The function ln(x-1) can be expressed as a Taylor Series of the form sum of c(subn)(x-2)^n. Calculate the formlua for c(subn).
We have:
Log(1+x) = sum from n = 1 to infinity of (-1)^(n+1) x^n/n
We can write:
Log(x-1) = Log[1+(x-2)] =
sum from n = 1 to infinity of
(-1)^(n+1) (x-2)^n/n
So, cn = (-1)^(n+1)/n