Chapter 9.1, Problem 72E

a.

To determine

To state: The first four nonzero terms and the general term for a power series for $f\left(x\right)$ centered at $t=0$ . Let $f\left(t\right)=\frac{4}{1+t^2}\text{ and }G\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ .

Answer to Problem 72E

The first four terms are $4,-4t^2,4t^4,-4t^6$ and the general term is $t_n={\left(-1\right)}^n\left(4t^{2n}\right)$ .

Explanation of Solution

Given information:

It is given that $f\left(t\right)=\frac{4}{1+t^2}\text{ and }G\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ .

Consider the given function: $f\left(t\right)=\frac{4}{1+t^2}\text{ and }G\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$

The sum of the infinite geometric series whose ratio is less than 1 is given by $S=\frac{a}{1-r}$ , where $a$ is the initial term and $r$ is the common ratio.

On comparing the function $f\left(x\right)\text{ with }\frac{a}{1-r}$ the value are $a=4\text{ and }r=-t^2$ .

Therefore, the first terms are: $4,-4t^2,4t^4,-4t^6$

The general term is: $t_n={\left(-1\right)}^n\left(4t^{2n}\right)$

b.

To determine

To state: The first four nonzero terms and the general term for a power series for $G\left(x\right)$ centered at $x=0$ . Let $f\left(t\right)=\frac{4}{1+t^2}\text{ and }G\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ .

Answer to Problem 72E

The first four terms are $4x,\frac{4}{3}{x}^3,\frac{4}{5}{x}^5,\frac{4}{7}{x}^7$ and the general term is $t_n={\left(-1\right)}^n\left(\frac{4}{2n+1}\right)x^{2n+1}$ .

Explanation of Solution

Given information:

It is given that $f\left(t\right)=\frac{4}{1+t^2}\text{ and }G\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ .

Consider the given function: $f\left(t\right)=\frac{4}{1+t^2}\text{ and }G\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$

From part (a), the series for the function $f\left(t\right)\text{ is }f\left(t\right)=4-4t^2+4t^4-4t^6+\cdots$

To find the power series $G\left(x\right)$ of integrate both sides of the power series with respect to x :

  $f\left(t\right)=4-4t^2+4t^4-4t^6+\cdots$

  ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}={\displaystyle \underset{0}{\overset{x}{\int }}\left(4-4t^2+4t^4-4t^6+\cdots \right)dt}$

The series will become:

  $G\left(x\right)=4x-\frac{4}{3}{x}^3+\frac{4}{5}{x}^5-\frac{4}{7}{x}^7+\cdots$

The first four terms are: $4x,\frac{4}{3}{x}^3,\frac{4}{5}{x}^5,\frac{4}{7}{x}^7$

The general term is: $t_n={\left(-1\right)}^n\left(\frac{4}{2n+1}\right)x^{2n+1}$

c.

To determine

To state: The interval of convergence of the power series in part (a).

Answer to Problem 72E

The resultant interval is $-1<t<1$ .

Explanation of Solution

Given information:

The given statement says to find the interval of convergence of the power series in part (a).

Consider the given function: $f\left(t\right)=\frac{4}{1+t^2}$

From part (a), the series is $4,-4t^2,4t^4,-4t^6$ and the value of the first term is $a=4$ and the common ratio is $r=-t^2$ .

A geometric series converges when $\left|r\right|<1$ . Now consider the common ratio of the given series to check if the series converges or not.

The common ratio is $r=-t^2$ . The series converges when $\left|-t^2\right|<1$ , that gives the interval of convergence as $-1<t<1$ .

d.

To determine

To state: The two numbers if the interval of convergence of the power series in part (b) is almost the same as the interval in part (c), but it includes two more numbers.

Answer to Problem 72E

The two numbers are $x=1\text{ and }x=-1$ .

Explanation of Solution

Given information:

The given statement says to find the two numbers if the interval of convergence of the power series in part (b) is almost the same as the interval in part (c), but it includes two more numbers..

The two numbers are $x=1\text{ and }x=-1$

which result in convergent series:

  $G\left(x\right)=4x-\frac{4}{3}{x}^3+\frac{4}{5}{x}^5-\frac{4}{7}{x}^7+\cdots$

Substitute the values $x=1\text{ and }x=-1$ ,

  $\begin{array}{c}G\left(1\right)=4\left(1\right)-\frac{4}{3}{\left(1\right)}^3+\frac{4}{5}{\left(1\right)}^5-\frac{4}{7}{\left(1\right)}^7+\cdots +{\left(-1\right)}^n\left(\frac{4}{2n+1}\right)+\cdots \\ =4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\cdots +{\left(-1\right)}^n\left(\frac{4}{2n+1}\right)+\cdots \end{array}$

Similarly,

  $\begin{array}{c}G\left(-1\right)=4\left(-1\right)-\frac{4}{3}{\left(-1\right)}^3+\frac{4}{5}{\left(-1\right)}^5-\frac{4}{7}{\left(-1\right)}^7+\cdots +{\left(-1\right)}^n\left(\frac{4}{2n+1}\right)+\cdots \\ =-4+\frac{4}{3}-\frac{4}{5}+\frac{4}{7}+\cdots +{\left(-1\right)}^n\left(\frac{4}{2n+1}\right)+\cdots \end{array}$