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VonMangoldt formula

  1. Aug 28, 2006 #1


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    Regarding the vonMongoldt formula:


    I cannot understand why the sum over the complex zeros exhibits discontinuities at powers of primes. I understand why [itex]\psi_0(x)[/tex] does as this results from the derivation of the formula from a Mellin transform (Perron's formula) and a function derived from Chebyshev's function which itself is discontinuous at powers of primes. But why should the sum over the complex zeros exhibit the same type of discontinuity? They seem unrelated. I'll be looking into it. Might take a while though.
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  3. Aug 28, 2006 #2


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    The explicit formula itself is the best explanation on why it has jumps at prime powers. This is really the most basic tool for going back and forth between primes and zeros.

    It's worth pointing out that this sum is only conditionally convergent, the order you put the zeros in does matter, so you wouldn't be able to assert continuity based on absolute convergence for example.
  4. Aug 28, 2006 #3


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    Thanks for replying Shmoe. Suppose though I looked at the sum in "complete isolation" without it's connection to Analytic Number Theory:

    [tex]\lim_{T\to\infty}\sum_{|\gamma|\leq T} \frac{x^{\rho}}{\rho}[/tex]

    for Zeta zeros (well then not complete isolation):


    You mean, there is no "dirrect" explanation of why this sum (and others like it) have discontinuities? Might be something interesting to look into, experiement with. For example, I can change the values slightly such as:

    [tex]\lim_{T\to\infty}\sum_{|\gamma|\leq T} \frac{x^{\rho+0.1}}{\rho+0.1}[/tex]

    and it still seems to exhibit discontinuities.

    Last edited: Aug 28, 2006
  5. Aug 30, 2006 #4
    - Perhaps it's something similar to a "Gibbs Phenomenon" that also happens with the Fourier series of certain function with discontinuities...
  6. Aug 30, 2006 #5


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    Hello Lokofer. I think the Gibbs Phenomenon is a property of Fourier Series in general and pertains to the "overshoot" and "undershoot" of the convergence of the series near a discontinuity. That however is not the reason the sum is discontinuous at prime powers.

    I think I know why it's discontinuous just not why at prime powers. No doubt the particular values of the zeta zeros somehow are causing the discontinuities there. I suspect no other values of [itex] \rho[/itex] in the sum will cause discontinuities at ALL prime powers and that there may exist a set of values that do so only at primes. Interesting problem . . .
  7. Aug 31, 2006 #6


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    If I write the sum as:

    \lim_{T\to\infty}\sum_{|\gamma|\leq T}\frac{x^{\rho}}{\rho}&=
    \left[\sum_{i=1}^{\infty}\frac{\sqrt{x}\left[Cos(t_i lnx)+2t_i Sin(t_i lnx)\right]}{1/4+t_i^2} \right] \\
    &=\sqrt{x}\left[\sum_{n=1}^{\infty}a_nCos(\omega_n ln(x))+\sum_{n=1}^{\infty}b_n Sin(\omega_n ln(x))\right]

    where the t's are the ordinates of the Zeta zeros above the real axis, I can begin to see why it exhibits "saw-tooth-like" discontinuities: The sum is actually a Fourier-like series in disquise which converges to the saw-tooth-like function which the sum represents!

    See guys . . . never a dull moment in the big-house.:biggrin:
    Last edited: Aug 31, 2006
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