21. About the roots of Riemann zeta function

The content of the article below is slightly different from the most recent article.Riemann zeta function and quantum theory, About roots of Riemann zeta function

 

 

In the previous article¹, the zeta function has the following relationship.

 

(2)   $\zeta(s)=-\frac{1}{2(s-1)} + \frac{p}{s-1}\zeta_1(s-1), b_0=-1/2$

 

Assume that a certain term $a_s$ is added to both sides and the reciprocal of the two forms an arithmetic sequences. In other words,

 

$\frac{1}{a_s}-\frac{p}{(s-1)a_s-1/2}=a$, $a$: constant.

 

=>$a(s-1)a_s^2-(s-p+a/2-1)a_s+1/2=0$

 

$a_s=\frac{s-p+a/2-1 \pm \sqrt{(s-p+a/2-1)^2-2a(s-1)}}{2a(s-1)}$.

 

Then, the equation (2) is modified as follows,

 

$\zeta(s)+ a_s =\frac{p}{s-1}(\zeta_1(s-1)+ a_{s-1})$

 

$\zeta(s)+ a_{s} =\frac{p}{s-1}(\zeta_1(s-1)+ a_{s-1})$

$\zeta(s-1)+ a_{s-1} =\frac{p}{s-2}(\zeta_1(s-2)+ a_{s-2})$

$\zeta(s-2)+ a_{s-2} =\frac{p}{s-3}(\zeta_1(s-3)+ a_{s-3})$

$\zeta(s-3)+ a_{s-3} =\frac{p}{s-4}(\zeta_1(s-4)+ a_{s-4})$

...

 

If $\zeta(s) \approx \zeta_1(s)$ by varying the value $a$,

 

(3) $\zeta(s)+ a_s =\frac{p^{s-2}}{(s-1)!}(\zeta(2)+a_2)$

 

If $\zeta(s)=0$, $s$ must satisfies $a_s =\frac{p^{s-2}}{(s-1)!}(\zeta(2)+a_2)$.

 

Above, the reciprocal of $a_s$ was assumed to be an arithmetic sequence, but in reality, it may be a different sequence.

Because $\zeta(s)->-a_s=1$ when $s->\infty$, $a < 0$ and $a$ must approach to $-1$. Since $\zeta(s)$ shows periodicity when $s$ is even and odd, $a_s$ is assumed to be related to trigonometric functions, and $a_s$ extended to complex numbers can be expressed like below.²

 

$a_s \approx -\frac{1 \pm \sqrt{1-\frac{2}{a(s-1)}}}{2}$

If $s>1$ then,

$a_s=-\frac{1}{2} \pm \frac{1}{2}\cos\theta=-\cos^2(\theta/2), -\sin^2(\theta/2), \theta=\frac{s}{(s-1)!+1 }\pi$

If $0<s<1$ and $\Gamma(s)=s$ then,

$a_s=-\frac{1}{2} \pm \frac{i}{2}\tan\theta=-\frac{e^{i\theta}}{2\cos\theta}, \theta=\frac{s}{s+1}\pi $

 

When $0<s<1$, $\zeta(s)$ can be changed as follows. Assume that $z=\sqrt{2}\cos(\theta/2)$ and $\Re(z) \leq 1$. Then,

$z^2=(1+\sqrt{1-\beta^2}), \beta= \sin\theta+\epsilon$

There is no inevitable reason to change $z^2$ like this. The reason is to simplify the equations and to approximate $\beta=\cos\theta$ when $s>1$.

Substituting $\beta$ with $1/\beta$ in order for $\beta$ to be shown greater or less than $1$ and swapping $\theta$ with $\pi/2-\theta$, $\beta \approx \cos\theta$ and

$z^2=1+i\tan\theta=\frac{\cos\theta+i\sin\theta}{\cos\theta}$,

$\cos\theta=\frac{\cos\theta+i\sin\theta}{z^2}$

Since the integral section and the sign of the differential operator change simultaneously, $\zeta(s)$ does not change. What to be careful here is that by changing $\beta$ to $1/\beta$, the $\cos\theta$ of the real number changes to the complex number $(\cos\theta+i\sin\theta)/z^2$.

Therefore, when $0<\Re(s)<1$,

(4)  $\zeta(s)=\int(\cos\theta)^{s-2}b_0(x)dx$

 

$=-2\int_{1}^{\infty}z^{1-2s}b_0(x)dz=2\int_{0}^{1}z^{2s-2}(zb_0(x))dz$

Here, $x=\cos\theta=\frac{1}{z^2}, dx=-2dz/z^3$. When replacing variable $x$ with $z$, $zb_0(x)$, not $b_0(x)$, converges.

 

$\zeta(s)$ can be described like below

 

$\zeta(s)=\frac{b'_0}{2s-1} + \frac{p'}{2s-1}\zeta_{1}(s-1)$.

 

$\zeta(s)+a_s=\frac{15(p'/2)^{s-3}}{8(s-1/2)!}\zeta(3)=\frac{15(p')^{s-3}}{\Gamma(s)}\zeta(3), \Gamma(s)=(2s-1)(2s-3)(2s-5)...1$. $b_0', p'$ are constants.³

 

Nontrivial zeros of $\zeta(s)$

 

When $0 < \Re(s) < 1$, $\Gamma(s)=2s-1$. If this is calculated by replacing $s$ with the complex number $1/2+it$,

 

(a) $\zeta(s)=\frac{15(p')^{s-3}}{2s-1}\zeta(3)-a_s=\frac{15(p')^{s-3}}{2ti}\zeta(3)+\frac{e^{i\theta_2}}{2\cos\theta_2}=0$

 

Therefore, all that remains is to find $t, \theta_2$ satisfies this equation. As you can see from the equation, if there is at least one root, there are countless roots.

