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3,270,725 | <p>Hello everyone I read on my notes this proposition: </p>
<p>Given a field <span class="math-container">$K$</span> and <span class="math-container">$R=K[T]$</span>, let <span class="math-container">$M$</span> be a (left) finitely generated <span class="math-container">$R$</span>-module; then <span class="math-container">$M$</span> is a torsion module if and only if <span class="math-container">$\dim_K(M)<\infty$</span>.</p>
<p>Since it has already been stated that <span class="math-container">$M$</span> is finitely generated, <span class="math-container">$\dim_K(M)$</span> must be something different from the number of generators of <span class="math-container">$M$</span>, then my question is: what does <span class="math-container">$\dim_K(M)$</span> mean?</p>
| hmakholm left over Monica | 14,366 | <p>Instead of dividing according to the sign of <span class="math-container">$x$</span>, give <span class="math-container">$\tan x$</span> a name such as <span class="math-container">$y$</span>, and ask yourself:</p>
<p>For which <span class="math-container">$y$</span> do we have <span class="math-container">$|y|>y$</span>?</p>
|
2,009,557 | <p>I am pretty sure this question has something to do with the Least Common Multiple. </p>
<ul>
<li>I was thinking that the proof was that every number either is or isn't a multiple of $3, 5$, and $8\left(3 + 5\right)$.</li>
<li>If it isn't a multiple of $3,5$, or $8$, great. You have nothing to prove.</li>
<li>But if it is divisible by one of them, I couldn't find a general proof that showed that it wouldn't be divisible by another one. Say $15$, it is divisible by $3$ and $5$, but not $8$.</li>
</ul>
| Amr | 434,994 | <ul>
<li>proof by strong induction for n,a,b.</li>
<li>proof hypotheses p(n) = 3*a + 5*b for all n >= 8 //any number greater than 7 consists of 3s and 5s.</li>
<li>base case p(8) holds because it consists of 3 and 5.</li>
<li>Inductive step: we assume p(n) holds for all n>=8 we must prove that p(n+1) holds: <br>
since p(n+1) = k+m <br>
since k,m<=n then p(k),p(m) holds for some 3*a + 5*b <br>
then p(n+1) = 3*a + 5*b + 3*a + 5*b <br>
we conclude that we can make change for any number greater than 7.
<br>
I suppose & hope this works. :)</li>
</ul>
|
3,764,030 | <p>If <span class="math-container">$|z| = \max \{|z-1|,|z+1|\}$</span>, then:</p>
<ol>
<li><span class="math-container">$\left| z + \overline{z} \right| =1/2$</span></li>
<li><span class="math-container">$z + \overline{z} =1$</span></li>
<li><span class="math-container">$\left| z + \overline{z} \right| =1$</span></li>
<li><span class="math-container">$z - \overline{z} = 5$</span></li>
</ol>
<p>I am totally confused that how to calculate maximum of <span class="math-container">$|z-1|$</span> and <span class="math-container">$|z+1|$</span>, I think they represent line <span class="math-container">$x=1$</span> and <span class="math-container">$x=-1$</span> and if so then how to calculate maximum.</p>
<p>Please! Guide me how to proceed.</p>
| Khosrotash | 104,171 | <p>Welcome to MSE. note that <span class="math-container">$|Z|,|z-1|,|z+1|$</span> are real numbers, so we can solve the equation, here split in two cases
<span class="math-container">$$ \max \{|z-1|,|z+1|\}=|z-1| or |z+1|\\
(1):|z| =|z-1|\to put \space z=x+iy\\|x+iy|=|x+iy-1|\\\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\\x^2+y^2=x^2+y^2-2x+1\to x=\frac12\\
(2):|z| =|z+1|\to put \space z=x+iy\\|x+iy|=|x+iy+1|\\\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2}\\x^2+y^2=x^2+y^2+2x+1\to x=-\frac12\\$$</span>you can see <span class="math-container">$Re(z)=\pm \frac 12 \to z=\pm \frac 12 +iy$</span> now
find <span class="math-container">$z+\bar{z}=(\pm \frac 12+iy )+(\pm \frac 12 -iy) =\pm1$</span> it seems <span class="math-container">$Z+\bar{Z}=1$</span> but when you put <span class="math-container">$x=\pm 0.5$</span> in the equation you will find, the question is wrong</p>
<p>let's see
<span class="math-container">$$|\pm 0.5 +iy| = \max \{|\pm 0.5 +iy-1|,|\pm 0.5 +iy+1|\}\\
\sqrt{(\pm0.5)^2+y^2}=\max \{\sqrt{(\pm 0.5+1)^2+y^2},\sqrt{(\pm0.5-1)^2+y^2} \}$$</span> but it is impossible <span class="math-container">$$\sqrt{(0.5)^2+y^2}<\sqrt{(1.5)^2+y^2}$$</span></p>
|
2,635,635 | <p>I have searched a lot and don't really understand the answers I've come across. So apologies in advance if I'm repeating a common question.</p>
<p>The problem is as follows: Distribute $69$ identical items across
$4$ groups where each groups needs to contain at least $5$ items.</p>
<p>The way I see the problem: $x_1 + x_2 + x_3 + x_4 = 69$, $x_i \geq 5$</p>
<p>Could this then be the same as: $x_1 + x_2 + x_3 + x_4 = 49$, $x_i \geq 0$, and therefore be solved using ${n + k -1} \choose {k - 1}$ = ${52}\choose{3}$ = $\frac{52!}{3!(52 - 3)!}$ = $\frac{52!}{3!49!}$</p>
<p>Or am I thinking completely wrong here?</p>
| Rohan Shinde | 463,895 | <p>You are completely right.</p>
<p>Distribute 5 items to each $x_i$ before using star and bars.
Hence we get to find non negative integral solutions of $$a+b+c+d=49$$
Hence the answer would be $\binom {52}{3}$</p>
|
1,353,015 | <p>Given a positive singular measure $\mu$ on $[-\pi,\pi]$, we define a singular inner function by</p>
<p>$$S(z)=\exp\left(-\int_{-\pi}^{\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}\,d\mu(\theta)\right).$$</p>
<p>It is stated in many different sources that the radial limits $\lim_{r\rightarrow1^{-}}S(re^{2\pi it})$ equal 0 for $\mu$-a.e. $t\in\overline{\text{supp}}(\mu)$. None of them give a proof, and most of them say it is "easy to see." I personally have not been able to figure it out. Can anyone either give me a proof or direct me to one?</p>
| zhw. | 228,045 | <p>Hint: For $\mu$-a.e. $x\in \text {supp} (\mu),$</p>
<p>$$\lim_{r\to 0^+}\frac{1}{2r}\int_{(x-r,x+r)}d\mu = \infty.$$</p>
<p>Show this implies that $P_r(\mu)(x) \to \infty,$ where $P_r$ is the Poisson kernel. Now what is the relationship between $|S(re^{ix})|$ and $P_r(\mu)(x)?$</p>
|
2,867,521 | <p>I am interested in calculating the following double summation:</p>
<p>$\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\frac{1}{4}^k \frac{1}{2}^{n-k-2}$</p>
<p>I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.</p>
<p>The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.</p>
| Andreas Blass | 48,510 | <p>No. Let the directed set $\Lambda$ consist of the natural numbers, ordered as usual, and infinitely many additional elements, incomparable with each other but each $<$ all of the natural numbers. Then a net indexed by $\Lambda$ converges iff the sequence of its values at the natural numbers converges, while the values at the additional elements are completely arbitrary. Those arbitrary values can be chosen to make the set of all the net's values plus the limit non-compact.</p>
|
2,867,521 | <p>I am interested in calculating the following double summation:</p>
<p>$\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\frac{1}{4}^k \frac{1}{2}^{n-k-2}$</p>
<p>I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.</p>
<p>The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.</p>
| Daniel Schepler | 337,888 | <p>Suppose the directed set you use as indices is $\mathbb{Z}$, and the net $\mathbb{Z} \to \mathbb{R}$ sends $n \mapsto 2^{-n}$. Then the net converges to 0; however, the image of the net along with its limit is $\{ 2^n \mid n \in \mathbb{Z} \} \cup \{ 0 \}$ which is unbounded so it cannot be compact or even relatively compact.</p>
|
2,193,779 | <p><a href="https://i.stack.imgur.com/d65g2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d65g2.png" alt="enter image description here"></a></p>
<p>Any idea where the missing 3^k+2 comes from?
(sorry for the format, this thing didn't allow me to post images)</p>
| LM2357 | 376,702 | <p>The last step is simply
$$(k+1)(3^{k+1})+\frac{(2k-1)(3^{k+1})+3}{4}$$
$$\frac{4(k+1)(3^{k+1})+{(2k-1)(3^{k+1})+3}}{4}$$
$$\frac{(3^{k+1})(4k+4+2k-1)+3}{4}$$
$$\frac{(3^{k+2})(2k+1)+3}{4}$$
They have simply taken the $3$ out
$$3\left(\frac{(3^{k+1})(2k-1)+1}{4}\right)$$</p>
|
362,250 | <p>Let <span class="math-container">$\nu$</span> be a <em>finite</em> Borel measure on <span class="math-container">$\mathbb{R}^n$</span> and define the shift operator <span class="math-container">$T_a$</span> on <span class="math-container">$L^p_{\nu}(\mathbb{R}^n)$</span> by <span class="math-container">$f\to f(x+a)$</span> for some fixed <span class="math-container">$a\in \mathbb{R}^n-\{0\}$</span>. Suppose moreover that
<span class="math-container">$\nu$</span> is absolutely continuous wrt the Lebesgue measure <span class="math-container">$m$</span> and let
<span class="math-container">$
\frac{d \nu}{dm}(x)= h(x).
$</span></p>
<p>In this case, can we obtain a bound on <span class="math-container">$\|T_{a}\|_{\mathrm{op}}$</span> in terms of <span class="math-container">$h$</span> and of <span class="math-container">$a$</span>? </p>
<p>Usually when <span class="math-container">$\nu$</span> is the Lebesgue measure then this is commonly known to be <span class="math-container">$1$</span>, but here, in the finite and dominated case I can't seem to find such a result...</p>
| Nik Weaver | 23,141 | <p>Well, for large <span class="math-container">$a$</span> the norm goes to infinity. Find a ball <span class="math-container">$B$</span> such that <span class="math-container">$\nu(B) > \nu(\mathbb{R}^n) - \epsilon$</span> and consider the characteristic function of <span class="math-container">$B$</span> shifted by <span class="math-container">$-a$</span>, for any <span class="math-container">$a$</span> greater than the radius of <span class="math-container">$B$</span>. Its <span class="math-container">$L^2$</span> norm is at most <span class="math-container">$\sqrt{\epsilon}$</span>, but after shifting by <span class="math-container">$a$</span> its norm is <span class="math-container">$> \sqrt{\nu(B)}$</span>.</p>
<p>For general <span class="math-container">$a$</span> it's just a matter of comparing <span class="math-container">$h$</span> and its shift by <span class="math-container">$a$</span>. The issue is if <span class="math-container">$f$</span> is the characteristic function of a tiny ball <span class="math-container">$B_1$</span> (tiny compared to <span class="math-container">$a$</span>), and <span class="math-container">$B_2$</span> is the shift of this ball by <span class="math-container">$a$</span>, then the ratio of the square roots of <span class="math-container">$s = \int_{B_2} h$</span> to <span class="math-container">$r = \int_{B_1} h$</span> gives a lower bound on the norm of the shift. Tiny balls are all we need to look at by a short argument using Lebesgue density. So the norm of the shift will be <span class="math-container">$\sqrt{\left\|\frac{h_a}{h}\right\|_\infty}$</span>, where <span class="math-container">$h_a$</span> is the shift of <span class="math-container">$h$</span> by <span class="math-container">$a$</span>. (Note that this could be infinite.)</p>
|
3,829,894 | <p>The problem is to find the parametric equation of the line that is tangent to the line of intersection between the plane <span class="math-container">$x+2y+3z=6$</span> and the surface <span class="math-container">$x^2+y^2=2$</span> and passes through the point <span class="math-container">$(1,1,1)$</span>.</p>
<p>How would I solve this problem?</p>
| gt6989b | 16,192 | <p>Why not reduce to
<span class="math-container">$$
\int_0^1 x^m (1-x)^n dx
$$</span>
and take it by parts, differentiating <span class="math-container">$(1-x)^n$</span> until it disappears?</p>
<p>E.g. if <span class="math-container">$u = (1-x)^n$</span> and <span class="math-container">$dv = x^m dx$</span> then <span class="math-container">$du = -n(1-x)^{n-1}dx$</span> and <span class="math-container">$v = \frac{x^{m+1}}{m+1}$</span> so <span class="math-container">$uv|_0^1 = 0$</span> and
<span class="math-container">$$
\begin{split}
I(m,n)
&= \int_0^1 x^m (1-x)^n dx \\
&= \frac{n}{m+1} \int_0^1 x^{m+1} (1-x)^{n-1} dx \\
&= \frac{n}{m+1} I(m+1,n-1)
\end{split}
$$</span>
and now just solve the recurrence:
<span class="math-container">$$
\begin{split}
I(m,n)
&= \frac{n}{m+1} I(m+1,n-1) \\
&= \frac{n}{m+1} \frac{n-1}{m+2} I(m+2,n-2) \\
&= \ldots \\
&= \frac{n!m!}{(m+n)!} I(m+n,0) \\
&= \frac{1}{\binom{m+n}{m}} \frac{1}{m+n+1}
\end{split}
$$</span></p>
|
3,829,894 | <p>The problem is to find the parametric equation of the line that is tangent to the line of intersection between the plane <span class="math-container">$x+2y+3z=6$</span> and the surface <span class="math-container">$x^2+y^2=2$</span> and passes through the point <span class="math-container">$(1,1,1)$</span>.</p>
<p>How would I solve this problem?</p>
| Henry Lee | 541,220 | <p>notice that:
<span class="math-container">$$I(n,k)=\int_0^1\frac{x^k(1-x)^n}{(1-x)^k}dx=\int_0^1 x^k(1-x)^{n-k}dx$$</span>
and we have a definition for the beta function as follows:
<span class="math-container">$$B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$</span>
and so:
<span class="math-container">$$I(n,k)=B(k+1,n-k+1)=\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}$$</span>
which works for non-integer values too :)</p>
|
1,862,807 | <blockquote>
<p>Show that for <span class="math-container">$x,y,z\in\mathbb{Z}$</span>, if <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are coprime and <span class="math-container">$z$</span> is nonzero, then <span class="math-container">$\exists n\in\mathbb{Z}$</span> such that <span class="math-container">$z$</span> and <span class="math-container">$y+xn$</span> are coprime.</p>
</blockquote>
<p>Not sure where to start on this one. I understand that coprime indicates that their GCD is 1, but I am somewhat confused how to proceed.</p>
| André Nicolas | 6,312 | <p>We give an elementary proof that does not use Dirichlet's Theorem. </p>
<p>Let $P$ be the product of the primes that divide $z$ but do not divide $x$. (Recall that an empty product is equal to $1$.) </p>
<p>Since $x$ and $P$ are relatively prime, there is a solution $n$ of the congruence
$$xn\equiv -y+1\pmod{P}.$$ We show this $n$ works, by showing that $y+xn$ is relatively prime to $z$. </p>
<p>Suppose to the contrary that $y+xn$ and $z$ are not relatively prime. Then there is a prime $p$ that divides both. We show that this is impossible by showing that if $p$ divides $z$, then $p$ cannot divide $y+xn$. </p>
<p>If $p$ divides $x$, then it cannot divide $y$, since $x$ and $y$ are coprime. So $p$ does not divide $x$, and therefore $p$ divides $P$. Thus
$xn+y\equiv 1\pmod{p}$. It follows that $p$ does not divide $y+xn$, and we are finished.</p>
<p><em>Remark</em>: It is now easy to see that $n'z+y$ and $z$ are relatively prime for $n'=n+tP$ with $t$ arbitrary, so in fact there are infinitely many $n$ that do the job.</p>
|
1,862,807 | <blockquote>
<p>Show that for <span class="math-container">$x,y,z\in\mathbb{Z}$</span>, if <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are coprime and <span class="math-container">$z$</span> is nonzero, then <span class="math-container">$\exists n\in\mathbb{Z}$</span> such that <span class="math-container">$z$</span> and <span class="math-container">$y+xn$</span> are coprime.</p>
</blockquote>
<p>Not sure where to start on this one. I understand that coprime indicates that their GCD is 1, but I am somewhat confused how to proceed.</p>
| darij grinberg | 586 | <p>This is an annoyingly nontrivial statement, despite its harmless sound.
I have two proofs lying around from a number theory homework set that
contained this exercise, so let me post them here; I am sorry for the
mismatching notations.</p>
<blockquote>
<p><strong>Problem 1.</strong>
Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> be three integers such that <span class="math-container">$\gcd\left( a,b\right) =1$</span> and
<span class="math-container">$c>0$</span>. Prove that there is an integer <span class="math-container">$x$</span> such that <span class="math-container">$\gcd\left( a+bx,c\right)
=1$</span>.</p>
</blockquote>
<p>My numbers <span class="math-container">$a, b, c, x$</span> correspond to the numbers <span class="math-container">$y, x, \left|z\right|, n$</span>
in the original post.</p>
<p>I will, in fact, solve a stronger problem:</p>
<blockquote>
<p><strong>Problem 2.</strong>
Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> be three integers such that <span class="math-container">$\gcd\left( a,b,c\right) =1$</span> and
<span class="math-container">$c>0$</span>. Prove that there is an integer <span class="math-container">$x$</span> such that <span class="math-container">$\gcd\left( a+bx,c\right)
=1$</span>.</p>
</blockquote>
<p>Problem 1 is a particular case of Problem 2, because if we have
<span class="math-container">$\gcd\left(a,b\right) = 1$</span>, then <span class="math-container">$\gcd\left(a,b,c\right) = 1$</span>
holds as well (since <span class="math-container">$\gcd\left(a,b,c\right)
= \gcd\left(\underbrace{\gcd\left(a,b\right)}_{=1},c\right)
= \gcd\left(1,c\right) = 1$</span>).
Thus, it suffices to solve Problem 2.</p>
<p><em>First solution to Problem 2.</em>
Let <span class="math-container">$\left( m_1, m_2, \ldots, m_r \right)$</span> be a list of all prime
divisors of <span class="math-container">$c$</span> (without multiplicities). Thus, the integers
<span class="math-container">$m_1, m_2, \ldots, m_r$</span> are distinct primes, and therefore are mutually
coprime.</p>
<p>If <span class="math-container">$p$</span> is any prime divisor of <span class="math-container">$c$</span>, then
<span class="math-container">\begin{equation}
\text{there exists some }
y\in \left\{ 0,1 \right\} \text{ such that } p\nmid a+by .
\label{2.3.p38.oneprime}
\tag{1}
\end{equation}</span></p>
<p>[<em>Proof of \eqref{2.3.p38.oneprime}:</em> Let <span class="math-container">$p$</span> be a
prime divisor of <span class="math-container">$c$</span>.
We must show that there exists some <span class="math-container">$y\in \left\{ 0,1 \right\}$</span> such
that <span class="math-container">$p\nmid a+by$</span>.</p>
<p>Assume the contrary. Thus, there exists no <span class="math-container">$y\in \left\{ 0,1 \right\}$</span> such
that <span class="math-container">$p\nmid a+by$</span>. In other words, every <span class="math-container">$y\in\left\{ 0,1 \right\}$</span>
satisfies <span class="math-container">$p\mid a+by$</span>. Applying this to <span class="math-container">$y=0$</span>, we obtain <span class="math-container">$p\mid a$</span>.
Also, <span class="math-container">$p \mid c$</span> (since <span class="math-container">$p$</span> is a prime divisor of <span class="math-container">$c$</span>).</p>
<p>But recall that every <span class="math-container">$y\in \left\{ 0,1 \right\}$</span> satisfies <span class="math-container">$p\mid a+by$</span>.
Applying this to <span class="math-container">$y=1$</span>, we obtain <span class="math-container">$p\mid a+b1=a+b$</span>. Now, both integers <span class="math-container">$a$</span> and
<span class="math-container">$a+b$</span> are divisible by <span class="math-container">$p$</span> (since <span class="math-container">$p\mid a$</span> and <span class="math-container">$p\mid a+b$</span>). Hence, their
difference <span class="math-container">$\left( a+b \right) - a = b$</span> is also divisible by <span class="math-container">$p$</span>. Thus, all
three integers <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> are divisible by <span class="math-container">$p$</span> (since <span class="math-container">$p \mid a$</span>
and <span class="math-container">$p \mid c$</span>). In other words, <span class="math-container">$p$</span> is a common divisor of <span class="math-container">$a$</span>, <span class="math-container">$b$</span>
and <span class="math-container">$c$</span>. Therefore, <span class="math-container">$p$</span> divides the greatest common divisor <span class="math-container">$\gcd\left(
a,b,c\right) $</span> of <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span>.
In other words, <span class="math-container">$p\mid \gcd\left( a,b,c \right) =1$</span>.
Hence, <span class="math-container">$p$</span> is either <span class="math-container">$1$</span> or <span class="math-container">$-1$</span>. This contradicts the fact that <span class="math-container">$p$</span> is prime.
This contradiction proves that our assumption was wrong; hence,
\eqref{2.3.p38.oneprime} is proven.]</p>
<p>Now, for every <span class="math-container">$i\in\left\{ 1,2,\ldots,r\right\} $</span>, there exists some
<span class="math-container">$y\in \left\{ 0,1 \right\}$</span> such that <span class="math-container">$m_i\nmid a+by$</span> (by
\eqref{2.3.p38.oneprime}, applied to <span class="math-container">$p=m_i$</span>). Let us denote this <span class="math-container">$y$</span> by
<span class="math-container">$a_{i}$</span>. Thus, for every <span class="math-container">$i\in\left\{ 1,2,\ldots,r\right\} $</span>, the integer
<span class="math-container">$a_{i}\in \left\{ 0,1 \right\}$</span> satisfies
<span class="math-container">\begin{equation}
m_{i}\nmid a+ba_{i}.
\label{2.3.p38.ai}
\tag{2}
\end{equation}</span></p>
<p>Now, recall that the integers <span class="math-container">$m_1, m_2, \ldots, m_r$</span> are mutually
coprime. Hence, the <a href="https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Theorem_statement" rel="nofollow noreferrer">Chinese Remainder Theorem</a>
(in its elementary form, stated for <span class="math-container">$r$</span> integers)
shows that the congruences
<span class="math-container">\begin{align}
\left\{
\begin{array}[c]{c}
x\equiv a_{1}\operatorname{mod}m_{1},\\
x\equiv a_{2}\operatorname{mod}m_{2},\\
\vdots\\
x\equiv a_{r}\operatorname{mod}m_{r}
\end{array}
\right.
\end{align}</span>
have a common solution. Let <span class="math-container">$z$</span> be such a solution. Thus, <span class="math-container">$z$</span> is an integer
and satisfies
<span class="math-container">\begin{equation}
\left\{
\begin{array}[c]{c}
z\equiv a_{1}\operatorname{mod}m_{1},\\
z\equiv a_{2}\operatorname{mod}m_{2},\\
\vdots\\
z\equiv a_{r}\operatorname{mod}m_{r}
\end{array}
\right. .
\label{2.3.p38.z}
\tag{3}
\end{equation}</span></p>
<p>Now, let me prove that
<span class="math-container">\begin{equation}
\gcd\left( a+bz,c\right) =1.
\label{2.3.p38.a+bz}
\tag{4}
\end{equation}</span></p>
<p>[<em>Proof of \eqref{2.3.p38.a+bz}:</em> Assume the contrary. Thus, <span class="math-container">$\gcd\left(
a+bz,c\right) >1$</span>. Hence, there exists a prime <span class="math-container">$p$</span> such that <span class="math-container">$p\mid\gcd\left(
a+bz,c\right) $</span>. Consider this <span class="math-container">$p$</span>.</p>
<p>We have <span class="math-container">$p\mid\gcd\left( a+bz,c\right) \mid c$</span>. Thus, <span class="math-container">$p$</span> is a prime divisor of
<span class="math-container">$c$</span> (since <span class="math-container">$p$</span> is a prime). In other words, <span class="math-container">$p\in\left\{ m_{1},m_{2}
,\ldots,m_{r}\right\} $</span> (since <span class="math-container">$\left( m_1, m_2, \ldots, m_r \right) $</span> is
a list of all prime divisors of <span class="math-container">$c$</span>). In other words, <span class="math-container">$p=m_{i}$</span> for some
<span class="math-container">$i\in\left\{ 1,2,\ldots,r\right\} $</span>. Consider this <span class="math-container">$i$</span>.</p>
<p>We have <span class="math-container">$m_{i}\nmid a+ba_{i}$</span> (by \eqref{2.3.p38.ai}). In other words,
<span class="math-container">$a+ba_{i}\not \equiv 0 \mod m_{i}$</span>. But <span class="math-container">$z\equiv a_{i}
\mod m_{i}$</span> (by \eqref{2.3.p38.z}). Hence, <span class="math-container">$a+b\underbrace{z}
_{\equiv a_{i}\mod m_{i}}\equiv a+ba_{i}\not \equiv
0\mod m_{i}$</span>. In other words, <span class="math-container">$m_{i}\nmid a+bz$</span>. This contradicts
<span class="math-container">$m_{i}=p\mid\gcd\left( a+bz,c\right) \mid a+bz$</span>. This contradiction proves that
our assumption was wrong. Thus, \eqref{2.3.p38.a+bz} is proven.]</p>
<p>Now that \eqref{2.3.p38.a+bz} is proven, we can immediately see that there is
an integer <span class="math-container">$x$</span> such that <span class="math-container">$\gcd\left( a+bx,c\right) =1$</span> (namely, <span class="math-container">$x=z$</span>).
Thus, Problem 1 is solved. <span class="math-container">$\blacksquare$</span></p>
<p><strong>Remark.</strong>
It is easy to see (using the uniqueness part of the Chinese Remainder Theorem)
that the number
of all <span class="math-container">$x\in\left\{ 0,1,\ldots,m_{1}m_{2}\cdots m_{r}-1\right\} $</span> satisfying
<span class="math-container">$\gcd\left( a+bx,c\right) =1$</span> is
<span class="math-container">\begin{align}
\prod\limits_{i=1}^{r}
\begin{cases}
m_{i}-1, & \text{if }m_{i}\nmid b;\\
m_{i}, & \text{if }m_{i}\mid b
\end{cases}
.
