Question

How do you derive the maclaurin series for `1/(1-x^2)`?

Answer 1

`sum_(n=0)^oox^(2n)=1+x^2+x^4+x^6...`

Convergent when `|x|<1`

Explanation:

If we look at the geometric series:
`sum_(n=0)^oox^n=1+x+x^2+x^3...=1/(1-x)`

This is basically the same as our expression, but we have an `x^2` instead. So to determine the series for our expression, we just replace `x` with `x^2`:
`1/(1-x^2)=sum_(n=0)^oo(x^2)^n=sum_(n=0)^oox^(2n)=1+x^2+x^4+x^6...`

So, our Maclaurin series is `sum_(n=0)^oox^(2n)=1+x^2+x^4+x^6...`

We know that the geometric series is convergent for `|x|<1`, so to determine when this series is convergent, we solve for when `|x^2|<1`:

`|x^2|<1`

`x^2` is always positive, so we don't need the absolute value:
`x^2<1`

`sqrt(x^2)< sqrt1`

`|x|<1`

As luck would have it, this series has the same radius of convergence as the geometric series.