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I'd like to simplify

$$\newcommand{\bigk}{\mathop{\huge\vcenter{\hbox{K}}}}B(s)=\prod_{p\in\mathbb{P}}\left(1+\bigk_{k=1}^{\infty }\frac{f_{k}(s)}{f_{k}(s)}\right)^{-1}$$ to something of the form $$B(s)=\bigk_{k=1}^{\infty }\frac{F_{k}(s)}{G_{k}(s)}$$

This is related to the question "Arithmetic of continued fractions, does it exist?".


Motivation:

Consider the following representation of a function $$ f(z)= 1+ \sum_{n=1}^{\infty}\left( \prod_{k=1}^{n}f_{k}(z) \right) $$ Using Euler's continued fraction formula, this function can have the following representation $$ f(z)= \cfrac{1}{ 1- \cfrac{f_{1}(z)}{ 1+f_{1}(z)- \cfrac{f_{2}(z)}{ 1+f_{2}(z)- \cfrac{f_{3}(z)}{ 1+f_{3}(z) - \ddots}}}} $$ Now consider the Jacobi theta function $$ \frac{1}{2}(\vartheta_{4}(0,x)+1)=1-x+x^{4}-x^{9}+x^{16}\cdots $$ and the fact that $$ n^{2}=\sum_{k=1}^{n}(2k-1) $$ We can represent the theta function as $$ \frac{1}{2}(\vartheta_{4}(0,x)+1)=1-x+x^{1+3}-x^{1+3+5}+x^{1+3+5+7}\cdots $$ or $$ \frac{1}{2}(\vartheta_{4}(0,x)+1)= 1+ \sum_{n=1}^{\infty}(-1)^{n}\left( \prod_{k=1}^{n}x^{2k-1} \right) $$ so $\frac{1}{2}(\vartheta_{4}(0,x)+1)$ has the following continued fraction representation $$ \frac{1}{2}(\vartheta_{4}(0,x)+1)= \cfrac{1}{ 1+ \cfrac{x}{ 1-x+ \cfrac{x^{3}}{ 1-x^{3}+ \cfrac{x^{5}}{ 1-x^{5} + \ddots}}}} $$ Now remember that the Riemann's zeta function can be represented by $$ \zeta(s)=\frac{A(s)}{B(s)} $$ with $$ B(s)=\prod_{p\in\mathbb{P}}\frac{1}{2}\left(\vartheta_{4}(0,\frac{1}{p^{z}})+1\right) $$ and $A(s)$ convergent for $\Re{(s)}>\frac{1}{4}$

We have than that $B(s)$ is an infinite product of continued fractions $$ B(s)=\prod_{p\in\mathbb{P}}\left(1+\bigk_{k=1}^{\infty }\frac{p^{-(2k-1)s}}{1-p^{-(2k-1)s}}\right)^{-1} $$

where $p\in\mathbb{P}$ are prime numbers.

Finally my question, could someone find a representation for $B(s)$ as a single continued fraction?

Thanks.

share|improve this question
    
What does K do? –  Alizter Dec 15 '13 at 19:25
    
@Alizter, Gauss notation –  Neves Dec 15 '13 at 20:24

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discard

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