Chapter 8.8, Problem 14E

(a)

To determine

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

Answer to Problem 14E

Taylor polynomials 1y1y63 $T_4(x)$

  $T_4(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi }{6}\right)-\frac{1}{4}{\left(x-\frac{\pi }{6}\right)}^2-\frac{\sqrt{3}}{12}{\left(x-\frac{\pi }{6}\right)}^3+\frac{1}{48}{\left(x-\frac{\pi }{6}\right)}^4$

Explanation of Solution

Given information:

  $f(x)=\sin x,a=\frac{\pi }{6},n=4,0\le x\le \frac{\pi }{3}$

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 $T_n(x)$

  $\begin{array}{l}{T}_4(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-a)}^3}{3!}+\frac{f^{\text{'}\text{'}\text{'}\text{'}}(a){(x-a)}^4}{4!}\\ f(x)=\sin x,a=\frac{\pi }{6}\\ f(\frac{\pi }{6})=\frac{1}{2}\\ f^{\text{'}}(\frac{\pi }{6})=\frac{\sqrt{3}}{2}\\ f^{\text{'}\text{'}}(\frac{\pi }{6})=\frac{-1}{2}\\ f^{\text{'}\text{'}\text{'}}(\frac{\pi }{6})=\frac{-\sqrt{3}}{2}\\ f^4(\frac{\pi }{6})=\frac{1}{2}\end{array}$

Taylor polynomials $T_4(x)$

  $T_4(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi }{6}\right)-\frac{1}{4}{\left(x-\frac{\pi }{6}\right)}^2-\frac{\sqrt{3}}{12}{\left(x-\frac{\pi }{6}\right)}^3+\frac{1}{48}{\left(x-\frac{\pi }{6}\right)}^4$

(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 14E

The approximation is accurate to within

  $0.000328$

Explanation of Solution

Given information:

  $f(x)=\sin x,a=\frac{\pi }{6},n=4,0\le x\le \frac{\pi }{3}$

Formula Used:

  $T_n(x)\to f(x)$ ???? ??&#??????????;&#??????????;

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 $a=\frac{\pi }{6}$ ?????? ??=??

  $\left|R_4(x)\right|\le \frac{M}{5!}{\left|x-\frac{\pi }{6}\right|}^5$

Where $\left|f^{(5)}(x)\right|\le M$

  $\begin{array}{l}{f}^4(x)=\cos x\\ \cos x\le M\end{array}$

With $x=0$

  $\begin{array}{l}1\le M\\ \left|R_4(x)\right|\le \frac{1}{5!}{\left|x-\frac{\pi }{6}\right|}^5\\ \left|R_4(x)\right|\le \frac{1}{5!}{\left|\frac{\pi }{6}\right|}^5\\ \left|R_4(x)\right|\le 0.000328\end{array}$

The approximation is accurate to within

  $0.000328$

(c)

To determine

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

Answer to Problem 14E

Error estimate from graphical method is

  $0.0003$

Explanation of Solution

Given information:

  $f(x)=\sin x,a=\frac{\pi }{6},n=4,0\le x\le \frac{\pi }{3}$

  $T_4(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi }{6}\right)-\frac{1}{4}{\left(x-\frac{\pi }{6}\right)}^2-\frac{\sqrt{3}}{12}{\left(x-\frac{\pi }{6}\right)}^3+\frac{1}{48}{\left(x-\frac{\pi }{6}\right)}^4$

Formula Used:

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

Calculation:

  $\begin{array}{l}\left|R_n(x)\right|=\left|f(x)-T_n(x)\right|\\ R_3(x)=\left|\sin x-\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi }{6}\right)-\frac{1}{4}{\left(x-\frac{\pi }{6}\right)}^2-\frac{\sqrt{3}}{12}{\left(x-\frac{\pi }{6}\right)}^3+\frac{1}{48}{\left(x-\frac{\pi }{6}\right)}^4\right)\right|\end{array}$