How can I solve this exponential recurrence relation?

Question

Does anyone know how to solve $a_{n+1}=1-Ce^{-a_n}$ explicitly for $a_n$ in terms of $n$ and $a_0$, where $C$ is constant?

Answer 1

I think I correct in writing \begin{equation} e^{-a_n} < 1-a_n+\frac{a_{n}^{2}}{2} \end{equation} For the purposes of ease, let $C=1$. Then \begin{align} 1-e^{-a_{n}} &< 1-\left(1-a_n \frac{a_n^{2}}{2}\right) \\ &= a_n-\frac{a_n^{2}}{2} \\ &=a_n\left(1-\frac{a_n}{2}\right) \end{align} Therefore \begin{align} a_n+1 &< a_n\left(1-\frac{a_n}{2}\right) \\ \implies \frac{a_n+1}{a_n} &< 1-\frac{a_n}{2} \end{align}

I wonder if at all these musings aid you?

Answer 2

Long story short, you won't have the explicit solution, but a decent approximation may be possible.

Now, as $C<1$, there is a limit value which can be expressed in terms of Lambert W function (specifically, $a_\infty=1+W(-{C\over e})$, see the picture below, that's $a_\infty$ as a function of $C$), and $a_n$ tends to that limit exponentially, i.e. $a_n\approx a_\infty+c_1\cdot c_2^{-n}$, where $c_1$ and $c_2$ are some constants.

As $C=1$, the limit is $0$, and $a_n$ tends to it in a slower manner, namely, as $a_n\approx{c_1\over n}$.

As $C>1$, there is no limit. $a_n$ tends to negative infinity, and pretty fast at that.