Loser66
  • Loser66
Problem is in comment
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
There is an "i" on the Gauss sums. It's a typo.
anonymous
  • anonymous
Let 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.

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anonymous
  • anonymous
Did I get everything right?
Loser66
  • Loser66
Yes, 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
  • anonymous
For 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 non-increasing 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
  • anonymous
... and so \(\sum a_nb_n\) converges for that case.
Loser66
  • Loser66
I got it,
Loser66
  • Loser66
Question: the sum we need prove is infinite sum, not stop at N. How to argue that part?
anonymous
  • anonymous
Slight 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
  • anonymous
The 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
  • anonymous
The 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
  • Loser66
Thank you so so so much. You make it simple.
anonymous
  • anonymous
You're welcome!
Loser66
  • Loser66
oh, I got it, nvm. :)

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