\end{align}</span>
Thus, if we let <span class="math-container">$\alpha_{i}$</span> be the multiplicity of the prime factor <span class="math-container">$m_{i}$</span>
in <span class="math-container">$c$</span> (so that <span class="math-container">$c=\prod\limits_{i=1}^{r}m_{i}^{\alpha_{i}}$</span>), then the number of all
<span class="math-container">$x\in\left\{ 0,1,\ldots,c-1\right\} $</span> satisfying <span class="math-container">$\gcd\left( a+bx,c\right) =1$</span>
is
<span class="math-container">\begin{align}
\prod\limits_{i=1}^{r}\left(
\begin{cases}
m_i-1, & \text{if }m_{i}\nmid b;\\
m_i, & \text{if }m_{i}\mid b
\end{cases}
m_i^{\alpha_i-1}\right)
\geq
\prod\limits_{i=1}^{r}\left(
\left(m_i - 1\right)
m_i^{\alpha_i-1}\right)
= \phi\left( c\right) .
\end{align}</span>
So not only do integers <span class="math-container">$x$</span> satisfying <span class="math-container">$\gcd\left( a+bx,c\right) =1$</span> exist, but
they are also at least as frequent as integers <span class="math-container">$x$</span> satisfying <span class="math-container">$\gcd\left(
x,c\right) =1$</span> (in an appropriate sense of "frequent").</p>
<p><em>Second solution to Problem 1.</em>
The following alternative solution was suggested by some of the students
in MIT 18.781 Spring 2016.</p>
<p>For any integer <span class="math-container">$n$</span>, let <span class="math-container">$\operatorname*{PF}n$</span> denote the set of all prime
factors of <span class="math-container">$n$</span>. This is a finite set when <span class="math-container">$n\neq0$</span>. Thus, in particular,
<span class="math-container">$\operatorname*{PF}c$</span> is a finite set.</p>
<p>Now, let
<span class="math-container">\begin{align}
x=\prod\limits_{p\in\left( \operatorname*{PF}c\right) \setminus\left(
\operatorname*{PF}a\right) }p.
\end{align}</span>
I claim that <span class="math-container">$\gcd\left( a+bx,c\right) =1$</span>. Obviously, once this is proven, the
problem will be solved.</p>
<p>So we must prove that <span class="math-container">$\gcd\left( a+bx,c\right) =1$</span>. Indeed, assume the
contrary. Thus, <span class="math-container">$\gcd\left( a+bx,c\right) >1$</span>. Hence, the integer <span class="math-container">$\gcd\left(
a+bx,c\right) $</span> has a prime divisor <span class="math-container">$q$</span>. Consider this <span class="math-container">$q$</span>.</p>
<p>We have <span class="math-container">$q\mid\gcd\left( a+bx,c\right) \mid c$</span> and thus <span class="math-container">$q\in\operatorname*{PF}
c$</span> (since <span class="math-container">$q$</span> is a prime). Moreover, <span class="math-container">$q\mid\gcd\left( a+bx,c\right) \mid a+bx$</span>.
Now, we must be in one of the following two cases:</p>
<p><em>Case 1:</em> We have <span class="math-container">$q\in\operatorname*{PF}a$</span>.</p>
<p><em>Case 2:</em> We have <span class="math-container">$q\notin\operatorname*{PF}a$</span>.</p>
<p>Let us consider Case 1 first. In this case, we have <span class="math-container">$q\in\operatorname*{PF}a$</span>.
Thus, <span class="math-container">$q$</span> is a prime divisor of <span class="math-container">$a$</span>. In particular, <span class="math-container">$q\mid a$</span>. Hence, both
integers <span class="math-container">$a$</span> and <span class="math-container">$a+bx$</span> are divisible by <span class="math-container">$q$</span>. Therefore, their difference
<span class="math-container">$\left( a+bx\right) -a=bx$</span> must also be divisible by <span class="math-container">$q$</span>. In other words,
<span class="math-container">$q\mid bx$</span>. Since <span class="math-container">$q$</span> is prime, this shows that <span class="math-container">$q\mid b$</span> or <span class="math-container">$q\mid x$</span> (because
if a prime divides a product of two integers, then it must divide at least
one of these two integers).</p>
<p>But <span class="math-container">$q \mid b$</span> is false. [<em>Proof:</em> Assume the contrary. Thus, we have
<span class="math-container">$q\mid b$</span>. Hence, <span class="math-container">$q$</span> divides all three integers <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span>.
Therefore, <span class="math-container">$q$</span> divides the gcd
<span class="math-container">$\gcd\left( a,b,c\right) $</span> of <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span>. In other words, <span class="math-container">$q\mid\gcd\left(
a,b,c\right) =1$</span>. Therefore, <span class="math-container">$q=1$</span> or <span class="math-container">$q=-1$</span>. But this contradicts the fact
that <span class="math-container">$q$</span> is a prime. This contradiction shows that our assumption was wrong,
qed.]</p>
<p>So we know that <span class="math-container">$q \mid b$</span> or <span class="math-container">$q \mid x$</span>, but we also know that <span class="math-container">$q \mid b$</span>
is false. Hence, we must have <span class="math-container">$q\mid x$</span>. Thus, <span class="math-container">$q\mid x=\prod\limits_{p\in\left(
\operatorname*{PF}c\right) \setminus\left( \operatorname*{PF}a\right) }p$</span>.</p>
<p>But <span class="math-container">$q$</span> is a prime. Hence, if <span class="math-container">$q$</span> divides a product of several integers, then
<span class="math-container">$q$</span> must divide one of the factors of the product (by a well-known property
of primes).
Since <span class="math-container">$q$</span> divides the product <span class="math-container">$\prod\limits_{p\in\left(
\operatorname*{PF}c\right) \setminus\left( \operatorname*{PF}a\right) }p$</span>,
we thus conclude that <span class="math-container">$q$</span> must divide one of the factors of this product. In
other words, <span class="math-container">$q\mid p$</span> for some <span class="math-container">$p\in\left( \operatorname*{PF}c\right)
\setminus\left( \operatorname*{PF}a\right) $</span>. Consider this <span class="math-container">$p$</span>.</p>
<p>We have <span class="math-container">$p\in\left( \operatorname*{PF}c\right) \setminus\left(
\operatorname*{PF}a\right) \subseteq\operatorname*{PF}c$</span>. Thus, <span class="math-container">$p$</span> is a
prime divisor of <span class="math-container">$c$</span>. Since <span class="math-container">$p$</span> is a prime, the only prime divisor of <span class="math-container">$p$</span> is
<span class="math-container">$p$</span>. Since <span class="math-container">$q$</span> is a prime divisor of <span class="math-container">$p$</span> (because <span class="math-container">$q$</span> is a prime and because
<span class="math-container">$q\mid p$</span>), this shows that <span class="math-container">$q=p$</span>. Thus, <span class="math-container">$q=p\in\left( \operatorname*{PF}
c\right) \setminus\left( \operatorname*{PF}a\right) $</span>, so that
<span class="math-container">$q\notin\operatorname*{PF}a$</span>. But this contradicts <span class="math-container">$q\in\operatorname*{PF}a$</span>.
Thus, we have obtained a contradiction in Case 1.</p>
<p>Let us now consider Case 2. In this case, we have <span class="math-container">$q\notin\operatorname*{PF}
a$</span>. Combining this with <span class="math-container">$q\in\operatorname*{PF}c$</span>, we obtain <span class="math-container">$q\in\left(
\operatorname*{PF}c\right) \setminus\left( \operatorname*{PF}a\right) $</span>.
Hence, <span class="math-container">$q$</span> is a factor in the product <span class="math-container">$\prod\limits_{p\in\left( \operatorname*{PF}
c\right) \setminus\left( \operatorname*{PF}a\right) }p$</span>. Therefore, <span class="math-container">$q$</span>
divides this product. Thus, <span class="math-container">$q\mid\prod\limits_{p\in\left( \operatorname*{PF}
c\right) \setminus\left( \operatorname*{PF}a\right) }p=x\mid bx$</span>. Hence,
both integers <span class="math-container">$bx$</span> and <span class="math-container">$a+bx$</span> are divisible by <span class="math-container">$q$</span>. Therefore, their
difference <span class="math-container">$\left( a+bx\right) -bx=a$</span> is also divisible by <span class="math-container">$q$</span>.
Consequently, <span class="math-container">$q$</span> is a prime divisor of <span class="math-container">$a$</span> (since <span class="math-container">$q$</span> is a prime). In other
words, <span class="math-container">$q\in\operatorname*{PF}a$</span>. But this contradicts <span class="math-container">$q\notin
\operatorname*{PF}a$</span>. Thus, we have obtained a contradiction in Case 2.</p>
<p>We have now found contradictions in both Cases 1 and 2. Since these two cases
cover all possibilities, this shows that we always get a contradiction. Thus,
our assumption was wrong, and therefore <span class="math-container">$\gcd\left( a+bx,c\right) =1$</span> is proven.
This solves Problem 2. <span class="math-container">$\blacksquare$</span></p>
|
1,355,901 | <p>Let $A$ be the set of all integers $x$ such that $x = 2k$ for some integer $k$</p>
<p>Let $B$ be the set of all integers $x$ such that $x = 2k+2$ for some integer $k$</p>
<p>Give a formal proof that $A = B$.</p>
| Jonathan Hebert | 120,932 | <p>You need to show us some effort in the future.</p>
<p>First, to show two sets are equal, we normally pick an element of the first set, show it is contained in the second, then pick an element in the second, and show it is contained in the first.</p>
<p>If we suppose $x \in A$, then $x=2k$ for some integer $k$. Since $x = 2k$, $x = 2(k-1)+2$, and since $k-1$ is an integer, $x \in B$. Thus, $A \subseteq B$.</p>
<p>I leave the proof of $B \subseteq A$ to you.</p>
|
2,451,092 | <p>I want to solve a Lagrange multiplier problem,</p>
<p>$$f(x,y) = x^2+y^2+2x+1$$
$$g(x,y)=x^2+y^2-16 $$</p>
<p>Where function $g$ is my constraint.
$$f_x=2x+2, \ \ \ f_y=2y, \ \ \ g_x=2x\lambda, \ \ \ g_y=2y\lambda$$</p>
<p>$$
\begin{cases}
2x+2=2x\lambda \\
2y=2y\lambda \\
x^2+y^2-16=0
\end{cases}
$$</p>
<p>See, this is a very nasty system of equations.
At any rate, I get $\lambda = 1$ because in this case, $y=0$. So I cannot do anything with this as far as algebra is concerned? How do I resolve a problem like this?</p>
| ashi | 299,991 | <p><img src="https://i.stack.imgur.com/0RK0z.jpg" alt="">
this method was taught in our class
Hope this could help you</p>
|
254,623 | <p>For example 100 is even and 100/2= 50 is also even</p>
<p>But 30 is also even but 30/2=15 is odd</p>
<p>Now let's say I have a number as large as 10^10000000000...</p>
<p>I want to know how many steps are involved in cutting this number in half. When the number is even, I divide it by 2. When it's odd, I subtract 1. I continue this process until I hit 0.</p>
<p>However, when the number is too high, I can't actually manipulate it directly (elaboration: too big to write out, and too big to fit into memory on a computer), so I am curious if there's a way for me to do this by just knowing the even/odd attributes along the chain.</p>
<p>I hope this makes sense!</p>
<p>Example:
If n=100, I have the following chain</p>
<p>100, 50, 25, 24, 12, 6, 3, 2, 1, 0</p>
<p>Which is a total of 9 "splitting steps" (10 if you count the original number)
And the following parities</p>
<p>Even, Even, Odd, Even, Even, Even, Odd, Even, Odd, Even</p>
<p>I am asking if, given n=10, there is a way to get this parity chain</p>
| Hagen von Eitzen | 39,174 | <p>Your sequence corresponds quirte directly with the binary representation of the original number: Starting form the least significant binary digit, each $0$ corresponds to "even, dide by two" and each $1$ corresponds to "odd, subtract one; even, divide by two".
For your example $n=100$, which is $1100100$ in binary we thus obtain</p>
<p><strong>Even,</strong> divide by two;
<strong>even,</strong> divide by two;
<strong>odd,</strong> subtract one, then <strong>even,</strong> divide by two;
<strong>even,</strong> divide by two;
<strong>even,</strong> divide by two;
<strong>odd,</strong> subtract one, then <strong>even,</strong> divide by two;
<strong>odd,</strong> subtract one. Finally zero is <strong>even.</strong></p>
<p>Thus determining your even/odd sequence is <em>equivalent</em> to determinig the binary representation of the given number.</p>
|
860,247 | <p>Simplify $$\frac{3x}{x+2} - \frac{4x}{2-x} - \frac{2x-1}{x^2-4}$$</p>
<ol>
<li><p>First I expanded $x²-4$ into $(x+2)(x-2)$. There are 3 denominators. </p></li>
<li><p>So I multiplied the numerators into: $$\frac{3x(x+2)(2-x)}{(x+2)(x-2)(2-x)} - \frac{4x(x+2)(x-2)}{(x+2)(x-2)(2-x)} - \frac{2x-1(2-x)}{(x+2)(x-2)(2-x)} $$</p></li>
</ol>
<p>I then tried 2 different approaches:</p>
<ol>
<li>Calculated it without eliminating the denominator into: $$\frac{-6x²-5x+2}{(x+2)(x-2)(2-x)}$$</li>
<li>Calculated it by multiplying it out to: $$\frac{-6x+2x²+2}{(x+2)(x-2)(2-x)}$$</li>
</ol>
<p>I can't seem to simplify them further and so they seem incorrect. Something I missed? Help! </p>
| Omran Kouba | 140,450 | <p>Note that
$$x^2\sqrt{x+x^2}=\frac{x^3+x^4}{\sqrt{x(1+x)}}$$
So, let us look for a polynomial $P(x)=a x^3+bx^2+cx+d$ such that the derivative
$(P(x)\sqrt{x(1+x)})^\prime$ is as close as we can to this function. An easy calculation shows that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{4 a x^4+(\frac{7 a }{2}+3 b)
x^3+(\frac{5 b }{2}+2 c)
x^2+(\frac{3 c }{2}+d )x+\frac{d}{2}}{\sqrt{x(1+x)}}
$$
So, choosing $a=\frac{1}{4}$, $b=\frac{1}{24}$, $c=-\frac{5}{96}$ and $d=\frac{5}{64}$ we see that
$$
\left(P(x)\sqrt{x(1+x)}\right)^\prime=\frac{ x^4+
x^3}{\sqrt{x(1+x)}}+\frac{5}{64}\cdot\frac{1}{2\sqrt{x(1+x)}}
$$
This reduces the considered integral to a simple one:
$$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}\Big]_0^1-\frac{5}{64}\int_0^1\frac{dx}{2\sqrt{x(1+x)}}$$
The last integral is easy since $\log(\sqrt{x}+\sqrt{1+x})$ is a primitive of the integrand.
Thus
$$\int_0^1x^2\sqrt{x(1+x)}dx=\Big[P(x)\sqrt{x(1+x)}-\frac{5}{64}\log(\sqrt{x}+\sqrt{1+x})\Big]_0^1$$
Finally,
$$\int_0^1x^2\sqrt{x(1+x)}dx=
\frac{61}{96 \sqrt{2}}-\frac{5}{64}
\log \left(1+\sqrt{2}\right).
$$</p>
|
1,378,536 | <p>Here is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known <em>nice real analysis tools</em> for calculating them (<em>including here all possible sources<br>
in literature that are publicaly available</em>). At some point I'll add my <em>real analysis</em> solution.<br>
It's a question for the informative purpose rather than finding solutions, the solution is optional.</p>
<p>Prove that</p>
<p>$$\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6}. $$</p>
<p><em>Here is a supplementary question</em></p>
<p>$$\int_{-1}^1 \frac{\log(1-x)}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{4}+\frac{\gamma }{6}+\frac{\log (2)}{6}-2 \log (A) $$</p>
<p>where $A$ is <a href="https://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant">Glaisher–Kinkelin constant</a>.</p>
<p>for the passionates of integrals, series and limits.</p>
| M.N.C.E. | 178,187 | <p><strong>Approach 1:</strong></p>
<p>For the first integral
\begin{align}
2\int^1_{0}\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}
&=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{(\ln{x}+\pi i)(1+x)^2}\tag1\\
&=-\frac{4}{\pi}\mathrm{Im}\int^1_{-1}\frac{{\rm d}x}{\ln{x}(1-x)^2}\tag2\\
&=\frac{4}{\pi}\mathrm{Im}\left[\int^\pi_0\frac{ie^{i\phi}}{i\phi(1-e^{i\phi})^2}\ {\rm d}\phi-\frac{\pi i}{2}\operatorname*{Res}_{z=1}\frac{1}{\ln{z}(z-1)^2}\right]\tag3\\
&=\frac{4}{\pi}\mathrm{Im}\left[\int^\pi_0\frac{{\rm d}\phi}{-4\phi\sin^2\left(\frac{\phi}{2}\right)}-\frac{\pi i}{2}\operatorname*{Res}_{z=1}\frac{1}{\ln{z}(z-1)^2}\right]\\
&=\frac{4}{\pi}\left(-\frac{\pi}{2}\right)\operatorname*{Res}_{z=1}\left[\frac{1}{(z-1)^3}+\frac{1}{2(z-1)^2}\color{red}{-}\frac{\color{red}{1}}{\color{red}{12}(z-1)}+\mathcal{O}(1)\right]\tag4\\
&=\frac{1}{6}
\end{align}
Explanation: <br/>
$(1)$: Substitute $x\mapsto \dfrac{1-x}{1+x}$. <br/>
$(2)$: Substitute $x\mapsto -x$ and extend the integration interval to $[-1,1]$. The integral over $[0,1]$ contributes nothing since it is purely real. <br/>
$(3)$: Change the path of integration to the unit semicircular arc in the upper half plane. We pick up the residue at $z=1$ in the process. <br/>
$(4)$: The first integral in the previous line is purely real; Expand $\displaystyle\frac{1}{\ln{z}(z-1)^2}$ as a Laurent series.</p>
<hr>
<p><strong>Approach 2:</strong></p>
<p>Here is another approach that avoids complex analysis.
\begin{align}
2\int^1_0\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}
&=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{\ln(-x)(1+x)^2}\\
&=\frac{4}{\pi}\int^\infty_0\int^1_0\frac{x^s\sin(\pi s)}{(1+x)^2}\ {\rm d}x\ {\rm d}s\\
&=\frac{4}{\pi}\sum^\infty_{k=0}(-1)^k\int^1_0\int^1_0\frac{x^{s+k}\sin(\pi s)}{(1+x)^2}\ {\rm d}x\ {\rm d}s\\
&=\frac{4}{\pi}\int^1_0\sin(\pi s)\int^1_0\frac{x^{s}}{(1+x)^3}\ {\rm d}x\ {\rm d}s\\
&=\frac{4}{\pi}\int^1_0\sin(\pi s)\left[-\frac{1}{8}-\frac{s}{4}+\frac{s(s-1)}{4}\left(\psi_0\left(\frac{s}{2}\right)-\psi_0\left(\frac{s-1}{2}\right)\right)\right]\ {\rm d}s\\
&=\frac{1}{\pi}\int^1_0\sin(\pi s)s(s-1)\left[\psi_0\left(\frac{s}{2}\right)-\psi_0\left(\frac{s+1}{2}\right)\right]\ {\rm d}s\\
&=\frac{2}{\pi}\int^1_0\ln\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{s+1}{2}\right)}\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s
\end{align}
Using the fact
$$\ln\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{s+1}{2}\right)}=\sum_{n\in2\mathbb{N}_0+1}\left[\frac{1}{n}\cos(n\pi s)+2\frac{\gamma+\ln(2n\pi)}{n\pi}\sin(n\pi s)\right]\ \ , \ \ 0<s\le1$$
in tandem with the results
\begin{align}
\int^1_0\sin((2n+1)\pi s)\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s
&=0\\
\int^1_0\cos((2n+1)\pi s)\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s
&=
\begin{cases}
\frac{\pi}{12}+\frac{1}{4\pi},& \text{if $n=0$}\\
\frac{2n+1}{4\pi}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right),& \text{if $n\in\mathbb{N}$}
\end{cases}
\end{align}
yields
\begin{align}
2\int^1_0\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}
&=\frac{2}{\pi}\left[\frac{\pi}{12}+\frac{1}{4\pi}+\frac{1}{4\pi}\sum^\infty_{n=1}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right)\right]=\frac{1}{6}
\end{align}</p>
<hr>
<p><strong>Supplementary Problem:</strong></p>
<p>Substituting $x\mapsto\tanh{x}$ then $2x\mapsto\ln{x}$,
\begin{align}
\int^1_{-1}\frac{\ln(1-x)}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x
&=\int^1_{-1}\frac{\ln{2}}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x+\int^\infty_{-\infty}\frac{\ln\left(\frac{1-\tanh{x}}{2}\right)}{(\pi^2+4x^2)\cosh^2{x}}\ {\rm d}x\\
&=\frac{\ln{2}}{6}-\int^\infty_{0}\frac{2\ln(1+x)}{(\pi^2+\ln^2{x})(1+x)^2}\ {\rm d}x\\
&=\frac{\ln{2}}{6}-\int^1_0\frac{1}{2\pi i}\int_Cf(z,x)\ {\rm d}z\ {\rm d}x
\end{align}
where $f(z,x)=\dfrac{2z}{(1+xz)(\ln{z}-\pi i)(1+z)^2}$ and $C$ is the keyhole contour deformed along the branch cut $[0,\infty]$. We note that
$$\frac{1}{(\ln{z}-\pi i)(1+z)^2}\sim_{-1}-\frac{1}{(z+1)^3}+\frac{1}{2(z+1)^2}+\frac{1}{12(z+1)}+\mathcal{O}(1)$$
and by the residue theorem, the contour integral is
\begin{align}
\frac{1}{2\pi i}\int_Cf(z,x)\ {\rm d}z
=&\ \operatorname*{Res}_{z=-1/x}f(z,x)+\operatorname*{Res}_{z=-1}f(z,x)\\
=&\ \frac{2\left(-\frac{1}{x}\right)}{x\left(\ln\left(-\frac{1}{x}\right)-\pi i\right)\left(1-\frac{1}{x}\right)^2}-\frac{1}{2}\left.\left(\frac{{\rm d}^2}{{\rm d}z^2}+\frac{{\rm d}}{{\rm d}z}\right)\frac{2z}{1+xz}\right|_{z=-1}-\frac{1}{6(1-x)}\\
=&\ \frac{2x}{(1-x)^3}+\frac{1}{(1-x)^2}-\frac{1}{6(1-x)}+\frac{2}{(1-x)^2\ln{x}}
\end{align}
Next, let
$$I(s)=\int^1_0\left(\frac{2x}{(1-x)^3}+\frac{1}{(1-x)^2}-\frac{1}{6(1-x)}+\frac{2}{(1-x)^2\ln{x}}\right)x^s\ {\rm d}x$$
such that the integral we seek is $\displaystyle\frac{\ln{2}}{6}-I(0)$. Differentiating twice and integrating by parts,
\begin{align}
I''(s)
&=1+\int^1_0\left(\frac{s^2x^{s-1}\ln^2{x}}{1-x}+\frac{2sx^{s-1}\ln{x}}{1-x}-\frac{x^s\ln^2{x}}{6(1-x)}\right)\ {\rm d}x\\
&=1-s^2\psi_2(s)-2s\psi_1(s)+\frac{1}{6}\psi_2(s+1)
\end{align}
and we may integrate back to obtain
\begin{align}
I'(s)
&=\frac{1}{2}+s-s^2\psi_1(s)+\frac{1}{6}\psi_1(s+1)\\
I(s)
&=L-s\ln(2\pi)-\frac{s}{2}+\frac{3s^2}{2}+\frac{1}{6}\psi_0(s+1)-s^2\psi_0(s)+2\ln{G(s+1)}
\end{align}
where
\begin{align}
L
&=\lim_{s\to\infty}\left(s\ln(2\pi)+\frac{s}{2}-\frac{3s^2}{2}-\frac{1}{6}\psi_0(s+1)+s^2\psi_0(s)-2\ln{G(s+1)}\right)\\
&=\lim_{s\to\infty}\left(s\ln(2\pi)+\frac{s}{2}-\frac{3s^2}{2}-\frac{1}{6}\ln{s}+s^2\ln{s}-\frac{s}{2}-\frac{1}{12}-s^2\ln{s}+\frac{3s^2}{2}-s\ln(2\pi)+\frac{1}{6}\ln{s}-2\zeta'(-1)+\mathcal{O}\left(\frac{1}{s}\right)\right)\\
&=-2\zeta'(-1)-\frac{1}{12}
\end{align}
Letting $s=0$ and noting that $\psi_0(1)=-\gamma$, we arrive at
$$\int^1_{-1}\frac{\ln(1-x)}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x=2\zeta'(-1)+\frac{\gamma}{6}+\frac{\ln{2}}{6}+\frac{1}{12}$$</p>
|
316,878 | <p>Why is the function not analytic in the complex plane? I believe it is analytic on real plane.</p>
<p>$e^{(-\frac{1}{z^2})}$ where $z\in\mathbb{C}$. </p>
<p>Well a complex function should be infinitely differentiable and should converge. this happens on real plane. But what happens in complex plane?</p>
| Mhenni Benghorbal | 35,472 | <p><strong>Hint:</strong> A function is analytic if and only if its Taylor series about $x_0$ converges to the function in some neighborhood for every $x_0$ in its domain.</p>
|
1,989,291 | <p>What is the closed form of the following:</p>
<p>$$\sum_{j=1}^n 3^{j+1}$$</p>
<p>I'm new to summations. Is it this?</p>
<p>$$\sum_{j=1}^n 3^{j} + \sum_{j=1}^n 3$$</p>
<p>Then using the closed form formula:</p>
<p>$$\frac{3^{n+1} - 1}{2} + 3n$$</p>
| MrYouMath | 262,304 | <p>Hint: This is the finite geometric series.</p>
<p>$$S_0=3\sum_{j=1}^{n}3^j=3S_n$$</p>
<p>We will only look at $S_n$ from now on. $S_n=3+3^2+3^3+\dots+3^n$ so multiply with 3 to get $3S_n=3^2+3^3+\dots+3^{n+1}$. Subtract both equations and notice the chancellations to get $S_n-3S_n=3-3^{n+1}$. Solve for $S_n$ and resubstitute into $S_0=3S_n$ to get your closed solutions.</p>
|
1,989,291 | <p>What is the closed form of the following:</p>
<p>$$\sum_{j=1}^n 3^{j+1}$$</p>
<p>I'm new to summations. Is it this?</p>
<p>$$\sum_{j=1}^n 3^{j} + \sum_{j=1}^n 3$$</p>
<p>Then using the closed form formula:</p>
<p>$$\frac{3^{n+1} - 1}{2} + 3n$$</p>
| Bernard | 202,857 | <p>No. Just factor out $3^2$ so as to obtain the standard geometric series:
$$ \sum_{j=1}^n 3^{j+1}=9 \sum_{j=0}^{n-1}3^j=9\frac{3^n-1}{3-1}=\frac{3^{n+2}-3^2}{2}. $$</p>
|
1,436,215 | <p>I'm using the following algorithm (in C) to find if a point lays within a given polygon</p>
<pre><code>typedef struct {
int h,v;
} Point;
int InsidePolygon(Point *polygon,int n,Point p)
{
int i;
double angle=0;
Point p1,p2;
for (i=0;i<n;i++) {
p1.h = polygon[i].h - p.h;
p1.v = polygon[i].v - p.v;
p2.h = polygon[(i+1)%n].h - p.h;
p2.v = polygon[(i+1)%n].v - p.v;
angle += Angle2D(p1.h,p1.v,p2.h,p2.v);
}
if (ABS(angle) < PI)
return(FALSE);
else
return(TRUE);
}
/*
Return the angle between two vectors on a plane
The angle is from vector 1 to vector 2, positive anticlockwise
The result is between -pi -> pi
*/
double Angle2D(double x1, double y1, double x2, double y2)
{
double dtheta,theta1,theta2;
theta1 = atan2(y1,x1);
theta2 = atan2(y2,x2);
dtheta = theta2 - theta1;
while (dtheta > PI)
dtheta -= TWOPI;
while (dtheta < -PI)
dtheta += TWOPI;
return(dtheta);
}
</code></pre>
<p>I found the algorithm while searching online for such an algorithm, <a href="http://bbs.dartmouth.edu/~fangq/MATH/download/source/Determining%20if%20a%20point%20lies%20on%20the%20interior%20of%20a%20polygon.htm" rel="nofollow">Point in Polygon algorithm</a>, Solution 2 (2D).</p>
<ol>
<li>The explanation in the link says it checks if the sum of angles is $2\pi$ but the algorithm (which works) checks it with $\pi$.</li>
<li>How are the angles even calculated?</li>
</ol>
| gammatester | 61,216 | <p>The angle is computed with the <code>atan2</code> function (see
<a href="https://en.wikipedia.org/wiki/Arctan2" rel="nofollow">https://en.wikipedia.org/wiki/Arctan2</a>). But the code is not computationally
robust. The mathematical idea may be OK, but since $\pi$ and $2\pi$ are not exactly
representable as floating point numbers, there will be uncertainties produced by using floating point arithmetic near the interesting sum $2\pi$. Testing equality of two floating point variables is a questionable concept in many situations (since the angle is in the interval $[-\pi, +\pi],\,$ the test $|\, . |<\pi\,$ is actually an equality test in disguise).</p>
<p>And there is a bug, if the sum happens to be $\pm\pi,$ the code should handle this separately.</p>
|
319,058 | <p>Denote <span class="math-container">$\square_m=\{\pmb{x}=(x_1,\dots,x_m)\in\mathbb{R}^m: 0\leq x_i\leq1,\,\,\forall i\}$</span> be an <span class="math-container">$m$</span>-dimensional cube.</p>
<p>It is all too familiar that <span class="math-container">$\int_{\square_1}\frac{dx}{1+x^2}=\frac{\pi}4$</span>.</p>
<blockquote>
<p><strong>QUESTION.</strong> If <span class="math-container">$\Vert\cdot\Vert$</span>stands for the Euclidean norm, then is this true?
<span class="math-container">$$\int_{\square_{2n-1}}\frac{d\pmb{x}}{(1+\Vert\pmb{x}\Vert^2)^n}
=\frac{\pi^n}{4^nn!}.$$</span></p>
</blockquote>
| Carlo Beenakker | 11,260 | <p>Yes, this is true, it follows from formulas in <A HREF="https://projecteuclid.org/download/pdf_1/euclid.em/1317758103" rel="noreferrer">Higher-Dimensional Box Integrals</A>, by Jonathan M. Borwein, O-Yeat Chan, and R. E. Crandall. </p>
<p><span class="math-container">\begin{align}
&\text{define}\;\;C_{m}(s)=\int_{[0,1]^m}(1+|\vec{r}|^2)^{s/2}\,d\vec{r},\;\;\text{we need}\;\;C_{2n-1}(-2n).\\
&\text{the box integral is}\;\;B_m(s)=\int_{[0,1]^m}|\vec{r}|^s\,d\vec{r},\\
&\text{related to our integral by}\;\;2n C_{2n-1}(-2n)=\lim_{\epsilon\rightarrow 0}\epsilon B_{2n}(-2n+\epsilon)\equiv {\rm Res}_{2n}.
\end{align}</span>
The box integral <span class="math-container">$B_n(s)$</span> has a pole at <span class="math-container">$n=-s$</span>, with "residue" <span class="math-container">${\rm Res}_n$</span>. Equation 3.1 in the cited paper shows that the residue is a piece (an <span class="math-container">$n$</span>-orthant) of the surface area of the unit <span class="math-container">$n$</span>- sphere, so it is readily evaluated,
<span class="math-container">\begin{align}
&{\rm Res}_{n}=\frac{1}{2^{n-1}}\frac{\pi^{n/2}}{\Gamma(n/2)},\\
&\text{hence}\;\;C_{2n-1}(-2n)=\frac{1}{2n}\frac{1}{2^{2n-1}}\frac{\pi^{n}}{\Gamma(n)}=\frac{1}{4^n}\frac{\pi^n}{n!}\;\;\text{as in the OP}.
\end{align}</span></p>
<p>More generally, we can consider</p>
<p><span class="math-container">$$C_{p-1}(-p)=\int_{[0,1]^{p-1}}(1+|\vec{r}|^2)^{-p/2}\,d\vec{r}=\frac{1}{p}\,{\rm Res}_p=\frac{1}{p} \frac{1}{2^{p-1}}\frac{\pi^{p/2}}{\Gamma(p/2)}. $$</span>
This formula holds for even <em>and</em> odd <span class="math-container">$p\geq 2$</span>, as surmised by <em>Gro-Tsen</em> in a comment.</p>
<hr>
<p>There are similarly remarkable hypercube integrals where this came from, for example</p>
<p><span class="math-container">$$\int_{[0,\pi/4]^k}\frac{d\theta_1 d\theta_2\cdots d\theta_k}{(1+1/\cos^2\theta_1+1/\cos^2\theta_2\cdots+1/\cos^2\theta_k)^{1/2}}=\frac{k!^2\pi^k}{(2k+1)!}.$$</span></p>
|
2,551,683 | <blockquote>
<p>A jet has a $5\%$ chance of crashing on any given test flight. Once it crashes the program will be halted. Find the probability that the program lasts less than three flights.</p>
</blockquote>
<p>The correct answer to this question is $0.1426$, but I can't figure out how to get it.</p>
<p>Here's my attempt at the question:
$p=0.05$</p>
<p>$q=0.95$</p>
<p>$x=0,1,2$</p>
<p>$$(0.05)(0.95)^{-1} + (0.05)(0.95)^0 + (0.05)(0.95)^1 = 0.1501$$ </p>
<p>I used the geometric probability method by using the formula $pq^{x-1}$ where $p$ is the probability of successes, $q$ is the probability of failures, and $x$ is the waiting time/how many events occurred before the success occurred.</p>
| Graham Kemp | 135,106 | <p>$pq^{x-1}$ is the probability for $x$ <em>trials until</em> the first success, where $x\in\{1,2,\ldots\}$.</p>
<p>$pq^x$ is the probability for $x$ <em>failures before</em> the first success, where $x\in\{0,1,2,\ldots\}$.</p>
<p>Unfortunately both distributions are called Geometric but it important to not confuse them.</p>
<p>If you use $X$ as the count of failures before the first successful crash, you need the second distribution. So the probability that there are two or fewer failed crashes is:</p>
<p>$$X\sim\mathcal{Geo}_0(0.05)\implies \mathsf P(X\leq 2) ~{= p(q^0+q^1+q^2)\\= 0.05(1+0.95+0.95^2) \\= 0.142625}$$</p>
|
1,791,673 | <p>I was wondering about this, just now, because I was trying to write something like:<br>
$880$ is not greater than $950$. <br>
I am wondering this because there is a 'not equal to': $\not=$ <br>
Not equal to is an accepted mathematical symbol - so would this be acceptable: $\not>$? <br>
I was searching around but I couldn't find any qualified sites that would point me in that direction.</p>
<p><br>
So, I would like to know if there are symbols for, not greater, less than, less than or equal to, greater than or equal to x. </p>
<p>Thanks for your help and time! </p>
| DomS | 518,038 | <p>I would just like to make it clear that ≮ is NOT the same as ≥</p>
<p>Here is an example: </p>
<p>1+²≮(1+)² is clearly not the same as 1+²≥(1+)²</p>
<p>Think of =-0.5 and =2 as examples to highlight this, because although when =-0.5, 1+²≥(1+)² but when =2, 1+²≱(1+)²</p>
<p>Therefore, think of ≮ as meaning "not greater than" and ≥ meaning "more than or equal to" but remember that they are not the same!</p>
|
177,515 | <p>From <a href="http://mitpress.mit.edu/algorithms/" rel="nofollow">Cormen et all</a>:</p>
<blockquote>
<p>The elements of a matrix or vectors are numbers from a number system, such as the real numbers , the complex numbers , or integers modulo a prime .</p>
</blockquote>
<p>What do they mean by <strong>integers modulo a prime</strong> ? I thought real numbers and complex numbers together make up all the elements of a matrix . Why did they put this additional one ?</p>
| Andrea Mori | 688 | <p>You can actually form matrices with entries in any <a href="http://en.wikipedia.org/wiki/Ring_%28mathematics%29" rel="nofollow">ring</a>, although sometimes you won't have the same nice properties.</p>
<p>The ring of integers modulo a prime, sometimes denoted $\Bbb F_p$, is the ring where you perform <a href="http://en.wikipedia.org/wiki/Modular_arithmetic" rel="nofollow">modular arithmetic</a> modulo $p$.</p>
<p>The reason to stick to a prime number (modular arithmetic can in fact be done modulo any natural number $N$) is that some nice properties are mantained, such as the possibility to find an inverse matrix whenever the determinant is not zero.</p>
|
951,332 | <p>the integer 220, 251 304 represent three consecutive perfect squares in base b. Determine the value of b.</p>
| Mark Bennet | 2,906 | <p>Hint: the differences between three consecutive squares are two consecutive odd integers.</p>
<blockquote class="spoiler">
<p> The first difference is $31$ so the second will be $33$ which means that $25+3=30$ and $b=8$</p>
</blockquote>
|
951,332 | <p>the integer 220, 251 304 represent three consecutive perfect squares in base b. Determine the value of b.</p>
| Jack D'Aurizio | 44,121 | <p>We have that $3b+1$ and $b^2-5b+3$ are two consecutive odd integers, hence $b^2-8b+2=2$, so the only possibility is $b=8$, and it fits.</p>
|
3,290,839 | <p>I want to prove that that given <span class="math-container">$f:R^2 \rightarrow R$</span> which is continuous with compact support s.t the integral of <span class="math-container">$f$</span> for every straight line <span class="math-container">$l$</span> is zero (<span class="math-container">$\int f(l(t))\mathrm{d}t=0$</span>) then <span class="math-container">$f$</span> is almost everywhere <span class="math-container">$0.$</span></p>
<p>Well I know how to proof it in case that <span class="math-container">$l = 1_{B(x,r)}$</span> is measurable and bounded with compact support (from other thread) , but that's not the case here.
Any idea? thanks!</p>
| Redundant Aunt | 109,899 | <p>Suppose there exists <span class="math-container">$p\in\mathbb{R}^2$</span> with <span class="math-container">$f(p)\neq 0$</span>, then WLOG (by replacing <span class="math-container">$f$</span> by <span class="math-container">$-f$</span> if necessary) we might assume <span class="math-container">$f(p)>0$</span>. But then by continuity of <span class="math-container">$f$</span> there exist <span class="math-container">$\varepsilon>0$</span> such that <span class="math-container">$f(x)>f(p)/2$</span> for all <span class="math-container">$x$</span> with <span class="math-container">$\|x-p\|\leq\varepsilon$</span>, which then gives that the integral of <span class="math-container">$f$</span> over every line contained in <span class="math-container">$B(p,\varepsilon)$</span> is strictly positive, contradiction.</p>
|
2,561,125 | <p>Hey having trouble finishing this question.</p>
<p>Prove by induction that $n^3 \le 2^n$ for all natural numbers $n\ge 10$.</p>
<p>This is what I have so far:</p>
<p>Base step: For $n = 10$ </p>
<p>$1000 \le 1024$</p>
<p>Assumption Step: For $n = k$</p>
<p>Assume $k^3 \le 2^k$</p>
<p>Induction step: For $n = (k+1)$</p>
<p>$(k+1)^3 \le 10^{k+1}$</p>
<p>$k^3 +3k^2 + 3k +1 \le 10^k*10$</p>
<p>Not really sure where to go from here</p>
| Robin Grajeda | 512,209 | <p>Define $$S=\left\{ n\in
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
,\text{ }n\geq 10\text{ }|\text{ }n^{3}\leq 2^{n}\right\} $$</p>
<p>For $n=10$</p>
<p>$$
10^{3}\leq 2^{10}
$$</p>
<p>Therefore $S\neq \varnothing .$</p>
<p>Suppose that $k^{3}\leq 2^{k}$. We want to show that for any positive intger $k$, $\left( k+1\right) \in
S.$ In other words $$\left( k+1\right) ^{3}\leq 2^{k+1}$$ </p>
<p>Using the inductive
hypothesis, we may write</p>
<p>$$
2k^{3}\leq 2\left( 2^{k}\right) =2^{k+1}
$$</p>
<p>All that remains to be shown is that $\left( k+1\right) ^{3}\leq 2k^{3}$ for
$k\geq 10.$ This is certainly true for $k>3,$ so the requirement is
satisfied. </p>
<p>Therefore, $S\subset
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
.$</p>
|
4,263,784 | <p>I'm trying to find the multiplicity of <span class="math-container">$z=0$</span> on <span class="math-container">$f(z)=z\cos(z)-\sin(z)$</span> using complex analysis.</p>
<p>I'm new to complex analysis and the argument principle/Rouché's theorem so I'm not quite sure where to start. I can prove how many zero's this function has but I'm not quite sure what theorem would help me determine the multiplicity of this root, can someone point me in the right direction or provide a proof?</p>
<p>Thank you</p>
| Henno Brandsma | 4,280 | <p>Given a cipher text of length <span class="math-container">$L$</span>, and plain texts of the same length <span class="math-container">$L$</span> are equally likely (as all keys are of the same length and random). Any attacker has no way of telling which is right.</p>
|
194,421 | <p>This is homework. The problem was also stated this way: </p>
<p>Let A be a dense subset of $\mathbb{R}$ and let x$\in\mathbb{R}$. Prove that there exists a decreasing sequence $(a_k)$ in A that converges to x.</p>
<p>I know:</p>
<p>A dense in $\mathbb{R}$ $\Rightarrow$ every point in $\mathbb{R}$ is either in A or a limit point of A.</p>
<p>If x is a limit point of A, then there is a sequence in A that converges to x. </p>
<p>What if $x\in A$?</p>
<p>Also, how can I know if the sequence is increasing or decreasing?</p>
| BaronVT | 39,526 | <p>There is a sequence converging to $x$, but you won't know it's decreasing - you'll have to construct it so this happens.</p>
<p>Try thinking about what would happen if this weren't true - then there would be an $a_0 \in A$ so that $(x,a_0) \cap A = \emptyset$... in other words this interval is full of points that are in $\mathbb{R}$, but not in $A$. Would such an $A$ still be dense?</p>
|
459,579 | <blockquote>
<p>Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$</p>
</blockquote>
<p>Doesn't this thing approaches 0 at the end? why does it approaches 1?</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Using <a href="http://www.proofwiki.org/wiki/Exponent_Combination_Laws" rel="nofollow">Exponent Combination Laws</a>,
$$a^m\cdot a^n\cdot a^p\cdots=a^{m+n+p+\cdot},$$</p>
<p>$$\displaystyle 3^9\cdot 3^3\cdot3\cdot 3^\frac13\cdots=3^{\left(3^2+3+1+\frac13+\cdots\right)}$$</p>
<p>Observe that the power of $3$ is an <a href="https://www.artofproblemsolving.com/Wiki/index.php/Geometric_sequence#Infinite_Geometric_Sequences" rel="nofollow">infinite</a> Geometric Series with the first Term $=9$ and common ratio $=\frac13<1$</p>
|
2,961,023 | <p>Is it allowed to solve this inequality <span class="math-container">$x|x-1|>-3$</span> by dividing each member with <span class="math-container">$x$</span>? What if <span class="math-container">$x$</span> is negative?</p>
<p>My textbook provides the following solution:</p>
<blockquote>
<p>Divide both sides by <span class="math-container">$x: $</span> <span class="math-container">$\frac { x | x - 1 | } { x } > \frac { - 3
} { x } ; \quad x \neq 0$</span></p>
<p>Simplify: <span class="math-container">$| x - 1 | > - \frac { 3 } { x } ; \quad x \neq 0$</span></p>
</blockquote>
<p>Edit: provided textbook's solution</p>
| nonuser | 463,553 | <p>For <span class="math-container">$x\geq0$</span> this inequality is always true. </p>
<p>Assume that <span class="math-container">$x<0$</span>, so <span class="math-container">$x=-y$</span> for some positive <span class="math-container">$y$</span> and we get <span class="math-container">$$y|\;\underbrace{y+1}_{>0}\;|<3\implies y(y+1)<3 ...$$</span></p>
|
2,475,507 | <blockquote>
<p>Find $f$ and $g$ such that domain $(f\circ g)=\mathbb{R}$ and domain $(g\circ f)=\emptyset$</p>
</blockquote>
<p>That's it, I can't think of any. </p>
<p>I've thought of $f(x)=-1$ and $g(x)=\sqrt{x}$, and then: $$f\big(g(x)\big)=-1$$ $$g\big(f(x)\big)=\sqrt{-1}$$ </p>
<p>Which would in principle satisfy it, but the thing is, in $f\circ g$, can I say that the domain is $\mathbb{R}$? Or is it the same as the domain of $g$?</p>
| Andreas Blass | 48,510 | <p>You already have a correct answer, from user334639, under the assumption that all the inputs and outputs of your functions are real numbers. On the other hand, if you're allowed to use some entity $Q$ that isn't a real number, then you can obtain an example. Let $g$ be the identity function on $\mathbb R$ and let $f$ be the constant function with domain $\mathbb R$ that maps all real numbers to $Q$. Then, $f(g(x))$ is defined (and equal to $Q$) whenever $x\in\mathbb R$, but $g(f(x))$ is never defined, because the only output of $f$ is $Q$, which is not in the domain of $g$.</p>
|
365,287 | <p>Let $([0,1],\mathcal{B},m)$ be the Borel sigma algebra with lebesgue measure and $([0,1],\mathcal{P},\mu)$ be the power set with counting measure. Consider the product $\sigma$-algebra on $[0,1]^2$ and product measure $m \times \mu$.</p>
<p>(1) Is $D=\{(x,x)\in[0,1]^2\}$ measurable?</p>
<p>(2) If so, what is $m \times \mu(D)$?</p>
<p>Edit: The product measure is defined on a rectangle by $m \times \mu(A\times B)=m(A)\mu(B)$ and on general set taking infimum over union of rectangles containing $D$. (Maybe we should use $0 \cdot \infty=0$)</p>
| Brian M. Scott | 12,042 | <p>Charles Wells, <em>The Handbook of Mathematical Discourse</em>; it’s available as a PDF <a href="https://www.abstractmath.org/Handbook/handbook.pdf" rel="nofollow noreferrer">here</a>. His site <a href="http://www.abstractmath.org/MM/MMIntro.htm" rel="nofollow noreferrer">abstractmath.org</a> may also be useful.</p>
|
2,541,991 | <p>I need to find a pair of dependent random variables $(X, Y)$ with covariance equal to $0.$ From this I gather:</p>
<p>$$0 = E((X-EX)(Y-EY)) = E \left(\left(X - \int_{-\infty}^\infty xf_X(x)\,dx\right) \left(Y - \int_{-\infty}^\infty xf_Y(x)\,dx \right)\right)$$</p>
<p>but what can I do now? How can I use the fact that they are dependent in the equation? Do you know of two such variables?</p>
| Remy | 325,426 | <p>Let $X$ be a random variable that takes on the values $1$ or $-1$ with equal probability. Let $Y$ be a random variable where $Y=0$ if $X=-1$, and $Y$ is $-1$ or $1$ with equal probability if $X=1$.</p>
<p>Then $X$ and $Y$ depend on each other, since if you know what $Y$ is then you know what $X$ is. However, their covariance is zero:</p>
<p>$$\begin{align*}
\operatorname{Cov}(X,Y)
&= E(XY)-E(X)E(Y)\\\\
&= 0
\end{align*}$$</p>
<p>since</p>
<p>$$\begin{align*}
E(XY)
&= (-1\cdot 0 \cdot P(X=-1, Y=0)) +(1\cdot -1 \cdot P(X=1, Y=-1))+(1\cdot 1 \cdot P(X=1, Y=1)) \\\\
&= 0-P(X=1, Y=-1)+P(X=1,Y=1)\\\\
&= 0
\end{align*}$$</p>
<p>and </p>
<p>$E(X)$ and $E(Y)$ are both zero.</p>
<p>This illustrates how random variables can be dependent but have no correlation, and thus no covariance.</p>
<p><a href="https://i.stack.imgur.com/UwAXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UwAXt.png" alt="enter image description here"></a></p>
<p>Correlation is a measure of linear dependence. It is possible for two random variables to be uncorrelated but nonlinearly dependent.</p>
|
2,331,961 | <p>Please suggest: </p>
<h1>Question 1)</h1>
<p>What are some reasonable assumptions regarding the limit of the Cumulative Distribution as the Variance grows to infinity.</p>
<p>$$
\lim_{\sigma\rightarrow\infty} F\left(t,\sigma\right) = \text{??}
$$</p>
<h1>Question 2)</h1>
<p>Also, is it a reasonable assumption to expect that the cumulative distribution will decrease in value with growing variance, as shown below?</p>
<p>$$\frac{\partial F(t,\sigma)}{\partial\sigma}\leq0$$</p>
<p>Here, $F(t,\sigma)$ are a family of distributions with parameter governed by $\sigma$, the variance. If helpful, we can make another simplifying assumption that $t\geq0$.</p>
<p>Please note, the questions are not specific to any particular distribution, but any general distribution and what are some valid forms of this limit and the behaviour of the distribution as variance increases.</p>
<p>Please point out any specific references on this topic and also list any additional assumptions made to arrive at any results.</p>
<p>Please let me know if the question is not clear or if you need any further information.</p>
| Michael Hardy | 11,667 | <p>One example is the normal distribution with expected value $0.$</p>
<p>Let $Z\sim N(0,1).$ Then $\sigma Z\sim N(0,\sigma^2).$ Let $F$ be the c.d.f. of $\sigma Z.$ Then
$$
F(t) = \Pr(\sigma Z\le t) = \Pr\left( Z \le \frac t \sigma \right).
$$
As $\sigma$ increases, then $t/\sigma\quad \begin{cases} \text{increases to } 0 & \text{if } t<0, \\ \text{decreases to } 0 & \text{if } t>0. \end{cases}$</p>
<p>Therefore $F(t)\quad \begin{cases} \text{increases to } 1/2 & \text{if } t<0, \\ \text{decreases to } 1/2 & \text{if }t>0. \end{cases}$</p>
|
2,331,961 | <p>Please suggest: </p>
<h1>Question 1)</h1>
<p>What are some reasonable assumptions regarding the limit of the Cumulative Distribution as the Variance grows to infinity.</p>
<p>$$
\lim_{\sigma\rightarrow\infty} F\left(t,\sigma\right) = \text{??}
$$</p>
<h1>Question 2)</h1>
<p>Also, is it a reasonable assumption to expect that the cumulative distribution will decrease in value with growing variance, as shown below?</p>
<p>$$\frac{\partial F(t,\sigma)}{\partial\sigma}\leq0$$</p>
<p>Here, $F(t,\sigma)$ are a family of distributions with parameter governed by $\sigma$, the variance. If helpful, we can make another simplifying assumption that $t\geq0$.</p>
<p>Please note, the questions are not specific to any particular distribution, but any general distribution and what are some valid forms of this limit and the behaviour of the distribution as variance increases.</p>
<p>Please point out any specific references on this topic and also list any additional assumptions made to arrive at any results.</p>
<p>Please let me know if the question is not clear or if you need any further information.</p>
| texmex | 238,328 | <p>Adding a few more examples for Question 2, in addition to the clarifications by Robert Israel and Michael Hardy.</p>
<h1>Example 1:</h1>
<p>The derivative of the cumulative distribution of a uniformly distributed
variable, having support $t\in\left[a,b\right]\;;0\leq a,b<\infty$,
with respect to the standard deviation (variance) is less than zero.
$$
F\left(t\right)=\left(\frac{t-a}{b-a}\right)\equiv F\left(t,\sigma_{D}\right)\propto\frac{\left(t-a\right)}{\sigma_{D}}\;\because\sigma_{D}^{2}=\frac{\left(b-a\right)^{2}}{12}
$$
$$
\frac{\partial F\left(t,\sigma_{D}\right)}{\partial\sigma_{D}}\propto-\frac{\left(t-a\right)}{\sigma_{D}^{2}}\Longrightarrow\frac{\partial F\left(t,\sigma_{D}\right)}{\partial\sigma_{D}}\leq0
$$</p>
<h1>Example 2:</h1>
<p>The derivative of the cumulative distribution of a log-normally distributed
variable, $\ln Y\sim N(\mu,\sigma^{2})$, with
respect to the variance is less than zero in the region where, $y>e^{\mu}$.
