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Let's say we have an infinite nested radical with random terms (positive integers) which can take on finitely many values in the range $(n_1,n_2)$. What would the distribution look like for these radicals if all terms appear with equal probability?

I naively expected it to look, well, random, even for $n_2=n_1+1$. However, this is not the case.

See how the distribution looks for infinite radicals with random terms which can be either $1$ or $2$ (number of terms is $300$ in each radical, size of the distribution is $700$, the histogram contains $250$ bins, and I used Mathematica RandomInteger):

enter image description here

It's obvious, that the numbers can't be smaller than the golden ratio:

$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=1.618034$$

And can't be larger than $2$:

$$2=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$

There are also two more 'attractors' for this distribution:

$$\sqrt{1+\sqrt{2+\sqrt{1+\sqrt{2+\cdots}}}}=1.710644$$

$$\sqrt{2+\sqrt{1+\sqrt{2+\sqrt{1+\cdots}}}}=1.926303$$

What I can't explain is the occurence of 'doublets' for each spectral line (which appear in any experiment I've run, for other pairs of numbers as well).

And how to find the size of the 'line broadening'? What does it depend on?

This is what the distribution looks like for terms which can take values $1,2,3$:

enter image description here


See also the thesis about nested radicals and Cantor sets here

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Let's write your nested radical as $r(x_1, x_2, \ldots) = \sqrt{x_1 + \sqrt{x_2 + \sqrt{x_3 + \ldots}}}$. Thus

$$r(x_1, x_2, \ldots) = \sqrt{x_1 + r(x_2, x_3, \ldots)}$$

In particular, since $r$ is an increasing function of each $r_i$, if the possible values are $1$ and $2$

$$\sqrt{r_1 + \phi} \le r(x_1, x_2, \ldots) \le \sqrt{r_1 + 2}$$

so that the distribution will be supported in the union of intervals $$[\sqrt{1+\phi},\sqrt{3}] \cup [\sqrt{2+\phi}, 2] \approx [1.618033988, 1.732050808] \cup [1.902113032, 2]$$

If you take the two possible values of $r_2$ into account,each of these intervals splits into two, and this continues; the end result is a somewhat distorted image of the Cantor set.

Similarly for your next example, but splitting each interval into three instead of two.

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  • $\begingroup$ Now I get it, thank you. So, if the size of the distribution becomes infinite, do I get an infinite number of 'infinitely thin' lines? The Cantor set has measure $0$, will this be the case for this distribution also? For any finite number of term values? $\endgroup$ – Yuriy S Mar 17 '16 at 22:01
  • $\begingroup$ If we use numbers 2 to 6, the intervals unite and cover the whole region. Thanks for the method, it's very useful $\endgroup$ – Yuriy S Mar 18 '16 at 0:13

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