1. ## Power Series

Question:

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

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

My attempt:

we know that

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

Therefore,

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

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

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

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

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

Am I correct?

2. Originally Posted by calc101

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

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

$\displaystyle \frac{x^2}{1-x^2} = \sum x^{4n}$
Nope. $\displaystyle x^2 \cdot x^{2n} = x^{2n+2}$

3. Also note that $\displaystyle x^{2n}+x^{2n+2}=(1+x^2)x^{2n}$.

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

$\displaystyle \sum_{n=0}^{\infty}(1+x^2)x^{2n}$ ?

5. Originally Posted by calc101
Question:

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

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

My attempt:

we know that

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

Therefore,

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

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

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

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

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

Am I correct?
It might be a little easier to recognize that

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