$$
F\left(t,\sigma\right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\left(\frac{\ln t-\mu}{\sigma}\right)}e^{-\frac{y^{2}}{2}}dy=\Phi\left(\frac{\ln t-\mu}{\sigma}\right)
$$</p>
<p>$$
\text{Var}\left(Y\right)=e^{2\mu+2\sigma^{2}}-e^{2\mu+\sigma^{2}}\;;\frac{\partial\text{Var}\left(Y\right)}{\partial\sigma}=2\sigma\left\{ 2e^{2\mu+2\sigma^{2}}-e^{2\mu+\sigma^{2}}\right\} >0
$$
$$
\frac{\partial F\left(t,\sigma\right)}{\partial\text{Var}\left(Y\right)}=\frac{\partial F\left(t,\sigma\right)}{\partial\sigma}\frac{\partial\sigma}{\partial\text{Var}\left(Y\right)}\propto e^{-\frac{\left(\ln t-\mu\right)^{2}}{2\sigma^{2}}}\left\{ -\left(\frac{\ln t-\mu}{\sigma^{2}}\right)\right\}
$$
$$\Longrightarrow\frac{\partial F\left(t,\sigma\right)}{\partial\text{Var}\left(Y\right)}\leq0\;;\text{ When }\left(\ln t-\mu\right)>0
$$</p>
<p>Perhaps simply written as,
$$
\frac{\partial F\left(t,\sigma\right)}{\partial\sigma}\propto e^{-\frac{\left(\ln t-\mu\right)^{2}}{2\sigma^{2}}}\left\{ -\left(\frac{\ln t-\mu}{\sigma^{2}}\right)\right\} \Longrightarrow\frac{\partial F\left(t,\sigma\right)}{\partial\sigma}\leq0\;;\text{ When }\left(\ln t-\mu\right)>0
$$</p>
|
1,151,726 | <p>The following question is from Fred H. Croom's book "Principles of Topology"</p>
<blockquote>
<blockquote>
<p>In <span class="math-container">$\mathbb{R}^n$</span>, let <span class="math-container">$R$</span> denote the set of points having only rational coordinates and <span class="math-container">$I$</span> its complement, the set of points having at least one irrational coordinate. Prove that</p>
</blockquote>
<ol>
<li>int<span class="math-container">$R$</span> = int<span class="math-container">$I$</span> = <span class="math-container">$\emptyset$</span>.</li>
<li><span class="math-container">$R^{'}$</span> = <span class="math-container">$I^{'}$</span> = <span class="math-container">$\mathbb{R}^n$</span>.</li>
<li>bdy<span class="math-container">$R$</span> = bdy<span class="math-container">$I$</span> = <span class="math-container">$\mathbb{R}^n$</span>.</li>
</ol>
</blockquote>
<p>I have an idea for the first part.</p>
<blockquote>
<blockquote>
<p>Part1: Since every open ball contains both a rational and irrational, it follows that <span class="math-container">$R$</span> as well as <span class="math-container">$I$</span> have no interior points. If <span class="math-container">$x$</span> were an interior point of <span class="math-container">$R$</span>, then there would exist <span class="math-container">$\delta >0$</span> such that <span class="math-container">$B(x,\delta) \subset R$</span>. However, we know this cannot happen since there is at least one irrational number in the ball. The same argument works for <span class="math-container">$I$</span>. Therefore, int<span class="math-container">$R$</span> = int<span class="math-container">$I$</span> = <span class="math-container">$\emptyset$</span>.</p>
</blockquote>
</blockquote>
<p>Does this make sense for the first part? I am rather confused on how to approach the second and third part. Any suggestions?</p>
<hr />
<p>I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.</p>
| user4894 | 118,194 | <p>Yes, your part 1 proof is good. </p>
<p>But now you're already 90% of the way to #2. If every ball contains points of both $R$ and $I$, what can you say about the set of limits points of $R$ and $I$, respectively?</p>
<p>And 3 is just another minor variation on this same theme. What's the definition of the boundary of a set? Isn't it exactly what you said about <em>every</em> ball in part 1?</p>
|
334,075 | <p>How do you compute $$\int_{0}^1 \frac{\arctan x }{1+x} dx$$</p>
| Santosh Linkha | 2,199 | <p>Using integration by parts
$$\int_0^1 \frac{\arctan x}{1+x} dx = \arctan(x) \ln(1+x)|_0^1 - \int_0^1 \frac{\ln (1+x)}{1+x^2}dx$$</p>
<p>The former part is $\displaystyle \frac{\pi}{
4} \ln 2 $ and the latter part is $\displaystyle \frac{\pi}{8} \ln 2$ which is answered <a href="https://math.stackexchange.com/questions/155941/evaluate-the-integral-int-01-frac-lnx1x21-dx">here</a>.</p>
|
334,075 | <p>How do you compute $$\int_{0}^1 \frac{\arctan x }{1+x} dx$$</p>
| robjohn | 13,854 | <p>$$
\begin{align}
\int_0^1\frac{\arctan(x)}{1+x}\,\mathrm{d}x
&=\int_0^{\pi/4}\frac{\theta}{1+\tan(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{1}\\[6pt]
&=\int_0^{\pi/4}\frac{\theta\,\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{2}\\[6pt]
&=\int_0^{\pi/4}\frac{(\frac\pi4-\theta)\,\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{3}\\[6pt]
&=\frac\pi8\int_0^{\pi/4}\frac{\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{4}\\[6pt]
&=\frac\pi8\int_0^{\pi/4}\frac{\sec^2(\theta)}{1+\tan(\theta)}\,\mathrm{d}\theta\tag{5}\\[6pt]
&=\frac\pi8\int_0^1\frac1{1+x}\mathrm{d}x\tag{6}\\[6pt]
&=\frac\pi8\Big[\log(1+x)\Big]_0^1\tag{7}\\[12pt]
&=\frac\pi8\log(2)\tag{8}
\end{align}
$$
$(1):$ $x=\tan(\theta)$</p>
<p>$(3):$ $\theta\mapsto\frac\pi4-\theta$</p>
<p>$(4):$ since $(2)=(3)$ we have $(3)=\frac{(2)+(3)}{2}$</p>
<p>$(6):$ $x=\tan(\theta)$</p>
|
164,002 | <p>When I am reading a mathematical textbook, I tend to skip most of the exercises.
Generally I don't like exercises, particularly artificial ones.
Instead, I concentrate on understanding proofs of theorems, propositions, lemmas, etc..</p>
<p>Sometimes I try to prove a theorem before reading the proof.
Sometimes I try to find a different proof.
Sometimes I try to find an example or a counter-example.
Sometimes I try to generalize a theorem.
Sometimes I come up with a question and I try to answer it. </p>
<p>I think those are good "exercises" for me.</p>
<p><strong>EDIT</strong>
What I think is a very good "excercise" is as follows:</p>
<p>(1) Try to prove a theorem before reading the proof.</p>
<p>(2) If you have no idea to prove it, take a look <strong>a bit</strong> at the proof.</p>
<p>(3) Continue to try to prove it.</p>
<p>(4) When you are stuck, take a look <strong>a bit</strong> at the proof.</p>
<p>(5) Repeat (3) and (4) until you come up with a proof.</p>
<p><strong>EDIT</strong>
Another method I recommend rather than doing "homework type" exercises:
Try to write a "textbook" on the subject.
You don't have to write a real one.
I tried to do this on Galois theory.
Actually I posted "lecture notes" on Galois theory on an internet mathematics forum.
I believe my knowledge and skill on the subject greatly increased.</p>
<p>For example, I found <a href="https://math.stackexchange.com/questions/131757/a-proof-of-the-normal-basis-theorem-of-a-cyclic-extension-field">this</a> while I was writing "lecture notes" on Galois theory.
I could also prove that any profinite group is a Galois group.
This fact was mentioned in Neukirch's algebraic number theory.
I found later that Bourbaki had this problem as an exercise.
I don't understand its hint, though.
Later I found someone wrote a paper on this problem.
I made other small "discoveries" during the course.
I was planning to write a "lecture note" on Grothendieck's Galois theory.
This is an attractive plan, but has not yet been started.</p>
<p><strong>EDIT</strong>
If you want to have exercises, why not produce them yourself?
When you are learning a subject, you naturally come up with questions.
Some of these can be good exercises. At least you have the motivation not given by others. It is not homework.
For example, I came up with the following question when I was learning algebraic geometry.
I found that this was a good problem.</p>
<p>Let $k$ be a field.
Let $A$ be a finitely generated commutative algebra over $k$.
Let $\mathbb{P}^n = Proj(k[X_0, ... X_n])$.
Determine $Hom_k(Spec(A), \mathbb{P}^n)$.</p>
<p>As I wrote, trying to find examples or counter-examples can be good exercises, too.
For example, <a href="https://math.stackexchange.com/questions/133790/an-example-of-noncommutative-division-algebra-over-q-other-than-quaternion-alg">this</a> is a good exercise in the theory of division algebras.</p>
<p><strong>EDIT</strong>
Let me show you another example of self-exercises.
I encountered the following problem when I was writing a "lecture note" on Galois theory.</p>
<p>Let $K$ be a field.
Let $K_{sep}$ be a separable algebraic closure of $K$.
Let $G$ be the Galois group of $K_{sep}/K$.</p>
<p>Let $A$ be a finite dimensional algebra over $K$.
If $A$ is isomorphic to a product of fields each of which is separable over $K$, $A$ is called a finite etale algebra.
Let $FinEt(K)$ be the category of finite etale algebra over $K$.</p>
<p>Let $X$ be a finite set.
Suppose $G$ acts on $X$ continuously.
$X$ is called a finite $G$-set.
Let $FinSets(G)$ be the category of finite $G$-sets.</p>
<p><em>Then $FinEt(K)$ is anti-equivalent to $FinSets(G)$.</em></p>
<p>This is a zero-dimensional version of the main theorem of Grothendieck's Galois theory.
You can find the proof elsewhere, but I recommend you to prove it yourself.
It's not difficult and it's a good exercise of Galois theory.
<em>Hint</em>: Reduce it to the the case that $A$ is a finite separable extension of $K$ and X is a finite transitive $G$-set.</p>
<p><strong>EDIT</strong>
If you think this is too broad a question, you are free to add suitable conditions.
This is a soft question.</p>
| benny rimmer | 35,048 | <p>I'm with the OP on this, I skip exercises too. </p>
<p>Here's the logic: A true understanding of maths is about being creative in its applications and not just the material initself. </p>
<p>Exercises, by definition, stifle creativity by presenting a sandbox in which to think. </p>
<p>It's a bit like Rocky 3, do you learn your craft by working out in a gym or by lifting logs down in the forest? </p>
<p>...and another thing, I reckon following examples in books leads to degenerate mathematics. They perpetuate particular styles of thinking about problems. </p>
|
2,313,060 | <p>$f(\bigcap_{\alpha \in A} U_{\alpha}) \subseteq \bigcap_{\alpha \in A}f(U_{\alpha})$</p>
<p>Suppose $y \in f(\bigcap_{\alpha \in A} U_{\alpha})$
$\implies f^{-1}(y) \in \bigcap_{\alpha \in A} U_{\alpha} \implies f^{-1}(y) \in U_{\alpha}$ for all $\alpha \in A$</p>
<p>$\implies y \in f (U_{\alpha})$ for all $\alpha \in A \implies y \in \bigcap_{\alpha \in A}f (U_{\alpha})$</p>
<p>$\bigcap_{\alpha \in A}f(U_{\alpha}) \subseteq f(\bigcap_{\alpha \in A} U_{\alpha})$</p>
<p>Suppose $y \in \bigcap_{\alpha \in A}f(U_{\alpha})\implies y \in f(U_{a}) $ for all $\alpha \in A$
$\implies f^{-1}(y)\in U_{a}$ for all $a\in A$</p>
<p>$ \implies f^{-1}(y)\in \bigcap_{a \in A}U_{a} \implies y \in f(\bigcap_{a \in A}(U_{a})$</p>
<p>Therefore</p>
<p>$f(\bigcap_{\alpha \in A} U_{\alpha}) \subseteq \bigcap_{\alpha \in A}f(U_{\alpha})$</p>
<p>Please let me know if my proof works, also I don't fully know how to do the following. Please give me some help. </p>
<p>Give an example of proper containment. Find a condition on f that would ensure equality.</p>
| Angina Seng | 436,618 | <p>There is no "$f^{-1}(y)$" in general. Although by definition $y$
is equal to $f(x)$ for some $x\in\bigcap U_\alpha$, that $x$ might
not be unique. So take an $x\in\bigcap U_\alpha$ and repeat your
argument with $x$ replacing "$f^{-1}(y)$".</p>
<p>You then attempt to prove the reverse inclusion: $\bigcap f(U_\alpha)
\subseteq f(\bigcap U_\alpha)$. But this is false. Consider just two $U_\alpha$, call them $U$ and $V$. They may be disjoint, yet there
are $x\in U$, $y\in V$ with $f(x)=f(y)$ so that $f(U)\cap f(V)\ne\emptyset$.</p>
|
2,313,060 | <p>$f(\bigcap_{\alpha \in A} U_{\alpha}) \subseteq \bigcap_{\alpha \in A}f(U_{\alpha})$</p>
<p>Suppose $y \in f(\bigcap_{\alpha \in A} U_{\alpha})$
$\implies f^{-1}(y) \in \bigcap_{\alpha \in A} U_{\alpha} \implies f^{-1}(y) \in U_{\alpha}$ for all $\alpha \in A$</p>
<p>$\implies y \in f (U_{\alpha})$ for all $\alpha \in A \implies y \in \bigcap_{\alpha \in A}f (U_{\alpha})$</p>
<p>$\bigcap_{\alpha \in A}f(U_{\alpha}) \subseteq f(\bigcap_{\alpha \in A} U_{\alpha})$</p>
<p>Suppose $y \in \bigcap_{\alpha \in A}f(U_{\alpha})\implies y \in f(U_{a}) $ for all $\alpha \in A$
$\implies f^{-1}(y)\in U_{a}$ for all $a\in A$</p>
<p>$ \implies f^{-1}(y)\in \bigcap_{a \in A}U_{a} \implies y \in f(\bigcap_{a \in A}(U_{a})$</p>
<p>Therefore</p>
<p>$f(\bigcap_{\alpha \in A} U_{\alpha}) \subseteq \bigcap_{\alpha \in A}f(U_{\alpha})$</p>
<p>Please let me know if my proof works, also I don't fully know how to do the following. Please give me some help. </p>
<p>Give an example of proper containment. Find a condition on f that would ensure equality.</p>
| egreg | 62,967 | <p>You are probably misled by the similar relation holding for $f^{-1}$:
$$
\bigcap_\alpha f^{-1}(U_\alpha)=
f^{-1}\Bigl(\bigcap_\alpha U_\alpha\Bigr)
$$
which <em>is</em> true and whose proof goes essentially like yours, with $f$ and $f^{-1}$ interchanged.</p>
<p>However, your assignment asks you to find an example of <em>proper containment</em>, which should make you suspect equality doesn't generally hold.</p>
<p>Indeed, it doesn't hold in general; it does when $f$ is injective.</p>
<hr>
<p>If $f\colon X\to Y$ and $V\subseteq Y$, then
$$
f^{-1}(V)=\{x\in X:f(x)\in V\}
$$
In other words, saying $x\in f^{-1}(U)$ is the same as saying $f(x)\in V$.</p>
<p>To the contrary, if $U\subseteq X$,
$$
f(U)=\{y\in Y:\text{there exists $x\in U$ with $y=f(x)$}\}
$$
You <strong><em>cannot</em></strong> say $y\in f(U)$ if and only if $f^{-1}(y)\in U$; what you can say, when you take $y\in f(U)$, is that there is $x\in U$ with $f(x)=y$.</p>
<p>Now try and fix your proof.</p>
|
939,725 | <p>Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.</p>
<p>I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out. </p>
| Pauly B | 166,413 | <p>Start with $a_n=\frac6{a_{n-1}-1}$, and replace $a_{n-1}$ with $\frac6{a_{n-2}-1}$. We obtain</p>
<p>$$a_n=\frac6{\frac6{a_{n-2}-1}-1}=\frac{6(1-a_{n-2})}{a_{n-2}-7}$$</p>
<p>Doing this again with $a_{n-2}=\frac6{a_{n-3}-1}$ and so forth, we get</p>
<p>$$a_n=\frac{6(7-a_{n-3})}{7a_{n-3}-13}=\frac{6(13-7a_{n-4})}{13a_{n-4}-55}=\frac{6(55-13a_{n-5})}{55a_{n-5}-133}$$</p>
<p>It would seem then, that</p>
<p>$$a_n=\frac{6(F_{k-1}-F_{k-2}a_{n-k})}{F_{k-1}a_{n-k}-F_k}$$</p>
<p>For some coefficients $F_k$. But what is $F_k$? To find out replace $a_{n-k}$ with $\frac6{a_{n-k-1}-1}$, then</p>
<p>$$a_n=\frac{6(F_{k-1}-F_{k-2}a_{n-k})}{F_{k-1}a_{n-k}-F_k}=\frac{6(F_{k-1}-F_{k-2}\frac6{a_{n-k-1}-1})}{F_{k-1}\frac6{a_{n-k-1}-1}-F_k}=\frac{6((6F_{k-2}+F_{k-1})-F_{k-1}a_{n-k-1})}{F_{k}a_{n-k-1}-(6F_{k-2}+F_{k-1})}=\frac{6(F_{k}-F_{k-1}a_{n-k-1})}{F_{k}a_{n-k-1}-F_{k+1}}$$</p>
<p>So it would seem that $F_k=F_{k-1}+6F_{k-2}$. By noting that $F_1=1,F_2=7$, we have the solution (I presume you can solve it yourself) $$F_k=\frac25(-2)^k+\frac353^k$$ </p>
<p>By plugging in $k=n$, we now have</p>
<p>$$a_n=\frac{6(F_{n-1}-F_{n-2}a_{0})}{F_{n-1}a_{0}-F_n}=\frac{6(F_{n-1}-2F_{n-2})}{2F_{n-1}-F_n}$$</p>
<p>which eventually simplifies to $$a_n=3-\frac{5}{4\left(-\frac23\right)^n+1}$$</p>
<p>which fits both $a_0=2$ and $a_n=\frac6{a_{n-1}-1}$, so the expression is indeed correct.</p>
|
1,075,879 | <p>I have to prove or disprove the following statement:</p>
<blockquote>
<p>If a group $G$ acts on a set $X$, then every subgroup $H$ of $G$ acts on the set $X$ as well, and every orbit of the action $G$ on $X$ is an union of orbits of the action $H$ on $X$.'</p>
</blockquote>
<p>But I have absolutely no clue what they mean with this question. The question is translated from Dutch, so I hope that I didn't make mistakes while translating it. What I don't understand is what the action $G$ on $X$ means and what a union of orbits should be.</p>
| user133281 | 133,281 | <p>In general, an action of a group $G$ on a set $X$ is a group homomorphism from $G$ to the group $S_X$ of permutations of the set $X$. This means that we send each group element $g \in G$ to some permutation of the elements in $X$, so each group element "acts" on $X$ by permuting its elements in some way. Usually, we write $g \cdot x$ to denote the element of $X$ that $x$ is sent to by the permutation in $S_X$ that corresponds to $g \in G$.</p>
<p>Then the <em>orbit</em> of a point $x \in X$ is the set of all points of $X$ that we can reach by applying elements of $g$ to $x$. In formula: $\text{Orb}(x) = G(x) = \{g \cdot x: g \in G\}$. This is a subset of $X$.</p>
<p>You are given a group $G$ with a subgroup $H$ and some action $G \to S_X$. Does this also give an action of $H$ on $X$? Yes, it does, because the composition of the homomorphisms $H \to G$ (inclusion) and $G \to S_X$ gives a homomorphism $H \to S_X$. Loosely speaking, since we can apply elements of $G$ to elements of $X$ we can do the same with elements of $H \subset G$. Since $H$ is a group under the group operation of $G$, this gives a well defined action of a group on a set.</p>
<p>Now we have two actions on $X$: one from $G$ and one from $H$. We also have orbits from both actions, partitioning $X$. (So we have two partitions of $X$: one into the orbits of the action of $G$, and one into the orbits of the action of $H$.) The question asks whether it is true that any orbit of $G$ (as a subset of $X$) is a union of orbits of $H$ (which are subsets of $X$ as well).</p>
<p>For instance, the dihedral group $D_4$ acts naturally on the vertices of a square (a set $X$ with $4$ points). $D_4$ has a cyclic subgroup $H$ with two elements, corresponding with reflecting in a (let's say horizontal) side of the square. The action of $D_4$ is transitive, so there is one orbit: $X$ itself. The action of $H$ has two orbits, one consisting of the upper two vertices and one consisting of the lower two vertices of the square. So, in this case, any $G$-orbit is a union of $H$-orbits: the only $G$-orbit $X$ is the union of the two $H$-orbits. The question is whether this holds for all group actions. </p>
<p>I hope this makes the question clear to you!</p>
<p>And to give you a hint to solve it: because $X$ is partitioned both by the $H$- and the $G$-orbits, it suffices to prove that any $H$-orbit is contained in only one $G$-orbit.</p>
|
1,075,879 | <p>I have to prove or disprove the following statement:</p>
<blockquote>
<p>If a group $G$ acts on a set $X$, then every subgroup $H$ of $G$ acts on the set $X$ as well, and every orbit of the action $G$ on $X$ is an union of orbits of the action $H$ on $X$.'</p>
</blockquote>
<p>But I have absolutely no clue what they mean with this question. The question is translated from Dutch, so I hope that I didn't make mistakes while translating it. What I don't understand is what the action $G$ on $X$ means and what a union of orbits should be.</p>
| Community | -1 | <p>As for the second part, recall that every <span class="math-container">$g\in G$</span> lays in some right coset of <span class="math-container">$H$</span> in <span class="math-container">$G$</span>. So, denoted with <span class="math-container">$R\subseteq G$</span> a complete set of coset representatives, for <span class="math-container">$x\in X$</span> we get:
<span class="math-container">\begin{alignat}{1}
\operatorname{Orb}_G(x) &= \{g\cdot x, g\in G\} \\
&= \{(hg_i)\cdot x, g_i\in R \text{ and } h\in H\} \\
&= \{h\cdot(g_i\cdot x), g_i\in R \text{ and } h\in H\} \\
&= \{h\cdot y_i, y_i\in Y(x)\subseteq \operatorname{Orb}_G(x) \text{ and } h\in H\} \\
&= \bigcup_{y_i\in Y(x)}\{h\cdot y_i, h\in H\} \\
&= \bigcup_{y_i\in Y(x)}\operatorname{Orb}_H(y_i) \\
\end{alignat}</span>
where <span class="math-container">$Y(x):=\{g_i\cdot x, g_i\in R\}$</span>. So, the claim is true.</p>
|
3,177,343 | <p>I have the following minimization problem in <span class="math-container">$x \in \mathbb{R}^n$</span></p>
<p><span class="math-container">$$\begin{array}{ll} \text{minimize} & \|x\|_2 - c^T x\\ \text{subject to} & Ax = b\end{array}$$</span></p>
<p>where <span class="math-container">$A \in \mathbb{R}^{m \times n}$</span> is right-invertible, <span class="math-container">$b \in \mathbb{R}^m$</span> and <span class="math-container">$c \in \mathbb{R}^n$</span>.</p>
<p>I tried to solve this using Lagrange multipliers, but am unable to find a closed form solution for <span class="math-container">$x$</span>, because the derivative of <span class="math-container">$\|x\|_2$</span> with respect to <span class="math-container">$x$</span>, which is <span class="math-container">$\frac{x}{\|x\|_2}$</span>, contains <span class="math-container">$\sqrt{x^Tx}$</span> in the denominator.</p>
<p>Any help would be appreciated.</p>
| Community | -1 | <p>Let <span class="math-container">$A^+$</span> be the Moore Penrose inverse of <span class="math-container">$A$</span>. Note that <span class="math-container">$x=x_0+u$</span> where <span class="math-container">$u$</span> varies in the image of the symmetric matrix <span class="math-container">$I_n-A^+A$</span>, a space <span class="math-container">$F$</span> of dimension <span class="math-container">$n-m$</span> (cf the Hans comment). Moreover <span class="math-container">$E=\ker(I_n-A^+A)$</span> is the orthogonal of <span class="math-container">$F$</span>; considering an orthonormal basis associated to the decomposition <span class="math-container">$E\bigoplus F$</span>, we may assume that </p>
<p><span class="math-container">$x_0=[a,0]^T,u=[0,x]^T,c=[w,v]^T$</span>. Then</p>
<p><span class="math-container">$g(x)=||x_0+u||-c^Tu=\sqrt{||a||^2+||x||^2}-v^Tx$</span>. Then</p>
<p><span class="math-container">$Dg_x=0$</span> iff <span class="math-container">$x/\sqrt{||a||^2+||x||^2 }=v$</span>. Then <span class="math-container">$x=\lambda v$</span> where <span class="math-container">$\lambda=\sqrt{||a||^2+\lambda ^2||v||^2}\geq 0$</span>. </p>
<p>Finally, if <span class="math-container">$||v||\geq 1$</span>, then no solution and no finite minimum. If <span class="math-container">$||v||<1$</span>, then </p>
<p><span class="math-container">$\lambda=\dfrac{||a||}{\sqrt{1-||v||^2}}$</span>.</p>
<p>It remains to show that the above <span class="math-container">$u$</span> is associated to the minimum of <span class="math-container">$g$</span>; that is your business.</p>
|
3,177,343 | <p>I have the following minimization problem in <span class="math-container">$x \in \mathbb{R}^n$</span></p>
<p><span class="math-container">$$\begin{array}{ll} \text{minimize} & \|x\|_2 - c^T x\\ \text{subject to} & Ax = b\end{array}$$</span></p>
<p>where <span class="math-container">$A \in \mathbb{R}^{m \times n}$</span> is right-invertible, <span class="math-container">$b \in \mathbb{R}^m$</span> and <span class="math-container">$c \in \mathbb{R}^n$</span>.</p>
<p>I tried to solve this using Lagrange multipliers, but am unable to find a closed form solution for <span class="math-container">$x$</span>, because the derivative of <span class="math-container">$\|x\|_2$</span> with respect to <span class="math-container">$x$</span>, which is <span class="math-container">$\frac{x}{\|x\|_2}$</span>, contains <span class="math-container">$\sqrt{x^Tx}$</span> in the denominator.</p>
<p>Any help would be appreciated.</p>
| greg | 357,854 | <p>The linear equation <span class="math-container">$Ax=b$</span> has the general solution
<span class="math-container">$$x = A^+b + Pw$$</span>
where <span class="math-container">$A^+$</span> is the Penrose inverse, <span class="math-container">$P=(I-A^+A)$</span> is the projector into the nullspace, and <span class="math-container">$w$</span> is an arbitrary vector. Using this, we can replace the constrained problem in terms of <span class="math-container">$x$</span> with an <strong>unconstrained</strong> problem in terms of <span class="math-container">$w$</span>.</p>
<p>For convenience, define the variables
<span class="math-container">$$\eqalign{
y &= Pc, \quad &Py = y \cr
x_b &= A^+b, \quad &x_w = Pw = Px, \quad
&x &= x_b + x_w \cr
\lambda_b^2 &= x_b:x_b, \quad &\lambda_w^2 = x_w:x_w, \quad
&\lambda^2 &= \lambda_b^2 + \lambda_w^2 \,= x:x \cr
}$$</span>
Note the following implications
<br><span class="math-container">$\quad P$</span> is ortho-projector <span class="math-container">$\implies P^2=P=P^T$</span>
<br><span class="math-container">$\quad A$</span> is right-invertible <span class="math-container">$\implies A^+=A^T(AA^T)^{-1}\,$</span> <strong>and</strong> <span class="math-container">$\,AA^+=I$</span>
<br><span class="math-container">$\quad AP\,=0\,\implies Ax=(AA^+b+0)=b$</span>
<br><span class="math-container">$\quad PA^+\!=0\implies Px=Pw\,\,$</span> <strong>and</strong> <span class="math-container">$\,\,x_b:x_w=0$</span>
<br><span class="math-container">$\quad\lambda^2\!=x\!:\!x\implies \lambda\,d\lambda = x:dx$</span></p>
<p>Start by calculating the gradient of the cost function.
