@mixedmath I wonder if he meant that $\sum f(a_n)$ also converges absolutely. I'm not sure if it's relevant, but if that's not required to converge absolutely, maybe things get messy.
@BalarkaSen Yup, you start by checking the characteristic polynom first, and if they match, you also need to check the geometric multiplexing if its $^?$ bigger than 1.
Monday - Tried to prove a theorem, Tuesday - Tried to prove the theorem, Wednesday - Tried to prove the theorem, Thursday - Tried to prove the theorem, Friday - Theorem turned out to be false.
@Balarka Yes, most of the issues were actually with my bounds over the double-integrals. I really need to be more careful there, giving too loose bounds ends up with false results. @Ilan checking
@Ilan I haven't check your computations, will if you want me to. But it's true - if they don't have the same polynom, they can't be similliar, as they don't have the same eigenvalues.
Two matrices are similliar <=> they have the same eigenvalues!!
To clarify, they are similliar *only if* they have the same eigenvalues, They can't be similiar if they don't have the same eigenvalues. They may or may not be similliar if they have the same eigenvalues.
@Chris'ssis however the thing inside the integral looks familiar (if I'm not confusing with something else) :D ... IDK (I have to fiddle with it to know)
@Chris'ssis But there are pretty elementary examples of moduli problem, which can be understood without any advanced stuff. For example, this elementary and relatively easy) problem arise from hyperbolic geometry : Can you draw a circle from a point in the xy-plane that doesn't intersect the y-axis?
@Chris'ssis maybe its just silly of me to think of that .. just because the expression has some $\binom{n}{k}$ and possibly has a combinatorial interpretation .. might not mean the divisibility has some combinatorial meaning too :)
Let us denote
$$A_m = \sum_{k=0}^m \binom{3m}{3k}(3n-1)^k.$$
Writing $z = \sqrt[3]{3n-1}$ and $\rho = e^{2\pi i/3}$, we find
$$3A_m = (1+z)^{3m} + (1+\rho z)^{3m} + (1+ \rho^2 z)^{3m}.$$
The sequence given by
$$3u_k = (1+z)^k + (1+\rho z)^k + (1+\rho^2 z)^k$$
satisfies the linear recurrence...
@Chris'ssis possibly .. a junior asked it to me .. I am still not aware of the source .. he won't tell .. :| (he puts me through testing problems some times :P )
Klein model is intuitively better understood, though. Take an open disk and the straightlines be the straightlines inside the disk. It's easy to see that this model satisfies all the axioms, except Playfair's.
@Vrouvrou you do realize that there is no unique kernel, don't you? All you have is a map $(a, b) \mapsto c$ for $a, b, c \in \Bbb Z$ and the map can send anything to pretty much anything.
@Vrouvrou the question is what is $i$? it's not sensible to talk about kernels if there is no homomorphism. what kind of homomorphism is it? group? ring?
@DanielFischer Well, all algebraists are currently out of town, so I grabbed you. Am I right in thinking that there is a unique galois extension of $\Bbb C(X)$ with given galois group?
What I want is to prove the uniqueness, by the way, not the inverse galois.
@skullpatrol Algebraists always think about groups and subgroups, so he doesn't get what's a set. We almost never work with N so that doesn't make sense to him too.
I am not sure about the function so I removed it. Usually, we call it "map" and function draws continuity stuffs.
@DanielFischer I was able to prove that non constant complex polynomials are closed maps. How would complementing help to prove that they're open maps?
@G.T.R Since they usually aren't injective, I don't think that complementing would help much here. Just look how the map locally looks like, the openness is an easy consequence of the local form.
@DanielFischer locally, it looks like a constant, and minus this constant it looks like zero. So WLOG I may look at what happens around the zero of a polynomial...
@DanielFischer If you ask Wolfram Alpha for the Laurent expansion of $f(z) = \frac{\log (\frac{1+z}{1-z})}{z\sqrt{1-z^{2}}}$ at infinity, it says that the leading order term is $\frac{\pi}{z^{2}}$. But when you ask it to compare $f(z)$ and $\frac{\pi}{z^{2}}$ for values of $z$ such that $|z|$ is large, the values always differ in sign. Is there a reason for that?
I'm going to propose it for a contest (don't use any software for this - if you wanna have fun) $$\int \frac{1}{(e^x-1)(e^{x+y}-1)} \ dx$$ What is the fastest way?
@DanielFischer When I did the calculation by hand, I restricted $\arg(1+z)$ to $- \pi < \arg(1+z) \le \pi$ and $\arg(1-z)$ to $ 0 \le \arg(1-z) < 2 \pi$ so that the branch cut would be on $[-1,1]$. Doing that, and being careful with signs, I got the leading order term to be $- \frac{\pi}{z^{2}}$. But I imagine different choices will lead to different expansions.
@RandomVariable Switching the branch of the square root gives a factor of $-1$, changing the branch of the logarithm gives a multiple of $2\pi i$ in the numerator. So the $z^{-2}$ term can be $$\frac{(2k+1)\pi}{z^2}$$ for any $k\in\mathbb{Z}$.
@BalarkaSen Let $G$ be a subgroup of $3$ elements of the multiplicative group of non-zero complex numbers with $a, b \in \mathbb{C}$. Compute $$\prod_{z\in\mathbb{G}}(a z+b)$$ (problems you often meet during high school)
@DanielFischer Someone asked about showing that $\int_{0}^{1} \frac{\log (\frac{1+x}{1-x})}{x\sqrt{1-x^{2}}} = \frac{\pi^{2}}{2}$. So I extended the interval of integration to $(-1,1)$ and integrated around a dogbone/dumbbell contour. That's where this comes from.
Let us denote
$$A_m = \sum_{k=0}^m \binom{3m}{3k}(3n-1)^k.$$
Writing $z = \sqrt[3]{3n-1}$ and $\rho = e^{2\pi i/3}$, we find
$$3A_m = (1+z)^{3m} + (1+\rho z)^{3m} + (1+ \rho^2 z)^{3m}.$$
The sequence given by
$$3u_k = (1+z)^k + (1+\rho z)^k + (1+\rho^2 z)^k$$
satisfies the linear recurrence...
