# Convergence of Riemann spectrum/Fourier transform of prime powers

Prime Numbers and the Riemann Hypothesis by Mazur and Stein makes use of an interesting function: $$\hat{\Phi}_{\le C}(\theta)=2\sum_{prime\:powers\:p^n\le C}p^{-n/2}\cdot log(p)\cdot cos(n\cdot log(p)\cdot \theta)$$

The function is basically a Fourier transform of symmetrized Dirac $$\delta$$ functions at the prime powers. Per the authors, the "spikes" of $$\hat{\Phi}_{\le C}(\theta)$$ "for large $$C$$ pinpoints the spectrum" $$\{\theta_1,\theta_2,\theta_3,... \}$$, i.e., the sequence in order of imaginary parts of the non-trivial $$\zeta$$-function zeros.

As an example, here's what I get with a C program and MATLAB to replicate Figure 32.7 in the book:

That matches the figure very well so I'm pretty sure the code is right. The red lines are at the locations of the the first six $$\theta$$'s

The first few $$\theta_i$$ are more or less at the negative peaks. However, increasing $$C$$ further doesn't really improve the situation:

Shouldn't the $$\theta_i$$ corresponding to the imaginary parts of the non-trivial $$\zeta$$-function zeros get more distinct as $$C$$ increases, not less? Or maybe there's a bug?

EDIT:

Per reuns suggestion of regularizing each term in the sum by $$p^{−a^2θ^2n/2}$$, here's a comparison of the result at $$\theta_1$$ using $$a=0$$ vs. $$a=0.1$$:

$$a=0$$:

$$a=0.1$$:

• The series $f(\theta) = 2\sum_{prime\:powers\:p^n}p^{-n/2}\cdot log(p)\cdot cos(n\cdot log(p)\cdot \theta)$ converges only in the sense of distributions, you need to look at the regularized version $f_a(\theta)= 2\sum_{prime\:powers\:p^n\le C}p^{-n/2}\cdot log(p)\cdot cos(n\cdot log(p)\cdot \theta) p^{- a^2 \theta^2 n/2}$ with $a$ small to make appear the convolution of the gaussian $|a|^{-1} e^{- \theta^2/(2a^2)}$ with the distribution $\sum_{t \text{ imaginary part of non trivial zeros}} \delta(\theta-t)$ en.wikipedia.org/wiki/Explicit_formulae_(L-function) Feb 22, 2019 at 0:37
• That regularization term is interesting--I'll try to grok what it's doing, but as a magic trick at the moment, given the right choice of $a$ it does seem to make the $\theta_i$ stand out (see edit above). Does $a$ need to be tuned empirically for each $\theta_i$? Feb 22, 2019 at 11:15

    N = 10^5; mu = zeros(1,N); mu(1)= 1; for n = 1:N,  mu(n+n:n:end) = mu(n+n:n:end)-mu(n); end;
Lambda = zeros(1,N); logn = log(1:N); for n = 1:N, Lambda(n:n:end) =  Lambda(n:n:end) + log(n)*mu(1:floor(N/n)); end;

Lambda12 = Lambda ./ (1:N).^(1/2);         % Lambda(p^k) = log(p)

% the raw Fourier transform of the OP
dx = 0.01; X = [0:dx:50]; f = zeros(1,length(X));
for l = 1:length(X), x = X(l); f(l) = sum(Lambda12 .* exp(-i*logn * x));end;

% substract the contribution of the pole at s=1/2 and the trivial zeros at s = -2k-1/2
h = (exp(log(N)*(1/2-i*X))-1)./(1/2-i*X); for k = 1:500, h = h - (exp(log(N)*(-2*k-1/2-i*X))- 1)./(-2*k-1/2-i*X); end; g = f-h;

% remove oscillations due to finite N
G = [g(end:-1:1),g]; K = 50; filter = ones(1,K)/K; filter = conv(filter,filter); filter = conv(filter,filter); F = conv(real(G),2*filter); F = F-F(5000);

plot(-F);


You can run it in https://octave-online.net/ (click 2 times on add 15 seconds during the execution)

The peaks are the non-trivial zeros of $$\zeta(s)$$, each one integrates to $$2\pi$$

• Neat--the code runs nicely on MATLAB. I gather you first get the Mobius function, then Mangoldt, and then the graph above is of $$f(\theta)=\sum_{c=1}^C \frac{\Lambda(c)-1}{\sqrt c}\cdot r(c)\cdot cos(\theta \log c)$$ Where $r$ is the regularization term. Is that it? Is this closely related to the $\Phi$ function above somehow? Feb 22, 2019 at 20:23
• One thing I noticed: the regularization term in your program doesn't do much. If it's set to 1 the output is pretty much the same. But if 1 isn't subtracted from $\Lambda$ the result looks pretty much like the $\Phi$ function above. So maybe that's really the key (removing the effect of the pole at 1 like the comment states)? Feb 23, 2019 at 1:43
• @JoeKnapp I cleaned up the code, can't do much better. Feb 23, 2019 at 5:24