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

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 an 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]

${\sqrt[ {4}]{{\frac {3+2{\sqrt[ {4}]{5}}}{3-2{\sqrt[ {4}]{5}}}}}}={\frac {{\sqrt[ {4}]{5}}+1}{{\sqrt[ {4}]{5}}-1}}={\tfrac 12}\left(3+{\sqrt[ {4}]5}+{\sqrt 5}+{\sqrt[ {4}]{125}}\right),$

${\sqrt {{\sqrt[ {3}]{28}}-{\sqrt[ {3}]{27}}}}={\tfrac 13}\left({\sqrt[ {3}]{98}}-{\sqrt[ {3}]{28}}-1\right),$

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

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

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

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:

${\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n+\cdots }}}}}}}}={\tfrac 12}\left(1+{\sqrt {1+4n}}\right).$