Convergence of this iterated logarithm: $\log(x + \log(x + \log(x + ... + \log(x + c)...)))$

Question

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?

See What is the maximum value of $\ln(x+\ln(x+\ln(x+\cdots)))-\ln x$? Convergence is yet to be shown rigorously. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Oct 01 '19 at 11:58

score 1 · Accepted Answer · 2019-10-01T14:15:13.310

1

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

$$y=\log(y+x').$$

edited Oct 01 '19 at 14:15

answered Oct 01 '19 at 12:03

$W$ has many branches, when the base is $e$ instead of an integer, the correct branch should be $W_{-1}(\cdot)$ for $x \ge 0$. – achille hui Oct 01 '19 at 12:10
@achillehui is there a $W$ function in base $b$ instead of $e$? It should be a solution to $e^{z\ln(b) +x} = e^x(z+x)$, isn't it? – Marcellus Oct 01 '19 at 12:37
Yves Daoust, thank you! What is puzzling me is the fact that $W_{-1}(-e^{x})$ is not a real number, but $L_n(x)$ is... – Marcellus Oct 01 '19 at 12:56
@Marcellus: why do you say it is not real ? – Oct 01 '19 at 13:48
I tested it for several values of $x$ on Wolfram Alpha and it always returns something like $W_{-1}(-e^x) \approx c_0 + c_1\cdot i$ with $c_1 \not = 0$. – Marcellus Oct 01 '19 at 14:04
@Marcellus: sorry there was a typo: $e^{-x}$. – Oct 01 '19 at 14:14

Convergence of this iterated logarithm: $\log(x + \log(x + \log(x + ... + \log(x + c)...)))$

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