<span class="math-container">$$\eqalign{
\phi &= \lambda - c:x \cr
d\phi &= d\lambda - c:dx = \Big(\frac{x}{\lambda}-c\Big):dx \cr
&= \Big(\frac{x-\lambda c}{\lambda}\Big):P\,dw \cr
&= \frac{Pw-\lambda y}{\lambda}:dw \cr
\frac{\partial\phi}{\partial w} &= \frac{Pw-\lambda y}{\lambda} \cr
}$$</span>
Setting the gradient to zero, we can solve for the vector <span class="math-container">$\,\,w=\lambda y$</span>. </p>
<p>Substitute this value of <span class="math-container">$w$</span> into the <span class="math-container">$\lambda$</span>-relationships.
<span class="math-container">$$\eqalign{
\lambda_w^2 &= Pw:Pw = \lambda Py:\lambda Py = \lambda y:\lambda y \cr
\lambda^2 &= \lambda_b^2 + \lambda_w^2 = \lambda_b^2 + \lambda^2(y:y) \cr
\lambda^2 &= \frac{\lambda_b^2}{1-y:y} = \frac{(A^+b):(A^+b)}{1-(Pc:Pc)} \cr
}$$</span>
Choosing the <em>real</em> positive square root (if it exists) completes the solution
<span class="math-container">$$\eqalign{
x &= A^+b + \lambda y \cr\cr
}$$</span>
<strong>NB:</strong> Many of the steps above utilize the trace/Frobenius product
<span class="math-container">$$A:B = {\rm Tr}(A^TB)$$</span></p>
|
82,765 | <p><strong>Bug introduced in 9.0 and persisting through 12.2</strong></p>
<hr />
<p>I get the following output with a fresh Mathematica (ver 10.0.2.0 on Mac) session</p>
<pre><code>FullSimplify[Exp[-100*(i-0.5)^2]]
(* 0. *)
Simplify[Exp[-100*(i-0.5)^2]]
(* E^(-100. (-0.5+i)^2) *)
</code></pre>
<p><code>FullSimplify</code> seems to be a bit overambitious and kills the expression completely. Is there anything that explains this behavior or is this simply a bug?</p>
<p>As suggested in the comments, I did an additional test:</p>
<pre><code>FullSimplify[Exp[-100*(i-1/2)^2]]
(* E^(-25 (1-2 i)^2) *)
</code></pre>
<p>Apparently, the float point math causes the problem.</p>
| David Zwicker | 26,656 | <p>The behavior seems to be a bug of Mathematica.
Here is an excerpt from an email I got from Wolfram after asking them about the problem:</p>
<blockquote>
<p>It does seem that the answer of FullSimplify is incorrect especially
since the exponential function is not identical to zero (or a
very-close-to-zero constant). Therefore, I filed a report with our
development team raising the issue [...]</p>
</blockquote>
|
4,441,034 | <blockquote>
<p>Consider function <span class="math-container">$f(x)$</span> whose derivative is continuous on the interval <span class="math-container">$[-3; 3]$</span> and the graph of the function <span class="math-container">$y = f'(x)$</span> is pictured below. Given that <span class="math-container">$g(x) = 2f(x) + x^2 + 4$</span> and <span class="math-container">$f(1) = -24$</span>, how many real roots of the equation <span class="math-container">$g(x) = 0$</span> are there on the interval <span class="math-container">$[-3; 3]$</span>?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/yMyHY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yMyHY.png" alt="enter image description here" /></a></p>
<p>[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)</p>
<p>By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]</p>
<p>Excuse my waterlogged, dropped-on-the-floor-too-many-times, memory-of-a-goldfish-with-dementia brick camera quality.</p>
<p>Anyhow, first of all, <span class="math-container">$g(1) = 2f(1) + 1^2 + 4 = -43$</span> and <span class="math-container">$$g'(x) = 0 \iff 2[f'(x) + x] = 0 \iff \left[ \begin{aligned} x &= -3\\ x &= 1\\ x&= 3 \end{aligned} \right.$$</span></p>
<p>From which, we can draw the table of variations (<em>in glorious Technicolor</em>).</p>
<p><a href="https://i.stack.imgur.com/gCTc2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gCTc2.png" alt="enter image description here" /></a></p>
<p>Now, we just need to know whether <span class="math-container">$2f(\pm 3) + 13$</span> are greater than zero. And this is where I have a problem. How am I supposed to know?</p>
<p>Perhaps it's through the values of <span class="math-container">$\displaystyle \int_1^{\pm 3}f'(x)\, \mathrm dx = f(\pm 3) - f(1)$</span>, which could be somewhat perceived in the graph itself, but I am not entirely sure.</p>
<p>Anyhow, thanks for reading, (and even more if you could help~)</p>
| heropup | 118,193 | <p>Let's look at an example. Say <span class="math-container">$n = 12$</span>. Then for <span class="math-container">$k \in \{1, \ldots, 12\}$</span>, we look at set of divisors of <span class="math-container">$k$</span>:</p>
<p><span class="math-container">$$\begin{array}{c|l|c}
k & d & d(k)\\
\hline
1 & \{1 \} & 1 \\
2 & \{1, 2 \} & 2\\
3 & \{1, 3 \} & 2 \\
4 & \{1, 2, 4 \} & 3 \\
5 & \{1, 5 \} & 2 \\
6 & \{1, 2, 3, 6 \} & 4 \\
7 & \{1, 7 \} & 2 \\
8 & \{1, 2, 4, 8 \} & 4 \\
9 & \{1, 3, 9 \} & 3 \\
10 & \{1, 2, 5, 10 \} & 4 \\
11 & \{1, 11 \} & 2 \\
12 & \{1, 2, 3, 4, 6, 12 \} & 6
\end{array}$$</span>
Now we want the sum to look like this:
<span class="math-container">$$\sum_{d=1}^n \sum_{k \in ?} 1$$</span>
So for <span class="math-container">$d = 1$</span>, we need to count the inner summand a total of <span class="math-container">$12$</span> times, one for each row in the table. But for <span class="math-container">$d = 2$</span>, we count only <span class="math-container">$6$</span> times, corresponding to <span class="math-container">$k \in \{2, 4, 6, 8, 10, 12\}$</span>. Similarly, for <span class="math-container">$d = 3$</span>, we count <span class="math-container">$4$</span> times, for <span class="math-container">$k \in \{3, 6, 9, 12\}$</span>. And now the pattern should be clear: for each <span class="math-container">$d$</span>, we count a total of <span class="math-container">$\lfloor n/d \rfloor$</span> times, for <span class="math-container">$$k \in \{d, 2d, \ldots, \lfloor n/d \rfloor d \}.$$</span> So the desired double sum might look like this:
<span class="math-container">$$S = \sum_{k=1}^n d(k) = \sum_{k=1}^n \sum_{d \mid k} 1 = \sum_{d=1}^n \sum_{k = 1}^{\lfloor n/d \rfloor} 1 = \sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor.$$</span></p>
|
4,441,034 | <blockquote>
<p>Consider function <span class="math-container">$f(x)$</span> whose derivative is continuous on the interval <span class="math-container">$[-3; 3]$</span> and the graph of the function <span class="math-container">$y = f'(x)$</span> is pictured below. Given that <span class="math-container">$g(x) = 2f(x) + x^2 + 4$</span> and <span class="math-container">$f(1) = -24$</span>, how many real roots of the equation <span class="math-container">$g(x) = 0$</span> are there on the interval <span class="math-container">$[-3; 3]$</span>?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/yMyHY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yMyHY.png" alt="enter image description here" /></a></p>
<p>[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)</p>
<p>By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]</p>
<p>Excuse my waterlogged, dropped-on-the-floor-too-many-times, memory-of-a-goldfish-with-dementia brick camera quality.</p>
<p>Anyhow, first of all, <span class="math-container">$g(1) = 2f(1) + 1^2 + 4 = -43$</span> and <span class="math-container">$$g'(x) = 0 \iff 2[f'(x) + x] = 0 \iff \left[ \begin{aligned} x &= -3\\ x &= 1\\ x&= 3 \end{aligned} \right.$$</span></p>
<p>From which, we can draw the table of variations (<em>in glorious Technicolor</em>).</p>
<p><a href="https://i.stack.imgur.com/gCTc2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gCTc2.png" alt="enter image description here" /></a></p>
<p>Now, we just need to know whether <span class="math-container">$2f(\pm 3) + 13$</span> are greater than zero. And this is where I have a problem. How am I supposed to know?</p>
<p>Perhaps it's through the values of <span class="math-container">$\displaystyle \int_1^{\pm 3}f'(x)\, \mathrm dx = f(\pm 3) - f(1)$</span>, which could be somewhat perceived in the graph itself, but I am not entirely sure.</p>
<p>Anyhow, thanks for reading, (and even more if you could help~)</p>
| PNT | 873,280 | <p>Let's make a table, the rows and columns are <span class="math-container">$1,2,3...,n$</span> and an element <span class="math-container">$(d,k)$</span> is <span class="math-container">$1$</span> if <span class="math-container">$d\mid k$</span> and <span class="math-container">$0$</span> otherwise,
<span class="math-container">\begin{array} {|r|r|}\hline & k & 1 & 2 & 3 & 4 & 5 \\ \hline d & * & * & * & * & * & * \\ \hline 1 & * & 1 & 1 & 1 & 1 & 1 \\ \hline 2 & * & 0 & 1 & 0 & 1 & 0 \\ \hline 3 & * & 0 & 0 & 1 & 0 & 0 \\ \hline 4 & * & 0 & 0 & 0 & 1 & 0 \\ \hline \end{array}</span></p>
<p>In the original sum, we fix a <span class="math-container">$k$</span> and count the divisors of <span class="math-container">$k$</span>, so if we want to swap the summation we would fix a <span class="math-container">$d$</span> and count how many <span class="math-container">$k\le n$</span> are there such that <span class="math-container">$d\mid k$</span>? The answer is <span class="math-container">$\lfloor n/d \rfloor $</span>
Hence <span class="math-container">$$\sum_{k=1}^nd(k)=\sum_{k=1}^n\sum_{d\mid k}^k1=\sum_{d=1}^n\sum_{k\le n\\ d\mid k}=\sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor $$</span></p>
|
3,830,231 | <p>I'm trying to prove the following 'covariance inequality'
<span class="math-container">$$
|\text{Cov}(x,y)|\le\sqrt{\text{Var}(x)}\sqrt{\text{Var}(y)}\,,
$$</span>
where covariance and variance are defined using discrete values,
<span class="math-container">$$
\text{Cov}(x,y) = \frac{1}{n-1}\sum_{i=1}^n \big[(x_i-\bar{x})(y_i-\bar{y})\big]\,,
$$</span>
<span class="math-container">$$
\text{Var}(x) = \frac{\sum_{i=1}^n(x_i-\bar{x})}{n-1}\,,
$$</span>
<span class="math-container">$$
\text{Var}(y) = \frac{\sum_{i=1}^n(y_i-\bar{y})}{n-1}\,.
$$</span></p>
<p>There are plenty of proofs of to be found online (such as <a href="https://www.probabilitycourse.com/chapter6/6_2_4_cauchy_schwarz.php" rel="nofollow noreferrer">this one</a>), however, they all either seem to be for continuous random variables, or just refer me to the Cauchy-Schwarz inequality, which I am aware of, but not sure how to apply to this particular proof. Basically, I am wondering if there is a way to prove this inequality using those above definitions.</p>
<p>I've tried substituting these definitions into the inequality above, but after expanding these summations and ridding of the <span class="math-container">$1/(n-1)$</span> on both sides, I'm left with a mess (as you can imagine) with summation terms on both sides, some in the absolute value, and some in the square root. I'm not sure if there's some algebraic mistake I'm making, some summation property I'm missing, or if substitution is just the wrong way to go about this proof.</p>
| User203940 | 333,294 | <p>First recall that</p>
<p><span class="math-container">$$ \text{Var}(x) = \frac{1}{n} \sum_{j=0}^{n-1} (x_i - \bar{x})^2.$$</span></p>
<p>See for example <a href="https://en.wikipedia.org/wiki/Variance#Discrete_random_variable" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Variance#Discrete_random_variable</a>.</p>
<p>Cauchy-Schwarz says that</p>
<p><span class="math-container">$$ \left( \sum_{j=1}^n u_j v_j \right)^2 \leq \left(\sum_{j=1}^n (u_j)^2 \right) \left( \sum_{j=1}^n (v_j)^2\right).$$</span></p>
<p>See for example <a href="https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality</a>.</p>
<p>Let's see how to apply it. First we write things out.</p>
<p><span class="math-container">$$ \text{Cov}(x,y)^2 = \left(\frac{1}{n} \sum_{j=0}^{n-1} (x_j - \bar{x}) (y_j - \bar{y}) \right)^2 = \frac{1}{n^2} \left(\sum_{j=0}^{n-1} (x_j - \bar{x}) (y_j - \bar{y}) \right)^2.$$</span></p>
<p>It now looks something like Cauchy-Schwarz. Let's apply it.</p>
<p><span class="math-container">$$ \text{Cov}(x,y)^2 \leq \frac{1}{n^2} \left(\sum_{j=0}^{n-1} (x_j-\bar{x})^2\right) \left( \sum_{j=0}^{n-1} (y_j - \bar{y})^2\right).$$</span></p>
<p>I can actually rewrite this as</p>
<p><span class="math-container">$$ \text{Cov}(x,y)^2 \leq \left(\frac{1}{n} \sum_{j=0}^{n-1} (x_j-\bar{x})^2\right) \left(\frac{1}{n} \sum_{j=0}^{n-1} (y_j - \bar{y})^2\right).$$</span></p>
<p>Note that this looks like our definition of variance. Applying that, we have</p>
<p><span class="math-container">$$ \text{Cov}(x,y)^2 \leq \text{Var}(x) \text{Var}(y).$$</span></p>
<p>Now take the square root of both sides.</p>
|
2,830,718 | <p>I want to estimate how many red balls in a box. Red, yellow, blue balls could be in the box. But I don't know how many of them are in the box.</p>
<p>What I did was randomly drawing 10 balls from the box and learned that there was no red ball.</p>
<p>(Edit: Assume the number of the balls in the box is a known finite number. Let's say 10,000)</p>
<p>Can I say the possibility of having at least one red ball is $$\left(\frac13\right)^{10}= 0.00169\ \%\ ?$$ (3 possible outcomes, and 10 observations)</p>
<p>I'm thinking I don't know the distribution of the colors of the balls, so I'm not sure this inference is reasonable or not. Thanks!</p>
| David K | 139,123 | <p>TL;DR: The answer to the title of your question is, "Yes, but not with anywhere near the kind of accuracy and confidence you seem to be asking for."</p>
<hr>
<p>Let's say I put $10000$ balls in a box, and exactly one of the balls is red.
You know the total number of balls but not how many there are of each color.</p>
<p>You draw $5000$ balls without replacement and look at each one to see if it is red.
I then ask you whether there was a red ball in the box before you started taking balls out, and you answer "yes" or "no."</p>
<p>If your criterion for "no red ball" is that none of the balls you drew were red, then there is a $50\%$ chance you will give me the wrong answer.</p>
<p>Now consider the case where you draw only $10$ balls. If you think getting only yellow and blue balls in the first ten is enough to say that there is <em>not</em> at least one red ball in the box, what's the probability that you'll give the wrong answer in the case where there actually <em>is</em> exactly one red ball in the box?</p>
<p>Note that it doesn't matter how many other colors there are. You'll get wrong answers just as likely when we fill the rest of the box with balls of $100$ different colors as when we use only yellow and blue balls in addition to the red ball.</p>
<hr>
<p>Trying to decide there are <em>no</em> individuals of type X in a large population is something you basically cannot resolve simply by examining a small sample.
The formula you give would be a reasonable one to use if the question you are trying to answer is, "Are at least $\frac23$ of the balls red?"
(And even then, I would not describe your result as the <em>probability</em> that the answer is "yes".)</p>
|
4,188,106 | <p>Let's say we have the following diagram
<span class="math-container">$$\require{AMScd}\begin{CD}
0 @>>> A @>>> B @>>> C @>>> 0\\
{} @V{\alpha}VV @V{\beta}VV @V{\gamma}VV {} \\
0 @>>> A' @>>> B' @>>> C' @>>> 0
\end{CD}$$</span>
where the top and bottom rows are short exact, so if <span class="math-container">$f: A \rightarrow B$</span> and <span class="math-container">$f': A' \rightarrow B'$</span> are injectives and <span class="math-container">$g:B \rightarrow C$</span>, <span class="math-container">$g': B' \rightarrow C'$</span> surjectives, then <span class="math-container">$\text{Im}(f)= \text{ker}(g)$</span> and <span class="math-container">$\text{Im}(f')= \text{ker}(g')$</span>. The short five lemma says that if the diagram is commutative and <span class="math-container">$\alpha$</span> and <span class="math-container">$\gamma$</span> are modules isomorphisms, then <span class="math-container">$\beta$</span> is an isomorphism.</p>
<p>Is it true that if <span class="math-container">$A \simeq A'$</span> and <span class="math-container">$C \simeq C'$</span> then there's an isomorphism <span class="math-container">$\beta :B \rightarrow B'$</span> such that the diagram is commutative? The hint is: let <span class="math-container">$A = A' = \mathbb{Z}_2$</span>, <span class="math-container">$C = C' = \mathbb{Z}_2 \oplus \mathbb{Z}_2$</span> and <span class="math-container">$B = B' = \mathbb{Z}_4 \oplus \mathbb{Z}_2$</span> with <span class="math-container">$\alpha = \gamma = id$</span> and
<span class="math-container">$$g(a \text{ mod }4, b\text{ mod }2) = (a\text{ mod }2,b\text{ mod }2) \\
g'(a\text{ mod }4, b\text{ mod }2) = (b\text{ mod }2,a\text{ mod }2).$$</span>
Find two injective morphisms <span class="math-container">$f,f': \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_2$</span> such that the top and bottom row are exact but there's no isomorphism <span class="math-container">$\beta: \mathbb{Z}_4 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_2$</span> such that the diagram is commutative.</p>
<p>So for my proof, I started by finding</p>
<p><span class="math-container">$$\text{ker}(g) = \text{ker}(g') = \left\{(0,0), (2,0) \right\},$$</span>
here I mean <span class="math-container">$(0,0) = ([0],[0])$</span> the corresponding classes. So, the functions <span class="math-container">$f,f'$</span> that satisfies the conditions are <span class="math-container">$f=f'$</span> such that <span class="math-container">$f(0) = (0,0), f(1) = (2,0)$</span> (because <span class="math-container">$\text{Im}(f)=\text{ker}(g)$</span>). Now, suppose that there is an isomorphism <span class="math-container">$\beta: \mathbb{Z}_4 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_2$</span> such that the diagram is commutative. Then <span class="math-container">$\beta \circ f = f' \circ \alpha$</span> i.e., <span class="math-container">$\beta \circ f = f$</span>. So I should prove that <span class="math-container">$g=g' \circ \beta$</span> doesn't hold. Now my teacher said that since there're only <span class="math-container">$4$</span> possible <span class="math-container">$\beta$</span> isomorphisms, this can be made by hand (it is easy if <span class="math-container">$\beta = id$</span> because <span class="math-container">$g \neq g'$</span>). My question is, is there a way to complete the proof without checking the condition for all possible isomorphisms <span class="math-container">$\beta$</span>? Thanks</p>
| Milten | 620,957 | <p>Let <span class="math-container">$(a,b)=\beta^{-1}(1,0)$</span>. Then <span class="math-container">$(a,b)$</span> has order <span class="math-container">$4$</span> in <span class="math-container">$B$</span>, so <span class="math-container">$a$</span> is odd. Therefore
<span class="math-container">$$
g(a,b) = (1,b) \ne (0,1) = g'\circ\beta(a,b).
$$</span></p>
|
2,550,568 | <p>Suppose we have an alphabet of $a$ letters and a word $w$ of length $r$. What is the probablity that $w$ will appear in a sequence of $n$ letters drawn at random from the given alphabet?</p>
<p>I have posted a general question since there seem to be a few of these questions appearing, and this is intended as a general form of question to which general answers can be given. Anyone who wants to add to what I have written - asymptotic of solutions in more detail, for example, or alternative methods - that would be great.</p>
<p>And are there any good references for this kind of problem.</p>
| Mark Bennet | 2,906 | <p>Let $a_k$ be the probability that the word appears in the first $k$ letters. We have $a_0=a_1= \dots =a_{r-1}=0$.</p>
<p>Either $w$ appears in the first $n-1$ letters chosen, or it appears for the first time at the $n^{th}$ digit. In this second case (excluding the case of overlaps - see below) the final $r$ digits are determined - probability $a^{-r}$. The word cannot appear in the first $n-r$ digits. In simple cases therefore we have $$p_n=p_{n-1}+a^{-r}(1-p_{n-r})$$.</p>
<hr>
<p>As an interlude, it is not always that simple. The word ONION taken from an alphabet of $26$ capital letters can appear overlapping $ONIONION$ - here two letters overlap and we have to exclude the word from the first $n-3$ letters and not just the first $n-5$ if we are to avoid double-counting. This case requires minor adjustments.</p>
<hr>
<p>We are left, then, with a simple recurrence, which is not homogeneous, but it is easy to see that $p_k=1$ is a particular solution.</p>
<p>The homogeneous equation is of the form $$p_n-p_{n-1}+a^{-r}p_{n-r}=0$$ and the auxiliary equation is $$x^r-x^{r-1}+a^{-r}=0$$</p>
<p>This can be solved using the usual methods for such equations.</p>
<p>The following is very sketchy.</p>
<p>The function $f(x)=x^r-x^{r-1}+a^{-r}$ has a local minimum at $x=1-\frac 1r$ and therefore the root with largest absolute value lies in the range $1-\frac 1r \lt \alpha \lt 1$. This will dominate the asymptotics which should be of the form $a_n=1-A\alpha^n$, with the probability tending to $1$ and $n$ increases.</p>
<p>Since $f(x)$ is a perturbation of $x^r-x^{r-1}$ we would expect one root near $x=1$ and other roots to have small absolute value (certainly for $a$ and $r$ reasonably large), so the asymptotics should give a good estimate.</p>
|
4,765 | <p>I have a grid made up of overlapping <span class="math-container">$3\times 3$</span> squares like so:</p>
<p><img src="https://i.stack.imgur.com/BaY9s.png" alt="Grid"></p>
<p>The numbers on the grid indicate the number of overlapping squares. Given that we know the maximum number of overlapping squares (<span class="math-container">$9$</span> at the middle), and the size of the squares (<span class="math-container">$3\times 3$</span>), is there a simple way to calculate the rest of the number of overlaps?</p>
<p>e.g. I know the maximum number of overlaps is <span class="math-container">$9$</span> at point <span class="math-container">$(2,2)$</span> and the square size is <span class="math-container">$3\times 3$</span> . So given point <span class="math-container">$(3,2)$</span> how can I calculate that there are <span class="math-container">$6$</span> overlaps at that point?</p>
| Alan D'Souza | 74,758 | <p>Try proving that for any irrational number $\alpha$, the set
$A=\left \{ a+b\alpha \mid a\in \mathbb{N},b\in \mathbb{Z} \right \}$ is dense in $\mathbb{R}$. Let $\alpha =\pi $.
Since the set $A$ is dense in $\mathbb{R}$, $\forall \; x\in \mathbb{R}$, there exists a sequence of terms $(z_{n})$ such that
$\lim_{n\rightarrow \infty }\left ( z_{n} \right )=x$.</p>
<p>$(z_{n})$ can be written as the sum of two sequences of integers. </p>
<p>$z_{n} = x_{n} + \pi y_{n}$. </p>
<p>Use the continuity of $\sin(x)$, to show that a sequence from the set $\left \{\sin (n)\mid n\in \mathbb{N} \right \}$ converges to every element of $[-1,1]$.</p>
|
1,198,373 | <p>I have a function of two variables, which I wish to check for monotonicity in the entire function domain. I cant find any formal definition of increasing or decreasing function for multi variable case. Can anybody please guide?
Thanks in advance.</p>
| Clement C. | 75,808 | <p>There is no general definiton (as mentioned in the comments, there is no total order on <span class="math-container">$\mathbb{R}^2$</span>, which would be required for a canonical definition of monotonicity of bivariate functions).</p>
<p>Two <em>possible</em> definitions, though: let <span class="math-container">$$f\colon (x,y)\in\mathbb{R}^2\mapsto f(x,y)$$</span>
be your function.</p>
<ul>
<li><p>First definition: <span class="math-container">$f$</span> is said to be <em>monotone</em> (non-decreasing) if for all fixed <span class="math-container">$x_0, y_0 \in \mathbb{R}$</span>, the two functions <span class="math-container">$f_{x_0}\colon y\in\mathbb{R}\mapsto f(x_0,y)$</span> and <span class="math-container">$f_{y_0}\colon x\in\mathbb{R}\mapsto f(x,y_0)$</span> are monotone (non-decreasing).
