This one's kinda easy (or at least Malik's algebra jumps its justification). Let say I have a group whose order is $91 = 7 * 13$. I know there exists only one normal 7-subgroup H and only one 13-subgroup K. Can I say that $H\cap K = \{e\}$?
@TedShifrin I can imply from that that the group of order 91 is always cyclic, because it can be seen as the product of both $H$ and $K$ (where both $H$ and $K$ are cyclic). Showing that there can't be a group by which $|G/Z(G)| = 91$. I might bother you later, but I think that's all.
@Hippalectryon and this one is more advanced ... $$\int_0^1 \tan(\pi x) \psi^{(-2)}(x)dx$$ where the integrand also contains the negapolygamma. These ones are deadly.
@TedShifrin That's kinda what I'm using. But the gist of the matter is that if $G/Z(G)$ is cyclic, $G$ is abelian and therefore $|G/Z(G)| = 1$. Is that a correct reasoning?
Is it true that the boundry of the set of continious functions $2\le f(x)<3, \forall x\in [0,1]$ in the space C[0,1] is the set of continious functions $2\le f(x)\le 3$, so that for some $a$ f(a)=3? Or am I wrong already here? I don't know if I can take some function arbitiraly close to 2, and how it will count exactly.
@Ted well, to be fair I am going faster than the others - I need to finish this within a semester, where the course is meant to be yearly. So he can't really elaborate with every question, as it is as if he is teaching two classes that way
@TedShifrin I have $n$ urns and $m$ balls. Let X be the variable that holds the number of balls in one given urn, we suppose that the distribution law is equiprobable. I have $P(X=k)=\binom{m}{k}\dfrac{(n-1)^{m-k}}{n^m}$. Now they ask for $P(X\ge k)$, is that $\sum_{i=k}^m P(X=i)$ ?
Seems that's fine, I think what was wrong was my explanation of that set of functions. It seemed as if I was saying I was raising some point to a different location, thus nullifying the continiuty.. Ach so, at least I get it now
@Ted precisely. $2\le f(x) \le 3$ so that there is some $a$, $f(a)=3$ is there. Question is whether the same. I can show nothing else is - if I take it to be smaller than $2$ for some value or larger than 3 for some value, I can take small enough epsilon to contradict it. And if there's no $f(a)=3$, it's in $S$.
Yes, too bad you never took upper level mechanics. The moments of inertia about the three axes of a box, unless the box has a square base, are $I_1<I_2<I_3$. Rotation about axis #2 is unstable. Didn't I ever show you this with tossing a book? This is something they should definitely do in 4700.
I am trying to work out the large $n$ asymptotics of $$S_n = \prod_{x=1}^{\lceil\frac{n}{\log_2{n} }\rceil} \left(\frac{1}{\sqrt{n}} + x\left(\frac{1}{n}-\frac{2}{n^\frac{3}{2}} \right)\right) .$$
Here is my attempt so far $$\prod_{x=1}^{k} (A + Bx) = \frac{B^k \Gamma(k+1+A/B)}{\Gamma(1+A/B)}.$$...
The mapping cylinder will be defined as $Z_f=X\times[0,1]\coprod Y/\sim$, where $\sim$ is defined by $(x,1)\sim f(x)$. Let $f:X\to Y$ a continuous map between topological spaces and the map $i:X\hookrightarrow Z_f$ identifies $X$ with $X\times \{0\}\subset Z_f$ is the inlusion. R is a ring. I wa...
@Chris'ssis Let $$\displaystyle T_k(q)=\int_0^\infty\dfrac{t^k\arctan(t)}{e^{2\pi q t}-1}$$ Show that $T_0(q)=\dfrac{1}2-\dfrac{\ln q}2+\dfrac{\ln q}{4q}+\dfrac{\ln\Gamma(q)}{2q}-\dfrac{\sqrt{2\pi}}{2q}$
What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
@Hippalectryon Honestly speaking, can you do that in a very easy way? Maybe you use some beta function, you add some fake parameters, make use of some generating function. Just curious about your way.
@evinda, so correct me if I'm wrong, but can we rewrite the expression $3x^2 + 5y^2 - 7z^2$ as $x^2 + y^2 + z^2$ given that all those coefficients are $1 \pmod{2}$?
I agree @Zach. It's boring at home, but I definitely needed to get away from coursework.
That certainly appears to be correct. If there is a solution in $\mathbb{Z}_p$, then there should be a solution in $\mathbb{Q}_p$.
How do you add and multiply in $\mathbb{Q}_p$ btw? Do you carry it out as you would in $\mathbb{Q}$ and then just mod out the numerator and denominator?
@KajHansen if D is any domain (no zero divisors), its field of fractions is comprised of the things of the form a/b (b not 0), with (ac)/(bc) identified with a/b (c not 0), and the operations (a/b)*(c/d)=(ac)/(bd) and a/c+b/d=(ad+bc)/(bd). this is exactly what we do to form Q out of Z.
@TedShifrin I have $n$ urns and $m$ balls. Let X be the variable that holds the number of balls in one given urn, we suppose that the distribution law is equiprobable. I have $P(X=k)=\binom{m}{k}\dfrac{(n-1)^{m-k}}{n^m}$. Now they ask for $P(X\ge k)$, is that $\sum_{i=k}^m P(X=i)$ ?
@anon @MikeMiller @KajHansen Now I want to check if the equation $3x^2+5y^2-7z^2=0$ has a solution in $\mathbb{Q}_3$.
Using this Lemma:
Let $p \neq 2$ be a prime,$ a,b$ and $c$ be pairwise coprime integers with $abc$ square-free and $p \mid a$, and $Q:ax^2+by^2+cz^2=0$ a quadratic form. Then there is a solution to $Q$ over $Q_p$ if and only if $\frac{-b}{c}$ is a square mod $p$.
I found that there is no solution in $\mathbb{Q}_5$.
What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
@MikeMiller Consider the following map $F:I\times S^1\to S^1$ where $F(t,e^{is})=e^{ist}e^{(1-t)i(s+\alpha)}$. This is a function that has $F(1,e^{is})=e^{is}=1_{S^1}$ and $F(0,e^{is})=e^{i(s+\alpha)}$, a rotation by $\alpha$ radians.
What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
