Posted by A. on .
Expand f(x)=ln (1+x/1x) 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/1x) and get 2/x^21
And then use the identity 1/1x= sum x^k
So then it will be Sum (2^k) (x^2k)

MATH 
Steve,
you have the right idea.
First, df/dx = 2/(1x^2), but I'd do it like this:
f(x) = ln(1+x)  ln(1x)
ln(1+x) = 0+xx^2/2+x^3/3x^4/4+x^5/5+...
ln(1x) = 0xx^2/2x^3/3x^4/4x^5/5...
subtract to get
0 + 2x + 2x^3/3 + 2x^5/5 + ...