(i.e., monotonicity wrt both projections)</p>
</li>
<li><p>Second definition: <span class="math-container">$f$</span> is said to be <em>monotone</em> (non-decreasing) if for all fixed <span class="math-container">$(x, y),(x', y') \in \mathbb{R}^2$</span>,
<span class="math-container">$$(x \leq x' \text{ and } y \leq y' ) \Rightarrow f(x,y) \leq f(x',y')$$</span>
(i.e., monotonicity wrt a partial order on <span class="math-container">$\mathbb{R}^2$</span>)</p>
</li>
</ul>
<p><strong>Edit:</strong> <em>I had originally written the two definitions are not equivalent. They are, as <a href="https://math.stackexchange.com/a/2488581/75808">Ij Huij's answer</a> below shows. I can't reconstruct what I had in mind at the time, but to be charitable it's probably either very contrived or wrong...</em></p>
|
1,198,373 | <p>I have a function of two variables, which I wish to check for monotonicity in the entire function domain. I cant find any formal definition of increasing or decreasing function for multi variable case. Can anybody please guide?
Thanks in advance.</p>
| Ij Huij | 495,418 | <p>I wanted to add this as a comment for the first answer, but I cannot put comments. The first and the second definition in this answer are equivalent.</p>
<p>If $f$ satisfies the first definition then it satisfies the second. We can see that by sandwiching $f(x',y)$ (or $f(x,y')$) between $f(x,y)$ and $f(x',y')$.</p>
<p>If $f$ satisfies the second definition, by once setting $x = x'$ and once setting $y = y'$ we can see that it is monotone with respect to both projections.</p>
<p>In general, let $\preceq$ be the elementwise inequality on $\mathbf{R}^n$ ($x \preceq y$ if and only if $x_i \leq y_i$ for all $i$). Now, a function $f: \mathbf{R}^n \mapsto \mathbf{R}$ is monotone with respect to this partial ordering when:</p>
<p>$$x \preceq y \Rightarrow f(x) \leq f(y).$$
This is equivalent to requiring that for all $x \in \mathbf{R}^n$
$$f(x) \leq f(x + \gamma e_i) \quad \text{for all } \gamma \geq 0 \text{ and }i \in\{1,2,...,n\}, $$
where $\{e_1,\ldots,e_n\}$ is the standard basis for $\mathbf{R}^n$.</p>
|
1,390,093 | <p>Let $G$ be act on $\Gamma$ with a fundamental domain $T$ where $T$ is tree. We construct <em>tree of groups</em> $(\mathcal{G},T)$ with the following structure:
$$\text{for every } v\in V(T),\,\,G_v=\operatorname{Stab}_G(v) $$</p>
<p>$$\text{for every } e\in E(T),\,\,G_e=\operatorname{Stab}_G(e) $$</p>
<p>Assume that $G_T$ is the direct limit of the system $(\mathcal{G},T)$. With using the universal property of the definition of the direct limit, we get the map $\phi\colon G_T\mapsto G$.</p>
<p>Now my question is:</p>
<p>If $\Gamma$ is connected, then why can we conclude that $\phi$ is a surjective map?</p>
| Lee Mosher | 26,501 | <p>We cannot conclude that $\phi$ is surjective, because it is not true in general. Here is a counterexample. </p>
<ul>
<li>$\Gamma = \mathbb{R}$ with vertex set $\mathbb{Z}$. </li>
<li>$G = \mathbb{Z}$ acting on $\mathbb{R}$ by translation. </li>
<li>$T = [0,1]$ is a fundamental domain. </li>
</ul>
<p>The stabilizer of each vertex $0$ and $1$ of $T$ is trivial, and the stablizer of the unique edge $[0,1]$ in $T$ is trivial. So $G_T$ is the trivial group. There is no map from the trivial group onto $\mathbb{Z}$.</p>
|
3,840,253 | <blockquote>
<p>How to show that <span class="math-container">$\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$</span>?</p>
</blockquote>
<p>My attempt:<br />
<span class="math-container">\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}</span>
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.</p>
| Michael Rozenberg | 190,319 | <p><span class="math-container">$$\frac{1}{\sin{x}}-\frac{1}{\sin\left(\frac{\pi}{3}+x\right)}+\frac{1}{\sin\left(\frac{\pi}{3}-x\right)}=$$</span>
<span class="math-container">$$=\frac{\sin\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{3}+x\right)+\sin{x}\left(\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}-x\right)\right)}{\sin{x}\sin\left(\frac{\pi}{3}+x\right)\sin\left(\frac{\pi}{3}-x\right)}=$$</span>
<span class="math-container">$$=\frac{\frac{1}{2}\left(\cos2x-\cos\frac{2\pi}{3}\right)+\sin{x}\cdot2\sin{x}\cos\frac{\pi}{3}}{\sin{x}\left(\frac{\sqrt3}{2}\cos{x}+\frac{1}{2}\sin{x}\right)\left(\frac{\sqrt3}{2}\cos{x}-\frac{1}{2}\sin{x}\right)}=$$</span>
<span class="math-container">$$=\frac{\frac{1}{2}-\sin^2x+\frac{1}{4}+\sin^2x}{\sin{x}\left(\frac{3}{4}\cos^2x-\frac{1}{4}\sin^2x\right)}=\frac{3}{\sin{x}(3(1-\sin^2x)-\sin^2x)}=\frac{3}{\sin3x}.$$</span></p>
|
4,082,588 | <blockquote>
<p><strong>Definition:</strong> <span class="math-container">$\beta X$</span> is the Stone-Čech compactification of <span class="math-container">$X$</span>.</p>
</blockquote>
<blockquote>
<p><strong>Theorem A:</strong> If <span class="math-container">$K$</span> is a compact Hausdorff space and <span class="math-container">$f\colon X \to K$</span> is<br />
continuous, there is a continuous <span class="math-container">$F: \beta X \to K$</span> such that <span class="math-container">$F \circ e = f$</span>, where <span class="math-container">$e\colon X\to\beta X$</span> is an embedding into a compact Hausdorff space.</p>
</blockquote>
<blockquote>
<p>Show that <span class="math-container">$\left|\beta\mathbb{N}\right|\geq\left|\beta\mathbb{Q}\right|$</span>.</p>
</blockquote>
<p>Let <span class="math-container">$f\colon\mathbb{N}\to\mathbb{Q}$</span> be a bijection. As <em>any</em> function from the discrete topology is continuous (<span class="math-container">$\mathbb{N}$</span> with the relative topology from <span class="math-container">$\mathbb{R}_\text{std.}$</span> is the discrete topology), we can enlarge the range. Therefore, we can enlarge the range to <span class="math-container">$\beta\mathbb{Q}$</span>, which is a compact Hausdorff space, so that <span class="math-container">$f\colon\mathbb{N}\to\beta\mathbb{Q}$</span> is continous. By <strong>Theorem A</strong> we can extend <span class="math-container">$f$</span> uniquely to to a continuous function
<span class="math-container">$\beta f\colon\beta\mathbb{N}\to\beta\mathbb{Q}$</span>.</p>
<p>I need to show that
<span class="math-container">$\beta f\colon\beta\mathbb{N}\to\beta\mathbb{Q}$</span> is surjective, i.e. <span class="math-container">$\beta f[\beta\mathbb{N}]=\beta\mathbb{Q}$</span>. As for any mapping <span class="math-container">$\beta f[\beta\mathbb{N}]\subseteq\beta\mathbb{Q}$</span>, it's enough to show that <span class="math-container">$\beta f[\beta\mathbb{N}]\supseteq\beta\mathbb{Q}$</span>.</p>
| Peluso | 884,108 | <p>This proof looks correct, you just need a little push.</p>
<p>We will use the following result:</p>
<p><strong>Lemma:</strong> If <span class="math-container">$X$</span> is a topological space, <span class="math-container">$D$</span> is a dense subset of <span class="math-container">$X$</span> and <span class="math-container">$A$</span> is a closed subset of <span class="math-container">$X$</span> such that <span class="math-container">$D\subseteq A$</span>, then we have the equality <span class="math-container">$A=X$</span>.</p>
<p>Since <span class="math-container">$\beta f$</span> extends the bijective function <span class="math-container">$f$</span>, we get the relations <span class="math-container">$$\mathbb{Q}=f[\mathbb{N}] \subseteq \beta f[\beta\mathbb{N}].$$</span></p>
<p>Now, since <span class="math-container">$\beta f$</span> is continuous, this tells us that <span class="math-container">$\beta f[\beta\mathbb{N}]$</span> is a compact space (therefore, a closed subset of <span class="math-container">$\beta\mathbb{Q}$</span>) that contains a dense subset of <span class="math-container">$\beta\mathbb{Q}$</span>. So, if we let <span class="math-container">$X=\beta\mathbb{Q}$</span>, <span class="math-container">$D=\mathbb{Q}$</span> and <span class="math-container">$A=\beta f[\beta\mathbb{N}]$</span>, then we can apply the lemma to get that <span class="math-container">$\beta f[\beta\mathbb{N}]$</span> must be exactly <span class="math-container">$\beta\mathbb{Q}$</span>; in other words, <span class="math-container">$\beta f$</span> is surjective.</p>
|
2,236,717 | <p>Let $S$ be a regular domain of characteristic $p>0$ with fraction field $K$. Assume that $K$ is $F$-finite, meaning that $K$ is a finite module over $K^p$. Does it follow that $S$ is also $F$-finite?</p>
<p>Diego</p>
| Takumi Murayama | 116,766 | <p>I believe this is false. [<a href="http://dx.doi.org/10.2140/ant.2016.10.1057" rel="nofollow noreferrer">Datta–Smith</a>, Ex. 4.5.1] give an example of a DVR that is not $F$-finite, whose fraction field is $\mathbf{F}_p(x,y)$.</p>
<p>A way to produce more examples is the following:</p>
<p><strong>Proposition</strong> [<a href="http://dx.doi.org/10.2140/ant.2016.10.1057" rel="nofollow noreferrer">Datta–Smith</a>, Prop. 2.6.1]<strong>.</strong> <em>A noetherian domain of positive characteristic is $F$-finite if and only if it is excellent and its fraction field is $F$-finite.</em></p>
<p>Thus, any non-excellent regular domain of positive characteristic with $F$-finite fraction field is a counterexample to your claim.</p>
<p>There are many examples of non-excellent regular domains in the literature: see [<a href="http://www.ams.org/mathscinet-getitem?mr=460307" rel="nofollow noreferrer">Nagata</a>, (E3.3)], [<a href="http://www.ams.org/mathscinet-getitem?mr=575344" rel="nofollow noreferrer">Matsumura</a>, (34.B)], and [<a href="http://www.cmls.polytechnique.fr/perso/orgogozo/travaux_de_Gabber/GTG/GTG.pdf" rel="nofollow noreferrer">Raynaud</a>, Exp. I, §11] for more examples. It was not clear to me, however, which of those existing examples have $F$-finite function field. On the other hand, combining Prop. 11.6 in Raynaud with Datta–Smith's proposition above gives a systematic way to construct non-excellent DVR's with $F$-finite fraction field.</p>
<p><strong>Edit.</strong> Datta–Smith recent posted <a href="https://arxiv.org/abs/1704.03628" rel="nofollow noreferrer">another preprint</a>, which discusses these questions in more detail. See §1.2 and §3 in particular.</p>
|
4,342,737 | <p>My question: If you throw a dice 5 times, what is the expected value of the square of the median of the 5 results?</p>
<p>A slightly modified question would be: If you throw a dice 5 times, what is the expected value of the median? The answer would be 3.5 by symmetry.</p>
<p>For the square, it seems to be that symmetry does not hold anymore. Is there a "smart" way to solve this problem?</p>
<p>If there isn't a smart way to solve the problem, if there a smart way to estimate the answer?</p>
| Masacroso | 173,262 | <p>Let <span class="math-container">$X_1,\ldots, X_n$</span> be i.i.d. r.v. If we order the previous list of r.v. by it values we get r.v. <span class="math-container">$X_{(1)},\ldots ,X_{(n)}$</span> named the ranks of the list <span class="math-container">$X_1,\ldots,X_n$</span>. Now, observe that the distribution of the <span class="math-container">$k$</span>-th rank of <span class="math-container">$n$</span> i.i.d. r.v. is given by</p>
<p><span class="math-container">$$
\Pr [X_{(k)}\leqslant c]=\Pr \left[\bigcup_{r\geqslant k}(\{X_{(j)}\leqslant c, j\leqslant r\}\cap \{X_{(j)}>c,j\geqslant r+1\})\right]\\
=\Pr \left[\bigcup_{r\geqslant k}\bigcup_{A\in \mathcal{A}_r}(\{X_j\leqslant c, j\in A\}\cap \{X_j>c,j\in A^\complement \})\right]\\
=\sum_{r\geqslant k}\binom{n}{r}F_{X_1}(c)^r(1-F_{X_1}(c))^{n-r}
$$</span></p>
<p>with the convention that <span class="math-container">$0^0:=1$</span> and where <span class="math-container">$\mathcal{A}_r:=\{B\subset \{1,\ldots,n\}: |B|=r\}$</span>. Now observe that the median, <span class="math-container">$M$</span>, of five i.i.d. r.v. is just <span class="math-container">$X_{(3)}$</span> and</p>
<p><span class="math-container">$$
\mathrm{E}[M^2]=\sum_{k= 1}^6 k^2p_M(k)=\sum_{k=1}^{6}k^2(F_M(k)-F_M(k-1))
$$</span></p>
<p>Therefore, putting all together and using a CAS we get the result <span class="math-container">$\mathrm{E}[M^2]=\frac{2207}{162}\approx 13.62$</span>.</p>
|
2,949,789 | <p>Suppose I have some function <span class="math-container">$V(x)=x+log(c)$</span>, where <span class="math-container">$x$</span> is a continuous random variable and <span class="math-container">$c$</span> a constant bounded on <span class="math-container">$[0,1]$</span>. I have some queries regarding the following:</p>
<p>i) May the above function be differentiated in the conventional manner w.r.t <span class="math-container">$x$</span> even though the resultant derivative will be random?</p>
<p>ii) How may I determine if <span class="math-container">$V(x)$</span> is differentiable?</p>
<p>iii) What is the derivative of <span class="math-container">$V(x)$</span>?</p>
<p>Note: <span class="math-container">$x$</span> is bounded on <span class="math-container">$[a,b]$</span></p>
<p>Any help would be greatly appreciated. Thank You</p>
| Community | -1 | <p><span class="math-container">$$\frac{dV(x)}{dx}=1$$</span></p>
<p>Randomness of <span class="math-container">$x$</span> does not matter. (And in this particular case, even the value of <span class="math-container">$x$</span> does not matter.)</p>
|
67,513 | <p>When processing a larger Dataset I came up do a point where I want to form a dataset with culumn heads from an intermediate structure. Here is an example of this structure:</p>
<pre><code>test = {<|"name" -> "alpha",
"group" -> "one"|> -> {<|"value" -> 459|>}, <|"name" -> "beta",
"group" -> "two"|> -> {<|"value" -> -338|>}, <|"name" -> "gamma",
"group" -> "two"|> -> {<|"value" -> 363|>}};
</code></pre>
<p>making a dataset:</p>
<pre><code>assoc = Association@test; Dataset@assoc
</code></pre>
<p>this gives:</p>
<p><img src="https://i.stack.imgur.com/9fH3r.png" alt="enter image description here"></p>
<p>Now I´m searching a way to make this a dataset with column-heads "name", "group" and "value". Maybe I´m blind for a moment, but I could not manage it??? Can anyone give me a hint?</p>
| Karsten 7. | 18,476 | <pre><code>test = {<|"name" -> "alpha", "group" -> "one"|> -> {<|"value" -> 459|>},
<|"name" -> "beta", "group" -> "two"|> -> {<|"value" -> -338|>},
<|"name" -> "gamma", "group" -> "two"|> -> {<|"value" -> 363|>}};
Association /@ Flatten /@ Normal[test /. Rule -> List] // Dataset
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/kDidv.png" alt="Dataset"></p>
</blockquote>
|
67,513 | <p>When processing a larger Dataset I came up do a point where I want to form a dataset with culumn heads from an intermediate structure. Here is an example of this structure:</p>
<pre><code>test = {<|"name" -> "alpha",
"group" -> "one"|> -> {<|"value" -> 459|>}, <|"name" -> "beta",
"group" -> "two"|> -> {<|"value" -> -338|>}, <|"name" -> "gamma",
"group" -> "two"|> -> {<|"value" -> 363|>}};
</code></pre>
<p>making a dataset:</p>
<pre><code>assoc = Association@test; Dataset@assoc
</code></pre>
<p>this gives:</p>
<p><img src="https://i.stack.imgur.com/9fH3r.png" alt="enter image description here"></p>
<p>Now I´m searching a way to make this a dataset with column-heads "name", "group" and "value". Maybe I´m blind for a moment, but I could not manage it??? Can anyone give me a hint?</p>
| kglr | 125 | <pre><code>Dataset[Join@@@({#,Join@@#2}&@@@test)]
(* or Dataset[Join@@@({#,Sequence@@#2}&@@@test)] *)
</code></pre>
<p><img src="https://i.stack.imgur.com/NMa7J.png" alt="enter image description here"></p>
|
74,188 | <blockquote>
<p>Let <span class="math-container">$a,c \in \mathbb R$</span> with <span class="math-container">$a \neq 0$</span>, and let <span class="math-container">$b \in \mathbb C$</span>. Define <span class="math-container">$$S=\{z\in \mathbb C: az\bar{z}+b\bar{z}+\bar{b}z+c=0\}.$$</span></p>
<p>a. Show that <span class="math-container">$S$</span> is a circle, if <span class="math-container">$|b|^2 > ac$</span>. Determine its centre and radius.<br />
b. What is <span class="math-container">$S$</span> if <span class="math-container">$a=0$</span> and <span class="math-container">$b \neq 0$</span>?</p>
</blockquote>
<p>How would I get started in this? I'm completely stuck. Most appreciated.</p>
| Gerry Myerson | 8,269 | <p>One way is to work your way backward to the problem from what you know about circles. A circle has a center and a radius. Let's call the center $w$, the radius, $r$. The circle is all the points $z$ whose distance from $w$ is $r$. The square of the distance between two points $u$ and $v$ in the complex plane is $|u-v|^2$. If $s$ is a complex number, then $|s|^2=s\overline s$. If you put all that together, you should come up with a formula looking very similar to the one in the problem, mostly just with different letters. </p>
|
4,046,532 | <p><strong>QUESTION 1:</strong> Let <span class="math-container">$f, g: S\rightarrow \mathbb{R}^m$</span> be differentiable vector-valued functions and let <span class="math-container">$\lambda\in \mathbb{R}$</span>. Prove that the function <span class="math-container">$(f+g):S\rightarrow \mathbb{R}^m$</span> is also differentiable.
<strong>PROOF:</strong> Let <span class="math-container">$X:U\rightarrow S$</span> be a parametrization of <span class="math-container">$S$</span>. Then <span class="math-container">$$(f+g)\circ X= f\circ X + g\circ X$$</span> which is differentiable, as a sum of differentiable functions.</p>
<p><strong>MY DOUBT IN QUESTION 1:</strong> How can I show in a step by step way that the sum above is distributive with the composition maps?</p>
<p><strong>QUESTION 2:</strong> Suppose that a surface <span class="math-container">$S$</span> is a union <span class="math-container">$S=\displaystyle\bigcup_{i\in I} S_i$</span>, where each <span class="math-container">$S_i$</span> is open. If <span class="math-container">$f:S\rightarrow \mathbb{R}^m$</span> is a map such that each <span class="math-container">$f|_{S_i}:S_i\rightarrow \mathbb{R}^m$</span> is differentiable, prove that <span class="math-container">$f$</span> is differentiable.</p>
<p><strong>MY ATTEMPT:</strong> We have already proved that open sets in surfaces are also surfaces. Defining <span class="math-container">$f|_{S_i}:S_i\rightarrow \mathbb{R}^m$</span> as <span class="math-container">$f_i:S_i\rightarrow \mathbb{R}^m$</span> such that <span class="math-container">$f=\sum f_i$</span>, considering <span class="math-container">$X:U\rightarrow S$</span> a parametrization of S (which is a surface), then we can write <span class="math-container">$(f_1+\cdots +f_n)\circ X= f_1\circ X+\cdots +f_n\circ X$</span> is differentiable, this is <span class="math-container">$f$</span> is differentiable.</p>
<p><strong>MY DOUBT IN QUESTION 2:</strong> I'm not sure if I can interpret the union as a sum. Would you help me?</p>
| Rafael | 894,475 | <p>Question 1. The definition of sum of functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span> is a function <span class="math-container">$(f+g):S\rightarrow \mathbb{R}^m$</span>, which can be described with a formula <span class="math-container">$(f+g)(s)=f(s)+g(s)$</span>. It is correct since <span class="math-container">$\mathbb{R}^m$</span> is a vector space. To show that functions <span class="math-container">$(f+g)\circ X$</span> and <span class="math-container">$f\circ X+g\circ X$</span> you need to show that they are pointwisely equal on <span class="math-container">$U$</span>. Indeed, <span class="math-container">\begin{equation*}((f+g)\circ X)(u)=(f+g)(X(u))=f(X(u))+g(X(u))=
(f\circ X)(u)+(g\circ X)(u)=(f\circ X+g\circ X)(u),\end{equation*}</span>
where I used the definition of composition of functions. Here you have no more than definitions.</p>
|
2,294,997 | <p>How to prove that $\displaystyle 0,02<\int_0^1 \frac{x^7}{(e^x+e^{-x})\sqrt{1+x^2}}dx<0,05$? I tried to use mean value theorems, but i failed.</p>
| Mark Viola | 218,419 | <p>HINT:</p>
<p>Note that $2\le e^x+e^{-x}\le e+e^{-1}$ and $x\le \sqrt{1+x^2}\le \sqrt{2}$.</p>
|
964,372 | <p>I have a general question.</p>
<p>If there is a matrix which is inverse and I multiply it by other matrixs which are inverse.
Will the result already be reverse matrix?</p>
<p>My intonation says is correct, but I'm not sure how to prove it.</p>
<p>Any ideas? Thanks.</p>
| Petite Etincelle | 100,564 | <p>When you have $n$ people, consider what you do with the last person.</p>
<p>If you let him alone, you have $A_{n-1}$ ways to group the other $n-1$ people.</p>
<p>If you group him with someone else, you have $n-1$ ways to do that. And after that you need to group the other $n-2$ people. That's where comes from $(n-1)A_{n-2}$</p>
|
1,828,042 | <p>This is my first question on this site, and this question may sound disturbing. My apologies, but I truly need some advice on this.</p>
<p>I am a sophomore math major at a fairly good math department (top 20 in the U.S.), and after taking some upper-level math courses (second courses in abstract algebra and real analysis, differential geometry, etc), I can say that I genuinely like math, and if I have A BIT chance to succeed, I will go to graduate school and choose math research as my career.</p>
<p>However, this is exactly the thing that I am afraid of. My grades on the courses are mediocre (my GPA for math courses is around 3.7), and for the courses I got A's, I had to work very hard, much harder than others to get the same result, and I often get confused in many of the classes, while the others understand the material quickly and could answer professor's questions, and at the same time I didn't even understand what the professor was really asking. I really wonder, if I have to work hard even on undergraduate courses, does that mean I am not naturally smart enough for more advanced math, especially compared to everyone else in my class? Can I even survive graduate level math if I even sometimes struggle with undergraduate courses? I always believe that adequate mathematicians could do well in their undergraduate courses easily. In my case, even if I work very hard, I forget definitions/theorems easily and then of course forget how to use them to solve problems.</p>
<p>Is it still worth to try if I am significantly behind the regular level and have to work hard even for undergraduate courses, providing that there are a lot of smart people who can understand them instantly. This feeling hurts me a lot, especially when I am struggling with something in math, I always feel I am a useless trash and ask myself why I am so stupid?</p>
<p>I thought about talking to my professors about this issue, but I find this too embarrassing to start. I am really afraid that if I ask them this question, they may tell me the truth in person that "you are really not smart enough to go to graduate school".</p>
<p>So how can I tell if it is still worth for me to think about this path, or I should realize that I have no chance to succeed and give up now? I appreciate encouraging comments, but please, please be honest on this case because it is really important for my future plan. Thanks again for your advice, and I am really grateful.</p>
| treble | 24,837 | <p>Hopefully I can answer before the tide of "talk to someone who knows you personally" and "this question is off topic" rolls in and your question is inevitably closed. It is true that you should talk to someone who knows you better, but I can give you some general advice that is better than just "follow your heart."</p>
<p>First, you should know that professors see students like yourself, who have a low opinion of their own ability, all of the time. We get questions like "am I cut out for this or that?" quite often. Your situation is not as unique as you might think, which, I hope, is comforting to you.</p>
<p>Here are some issues to consider.</p>
<p>Not all graduate schools are the same, and not all career paths are the same. As a Ph.D. in mathematics, you might become: A career lecturer at a large public university, a research mathematician, a professor at a small liberal arts college, a teacher in the math dept. of a private high school, a simulation/modeling expert in thermodynamics for a large oil conglomerate, a control systems analyst for Boeing, a data analyst for Facebook, a quantitative analyst at a large wall street firm, a mathematical ecologist for the united states government, etc. etc. etc. All of these areas are well served by mathematics Ph.D.s but require hugely different <em>skill sets</em> that you develop while you are in graduate school, and after you leave graduate school.</p>
<p>There are many different graduate schools. Some you may be well prepared for, others you may not be. At some well-known graduate schools, we have people take real analysis and other upper level undergraduate courses during their first year of study. There is no shame in this! The point is whether you love what you are doing and whether or not you are making <em>consistent, quantifiable progress.</em> If you are working really hard and not getting anywhere, that is bad and you should consider a career shift. If you are working hard and making progress and love what you are doing, then for goodness sake keep doing it!</p>
<p>Here is how you should approach your professors. Ask direct questions that have answers. Here are some examples: Based on my performance in your class, do you feel that I would be a successful first year student in graduate school 3 years from now, supposing that I make the same level of progress that I've been doing? If so, do you have recommendations for specific schools? Your professor might hear this question and go off on a tangent, or become alerted to something else in your tone, etc... and give you more or less advice than you wanted. But the fact that you asked a specific question is much better than "Am I cut out for grad school?", which is a question that essentially has no answer.</p>
<p>In sum, your education is not wasted. <em>Keep learning.</em> You will thank yourself in 10 years.</p>
|
2,165,296 | <p>Can every separable Banach space be isometrically embedded in $l^2$ ? Or at least in $l^p$ for some $1\le p<\infty$ ? </p>
<p>I only know that any separable Banach space is isometrically isomorphic to a linear subspace of $l^{\infty}$.</p>
<p>Please help . Thanks in advance </p>
| user90369 | 332,823 | <p>$\displaystyle \frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n\enspace$ for $\enspace-1<x<1$ , proof by multiplication with $1-x$ . </p>
<p>One derivation for $x$ gives $\enspace\displaystyle \frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}$ . </p>
<p>With $x:=-2t$ and therefore $\enspace -\frac{1}{2}< t< \frac{1}{2}$ follows:</p>
<p>$$\displaystyle \frac{1}{(1+2t)^2}=\sum\limits_{n=1}^\infty n(-2t)^{n-1}$$ </p>
|
1,212,000 | <p>I was trying to solve this square root problem, but I seem not to understand some basics. </p>
<p>Here is the problem.</p>
<p>$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2$$</p>
<p>The solution is as follows:</p>
<p>$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^2 = \Bigg(\frac{3}{2} - \sqrt{2} - 1 + \sqrt{2}\Bigg)^2 = \bigg(\frac{1}{2}\bigg)^2 = \frac{1}{4}$$</p>
<p>Now, what I don't understand is how the left part of the problem becomes:
$$\frac{3}{2} - \sqrt{2}$$</p>
<p>Because I thought that $$\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2}$$
equals to $$\bigg(\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2\bigg)^{\frac{1}{2}}$$
Which becomes $$\sqrt{2} - \frac{3}{2}$$</p>
<p>But as you can see I'm wrong. </p>
<p>I think that there is a step involving absolute value that I oversee/don't understand.
