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

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

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

oldrin.bataku
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0then, the question is wrong?

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

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

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

oldrin.bataku
 one year 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
 one year 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
 one year 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
 one year 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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