 \Leftrightarrow \bar{\zeta}=\frac{1}{\zeta}$.</p>
<p>$|z|=|w|=1 \Leftrightarrow \bar{z}=\frac{1}{z} \text{and } \bar{w}=\frac{1}{w} \Leftrightarrow \frac{\bar{z}+\bar{w}}{1+\bar{z}\bar{w}}=\frac{\frac{1}{z}+\frac{1}{w}}{1+\frac{1}{z} \frac{1}{w}} \Leftrightarrow \overline{ \left(\frac{z+w}{1+zw}\right)}=\frac{z+w}{1+zw} \Leftrightarrow \frac{z+w}{1+zw} \in \mathbb R$ </p>
|
427,663 | <p>If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $</p>
<p>i found one link that had a similar problem.
<a href="https://math.stackexchange.com/questions/343982/prove-if-z-1-and-w-1-then-1-zw-neq-0-and-z-w-over-1">Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$</a></p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>As $z \bar z=|z|^2=1$ and similarly for $w,$</p>
<p>$$\frac{z+w}{1+zw}=\frac{\frac1{\bar z}+\frac1{\bar w}}{1+\frac1{\bar z}\frac1{\bar w}}=\frac{\bar w+\bar z}{1+\bar w\bar z}$$</p>
<p>$$\text{Using }\frac ab=\frac cd=\frac{a+c}{b+d},$$</p>
<p>$$\frac{z+w}{1+zw}=\frac{\bar w+\bar z}{1+\bar w\bar z}=\frac{z+w+\bar w+\bar z}{1+zw+1+\bar w\bar z}=\frac{(z+\bar z)+(w+\bar w)}{2+(zw+\overline{wz})}$$ as $\bar z\cdot \bar w=\overline{z\cdot w}$</p>
<p>Observe that each pair within parentheses is real</p>
|
1,644,715 | <p>Define a locally lipschitz and nonnegative function $f\colon\mathbb{R}^n\to\mathbb{R}$. Let $M\in\mathbb{R}^{n\times n}$ and $\eta>0\in\mathbb{R}$. Consider the function $h\colon\mathbb{R}^n\to\mathbb{R}^n$ defined as </p>
<p>$$h(\mathbf{x})=
\begin{cases}
\frac{1}{\Vert M\mathbf{x}\Vert}M\mathbf{x}, &\text{if }f(\mathbf{x})\Vert M\mathbf{x}\Vert\ge\eta,\\
\frac{f(\mathbf{x})}{\eta}M\mathbf{x}, &\text{if }f(\mathbf{x})\Vert M\mathbf{x}\Vert<\eta.
\end{cases}$$</p>
<p>Show $h$ is lipschitz on any compact subset $\mathcal{D}\subseteq\mathbb{R}^n$.</p>
<hr>
<p>Let $\mathbf{x},\mathbf{y}\in\mathcal{D}$, then $h$ is Lipschitz on $\mathcal{D}\subseteq\mathbb{R}^n$ if
$$\Vert h(\mathbf{x})-h(\mathbf{y})\Vert\le L\Vert \mathbf{x}-\mathbf{y}\Vert$$</p>
<p>for some Lipschitz constant $L>0\in\mathbb{R}$.</p>
<hr>
<p>What is the best plan of attack for this? I initially split this into two cases when $f(\mathbf{x})\Vert M\mathbf{x}\Vert\ge\eta$ and when $f(\mathbf{x})\Vert M\mathbf{x}\Vert<\eta$. I then tried to write $\Vert h(\mathbf{x})-h(\mathbf{y})\Vert$ in terms of a const. multiplied by $\Vert\mathbf{x}-\mathbf{y}\Vert$ then we can take $L=$ const. but I come to difficulties since $f$ is only locally lipschitz.</p>
| copper.hat | 27,978 | <p>Note that if $x$ is such that $f(x)=0$ or $Mx=0$ then there is a neighbourhood $U$ of $x$ such that $f(y) \|My\| < \eta$ for $y \in U$. Hence $h$ is locally Lipschitz in a neighbourhood of any point such that $f(x)=0$ or $Mx=0$.</p>
<p>Suppose $x$ is such that $f(x) \neq 0$ and $Mx \neq 0$.
Let $\phi(y) = \min({1 \over \|My\|}, {f(y) \over \eta})$, this is well defined
in a neighbourhood of $x$ and furthermore it is locally Lipschitz in this
neighbourhood (since $f$ is locally Lipschitz and the function $t \mapsto {1 \over t}$ is locally Lipschitz and compositions of locally Lipschitz functions is again locally Lipschitz). Since $h = f \cdot \phi$ it follows
that $h$ is locally Lipschitz (since products of locally Lipschitz functions
are locally Lipschitz).</p>
<p>A slightly more involved approach would be to use the Rademacher theorem.</p>
|
2,835,177 | <p>For example 6,</p>
<p>convert to the base 2 = 110</p>
<p>number of 1's is 2(prime), number of zero's is 1(prime - 1)</p>
<p>or 496 = (111110000)2</p>
<p>5(prime) times 1 and 4 times 0</p>
<p>Is this always correct?</p>
| Graham Kemp | 135,106 | <p>Use the Law of Total Probability: for example $d6, d10$ the results of independen six and ten sided dice.</p>
<p>$$\begin{align}\mathsf P(d6>d10) &= \mathsf P(d10>6)\mathsf P(d6>d10\mid d10>6)+\mathsf P(d10\leq 6)\mathsf P(d6>d10\mid d10\leq 6) \\ &=\tfrac 4{10}\cdot 1+\tfrac 6{10}\cdot\mathsf P(d6>d10\mid d10\leq 6)\\ &=\tfrac 2{5}+\tfrac 3 5\cdot\mathsf P(d6>d10\mid d10\leq 6)\end{align}$$</p>
<p>All that is left is to evaluate that last term</p>
<p>Hints: $1=\mathsf P(d6>d10\mid d10\leq 6)+\mathsf P(d6=d10\mid d10<6)+\mathsf P(d6<d10\mid d10\leq 6)\\\mathsf P(d6>d10\mid d10\leq 6)=\mathsf P(d6<d10\mid d10\leq 6)$</p>
<p>Also, when given the condition that it is at most 6, the distribution of a 10 sided die is identical to the distribution of a six sided die .</p>
<p>Extend this principle to account for selections from any two independent uniform discrete distributions from non-identical supports.</p>
|
54,395 | <p>A generic question: </p>
<p>are there any spectral techniques to estimate the genus of a graph? I am interested in complete balance multipartite graph.</p>
| David E Speyer | 297 | <p>I believe that this is the subject of Jon Kelner's paper <a href="http://math.mit.edu/~kelner/Publications/Docs/LowGenusJournal.pdf" rel="nofollow">Spectral partitioning, eigenvalue bounds, and circle packings for graphs of bounded genus</a>. In particular, he proves a lower bound of the form $O(g/n)$ for $\lambda_2$, resolving a conjecture of Spielman and Teng.</p>
|
341,946 | <p>Using $n$-th Taylor polynomial for $f_1(x)=\frac{1}{1-x}$ with center in $0$, find $4$-th derivative of $f_2(x)=\frac{1+x+x^2}{1-x+x^2}$ in the point $0$ without calculating it's $1$,$2$ or $3$ derivative. </p>
<p>I'm looking for hints, it's a homework. I've tried using the Taylor expansion with Lagrange rest, but it won't work because of restriction of calculating $1-3$ derivatives. Also I don't see connection between $f_1$ and $f_2$</p>
<p>Thanks in advance for help!</p>
| Chris Godsil | 16,143 | <p>This an answer to the version of the question given in Mike's comment. Note first that for any polynomials $p$ and $q$ we have $(xp,q)=(p,xq)$. Second, note that $P_n$ is orthogonal to all polynomials of degree less than $n$. </p>
<p>So if $m \le n-2$, then
$$
(xP_n,P_m) = (P_n,xP_m) = 0.
$$</p>
<p>But we know that $xP_n$ must be a linear combination of $P_0,\ldots,P_n,P_{n+1}$
(because this holds for any polynomial of degree at most $n+1$), and the coefficient of $P_m$ in this expansion if $(xP_,P_m)/(P_m,P_m)$. Therefore $xP_n$ is a linear combination of $P_{n-1}$, $P_n$ and $P_{n+1}$.</p>
<p>Finally note that this works for any family of orthogonal polynomials, not just Legendre polynomials.</p>
|
747,286 | <p>Pardon my ignorance. I don't know enough calculus to understand this. I assume this is a very easy question for this amazing site. I saw this on the The Theory of Riemann Zeta Function Book.</p>
<p>$$\sum_{n=2}^{\infty}\pi (n)\int_{n}^{n+1}\frac{s}{x(x^{s}-1)}dx =s\int_{2}^{\infty}\frac{\pi (x)}{x(x^{s}-1)}dx$$</p>
<p>where $\pi(n)$ is the prime counting function.</p>
<p>Can someone explain to me how did that sum symbol get lost?</p>
| ploosu2 | 111,594 | <p>Linearity of integral and $\pi$ is constant $\pi (n)$ on the interval $[n, n+1)$.</p>
<p>EDIT: Also: the set $[2, \infty)$ is broken to intervals $[n, n+1],$ that's not called linearity... (is it additivity?)</p>
|
747,286 | <p>Pardon my ignorance. I don't know enough calculus to understand this. I assume this is a very easy question for this amazing site. I saw this on the The Theory of Riemann Zeta Function Book.</p>
<p>$$\sum_{n=2}^{\infty}\pi (n)\int_{n}^{n+1}\frac{s}{x(x^{s}-1)}dx =s\int_{2}^{\infty}\frac{\pi (x)}{x(x^{s}-1)}dx$$</p>
<p>where $\pi(n)$ is the prime counting function.</p>
<p>Can someone explain to me how did that sum symbol get lost?</p>
| Brian | 114,928 | <p>First, remember that if $a<b<c$, then $\int_{a}^{c}f\left(x\right)\;dx=\int_{a}^{b}f\left(x\right)\;dx+\int_{b}^{c}f\left(x\right)\;dx$. Also, since $\pi$ is constant on $\left[n,n+1\right)$, you can push it into the integral as $\pi\left(x\right)$. Then, if you look at the partial sums, at the $n$th term, you have</p>
<p>$$\pi\left(2\right)\int_{2}^{3}\frac{s}{x\left(x^{s}-1\right)}dx+\pi\left(3\right)\int_{3}^{4}\frac{s}{x\left(x^{s}-1\right)}dx+\cdots+\pi\left(n\right)\int_{n}^{n+1}\frac{s}{x\left(x^{s}-1\right)}dx=\\s\int_{2}^{3}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx+s\int_{3}^{4}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx+\cdots+s\int_{n}^{n+1}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx=\\s\int_{2}^{n+1}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx.$$</p>
<p>The sum is just
$$\lim_{n\to\infty}s\int_{2}^{n+1}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx=s\int_{2}^{\infty}\frac{\pi\left(x\right)}{x\left(x^{s}-1\right)}dx.$$</p>
|
1,717,775 | <p>I have defined, for a probability measure $\eta$ we have the fourier transform as $\hat{\eta} = \int e^{itx} \ d\eta(x)$, and for a function $h: \mathbb{R} \to \mathbb{R}$ we have that the fourier transform of $h$ is given by $\hat{h}(t) = \int e^{itx} h(x) \ dx$.</p>
<p>I then am told that the fourier transform of a probability measure, is the fourier transform of its density if it has one. </p>
<p>I don't really understand this statement - what is the density of a measurable function? I thought random variables are those to have densities. Any explanation please, thanks.</p>
| Henry | 313,321 | <p>Since you are dealing with non-normal distributions, shouldn't you be using tests for non-normally distributed data? Also, when the sample size is less than 30, you would usually conduct a t-test. In this case, the t-test equivalents for non-normally distributed data are the following:Mann-Whitney test, Mood’s median test, and Kruskal-Wallis test.</p>
|
1,717,775 | <p>I have defined, for a probability measure $\eta$ we have the fourier transform as $\hat{\eta} = \int e^{itx} \ d\eta(x)$, and for a function $h: \mathbb{R} \to \mathbb{R}$ we have that the fourier transform of $h$ is given by $\hat{h}(t) = \int e^{itx} h(x) \ dx$.</p>
<p>I then am told that the fourier transform of a probability measure, is the fourier transform of its density if it has one. </p>
<p>I don't really understand this statement - what is the density of a measurable function? I thought random variables are those to have densities. Any explanation please, thanks.</p>
| grand_chat | 215,011 | <p>Try computing confidence intervals for samples from distributions that deviate more from normality. For example, try generating observations from a Poisson distribution with $\lambda=.001$. You'll find that your $z$-theory intervals will not have the advertised coverage.</p>
<p>In general, you'll find that the normal theory works pretty well if the distribution you're sampling from is reasonably symmetric, so aim for samples from non-symmetric distributions, or distributions that are almost a point mass, such as Poisson with tiny $\lambda$. Keep away from the uniform distribution; the normal approximation kicks in pretty quickly no matter how you scale it.</p>
<p>Knowing the variance takes some of the 'noise' out of the confidence interval. So yes, this will improve coverage over using an estimator for the standard deviation.</p>
|
4,553,447 | <p>In order to create a Monte-Carlo Simulation for Mertons Jump Diffusion Model I need to simulate a random variable <span class="math-container">$Y$</span>, which is log-normally distributed. The mean of <span class="math-container">$Y$</span> is supposed to be <span class="math-container">$m$</span> and the variance of the logarithm of <span class="math-container">$Y$</span> should be <span class="math-container">$v^2$</span>.
