Metallic mean

The metallic mean (also metallic ratio, metallic constant, or noble mean) of a natural number n is a positive real number, denoted here S n , {\displaystyle S_{n},} that satisfies the following equivalent characterizations:

• the unique positive real number x {\displaystyle x} such that x = n + 1 x {\textstyle x=n+{\frac {1}{x}}}
• the positive root of the quadratic equation x 2 − n x − 1 = 0 {\displaystyle x^{2}-nx-1=0}
• the number n + n 2 + 4 2 = 2 n 2 + 4 − n {\textstyle {\frac {n+{\sqrt {n^{2}+4}}}{2}}={\frac {2}{{\sqrt {n^{2}+4}}-n}}}
• the number whose expression as a continued fraction is [ n ; n , n , n , n , … ] = n + 1 n + 1 n + 1 n + 1 n + ⋱ {\displaystyle [n;n,n,n,n,\dots ]=n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+\ddots \,}}}}}}}}}

Metallic means are (successive) derivations of the golden (n = 1 {\displaystyle n=1}) and silver ratios (n = 2 {\displaystyle n=2}), and share some of their interesting properties. The term "bronze ratio" (n = 3 {\displaystyle n=3}) (cf. Golden Age and Olympic Medals) and even metals such as copper (n = 4 {\displaystyle n=4}) and nickel (n = 5 {\displaystyle n=5}) are occasionally found in the literature.

In terms of algebraic number theory, the metallic means are exactly the real quadratic integers that are greater than 1 {\displaystyle 1} and have − 1 {\displaystyle -1} as their norm.

The defining equation x 2 − n x − 1 = 0 {\displaystyle x^{2}-nx-1=0} of the nth metallic mean is the characteristic equation of a linear recurrence relation of the form x k = n x k − 1 + x k − 2 . {\displaystyle x_{k}=nx_{k-1}+x_{k-2}.} It follows that, given such a recurrence the solution can be expressed as

x k = a S n k + b ( − 1 S n ) k , {\displaystyle x_{k}=aS_{n}^{k}+b\left({\frac {-1}{S_{n}}}\right)^{k},}

where S n {\displaystyle S_{n}} is the nth metallic mean, and a and b are constants depending only on x 0 {\displaystyle x_{0}} and x 1 . {\displaystyle x_{1}.} Since the inverse of a metallic mean is less than 1, this formula implies that the quotient of two consecutive elements of such a sequence tends to the metallic mean, when k tends to the infinity.

For example, if n = 1 , {\displaystyle n=1,} S n {\displaystyle S_{n}} is the golden ratio. If x 0 = 0 {\displaystyle x_{0}=0} and x 1 = 1 , {\displaystyle x_{1}=1,} the sequence is the Fibonacci sequence, and the above formula is Binet's formula. If n = 1 , x 0 = 2 , x 1 = 1 {\displaystyle n=1,x_{0}=2,x_{1}=1} one has the Lucas numbers. If n = 2 , {\displaystyle n=2,} the metallic mean is called the silver ratio, and the elements of the sequence starting with x 0 = 0 {\displaystyle x_{0}=0} and x 1 = 1 {\displaystyle x_{1}=1} are called the Pell numbers.

Geometry

The defining equation x = n + 1 x {\textstyle x=n+{\frac {1}{x}}} of the nth metallic mean induces the following geometrical interpretation.

Consider a rectangle such that the ratio of its length L to its width W is the nth metallic ratio. If one remove from this rectangle n squares of side length W, one gets a rectangle similar to the original rectangle; that is, a rectangle with the same ratio of the length to the width (see figures).

Some metallic means appear as segments in the figure formed by a regular polygon and its diagonals. This is in particular the case for the golden ratio and the pentagon, and for the silver ratio and the octagon; see figures.

