Results 1 to 4 of 4

Math Help - Zeta function: A recurrence relation

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    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 \sum_{j=1}^{n-1} \zeta(2j) \zeta(2n-2j) = \left(n + \frac{1}{2} \right) \zeta(2n), \ \ n \geq 2.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    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 :
     \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
     \pi \cot{\pi x} = \frac{1}{x} - \sum_{k=1}^{\infty} \frac{2x}{k^2 - x^2}
    Then  1 - \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2} = \pi x \cot{\pi x}

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

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

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

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

    Thus , from the third step , i got

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

    Square both ,
     [ 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
     {\pi}^2 {x}^2 ( \csc^2{\pi x} - 1 )
     = - {\pi}^2 {x}^2 [ \frac{1}{\pi} ( \cot{\pi x} )' +1 ]

    From (1) , i got
     [ 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) ]^2 = <br />
1 + 2 \sum_{r=1}^{\infty}(2r-1) x^{2r} \zeta(2r) - {\pi}^2 {x}^2


    Then compare the coefficient of  x^{2n} ,
    the identity  \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  x^2
     \zeta(2) = \frac{{\pi}^2}{6}

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

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

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

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

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

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

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

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

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

     \zeta(18) = \frac{43867{\pi}^{18}}{38979295480125}
    Last edited by simplependulum; May 25th 2009 at 07:37 PM. Reason: mistakes
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by simplependulum View Post
    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 :
     \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
     \pi \cot{\pi x} = \frac{1}{x} - \sum_{k=1}^{\infty} \frac{2x}{k^2 - x^2}
    Then  1 - \sum_{k=1}^{\infty} \frac{2 x^2}{k^2 - x^2} = \pi x \cot{\pi x}

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

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

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

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

    Thus , from the third step , i got

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

    Square both ,
     [ 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
     {\pi}^2 {x}^2 ( \csc^2{\pi x} - 1 )
     = - {\pi}^2 {x}^2 [ \frac{1}{\pi} ( \cot{\pi x} )' +1 ]

    From (1) , i got
     [ 1 - 2 \sum_{r=1}^{\infty} x^{2r} \zeta(2r) ]^2 = <br />
1 + 2 \sum_{r=1}^{\infty}(2r-1) x^{2r} \zeta(2r) - {\pi}^2 {x}^2


    Then compare the coefficient of  x^{2n} ,
    the identity  \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  x^2
     \zeta(2) = \frac{{\pi}^2}{6}

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

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

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

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

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

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

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

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

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

     \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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Unfortunately , the value of  \zeta(10) is  \frac{{\pi}^{10}}{93555} but not 93455
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Recurrence relation
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: December 8th 2011, 10:32 AM
  2. recurrence relation
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 15th 2009, 06:20 PM
  3. Riemann Zeta function - Zeta(0) != Infinity ??
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 8th 2009, 12:50 AM
  4. Recurrence Relation
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: January 13th 2009, 03:37 PM
  5. Recurrence Relation
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: June 2nd 2007, 11:14 PM

Search Tags


/mathhelpforum @mathhelpforum