3
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Let the matrix $A$ be the lower triangular matrix defined as: $$A=\text{If } n \bmod k=0 \text{ then } \frac{1}{n^c} \text{ else } 0$$ and let the matrix $B$ be the upper triangular matrix defined as: $$B=\text{If } k \bmod n=0 \text{ then } \frac{n \cdot \mu (n)}{k^s} \text{ else } 0$$
where $\mu(n)$ is the Möbius function.

Then the matrix product of the matrices $A$ and $B$ is the symmetric matrix $T$ starting:

$$T=A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which has the Dirichlet generating function:

$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$

For a while consider the three limits of the generating function:

$(1)$
$$\lim_{c\to \infty } \, \frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}=\zeta (s)$$

$(2)$
For $s$ equal to a Riemann zeta zero: $$\lim_{c\to 1 } \, \frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}=0$$ $(3)$
while for $s$ not equal to a Riemann zeta zero: $$\lim_{c\to 1 } \, \frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}=\infty$$

So thereby there is/exists a continuous transformation through the value of the variable $c$ between points $t$ such that:

$$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$$ and: $$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) \neq 0$$

and points $t$ such that (both):

$$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$$ and: $$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$$

or in other words we can push the Franca-LeClair points (or complementary Gram points) to coincide with the Riemann zeta zeros by letting $c\to 1$.

The plot for: $c= 1+\frac{1}{100}$
$$f(t)=\frac{\zeta (\frac{1}{2}+it) \zeta (c)}{\zeta (\frac{1}{2}+it+c-1)}$$

zeta zero spectrum

illustrates the accentuation of the zeta zeros.

There is a Fourier like series of this spectrum over here For $c=1$: $$f(t)=\sum\limits_{n=1}^{\infty} \frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(\frac{1}{2}+it-1)}}}{n^c}$$

Fourierlikespectrum

and that works better for finding points $t$ such that the real part of the Riemann zeta function is zero, iteratively, but I find the Dirichlet generating function more interesting because it is smoother and accentuates the Riemann zeta zeros more.

The Franca-LeClair equation (131) at page 37 in their arXiv paper is:

$(4)$ $$\frac{y_n}{2\pi}\log\left(\frac{y_n}{2\pi e}\right)+\frac{1}{\pi}\lim_{\delta\to 0^{+}} \, \arg \left(\zeta \left(\delta+i y_n+\frac{1}{2}\right)\right)=n-\frac{11}{8}$$

The Franca-LeClair formula (163) at page 47:

$(5)$ $$y_n = 2 \pi \exp (1) \exp \left(W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)\right)$$

We then solve $(4)$ for $n$ and input the resulting expression into $(5)$ which gives as at mathematics stack exchange at this question for $k=\frac{1}{2}$ the relationship:

$(6)$ $$y_n=2 \pi \exp (1) \exp \left(W\left(\frac{\frac{y_n}{2 \pi }\log \left(\frac{y_n}{2 \pi e}\right) -k+n-\frac{\vartheta (y_n)}{\pi }}{\exp (1)}\right)\right)$$

The above can be iterated and converges to the Franca-LeClair points. With Franca-LeClair points (or complementary Gram points) as already said I mean points $t$ such that:$\Re\left(\zeta \left(i t+\frac{1}{2}\right)\right) = 0$ and:$\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right) \neq 0$.

What remains is to include the Dirichlet generating function $\frac{\zeta (s) \zeta (c)}{\zeta (c+s-1)}$ to push the Franca-LeClair points closer to the Riemann zeta zeros. Since the Franca-LeClair equation (131) already corresponds to $\zeta(s)$, we only need to add the part $\frac{\zeta (c)}{\zeta (c+s-1)}$ to $(6)$. This we include as:

$$\arg \left(\frac{\zeta (c)}{\zeta (c+s-1)}\right)$$

or on the critical line:

$$\arg \left(\frac{\zeta (c)}{\zeta (\frac{1}{2}+it+c-1)}\right)$$

The right hand side in $(6)$ for $k=\frac{1}{2}$ then becomes:
$(7)$

$$t = 2 \pi e \exp \left(W\left(\frac{\frac{\arg \left(\frac{\zeta (c)}{\zeta \left(c+i t+\frac{1}{2}-1\right)}\right)}{\pi }-k+n-\frac{\vartheta (t)}{\pi }+\frac{t \log \left(\frac{t}{2 \pi e}\right)}{2 \pi }-1}{e}\right)\right)$$

Plots of the family of functions $f(t,n,k,c)$ for $k=\frac{1}{2}$, $c = 1 + 1/40$ and $n=1,2,3...12$: $$f(t,n,k,c)=2 \pi e \exp \left(W\left(\frac{\frac{\arg \left(\frac{\zeta (c)}{\zeta \left(c+i t+\frac{1}{2}-1\right)}\right)}{\pi }-k+n-\frac{\vartheta (t)}{\pi }+\frac{t \log \left(\frac{t}{2 \pi e}\right)}{2 \pi }-1}{e}\right)\right)$$

family of functions passing through the line x at zeta zeros

Black dots are at $t$ equal to imaginary parts of zeta zeros.

Plots of the sharpened family of functions $f(t,n,k,c)$ for $k=\frac{1}{2}$, $c = 1 + 1/10^{11}$ and $n=1,2,3...12$:

sharpened family of functions passing through the line x at zeta zeros

Is this proof enough that for $k=\frac{1}{2}$ the imaginary part of the $n$-th Riemann zeta zero is the point $t$ such that: $$t = \lim_{c\to 1} \, 2 \pi e \exp \left(W\left(\frac{\frac{\arg \left(\frac{\zeta (c)}{\zeta \left(c+i t+\frac{1}{2}-1\right)}\right)}{\pi }-k+n-\frac{\vartheta (t)}{\pi }+\frac{t \log \left(\frac{t}{2 \pi e}\right)}{2 \pi }-1}{e}\right)\right)$$ ?

$W(z)$ is the Lambert W function.

Mathematica program to verify the claim:

(* start *)
Clear[k, c, n, t];
k = 1/2;
c = 1 + 1/10^100;
n = Range[50];
t = Im[ZetaZero[n]]
N[2*Pi*E*E^
   LambertW[((t/(2*Pi))*Log[t/(2*Pi*E)] + 
       Arg[Zeta[c]/Zeta[1/2 + I*t + c - 1]]/Pi - k + n - 1 - 
       RiemannSiegelTheta[t]/Pi)/E], 110]
% - t
Print["The differences above are zero which verifies the conjecture."]
(* end *)
$\endgroup$
  • $\begingroup$ Come on... Do you understand that $(s-1) \zeta(s)$ is analytic and $=1$ at $s=1$ and if $\zeta(s)$ has a simple zero at $\rho $ then $\frac{s-1}{\zeta(s-1+\rho)}$ is analytic $= \frac{1}{\zeta'(\rho)}$ as $s= 1$ ? And do you understand you need to explain your question in less than 10 lines ? $\endgroup$ – reuns Jun 24 '17 at 19:31
  • 2
    $\begingroup$ Also there is no França-Leclair asymptotic. The asymptotic is thisone known 1 century ago. $\endgroup$ – reuns Jun 24 '17 at 19:33

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