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Loser66

  • one year ago

Complex analysis hard problem. Please, help

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  1. Loser66
    • one year ago
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  2. Loser66
    • one year ago
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    In 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.

  3. Loser66
    • one year ago
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    \(\sum_{n =1}^\infty \dfrac{z^{n^2}}{n}=\sum_{n=1}^N \dfrac{z^{n^2}}{n}+\sum_{n=N+1}^\infty \dfrac{z^{n^2}}{n}\)

  4. Loser66
    • one year ago
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    I 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.

  5. anonymous
    • one year ago
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    there is a typo in the Gauss sums, it is indeed \(e^{2\pi in^2/N}\)

  6. anonymous
    • one year ago
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    in this case, it's obvious that the only one of those expressions with a finite limit as \(N\to\infty\) as the case where \(N\equiv 2\)

  7. anonymous
    • one year ago
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    the point they're trying to make is that this lacunary series converges on only part of the unit circle

  8. Loser66
    • one year ago
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    Please, guide me more, on the Gauss sums, it doesn't have /n there, only \(z^{n^2}\), how to link them?

  9. anonymous
    • one year ago
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    consider 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.

  10. anonymous
    • one year ago
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    oops, that should be \(\ge\) but you see what i'm saying

  11. Loser66
    • one year ago
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    Wow!! Thank you so much. I need time to digest it.

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