The problem I have is that I can only specify either both the mean and variance of the logarithm of <span class="math-container">$Y$</span> <strong>or</strong> both the mean and variance of <span class="math-container">$Y$</span> directly, via the following formulae (see <a href="https://en.wikipedia.org/wiki/Log-normal_distribution#Generation_and_parameters" rel="nofollow noreferrer">Wikipedia</a>):</p>
<p><span class="math-container">$$
\mu=\ln \left(\frac{m^2}{\sqrt{m^2+v^2}}\right) \text { and } \sigma^2=\ln \left(1+\frac{v^2}{m^2}\right).
$$</span>
<span class="math-container">$Y \sim \mathcal{LN}( \ln \left(\frac{k^2}{\sqrt{k^2+v^2}}\right),v^2)$</span> (as a combination of the two) unfortunately does not generate the desired distribution. I need a way to specify <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma$</span> of the log-normal distribution that satisfies the above criterium.</p>
<p><strong>Edit:</strong> I accidentally wrote that the variance of the logarithm of Y needs to be equal to <span class="math-container">$v$</span> instead of <span class="math-container">$v^2$</span>. Thanks to heropup for the comment.</p>
| Narasimham | 95,860 | <p>From trig figure directly</p>
<p><a href="https://i.stack.imgur.com/Yq9Vt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yq9Vt.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ \text{L is ladder length},~x = L\sin \theta~, \frac{dx}{d\theta} =L\cos \theta=50\cos 45^{\circ} \tag 1$$</span></p>
<p>And if you want the rate for any other <span class="math-container">$x,$</span> there is the circle relation</p>
<p><a href="https://i.stack.imgur.com/1HsJJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1HsJJ.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ x^2+ \left( \frac{dx}{d\theta}\right)^2 =L^2 \tag 2 $$</span></p>
<p>obtained by eliminating <span class="math-container">$\theta$</span> in equation 1)</p>
|
4,553,447 | <p>In order to create a Monte-Carlo Simulation for Mertons Jump Diffusion Model I need to simulate a random variable <span class="math-container">$Y$</span>, which is log-normally distributed. The mean of <span class="math-container">$Y$</span> is supposed to be <span class="math-container">$m$</span> and the variance of the logarithm of <span class="math-container">$Y$</span> should be <span class="math-container">$v^2$</span>.
The problem I have is that I can only specify either both the mean and variance of the logarithm of <span class="math-container">$Y$</span> <strong>or</strong> both the mean and variance of <span class="math-container">$Y$</span> directly, via the following formulae (see <a href="https://en.wikipedia.org/wiki/Log-normal_distribution#Generation_and_parameters" rel="nofollow noreferrer">Wikipedia</a>):</p>
<p><span class="math-container">$$
\mu=\ln \left(\frac{m^2}{\sqrt{m^2+v^2}}\right) \text { and } \sigma^2=\ln \left(1+\frac{v^2}{m^2}\right).
$$</span>
<span class="math-container">$Y \sim \mathcal{LN}( \ln \left(\frac{k^2}{\sqrt{k^2+v^2}}\right),v^2)$</span> (as a combination of the two) unfortunately does not generate the desired distribution. I need a way to specify <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma$</span> of the log-normal distribution that satisfies the above criterium.</p>
<p><strong>Edit:</strong> I accidentally wrote that the variance of the logarithm of Y needs to be equal to <span class="math-container">$v$</span> instead of <span class="math-container">$v^2$</span>. Thanks to heropup for the comment.</p>
| zeeshan khan | 1,107,512 | <p>@Narasimham did a nice job but I'll give another answer to this</p>
<p><span class="math-container">$$x/50 =tan(θ) $$</span>
<span class="math-container">$$x = 50 [tan(θ)]$$</span>
<span class="math-container">$$dx/dθ = 50 sec^2(θ)$$</span>
<span class="math-container">$$put θ= 45$$</span>
<span class="math-container">$$dx/dθ= 50sec^2(45)$$</span>
<span class="math-container">$$dx/dθ= 50(sqrt2)^2 = 50 (2) = 100$$</span></p>
|
4,553,447 | <p>In order to create a Monte-Carlo Simulation for Mertons Jump Diffusion Model I need to simulate a random variable <span class="math-container">$Y$</span>, which is log-normally distributed. The mean of <span class="math-container">$Y$</span> is supposed to be <span class="math-container">$m$</span> and the variance of the logarithm of <span class="math-container">$Y$</span> should be <span class="math-container">$v^2$</span>.
The problem I have is that I can only specify either both the mean and variance of the logarithm of <span class="math-container">$Y$</span> <strong>or</strong> both the mean and variance of <span class="math-container">$Y$</span> directly, via the following formulae (see <a href="https://en.wikipedia.org/wiki/Log-normal_distribution#Generation_and_parameters" rel="nofollow noreferrer">Wikipedia</a>):</p>
<p><span class="math-container">$$
\mu=\ln \left(\frac{m^2}{\sqrt{m^2+v^2}}\right) \text { and } \sigma^2=\ln \left(1+\frac{v^2}{m^2}\right).
$$</span>
<span class="math-container">$Y \sim \mathcal{LN}( \ln \left(\frac{k^2}{\sqrt{k^2+v^2}}\right),v^2)$</span> (as a combination of the two) unfortunately does not generate the desired distribution. I need a way to specify <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma$</span> of the log-normal distribution that satisfies the above criterium.</p>
<p><strong>Edit:</strong> I accidentally wrote that the variance of the logarithm of Y needs to be equal to <span class="math-container">$v$</span> instead of <span class="math-container">$v^2$</span>. Thanks to heropup for the comment.</p>
| Narasimham | 95,860 | <p>If the base is of <span class="math-container">$50$</span> meters length, then the figure should be scaled up by a magnification factor <span class="math-container">$\sqrt2.$</span></p>
<p>From trig figure directly</p>
<p><a href="https://i.stack.imgur.com/YVq4r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YVq4r.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ \text{LL is ladder length} = 50\sqrt 2,~x = LL\sin \theta~, \frac{dx}{d\theta} =LL\cos \theta\tag 1$$</span></p>
<p>And if you want the rate for any other <span class="math-container">$x,$</span> there is the circle relation</p>
<p><span class="math-container">$$ x^2+ \left( \frac{dx}{d\theta}\right)^2 =LL^2 \tag 2 $$</span></p>
<p>obtained by eliminating <span class="math-container">$\theta$</span> in equations 1)</p>
<p><a href="https://i.stack.imgur.com/YekHs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YekHs.png" alt="enter image description here" /></a></p>
|
89,666 | <p>As I am trying to get the solution of the simultaneous ordinary differential equations </p>
<pre><code> b1'[z]-1I*beta1*b1[z]-C1*b2[z]==0,
b2'[z]-1I*beta2*b2[z]+C1*b1[z]==0
</code></pre>
<p>with boundary conditions </p>
<pre><code>b1[1.581825567*10^-6]==0.876212, b2[1.581825567*10^-6]==0.481925
</code></pre>
<p>By running the following command</p>
<pre><code>S = DSolve[{b1'[z]-1I*beta1*b1[z]-C1*b2[z]==0, b2'[z]-1I*beta2*b2[z]+C1*b1[z]==0,
b1[1.581825567*10^-6]==0.876212, b2[1.581825567*10^-6]==0.481925,
(b1[z])^2 (b2[z])^2==1}, {b1, b2}, z]
</code></pre>
<p>I am getting the values of <code>b1[z]</code> and <code>b2[z]</code>. But my main problem is that whatever the values of <code>b1[z]</code> and <code>b2[z]</code>, I am getting after solving the above equation, the sum of there values i.e.<code>|b1|^2+|b2|^2</code> should be less then or equal to 1, i.e., <code>|b1|^2+|b2|^2 <= 1</code> (which is my third boundary condition); so please can anyone suggest me that how I will run the above equation with this third boundary condition <code>|b1|^2+|b1|^2 <=1</code>. So that I can get those values of <code>b1[z]</code> and <code>b2[z]</code> which are satisfying our third boundary condition i.e.<code>|b1|^2+|b2|^2</code>. Even I had tryied to run this programme with my third boundary condition. But it is showing some error whose snapshot I am attaching with this problem so please suggest me the right answer for the same problem. And please I have request just do no put simple comment, if it is possible try this problem in mathematica and then send to me.</p>
<p><a href="https://i.stack.imgur.com/FlqDj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FlqDj.png" alt="enter image description here"></a></p>
| m_goldberg | 3,066 | <p>We have seen this problem posted by you before. This time you don't seem to understand that you can't have more equations than you have dependent variables. You have three equations </p>
<pre><code>b1'[z] - 1 I*beta1*b1[z] - C1*b2[z] == 0
b2'[z] - 1 I*beta2*b2[z] + C1*b1[z] == 0
(b1[z])^2 + (b2[z])^2 == 1
</code></pre>
<p>and only two dependent variables, <code>b1</code> and <code>b2</code>. That one of the equations is not a <em>differential</em> equation doesn't mean it isn't a member of the system.</p>
<p>You may think as the last equation as a constraint, but <code>DSolve</code> doesn't see it that way.</p>
|
325,236 | <blockquote>
<p>Is there a bijection between $\mathbb N$ and $\mathbb N^2$?</p>
</blockquote>
<p>If I can show $\mathbb N^2$ is equipotent to $\mathbb N$, I can show that $\mathbb Q$ is countable. Please help. Thanks,</p>
| Dhruv Ranganathan | 42,999 | <p>You're trying to find a way to "count" the things in $\mathbb N^2$ in a manner that gets you to any pair $(a,b)$ in some finite time. Notice that "counting" something is essentially finding this bijection - there is a "first" pair, and a "second pair" and so on.</p>
<p>There are many ways to do this, but most of them come down to ordering the set $\mathbb N^2$ in some meaningful way. </p>
<p>Hint: There are countably many diagonals, i.e. there is a "first diagonal" a "second diagonal" and so on. Does this work? </p>
|
1,902,407 | <p>Expand given function $f$ as Taylor series around $c=3$
$$f(x) = \frac{x-3}{(x-1)^2}+\ln{(2x-4)} $$</p>
<p>and find out open interval at what that series is convergent. What is radius of convergence?</p>
<p>This is what i have so far. We know that $\ln{(1+x)} = \sum_{n=1}^{\infty} \frac{x^n}{n}$ and $\frac{1}{1-x}=\sum_{n=1}^\infty x^n ,|x| \lt 1$</p>
<p>We can rewrite $\ln{(2x-4)}$ as $\ln{(-\frac{1}{4})}+\ln{(-x/2+1)}$ and then expand last expression, but i don't know what to do with $\ln{(-1/4)}$. We can can split given fraction onto partial fractions as $\frac{x-3}{(x-1)^2} = \frac{1}{x-1}-\frac{2}{(x-1)^2}$. First fraction we can expand easily, but i don't know how to expand fraction with binomial as denominator.</p>
| Robert Z | 299,698 | <p>Try 3 and 11 by using <a href="https://en.wikipedia.org/wiki/Divisibility_rule" rel="nofollow">divisibility criteria</a>.
Note that $10^{100}-1$ is a sequence of 100 digits all equal to $9$.</p>
|
3,799,421 | <p>While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
<span class="math-container">$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$$</span>
where <span class="math-container">$R$</span> and <span class="math-container">$x$</span> are positive constants.
Can this integral be solved using substitution?</p>
| Claude Leibovici | 82,404 | <p><span class="math-container">$$\frac{1}{\sqrt{R^2+x^2-2 R x \cos (\theta )}}=\frac{1}{\sqrt{R^2+x^2}}\frac{1}{\sqrt{1-k\cos (\theta )}}$$</span> with <span class="math-container">$k=\frac{2Rx}{R^2+x^2}$</span>.
