Question

find the taylor expansion around a singular point from a function f(z) = [exp(z)/(z-1)^{3}]!

Answer 1

find the taylor expansion around a singular point from a function
\[f(z) = \frac{ \exp (z) }{ (z-1)^{3} } !\]

Answer 2

@oldrin.bataku @UnkleRhaukus @sirm3d @hartnn

Answer 3

@hba

Answer 4

Do you know what a singular point is?

Answer 5

@gerryliyana the point is that your radius of convergence is the distance your function is from its nearest singular point, where the function has a singularity (see: http://en.wikipedia.org/wiki/Mathematical_singularity#Complex_analysis ).

I'm going to presume that stray \(!\) is a typo.

Answer 6

then, the question is wrong?

Answer 7

@gerryliyana no, but it won't converge I think.

Answer 8

with a radius of convergence of 0

Answer 9

ah ok.., then What form of expansion ??

Answer 10

Watch mate... we know \(\exp(z)\) is entire, but \(\frac1{(z-1)^3}\) is not -- it has a pole at \(z=1\). So we've found a singularity. Now let's try constructing a Taylor expansion around \(z=1\):$$\exp(z)=\sum_{n=0}^\infty\frac{e}{n!}(z-1)^n\\\frac{\exp(z)}{(z-1)^3}=\sum_{n=0}^\infty\frac{e}{n!}(z-1)^{n-3}$$oops, that's not a Taylor series, though -- it's a Laurent series.

Answer 11

yeah..., it's laurent.., i'm sorry, paper told us laurent., :). Thank you oldrin

Answer 12

@gerryliyana if you want it int he proper Laurent form rewrite:$$\frac{\exp(z)}{(z-1)^3}=\sum_{n=-3}^\infty\frac{e}{(n+3)!}(z-1)^n$$

Answer 13

ok.., thank you.., hei have a facebook ?? add me on fb at https://www.facebook.com/gerryliyana