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Math Help - Euler product of a Dirichlet function

  1. #1
    Super Member Deadstar's Avatar
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    Euler product of a Dirichlet function

    I'm so close to this answer but just can't quite get it 100%...

    Show that if f(n) is a strongly multiplicative function then the Euler product of its Dirichlet function D_f (s) is of the form

    \prod_p \left ( 1 - \frac{f(p)}{p^s} \right )^{-1}

    What I know...

    Since f(n) is strongly multiplicative... It can be written as,

    \prod_p f(p) where p is a prime number. (1)

    The Dirichlet series I'm after is of the form,

    \sum_{n=1}^{\infty} f(n) n^{-s}.

    Now here's where I kinda get lost. I'll write down exactly what I've been writing down as my answer.

    \sum_{n=1}^{\infty} f(n) n^{-s} = \frac{1}{1 - \frac{f(n)}{n^s}} by the geometric series.

    So by (1) we can write this as...

    \prod_p \left (1-\frac{f(p)}{p^s} \right )^{-1}

    Is this right? Something about it feels wrong, like I'm not being very rigorous or that I've missed a step...
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    I'm so close to this answer but just can't quite get it 100%...

    Show that if f(n) is a strongly multiplicative function then the Euler product of its Dirichlet function D_f (s) is of the form

    \prod_p \left ( 1 - \frac{f(p)}{p^s} \right )^{-1}

    What I know...

    Since f(n) is strongly multiplicative... It can be written as,

    \prod_p f(p) where p is a prime number. (1)

    The Dirichlet series I'm after is of the form,

    \sum_{n=1}^{\infty} f(n) n^{-s}.

    Now here's where I kinda get lost. I'll write down exactly what I've been writing down as my answer.

    \sum_{n=1}^{\infty} f(n) n^{-s} = \frac{1}{1 - \frac{f(n)}{n^s}} by the geometric series.

    So by (1) we can write this as...

    \prod_p \left (1-\frac{f(p)}{p^s} \right )^{-1}

    Is this right? Something about it feels wrong, like I'm not being very rigorous or that I've missed a step...


    It looks just fine to me: since the function is completely (or totally or, as you say, strong) multiplicative, it's easy to see that \sum^\infty_{n=1}f(x)=\prod^{}_{p,\,a\,\,prime}\le  ft(1-f(p)\right)^{-1}, so its Dirichlet series \sum^\infty_{n=1}\frac{f(n)}{n^s} fulfills what you wrote.

    Tonio
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