<span class="math-container">$$\int \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 2 {\sqrt{{1-k}}} F\left(\frac{\theta }{2}|\frac{2 k}{k-1}\right)$$</span>
<span class="math-container">$$\int_0^{2\pi} \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)$$</span></p>
<p>If <span class="math-container">$k$</span> is small, you can use the expansion
<span class="math-container">$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi \left(1+\frac{3 k^2}{16}+\frac{105 k^4}{1024}+\frac{1155 k^6}{16384} \right)+O\left(k^8\right)$$</span></p>
<p>If you want a much better approximation, you could use the Padé approximant
<span class="math-container">$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi\frac{1-\frac{497 }{576}k^2+\frac{3835}{36864}k^4 } {1-\frac{605 }{576}k^2+\frac{7315 }{36864}k^4 }$$</span></p>
|
2,202,498 | <p>Test whether the series $\sum_{n=1}^\infty\frac{1+\sin(n)}{2^n}$ is converges or diverges. I used the limit comparison method in order to test its convergence. I chose $a_n = \frac{1+\sin(n)}{2^n}$ and $b_n = \frac{1}{2^n}$.</p>
<p>$$
\lim_{n \rightarrow \infty}\frac{\frac{1+\sin(n)}{2^n}}{\frac{1}{2^n}} = \lim_{n \rightarrow \infty}({1+\sin(n)}) = \infty
$$</p>
<p>However, WolframAlpha suggests that the series is in fact convergent. If this is the case, then is the way I set up the limit comparison test incorrect or is there another test that proves this series is convergent? Any help would be appreciated!</p>
| AlexT | 428,860 | <p>You have a map $\phi: \mathbb{Z_7} \rightarrow M_2(\mathbb{Z_7})$ such that $\ f \mapsto f \begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix}$. Basically you evaluate all possible linear combinations of the matrix $\begin{pmatrix} 0 & 2 \\ 2 & 2 \end{pmatrix}$, which we'll call $A$ for brevity, as f varies in $\mathbb{Z_7}[x]$.</p>
<p>When does $\phi$ map a polynomial to $0$? It happens exactly when $f\in I(A)$, where $I(A)=\{p\in\mathbb{Z_7}[x] | p(A)=0\}$. We denote this set $I(A)$ because it is also an ideal in the ring $\mathbb{Z_7}[x]$. We would like to find a polynomial that generates $I$ - we'll call it $m(t)$. To find it we first calculate the characteristic polynomial of $A$, which is $p(t)=det(A-tI)$. Hamilton-Cayley's Theorem tells us that $p(t) \in I(A)$, so we know that $m(t) | p(t)$. We also know that the roots of both $p(t)$ and $m(t)$ are all the eigenvalues of A, which may eventually appear in $m(t)$ with a smaller algebraic multiplicity. However, we see that $p(t)=det(A-tI)= t^2-2t-4 \equiv t^2 +5t +3 \ (7)$ has two distinct roots, thus it must coincide with $m(t)$ for they are both of degree 2 and monic (we usually choose $m(t)$ with leading coefficient equal to 1).</p>
|
1,137,914 | <p>Can Someone please help me prove that the number 1729 is a Pseudoprime?</p>
<p>So a pseudoprime is a composite $n$ such that $n |(2^n − 2)$. And every prime
number also has this property.</p>
| Mikko Korhonen | 17,384 | <p>Suppose that $G$ has a composition series. If you take any series (proper inclusions)</p>
<p>$$1 < N_1 < \cdots < N_t < N_{t+1} = G$$</p>
<p>with each $N_i$ normal in $G$, by Jordan-Hölder the length of this series is at most the composition length of $G$. Hence we can find a minimal normal subgroup $N \neq 1$ of $G$. Then we find a chief series of $G$ by considering $G/N$ and applying induction on the composition length of $G$.</p>
|
262,985 | <p>$Ax=0$, $A$ has $m$ rows and $n$ columns, $m \le n$, all entries of $x$ are non-negative.</p>
<p>What should $A$ satisfy to guarantee the equation set have only zero solution?</p>
| Dima Pasechnik | 11,100 | <p>If a nonzero $x\geq 0$ is a solution to $Ax=0$ then $\sum_j x_j=a>0$, and $x$ is also a solution to the system
$$A x=0,\quad 0\leq x,\quad e^\top x\leq a,\qquad\qquad\qquad\qquad(*) $$
where $e$ denotes the all-1 vector.
It is a standard application of Farkas lemma in an appropriate form (one might want to use instead $e^\top x=a$) to check whether $(*)$ has a solution. Or, indeed, you may maximise $\sum_j x_j$ subject to $(*)$ and see whether it has maximum $a$ or $0$.</p>
|
4,174,481 | <p>I'm going through a proof in which at one point, the authors make the following statement :</p>
<blockquote>
<p>Let <span class="math-container">$(X,d)$</span> be a complete metric space and <span class="math-container">$f:X\to X$</span> and <span class="math-container">$f_m:X\to X$</span> Lipschitz with <span class="math-container">$lip(f)<1$</span> and <span class="math-container">$lip(f_m)<1\; \forall m\in\mathbb{N}$</span>. Let <span class="math-container">$x^*_m$</span> the fixed point of <span class="math-container">$f_m$</span> and <span class="math-container">$x^*$</span> the fixed point of <span class="math-container">$f$</span>.
If for each <span class="math-container">$x\in X$</span> we have that<br />
<span class="math-container">$$\lim_{m\to+\infty} d(f_m(x),f(x)) = 0, $$</span> it follows that <span class="math-container">$$\lim_{m\to+\infty} d(x^*_m,x^*) = 0 .$$</span></p>
</blockquote>
<p>I tried to figure out how they came up with this by an argument along the lines of
<span class="math-container">\begin{align*}d(x^*_m,x^*) &= d(f_m(x^*_m),f(x^*)) \leq d(f_m(x^*_m),f(x_m^*))+ d(f(x_m^*),f(x^*))\\&\leq d(f_m(x^*_m),f(x_m^*))+lip(f)d(x_m^*,x^*).
\end{align*}</span></p>
<p>Then
<span class="math-container">$$0\leq (1-lip(f))d(f(x_m^*),f(x^*))\leq d(f_m(x^*_m),f(x_m^*)).$$</span>
But now we can't apply the first limit because <span class="math-container">$x_m^*$</span> depends on <span class="math-container">$m$</span>.</p>
<p>Also another thing I tried is by using the fact that <span class="math-container">$$\lim_{n
\to+\infty}d(f^{\circ n}(x),x^*)=0 \; \forall x\in X$$</span>
Where <span class="math-container">$f^{\circ n} = \underbrace{f\circ f\circ \dots \circ f}_{\text {n times}}.$</span></p>
<p>Then for a fixed <span class="math-container">$x\in X$</span>
<span class="math-container">\begin{equation*}d(x^*_m,x^*)\leq d(x^*_m,f^{\circ n}_m(x)) + d(f^{\circ n}_m(x),f^{\circ n}(x))+ d(f^{\circ n}(x),x^*),\; \forall n\in\mathbb{N}
\end{equation*}</span></p>
<p>Now let <span class="math-container">$\varepsilon > 0$</span> and set <span class="math-container">$N$</span> big enough so that <span class="math-container">$d(f^{\circ N}(x),x^*)<\varepsilon/3$</span> and <span class="math-container">$d(f_m^{\circ N}(x),x_m^*)< \varepsilon/3$</span>.</p>
<p>Then <span class="math-container">\begin{align*}
d(x^*_m,x^*) &\leq d(x^*_m,f^{\circ N}_m(x)) + d(f^{\circ N}_m(x),f^{\circ N}(x))+ d(f^{\circ N}(x),x^*)\\&<2\varepsilon/3 + d(f^{\circ N}_m(x),f^{\circ N}(x)).
\end{align*}</span></p>
<p>Now it can be shown that <span class="math-container">$\lim\limits_{m\to+\infty} d(f^{\circ N}_m(x),f^{\circ N}(x)) =0$</span>, so there is some <span class="math-container">$M$</span> such that <span class="math-container">$d(f^{\circ N}_m(x),f^{\circ N}(x)) <\varepsilon/3\; \forall m>M$</span>, from which we get <span class="math-container">$$d(x^*_m,x^*)<\varepsilon.$$</span></p>
<p>But I feel like something is not right in my second attempt.</p>
<p>Is something missing from the statement?</p>
| Son Gohan | 865,323 | <p>I have a strong feeling that one needs uniform convergence and not only pointwise convergence to conclude this. Let me give you a partial answer (I'll think about a counterexample) when we assume <span class="math-container">$$\lim_{m\to\infty} \sup_{x \in X} d(f_m(x),f(x)) = 0 $$</span></p>
<p>Indeed note that under this assumption we can write:</p>
<p><span class="math-container">$$d(x^*_m,x^*) = d(f_m(x^*_m),f(x^*)) \leq \sup_{y \in X} d(f_m(y),f(y)) \xrightarrow{m \rightarrow \infty} 0$$</span></p>
<p>In your last attemp you cannot say "set <span class="math-container">$N$</span> big enough so that <span class="math-container">$d(f_m^{\circ N}(x),x_m^*)< \varepsilon/3$</span>" since to write this passage you need uniform convergence (as in the comment is kindly pointed out - I slipped on the dependence of <span class="math-container">$N$</span> on <span class="math-container">$m$</span>, as written in the comment below that does not allow to pass to the limit).</p>
|
3,789,429 | <blockquote>
<p><strong>Problem</strong>:
Can an <span class="math-container">$f$</span> function be created where:<span class="math-container">$$f\colon\mathbb Q_{+}^{*}\to \mathbb Q_{+}^{*}$$</span>
The function is defined on the set of fully positive rational numbers and is achieved:
<span class="math-container">$\forall(x,y)\in \mathbb Q_{+}^{*}\times\mathbb Q_{+}^{*},f(xf(y))=\frac{f(f(x))}{y}$</span></p>
</blockquote>
<p>This question is similar to one of the Olympiad questions that I was very passionate about and used several ideas to solve this problem, but I did not arrive at any result from one of them by using the basic theorem in arithmetic that states that there is a corresponding application between <span class="math-container">$(\mathbb Q_{+}^{*})$</span>and <span class="math-container">$(\mathbb Z^{\mathbb N})$</span> where:
<span class="math-container">$$\left\{\mathbb Z^{\mathbb N} =\text{ A set of stable sequences whose values are set in} \quad\mathbb Z\right\}$$</span>
This app is defined like this
<span class="math-container">$$\varphi\colon\mathbb Z^{\mathbb N}\to \mathbb Q_{+}^{*} ,(\alpha_n)_{n\in\mathbb N}\longmapsto \prod_{n\in\mathbb N} P_n^{\alpha_n}$$</span>
Where:<span class="math-container">$$\mathbb P=\left\{P_k:k\in\mathbb N\right\}\text{ is the set of prime numbers} $$</span>
And put <span class="math-container">$x=\prod_{n\in\mathbb N}P_n^{\alpha_n},\quad y=\prod_{n\in\mathbb N }P_n^{\beta_n},\text{and}\quad $</span>
<span class="math-container">$$f(\prod_{n\in\mathbb N}P_n^{\alpha_n})=\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)$$</span></p>
<p>Found in the latter<span class="math-container">$$f(xf(y))=\frac{\left(\prod_{n\in\mathbb N}P_{2n}^{\alpha_{2n+1}}\right)\left(\prod_{n\in\mathbb N}P_{2n+1}^{-\alpha_{2n}}\right)}{\left(\prod_{n\in\mathbb N}P_n^{\beta_{2n}}\right)}=\frac{f(x)}{y}$$</span>
However, this did not help me create this method</p>
<p>I need an idea or suggestion to solve this problem if possible and thank you for your help</p>
<blockquote>
<p>Note: <span class="math-container">$(\alpha_n)_{n\in\mathbb N}\quad \text{is a stable sequence}\leftrightarrow \forall n\in\mathbb N ,\exists n_0\in\mathbb N :\left( n\geq n_0 \quad \alpha_{n}=0\right) $</span></p>
</blockquote>
| freakish | 340,986 | <p>For <span class="math-container">$x=1$</span> we get <span class="math-container">$f(f(y))=f(f(1))/y$</span>. For <span class="math-container">$y=1$</span> we get <span class="math-container">$f(xf(1))=f(f(x))$</span>. And so if we replace <span class="math-container">$y$</span> with <span class="math-container">$x$</span> in the first equation and compare both we get</p>
<p><span class="math-container">$$f(f(1))/x=f(xf(1))$$</span></p>
<p>We now replace <span class="math-container">$x$</span> with <span class="math-container">$x\cdot f(1)^{-1}$</span> and we get</p>
<p><span class="math-container">$$f(x)=\frac{f(f(1))\cdot f(1)}{x}$$</span></p>
<p>We conclude that our function is of the form:</p>
<p><span class="math-container">$$f(x)=\frac{c}{x}$$</span></p>
<p>for some constant <span class="math-container">$c\in\mathbb{Q}^*_+$</span>. Note that <span class="math-container">$f(f(x))=x$</span>. We now evaluate both sides of your original equation:</p>
<p><span class="math-container">$$f(xf(y))=f(x\cdot\frac{c}{y})=\frac{c}{x\cdot\frac{c}{y}}=\frac{y}{x}$$</span>
<span class="math-container">$$\frac{f(f(x))}{y}=\frac{x}{y}$$</span></p>
<p>These are clearly different, e.g. when <span class="math-container">$x=1,y=2$</span>, regardless of <span class="math-container">$c$</span>. Hence no such function exists.</p>
|
137,915 | <blockquote>
<p>Given two continuously differentiable functions $f,\
g:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x_0) = g(x_0)$ and
$f'(x_0) < g'(x_0)$, there exists $\epsilon > 0$ such that $f(x) <
g(x)$ for $x \in (x_0, x_0 + \epsilon)$.</p>
</blockquote>
