[Properties near limiting value in Riemann zeta function]
This article is about $b_0(x)$ in the previous article1.
(1) $b_0(x)=(\frac{x}{e^{px}-1})'=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(px)^n}{n!(n+1)(n+2)}$
(2) $b_0(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(px)^n}{(n+1)(n+2)}$
When the equation (1), (2) is used, the following difference occurs.
(1)-1 $(x^2(xb_0(x))')'=-1+px-(px)^2/2!+(px)^3/3!+...=-e^{-px}$
(2)-1 $(x^2(xb_0(x))')'=-1+px-(px)^2+(px)^3+...=-\frac{1}{px+1}$
In both cases, the value is same when $x=0$. The difference after multiplying $x^{s-1}$ and integrating is follows.2
(1)-2 $x^2(xb_0(x))'=>-\int x^{s-1}e^{-px}dx=-\frac{(s-1)!}{p^{n-1}}\int x^{s-n}e^{-px}dx$
($0 \leq s-n < 1$)
(2)-2 $x^2(xb_0(x))'=>-\int \frac{x^{s-1}}{px+1}=-\frac{1}{p^s}(p_1+ p_2+\sum_{k=0}^{s-3}(-1)^k\frac{{}_{s-1}C_k}{s-k-1})$
It is assumed that $p_1, p_2$3 are integral constant and $s$ is a natural number.
If [(1)-2]=$f(s)$[(2)-2],
(3) $\frac{(s-1)!}{p^{n-1}}\int x^{s-n}e^{-px}dx =\frac{f(s)}{p^s}(p_1+ p_2+\sum_{k=0}^{s-3}(-1)^k\frac{{}_{s-1}C_k}{s-k-1})$
$\int x^{s-n}e^{-px}dx=\frac{f(s)}{(s-1)!p^{s-n+1}}(p_1+ p_2+\sum_{k=0}^{s-3}(-1)^k\frac{{}_{s-1}C_k}{s-k-1})$
Therefore, it is possible to assume that
(4) $\zeta(s-n)=\int x^{s-n}(\frac{1/x-1}{e^{p(1/x-1)}-1})'dx=\frac{F(s)}{(s-1)!p^{s-n+1}}(p_1+ p_2+\sum_{k=0}^{s-3}(-1)^k\frac{{}_{s-1}C_k}{s-k-1})$
And considering $-a_s \geq 0$,
(5) $\frac{F(s)}{(s-1)!p^{s-n+1}}(p_2+\sum_{k=0}^{s-3}(-1)^k\frac{{}_{s-1}C_k}{s-k-1})=-a_s$
If $F(s)=F_0p^{2s-1}$ and $n=0$ and $F_0$ is constant,
(6) $a_s=-\frac{F_0p^{s-2}}{(s-1)!}(p_2+\sum_{k=0}^{s-3}\frac{(-1)^k {}_{s-1}C_k}{(s-k-1)})$
Here, $s-3$ varies depending on how $b_0(x)$ is treated. In summary, the question is whether $f(s)$ or $F(s)$ can be expressed as a relatively simple function like the equation (6). And this is because $x$ has a limiting value at a certain point. This would not be the case if it were a differentiable function.
When $s$ is near $1$, $a_s$ can be written in two ways from the equation (a) in the article 21:4
(7) $-a_s=\frac{e^{i\theta}}{2\cos\theta}, \frac{e^{i\theta}}{2\sin\theta}$
If you quote this article, please cite the source.
* 2024.02.20
The reason I left the old file in the attached file is because someone might have referenced it.
* 2024.02.11
[Vacant integral, Generating function. 2024.02.09]
If $f(x)=\int_{x}^{\infty} \sum a_n(x+1)^ndx=[\sum \frac{a_n(x+1)^{n+1}}{n+1}]_{x}^{\infty}=- \sum \frac{a_n(x+1)^{n+1}}{n+1}$, then $f(x)$ is a generating function because $f(0)=-\sum \frac{a_n}{n+1}$. If neglecting constant term when integrating, $f(0)=0$.
If $f(x)=\int_{x}^{ \infty }\sum\frac{b_ndx}{ax+1}=[\sum\frac{b_n}{a}\ln(ax+1)]_{x}^{ \infty }, [\sum\frac{b_n}{a}\ln(x+1/a)]_{x}^{ \infty }$, $f(0)=0, \sum\frac{b_n\ln a}{a}$ both are possible.
Above two cases are assumed that $f(\infty)=0$. Normally, the constant after integration is ignored in definite integrals, but when one side of the integral section goes to $\infty$, the handling of the constant becomes a bit ambiguous. Therefore, 'Vacant integral' or 'Generating function' refers to these cases. this is about $p_1, p_2$.
[Theorem 1. 2024.07.10]
All Taylor series can be expressed as the sum of infinite sequences in a specific interval5. The condition for this equation to be established is $|ax| < 1$.
If $f(x)=\frac{1}{1+ax}$,
$f(x)=1-ax+(ax)^2-(ax)^3+...$ in Taylor series, and
If $f(x)=e^{-px}$, there is a constant $a$ that satisfies
$f(x)=\frac{1}{1+ax}, x-\epsilon < x < x + \epsilon$.
