Chapter 10, Problem 57RE

(a)

To determine

The first three non-zero terms and the general term of the Taylor series by $f\left(x\right)=5\sin \left(x/2\right)$

Answer to Problem 57RE

The first three non-zero terms of the series are $T_0=\frac{5x}{2}{T}_1=\frac{5x^3}{48}{T}_2=\frac{x^5}{768}$

Explanation of Solution

Given information:

The given function is

  $f\left(x\right)=5\sin \left(x/2\right)$

Formula used:

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}+\mathrm{....}$

Calculation:

  $f\left(x\right)=5\sin \left(x/2\right)$ series generated by function sin x is given by

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}+\mathrm{....}$

Replacing x by x/2 and multiplying whole series by 5

  $\begin{array}{l}f\left(x\right)=5\left[x/2-\frac{{\left(x/2\right)}^3}{3!}+\frac{{\left(x/2\right)}^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{{\left(x/2\right)}^{2n+1}}{\left(2n+1\right)!}+\mathrm{....}\right]\\ 5\sin \left(x/2\right)=\frac{5x}{2}-\frac{5x^3}{48}+\frac{x^5}{768}-\mathrm{....}+{\left(-1\right)}^n\frac{5}{\left(2n+1\right)!}{\left(\frac{x}{2}\right)}^{2n+1}+\mathrm{......}\end{array}$

Therefore, the first three non-zero terms of the series are $T_0=\frac{5x}{2}{T}_1=\frac{5x^3}{48}{T}_2=\frac{x^5}{768}$

The general term of the series is given by $T_n={\left(-1\right)}^n\frac{5}{\left(2n+1\right)!}{\left(\frac{x}{2}\right)}^{2n+1}$

Conclusion:

The first three non-zero terms of the series are $T_0=\frac{5x}{2}{T}_1=\frac{5x^3}{48}{T}_2=\frac{x^5}{768}$

(b)

To determine

The interval of convergence

Answer to Problem 57RE

The series converges for all real values of x.

Explanation of Solution

Given information:

Since the series is,

  $S={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{5}{\left(2n+1\right)!}{\left(\frac{x}{2}\right)}^{2n+1}$

Formula used:

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\left|\frac{5{\left(-1\right)}^{n+1}}{\left(2n+3\right)!}{\left(\frac{x}{2}\right)}^{2n+3}\right|\left|\frac{\left(2n+1\right)!}{5{\left(-1\right)}^n}{\left(\frac{2}{x}\right)}^{2n+1}\right|$

Calculation:

  $S={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{5}{\left(2n+1\right)!}{\left(\frac{x}{2}\right)}^{2n+1}$

By using ratio test,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\left|\frac{5{\left(-1\right)}^{n+1}}{\left(2n+3\right)!}{\left(\frac{x}{2}\right)}^{2n+3}\right|\left|\frac{\left(2n+1\right)!}{5{\left(-1\right)}^n}{\left(\frac{2}{x}\right)}^{2n+1}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{-{\left(\frac{x}{2}\right)}^2}{\left(2n+2\right)\left(2n+3\right)}\right|\\ =\left|-\frac{x^2}{4}\right|\underset{n\to \infty }{\lim }\frac{1}{\left(2n+2\right)\left(2n+3\right)}\\ =\left|-\frac{x^2}{4}\right|.0\\ \underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0\end{array}$

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$

Therefore, the series converges for all real values of x.

Conclusion:

The series converges for all real values of x.

(c)

To determine

The minimum number of terms for f(x) in the interval (-2,2)

Answer to Problem 57RE

The Maximum error is $\frac{5}{2^{n+1}},\frac{{\left|2\right|}^{n+1}}{\left(n+1\right)!}=\frac{5}{\left(n+1\right)!}$

Explanation of Solution

Given information:

The given function is

  $f\left(x\right)=5\sin \left(x/2\right)$

Formula used:

  $5\sin \left(x/2\right)=\frac{5x}{2}-\frac{5x^3}{48}+\frac{x^5}{768}-\mathrm{....}+{\left(-1\right)}^n\frac{5}{\left(2n+1\right)!}{\left(\frac{x}{2}\right)}^{2n+1}+\mathrm{....}$

Calculation:

  $f\left(x\right)=5\sin \left(x/2\right)$

  $5\sin \left(x/2\right)=\frac{5x}{2}-\frac{5x^3}{48}+\frac{x^5}{768}-\mathrm{....}+{\left(-1\right)}^n\frac{5}{\left(2n+1\right)!}{\left(\frac{x}{2}\right)}^{2n+1}+\mathrm{....}$

Therefore, all values of the function $f^{\left(n\right)}\left(x\right)$ is bounded by $\frac{5}{2^n}$ for all x $n=1,2,\mathrm{3....}$

  $M=5,r=\frac{1}{2}$

Which gives $\left|R_n\left(x\right)\right|\le \frac{5}{2^{n+1}}\frac{{\left|x\right|}^{n+1}}{\left(n+1\right)!}$

Therefore,

Maximum error is $\frac{5}{2^{n+1}},\frac{{\left|2\right|}^{n+1}}{\left(n+1\right)!}=\frac{5}{\left(n+1\right)!}$

So, to make this error less than 0.1, requires $n\ge 4$

So, two terms (up through degree four) are needed.

Conclusion:

The Maximum error is $\frac{5}{2^{n+1}},\frac{{\left|2\right|}^{n+1}}{\left(n+1\right)!}=\frac{5}{\left(n+1\right)!}$