Nested radical

 ( Algebra ) 

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

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

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

$$\backslash sqrt3\{2+\backslash sqrt\{3\}+\backslash sqrt3\{4\}\backslash \; \}.$$

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

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

= \sqrt{3},\quad}}
     

$$\backslash sqrt3\{\backslash sqrt3\{2\}\; -\; 1\}\; =\; \backslash frac\{1\; -\; \backslash sqrt3\{2\}\; +\; \backslash sqrt3\{4\}\}\{\backslash sqrt3\{9\}\}\; \backslash ,.$$

Another simple example,

$$\backslash sqrt3\{\backslash sqrt\{2\}\}\; =\; \backslash sqrt6\{2\}$$

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

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Two nested square roots In the case of two nested square roots, the following theorem completely solves the problem of denesting. In this section the use of the radical sign denotes, as usual, the positive square root of the radicand, which is supposed to be a positive real number,

One has two nested square roots with an expression of the form $$\backslash sqrt\{\backslash frac\{\backslash alpha\; +\; \backslash beta\backslash sqrt\; r\}\{\backslash gamma\; +\; \backslash delta\backslash sqrt\; r\}\},$$ where all variables denote rational numbers. By rationalizing the fraction, one gets an expression of the form $$\backslash sqrt\{a\backslash pm\; b\backslash sqrt\; r\},$$ where are and . By putting , the problem is reduced to denest an expression of the form $$\backslash sqrt\{a\backslash pm\backslash sqrt\; c\}.$$

&=\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},

The theorem will be proved in two separate steps. Firstly, one has a denested formula of the form if and only if . This will be done with elementary algebraic computation. The second step consists to prove that more complicated denested formulas, such as , can always be reduced to the form given in the theorem. This involves basic Galois theory.

Proof that the given denesting formula is correct: If and , all radicands are positive in the given formulas. This is almost immediate for the left-hand sides. For the right-hand sides, this results from the positiveness of the sum and the product of the two radicands that are respectively and . Then, since two positive numbes are equal if and only if their squares are equal, it suffices to square everything and apply the binomial formula.

Proof that, if denesting is possible into a sum or difference of square roots, then and . By squaring, the equation $$\backslash sqrt\{a\backslash pm\backslash sqrt\{c\}\}\; =\; \backslash sqrt\{x\}\backslash pm\backslash sqrt\{y\}$$ implies $$a\backslash pm\backslash sqrt\{c\}=x+y\backslash pm\; 2\backslash sqrt\{xy\}.$$ This implies that is an element of the field , and can thus be uniquely written , with and rational numbers. One has since, otherwise one would have an irrational number equal to a rational number. One has $$xy=(\backslash alpha+\backslash beta\backslash sqrt\; c)^2=\backslash alpha^2+c\backslash beta^2+2\backslash alpha\backslash beta\backslash sqrt\; c.$$ As is a rational number and , one has . It results that and $$a+\backslash sqrt\{c\}=x+y\backslash pm\; 2\backslash beta\; \backslash sqrt\; c.$$ Thus , , and . This implies by Vieta's formula that and are the two roots ot the equation $$X^2-aX+\backslash frac\; c4.$$ As these roots must be rational and positive, this implies that and is the square of a rational number. Then, the quadratic formula provides the solution as given in the theorem.

Other forms of denesting:

A denesting of the form $$\backslash sqrt\{a+\backslash sqrt\{c\}\}\; =\; \backslash alpha\backslash sqrt\{r\}+\backslash beta\backslash sqrt\{s\}$$ can be reduced to the above form: setting and , one gets $$\backslash sqrt\{a+\backslash sqrt\{c\}\}\; =\; \backslash pm\backslash sqrt\{x\}\backslash pm\backslash sqrt\{y\}.$$ Since the left member of the equality is positive, at least one of the "" must be positive, and one gets the above form after exchanging, if needed, and .

If $\backslash sqrt\{a\backslash pm\backslash sqrt\{c\}\; \}\; \backslash in\; \backslash Q(\backslash sqrt\; c),$ there are rational numbers and such that $$\backslash sqrt\{a\backslash pm\; \backslash sqrt\{c\}\}\; =\; \backslash alpha\; +\; \backslash beta\; \backslash sqrt\; c=\backslash pm\; \backslash sqrt\{\backslash alpha^2\}\backslash pm\; \backslash sqrt\{\backslash beta^2\; c\}.$$ So, the theorem applies to test whether an element of has a square root in , and, if it has one, to compute it.

A general denesting formula would be an expression of $\backslash sqrt\{a\backslash pm\backslash sqrt\{c\}\}$ as an element of a field , where are rational numbers. One may suppose that the are distinct , by field extension with the square roots of the prime numbers that appear as factors of the numerator or the denominator of some , and then removing the original , since they can be expressed as product and quotients of the new square roots. If the are distinct , is a Galois extension of with as Galois group. An element of this Galois groupe is a -tuple of 0 and 1, and the corresponding automorphism consists to change the signs of the such the th element of the tuple is 1.

