# Bessel Function of order 1

• January 29th 2011, 08:56 AM
deuteronomy
Bessel Function of order 1
The function

$J_{1} (x) = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{2^{2n+1} (n!)(n+1)!}$

is called a Bessel function of order 1. Show that it satisfies the differential equation

$x^{2}$ $J''_{1}(x)$ $+xJ'_{1}(x)$ $+ (x^{2}$ $- 1)J_{1}(x) = 0$

I have tried differentiating the series and plugging in but I cannot get the right answer.
• January 29th 2011, 09:46 AM
FernandoRevilla
We can check your computations.

Fernando Revilla
• January 30th 2011, 12:05 AM
chisigma
May be that a more confortable way is to valuate what are the solution of the DE...

$\displaystyle y^{''} + \frac{y^{'}}{x} + (1-\frac{\nu^{2}}{x^{2}}) =0$ (1)

... that can be expressed in the form...

$\displaystyle y(x)= x^{\nu}\ \sum_{k=0}^{\infty} a_{k}\ x^{2k}$ (2)

With a little of patience You find that is...

$\displaystyle y(x)= a_{0}\ x^{\nu}\ \{1+ \sum_{k=1}^{\infty} (-1)^{k}\ \frac {(\frac{x}{2})^{2 k}}{k!\ (1+\nu)\ (2+\nu)...(k+\nu)}\}$ (3)

Now You set in (3) $\nu=1$ and $a_{0}=1$ and Youn obtain $J_{1}(x)$...

Kind regards

$\chi$ $\sigma$