<p>The above result is not too difficult to prove, but I was wondering if the condition that the functions be <strong>continuously</strong> differentiable is absolutely necessary. I have not taken much analysis, but I'm wondering if the fact that the derivative exists at $x_0$ would be enough to prove this result without needing the derivative to be continuous. The reason I ask this is because it seems that the existence of a derivative function already implies it must satisfy some quite stringent requirements (i.e. Darboux's Theorem), perhaps these are enough?</p>
| Siminore | 29,672 | <p>If $f$ and $g$ are differentiable at $x_0$, there exist two functions $\omega_f$ and $\omega_g$ such that $\omega_f$ and $\omega_g$ are continuous at $x_0$ and $$f(x)=f(x_0)+\omega_f(x)(x-x_0), \quad g(x)=g(x_0)+\omega_g (x)(x-x_0)$$ for all $x$ (in some neighborhood of $x_0$). From the assumptions you get $\omega_f(x_0) < \omega_g(x_0)$ and there fore $f(x)-g(x)=\left(\omega_f(x)-\omega_g(x)\right)(x-x_0)$. You can immediately conclude that $f(x)<g(x)$ in a right neighborhood of $x_0$.</p>
|
2,491,560 | <p>Given a sequence $(s_n)$ in $\mathbb R$ such that
$$\lim \limits_{n \to \infty}( s_{n+1}-s_n)=0,$$
I am asked to prove
$(s_n)$ converges. </p>
<p>I know all Cauchy sequences converge in $\mathbb R^k$. So I want to prove that $(s_n)$ is Cauchy. </p>
<p>I am stuck as to how to show the given sequence is a Cauchy. Thank you.</p>
| José Carlos Santos | 446,262 | <p>Another counter-example is: $s_n=\log n$. Then$$\lim_{n\to\infty}\bigl(\log(n+1)-\log(n)\bigr)=\lim_{n\to\infty}\log\left(1+\frac1n\right)=0.$$</p>
|
1,826,435 | <p>I was given a problem:
(Edited) </p>
<blockquote>
<p>Show $$ S=\bigl\{S_p:p\in\mathbb{P}\bigr\}\cup\bigl\{\{1\}\bigr\} $$
where $\mathbb{P}$ is the set of prime numbers, and $S = \{n \in
\mathbb{N}: n \text{ is a multiple of } p\}$ is a subbasis on
$\mathbb{N}$</p>
</blockquote>
<p>The easiest way to do this in my opinion to show that the $\mathcal{S}$ covers $\mathbb{N}$</p>
<p>i.e. $\bigcup S= \mathbb{N}$</p>
<p>Does this actually cover $\mathbb{N}$?</p>
<p>Take $n' \in \mathbb{N}$, then we need to show that $n' \in\bigcup S$</p>
<p>Is there some result that all natural number are multiple of primes? Without knowing that I don't see how I can show that $\mathcal{S}$ covers $\mathbb{N}$</p>
| 5xum | 112,884 | <p>It's actually really easy to prove every number has a prime divisor, and the proof is rather nice:</p>
<hr>
<p>Take a number $n\in\mathbb N$. If $n$ is prime, then $n=1\cdot n$ and $n$ is a multiple of a prime. </p>
<p>If $n$ is not prime, then $n$ has a divisor that is not $1$ and is not $n$ (because otherwise it would be prime). Call that divisor $n_1$.</p>
<p>If $n_1$ is prime, then $n$ is a multiple of $n_1$, therefore a multiple of a prime. If not, $n_1$ has a divisor that is not $1$ and not $n_1$. Call that divisor $n_2$.</p>
<p>Continue.</p>
<hr>
<p>Now, the process above either stops, or it does not. If it stops, it stops after it finds a divisor for $n$, so that's OK.</p>
<p>Now, if it doesn't stop, it produces an infinite sequence of numbers $n>n_1>n_2>n_3\dots > 1$, which is of course impossible (there can be only $n-1$ distinct numbers between $n$ and $1$!)</p>
<p>Conclusion: the process above stops. Therefore, $n$ has a prime divisor, or in other words, $n$ is a multiple of a prime.</p>
|
1,379,878 | <blockquote>
<p>Let $M_n(\mathbb{C})$ denote the vector space over $\mathbb{C}$ of all $n\times n$ complex matrices. Prove that if $M$ is a complex $n\times n$ matrix then $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.</p>
</blockquote>
<p>My Try:</p>
<p>I proved that $C(M)$ is a subspace. But how can I show that it is of dimension at least $n$. No idea how to do it. I found similar questions posted in MSE but could not find a clear answer. So, please do not mark this as duplicate.</p>
<p>Can somebody please help me how to find this? </p>
<p>EDIT: Non of the given answers were clear to me. I would appreciate if somebody check my try below:</p>
<p>If $J$ is a Jordan Canonical form of $A$, then they are similar. Similar matrices have same rank. $J$ has dimension at least $n$. So does $A$. Am I correct?</p>
| Community | -1 | <p>Let <span class="math-container">$K$</span> be a field, <span class="math-container">$A\in M_n(K)$</span> and <span class="math-container">$C(A)$</span> be the commutant of <span class="math-container">$A$</span>. Clearly, if <span class="math-container">$A,B$</span> are similar, then <span class="math-container">$\dim(C(A))=\dim(C(B))$</span>.</p>
<p>Definition. <span class="math-container">$U\in M_n(K)$</span> is said to be cyclic if there is <span class="math-container">$u\in K^n$</span> s.t. <span class="math-container">$\{u,Uu,\cdots,U^{n-1}u\}$</span> is a basis of <span class="math-container">$K^n$</span>. Remark that, when <span class="math-container">$U$</span> is cyclic, the matrices <span class="math-container">$I,U,\cdots,U^{n-1}$</span> are linearly independent and, therefore, <span class="math-container">$\dim(C(U))\geq n$</span>.</p>
<p>The key is the following proposition</p>
<p>Proposition. If <span class="math-container">$A\in M_n(K)$</span> is not cyclic, then there are two complementary proper subspaces of <span class="math-container">$K^n$</span> that are <span class="math-container">$A$</span>-invariant. </p>
<p>Proof. Let <span class="math-container">$m_A=p_1^{u_1}\cdots p_k^{u_k}$</span> be the decomposition of the minimal polynomial of <span class="math-container">$A$</span> in irreducibles. If <span class="math-container">$k>1$</span>, then, according to the kernels theorem, <span class="math-container">$K^n=\oplus_i \ker(p_i^{u_i})$</span> and we are done. Then we may assume that <span class="math-container">$m_A=p^u$</span> where <span class="math-container">$p$</span> is irreducible of degree <span class="math-container">$d$</span>. If <span class="math-container">$\mathcal{B}$</span> is a basis of <span class="math-container">$K^n$</span>, then there is <span class="math-container">$e\in\mathcal{B}$</span> s.t. <span class="math-container">$m_A$</span> is its minimal polynomial; thus <span class="math-container">$\{e,Ae,\cdots,A^{ud-1}e\}$</span> is a free system and spans <span class="math-container">$E_1$</span>, the first <span class="math-container">$A$</span>-invariant subspace. Note that, if <span class="math-container">$A$</span> is not cyclic, then <span class="math-container">$ud<n$</span>. </p>
<p>EDIT. Now <span class="math-container">$A$</span> induces an endomorphism of <span class="math-container">$K^n/E_1$</span> which can be represented (as a first step) by <span class="math-container">$\pi\circ A_{|U}$</span> where <span class="math-container">$K^n=E_1\oplus U$</span> and <span class="math-container">$\pi$</span> is the associated projection onto <span class="math-container">$U$</span>; let <span class="math-container">$m_1=p^v,v\leq u$</span> be its minimal polynomial; we consider a basis <span class="math-container">$\mathcal{B}_1$</span> of <span class="math-container">$\ker(m_1(A))$</span>; by construction, there is <span class="math-container">$e_1\in \mathcal{B}_1$</span> s.t. <span class="math-container">$m_1$</span> is its minimal polynomial and s.t. <span class="math-container">$\{e_1,Ae_1,\cdots,A^{vd-1}e_1\}$</span> spans <span class="math-container">$E_2$</span>, a second <span class="math-container">$A$</span>-invariant subspace which is a direct sum with <span class="math-container">$E_1$</span>; and so on...</p>
<p>Corollary. <span class="math-container">$\dim(C(A))\geq n$</span>.</p>
<p>Proof. We proceed by recurrence over <span class="math-container">$n$</span>. If <span class="math-container">$A$</span> is cyclic, then, according to the above remark, we are done. Otherwise, according to Proposition, there is <span class="math-container">$p\in \left\{1,\cdots,n-1\right\}$</span> s.t. <span class="math-container">$A$</span> is similar to a matrix in the form <span class="math-container">$B=\operatorname{diag}(U,V)$</span> where <span class="math-container">$U\in M_p(K),V\in M_{n-p}(K)$</span>. Note that <span class="math-container">$C_1=\{\operatorname{diag}(X,Y)|X\in M_p(K),Y\in M_{n-p}(K),UX=XU,YV=VY \}$</span> is a subspace of <span class="math-container">$C(B)$</span>. By the hypothesis of recurrence, <span class="math-container">$\dim(C_1)\geq p+n-p=n$</span> and we are done.</p>
|
1,183,643 | <p>Given a integer $h$</p>
<blockquote>
<p>What is $N(h)$ the number of full binary trees of height less than $h$?</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/XcNVi.jpg" alt="enter image description here"></p>
<p>For example $N(0)=1,N(1)=2,N(2)=5, N(3)=21$(As pointed by <a href="https://math.stackexchange.com/users/212738/travisj">TravisJ</a> in his partial answer) I can't find any expression of $N(h)$ neither a reasonable upper bound.</p>
<p><strong>Edit</strong> In a full binary tree (sometimes called proper binary tree) every node other than the leaves has two children.</p>
| TravisJ | 212,738 | <p>(Partial Answer) First, $N(3)=21$. Second, and probably more useful, you can construct all the trees of height $h$ by adding two children to every non-empty subset of lowest vertices on a tree of height $h-1$. So, if at height $h-1$ you have trees $T_{1},...,T_{n}$ and tree $T_{i}$ has $v_{i}$ vertices on the lowest level, then the number of trees on level $h$ is </p>
<p>$$\sum_{i=1}^{n} (2^{v_{i}}-1).$$</p>
<p>Unfortunately, it seems to be not clear how to count how many leaves are at the lowest level in the newly constructed trees. This does provide an algorithm for constructing all such trees though.</p>
|
1,183,643 | <p>Given a integer $h$</p>
<blockquote>
<p>What is $N(h)$ the number of full binary trees of height less than $h$?</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/XcNVi.jpg" alt="enter image description here"></p>
<p>For example $N(0)=1,N(1)=2,N(2)=5, N(3)=21$(As pointed by <a href="https://math.stackexchange.com/users/212738/travisj">TravisJ</a> in his partial answer) I can't find any expression of $N(h)$ neither a reasonable upper bound.</p>
<p><strong>Edit</strong> In a full binary tree (sometimes called proper binary tree) every node other than the leaves has two children.</p>
| Ross Millikan | 1,827 | <p>You can define $T(h,b)$ as the number of trees of height $h$ with $b$ leaves at the bottom. Your $N(h)$ is then the sum over $b$ of $T(h,b)$ To get a tree of height $h+1$ and $b$ leaves on the bottom, you pick a tree of height $h$ and at least $b/2$ leaves on the bottom, then pick $b/2$ leaves to hang new leaves from. This gives a recurrence $$T(h+1,b)=\sum_{k=b/2}^{2^h}T(h,k){k \choose b/2}$$
We have $T(3,2)=8, T(3,4)=8, T(3,6)=4, T(3,8)=1$, giving the total of $21$ cited by TravisJ. Then $T(4,2)=8 \cdot 2 + 8 \cdot 4 + 4 \cdot 6 + 1 \cdot 8=80, T(4,4)=8\cdot 1+8\cdot 6 + 4 \cdot 15+1\cdot 28=144, T(4,6)=8\cdot 4+4 \cdot 20+1\cdot 56=168,T(4,8)=8\cdot 1+4\cdot 15+1\cdot 70=138,T(4,10)=4\cdot 6+1\cdot 56=80,T(4,12)=4\cdot 1+1\cdot 28=32,T(4,14)=8,T(4,16)=1$ for a total of $N(4)=651$ The sequence is given in <a href="http://oeis.org/A001699" rel="nofollow">OEIS A001699</a> where it is said to approach $1.5028368...^{(2^n)}$ and some references are given.</p>
|
1,986,007 | <p>What is the intuitive explanation of the Inverse Function Theorem (and generalized to multiple dimensions)?</p>
| Christian Blatter | 1,303 | <p>If you have $n\geq1$ independent variables $x_k$ and $n$ functions $$f_i:\quad U\to{\mathbb R},\quad (x_1,\ldots, x_n)\mapsto y_i:=f_i(x_1,\ldots, x_n)\qquad(1\leq i\leq n)$$ defined in a neighborhood $U$of ${\bf0}$, whereby ${\bf f}({\bf 0})={\bf 0}$, then under certain "technical assumptions", this setup defines the $x_k$ as functions of the $y_i$: There are $n$ functions
$$g_k:\quad V\to{\mathbb R},\quad (y_1,\ldots, y_n)\mapsto x_k:=g_k(y_1,\ldots, y_n)\qquad(1\leq k\leq n)$$
defined in a neighborhood $V$of ${\bf0}$, whereby ${\bf g}({\bf 0})={\bf 0}$, such that
$$y_i:=f_i(x_1,\ldots, x_n)\quad(1\leq i\leq n)\quad\Leftrightarrow \quad x_k:=g_k(y_1,\ldots, y_n)\quad(1\leq i\leq n)\ .$$</p>
<p>The essential point of the theorem is that it guarantees the existence of such functions $g_k$ and their properties even if you are not able to solve the system
