Closed form expression of a recursive relation

Question

Consider the function $f \colon (0, +\infty) \rightarrow \mathbb{R} \colon x \mapsto 1/2(1 - e^{-1/x})$. Consider the recursive relation $$ \left \{ \begin{array}{ll} g_0(x) = f(x) & \\ g_k(x) = g_{0}(g_{k - 1}(x)) & \forall k \geq 1 \end{array} \right . $$ I would like to find a closed form for the function $g_k(x)$. Alternatively, it would be interesting to understand the asymmtotic behavious of the functions $g_k(x)$, i.e. find $$ \lim_{x\to + \infty} g_k(x) $$

Answer 1

The first iterate is asymptotic to

$$\frac1{2x},$$ which obviously tends to zero.

The second iterate is thus asymptotic to

$$\frac{1-e^{-2x}}2$$ and quickly tends to $\dfrac 12$.

Hence the next iterates are virtually constant and equal to the iterates from $\dfrac12$, which linearly tend to the fixed point, solution of

$$\frac{1-e^{-1/x}}2=x.$$

Notice that the iterates starting from $x=0$ are $\dfrac12$, then $f\left(\dfrac12\right)\cdots$ so that except for the first, all iterates are very flat and vary only in the interval between terms iterated from $0$.

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An engineer would say "$g_k(x)$ is $0.45$".