The Maximal Number of Transverse Self-Intersections of Geodesics ...

µ 2 3
As an example, we see that a 2 =
3 5
µ 5 8
,a 3 =
8 13
, all Fibonacci numbers.
Definition 3.4 Let φ = 1+√ 5
, represent the golden ratio.
2
µ
µ
Fn+1
Fn−1
Lemma 3.5 The lim n→∞ = φ, lim n→∞ = 1 φ
F n F n
F n−1
+ F n+1
− φ monotonicallydecreaseto2φ − 1 as n →∞.
F n F n
= φ − 1 and the values of
Lemma 3.6 If W = a n 2 b n 2 , then all cyclic permutations W p1 = a k b n 2 a m and W p2 = b k a n 2 b m
of W have symmetric roots.
Proof. We look at equations of the form W p1 = a k b n 2 a m and W p2 = b k a n 2 b m . Since n
is even, we know that k + m = n , assuming these equations are cyclic permutations of W .
2
Then µ we perform the µ following matrix µ multiplication:
F2k−1 F 2k F2n−1 −F 2n F2m−1 F 2m
=
µ
F 2k F 2k+1 −F 2n F 2n+1 F 2m F 2m+1
F2m (F 2k F 2n+1 − F 2n F 2k−1 )+F 2m−1 (−F 2k F 2n + F 2k−1 F 2n−1 )
F 2m (−F 2k F 2n + F 2k+1 F 2n+1 )+F 2m−1 (F 2k F 2n−1 − F 2n F 2k+1 )
F 2m (−F 2k F 2n + F 2k−1 F 2n−1 )+F 2m+1 (F 2k F 2n+1 − F 2n F 2k−1 )
µ
F 2m (F 2k F 2n−1 −
µ
F 2n F 2k+1 )+F 2m+1
µ
(−F 2k F 2n + F 2k+1 F
2n+1 )
F2k−1 −F 2k F2n−1 F 2n F2m−1 −F 2m
=
µ
−F 2k F 2k+1 F 2n F 2n+1 −F 2m F 2m+1
F2m (F 2k F 2n+1 − F 2n F 2k−1 )+F 2m−1 (−F 2k F 2n + F 2k−1 F 2n−1 )
F 2m (F 2k F 2n − F 2k+1 F 2n+1 )+F 2m−1 (−F 2k F 2n−1 + F 2n F 2k+1 )
F 2m (F 2k F 2n − F 2k−1 F 2n−1 )+F 2m+1 (−F 2k F 2n+1 + F 2n F 2k−1 )
.
F 2m (F 2k F 2n−1 − F 2n F 2k+1 )+F 2m+1 (−F 2k F 2n + F 2k+1 F 2n+1 )
Since the only changes in the matrices are the signs of the upper right and lower left
entries, µ let us assume that the entries are
µ
µ
α β
α β
α −β
, without loss of generality. So W
γ δ
p1 = ,andW
γ δ
p2 =
.
−γ δ
By performing our transformation, we seeq
T (W p1 )=z has roots at − (δ − α) ± (δ − α) 2 +4γβ
, while T (W p2 )=z has roots at
q
2γ
− (δ − α) ± (δ − α) 2 +4γβ
. Since the only difference in these two roots is the lower
−2γ
term’s sign, they are symmetric.
Lemma 3.7 Words with n even, n ≥ 4 of the form W = a n 2 b n 2
[1, ∞).
have roots in [0, 1] and
12