How to Prove the Bessel Function Identity J₂(x) = (2/x)J₁(x) - J₀(x)?

[m0nk3y]

I have been working on this for some hours now. How can I prove
Jsub2(x) = (2/x)*Jsub1(x) - Jsub0(x)??

 

[Astronuc]

Well, the thing to do would be to prove the general recursive relationship,

J_{n+1}(x)\,=\,\frac{2n}{x}J_n(x)\,-\,J_{n-1}(x)


using the identity
\frac{d}{dx}[x^n{J_n(x)}]\,=\,x^n{J_{n-1}(x)}

and Bessel's equation.


With what identities or properties of Bessel's function of the first kind is one familiar?

 

[m0nk3y]

We are using Bessel Equations of the first kind.

So I did the recursion relation for 2, 1, and zero.
Like for 2nd i got
p=2
n=0 to inf
(1/(n-1)(n+1+2))(x/2)^(2n+2)
i started doing n=0 though 3 to see a pattern but my TA told us not to do that and just to just the general part which it what i wrote about for Jsub2

I don't know if I am explaining myself right.
Sorry if I am not, and thank you for replying :)

 

[Astronuc]

It seems that one was trying to do it inductively, but the TA indicated that one should start with the infinite series for J₂(x) and manipulate the terms and show that they can be rearranged to give (2/x)J₁(x) - J₀(x), which is fine.

I would recommend that at some point try to prove the general recursive relationship, i.e.

J_n+1(x) = (2n/x)J_n(x) - J_n-1(x), using the method proposed by the TA.

 

[Astronuc]

It occurred to me that one could use the definition of J_p(x)

J_p(x)\,=\,\sum_{m=0}^\infty \frac{(-1)^m(\frac{x}{2})^{2m+p}} {m!\,{\Gamma(1+m+p)}}

then using p = n+1 and p = n-1, show

J_{n+1}(x)\,+\,J_{n-1}(x)\,=\,\frac{2n}{x}J_n(x),

which is just a reordering of terms.


This book might prove useful: Special Functions