The answers to your question is: yes!...
The use of instead of in Taylor expansion of is correct, so that is...
(1)
In this case the function expressed in (1) as Laurent series is said to have an essential singularity in ...
Kind regards
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?
thanks in advance
Thanks for the reply.
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?
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?
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 )