# Do infinitely nested radicals have any applications?

There is a simple necessary and sufficient condition for a continued radical of the form $$\sqrt{a_1 + \sqrt{a_2 + \dotsc}}$$ to converge (where all terms $$a_1, a_2$$ etc. are nonnegative). Namely, that the sequence $$n \mapsto a_n^{2^{-n}}$$ should be bounded. This is known as Herschfeld's Convergence Theorem (though it was discovered independently of Herschfeld by Paul Wiernsberger thirty years before).

Lately, I've developed a constructive proof of this theorem, which you can find here.

It leads me to wonder: Is there any actual use for continued square roots, apart from studying them for their own sake? Googling doesn't show up anything.

• may be, it has some use in galois theory, typically kummer theory, or serves as an example for the applications of galois theory – vidyarthi Apr 28 at 17:41
• It might be useful to look through the 98 page (unpublished?) manuscript A chronology of continued square roots and other continued compositions, through the year 2016 by Dixon J. Jones. – Dave L Renfro Apr 28 at 18:05
• Here's a paper (behind paywall, sorry: JSTOR still earns money on 45 year old papers) on the case of complex iterated radicals. – YCor Apr 29 at 7:46
• If it was discovered 30 years before Herschfeld, would you mention by who? – YCor Apr 29 at 8:01

The first application goes back to Archimedes. Let me explain how. $$\underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n}\to 2.$$ The question is how fast. It turns out that the rate of convergence can be caught quite precisely from the formula: $$2^{n+2}\cdot\sqrt{2-{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}}\to 2\pi.$$ This formula has an easy geometric interpretation: the expression on the left hand side equals the circumference of the regular $$2^{n+2}$$-gon inscribed in the unit circle. In fact, Archimedes used this approach to find the approximate value of $$\pi$$. The method was mastered later by Ludolph van Ceulen who published in 1596 the first 20-decimals of $$\pi$$.

• The method from my paper gives that $2 - \underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n} \leq 2^{2^{-n}}(2^{2^{-n}} - 1)$, but this is quite pessimistic – ogogmad Apr 28 at 22:21
• @ogogmad I think you should provide coordinates for your paper in the statement of the question. Otherwise nobody knows what you are talking about. – Piotr Hajlasz Apr 29 at 1:45
• I'm not sure if it's clear, but the general idea is described in the section Overview and Strategy... – ogogmad Apr 29 at 9:43

The following is not an answer.

In a comment to @PiotrHajlasz's answer, I say that a method from my paper shows that $$2-\underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n} \leq 2^{2^{-n}}(2^{2^{-n}} - 1)$$. You can find a description of this in the subsection Overview and Strategy of my proof. But in case it's not clear, I describe how to do this below:

The general idea is that given $$U \geq L$$ and a non-negative sequence $$(a_i)_i$$, we have that \begin{align} &\sqrt{a_1 + \sqrt{a_2 + \dotsb \sqrt{a_{n-1} + \sqrt{U}}}} - \sqrt{a_1 + \sqrt{a_2 + \dotsb \sqrt{a_{n-1} + \sqrt{L}}}} \leq \sqrt{0 + \sqrt{0 + \dotsc\sqrt{0 + \sqrt U}}} - \sqrt{0 + \sqrt{0 + \dotsc\sqrt{0 + \sqrt L}}} = U^{1/2^n} - L^{1/2^n}. \end{align} In other words, we can upper bound the difference between an upper bound and lower bound by driving the terms $$a_i$$ down to $$0$$.

In the case of $$\sqrt{2 + \sqrt{2 + \dotsb}}$$, we have an upper bound in the form of $$\sqrt{2 + \sqrt{2 + \dotsb \sqrt{2 + \sqrt{\color{red} 4}}}}=2$$, and a lower bound in the form of $$\sqrt{2 + \sqrt{2 + \dotsb \sqrt{2 + \sqrt{\color{red} 2}}}}$$. The difference between upper and lower bound can thus be increased to $$4^{1/2^n} - 2^{1/2^n} = 2^{1/2^n} (2^{1/2^n} - 1)$$, which clearly goes to zero.