Q. 44 - 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 $F=\int f$ that satisfies the specified initial condition

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

Step-by-Step Solution

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

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,

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

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 $f\left(x\right)$

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

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

This implies that:

 $C=\mathrm{\pi }$

Therefore,

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