Question

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

Answer 1

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`.