Chapter 8.7, Problem 28E

To determine

To obtain the Maclaurin series for the function: $f(x)=e^x+2e^{-x}$ .

Answer to Problem 28E

The Maclaurinseries expansion of $e^x+2e^{-x}=3-\frac{x}{1!}+3\cdot \frac{x^2}{2!}-\frac{x^3}{3!}+\mathrm{........}$

Explanation of Solution

Given information: $f(x)=e^x+2e^{-x}$

Formula used: The Maclaurin expansion of the function:

  $e^x={\displaystyle \sum _{n=0}^{\infty }\left(\frac{x^n}{n!}\right)}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\mathrm{........}$

Calculation : Replace $x\text{to}-x$ in the second term, the above expression becomes as:

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

Hence, the Maclaurin expansion of the function:

  $e^x+2e^{-x}=3-\frac{x}{1!}+3\cdot \frac{x^2}{2!}-\frac{x^3}{3!}+\mathrm{........}$