Chapter 9, Problem 60RE

$\text{(a)}$

To determine

To find: The first four terms and the general term of the Taylor series.

Answer to Problem 60RE

The answer: $f(x)=1-(x-3)+{(x-3)}^2-{(x-3)}^3+\dots +{(-1)}^n{(x-3)}^n+\dots$

Explanation of Solution

Given information:

The equation is $f(x)=1/(x-2)$ .

The Taylor series generated by $f(x)$ at $x=3$ .

Calculation:

To use the general power rule,

  $\begin{array}{}$ f^{\prime }(x)=(-1){(x-2)}^{-2}$$ f^{″}(x)=(-1)(-2){(x-2)}^{-3}$$ ={(-1)}^2(2!){(x-2)}^{-3}$$ f^{‴}(x)=(-1)(-2)(-3){(x-2)}^{-4}$$ ={(-1)}^3(3!){(x-2)}^{-4}$$ \mathrm{...}$$ f^{(n)}(x)={(-1)}^n(n!){(x-2)}^{-(n+1)}$$ ={(-1)}^n\frac{n!}{{(x-2)}^{n+1}}$\end{array}$

Evaluate the function and the derivatives at $x=3$ by substitution into their formulas, thus

  $\begin{array}{c}f(3)=1\\ $ f^{\prime }(3)=(-1){(3-2)}^{-2}$$ =-1$$ f^{″}(3)={(-1)}^2(2!){(3-2)}^{-3}$$ =2$$ f^{‴}(3)={(-1)}^3(3!){(3-2)}^{-4}$$ =-6$$ \mathrm{...}$$ f^{(3)}(x)={(-1)}^n\frac{n!}{{(3-2)}^{n+1}}$$ ={(-1)}^{n}n!$\end{array}$

To use the results of the previous step and the definition of Taylor series, obtain

  $\begin{array}{}$ f(x)=1-(x-3)+\frac{2}{2!}{(x-3)}^2-\frac{6}{3!}{(x-3)}^3+\dots +{(-1)}^n\frac{n!}{n!}{(x-3)}^n+\dots $$ =1-(x-3)+{(x-3)}^2-{(x-3)}^3+\dots +{(-1)}^n{(x-3)}^n+\dots $\end{array}$

Therefore, the required first four terms and the general term of the Taylor series is

  $f(x)=1-(x-3)+{(x-3)}^2-{(x-3)}^3+\dots +{(-1)}^n{(x-3)}^n+\dots$

$\text{(b)}$

To determine

To find: The first four terms and the general term of the series generated by $\ln |x-2|$ at $x=3$ .

Answer to Problem 60RE

The answer: $f(x)=1-(x-3)+{(x-3)}^2-{(x-3)}^3+\dots +{(-1)}^n{(x-3)}^n+\dots$

Explanation of Solution

Given information:

The equation is $f(x)=1/(x-2)$ .

Calculation:

By use

Theorem2, i.e. the Term-by-Term Integration from Section 10.1, obtain at $x=3$ .

So,

  

  $=(x-3)-\frac{{(x-3)}^2}{2}+\frac{{(x-3)}^3}{3}-\frac{{(x-3)}^4}{4}+\dots {(-1)}^n\frac{{(x-3)}^{n+1}}{n+1}+\dots$

Therefore, the required first four terms and the general term of the series generated by $\ln |x-2|$ at $x=3$ is:

  $f(x)=1-(x-3)+{(x-3)}^2-{(x-3)}^3+\dots +{(-1)}^n{(x-3)}^n+\dots$

$(c)$

To determine

To find: The number differs from $\text{In }(3/2)$ by less than $0.05$ .

Answer to Problem 60RE

The answer: An appropriate value can be $0.375$ .

Explanation of Solution

Given information:

The equation is $f(x)=1/(x-2)$ .

Calculation:

First evaluate the value of $\ln (3/2)$ , it is approximately $0.4055$ . A number $\text{c}$ differs from this value less than $0.05$ .if $0.03555<c<0.4555$ .

Evaluate $P_n(x)$ at $x=3/2$ for some values of $n$ .

Thus,

  $\begin{array}{c}{P}_1(x)=\left(\frac{3}{2}-3\right)\\ =-\frac{3}{2}\end{array}$

And,

  $\begin{array}{c}{P}_2(x)=\left(\frac{3}{2}-3\right)+\frac{{\left(\frac{3}{2}-3\right)}^2}{2}\\ =-\frac{3}{2}\\ =-0.375\end{array}$

Therefore, the required appropriate value can be $0.375$ .