Asymptotic expansion of maximum of function.

Question

Denoted $M(n)$ by maximum of $$e^{-t}-\left(1-\frac{t}{n}\right)^n \ \ \ t\in [0,n]$$ how to calculate the asymptotic expansion of $M(n)$ as $n\rightarrow \infty$?

Answer 1

If we consider the function

$$f(t)=e^{-t}-\left(1-\frac{t}{n}\right)^n $$ its first derivative is given by $$f'(t)=-e^{-t}+\left(1-\frac{t}{n}\right)^{n-1} $$ It cancels at a value $$t_*=n+(n-1)\, W\left(-\frac{ n}{n-1}e^{-\frac{n}{n-1}}\right)$$ where $W(.)$ is Lambert function.

So $M(n)=f(t_*)$.

Using successive Taylor expansions, we have $$e^{-\frac{n}{n-1}}=\frac{1}{e}-\frac{1}{e n}-\frac{1}{2 e n^2}-\frac{1}{6 e n^3}+O\left(\frac{1}{n^4}\right)$$ $$-\frac{ n}{n-1}e^{-\frac{n}{n-1}}=-\frac{1}{e}+\frac{1}{2 e n^2}+\frac{2}{3 e n^3}+O\left(\frac{1}{n^4}\right)$$ $$ W\left(-\frac{ n}{n-1}e^{-\frac{n}{n-1}}\right)=-1+\frac{1}{n}+\frac{1}{3 n^2}-\frac{37}{72n^3}+O\left(\frac{1}{n^4}\right)$$ $$t_*=2-\frac{2}{3 n}-\frac{61}{72 n^2}+O\left(\frac{1}{n^3}\right)$$

$$n \log\left(1-\frac{t_*}{n}\right)=-2-\frac{4}{3 n}-\frac{35}{72 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\left(1-\frac{t_*}{n}\right)^n=\frac{1}{e^2}-\frac{4}{3 e^2 n}+\frac{29}{72 e^2 n^2}+O\left(\frac{1}{n^3}\right)$$ from which $$M(n)=f(t_*)=\frac{2}{e^2 n}+\frac{2}{3 e^2 n^2}+O\left(\frac{1}{n^3}\right)=\frac{2}{e^2 n}\left(1+\frac{1}{3 n}+O\left(\frac{1}{n^2}\right)\right)$$

So, your result is $$\color{blue}{M(n)=\frac{2}{e^2 n}+O\left(\frac{1}{n^2}\right)}$$

It has been checked numerically.