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I Series expansions of log()

  1. Feb 18, 2017 #1

    see attached PdF file for my question concerning serie expansion of log(z) at infinity.

    Thank you
    Belgium 12

    Attached Files:

  2. jcsd
  3. Feb 18, 2017 #2


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    Staff Emeritus
    Science Advisor

    So the usual expansion for [itex]log(1+x)[/itex] when [itex]x[/itex] is small is given by:

    [itex]log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...[/itex]

    We also know that [itex]log(1+z) = log(z (1 + \frac{1}{z})) = log(z) + log(1 + \frac{1}{z})[/itex]

    So you can put the two together (replacing [itex]x[/itex] by [itex]\frac{1}{z}[/itex]):

    [itex]log(1+z) = log(z) + \frac{1}{z} - \frac{1}{2 z^2} + \frac{1}{3 z^3} ...[/itex]​
  4. Feb 18, 2017 #3


    Staff: Mentor

    Instead of posting an image of barely legible writing, please learn to use LaTeX. Everything you wrote can be entered directly into the input pane here. See our tutorial on LaTeX here: https://www.physicsforums.com/help/latexhelp/

    Here are some examples of expressions you wrote and how they appear in LaTeX:
    ##\log(\frac{z - 1}{z})##
    Script for the above: ##\log(\frac{z - 1}{z})##

    $$\sum_{q \ge 1}^{\infty} \frac{(-1)^q}{qz^q}$$
    Script for the above: $$\sum_{q \ge 1}^{\infty} \frac{(-1)^q}{qz^q}$$

    (Inline version of the above would be ##\sum_{q \ge 1}^{\infty} \frac{(-1)^q}{qz^q} ##.)

    BTW, in the summation that appears twice, it's very difficult to tell that the denominator is ##qz^q##. What you wrote--twice--looks like ##qzq##. The only clue that the exponent is q is that this letter appears slightly raised.
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