High School Mathematics

Consider the function, $f(x)=cosx$

Euler factorised $\frac{sinx}{x}$ as $\frac{sinx}{x}=\left ( 1-\frac{x}{\pi} \right )\left ( 1+\frac{x}{\pi} \right ) \left ( 1-\frac{x}{2\pi} \right ) \left (1+\frac{x}{2\pi} \right )...$

Thus, $\frac{sinx}{x}=\prod_{r=1}^{\infty}\left ( 1-\frac{x^2}{(r\pi)^2} \right )$

Similarly, the roots of $cosx$ are $\frac{\pi}{2},\frac{-\pi}{2},\frac{3\pi}{2},\frac{-3\pi}{2}...$

We can then write, $f(x)=cosx=\prod_{r=1}^{\infty}\left ( 1-\frac{x^2}{(\frac{(2r-1)\pi}{2})^2} \right )=\prod_{r=1}^{\infty}\left ( 1-\left (\frac{2x}{(2r-1)\pi} \right )^2 \right )$

Now, let $x=\frac{\pi y}{2}$

We have: $f(\frac{\pi y}{2})=\prod_{r=1}^{\infty}\left ( 1-\left (\frac{y}{(2r-1)} \right )^2 \right )$

A little fiddling around should convince you that this is indeed the expression given in the question.

Thus, $L(x)=cos(\frac{\pi x}{2})$ .

To find the coefficient of $x^{48}$ , note that the Taylor series for cosx is:

$cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...\frac{x^{48}}{48!}...$

The rest is trivial.

$L(x)=cos(\pi x/2)$