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Laurent series

  1. Jul 22, 2014 #1
    Hello.
    I need explanation about this Laurent series.

    The question is:
    Let {##z\inℂ|0<|z|##}, expand ##\frac{e^{z^2}}{z^3}## where the centre z=0 into Laurent series.

    And the solution is:
    $$\frac{e^{z^2}}{z^3}=\sum_{n=0}^{\infty}\frac{\frac{(z^2)^n}{n!}}{z^3}=\sum_{n=0}^{\infty}\frac{z^{2n-3}}{n!}$$

    I don't understand the solution because isn't the formula for Laurent series
    $$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$

    where
    $$a_n=\frac{1}{2\pi{i}}\oint_{c}^{}\frac{f(z^*)}{(z-z_0)^{n+1}}dz^*$$

    $$b_n=\frac{1}{2\pi{i}}\oint_{c}^{}(z-z_0)^{n-1}f(z^*)dz^*$$
     
  2. jcsd
  3. Jul 22, 2014 #2
    Why is this a laurent series? This looks like a maclaurin expansion of the exponential function.
     
  4. Jul 22, 2014 #3

    SammyS

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    Does a MacLaurin expansion have negative exponents?
     
  5. Jul 22, 2014 #4
    No it doesn't but that definitely is a power series of the exponential function. Is Laurent series the same as power series? I am confused.
     
  6. Jul 22, 2014 #5

    Ray Vickson

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    Just write out the first 3 or 4 terms of your series. Does that not look like a Laurent series to you?
     
  7. Jul 22, 2014 #6
    Aha, now I see it. So, can I conclude that a Taylor/maclaurin series is a Laurent series with only positive exponents?
     
  8. Jul 23, 2014 #7

    HallsofIvy

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    Yes, of course. But what does that have to do with this problem? This series has negative powers.
     
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