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  1. #1
    MHF Contributor chiph588@'s Avatar
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    Euler Product

    Show $\displaystyle \displaystyle{\prod_{p\equiv1\;(4)}\frac{p^3+1}{p^ 3-1} = \frac{105}{4\pi^3}\zeta(3)} $ and $\displaystyle \displaystyle{\prod_{p\equiv3\;(4)}\frac{p^3+1}{p^ 3-1} = \frac{28}{\pi^3}\zeta(3)} $.
    Last edited by chiph588@; Jul 16th 2010 at 12:47 PM.
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  2. #2
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    Quote Originally Posted by chiph588@ View Post
    Show $\displaystyle \prod_{p\equiv1\;(4)}\frac{p^3+1}{p^3-1} = \frac{105}{4\pi^3}\zeta(3) $ and $\displaystyle \prod_{p\equiv3\;(4)}\frac{p^3+1}{p^3-1} = \frac{28}{\pi^3}\zeta(3) $.


    I consider

    $\displaystyle \ F(s) = \big{ \frac{1}{1^s} - \frac{1}{3^s} + \frac{1}{5^s} - ... } = \displaystyle{\sum_{k=0}^{\infty} } \frac{(-1)^k}{(2k+1)^s}$ which is similar to zeta function .

    I find that it can be ' factorised ' in this form :


    $\displaystyle \left( \displaystyle{\prod_{p\equiv 1(4)}} ( 1 - \frac{1}{p^s} ) \displaystyle{\prod_{p\equiv 3(4)}} ( 1 + \frac{1}{p^s} ) \right)^{-1} ~~~~~ (1) $

    The reason is when we expand the product from RHS we obtain something like :


    $\displaystyle \displaystyle{ \prod_{1(4)}} ( 1 + \frac{1}{p^s } + ... ) \prod_{3(4)} ( 1 - \frac{1}{p^s } + ... ) $

    For a number $\displaystyle n $ if it is $\displaystyle 1 \bmod{4} $ , no matter how many primes $\displaystyle 3\bmod{4}$ it contains , the number must be even , and for $\displaystyle n \equiv 3 \bmod{4} $ , we conclude that the number of primes $\displaystyle 3\bmod{4} $ as factors must be odd , so we have this formula (1) .



    Together with $\displaystyle \zeta(s) = \left( \displaystyle{\prod_p} ( 1 - \frac{1}{p^s} ) \right) ^{-1} $

    $\displaystyle = \frac{1}{1 - \frac{1}{2^s} } \left( \displaystyle{\prod_{p\equiv 1(4)}} ( 1 - \frac{1}{p^s} ) \displaystyle{ \prod_{p \equiv 3(4) }} ( 1 - \frac{1}{p^s} ) \right) ^{-1} $

    We obtain , by division :

    $\displaystyle \boxed{ \prod_{p \equiv 3(4) } \frac{p^s+1 }{ p^s - 1 } = ( 1 - \frac{1}{2^s} ) ~ \frac{ \zeta(s) }{ F(s) } } $

    Also , we have

    $\displaystyle \left( \frac{ 2^s + 1 }{2^s - 1 } \right) \displaystyle{ \prod_{p \equiv 1(4) }} \frac{ 1 + \frac{1}{p^s} }{ 1 - \frac{1}{p^s} } \displaystyle{\prod_{p \equiv 3(4) }} \frac{ 1 + \frac{1}{p^s} }{ 1 - \frac{1}{p^s} } $

    $\displaystyle = \frac{ \zeta^2(s) }{\zeta(2s ) }$ so we obtain

    $\displaystyle \boxed{ \prod_{p \equiv 1(4) } \frac{ p^s + 1 }{ p^s - 1 } = \frac{1}{ 1 + \frac{1}{2^s } } ~ \frac{ \zeta(s) F(s) }{ \zeta(2s) }}$


    I don't know the value of $\displaystyle F(3) $ but from your writing , it should be $\displaystyle \frac{\pi^3}{32} $ and if I remember correctly , $\displaystyle \zeta(6) = \frac{\pi^6}{945} $ we should have

    $\displaystyle \displaystyle{ \prod_{p \equiv 1(4) }} \frac{ p^3 + 1 }{ p^3 - 1 } } = \frac{8}{9} ~ \frac{\pi^3}{32 } ~ \frac{945}{\pi^6} ~ \zeta(3) $

    $\displaystyle = \frac{ 105}{4 \pi^3 } ~ \zeta(3) $ , it looks ok to me ...

    Remarks : I've read a book by Euler , called Introduction to Analysis of the Infinite , in this book he really introduced a method of evaluating $\displaystyle F(2n-1) $ ( of course , in the mean time , he evaluated $\displaystyle \zeta(2n) $ , too ) but i forgot this method , recently i got another method using Fourier Analysis , I guess you experts are using this skilfully .
    Last edited by simplependulum; Jul 16th 2010 at 12:45 AM.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Very good!

    It can be shown $\displaystyle F(2k+1)=\displaystyle{{{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k!)}}} $, where $\displaystyle E_{n} $ are the Euler Numbers.

    -------

    Also if anyone didn't follow the first step, $\displaystyle F(s)=\displaystyle{\sum_{k=0}^{\infty} } \frac{(-1)^k}{(2k+1)^s} = \displaystyle{\prod_p} \left(1+(-1)^{(p-1)/2}p^{-s}\right)^{-1} = \left( \displaystyle{\prod_{p\equiv 1(4)}} ( 1 - p^{-s}) \displaystyle{\prod_{p\equiv 3(4)}} ( 1 + p^{-s} ) \right)^{-1} $.

    The second equality comes from the fact that $\displaystyle F(s) = \displaystyle{\sum_{n=0}^{\infty} } \frac{\chi(n)}{n^s} $, where $\displaystyle \chi(n)=\begin{cases} (1)^{(n-1)/2} ,\;\;\text{if }n\text{ is odd}\\0,\;\;\text{if }n\text{ is even}\end{cases} $

    Now observe $\displaystyle \chi(n) $ is completely multiplicative so we can transform $\displaystyle F(s) $ into its Euler Product representaion.
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  4. #4
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    Oh , is $\displaystyle \chi(n)$ the Dirichlet function ? I first saw it in the later part of a book about elementary number theory .
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by simplependulum View Post
    Oh , is $\displaystyle \chi(n)$ the Dirichlet function ? I first saw it in the later part of a book about elementary number theory .
    Sort of... it's not the but rather a Dirichlet character.

    The link I provided goes in depth and has tables for you to see, but this particular $\displaystyle \chi(n) $ is the only non-principle character modulo $\displaystyle 4 $.
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