Chapter 8.5, Problem 29E

a.

To determine

To find: the domain of the function.

Answer to Problem 29E

  $x\in (-\infty ,\infty )$ .

Explanation of Solution

Given information: The function $J_1(x)={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{n!(n+1)!2^{2n+1}}}$ is called the Bessel Function

of order 1.

Calculation:

  $\begin{array}{c}{J}_1(x)={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{n!(n+1)!2^{2n+1}}}.\\ a_n=\frac{{(-1)}^n{x}^{2n+1}}{n!(n+1)!2^{2n+1}}.\\ a_{n+1}=\frac{{(-1)}^{n+1}{x}^{2(n+1)+1}}{(n+1)!(n+2)!2^{2(n+1)+1}}\\ a_{n+1}=\frac{{(-1)}^n(-1)x^{2n+1}.x^2}{(n+1)n!(n+2)(n+1)!2^{2n+1}{.2}^2}\\ \frac{a_{n+1}}{a_n}=\frac{{(-1)}^n(-1)x^{2n+1}.x^2}{(n+1)n!(n+2)(n+1)!2^{2n+1}{.2}^2}.\frac{n!(n+1)!2^{2n+1}}{{(-1)}^n{x}^{2n+1}}\\ \frac{a_{n+1}}{a_n}=\frac{-x^2}{(n+1)(n+2)2^2}.\\ \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{\lim }\frac{-x^2}{(n+1)(n+2)2^2}\\ =\frac{-x^2}{\infty }\\ =0=L\end{array}$

From the ratio test $L=0<1$ , Therefore have convergence.

So given series converges for all values of x , therefore the domain of x is . $x\in (-\infty ,\infty )$

b.

To determine

To graph: the first several partial sums on a common screen.

Answer to Problem 29E

Explanation of Solution

Given information: The function $J_1(x)={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{n!(n+1)!2^{2n+1}}}$ is called the Bessel Function

of order 1.

Calculation:

The graphs for the first 6 values of n are shown below by using graphing utility.

  

c.

To determine

To observe: how partial sums approximate $J_1$ .

Answer to Problem 29E

Explanation of Solution

Given information: The function $J_1(x)={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{n!(n+1)!2^{2n+1}}}$ is called the Bessel Function

of order 1.

Calculation:

The graphs for the first 6 values of n are shown below together with the Bessel function by using graphing utility.

  

See how the graphs converge for $-1\le x\le 1$ .