$$y_i=f_i(x_1,\ldots, x_n)\qquad(1\leq i\leq n)$$
explicitly for the unknowns $x_k$.</p>
|
173,360 | <p>From my lecture notes: "The notation $\mathbb T$ will be used for the additive circle and $S^1$ for the multiplicative circle."</p>
<p>What I understand: As a topological group, $S^1$ has the subspace topology of $\mathbb R^2$ and multiplication is defined as $(e^{ia}, e^{ib}) \mapsto e^{i(a + b)}$. </p>
<p>My guess is that $\mathbb T$ as an additive group should then be something like $(a,b) \mapsto (a + b) \mod 1$. The problem with that is that the space would look like $[0,1)$ but that's not compact. </p>
<p>But I'm confused: "mod 1" seems to be the same as $\mathbb R / \mathbb Z$ which is $S^1$. But I can't add complex numbers on the unit circle and stay on the unit circle. </p>
<p>So: What's $\mathbb T$? How are elements in it added?</p>
| Alex Becker | 8,173 | <p>In my experience the two are used interchangeably, as the two groups are isomorphic, even as topological groups, and this isomorphism (as topological groups) is unique! You are right that "$\bmod 1$" is the same as $\mathbb R/\mathbb Z$, which is best viewed as $[0,1]$ with the endpoints identified. The only differences I can see are one of notation (elements of $\mathbb T$ are added while those in $S^1$ are multiplied) and that $S^1$ is more naturally considered a subset of $\mathbb C$.</p>
|
3,028,385 | <p>How to determine if there is a surjective homomorphism from <span class="math-container">$U(15)$</span> to <span class="math-container">$\mathbb{Z}_4$</span>? <span class="math-container">$U(15)$</span> is the multiplicative group of positive integers less than <span class="math-container">$n$</span> that are coprime with <span class="math-container">$n$</span>. </p>
<p>I already know <span class="math-container">$U(15)$</span> is not cyclic with 3 elements of order 2, 4 elements of order 4. <span class="math-container">$U(15)=\{1,2,4,7,8,11,13,14\}$</span>. While <span class="math-container">$\mathbb{Z}_4$</span> is obviously cyclic.</p>
| Mauro ALLEGRANZA | 108,274 | <p>A formula is an <em>expression</em> of the language of set theory built up acoording to the <a href="https://en.wikipedia.org/wiki/First-order_logic#Syntax" rel="nofollow noreferrer">rules of the syntax</a>. </p>
<p>Examples : <span class="math-container">$∃y \ ∀x \ ¬(x ∈ y), ∀x \ ¬(x ∈ \emptyset)$</span>.</p>
<p>A formula can be a <em>sentence</em>, i.e. without <em>free variables</em> (like the two previous examples) ore an <em>open one</em>, like e.g. : <span class="math-container">$(x∈y)$</span>.</p>
<p>A sentence has a definite truth value : <span class="math-container">$∀x \ ¬(x ∈ \emptyset)$</span> means "the <em>empty set</em> has no elements" and it is true in set theory.</p>
<p>An open formula, like <span class="math-container">$(x∈y)$</span> has not a definite truth value; its truth value depends on the "reference" assigned to the variables.</p>
<p>Consider some simple arithmetical examples : <span class="math-container">$\forall n (n \ge 0)$</span> is true in <span class="math-container">$\mathbb N$</span>.</p>
<p>Consider <span class="math-container">$(n > 0)$</span> instead : it is false if <span class="math-container">$n$</span> denotes <span class="math-container">$0$</span> and is true otherwise.</p>
<hr>
<p>For a formal definition, see the post : <a href="https://math.stackexchange.com/questions/802963/in-mathematical-logic-what-is-a-language">In Mathematical Logic, What is a Language?</a></p>
|
3,028,385 | <p>How to determine if there is a surjective homomorphism from <span class="math-container">$U(15)$</span> to <span class="math-container">$\mathbb{Z}_4$</span>? <span class="math-container">$U(15)$</span> is the multiplicative group of positive integers less than <span class="math-container">$n$</span> that are coprime with <span class="math-container">$n$</span>. </p>
<p>I already know <span class="math-container">$U(15)$</span> is not cyclic with 3 elements of order 2, 4 elements of order 4. <span class="math-container">$U(15)=\{1,2,4,7,8,11,13,14\}$</span>. While <span class="math-container">$\mathbb{Z}_4$</span> is obviously cyclic.</p>
| Alberto Takase | 146,817 | <p>Put simply in the language of set theory we start with atomic formulas
<span class="math-container">$$(x\in y)$$</span>
or
<span class="math-container">$$(x=y)$$</span>
where <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are variables.</p>
<p>Then we expand the definition of formulas to so that it is closed under <span class="math-container">$$\neg(\cdot)$$</span> and <span class="math-container">$$(\cdot)\wedge(\cdot)$$</span> and <span class="math-container">$$(\exists x)(\cdot)$$</span> where <span class="math-container">$x$</span> is a variable.</p>
<p>Then we introduce the following notations.</p>
<p><span class="math-container">$$(\varphi\vee\psi)\equiv\neg(\neg\varphi\wedge\neg\psi)$$</span>
<span class="math-container">$$(\varphi\Rightarrow\psi)\equiv\neg\varphi\vee\psi$$</span>
<span class="math-container">$$(\varphi\Leftrightarrow\psi)\equiv(\varphi\Rightarrow\psi)\wedge(\psi\Rightarrow\varphi)$$</span>
<span class="math-container">$$(\forall x)\varphi\equiv\neg(\exists x)\neg\varphi$$</span></p>
<p>We take variables, punctuations, <span class="math-container">$=$</span>, <span class="math-container">$\in$</span>, <span class="math-container">$\neg$</span>, <span class="math-container">$\wedge$</span>, <span class="math-container">$\exists$</span> as primitive notions. That is to say, they are undefined symbols which are informally trying to capture the notion of variables, punctuations, equality, membership, negation, conjunction, existential quantifier. Establishing a collection of axioms is how we try to capture desired notions.</p>
<p>Let me speak in the language of set theory:</p>
<p><span class="math-container">$$(\exists x)(\neg(x=x))$$</span></p>
<p>What I just said is false because of established axioms of set theory; most popular being <span class="math-container">$\mathsf{ZFC}$</span>.</p>
<p>Let <span class="math-container">$\varphi$</span> be an arbitrary formula. Then certain variables within <span class="math-container">$\varphi$</span> are "free." We typically denote those variables by writing <span class="math-container">$\varphi(x_1,\ldots,x_n)$</span> instead of simply <span class="math-container">$\varphi$</span>.</p>
|
1,178,969 | <p>$$\sum_{k=1}^{\infty}\ln\frac{k+5}{k+4}$$</p>
<p>$$
\frac{\ln\frac{k+6}{k+5}}{\ln\frac{k+5}{k+4}}
=\cdots
$$</p>
<p>(ln(k+6/k+5)/ln(k+5/k+4))</p>
<p>ln((k+6/k+5)*(k+4/k+5))</p>
<p>simplify</p>
<p>lim as k approaches infinity of ln(kˆ2 +10k +24/ kˆ2 +10k+25)= ln(1) = 0.</p>
<p>why doesn't the series converge?</p>
| Claude Leibovici | 82,404 | <p>Concerning the ratio test, if you consider $$a_k=\ln\frac{k+5}{k+4}$$ then $$\frac{a_{k+1}}{a_k}=\frac{\log \left(\frac{k+6}{k+5}\right)}{\log \left(\frac{k+5}{k+4}\right)}=\frac{\log \left(1+\frac{1}{k+5}\right)}{\log \left(1+\frac{1}{k+4}\right)}$$ Now, use that, for small values of $y$, $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ Make $y=\frac{1}{k+5}$ for the numerator, $y=\frac{1}{k+4}$ for the denominator, use the long division to arrive to $$\frac{a_{k+1}}{a_k}=1-\frac{1}{k}+\frac{11}{2 k^2}+O\left(\left(\frac{1}{k}\right)^3\right)$$</p>
|
2,476,267 | <p>Here's a problem presented in my proofs class that I cannot for the life of me figure out. Apparently the answer is P2 according to the answer book but I have no idea how Help is much appreciated, thanks in advance!</p>
<p>Anna wrote down the following 3 statements, denoted P1, P2, P3, on a blank piece of paper:</p>
<p>P1 : There is exactly one FALSE statement written on this piece of paper.</p>
<p>P2 : There are exactly two FALSE statements written on this piece of paper .</p>
<p>P3 : There are exactly three FALSE statements written on this piece of paper.</p>
<p>One of the above statements is TRUE. What statement is TRUE?</p>
| Alex B. | 492,378 | <p>Suppose that P1 is True. Then P2 and P3 must be false, a contradiction to P1. </p>
<p>If you follow the same reasoning for each statement, you will find your answer.</p>
|
1,160,726 | <p>I am trying to figure out why the equation,</p>
<p>$2x^2\:-5y^2\:=\:1$</p>
<p>has no solutions. I have tried taking the equation mod $2$,$4$, and $5$, but when I do this I am not sure what I am supposed to look for. I would really appreciate an explanation on how to solve this.</p>
<p>Thanks!</p>
| Bill Dubuque | 242 | <p>${\rm mod}\ 5\!:\ 1 \equiv 2x^2\overset{(\ \ )^{\large 2}}\Rightarrow 1\equiv 4x^4\overset{\rm Fermat}\equiv 4\,\Rightarrow\, 5\mid 4-1\,\Rightarrow\!\Leftarrow$</p>
|
18,174 | <p>I am an undergraduate secondary math education major. In <span class="math-container">$2$</span> weeks I have to give a <a href="https://www.mathmammoth.com/lessons/number_talks.php" rel="nofollow noreferrer">Number Talk</a> in my math ed class on the problem "<span class="math-container">$3.9$</span> times <span class="math-container">$7.5$</span>". I need to come up with as many different solution methods as possible. </p>
<p>Here is what I have come up with so far:</p>
<ol>
<li><p>The most common way: multiply the two numbers "vertically", ignoring the decimal, to get <span class="math-container">$2925$</span>:
<span class="math-container">\begin{array}
{}\hfill {}^6{}^439\\
\hfill \times\ 75 \\\hline
\hfill {}^1 195 \\
\hfill +\ 273\phantom{0} \\\hline
\hfill 2925
\end{array}</span> Since there are two numbers that are to the right of the decimal, place the decimal after the <span class="math-container">$9$</span> to get the answer <span class="math-container">$29.25$</span>.</p></li>
<li><p>Write both numbers as improper fractions: <span class="math-container">$$3.9= \dfrac{39}{10}$$</span> and <span class="math-container">$$7.5=\dfrac{75}{10}$$</span>Then multiply <span class="math-container">$$\dfrac{39}{10}\cdot\dfrac{75}{10}$$</span> to get <span class="math-container">$\dfrac{2925}{100}$</span> which simplifies to 29.25.</p></li>
<li><p>Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.</p></li>
<li><p>Since <span class="math-container">$3.9$</span> is very close to <span class="math-container">$4$</span>, we could instead do <span class="math-container">$4\cdot7.5=30$</span> and then subtract <span class="math-container">$0.1\cdot7.5=0.75$</span> to get <span class="math-container">$30 - 0.75=29.25$</span></p></li>
<li><p>Similarly, since <span class="math-container">$7.5$</span> rounds up to <span class="math-container">$8$</span>, we can do <span class="math-container">$3.9\cdot 8=31.2$</span> and then subtract .<span class="math-container">$5\cdot 3.9=1.95$</span> to get <span class="math-container">$31.2-1.95=29.25$</span> </p></li>
</ol>
<p>Are there any other possible methods the students might use? (<strong>Note:</strong> they are junior college math ed students.) Thanks!</p>
| Adam | 4,791 | <p>Combining your last two methods: <span class="math-container">$3.9*7.5 = 4*8 - 0.1*8 - 4 * 0.5 + 0.1*0.5$</span>, which can be thought of as computing the area of the big rectangle below, cutting off the two extra strips along the edge, then adding back in a copy of the corner piece since you have removed it twice, instead of once.</p>
<p><a href="https://i.stack.imgur.com/NtDJV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/NtDJV.png" alt="rectangles"></a></p>
|
946,738 | <p>I ran into a nice question from one book in Discrete Mathematics. I want to someone lean me how solve such a problem, because I prepare for entrance exam. </p>
<blockquote>
<blockquote>
<p>if the time is "Wednesday 4 afternoon", after $47^{74}$ hours, we are in what hours? and what day? </p>
</blockquote>
</blockquote>
<p>Thanks to all. </p>
| paw88789 | 147,810 | <p>Since $47^{74}$ hours is more than a googol ($10^{100}$) years, by all estimates, the sun will be long gone, and we won't have days anymore.</p>
<p>But if you ignore all that, you should try to reduce $47^{74} \pmod{168}$ (the number of hours in a week).</p>
|
