Question

What is the power series representation of `ln((1+x)/(1-x))`?

Answer 1

`ln((1+x)/(1-x)) =2x^3/3+2x^5/5+2x^7/7 ... = 2sum_(n=1)^oox^(2n+1)/(2n+1)`

Explanation:

I would use the following

1. The log rule; `log(A/B) = logA-logB`
2. The known power series :
   `ln(1+x) = 1-x^2/2+x^3/3-x^4 ...=sum_(n=1)^oo(-1)^(n+1)x^n/n`

So:
`\ \ \ \ \ ln((1+x)/(1-x)) =ln(1+x)-ln(1-x)`
`:. ln((1+x)/(1-x)) ={1-x^2/2+x^3/3-x^4 + ...} - {1-(-x)^2/2+(-x)^3/3-(-x)^4 + ...}`
`:. ln((1+x)/(1-x)) ={1-x^2/2+x^3/3-x^4 + ...} - {1-x^2/2-x^3/3-x^4 + ...}`
`:. ln((1+x)/(1-x)) =1-x^2/2+x^3/3-x^4 + ... - 1+x^2/2+x^3/3+x^4 + ...`
`:. ln((1+x)/(1-x)) =2x^3/3+2x^5/5+2x^7/7 ... = 2sum_(n=1)^oox^(2n+1)/(2n+1)`