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Riemann's zeta function is a function with many faces, I mean representations. I recently derived this one, bellow, as a continued fraction over prime numbers.

$$ \zeta(s)=1 +\cfrac{\frac{1}{2^{s}}}{1-\frac{1}{2^{s}} -\cfrac{\frac{2^{s}-1}{3^{s}}}{1+\frac{2^{s}-1}{3^{s}} -\cfrac{\frac{3^{s}-1}{5^{s}}}{1+\frac{3^{s}-1}{5^{s}} -\cfrac{\frac{5^{s}-1}{7^{s}}}{1+\frac{5^{s}-1}{7^{s}} -\cfrac{\frac{7^{s}-1}{11^{s}}}{1+\frac{7^{s}-1}{11^{s}} -\ddots}}}}} $$

... and I'd like to know if this is known in the literature and if so I'd appreciate to have references about it.


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"Found" in the sense of derived it yourself, or found it in a written source somewhere? If the latter, you might include that source here. –  Jack M Apr 8 '14 at 11:50
@Jack M, I derived it myself. –  Neves Apr 8 '14 at 11:54
Please show your derivation. Thank you. –  marty cohen Apr 11 '14 at 2:46
@marty, ok, I'll do that. –  Neves Apr 11 '14 at 6:23
@FredKline, Thanks, but I already knew this one... and none of the others is "over primes". –  Neves Apr 12 '14 at 6:14

2 Answers 2

up vote 5 down vote accepted

The continued fraction representation above had its origins on another problem I was working on sometime ago.

It's based on a very simple way of looking at the Euler's product representation of $\frac{1}{\zeta(s)}$. Interestingly it applies to every infinite product.

And this is as follows

$$ \frac{1}{\zeta(s)}=\left(1-\frac{1}{2^s}\right)-\left(1-\frac{1}{2^s}\right)\frac{1}{3^s}-\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\cdots $$

From here its easy to derive the above continued fraction using Euler's continued fraction formula.

And thats it, It's nice and eventually a new thing.


Just to make it clear, note that $$ \begin{align*} \frac{1}{\zeta(s)}&=\left(1-\frac{1}{2^s}\right)\left[\left(1-\frac{1}{3^s}\right)-\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\ &=\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left[\left(1-\frac{1}{5^s}\right)-\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\ &\vdots\\ &=\prod_{p\in\mathbb{P}}\left(1-\frac{1}{p^{s}}\right) \end{align*} $$ where $\mathbb{P}$ is the set of the prime numbers.


To derive the continued faction just put $\frac{1}{\zeta(s)}$ in the form $$ \frac{1}{\zeta(s)}=1-\frac{1}{2^s}\left(1+\frac{2^s-1}{3^s}\left(1+\frac{3^s-1}{5^s}\left(1+\frac{5^s-1}{7^s}\left(1+\frac{7^s-1}{11^s}\left(1+\ddots\right ) \right ) \right ) \right ) \right) $$ and then just apply the Euler continued fraction formula.

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Found this: arxiv.org/abs/1003.4015 –  Fred Kline Apr 15 '14 at 3:50

By using Mathematica to simplify the first 7 primes, we get: $$\frac{510510^s}{\left(2^s-1\right) \left(3^s-1\right) \left(5^s-1\right) \left(7^s-1\right) \left(11^s-1\right) \left(13^s-1\right) \left(17^s-1\right)},$$ which is equivalent to: $$\prod _{p\text{ prime}} \frac{p^s}{p^s-1} = \zeta(s).$$ Product does not converge when $s=1.$

Scroll down to Euler product formula (2nd paragraph).
When $s=1\text{, }\frac{1}{1-\frac{1}{p^s}}$ simplifies to $\frac{p}{p-1}.$ When $s>1,$ there is no simplification.

Neves's formula puts the exponents back onto the primes when it is simplified, $\frac{p^s}{p^s-1}.$

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Thats $\zeta(s)$ in the Euler's product form... –  Neves Apr 11 '14 at 6:18
@Neves, we crossed paths. My last edit explains. –  Fred Kline Apr 11 '14 at 7:03

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