Chapter 10.2, Problem 13E

To determine

To find:The Taylor series generated by the function at the given point,

  $f(x)=\frac{1}{x+1},x=2$

Answer to Problem 13E

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n}{3^{n+1}}}{(x-2)}^n$

Explanation of Solution

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

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, $f(x)=\frac{1}{x+1},x=2$ . Then,

  $\begin{array}{l}f(2)={\left[\frac{1}{x+1}\right]}_{x=2}=\frac{1}{3}\\ f\text{'}(2)={\left[-\frac{1}{{(x+1)}^2}\right]}_{x=2}=-\frac{1}{3^2}=-\frac{1!}{3^2}\\ f\text{'}\text{'}(2)={\left[\frac{2}{{(x+1)}^3}\right]}_{x=2}=\frac{2}{3^3}=\frac{2!}{3^3}\\ f\text{'}\text{'}\text{'}(2)={\left[-\frac{6}{{(x+1)}^4}\right]}_{x=2}=-\frac{6}{3^4}=-\frac{3!}{3^4}\\ f^{(4)}(2)={\left[\frac{24}{{(x+1)}^5}\right]}_{x=2}=\frac{24}{3^5}=\frac{4!}{3^5}\end{array}$

Similarly, $f^{(k)}(2)={(-1)}^k\frac{k!}{3^{k+1}}$

The Taylor series generated by $f(x)=\frac{1}{x+1}$ at $x=2$ is given by,

  $\begin{array}{}$ \frac{1}{3}-\frac{1}{3^2}(x-2)+\frac{2!}{3^3}\cdot \frac{1}{2!}{(x-2)}^2-\frac{3!}{3^4}\cdot \frac{1}{3!}{(x-2)}^3+\frac{4!}{3^5}\cdot \frac{1}{4!}{(x-2)}^4-\mathrm{...}+{(-1)}^k\cdot \frac{k!}{3^{k+1}}\cdot \frac{1}{k!}{(x-2)}^k+\mathrm{...}$$ =\frac{1}{3}-\frac{1}{3^2}(x-2)+\frac{1}{3^3}{(x-2)}^2-\frac{1}{3^4}{(x-2)}^3+\frac{1}{3^5}{(x-2)}^4-\mathrm{...}+{(-1)}^k\frac{1}{3^{k+1}}{(x-2)}^k+\mathrm{...}$$ ={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n}{3^{n+1}}}{(x-2)}^n$\end{array}$