I want to expand $\exp(-x)$ in a series centered at infinity, i.e. ,
$\exp(-x)=\sum_{i=-\infty}^{\infty}b_n (x-\infty)^n$
Obviously, this does not make sense, so what I did is:
We define $z=1/x$ such that now the expansion is centered at zero, i.e.,
$\exp(-1/z)=\sum_{i=-\infty}^{\infty}a_n z^n$
where
$a_n=\frac{1}{2 \pi \imath }\oint_C dz \frac{e^{-1/z}}{z^{n+1}}$,
and the contour C is taken in any way such that contains the origin.
Then, for $n\leq 1$, the function integrating is analytic in all complex space. Thus, the integral is zero. However, for all other $n$ I used the residue theorem and obtained
$a_n=\lim_{z\rightarrow 0} \frac{1}{n!} \frac{d^n}{dz^n}(\exp(-1/z) ) $
$ = \lim_{z\rightarrow 0} \frac{1}{n!} e^{-1/z}P_{2n}(\frac{1}{z}) $
here $P_{2n}(x)$ is a polinomial of order $2n$. But this result says that for all $n$ finite $a_n=0$, but I'm not sure if this is correct, and if it is then how would the Laurent expansion would be?
So, how could I express $\exp(-x)$ in a series centered at infinity, so that the first term of the series is the heaviest?
This is for an approximation I'm doing where the function I'm approximating is $exp(-N\sigma)$ ($N$ being number of particles, which is a big number), and I want to be able to make first order approx, 2nd order approx, and so.
Thanx!
Best Answer
As you have written, the functions $u$ and $v$ are not defined at zero and hence they are not differentiable at origin. Well, only thing that we can say is that $\exp({-1/z})$ is analytic at any point $z_{0}\in C-\{0\}$ and at $z_{0}=0$ the function has an essential isolated singularity. For more details see "Comlex Variables and Applications by Brown and Churchill", Chapters 2 and 5 (sec 62).