# Power Series

• Jul 19th 2009, 03:44 PM
calc101
Power Series
Question:

Find a power series representation for the function and determine the interval of convergence:

$f(x) = \frac {1+x^2}{1-x^2}$

My attempt:

we know that

$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

Therefore,

$\frac{1}{1-x^2} = \sum x^{2n}$

Now, if we multiply by $x^2$ we get:

$\frac{x^2}{1-x^2} = \sum x^{4n}$

Then we add: $\frac{1}{1-x^2}$

So, $\sum_{n=0}^{\infty} x^{2n} + x^{4n}$

Am I correct?
• Jul 19th 2009, 03:53 PM
o_O
Quote:

Originally Posted by calc101

$\frac{1}{1-x^2} = \sum x^{2n}$

Now, if we multiply by $x^2$ we get:

$\frac{x^2}{1-x^2} = \sum x^{4n}$

Nope. $x^2 \cdot x^{2n} = x^{2n+2}$
• Jul 19th 2009, 04:00 PM
Plato
Also note that $x^{2n}+x^{2n+2}=(1+x^2)x^{2n}$.
• Jul 19th 2009, 04:25 PM
calc101
Quote:

Originally Posted by Plato
Also note that $x^{2n}+x^{2n+2}=(1+x^2)x^{2n}$.

$\sum_{n=0}^{\infty}(1+x^2)x^{2n}$ ?
• Jul 19th 2009, 04:30 PM
Jester
Quote:

Originally Posted by calc101
Question:

Find a power series representation for the function and determine the interval of convergence:

$f(x) = \frac {1+x^2}{1-x^2}$

My attempt:

we know that

$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

Therefore,

$\frac{1}{1-x^2} = \sum x^{2n}$ (**)

Now, if we multiply by $x^2$ we get:

$\frac{x^2}{1-x^2} = \sum x^{4n}$

Then we add: $\frac{1}{1-x^2}$

So, $\sum_{n=0}^{\infty} x^{2n} + x^{4n}$

Am I correct?

It might be a little easier to recognize that

$
\frac{1+x^2}{1-x^2} = \frac{2}{1-x^2} - 1.
$
then use (**) above