Q. 64 - Calculus

Question

Exercise 64-68 concern with the bessel function.

What is the interval for convergence for $J_0\left(x\right)?$

Step-by-Step Solution

Answer

The series converges for every x .

Step 1.Given information

We have to find out the interval for convergence for $J_p\left(x\right)$

Step 2. Representation of function

We denote the the given function as 

$J_p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{k!(k+p)!2^{2k+p}}{x}^{2k+p}$

Step 3. Finding the interval of convergence for given function

$\begin{array}{r}{J}_0\left(x\right)=\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{k!(k+0)!2^{2k+0}}{x}^{2k+0}\\ =\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{(k!{)}^2{2}^{2k}}{x}^{2k}\end{array}$

For finding the convergence of the function we will do the ratio test for absolute convergence we will assume $b_k=\frac{(-1{)}^k}{(k!{)}^2{2}^{2k}}{x}^{2k}$ therefore the next term will be $b_{k+1}=\frac{(-1{)}^{k+1}}{\left[\right(k+1)!{]}^2{2}^{2(k+1)}}{x}^{2(k+1)}$ implies that

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{\left[\right(k+1)!{]}^2{2}^{2(k+1)}}{}{x}^{2(k+1)}\right|\frac{(-1{)}^k}{(k!{)}^2{2}^{2k}{x}^{2k}}\mid \\ =\underset{k\to \mathrm{\infty }}{lim} \left|-\frac{1}{(k+1{)}^{2}4}{x}^2\right|\end{array}$

Now we will be evaluating the limit $k\to \infty$

so, $\underset{k\to \mathrm{\infty }}{lim} \left|x^2\right|\frac{-1}{(k+1{)}^2}=0$ zero is the value no matter what is the value of x .

Step 4. The convergence interval is

Hence by the ratio test it is clear that this series absolutely converges for every value of x