Q.25 - Calculus

Question

In Exercises 25-30, find Maclaurin series for the given pairs of functions, using these steps:

(a) Use substitution and/or multiplication and the Maclaurin series for $\frac{1}{1-x}$ to find the Maclaurin series for the given function. Also, provide the interval of convergence for the series you found.

(b) Use Theorem 8.11 and your answer from part (a) to find the Maclaurin series for the given function. Also, provide the interval of convergence for the series you found.

25. (a) $\frac{1}{1+x^2}$  (b) $\frac{x}{{\left(1+x^2\right)}^2}$

Step-by-Step Solution

Answer

(a)The Maclaurin series of $\frac{1}{1+x^2}$ is $\sum _{k=0}^{\infty }(-1{)}^k{x}^{2k}$

The interval of convergence is  $[-1,1]$

(b)The Maclaurin series of  $\frac{x}{{\left(1+x^2\right)}^2}$ is $\sum _{k=0}^{\infty }(-1{)}^k·2k\left(x^{2k-1}\right)$

Part (a) Step 1: Given Information

Given equation : $\frac{1}{1+x^2}$ 

Part (a): Step 2 : Simplification

To find the Maclaurin series for the function$f\left(x\right)=\frac{1}{1+x^2}$, then the $x$ will be substituted by $-x^2$  in the Maclaurin series of $\frac{1}{1-x}$.

So,

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

Considering $b_k=(-1{)}^k{x}^{2k}$ ,to find the interval convergence,  the ratio test for the absolute. convergence is to be used. So, $b_{k+1}=(-1{)}^{k+1}{x}^{2(k+1)}$.

In this way,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{(-1{)}^{k+1}{x}^{2k+2}}{(-1{)}^k{x}^{2k}}\right| \\ =\underset{x\to \infty }{\lim }\left|-x^2\right| \end{gathered}$$

when $\left|-x^2\right|<1$.

The series will converge based on the ratio test of absolute convergence.

As a result, the convergence interval will contain the value $(-1,1)$. We examine the behavior of the series at the interval's endpoints because it is a finite interval.

The series will be $\sum _{k=0}^{\infty }(-1{)}^k(-1{)}^{2k}=\sum _{k=0}^{\infty }(-1{)}^k$ when $x=-1$.

The series will be  $\sum _{k=0}^{\infty }(-1{)}^k(1{)}^{2k}=\sum _{k=0}^{\infty }(-1{)}^k$ when $x=1$

Both the series will converge.

As a result, the series $\frac{1}{1+x^2}$ has a convergence interval of $\mathbf{[}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{]}$.

Part (b): Step 1 : Given information

Given equation : $\frac{x}{{\left(1+x^2\right)}^2}$

Part (b): Step 2 : Simplification

The first function will be derivated. So,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{1+x^2}\right) \\ =\frac{\left(1+x^2\right)0-1\left(2x\right)}{{\left(1+x^2\right)}^2} \\ =\frac{-2x}{{\left(1+x^2\right)}^2} \end{gathered}$$

the Maclaurin series for $\frac{-2x}{{\left(1+x^2\right)}^2}$ is:

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

This means that the Maclaurin series for $\frac{x}{{\left(1+x^2\right)}^2}$ is

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