- #1
cpt_carrot
- 29
- 4
The Bessel function can be written as a generalised power series:
<br /> J_m(x) = \sum_{n=0}^\infty \frac{(-1)^n}{ \Gamma(n+1) \Gamma(n+m+1)} ( \frac{x}{2})^{2n+m}<br />
Using this show that:
<br /> \sqrt{\frac{ \pi x}{2}} J_{1/2}(x)=\sin{x}
where
<br /> \Gamma(p)=\int_{0}^{\infty} x^{p-1}e^{-x}dx<br />
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
)
<br /> J_m(x) = \sum_{n=0}^\infty \frac{(-1)^n}{ \Gamma(n+1) \Gamma(n+m+1)} ( \frac{x}{2})^{2n+m}<br />
Using this show that:
<br /> \sqrt{\frac{ \pi x}{2}} J_{1/2}(x)=\sin{x}
where
<br /> \Gamma(p)=\int_{0}^{\infty} x^{p-1}e^{-x}dx<br />
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
)
