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Thread: More on the zeta function....

  1. #1
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    More on the zeta function....

    I cannot see why the folowing is correct:
    $\displaystyle \zeta(s)^{-1}= \Sigma {\mu(n) \over n^s}$
    where $\displaystyle \mu$ is the Mobius function....
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  2. #2
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    Quote Originally Posted by bigdoggy View Post
    I cannot see why the folowing is correct:
    $\displaystyle \zeta(s)^{-1}= \Sigma {\mu(n) \over n^s}$
    where $\displaystyle \mu$ is the Mobius function....
    well, by the Euler product formula or just a simple observation we have: $\displaystyle \zeta(s)=\prod_p \left(1+ \frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots \right)=\prod_p \frac{1}{(1- \frac{1}{p^s})},$ where the product is over all (distinct) primes.

    thus $\displaystyle \frac{1}{\zeta(s)}=\prod_p \left(1-\frac{1}{p^s} \right)=\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s},$ by the definition of $\displaystyle \mu(n)$ and after expanding the product.
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  3. #3
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    Quote Originally Posted by bigdoggy View Post
    I cannot see why the folowing is correct:
    $\displaystyle \zeta(s)^{-1}= \Sigma {\mu(n) \over n^s}$
    where $\displaystyle \mu$ is the Mobius function....
    If s > 1,

    $\displaystyle \sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_p(1-p^{-s})=\frac{1}{\zeta(s)}. $
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