Chapter 8.8, Problem 17E

(a)

To determine

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

Answer to Problem 17E

Taylor polynomials $T_4(x)$

  $T_4(x)={(x)}^2-\frac{{(x)}^4}{6}$

Explanation of Solution

Given information:

  $f(x)=x\sin x,a=0,n=4,-1\le x\le 1$

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 $T_n(x)$

  $\begin{array}{l}f(x)=x\sin x,a=0,n=4,-1\le x\le 1\\ T_4(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-a)}^3}{3!}+\frac{f^{\text{'}\text{'}\text{'}\text{'}}(a){(x-a)}^4}{4!}\\ f(x)=x\sin x,a=0\\ f(0)=0\\ f^{\text{'}}(0)=0\\ f^{\text{'}\text{'}}(0)=2\\ f^{\text{'}\text{'}\text{'}}(0)=0\\ f^4(0)=-4\end{array}$

Taylor polynomials $T_4(x)$

  $\begin{array}{l}{T}_4(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-a)}^3}{3!}+\frac{f^{\text{'}\text{'}\text{'}\text{'}}(a){(x-a)}^4}{4!}\\ T_4(x)=f(0)+\frac{f^{\text{'}}(0)}{1!}(x)+\frac{f^{\text{'}\text{'}}(0){(x)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(0){(x)}^3}{3!}+\frac{f^{\text{'}\text{'}\text{'}\text{'}}(0){(x)}^4}{4!}\\ T_4(x)={(}^x-\frac{{(x)}^4}{6}\end{array}$

(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 17E

The approximation is accurate to within

  $0.00833$

Explanation of Solution

Given information:

  $f(x)=x\sin x,a=0,n=4,-1\le x\le 1$

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 $a=0$ ?????? ??=??

  $\left|R_4(x)\right|\le \frac{M}{5!}{\left|x\right|}^4$

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

  $f^5(x)=5\sin x+x\cos x$

  $f^5(0)=0$ so, the corresponding Taylor term is $0$ and does not affect the series

  $f^6(x)=6\cos x+x\sin x$

  $\left|f^6(0)\right|\le M$

With $x=0$

  $6\le M$

  $\begin{array}{l}\left|R_4(x)\right|\le \frac{6}{6!}{\left|1\right|}^6\\ \left|R_4(x)\right|\le \frac{6}{6!}\\ \left|R_4(x)\right|\le 0.00833\end{array}$

The approximation is accurate to within

  $0.00833$

(c)

`

To determine

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

Answer to Problem 17E

Error estimate from graphical method is

  $0.0082$

Explanation of Solution

Given information:

  $\begin{array}{l}f(x)=x\sin x,a=0,n=4,-1\le x\le 1\\ T_4(x)={(}^x-\frac{{(x)}^4}{6}\end{array}$

Formula Used:

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

Calculation:

  $\begin{array}{l}\left|R_3(x)\right|=\left|f(x)-T_3(x)\right|\\ \left|R_3(x)\right|=\left|x\sin x-\left(x^2-\frac{x^4}{6}\right)\right|\end{array}$