Some fun with special infinite nested radicals

Question

Let us define the following functions:

$$f_n(x)=\sqrt{x^{n}-\sqrt{x^{n+1}- \sqrt{x^{n+2}-\cdots}}} $$ $$g_n(x)=\sqrt{x^{n}+\sqrt{x^{n+1}+ \sqrt{x^{n+2}+\cdots}}} $$

with $f(x)=f_1(x)$ and $g(x)=g_1(x)$. Very little is known about $f(x)$ and $g(x)$, except:

• The nested radical converges in both cases if $x > 1$
• $\lim_{x\rightarrow 1^+} f(x) = \frac{-1+\sqrt{5}}{2}$ and $g(1) = \frac{1+\sqrt{5}}{2}$
• $f(4)=1$ and $g(4)=3$

Let us now focus on the case where $x>1$ is an integer. The goal here is to obtain more advanced, interesting results about these nested radicals, maybe even a closed form or some asymptotic formulas.

1. Integer part of the infinite nested radicals

Let $\lfloor\cdot\rfloor$ denotes the integer part function. We have:

$$\lfloor f_{2n}(x) \rfloor = x^{n}-\phi(x)\\ \mbox{ } \lfloor g_{2n}(x) \rfloor = x^{n}+\psi(x)$$

with $\phi(x)=k$ if $x\in A_k$, $\psi(x)=k$ if $x\in B_k$. The sets $A_k, B_k$ are as follows:

• $A_1=[2,5[, A_2=[5,15[, A_3= [15,33[, A_4=[33,61[, A_5=[61,96[, \cdots$
• $B_0=[2,4[, B_1=[4,17[, B_2= [17,38[, B_3=[38,67[, B_4=[67,104[, \cdots$

2. Fractional part of the infinite nested radicals

Let $\{\cdot\}$ denotes the fractional part function. We seem to have:

$$\lim_{n\rightarrow\infty} \{ f_{2n}(x)\}=1-\Big\{\frac{\sqrt{x}}{2}\Big\}\\ \lim_{n\rightarrow\infty} \{ g_{2n}(x)\}=\Big\{\frac{\sqrt{x}}{2}\Big\} $$ Also, the roots of $\{g_1(x)\}$ have a very peculiar quadratic distribution. The first few ones, for $x>0$, are $\rho_1=4.0000$, $\rho_2=7.3370$, $\rho_3=11.6689$, $\rho_4=16.9982$, $\rho_5=23.3260$, $\rho_6=30.6526$, $\rho_7=38.9787$. Furthermore, it seems that

$$\lim_{k\rightarrow\infty} (\rho_{k+2}-2\rho_{k+1} +\rho_k) = 1.$$

Finally, values of $\{g_1(x)\}$ for large successive integers $x$ lying between two successive roots of $\{g_1(\cdot)\}$ tend to be equally spaced as $x\rightarrow\infty$. See table below.

3. My question

Actually a few related questions. Feel free to answer the one(s) you are most interested in.

• Many of my results are experimental (thus I often use the word "it seems"). Can you prove some of them?

• Obtain an explicit closed form for all sets $A_k,B_k$ used in the definition of $\phi(x)$ and $\psi(x)$ in section 1. Not sure if it is easy or not.

• We focused on $n$ even. What happens for $n$ odd? Do we have interesting results? For instance, if $x=2$, the successive values of $\lfloor g_{2n+1}(x)\rfloor$ are $2, 3, 6, 12, 23, 45, 91, 181, 362, \cdots$ (for $n=0, 1, \cdots$). I did a reverse lookup on that sequence (see here) but it did not return any result despite the semi-obvious pattern.

• What happens if $x$ is not an integer? Any interesting pattern or result?

• Can you derive even more intriguing insights from the empirical results I presented here?

• It looks like the larger $x$, the faster my limits are converging. Worth exploring.

Answer 1

I am focusing here on $g(x)$, with $x$ a strictly positive integer. All the results below have been obtained empirically. A proof (or rebuttal) would be welcome. Again, $\{ \cdot \}$ represents the floor function.

Let $b_k=4k^2 + k - 1$. We have $B_k=[b_k,b_{k+1}[$ if $k>0$, and $B_0=[2, 4[$. Thus we now have a closed form for $\psi(x)$ and thus for $\lfloor g_{2n}(x)\rfloor$, regardless of $x$ and $n$, assuming $x$ is an integer. In particular, for $x>0$, we have:

$$\lfloor g_{2n}(x)\rfloor = x^{n}+\psi(x), \mbox{ with } \psi(x)=\Big\lfloor \small \frac{-1+\sqrt{17+16x}}{8}\Big\rfloor.$$

It also works for $n=0$. Let $\eta(x)=\psi(x) -\lfloor\sqrt{x}/2\rfloor$. It is equal either to zero (for most $x$'s) or one (for $x=16, 36, 37$, $64, 65, 66,\cdots$). Also, for $x<16$, we have the following approximation: $$\{ g_{0}(x)\} \approx \Big\{\frac{\sqrt{x}}{2}\Big\}+\frac{2-\sqrt{5}}{6} (x-4)$$

resulting in

$$g_0(x)=\lfloor g_0(x)\rfloor + \{ g_{0}(x)\} \approx \frac{\sqrt{x}}{2} +\frac{2-\sqrt{5}}{6} (x-4) + 1.$$

The approximation is exact if $x=1$ or $x=4$. It is also pretty good even if $x$ is not an integer. Note that $g(x)=g_1(x)=g_0^2(x)-1$.

Another potentially interesting result is this:

$$\lim_{n\rightarrow\infty} \frac{ \{ g_{2n}(x)\}-\{\frac{\sqrt{x}}{2}\} }{ \{ g_{2n+2}(x)\}-\{\frac{\sqrt{x}}{2}\} } = x.$$

Asymptotics and distribution of roots

A better approximation to $g_0(x)$, especially for large $x$, is the only real, positive solution of the equation $(y^2−1)^2−y=2x−1$ with respect to $x$. This approximation is also exact for $x=1$ and $x=4$ and it works for non-integer values of $x$. For large $x$, we have the following asymptotic expansion for $g(x)=g_1(x)$:

$$g(x) =\sqrt{2}\cdot\Big(\sqrt{x}+\frac{1}{8}-\frac{5}{128 \sqrt{x}}+O\Big(\frac{1}{x}\Big)\Big) .$$

The above formula is easy to derive (see Mathematica computation here) and is particularly useful to study the roots of $\{ g(x)\}$. If $x$ is very large, $x$ is a root of $\{g(x)\}$ if and only if $\sqrt{2} (\sqrt{x} + \frac{1}{8})$ is very close to an integer. Since the first root is $\rho_1=4$, an (excellent) approximation to the $k$-th root $\rho_k$ is the value of $x$ satisfying $\sqrt{2} (\sqrt{x} + \frac{1}{8})=k+2$. In order words, $$\rho_k = \frac{(k+2)^2}{2} -\frac{k+2}{4\sqrt{2}}+\frac{1}{64} + O\Big(\frac{1}{k}\Big).$$

So, $g(\rho_k)=k+2$ and thus $\{g(\rho_k)\}=0$ and there is no other root beyond those discussed here. Note that using my approximation, we have $\rho_1\approx 3.9853$ while the exact value is $4$. The larger $k$, the better the approximation since the error term is of the order $1/k$ and thus tends to zero as $k\rightarrow\infty$.