Powers

Denoting by S m {\displaystyle S_{m}} the metallic mean of m one has

S m n = K n S m + K n − 1 , {\displaystyle S_{m}^{n}=K_{n}S_{m}+K_{n-1},}

where the numbers K n {\displaystyle K_{n}} are defined recursively by the initial conditions K0 = 0 and K1 = 1, and the recurrence relation

K n = m K n − 1 + K n − 2 . {\displaystyle K_{n}=mK_{n-1}+K_{n-2}.}

Proof: The equality is immediately true for n = 1. {\displaystyle n=1.} The recurrence relation implies K 2 = m , {\displaystyle K_{2}=m,} which makes the equality true for k = 2. {\displaystyle k=2.} Supposing the equality true up to n − 1 , {\displaystyle n-1,} one has

S m n = m S m n − 1 + S m n − 2 (defining equation) = m ( K n − 1 S n + K n − 2 ) + ( K n − 2 S m + K n − 3 ) (recurrence hypothesis) = ( m K n − 1 + K n − 2 ) S n + ( m K n − 2 + K n − 3 ) (regrouping) = K n S m + K n − 1 (recurrence on K n ) . {\displaystyle {\begin{aligned}S_{m}^{n}&=mS_{m}^{n-1}+S_{m}^{n-2}&&{\text{(defining equation)}}\\&=m(K_{n-1}S_{n}+K_{n-2})+(K_{n-2}S_{m}+K_{n-3})&&{\text{(recurrence hypothesis)}}\\&=(mK_{n-1}+K_{n-2})S_{n}+(mK_{n-2}+K_{n-3})&&{\text{(regrouping)}}\\&=K_{n}S_{m}+K_{n-1}&&{\text{(recurrence on }}K_{n}).\end{aligned}}}

End of the proof.

One has also [citation needed]

K n = S m n + 1 − ( m − S m ) n + 1 m 2 + 4 . {\displaystyle K_{n}={\frac {S_{m}^{n+1}-(m-S_{m})^{n+1}}{\sqrt {m^{2}+4}}}.}

The odd powers of a metallic mean are themselves metallic means. More precisely, if n is an odd natural number, then S m n = S M n , {\displaystyle S_{m}^{n}=S_{M_{n}},} where M n {\displaystyle M_{n}} is defined by the recurrence relation M n = m M n − 1 + M n − 2 {\displaystyle M_{n}=mM_{n-1}+M_{n-2}} and the initial conditions M 0 = 2 {\displaystyle M_{0}=2} and M 1 = m . {\displaystyle M_{1}=m.}

Proof: Let a = S m {\displaystyle a=S_{m}} and b = − 1 / S m . {\displaystyle b=-1/S_{m}.} The definition of metallic means implies that a + b = m {\displaystyle a+b=m} and a b = − 1. {\displaystyle ab=-1.} Let M n = a n + b n . {\displaystyle M_{n}=a^{n}+b^{n}.} Since a n b n = ( a b ) n = − 1 {\displaystyle a^{n}b^{n}=(ab)^{n}=-1} if n is odd, the power a n {\displaystyle a^{n}} is a root of x 2 − M n − 1 = 0. {\displaystyle x^{2}-M_{n}-1=0.} So, it remains to prove that M n {\displaystyle M_{n}} is an integer that satisfies the given recurrence relation. This results from the identity

a n + b n = ( a + b ) ( a n − 1 + b n − 1 ) − a b ( a n − 2 + a n − 2 ) = m ( a n − 1 + b n − 1 ) + ( a n − 2 + a n − 2 ) . {\displaystyle {\begin{aligned}a^{n}+b^{n}&=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+a^{n-2})\\&=m(a^{n-1}+b^{n-1})+(a^{n-2}+a^{n-2}).\end{aligned}}}

This completes the proof, given that the initial values are easy to verify.

