A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Loser66

  • one year ago

Problem is in comment

  • This Question is Closed
  1. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  2. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    There is an "i" on the Gauss sums. It's a typo.

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did I get everything right?

  5. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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, \[\left|\sum_{n=1}^N\exp\dfrac{2i\pi n^2}{N}\right|=\left|\sum_{n=1}^N0\right|=0\le M\]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ... and so \(\sum a_nb_n\) converges for that case.

  8. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got it,

  9. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Question: the sum we need prove is infinite sum, not stop at N. How to argue that part?

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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, \(\sum\limits_{n=1}^Nb_n\).

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  13. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so so so much. You make it simple.

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You're welcome!

  15. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, I got it, nvm. :)

  16. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.