Question

Annie throws a rock into the water near her camp on a very calm morning and watches the ripples spread from the point where the rock splashed. She knows that those ripples can be described radially by Bessel functions. The

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

Short answer

It is proved that \(xJ_{0}\left ( x \right )=\frac{d}{dx}\left ( xJ_{1}\left ( x \right ) \right )\).

Step-by-step solution

Step 1. Given Information

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

Step 2. Show

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