Zeta(3) and Euler's formula

  • Thread starter camilus
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  • #1
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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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  • #2
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The step going from the second to last line, to the last line.
 
  • #3
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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!!!
 
  • #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
 
  • #5
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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
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.
 
  • #6
lurflurf
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Yes that is a geometric series expansion.
 
  • #7
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What about the negative one?
 
  • #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).
 
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