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gerryliyana

  • 3 years ago

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

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

  2. gerryliyana
    • 3 years ago
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    @oldrin.bataku @UnkleRhaukus @sirm3d @hartnn

  3. gerryliyana
    • 3 years ago
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    @hba

  4. oldrin.bataku
    • 3 years ago
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    Do you know what a singular point is?

  5. oldrin.bataku
    • 3 years ago
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    @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.

  6. gerryliyana
    • 3 years ago
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    then, the question is wrong?

  7. oldrin.bataku
    • 3 years ago
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    @gerryliyana no, but it won't converge I think.

  8. oldrin.bataku
    • 3 years ago
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    with a radius of convergence of 0

  9. gerryliyana
    • 3 years ago
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    ah ok.., then What form of expansion ??

  10. oldrin.bataku
    • 3 years ago
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    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.

  11. gerryliyana
    • 3 years ago
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    yeah..., it's laurent.., i'm sorry, paper told us laurent., :). Thank you oldrin

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

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

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