Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question
    
"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 at 11:50
3  
@Jack M, I derived it myself. –  Neves Apr 8 at 11:54
    
Please show your derivation. Thank you. –  marty cohen Apr 11 at 2:46
    
@marty, ok, I'll do that. –  Neves Apr 11 at 6:23
2  
@FredKline, Thanks, but I already knew this one... and none of the others is "over primes". –  Neves Apr 12 at 6:14

2 Answers 2

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.

EDIT

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.

EDIT

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.

share|improve this answer
    
Found this: arxiv.org/abs/1003.4015 –  Fred Kline Apr 15 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, $\frac{p^s}{p^s-1}.$

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.