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:

$\sqrt{5-2\sqrt{5}\ }$

which arises in discussing the regular pentagon;

$\sqrt{5+2\sqrt{6}\ },$

or more complicated ones such as:

$\sqrt[3]{2+\sqrt{3}+\sqrt[3]{4}\ }.$

## Denesting nested radicals

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

$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}\,,$
$\sqrt{5+2\sqrt{6}} = \sqrt{2}+\sqrt{3},$
$\sqrt[3]{\sqrt[3]{2} - 1} = \frac{1 - \sqrt[3]{2} + \sqrt[3]{4}}{\sqrt[3]{9}} \,.$

Rewriting a nested radical in this way is called denesting. This process is generally considered a difficult problem, although a special class of nested radical can be denested by assuming it denests into a sum of two surds:

$\sqrt{a\pm b \sqrt{c}\ } = \sqrt{d}\pm\sqrt{e}.$

Squaring both sides of this equation yields:

$a\pm b \sqrt{c} = d + e \pm 2 \sqrt{de}.$

This can be solved by finding two numbers such that their sum is equal to a and their product is b2c/4, or by equating coefficients of like terms—setting rational and irrational parts on both sides of the equation equal to each other. The solutions for e and d can be obtained by first equating the rational parts:

$a = d + e,$

which gives

$d = a - e,$
$e = a - d.$

For the irrational parts note that

$b\sqrt{c} = 2\sqrt{de},$

and squaring both sides yields

$b^2 c = 4de.$

By plugging in ad for e one obtains

$b^2 c = 4(a-d)d = 4ad - 4d^2.$

Rearranging terms will give a quadratic equation which can be solved for d using the quadratic formula:

$4d^2 - 4ad + b^2 c = 0,$
$d=\frac{a \pm \sqrt {a^2-b^2c}}{2}.$

Since a = d+e, the solution e is the algebraic conjugate of d. If we set

$d=\frac{a + \sqrt {a^2-b^2c}}{2},$

then

$e=\frac{a - \sqrt {a^2-b^2c}}{2}.$

However, this approach works for nested radicals of the form $\sqrt{a\pm b \sqrt{c}\ }$ if and only if $\sqrt{a^2 - b^2c}$ is a rational number, in which case the nested radical can be denested into a sum of surds.

In some cases, higher-power radicals may be needed to denest the nested radical.

### Some identities of Ramanujan

Srinivasa Ramanujan demonstrated a number of curious identities involving denesting of radicals. Among them are the following:[1]

$\sqrt[4]{\frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}}} = \frac{ \sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}=\tfrac12\left(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125}\right),$
$\sqrt{ \sqrt[3]{28} - \sqrt[3]{27}} = \tfrac13\left(\sqrt[3]{98} - \sqrt[3]{28} -1\right),$
$\sqrt[3]{ \sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}} } = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}},$
$\sqrt[3]{\ \sqrt[3]{2}\ - 1}= \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}.$ [2]

Other odd-looking radicals inspired by Ramanujan include:

$\sqrt[4]{49 + 20\sqrt{6}} + \sqrt[4]{49 - 20\sqrt{6}} = 2\sqrt{3},$
$\sqrt[3]{\left(\sqrt{2}+ \sqrt{3}\right)\left(5 - \sqrt{6}\right) + 3\left(2\sqrt{3} + 3\sqrt{2}\right)} = \sqrt{10 - \frac{13 - 5\sqrt{6}}{5 + \sqrt{6}}}.$

### Landau's algorithm

In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested.[3] Earlier algorithms worked in some cases but not others.

## In trigonometry

In trigonometry, the sines and cosines of many angles can be expressed in terms of nested radicals. For example,

$\sin\frac{\pi}{60}=\sin 3^\circ=\tfrac{1}{16} \left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]\,$

and

$\sin\frac{\pi}{24}=\sin 7.5^\circ=\tfrac{1}{2} \sqrt{2-\sqrt{2+\sqrt3}} = \tfrac{1}{2} \sqrt{2 - \tfrac{1 + \sqrt3}{\sqrt2}} .$

## 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

$x^3+px+q=0,$

whose general solution for one of the roots is

$x=\sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}};$

here the first cube root is defined to be any specific cube root of the radicand, and the second cube root is defined 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 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

$x^3-7x+6=0,$

which has the rational solutions 1, 2, and —3. The general solution formula given above gives the solutions

$x=\sqrt[3]{-3+\frac{10\sqrt{3}i}{9}} + \sqrt[3]{-3-\frac{10\sqrt{3}i}{9}} .$

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

$x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$

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

$x = \sqrt{2+x}.$

If we solve this equation, we find that x = 2 (the second solution 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 n > 0, then:

$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \tfrac12\left(1 + \sqrt {1+4n}\right)$

and is the real root of the equation x2 − x − n = 0. For n = 1, this root is the golden ratio φ, approximately equal to 1.618. The same procedure also works to get that

$\sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}} = \tfrac12\left(-1 + \sqrt {1+4n}\right).$

and is the real root of the equation x2 + x − n = 0. For n = 1, this root is the reciprocal of the golden ratio Φ, which is equal to φ − 1. This method will give a rational x value for all values of n such that

$n = x^2 + x. \,$

Ramanujan posed this problem to the 'Journal of Indian Mathematical Society':

$? = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \,$

This can be solved by noting a more general formulation:

$? = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}} \,$

Setting this to F(x) and squaring both sides gives us:

$F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}} \,$

Which can be simplified to:

$F(x)^2 = ax+(n+a)^2 +xF(x+n) \,$

It can then be shown that:

$F(x) = {x + n + a} \,$

So, setting a =0, n = 1, and x = 2:

$3 = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \,$

Ramanujan stated this radical in his lost notebook

$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$

(The repeating pattern of the signs is $(+,+,-,+)$

#### In Viète's expression for pi

Viète's formula for pi, the ratio of a circle's circumference to its diameter, is

$\frac2\pi= \frac{\sqrt2}2\cdot \frac{\sqrt{2+\sqrt2}}2\cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots.$

### Cube roots

In certain cases, infinitely nested cube roots such as

$x = \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\cdots}}}}$

can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation

$x = \sqrt[3]{6+x}.$

If we solve this equation, we find that x = 2. More generally, we find that

$\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\cdots}}}}$

is the real root of the equation x3 − x − n = 0 for all n > 0. For n = 1, this root is the plastic number ρ, approximately equal to 1.3247.

The same procedure also works to get

$\sqrt[3]{n-\sqrt[3]{n-\sqrt[3]{n-\sqrt[3]{n-\cdots}}}}$

as the real root of the equation x3 + x − n = 0 for all n and x where n > 0 and |x| ≥ 1.