155,897 | <p>I would like to exclude non-western characters and words from a text file. I do not know how to insert the text file here, but I suppose you can do without it. All your suggestions will be much appreciated.</p>
<p>Update: I have used:</p>
<pre><code>Alphabet["Russian"]
Select[dict21, Not@StringContainsQ[#, Alternatives @@ dict24Russion] &]
</code></pre>
<p>The problem is that there are several alphabets in the text (even unknown). There must be some solution of kind "include only <code>Alphabet[]</code>". What do you think?</p>
| b3m2a1 | 38,205 | <p>Here's a better <code>StringDelete</code> than my first method, courtesy of Alexey Popkov:</p>
<pre><code>text = Import["https://sv.wikipedia.org/wiki/Kategori:Sjötermer"];
pattern =
Except[
Join[Alphabet["Swedish"],
ToUpperCase@Alphabet["Swedish"], {PunctuationCharacter,
WhitespaceCharacter}
]
];
{timing, result} = StringDelete[text, pattern] // RepeatedTiming;
{timing, StringTake[result, 50]}
{0.00033, "Hjälp Kategori:Sjötermer
Från Wikipedia
"}
</code></pre>
<p>Note how fast it is. And note that all it leaves behind are the Swedish diacritics:</p>
<pre><code>StringCases[result,
(Except[WhitespaceCharacter]?(Not@*PrintableASCIIQ))]
{"ä", "ö", "å", "ö", "ä", "å", "ö", "ä", "ä", "å", "ä", "ö", "å", \
"ö", "å", "ä", "ö", "ö", "ö", "ö", "ö", "ä", "ö", "ä", "ö", "å", "ä", \
"ö", "å", "ä", "å", "å", "å", "å", "ä", "ä", "ö", "ö", "ö", "ö", "ö", \
"ö", "å", "å", "ä", "å", "ö", "å", "ö", "ö", "ö", "ä", "ä", "ä", "ö", \
"å", "ä", "ä", "å", "å", "å", "ö", "ä", "ä", "ä", "ä", "å", "ä", "ä", \
"ä", "ä", "ä", "ä", "ä", "ä", "ö", "ö", "ä", "ä", "å", "ö", "ö", "å", \
"ö", "ö", "ä", "ö", "ä", "ä", "å", "ä", "ä", "ä", "å", "", "ä", "ä", \
"ä", "ä", "ö", "å", "ä", "ä", "ö", "å"}
</code></pre>
<p>We can compile it as a regex too, but unsurprisingly we get no noticeable boost:</p>
<pre><code>regex =
RegularExpression@
StringPattern`PatternConvert[pattern][[1]];
StringDelete[text, regex] // RepeatedTiming // First
0.00032
</code></pre>
<hr>
<p>Here's a way that works by matching only against <code>LetterCharacter</code>:</p>
<pre><code>pattern2 =
Except[
Join[Function[Join[#, ToUpperCase[#]]]@Alphabet["Swedish"]],
LetterCharacter
];
{timing2, result2} = StringDelete[text, pattern2] // RepeatedTiming;
timing2
0.00042
</code></pre>
<hr>
<p>Here's a <a href="http://reference.wolfram.com/language/ref/StringCases.html" rel="nofollow noreferrer"><code>StringCases</code></a> way:</p>
<pre><code>StringJoin@
StringCases[
Import["https://sv.wikipedia.org/wiki/Kategori:Sjötermer"],
(Alternatives @@
Join[Alphabet["Swedish"], ToUpperCase@Alphabet["Swedish"]]) |
PunctuationCharacter | WhitespaceCharacter
]
</code></pre>
<p>Here's an updated regex base method which is faster than the <code>StringCases</code> way:</p>
<pre><code>regex =
RegularExpression[
StringPattern`PatternConvert[(
RegularExpression["[A-z]+"] |
Alternatives @@
Function[Join[#, ToUpperCase[#]]]@
DeleteCases[Alphabet["Swedish"],
Alternatives @@ Alphabet[]]
) | PunctuationCharacter | WhitespaceCharacter
][[1]]
];
StringJoin@
StringCases[
text,
regex
]
</code></pre>
<p>Note that this is much faster than the older method and gives similar results</p>
<pre><code>{timing1, result1} =
StringJoin@
StringCases[
text,
regex
] // AbsoluteTiming;
{timing1, StringTake[result1, 100]}
{timing1, result1} =
StringJoin@
StringCases[
text,
regex
] // AbsoluteTiming;
{timing1, StringTake[result1, 100]}
{0.000769, "Hjälp Kategori:Sjötermer
Från Wikipedia
Hoppa till: navigering , sök
Termerna har"}
</code></pre>
|
3,700,575 | <p>I have to find Taylor series at <span class="math-container">$a=1$</span> for </p>
<p><span class="math-container">$
f(x)=\begin{cases}
\frac{e^{x}-e}{x-1},\quad &\text{if } x\ne1\\
e,\quad &\text{if } x=1\\
\end{cases}
$</span></p>
<p>I haven't found Taylor series for such functions before and I also don't know how to find n-th derivative of this function. Can anyone help me with this?</p>
| Kavi Rama Murthy | 142,385 | <p>Using the series expansion of <span class="math-container">$e^{x-1}$</span> we get <span class="math-container">$f(x)=e \frac {e^{x-1}-1}{x-1}=e\sum\limits_{k=1}^{\infty} \frac {(x-1)^{k}} {(x-1)k!} =\sum\limits_{k=1}^{\infty}\frac { e{(x-1)^{k-1}}} {k!}=\sum\limits_{k=0}^{\infty}\frac { e{(x-1)^{k}}} {(k+1)!}$</span></p>
|
3,700,575 | <p>I have to find Taylor series at <span class="math-container">$a=1$</span> for </p>
<p><span class="math-container">$
f(x)=\begin{cases}
\frac{e^{x}-e}{x-1},\quad &\text{if } x\ne1\\
e,\quad &\text{if } x=1\\
\end{cases}
$</span></p>
<p>I haven't found Taylor series for such functions before and I also don't know how to find n-th derivative of this function. Can anyone help me with this?</p>
| vin92 | 795,021 | <p>First I compute </p>
<p><span class="math-container">$$f'(x)=\dfrac{e^x-f(x)}{x-1}.$$</span>
And </p>
<p><span class="math-container">$$
f''(x)=\dfrac{e^x-2f'(x)}{x-1}.$$</span></p>
<p>So that I find a pattern</p>
<p><span class="math-container">$$
f'''(x)=\dfrac{[e^x-2f''(x)](x-1)-[e^x-2f'(x)]}{(x-1)^2}=\dfrac{[e^x-2f''(x)](x-1)-f''(x)(x-1)}{(x-1)^2}=\dfrac{e^x-3f''(x)}{x-1}.$$</span></p>
<p>And I can give a general formul for the <span class="math-container">$n$</span>-th derivative</p>
<p><span class="math-container">$$
f^{(n)}(x)=\dfrac{e^x-nf^{n-1}(x)}{x-1}.$$</span></p>
<p>In the Taylor series, this terms appear evaluated at <span class="math-container">$x=1$</span>. Then <span class="math-container">$f(1)=e$</span> but <span class="math-container">$f''(1)=0$</span>, <span class="math-container">$f'''(1)=0$</span> etc so we have to make sense of <span class="math-container">$\dfrac{0}{0}$</span> limits. We can use Hôpital for this. Then</p>
<p><span class="math-container">$$
\lim_{x\rightarrow 1}f^{(n)}(x)=\lim_{x\rightarrow 1}\dfrac{e^x-nf^{n-1}(x)}{x-1}=\dfrac{0}{0}\rightarrow \lim_{x\rightarrow 1} [e^x-nf^{(n)}(x)]=e-n \lim_{x\rightarrow 1} f^{(n)}(x)$$</span></p>
<p>Now, since all this limits must tend to <span class="math-container">$0$</span> (differentiable),</p>
<p><span class="math-container">$$e-n \lim_{x\rightarrow 1} f^{(n)}(x)=0 \Rightarrow \lim_{x\rightarrow 1} f^{(n)}(x)=\dfrac{e}{n} \quad n\geq 1.$$</span></p>
<p>And I get for the Taylor series</p>
<p><span class="math-container">$$f(x)=
\begin{cases}
e+\displaystyle\sum_{n=1}^{\infty} \dfrac{e}{n! }\dfrac{(x-1)^n}{n}\quad x\neq 1\\\\
e \quad x=1
\end{cases}.$$</span></p>
|
3,381,543 | <p>Let <span class="math-container">$X \sim \text{Poisson}(\lambda)$</span>. If <span class="math-container">$P(X = 1) = 0.1$</span> and <span class="math-container">$P(X = 2) = 0.2$</span>, what is the value of <span class="math-container">$P(X = 3)?$</span></p>
<p>I've tried to expand the given equalities, finding <span class="math-container">$\lambda = 4$</span>, which is clearly wrong if you try to evaluate <span class="math-container">$P(X = 1)$</span> or <span class="math-container">$P(X = 2)$</span>. I'm stuck at this point; I don't know how to know the value of <span class="math-container">$\lambda$</span> given that</p>
<p><span class="math-container">\begin{cases}e^{-\lambda}\cdot\lambda &= 0.1\\
e^{-\lambda}\cdot\lambda^2 &= 0.4
\end{cases}</span></p>
| Ongky Denny Wijaya | 556,947 | <p>Given <span class="math-container">$X\sim \text{POI}(\lambda)$</span>, so <span class="math-container">$X$</span> have p.d.f.
<span class="math-container">$$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}, x=0,1,\ldots.$$</span>
So, we have
<span class="math-container">\begin{eqnarray}
P(X=1)=0.1\iff \dfrac{e^{-\lambda}\lambda}{1}=0.1\\
P(X=2)=0.2\iff \dfrac{e^{-\lambda}\lambda^2}{2}=0.2.
\end{eqnarray}</span></p>
<p>Now we have <span class="math-container">$e^{-\lambda}\lambda=0.1$</span> and <span class="math-container">$e^{-\lambda}\lambda^2=0.4$</span>. Divide two last equations,
<span class="math-container">$$\dfrac{e^{-\lambda}\lambda^2}{e^{-\lambda}\lambda}=\dfrac{0.4}{0.1}=4,$$</span>
we have <span class="math-container">$\lambda=4$</span>. So,
<span class="math-container">$$P(X=3)=\dfrac{e^{-\lambda}\lambda^3}{3!}=\dfrac{e^{-4}4^3}{6}=0.1953668148.$$</span></p>
|
3,068,381 | <p>I calculated, using Mathematica, that for <span class="math-container">$4\leq k \leq 100$</span>,
<span class="math-container">$$ \sum_{j=k}^{2k} \sum_{i=j+1-k}^j (-1)^j 2^{j-i} \binom{2k}{j} S(j,i) s(i,j+1-k) = 0,$$</span>
where <span class="math-container">$s(i,j)$</span> and <span class="math-container">$S(i,j)$</span> are Stirling numbers of the first and second kinds, respectively.</p>
<p>Here the code:</p>
<blockquote>
<p>F[k_] := Sum[(-1)^j 2^(j - i) Binomial[2 k, j] StirlingS2[j,
i] (StirlingS1[i, j + 1 - k]), {j, k, 2 k}, {i, j - k + 1, j}];</p>
<p>Table[F[k], {k, 4, 100}]</p>
</blockquote>
<p>How do I prove it holds for all <span class="math-container">$k \geq 4$</span> ?</p>
| Marko Riedel | 44,883 | <p>It looks like my first response interprets the problem statement to
use unsigned Stirling numbers of the first kind. We find for signed
ones,</p>
<p><span class="math-container">$$S_n = (-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j
(-1)^k 2^{j-k} {2n\choose j} {j\brace k}
{k\brack j+1-n}.$$</span></p>
<p>With the usual EGFs we get</p>
<p><span class="math-container">$$(-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j
(-1)^k 2^{j-k} {2n\choose j} j! [z^j] \frac{(\exp(z)-1)^k}{k!}
\\ \times k! [w^k]
\frac{1}{(j+1-n)!} \left(\log\frac{1}{1-w}\right)^{j+1-n}.$$</span></p>
<p>Now we have</p>
<p><span class="math-container">$${2n\choose j} j! \frac{1}{(j+1-n)!}
= \frac{(2n)!}{(2n-j)! \times (j+1-n)!}
= \frac{(2n)!}{(n+1)!} {n+1\choose j+1-n}.$$</span></p>
<p>This yields for the sum</p>
<p><span class="math-container">$$(-1)^{n+1} \frac{(2n)!}{(n+1)!}
\sum_{j=n}^{2n} {n+1\choose j+1-n} 2^j
\\ \times [z^j] \sum_{k=j+1-n}^j (-1)^k 2^{-k}
(\exp(z)-1)^k [w^k]
\left(\log\frac{1}{1-w}\right)^{j+1-n}
\\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n
\sum_{j=0}^{n} {n+1\choose j+1} 2^j
\\ \times [z^{n+j}] \sum_{k=j+1}^{j+n} (-1)^k 2^{-k}
(\exp(z)-1)^k [w^k]
\left(\log\frac{1}{1-w}\right)^{j+1}.$$</span></p>
<p>Observe that <span class="math-container">$(\exp(z)-1)^k = z^k + \cdots$</span> and hence we may extend
the inner sum beyond <span class="math-container">$j+n$</span> due to the coefficient extractor
<span class="math-container">$[z^{n+j}].$</span> We find</p>
<p><span class="math-container">$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n
\sum_{j=0}^{n} {n+1\choose j+1} 2^j
\\ \times [z^{n+j}] \sum_{k\ge j+1} (-1)^k 2^{-k}
(\exp(z)-1)^k [w^k]
\left(\log\frac{1}{1-w}\right)^{j+1}.$$</span></p>
<p>Furthermore note that <span class="math-container">$\left(\log\frac{1}{1-w}\right)^{j+1} = w^{j+1}
+\cdots$</span> so that the coefficient extractor <span class="math-container">$[w^k]$</span> covers the entire
series, producing</p>
<p><span class="math-container">$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n
\sum_{j=0}^{n} {n+1\choose j+1} 2^j [z^{n+j}]
\left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}.$$</span></p>
<p>Working with formal power series we are justified in writing</p>
<p><span class="math-container">$$[z^{n+j}] \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}
= [z^{n-1}] \frac{1}{z^{j+1}}
\left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}$$</span></p>
<p>because the logarithmic term starts at <span class="math-container">$(-1)^{j+1} z^{j+1}/2^{j+1}.$</span>
To see this write</p>
<p><span class="math-container">$$-\frac{\exp(z)-1}{2}
+ \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2}
- \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3}
\pm \cdots$$</span></p>
<p>We continue</p>
<p><span class="math-container">$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\sum_{j=0}^{n} {n+1\choose j+1} 2^{j+1}
\frac{1}{z^{j+1}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}
\\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\sum_{j=1}^{n+1} {n+1\choose j} 2^{j}
\frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j}.$$</span></p>
<p>The term for <span class="math-container">$j=0$</span> in the sum is one and hence only contributes to
<span class="math-container">$n=1$</span> so that we may write</p>
