Answer: Power series from the geometric series Use the

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

ISBN: 9780321570567 2

Solution for problem 22RE Chapter 9

Calculus: Early Transcendentals | 1st Edition

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Problem 22RE

Power series from the geometric series Use the geometric I series ,for |x|<1 to determine the Maclanrin series and the interval of convergence for the following functions.

Step-by-Step Solution:

Solution 22RE

Step 1:

In this problem we have to use the geometric series $\sum\limits_{k\ =\ 0}^{\infty{}}{x}^{k}=\frac{1}{1\ -\ x}$ for |x|<1, to determine the Maclaurin series of $f(x)=\frac{1}{1\ +\ {x}^{3}}$ and also we have to find the interval of convergence

We have $f(x)=\frac{1}{1\ +\ {x}^{3}}$

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

Using geometric series $\sum\limits_{k\ =\ 0}^{\infty{}}{x}^{k}=\frac{1}{1\ -\ x}$ for |x| < 1 we get,

$\frac{1}{1\ -\ (-{x}^{3})}=\sum\limits_{k\ =\ 0}^{\infty{}}(-{x}^{3}{)}^{k}=\sum\limits_{k\ =\ 0}^{\infty{}}(-1{)}^{k}({x}^{3k})$

Thus the Maclaurin series of $\frac{1}{1\ +\ {x}^{3}}$ is given by $\sum\limits_{k\ =\ 0}^{\infty{}}(-1{)}^{k}({x}^{3k})$

Notice that we replaced both the x in the geometric series with ${x}^{3}$ to get the required Maclaurin series and so we will do the same in the interval of convergence.