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Calculating Riemann Zeta function

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Using method of Euler, calculate [tex]\zeta(4)[/tex], the Riemann Zeta function of 4th order.

    2. Relevant equations

    [tex]\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}[/tex]

    Finding [tex]\zeta(2)[/tex]:

    [tex]\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^s}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...[/tex]

    [tex]\sin{x}=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5+...[/tex]
    [tex]\frac{\sin{x}}{x}=1-\frac{1}{3!}x^2+\frac{1}{5!}x^4+...[/tex]

    [tex]\frac{\sin{x}}{x}=\left(1-\left(\frac{x}{\pi}\right)^2\right)\left(1-\left(\frac{x}{2\pi}\right)^2\right)\left(1-\left(\frac{x}{3\pi}\right)^2\right)\left(1-\left(\frac{x}{4\pi}\right)^2\right)\left(1-\left(\frac{x}{5\pi}\right)^2\right)...[/tex]

    From the above, the coefficients of [tex]x^2[/tex] are:

    [tex]\frac{1}{\pi^2}+\frac{1}{2^2\pi^2}+\frac{1}{3^2\pi^2}+\frac{1}{4^2\pi^2}+\frac{1}{5^2\pi^2}+...[/tex]

    Now equate these coefficients to [tex]x^2[/tex] in the sine function series:

    [tex]\frac{1}{\pi^2}+\frac{1}{2^2\pi^2}+\frac{1}{3^2\pi^2}+\frac{1}{4^2\pi^2}+\frac{1}{5^2\pi^2}+...=\frac{1}{3!}[/tex]
    [tex]\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...=\frac{\pi^2}{3!}[/tex]
    [tex]\zeta(2)=\frac{\pi^2}{6}[/tex]

    3. The attempt at a solution

    I get the following for the coefficients of [tex]x^4[/tex]:

    [tex]\frac{1}{2^2\pi^4}+\frac{1}{3^2\pi^4}+\frac{1}{2^23^2\pi^4}+\frac{1}{4^2\pi^4}+\frac{1}{2^24^2\pi^4}+\frac{1}{3^24^2\pi^4}+...=\frac{1}{\pi^4}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^23^2}+\frac{1}{4^2}+\frac{1}{2^24^2}+\frac{1}{3^24^2}+...\right)[/tex]

    The problem is, how do I get [tex]\zeta(4)=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+...[/tex] out of that sum?
     
    Last edited: Dec 8, 2008
  2. jcsd
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