Chapter 9.2, Problem 28E

(a)

To determine

The first four non zero terms and general term of series.

Answer to Problem 28E

The first four non zero terms of Taylor series for $f\left(x\right)$ is $f\left(x\right)=1+3x+\frac{9x^2}{2}+\frac{9x^3}{2}$ .

General term of power series is $=1+3x+\frac{9x^2}{2}+\frac{9x^3}{2}+\mathrm{..........}$

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ .

The value of

  $\begin{array}{l}{a}_0=1\\ a_n=\left(\frac{3}{n}\right)a_{n-1}\end{array}$

The general term of power series is

  $\begin{array}{c}={\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}\\ =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\mathrm{.......}\\ =a_0+\frac{3}{1}{a}_{0}x+\frac{3}{2}\times \frac{3}{1}{a}_0{x}^2+\frac{3}{3}\times \frac{9}{2}{a}_0{x}^3+\mathrm{........}\\ =1+3x+\frac{9x^2}{2}+\frac{9x^3}{2}+\mathrm{..........}\end{array}$

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}}{\left(x-a\right)}^n\\ =f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)+\frac{f\text{'}\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(a\right)}{3!}{\left(x-a\right)}^3+\mathrm{............}\end{array}$

(b)

To determine

The function represented by power series.

Answer to Problem 28E

The power series represent the function $f\left(x\right)=e^{3x}atx=a=0$ .

Explanation of Solution

Given:

General term of power series is $=1+3x+\frac{9x^2}{2}+\frac{9x^3}{2}+\mathrm{..........}$

The Taylor series for $f\left(x\right)$ at $x=a=0$ is

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(0\right)}{n!}}{\left(x\right)}^n\\ =f\left(0\right)+f\text{'}\left(0\right)\left(x\right)+\frac{f\text{'}\text{'}\left(0\right)}{2!}{\left(x\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{\left(x\right)}^3+\mathrm{............}\end{array}$

On comparing the general term with Taylor series.

  $\begin{array}{l}f\left(0\right)=1\\ f^{\text{'}}\left(0\right)=3\\ \frac{f^{\text{'}\text{'}}\left(0\right)}{2!}=\frac{9}{2}\Rightarrow f^{\text{'}\text{'}}\left(0\right)=3^2\\ \frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}=\frac{9}{2}=\frac{27}{6}\Rightarrow f^{\text{'}\text{'}\text{'}}\left(0\right)=3^3\end{array}$

Taylor series expansion of $F\left(x\right)=e^{x}atx=a=0$ is

  $\begin{array}{c}F\left(x\right)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\mathrm{..........}\\ \text{Replace }x\text{ with 3}x\\ F\left(x\right)=e^{3x}=1+3x+\frac{{\left(3x\right)}^2}{2!}+\frac{{\left(3x\right)}^3}{3!}+\mathrm{..........}\\ =e^{3x}=1+3x+\frac{9x^2}{2}+\frac{27x^3}{3!}+\mathrm{..........}\\ =e^{3x}=1+3x+\frac{9x^2}{2}+\frac{9x^3}{2}+\mathrm{..........}\end{array}$

The power series represent the function $f\left(x\right)=e^{3x}atx=a=0$ .

(c)

To determine

The exact value of the function $f^{\text{'}}\left(1\right)$ .

Answer to Problem 28E

The exact value of the function $f^{\text{'}}\left(1\right)$ is $f^{\text{'}}\left(1\right)=60.26$ .

Explanation of Solution

Given:

The function of the power series is $f\left(x\right)=e^{3x}$

On differentiating the given function.

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}{e}^{3x}\\ f^{\text{'}}\left(x\right)=3e^{3x}\\ f^{\text{'}}\left(1\right)=3e^3\\ f^{\text{'}}\left(1\right)=60.26\end{array}$