# Bessel Function of order 1

$\displaystyle$J_{1} (x) = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{2^{2n+1} (n!)(n+1)!}$$is called a Bessel function of order 1. Show that it satisfies the differential equation \displaystyle x^{2}$$\displaystyle J''_{1}(x) \displaystyle +xJ'_{1}(x) \displaystyle + (x^{2}\displaystyle - 1)J_{1}(x) = 0$I have tried differentiating the series and plugging in but I cannot get the right answer. • Jan 29th 2011, 09:46 AM FernandoRevilla We can check your computations. Fernando Revilla • Jan 30th 2011, 12:05 AM chisigma May be that a more confortable way is to valuate what are the solution of the DE...$\displaystyle \displaystyle y^{''} + \frac{y^{'}}{x} + (1-\frac{\nu^{2}}{x^{2}}) =0 $(1) ... that can be expressed in the form...$\displaystyle \displaystyle y(x)= x^{\nu}\ \sum_{k=0}^{\infty} a_{k}\ x^{2k}$(2) With a little of patience You find that is...$\displaystyle \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)}\}$(3) Now You set in (3)$\displaystyle \nu=1$and$\displaystyle a_{0}=1$and Youn obtain$\displaystyle J_{1}(x)$... Kind regards$\displaystyle \chi\displaystyle \sigma\$