- Source: Metallic mean
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
,
{\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
{\textstyle {\frac {n+{\sqrt {n^{2}+4}}}{2}}}
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
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,
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 complicate. If n is a positive even integer then
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,
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}
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 primitive Pythagorean triangles.
Relation to Pythagorean triples
Metallic means are precisely represented by some primitive Pythagorean triples, a2 + b2 = c2, with positive integers a < b < c.
In a primitive Pythagorean triple, if the difference between hypotenuse c and longer leg b 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
n
{\displaystyle S_{n}}
as follows :
S
n
=
cot
α
4
{\displaystyle S_{n}=\cot {\frac {\alpha }{4}}}
where α is the smaller acute angle of the Pythagorean triangle and the metallic mean index is
n
=
2
cot
α
2
=
2
a
c
−
b
=
2
c
+
b
c
−
b
.
{\displaystyle n=2\cot {\frac {\alpha }{2}}={\frac {2a}{c-b}}=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.
Numerical values
See also
Constant
Mean
Ratio
Plastic ratio
Notes
References
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.
Kata Kunci Pencarian:
- Besi
- Hardcore punk
- Metallic mean
- Golden ratio
- Pisot–Vijayaraghavan number
- Rectifier (neural networks)
- Pisano period
- List of mathematical constants
- Generalizations of Fibonacci numbers
- Atomic radius
- Opacity
- Vera W. de Spinadel
