Laurent series of e^(1/z)

In summary, the conversation discusses the classification of singularities of the function e^1/z and finding its Laurent series. It is determined that there is a singularity at z=0 and the order of the pole needs to be found. The general expression for the expansion of an exponential is used, but it leads to an inconsistent first term. There is a suggestion to consider a new variable for 1/z, and it is explained how the order of the pole can be determined. It is concluded that the function has an essential singularity and the Laurent series is 1-1/z+1/(2!z^2)+...
  • #1
247
11

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 :)
 
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  • #2
Physgeek64 said:

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
stevendaryl said:
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
Steven didn't conclude it. He wrote If ... then ... !
 
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  • #5
Physgeek64 said:
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
Physgeek64 said:
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 :)
stevendaryl said:
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
BvU said:
Steven didn't conclude it. He wrote If ... then ... !

My mistake, I misinterpreted his response. Thank you for clarifying :)
 

1. What is a Laurent series?

A Laurent series is a representation of a complex function as a sum of infinitely many terms, with each term being a power of z.

2. How is a Laurent series different from a Taylor series?

A Taylor series represents a complex function as a sum of infinitely many terms, with each term being a power of z. However, a Taylor series can only be used for functions that are analytic at a given point, while a Laurent series can be used for functions that have singularities or poles at a given point.

3. What is the Laurent series of e^(1/z)?

The Laurent series of e^(1/z) is given by Σ(n=0 to ∞) (1/z)^n/n! = 1 + 1/z + 1/(2!z^2) + 1/(3!z^3) + ...

4. How is the Laurent series of e^(1/z) useful?

The Laurent series of e^(1/z) is useful in complex analysis, as it allows us to extend the concept of a power series to functions that have singularities or poles. It can also be used to evaluate complex integrals and solve differential equations.

5. Can the Laurent series of e^(1/z) be used to approximate the function?

Yes, the Laurent series of e^(1/z) can be used to approximate the function for values of z near the origin. However, as z gets further away from the origin, the approximation becomes less accurate due to the presence of singularities in the series.

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