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Laurent series of e^(1/z)

  • Thread starter Physgeek64
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Homework Statement


Cassify the singularities of [itex]e^\frac{1}{z} [/itex] and find the Laurent series

Homework Equations


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

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 :)
 

Answers and Replies

  • #2
stevendaryl
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Homework Statement


Cassify the singularities of [itex]e^\frac{1}{z} [/itex] and find the Laurent series

Homework Equations


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

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.
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?
 
  • #3
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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?
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 :)
 
  • #4
BvU
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Steven didn't conclude it. He wrote If ... then ... !
 
  • #5
stevendaryl
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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]
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.
 
  • #6
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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 :)
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.
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 :)
 
  • #7
247
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Steven didn't conclude it. He wrote If ... then ... !
My mistake, I misinterpreted his response. Thank you for clarifying :)
 

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