Thread: Bessel function ??

n : integer .

Originally Posted by amro05

n : integer .

What have you tried?

Originally Posted by amro05

n : integer .

You need to mess around with the series for the Bessel function of the first kind:

$\displaystyle J_n(x) = \displaystyle\sum\limits_{m=0}^{\infty}\dfrac{(-1)^m x^{2m+n}}{2^{2m+n}m!\Gamma(m+n+1)}$

Clearly,

$\displaystyle J_{-n}(x) = \displaystyle\sum\limits_{m=0}^{\infty}\dfrac{(-1)^m x^{2m-n}}{2^{2m-n}m!\Gamma(m-n+1)}$

Now, make a substitution. Let $\displaystyle m=m^{\prime}+n$. Then,

$\displaystyle \begin{aligned}\sum\limits_{m=0}^{\infty}\dfrac{(-1)^m x^{2m-n}}{2^{2m-n}m!\Gamma(m-n+1)} &=\sum\limits_{m^{\prime}+n=0}^{\infty}\dfrac{(-1)^{m^{\prime}+n} x^{2m^{\prime}+n}}{2^{2m^{\prime}+n}m!\Gamma(m^{\p rime}+n+1)}\\ &= \sum\limits_{m^{\prime}=-n}^{-1}\dfrac{(-1)^{m^{\prime}+n} x^{2m^{\prime}+n}}{2^{2m^{\prime}+n}m!\Gamma(m^{\p rime}+n+1)}+\sum\limits_{m^{\prime}=0}^{\infty}\df rac{(-1)^{m^{\prime}+n} x^{2m^{\prime}+n}}{2^{2m^{\prime}+n}m!\Gamma(m^{\p rime}+n+1)}\end{aligned}$

A couple observations need to be made to finish it up.

Can you proceed?

Originally Posted by Chris L T521

You need to mess around with the series for the Bessel function of the first kind:

$\displaystyle J_n(x) = \displaystyle\sum\limits_{m=0}^{\infty}\dfrac{(-1)^m x^{2m+n}}{2^{2m+n}m!\Gamma(m+n+1)}$

Clearly,

$\displaystyle J_{-n}(x) = \displaystyle\sum\limits_{m=0}^{\infty}\dfrac{(-1)^m x^{2m-n}}{2^{2m-n}m!\Gamma(m-n+1)}$

Now, make a substitution. Let $\displaystyle m=m^{\prime}+n$. Then,

$\displaystyle \begin{aligned}\sum\limits_{m=0}^{\infty}\dfrac{(-1)^m x^{2m-n}}{2^{2m-n}m!\Gamma(m-n+1)} &=\sum\limits_{m^{\prime}+n=0}^{\infty}\dfrac{(-1)^{m^{\prime}+n} x^{2m^{\prime}+n}}{2^{2m^{\prime}+n}m!\Gamma(m^{\p rime}+n+1)}\\ &= \sum\limits_{m^{\prime}=-n}^{-1}\dfrac{(-1)^{m^{\prime}+n} x^{2m^{\prime}+n}}{2^{2m^{\prime}+n}m!\Gamma(m^{\p rime}+n+1)}+\sum\limits_{m^{\prime}=0}^{\infty}\df rac{(-1)^{m^{\prime}+n} x^{2m^{\prime}+n}}{2^{2m^{\prime}+n}m!\Gamma(m^{\p rime}+n+1)}\end{aligned}$

A couple observations need to be made to finish it up.

Can you proceed?

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yes I think I got it ( thank you very very mach )



because the Gamma function of the negative integer = infinity
so the first part = zero
and the 2.nd part ( will be ) = (-1)^n Jn(x) ( but need some work )

is it correct ??

Originally Posted by amro05

is it correct ??

Looks good to me! (but note your sum should start at zero..you forgot to put that in.)