Metallic mean

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

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

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

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

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

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

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

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

End of the proof.

One has also ^{[citation needed]}

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

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

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

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

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

In particular, one has

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

and, in general,^{[citation needed]}

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

where

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

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

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

Additionally,^{[citation needed]}

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

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

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

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

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

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

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

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

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

is any root of the equation, one has

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

The silver mean of m is also given by the integral^[4]

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

Another form of the metallic mean is^[4]

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

A tangent half-angle formula gives $${\displaystyle \cot \theta ={\frac {\cot ^{2}{\frac {\theta }{2}}-1}{2\cot {\frac {\theta }{2}}}}}$$ which can be rewritten as $${\displaystyle \cot ^{2}{\frac {\theta }{2}}-(2\cot \theta )\cot {\frac {\theta }{2}}-1=0\,.}$$ That is, for the positive value of ${\textstyle \cot {\frac {\theta }{2}}}$, the metallic mean $${\displaystyle S_{2\cot \theta }=\cot {\frac {\theta }{2}}\,,}$$ which is especially meaningful when ${\textstyle 2\cot \theta }$ is a positive integer, as it is with some 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.^[5]

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

• Constant
• Mean
• Ratio
• Plastic ratio