Q. 50 - 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.

$\left(\mathrm{a}\right) \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \left(\mathrm{b}\right) \mathrm{F}\left(0\right)=0$

Step-by-Step Solution

Answer

Part (a) :

${\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} {\mathrm{x}}^{6\mathrm{k}+7}$

Part (b) :

$\mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} \left(\frac{{\mathrm{x}}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right)$

Part (a) Step 1. Given information

Let us consider the given function $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right)$

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

$$\begin{gathered} \mathrm{The} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} \mathrm{g}\left(\mathrm{x}\right)={\tan }^{-1}\mathrm{x} \mathrm{is}: \\ {\tan }^{-1}\mathrm{x}=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\mathrm{x}}^{2\mathrm{k}+1} \\ \mathrm{So}, \mathrm{the} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \mathrm{can} \mathrm{be} \mathrm{found} \mathrm{by} \mathrm{substituting} \mathrm{x} \mathrm{by} 3{\mathrm{x}}^3 \\ {\tan }^{-1}\left(3{\mathrm{x}}^3\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3{\mathrm{x}}^3\right)}^{2\mathrm{k}+1} = \sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} {\mathrm{x}}^{6\mathrm{k}+3} \\ \mathrm{Now}, \mathrm{the} \mathrm{Maclaurin} \mathrm{series} \mathrm{for} {\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right) \mathrm{is}: \\ {\mathrm{x}}^4 {\tan }^{-1}\left(3{\mathrm{x}}^3\right)={\mathrm{x}}^4\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} {\mathrm{x}}^{6\mathrm{k}+3} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} {\mathrm{x}}^{6\mathrm{k}+7} \end{gathered}$$

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

$$\begin{gathered} \mathrm{F}\left(\mathrm{x}\right) = \int \left(\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} {\mathrm{x}}^{6\mathrm{k}+7}\right)\mathrm{dx} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} \int {\mathrm{x}}^{6\mathrm{k}+7} \mathrm{dx} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} \left(\frac{{\mathrm{x}}^{6\mathrm{k}+7+1}}{6\mathrm{k}+7+1}\right)+\mathrm{C} \\ =\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} \left(\frac{{\mathrm{x}}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right)+\mathrm{C} \\ \mathrm{Since} \mathrm{the} \mathrm{initial} \mathrm{condition} \mathrm{is} \mathrm{F}\left(0\right)=0; \\ \mathrm{This} \mathrm{implies} \mathrm{that}, \\ \mathrm{C}=0 \\ \mathrm{Therefore}, \\ \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\infty }\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\left(3\right)}^{2\mathrm{k}+1} \left(\frac{{\mathrm{x}}^{6\mathrm{k}+8}}{6\mathrm{k}+8}\right) \end{gathered}$$