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Thread: The order of the poles of cos(z)/z^2 and sin(z)/z^2

  1. #1
    CSM
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    The order of the poles of cos(z)/z^2 and sin(z)/z^2

    $\displaystyle g(z)=\frac{sin(z)}{z^2}$and $\displaystyle f(z)=\frac{cos(z)}{z^2}$ both have a pole at $\displaystyle z=0$

    From the Laurent series expansion I deduced that for $\displaystyle f(z)$ this pole is of order one, while for $\displaystyle g(z)$ the pole is order two.

    Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)
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    MHF Contributor FernandoRevilla's Avatar
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    Re: The order of the poles of cos(z)/z^2 and sin(z)/z^2

    Quote Originally Posted by CSM View Post
    Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)

    The series expansion of $\displaystyle \sin z$ and $\displaystyle \cos z$ around $\displaystyle z=0$ are well known, so we find immediately the order of the pole: $\displaystyle \sin z/z^2=1/z+\ldots$ (simple) and $\displaystyle \cos z/z^2=1/z^2+\ldots$ (double) .
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    MHF Contributor chisigma's Avatar
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    Re: The order of the poles of cos(z)/z^2 and sin(z)/z^2

    Quote Originally Posted by CSM View Post
    $\displaystyle g(z)=\frac{sin(z)}{z^2}$and $\displaystyle f(z)=\frac{cos(z)}{z^2}$ both have a pole at $\displaystyle z=0$

    From the Laurent series expansion I deduced that for $\displaystyle f(z)$ this pole is of order one, while for $\displaystyle g(z)$ the pole is order two.

    Was there any faster way to have noticed this (because first I thought they were of order 2 for both, since there was a square sign in the denominator....)
    If You know that $\displaystyle f(z)$ has a pole in $\displaystyle z=a$ but You don't know the Laurent expansion of $\displaystyle f(z)$ around $\displaystyle z=a$ , then the order of the pole in the value of n for which...

    $\displaystyle \lim_{z \rightarrow a} (z-a)^{n}\ f(z)= \lambda$ (1)

    ... where is $\displaystyle \lambda \ne 0$. An interesting 'counterexample' is the function $\displaystyle \ln z$ , for which $\displaystyle z=0$ is a 'singularity'. If we try to apply (1) in order to determine the 'order of the pole' we discover that...

    $\displaystyle \lim_{z \rightarrow 0} z^{n}\ \ln z = \begin{cases} -\infty &\phantom{|}\textrm{if }n=0\\ 0 &\phantom{|}\textrm{if } n>0 \end{cases}$ (2)

    ... so that no value on n satisfies (1). The reason is that $\displaystyle z=0$ is a singularity of the function $\displaystyle \ln z$ but isn't a 'pole'...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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