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 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 a − e for d one obtains

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

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

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

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

The solution d is the algebraic conjugate of e. If

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

then

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

However, this approach works for nested radicals of the form ${\sqrt {a+b{\sqrt {c}}\ }}$ if and only if ${\sqrt {a^{2}-b^{2}c}}$ 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

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}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right),

{\sqrt {{\sqrt[{3}]{28}}-{\sqrt[{3}]{27}}}}={\tfrac {1}{3}}\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

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

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 }}}}}}}}={\tfrac {1}{2}}\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 }}}}}}}}={\tfrac {1}{2}}\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