Excluding the already considered case $\backslash sqrt\{a\backslash pm\backslash sqrt\{c\}\; \}\; \backslash in\; \backslash Q(\backslash sqrt\; c),$ the field is a subfield of of degree four over of . It follows that it is a galois extension of , with the Klein four-group as its Galois group. The automorphisms in this Galois group are the identity, two involutions and whose square are the identity, and their product . The elements of that are fixed by for a quadratic extension of , which is thus of the form , where . Similarly, the elements fixed b generate a field , and one has .

This shows that when denesting is possible, it is possible with at most two radicals.

As for every element of , one may write uniquely $$\backslash sqrt\{a\backslash pm\backslash sqrt\{c\}\; \}=\backslash alpha\; +\backslash beta\backslash sqrt\; x\; +\backslash gamma\; \backslash sqrt\; y+\backslash delta\backslash sqrt\; x\; \backslash sqrt\; y,$$ with . There exists an automorphism that changes the sign of the left member. It must change the signs of all nonzero terms of the right member|. It follows that and at least one of must be zero. Indeed, if and , one must have , because the change of sign of the two first radicals implies that the sign of their product is not changed.

So, setting , and, if needed, permuting and renaming the variables, we are reduced to a denesting into a sum or difference of two square roots that has already been considered.

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

$$\backslash sqrt4\{\backslash frac\{3\; +\; 2\; \backslash sqrt4\{5\}\}\{3\; -\; 2\; \backslash sqrt4\{5\}\}\}\; =\; \backslash frac\{\; \backslash sqrt4\{5\}\; +\; 1\}\{\backslash sqrt4\{5\}\; -\; 1\}=\backslash tfrac12\backslash left(3+\backslash sqrt45+\backslash sqrt5+\backslash sqrt4\{125\}\backslash right),$$

$$\backslash sqrt\{\; \backslash sqrt3\{28\}\; -\; \backslash sqrt3\{27\}\}\; =\; \backslash tfrac13\backslash left(\backslash sqrt3\{98\}\; -\; \backslash sqrt3\{28\}\; -1\backslash right),$$

$$\backslash sqrt3\{\; \backslash sqrt5\{\backslash frac\{32\}\{5\}\}\; -\; \backslash sqrt5\{\backslash frac\{27\}\{5\}\}\; \}\; =\; \backslash sqrt5\{\backslash frac\{1\}\{25\}\}\; +\; \backslash sqrt5\{\backslash frac\{3\}\{25\}\}\; -\; \backslash sqrt5\{\backslash frac\{9\}\{25\}\},$$

and

- \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}.\quad}}
     

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Landau's algorithm In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested and denesting them when possible.

Susan Landau (1992). 9780818619823, SIAM. ISBN 9780818619823

Earlier algorithms worked in some cases but not others. Landau's algorithm is based on field theory, Galois theory, and polynomial factorization over algebraic field extensions. It runs in exponential time with respect to the depth of nested radicals.

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In trigonometry In trigonometry, the sines and cosines of many angles can be expressed in terms of nested radicals. For example, $$\backslash sin\backslash frac\{\backslash pi\}\{60\}=\backslash sin\; 3^\backslash circ=\backslash frac\{1\}\{16\}\; \backslash left2(1-\backslash sqrt3)\backslash sqrt\{5+\backslash sqrt5\}+\backslash sqrt2(\backslash sqrt5-1)(\backslash sqrt3+1)\backslash right$$

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

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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 $$x=\backslash sqrt3\{-\{q\backslash over\; 2\}+\; \backslash sqrt\}\; +\backslash sqrt3\{-\{q\backslash over\; 2\}-\; \backslash sqrt\}.$$

In the case in which the cubic has only one real root, the real root is given by this expression with the 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 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 $$x=\backslash sqrt3\{-3+\backslash frac\{10i\backslash sqrt\{3\}\}\{9\}\}\; +\; \backslash sqrt3\{-3-\backslash frac\{10i\backslash sqrt\{3\}\}\{9\}\}\; .$$

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.

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Infinitely nested radicals

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Square roots Under certain conditions infinitely nested square roots such as $$x=\; \backslash sqrt\{2+\backslash sqrt\{2+\backslash sqrt\{2+\backslash sqrt\{2+\backslash cdots\}\}\}\}$$

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

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

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 $$\backslash sqrt\{n+\backslash sqrt\{n+\backslash sqrt\{n+\backslash sqrt\{n+\backslash cdots\}\}\}\}\; =\; \backslash tfrac\{1\}\{2\}\backslash left(1\; +\; \backslash sqrt\; \{1+4n\}\backslash right)$$

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 , $$\backslash sqrt\{n-\backslash sqrt\{n-\backslash sqrt\{n-\backslash sqrt\{n-\backslash cdots\}\}\}\}\; =\; \backslash tfrac12\backslash left(-1\; +\; \backslash sqrt\; \{1+4n\}\backslash right),$$ which is the positive root of the equation .

