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Q. 64

Expert-verified
Found in: Page 681

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Exercise 64-68 concern with the bessel function.What is the interval for convergence for ${J}_{0}\left(x\right)?$

The series converges for every x .

See the step by step solution

## 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

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

## Step 3. Finding the interval of convergence for given function

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

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

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

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

so, $\underset{k\to \mathrm{\infty }}{lim} \left|{x}^{2}\right|\frac{-1}{\left(k+1{\right)}^{2}}=0$ 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