Chapter 10.1, Problem 72E

(a)

To determine

The first four non-zero terms and the general form of power series of f(t).

Answer to Problem 72E

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

Explanation of Solution

Given information:

The given functions are

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

Formula used:

  $S=\frac{a}{1-r}$

Calculation:

The sum of the infinite geometric series whose ratio is less than 1. $S=\frac{a}{1-r}$ where, a is initial term and r is the common ratio. Computing $f\left(x\right)$ with $\frac{a}{1-r}$ get $a=4$ and $r=-t^2$ . Therefore, first four terms are $4,-4t^2,4t^4,4t^6$ and the general term, $t_n={\left(-1\right)}^n\left(4t^{2n}\right)$

Conclusion:

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

(b)

To determine

The first four non-zero terms and the general form of power series of f(t).

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 $t_n={\left(-1\right)}^n\left(\frac{4}{2n+1}\right)x^{2n+1}$

Explanation of Solution

Given information:

The given functions are

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

Formula used:

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

Calculation:

  $f\left(t\right)=4-4t^2+4t^4-4t^6+\mathrm{......}$ to find the power series, integrate both the side with respect to x

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

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

Conclusion:

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 $t_n={\left(-1\right)}^n\left(\frac{4}{2n+1}\right)x^{2n+1}$

(c)

To determine

The internal of convergence of the power series.

Answer to Problem 72E

The interval of convergence as $-1<r<1$

Explanation of Solution

Given information:

The given functions are

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

Formula used:

  $\left|-t^2\right|<1$

Calculation:

  $f\left(t\right)=\frac{4}{1+t^2}a=4$ and $r=-t^2$ , where, a is initial term and r is the common ratio. The series converges when $\left|-t^2\right|<1$ , that gives the interval of convergence as $-1<r<1$

Conclusion:

The interval of convergence as $-1<r<1$

(d)

To determine

The internal of convergence of the power series.

Answer to Problem 72E

The value is G(1) and G(-1)

Explanation of Solution

Given information:

The given functions are

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

Formula used:

  $G\left(1\right)=4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}-\mathrm{.....}+{\left(-1\right)}^n\left(\frac{4}{2n+1}\right)+\mathrm{....}$

Calculation:

The two numbers are $x=1\text{and}x=-1$ which results in convergent series.

  $G\left(1\right)=4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}-\mathrm{.....}+{\left(-1\right)}^n\left(\frac{4}{2n+1}\right)+\mathrm{....}$ and

  $G\left(-1\right)=-4+\frac{4}{3}-\frac{4}{5}+\frac{4}{7}-\mathrm{.....}+{\left(-1\right)}^{n-1}\left(\frac{4}{2n+1}\right)+\mathrm{......}$ respectively.

Conclusion:

The value is G(1) and G(-1)