<p><span class="math-container">$$-[[n=1]]
+ (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\sum_{j=0}^{n+1} {n+1\choose j} 2^{j}
\frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j}
\\ = -[[n=1]]
+ (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\left(1+\frac{2}{z}
\log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1}.$$</span></p>
<p>Finally observe that</p>
<p><span class="math-container">$$\left(1+\frac{2}{z}
\log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1}
\\ = \left(1+\frac{2}{z}
\left( -\frac{\exp(z)-1}{2}
+ \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2}
- \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3}
\pm \cdots \right)\right)^{n+1}
\\ = \left( -\frac{1}{4} z - \cdots \right)^{n+1}$$</span></p>
<p>and furthermore</p>
<p><span class="math-container">$$[z^{n-1}] \left((-1)^{n+1}
\frac{1}{4^{n+1}} z^{n+1} + \cdots \right) = 0$$</span></p>
<p>which is the claim.</p>
|
220,170 | <p>My question is rather philosophical : without using advanced tools as Perlman-Thurston's geometrisation, how can we get convinced that the class of closed oriented $3$-manifolds is large and that simple invariants as Betti number are not even close to classify ?</p>
<p>For example i would start with :</p>
<ol>
<li><p>If $S_g$ is the closed oriented surface of genus $g$, the family $S_g \times S^1$ gives an infinite number of non pairwise homeomorphic $3$-manifolds.</p></li>
<li><p>Mapping tori of fiber $S_g$ gives as much as non-diffeomorphic $3$-manifolds as conjugacy classes in the mapping class group of $S_g$ which can be shown to be large using the symplectic representation for instance.</p></li>
</ol>
<p>I think that I would like also say that Heegaard splittings give rise to a lot of different $3$-manifolds which are essentially different, but I don't know any way to do this. </p>
<p>So if you know a nice construction which would help understanding the combinatorial complexity of three manifolds, please share it :) </p>
| Marc Kegel | 84,120 | <p>I would explain it like this:</p>
<p>Start with a framed link in $S^3$ and do <a href="https://en.wikipedia.org/wiki/Dehn_surgery" rel="nofollow noreferrer">Dehn-surgery</a> along it to get a new $3$-manifold. In fact one can get every compact oriented $3$-manifold in this way as mentioned earlier.</p>
<p>To distinguish this $3$-manifolds one can use the fundamental group. Actually the fundamental group can distinguish <a href="https://mathoverflow.net/questions/204245/3-manifolds-with-isomorphic-fundamental-groups">almost</a> all $3$-manifolds. It is easy to give a presentation of the fundamental group out of a surgery diagram:</p>
<p>First use the <a href="https://en.wikipedia.org/wiki/Wirtinger_presentation" rel="nofollow noreferrer">Wirtinger presentation</a> of the link exterior and then add a new realtion (of the form $p\mu+q\lambda=1$) for every surgery.</p>
<p>Of course its difficult to distinguish two such presentations in general. But to convince someone that there are many different should not be hard. For example one can do the following:</p>
<ul>
<li>make this groups abelian and use the classification of abelian groups.</li>
<li>compare the orders of the groups (this is how Poincaré proved the Poincaré sphere to be not $S^3$).</li>
<li>show that some of this groups are non-abelian (for example by finding surjective group homomorphisms in non-abelian groups) others are abelian.</li>
<li>compare orders of the elements.</li>
</ul>
|
2,559,623 | <p>The question is as follows:<p>
Calculate $\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx$.<p></p>
<p>$\textbf{Some ideas:}$<p>
We can use the fact that $\sin(\frac{x}{n}) \simeq \frac{x}{n} $.But then we find that <p>
$\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx \simeq \lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \times \frac{x}{n}}{1 + x^2} = \lim_{n \to +\infty} \int_{0}^{n} \frac{ x }{1 + x^2} dx $<p>
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{1}{2}\int_{0}^{n} \frac{ 2x }{1 + x^2} dx $<p>
$ \hspace{9.1cm} \text{take } x^2=y$<p>
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{1}{2}\int \frac{ dy }{1 + y} $<p>
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(y)}{2} $<p>
$\hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(x^2)}{2} \mid_{0}^{n}$<p>
$\hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(n^2)}{2} = +\infty$<p>
But someone said me that the final result should be $\frac{\pi}{2}$?<p>
Can you please let me know where is my mistake?<p>
Thanks!</p>
| Deepak | 151,732 | <p>Easiest to solve geometrically with a quick Argand diagram sketch. The first locus is a ray in the first quadrant making an angle with the horizontal real axis of $\frac{\pi}{6}$ with a "hole" at the origin (since the origin is excluded from $C_1$). The second locus $C_2$ is a circle of undetermined radius but centered at $(0,2\sqrt 3)$ on the vertical imaginary axis.</p>
<p>For a particular value of $r$ (which is geometrically the radius of the circle of the second locus), the circle will be just tangent to the ray of the first locus. Using trigonometry, you should be able to determine this minimal value of $r$ to be $2\sqrt 3 \cos \frac{\pi}{6} = 3$. Any value of $r$ smaller than this will give a circle that doesn't contact the ray at all.</p>
<p>For a certain higher value of $r$, the circle will just pass through the origin (which is not on $C_1$ - it's a "hole" at the origin along the ray). For higher values of $r$, the circle will intersect the ray once in the first quadrant and not pass through the origin. This maximal value of $r$ is simply $2\sqrt 3$.</p>
<p>The bounds are both exclusive. So the required range is $3 < r < 2\sqrt 3$.</p>
|
2,535,111 | <p>Let $n$ be a natural number, and let $1\leq j,k\leq n$ be two randoms.
Show that if $k\mid n$ ($k$ divides $n$) then the probabilty of
$k$ also dividing $j$ is $\frac{1}{k}$, that is $\text{P}\left(k\mid j\right)=\frac{1}{k}$.</p>
<p>One way i was thinking about is to look straight forward for
$$\text{P}\left(k\mid j\,\biggl|\,k\mid n\right)=\frac{\text{P}\left(k\mid j\,\cap\,k\mid n\right)}{\text{P}\left(k\mid n\right)}
$$But i dont really know how to find any of those on the right side
(Any help with that?)</p>
<p>Other way i was thinking about is that if $k\mid n$ then there is
an $s\in\mathbb{N}$ such that $n=k\cdot s$, so we can divide $\left\{ 1,\ldots,n\right\} $
to $s$ distinct sets or to $k$ distinct sets in these ways for example:
$$\begin{aligned}(1)\quad & \left\{ 1,\ldots,k\right\} ,\left\{ k+1,\ldots,2k\right\} ,\dots,\left\{ \left(s-1\right)k+1,\ldots,sk\right\} \\
(2)\quad & \left\{ 1,k+1,\ldots,\left(s-1\right)k+1\right\} ,\left\{ 2,k+2,\ldots,\left(s-1\right)k+2\right\} ,\ldots,\left\{ k,2k,\ldots sk\right\}
\end{aligned}
$$Somehow i think that using the 2nd way would be more helpful. I know
that the probabilty of choosing a random $j$ from $\left\{ 1,\ldots,n\right\} $
is $\frac{1}{n}$, but how do i find the probabilty that it is also
from the last set $\left\{ k,2k,\ldots,sk\right\} $ ?</p>
| ℋolo | 471,959 | <p>By symmetry I can tell that it is on the line $x=y=0$</p>
<p>Now to get $z$ we need to find few things:</p>
<p>the volume of the shape is: $\frac23\pi r^3=\frac23\pi$</p>
<p>Now the integral: we are searching for an integral of $dz$ so we need a function that "have" the change of $z$ in it, let's take the function of the radius of a circle parallel to $z=0$. By Pythagorean theorem $z^2+r^2=1\implies r(z)=\sqrt{1-z^2}$ now the area of the circle is $\pi r^2=\pi (1-z^2)$, we need to add the thickness of the circle, so we get $\pi (1-z^2)dz$</p>
<p>Now there is one last thing before the integration, we need to multiply the circle by its "hight", by how far it is from the circle at $z=0$, because all the circles are parallel this "hight" is just $z$ so we get $\int_0^1\pi(1-z^2)z\, dz=\frac14\pi$</p>
<p>Now the last part, the average of a shape in $\Bbb R^3$ is the integral/the volume so we get $\frac{\frac{\pi}4}{\frac{2\pi}3}=\frac38$ and because I know that $x=y=0$ I have the point $(0,0,\frac38)$</p>
|
1,576,561 | <p>I'd like to show that $$\sum\limits_{n = 1}^\infty {{{{x^{n + 1}}} \over {n(n + 1)}}} $$ absolutely converges for $|x| < 1$</p>
| Community | -1 | <p>$$\sum\limits_{n = 1}^\infty \left|{{{{x^{n + 1}}} \over {n(n + 1)}}}\right|<\sum\limits_{n = 1}^\infty {{{{|x|^{n + 1}}}} }=\lim_{n\to\infty}x^2\frac{1-|x|^n}{1-|x|}.$$</p>
|
3,284,729 | <p>I was wondering how to calculate the following:
I know the average chance of death per year for a certain age.
What is the average chance of death for a 5 year period then?</p>
<p>Is it simply additive?
The chances are listed here (jaar is the dutch word for Year)</p>
<p><a href="https://i.stack.imgur.com/09wt9.png" rel="nofollow noreferrer">Chance Table</a></p>
| heropup | 118,193 | <p>What you have is an excerpt of a life table, which is used in the actuarial sciences. The entries you have are ages <span class="math-container">$x$</span>, and the corresponding probabilities <span class="math-container">$q_x$</span> of death within one year for a life aged <span class="math-container">$x$</span>.</p>
<p>To obtain the the probability of death within <span class="math-container">$5$</span> years, denoted <span class="math-container">${}_5 q_x$</span>, you would observe that it is the complementary probability, survival for at least <span class="math-container">$5$</span> years, called <span class="math-container">${}_5 p_x$</span>, that can be expressed as a product of the individual annual survival probabilities; that is to say,</p>
<p><span class="math-container">$${}_5 p_x = (p_x)(p_{x+1})(p_{x+2})(p_{x+3})(p_{x+4}).$$</span> Since <span class="math-container">$q_x + p_x = 1$</span> for any age <span class="math-container">$x$</span>, we can then write <span class="math-container">$${}_5 q_x = 1 - {}_5 p_x = 1 - (1 - q_x)(1 - q_{x+1})(1 - q_{x+2})(1 - q_{x+3})(1 - q_{x+4}).$$</span> In your table, for someone aged <span class="math-container">$59.5$</span>, the <span class="math-container">$5$</span>-year failure (death) probability is <span class="math-container">$${}_5 q_{59.5} = 1 - (1 - 0.005380)(1 - 0.005980)(1 - 0.006320)(1 - 0.006800)(1 - 0.007120) = 0.031204.$$</span> However, for a life aged <span class="math-container">$60.5$</span>, <span class="math-container">$${}_5 q_{60.5} = 0.0335904.$$</span> A more complete table would allow you to make projections for survival beyond age <span class="math-container">$64.5$</span>.</p>
|
214,705 | <p>If $a(x) + b(x) = x^6-1$ and $\gcd(a(x),b(x))=x+1$ then find a pair of polynomials of $a(x)$,$b(x)$.</p>
<p>Prove or disprove, if there exists more than 1 more distinct values of the polynomials.</p>
| Brian M. Scott | 12,042 | <p>Since $\gcd\{a(x),b(x)\}=x+1$, there are polynomials $p(x)$ and $q(x)$ such that $a(x)=(x+1)p(x)$ and $b(x)=(x+1)q(x)$. </p>
<p>$$x^6-1=(x+1)\left(x^5-x^4+x^3-x^2+x-1\right)\;,$$</p>
<p>so </p>
<p>$$p(x)+q(x)=x^5-x^4+x^3-x^2+x-1\;.\tag{1}$$</p>
<p>You can easily find several pairs of relatively prime polynomials $p(x)$ and $q(x)$ that satisfy $(1)$, and they give you pairs $a(x),b(x)$.</p>
|
1,914,536 | <p><strong>Problem</strong>: Solve $3^x \equiv 2 \pmod{29}$ using Shank's Baby-Step Giant-Step method.</p>
<p>I choose $k=6$ and calculated $3^i \pmod {29}$ for $i=1,2,...,6$.
$$3^1 \equiv 3 \pmod {29}$$
$$3^2 \equiv 9 \pmod {29}$$
$$3^3 \equiv 27 \pmod {29}$$
$$3^4 \equiv 23 \pmod {29}$$
$$3^5 \equiv 11 \pmod {29}$$
$$3^6 \equiv 4 \pmod {29}$$</p>
<p>Then I have calculated $3^{-1} \equiv 10 \pmod {29}$ and started calculating second list:</p>
<p>$$2 \cdot 3^{-6} \equiv 2 \cdot 10^6 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-12} \equiv 2 \cdot {3^{-6}}^2 \equiv 2 \cdot 15^2 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-18} \equiv 2 \cdot {3^{-12}}^2 \equiv 2 \cdot 22^2 \equiv 2 \cdot 20 = 40 \equiv 11\pmod {29}$$</p>
<p>And now I can stop. I can see that:
$$ 3^5 \equiv 11 \pmod {29}\ and\ 2 \cdot 3^{-18} \equiv 11 \pmod {29}$$
Therefore $x = 5 + 18 = 23$</p>
<p>But when I plugin $x=23$ above I get that $3^{23} \equiv 8 \pmod {29}$.
