Question

How do you find the maclaurin series expansion of `f(x) = 1 / (1-x)`?

Answer 1

The simplest way is to use the geometric series formula `a+ar+ar^2+ar^3+cdots=a/(1-r)` when `|r| < 1` to get `f(x)=1/(1-x)=1+x+x^2+x^3+x^4+cdots` for `|x| < 1`. You can also use the general formula for Maclaurin series expansions as shown below.

Explanation:

Given `f(x)=1/(1-x)=(1-x)^(-1)`, if we want to use the general formula, we must compute:

`f'(x)=-(1-x)^{-2} * (-1)=(1-x)^{-2}`

`f''(x)=-2(1-x)^{-3} * (-1) = 2(1-x)^{-3}`

`f'''(x)=-6(1-x)^{-4) * (-1)=6(1-x)^{-4}`, etc...

and `f(0)=1`, `f'(0)=1`, `f''(0)=2=2!`, `f'''(0)=6=3!`, etc...

Therefore

`f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+cdots`

`=1+x+x^2+x^3+x^4+cdots`

The Ratio Test can also be used to confirm this converges for `|x|<1`.