91Ó°ÊÓ

The given Bessel function of order p is \(J_{p}\left ( x \right )=\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k+p}}{k!\left ( k+p \right )!2^{2k+p}}\). We have to show that \(xJ_{0}\left ( x \right )=\frac{d}{dx}\left ( xJ_{1}\left ( x \right ) \right )\).

If we put \(p=0\) in Bessel's function then we get,

\(J_{0}\left ( x \right )=\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k}}{\left ( k! \right )^{2}2^{2k}}\).

Thus,

\(xJ_{0}\left ( x \right )=x\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k}}{\left ( k! \right )^{2}2^{2k}}\)

\(xJ_{0}\left ( x \right )=\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k+1}}{\left ( k! \right )^{2}2^{2k}}\)

Similarly, when we put \(p=1\) in Bessel's function then we get,

\(J_{1}\left ( x \right )=\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k+1}}{k!\left ( k+1 \right )!2^{2k+1}}\).

Thus,

\(xJ_{1}\left ( x \right )=x\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k+1}}{k!\left ( k+1 \right )!2^{2k+1}}\)

\(xJ_{1}\left ( x \right )=\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}x^{2k+2}}{k!\left ( k+1 \right )!2^{2k+1}}\).

Now, let's find the derivative

\(\begin{aligned}\frac{d}{d x}\left[x J_1(x)\right] &=\frac{d}{d x}\left[\sum_{k=0}^{\infty} \frac{(-1)^k x^{2 k+2}}{k !(k+1) ! 2^{2 k+1}}\right] \\&=\sum_{k=0}^{\infty} \frac{(-1)^k}{k !(k+1) ! 2^{2 k+1}} \frac{d}{d x}\left[x^{2 k+2}\right] \\&=\sum_{k=0}^{\infty} \frac{(-1)^k}{k !(k+1) ! 2^{2 k+1}}(2 k+2) x^{2 k+1} \\&=\sum_{k=0}^{\infty} \frac{(-1)^k}{(k !)^2 2^{2 k}} x^{2 k+1}\end{aligned}\)

Hence, \(xJ_{0}\left ( x \right )=\frac{d}{dx}\left ( xJ_{1}\left ( x \right ) \right )\).