Let $x$ be a positive real number. Consider the sum $\sum \sin(p_i x)$ taken over all primes $p_i$ from 2 till $n$. Call this function $f(n,x)$.

Can we give good upper and lower bounds of $f(n,x)$ for fixed $n$ or $x$? Is the best known upper bound of $f(n,x)$ for sufficiently large $n$:

  1. approximately equal to $\sqrt{2n+3\sin^2(x)}$ ?

  2. equivalent to GRH ?

  3. a consequence of GRH ?

Is it useful to consider taking the power series of $sin(x)$ to help compute an upper bound? It seems natural considering we have an approximation for $\sum$ $p_i^k$. (see my other recent question Generalized PNT ?")

But when we do that, we seem to arrive at a too small upper bound. (or at least when I try it). Perhaps it's just a bad idea to try taylor series for something that grows slower than polynomial ... here ? ( or in general ? )

The analogue question for $cos$.

  • $\begingroup$ it seems greetings are not done here ( considering the edit ). Lots of rules on this site imho. $\endgroup$ – mick Sep 4 '12 at 15:34
  • $\begingroup$ The maximum of $\cos$ is easy: $f(n,2\pi k) = \pi(n)$ $\endgroup$ – Hurkyl Sep 4 '12 at 16:36
  • $\begingroup$ @mick: The editor probably removed it because it's not part of the question proper. Don't think too much of it. People edit others' questions fairly often. $\endgroup$ – Snowball Sep 4 '12 at 17:09
  • $\begingroup$ @Hurkyl That is a very simple answer. Its an upperbound for only a countable amount of fixed x. If one replaces 2 $\pi$ with another rational multiple of \pi one might see why i consider a connection with GRH. $\endgroup$ – mick Sep 4 '12 at 17:56

You are looking for bounds on Exponential sums over prime numbers. There are lots of articles on this subject matter, since as you mention, they can be very useful in analytic number theory. (The circle method for example, or more simply when looking at Fourier transform of the indicator function of the primes) Take a look at this paper by Goldston, it looks at results regarding the sum $$\sum_{p\leq x} e(p\alpha)\log p $$ where $e(p\alpha)=e^{2\pi \alpha i}$. The $\log p$ is put in because it is natural to weight the primes this way, it can be removed with partial summation, and since $e^{2\pi \alpha i}=\cos(2\pi \alpha)+i\sin(2\pi \alpha)$ you can recover your sums above from these results.

Hope that helps,

  • $\begingroup$ You forgot to type a 2 in the sin. I could not edit it. Its not a big deal but just in case someone reads this who is not good with complex numbers or so. $\endgroup$ – mick Sep 4 '12 at 19:07
  • $\begingroup$ Although this was a good and usefull answer , it also the only answer and i hoped for more and better ones. Do i have too accept the best answers ? I dont want to be unthankfull , but many questions remain ... $\endgroup$ – mick Sep 8 '12 at 21:26
  • $\begingroup$ @Mick: An enormous amount can be said about exponential sums over primes, and there already exists expositions online, or entire chapters of textbooks dedicated to the subject. $\endgroup$ – Eric Naslund Sep 8 '12 at 21:36

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