Magnitude (astronomy)

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:

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

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

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

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

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

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In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested. 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:

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

The same procedure also works to get

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

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

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

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

${\sqrt {5+{\sqrt {5+{\sqrt {5-{\sqrt {5+{\sqrt {5+{\sqrt {5+{\sqrt {5-\cdots }}}}}}}}}}}}}}={\frac {2+{\sqrt {5}}+{\sqrt {15-6{\sqrt {5}}}}}{2}}$

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. For n = 1, this root is the plastic number ρ, approximately equal to 1.3247.

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.

See also

• Sum of radicals
• Spiral of Theodorus
• Viète's formula

References

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