Q. 40 - Calculus

Question

In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.

$x^2{e}^x$

Step-by-Step Solution

Answer

Ans: The Maclaurin series of the function  $f\left(x\right)=\sum _{k=2}^{\infty }\frac{2^{k-1}}{k!}{x}^k$

Step 1. Given information:

$f\left(x\right)=x^2{e}^x$

Step 2. Finding the general form Maclaurin series:

Since for any function $f$ with derivatives of all orders at the point $x_0=0$, then the Maclurin series is

$f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}}\left(0\right)}{3!}{x}^3+\frac{f^{\text{'}\text{'}\text{'}\text{'}}\left(0\right)}{4!}{x}^4+\dots$

Or, we can write the general form Maclurin series of the function $f$ is

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n$

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the function$f\left(x\right)=x^2{e}^x$

$\begin{array}{|cccc|}\hline n& f^n\left(x\right)& f^n\left(0\right)& \frac{f^n\left(0\right)}{n!}\\ 0& x^2{e}^x& 0& 0\\ 1& \left(x^2+2x\right)e^x& 0& 0\\ 2& \left(x^2+2x+2\right)e^x& 2& \frac{2}{2!}\\ 3& \left(x^2+4x+4\right)e^x& 4& \frac{4}{3!}\\ 4& \left(x^2+6x+8\right)e^x& 8& \frac{8}{4!}\\ ·& ·& ·& ·\\ k& \left[x^2+2(k-1)x+2^{k-1}\right]e^x& 2^{k-1}& \frac{2^{k-1}}{k!}\\ \hline\end{array}$

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the function $f\left(x\right)=x^2{e}^x$ is $0+0·x+\frac{2}{2!}{x}^2+\frac{4}{3!}{x}^3+\frac{8}{4!}{x}^4+\dots +\frac{2^{k-1}}{k!}·x^k$

Or, we can write as

$f\left(x\right)=\sum _{k=2}^{\infty }\frac{2^{k-1}}{k!}{x}^k$