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Maclaurin and Laurent series.

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    I know the sum of the Laurent series (around x=2) is equal to
    But I can't find what the series is from this information alone.

    2. Relevant equations
    In the textbook, you have (for -1 < x < 1):
    [tex]\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n[/tex]
    and for |x|>1 I know (but have no idea how to deduce) that
    [tex]\frac{1}{1+x} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{x^n}[/tex]

    I just don't know how I can use this information to find the sum for 1/(x+3).

    3. The attempt at a solution
    I am sorry, but I don't want to further destroy my confidence by reliving my pathetic attempts to finding the solution. :D
  2. jcsd
  3. Oct 22, 2008 #2
    Yes you can, you just use the geometric series

    [tex] \frac{1}{x+3}=\frac{1}{3-(-x)} =\frac{1}{3(1- \frac{-x}{3})}[/tex]

    Now use u = -x/3 and employ the geometric series for |x|<1.
  4. Oct 22, 2008 #3
    Thank you! A clever little move there.

    I'll be sure to include you in my 'Thank You' speech when I accept my Fields medal. ;)
  5. Oct 22, 2008 #4


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    Homework Helper

    If you want the Laurant series around x=2, you want a series of powers of (x-2). You might want to rearrange the form a bit before you do the geometric series trick.
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