# series expansion of ln(x)

by jbowers9
Tags: expansion, series
 P: 70 How do you go about deriving the series expansion of ln(x)? 0 < x I got the representation at math.com but i'd still like to know how they got it. It's been a while since i did calc. iii. Thanks. John
 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.
 P: 70 I found the following link at math.com http://www.math.com/tables/expansion/log.htm I derived the first expression in the link and the one you gave the following way: 1/1-x = 1 + x + x^2 + x^3 + ... let x = -x and integrate 1/1+x = 1 - x + x^2 - x^3 int {1/1+x dx} = x - x^2/2 + x^3/3 - x^4/4 ... = ln(1+x) let x = x-1 ln(x) = (x-1) - (x-1)^2/2 + (x-1)^3/3 ... 0 < |x| < 1 The above expression is invalid at x=0. How are the three expressions following the first in the link derived?

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