Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q. 66

Expert-verified

Calculus

Found in: Page 681

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

The second-order differential equation
$x^{2} y^{''} + x y^{'} + (x^{2} - p^{2}) = 0$
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 $J_{p} (x)$ . It may be shown that $J_{p} (x)$ is given by the following power series in x :
$J_{p} (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{k! (k + p)! 2^{2 k + p}} x^{2 k + p}$
What is the interval of convergence for $J_{1} (x)$ ?

The series is converges for all values of x.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

An expression is given as $J_{p} (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{k! (k + p)! 2^{2 k + p}} x^{2 k + p}$

Step 2. Interval of convergence

The $J_{1} (x)$ is

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

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

Assume that $b_{k} = \frac{(- 1)^{k}}{k! (k + 1)! 2^{2 k + 1}} x^{2 k + 1}$

$b_{k + 1} = \frac{(- 1)^{k + 1}}{(k + 1)! (k + 1 + 1)! 2^{2 (k + 1) + 1}} x^{2 (k + 1) + 1} = \frac{(- 1)^{k + 1}}{(k + 1)! (k + 2)! 2^{2 k + 3}} x^{2 k + 3}$

It implies that

role="math" localid="1649407841088" $\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{\frac{(- 1)^{k + 1}}{(k + 1)! (k + 2)! 2^{2 k + 3}} x^{2 k + 3}}{\frac{(- 1)^{k}}{k! (k + 1)! 2^{2 k + 1}} x^{2 k + 1}}| = \lim_{k \to \infty} |- \frac{1}{(k + 1) (k + 2) 4} x^{2}|$

Calculate for k tending to infinity,

$\lim_{k \to \infty} |x^{2}| \frac{- 1}{(k + 1) (k + 2) 4} = 0$

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

Most popular questions for Math Textbooks

What is the definition of an odd function? An even function?

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$

$e^{x}$

Is it possible for a power series to have $(0, \infty)$ as its interval converge? Explain your answer.

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

$\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}^{}$

Find the interval of convergence for power series: $\sum_{k = 0}^{\infty} \frac{{(- 2)}^{k}}{k!} x^{k}$

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free