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

Question

Ezra Adams · Accepted Answer

First, note that #\frac{1}{(1+x)^2}=(1+x)^(-2)=\frac{d}{dx}(-(1+x)^{-1})=\frac{d}{dx}(-\frac{1}{1-(-x)})#.   Now use the power series expansion #\frac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots#, which converges for #|x|<1#, multiply everything by #-1#, and replace all the "#x#'s" with "#-x#'s to get #-\frac{1}{1-(-x)}=-1+x-x^2+x^3-x^4+\cdots#, which converges for #|-x|<1 \Leftrightarrow |x|<1#. Finally, differentiate this term-by-term (which is justified in the interior of the interval of convergence) to get #\frac{1}{(1+x)^{2}}=\frac{d}{dx}(-1+x-x^2+x^3-x^4+\cdots)# #=1-2x+3x^{2}-4x^{3}+\cdots#. This also converges for #|x|<1#.

Scarlett Allison · Answer

undefined To find a power series representation for $ f(x) = \frac{1}{{(1 + x)}^2} $, you can start by differentiating the geometric series representation $ \frac{1}{1 - (-x)} = \sum_{n=0}^{\infty} (-x)^n $. Then, multiply the result by $ \frac{d}{dx}(1 + x) $ and simplify. This will give you the power series representation for $ f(x) $.

The geometric series representation is $ \frac{1}{{(1 + x)}^2} = \sum_{n=0}^{\infty} (-1)^n x^n $.