Q. 49 - Calculus

Question

In Exercises 41–50, find Maclaurin series for the given pairs of functions, using these steps:

(a) Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .

(b) Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative     $\mathrm{F}=\int \mathrm{f}$ that satisfies the specified initial condition

$\left(\mathrm{a}\right) \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^{2 }} \left(\mathrm{b}\right) \mathrm{F}\left(0\right)=1$

Step-by-Step Solution

Answer

Part (a) :

${\mathrm{x}}^2{\mathrm{e}}^{-3{\mathrm{x}}^2}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left(\mathrm{x}\right)}^{2\mathrm{k}+2}$

Part (b) :

$\mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{{\left(\mathrm{x}\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right)+1$

Part (a) Step 1. Given information

Let us consider the given function $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2 {\mathrm{e}}^{-3{\mathrm{x}}^{2 }}$

Part (a) Step 2. Maclaurin series for given function

$$\begin{gathered} \mathrm{The} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}} \mathrm{is}: \\ {\mathrm{e}}^{\mathrm{x}}=\sum _{\mathrm{k}=0}^{\infty }\frac{1}{\mathrm{k}!}{\mathrm{x}}^{\mathrm{k}} \\ \mathrm{So}, \mathrm{the} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} {\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{can} \mathrm{be} \mathrm{found} \mathrm{by} \mathrm{substituting} \mathrm{x} \mathrm{by} -3{\mathrm{x}}^2 \\ {\mathrm{e}}^{-3{\mathrm{x}}^2}= \sum _{\mathrm{k}=0}^{\infty }\frac{1}{\mathrm{k}!}{\left(-3{\mathrm{x}}^2\right)}^{\mathrm{k}} \\ = \sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left(\mathrm{x}\right)}^{2\mathrm{k}} \\ \mathrm{Now}, \mathrm{the} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} {\mathrm{x}}^2{\mathrm{e}}^{-3{\mathrm{x}}^2} \mathrm{is}: \\ {\mathrm{x}}^2{\mathrm{e}}^{-3{\mathrm{x}}^2}={\mathrm{x}}^2\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left(\mathrm{x}\right)}^{2\mathrm{k}} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left(\mathrm{x}\right)}^{2\mathrm{k}+2} \end{gathered}$$

Part (b) Step 1. Given information

Let us consider the given function $\mathrm{F}=\int \mathrm{f}$

Part (b) Step 2. Maclaurin series for the antiderivative

$$\begin{gathered} \mathrm{F}\left(\mathrm{x}\right) = \int \left(\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}{\left(\mathrm{x}\right)}^{2\mathrm{k}+2}\right)\mathrm{dx} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\int {\left(\mathrm{x}\right)}^{2\mathrm{k}+2} \mathrm{dx} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{{\left(\mathrm{x}\right)}^{2\mathrm{k}+2+1} }{2\mathrm{k}+2+1}\right)+\mathrm{C} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{{\left(\mathrm{x}\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right)+\mathrm{C} \\ \mathrm{Since} \mathrm{the} \mathrm{initial} \mathrm{condition} \mathrm{is} \mathrm{F}\left(0\right)=1; \\ \mathrm{This} \mathrm{implies} \mathrm{that}, \\ \mathrm{C}=1 \\ \mathrm{Therefore}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}} 3^{\mathrm{k}}}{\mathrm{k}!}\left(\frac{{\left(\mathrm{x}\right)}^{2\mathrm{k}+3} }{2\mathrm{k}+3}\right)+1 \end{gathered}$$