Nested radical

In algebra, a nested radical is a radical expression that contains 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,

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

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

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:

$\sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e},$

Squaring both sides:

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

This can be solved by using the quadratic formula and setting rational and irrational parts on both sides of the equation equal to each other. The solutions for e and d are:

d = a − e

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

The solution d is the algebraic conjugate of e. If

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

then

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

However, this approach works for nested radicals of the form

$\sqrt{a+b \sqrt{c}\ }$

if and only if

$\sqrt{a^2 - b^2c}$

is an integer, 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.

Landau's algorithm

Main article: Landau's algorithm

In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested.^[1] Earlier algorithms worked in some cases but not others.

Some identities of Ramanujan

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

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

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

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

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:

$3 = \sqrt{1+2\sqrt{1+3 \sqrt{1+\cdots}}}. \,$ (Contributed by RSC)

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:

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

The same procedure also works to get

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

This method will give a rational x value for all values of n such that

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

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.

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. This root is the plastic number ρ, approximately equal to 1.3247. (Contributed by RSC)

See also

• Sum of radicals

References and external links

1. ^ S. Landau, "Simplification of Nested Radicals", SIAM Journal of Computation, volume 21 (1992), pages 85–110.[1]
2. ^ "A note on 'Zippel Denesting'", Susan Landau, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.35.5512&rep=rep1&type=pdf

• Susan Landau, "How to Tangle with a Nested Radical." Mathematical Intelligencer, 16, 49–55, 1994.
• Simplifcation of Nested Radicals, by Susan Landau
• Decreasing the Nesting Depth of Expressions Involving Square Roots
• Simplifying Square Roots of Square Roots
• Weisstein, Eric W., "Square Root" from MathWorld.
• Weisstein, Eric W., "Nested Radical" from MathWorld.