 

The reason why there are no roots when the real part of $s$ is not $1/2$ is as follows.

 

If $p'=|p'|e^{ic}$, the equation (a) becomes like

$ d|p'|^{s-3}e^{i(c(s-3)-\theta_2)}/(2s-1)+1=0, d=30\zeta(3)\cos\theta_2 $

The minimum condition for this to be true is that the size, $d|p'|^{s-3}=1-2s$. The $s$ that satisfies this does not exist if $1/2 < \Re(s) <1$.⁴

 

If $\zeta(s)=\frac{1}{p_0}\int{x^{s-2}b_0(x)}dx$ considering the actual zeta function value, $p$ is changed to $p/p_0$ in the equation (3), but it does not affect the result.

 

If you quote this article, please cite the source.

About_roots_of_Riemann_zeta_function_20240102.pdf

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[Another mothod. 2024.01.04]

 

$b_0(z^2)$ can be approximated in the form of a normal distribution. If you simply regard it as the result of the definite integration, it can be written as follows.

 

(b)   $b_0(1/z^2)/z \approx e^{-(z-1)^2/2\sigma^2}$, $\sigma$: a function of $s, p$

 

Using this approximation, $\zeta(s)$ can be expressed like belows.

 

$\zeta(s)=\int z^{2s-2}e^{-(z-1)^2/2\sigma^2}dz$

 

$=\frac{1}{2s-1}+\frac{1}{\sigma^2(2s-1)}\int z^{2s-1}(z-1)e^{-(z-1)^2/2\sigma^2}dz$

 

$=\frac{1}{2s-1}+ \frac{1}{\sigma^2(2s-1)}(\zeta(s+1)-\zeta(s+1/2))$

 

$\zeta(s)$ can be obtained in a similar way as in the equation (3). The difference here is that while the $\zeta(s)$ can be expressed exactly as an integral, the $\sigma$ value must be expressed as $s, p$.

 

The gamma function, $\Gamma(s)$ can be expressed like belows.

 

$\Gamma(s)=\frac{1}{p}\int z^{2s}e^{-z^2/2\sigma^2}dz=\frac{2}{p}\sigma^2(s-1/2)\int z^{2s-2}e^{-z^2/2\sigma^2}$

 

$=\frac{2}{p}\sigma^2(s-1/2)\Gamma(s-1)=(\frac{2}{p}\sigma^2)^{s}(s-1/2)!\int z^{\alpha}e^{-z^2/2\sigma^2}dz$

 

($s \geq 0, 0 \leq \alpha <1$).

 

When $s$ is between $0$ and $1/2$, $(\frac{2}{p}\sigma^2)^{s}(s-1/2)!=s\sigma^2/p $.

 

[Calculation of $b_n$ and Relation between $\zeta_s(s)$ and $\zeta(2s)$. 2024.02.11]

 

If $b_n(x)=\frac{f_n(x)}{(e^x-1)^n}$ then,

 

$b_{n+1}(x)=\frac{f'_n(x)(e^x-1)-ne^xf_n(x)}{(e^x-1)^{n+1}}=\frac{f_{n+1}(x)}{(e^x-1)^{n+1}} $

 

$f'_n(x)(e^x-1)-ne^xf_n(x) =f_{n+1}(x)$

 

$f'_n(x)=\frac{f_{n+1}(x)+ne^xf_n(x)}{e^x-1}$.

 

Using L'Hopital's theorem,

 

$f'_{n}(\epsilon)=-(n-2)f'_{n-1}(\epsilon)-(n-1)f_{n-1}(\epsilon) , (\epsilon \approx 0)$

 

If $f_{n-1}(\epsilon)$ is negligible compared to $f'_{n-1}(\epsilon)$,

 

$f'_{n}(\epsilon)=-(n-2)f'_{n-1}(\epsilon)$

 

By using L'Hopital's theorem repeatly.

 

(c)   $ b_n=b_n(0)=\frac{-(n-2)f'_{n-1}(\epsilon)}{n(e^\epsilon-1)^{n-1}}=...=\frac{(-1)^{n-1}}{n(n-1)}$

 

Therefore,

 

(d)   $\zeta(2s)=\frac{b_0}{2s-1}+\frac{b_1(-p)^1}{(2s-1)(2s-2)}+...+\frac{b_{s-1}(-p)^{s-1}}{(2s-1)(2s-2)...(s-2)}+\frac{(-p)^s}{(2s-1)(...)(s-1)}\zeta_s(s)$.

($\zeta_s(s)=\int x^{s-2}b_s(x)dx$)

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^{1. Refer to Riemann zeta function and quantum mechanics, Bernoulli sequence and Riemann zeta function, Potential and Riemann zeta function.}

^{2. If $2/a \approx (s-1)\sin^2\theta$, $1/a$ can be regarded as arithmetic sequence. The remaining problem will be to identify the exact integral relationship $\int x^{s-3}(b_1(x)-b_0(x))dx=-\cos^2(\theta_{s-1}/2)+\frac{1}{2p}+\frac{s-1}{p}\cos^2(\theta_s/2)$. More detailed explanation is in Supplementary explanation of article 21}

^{3. If $(s-1)!$ is changed to $(s-1/2)!$ in $\theta$ of $a_s$, the value of $a_3$ does not become $0$, but it is ignored here.}

^{4. Assuming that $\theta=s\pi/(s+1)$ changes to $\theta_2=s\pi/(s+1/2)$ when $1/2<s<1$ in the real number range, $\pi/3< \theta_2 < \pi/2$.}