On continued fractions and diophantine approximation in power ...

162 W. M. Schmidt
10. The Mills–Robbins continued fraction. Let Ω = ∅, Ω1 = X,
and for n ≥ 2 let Ωn be the finite sequence of polynomials
Ωn = Ωn−1, −X, Ω (3)
n−2 , −X, Ωn−1,
where commas signify juxtaposition of sequences, and Ω (3)
k is obtained by
cubing every element of Ωk. Since Ωn−1 appears as the initial segment of
Ωn, we may define Ω∞ as the infinite sequence having each Ωn as an initial
segment. Set
α = [0, Ω∞].
Mills and Robbins [16] had conjectured and Buck and Robbins [4] proved
the following intriguing theorem:
Suppose k = F3. Then α is the unique root in k((X −1 )) with
(10.1) α 4 + α 2 − Xα + 1 = 0.
Lasjaunias [8] gave another proof in a wider setting. Here we will rearrange
ideas of [4] to give a rather short argument.
Writing
Pn−1 Pn
Nn =
(n = −1, 0, 1, . . .),
we have
N−1 =
Qn−1 Qn
0 1
0 1
, Nn = Nn−1
1 0
1 An
(n = 0, 1, . . .)
where Pl = Pl(A0, . . . , Al), Ql = Ql(A0, . . . , Al) (l = 0, 1, . . .).
In the case when A0, A1, . . . is the sequence Ω∞, set M0 = 1 0
, M1 =
0 1
0 1
,
1 X
0 1
(10.2) Mn = Mn−1
M
1 −X
(3)
0 1
n−2
Mn−1 (n ≥ 2),
1 −X
where M (3)
l
is obtained from Ml by cubing each entry. Then
Mn = N r(n)
where r(n) is the number of elements of Ωn. Since Mn is symmetric, we
may write
(10.3)
Rn
Mn =
Sn
,
Sn Tn
and notice that Rn/Sn, Sn/Tn are consecutive convergents of the continued
fraction of α. Therefore
lim
n→∞ Rn/Sn = lim
n→∞ Sn/Tn = α, lim
n→∞ Rn/Tn = α 2 .
All this is already in [4].