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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 $$R(b\_k,\; \backslash ldots,\; b\_1)=\backslash frac\{b\_k\}\{2\}\; \backslash sqrt\{2+b\_\{k-1\}\; \backslash sqrt\{2+b\_\{k-2\}\; \backslash sqrt\{2+\backslash cdots+b\_2\; \backslash sqrt\{2+x\}\}\}\}$$

where $x=2\; \backslash sin(\backslash pi\; b\_1/4)$ with $b\_1$ in −2,2 and $b\_i\backslash in\; \backslash \{-1,0,1\backslash \}$ for $i\; \backslash neq\; 1$, are such that $R(b\_k,\; \backslash ldots,\; b\_1)=\backslash cos\; \backslash theta$ for $$\backslash theta=\backslash left(\backslash frac\{1\}\{2\}-\backslash frac\{b\_k\}\{4\}-\backslash frac\{b\_k\; b\_\{k-1\}\}\{8\}-\backslash frac\{b\_k\; b\_\{k-1\}\; b\_\{k-2\}\}\{16\}-\backslash cdots-\backslash frac\{b\_k\; b\_\{k-1\}\; \backslash cdots\; b\_1\}\{2^\{k+1\}\}\backslash right)\; \backslash pi\; .$$

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

If $x\; \backslash geq\; 2$, then

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

where $b\_1\backslash neq\; 2$.

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Ramanujan's infinite radicals Ramanujan posed the following problem to the Journal of Indian Mathematical Society:

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

This can be solved by noting a more general formulation: $$?\; =\; \backslash sqrt\{ax+(n+a)^2\; +x\backslash sqrt\{a(x+n)+(n+a)^2+(x+n)\; \backslash sqrt\{\backslash mathrm\{\backslash cdots\}\}\}\}.$$

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

which can be simplified to $$F(x)^2\; =\; ax\; +\; (n+a)^2\; +\; x\; F(x+n)\; .$$

It can be shown that

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

satisfies the equation for $F(x)$, 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 $F(x)$.

So, setting , , and , we have $$3\; =\; \backslash sqrt\{1+2\backslash sqrt\{1+3\; \backslash sqrt\{1+\backslash cdots\}\}\}.$$ Ramanujan stated the following infinite radical denesting in his lost notebook: $$\backslash sqrt\{5+\backslash sqrt\{5+\backslash sqrt\{5-\backslash sqrt\{5+\backslash sqrt\{5+\backslash sqrt\{5+\backslash sqrt\{5-\backslash cdots\}\}\}\}\}\}\}=\backslash frac\{2+\backslash sqrt\{5\}+\backslash sqrt\{15-6\backslash sqrt\{5\}\}\}\{2\}.$$ The repeating pattern of the signs is $(+,+,-,+).$

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

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Cube roots In certain cases, infinitely nested cube roots such as $$x\; =\; \backslash sqrt3\{6+\backslash sqrt3\{6+\backslash sqrt3\{6+\backslash sqrt3\{6+\backslash cdots\}\}\}\}$$ can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation $$x\; =\; \backslash sqrt3\{6+x\}.$$

If we solve this equation, we find that . More generally, we find that $$\backslash sqrt3\{n+\backslash sqrt3\{n+\backslash sqrt3\{n+\backslash sqrt3\{n+\backslash cdots\}\}\}\}$$ 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

$$\backslash sqrt3\{n-\backslash sqrt3\{n-\backslash sqrt3\{n-\backslash sqrt3\{n-\backslash cdots\}\}\}\}$$

as the real root of the equation for all .

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Herschfeld's convergence theorem An infinitely nested radical $\backslash sqrt\{a\_1\; +\; \backslash sqrt\{a\_2\; +\; \backslash dotsb\}\}$ (where all $a\_i$ are nonnegative) converges if and only if there is some $M\; \backslash in\; \backslash mathbb\; R$ such that $M\; \backslash geq\; a\_n^\{2^\{-n\}\}$ for all $n$, or in other words

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Proof of "if" We observe that Moreover, the sequence $\backslash left(\backslash sqrt\{a\_1\; +\; \backslash sqrt\{a\_2\; +\; \backslash dotsc\; \backslash sqrt\{a\_n\}\}\}\backslash right)$ is monotonically increasing. Therefore it converges, by the monotone convergence theorem.

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Proof of "only if" If the sequence $\backslash left(\backslash sqrt\{a\_1\; +\; \backslash sqrt\{a\_2\; +\; \backslash cdots\; \backslash sqrt\{a\_n\}\}\}\backslash right)$ converges, then it is bounded.

However, $a\_n^\{2^\{-n\}\}\backslash le\backslash sqrt\{a\_1\; +\; \backslash sqrt\{a\_2\; +\; \backslash cdots\; \backslash sqrt\{a\_n\}\}\}$, hence $\backslash left(a\_n^\{2^\{-n\}\}\backslash right)$ is also bounded.

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See also

• Exponentiation
• Sum of radicals

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Further reading

• Decreasing the Nesting Depth of Expressions Involving Square Roots
• Simplifying Square Roots of Square Roots

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