Chapter 8.8, Problem 12E

(a)

To determine

To Find Taylor’s polynomial with degree n at the number a

Answer to Problem 12E

Taylor polynomials $T_2(x)$

  $T_2(x)=1-2(x-1)+3{(x-1)}^2$

Explanation of Solution

Given information:

  $f(x)=x^{-2},a=1,n=2,0.9\le x\le 1.1$

Formula Used:

Taylor Polynomials

  $T_n(x)={\displaystyle \sum _{i=0}^n\frac{f^{(i)}(a)}{i!}{(x-a)}^i}=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}\dots +\frac{f^n(a)}{n!}{(x-a)}^n$

Calculation:

Taylor polynomials $T_n(x)$

  $\begin{array}{l}f(x)=x^{-2},a=1,n=2,0.9\le x\le 1.1\\ T_2(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}\\ f(x)=x^{-2},a=1\\ f(1)=1\\ f^{\text{'}}(1)=-2\\ f^{\text{'}\text{'}}(1)=-2\\ T_2(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}\\ T_2(x)=f(1)+\frac{f^{\text{'}}(1)}{1!}(x-1)+\frac{f^{\text{'}\text{'}}(1){(x-1)}^2}{2!}\\ T_2(x)=1-2(x-1)+3{(}^x\end{array}$

Taylor polynomials $T_2(x)$

  $T_2(x)=1-2(x-1)+3{(x-1)}^2$

(b)

To determine

To Find Use Taylor’s Inequality to estimate the accuracy of the approximation

  $f(x)\approx T_n(x)$

Answer to Problem 12E

The approximation is accurate to within

  $0.00677$

Explanation of Solution

Given information:

  $f(x)=x^{-2},a=1,n=2,0.9\le x\le 1.1$

Formula Used:

  $T_n(x)\to f(x)$ as $n\to \infty$

And $T_n(x)$ can be used as an approximation to

  $f:f(x)=T_n(x)$

Taylor’s Inequality, which says that if

  $\left|f^{(n+1)}(x)\right|\le M$ , then

  $\left|R_n(x)\right|\le \frac{M}{(n+1)!}{\left|x-a\right|}^{n+1}$

Calculation:

Taylor’s Inequality with $a=4$ and n = 2

  $\left|R_2(x)\right|\le \frac{M}{3!}{\left|x-a\right|}^3$

Where $\left|f^3(x)\right|\le M$ because $x\le 4.2$

  $\begin{array}{l}\left|f^3(x)\right|\le \left|\frac{24}{{\left(0.9\right)}^5}\right|\\ \left|f^3(x)\right|\le 40.6442\\ \left|R_2(x)\right|\le \frac{40.6442}{3!}{\left|1.1-1\right|}^3\\ \left|R_2(x)\right|\le 0.00677\end{array}$

The approximation is accurate to within

  $0.00677$

(c)

To determine

To check the result by graphing $\left|R_n(x)\right|$

Answer to Problem 12E

Error estimate from graphical method is

  $0.00677$

Explanation of Solution

Given information:

  $f(x)=x^{-2},a=1,n=2,0.9\le x\le 1.1$

  $T_2(x)=1-2(x-1)+3{(x-1)}^2$

Formula Used:

  $\left|R_n(x)\right|=\left|f(x)-T_n(x)\right|$

Calculation:

  $\begin{array}{l}\left|R_2(x)\right|=\left|f(x)-T_2(x)\right|\\ R_2(x)=\left|x^{-2}-1+2(x-1)-3{(x-1)}^2\right|\end{array}$