Nested 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

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

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

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

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

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

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

Another simple example,

$$\sqrt[3]{\sqrt{2}} = \sqrt[6]{2}$$

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.

If $a$ and $c$ are rational numbers and $c$ is not the square of a rational number, there are two rational numbers $x$ and $y$ such that $$\sqrt{a+\sqrt{c}} = \sqrt{x}\pm\sqrt{y}$$ if and only if $$a^2-c~$$ is the square of a rational number $d$.

If the nested radical is real, $x$ and $y$ are the two numbers $$\frac{a+d}2~$$ and $$~\frac{a-d}2~,~$$ where $$~d=\sqrt{a^2-c}~$$ is a rational number.

In particular, if $a$ and $c$ are integers, then $2x$ and $2y$ are integers.

This result includes denestings of the form $$\sqrt{a+\sqrt{c}}=z\pm\sqrt{y}~,$$ as $z$ may always be written $$z=\pm\sqrt{z^2},$$ 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 $$\sqrt{a+\sqrt{c}} = \alpha+ \beta\sqrt{x}+\gamma\sqrt{y}+\delta\sqrt x\sqrt y~.$$ However, Galois theory implies that either the left-hand side belongs to $$\mathbb Q(\sqrt c),$$ or it must be obtained by changing the sign of either $$\sqrt x,$$ $$\sqrt y,$$ or both. In the first case, this means that one can take $x = c$ and $$\gamma=\delta=0.$$ In the second case, $$\alpha$$ and another coefficient must be zero. If $$\beta=0,$$ one may rename $xy$ as $x$ for getting $$\delta = 0.$$ Proceeding similarly if $$\alpha=0,$$ it results that one can suppose $$\alpha = \delta = 0.$$ This shows that the apparently more general denesting can always be reduced to the above one.

Proof: By squaring, the equation $$\sqrt{a+\sqrt{c}} = \sqrt{x}\pm\sqrt{y}$$ is equivalent with $$a+\sqrt{c}=x+y\pm 2\sqrt{xy},$$ and, in the case of a minus in the right-hand side,

(square roots are nonnegative by definition of the notation). As the inequality may always be satisfied by possibly exchanging $x$ and $y$, solving the first equation in $x$ and $y$ is equivalent with solving $$a+\sqrt{c}=x+y\pm 2\sqrt{xy}.$$

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

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

Some identities of Ramanujan
Srinivasa Ramanujan demonstrated a number of curious identities involving nested radicals. Among them are the following:

$$ \sqrt[4]{\frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}}} = \frac{ \sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}=\tfrac12\left(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125}\right),$$

$$ \sqrt{ \sqrt[3]{28} - \sqrt[3]{27}} = \tfrac13\left(\sqrt[3]{98} - \sqrt[3]{28} -1\right), $$

$$ \sqrt[3]{ \sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}} } = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}, $$

and

Landau's algorithm
In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested. 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.

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

and $$\sin\frac{\pi}{24}=\sin 7.5^\circ=\frac{1}{2} \sqrt{2-\sqrt{2+\sqrt3}} = \frac{1}{2} \sqrt{2 - \frac{1 + \sqrt3}{\sqrt2}} .$$ The last equality results directly from the results of.

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

In the case in which the cubic has only one real root, the real root is given by this expression with the radicands 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 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.

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 $a > 0$ (the second solution $x = 2$ doesn't apply, under the convention that the positive square root is meant). This approach can also be used to show that generally, if $x = −1$, then $$ \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \tfrac{1}{2}\left(1 + \sqrt {1+4n}\right) $$

and is the positive root of the equation $n > 0$. For $x^{2} − x − n = 0$, this root is the golden ratio $n = 1$, approximately equal to 1.618. The same procedure also works to obtain, if $φ$, $$ \sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}} = \tfrac12\left(-1 + \sqrt {1+4n}\right),$$ which is the positive root of the equation $n > 0$.

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 sines and cosines. For example, it has been shown that nested square roots of 2 as $$R(b_k, \ldots, b_1)=\frac{b_k}{2} \sqrt{2+b_{k-1} \sqrt{2+b_{k-2} \sqrt{2+\cdots+b_2 \sqrt{2+x}}}}$$

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

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

If $$x \geq 2$$, then $$\begin{aligned} R_k(x) &=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}} \\ &=\left(\frac{x+\sqrt{x^2-4}}{2}\right)^{1 / 2^k}+\left(\frac{x+\sqrt{x^2-4}}{2}\right)^{-1 / 2^k} \end{aligned}$$

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

where $$b_1\neq 2$$.

Ramanujan's infinite radicals
Ramanujan posed the following 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 $x^{2} + x − n = 0$ 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 $$F(x)^2 = ax + (n+a)^2 + x F(x+n) .$$

It can then be shown that, assuming $$F$$ is analytic,

$$F(x) = {x + n + a}. $$

So, setting $F(x)$, $a = 0$, and $n = 1$, we have $$3 = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. $$ Ramanujan stated the following infinite radical denesting 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 $$(+,+,-,+).$$

Viète's expression for $\pi$
Viète's formula for π, the ratio of a circle's circumference to its diameter, is $$\frac{2}\pi= \frac{\sqrt2}{2}\cdot \frac{\sqrt{2+\sqrt2}}2\cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots.$$

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 positive real root of the equation $x = 2$ for all $x^{3} − x − n = 0$. For $n > 0$, this root is the plastic ratio 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 $n = 1$ for all $x^{3} + x − n = 0$.

Herschfeld's convergence theorem
An infinitely nested radical $$\sqrt{a_1 + \sqrt{a_2 + \dotsb}}$$ (where all $$a_i$$ are nonnegative) converges if and only if there is some $$M \in \mathbb R$$ such that $$M \geq a_n^{2^{-n}}$$ for all $$n$$, or in other words $ \sup a_n^{2^{-n}} <+\infty. $

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

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

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