Chapter 8.7, Problem 68E

a.

To determine

To prove: The Function

  $f(x)=\left\{\begin{array}{l}{e}^{\frac{-1}{x^2}}\text{ if }x\ne 0\\ 0\text{ if x=0}\end{array}\right\}$

Is not equal to its Maclaurin Series

Explanation of Solution

Given information:

The Function

  $f(x)=\left\{\begin{array}{l}{e}^{\frac{-1}{x^2}}\text{ if }x\ne 0\\ 0\text{ if x=0}\end{array}\right\}$

Formula used:

Maclaurin Series

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

Maclaurin series for $e^x$

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

Proof:

Use, Maclaurin series for $e^x$

  $e^{\frac{-1}{x^2}}={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\frac{-1}{x^2}\right)}^n}{n!}}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!x^{2n}}}$ ???? ??&#??2260;0

However, looking at the definition of a Maclaurin series

  $\begin{array}{l}f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n(0)x^n}{n!}}\\ f(x)=f(0)+f^{\text{'}}(0)x+\frac{f^{\text{'}\text{'}}(0)x^2}{2!}+\mathrm{..}\\ f(0)=0\end{array}$

Every derivative of $f(x)$ at $x=0$ is $0$ , and all the McLaurin terms are $0$ , making the whole series $0$

(b)

To determine

To graph: the function $f(x)=\left\{\begin{array}{l}{e}^{\frac{-1}{x^2}}\text{ if }x\ne 0\\ 0\text{ if }x=0\end{array}\right\}$

Explanation of Solution

Given information: The Function

  $f(x)=\left\{\begin{array}{l}{e}^{\frac{-1}{x^2}}\text{ if }x\ne 0\\ 0\text{ if }x=0\end{array}\right\}$

Graph:

Use graphing utility to graph the function,