So could you please explain by which property or rule of square root is this problem solved? </p>
<p>Thanks in advance</p>
| Community | -1 | <p>This is a mistake that might have been avoided by being aware of the numerical values of some of the expressions involved. A very rough approximation for $\sqrt{2}$ is $1.41$. Clearly $\frac{3}{2} = 1.5$. Then $\sqrt{2} - \frac{3}{2} \approx -0.09$, and that squares to $0.0081$, and the square root of that is $0.09$, not $-0.09$.</p>
<p>At no point in this problem is it actually necessary to take the square root of a negative number. But the possibility of going down that path, even if erroneously, would perhaps have alerted you that some things in this problem should be evaluated rather than canceled. Thus, $$\left(\sqrt{2} - \frac{3}{2}\right)^2 = \left(-\frac{3}{2} + \sqrt{2}\right)^2 = \frac{17}{4} - 3 \sqrt{2} \neq -\frac{3}{2} + \sqrt{2}.$$</p>
<p>I'm not suggesting that you always run through problems with crude numerical approximations. But it can be a good way to check you're on the right track.</p>
|
2,661,210 | <p>Let $a_{1}, \dots, a_{n}$ be real numbers not all zero; let $b_{1},\dots, b_{n}$ be real numbers; let $\sum_{1}^{n}b_{i} \neq 0$. Then does there exist real numbers $w_{1},\dots, w_{n} > 0$ such that
$$
\frac{\sum_{1}^{n}w_{i}a_{i}}{\sum_{1}^{n}w_{i}b_{i}} > \frac{\sum_{1}^{n}a_{i}}{\sum_{1}^{n}b_{i}}?
$$
Some function theory results seem prominent. But it seems that perhaps such a result is not in my current set of working knowledge. </p>
| mucciolo | 222,084 | <p>$\DeclareMathOperator{\spn}{span}\DeclareMathOperator{\img}{Im}$There is really some linear algebra hidden there. Note that $\sum_{i=1}^{n}w_{i}a_{i}$ is a linear functional on $\mathbb{R}^n$ with respect to $w$. That is, $\alpha: \mathbb{R}^n \to \mathbb{R} : (w_1, \cdots, w_n) \mapsto \sum_{i=1}^{n}w_{i}a_{i}$ is linear. Likewise, let $\beta: \mathbb{R}^n \to \mathbb{R} : (w_1, \cdots, w_n) \mapsto \sum_{i=1}^{n}w_{i}b_{i}$. Also let $w = (w_1, \cdots, w_n)$ and $f : D \to \mathbb{R} : w \mapsto \frac{\alpha(w)}{\beta(w)}$. Thus your question in this linear algebra framework is</p>
<blockquote>
<p>Let $\alpha, \beta \in \mathcal{L}(\mathbb{R}^n, \mathbb{R})$ such that $\alpha \neq 0$ and $\beta(\vec{1}) \neq 0 $. Then there exists $w \in \mathbb{R}^n$ such that $$f(w) > f(\vec{1})? \tag{1}$$</p>
</blockquote>
<p>The domain of $f$ (undetermined so far) is simply all $w \in \mathbb{R}^n$ such that $\beta(w) \neq 0$, that is, $D = \mathbb{R}^n \setminus \ker(\beta)$. By hypothesis we have $\beta(\vec{1}) \neq 0$, therefore $D = \mathbb{R}^n \setminus \spn(\vec{1})^{\perp}$.</p>
<p>Furthermore, an additional condition is required. It is necessary that $\alpha$ and $\beta$ are linearly independent, meaning that must <em>not</em> exist a $\lambda \in \mathbb{R}$ such that $\alpha = \lambda \beta$, because otherwise $f$ would be constant and $(1)$ would never be satisfied. This is equivalent to require $\ker(\alpha) \neq \ker(\beta) = \spn(\vec{1})^{\perp}$.</p>
<p>Now, since $\img(\alpha) = \img(\beta) = \mathbb{R}$ and for every sequence $(x_n)_{n \in \mathbb{N}}$ in $D$ converging to a non-null vector of $\ker(\beta)$ we have that $\lim\limits_{n \to \infty} |f(x_n)| = \infty$, we can construct $w$ that satisfies $(1)$ as following.</p>
<p>If $f(1) = 0$ just take any $w \in f^{-1}(\mathbb{R}_{>0})$. Otherwise if $f(1) > 0$ consider $A = f^{-1}(\mathbb{R}_{>0})$ or $A = f^{-1}(\mathbb{R}_{<0})$ if $f(1) < 0$. Take any non-null element of $\ker(\beta) \cap \partial A$, say $\ell$, and a sequence $(x_n)_{n \in \mathbb{N}}$ in $A$ converging to $\ell$. Because $\lim\limits_{n \to \infty} |f(x_n)| = \infty$, there exists $N \in \mathbb{N}$ such that for every $n > N$ it follows that $f(x_n) > f(1)$. As we wanted to show.</p>
<p>(In fact, since without loss of generality we can substitute that sequence for a curve, there exists an uncountable number of such $w$'s.)</p>
<p>For example, take $\alpha(x,y) = 4x-2y$ and $\beta(x,y) = x+y$. So $f(1) = 1$. Thus
$$A = f^{-1}(\mathbb{R}_{>0}) = \{ (x+y, x-y) : x,y \in \mathbb{R}_{>0} \}$$
$$\partial A = \{ (x, x) : x \in \mathbb{R}_{\ge 0} \} \cup \{ (x, -x) : x \in \mathbb{R}_{\ge 0} \}$$
and
$$\ker(\beta) = \{ (x, -x) : x \in \mathbb{R} \}$$</p>
<p>Then
$$ \ker(\beta) \cap \partial A = \{ (x, -x) : x \in \mathbb{R}_{\ge 0} \} $$</p>
<p>Now take $\ell = (1, -1) \in \ker(\beta) \cap \partial A$, $x_0 = (2,0)$ and $x_n = \frac{1}{n} x_0 + (1-\frac{1}{n}) \ell = (\frac{1}{n} + 1, \frac{1}{n} - 1)$. So for all $n \in \mathbb{N}$
$$f(x_n) = \frac{\frac{4}{n} + 4 - \frac{2}{n} + 2}{\frac{2}{n}} = \frac{\frac{2}{n} + 6}{\frac{2}{n}} = \frac{\frac{1}{n} + 3}{\frac{1}{n}} = 1 + 3n > 1 = f(\vec{1})$$</p>
|
4,629,922 | <p>I'm reading about the method of two-timing in section 7.6 of <em>Nonlinear Dynamics and Chaos</em> by Strogatz, and I have some questions about how to make this concept rigorous. In this section the book considers equations of the form
<span class="math-container">$$ \hspace{4.5cm} \ddot{x} + x + \epsilon h(x,\dot{x}) = 0 \hspace{4.5cm} (1) $$</span>
where <span class="math-container">$0 \leq \epsilon \ll 1$</span>, <span class="math-container">$x: \mathbb{R} \to \mathbb{R}$</span>, and <span class="math-container">$h: \mathbb{R}^2 \to \mathbb{R}$</span> is an arbitrary smooth function. The section first covers regular perturbation theory, which I feel okay with. But then author gets to the method of two-timing for ODEs that exhibit multiple time scales, and this is where the explanation starts to feel a little hand-wavey. Here is Strogatz's description of the method:</p>
<blockquote>
<p>To apply two-timing to (1), let <span class="math-container">$\tau = t$</span> denote the fast <span class="math-container">$O(1)$</span> time, and let <span class="math-container">$T = \epsilon t$</span> denote the slow time. We'll treat these two times as if they were <em>independent</em> variables. In particular, functions of the slow time <span class="math-container">$T$</span> will be regarded as <em>constants</em> on the fast time scale <span class="math-container">$\tau$</span>. It's hard to justify this idea rigorously, but it works. [Arg! Show me the rigor!]...Next we turn to the mechanics of the method. We expand the solution of (1) as a series
<span class="math-container">$$ \hspace{3cm} x(t,\epsilon) = x_0(\tau, T) + \epsilon x_1(\tau, T) + O(\epsilon^2). \hspace{3cm} (2) $$</span>
The time derivatives in (1) are transformed according to the chain rule:
<span class="math-container">\begin{align*}
\hspace{3cm} \dot{x} = \frac{dx}{dt} = \frac{\partial x}{\partial \tau} + \frac{\partial x}{\partial T}
\frac{\partial T}{\partial t} = \frac{\partial x}{\partial \tau} + \epsilon \frac{\partial x}{\partial T} \hspace{3cm} (3)
\end{align*}</span>
A subscript notation for differentiation is more compact; thus we write (3) as
<span class="math-container">$$ \hspace{5.1cm} \dot{x} = \partial_{\tau} x + \epsilon \partial_T x. \hspace{5.1cm} (4) $$</span></p>
<p>After substituting (2) into (4) and collecting powers of <span class="math-container">$\epsilon$</span>, we find
<span class="math-container">$$ \hspace{3.3cm} \dot{x} = \partial_{\tau} x_0 + \epsilon (\partial_T x_0 + \partial_{\tau} x_1) + O(\epsilon^2). \hspace{3.3cm} (5) $$</span>
Similarly,
<span class="math-container">$$ \hspace{3cm} \ddot{x} = \partial_{\tau \tau} x_0 + \epsilon(\partial_{\tau \tau} x_1 + 2 \partial_{T \tau} x_0) + O(\epsilon^2). \hspace{3cm} (6) $$</span></p>
</blockquote>
<p><strong>My main question</strong>: How can we make the above ideas rigorous? <br/></p>
<p>My attempt is as follows: Given the ODE (1), we first <em>assume</em> that the solution <span class="math-container">$x(t;\epsilon)$</span> can be expressed as
<span class="math-container">$$ x(t;\epsilon) = X(t,\epsilon t)$$</span></p>
<p>for some function <span class="math-container">$X: \mathbb{R}^2 \to \mathbb{R}$</span>. Now let <span class="math-container">$\tau, T: \mathbb{R} \to \mathbb{R}$</span> be functions defined by <span class="math-container">$\tau(t) := t$</span> and <span class="math-container">$T(t;\epsilon) := \epsilon t$</span> for all <span class="math-container">$t \in \mathbb{R}, \epsilon > 0$</span>. We then have <span class="math-container">$X(\tau(t),T(t)) = x(t;\epsilon)$</span> for all <span class="math-container">$t \in \mathbb{R}, \epsilon > 0$</span>. The chain rule computation is then straightforward:<br />
<span class="math-container">\begin{align*}
\dot{X} := \frac{dX}{dt} &= \frac{\partial X}{\partial \tau} \frac{d\tau}{dt} + \frac{\partial X}{\partial T} \frac{dT}{dt} = \frac{\partial X}{\partial \tau} + \frac{\partial X}{\partial T} \epsilon.
\end{align*}</span>
The subsequent equations (4)-(6) should now be the same, just with <span class="math-container">$x$</span> replaced by <span class="math-container">$X$</span>. (Is my interpretation correct so far?) <br/></p>
<p>The Taylor expansion part is where I'm a bit perplexed: It seems that (2) is a Taylor expansion of the function <span class="math-container">$\epsilon \mapsto x(t; \epsilon)$</span>, or equivalently, <span class="math-container">$\epsilon \mapsto X(\tau(t), T(t))$</span>. Now if we had some arbitrary smooth function <span class="math-container">$f: \mathbb{R}^2 \to \mathbb{R}$</span> indexed by some parameter <span class="math-container">$\epsilon$</span>, then I agree that the map <span class="math-container">$\epsilon \mapsto f(x,y; \epsilon)$</span> has some expansion
<span class="math-container">$$ f(x,y; \epsilon) = f_0(x,y) + f_1(x,y) \epsilon + f_2(x,y) \epsilon^2 + \cdots, $$</span></p>
<p>(provided that it is smooth), where the coefficient functions <span class="math-container">$f_0, f_1,\ldots$</span> do not depend on <span class="math-container">$\epsilon$</span>. But my problem with (2) is that the "coefficients" of the Taylor series, namely <span class="math-container">$x_0(\tau,T), x_1(\tau,T),\ldots$</span> are functions of <span class="math-container">$\epsilon$</span>, since <span class="math-container">$T$</span> depends on <span class="math-container">$\epsilon$</span>. How then is this a legitimate Taylor expansion? <br/></p>
<p>Also, an additional question: Is there a way to know a prior if the solution <span class="math-container">$x = x(t;\epsilon)$</span> to some ODE <span class="math-container">$\dot{x} = f(x)$</span> can be expressed in the form <span class="math-container">$x(t;\epsilon) = X(t, \epsilon t)$</span>? Or do we generally make such an assumption based on intuition? In one of the examples that Strogatz uses, the ODE has the analytic solution <span class="math-container">$x(t) = (1-\epsilon^2)^{-1/2} e^{-\epsilon t} \cos((1-\epsilon^2)^{1/2} t)$</span>, so clearly <span class="math-container">$X(t_1, t_2) := (1-\epsilon^2)^{-1/2} e^{-t_2} \cos((1-\epsilon^2)^{1/2} t_1)$</span> satisfies the requirement. But is there a way to see this a priori (i.e., if we couldn't analytically solve the ODE)?</p>
<p>Any insights would be greatly appreciated. This two-timing concept is rather confusing to me, and I think seeing the details spelled out very rigorously would help my understanding.</p>
| eyeballfrog | 395,748 | <p>So firstly, they really should be writing <span class="math-container">$X(t, \epsilon t; \epsilon)$</span>. Without that the expansion in terms of <span class="math-container">$x_i(\tau, T)$</span> doesn't quite make sense. But as you note, it <em>does</em> make sense that for any function <span class="math-container">$X(\tau, T; \epsilon)$</span>, there exists some sequence of <span class="math-container">$x_i$</span> such that
<span class="math-container">$$
X(\tau, T;\epsilon) = x_0(\tau, T) + \epsilon x_1(\tau, T) + \epsilon^2x_2(\tau, T) + ...
$$</span></p>
<p>So here's what we're going to do with this. We substitute this series into the differential equation using <span class="math-container">$x(t;\epsilon) = X(t, \epsilon t;\epsilon)$</span>, then collect orders of <span class="math-container">$\epsilon$</span>. This will give some series of relations among the <span class="math-container">$x_i$</span> and their <span class="math-container">$\tau$</span> and <span class="math-container">$T$</span> derivatives when evaluated at <span class="math-container">$\tau = t$</span>, <span class="math-container">$T = \epsilon t$</span>. Then we add the additional requirement that the <span class="math-container">$x_i$</span> and their derivatives satisfy those relations for <em>every</em> <span class="math-container">$\tau, T$</span>, which we can do because any such set of <span class="math-container">$x_i$</span> will of course satisfy the relations at <span class="math-container">$\tau = t, T = \epsilon t$</span>. And now we have a set of simpler differential equations entirely in terms of <span class="math-container">$\tau$</span> and <span class="math-container">$T$</span>, with no <span class="math-container">$\epsilon$</span>s or <span class="math-container">$t$</span>s around. If things have gone right, these can then be solved recursively order-by-order.</p>
|
4,629,922 | <p>I'm reading about the method of two-timing in section 7.6 of <em>Nonlinear Dynamics and Chaos</em> by Strogatz, and I have some questions about how to make this concept rigorous. In this section the book considers equations of the form
<span class="math-container">$$ \hspace{4.5cm} \ddot{x} + x + \epsilon h(x,\dot{x}) = 0 \hspace{4.5cm} (1) $$</span>
where <span class="math-container">$0 \leq \epsilon \ll 1$</span>, <span class="math-container">$x: \mathbb{R} \to \mathbb{R}$</span>, and <span class="math-container">$h: \mathbb{R}^2 \to \mathbb{R}$</span> is an arbitrary smooth function. The section first covers regular perturbation theory, which I feel okay with. But then author gets to the method of two-timing for ODEs that exhibit multiple time scales, and this is where the explanation starts to feel a little hand-wavey. Here is Strogatz's description of the method:</p>
<blockquote>
<p>To apply two-timing to (1), let <span class="math-container">$\tau = t$</span> denote the fast <span class="math-container">$O(1)$</span> time, and let <span class="math-container">$T = \epsilon t$</span> denote the slow time. We'll treat these two times as if they were <em>independent</em> variables. In particular, functions of the slow time <span class="math-container">$T$</span> will be regarded as <em>constants</em> on the fast time scale <span class="math-container">$\tau$</span>. It's hard to justify this idea rigorously, but it works. [Arg! Show me the rigor!]...Next we turn to the mechanics of the method. We expand the solution of (1) as a series
<span class="math-container">$$ \hspace{3cm} x(t,\epsilon) = x_0(\tau, T) + \epsilon x_1(\tau, T) + O(\epsilon^2). \hspace{3cm} (2) $$</span>
The time derivatives in (1) are transformed according to the chain rule:
<span class="math-container">\begin{align*}
\hspace{3cm} \dot{x} = \frac{dx}{dt} = \frac{\partial x}{\partial \tau} + \frac{\partial x}{\partial T}
\frac{\partial T}{\partial t} = \frac{\partial x}{\partial \tau} + \epsilon \frac{\partial x}{\partial T} \hspace{3cm} (3)
\end{align*}</span>
A subscript notation for differentiation is more compact; thus we write (3) as
<span class="math-container">$$ \hspace{5.1cm} \dot{x} = \partial_{\tau} x + \epsilon \partial_T x. \hspace{5.1cm} (4) $$</span></p>
<p>After substituting (2) into (4) and collecting powers of <span class="math-container">$\epsilon$</span>, we find
<span class="math-container">$$ \hspace{3.3cm} \dot{x} = \partial_{\tau} x_0 + \epsilon (\partial_T x_0 + \partial_{\tau} x_1) + O(\epsilon^2). \hspace{3.3cm} (5) $$</span>
Similarly,
<span class="math-container">$$ \hspace{3cm} \ddot{x} = \partial_{\tau \tau} x_0 + \epsilon(\partial_{\tau \tau} x_1 + 2 \partial_{T \tau} x_0) + O(\epsilon^2). \hspace{3cm} (6) $$</span></p>
</blockquote>
<p><strong>My main question</strong>: How can we make the above ideas rigorous? <br/></p>
<p>My attempt is as follows: Given the ODE (1), we first <em>assume</em> that the solution <span class="math-container">$x(t;\epsilon)$</span> can be expressed as
<span class="math-container">$$ x(t;\epsilon) = X(t,\epsilon t)$$</span></p>
<p>for some function <span class="math-container">$X: \mathbb{R}^2 \to \mathbb{R}$</span>. Now let <span class="math-container">$\tau, T: \mathbb{R} \to \mathbb{R}$</span> be functions defined by <span class="math-container">$\tau(t) := t$</span> and <span class="math-container">$T(t;\epsilon) := \epsilon t$</span> for all <span class="math-container">$t \in \mathbb{R}, \epsilon > 0$</span>. We then have <span class="math-container">$X(\tau(t),T(t)) = x(t;\epsilon)$</span> for all <span class="math-container">$t \in \mathbb{R}, \epsilon > 0$</span>. The chain rule computation is then straightforward:<br />
<span class="math-container">\begin{align*}
\dot{X} := \frac{dX}{dt} &= \frac{\partial X}{\partial \tau} \frac{d\tau}{dt} + \frac{\partial X}{\partial T} \frac{dT}{dt} = \frac{\partial X}{\partial \tau} + \frac{\partial X}{\partial T} \epsilon.
\end{align*}</span>
The subsequent equations (4)-(6) should now be the same, just with <span class="math-container">$x$</span> replaced by <span class="math-container">$X$</span>. (Is my interpretation correct so far?) <br/></p>
<p>The Taylor expansion part is where I'm a bit perplexed: It seems that (2) is a Taylor expansion of the function <span class="math-container">$\epsilon \mapsto x(t; \epsilon)$</span>, or equivalently, <span class="math-container">$\epsilon \mapsto X(\tau(t), T(t))$</span>. Now if we had some arbitrary smooth function <span class="math-container">$f: \mathbb{R}^2 \to \mathbb{R}$</span> indexed by some parameter <span class="math-container">$\epsilon$</span>, then I agree that the map <span class="math-container">$\epsilon \mapsto f(x,y; \epsilon)$</span> has some expansion
<span class="math-container">$$ f(x,y; \epsilon) = f_0(x,y) + f_1(x,y) \epsilon + f_2(x,y) \epsilon^2 + \cdots, $$</span></p>
<p>(provided that it is smooth), where the coefficient functions <span class="math-container">$f_0, f_1,\ldots$</span> do not depend on <span class="math-container">$\epsilon$</span>. But my problem with (2) is that the "coefficients" of the Taylor series, namely <span class="math-container">$x_0(\tau,T), x_1(\tau,T),\ldots$</span> are functions of <span class="math-container">$\epsilon$</span>, since <span class="math-container">$T$</span> depends on <span class="math-container">$\epsilon$</span>. How then is this a legitimate Taylor expansion? <br/></p>
<p>Also, an additional question: Is there a way to know a prior if the solution <span class="math-container">$x = x(t;\epsilon)$</span> to some ODE <span class="math-container">$\dot{x} = f(x)$</span> can be expressed in the form <span class="math-container">$x(t;\epsilon) = X(t, \epsilon t)$</span>? Or do we generally make such an assumption based on intuition? In one of the examples that Strogatz uses, the ODE has the analytic solution <span class="math-container">$x(t) = (1-\epsilon^2)^{-1/2} e^{-\epsilon t} \cos((1-\epsilon^2)^{1/2} t)$</span>, so clearly <span class="math-container">$X(t_1, t_2) := (1-\epsilon^2)^{-1/2} e^{-t_2} \cos((1-\epsilon^2)^{1/2} t_1)$</span> satisfies the requirement. But is there a way to see this a priori (i.e., if we couldn't analytically solve the ODE)?</p>
<p>Any insights would be greatly appreciated. This two-timing concept is rather confusing to me, and I think seeing the details spelled out very rigorously would help my understanding.</p>
| Lutz Lehmann | 115,115 | <p>Already the simple perturbation series is a deliberate decomposition of the solution that gets its justification from its result, if it works. Of course one can make a theory of and for classes of perturbation problems where it works.</p>
<p>In the multiple time-scale expansion, the calculator gives themselves even more freedom to distribute terms in the solution. This allows to calculate directly perturbations in the time scale that otherwise are a consequence of careful post-processing.</p>
<p>For instance in <span class="math-container">$\ddot x+x+ϵx^3=0$</span> one knows that the solutions will be closed and periodic. However the simple perturbation series will produce resonance terms that are trigonometric functions with polynomial coefficients, so the perturbation series extended to <span class="math-container">$\Bbb R$</span> will diverge from a closed curve. See <a href="https://math.stackexchange.com/questions/2013417/help-with-nonlinear-ode">Help with nonlinear ODE</a></p>
<p>This can be corrected by having also a perturbation series for the frequency, see some answers in <a href="https://math.stackexchange.com/questions/3961162/help-with-understanding-duffings-oscillator">Help with understanding Duffing's oscillator</a>, <a href="https://math.stackexchange.com/questions/1474995/find-an-approximate-solution-up-to-the-order-of-epsilon">find an approximate solution, up to the order of epsilon</a></p>
<p>However in a multi-scale expansion one can use the second (third,...) scale to set the combined coefficients of the resonance terms to zero, resulting in a systematic perturbation of the time scale. I found no direct examples for the above equation, but frequently the Mathieu equation is analyzed using multiple time scales, see <a href="https://math.stackexchange.com/questions/890843/uniform-perturbative-solutions-to-the-mathieu-equation">Uniform perturbative solutions to the Mathieu equation</a>, <a href="https://math.stackexchange.com/questions/3497182/method-of-multiple-scales-on-mathieus-equation">Method of multiple scales on Mathieu's equation</a>, or the nearly circular case of the van der Pol oscillator (incomplete treatment, detail question) <a href="https://math.stackexchange.com/questions/4094687/why-is-sin3-tau-phi-not-a-secular-term-in-the-context-of-the-van-der-po?rq=1">Why is $\sin^3(\tau + \phi)$ not a Secular term in the context of the van der Pol oscillator</a></p>
|
2,220 | <p>Some of you (myself included) might remember how as a new user you struggle with finding stuff to answer, and hope to have these answers upvoted and accepted...</p>
<p>You really want that. You want to write comments, to the least but that requires 50 reputation.</p>
<p>As a result you look for old questions, possible with very good answers that were accepted, and write your own answer.</p>
<p>Now there is nothing wrong with adding more into threads like that, but it turns out that the answers added by some new users are of... "low quality content".</p>
<p>In his answer to my previous question Qiaochu said that the community should discuss and decide what is low quality and how to treat it. Since I'm starting to feel flooded with old questions, I would like to discuss this right now, so in the future we can judge what should we do about it.</p>
<p>So, what is "low quality" content, and how should we deal with it?</p>
| Jeff Atwood | 153 | <p>The general guidance I give is twofold. Ask yourself …</p>
<ol>
<li><p>Could a student (of x skill level) learn anything useful/practical from this answer?</p></li>
<li><p>Would I be embarrassed to be associated with this answer?</p></li>
</ol>
<p>(Beyond that of course if it's a purely duplicate answer, it is not of any value.)</p>
<p>The page to monitor your new users is here:</p>
<p><a href="https://math.stackexchange.com/review">https://math.stackexchange.com/review</a></p>
<p>Almost all the tabs on that page are of critical importance to monitor incoming content by new users, and the more eyes on them, the better. This works both ways, to welcome great new users to your community and perhaps .. er .. discourage .. the not-so-great new users. :)</p>
<p>Do not hesitate to flag anything that you are uncomfortable with, and as always, vote honestly on quality.</p>
|
3,212,499 | <p>I'm struggling to find a solution to this exercise:</p>
<blockquote>
<p>Consider a set of 65 girls and a set of 5 boys. Prove that there are 3
girls and 3 boys such that either every girl knows every boy or no
girl knows any of the boys.</p>
</blockquote>
<p>I know I should use the Ramsey Theorem but I have no idea on how to apply it.</p>
<p>The only solution I came out with is that there are 8 girls all knowing or not knowing 3 boys. However, this doesn't sound right to me because the exercise clearly talks about 3 girls.</p>
| Empy2 | 81,790 | <p>You only need 41 girls. Then either 21 girls each know 3 boys, or 21 each don't know 3 boys. There are ten trios of 3 boys, so one of the trios is known/not known to 3 girls.</p>
|
2,725,019 | <blockquote>
<p>Define $S := \{x ∈ \mathbb Q : x^2 ≤ 2\}$. Prove that $a:=\inf \{S\}$
satisfies $a^2 = 2$.</p>
</blockquote>
<p>Since a is a lower bound for S, we have $a^2\le 2$. if $a^2\neq 2$ then $a^2 < 2$ and we may set $\epsilon:= a^2 − 2 > 0$ and then I am not sure how to show it satisfies $a^2 = 2$.</p>
| felipeh | 73,723 | <p>It is possible to flesh out the reasoning in your question to solve a more general question. Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function, and consider the set
$$
S_f = \{x\in\mathbb{Q} \,:\, f(x) \leq 0\}.
