Chapter 10.2, Problem 8E

To determine

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

Answer to Problem 8E

The first three nonzero terms are, $T_0=7x,T_1=7x^2\text{and}{T}_2=\frac{7x^3}{2!}$

The general term is, $T_n=\frac{7x^{n+1}}{n!}$

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

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=7x{e}^x$

Calculation:

As Maclaurin series of $f\left(x\right)=e^x$ is

  $f\left(x\right)=e^x=1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}+\mathrm{..}$

So for Maclaurin series of $f\left(x\right)=7x{e}^x$ multiply by 7x .

So, the series is

  $\begin{array}{l}f\left(x\right)=7x\left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}+\mathrm{..}\right)\\ =7x+7x^2+\frac{7x^3}{2!}+\mathrm{...}+\frac{7x^{n+1}}{n!}+\mathrm{..}\end{array}$

So, the first three nonzero terms are,

  $T_0=7x,T_1=7x^2\text{and}{T}_2=\frac{7x^3}{2!}$

And the general term of the series is

  $T_n=\frac{7x^{n+1}}{n!}$

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{7x^{n+1}}{n!}\times \frac{\left(n-1\right)!}{7x^n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{x}{n}\right|\\ =\left|x\right|\underset{n\to \infty }{\lim }\frac{1}{n}\\ =0\end{array}$

As it is very less than 1.

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