Chapter 9, Problem 39RE

To determine

To find: The Taylor series of the function $-\frac{1}{x}\text{ at }a=3$ .

Answer to Problem 39RE

  $-\frac{1}{3}+\frac{1}{9}\left(x-3\right)-\frac{1}{27}{\left(x-3\right)}^2+\frac{1}{81}{\left(x-3\right)}^3$

Explanation of Solution

Given information:

  $-\frac{1}{x}\text{ at }a=3$

Concept Used:

Taylor Series of $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n}$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n}=f\left(a\right)+\frac{f\text{'}\left(a\right)}{1!}\left(x-a\right)+\frac{f\text{'}\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+\mathrm{...}$

Calculation:

Consider the function, $f\left(x\right)=-\frac{1}{x}$ .

Now find the derivative and their values at $a=-1$ .

  $\begin{array}{cccc}f\left(x\right)=-\frac{1}{x}& & & f\left(3\right)=-\frac{1}{3}\\ f\text{'}\left(x\right)=\frac{1}{x^2}& & & f\text{'}\left(3\right)=\frac{1}{9}\\ f\text{'}\text{'}\left(x\right)=-\frac{2}{x^3}& & & f\text{'}\text{'}\left(3\right)=-\frac{2}{27}\\ f\text{'}\text{'}\text{'}\left(x\right)=\frac{6}{x^4}& & & f\text{'}\text{'}\text{'}\left(3\right)=\frac{2}{27}\\ ⋮& & & ⋮\end{array}$

Substitute the values into the Taylor series and yields,

  $\begin{array}{c}-\frac{1}{x}=\frac{-\frac{1}{3}}{0!}{\left(x-\left(3\right)\right)}^0+\frac{\frac{1}{9}}{1!}{\left(x-\left(3\right)\right)}^1+\frac{-\frac{2}{27}}{2!}{\left(x-\left(3\right)\right)}^2+\frac{\frac{2}{27}}{3!}{\left(x-\left(3\right)\right)}^3+\mathrm{...}\\ =-\frac{1}{3}+\frac{1}{9}\left(x-3\right)-\frac{1}{27}{\left(x-3\right)}^2+\frac{1}{81}{\left(x-3\right)}^3\mathrm{...}\end{array}$

Thus, the first four non zero terms of the series is $-\frac{1}{3}+\frac{1}{9}\left(x-3\right)-\frac{1}{27}{\left(x-3\right)}^2+\frac{1}{81}{\left(x-3\right)}^3$ .