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gerryliyana
 2 years ago
find the taylor expansion around a singular point from a function f(z) = [exp(z)/(z1)^{3}]!
gerryliyana
 2 years ago
find the taylor expansion around a singular point from a function f(z) = [exp(z)/(z1)^{3}]!

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gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0find the taylor expansion around a singular point from a function \[f(z) = \frac{ \exp (z) }{ (z1)^{3} } !\]

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku @UnkleRhaukus @sirm3d @hartnn

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2Do you know what a singular point is?

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2@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.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0then, the question is wrong?

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2@gerryliyana no, but it won't converge I think.

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2with a radius of convergence of 0

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0ah ok.., then What form of expansion ??

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2Watch mate... we know \(\exp(z)\) is entire, but \(\frac1{(z1)^3}\) is not  it has a pole at \(z=1\). So we've found a singularity. Now let's try constructing a Taylor expansion around \(z=1\):$$\exp(z)=\sum_{n=0}^\infty\frac{e}{n!}(z1)^n\\\frac{\exp(z)}{(z1)^3}=\sum_{n=0}^\infty\frac{e}{n!}(z1)^{n3}$$oops, that's not a Taylor series, though  it's a Laurent series.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0yeah..., it's laurent.., i'm sorry, paper told us laurent., :). Thank you oldrin

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.2@gerryliyana if you want it int he proper Laurent form rewrite:$$\frac{\exp(z)}{(z1)^3}=\sum_{n=3}^\infty\frac{e}{(n+3)!}(z1)^n$$

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0ok.., thank you.., hei have a facebook ?? add me on fb at https://www.facebook.com/gerryliyana
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