Chapter 9.2, Problem 12E

To determine

To find: To find the general term of the Maclaurin series.

Answer to Problem 12E

  $T_n={(-1)}^n\frac{2^n{x}^n}{n!}$ and series always converges for all real value of x.

Explanation of Solution

Given:

The given series $f(x)=e^{-2x}$ .

Formula used:

  $e^x=1+x+\frac{x^2}{2!}+\mathrm{......}+\frac{x^n}{n!}+\mathrm{...}$

Calculation:

Given the given function is,

  $f(x)=e^{-2x}$

  $\begin{array}{l}{e}^{2x}=1+2x+\frac{2x^2}{2!}+\mathrm{......}+\frac{2x^n}{n!}+\mathrm{...}\\ $ e^{-2x}=1-2x+\frac{4x^2}{2!}+\mathrm{......}+{(-1)}^n\frac{2^n{x}^n}{n!}+\mathrm{...}$\end{array}$

  $T_n={(-1)}^n\frac{2^n{x}^n}{n!}$

The first three non- zero terms of the series are $T_1=1,T_2=-2x,T_3=\frac{4x^3}{2!}$

The interval of convergence use the ratio test.

  $\begin{array}{l}$ \underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }{(-1)}^n\frac{2^n{x}^n}{n!}\frac{n-1!}{2^{n-1}{x}^{n-1}{(-1)}^{n-1}}$\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left(\frac{{(2x)}^2}{n}\right)\\ \underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=4x^2\underset{n\to \infty }{\lim }\left(\frac{1}{n}\right)\\ \underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=4x^2.0\\ \underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=0\end{array}$

So the ratio tells us that if $L<1$ the series will converge if $L>1$ the series will diverge and if $L=1$ . We don’t know that what will happen so we have $\left|x\right|<1$ , $-1<x<1$ . The way to determine the convergence at end points is to simply plug them into the power series and see if the convergence or divergence using any test necessary.

The series converges for all real value of $x$ .