So where am I wrong?</p>
| Darío A. Gutiérrez | 353,218 | <p>$$3^x \equiv 2 (\text{ mod } 29)$$
$\varphi(29) = 28$<br>
$m = \lceil \sqrt{28} \rceil = 6$</p>
<p>$\begin{array}{c|ccccc}
j& 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
3^j & 1 & 3& 9 & 27 & 23 & \color{#f00}{11}
\end{array}$
<br><br>
$3^{-6}= 3^{28 -6 } = 3^{22} = 22$</p>
<p>$\begin{array}{c|ccccc}
i & 0 & 1 & 2\\
\hline
2\cdot 22^i & 2 & 15 & \color{#f00}{11}
\end{array}$
<br><br>
Also... $$x = i\cdot m + j = 2\cdot 6 + 5 = 17 $$</p>
<blockquote>
<p>$$3^{17} \equiv 2 (\text{ mod } 29)$$</p>
</blockquote>
<p>See <a href="https://en.wikipedia.org/wiki/Baby-step_giant-step" rel="nofollow noreferrer">Baby-step Giant-step</a></p>
|
2,842,481 | <p>If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?</p>
<p>I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appreciated!</p>
| dxiv | 291,201 | <p>Hint: suppose $\,c, x \gt 0\,$, then $\,u = x + \dfrac{c}{x} \ge 2 \sqrt{c}\,$ by AM-GM with equality iff $\,x = \sqrt{c}\,$, and:</p>
<p>$$
f(x)=\frac{x^2-3x+c}{x^2+3x+c} \color{red}{\cdot \frac{\;\;\cfrac{1}{x}\;\;}{\cfrac{1}{x}}}=\dfrac{x+ \cfrac{c}{x}-3}{x+ \cfrac{c}{x}+3} = \dfrac{u-3}{u+3}=1 - \cfrac{6}{u+3} \ge 1 - \frac{6}{2 \sqrt{c}+3}
$$</p>
<p>The RHS gives a lower bound, which is actually a minimum since it is attained for $\,x = \sqrt{c}\,$.</p>
<p>For it to equal $\,\dfrac{1}{7}\,$, it follows that $\displaystyle\,1 - \frac{6}{2 \sqrt{c}+3}=\frac{1}{7} \iff 2 \sqrt{c} +3 = 7 \iff c = 4\,$.</p>
<p>What remains to be filled in is the details to cover the rest of cases, which are fairly straightforward to work out, for example $\,f(-x) = \dfrac{1}{f(x)}\,$.</p>
|
40,618 | <p>Let $f_{=}$ be a function from $\mathbb{R}^{2}$ be defined as follows:
(1) if $x = y$ then $f_{=}(x,y) = 1$;
(2) $f_{x,y} = 0$ otherwise.</p>
<p>I would like to have a proof for / a reference to a textbook proof of the following theorem (if it indeed is a theorem):</p>
<p>$f_{=}$ is uncomputable even if one restricts the domain of $f_{=}$ to a proper subset of $\mathbb{R}^{2}$, viz. the set of the computable real numbers</p>
<p>Thanks!</p>
| slimton | 9,355 | <p>Suppose that $f_=$ is computable when restricted to computable real numbers, which means that there exists a Turing machine that, given as input the encoding of two Turing machines $M_1$ and $M_2$ that compute the fractional digits of two computable real numbers $r_1$ and $r_2$ in $[0,1]$, produces $1$ if $r_1 = r_2$ and $0$ otherwise. I will use this assumption to show that the Halting problem is also computable, which is impossible. </p>
<p>Given a Turing machine $M$ and an input $x$ for which we want to know if $M$ on input $x$ halts or not, let $M_x$ be the Turing machine that acts as follows: given an integer $i$ as input, $M_x$ starts a simulation of $M$ on input $x$ for up to $i$ steps, and if the simulation does not halt within that number of steps, it outputs $0$ and otherwise it outputs $1$. By definition, $M_x$ computes the digits of a computable real number (more precisely, it computes the $i$-th digit for every given $i$). Moreover, that real number is $0$ if $M$ on input $x$ does not halt, and the real number $0.0\cdots 011 \cdots = 2^{-k}$ otherwise for some $k \geq 1$. In other words, $M_x$ computes the real number $0$ if and only if $M$ on input $x$ does not halt. To complete the argument, note that $0$ is a computable real number, so if you could tell whether two computable real numbers are equal you would also be able to tell if $M$ on input $x$ halts or not.</p>
|
40,618 | <p>Let $f_{=}$ be a function from $\mathbb{R}^{2}$ be defined as follows:
(1) if $x = y$ then $f_{=}(x,y) = 1$;
(2) $f_{x,y} = 0$ otherwise.</p>
<p>I would like to have a proof for / a reference to a textbook proof of the following theorem (if it indeed is a theorem):</p>
<p>$f_{=}$ is uncomputable even if one restricts the domain of $f_{=}$ to a proper subset of $\mathbb{R}^{2}$, viz. the set of the computable real numbers</p>
<p>Thanks!</p>
| Carl Mummert | 5,442 | <p>The main difficulty in finding a reference for this is that it's so well known :). The fact that equality of reals is only (negatively) semidecidable is a basic and important result in both computable analysis and constructive analysis. </p>
<p>The underlying phenomenon here <em>is</em> about continuity. As Gerald Edgar says, the equality function is not continuous (in particular, it's not sequentially continuous). The proof that slimton presents shows not only that it's discontinuous, but that it's <em>effectively discontinuous</em>: we can make an effective sequence of effective reals that witnesses the discontinuity.</p>
<p>This is closely related to the type-2 functional $E\colon \{0,1\}^\omega \to \{0,1\}$ defined such that $E(f) = 1 \leftrightarrow (\exists k)(f(k) = 1)$. This functional is not computable. </p>
<p>If you look more deeply at slimton's proof, you see that he actually proves that if you had a uniform way to test equality of reals, then you would have a uniform way to compute $E$. In particular the problem of computing equality of computable reals is no easier than that of computing $E$ on computable reals. It can be shown with only a little more work that these are equivalent problems. </p>
<p>This phenomenon is a particular instance of a general phenomenon first studied by Grilliot [1] and now called <em>Grilliot's trick</em>: a functional $\Phi$ is effectively discontinuous if and only if $E$ is computable from $\Phi$. In particular, no effectively discontinuous functional is computable. </p>
<p>1: Thomas J. Grilliot, "On Effectively Discontinuous Type-2 Objects", <em>Journal of Symbolic Logic</em> v. 36, n. 2 (Jun., 1971), pp. 245-248. <a href="http://www.jstor.org/stable/2270259">http://www.jstor.org/stable/2270259</a></p>
|
1,456,262 | <p>Show that the equation to a circle with center at $z_0$ and radius $r$ can be written as $$|z-z_0| = r$$ or as $$z\bar{z} - z\bar{z_0} - \bar{z}z_0 + |z_0|^2 = r^2$$</p>
<p>I let $z_0 = (x_0 + iy_0) = (x_0,y_0)$. Now I have $$(x-x_0)^2 + (y-y_0)^2 = r^2$$</p>
<p>I know $x=\frac{1}{2}(z+\bar{z})$ and $y = \frac{1}{2i}(z-\bar{z})$.</p>
<p>I'm not really sure where to go from here. I seem to be having issues with these types of geometric characterization problems.</p>
| Urgje | 95,681 | <p>Subtract the second equation from the first. Then
\begin{equation*}
x^{\prime }(t)-y^{\prime }(t)=x(t)-y(t)+4t-e^{t}
\end{equation*}
from which $u(t)=x(t)-y(t)$ can be found. Next
\begin{equation*}
x^{\prime }(t)=2x(t)-y(t)+4t=x(t)+u(t)+4t
\end{equation*}
from which $x(t)$ can be obtained.</p>
|
1,729,220 | <p>Let</p>
<p>$$f(x)=\frac{1}{x^2+3x+2}$$</p>
<p>I must find $$\lim_{n\to\infty}\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)$$</p>
<p>How should I proceed?</p>
| Mark Viola | 218,419 | <p>Let $f(x)=\frac{1}{x^2+3x+2}$. Then, using partial fraction expansion, we find that</p>
<p>$$f(x)=\frac{1}{x+1}-\frac{1}{x+2}$$</p>
<p>The $i$'th derivative of $f(x)$ is given by </p>
<p>$$f^{(i)}(x)=(-1)^i\, i!\left(\frac{1}{(x+1)^{i+1}}-\frac{1}{(x+2)^{i+1}}\right)$$</p>
<p>Therefore, we find that</p>
<p>$$\begin{align}
\sum_{i=0}^n \frac{(-1)^i}{i!}f^{(i)}(1)&=\sum_{i=0}^n\left(\frac{1}{2^{i+1}}-\frac{1}{3^{i+1}}\right)\\\\
&=\left(1-\frac{1}{2^{n+1}}\right)-\left(\frac12-\frac{1/2}{3^{n+1}}\right)\\\\
&\to \frac12\,\,\text{as}\,\,n\to \infty
\end{align}$$</p>
|
1,797,245 | <p>How does one show that $I_n = \int\limits_0^1 x^n e^x dx$ is decreasing?</p>
<p>The best I came up with is this: $I_{n-1} - I_n= \int\limits_0^1 e^xx^{n-1}(1-x)dx$, but how do we go from here?</p>
<p>I'd appreciate some hints.</p>
| Jack D'Aurizio | 44,121 | <p>You may prove much more than monotonicity: since $e^x$ is bounded between $1$ and $e$ on $[0,1]$,</p>
<p>$$ \int_{0}^{1} x^n(1-x)\,dx\leq \int_{0}^{1}e^x x^n(1-x)\,dx \leq e\int_{0}^{1} x^n(1-x)\,dx \tag{1}$$
hence:</p>
<blockquote>
<p>$$ \frac{1}{(n+1)(n+2)}\leq \left(I_n-I_{n+1}\right)\leq\frac{e}{(n+1)(n+2)}.\tag{2} $$</p>
</blockquote>
<p>The inequality in the RHS of $(2)$ is tighter since $g(x)=(n+1)(n+2)x^n(1-x)\mathbb{1}_{(0,1)}(x)$ is the PDF of a random variable that converges towards $X=1$.</p>
|
3,528,946 | <p><a href="https://i.stack.imgur.com/JG89d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JG89d.png" alt="enter image description here"></a></p>
<p>This is the example from Friedberg's Linear Algebra book. It is giving an example about Null space (denoted by <span class="math-container">$N(T)$</span>), and range, denoted by <span class="math-container">$R(T)$</span>. I don't understand is why is <span class="math-container">$N(T_{0})$</span>=V not <span class="math-container">$0$</span>. I thought since <span class="math-container">$N(T)$</span>={<span class="math-container">$x \in V: T(x)=0$</span>}, it should always be <span class="math-container">$0$</span>, why not?</p>
| egreg | 62,967 | <p>If <span class="math-container">$\lambda=0$</span> the statement may be true or false, depending on the matrix <span class="math-container">$A$</span>.</p>
<p>The easiest example is for <span class="math-container">$A=0$</span> a null matrix: the only eigenvalue of both <span class="math-container">$A^T\!A$</span> and <span class="math-container">$AA^T$</span> is <span class="math-container">$0$</span> and every nonzero vector is an eigenvector.</p>
<p>For a case when the statement is false, consider <span class="math-container">$A=[1\ 0]$</span>. Then
<span class="math-container">$$
AA^T=[1],\qquad A^TA=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
$$</span>
so <span class="math-container">$0$</span> is not even an eigenvalue of <span class="math-container">$AA^T$</span>, but it is of <span class="math-container">$A^T\!A$</span>.</p>
<p>If you consider the <em>full</em> statement, that is,</p>
<blockquote>
<p>if <span class="math-container">$0$</span> is an eigenvalue of both <span class="math-container">$A^T\!A$</span> and <span class="math-container">$AA^T$</span> and <span class="math-container">$v$</span> is an eigenvector of <span class="math-container">$A^T\!A$</span> relative to <span class="math-container">$0$</span>, then <span class="math-container">$Av$</span> is an eigenvector of <span class="math-container">$AA^T$</span> relative to <span class="math-container">$0$</span>,</p>
</blockquote>
<p>then this statement is indeed false, because the assumption is that <span class="math-container">$A^T\!Av=0$</span>, so <span class="math-container">$v^T\!A^T\!Av=0$</span> as well, which implies <span class="math-container">$Av=0$</span>.</p>
|
1,628,386 | <p>I had following limit of two variables as a problem on my calculus test. How does one show whether the limit below exists or does not exist? I think it does not exist but I was not able to show that rigorously. There was a hint reminding that $\lim_{t\to 0}\sin t / t=1$.</p>
<p>$$\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$$</p>
| محمد حسین فصیحی هرندی | 417,896 | <p>$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2sin^2(y)}{x^4+siny^4})$$
If Dividing by $y^4$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2sin(y)sin(y)}{y^2yy}}{\frac{x^4}{y^4}+\frac {sin(y^4)}{y^4}})$$
Know:$\lim_{t\to0}\frac{sint}{t}=1$</p>
<p>$(y^4=t)$,$(y=t)$</p>
<p>$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2}{y^2}}{\frac{x^4}{y^4}+1})$$</p>
<p>$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2y^2}{x^4+y^4})$$</p>
<p>Way $y=x$</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{3x^4}{4x^4}=\frac{3}{4}$$
Way $Y=0$ limit hint to 0</p>
<p>Not exist</p>
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