Nested radical

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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[edit]

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 of this equation yields:

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 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 ae for d one obtains

 b^2 c = 4(a-e)e = 4ae - 4e^2.

Rearranging terms will give an quadratic equation which can be solved for e:

 4e^2 - 4ae + b^2 c = 0,
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},


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.

Some identities of Ramanujan[edit]

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:

 \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[edit]

Main article: 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.

Infinitely nested radicals[edit]

Square roots[edit]

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


Cube roots[edit]

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


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


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

See also[edit]


  1. ^ "A note on 'Zippel Denesting'", Susan Landau,
  3. ^ Landau, Susan (1992). "Simplification of Nested Radicals". Journal of Computation (SIAM) 21: 85–110. doi:10.1109/SFCS.1989.63496. CiteSeerX: 

Further reading[edit]