After integration, the value is the same.
If $f(x)=x^{s-1}e^{-px}$, there is the function $F(s)$ that satisfies the below.
$\int f(x)dx=F(s)\int \frac{x^{s-1}}{ax+1}dx$
Here, $F(s)$ may be the simple function.
This is also true when $x$ is a complex number.
[Convergent, Divergent Taylor(or Maclaurin) Series. 2024.02.14]
$1/(ax+1)$ can be expressed in two types of infinite series as follows. One case is when $ax$ is greater than 1 and the other is when $ax$ is less than 1.
(10) $1-ax+(ax)^2+...$
(11) $\frac{1}{ax}(1-1/ax+1/(ax)^2+...)$ or $\sum a_k(ax-n)^{k}=\sum b_k\cos^k(ax-n) $
If $ax$ is greater than 1, equation (10) is incorrect. Although the easy form was shown above, it is thought that such incorrect infinite series expressions often appear in more complex functions. For example, this is the usual case in trigonometric functions($\tan ax, \cot ax, \csc ax,...$). That is, Taylor series of these functions must appear differently when $ax$ is near 1. This may be the reason why the Bernoulli sequence cannot be related to zeta functions other than even numbers. The infinite series is expressed as a polynomial of $x$ for convenience, but it is not necessary.
[An algebric function into a transcendental function. 2024.07.10]
- Analytic extension of $\gamma, \zeta$ function.
$(1+x)^n={}_{n}C_{0}+ {}_{n}C_{1}x+ {}_{n}C_{2}x^2+...+ {}_{n}C_{n}x^n$
When $x=1$, $2^n= {}_{n}C_{0}+ {}_{n}C_{1}+ {}_{n}C_{2}+...+ {}_{n}C_{n}$
One thing to check is that the right side is an polynomial of $n$, while the left side is an exponential function of 2. If you substitute $ix/n$ in place of $x$, you can see that it can be approximated as $e^{ix}$ and replaced with $\sin x, \cos x$ function. Also, combinations, algebraically factorials are used here. This shows that it is possible to express transcendental functions as factorial or gamma functions. Therefore, it needs to be more concrete theory.
The binomial theorem is expressed as $(1+x)^n$ and the exponential function is expressed as $(1+x/n)^n$.
If $\gamma(n, x)= (1+x)(2+x)(3+x)...(n-1+x)/n!$,
$=\frac{1}{n}(1+\sum x/n_i+\sum x^2/n_i\cdot n_j+...+\sum x^{n-1}/(n-1)!)$.
When $0 \leq x \leq 1$, $1/n \leq \gamma(n, x) \leq 1$ and $(n-1)! \leq \gamma(n, x)n! \leq n!$
Because $\gamma(n,-x)=(1-x)(2-x)(3-x)...(n-1-x)/n!$,
(12) $\gamma(n,-x)\gamma(n,x)=(1^2-x^2)(2^2-x^2)(3^2-x^2)...((n-1)^2-x^2)/(n!)^2$
When $n$ goes to $\infty$, $\gamma(n,x)$ can be written as $\gamma(x)$6, and
(13) $\gamma(-x)\gamma(x) = \sin\pi x, |x| \geq 1$
Equation.13 is valid because equation.12 , $\gamma(n,x)=0$ has the roots when $x$ is an integer. When $x$ is 0, $\sin\pi x$ is 0, but the absolute value of the $x$ domain is set to a value greater than 1. The reason is that considering the case less than 1, it must be in the form $\gamma(n,x)=x(1+x)...$.
It can be also thought about prime numbers rather than integers, in this case:
(14) $\gamma_p(-x) \gamma_p(x)=q(x)(p_1^2-x^2)(p_2^2-x^2)...(p_{n-1}^2-x^2)/(p_1p_2...p_n)^2$
$ \approx \sin\pi p(x)$
($p_i$: prime number, $p(x)=\frac{(x-1)!+1}{x}, q(x)= k(q_1^2-x^2)(q_2^2-x^2)..., k:$ constant)
When $x$ is a real number, $\sin\pi p(x)=0$ also holds true when $x$ is not a prime number. $q(x)$ is the product of those terms. It needs to be defined for $p(x)$ when $x$ is negative as follows.
$p(x)=\frac{(x+1)!-1}{x}, (x+1)!=-(-x-1)!, x \leq -1$
$\gamma(n, x)$ function can be expanded and written as follows.
