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Q. 66

Expert-verified
Found in: Page 681

### Calculus

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

# The second-order differential equation ${x}^{2}{y}^{\text{'}\text{'}}+x{y}^{\text{'}}+\left({x}^{2}-{p}^{2}\right)=0$where p is a non-negative integer, arises in many applications in physics and engineering, including one model for the vibration of a beaten drum. The solution of the differential equation is called the Bessel function of order p, denoted by ${J}_{p}\left(x\right)$. It may be shown that ${J}_{p}\left(x\right)$is given by the following power series in x :${J}_{p}\left(x\right)=\sum _{k=0}^{\infty }\frac{\left(-1{\right)}^{k}}{k!\left(k+p\right)!{2}^{2k+p}}{x}^{2k+p}$What is the interval of convergence for ${J}_{1}\left(x\right)$?

The series is converges for all values of x.

See the step by step solution

## Step 1. Given information

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

## Step 2. Interval of convergence

The ${J}_{1}\left(x\right)$ is

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

We have to do ratio test first for the absolute convergence,

Assume that ${b}_{k}=\frac{\left(-1{\right)}^{k}}{k!\left(k+1\right)!{2}^{2k+1}}{x}^{2k+1}$

${b}_{k+1}=\frac{\left(-1{\right)}^{k+1}}{\left(k+1\right)!\left(k+1+1\right)!{2}^{2\left(k+1\right)+1}}{x}^{2\left(k+1\right)+1}\phantom{\rule{0ex}{0ex}}=\frac{\left(-1{\right)}^{k+1}}{\left(k+1\right)!\left(k+2\right)!{2}^{2k+3}}{x}^{2k+3}$

It implies that

role="math" localid="1649407841088" $\underset{k\to \infty }{\mathrm{lim}}\left|\frac{{b}_{k+1}}{{b}_{k}}\right|=\underset{k\to \infty }{\mathrm{lim}}\left|\frac{\frac{\left(-1{\right)}^{k+1}}{\left(k+1\right)!\left(k+2\right)!{2}^{2k+3}}{x}^{2k+3}}{\frac{\left(-1{\right)}^{k}}{k!\left(k+1\right)!{2}^{2k+1}}{x}^{2k+1}}\right|\phantom{\rule{0ex}{0ex}}=\underset{k\to \infty }{\mathrm{lim}}\left|-\frac{1}{\left(k+1\right)\left(k+2\right)4}{x}^{2}\right|$

Calculate for k tending to infinity,

$\underset{k\to \infty }{\mathrm{lim}}\left|{x}^{2}\right|\frac{-1}{\left(k+1\right)\left(k+2\right)4}=0$

Limit is zero independently of x. So the series is converges for all values of x.