Q. 33 - Calculus

Question

Find Maclaurin series for the given pairs of functions, using these steps: (a) Use substitution in the appropriate Maclaurin series to find the Maclaurin series for the given function. (b) Use Theorem 8.11 and your answer from part (a) to find the Maclaurin series for the given function. (c) Find the Maclaurin series for the function in (b), using multiplication and substitution with the appropriate Maclaurin series. Compare your answers from (b) and (c).

(a) $e^{-x^2}$

(b) $x{e}^{-x^2}$

Step-by-Step Solution

Answer

The Maclaurin series for the given functions are, (a) $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k}\end{array}$ and (b) $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k+1}\end{array}$. 

Step 1. Given information.

Consider the given functions are (a) $e^{-x^2}$ and (b) $x{e}^{-x^2}$.

Step 2. Use Formula for Maclaurin series of e x .

The Maclaurin series for $e^x$ is $\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$.

Step 3. Find Maclaurin series for e - x 2 .

Substitute $-x^2$ for $x$ into the Maclaurin series of $e^x$.

$\begin{array}{rcl}{e}^{-x^2}& =& \sum _{k=0}^{\infty }\frac{1}{k!}{\left(-x^2\right)}^k\\ & =& \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k}\end{array}$

Step 4. Apply theorem.

Differentiation of a power series is used if $\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k$ is power series in $\left(x-x_0\right)$ that converges to a function on an interval I, then derivative of function can be written as,

$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left(\sum _{k=0}^{\infty }{a}_k{\left(x-x_0\right)}^k\right)\\ & =& \sum _{k=0}^{\infty }\frac{d}{dx}\left(a_k{\left(x-x_0\right)}^k\right)\\ & =& \sum _{k=0}^{\infty }k{a}_k{\left(x-x_0\right)}^{k-1}\end{array}$

  

Step 5. Find differentiation of power series

Apply theorem into the power series $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k}\end{array}$.

$\begin{array}{rcl}\frac{d}{dx}\left(\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k}\right)& =& \sum _{k=0}^{\infty }\frac{d}{dx}\left(\frac{{\left(-1\right)}^k}{k!}{x}^{2k}\right)\\ & =& \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}\frac{d}{dx}\left(x^{2k}\right)\\ & =& \sum _{k=0}^{\infty }\frac{2k{\left(-1\right)}^k}{k!}{x}^{2k-1}\\ & =& \sum _{k=1}^{\infty }\frac{2{\left(-1\right)}^k}{\left(k-1\right)!}{x}^{2k-1}\\ & =& \sum _{k=0}^{\infty }\frac{2{\left(-1\right)}^k}{k!}{x}^{2k+1}\end{array}$

Step 6. Find Maclaurin series for x e - x 2 .

It can be observed that $-\frac{1}{2}\frac{d}{dx}\left(e^{-x^2}\right)=x{e}^{-x^2}$.

The Maclaurin series for $\frac{d}{dx}\left(e^{-x^2}\right)$ is $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{2{\left(-1\right)}^k}{k!}{x}^{2k+1}\end{array}$.

Multiply $\frac{1}{2}$ into the Maclaurin series of $\frac{d}{dx}\left(e^{-x^2}\right)$ to find the Maclaurin series of $x{e}^{-x^2}$.

$\begin{array}{rcl}x{e}^{-x^2}& =& \frac{1}{2}\left(\sum _{k=0}^{\infty }\frac{2{\left(-1\right)}^k}{k!}{x}^{2k+1}\right)\\ & =& \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k+1}\end{array}$

 

The Maclaurin series obtained in part (b) is $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{2{\left(-1\right)}^k}{k!}{x}^{2k+1}\end{array}$ where as the Maclaurin series obtained in part (c) is $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{k!}{x}^{2k+1}\end{array}$.