Full text of DOI:10.1112/S0024609302001455

f
C
Bull. London Math. Soc. 35 (2003) 47–54
2003 London Mathematical Society
DOI: 10.1112/S0024609302001455
GENERAL FORMS FOR MINIMAL SPECTRAL VALUES FOR
A CLASS OF QUADRATIC PISOT NUMBERS
PETER BORWEIN and KEVIN G. HARE
Abstract
This paper studies the spectrum that results when all height one polynomials are evaluated at a Pisot
˝
´
number. This continues the research theme initiated by Erdos, Joo and Komornik in 1990. Of particular
interest is the minimal non-zero value of this spectrum. Formally, this value is denoted as l1(q), and this
definition is extended to all height m polynomials as
lm(q) := inf(|y| : y = ꢀ0 + ꢀ1q1 + . . . + ꢀnqn, ꢀi ∈ Z, |ꢀi| 6 m, y = 0).
A recent result in 2000, of Komornik, Loreti and Pedicini gives a complete description of lm(q) when q
is the Golden ratio. This paper extends this result to include all unit quadratic Pisot numbers. A main
theorem is as follows.
Theorem. Let q be a quadratic Pisot number that satisfies a polynomial of the form p(x) = x2 −ax±1,
with conjugate r. Let q have convergents {Ck/Dk} and let k be the maximal integer such that
1
|Dkr − Ck| 6 m
;
1 − |r|
then
lm(q) = |Dkq − Ck|.
A value related to l(q) is a(q), the minimal non-zero value when all ±1 polynomials are evaluated at
q. Formally, this is
a(q) := inf(|y| : y = ꢀ0 + ꢀ1q2 + . . . + ꢀnqn, ꢀi = ±1, y = 0).
An open question concerning how often a(q) = l(q) is also answered in this paper.
1. Introduction
˝
Erdos, Joo and Komornik in 1990 [6] initiated the study of the spectra resulting
from evaluating certain classes of polynomials at values q > 1. Recall the following.
´
Definition 1. A Pisot number is a real algebraic integer, all of whose conjugates
are of modulus strictly less than 1. A Pisot polynomial is the minimal polynomial of
a Pisot number.
It is known that if we evaluate a height m polynomial at a Pisot number, then
it is either zero, or bounded away from zero [3, 8, 11]. Thus the infimums of these
˝
spectra being studied are greater than zero for all Pisot numbers. To this end Erdos,
m
´
Joo and Joo [5] define these infimums as l (q).
´
Received 6 March 2001; final revision 23 October 2001.
2000 Mathematics Subject Classification 11Y60, 11Y40 (primary).
Research of K. G. Hare supported by MITACS and by NSERC of Canada, and P. Borwein supported
by MITACS and by NSERC of Canada.

48
peter borwein and kevin g. hare
Definition 2. Define lm(q) as
lm(q) := inf(|y| : y = ꢀ0 + ꢀ1q1 + . . . + ꢀnqn, ꢀi ∈ Z, |ꢀi| 6 m, y = 0).
We typically denote l(q) := l1(q).
A related area of interest is the case where the class of polynomials is restricted
to polynomials with ±1 coefficients [1, 14]. The minimal value in this case is defined
as a(q).
Definition 3. Define a(q) as
a(q) := inf(|y| : y = ꢀ0 + ꢀ1q1 + . . . + ꢀnqn, ꢀi = ±1, y = 0).
An open question concerning when a(q) = l(q) is answered in Section 5. For
additional history of problems relating to lm(q) and a(q), see [1, 12].
Specific values of lm(q) are calculated for some Pisot numbers q. If q is the Pisot
number that satisfies q3 − q2 − 1, then l(q) = q2 − 2 [13]. If q is the Pisot number
satisfying qn − qn−1 − . . . − 1 then l(q) = 1/q [5, 13]. If q is the Golden ratio (the
greater root of x2 − x − 1) then l2(q) = q3 − 2q2 + 2q − 2 = 2q − 3 (this corrects a
misprint in [3], which used the notation lim inf(u(n2)) for l2(q)). For general m, and
q the Golden ratio, all lm(q) are known. If Fk is the kth Fibonacci number (F0 = 0,
F1 = 1, Fn = Fn−1 + Fn−2), and qk−2 < m 6 qk−1 then lm(q) = |Fkq − Fk+1| [13].
In [1] an algorithm is given to calculate lm(q) for any Pisot number q and any
integer m, limited only by the memory of the computer. Although this method can
make calculations for any given q and m, it obviously only solves the problem for
specific examples. A tabulation of other lm(q) for various m and q, based on these
methods of calculation, is found at [10]. Upon examination of these tables, we find
another pattern, similar to that of lm(q) when q is the Golden ratio. This pattern
concerns lm(q) for unit quadratic Pisot numbers. The class of Pisot numbers we
consider is as follows.
Definition 4. Let P be the set of unit quadratic Pisot numbers. This is easily seen
to be the appropriate roots of polynomials of the form x2 − rx − 1 for r = 1, 2, 3, . . . ,
and x2 − rx + 1 for r = 3, 4, 5, . . . .
Let q ∈ P. This paper shows that lm(q) = |Dq − C| when C/D is a convergent of
the continued fraction of q. A better description of which convergent lm(q) is equal
to is given in Theorem 1, in Section 2.
2. A description of lm(q)
In this section we give a description of lm(q) for all q ∈ P. First though, we need a
few lemmas and definitions. Here and throughout, let a and b be two fixed integers.
Definition 5. We define the sequences {An}∞ and {Bn}∞ as
n=0
n=0
(1) A0 = 0, A1 = 1, An = aAn−1 + bAn−2
;
.
(2) B0 = 1, B1 = 0, Bn = aBn−1 + bBn−2
Lemma 1. Using the notation of Definition 5,
ꢀꢁ
ꢂꢃ
An An−1
Bn Bn−1
det
= (−b)n−1
.

