# How do you use differentiation to find a power series representation for f(x)=1/(1+x)^2?

##### 1 Answer
Apr 3, 2015

First, note that $\setminus \frac{1}{{\left(1 + x\right)}^{2}} = {\left(1 + x\right)}^{- 2} = \setminus \frac{d}{\mathrm{dx}} \left(- {\left(1 + x\right)}^{- 1}\right) = \setminus \frac{d}{\mathrm{dx}} \left(- \setminus \frac{1}{1 - \left(- x\right)}\right)$.

Now use the power series expansion $\setminus \frac{1}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \setminus \cdots$, which converges for $| x | < 1$, multiply everything by $- 1$, and replace all the "$x$'s" with "$- x$'s to get

$- \setminus \frac{1}{1 - \left(- x\right)} = - 1 + x - {x}^{2} + {x}^{3} - {x}^{4} + \setminus \cdots$, which converges for $| - x | < 1 \setminus \Leftrightarrow | x | < 1$.

Finally, differentiate this term-by-term (which is justified in the interior of the interval of convergence) to get

$\setminus \frac{1}{{\left(1 + x\right)}^{2}} = \setminus \frac{d}{\mathrm{dx}} \left(- 1 + x - {x}^{2} + {x}^{3} - {x}^{4} + \setminus \cdots\right)$

$= 1 - 2 x + 3 {x}^{2} - 4 {x}^{3} + \setminus \cdots$.

This also converges for $| x | < 1$.