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{{Short description|Mathematical expression with inner radical}}
In [[algebra]], a '''nested radical''' is a [[radical expression]] (one containing a square root sign, cube root sign, etc.) that contains (nests) another radical expression. Examples include

<math display="block">\sqrt{5-2\sqrt{5}\ },</math>

which arises in discussing the [[pentagon|regular pentagon]], and more complicated ones such as

<math display="block">\sqrt[3]{2+\sqrt{3}+\sqrt[3]{4}\ }.</math>

== Denesting==

Some nested radicals can be rewritten in a form that is not nested. For example,

<math display=block>\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}\,,</math>

{{bi|left=1.6|1=<math>\sqrt2+\sqrt{5-2\sqrt{6}} = \sqrt{3},\quad</math><ref>{{citation
 | last = Scheinerman | first = Edward R.
 | doi = 10.2307/2589344
 | issue = 6
 | journal = American Mathematical Monthly
 | jstor = 2589344
 | mr = 1766736
 | pages = 489–499
 | title = When close enough is close enough
 | volume = 107
 | year = 2000}}</ref>}}

<math display=block>\sqrt[3]{\sqrt[3]{2} - 1} = \frac{1 - \sqrt[3]{2} + \sqrt[3]{4}}{\sqrt[3]{9}} \,.</math>

Another simple example,

<math display=block>\sqrt[3]{\sqrt{2}} = \sqrt[6]{2}</math>

Rewriting a nested radical in this way is called '''denesting'''. This is not always possible, and, even when possible, it is often difficult.

==Two nested square roots==
In the case of two nested square roots, the following theorem completely solves the problem of denesting.<ref>{{cite book | last=Euler | first=Leonhard | title= Elements of algebra | publisher= Springer Science & Business Media | year= 2012 | at=Chapter VIII}}</ref>

If {{mvar|a}} and {{mvar|c}} are [[rational number]]s and {{mvar|c}} is not the square of a rational number, there are two rational numbers {{mvar|x}} and {{mvar|y}} such that
<math display="block">\sqrt{a+\sqrt{c}} = \sqrt{x}\pm\sqrt{y}</math>
if and only if <math>a^2-c~</math> is the square of a rational number {{mvar|d}}.

If the nested radical is real, {{mvar|x}} and {{mvar|y}} are the two numbers
<math display="block">\frac{a+d}2~</math> and <math>~\frac{a-d}2~,~</math> where <math>~d=\sqrt{a^2-c}~</math> is a rational number.

In particular, if {{mvar|a}} and {{mvar|c}} are integers, then {{math|2''x''}} and {{math|2''y''}} are integers.

This result includes denestings of the form
<math display="block">\sqrt{a+\sqrt{c}}=z\pm\sqrt{y}~,</math>
as {{mvar|z}} may always be written <math>z=\pm\sqrt{z^2},</math> and at least one of the terms must be positive (because the left-hand side of the equation is positive).

A more general denesting formula could have the form
<math display="block">\sqrt{a+\sqrt{c}} = \alpha+ \beta\sqrt{x}+\gamma\sqrt{y}+\delta\sqrt x\sqrt y~.</math>
However, [[Galois theory]] implies that either the left-hand side belongs to <math>\mathbb Q(\sqrt c),</math>  or it must be obtained by changing the sign of either <math>\sqrt x,</math> <math>\sqrt y,</math> or both. In the first case, this means that one can take {{math|1=''x'' = ''c''}} and <math>\gamma=\delta=0.</math> In the second case, <math>\alpha</math> and another coefficient must be zero. If <math>\beta=0,</math> one may rename {{math|''xy''}} as {{mvar|x}} for getting <math>\delta = 0.</math> Proceeding similarly if <math>\alpha=0,</math> it results that one can suppose <math>\alpha = \delta = 0.</math> This shows that the apparently more general denesting can always be reduced to the above one.

