I want to prove that the Bessel function:
- $$J_\nu(z)=\sum_{k=0}^{\infty}\frac{(-1)^k(z/2)^{2k+\nu}}{k!\Gamma(\nu+k+1)}$$
satisfies the Bessel differential equation:
- $$\frac{d^2y}{dz^2}+\frac{1}{z}\frac{dy}{dz}+\left(1-\frac{\nu^2}{z^2}\right)y=0$$
The first thing that I did was take the derivative and second derivative of $J_\nu(z)$ to get:
- $$J_\nu(z)=\sum_{k=0}^{\infty}\frac{(-1)^k(z/2)^{2k+\nu}}{k!(\nu+k)!}$$
- $$J_\nu^{'}(x)=\sum_{k=0}^{\infty}\frac{(-1)^k(2k+\nu)(z/2)^{2k+\nu-1}}{2k!(\nu+k)!}$$
- $$J_\nu^{''}(x)=\sum_{k=0}^{\infty}\frac{(-1)^k(2k+\nu)(2k+\nu-1)(z/2)^{2k+\nu-2}}{4k!(\nu+k)!}$$
After that I simply let $y=J_\nu(x)$ and plugged in the respective derivatives. After simplifying, I get:
- $$\sum_{k=0}^{\infty}\frac{(-1)^k(z/2)^{2k+\nu}}{k!(\nu+k)!}\left[\frac{1}{4}(2k+\nu)(k+\nu-1)+\frac{1}{2z}(2k+\nu)+\left(1-\frac{\nu^2}{z^2}\right)\right]=0$$
This is where I am now stuck. I'm unsure how to prove that the left side of the equation equals zero. I've tried expanding the summation, where the first two terms are:
- $$\left(\frac{(z/2)^\nu}{\nu!}\right)\left[\frac{1}{4}\nu(\nu-1)+\frac{1}{2z}\nu+\left(1-\frac{\nu^2}{z^2}\right)\right]-\left(\frac{(z/2)^{2+\nu}}{1!(\nu+1)!}\right)\left[\frac{1}{4}(2+\nu)(2+\nu-1)+\frac{1}{2z}(2+\nu)+\left(1-\frac{\nu^2}{z^2}\right)\right]+...=0$$
I still can't figure out how to get this expanded left side to equal zero. Should I try to prove that the summation I have in equation 6 converges to zero? Or am I not seeing terms that I am supposed to cancel out in equation 6 or 7? Or am I completely off the mark? Any advice would be greatly appreciated.
I doubt you’ll be happy with this, but you’ll want to do a bit more multiplying out: You want to expose the powers of $z$ explicitly. Assuming your work above to be correct, you’ll end up seeing that in order for the lowest order coefficients to cancel — and the final solution to be nontrivial — certain restrictions on $\nu$ will be required.