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Apr
14
answered Generators of a subgroup of the modular group
Apr
9
answered The leaf of the foliation has countable basis
Mar
18
awarded  Altruist
Mar
16
comment Existence of an holomorphic function
Dear George, your comment is very interesting. If you could flesh it out as a proper answer (and provide a reference for the theorem you're quoting), I would be more than happy to give you the +100 bounty.
Mar
13
awarded  Enlightened
Mar
13
awarded  Nice Answer
Mar
11
comment Existence of an holomorphic function
OK, so the question has been asked (and solved) on Math Overflow: mathoverflow.net/questions/4434/… ; they noted the power series approach was necessarily very limited, like I did, but someone gave a satisfying answer based on contour integration.
Mar
11
comment Existence of an holomorphic function
Not so sure about that. To me, that would be the equivalent of summoning Satan because your little brother is annoying. Assuming you get an answer (which is not likely), it would probably solve the problem, but it's not the wisest option.
Mar
11
comment Existence of an holomorphic function
In this article (archive.numdam.org/article/SHC_1951-1952__4__A20_0.pdf), which is quoted by Forster, J.-P. Serre says that this result is « un théorème classique sur les fonctions entières. » Thank you for the precision, mate!
Mar
11
answered Existence of an holomorphic function
Mar
10
awarded  Investor
Mar
9
revised Sum of Products of all Combinations
added 2 characters in body
Mar
9
comment Existence of an holomorphic function
Could you explain your proof? If you try and do it the ugly way with power series, there is an obvious candidate (which comes from the fact that Bernoulli polynomials, [more precisely, $P_n(z) = \frac 1{n+1} B_{n+1}(z)$] satisfy $P_n(z+1) - P_n(z) = z^n$) but, unless I'm mistaken, having $a_n = O(R^{-n})$ for all $R > 0$ is not enough for $\sum_{n=0}^{+\infty} a_n P_n$ to converge to a well-defined element in $\mathbb C[[z]]$. I must say this confuses me: I would have thought that, if a solution existed, it would be obtainable in that way...
Mar
9
comment Find an infinite power series of the form an$z^n$ with radius of convergence 1 that converges for z such that |z|=1 except when z = z1, z2, …zm.
The technical term is Cauchy Product: en.wikipedia.org/wiki/Cauchy_product ; basically, it is the same formula as the one you use to multiply two polynomials. If you do that to all your power series, you will get the example you're looking for.
Mar
9
comment Find an infinite power series of the form an$z^n$ with radius of convergence 1 that converges for z such that |z|=1 except when z = z1, z2, …zm.
No, what you have to do is to multiply all these power series. Do you know how to do that?
Mar
9
comment Find an infinite power series of the form an$z^n$ with radius of convergence 1 that converges for z such that |z|=1 except when z = z1, z2, …zm.
Yes. So now you have power series $\sum \frac{(\overline{z_i} z)^n}{n}$, each only diverging on the point $z = z_i$. How to combine them so to get a power series diverging on all of these points?
Mar
9
comment Find an infinite power series of the form an$z^n$ with radius of convergence 1 that converges for z such that |z|=1 except when z = z1, z2, …zm.
Well, either will do. But it is probably more simple to use it in exponential form. Now, the only thing you have to do is to modify the $\sum \frac{z^n}n$ example a bit to produce the example you're looking for. First step: how to make an example of a power series converging for all complex numbers of modulus 1 except for $z=z_1$?
Mar
9
revised Find an infinite power series of the form an$z^n$ with radius of convergence 1 that converges for z such that |z|=1 except when z = z1, z2, …zm.
added 10 characters in body
Mar
9
comment Find an infinite power series of the form an$z^n$ with radius of convergence 1 that converges for z such that |z|=1 except when z = z1, z2, …zm.
Do you know an example of power series of radius of convergence 1 which converges for all $z$ s.t. $|z| = 1$, except for $z = 1$?
Mar
6
awarded  Excavator