Q.26 - Calculus

Question

Find Maclaurin series for the given pairs of functions, using these steps:

(a) Use substitution and/or multiplication and the Maclaurin series for  to find the Maclaurin series for the given function. Also, provide the interval of convergence for the series you found.

(b) Use Theorem  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.

(a) $\frac{1}{1-x^3}$

(b) $\frac{x^2}{{\left(1-x^3\right)}^2}$

Step-by-Step Solution

Answer

Part (a)The Maclaurin series of  $\frac{1}{1-x^3}$ is $\sum _{i=0}^{\infty }{x}^3$

The interval of convergence is  $\left(-1,1\right)$

Part (b)The Maclaurin series of  $\frac{x^2}{{\left(1-x^3\right)}^2}$ is $\sum _{k=0}^{\infty }k\left(x^{3k-1}\right)$

Part (a) Step 1:The given Description

The Maclaurin series for the function  is .

The function to find the Maclaurin series is $\frac{1}{1-x^3}$ 

Part (a) Step:2 Calculation

To find the Maclaurin series for the function  $\frac{1}{1-x^3}$ , then the $x^3$  will be substituted bin the Maclaurin series of $\frac{1}{1-x}$ .

$$\begin{gathered} \frac{1}{1-x^3}=\sum _{k=0}^{\infty }{\left(x^3\right)}^k \\ \frac{1}{1-x^3}=\sum _{i=0}^{\infty }{x}^3 \end{gathered}$$

Considering $b_k=x^{3k}$ ,to find the interval convergence,  the ratio test for the absolute convergence is to be used.So $b_{k+1}=x^{3(k+1)}$,

 .

In this way,

$$\begin{gathered} \underset{x\to x}{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{x^{3k+3}}{x^{3k}}\right| \\ =\underset{k\to \infty }{\lim }\left|x^3\right| \end{gathered}$$

when $-1<x<1$The series will converge based on the ratio test of absolute convergence.

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

The series will be $\frac{1}{1-x^3}=\sum _{i=0}^{\infty }(-1{)}^{3k}$ when  $x=-1$.

The series will converge. 

The series will be  $\frac{1}{1-x^3}=\sum _{k=0}^{\infty }(1{)}^{3k}$ when $x=1$

The series will diverge.

As a result, the series $\frac{1}{1-x^3}$ has a convergence interval of $\left(-1,1\right)$.

Part (b) Step 3:The Maclaurin series is 3 x 2 1 - x 3 2

The functions to find the Maclaurin series is 

Part (b) Step 4:Calculation

The first function will be derivated .So, 

The Maclaurin series for  is:

This means that the Maclaurin series for  is