Solving Bessel Function for Sin: $\sqrt{\frac{\pi x}{2}} J_{1/2}(x) = \sin{x}$

[cpt_carrot]

The Bessel function can be written as a generalised power series:
$$J_m(x) = \sum_{n=0}^\infty \frac{(-1)^n}{ \Gamma(n+1) \Gamma(n+m+1)} ( \frac{x}{2})^{2n+m}$$

Using this show that:
$$\sqrt{\frac{ \pi x}{2}} J_{1/2}(x)=\sin{x}$$

where
$$\Gamma(p)=\int_{0}^{\infty} x^{p-1}e^{-x}dx$$
and therefore
$$\Gamma(p+1)=p\Gamma(p)$$

Wea re also given that:
$$\Gamma(3/2)=\frac{\sqrt{\pi}}{2}$$

My answer so far goes something like:
We are obviously trying to get the series expansion for the Bessel function into the form of the Taylor series for sin:
$$\sin{x} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

Simplyfing the expression for J by replacing m by 1/2 and Gamma(n+1) by n! I ca get reasonably close to the Taylor series but I'm having trouble getting rid of the Gamma(n+3/2)
Any help would be much appreciated!
Also as this is my first post: Hello Everybody
(And I hope the LaTex works :tongue2: )

 

[benorin]

Use the Legendre duplication formula to simplify.