Chapter 10, Problem 37RE

To determine

To find:The first four nonzero terms and the general term of the Taylor series generated by $f$ at $x=a$ .

  $f(x)=\frac{1}{3-x},a=2$

Answer to Problem 37RE

The first four nonzero terms of the Taylor series is $1+(x-2)+{(x-2)}^2+{(x-2)}^3$.

The general term of the Taylor series is ${(x-2)}^n$ .

Explanation of Solution

Given information: $f(x)=\frac{1}{3-x},a=2$

Calculation:The Taylor series generated by $f$ at $x=a$ is,

  $f(a)+f\text{'}(a)(x-a)+\frac{f\text{'}\text{'}(a)}{2!}{(x-a)}^2+\mathrm{...}+\frac{f^{(n)}(a)}{n!}{(x-a)}^n+\mathrm{...}$

Here, $f(x)=\frac{1}{3-x},a=2$ . Then,

  $\begin{array}{l}f(2)={\left[\frac{1}{3-x}\right]}_{x=2}=1\\ f\text{'}(2)={\left[\frac{1}{{(3-x)}^2}\right]}_{x=2}=1=1!\\ f\text{'}\text{'}(2)={\left[\frac{2}{{(3-x)}^3}\right]}_{x=2}=2=2!\\ f\text{'}\text{'}\text{'}(2)={\left[\frac{2\cdot 3}{{(3-x)}^4}\right]}_{x=2}=6=3!\end{array}$

The first four nonzero terms of the Taylor series generated by $f(x)=\frac{1}{3-x}$ at $x=2$ is,

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

The general term of the Taylor series is ${(x-2)}^n$ .