Table Of Content√
DIFFERENT BASES IN INVESTIGATION OF 3 2
MITJA LAKNER, PETER PETEK, MARJETA SˇKAPIN RUGELJ
Abstract. Thepresentpaperisinasenseacontinuationof[11],
it relies on the notation and some results. The problem tackled
in both papers is the nature of the continued fraction expansion
√
of 32: are the partial quotients bounded or not. Numerical ex-
1
periments suggest an even stronger result on the lines of Kuzmin
1
0 statistics. Hereweapplydifferent√setsofbasesforthevectorspace
2 V, where the adjunction ring Z[32] lives. And as a result we get
n a criterion for continued fraction convergents in terms of the coef-
a ficient vector from a lattice.
J
Key words: bases, cubic root, continued fractions.
1
Mathematics Subject Classification (2000): 11A55, 11K50, 11R16.
3
]
T
N
h. 1. Introduction
t
a
m Stability of invariant circles in K.A.M. theory depends on the respec-
√
[ tiverotationnumber. Themoststablecirclehasφ = −1+ 5, thegolden
2
1 mean ratio, as its rotation number, all partial quotients equal bi = 1.
v As in [8] boundness of partial quotients would be a bonus in represen-
7
tation on a computer. Here however, experiments strongly suggest the
6
8 opposite. Also cubic irrationals are interesting in studying quasiperi-
√
5 odic motion [3], [9]. Here we investigate 3 2 and its adjunction ring. It
. √
1 is a common belief that the partial quotients of 3 2 are not bounded,
0
supported by extensive computations, but no proof.
1
1
Even more, computations suggest that their relative frequencies in the
:
v
limit obey the Kuzmin law P(b = k) = log (k+1)2. In [12] several
Xi n 2 k(k+2) √ √ √
algebraicnumberswereusedincomputations, amongthem 3 2, 4 2, 5 2
r √
a and good accordance was found with Kuzmin’s statistics, for 3 2 even
too good. So later in [6] and [2] larger samples were taken and the
anomaly seemed to disappear. We played the same game, only we had
theadvantageofmoresophisticalcomputationtoolsthatevolvedinthe
Date: February 1, 2011.
M. Lakner, M. Sˇkapin Rugelj: University of Ljubljana, Faculty of Civil and
Geodetic Engineering, Jamova 2, 1000 Ljubljana, Slovenia.
P. Petek: University of Ljubljana, Faculty of Education, Kardeljeva ploˇsˇcad 16,
1000 Ljubljana, Slovenia.
E-mails: [email protected], [email protected], [email protected].
1
2 MITJA LAKNER, PETER PETEK, MARJETA SˇKAPIN RUGELJ
years in between. The experimental results supporting of the stronger
hypothesis instigated us to try towards some theoretical results.
However questions about ”big” partial quotients still linger and tease
us. Though some explanation was given in a special case, using elliptic
modular functions [13], [4].
In this paper we work towards the proof of
√
Hypothesis. The partial quotients of 3 2 are not bounded.
denoted as Hypothesis B in [11]. Only a very limited partial result
is given, helping to recognize possible convergents and estimating the
next partial quotient.
2. Adjunction ring and ambient vector space V
For convenience of the reader we repeat some definitions and notations
from [11].
√ √
In the adjunction ring we have the unit ρ = 1 + 3 2 + 3 4 and its
√
inverse σ = −1+ 3 2, ρσ = 1. We span the 3-dim space V on ρ,1,σ.
√
Obviously the continued fraction expansions for 3 2 and σ differ only
in the starting partial quotient, so we rather consider approximations
to σ. And in order to find ever better approximations, we construct
series of vector space bases for V.
√ √ √
Multiplicative norm is defined in Z[ 3 2]. Let x = a+b 3 2+c 3 4, its
norm is
(1) N(x) = a3 +2b3 +4c3 −6abc = x·x(cid:48) ·x(cid:48)(cid:48)
with
√ √
x(cid:48) = a+ωb 3 2+ω2c 3 4,
√ √
x(cid:48)(cid:48) = a+ω2b 3 2+ωc 3 4
and ω = e2πi a third root of one.