$$
If $S_f$ is nonempty and $S_f\not=\mathbb{R}$, then $x^* = \inf S_f$ satisfies $f(x^*) = 0$. To apply this to your question, simply set $f(x) = x^2-2$.</p>
<p>To prove it, let $y^* = f(x^*)$. Suppose $y^*>0$, and let $\epsilon=y^*/2$. Because $f$ is continuous we can find $\delta$ such that for every $x\in[x^*, x^*+\delta]$,
$$|f(x)-f(x^*)|<\epsilon.$$<br>
In particular, $f(x) > y^*/2 > 0$. In particular, $[x^*,x^*+\delta]\cap S_f = \emptyset$, so $x^*+\delta$ is a larger lower bound for $S_f$. But $x^*$ is the greatest lower bound for $S_f$, so this is a contradiction. We conclude that $f(x^*)\leq 0$.</p>
<p>On the other hand, if $y^*<0$ we let $\epsilon = -y^*/2$ and again choose $\delta$ such that for every $x\in[x^*-\delta,x^*]$,
$$|f(x)-f(x^*)|<\epsilon.$$
In particular, for such $x$, $f(x) < y^*/2 < 0$. Because $\mathbb{Q}$ is dense, we can find a rational number $q\in [x^*-\delta, x^*]$ such that $f(q) < 0$. This implies that $x^*$ is not a lower bound for $S_f$, and thus cannot be the infimum. This is a contradiction, so we must have $f(x^*) \geq 0$.</p>
<p>We conclude that $f(x^*)=0$, which is what we wanted to show.</p>
|
1,072,669 | <ul>
<li><p>Let <span class="math-container">$C_1,C_2,C_3,C_4$</span> be compact convexes of <span class="math-container">$\mathbb{R}^2$</span> such that <span class="math-container">$C_1\cap C_2\cap C_3\neq\emptyset,C_1\cap C_2\cap C_4\neq\emptyset,C_1\cap C_3\cap C_4\neq\emptyset,C_2\cap C_3\cap C_4\neq\emptyset$</span>.</p>
<p>Show that <span class="math-container">$C_1\cap C_2\cap C_3\cap C_4\neq\emptyset$</span></p>
</li>
<li><p>Let <span class="math-container">$C_1,\dots,C_n$</span> be compact convexes of <span class="math-container">$\mathbb{R}^m,(m,n)\in\mathbb{N}^2$</span>. We suppose that <span class="math-container">$\displaystyle\forall 1\le j\le n,\bigcap_{1\le i\le n,i\neq j} C_i\neq\emptyset$</span>.</p>
<p>Do we have <span class="math-container">$\displaystyle\bigcap_{1\le i\le n} C_i\neq\emptyset$</span> ?</p>
</li>
</ul>
<hr />
<p>The first one seems quite logical by drawing a graph, but I can't find a proper mathematical 'rigorous' proof. I don't know what to do for the second.</p>
| Valerii Sokolov | 25,856 | <p>For second question, let $m=2, n=3$. We can easily draw tree circles each pair of which intersects and all three of them have empty intersection. So the answer is negative.</p>
<p>As for the first one, a convex set $X$ has, by definition, a property that for any two points $u, v \in X$ point $au + (1-a)v$ belongs to $X$ for any $a \in [0, 1]$. In any words, whole segment between $u$ and $v$ lays in $X$.</p>
<p>Let us take four points $p_1 \in C_2 \cap C_3 \cap C_4$, $p_2 \in C_1 \cap C_3 \cap C_4$, $p_3 \in C_1\cap C_2 \cap C_4$ and $p_4 \in C_1 \cap C_2 \cap C_3$.</p>
<p>From the definition of convex set, $[p_1, p_2] \subset C_3 \cap C_4$, $[p_1, p_3] \subset C_2 \cap C_4$, $[p_1, p_4] \subset C_2 \cap C_3$, $[p_2, p_3] \subset C_1 \cap C_4$, $[p_2, p_4] \subset C_1 \cap C_3$ and $[p_3, p_4] \subset C_1 \cap C_2$.</p>
<p>Next we consider the convex hull of these points. It can be a single point, a segment, a triangle or a quadrangle.</p>
<ul>
<li>If it is only one point - it is in the intersection of all sets</li>
<li>If it is a segment, then all its points belongs, we can assume w.l.o.g we can assume that its ends are $p_1$ and $p_2$. So $C_1\cap C_2 \supset [p_3, p_4] \subset [p_1, p_2] \subset C_3 \cap C_4$ which means that intersection is not empty.</li>
<li>For a triangle (w.l.o.g.) on $p_1, p_2, p_3$ and $p_4$ is inside of the triangle. It is easy to see that triangle's vertexes are in $C_4$, so the whole triangle belongs to $C_4$ including $p_4$ which also belongs to three other sets.</li>
<li>If the convex hull is a quadrangle, then again, w.l.o.g. we can assume that $[p_1, p_2]$ and $[p_3, p_4]$ are its diagonals. And since the quadrangle is convex, the diagonals intersects. Again it is easy to verify that intersection point belongs to all sets.</li>
</ul>
<p>q.e.d.</p>
|
3,669,269 | <p>I have learned that I can compute the moments of a random variable with this formula
<span class="math-container">$$\mu_n=\mbox{E}(X-\mbox{E}X)^n$$</span>
However, for the moment of order <span class="math-container">$1$</span> I can not use this, since I get <span class="math-container">$\mbox{E}(X-\mbox{E}X)=0$</span>. Then is this formula not correct?</p>
| Community | -1 | <p>Indeed, the first order central moment is always zero and is not used.</p>
<p>Instead you can compute the first order <em>absolute</em> central moment,</p>
<p><span class="math-container">$$E(|X-E(X)|)$$</span> which is a measure of the spread, like the variance. It is not as efficient, but is more robust.</p>
|
3,278 | <h3>What are Community Promotion Ads?</h3>
<p>Community Promotion Ads are community-vetted advertisements that will show up on the main site, in the right sidebar. The purpose of this question is the vetting process. Images of the advertisements are provided, and community voting will enable the advertisements to be shown.</p>
<h3>Why do we have Community Promotion Ads?</h3>
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|
3,278 | <h3>What are Community Promotion Ads?</h3>
<p>Community Promotion Ads are community-vetted advertisements that will show up on the main site, in the right sidebar. The purpose of this question is the vetting process. Images of the advertisements are provided, and community voting will enable the advertisements to be shown.</p>
<h3>Why do we have Community Promotion Ads?</h3>
<p>This is a method for the community to control what gets promoted to visitors on the site. For example, you might promote the following things:</p>
<ul>
<li>the site's twitter account</li>
<li>useful tools or resources for the mathematically inclined</li>
<li>interesting articles or findings for the curious</li>
<li>cool events or conferences</li>
<li>anything else your community would genuinely be interested in</li>
</ul>
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[2]: http://clickthrough-url
</code></pre>
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|
3,278 | <h3>What are Community Promotion Ads?</h3>
<p>Community Promotion Ads are community-vetted advertisements that will show up on the main site, in the right sidebar. The purpose of this question is the vetting process. Images of the advertisements are provided, and community voting will enable the advertisements to be shown.</p>
<h3>Why do we have Community Promotion Ads?</h3>
<p>This is a method for the community to control what gets promoted to visitors on the site. For example, you might promote the following things:</p>
<ul>
<li>the site's twitter account</li>
<li>useful tools or resources for the mathematically inclined</li>
<li>interesting articles or findings for the curious</li>
<li>cool events or conferences</li>
<li>anything else your community would genuinely be interested in</li>
</ul>
<p>The goal is for future visitors to find out about <em>the stuff your community deems important</em>. This also serves as a way to promote information and resources that are <em>relevant to your own community's interests</em>, both for those already in the community and those yet to join. </p>
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[1]: http://image-url
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</code></pre>
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<h3>Image requirements</h3>
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<li>Must be hosted through our standard image uploader (imgur)</li>
<li>Must be GIF or PNG</li>
<li>No animated GIFs</li>
<li>Absolute limit on file size of 150 KB</li>
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3,278 | <h3>What are Community Promotion Ads?</h3>
<p>Community Promotion Ads are community-vetted advertisements that will show up on the main site, in the right sidebar. The purpose of this question is the vetting process. Images of the advertisements are provided, and community voting will enable the advertisements to be shown.</p>
<h3>Why do we have Community Promotion Ads?</h3>
<p>This is a method for the community to control what gets promoted to visitors on the site. For example, you might promote the following things:</p>
<ul>
<li>the site's twitter account</li>
<li>useful tools or resources for the mathematically inclined</li>
<li>interesting articles or findings for the curious</li>
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<li>anything else your community would genuinely be interested in</li>
</ul>
<p>The goal is for future visitors to find out about <em>the stuff your community deems important</em>. This also serves as a way to promote information and resources that are <em>relevant to your own community's interests</em>, both for those already in the community and those yet to join. </p>
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<ol>
<li><p>All answers should be in the exact form of:</p>
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[1]: http://image-url
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</code></pre>
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<h3>Image requirements</h3>
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|
1,454,919 | <p>I am trying to understand derivative and I want to know intuitive and rigorous definitions for a curve and if derivative is lmited only to curves or not..</p>
| Surb | 154,545 | <p>A curve is the graph of an application \begin{align*}\gamma :A\subset \mathbb R&\longrightarrow \mathbb R^n\\ t&\mapsto (\alpha_1(t),...,\alpha_n(t))\end{align*}</p>
<p>where \begin{align*}
\alpha_i:A&\longrightarrow \mathbb R\\
t&\mapsto \alpha_i(t).
\end{align*}</p>
<p>And yes, derivative are limited to curve. For a surface or any other space... you will talk about differential and not derivative. For a curve, the derivative and the differential coincident. </p>
|
1,820,036 | <p>I'd be thankful if some could explain to me why the second equality is true...
I just can't figure it out. Maybe it's something really simple I am missing?</p>
<blockquote>
<p>$\displaystyle\lim_{\epsilon\to0}\frac{\det(Id+\epsilon H)-\det(Id)}{\epsilon}=\displaystyle\lim_{\epsilon\to0}\frac{1}{\epsilon}\left[\det \begin{pmatrix}
1+\epsilon h_{11} & \epsilon h_{12} &\cdots & \epsilon h_{1n} \\
\epsilon h_{21} & 1+\epsilon h_{22} &\cdots \\
\vdots & & \ddots \\
\epsilon h_{n1} & & &1+\epsilon h_{nn}
\end{pmatrix}-1\right]$</p>
<p>$\qquad\qquad\qquad\qquad\qquad\qquad=\displaystyle\sum_{i=1}^nh_{ii}=\text{trace}(H)$</p>
</blockquote>
| Laurent.C | 346,533 | <p>actually, you're trying to calculate the differential $\phi$ of the function $\det : M_n(\mathbb{R}) \rightarrow \mathbb{R}$ at $I_n$, which is defined as
$$\forall H \in M_n(\mathbb{R}), \phi(H) = \lim\limits_{t \to 0} \dfrac{\det(I_n+tH)-\det(I_n)}{t}=
\dfrac{d}{dt}_{t=0} \det(I_n+tH)
$$</p>
<p>First, since $\det$ is a polynomial, it is differentiable at any point and $\phi$ is a linear form on $M_n(\mathbb{R})$. Therefore, it suffices to calculate its value on the canonical basis of $M_n(\mathbb{R})$.</p>
<ul>
<li>For $H=E_{i,i}$, $\det(I_n+tE_{i,i})=1+t$, so $\phi(E_{i,i})=1$.</li>
<li>For $H=E_{i,j},$ where $i \neq j, \det(I_n+tE_{i,j})=1$, so $\phi(E_{i,j})=0$.</li>
</ul>
<p>So, $\phi=\textrm{Trace}$, since these two linear forms coincide on a basis of $M_n(\mathbb{R})$.</p>
|
2,639,013 | <p>Let $V$ a vector space over field $\mathbb{K}$ with inner product and let $U$ and $W$ subspaces of $V$ so that $U \subseteq W^{\perp}$ and $V = W+U$. Show that $U=W^{\perp}$.</p>
<p>I try this approach:
Let $w \in W+U$ so $w \in U$ and $\langle w,v \rangle = 0$ for all $v \in W$. In particular $\langle w,w \rangle = 0$ implies $w=0$ and direct sum $V = W \oplus U$.</p>
<p>My question is, can I compare $V = W\oplus U$ with decomposition $V = W\oplus W^{\perp}$ and conclude $U=W^{\perp}$?</p>
| carmichael561 | 314,708 | <p>Note that $A=nI_n+vv^T$, where $v=(1,\dots,1)^T$. Using the <a href="https://en.wikipedia.org/wiki/Matrix_determinant_lemma" rel="nofollow noreferrer">matrix determinant lemma</a>, we have
$$ \det(nI_n+vv^T)=\Big(1+\frac{1}{n}v^Tv\Big)\det(nI_n)=2n^n$$</p>
|
2,639,013 | <p>Let $V$ a vector space over field $\mathbb{K}$ with inner product and let $U$ and $W$ subspaces of $V$ so that $U \subseteq W^{\perp}$ and $V = W+U$. Show that $U=W^{\perp}$.</p>
<p>I try this approach:
Let $w \in W+U$ so $w \in U$ and $\langle w,v \rangle = 0$ for all $v \in W$. In particular $\langle w,w \rangle = 0$ implies $w=0$ and direct sum $V = W \oplus U$.</p>
<p>My question is, can I compare $V = W\oplus U$ with decomposition $V = W\oplus W^{\perp}$ and conclude $U=W^{\perp}$?</p>
| Joca Ramiro | 264,635 | <p>Writing $A=nI_n+U$, where $U$ is the matrix with all elements equal to one, the eigenvalues $\lambda$ are solutions to the characteristic polynomial $\det(U-(\lambda-n)I_n)=\det(U-\lambda'I_n)$, where $\lambda'=\lambda-n$. In this <a href="https://math.stackexchange.com/questions/217521/what-are-the-eigenvalues-of-matrix-that-have-all-elements-equal-1">question</a> was shown that $U$ has all eigenvalues $\lambda'$ equal to $0$, except for one that is equal to $n$. Then $A$ has all eigenvalues $\lambda$ equal to $n$, except for one that is equal to $2n$. As the determinant is the product of eigenvalues, $\det(A)=n^{n-1}\cdot 2n=2n^n$.</p>
|
244,115 | <p>I create a very large output</p>
<pre><code> D[x^100*E^(2*x^5)*Cos[x^2], {x, 137}]
</code></pre>
<p>I want to assign this to a function as</p>
<pre><code> f[x_]:= {very large output from previous command}
</code></pre>
<p>This allows me to evaluate that output for various values of <span class="math-container">$x$</span>.</p>
<p>I tried</p>
<pre><code> Function[x, Evaluate[%11]]
</code></pre>
<p>But it seems cumbersome and I wasn't sure I was getting correct results from it.</p>
<p>Is there a clean way to do this?</p>
| Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
f1[x_] = D[x^100*E^(2*x^5)*Cos[x^2], {x, 137}];
f2[x_] = D[x^100*E^(2*x^5)*Cos[x^2], {x, 137}] // Simplify;
</code></pre>
<p>While defining <code>f2</code> takes much longer,</p>
<pre><code>LeafCount /@ {f1[x], f2[x]}
(* {4274535, 6386} *)
AbsoluteTiming[#[3.]] & /@ {f1, f2}
(* {{1.07832, -8.90666869434476*10^664},
{0.002447, -8.906668694344757*10^664}} *)
</code></pre>
|
2,495,918 | <p>A triangle is formed by the lines $x-2y-6=0$ , $ 3x−y+6=0$, $7x+4y−24=0$.</p>
<p>Find the equation of the line that bisects the inner angle of the triangle that is facing the side $7x+4y−24=0$.</p>
<p>I tried to find the intersect point of three equations by put them equal two by two. However, I don't know what to do next. could someone help me, please?</p>
| GAVD | 255,061 | <p>HINT: if point $P(x,y)$ is in the bisector of angle between two lines $x-2y-6=0$ and $3x-y+6=0$, then the distances from this point to each lines are equal. So, you get two lines which are bisector. Check which is the inner bisector.</p>
<p>HINT 2: See this <a href="https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line" rel="nofollow noreferrer">link</a> for distance between point $P(x_0,y_0)$ and the line $ax+by+c=0$. Using this formula, you have distance from $P(x_0,y_0)$ to two lines. Solve the system of equations, you get $P$. </p>
|
145,429 | <p>I have expression like this:</p>
<pre><code>expr = xuyz;
</code></pre>
<p>then</p>
<pre><code>Head[expr] = xuyz
</code></pre>
<p>But I wanted the product of four factors, so it should have been written as
<code>x*u*y*z</code> or <code>x u y z</code>, because Mathematica understands multiplication of four single-character variables in these two ways only.</p>
<p>So I need to detect whether or not an expression is written in the right way. By "right" I mean all variables should are required to be single-character, all factors to be such variables or function thereof, and all terms to be products of such factors . </p>
<p>I have tried with:</p>
<pre><code>FreeQ[expr, Times].
</code></pre>
<p>But it doesn't help, because I can have the situation</p>
<pre><code>xu*y
</code></pre>
<p>which is not right for me, because <code>xu</code> and <code>y</code> are are the factors and I want <code>x</code>, <code>u</code> and <code>y</code> to be the factors. </p>
<p>So is there any way to detect this problem?</p>
| Coolwater | 9,754 | <p>This returns <code>True</code> because of <code>xu</code> and/or <code>gh</code></p>
<pre><code>Max[StringLength[ToString /@ DeleteCases[Level[xu*y/gh, {-1}], _?NumericQ]]] > 1
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
145,429 | <p>I have expression like this:</p>
<pre><code>expr = xuyz;
</code></pre>
<p>then</p>
<pre><code>Head[expr] = xuyz
</code></pre>
<p>But I wanted the product of four factors, so it should have been written as
<code>x*u*y*z</code> or <code>x u y z</code>, because Mathematica understands multiplication of four single-character variables in these two ways only.</p>
<p>So I need to detect whether or not an expression is written in the right way. By "right" I mean all variables should are required to be single-character, all factors to be such variables or function thereof, and all terms to be products of such factors . </p>
<p>I have tried with:</p>
<pre><code>FreeQ[expr, Times].
</code></pre>
<p>But it doesn't help, because I can have the situation</p>
<pre><code>xu*y
</code></pre>
<p>which is not right for me, because <code>xu</code> and <code>y</code> are are the factors and I want <code>x</code>, <code>u</code> and <code>y</code> to be the factors. </p>
<p>So is there any way to detect this problem?</p>
| m_goldberg | 3,066 | <p>I haven't thoroughly tested the following function, but at least it might be a good starting point for you to solve your problem.</p>
<pre><code>singleCharVarsQ[expr_] :=
AllTrue[
StringLength /@ SymbolName /@
Cases[expr, Except[_?NumericQ, _?AtomQ], ∞], # == 1 &]
singleCharVarsQ[x y z + x Sqrt[x^2 + y^2] - Cos[xy]]
</code></pre>
<blockquote>
<p><code>False</code></p>
</blockquote>
<pre><code>singleCharVarsQ[x y z + x Sqrt[x^2 + y^2] - Cos[x y]]
</code></pre>
<blockquote>
<p><code>True</code></p>
</blockquote>
|
145,429 | <p>I have expression like this:</p>
<pre><code>expr = xuyz;
</code></pre>
<p>then</p>
<pre><code>Head[expr] = xuyz
</code></pre>
<p>But I wanted the product of four factors, so it should have been written as
<code>x*u*y*z</code> or <code>x u y z</code>, because Mathematica understands multiplication of four single-character variables in these two ways only.</p>
<p>So I need to detect whether or not an expression is written in the right way. By "right" I mean all variables should are required to be single-character, all factors to be such variables or function thereof, and all terms to be products of such factors . </p>
<p>I have tried with:</p>
<pre><code>FreeQ[expr, Times].
</code></pre>
<p>But it doesn't help, because I can have the situation</p>
<pre><code>xu*y
</code></pre>
<p>which is not right for me, because <code>xu</code> and <code>y</code> are are the factors and I want <code>x</code>, <code>u</code> and <code>y</code> to be the factors. </p>
<p>So is there any way to detect this problem?</p>
| kglr | 125 | <pre><code>f1 = Max@StringLength[SymbolName /@ Variables@#] == 1 &;
f1 /@ {xu*y, 3 x + w z^2}
</code></pre>
<blockquote>
<p>{False, True}</p>
</blockquote>
|
1,393,822 | <p>For a complex number $w$, or $a+bi$, is there a specific term for the value $w\overline{w}$, or $a^2+b^2$?</p>
| Barry Cipra | 86,747 | <p>$${n^{2.5}+(n+1)^{2.5}\over5}={n^{2.5}\over5}\left(1+\left(1+{1\over n}\right)^{2.5} \right)$$</p>
<p>and</p>
<p>$$\left(1+{1\over n}\right)^{2.5}\approx1+{2.5\over n}$$</p>
<p>so</p>
<p>$${n^{2.5}+(n+1)^{2.5}\over5}\approx{n^{2.5}\over5}\left(2+{2.5\over n}\right)={2\over5}n^{2.5}+{1\over2}n^{1.5}$$</p>
|
371,318 | <p>The original problem was to consider how many ways to make a wiring diagram out of $n$ resistors. When I thought about this I realized that if you can only connect in series and shunt. - Then this is the same as dividing an area with $n-1$ horizontal and vertical lines. When each line only divides one of the current area sections into two smaller ones.</p>
<p>This is also the same as the number of ways to make a set of $n$ (and only $n$) rectangles into a bigger rectangle. If the rectangles can be drawn by dividing the big rectangle, line by line, into the set of rectangles without lose endpoints of the line. - Can someone come to think of "a expression of $n$" which equals this amount, independent of the order of the rectangles or position?</p>
<p>(It is only the relations between the area sections that matters and not left or right, up or down. However dividing an area with a horizontal line is not the same as dividing it with a vertical line.)</p>
| Don Mintz | 151,541 |
<p>For $x$ > 0, define an
infinite number by the divergent geometric series:
$\displaystyle\sum_{i=0}^{n\rightarrow\infty} \left(\frac{x+1}{x}\right)^i $ </p>
<p>and define an infinitesimal number as the difference between a convergent geometric series and its sum:</p>
<p>$ x+1 -\displaystyle\sum_{i=0}^{n\rightarrow\infty} \left(\frac{x}{x+1}\right)^i$</p>
<p>If the x is the same in both the infinity and the infinitesimal their product will converge to the finite number x(x+1)
as n increases without bound. If the x in the infinity is smaller than the x in the infinitesimal their product in the limit will be an infinity. If the x in the infinity is larger than the x in the infinitesimal their product in the limit will be an infinitesimal.</p>
<p>Division of infinity by infinity as defined by these divergent geometric series will result
in the limit (1) an infinity if the numerator has a smaller x, (2) an infinitesimal if the
numerator has a larger x, (3) the finite value 1 if numerator and denominator have the same x.</p>
|
498,694 | <p>So, I'm learning limits right now in calculus class.</p>
<p>When $x$ approaches infinity, what does this expression approach?</p>
<p>$$\frac{(x^x)}{(x!)}$$</p>
<p>Why? Since, the bottom is $x!$, doesn't it mean that the bottom goes to zero faster, therefore the whole thing approaches 0?</p>
| Vaidyanathan | 86,594 | <p>You may use Stirlings approximation to do this.</p>
<p>Let $$y = x^x/x! $$, then $$ln(y) = ln((x^x/x!)) = ln(x^x) - ln(x!)$$</p>
<p>The last term on the right hand side can be expanded as $$ ln(x!) = xln(x) - x $$ for large n. As a side note, this is a typical expansion used in physics.</p>
<p>Substituting back and taking limits, $$ \lim_{x \to +\infty} ln(y) = xln(x) - (xln(x) -x )$$</p>
<p>$$ \lim_{x \to +\infty}\Rightarrow ln(y) = \lim_{x \to +\infty}x = \infty $$.</p>
<p>Stirlings approximation induces an approximation of O(log(x)) which goes as $$ \sqrt{2\pi x}$$ in the first term. If this is included and reworked this is will be result as @alex suggested</p>
|
498,694 | <p>So, I'm learning limits right now in calculus class.</p>
<p>When $x$ approaches infinity, what does this expression approach?</p>
<p>$$\frac{(x^x)}{(x!)}$$</p>
<p>Why? Since, the bottom is $x!$, doesn't it mean that the bottom goes to zero faster, therefore the whole thing approaches 0?</p>
| N. S. | 9,176 | <p><strong>Hint</strong>
$$\frac{x^x}{x!} > \frac{x}{1}$$</p>
|
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