Engineering Mathematics: Apr 11, 2011

firstly , I recommend you to read these Riemann zeta function , Euler product to have fundamental understanding of the Riemann zeta function.

RZ.ID.1:

$\frac{\zeta (s)}{\zeta (2s)}= \prod (1+p^{-s})$

p is a prime number

Proof

given

$\zeta (s)= \prod (1-p^{-s})^{-1}$

(1)

$\frac{\zeta (s)}{\zeta (2s)}= \frac{\prod (1-p^{-s})^{-1} }{\prod (1-p^{-2s})^{-1} }$

(2)

$\frac{\zeta (s)}{\zeta (2s)}= \frac{\prod (1-p^{-s})^{-1} }{\prod (1+p^{-s})^{-1}(1-p^{-s})^{-1} }$

(3)

cancelling out the similarities, therefore

$\frac{\zeta (s)}{\zeta (2s)}= \prod (1+p^{-s})$

(4)

RZ.ID.2:

$\frac{\zeta ^{2}(s)}{\zeta (2s)}= \prod \left ( \frac{1+p^{-s}}{1-p^{-s}} \right )$

Proof

derived by multiplying (1) and (4)

RZ.ID.3:

$\zeta' (s)=\sum _{n=1}^{\infty }\frac{-\ln n}{n^{s}}$

proof:

$\zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^{s}}=\sum _{n=1}^{\infty }e^{-\ln n^{s}}=\sum _{n=1}^{\infty }e^{-s\ln n}$

by differentiating the above equation

$\zeta' (s)=\sum _{n=1}^{\infty }\frac{-\ln n}{n^{s}}$

RZ.ID.4:

$\frac{1}{1-2^{1-s}}\sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n^{s}}=\zeta (s)$

Proof

$\sum _{n=1}^{\infty }\frac{(-1)^{n}}{n^{s}}+\sum _{n=1}^{\infty }\frac{1}{n^{s}}=2\sum _{n=2,4,6,\cdot \cdot \cdot }^{\infty }\frac{1}{n^{s}}$

(1)

replacing

$n=2k$

therefore,

$2\sum _{n=2,4,6,\cdot \cdot \cdot }^{\infty }\frac{1}{n^{s}}= 2\sum _{k=1}^{\infty }\frac{1}{(2k)^{s}}= 2^{1-s}\sum _{k=1}^{\infty }\frac{1}{k^{s}}=2^{1-s}\zeta (s)$

(2)

from (1) and (2)

$\sum _{n=1}^{\infty }\frac{(-1)^{n}}{n^{s}}+\zeta (s)=2^{1-s}\zeta (s)$

$\sum _{n=1}^{\infty }\frac{(-1)^{n}}{n^{s}}=(2^{1-s}-1)\zeta (s)$

$\frac{1}{2^{1-s}-1}\sum _{n=1}^{\infty }\frac{(-1)^{n}}{n^{s}}=\zeta (s)$

multiplying and dividing by -1

$\frac{1}{1-2^{1-s}}\sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n^{s}}=\zeta (s)$

Monday, April 11, 2011

Riemann zeta function