# Partial Fourier of von Mangoldt error estimation?

Through some properties of von Mangoldt we can extract Riemann zeta zeros alike Fourier transform but without going beyond scale 1 as this hides prime numbers in the process. This is a flat part of one of the possible options:

$\displaystyle f(x)=1-\frac{\ln(x)}{2}+\sum\limits_{n=1}^{M}\left(\Lambda(n)-\frac{n}{M}\right)\frac{\cos(x\ln(n))}{\sqrt{n}}$

M=600 (It is necessary to increase M in order to get a higher accuracy of larger Riemann zeta zeros, as one would expect. Increasing M, however, does not necessarily increase the accuracy of the positions of smaller zeros, as one would expect.)

For small values of M, $f(x)$ is behaving nicer than taking, for example, purely $\Lambda(n)$. If you take $\Lambda(n)-1$ (instead of $\Lambda(n)-\frac{n}{M}$) the function does not behave as smoothly and it is not clear if the error term remains bounded.

This is a great tool as it illustrates nicely the connection between Riemann zeta zeros displayed in this partial Fourier transform and Riemann zeta pole at 1. For example, this is what is the part of the illustration.

We can find from explicit formulae that it is formally:

$\displaystyle \Lambda(n)=\lim_{\epsilon \to 0}\int_{n-\epsilon}^{n+\epsilon}\sum\limits_{\rho}\frac{1}{x^{\rho}} dx$

$\displaystyle \sum\limits_{n=1}^{\infty} \frac{\Lambda(n)}{n^s} =\lim_{\epsilon \to 0}\int_{n-\epsilon}^{n+\epsilon}\sum\limits_{n=1}^{\infty}\sum\limits_{\rho}\frac{1}{n^{s}x^{\rho}} dx$

$\displaystyle \sum\limits_{n=1}^{\infty} \frac{\Lambda(n)}{n^s} =2\epsilon\sum\limits_{n=1}^{\infty}\sum\limits_{\rho}\frac{1}{n^{s}n(\epsilon)^{\rho}} \approx 2\epsilon\sum\limits_{\rho}\zeta(s+\rho)$

If you take $s \to \frac{1}{2}+\rho_{k}$ this last is formally saying that we can expect a sharp drop to the left at each Riemann zeta zero because to the right we have a pole at 1 and Riemann zeta zeros are symmetrical. $f(x)$ in some form might be the answer in what sense the above formal derivation that ignores much of the convergence works.

Another thing is that first Riemann zeta zero can be approximated by simply minimizing

$\displaystyle \frac{\ln(2)\cos(x\ln(2))}{\sqrt{2}}+\frac{\ln(3)\cos(x\ln(3))}{\sqrt{3}}$

around 14, and so on. All this because primes are accessible in the expression.

It seems that $f(x)$ never reaches more than $O(1)$, outside Riemann zeta zeros and region that is initially x < 10 and is slowing moving to the right, (it seems it stays within -5 and 5), for M increased as one is pleased, but I cannot reach any hint of proving it.

Does it look more obvious to someone else?

(In general for the expression above the unbounded oscillations seen to the left are slowly overwhelming the zeros peaks. Still, even for M one million, this will start engulfing only the first zero. The part staying to the right seems bounded all along. Is this visible smoothness anything substantial or should we be happy with $\Lambda(n)-1$ and consider the above useful for small zeros only? That is the question behind the question.)

• For a fixed $M$, the function $f(x)$ equals $-\frac12\ln x$ plus a bounded function of $x$. Therefore it will tend to $-\infty$ as $x$ increases, and certainly not itself stay bounded. – Greg Martin Jul 16 '17 at 18:13
• Under the RH $|x-\psi(x)| < C x^{1/2}\log^2 x$ so that $$\frac{1}{s}\sum_{n=1}^N \frac{1-\Lambda(n)}{n^{s}}= \frac{N-\psi(N)}{s N^s}+\sum_{n=1}^{N-1} (n-\psi(n))\frac{n^{-s}-(n+1)^{-s}}{s}$$ converges to $\frac{\zeta'(s)}{s\zeta(s)}+\frac{\zeta(s)}{s}$ uniformly for $\Re(s) \ge \sigma > 1/2$. You can also look at things like $\sum_{n=1}^\infty \frac{1-\Lambda(n)}{n^{1/2+it}} e^{- \pi (\log n)^2/N^2}$ which converges to $[\frac{\zeta'}{\zeta}+\zeta](1/2+it)$ in the sense of distributions – reuns Jul 16 '17 at 19:41
• Similar to this second came into the picture by replacing $\Lambda(n)-\frac{n}{M}$ with simpler $\Lambda(n)-1$. This second however gives far worse error (even though spikes are there), looks like O(n), but at least it is only O(n). Thanks. – user318107 Jul 16 '17 at 20:08
• @alex.peter I told you need to take $\Re(s) > 1/2$ to obtain the convergence. Also you can take $C = 1$ obtaining an effective bound for $\left|\frac{\zeta'(s)}{s \zeta(s)}+\frac{\zeta(s)}{s}-\frac{1}{s} \sum_{n=1}^N \frac{1-\Lambda(n)}{n^s}\right|$ (for $\Re(s) > 1/2$ !!!) using that $|\frac{n^{-s}-(n+1)^{-s}}{s}| = |\int_n^{n+1} y^{-s-1}dy| \le |n^{-s-1}|$ – reuns Jul 16 '17 at 21:22
• Just to add a reference from the literature (I am aficionado, thus I am no able to read it, but I know that has a high quality). You can see a different representation of the Mangoldt function due to Jesús Guillera here, and if I understand well the corresponding analysis referenced by the professor as a paper of this 2017, from arXiv. Good luck. – user243301 Jul 17 '17 at 0:03

Under the RH $|\psi(x)-x| \le (\log x)^2 x^{1/2}$ so by summation by parts for $\Re(s) > 1/2$
$$\frac{1}{|s|}\ \left|\frac{\zeta'(s)}{ \zeta(s)}+\zeta(s)-\sum_{n=1}^{N+1} \frac{1-\Lambda(n)}{n^s}+\frac{\psi(N\!+\!1)\!-\!N\!-\!1}{N^s}\right| = \left| \sum_{n=N+1}^\infty (n\!-\!\psi(n))\int_n^{n+1} y^{-s-1}dy\right|$$ $$\textstyle\le \int_N^\infty (\log y)^2y^{1/2-\sigma-1}dy = \frac{d^2}{d^2\sigma} \frac{N^{1/2-\sigma}}{\sigma-1/2}=\frac{N^{1/2-\sigma}}{\sigma-1/2}\left((\log N)^2-\frac{2 (\log N)^2}{\sigma-1/2}+\frac{2}{(\sigma-1/2)^2}\right)$$
Also since $\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1} = \sum_\rho \frac{1}{s-1/2-i\Im(\rho)}- \log 2\pi + 2\frac{\Gamma'(s/2)}{\Gamma'(s/2)}$, you can use the preceding to detect the number of non-trivial zeros in the neighborhood of $s$.