3
√
Division in general leads to the corresponding field Q[ 3 2]. Carrying
out the rationalization of the denominator as in
√ √
1 x(cid:48) ·x(cid:48)(cid:48) a2 −2bc+(2c2 −ab) 3 2+(b2 −ac) 3 4
= =
x N(x) N(x)
√ √ √
gives the elements of Q[ 3 2] in the form a + b 3 2 + c 3 4 with a,b,c
rational fractions.
3. The ambient vector space V
√ √
As already mentioned, instead of (1, 3 2, 3 4) we use the algebraic basis
√
(ρ,1,σ) and elements of Z[ 3 2] are expressed as
w = x·ρ+y ·1+z ·σ
√
DIFFERENT BASES IN INVESTIGATION OF 32 3
or changing the basis
√ √
a+b 3 2+c 3 4 = c·ρ+(a−2c+b)·1+(b−c)·σ.
Now, let V = R3 be the 3-dimensional space endowed with the usual
scalar product (cid:104)a,b(cid:105) and cross product a×b. Vectors can be written
as triplets
V = {v = (x,y,z);x,y,z ∈ R}
and define a linear mapping
√
η : Z[ 3 2] → V
by η(x·ρ+y·1+z·σ) = (x,y,z), the resulting image consisting of all
√
vectors with integer entries, multiplication inherited from Z[ 3 2].
Taking into account
√ √
x·ρ+y ·1+z ·σ = (x+y −z)+(x+z) 3 2+x 3 4
we can define the norm function in the whole V:
N(cid:101)(x,y,z) = (x+y −z)3 +2(x+z)3 +4x3 −6(x+y −z)(x+z)x
Multiplication with σ will prove very important and we observe
η(σ ·w) = Sη(w)
where
0 0 1
S = 1 0 −3.
0 1 −3
which, by the way, represents a hyperbolic toral authomorphism [5].
4. Interplay of different bases in V
The term basis comes in several ways in mathematics. Our discussion
needs it in two appearances:
• as the basis of a number system (ρ in our case),
• as the basis of a vector space (different bases of V here).
The first usage figures in our paper [11], here we are concerned with
the second one.
We can think of the basis B = ((1,0,0),(0,1,0),(0,0,1)) as the canon-
0
ical one. And we shall also write B = (s ,s ,s ) as we shall denote
0 −1 0 1
η(σj) = s , and noting Ss = s we have
j j j+1
s = (0,1,0),
0
s = (0,0,1),
1
s = (1,−3,−3),
2
s = (−3,10,6),
3
4 MITJA LAKNER, PETER PETEK, MARJETA SˇKAPIN RUGELJ
s = (6,−21,−8),
4
s = (−8,30,3),
5
s = (3,−17,21),
6
and we may also need the ones with negative indices
s = (1,0,0),
−1
s = (3,3,1),
−2
s = (12,10,3),
−3
s = (46,39,12),
−4
s = (177,150,46),
−5
s = (681,577,177),
−6
as well as the inverse matrix
3 1 0
S−1 = 3 0 1,
1 0 0
which we also meet in the Jacobi-Perron algorithm [1].
Further we define the series of bases B = (s ,s ,s ) for all integer
j j−1 j j+1
j. These are good bases for our purposes as
Lemma 4.1. Elements of Z3 have integer coefficients in each basis B .
j
Proof. Let v be an element from Z3. If we multiply expansion
v = αs +βs +γs
j−1 j j+1
by integer element matrix S−j, we get vector with integer components
S−jv = αs +βs +γs = (α,β,γ).
−1 0 1
(cid:3)
Besides the series of bases B we also define the conjugate series B∗ in
j j
the following manner.
For start s∗ = (1,0,0) and with the adjoint matrix
0
0 1 0
S∗ = 0 0 1
1 −3 −3
we define vectors s∗ = S∗s∗ for positive and negative indices. So we
j+1 j
have
s∗ = (46,12,3),
−3
s∗ = (12,3,1),
−2
s∗ = (3,1,0),
−1
√
DIFFERENT BASES IN INVESTIGATION OF 32 5
s∗ = (1,0,0),
0
s∗ = (0,0,1),
1
s∗ = (0,1,−3),
2
s∗ = (1,−3,6).
3
Remark 4.2. Comparing with the vectors s = (x ,y ,z ) we see that
j j j j
s∗ = (x ,x ,x ) = (z ,z ,z ).
j j−1 j j+1 j−2 j−1 j
ThebasesseriesB∗ = (s∗ ,s∗,s∗ )againprovidingintegercoefficients
√ j j−1 j j+1
for Z[ 3 2]. The two bases series shall be useful in further computations.
