Series expansion of ln(x)

by jbowers9
Tags: expansion, series
 HW Helper P: 1,391 You expand it as you would any other function, you just have to be careful which point you expand it about. Typically the function is expanded about the point x = 1. The series expansion is $$\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$$, so for $x_0 = 1$ and f(x) = lnx, the first few terms are $$\ln x_0 + \frac{1}{x_0}(x - x_0) - \frac{1}{2}\frac{1}{x_0^2}(x - x_0)^2 + ...$$ $$= (x - 1) - \frac{1}{2}(x - 1)^2 + ...$$ This can be made to look a little cleaner by letting u = x - 1.