Chapter 8.8, Problem 11E

(a)

To determine

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

Answer to Problem 11E

Taylor polynomials $T_2(x)$

  $\begin{array}{l}{T}_2(x)=2+\frac{1}{4}(x-4)-\frac{{(x-4)}^2}{64}\\ \end{array}$

Explanation of Solution

Given information:

  $f(x)=\sqrt{x},a=4,n=2,4\le x\le 4.2$

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 $T_n(x)$

  $\begin{array}{l}f(x)=\sqrt{x},a=4,n=2,4\le x\le 4.2\\ T_2(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}\\ f(x)=\sqrt{x},a=4\\ f(4)=0\\ f^{\text{'}}(4)=\frac{1}{4}\\ f^{\text{'}\text{'}}(4)=\frac{-1}{32}\\ T_2(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}\\ T_2(x)=f(4)+\frac{f^{\text{'}}(4)}{1!}(x-4)+\frac{f^{\text{'}\text{'}}(4){(x-4)}^2}{2!}\\ T_2(x)=2+\frac{1}{4}(x-4)-\frac{{(x-4)}^2}{64}\\ \end{array}$

Taylor polynomials $T_2(x)$

  $\begin{array}{l}{T}_2(x)=2+\frac{1}{4}(x-4)-\frac{{(x-4)}^2}{64}\\ \end{array}$

(b)

To determine

To FindUse Taylor’s Inequality to estimate the accuracy of the approximation

  $f(x)\approx T_n(x)$

Answer to Problem 11E

The approximation is accurate to within

  $1.5625\times {10}^{-5}$

Explanation of Solution

Given information:

  $f(x)=\sqrt{x},a=4,n=2,4\le x\le 4.2$

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

  $\begin{array}{l}\left|f^3(x)\right|\le \frac{3}{8{\left(4\right)}^{\frac{5}{2}}}\\ \left|f^3(x)\right|\le 0.011718\\ \left|R_2(x)\right|\le \frac{0.011718}{3!}{\left|4.2-4\right|}^3\\ \left|R_2(x)\right|\le 1.5625\times {10}^{-5}\end{array}$

The approximation is accurate to within

  $1.5625\times {10}^{-5}$

(c)

To determine

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

Answer to Problem 11E

Error estimate from graphical method is

  $1.5625\times {10}^{-5}$

Explanation of Solution

Given information:

  $f(x)=\sqrt{x},a=4,n=2,4\le x\le 4.2$

  $\begin{array}{l}{T}_2(x)=2+\frac{1}{4}(x-4)-\frac{{(x-4)}^2}{64}\\ \end{array}$

Formula Used:

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

Calculation:

  $\begin{array}{l}\left|R_2(x)\right|=\left|f(x)-T_2(x)\right|\\ R_2(x)=\sqrt{x}-2-\frac{\left(x-4\right)}{4}+\frac{{\left(x-4\right)}^2}{64}\end{array}$