The action of linear transformation S is best understood in terms of
its eigenvalues and eigenspaces. One eigenvalue is real, smaller then 1,
and two are complex conjugate greater than 1
√
λ = σ = −1+ 3 2,
1
√ √ √ √
λ = σ(cid:48) = −1+ω 3 2 = −1− 32 + i 3 3 2,
2 2 2
√ √ √ √
λ = σ(cid:48)(cid:48) = −1+ω2 3 2 = −1− 32 − i 3 3 2.
3 2 2
The eigenvectors being h and g±ik where
√ √ √ √ √
.
h = 1( 3 2,2−2 3 2+ 3 4,− 3 2+ 3 4) = (0.209987,0.177926,0.05458),
6
√ √ √ √ √
.
g = 1 (− 3 2,4+2 3 2− 3 4, 3 2− 3 4) = (−0.104993,0.411037,−0.02729),
12
√ √ √ √ √ √ .
k = 3( 3 2,−2 3 2− 3 4,− 3 2− 3 4) = (0.181854,−0.592829,−0.410976).
12
We can also compute the rotation angle
√ √
3 3 2 .
θ = π −arctan √ = 146.20
2+ 3 2
and after some computation we can express
s = σjh+2ρj(gcos(jθ)−ksin(jθ)),
j 2
and
√
Sg = ρ(gcosθ−ksinθ),
√
Sk = ρ(gsinθ+kcosθ).
Thenorm(1)takeszerovalueontheunionoftheeigenplaneP spanned
by vectors g, k and the eigenline of h. Except for the origin, there is
no rational point (x,y,z) of zero norm.
The basic vectors s with increasing positive j are approaching the in-
j
variantplaneandfornegativej beingalmostcolineartotheeigenvector
h.
On the other hand we can construct the eigenbasis B = (h,g,k) and
e
the conjugate eigenbasis.
6 MITJA LAKNER, PETER PETEK, MARJETA SˇKAPIN RUGELJ
To make the conjugate eigenbasis, we compute the vector products
h∗ = g×k,
g∗ = h×k,
k∗ = h×g
and they constitute the conjugate eigenbasis B∗
e
√ √ √ √ .
h∗ = − 3(1+ 3 2+ 3 4,1,−1+ 3 2) = (−0.185104,−0.0481125,−0.0125055),
36
√ √ √ √ .
g∗ = − 3(−2+ 3 2+ 3 4,−2,2+ 3 2) = (−0.0407668,0.096225,−0.156843),
36
√ √ √
.
k∗ = 1 ( 3 2− 3 4,0, 3 2) = (−0.02729,0,0.104993).
12
Later we shall need also the mixed product
√
− 3 .
[h,g,k] = = −M = −0.0481125.
36
The plane P∗ of the vectors g∗,k∗ is the invariant plane of S∗.
Lemma 4.3. The vectors h∗,g∗ ± ik∗ are eigenvectors of the matrix
S∗.
Proof. First we find the scalar products:
√
(cid:104)S∗h∗,g(cid:105) = (cid:104)g×k,Sg(cid:105) = (cid:104)g×k, ρ(gcosθ−ksinθ)(cid:105) = 0,
√
(cid:104)S∗h∗,k(cid:105) = (cid:104)g×k,Sk(cid:105) = (cid:104)g×k, ρ(gsinθ+kcosθ)(cid:105) = 0.
So it is clear S∗h∗ is orthogonal to both g and k and therefore colinear
to h∗ itself, therefore an eigenvector with a real eigenvalue, the only
one being σ, therefore S∗h∗ = σh∗. The other two cases demand a
little more work to tell apart the two complex eigenvalues, of course
unless we want to go into direct computation. (cid:3)
We can also express the vectors s∗ in terms of the conjugate eigenbasis
j
√ √ (cid:16) π π (cid:17)
s∗ = −2 3 3 2 σjh∗ +ρj(g∗cos(jθ− )−k∗sin(jθ− ))
2
j 3 3
and infer a connection between the basis and conjugate basis via the
matrix
3 1 0
T = 1 0 0,
0 0 1
namely
s∗ = Ts
j j
and also connecting the eigenbasis with conjugate eigenbasis
√ √
Th = −2 3 3 2h∗,
√ √ √
Tg = 332(−g∗ − 3k∗),
2
√
DIFFERENT BASES IN INVESTIGATION OF 32 7
√ √ √
Tk = 332( 3g∗ −k∗).
