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

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

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
@oldrin.bataku @UnkleRhaukus @sirm3d @hartnn
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
Do you know what a singular point is?
 one year ago

oldrin.bataku Group TitleBest 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.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
then, the question is wrong?
 one year ago

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

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
with a radius of convergence of 0
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
ah ok.., then What form of expansion ??
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
Watch 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.
 one year ago

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

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

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