Chapter 8.5, Problem 30E

a.

To determine

To find: the domain of the Airy function.

Answer to Problem 30E

  $x\in (-\infty ,\infty )$ .

Explanation of Solution

Given information: The function $A(x)=1+\frac{x^3}{2.3}+\frac{x^6}{\mathrm{2.3.5.6}}+\frac{x^9}{\mathrm{2.3.5.6.8.9}}+\mathrm{...}$ is called the

Airy Function.

Calculation:

  $\begin{array}{c}A(x)=1+\frac{x^3}{2.3}+\frac{x^6}{\mathrm{2.3.5.6}}+\frac{x^9}{\mathrm{2.3.5.6.8.9}}+\mathrm{...}\\ \text{The}\text{given}\text{series}\text{can}\text{be}\text{written}\text{as:}\\ A(x)={\displaystyle \sum _{n=0}^{\infty }\frac{x^{3n}}{(3n)!}}.\\ a_n=\frac{x^{3n}}{(3n)!}.\\ a_{n+1}=\frac{x^{3(n+1)}}{3(n+1))!}\\ a_{n+1}=\frac{x^{3n+3}}{(3n+3)!}\\ \frac{a_{n+1}}{a_n}=\frac{{(x)}^{3n}.x^3}{(3n+3)(3n+2)(3n+1)(3n)!}.\frac{(3n)!}{x^{3n}}\\ \frac{a_{n+1}}{a_n}=\frac{x^3}{(3n+3)(3n+2)(3n+1)}.\\ \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{\lim }\frac{x^3}{(3n+3)(3n+2)(3n+1)}.\\ =\frac{x^3}{\infty }\\ =0=L\end{array}$

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 30E

Explanation of Solution

Given information: The function $A(x)=1+\frac{x^3}{2.3}+\frac{x^6}{\mathrm{2.3.5.6}}+\frac{x^9}{\mathrm{2.3.5.6.8.9}}+\mathrm{...}$ is called the

Airy Function.

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 $A$ .

Answer to Problem 30E

Explanation of Solution

Given information: The function $A(x)=1+\frac{x^3}{2.3}+\frac{x^6}{\mathrm{2.3.5.6}}+\frac{x^9}{\mathrm{2.3.5.6.8.9}}+\mathrm{...}$ is called the

Airy Function.

Calculation:

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