Chapter 9, Problem 57RE

a.

To determine

To find: first 3 terms of taylor series

Answer to Problem 57RE

  $-\frac{5x^3}{48}+\frac{5x^5}{3840}+\frac{5x^7}{645120}$

Explanation of Solution

Given: $f(x)=5\sin (x/2)$

Concept Used: $\sin x={\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}}$

Calculation:

  $\begin{array}{c}\sin (x)=-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\\ 5\sin (x/2)=\frac{-5{(x/2)}^3}{6}+\frac{5{(x/2)}^5}{120}-\frac{5{(x/2)}^7}{5040}\\ =-\frac{5x^3}{48}+\frac{5x^5}{3840}+\frac{5x^7}{645120}\end{array}$

Conclusion: the first 3 terms are $-\frac{5x^3}{48}+\frac{5x^5}{3840}+\frac{5x^7}{645120}$

b.

To determine

To find: the interval in which the function is convergent.

Answer to Problem 57RE

Converging on all real number

Explanation of Solution

Given: $f(x)=5\sin (x/2)$

Concept Used: the value of sin is always less than 1

Calculation:

  $\left|\sin x\right|<1$ therefore the series is convergent on all x

Conclusion: the series is absolutely convergent.

c.

To determine

To find: minimum terms required to not exceed 0.1 in magnitude

Answer to Problem 57RE

2

Explanation of Solution

Given: interval is (-2,2)

Concept Used: alternating series estimation theorem

Calculation:

Approximating to one term

  $\begin{array}{c}5\sin (x/2)=-5/48\\ =-0.1041\end{array}$

2 terms

  $\begin{array}{c}5\sin (x/2)=-5/48+5/3840\\ =-0.1028\end{array}$

3 terms

  $\begin{array}{c}5\sin (x/2)=-5/48+5/3840-5/645120\\ =-0.1028\end{array}$

The series has been stable to 4 decimals

Conclusion: minimum 2 terms is required to not exceed a 0.1 error