I have a growth model such as:
$$G(n)= (1-1/N)(1-2/N)...(1-n/N) \text{, with } n<N$$
As analytical approximation of $G(n)$, I numerically found that $1-\exp(-\exp(1.71 (1-1.05 n/\sqrt{N}))$ which is the Gompertz function was sufficient for my application.
N ranges from 5e4 to 5e7, and n ranges from 1 to $5 \sqrt{N}$
But is there a mathematical evaluation of this approximation?
Is there another analytical approximation of $G(n)$?
the accuracy can be lower when n approaches N
The Taylor expansion given by Claude is very close to the Gompertz approximation and works
for the physical point of view G(n) is a sigmoid curve with its inflexion point close, or equal, to $\sqrt{N}$, it is why I was hoping that there was a sigmoid curve as a function of $n/\sqrt{N}$ which makes more sense regarding the physics
in fact 1-G(n) corresponds to the probability to draw at least twice the same balls after n random drawings with replacing the drawn ball in the box
Thanks
Stephan
1 Answer
$$G_n=\prod_{i=1}^n \left(1-\frac i N \right)\implies \log(G_n)=\sum_{i=1}^n \log\left(1-\frac i N \right)$$ Assuming that $n \ll N$, we can use the Taylor expansion for the logarithm which makes $$\log\left(1-\frac i N \right)=-\frac{i}{N}-\frac{i^2}{2 N^2}-\frac{i^3}{3 N^3}+O\left(\frac{1}{N^4}\right)$$ and make the summation over $i$ to get $$\log(G_n)=-\frac{n (n+1)}{2 N}-\frac{n (n+1) (2 n+1)}{12 N^2}+O\left(\frac{1}{N^3}\right)$$
So, an approximation could be $$G_n \sim \exp\left(-\frac{n (n+1)}{2 N}\left(1+\frac {2n+1}{6N}+\frac{n (n+1)}{6 N^2} \right)\right)$$
Let us try for $N=100$ $$\left( \begin{array}{ccc} n & \text{approximation} &\text{exact} \\ 5 & 0.858280 & 0.858278 \\ 10 & 0.565379 & 0.565341 \\ 15 & 0.281732 & 0.281592 \\ 20 & 0.104538 & 0.104320 \\ 25 & 0.0283974 & 0.0282133 \\ 30 & 0.00554519 & 0.00545377 \\ 35 & 0.000763345 & 0.000735507 \\ 40 & 0.0000725573 & 0.0000672946 \end{array} \right)$$
