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can we say that essential singularities has order like poles ? I mean ;

we know e^(1/z) has an essential singularity at z=0 then do I need to say e^(1/z^3) has essential singularity at z=0 with order 3 ?

thanks

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No. The term order only applies to removable singularities. Your functions both have "infinite order" poles; $e^{1/z^3}$ certainly does not have a singularity of order 3 at 0. –  rotskoff Jun 3 '12 at 23:28
    
it does not really give any computational advantages, as the behaviour of the function remains equally "bad" in the neighbourhood of the essential singularity –  Valentin Jun 4 '12 at 0:33

2 Answers 2

It depends on what you mean by order. Usually, we define the order of a pole $a$ of $f(z)$ as the unique integer between $\alpha$ and $\alpha+1$, where $\alpha\in \mathbb{R}$ such that $\lim_{z\rightarrow a}f(z)(x-a)^\alpha=\infty$ and $\lim_{z\rightarrow a}f(z)(x-a)^{\alpha+1}=0$. In a way, it's the transition point where multiplying by a factor of $(x-a)$ makes the limit go from $\infty$ to zero (the story is reversed if $a$ is a zero of $f$).

By this definition of order of a pole, you cannot rightfully say that $e^{1/z^3}$ has an essential singularity of order $3$ since it will not satisfy the above definition.

However, if you look at the power series of $e^{1/z^3}=1/z^3+\ldots$, you can try to define the 'order' of the essential singularity as the lowest order term of the negative Laurant expansion part, in this case $z^{-3}$. Perhaps someone else can comment on the usefulness of this definition for essential singularities. Of course one characterization of an essential singularity is that the negative part of the Laurent expansion has infinitely many terms.

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You can move the singularity to $\infty$ and use the same notions of order and type as for entire functions. The study of entire functions is, to some extent, the study of an isolated essential singularity.

In this sense, $e^{1/z^3}$ has order $3$ at $z=0$, which agrees with your intuition.

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