posted by A. .
Expand f(x)=ln (1+x/1-x) in a Taylor Series about x=0. You must express your answer using summation notation.
---This is what I tried to do---
So I was thinking of taking the derivative of ln(1+x/1-x) and get 2/x^2-1
And then use the identity 1/1-x= sum x^k
So then it will be Sum (-2^k) (x^2k)
you have the right idea.
First, df/dx = -2/(1-x^2), but I'd do it like this:
f(x) = ln(1+x) - ln(1-x)
ln(1+x) = 0+x-x^2/2+x^3/3-x^4/4+x^5/5+...
ln(1-x) = 0-x-x^2/2-x^3/3-x^4/4-x^5/5-...
subtract to get
0 + 2x + 2x^3/3 + 2x^5/5 + ...