''Proof'': By squaring, the equation 
<math display="block">\sqrt{a+\sqrt{c}} = \sqrt{x}\pm\sqrt{y}</math>
is equivalent with 
<math display="block">a+\sqrt{c}=x+y\pm 2\sqrt{xy},</math>
and, in the case of a minus in the right-hand side, 
{{block indent | em = 1.5 | text = {{math|''{{abs|x}}'' ≥ ''{{abs|y}}''}},}}
(square roots are nonnegative by definition of the notation). As the inequality may always be satisfied by possibly exchanging {{mvar|x}} and {{mvar|y}}, solving the first equation in {{mvar|x}} and {{mvar|y}} is equivalent with solving  
<math display="block">a+\sqrt{c}=x+y\pm 2\sqrt{xy}.</math>

This equality implies that <math>\sqrt{xy}</math> belongs to the [[quadratic field]] <math>\mathbb Q(\sqrt c).</math> In this field every element may be uniquely written <math>\alpha +\beta\sqrt c,</math> with <math>\alpha</math> and <math>\beta</math> being rational numbers. This implies that <math>\pm 2\sqrt{xy}</math> is not rational (otherwise the right-hand side of the equation would be rational; but the left-hand side is irrational). As {{mvar|x}} and {{mvar|y}} must be rational, the square of <math>\pm 2\sqrt{xy}</math> must be rational. This implies that <math>\alpha=0</math> in the expression of <math>\pm 2\sqrt{xy}</math> as <math>\alpha +\beta\sqrt c.</math> Thus
<math display="block">a+\sqrt{c}=x+y +\beta\sqrt{c}</math>
for some rational number <math>\beta.</math>
The uniqueness of the decomposition over {{math|1}} and <math>\sqrt c</math> implies thus that the considered equation is equivalent with
<math display="block">a= x+y\quad \text{and}\quad \pm 2\sqrt{xy} = \sqrt c.</math>
It follows by [[Vieta's formulas]] that {{mvar|x}} and {{mvar|y}} must be roots of the [[quadratic equation]]
<math display="block">z^2-az+\frac c4 = 0~;</math>
its <math>~\Delta = a^2-c = d^2 > 0~</math> ({{math|≠ 0}}, otherwise {{mvar|c}} would be the square of {{mvar|a}}), hence {{mvar|x}} and {{mvar|y}} must be
<math display="block">\frac{a+\sqrt{a^2-c}}{2}~</math> and <math>~\frac{a-\sqrt{a^2-c}}{2}~.</math>
Thus {{mvar|x}} and {{mvar|y}} are rational if and only if <math>d=\sqrt{a^2-c}~</math> is a rational number.

For explicitly choosing the various signs, one must consider only positive real square roots, and thus assuming {{math|''c'' > 0}}. The equation <math>a^2=c+d^2</math> shows that {{math|''{{abs|a}}'' > {{sqrt|''c''}}}}. Thus, if the nested radical is real, and if denesting is possible, then {{math|''a'' > 0}}. Then the solution is
<math display="block">\begin{align}
\sqrt{a+\sqrt{c}}&=\sqrt{\frac{a+d}2}+\sqrt{\frac{a-d}2},\\[6pt]
\sqrt{a-\sqrt{c}}&=\sqrt{\frac{a+d}2}-\sqrt{\frac{a-d}2}.
\end{align}</math>

== Some identities of Ramanujan ==

[[Srinivasa Ramanujan]] demonstrated a number of curious identities involving nested radicals.  Among them are the following:<ref>{{cite web | title = A note on 'Zippel Denesting' | first = Susan | last = Landau | authorlink = Susan Landau |date=July 16, 1993| citeseerx = 10.1.1.35.5512 | url = https://citeseerx.ist.psu.edu/pdf/da359f4651fd157a1aa47e21e3ea9284bd5e7282 | access-date = August 23, 2023}}</ref>

<math display=block> \sqrt[4]{\frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}}} = \frac{ \sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}=\tfrac12\left(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125}\right),</math>

<math display=block> \sqrt{ \sqrt[3]{28} - \sqrt[3]{27}} = \tfrac13\left(\sqrt[3]{98} - \sqrt[3]{28} -1\right), </math>

<math display=block> \sqrt[3]{ \sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}} } = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}, </math>

and
{{bi|left=1.6|1=<math>\sqrt[3]{\ \sqrt[3]{2}\ - 1}= \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}.\quad</math><ref>{{cite journal |last1=Berndt |first1=Bruce |last2=Chan |first2=Heng |last3=Zhang |first3=Liang-Cheng |title=Radicals and units in Ramanujan's work |journal=Acta Arithmetica |date=1998 |volume=87 |issue=2 |pages=145–158 |url=https://faculty.math.illinois.edu/~berndt/articles/radicals.pdf|doi=10.4064/aa-87-2-145-158 |doi-access=free }}</ref>}}

