Laurent series

• Apr 2nd 2009, 10:29 PM
krindik
Laurent series
Hi i'm new to this forum. I have a question with Laurent series.
Found below in an Eng Math text

z^2 * exp(1/z)

= z^2 * (expansion of exp(z), replacing z with 1/z )

However, as I understand eventhough exp(z) is entire (analytic
everywhere) exp(1/z) is not. so can the expansion of exp(z)
1 + z/1! + (z^2)/2! + ...

replacing z by 1/z be valid here?

• Apr 2nd 2009, 11:48 PM
chisigma

The use of $\frac{1}{z}$ instead of $z$ in Taylor expansion of $e^{z}$ is correct, so that is...

$f(z)=z^{2}\cdot e^{\frac{1}{z}}= z^{2}\cdot \sum_{n=0}^{\infty} \frac{1}{z^{n}\cdot n!}$ (1)

In this case the function $f(*)$ expressed in (1) as Laurent series is said to have an essential singularity in $z=0$...

Kind regards

$\chi$ $\sigma$
• Apr 3rd 2009, 03:06 AM
krindik
Anyway, I was wondering how could Taylor expansion be used here if there's a singularity at z=0 ?
If we use the Taylor expansion for exp(z) for the solution, replacing z with 1/z aren't we assuming that exp(1/z) is also analytic at z=0?

Though Laurent expansion allows for singularities how does it relate here?
• Apr 3rd 2009, 03:35 AM
Opalg
Quote:

Originally Posted by krindik
I was wondering how could Taylor expansion be used here if there's a singularity at z=0 ?
If we use the Taylor expansion for exp(z) for the solution, replacing z with 1/z aren't we assuming that exp(1/z) is also analytic at z=0?

Though Laurent expansion allows for singularities how does it relate here?

That series for $z^2e^{1/z}$ is a Laurent series, not a Taylor series (because it includes negative powers of z). In general, a Laurent series converges in an annulus $m<|z|. In this case, m=0 and M=∞.
• Apr 3rd 2009, 03:51 AM
krindik
Thanks.

exp(z) = 1 + z + z/2! + ...

Above expansion is based on the property that exp(z) is analytic. That's how the Taylor expansion was written for exp(z)

But then, how could be used for Laurents expansion as exp(1/z) is not analytic. ie. we may be able arrive at the solution by using the formulas for Laurent expansion. But can we take the assumptions for Taylor expansion to this?
• Apr 3rd 2009, 05:06 AM
HallsofIvy
The point is that it is valid everywhere except at x= 0. That doesn't have to be said explicitely because the function itself does not exist at x= 0.
• Apr 3rd 2009, 01:45 PM
krindik
That's precisely my point.
Since it doesn't exist for z=0 how can u expand it with Taylor series?

I'd really appreciate if you would let me know how the derivation for exp(z) came from.

In the text I read it is from Taylor series
i.e.
f(z) = sum(An * (z-z0)^n) : n=0, ... infinity
where
f(z) = exp(z)
An = (n th differential of f at z0) / n!

Only then
exp(z) = 1 + z + z^2/2! + ....
is derived.

My question is how can we use the above derivation for exp(1/z) as we have assumed the analyticity of the function?

Thanks

(btw, how do u write the math symbols? the sigma symbol on the tool bar inserts only [tex] tags )