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 Post subject: An infinite product
PostPosted: Mon May 15, 2017 2:28 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 155
Location: Melbourne, Australia
Prove that

$$\prod_{n=1}^{\infty} \left ( 1 + \frac{x^2}{n^2+n-1} \right ) = \frac{1}{\cos \left ( \frac{\pi \sqrt{5}}{2} \right )} \frac{\cos \left ( \frac{\pi \sqrt{5-4x^2}}{2} \right )}{1-x^2}$$

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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 Post subject: Re: An infinite product
PostPosted: Thu May 25, 2017 8:10 am 

Joined: Thu Dec 10, 2015 1:58 pm
Posts: 59
Location: India
Using the infinite product for cosines: $\displaystyle \cos (\pi x) = \prod\limits_{n=1}^{\infty} \left(1-\frac{x^2}{\left(n-\frac{1}{2}\right)^2}\right)$,

\begin{align*}\prod\limits_{n=1}^{\infty} \left(1 + \frac{x^2}{n^2+n-1}\right) &= \prod\limits_{n=2}^{\infty} \frac{\left(n-\frac{1}{2}\right)^2 - \frac{5}{4} + x^2}{\left(n-\frac{1}{2}\right)^2 - \frac{5}{4}} \\&= \frac{1}{(1-x^2)}\prod\limits_{n=1}^{\infty}\frac{\left(1 - \frac{\frac{5}{4} - x^2}{\left(n-\frac{1}{2}\right)^2}\right)}{1 - \frac{\frac{5}{4}}{\left(n-\frac{1}{2}\right)^2}} \\&= \frac{1}{1-x^2}\frac{\cos \left(\frac{\pi\sqrt{5 - 4x^2}}{2}\right)}{\cos \left(\frac{\pi\sqrt{5}}{2}\right)}\end{align*}


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