== Landau's algorithm ==
{{expand section|date=February 2015}}
In 1989 [[Susan Landau]] introduced the first [[algorithm]] for deciding which nested radicals can be denested.<ref>{{cite book | citeseerx = 10.1.1.34.2003 | chapter = Simplification of Nested Radicals | authorlink = Susan Landau | first = Susan | last = Landau | title = 30th Annual Symposium on Foundations of Computer Science | publisher = [[Society for Industrial and Applied Mathematics|SIAM]] | volume = 21 | year = 1992 | pages = 85–110 | doi = 10.1109/SFCS.1989.63496 | isbn = 978-0-8186-1982-3 | s2cid = 29982884 }}</ref> Earlier algorithms worked in some cases but not others. Landau's algorithm involves complex [[roots of unity]] and runs in [[exponential time]] with respect to the depth of the nested radical.<ref>{{Cite journal|last=Gkioulekas|first=Eleftherios|date=2017-08-18|title=On the denesting of nested square roots | url=https://www.tandfonline.com/doi/full/10.1080/0020739X.2017.1290831|journal=International Journal of Mathematical Education in Science and Technology| language=en| volume=48| issue=6| pages=942–953| doi=10.1080/0020739X.2017.1290831 | bibcode=2017IJMES..48..942G | s2cid=9737528 | issn=0020-739X}}</ref>

==In trigonometry==
{{Main|Exact trigonometric values}}

In [[trigonometry]], the [[sine and cosine|sines and cosines]] of many angles can be expressed in terms of nested radicals. For example,
<math display=block>\sin\frac{\pi}{60}=\sin 3^\circ=\frac{1}{16} \left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]</math>

and
<math display=block>\sin\frac{\pi}{24}=\sin 7.5^\circ=\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt3}} = \frac{1}{2} \sqrt{2 - \frac{1 + \sqrt3}{\sqrt2}} .</math>
The last equality results directly from the results of {{slink||Two nested square roots}}.

==In the solution of the cubic equation==

Nested radicals appear in the [[algebraic solution]] of the [[cubic equation]]. Any cubic equation can be written in simplified form without a quadratic term, as

<math display="block">x^3+px+q=0,</math>

whose general solution for one of the roots is
<math display=block> x=\sqrt[3]{-{q\over 2}+ \sqrt{{q^2\over 4}+{p^3 \over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^2\over 4}+{p^3\over 27}}}.</math>

In the case in which the cubic has only one real root, the real root is given by this expression with the [[radicand]]s of the cube roots being real and with the cube roots being the real cube roots. In the case of three real roots, the square root expression is an imaginary number; here any real root is expressed by defining the first cube root to be any specific complex cube root of the complex radicand, and by defining the second cube root to be the [[complex conjugate]] of the first one. The nested radicals in this solution cannot in general be simplified unless the cubic equation has at least one [[rational number|rational]] solution. Indeed, if the cubic has three irrational but real solutions, we have the ''[[casus irreducibilis]]'', in which all three real solutions are written in terms of cube roots of complex numbers. On the other hand, consider the equation

<math display="block">x^3-7x+6=0,</math>

which has the rational solutions 1, 2, and&nbsp;−3. The general solution formula given above gives the solutions
<math display=block>x=\sqrt[3]{-3+\frac{10\sqrt{3}i}{9}} + \sqrt[3]{-3-\frac{10\sqrt{3}i}{9}} .</math>

For any given choice of cube root and its conjugate, this contains nested radicals involving complex numbers, yet it is reducible (even though not obviously so) to one of the solutions 1, 2, or –3.

== Infinitely nested radicals ==

=== Square roots ===

Under certain conditions infinitely nested square roots such as
<math display=block> x= \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}} </math>

represent rational numbers. This rational number can be found by realizing that ''x'' also appears under the radical sign, which gives the equation

<math display="block"> x = \sqrt{2+x}. </math>

If we solve this equation, we find that {{math|1=''x'' = 2}} (the second solution {{math|1=''x'' = −1}} doesn't apply, under the convention that the positive square root is meant). This approach can also be used to show that generally, if {{math|''n'' > 0}}, then
<math display=block> \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \tfrac{1}{2}\left(1 + \sqrt {1+4n}\right) </math>

and is the positive root of the equation {{math|1=''x''<sup>2</sup> − ''x'' − ''n'' = 0}}. For {{math|1=''n'' = 1}}, this root is the [[golden ratio]] {{math|φ}}, approximately equal to 1.618. The same procedure also works to obtain, if {{math|''n'' > 0}},
<math display=block> \sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}} = \tfrac12\left(-1 + \sqrt {1+4n}\right),</math>
which is the positive root of the equation {{math|1=''x''<sup>2</sup> + ''x'' − ''n'' = 0}}.

==== Nested square roots of 2 ====
The nested square roots of 2 are a special case of the wide class of infinitely nested radicals. There are many known results that bind them to [[Trigonometric functions|sines and cosines]]. For example, it has been shown that nested square roots of 2 as<ref name=":0">{{Cite journal|last=Servi|first=L. D.|date=April 2003|title=Nested Square Roots of 2 | url=https://www.tandfonline.com/doi/full/10.1080/00029890.2003.11919968|journal=The American Mathematical Monthly | language=en|volume=110|issue=4|pages=326–330|doi=10.1080/00029890.2003.11919968|s2cid=38100940 |issn=0002-9890}}</ref>
<math display=block>R(b_k, \ldots, b_1)=\frac{b_k}{2} \sqrt{2+b_{k-1} \sqrt{2+b_{k-2} \sqrt{2+\cdots+b_2 \sqrt{2+x}}}}</math>

where <math>x=2 \sin(\pi b_1/4)</math> with <math>b_1</math> in [−2,2] and <math>b_i\in \{-1,0,1\}</math> for <math>i \neq 1</math>, are such that <math>R(b_k, \ldots, b_1)=\cos \theta</math> for
<math display=block>\theta=\left(\frac{1}{2}-\frac{b_k}{4}-\frac{b_k b_{k-1}}{8}-\frac{b_k b_{k-1} b_{k-2}}{16}-\cdots-\frac{b_k b_{k-1} \cdots b_1}{2^{k+1}}\right) \pi .</math>

This result allows to deduce for any <math>x \in [-2,2]</math> the value of the following infinitely nested radicals consisting of k nested roots as
<math display=block>R_k(x)=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}.</math>

If <math>x \geq 2</math>, then<ref>{{Cite journal|last=Nyblom|first=M. A.|date=November 2005|title=More Nested Square Roots of 2|url=https://www.tandfonline.com/doi/full/10.1080/00029890.2005.11920256|journal=The American Mathematical Monthly | language=en | volume=112 | issue=9 | pages=822–825 | doi=10.1080/00029890.2005.11920256 | s2cid=11206345 |issn=0002-9890}}</ref>
<math display=block>\begin{aligned}
R_k(x) &=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}} \\
&=\left(\frac{x+\sqrt{x^2-4}}{2}\right)^{1 / 2^k}+\left(\frac{x+\sqrt{x^2-4}}{2}\right)^{-1 / 2^k}
\end{aligned}</math>

These results can be used to obtain some nested square roots representations of <math>\pi</math> . Let us consider the term <math>R\left(b_{k}, \ldots, b_{1}\right)</math> defined above. Then<ref name=":0" />
<math display=block>\pi=\lim _{k \rightarrow \infty}\left[\frac{2^{k+1}}{2-b_{1}} R(\underbrace{1,-1,1,1, \ldots, 1,1, b_1}_{k \text { terms }})\right]</math>

where <math>b_1\neq 2</math>.

====Ramanujan's infinite radicals====

[[Srinivasa Ramanujan|Ramanujan]] posed the following problem to the ''Journal of Indian Mathematical Society'':

<math display="block">? = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. </math>

This can be solved by noting a more general formulation:
<math display=block>? = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}}. </math>

Setting this to {{math|''F''(''x'')}} and squaring both sides gives us
<math display=block>F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}, </math>

which can be simplified to
<math display="block">F(x)^2 = ax + (n+a)^2 + x F(x+n) .</math>

It can be shown that

<math display="block">F(x) = {x + n + a}</math>

satisfies the equation for <math>F(x)</math>, so it can be hoped that
it is the true solution.  For a complete proof, we would need to show
that this is indeed the solution to the equation for <math>F(x)</math>.

So, setting {{math|1=''a'' = 0}}, {{math|1=''n'' = 1}}, and&nbsp;{{math|1=''x'' = 2}}, we have
<math display=block>3 = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. </math>
Ramanujan stated the following infinite radical denesting in his [[Ramanujan's lost notebook|lost notebook]]:
<math display=block>\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}.</math>
The repeating pattern of the signs is <math>(+,+,-,+).</math>

====Viète's expression for {{pi}}====

[[Viète's formula]] for [[pi|{{pi}}]], the ratio of a circle's circumference to its diameter, is
<math display=block>\frac{2}\pi=
\frac{\sqrt2}{2}\cdot
\frac{\sqrt{2+\sqrt2}}2\cdot
\frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots.</math>

=== Cube roots ===
{{Unreferenced section|date=November 2018}}
In certain cases, infinitely nested cube roots such as
<math display="block"> x = \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\cdots}}}} </math>
can represent rational numbers as well.  Again, by realizing that the whole expression appears inside itself, we are left with the equation
<math display="block"> x = \sqrt[3]{6+x}. </math>

If we solve this equation, we find that&nbsp;{{math|1=''x'' = 2}}. More generally, we find that
<math display="block"> \sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\cdots}}}}</math>
is the positive real root of the equation {{math|1=''x''<sup>3</sup> − ''x'' − ''n'' = 0}} for all&nbsp;{{math|''n'' > 0}}. For {{math|1=''n'' = 1}}, this root is the [[plastic ratio]] ''ρ'', approximately equal to 1.3247.

The same procedure also works to get

<math display="block"> \sqrt[3]{n-\sqrt[3]{n-\sqrt[3]{n-\sqrt[3]{n-\cdots}}}} </math>

as the real root of the equation {{math|1=''x''<sup>3</sup> + ''x'' − ''n'' = 0}} for all {{math|''n'' > 1}}.

=== Herschfeld's convergence theorem===
An infinitely nested radical <math>\sqrt{a_1 + \sqrt{a_2 + \dotsb}}</math> (where all <math>a_i</math> are [[nonnegative]]) converges if and only if there is some <math>M \in \mathbb R</math>  such that <math>M \geq a_n^{2^{-n}}</math> for all <math>n</math>,<ref>{{Cite journal|last=Herschfeld|first=Aaron|date=1935|title=On Infinite Radicals|journal=The American Mathematical Monthly | volume=42|issue=7|pages=419–429|doi=10.2307/2301294|jstor=2301294|issn=0002-9890}}</ref> or in other words <math display=inline> \sup a_n^{2^{-n}} <+\infty. </math>

==== Proof of "if" ====
We observe that 
<math display=block>\sqrt{a_1 + \sqrt{a_2 + \dotsb}} \leq \sqrt{M^{2^1} + \sqrt{M^{2^2} + \cdots}} = M\sqrt{1 + \sqrt{1 + \dotsb}}<2M.</math>
Moreover, the sequence <math>\left(\sqrt{a_1 + \sqrt{a_2 + \dotsc \sqrt{a_n}}}\right)</math> is monotonically increasing. Therefore it converges, by the [[monotone convergence theorem]].

==== Proof of "only if" ====
If the sequence <math>\left(\sqrt{a_1 + \sqrt{a_2 + \cdots \sqrt{a_n}}}\right)</math> converges, then it is bounded.

However, <math>a_n^{2^{-n}}\le\sqrt{a_1 + \sqrt{a_2 + \cdots \sqrt{a_n}}}</math>, hence <math>\left(a_n^{2^{-n}}\right)</math> is also bounded.

==See also==
*[[Exponentiation]]
*[[Sum of radicals]]

== References ==

{{reflist}}

===Further reading ===
* {{cite journal | authorlink = Susan Landau | first = Susan | last = Landau | title = How to Tangle with a Nested Radical | journal = [[Mathematical Intelligencer]] | volume = 16 | issue = 2 | pages = 49–55 | year = 1994 | doi=10.1007/bf03024284| s2cid = 119991567 }}
*
* [https://web.archive.org/web/20060312035635/http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf Decreasing the Nesting Depth of Expressions Involving Square Roots]
* [http://www.cybertester.com/data/denest.pdf Simplifying Square Roots of Square Roots]
* {{mathworld|urlname=SquareRoot|title=Square Root}}
* {{mathworld|urlname=NestedRadical|title=Nested Radical}}

[[Category:Algebra]]