minimal spectral values of quadratic pisot numbers
49
Proof.
ꢀꢁ
ꢂꢃ
ꢀꢁ
ꢂꢃ
An An−1
Bn Bn−1
aAn−1 + bAn−2 An−1
aBn−1 + bBn−2 Bn−1
det
= det
ꢀꢁ
ꢂꢃ
An−1 An−2
= −b det
Bn−1 Bn−2
.
.
.
ꢀꢁ
ꢂꢃ
1
0
0
1
= (−b)n−1 det
= (−b)n−1
.
q
Lemma 2. For n > 0 we have
xn ≡ Anx + Bn (mod x2 − ax − b).
Proof. A simple induction argument is used.
q
Lemma 3. Let q and r be the two roots of x2 − ax − b. Then
An =
1
1
qn +
rn
q − r
r − q
and
r
q
Bn =
qn +
rn.
r − q
Proof. This is a standard result from recurrence relations; see for example [9].
q − r
q
By combining Lemmas 2 and 3 we get the following.
Lemma 4. Let q and r be the two roots of x2 − ax − b. Then
1
1
xn ≡
qn(x − r) +
rn(x − q) (mod x2 − ax − b).
q − r
r − q
The next result is well known in the literature on continued fractions, see for
example [4].
Lemma 5. Let q be a real number, and m be some integer, then the best approxi-
mation to q by C/D, where 0 < D 6 m, is a convergent Cn/Dn of the continued
fraction of q, for n maximal such that Dn 6 m.
One property of a continued fraction C/D is that |D0q − C0| > |Dq − C| for all
0 < D0 < D. The next lemma is reminiscent of those lemmas in [13, Section 3], but
the presentation is different. For this lemma we need the following definition.
P
n
Definition 6. Define R[x] = {p ∈ R[x] : H(p) 6 1}. Here H( i=0 aixi) =
ˆ
max{|ai| : i = 0, . . . , n}.
Lemma 6. Let p(x) = x2 − ax ± 1 be associated with some q ∈ P. Let m be such
0
0
ˆ
that |An| 6 2m|Bn| for all n > 2. Let y ∈ mR[x] such that y ≡ cx + d (mod p(x))
ˆ
where c, d ∈ Z. Then there exists a y ∈ mR[x] ∩ Z[x] such that y ≡ cx+d (mod p(x)).

