Chapter 8.8, Problem 26E

To determine

To Prove: The fifth degree Taylor polynomial approximates $f\left(5\right)$ with error less than 0.0002.

Explanation of Solution

Given:

The given function is $f^n\left(4\right)=\frac{{\left(-1\right)}^{n}n!}{3^n\left(n+1\right)}$ .

The Taylor series of f centred at 4 converges to $f\left(x\right)$ for all the values of $x$ in the interval of congruence.

Calculation:

Consider the given function is $f^n\left(4\right)=\frac{{\left(-1\right)}^{n}n!}{3^n\left(n+1\right)}$

Then,

  $\frac{f^n\left(4\right)}{n!}=\frac{{\left(-1\right)}^n}{3^n\left(n+1\right)}$

Then, the series around 4 is of the form $f\left(4\right)=0$ then,

  $\begin{array}{c}\frac{f^{\prime }\left(4\right)}{1!}=\frac{{\left(-1\right)}^1}{3^1\left(1+1\right)}\\ =\frac{-1}{6}\end{array}$

From above, in the same way,

  $\begin{array}{l}\frac{f^{\prime }\left(4\right)}{2!}=\frac{1}{27}\\ \frac{f^{″}\left(4\right)}{3!}=\frac{1}{108}\\ \frac{f^4\left(4\right)}{4!}=\frac{1}{405}\\ \frac{f^5\left(4\right)}{5!}=-\frac{1}{1458}\end{array}$

Thus,

  $f\left(x\right)=0-\frac{1}{6}\left(x-4\right)+\frac{1}{27}{\left(x-4\right)}^2-\frac{1}{108}{\left(x-4\right)}^3+\frac{1}{405}{\left(x-4\right)}^4+\mathrm{...}+\frac{{\left(-1\right)}^n}{3^n\left(n+1\right)}{\left(x-4\right)}^n+\mathrm{..}$

Consider the inequality as, $\left|f^{n+1}\left(x\right)\right|\le M$ then, $\left|R_{n}x\left(x\right)\right|=\left|f\left(x\right)-T_n\left(x\right)\right|\le \frac{M}{\left(n+1\right)!}{\left|x-a\right|}^{n+1}$ .

Thus,

  $\begin{array}{c}{f}^4\left(4\right)=\frac{6!}{3^6\left(6+1\right)}\\ =0.1410935\end{array}$

Thus, the value of the fifth polynomial is,

  $\begin{array}{l}\left|R_5\left(5\right)\right|=\left|f\left(5\right)-T_5\left(5\right)\right|\le \frac{0.1410935}{6!}{\left|5-4\right|}^6\\ \left|f\left(5\right)-T_5\left(5\right)\right|\le 0.0001959632<0.0002\end{array}$

Thus, the value of the fifth polynomial is proved.