In this biology textbook I found the following approximation:
$$\frac{1}{2N}\left( 1-\frac{1}{2N} \right)^t ≈ \frac{1}{2N}e^{\frac{-t}{2N}}$$
Can you help me to understand this approximation and help me to understand what assumption are needed for this approximation to be useful.
On
Let us compare Taylor expansions of $\left( 1-\frac{1}{x} \right)^t$ and $e^{\frac{-t}{x}}$ around $t=0$. For the first one, we have $$\left( 1-\frac{1}{x} \right)^t \simeq 1+t \log \left(1-\frac{1}{x}\right)+\frac{1}{2} t^2 \log ^2\left(1-\frac{1}{x}\right)+\frac{1}{6} t^3 \log ^3\left(1-\frac{1}{x}\right)+O\left(t^4\right)$$ while for the second one, we have $$e^{\frac{-t}{x}}\simeq 1-\frac{t}{x}+\frac{t^2}{2 x^2}-\frac{t^3}{6 x^3}+O\left(t^4\right)$$ Now, in the first one, assume that $x$ is large compared to $1$ and use the Taylor expansion of $\log(1+y)$ for small values of $y$. Replace $y$ by $1/x$.
I am sure that you can take from here and conclude about the validity of the approximation.
One way to define $e^x$ is via $$ e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n $$
By re-arranging your forumula a bit, you get $$ \frac{1}{2N}\left(1 - \frac{1}{2N}\right)^t = \frac{1}{2N}\underbrace{\left(1 + \frac{\frac{-t}{2N}}{t}\right)^t}_{\text{Compare to $e^x$ for $x=\frac{-t}{2N}$}} \approx \frac{1}{2N}e^{-\frac{-t}{2N}} \text{,} $$ at least for large enough $t$.