Chapter 8.8, Problem 18E

(a)

To determine

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

Answer to Problem 18E

Taylor polynomials $T_3(x)$

  $T_3(x)=(x-1)+\frac{{(x-1)}^2}{2}-\frac{{(x-1)}^3}{6}$

Explanation of Solution

Given information:

  $f(x)=x\ln x,a=1,n=3,0.5\le x\le 1.5$

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:

  $f(x)=x\ln x,a=1,n=3,0.5\le x\le 1.5$

Taylor polynomials $T_n(x)$

  $\begin{array}{l}{T}_3(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-a)}^3}{3!}\\ f(x)=x\ln x,a=1\\ f(1)=0\\ f^{\text{'}}(1)=1\\ f^{\text{'}\text{'}}(1)=1\\ f^{\text{'}\text{'}\text{'}}(1)=-1\\ T_3(x)=f(a)+\frac{f^{\text{'}}(a)}{1!}(x-a)+\frac{f^{\text{'}\text{'}}(a){(x-a)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-a)}^3}{3!}\\ T_3(x)=f(1)+\frac{f^{\text{'}}(1)}{1!}(x-1)+\frac{f^{\text{'}\text{'}}(1){(x-1)}^2}{2!}+\frac{f^{\text{'}\text{'}\text{'}}(a){(x-1)}^3}{3!}\\ T_3(x)=(x-1)+\frac{{(x-1)}^2}{2}-\frac{{(x-1)}^3}{6}\end{array}$

Taylor polynomials $T_3(x)$

  $T_3(x)=(x-1)+\frac{{(x-1)}^2}{2}-\frac{{(x-1)}^3}{6}$

(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 18E

The approximation is accurate to within

  $0.0416$

Explanation of Solution

Given information:

  $f(x)=x\ln x,a=1,n=3,0.5\le x\le 1.5$

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 $a=1$ ?????? ??=??

  $\left|R_3(x)\right|\le \frac{M}{4!}{\left|x-1\right|}^4$

Where $\left|f^4(x)\right|\le M$

  $\begin{array}{l}\left|\frac{2}{{(x)}^3}\right|\le M\\ 16\le M\\ \left|R_3(x)\right|\le \frac{16}{4!}{\left|1.5-1\right|}^4\\ \left|R_3(x)\right|\le 0.0416\end{array}$

The approximation is accurate to within

  $0.0416$

(c)

`

To determine

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

Answer to Problem 18E

Error estimate from graphical method is

  $0.0416$

Explanation of Solution

Given information:

  $f(x)=x\ln x,a=1,n=3,0.5\le x\le 1.5$

  $T_3(x)=(x-1)+\frac{{(x-1)}^2}{2}-\frac{{(x-1)}^3}{6}$

Formula Used:

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

Calculation:

  $\begin{array}{l}\left|R_3(x)\right|=\left|f(x)-T_3(x)\right|\\ \left|R_3(x)\right|=\left|x\ln x-\left(\left(x-1\right)+\frac{{(x-1)}^2}{2}-\frac{{(x-1)}^3}{6}\right)\right|\end{array}$