Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spacing of Riemann's Zeros

  1. Apr 29, 2007 #1
    I've just been reading through my number theory lecture notes and I noticed this line of reasoning:

    ***Assuming the Rieman hypothesis***

    [tex]\sum_{\gamma > 0}\sin(\gamma \log x) \frac{-(1/2)^2}{\gamma((1/2)^2 + \gamma^2)} = O(1)[/tex]

    by comparisson with [tex]\sum_{n \ge 1}\frac{1}{n^2}[/tex].

    Let me explain the context a little. Importantly the gamma are the non-trivial zeros of the zeta function in the upper half plane. (The goal is to obtain an expression which demonstrates a connection between the Von-Mangolt explicit formula and Fourier series if anyone's interested but that is really not important for my question.)

    Now as far as I can see the statement is true by considering absolute convergence etc. but surely this requires also that the spacing of the zeros on the line Re(z) = 1/2 is sufficenty sparse so that the comparison is valid.

    There is no mention anywhere else in the course of this being a fact. Is it true?

    How much do we know about properties of the distribution? I suppose the above question concerns the existence of a lower bound on the spacing.
    Last edited: Apr 29, 2007
  2. jcsd
  3. Apr 29, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It sounds wrong, but I've only looked at small zeroes.
  4. Apr 29, 2007 #3


    User Avatar

    If in V.Mangoldt formula you put x=exp(u) and differentiate respect to 'u' setting [tex] \rho _{n} = 1/2+it_{n} [/tex] you can find an expression for:

    [tex] \sum_{n} exp(it_{n}) [/tex] similar to a kind of sum over 'frecuencies' where the 't'-s are just the imaginary part of NOn-trivial roots.
  5. May 1, 2007 #4
    Ok to simplify the situation a little, it has been claimed to me that

    [tex]\sum_{\gamma > 0}\frac{1}{\gamma^2}[/tex]

    is convergent, where, as usual, the gamma are just the imaginary parts of the Riemann zerosin the critical strip.

    I think this is false based on the following heuristic arguament:

    The number of Riemann zeros in critical strip between 0 and iT is known to be asyptotic to Tlog(T)/2pi. It follows then that the average density becomes unbounded as we go further up the strip, and hence the series must diverge.

    If I am correct, how might I go about proving this?
    Last edited: May 1, 2007
  6. May 1, 2007 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    By finding an asymptotic upper bound on the n-th zero, and thus a lower bound on the series. If the lower bound on the series is infinity...

    Are you sure that gamma is the imaginary part, and not the actual value of the zero? You said otherwise in your OP.
    Last edited: May 2, 2007
  7. May 1, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    The average density tending to infinity does not imply the series will diverge. How fast it tends to infinity is important. Here, you only have log(T) times as many zeros up to T as you do integers (ignoring constants and lower order terms), which really isn't that much, and not enough to make this diverge.

    Try bounding the tail of the series by looking at

    [tex]\sum_{T+t \geq\gamma > T}\frac{1}{\gamma^2}[/tex]

    and smacking it around with partial summation and your asymptotic for the number of zeros (really the bound O(Tlog(T)) for the number of zeros will do). Let "t" go to infinity and you should be able to get something that shows the tail tends to 0 as T->infinity, so the original series converges.

    For Hurkyl's question about gamma, it doesn't really matter here if it's the imaginary part or the whole zero for the convergence of this series. The real part will be between 0 and 1, while the imaginary part is tending to infinity, so the real part doesn't do much. However, the usual convention is little gamma standing for the imaginary part, so that's likely what they were talking about.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook