I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.

I'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets.

$$2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$$

$$1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$$

$$0=\sqrt{2 - \sqrt{2 + \sqrt{2 + \dots}}}=-(+)$$

$$\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$$

$$\frac{1}{\phi}=\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \dots}}}}=(-+)$$

In general, the expansion can be found by a very easy algorithm:

  • take any number in $(0,2)$, square it
  • if the result $>2$ write $+$, if the result $<2$ write $-$
  • subtract $2$ from the result, square, repeat

If on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.


$$\pi-2=--+-++-+-+++++++-+-+---------+-+--+--+--+++---++++ \dots=1.141592653589793 \dots$$

Basically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\pi$. But considering the expansion can be coded as binary, it's not so bad.

The convergence plot, and two binary plots for this $50$ terms can be seen below:

enter image description here

$$e-1=+-----+++-++-+---++-++++-+---++-+++-++++-++++---++ \dots=1.71828182845905 \dots$$

enter image description here

Do you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?

Is number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?


Now that I think about it, we can use the general expansion for $x \in [0,a]$:

$$x=\left(a \pm \left(a \pm \left(a \pm \dots \right)^p \right)^p \right)^p$$


For example:

$$\frac{1}{4}=\left(\frac{1}{4} + \left(\frac{1}{4} + \left(\frac{1}{4} + \dots \right)^2 \right)^2 \right)^2$$

$$\frac{3}{4}-\frac{\sqrt{2}}{2}=\left(\frac{1}{4} - \left(\frac{1}{4} - \left(\frac{1}{4} - \dots \right)^2 \right)^2 \right)^2$$


However, this case $p=2,~~~a=\frac{1}{4}$ is not just a random example, it's the only rational expansion of this kind. So I would say it's more important than the titular root expansion.


An interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344

Yuriy S
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    Wow. Great question, +1 –  Apr 06 '16 at 21:33
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    Beautiful idea. One could make a binary-like number system based on it. Let a "bit" code 0 for minus or 1 for plus. I wonder if it could have unique representation contrary to the usual way to write numbers, for example $0.1b = 0.011\dots b = 1/2$. – mathreadler Apr 07 '16 at 08:01
  • I wonder if there is some class of numbers which have periodic "number expansions" in that number system. Like the rationals have for ordinary number systems using the division algorithm. – mathreadler Apr 07 '16 at 08:27
  • @mathreadler, that would be a subset of algebraic numbers – Yuriy S Apr 07 '16 at 08:30
  • the $50$ bits gives $50/\log_{10}(2) \approx 15$ decimals which is just one or two bits short of mantissa of a ordinary double precision floating point number system. – mathreadler Apr 07 '16 at 15:36
  • Next question: if you have representations of two numbers in this way, how hard is it to add and subtract them, and how can you detect "overflow" and "underflow"? – marty cohen Apr 08 '16 at 00:32
  • @martycohen, just as with continued fractions, we will have to compute the numbers to the required (or double) precision, and then compute the sum or difference. In fact, nested radicals are even worse, because they allow no forward recursion. This expansion is just a curiosity, I see no application so far – Yuriy S Apr 08 '16 at 11:20
  • We can use it to represent irrational numbers exactly with finite number of bits. Just a few bits to represent or store an irrational number like $\phi$ is not very much compared to the many tens of bits storing a floating point number. – mathreadler Apr 08 '16 at 12:41
  • Which irrational numbers? How about $\sqrt[3]{2}$? – marty cohen Apr 08 '16 at 18:38
  • Even the case of except 2, it seem this expansion holds. – Takahiro Waki Jan 13 '17 at 03:53
  • All the exact solutions in the general case can be found here: http://viXra.org/abs/2004.0456 – Γιώργος Πλούσος Apr 21 '20 at 14:35

3 Answers3


Here is a possible explanation. Let $\alpha \in [0, \pi/2]$ and define $\epsilon_1, \epsilon_2, \cdots$ by $ \epsilon_i = \operatorname{sgn}( \cos ( 2^i \alpha )) \in \{-1, 1\}$. Here, we take the convention that $\operatorname{sgn}(0) =1 $. Then applying the identity $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt{2 + 2\cos(2\theta)}$ repeatedly, we have

$$ 2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ \cdots + \epsilon_n \sqrt{2 + \smash[b]{2\cos(2^{n+1} \alpha)} }}}}. $$

This can be used to show that, with an appropriate definition of infinite nested radical, the following identity

$$ 2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ 2 + \cdots }}} $$

is true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\cos\alpha$, then

  • $\epsilon_1 = \operatorname{sgn}(2\cos (2\alpha)) = \operatorname{sgn}(x^2 - 2)$,
  • $\epsilon_2 = \operatorname{sgn}(2\cos (4\alpha)) = \operatorname{sgn}((x^2 - 2)^2 - 2)$,

and likewise. This explains why signs are determined by OP's algorithm.

Sangchul Lee
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    Wow, this is a great explanation, which also shows where $2$ comes from – Yuriy S Apr 06 '16 at 22:47
  • @YuriyS, Thank you! And yes, this explains why the range is $[0, 2]$. Unfortunately this seems not work for other cases with $a > 2$ instead of $2$, and I even suspect that the analogous claim may be false. (For an an analogous situation, we can in fact check this. Numbers of the form $\pm \frac{1}{2} \pm \frac{1}{4} \pm \frac{1}{8} \pm \cdots $ represents any real number in $[-1, 1]$, but the range of $\pm \frac{1}{3} \pm \frac{1}{9} \pm \frac{1}{27} \pm \cdots $ is the (shifted) Cantor set.) – Sangchul Lee Apr 06 '16 at 22:59
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    Nice answer (+1). This is theme of problems 183 - 185 in the classic *[Problems and Theorems in Analysis I](http://www.amazon.com/Problems-Theorems-Analysis-Functions-Mathematics/dp/3540636404?ie=UTF8&keywords=Polya%20szegO&qid=1460217036&ref_=sr_1_1&sr=8-1)* by G. Polya and G. Szegö – epi163sqrt Apr 09 '16 at 16:02
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    This is incredible. – 6005 Jul 26 '16 at 21:08

Peculiar observation

If we define a binary number $b = b_1b_2\cdots b_n$ with digits mapped to the symbols like this: $$b_k = \begin{cases}0 \text{ if } (-) \text{ at position } k\\1 \text{ if } (+) \text{ at position } k\end{cases}$$ Then if we run the algorithm proposed in the question, looping

x(k) = x(k-1)^2-2;
b(k) = (x(k)>0);

the vector b will get logical values corresponding to bits 1 and 0 of the binary number above and we can calculate it for the linear space of $x\in[0,1]$. If we do this we can then calculate each number as the scalar product $$[1/2,1/4,\cdots,1/2^k]b$$ and if we then plot it, it will look like

enter image description here Which is kind of a peculiar plot having a bit of a discontinuous and fractal structure. I think the largest discontinuity is around $x = \sqrt{1/2}$ but I have no theoretical explanation why..

edit as pointed out by Sangchul Lee this seems similar to Tent Map

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    Good observation. I am not expert of symbolic dynamics, but this is related to the [*tent map orbits*](https://en.m.wikipedia.org/wiki/Tent_map#/media/File%3ATent-map.png). Basically this is because the string of signs are produced by the doubling transform on the circle via the correspondance I explained in my answer. – Sangchul Lee Apr 08 '16 at 02:58

Your algorithm pretty much shows that there exists an expansion for every number in $[0,2]$.

If we replace $2$ by $a>2$ (and keep the square root), we will fail because we need that squaring a number from the interval $[u,v]$ produces a number that is either in $a+[u,v]$ or in $a-[u,v]$. So we must have $u=0$ and $v\ge a$ and $v^2\le a+v$. The last two imply $a\le v\le 2$.

If we additionally switch to $k$th roots, the condition becomes that $v\ge a$ and $v^k\le a+v$, hence $a\le v\le\sqrt[k-1]2$.

Hagen von Eitzen
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