Monday Math 99

Find the infinite product $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}\right)$ .

The key is to recall the infinite product for sine:
$\sin\left(\pi{z}\right)=\pi{z}\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$ .
Thus,
$\frac{\sin\left(\pi{z}\right)}{\pi{z}}=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$ ,
and letting z=ıx, then
$\frac{\sin\left(\imath\pi{x}\right)}{\imath\pi{x}}=\prod_{n=1}^{\infty}\left(1-\frac{(\imath{x})^2}{n^2}\right)=\prod_{n=1}^{\infty}\left(1+\frac{x^2}{n^2}\right)$ .

To find the sine of an imaginary number, we use $\sin{z}=\frac{e^{\imath{z}}-e^{-\imath{z}}}{2\imath}$ ; substituting z=ıx,
$\begin{eqnarray}\sin(\imath{x})&=&\frac{e^{\imath(\imath{x})}-e^{-\imath(\imath{x})}}{2\imath}\\&=&\frac{e^{-x}-e^{x}}{2\imath}\\&=&\imath\frac{e^{x}-e^{-x}}{2}\\\sin(\imath{x})&=&\imath\sinh{x}\end{eqnarray}$ ,
so
$\frac{\sin\left(\imath\pi{x}\right)}{\imath\pi{x}}=\frac{\imath\sinh\left(\pi{x}\right)}{\imath\pi{x}}=\frac{\sinh\left(\pi{x}\right)}{\pi{x}}$ ,
and thus
$\prod_{n=1}^{\infty}\left(1+\frac{x^2}{n^2}\right)=\frac{\sinh\left(\pi{x}\right)}{\pi{x}}$ .
Letting x=1 then gives us our answer
$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}\right)=\frac{\sinh\pi}{\pi}=3.67607791\ldots$ .

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Tags: Hyperbolic Sine, Infinite Product, Math, Monday Math, Sine Product Formula

This entry was posted on November 30, 2009 at 12:32 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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Monday Math 99

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