Nested radical explained

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

$$\backslash sqrt,$$

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

$$\backslash sqrt[3].$$

Denesting

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

$$\backslash sqrt\; =\; 1+\backslash sqrt\backslash ,,$$

$$\backslash sqrt[3]\; =\; \backslash frac\; \backslash ,.$$

Another simple example,

$$\backslash sqrt[3]\; =\; \backslash sqrt[6]$$

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.^[1]

If and are rational numbers and is not the square of a rational number, there are two rational numbers and such that$$\backslash sqrt\; =\; \backslash sqrt\backslash pm\backslash sqrt$$if and only if

a ^{2-c~} is the square of a rational number .

If the nested radical is real, and are the two numbers$$\backslash frac2~$$ and

~ (a - d)/(2 ~,~) where ~d = \sqrt{a ^{2-c}~} is a rational number.

In particular, if and are integers, then and are integers.

This result includes denestings of the form$$\backslash sqrt=z\backslash pm\backslash sqrt~,$$as 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$$\backslash sqrt\; =\; \backslash alpha+\; \backslash beta\backslash sqrt+\backslash gamma\backslash sqrt+\backslash delta\backslash sqrt\; x\backslash sqrt\; y~.$$However, Galois theory implies that either the left-hand side belongs to

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 and \gamma = \delta =0 . In the second case, \alpha and another coefficient must be zero. If \beta =0 , one may rename as 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 $$\backslash sqrt\; =\; \backslash sqrt\backslash pm\backslash sqrt$$is equivalent with $$a+\backslash sqrt=x+y\backslash pm\; 2\backslash sqrt,$$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 and, solving the first equation in and is equivalent with solving $$a+\backslash sqrt=x+y\backslash pm\; 2\backslash sqrt.$$

This equality implies that

\sqrt{xy} belongs to the quadratic field 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 and 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$$a+\backslash sqrt=x+y\; +\backslash beta\backslash sqrt$$for some rational number \beta. The uniqueness of the decomposition over and \sqrt c implies thus that the considered equation is equivalent with$$a=\; x+y\backslash quad\; \backslash text\backslash quad\; \backslash pm\; 2\backslash sqrt\; =\; \backslash sqrt\; c.$$It follows by Vieta's formulas that and must be roots of the quadratic equation$$z^2-az+\backslash frac\; c4\; =\; 0~;$$its ~\Delta = a ^{2-c} = d ^{2} > 0 ~ (otherwise would be the square of), hence and must be$$\backslash frac~$$ and ~ (a - \sqrt{a ^{2-c})/() }~.Thus and 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 . The equation

a ^{2=c+d} ^{2} shows that . Thus, if the nested radical is real, and if denesting is possible, then . Then the solution is

Some identities of Ramanujan

Srinivasa Ramanujan demonstrated a number of curious identities involving nested radicals. Among them are the following:^[2]

$$\backslash sqrt[4]\; =\; \backslash frac=\backslash tfrac12\backslash left(3+\backslash sqrt[4]5+\backslash sqrt5+\backslash sqrt[4]\backslash right),$$

$$\backslash sqrt\; =\; \backslash tfrac13\backslash left(\backslash sqrt[3]\; -\; \backslash sqrt[3]\; -1\backslash right),$$

$$\backslash sqrt[3]\; =\; \backslash sqrt[5]\; +\; \backslash sqrt[5]\; -\; \backslash sqrt[5],$$

and

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. Landau's algorithm involves complex roots of unity and runs in exponential time with respect to the depth of the nested radical.^[4]

In trigonometry

See main article: Exact trigonometric values.

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

and$$\backslash sin\backslash frac=\backslash sin\; 7.5^\backslash circ=\backslash frac\; \backslash sqrt\; =\; \backslash frac\; \backslash sqrt\; .$$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$$x=\backslash sqrt[3]\; +\backslash sqrt[3].$$

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$$x=\backslash sqrt[3]\; +\; \backslash sqrt[3]\; .$$

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$$x=\; \backslash sqrt$$

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

$$x\; =\; \backslash sqrt.$$

If we solve this equation, we find that (the second solution doesn't apply, under the convention that the positive square root is meant). This approach can also be used to show that generally, if, then$$\backslash sqrt\; =\; \backslash tfrac\backslash left(1\; +\; \backslash sqrt\; \backslash right)$$

and is the positive root of the equation . For, this root is the golden ratio, approximately equal to 1.618. The same procedure also works to obtain, if,$$\backslash sqrt\; =\; \backslash tfrac12\backslash left(-1\; +\; \backslash sqrt\; \backslash right),$$which is the positive root of the equation .