2
We shall also need some scalar and cross products of the basis vectors.
Lemma 4.4. The scalar products of basis vectors are as follows:
(cid:104)s∗,s (cid:105) = (cid:104)s∗,s (cid:105)
n k 0 k+n
with
(cid:104)s∗,s (cid:105) = 3,
0 −2
(cid:104)s∗,s (cid:105) = 1,
0 −1
(cid:104)s∗,s (cid:105) = 0,
0 0
(cid:104)s∗,s (cid:105) = 0,
0 1
(cid:104)s∗,s (cid:105) = 1,
0 2
(cid:104)s∗,s (cid:105) = −3.
0 3
Proof. Since
(cid:104)s∗,s (cid:105) = (cid:104)S∗ns∗,s (cid:105) = (cid:104)s∗,Sns (cid:105) = (cid:104)s∗,s (cid:105)
n k 0 k 0 k 0 k+n
we only need to read off the first component of s as s∗ = (1,0,0). (cid:3)
j 0
Lemma 4.5. For two consecutive basis vectors we have the cross prod-
uct
s ×s = s∗
−j −j+1 j
and if we jump by one index
s ×s = −s∗ +3s∗.
−j−1 −j+1 j−1 j
Proof. Setting unknown coefficients α,β,γ
s ×s = αs∗ +βs∗ +γs∗
−j −j+1 j−1 j j+1
and taking scalar products in turn with s ,s ,s we get
−j−1 −j −j+1
1 = α·3+β ·1+γ ·0
0 = α·1+β ·0+γ ·0
0 = α·0+β ·0+γ ·1
and from here α = γ = 0 and β = 1. Likewise we prove the second
formula. (cid:3)
8 MITJA LAKNER, PETER PETEK, MARJETA SˇKAPIN RUGELJ
j a |a(cid:48)|
8 (20,-69,-33) 5.34
9 (-9,27,20) 1.68
10 (0,-7,-9) 0.39
11 (-7,-9,0) 0.28
12 (-30,-21,-7) 0.65
13 (-111,-97,-30) 1.89
Table 1. Appropriate vector
5. The shortest coefficient vector and convergents
The vector (0,p,−q) = η(p−qσ) can as any vector be expanded in any
basis B and with integer coefficients.
j
In [11], using ρ basis, we expressed (0,p,−q) with a wider choice of vec-
tors, but coefficients being limited to 0,1,2 or 3: (0,p,−q) = (cid:80)n a s .
j=k j j
Whereas here, since each time only 3 vectors form the basis, we allow
all integer coefficients.
Should p be a convergent to σ, we can control the size of the these
q
coefficients, provided j has been chosen appropriately.
Definition 5.1. Let p be a convergent to σ and a = (a ,a ,a ) the
q 1 2 3
coordinates of η(p − qσ) = (0,p,−q) = a s + a s + a s in the
1 j−1 2 j 3 j+1
basis B . This basis is called appropriate for the convergent, when the
j
vector a(cid:48) = ρ−ja is the shortest for some positive integer j.
4
Example 5.2. Fromthetableofconvergents[11]wetakep = 1251,q =
4813 that is just preceding the relatively big partial quotient b = 14,
11
so that |δ| < 1 in the estimate p−qσ = δ.
14 q
Which j take, to make the vector a(cid:48) shortest? Here are some results in
the Table 1.
And we see that the appropriate j and B to give the shortest a(cid:48) is
j
j = 11.
Theorem 5.3. Let p be a convergent to σ, B its appropriate basis.
q j
Then for its reduced coefficient vector we have
|a(cid:48)| < 2.01.
