essential singularity

**Kat-M** · April 24th 2010, 08:24 PM

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.

**chisigma** · April 24th 2010, 09:26 PM

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

$\chi$ $\sigma$

**Kat-M** · April 25th 2010, 11:20 AM

the exponent of 2 is not -2j. it is $-2^j$ . so it wont be geometric.

**chisigma** · April 25th 2010, 11:38 PM

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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