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Loser66
 one year ago
Problem is in comment
Loser66
 one year ago
Problem is in comment

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0There is an "i" on the Gauss sums. It's a typo.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me just retype the problem here, MSWord always gives me trouble. \[\sum_{n=1}^N\exp\frac{2i\pi n^2}{N}=\begin{cases} \sqrt N&\text{if }N\equiv1\mod4\\ 0&\text{if }N\equiv 2\mod4\\ i\sqrt N&\text{if }N\equiv3\mod4\\ (1+i)\sqrt N&\text{if }N\equiv0\mod4 \end{cases}\] and you want to prove that \(\sum\limits_{n\ge1}\dfrac{z^{n^2}}{n}\) converges for \(z=\exp\dfrac{2i\pi}{N}\) if \(N\equiv2\mod4\) and diverges otherwise.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Did I get everything right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it is right. My prof said that we need use Dirichlet test to attack it. From part 1, we have \(\sum_{n\geq 1} \dfrac{z^{n^2}}{n} \) converges with radius R =1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the case \(N\equiv2\mod4\), you have \(z=\exp\dfrac{2i\pi}{N}\) and \[\sum_{n=1}^N\frac{z^{n^2}}{n}=\sum_{n=1}^N\frac{\exp\dfrac{2i\pi n^2}{N}}{n}\] Call \(a_n=\dfrac{1}{n}\) and \(b_n=\exp\dfrac{2i\pi n^2}{N}\). To satisfy Dirichlet's test, you must (1) \(a_n\) a nonincreasing and converging to \(0\), and (2) \(\left\sum\limits_{n=1}^Nb_n\right\le M\) for some constant \(M\) and all \(N\). The first condition is clearly satisfied. Meanwhile, \[\left\sum_{n=1}^N\exp\dfrac{2i\pi n^2}{N}\right=\left\sum_{n=1}^N0\right=0\le M\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0... and so \(\sum a_nb_n\) converges for that case.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question: the sum we need prove is infinite sum, not stop at N. How to argue that part?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Slight typo above: In that last line, it should just be \(0\), not the sum of \(0\)s. The procedure is similar for the other case. Set \(N\equiv1\mod4\), use the same \(a_n\), and replace \(b_n\) accordingly. You have \[\left\sum_{n=1}^N\exp\dfrac{2i\pi n^2}{N}\right=\left\sqrt N\right=\sqrt N\] which is unbounded, so the second required condition is not met, and the series diverges.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The test is designed to check the convergence of the series \(\sum\limits_{n=1}^\infty a_nb_n\), while one of the conditions for the test is to check the boundedness of a partial sum, \(\sum\limits_{n=1}^Nb_n\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The case for \(N\equiv3\mod4\) is almost identical as for \(N\equiv 1\mod4\). The partial sum of \(b_n\) is equal to \(i\sqrt N=\sqrt N\), and you get the same conclusion. For \(N\equiv0\mod4\), you have \((1+i)\sqrt N=\sqrt{2N}\), etc.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so so so much. You make it simple.
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