In particular, one has

S m 3 = S m 3 + 3 m S m 5 = S m 5 + 5 m 3 + 5 m S m 7 = S m 7 + 7 m 5 + 14 m 3 + 7 m S m 9 = S m 9 + 9 m 7 + 27 m 5 + 30 m 3 + 9 m S m 11 = S m 11 + 11 m 9 + 44 m 7 + 77 m 5 + 55 m 3 + 11 m {\displaystyle {\begin{aligned}S_{m}^{3}&=S_{m^{3}+3m}\\S_{m}^{5}&=S_{m^{5}+5m^{3}+5m}\\S_{m}^{7}&=S_{m^{7}+7m^{5}+14m^{3}+7m}\\S_{m}^{9}&=S_{m^{9}+9m^{7}+27m^{5}+30m^{3}+9m}\\S_{m}^{11}&=S_{m^{11}+11m^{9}+44m^{7}+77m^{5}+55m^{3}+11m}\end{aligned}}}

and, in general,[citation needed]

S m 2 n + 1 = S M , {\displaystyle S_{m}^{2n+1}=S_{M},}

where

M = ∑ k = 0 n 2 n + 1 2 k + 1 ( n + k 2 k ) m 2 k + 1 . {\displaystyle M=\sum _{k=0}^{n}{{2n+1} \over {2k+1}}{{n+k} \choose {2k}}m^{2k+1}.}

For even powers, things are more complicated. If n is a positive even integer then[citation needed]

S m n − ⌊ S m n ⌋ = 1 − S m − n . {\displaystyle {S_{m}^{n}-\left\lfloor S_{m}^{n}\right\rfloor }=1-S_{m}^{-n}.}

Additionally,[citation needed]

1 S m 4 − ⌊ S m 4 ⌋ + ⌊ S m 4 − 1 ⌋ = S ( m 4 + 4 m 2 + 1 ) {\displaystyle {1 \over {S_{m}^{4}-\left\lfloor S_{m}^{4}\right\rfloor }}+\left\lfloor S_{m}^{4}-1\right\rfloor =S_{\left(m^{4}+4m^{2}+1\right)}}

1 S m 6 − ⌊ S m 6 ⌋ + ⌊ S m 6 − 1 ⌋ = S ( m 6 + 6 m 4 + 9 m 2 + 1 ) . {\displaystyle {1 \over {S_{m}^{6}-\left\lfloor S_{m}^{6}\right\rfloor }}+\left\lfloor S_{m}^{6}-1\right\rfloor =S_{\left(m^{6}+6m^{4}+9m^{2}+1\right)}.}

For the square of a metallic ratio we have:S m 2 = [ m m 2 + 4 + ( m + 2 ) ] / 2 = ( p + p 2 + 4 ) / 2 {\displaystyle S_{m}^{2}=[m{\sqrt {m^{2}+4}}+(m+2)]/2=(p+{\sqrt {p^{2}+4}})/2}

where p = m m 2 + 4 {\displaystyle p=m{\sqrt {m^{2}+4}}} lies strictly between m 2 + 1 {\displaystyle m^{2}+1} and m 2 + 2 {\displaystyle m^{2}+2}. Therefore

S m 2 + 1 < S m 2 < S m 2 + 2 {\displaystyle S_{m^{2}+1}<S_{m}^{2}<S_{m^{2}+2}}

Generalization

One may define the metallic mean S − n {\displaystyle S_{-n}} of a negative integer −n as the positive solution of the equation x 2 − ( − n ) x − 1. {\displaystyle x^{2}-(-n)x-1.} The metallic mean of −n is the multiplicative inverse of the metallic mean of n:

S − n = 1 S n . {\displaystyle S_{-n}={\frac {1}{S_{n}}}.}

Another generalization consists of changing the defining equation from x 2 − n x − 1 = 0 {\displaystyle x^{2}-nx-1=0} to x 2 − n x − c = 0 {\displaystyle x^{2}-nx-c=0}. If

R = n ± n 2 + 4 c 2 , {\displaystyle R={\frac {n\pm {\sqrt {n^{2}+4c}}}{2}},}

is any root of the equation, one has

R − n = c R . {\displaystyle R-n={\frac {c}{R}}.}

The silver mean of m is also given by the integral

S m = ∫ 0 m ( x 2 x 2 + 4 + m + 2 2 m ) d x . {\displaystyle S_{m}=\int _{0}^{m}{\left({x \over {2{\sqrt {x^{2}+4}}}}+{{m+2} \over {2m}}\right)}\,dx.}

