3
$\begingroup$

In this book on page 82 I found an estimate $$\sum_{j=1}^{n-2}\frac{\sin(k+1)\theta_{j}}{\sin\theta_{j}}=O_{\epsilon}\left( p^{k\epsilon}\right)$$ as $k$ goes to infinity, for all $\epsilon>0$. Here $p$ is a prime number and $\theta_{1},...,\theta_{n-2}$ are complex numbers. After this estimate, the authors says: The last clearly implies that $\theta_{j}$ $\left(1\leq\theta_{j}\leq n-2\right)$ are real numbers. Why is this clearly true?

P.S. I hope this edited question is now ok.

$\endgroup$
  • $\begingroup$ Nothing is «clearly» true. $\endgroup$ – Mariano Suárez-Álvarez Jan 2 '14 at 19:17
  • $\begingroup$ I can't understand that part. So, if someone is clear how the author got the result, I would appreciate if you could write it. $\endgroup$ – Alem Jan 2 '14 at 19:26
  • 10
    $\begingroup$ Keep in mind that $|\sin z|$ can get large if $z =x+iy$ is complex (as large as $e^{|y|}$). So if any of the $\theta_j$ are complex, you can find values of $k$ for which the LHS grows exponentially in $k$ (this is obvious if there is a single $\theta_j$ whose imaginary part is largest, and you'll need some argument if there are several with the same largest imaginary part). Hope that helps. $\endgroup$ – Lucia Jan 2 '14 at 19:29
  • $\begingroup$ I really can't see the details. $\endgroup$ – Alem Jan 2 '14 at 19:56
  • 4
    $\begingroup$ @Alem In the future, when you post questions, please try to make the title informative. I suspect that the title that you used is a reason that people have voted to close your question. $\endgroup$ – Joe Silverman Jan 3 '14 at 0:01
6
$\begingroup$

A much more detailed writeup of this is is Davidoff/Sarnak/Vallette, see page 127 and on, especially p. 130.

$\endgroup$
  • $\begingroup$ @AlainValette I will take you up on it :) $\endgroup$ – Igor Rivin Jan 3 '14 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.