Metallic mean

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

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

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

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

Metallic means are (successive) derivations of the golden ($n=1$) and silver ratios ($n=2$), and share some of their interesting properties. The term "bronze ratio" ($n=3$) (Cf. Golden Age and Olympic Medals) and even metals such as copper ($n=4$) and nickel ($n=5$) 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 $1$ and have $-1$ as their norm.

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

$x_k=a{S}_n^k+b{\left(\frac{-1}{S_n}\right)}^k,$

where $S_n$ is the nth metallic mean, and a and b are constants depending only on $x_0$ and $x_1.$ 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 $n=1,$ $S_n$ is the golden ratio. If $x_0=0$ and $x_1=1,$ the sequence is the Fibonacci sequence, and the above formula is Binet's formula. If $n=1,x_0=2,x_1=1$ one has the Lucas numbers. If $n=2,$ the metallic mean is called the silver ratio, and the elements of the sequence starting with $x_0=0$ and $x_1=1$ are called the Pell numbers.

Geometry

File:Gold, silver, and bronze rectangles.svg

If one removes n largest possible squares from a rectangle with ratio length/width equal to the nth metallic mean, one gets a rectangle with the same ratio length/width (in the figures, n is the number of dotted lines).

File:Pentagram-phi.svg

File:Silver ratio octagon.svg

Golden ratio within the pentagram (φ = red/ green = green/blue = blue/purple) and silver ratio within the octagon.

The defining equation $x=n+\frac{1}{x}$ 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 $S_m$ 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

$K_n=m{K}_{n-1}+K_{n-2}.$

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

$\begin{array}{cccc}{S}_m^n& =m{S}_m^{n-1}+S_m^{n-2}& & \text{(defining equation)}\\ & =m(K_{n-1}{S}_n+K_{n-2})+(K_{n-2}{S}_m+K_{n-3})& & \text{(recurrence hypothesis)}\\ & =(m{K}_{n-1}+K_{n-2})S_n+(m{K}_{n-2}+K_{n-3})& & \text{(regrouping)}\\ & =K_n{S}_m+K_{n-1}& & \text{(recurrence on }{K}_n).\end{array}$

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 $M_n=m{M}_{n-1}+M_{n-2}$ 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

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

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

In particular, one has

$\begin{array}{cc}{S}_m^3& =S_{m^3+3m}\\ S_m^5& =S_{m^5+5m^3+5m}\\ S_m^7& =S_{m^7+7m^5+14m^3+7m}\\ S_m^9& =S_{m^9+9m^7+27m^5+30m^3+9m}\\ S_m^{11}& =S_{m^{11}+11m^9+44m^7+77m^5+55m^3+11m}\end{array}$

and, in general,

$S_m^{2n+1}=S_M,$

where

$M=\sum _{k=0}^n\frac{2n+1}{2k+1}(\genfrac{}{}{0ex}{}{n+k}{2k})m^{2k+1}.$

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

$S_m^n-\lfloor S_m^n\rfloor =1-S_m^{-n}.$

Additionally,

$\frac{1}{S_m^4-\lfloor S_m^4\rfloor }+\lfloor S_m^4-1\rfloor =S_{(m^4+4m^2+1)}$

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

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

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

$S_{m^2+1}<S_m^2<S_{m^2+2}$

Generalization

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

$S_m=\underset{0}{\overset{m}{\int }}(\frac{x}{2\sqrt{x^2+4}}+\frac{m+2}{2m})dx.$

Another form of the metallic mean is

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

Relation to half-angle cotangent

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

Relation to Pythagorean triples

File:AgamRiaSanga.png

Metallic Ratios in Primitive Pythagorean Triangles

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 $$\tan \frac{\alpha }{2}=\frac{c-b}{a}.$$ When c − b ∈ {1, 2, 8}, we will always get that $$n=2\cot \frac{\alpha }{2}=\frac{2a}{c-b}$$ is an integer and that $$\cot \frac{\alpha }{4}=S_n,$$ 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

First metallic means
n	Ratio	Value	Name
$n$	$\frac{n+\sqrt{4+n^2}}{2}=\frac{n}{2}+\sqrt{1+{\left(\frac{n}{2}\right)}^2}$
0	$\frac{0+\sqrt{4}}{2}=0+\sqrt{1}$	1
1	$\frac{1+\sqrt{5}}{2}$	1.618033988...	Golden
2	$\frac{2+\sqrt{8}}{2}=1+\sqrt{2}$	2.414213562...	Silver
3	$\frac{3+\sqrt{13}}{2}$	3.302775637...	Bronze
4	$\frac{4+\sqrt{20}}{2}=2+\sqrt{5}$	4.236067977...	Copper
5	$\frac{5+\sqrt{29}}{2}$	5.192582403...	Nickel
6	$\frac{6+\sqrt{40}}{2}=3+\sqrt{10}$	6.162277660...
7	$\frac{7+\sqrt{53}}{2}$	7.140054944...
8	$\frac{8+\sqrt{68}}{2}=4+\sqrt{17}$	8.123105625...
9	$\frac{9+\sqrt{85}}{2}$	9.109772228...
10	$\frac{10+\sqrt{104}}{2}=5+\sqrt{26}$	10.099019513...

Relation to Aperiodic Order

The $k$-th metallic mean serves as the inflation ratio for one-dimensional substitution tilings, such as $a\to a^{k}b$ and $b\to a$. 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

Further reading

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

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.