More on the zeta function....

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March 1st 2010, 11:01 AM
bigdoggy

More on the zeta function....

I cannot see why the folowing is correct:
$\zeta(s)^{-1}= \Sigma {\mu(n) \over n^s}$
where $\mu$ is the Mobius function....
March 1st 2010, 12:12 PM
NonCommAlg

Quote:

Originally Posted by bigdoggy

I cannot see why the folowing is correct:
$\zeta(s)^{-1}= \Sigma {\mu(n) \over n^s}$
where $\mu$ is the Mobius function....

well, by the Euler product formula or just a simple observation we have: $\zeta(s)=\prod_p \left(1+ \frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots \right)=\prod_p \frac{1}{(1- \frac{1}{p^s})},$ where the product is over all (distinct) primes.

thus $\frac{1}{\zeta(s)}=\prod_p \left(1-\frac{1}{p^s} \right)=\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s},$ by the definition of $\mu(n)$ and after expanding the product.
March 1st 2010, 12:15 PM
aliceinwonderland

Quote:

Originally Posted by bigdoggy

I cannot see why the folowing is correct:
$\zeta(s)^{-1}= \Sigma {\mu(n) \over n^s}$
where $\mu$ is the Mobius function....

If s > 1,

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

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