My title - Departamento de Matemática da Universidade do Minho

10 STATISTICAL DESCRIPTION OF ORBITS 72
Gauss map.
of the form
Any irrational real number x ∈ ]0, 1] has a unique continued fraction representation
x = [0; a 1 , a 2 , a 3 , . . . ] =
a 1 +
1
a 2 +
1
1
a 3 + 1
. ..
where the a n are nonnegative integers. The equality sign and the “infinite fraction” above mean
that the sequence of finite continued fractions
1
r n = [0; a 1 , a 2 , . . . , a n ] =
1
a 1 +
1
a 2 +
. .. 1 +
a n
which are called “convergents”, do converge to x as n → ∞. The sequence of rationals r n (any
such r n provide the best rational approximation for x with denominator less or equal than that
of r n , as you may have been teached in a course on number theory) is inductively constructed as
follows. First, observe that if a 1 = [1/x] and x 1 = 1/x − a 1 we may write
1
x =
a 1 + x 1
with x 1 ∈ [0, 1]. Then, since x 1 ≠ 0, for otherwise x would be rational, we may define a 2 = [1/x 1 ]
and x 2 = 1/x 1 − a 2 to get
1
x =
1
a 1 +
a 2 + x 2
Inductively, we see that
1
x =
1
a 1 +
1
a 2 +
... + 1
a n+xn
where x n = 1/x n−1 − a n and a n = [1/x n−1 ]. This amounts to say that the sequence (x n ) is the
trajectory of x under the Gauss map g : ]0, 1] → ]0, 1], defined as
x ↦→ 1/x − [1/x]
Observe that g is not defined at the origin, hence to iterate g we need to avoid all the preimages
of 0, which are the rationals. This is not a problem if we want to study the statistical properties
of g with respect to Lebesgue measure, since rationals form a subset of zero measure. The Gauss
map admits an absolutely continuous invariant measure µ = ρdx, defined as
µ (A) = 1
log 2 ·
∫
A
1
1 + x dx
for any Borel subset A ⊂ ]0, 1]. The denominator log 2 is there to normalize the measure, so we
just have to check the invariance criterium for the density ρ (x) = 1/ (1 + x). Since any x ′ ∈ ]0, 1]
has one preimage x k = 1/ (x ′ + k) in each interval ]1/ (k + 1) , 1/k], we compute
∑
x∈g −1 {x ′ }
ρ (x)
|det g ′ (x)|
x 2 k
= ∑ 1 + x k
k≥1
= ∑ ( )
1
x ′ + k − 1
x ′ + k + 1
k≥1
=
1
1 + x ′ = ρ (x′ )