## gerryliyana Group Title find the taylor expansion around a singular point from a function f(z) = [exp(z)/(z-1)^{3}]! one year ago one year ago

1. gerryliyana Group Title

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

2. gerryliyana Group Title

@oldrin.bataku @UnkleRhaukus @sirm3d @hartnn

3. gerryliyana Group Title

@hba

4. oldrin.bataku Group Title

Do you know what a singular point is?

5. oldrin.bataku Group Title

@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 Group Title

then, the question is wrong?

7. oldrin.bataku Group Title

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

8. oldrin.bataku Group Title

with a radius of convergence of 0

9. gerryliyana Group Title

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

10. oldrin.bataku Group Title

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 Group Title

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

12. oldrin.bataku Group Title

@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 Group Title