Metallic mean

First metallic means
N	Ratio	Value	Name
0	0 + √4/2	1
1	1 + √5/2	1.618033989	Golden
2	2 + √8/2	2.414213562	Silver
3	3 + √13/2	3.302775638	Bronze
4	4 + √20/2	4.236067978
5	5 + √29/2	5.192582404
6	6 + √40/2	6.162277660
7	7 + √53/2	7.140054945
8	8 + √68/2	8.123105626
9	9 + √85/2	9.109772229
10	10+ √104/2	10.099019513

Golden ratio within the pentagram and silver ratio within the octagon.

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

• the unique positive real number $x$ such that ${\textstyle x=n+{\frac {1}{x}}}$
• the positive root of the quadratic equation $x^{2}-nx-1=0$
• the number ${\textstyle {\frac {n+{\sqrt {n^{2}+4}}}{2}}}$
• the number whose expression as a continued fraction is

  $[n;n,n,n,n,\dots ]=n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+\ddots \,}}}}}}}}$

Metallic means are generalizations of the golden ratio ($n=1$) and silver ratio ($n=2$), and share some of their interesting properties. The term "bronze ratio" ($n=3$), and terms using other metals names (such as copper or nickel), are occasionally used to name subsequent metallic means.

As the golden ratio is connected to the pentagon (first diagonal/side), the silver ratio is connected to the octagon (second diagonal/side). As the golden ratio is connected to the Fibonacci numbers, the silver ratio is connected to the Pell numbers, and the bronze ratio is connected to OEIS: A006190. Each Fibonacci number is the sum of the previous number times one plus the number before that, each Pell number is the sum of the previous number times two and the one before that, and each "bronze Fibonacci number" is the sum of the previous number times three plus the number before that. Taking successive Fibonacci numbers as ratios, these ratios approach the golden mean, the Pell number ratios approach the silver mean, and the "bronze Fibonacci number" ratios approach the bronze mean.

Powers

If one removes the largest possible square from the end of a gold rectangle one is left with a gold rectangle. If one removes two from a silver, one is left with a silver. If one removes three from a bronze, one is left with a bronze. Examine the dotted lines representing the boundaries of the perfect squares within each rectangle, and note that the number of said dotted lines always equals N.

Gold, silver, and bronze ratios within their respective rectangles.

Denoting by $S_{m}$ the metallic mean of m one has

$S_{m}^{n}=K_{n}S_{m}+K_{n-1},$

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

$K_{n}=mK_{n-1}+K_{n-2}.$

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

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

End of the proof.

One has also

$K_{n}={\frac {S_{m}^{n+1}-(m-S_{m})^{n+1}}{\sqrt {m^{2}+4}}}.$

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

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

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

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

In particular, one has

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

and, in general,

$S_{m}^{2n+1}=S_{M},$

where

$M=\sum _{k=0}^{n}{{2n+1} \over {2k+1}}{{n+k} \choose {2k}}m^{2k+1}.$

For even powers, things are more complicate. If n is a positive even integer then

${S_{m}^{n}-\left\lfloor S_{m}^{n}\right\rfloor }=1-S_{m}^{-n}.$

Additionally,

${1 \over {S_{m}^{4}-\left\lfloor S_{m}^{4}\right\rfloor }}+\left\lfloor S_{m}^{4}-1\right\rfloor =S_{\left(m^{4}+4m^{2}+1\right)}$

${1 \over {S_{m}^{6}-\left\lfloor S_{m}^{6}\right\rfloor }}+\left\lfloor S_{m}^{6}-1\right\rfloor =S_{\left(m^{6}+6m^{4}+9m^{2}+1\right)}.$

A golden triangle. The ratio a:b is equivalent to the golden ratio φ. In a silver triangle this would be equivalent to δ_S.