50
peter borwein and kevin g. hare
Proof. First it is worth noting that we can always find an m such that |An| 6 m|Bn|
as limn→∞ |An/Bn| = |1/r| < ∞.
By assuming that y ≡ cx + d (mod p(x)) with c, d ∈ Z, we show here how to find
0
0
0
ˆ
y such that y ≡ y (mod p(x)) and y ∈ Z[x] ∩ mR[x]. Let
y = anxn + . . . + a0.
If an ∈ Z then continue inductively on an−1xn−1 + . . . + a0. If a0 ∈ Z then continue
inductively on anxn−1 +. . .+a1. If neither a0 nor an is an integer, then use the identity
xn − Anx − Bn ≡ 0 (mod p(x)) along with the fact that |An| 6 2m|Bn| to solve for α
where an + α ∈ Z or a0 − αBn ∈ Z, and |an + α|, |a1 − Anα|, |a0 − Bnα| 6 m. By solving
for α such that a0 − αBn ≡ da0e or ba0c we get two possible values, one negative
and one positive. The possible values have the property that they differ by less than
1/|Bn|. Similarly by solving for α such that an + α = dane or banc we again get two
possible values, one negative and one positive. Let αl be the maximal of the two
negative values, and let αu be the minimal of the two positive values. If α is equal to
either αl or αu either an + α ∈ Z or a0 − αBn ∈ Z and further |an + α|, |a0 − Bnα| 6 m.
Lastly, by noticing that αu − αl 6 1|Bn|, we see that for at least one of these two
values |a1 − α1A1| 6 m. Continue inductively on anxn + . . . + a0 + α(xn − Anx − Bn).
By repeated application of this we see that y0 is such that all of the ai are
integers with the possible exception of two consecutive terms, aj and aj−1. Notice
that anxn + . . . + aj+1xj+1 + aj−2xj−2 + . . . + a0 ≡ c0x + d0 for some d0, c0 ∈ Z. Thus we
j−1
see that ajxj + a
x
≡ (c − c0)x + (d − d0), where c − c0, d − d0 ∈ Z. By Lemma 2
j−1
j−1
we know that ajxj + a
x
≡ ajAjx + aj−1Aj−1x + ajBj + aj−1Bj−1. Thus we have
j−1
ꢁ
By noticing that the determinant of
ꢂ ꢁ
ꢂ
ꢁ
ꢂ
Aj Aj−1
aj
c − c0
d − d0
=
.
Bj Bj−1
aj−1
ꢁ
ꢂ
Aj Aj−1
Bj Bj−1
is ±1, we see that the inverse of this matrix is integral, and thus aj, aj−1 ∈ Z.
q
What is interesting here is that this proof is constructive, and a computer algorithm
can be designed from this. This is described in Section 3.
Theorem 1. Let q ∈ P satisfy a polynomial of the form p(x) = x2 − ax ± 1, with
conjugate r. Let q have convergents {Ck/Dk} and let k be the maximal integer such
that
1
|Dkr − Ck| 6 m
;
1 − |r|
lm(q) = |Dkq − Ck|.
then
It is worth noting that when q is the Golden ratio then the result is equivalent to
[13, Theorem 3.1].
Proof of Theorem 1. The case where |An| > 2m|Bn| satisfies the theorem trivially
with lm(q) = 1. Recall from Lemma 4 that
1
1
xn ≡
qn(x − r) +
rn(x − q).
q − r
r − q

minimal spectral values of quadratic pisot numbers
It follows from this that if
51
X
ˆ
R[x] ≡
αnxn (mod p(x))
for |αn| 6 1 then
ꢀ
ꢃ
X
1
1
ˆ
R[x] ≡
αn
qn(x − r) +
rn(x − q)
(mod p(x))
q − r
r − q
≡ v(x − r) + w(x − q) (mod p(x))
where v ∈ R and |w| 6 1/(q − r)(1 − |r|).
Consider the continued fraction of q, {Ck/Dk}. Lemma 5 indicates that the best
ˆ
linear terms are of the form Dkq − Ck. Lemma 6 indicates that if Dkx − Ck ∈ mR[x]
ˆ
then there exists a y ∈ Z[x] ∩ R[x] (mod p(x)) such that y ≡ Dkx−Ck (mod p(x)). It
m
ˆ
follows that l (q) = Dkq − Ck when Dkx − Ck ∈ mR[x] (mod p(x)) with k maximal.
ˆ
Write Dkx − Ck as v(x − r) + w(x − q). As Dkx − Ck ∈ mR[x] (mod p(x)) we have
|w| 6 m/((q − r)(1 − |r|)). Thus
|Dkr − Ck| = |v(r − r) + w(r − q)|
= |w(q − r)|
m
6
(q − r)
(q − r)(1 − |r|)
m
6
.
1 − |r|
This is the desired result.
q
Corollary 1. Define Fn = rFn−1 + Fn−2 with F0 = 0 and F1 = 1 and q the Pisot
root of x2 − rx − 1. If qk−1(q − 1) 6 m < qk(q − 1) then lm(q) = |Fkq − Fk+1|.
Proof. It is easy to verify that {Fk+1/Fk} are the continued fractions of q. A
simple calculation shows that
1
Fk =
(qk − rk)
q − r
and that r = −1/q. This yields lm(q) = |Fkq − Fk+1| when
m
|Fkr − Fk+1| 6
,
,
,
1 − |r|
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
1
m
((qk − rk)r − (qk+1 − rk+1)) 6
ꢄ
ꢄ
q − r
1 − |r|
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
1
m
(qkr − qk+1) 6
ꢄ
ꢄ
q − r
1 − |r|
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
ꢄ
r − q
q − r
m
qk
6
,
,
1 − |r|
m
qk 6
1 − |r|
(q − 1)qk−1 6 m,
and the result follows.
q