Proof. We can write p = qσ + δ with |δ| < 1
q
√
(0,p,−q) = ps −qs = p(h+2g)−q(σh+2 ρ(gcosθ−ksinθ))
0 1
√
DIFFERENT BASES IN INVESTIGATION OF 32 9
and when we rearrange the terms
δ √ √ √ 2δ
(0,p,−q) = h+q 3 2 3( 3g+k)+ g,
q q
we see what happens to either term under action of S−j with grow-
ing j. The first term grows exponentially with ρj in the direction of
the eigenvector h, the second decreases with ρ−j and rotates in the
2
eigenplane, and the last term also decreases
δ √ √ π π
a = S−j(0,p,−q) = ρj h+ρ−jq2 3 2 3(gcos(jθ+ )+ksin(jθ+ ))
2
q 6 6
2δ
(2) + ρ−j (gcos(jθ)+ksin(jθ)).
2
q
But if we put K = qρ−3j we get
j 4
δ √ √ π π
a = S−j(0,p,−q) = ρj h+ρjK 2 3 2 3(gcos(jθ+ )+ksin(jθ+ ))
4 4 j
K 6 6
j
2δ
+ ρ−5j (gcos(jθ)+ksin(jθ)).
2
K
j
First we represent the vector a(cid:48) = ρ−ja = a(cid:48)(cid:48) +a(cid:48)(cid:48)(cid:48) with
4
δ √ √
a(cid:48)(cid:48) = h+2K 3 2 3(gcosα +ksinα ),
j j j
K
j
2δ
a(cid:48)(cid:48)(cid:48) = ρ−5j (gcosjθ+ksinjθ),
4
K
j
where α = jθ+ π. As for a(cid:48)(cid:48) we write the square of its norm as a sum
j 6
of three terms
δ2|h|2 √ √ √
|a(cid:48)(cid:48)|2 = +12 3 4K2|gcosα +ksinα |2 +4δ 3 2 3(cid:104)h,gcosα +ksinα (cid:105)
K2 j j j j j
j
= T +T +T .
1 2 3
The last term is estimated independently of K
j
√ √
(3) |T | < 4·1· 3 2 3max(cid:104)h,gcosα+ksinα(cid:105) < 0.894896.
3
α∈R
Denote
δ2|h|2 √
F(K ,α ) = T +T = +12 3 4K2|gcosα +ksinα |2
j j 1 2 K2 j j j
j
andtoeliminatethedependenceonα andδ wedefineanotherfunction
j
|h|2 √
G(K ) = +12 3 4K2max|gcosα+ksinα|2
j K2 j α∈R
j
10 MITJA LAKNER, PETER PETEK, MARJETA SˇKAPIN RUGELJ
or inserting the numerical values
0.07873129
G(K ) < +12.95559953K2 = H(K ),
j K2 j j
j
observing that F(K ,α ) < H(K ). The values of both functions de-
j j j
pend only on the choice of j, the variable K assumes discrete values
j
from a geometric series as K = qρ−3j. The function H(x) = a +bx2
j 4 x2
as a function of continous variable x > 0 features just one minimum at
√
x = (cid:112)4 a with value H(x ) = 2 ab, but the discrete variable K shall
0 b 0 j
almost certainly miss this minimum point. We shall further denote by
x(cid:48) the unique solution to the equation H(x(cid:48)) = H(x(cid:48)ρ−3). Indeed the
4
equation H(x(cid:48)) = H(x(cid:48)ρ−3) reads
4
a a
+bx2 = +bx2ρ−3
2
x2 x2ρ−3
2
and we can easily solve it
(cid:114)
a
x(cid:48) = 4 ρ38 = x0ρ38,
b
with the value
√
H(x(cid:48)) = ab(ρ3 +ρ−3).
4 4
However within the interval [x(cid:48)ρ−3,x(cid:48)] there is exactly one K and this
4 j
defines also the choice of j.
To determine j we have
x(cid:48)ρ−3 < K < x(cid:48),
4 j
x(cid:48)ρ−3 < qρ−3j < x(cid:48).
4 4
Taking logarithms
3 3j
lnx(cid:48) − lnρ < lnq − lnρ < lnx(cid:48)
4 4
and dividing by (−3 lnρ) we get
4
4lnx(cid:48) 4lnq 4lnx(cid:48)
(4) 1− > j − > − ,
3lnρ 3lnρ 3lnρ
4(lnq −lnx(cid:48)) 4(lnq −lnx(cid:48))
1+ > j > ,
3lnρ 3lnρ
(cid:20) 4(lnq −lnx(cid:48))(cid:21)
j = 1+ ∈ N.
3lnρ
Thus we have
F(K ,α ) < H(K ) < H(x(cid:48)).
j j j
