I have noticed an interesting property related to the Gibbs phenomenon for the Fourier transform of the zeta zeros in Riemann's explicit formula, namely that the rate at which $r\rightarrow 2 $ in the interval $[2,3]$

enter image description here

where $r$ is the point at which

$$\operatorname{li}(x)-\sum_{\rho}\operatorname{li}(x^\rho)-\log 2+\int_{x}^{\infty}(dt)/(t(t^2-1)\log t)=1$$

for partial sums of $\sum_{\rho}\operatorname{li}(x^\rho).$

From initial observations, it seems that for each sucessive zero added, $r-2\sim C/\operatorname{li}(n)$ for some $C<1/2.$

Much the same results can be achieved with finding $r$ for the partials sums at the points where

$$n-\sum_{\gamma}^{}\dfrac{2\log n\sin(\gamma \log n )}{\gamma\sqrt{n}} = 5/2$$

where $\gamma=$ imaginary parts of zeta zeros.

Does this suggest that the zeta zeros and the prime powers are in some sort of one to one correspondance?

Resposted to MO here

  • $\begingroup$ I don't see where prime powers appear anywhere? $\endgroup$ – anon Jul 3 '15 at 15:22
  • $\begingroup$ @anon that is precisely my point - and yet $r\rightarrow 2$ proportionally to the rate growth of the prime powers. Note this seems to be independent of the logarithmic integral from which they are being taken. $\endgroup$ – martin Jul 5 '15 at 8:35
  • $\begingroup$ Do you mean inversely proportionally? So, you're saying $r_n-2$ is $\sim$ to the reciprocal of the number of prime powers $\le n$? Aren't the number of primes and prime powers less than $n$ asymptotically proportional? If so, why are you talking about prime powers instead of primes, or anything else with the same growth rate? As I understand it, $n$, $\log n$, $\log\log n$ and so on and their powers/products/exponentials appear ubiquitously in analytic number theory's asymptotics. Anyway, could you explain what $r$ is supposed to be more intuitively? $\endgroup$ – anon Jul 5 '15 at 18:34

Your Answer

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

Browse other questions tagged or ask your own question.