Does this normalization of the Riemann zeta function make sense?

Question

For $c=0$ the following should be true for the $n$-th Gram point: $$\frac{x}{2 \pi e}\log\left(\frac{x}{2 \pi e}\right) = \frac{x}{2 \pi e}\log \left(\frac{x}{2 \pi e}\right) + \frac{-c+n}{e}-\frac{\vartheta (x)}{\pi e}$$ and for $c=\frac{1}{2}$ it should be true for the $n$-th complementary Gram point. $\vartheta (x)$ is the Riemann-Siegel theta function.

That equation can be solved for $x$ like this:

$$x=2 \pi e \exp \left(W\left(\frac{-c+n-\frac{\vartheta (x)}{\pi }+\frac{x \log \left(\frac{x}{2 \pi e}\right)}{2 \pi }}{e}\right)\right)$$

where $W$ is the Lambert W function.

Then name the right hand side the function $f(x)$: $$f(x)=2 \pi e \exp \left(W\left(\frac{-c+n-\frac{\vartheta (x)}{\pi }+\frac{x \log \left(\frac{x}{2 \pi e}\right)}{2 \pi }}{e}\right)\right)$$

and iterate it as many times as computer power allows. This iteration is the normalizing function that we call $g(n,x)$:

$$g(n,x)=f(...f(f(f(f(x))))...)$$

This should allow us to cut the real part of the Riemann zeta function:

$$\Re\left(\zeta \left(\frac{1}{2}+i g(n,1)\right)\right)$$

at complementary Gram points $c=\frac{1}{2}$ into a family of functions with intervals $n=\left[k,k+1\right]$ where $k=0,1,2,3,4,...,100$. The seed point is $x=1$. In case I did not say this correctly, this how it is written in Mathematica:

Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]

Plot of the superimposed real parts:

where the end points at $0$ and $1$ are complementary Gram points.

The average of the 101 curves looks like this:

This looks very much like the function $1+\cos (x+\pi )$ on the interval $\left[0,2\pi\right]$:

Mathematica code for the normalized real parts:

Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
   E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n - 
        RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
  Table[ListLinePlot[
    Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], 
    DataRange -> {0, 1}, PlotRange -> {-1.3, 6.4}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
   E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n - 
        RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
  ListLinePlot[
   Sum[Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0, 
      100}]/101, DataRange -> {0, 1}], k]]

We can do the same for the imaginary part of Riemann zeta:

$$\Im\left(\zeta \left(\frac{1}{2}+i g(n,1)\right)\right)$$

Again we look at the family of functions between complementary Gram points $c=\frac{1}{2}$ with intervals $n=\left[k,k+1\right]$ where $k=0,1,2,3,4,...,100$. The seed point is $y=1$. In Mathematica:

Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]

The plot of the superimposed imaginary parts looks like this:

where the middle common point at $\frac{1}{2}$ is at a complementary Gram point.

The average of the 101 curves is:

This again looks like $\sin (x)$ on the interval $\left[0,2\pi\right]$:

Would such a normalization with the function $g(n,x)$, be any easier to study, than the Riemann zeta function directly?

Mathematica code for the normalized imaginary parts:

Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
   E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n - 
        RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
  Table[ListLinePlot[
    Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], 
    DataRange -> {0, 1}, PlotRange -> {-5, 5}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
   E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n - 
        RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
  ListLinePlot[
   Sum[Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0, 
      100}]/101, DataRange -> {0, 1}], k]]

Answer 1

I don't know what you want to say with this iteration.

---

See this. Defining the arguments correctly (zero on $s > 1$, continuous on $\Re(s) > 1$, and extended by continuous variation on horizontal lines to $\Re(s) \le 1$) we can express the number of non-trivial zeros $\zeta(\rho) = 0, \Im(\rho) \in (0,t]$ with $$ N(t) =\frac{\text{arg }(Z(t))}{\pi} = \frac{\vartheta(t)}{\pi}+ \frac{\text{arg }(\zeta(1/2+it))}{\pi} = \frac{t}{2\pi} \log(\frac{t}{2\pi e}) + 7/8+\mathcal{O}(1/t) + \mathcal{O}(\log t)$$ where $\frac{t}{2\pi} \log(\frac{t}{2\pi e}) + 7/8+\mathcal{O}(1/t)$ comes from the Stirling approximation applied to $\vartheta(t)$, and $ \mathcal{O}(\log t)$ comes from a bound for $\text{arg }(\zeta(1/2+it))$.

Assuming the Riemann hypothesis and that $\text{arg }(\zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$\rho_n \approx 1/2+i \vartheta^{-1}(n)$$