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Laurent series of e^(1/z)

  1. May 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Cassify the singularities of [itex]e^\frac{1}{z} [/itex] and find the Laurent series

    2. Relevant equations
    [itex]e^\frac{1}{x} [/itex]=[itex]\sum \frac{(\frac{1}{x})^n}{n!}[/itex]

    3. The attempt at a solution
    Theres a singularity at z=0, but I need to find the order of the pole

    So using the general expression for the expansion of an exponential:
    [itex]e^\frac{1}{z} [/itex]=[itex]\sum \frac{(\frac{1}{z})^n}{n!}[/itex] but this leads to a 1 as the first term, which is obviously not consistent.

    I also tried considering re-defining a new variable for [itex]\frac{1}{z} [/itex], but I'm not really sure how to proceed from here

    Many thanks :)
     
  2. jcsd
  3. May 20, 2016 #2

    stevendaryl

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    If the order of the pole for [itex]f(z)[/itex] at [itex]z=0[/itex] is [itex]n[/itex], then that means that [itex]f(z) \propto \frac{1}{z^n}[/itex] near [itex]z=0[/itex], which in turn means that [itex]z^n f(n)[/itex] remains finite as [itex]z \rightarrow 0[/itex]. So in the case of [itex]e^{\frac{1}{z}}[/itex], do you think there is some number [itex]n[/itex] that would work?
     
  4. May 21, 2016 #3
    Im sorry I don't see how you concluded that [itex]f(z) \propto \frac{1}{z^n}[/itex] near [itex]z=0[/itex], since would the expansion for [itex]f(z) =1-\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+...[/itex]

    Thank you :)
     
  5. May 21, 2016 #4

    BvU

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    Steven didn't conclude it. He wrote If ... then ... !
     
  6. May 21, 2016 #5

    stevendaryl

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    Let me quote from Wikipedia (https://en.wikipedia.org/wiki/Singularity_(mathematics)#Complex_analysis)

    • The point a is a removable singularity of f if there exists a holomorphic function g defined on all of U such that f(z) = g(z) for all z in U \ {a}. The function g is a continuous replacement for the function f.
    • The point a is a pole or non-essential singularity of f if there exists a holomorphic function g defined on U with g(a) nonzero, and a natural number n such that f(z) = g(z) / (za)n for all z in U \ {a}. The number n here is called the order of the pole. The derivative at a non-essential singularity itself has a non-essential singularity, with n increased by 1 (except if n is 0 so that the singularities are removable).
    • The point a is an essential singularity of f if it is neither a removable singularity nor a pole. The point a is an essential singularity if and only if the Laurent series has infinitely many powers of negative degree.
     
  7. May 21, 2016 #6
    Ah okay, so in this case we have an essential singularity?

    Would I be okay in thinking that the Laurent series is [itex]f(z) =1-\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+...[/itex] ? Or have I oversimplified the problem

    Thank you :)
     
  8. May 21, 2016 #7
    My mistake, I misinterpreted his response. Thank you for clarifying :)
     
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