Register to reply 
Series expansion of ln(x) 
Share this thread: 
#1
Dec807, 11:15 AM

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 


#2
Dec807, 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. 


#3
Dec907, 12:21 PM

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: 1/1x = 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 = x1 ln(x) = (x1)  (x1)^2/2 + (x1)^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? 


Register to reply 
Related Discussions  
The Taylor series expansion for sin about z_0 = (pi/2)  Calculus & Beyond Homework  1  
Series expansion  Calculus & Beyond Homework  3  
Series expansion..  Calculus  4  
Taylor Series Expansion  Calculus & Beyond Homework  2  
Power series expansion  Calculus  2 