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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)

  • MATH -

    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 + ...

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