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Series expansion of ln(x)

  1. Dec 8, 2007 #1
    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. jcsd
  3. Dec 8, 2007 #2

    Mute

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    Homework Helper

    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.
     
  4. Dec 9, 2007 #3
    ln(x) series

    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?
     
    Last edited: Dec 9, 2007
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