series expansion of ln(x)

by jbowers9
Tags: expansion, series
jbowers9 is offline
Dec8-07, 11:15 AM
P: 70
How do you go about deriving the series expansion of ln(x)?
0 < x
I got the representation at but i'd still like to know how they got it. It's been a while since i did calc. iii. Thanks.
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Mute is offline
Dec8-07, 11:23 AM
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 [tex]\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x - x_0)^n[/tex], so for [itex]x_0 = 1[/itex] and f(x) = lnx, the first few terms are

[tex]\ln x_0 + \frac{1}{x_0}(x - x_0) - \frac{1}{2}\frac{1}{x_0^2}(x - x_0)^2 + ...[/tex]

[tex] = (x - 1) - \frac{1}{2}(x - 1)^2 + ...[/tex]

This can be made to look a little cleaner by letting u = x - 1.
jbowers9 is offline
Dec9-07, 12:21 PM
P: 70
I found the following link at

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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