Chapter 8: Infinite Sequences and Series

Section 8.5: Taylor Series

Example 8.5.16

$$ Obtain the Maclaurin series for $\frac{1-x^2}{1\+x^2}$ from appropriate geometric series. $$ Solution $$ Write $\frac{1-x^2}{1\+x^2}$ as $\frac{1}{1\+x^2}-\frac{x^2}{1\+x^2}$ and note that, by the sum formula for the geometric series, $\frac{1}{1\+x^2}\=\sum _{n\=0}^{\infty }{\left(-x^2\right)}^n$. Thus,     $\frac{1}{1\+x^2}-\frac{x^2}{1\+x^2}\=\sum _{n\=0}^{\infty }{\left(-x^2\right)}^n-x^2 \sum _{n\=0}^{\infty }{\left(-x^2\right)}^n\=1-2x^2\+2x^4-2x^6\+2x^8-2 x^{10}\+\cdots$  $$$$

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