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[tex]\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p}\left(1-p^{-s}\right)^{-1}[/tex]

- Thread starter amcavoy
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- #1

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[tex]\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p}\left(1-p^{-s}\right)^{-1}[/tex]

- #2

shmoe

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If the coefficients of your Dirichlet series is a multiplicative function f, that isapmcavoy said:

[tex]\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p}\left(1-p^{-s}\right)^{-1}[/tex]

[tex]\sum_{n=1}^\infty f(n)n^{-s}[/tex]

then you can write this as an Euler product

[tex]\prod_{p}(1+f(p)p^{-s}+f(p^2)p^{-2s}+\ldots)[/tex]

where the product is over the primes (this is assuming you have absolute convergence of both product and sum). You can think of this as the fundamental theorem of arithmetic in an analytic form. There are plenty of interesting examples of this, powers of Zeta, Dirichlet L-functions, and anything that gets the name "L-function" is usually assumed to satisfy some form of this (as well as many other properties).

For more general sums and products you can still use exponentiation and logarithms to convert from one to another, again being careful with convergence issues if any.

- #3

SGT

Exponentials turn sums into products, while logarithms turn products into sums. So:apmcavoy said:

[tex]\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p}\left(1-p^{-s}\right)^{-1}[/tex]

[tex]exp(\sum_{n=1}^{\infty}\frac{1}{n^{s}})=\prod_{n=1}^{\infty}exp(\frac{1}{n^{s}})[/tex]

You must now find [tex]p[/tex] such that

[tex]\left(1-p^{-s}\right)^{-1} = exp(\frac{1}{n^{s}})[/tex]

- #4

shmoe

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Although it wasn't mentioned, the product in the orignal post is almost surely a product over all the primes (it's the Euler product form of the Riemann Zeta function. The terms won't match up via exponentiation like this.SGT said:Exponentials turn sums into products, while logarithms turn products into sums. So:

[tex]exp(\sum_{n=1}^{\infty}\frac{1}{n^{s}})=\prod_{n=1}^{\infty}exp(\frac{1}{n^{s}})[/tex]

You must now find [tex]p[/tex] such that

[tex]\left(1-p^{-s}\right)^{-1} = exp(\frac{1}{n^{s}})[/tex]

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