Chapter 10.2, Problem 12E

To determine

To find: First three nonzero term, general term and interval of convergence of the given function.

Answer to Problem 12E

The first three nonzero terms are, $T_0=1,T_1=-2x\text{and}{T}_2=\frac{4x^2}{2!}$

The general term is, $T_n={\left(-1\right)}^n\frac{2^n{x}^n}{n!}$

And the series will converge for all real values of x .

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=e^{-2x}$

Calculation:

As, the Maclaurin series of $e^x$ is

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

So for Maclaurin series of $f\left(x\right)=e^{-2x}$ replace x by - 2x

  $\begin{array}{l}{e}^{-2x}=1+\left(-2x\right)+\frac{{\left(-2x\right)}^2}{2!}+\mathrm{...}+\frac{{\left(-2x\right)}^n}{n!}+\mathrm{..}\\ =1-2x+\frac{4x^2}{2!}-\frac{8x^3}{3!}+\mathrm{...}+{\left(-1\right)}^n\frac{2^n{x}^n}{n!}+\mathrm{...}\end{array}$

So, the first three nonzero terms are,

  $T_0=1,T_1=-2x\text{and}{T}_2=\frac{4x^2}{2!}$

And the general term of the series is

  $T_n={\left(-1\right)}^n\frac{2^n{x}^n}{n!}$

Now, for determining the interval of convergence use ration test. So,

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

As it is always less than 1.

Hence, the series will converge for all real values of x .