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Loser66
 one year ago
Complex analysis hard problem. Please, help
Loser66
 one year ago
Complex analysis hard problem. Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0In the previous part, we have the series converges with the radius of convergence is R =1. But I don't know how to attack this problem.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(\sum_{n =1}^\infty \dfrac{z^{n^2}}{n}=\sum_{n=1}^N \dfrac{z^{n^2}}{n}+\sum_{n=N+1}^\infty \dfrac{z^{n^2}}{n}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I believe that in the Gauss sums, we have an "i" on the exponent. That is \(\sum e^{2i\pi n^2/N}\). If it is so, we can see the link between them. Like if \(z = e^{2i\pi /N}\) then \(z^{n^2} = e^{2 i \pi n^2/N}\) which is the summand of the Gauss sums.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is a typo in the Gauss sums, it is indeed \(e^{2\pi in^2/N}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in this case, it's obvious that the only one of those expressions with a finite limit as \(N\to\infty\) as the case where \(N\equiv 2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the point they're trying to make is that this lacunary series converges on only part of the unit circle

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Please, guide me more, on the Gauss sums, it doesn't have /n there, only \(z^{n^2}\), how to link them?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider that $$\begin{align*}\sum_{n=kN+1}^{(k+1)N} e^{2\pi in^2/N}&=\sum_{n=1}^N e^{2\pi i(n+kN)^2/N}\\&=\sum_{n=1}^N e^{2\pi i(n^2+k^2N^2+2nkN)/N}\\&=\sum_{n=1}^N e^{2\pi in^2/N}\cdot e^{2\pi i(k^2N+2nk)}\\&=\sum_{n=1}^Ne^{2\pi in^2/N}\end{align*}$$so we can rewrite our infinite sum thusly: $$\sum_{n=0}^\infty\frac{e^{2\pi in^2/N}}n=\sum_{k=0}^\infty\sum_{n=kN+1}^{(k+1)N}\frac{e^{2\pi in^2/N}}n$$ and consider that we can bound this like so: $$\sum_{n=kN+1}^{(k+1)N}\frac{e^{2\pi in^2/N}}n\ge\frac1{(k+1)N}\sum_{n=kN+1}^{(k+1)N}e^{2\pi in^2/N}$$and substituting our simplified expression for the sums on \(kN+1,\dots,(k+1)N\) we see: $$\sum_{n=kN+1}^{(k+1)N}\frac{e^{2\pi in^2/N}}n\ge\frac1{(k+1)N}\sum_{n=1}^Ne^{2\pi in^2/N}$$ now suppose \(\sum\limits_{n=1}^Ne^{2\pi in^2/N}=a\sqrt{N}\) for some constant \(a\). we see: $$\sum_{n=kN+1}^{(k+1)N}\frac{e^{2\pi in^2/N}}n\ge\frac{a}{(k+1)\sqrt{N}}$$ and so it follows that $$\sum_{n=1}^\infty\frac{e^{2\pi in^2/N}}n>\frac{a}{\sqrt{N}} \sum_{k=0}^\infty\frac1{k+1}$$ clearly the term on the right, being the harmonic series, will diverge if \(a\) is nonzero, and it will only converge in the case that \(a\) *is* zero. now recall that \(a=0\) iff \(N\equiv2\pmod 4\), so we're done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, that should be \(\ge\) but you see what i'm saying

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Wow!! Thank you so much. I need time to digest it.
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