Generalization

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

$S_{-n}={\frac {1}{S_{n}}}.$

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

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

is any root of the equation, one has

$R-n={\frac {c}{R}}.$

The silver mean of m is also given by the integral

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

Another form of the metallic mean is

${\frac {n+{\sqrt {n^{2}+4}}}{2}}=e^{\operatorname {arsinh(n/2)} }.$

Trigonometric expressions

N	Trigonometric expression	Associated regular polygon
1	$2\cos {\frac {\pi }{5}}$	pentagon
2	$\tan {\frac {3\pi }{8}}$	octagon
3	$8\cos {\frac {\pi }{13}}\cos {\frac {3\pi }{13}}\cos {\frac {4\pi }{13}}$	tridecagon
4	$8\cos ^{3}{\frac {\pi }{5}}$	pentagon
5	$128\cos {\frac {\pi }{29}}\cos {\frac {4\pi }{29}}\cos {\frac {5\pi }{29}}\cos {\frac {6\pi }{29}}\cos {\frac {7\pi }{29}}\cos {\frac {9\pi }{29}}\cos {\frac {13\pi }{29}}$	29-gon
6	${\frac {\sin {\frac {21\pi }{40}}}{\sin {\frac {7\pi }{8}}-\sin {\frac {5\pi }{8}}\sin {\frac {\pi }{40}}\csc {\frac {11\pi }{40}}-\sin {\frac {19\pi }{20}}\sin {\frac {9\pi }{40}}\csc {\frac {7\pi }{10}}}}$	tetracontagon
7
8	${\frac {\sin {\frac {6\pi }{17}}\sin {\frac {7\pi }{17}}\sin {\frac {3\pi }{17}}\sin {\frac {12\pi }{17}}}{\sin {\frac {\pi }{17}}\sin {\frac {9\pi }{17}}\sin {\frac {4\pi }{17}}\sin {\frac {2\pi }{17}}}}$	heptadecagon
9
10

Geometric construction

The metallic mean for any given integer $N$ can be constructed geometrically in the following way. Define a right triangle with sides $A$ and $B$ having lengths of $1$ and $N/2$, respectively. The $N$th metallic mean $M$ is simply the sum of the length of $B$ and the hypotenuse, $H$.

For $N=1$,

N = 1

$H={\sqrt {(N/2)^{2}+1^{2}}}={\sqrt {5/4}}=1.1180339...$

and so

$M=B+H=1/2+1.1180339...=1.6180339...=$ φ.

Setting $N=2$ yields the silver ratio.

N = 2

$H={\sqrt {(2/2)^{2}+1^{2}}}={\sqrt {2}}=1.4142135...$

Thus

$M=B+H=2/2+1.4142135...=2.4142135...$

Likewise, the bronze ratio would be calculated with $N=3$ so

N = 3

$H={\sqrt {(3/2)^{2}+1^{2}}}={\sqrt {13/4}}=1.8027756...$

yields

$M=B+H=3/2+1.8027756...=3.3027756...$

Non-integer arguments sometimes produce triangles with a mean that is itself an integer. Examples include N = 1.5, where

N = 1.5

$H={\sqrt {(1.5/2)^{2}+1^{2}}}={\sqrt {1.5625}}=1.25$

and

$M=B+H=1.5/2+1.25=2$

which is simply a scaled-down version of the 3–4–5 Pythagorean triangle.

Relation to Pythagorean triples

Metallic Ratios in Primitive Pythagorean Triangles

Metallic means are precisely represented by primitive Pythagorean triples.

In a primitive Pythagorean triple, if the difference between hypotenuse and longer leg is 1, 2 or 8, such Pythagorean triple accurately represents one particular metallic mean. The cotangent of the quarter of smaller acute angle of such Pythagorean triangle equals the precise value of one particular metallic mean.

Consider a primitive Pythagorean triple (a,b,c) in which a < b < c and c - b ∈ {1, 2, 8}. Such Pythagorean triangle (a,b,c) yields the precise value of a particular metallic mean $S_{m}$ as follows :

$S_{m}=cot\left({\frac {\theta }{4}}\right)$

where θ is the smaller acute angle of the Pythagorean triangle

and $m=2{\sqrt {\frac {c+b}{c-b}}}$

For example, the primitive Pythagorean triple 20-21-29 incorporates the 5th metallic mean. Cotangent of the quarter of smaller acute angle of the 20-21-29 Pythagorean triangle yields the precise value of the 5th metallic mean. Similarly, the Pythagorean triangle 3-4-5 represents the 6th metallic mean. Likewise, the Pythagorean triple 12-35-37 gives the 12th metallic mean, the Pythagorean triple 52-165-173 yields the 13th metallic mean, and so on.

See also

• Constant
• Mean
• Ratio
• Plastic ratio

Further reading

• Stakhov, Alekseĭ Petrovich (2009). The Mathematics of Harmony: From Euclid to Contemporary Mathematics and Computer Science, p. 228, 231. World Scientific. ISBN 9789812775832.

External links

• Cristina-Elena Hrețcanu and Mircea Crasmareanu (2013). "Metallic Structures on Riemannian Manifolds", Revista de la Unión Matemática Argentina.
• Rakočević, Miloje M. "Further Generalization of Golden Mean in Relation to Euler's 'Divine' Equation", Arxiv.org.