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Laurent series expansion of ℘(z).

  1. May 20, 2017 #1
    1. The problem statement, all variables and given/known data
    2nd equality.png

    2. Relevant equations


    3. The attempt at a solution

    Hi,

    I am trying to understand the 2nd equality .

    I thought perhaps it is an expansion of ##(1-\frac{z}{w})^{-2}## (and then the ##1## cancels with the ##1## in ##( (1-\frac{z}{w})^{-2}) -1 ) ##) in the form ##(1-x)^{-2}##, however this doesn't give me the right answer, can someone please help explain how they got from line ##1## to ##2## ?

    Many thanks in advance.
     
  2. jcsd
  3. May 20, 2017 #2

    Ray Vickson

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    I get the right answer when I expand ##(1 - (z/w))^{-2}## in a Maclauren series about ##z = 0##, or use the binomial expansion involving the "negative binomial" ##C^{-2}_n##. It sounds like that is what you did too, but got the wrong answer for reasons you do not explain, using work that you do not show.
     
  4. May 20, 2017 #3
    oh okay, thank you for your reply.
    So for ##(1+x)^{-2}## where ##x= \frac{z}{w}## is of the form ##1+nx+\frac{n(n+1)x^2}{1.2}+\frac{n(n+1)(n+2)x^3}{1.2.3}+... ##I don't know how to express this as a summation over ##n## unless it simplifies somehow for the case here ##n=2##


    you know me well...
     
  5. May 21, 2017 #4
    What does called ?
     
  6. May 21, 2017 #5

    Ray Vickson

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    Your expansion is wrong. It should be
    $$(1-t)^{-2} = 1 + 2 t + 3t^2 + 4 t^3 + \cdots = \sum_{n=0}^{\infty} (n+1)t^n.$$
    Note that ##(1-t)^{-2} = \sum_n (-1)^n C(-2,n) t^n##, and ##C(-2,n) = (-1)^n (n+1).##

    Of course, you can also get it by looking at
    $$(1-t)^{-2} = \frac{d}{dt} (1-t)^{-1} = \frac{d}{dt} (1 + t + t^2 + \cdots)$$
     
    Last edited: May 21, 2017
  7. May 21, 2017 #6
    Apologies it should have read ##(1-x)^{-2}## , wrong sign, but plugging in ##n=2## gives cancellations to give the same as your expansion there
     
  8. Jun 22, 2017 #7
    Can I just ask, for general ##n## there is no simplified general expression as a summation ?

    For example considering ##n=3## the most I can simplify this to is:


    ##1+3x+\frac{(3)(4)x^2}{(2)(1)}+\frac{(3)(4)(5)x^3}{(3)(2)(1)}+\frac{(3)(4)(5)(6)x^4}{(4)(3)(2)(1)}+\frac{(3)(4)(5)(6)(7)x^5}{(5)(4)(3)(2)(1)}+...##
    ##=1+3x+6x^2+\frac{(4)(5)x^3}{(2)(1)}+\frac{(5)(6)x^4}{(2)(1)}+\frac{(6)(7)x^5}{(2)(1)}##
    ##=1+3x+6x^2+\sum\limits_{k=3}^{k=\infty} \frac{(k+1)(k+1)x^k}{2} ##

    so for a general expression the ##x^{n-1}## terms will not be included in some general summation? (for ##n## integer)


    Is there a more simplified way to write this or not?

    Many thanks.
     
    Last edited: Jun 22, 2017
  9. Jun 22, 2017 #8

    Dick

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