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The following infinite nested radical $$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$$ is known to converge to $\phi=\displaystyle\frac{\sqrt{5}+1}{2}$.

It is also known that the similar infinite nested radical $$\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\cdots}}}}$$ converges to $\displaystyle\frac{1}{t_3-1}=1.19148...$, where $t_3=1.83929...$ is the tribonacci constant, corresponding to the ratio to which adjacent tribonacci numbers tend. Note the similarities with the previous case, since $\displaystyle\frac{1}{\phi-1}=\phi$.

Interestingly, assuming that the degree of the radicals is $n$ and that the constant term under the radicals is $1/(n-1)$, the relationship between this type of nested radicals and the n-bonacci constants is valid even for the degenerate 1-bonacci numbers (whose sequence reduces to a series of 1). In this case, the radicals are canceled out, the constant term becomes $1/0=\infty$, and the result is $\infty$, which is in accordance with $\frac{1}{1-1}=\infty$.

Based on these considerations, it could be hypothesized that similar nested radicals may exist for higher order n-bonacci numbers. For example, for $n=4$, we could expect that a nested radical giving $\displaystyle\frac{1}{t_4-1}=1.07809...$, where $t_4=1.92756...$ is the so-called tetranacci constant (limit of the ratio between adjacent 4-bonacci numbers), might be

$$\sqrt[4]{\frac{1}{3}+\sqrt[4]{\frac{1}{3}+\sqrt[4]{\frac{1}{3}+\sqrt[4]{\frac{1}{3}+\cdots}}}}$$

Unfortunately, in this case the radical seems not to work (it gives $1.09279...$ instead of the expected $1.07809...$). Also, I was not able to find any rational number that, inserted in this nested radical as the constant term, satisfies the equivalence (the numerical value that works is $k=0.27282...$). Similar results are obtained for higher values of $n$. Which is the reason for this? Is there any way to prove or disprove the existence of similar nested radicals linked to n-bonacci numbers for $n>3$?

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  • $\begingroup$ Here the first few solutions. $\endgroup$ – Nikolaj-K Jul 10 '14 at 15:03
  • $\begingroup$ @Anatoly: You are partially right, but its nested radical does not involve $\frac{1}{3}$. See my answer below. $\endgroup$ – Tito Piezas III Dec 11 '14 at 5:20
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This is equivalent to $x^n=\dfrac1{n-1}+x$ , which means solving a polynomial of the n-th degree in x. The generalized Fibonacci numbers fulfill the general equation $x^{−n}+x=2$, so there is no reason to expect any universally valid connection between the two. See also strong law of small numbers.

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  • $\begingroup$ I know that the radical converges to 1.09279, as I wrote in the question. But the expected result, according to what works for $n\leq3$, should be $1/(t_4-1)=1.0780...$. It does not work. $\endgroup$ – Anatoly Jul 10 '14 at 15:10
  • $\begingroup$ @Anatoly: Solving $x^4=\dfrac13+x$ yields $1.09279$ as the only real and positive solution, so I don't see the problem. $\endgroup$ – Lucian Jul 10 '14 at 15:19
  • $\begingroup$ Ok, but the question does not refer to the result $1.09279$ of the nested radical, which is clearly equal to the root of the fourth-grade equation. The question is why this number is not related to the tetranacci constant via the expected relation $1/(t_4-1)$. This relation works for $n=1,2,3$, but not for $n\geq4$. $\endgroup$ – Anatoly Jul 10 '14 at 15:33
  • $\begingroup$ @Anatoly: The numbers you speak of fulfill the general equation $x^{-n}+x=2$, so there is no reason to expect any universally-valid connection between the two. This reminds me of the so-called strong law of small numbers. $\endgroup$ – Lucian Jul 10 '14 at 15:45
  • $\begingroup$ Thanks for your comment on the general equation of n-bonacci numbers. I agree that a universally valid connection is unlikely, even hypothesizing that for some values of $n>3$ there exist more complex forms of radicals that work. For the sake of clarity, could you update your answer by focusing it on the correct issue, as clarified in these comments? I will then be happy to accept it! $\endgroup$ – Anatoly Jul 10 '14 at 18:04
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(Edit: Generalized previous answer.)

Actually, you are partially right. There is a general approach, but it works in radicals only for $n \leq 5$. Given,

$$x = \sqrt[n]{b+a\sqrt[n]{b+a\sqrt[n]{b+a\sqrt[n]{b+\dots}}}}$$

Raise to the $n$th power,

$$x^n = b+a\sqrt[n]{b+a\sqrt[n]{b+a\sqrt[n]{b+\dots}}}$$

Or,

$$x^n = b+ax\tag1$$

The general eqn of degree $n\leq5$ can be transformed, in radicals, to this form with the case $n=5$ called the Bring quintic. Hence, given the positive real root of $n$-nacci eqn, $y^{-n}+y = 2$, then,

$n = 2$: Fibonacci constant $y$:

$$y = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}} = 1.618033\dots$$

$n = 3$: Tribonacci constant $y$:

$$y(3-y) = \sqrt[3]{14-2\sqrt[3]{14-2\sqrt[3]{14-2\sqrt[3]{14-\dots}}}} = 2.134884\dots$$

$n = 4$: Tetranacci constant $y$:

$$y(3-y) = \sqrt[4]{41-11\sqrt[4]{41-11\sqrt[4]{41-11\sqrt[4]{41-\dots}}}} = 2.06719\dots$$

$n = 5$: Pentanacci constant $y$:

$$\text{doable, but complicated}$$

You can still do so for $n=5$, but the LHS will be more complicated and $a,b$ will be algebraic numbers with square roots and cube roots. But if you are curious how to transform the general quintic into the form $(1)$, see this post.

P.S. Let me guess. You found that other nested radical for the tribonacci constant in this website?

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