(15) $\gamma(n^s, x^s)= (1+x^s)(1+(x/2)^s)(1+(x/3)^s)...(1+(x/n-1)^s)/n^s$
$\gamma(n^s, x^s)\gamma(n^s, -x^s)$
$= (1-x^{2s})(1-(x/2)^{2s})...(1-(x/(n-1))^{2s})/n^s \approx \sin\pi x(1+ f_{2s}(x))$
The reason the $f_{2s}(x)$ term was added is because when the left side $1-(x/n)^{2s}$ is 0, there must be $2s$ roots. For prime numbers, it can be written:
(16) $q(x)(1-x^{2s})(1-(x/p_1)^{2s})(1-(x/p_2)^{2s})...(1-(x/p_{n-1})^{2s})/p_n^{2s}$
$\approx q(x)(1-x^{2s})/\zeta_e(2s, x) = \sin\pi p(x)(1+ g_{2s}(x))$
($1/\zeta_e(2s, x)= (1-(x/p_1)^{2s})(1-(x/p_2)^{2s})...(1-(x/p_{n-1})^{2s})....$)
When $x$ approaches 1, $\zeta_e(2s, 1)=\zeta(2s)$.
Assuming $ \sin\pi p(x) $ is the real part $(e^{i\pi p(x)}-1)/i$ to extend to complex function, $\zeta_e(2s,x)$ can be written like this.
(17) $\zeta_e(2s, x)= \frac{q(x)(1-x^{2s})i}{(e^{i\pi p(x)}- 1)(1+ g_{2s}(x))}$
$p(x)= 1$ when $x$ is 2, 3, and $\sin^2\pi p(x)= \sin^2\frac{\pi}{x}$ when $x$ is natural number, but not prime number. If this holds true for real numbers near 1, 2 and 3, we can write $ \sin^2\pi p(x) \approx \sin^2(\pi /p(x)) $. In particular, if $(x-1)!=x$ when $1 \leq x \leq 3$ ,
$ \sin^2\pi p(x) \approx \sin^2\frac{\pi x}{x+1}$
Extending to complex variable, if $x^{2s}=1$,
$x=\cos\frac{\pi}{s}+i\sin\frac{\pi}{s}=e^{\pi i/s}$
If you quote this article, please cite the source.
Furthermore, If $\zeta(s, t)=\int_t^1 t^{s-2}b_0(t)dt$,
There may be the relation of $p(x)/\pi=((x-1)!+1)/x=1/t$. Refer to Coordinate transform and derivative.
[Comparison with the Laplace transform]
The Laplace transform is used to simplify most natural phenomena expressed as exponential functions of $e$. Due to the nature of the exponential function, it takes advantage of the characteristic that if it is differentiated once or twice, it returns to a form similar to the original function. The integral form that appears in the process of calculating the zeta function in this article can be written in general form as follows.
(8) $A(s)=\int x^{s-1}f(x)dx, f(x)=\sum a_ne^{-px}$
When $f(x)=e^{-x}$, $A(s)=\Gamma(s)$. Functions expressed as irrational numbers, such as transcendental numbers, are usually expressed as integers or infinite series of natural numbers. This transformation seems to have the characteristic of expressing the function as a finite series by making the variable $x$ of these transcendental functions into the function of $\Gamma(x)$.
For example,
$\sin(g(\Gamma(x)))=\sum_{k=0}^{n}c_kx^k, x=( e^x-1 )\sum (e^x-1)^{n}/(n+1)$
[Expressing $e^x, \cos x, \sin x$ with a combination ${}_nC_k$]
$e^x \approx (1+\frac{x}{n})^n$ when $n$ is large enough.
(9) $(1+\frac{ix}{n})^{n}=\cos x + i\sin x$
$=\sum (-1)^k{}_nC_{2k}(x/n)^{2k}+i\sum (-1)^{k}{}_nC_{2k-1}(x/n)^{2k-1}$
If $x=n/((n-1)!+1)$ or the other function of $n$, the equation (9) becomes the series of $n$.
1. Refer to Supplementary explanation of article 21.
2. By substituting $x=t^2$, you can also calculate $2\int t^{2s-1}e^{-pt^2}dt$ in (1)-2, and by substituting $x=\tan^2\theta/p$, $\frac{2}{p^s}\int(\tan\theta)^{2s-1}d\theta$ in (2)-2.
3. $p_2$ may be not a constant, but $(-1)^{s-2}(s-1)$.
4. The first appears to refer to the creation of mass and the second to the generation of charge when the certain value $m_p$ is multiplied as $-m_pa_s$. And $m_p$ is related with $F_0, p_1, p_2$.
5. I don't know if it has an official designation, but it can be called a divergent Taylor series or a convergent Taylor series. A convergent Taylor series converges to a certain value unless $x$ is infinity, but a divergent Taylor series diverges to infinity at a certain value or interval. The convergence Taylor series has the characteristic that as the degree of $x$ increases, the term or sum of neighboring terms converges to $0$. It can be replaced with other functions besides $1/(1+ax)$, but the value of $a$ must remain after integration, it must be as simple as possible.
6. Refer to Euler's proof, Cauchy's proof.
'Mathematics and Science' 카테고리의 다른 글
| [Book Info] General Relativistic Quantum Theory (2) | 2024.10.22 |
|---|---|
| 33. Verification of Riemann zeta function (0) | 2024.03.16 |
| 29. [Book Info] Riemann zeta function and quantum theory (0) | 2024.01.23 |
| 28. Coordinate transform and derivative (0) | 2024.01.22 |
| 27. Trigonometric representation of Gamma function (0) | 2024.01.14 |