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^[5] $$R(b\_k,\; \backslash ldots,\; b\_1)=\backslash frac\; \backslash sqrt$$

where

x =2 \sin(\pi b _{1/4)} with b _{1} in [−2,2] and b _{i\in} \{ -1 , 0 , 1 \} for i ≠ 1 , are such that R(b _{k,} \ldots, b _{1)=\cos} \theta for$$\backslash theta=\backslash left(\backslash frac-\backslash frac-\backslash frac-\backslash frac-\backslash cdots-\backslash frac\backslash right)\; \backslash pi\; .$$

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$$R\_k(x)=\backslash sqrt.$$

If

x \geq 2 , then^[6]

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$$\backslash pi=\backslash lim\; \_\backslash left[\backslash frac\{2^\{k+1\}\}\{2-b\_\{1\}\}\; R(\backslash underbrace\{1,-1,1,1,\; \backslash ldots,\; 1,1,\; b\_1\}\_\{k\; \backslash text\; \{\; terms\; \}\})\backslash right]$$

where

b _{1 ≠} 2 .

Ramanujan's infinite radicals

Ramanujan posed the following problem to the Journal of Indian Mathematical Society:

$$?\; =\; \backslash sqrt.$$

This can be solved by noting a more general formulation:$$?\; =\; \backslash sqrt.$$

Setting this to and squaring both sides gives us$$F(x)^2\; =\; ax+(n+a)^2\; +x\backslash sqrt,$$

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,

$$F(x)\; =\; .$$

So, setting,, and , we have$$3\; =\; \backslash sqrt.$$Ramanujan stated the following infinite radical denesting in his lost notebook:$$\backslash sqrt=\backslash frac.$$The repeating pattern of the signs is

( + , + , - , + ).

Viète's expression for

Viète's formula for , the ratio of a circle's circumference to its diameter, is$$\backslash frac\backslash pi=\backslash frac\backslash cdot\backslash frac2\backslash cdot\backslash frac2\backslash cdots.$$

Cube roots

In certain cases, infinitely nested cube roots such as$$x\; =\; \backslash sqrt[3]$$can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation$$x\; =\; \backslash sqrt[3].$$

If we solve this equation, we find that . More generally, we find that$$\backslash sqrt[3]$$is the positive real root of the equation for all . For, this root is the plastic ratio ρ, approximately equal to 1.3247.

The same procedure also works to get

$$\backslash sqrt[3]$$

as the real root of the equation for all .

Herschfeld's convergence theorem

An infinitely nested radical

\sqrt{a _{1} + \sqrt{a _{2} + ... b}} (where all a _{i} are nonnegative) converges if and only if there is some M \in R such that M \geq a ^{2 ^{-n}} _{n} for all n ,^[7] or in other words

Proof of "if"

We observe that Moreover, the sequence

\left(\sqrt{a _{1} + \sqrt{a _{2} + ... c \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} + … \sqrt{a _{n}}}\right)} converges, then it is bounded.

However,

a ^{2 ^{-n}} _{n} \le\sqrt{a _{1} + \sqrt{a _{2} + … \sqrt{a _{n}}}} , hence \left(a ^{2 ^{-n}} _{n} \right) is also bounded.

See also

• Exponentiation
• Sum of radicals

References

Further reading

• Susan Landau . Susan . Landau . How to Tangle with a Nested Radical . . 16 . 2 . 49–55 . 1994 . 10.1007/bf03024284. 119991567 .
  • Decreasing the Nesting Depth of Expressions Involving Square Roots
• Simplifying Square Roots of Square Roots

Notes and References

1. Book: Euler, Leonhard . Elements of algebra . Springer Science & Business Media . 2012 . Chapter VIII.
2. Web site: A note on 'Zippel Denesting' . Susan . Landau . Susan Landau . July 16, 1993. 10.1.1.35.5512 . August 23, 2023.
3. Book: Landau, Susan . Susan Landau

   . 10.1.1.34.2003 . Simplification of Nested Radicals . Susan Landau . 30th Annual Symposium on Foundations of Computer Science . . 21 . 1992 . 85–110 . 10.1109/SFCS.1989.63496 . 978-0-8186-1982-3 . 29982884 .

4. Gkioulekas. Eleftherios. 2017-08-18. On the denesting of nested square roots . International Journal of Mathematical Education in Science and Technology. en. 48. 6. 942–953. 10.1080/0020739X.2017.1290831 . 2017IJMES..48..942G . 9737528 . 0020-739X.
5. Servi. L. D.. April 2003. Nested Square Roots of 2 . The American Mathematical Monthly . en. 110. 4. 326–330. 10.1080/00029890.2003.11919968. 38100940 . 0002-9890.
6. Nyblom. M. A.. November 2005. More Nested Square Roots of 2. The American Mathematical Monthly . en . 112 . 9 . 822–825 . 10.1080/00029890.2005.11920256 . 11206345 . 0002-9890.
7. Herschfeld. Aaron. 1935. On Infinite Radicals. The American Mathematical Monthly . 42. 7. 419–429. 10.2307/2301294. 2301294. 0002-9890.