1
$\begingroup$

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.

enter image description here

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.

$\endgroup$
2
  • $\begingroup$ The case $n$ odd can be derived from $n$ even, via the formula $g_{2n+1}(x) = \sqrt{x^{2n+1}+g_{2n+2}(x)}$. $\endgroup$ – Vincent Granville Oct 9 '20 at 15:49
  • $\begingroup$ Yes. I also stated that $f$ converges when $x>1$, however I have no proof for this, though the result sounds easy to prove, albeit a bit more tricky than for $g$. $\endgroup$ – Vincent Granville Oct 9 '20 at 19:07
0
$\begingroup$

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 $k\rightarrow\infty$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.