Another form of the metallic mean is

n + n 2 + 4 2 = e a r s i n h ( n / 2 ) . {\displaystyle {\frac {n+{\sqrt {n^{2}+4}}}{2}}=e^{\operatorname {arsinh(n/2)} }.}

Relation to half-angle cotangent

A tangent half-angle formula gives cot ⁡ θ = cot 2 ⁡ θ 2 − 1 2 cot ⁡ θ 2 {\displaystyle \cot \theta ={\frac {\cot ^{2}{\frac {\theta }{2}}-1}{2\cot {\frac {\theta }{2}}}}} which can be rewritten as cot 2 ⁡ θ 2 − ( 2 cot ⁡ θ ) cot ⁡ θ 2 − 1 = 0 . {\displaystyle \cot ^{2}{\frac {\theta }{2}}-(2\cot \theta )\cot {\frac {\theta }{2}}-1=0\,.} That is, for the positive value of cot ⁡ θ 2 {\textstyle \cot {\frac {\theta }{2}}}, the metallic mean S 2 cot ⁡ θ = cot ⁡ θ 2 , {\displaystyle S_{2\cot \theta }=\cot {\frac {\theta }{2}}\,,} which is especially meaningful when 2 cot ⁡ θ {\textstyle 2\cot \theta } is a positive integer, as it is with some Pythagorean triangles.

Relation to Pythagorean triples

For a primitive Pythagorean triple, a2 + b2 = c2, with positive integers a < b < c that are relatively prime, if the difference between the hypotenuse c and longer leg b is 1, 2 or 8 then the Pythagorean triangle exhibits a metallic mean. Specifically, the cotangent of one quarter of the smaller acute angle of the Pythagorean triangle is a metallic mean.

More precisely, for a primitive Pythagorean triple (a, b, c) with a < b < c, the smaller acute angle α satisfies tan ⁡ α 2 = c − b a . {\displaystyle \tan {\frac {\alpha }{2}}={\frac {c-b}{a}}\,.} When c − b ∈ {1, 2, 8}, we will always get that n = 2 cot ⁡ α 2 = 2 a c − b {\displaystyle n=2\cot {\frac {\alpha }{2}}={\frac {2a}{c-b}}} is an integer and that cot ⁡ α 4 = S n , {\displaystyle \cot {\frac {\alpha }{4}}=S_{n}\,,} the n-th metallic mean.

The reverse direction also works. For n ≥ 5, the primitive Pythagorean triple that gives the n-th metallic mean is given by (n, n2/4 − 1, n2/4 + 1) if n is a multiple of 4, is given by (n/2, (n2 − 4)/8, (n2 + 4)/8) if n is even but not a multiple of 4, and is given by (4n, n2 − 4, n2 + 4) if n is odd. For example, the primitive Pythagorean triple (20, 21, 29) gives the 5th metallic mean; (3, 4, 5) gives the 6th metallic mean; (28, 45, 53) gives the 7th metallic mean; (8, 15, 17) gives the 8th metallic mean; and so on.

Numerical values

Relation to Aperiodic Order

The k {\displaystyle k}-th metallic mean serves as the inflation ratio for one-dimensional substitution tilings, such as a → a k b {\displaystyle a\to a^{k}b} and b → a {\displaystyle b\to a}. These sequences exhibit long-range aperiodic order. By applying an interval removal process to these tilings, one can construct self-similar Cantor sets where the Hausdorff dimension is determined by the metallic mean scaling factor.

See also

• Constant
• Mean
• Ratio
• Plastic ratio

Notes

Further reading

• Stakhov, Alekseĭ Petrovich (2009). The Mathematics of Harmony: From Euclid to Contemporary Mathematics and Computer Science, p.228, 231. World Scientific. ISBN9789812775832.

External links

• Cristina-Elena Hrețcanu and Mircea Crasmareanu (2013). "", Revista de la Unión Matemática Argentina.
• Rakočević, Miloje M. "", Arxiv.org.