Chapter 9.2, Problem 22E

(a)

To determine

To find:The third order Taylor polynomial for $f$ at $x=0$ and the value at $f\left(0.2\right)$ .

Answer to Problem 22E

The third order Taylor polynomial is $f\left(x\right)=4+5x-4x^2+x^3$ .

The value of $f\left(0.2\right)$ is $f\left(0.2\right)=4.848$ .

Explanation of Solution

Given:

The value of function at $x=0$ is $f\left(0\right)=4$ .

The value of first derivative of function at $x=0$ is $f\text{'}\left(0\right)=5$ .

The value of second derivative of function at $x=0$ is $f\text{'}\text{'}\left(0\right)=-8$ .

The value of third derivative of function at $x=0$ is $f\text{'}\text{'}\text{'}\left(0\right)=6$ .

Concept Used:

Taylor polynomial is given by:

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

Put $a=0$ in general expression of Taylor polynomial.

  $\begin{array}{c}f\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)\left(x-0\right)+\frac{f\text{'}\text{'}\left(0\right)}{2!}{\left(x-0\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{\left(x-0\right)}^3\\ =4+5x+\frac{-8}{2!}{x}^2+\frac{6}{3!}{x}^3\\ =4+5x+\frac{-8}{2\times 1}{x}^2+\frac{6}{3\times 2\times 1}{x}^3\\ =4+5x-4x^2+x^3\mathrm{....................}\left(1\right)\end{array}$

The value of $f\left(0.2\right)$ is obtained if $x=0.2$ in equation $\left(1\right)$ is:

  $\begin{array}{c}f\left(0.2\right)=4+5\left(0.2\right)-4{\left(0.2\right)}^2+{\left(0.2\right)}^3\\ =4+1-4\times 0.04+0.008\\ =5-0.16+0.008\\ =4.848\end{array}$

Conclusion:

The third order Taylor polynomial is $f\left(x\right)=4+5x-4x^2+x^3$ .

The value of $f\left(0.2\right)$ is $f\left(0.2\right)=4.848$ .

(b)

To determine

To find: The second order Taylor polynomial for derivative of $f$ at $x=0$ and the value at $f\text{'}\left(0.2\right)$ .

Answer to Problem 22E

The second order Taylor polynomial for derivative of $f$ at $x=0$ is $f\text{'}\left(x\right)=3x^2-8x+5$ .

The value of $f\text{'}\left(0.2\right)$ is $f\text{'}\left(0.2\right)=3.52$ .

Explanation of Solution

Given:

The third order Taylor polynomial is $f\left(x\right)=4+5x-4x^2+x^3$

The derivative of Taylor polynomial is given by:

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(4+5x-4x^2+x^3\right)\\ f\text{'}\left(x\right)=\frac{d}{dx}\left(4\right)+5\times \frac{d}{dx}\left(x\right)-4\times \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x^3\right)\\ f\text{'}\left(x\right)=0+5-4\left(2x\right)+\left(3x^2\right)\\ f\text{'}\left(x\right)=3x^2-8x+5\mathrm{................}\left(2\right)\end{array}$

Put the value $x=0.2$ in the equation $\left(2\right)$ .

  $\begin{array}{c}f\text{'}\left(0.2\right)=3{\left(0.2\right)}^2-8\left(0.2\right)+5\\ =3\times 0.04-1.6+5\\ =0.12-1.6+5\\ =3.52\end{array}$

Conclusion:

The second order Taylor polynomial for derivative of $f$ at $x=0$ is $f\text{'}\left(x\right)=3x^2-8x+5$ .

The value of $f\text{'}\left(0.2\right)$ is $f\text{'}\left(0.2\right)=3.52$ .