# Riemann zeta zeros Fourier like divergent square wave. Can you complete this analogy?

The question is to complete this analogy:

$$\left|Z(t)\right|=\left|\zeta \left(\frac{1}{2}+i t\right)\right| \tag{1}$$

is to:

$$Z(t)=e^{i \vartheta (t)} \zeta \left(\frac{1}{2}+i t\right) \tag{2}$$

as:

$$\left|f(t)\right|=\left|\sum\limits_{n=1}^{n=k} \frac{1}{n} \zeta(1/2+i t)\sum\limits_{d|n}\frac{\mu(d)}{d^{(1/2+i t-1)}}\right| \tag{3}$$

is to what? $$f(t) = \text{?} \tag{4}$$

I am asking this because $$f(t)$$ divided by Riemann Siegel theta and the $$k$$-th Harmonic number:

$$f_{\vartheta(t)}(t)=\frac{\text{sgn} (Z(t))\left|\sum\limits_{n=1}^{n=k} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n}\frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}\right|}{g(t)+H_{\text{k}}}$$

where:

$$g(t)=\frac{\partial \vartheta (t)}{\partial t}$$

and where $$\vartheta(t)$$ is the Riemann-Siegel theta function,

has a nice plot:

The blue graph passes through the Riemann zeta zeros, like the Riemann-Siegel zeta function. However, the square wave $$f(t)$$ is divergent despite the appearance of the plot.

$$Z(t)$$ is the Riemann-Siegel zeta function.

• $P_k(s)=\sum_{n=1}^k n^{-1} \sum_{d |n} \mu(d) d^{1-s}$ is a Dirichlet polynomial with real coefficients so $P_k(s) e^{-i h_k(s)/2}$ is real for $\Re(s) = 1/2$ where $h_k(s) = \log P_k(s) -\log P_k(1-s)$ is analytic on $\Re(s) = 1/2$. – reuns Nov 17 '18 at 10:49
• What $f_{\vartheta}(t)$ means to you – reuns Nov 17 '18 at 10:56
• $f_{\vartheta}(t)$ means that the function is divided by Riemann Siegelt theta. – Mats Granvik Nov 17 '18 at 11:36