# Thread: Zeta function: A recurrence relation

1. ## Zeta function: A recurrence relation

I hope you haven't got tired of my questions yet! The good news is that I won't be posting new problems for a while! haha

Anyway, apparently this problem was first appeared in the American Mathematical Monthly. Some of you might have seen it:

Prove that $\displaystyle \sum_{j=1}^{n-1} \zeta(2j) \zeta(2n-2j) = \left(n + \frac{1}{2} \right) \zeta(2n), \ \ n \geq 2.$

2. NoNo it isn't a good news , i am interested in your ' techniques of integration' and waiting for your new posts .

i have a method but it's quite complicated , hope you dont mind it .

i first considered the infinite product of sine function :
$\displaystyle \sin{\pi x } = \pi x \prod_{k=1}^{\infty}( 1 - \frac{x^2}{k^2} )$

Then take natural log. for each side , followed by differentiation , i got
$\displaystyle \pi \cot{\pi x} = \frac{1}{x} - \sum_{k=1}^{\infty} \frac{2x}{k^2 - x^2}$
Then $\displaystyle 1 - \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2} = \pi x \cot{\pi x}$

Consider $\displaystyle \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2}$

$\displaystyle = \sum_{k=1}^{\infty} 2 ( \frac{x^2}{k^2} )( \frac{1}{ 1- \frac{x^2}{k^2}} )$

$\displaystyle = 2 \sum_{r=1}^{\infty} \sum_{k=1}^{\infty} ( \frac{x^2}{k^2} )^r$

$\displaystyle = 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r)$

Thus , from the third step , i got

$\displaystyle 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) = \pi x \cot{\pi x}$ --- (1)

Square both ,
$\displaystyle [ 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) ]^2 = [ \pi x \cot{\pi x}]^2$

just consider the right hand side , it becomes
$\displaystyle {\pi}^2 {x}^2 ( \csc^2{\pi x} - 1 )$
$\displaystyle = - {\pi}^2 {x}^2 [ \frac{1}{\pi} ( \cot{\pi x} )' +1 ]$

From (1) , i got
$\displaystyle [ 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) ]^2 = 1 + 2 \sum_{r=1}^{\infty}(2r-1) x^{2r} \zeta(2r) - {\pi}^2 {x}^2$

Then compare the coefficient of $\displaystyle x^{2n}$ ,
the identity $\displaystyle \sum_{j=1}^{n-1} \zeta(2j) \zeta(2n-2j) = \left(n + \frac{1}{2} \right) \zeta(2n), \ \ n \geq 2.$ is found

Also compare the coefficient of $\displaystyle x^2$
$\displaystyle \zeta(2) = \frac{{\pi}^2}{6}$

As we have the identity , the value of $\displaystyle \zeta(2n)$ can be easily found , here is my result :

$\displaystyle \zeta(2) = \frac{{\pi}^2}{6}$

$\displaystyle \zeta(4) = \frac{{\pi}^4}{90}$

$\displaystyle \zeta(6) = \frac{{\pi}^6}{945}$

$\displaystyle \zeta(8) = \frac{{\pi}^8}{9450}$

$\displaystyle \zeta(10) = \frac{{\pi}^{10}}{93555}$

$\displaystyle \zeta(12) = \frac{691{\pi}^{12}}{638512875}$

$\displaystyle \zeta(14) = \frac{2{\pi}^{14}}{18243225}$

$\displaystyle \zeta(16) = \frac{3617{\pi}^{16}}{325641566250}$

$\displaystyle \zeta(18) = \frac{43867{\pi}^{18}}{38979295480125}$

3. Originally Posted by simplependulum
NoNo it isn't a good news , i am interested in your ' techniques of integration' and waiting for your new posts .

i have a method but it's quite complicated , hope you dont mind it .

i first considered the infinite product of sine function :
$\displaystyle \sin{\pi x } = \pi x \prod_{k=1}^{\infty}( 1 - \frac{x^2}{k^2} )$

Then take natural log. for each side , followed by differentiation , i got
$\displaystyle \pi \cot{\pi x} = \frac{1}{x} - \sum_{k=1}^{\infty} \frac{2x}{k^2 - x^2}$
Then $\displaystyle 1 - \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2} = \pi x \cot{\pi x}$

Consider $\displaystyle \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2}$

$\displaystyle = \sum_{k=1}^{\infty} 2 ( \frac{x^2}{k^2} )( \frac{1}{ 1- \frac{x^2}{k^2}} )$

$\displaystyle = 2 \sum_{r=1}^{\infty} \sum_{k=1}^{\infty} ( \frac{x^2}{k^2} )^r$

$\displaystyle = 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r)$

Thus , from the third step , i got

$\displaystyle 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) = \pi x \cot{\pi x}$ --- (1)

Square both ,
$\displaystyle [ 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) ]^2 = [ \pi x \cot{\pi x}]^2$

just consider the right hand side , it becomes
$\displaystyle {\pi}^2 {x}^2 ( \csc^2{\pi x} - 1 )$
$\displaystyle = - {\pi}^2 {x}^2 [ \frac{1}{\pi} ( \cot{\pi x} )' +1 ]$

From (1) , i got
$\displaystyle [ 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) ]^2 = 1 + 2 \sum_{r=1}^{\infty}(2r-1) x^{2r} \zeta(2r) - {\pi}^2 {x}^2$

Then compare the coefficient of $\displaystyle x^{2n}$ ,
the identity $\displaystyle \sum_{j=1}^{n-1} \zeta(2j) \zeta(2n-2j) = \left(n + \frac{1}{2} \right) \zeta(2n), \ \ n \geq 2.$ is found

Also compare the coefficient of $\displaystyle x^2$
$\displaystyle \zeta(2) = \frac{{\pi}^2}{6}$

As we have the identity , the value of $\displaystyle \zeta(2n)$ can be easily found , here is my result :

$\displaystyle \zeta(2) = \frac{{\pi}^2}{6}$

$\displaystyle \zeta(4) = \frac{{\pi}^4}{90}$

$\displaystyle \zeta(6) = \frac{{\pi}^6}{945}$

$\displaystyle \zeta(8) = \frac{{\pi}^8}{9450}$

$\displaystyle \zeta(10) = \frac{{\pi}^{10}}{93455}$

$\displaystyle \zeta(12) = \frac{691{\pi}^{12}}{638512875}$

$\displaystyle \zeta(14) = \frac{2{\pi}^{14}}{18243225}$

$\displaystyle \zeta(16) = \frac{3617{\pi}^{16}}{325641566250}$

$\displaystyle \zeta(18) = \frac{43867{\pi}^{18}}{38979295480125}$
nice! it's actually simpler than the solution i know! i haven't checked all the calculations at the end of your solution but it looks ok to me.

4. Unfortunately , the value of $\displaystyle \zeta(10)$ is $\displaystyle \frac{{\pi}^{10}}{93555}$ but not 93455