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 $F=\int f$that satisfies the specified initial condition

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

Short answer

Part (a) ${\tan }^{-1}\left(x^2\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{4k+2}$

Part (b)$F\left(x\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left(\frac{x^{4k+3}}{4k+3}\right)+\mathrm{\pi }$

Step-by-step solution

Part (a) Step 1. Given information

Let us consider the given function$f\left(x\right)={\tan }^{-1}\left(x^2\right)$

Part (a) Step 2. Use substitution and/or multiplication and the appropriate Maclaurin series to find the Maclaurin series for the given function f .

The maclaurin series for $g\left(x\right)={\tan }^{-1}\left(x\right)$is :

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

So,the maclaurin series for $f\left(x\right)={\tan }^{-1}\left(x^2\right)$can be founded by substituting $x$by$x^2$

Thus,

role="math" localid="1650696937207" $$\begin{gathered} {\tan }^{-1}\left(x^2\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{\left(x^2\right)}^{2k+1} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{4k+2} \end{gathered}$$

Part (b) Step 1. Given information

Let us consider the given function$F=\int f\left(x\right)$

Part (b) Step 2.  Use Theorem 8.12 and your answer from part (a) to find the Maclaurin series for the antiderivative F=∫f that satisfies the specified initial condition 

Put the value of function $f\left(x\right)$

localid="1650696832847" $$\begin{gathered} F\left(x\right)=\int \sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{4k+2}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\int x^{4k+2}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left(\frac{x^{4k+2+1}}{4k+2+1}\right)+C \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left(\frac{x^{4k+3}}{4k+3}\right)+C \end{gathered}$$

Since,the initial condition is$F\left(0\right)=\mathrm{\pi }$

This implies that:

$C=\mathrm{\pi }$

Therefore,

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