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Math Help - essential singularity

  1. #1
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    essential singularity

    prove that \Sigma_{j=1} ^\infty 2^{-2^j} z^{-j} converges for z \neq 0 and defines a function which has an essential singularity at P=0.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The series...

     \varphi(z)= \sum_{j=1}^{\infty} \frac{1}{(4 z)^{j}} (1)

    ... is a Laurent series with an infinite numers of negative powers of z, so that in z=0 is has an essential singularity. If we consider (1) as a 'geometric series' we have...

    \varphi(z) = \frac{\frac{1}{4z}}{1 - \frac{1}{4z}} = \frac{1}{4 z-1} (2)

    ... and the (1) converges for |z|>\frac{1}{4} , not symply for z\ne 0 ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 24th 2010 at 10:01 PM.
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  3. #3
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    the exponent of 2 is not -2j. it is -2^j. so it wont be geometric.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Very sorry for my misreading! ...

    According to the 'Stirling approximation' is...

    \ln \frac{1}{j!} \approx -j\cdot \ln j + j > -2^{j}\cdot \ln 2 = \ln 2^{-2^{j}}, \forall j>0 (1)

    ... so that is...

     \frac{1}{j!} > 2^{-2^{j}}, \forall j>0 (2)

    Now the series \sum_{j=1}^{\infty} \frac{1}{j!\cdot z^{j}} absolutely converges \forall z\ne 0 and therefore the same is for the series \sum_{j=1}^{\infty} \frac{2^{-2^{j}}}{z^{j}}...

    Kind regards

    \chi \sigma
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