Here, all the logarithms are in some base $b \in \mathbb{N}$, with $b \ge 2$.
Let $x$ be a real number and $c$ some small real constant (for instance, we can consider $0 \le c \le 10^{9}$).
Let $L_1(x) = \log(x + c)$ and for $n \in \mathbb{N}_{> 1}$, define
$$L_n(x) = \log(x + L_{n-1}(x)).$$
Thus, $L_n(x)$ is basically applying the $\log$ function $n$ times and putting that extra constant $c$ at the end. For instance, $L_3(x) = \log(x + \log(x + \log(x + c)))$.
So, the question is, considering fixed values of $x$ and $c$, is there a known easy-to-compute formula or even a easy-to-compute approximation for $L_n(x)$? Do you know if $L_n(x)$ converges to some known simple function as $n$ increases?
Best Answer
If the iterations converge, they converge to a $z$ such that
$$z=\log(z+x).$$
Taking the antilogarithm and multiplying by $-e^x$,
$$-e^{z+x}=-e^x(z+x)$$
or
$$-(z+x)e^{-(z+x)}=-e^{-x}$$
which is solved by the Lambert $W$ function,
$$-z-x=W(-e^{-x}).$$
As the derivative of the iterated function is
$$\frac1{x+z},$$ convergence is guaranteed where $x+z>1$. In particular, it converges with $x\ge1,z>0$.
The discussion was based on the natural logarithm , but you can generalize to other bases as follows. Let
$$z=\frac{y-\log(\log b)}{\log b}.$$
Then
$$\log_b(z+x)=\frac{\log(y-\log(\log b)+x\log b)}{\log b}-\frac{\log(\log b)}{\log b},$$
so that by setting
$$x':=x\log b-\log(\log b)$$ the recurrence turns to
$$y=\log(y+x').$$