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Zeta(3) and Euler's formula

  1. Jun 11, 2013 #1
    Hi everyone. I'm trying to understand the step where they wrote

    1/2 ∏1/(1+p^-3) =1/2 Ʃ(-1)^ord(k)/k^3

    How can I see this? I know the Euler product formula, but it has a negative sign before the p^-3, where here we have a + sign.

    Thanks for the help.
     

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  3. Jun 11, 2013 #2
    The step going from the second to last line, to the last line.
     
  4. Jun 11, 2013 #3
    I need help understanding this equality:

    [itex]\prod_{p-prime} \frac{1}{1+\frac{1}{p^3}}= \sum_{k=1}^\infty \frac{(-1)^{\sum_p ord_p(k)}}{k^3}[/itex]


    Any help is greatly appreciated!!!
     
  5. Jun 11, 2013 #4

    lurflurf

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    We know
    $$\prod_\mathbb{P} (1-p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} n^{-s}$$

    with the minus sign it is a little more complicated

    $$\prod_\mathbb{P} (1+p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} (-p)^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} (-1)^{\sum ord_p(k)}n^{-s}$$

    Where the ord_p(k) makes sure we get the right sign (it counts the minuses), clearly

    $$\left( \prod_\mathbb{P} (1-p^{-s})^{-1} \right) \left( \prod_\mathbb{P} (1+p^{-s})^{-1} \right) =\zeta(2n)$$

    but instead of using that your link makes a simple estimate
     
  6. Jun 11, 2013 #5
    Should [itex]\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1}[/itex] have that ^(-1) after it? Or am I missing something..? Are you rewriting 1/(1-p^-s) using geometric series?

    Anyways, thanks that was very helpful, I'm looking into the proofs of the product formula via this route.
     
  7. Jun 11, 2013 #6

    lurflurf

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    Yes that is a geometric series expansion.
     
  8. Jun 11, 2013 #7
    What about the negative one?
     
  9. Jun 12, 2013 #8

    lurflurf

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    $$\prod_\mathbb{P} (1+p^{-s})^{-1}=
    \prod_\mathbb{P} (1-(-p^{-s}))^{-1}= \\
    \prod_\mathbb{P} \left( 1-p^{-s}+p^{-2s}-p^{-3s}+...+(-1)^k p^{-ks}+... \right) ^{-1} = \sum_{\mathbb{Z}>0} a_n n^{-s}=\frac{\zeta(2n)}{\zeta(n)}$$
    a_n being in{-1,1}
    In the sum each term is positive or negative and we can determine which using
    ord_p(k)
    it is a bit tedious to compute, which is why the link estimates
    here is the wikipedia
    [PLAIN]http://en.wikipedia.org/wiki... one way (fundamental theorem of arithmetic).
     
    Last edited by a moderator: May 6, 2017
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