Chapter 8.7, Problem 36E

To determine

To find the Maclaurin series of $f(x)=e^{-x^2}+\cos x$ and then sketch the graph of the taylor’s polynomial and the function.

Answer to Problem 36E

The Maclaurin expansion of the function is

  $e^{-x^2}+\cos x=2-\frac{3}{2}{x}^2+\frac{13}{24}{x}^4-\frac{121}{720}{x}^6+\mathrm{.....}$

Explanation of Solution

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

Formula used: The Maclaurin expansion of the function:

  $\begin{array}{l}\cos x={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\left(\frac{x^{2n}}{(2n)!}\right)}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\mathrm{........}\\ e^x={\displaystyle \sum _{n=0}^{\infty }\left(\frac{x^n}{n!}\right)}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\mathrm{........}\end{array}$

Calculation : The Maclaurin expansion of the function becomes as:

  $\begin{array}{l}{e}^{-x^2}+\cos x=\left(1+(-x^2)+\frac{{(-x^2)}^2}{2!}+\frac{{(-x^2)}^3}{3!}+\frac{{(-x^2)}^4}{4!}+\mathrm{........}\right)+\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\mathrm{........}\right)\\ =\left(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}+\mathrm{........}\right)+\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\mathrm{........}\right)\\ =2-\frac{3}{2}{x}^2+\frac{13}{24}{x}^4-\frac{121}{720}{x}^6+\mathrm{.....}\end{array}$

Graph:Using graphingutility to sketch the graph.

Enter the expression: $f(x)=e^{-x^2}+\cos x$ and $g(x)=2-\frac{3}{2}{x}^2+\frac{13}{24}{x}^4$ in the input box of the graphing utility. The sketch of the graph is:

  

From the above graph, it is concluded that both the function and the Taylor’s polynomial having same origin. Also, the graph of both the functions are same upto certain value.

Hence, the Maclaurin expansion of the function:

  $e^{-x^2}+\cos x=2-\frac{3}{2}{x}^2+\frac{13}{24}{x}^4-\frac{121}{720}{x}^6+\mathrm{.....}$

Also, the radius of convergence is $R=\infty$