Let $b>0$, how to expand $\exp(-bx)$ when $1<x<B$?
I am seeking a series expansion like the following ($n_0> 0$): $$\exp(-bx)=\sum_{n=n_0}^{\infty}\frac{a_n}{x^n} $$
EDIT:
@GerryMyerson: Thanks for question.
I now think that Taylor series will be sufficient as long as I keep enough terms.
$$\exp(-bx)\approx\sum_{n=0}^{N+1}(-b)^n\frac{x^n}{n!} $$
where $N$ is determined by $$\frac{B^{N+1}}{(N+1)!}\approx \frac{B^{N}}{(N/e)^N}\frac{B}{\sqrt{2\pi N}}<<1$$
Thus we may require $$N/e>>B,\qquad 2\pi N>>B^2$$ or $$N>>\max\left(e B, \frac{B^2}{2\pi}\right)$$