52
peter borwein and kevin g. hare
Table 1. Relation between lm(q) and Fk−1q − Fk
lm(q)
q2 − q − 1
[1, 1]
q2 − 2q − 1
q2 − 3q − 1
q2 − 4q − 1
q2 − 5q − 1
|F0q − F1|
|F1q − F2|
|F2q − F3|
|F3q − F4|
|F4q − F5|
|F5q − F6|
|F6q − F7|
|F7q − F8|
|F8q − F9|
[1, 1]
[2, 3]
[1, 2]
[3, 7]
[1, 3]
[4, 13]
[1, 4]
[5, 21]
[22, 113]
[114, 586]
[587, 3048]
[3049, 15 826]
[15 827, 82 183]
[82 184, 426 742]
[426 743, 2 215 893]
[4, 8]
[9, 19]
[8, 25]
[26, 82]
[14, 58]
[59, 245]
[2, 2]
[3, 4]
[5, 6]
[7, 11]
[12, 17]
[18, 29]
[20, 48]
[49, 115]
[116, 280]
[281, 675]
[676, 1632]
[83, 274]
[275, 904]
[905, 2989]
[2990, 9871]
[9872, 32 605]
[246, 1042]
[1043, 4413]
[4414, 18 698]
[18 699, 79 205]
[79 206, 335 521]
Table 2. Relation between lm(q), Ek−1q − Ek and Gk−1q − Gk
lm(q)
q2 − 3q + 1
q2 − 4q + 1
q2 − 5q + 1
q2 − 6q + 1
q2 − 7q + 1
[1, 5]
|G0q − G1|
|E0q − E1|
|G1q − G2|
|E1q − E2|
|G2q − G3|
|E2q − E3|
|G3q − G4|
|E3q − E4|
[1, 1]
[1, 2]
[1, 2]
[3, 3]
[1, 3]
[4, 4]
[5, 23]
[24, 28]
[29, 135]
[136, 164]
[165, 791]
[2, 2]
[3, 4]
[5, 6]
[7, 11]
[12, 17]
[3, 7]
[8, 10]
[11, 27]
[28, 38]
[39, 103]
[4, 14]
[15, 18]
[19, 68]
[69, 87]
[88, 329]
[6, 34]
[35, 40]
[41, 234]
[235, 275]
[276, 1609]
Table 1 gives the ranges that m is in, for lm(q) = Fk−1q − Fk.
With a proof similar to that of Corollary 1, we get the following.
Corollary 2. Define En = rEn−1 − En−2 with E0 = 0 and E1 = 1. Define Gn =
rGn−1 − Gn−2 with G0 = 1 and G1 = 1. Let q be the Pisot root of x2 − rx + 1. If
q
q
k−3(q − 1)2 6 m < qk−2(q − 1) then lm(q) = |Gk−1q − Gk| and if qk−2(q − 1) 6 m <
k−2(q − 1)2 then lm(q) = |Ek−1q − Ek|.
Table 2 gives the ranges of m for lm(q) = |Gk−1q − Gk| and lm(q) = |Ek−1q − Ek|.
3. Finding the height m polynomials
For q ∈ P with minimal polynomial p(x), we have lm(q) = |Dq − C| for some
integers C and D where C/D is a convergent of q. We are interested in finding the
particular height m polynomial that lm(q) relates to. We notice that Lemma 6 can
ˆ
be implemented as an algorithm. Thus it is sufficient to find a t(x) ∈ mR[x] such
that t(x) ≡ Dx − C (mod p(x)). For this we can use the simplex method [15]. Write
!
n
n+2
X
X
Dx − C +
akxk p(x) =
bkxk
k=0
k
for unknowns ak. We wish −m 6 bk 6 m for all k = 0, . . . , (n + 2), so for the correct
value of n we minimize for h with
−h 6 bk 6 h (1)
and solve for the ak. A careful calculation can yield the minimal value for n that

minimal spectral values of quadratic pisot numbers
53
works as
ꢅ
ꢀ
ꢃ
ꢆ
m + |Dr − C| |r| − |Dr − C|
m|r|
n = ln
(ln(|r|))−1
.
(2)
Using the simplex method in this way works for any polynomial, though the value
for n given in equation (2) is specifically for q ∈ P. Thus if we wish to implement
this algorithm for polynomials that come from some q ∈/ P, we can simply take n
increasing until we find one that works. We now do an example.
Example 1. Let q be the root of q2 − 3q + 1. A simple calculation demonstrates
that l7(q) = 5q − 13. Using equation (2) the minimal value for n is 3. Minimizing h
with respect to the constraints in equation (1) as n = 3 gives h = 305/44 < 7. This
gives a polynomial of
17
4
305
44
305
44
305
44
305
44
305
44
x5 −
x4 −
x3 −
x2 −
x −
.
Here we use the techniques in Lemma 6 iteratively. Notice that at any step, only
three coefficients are altered.
17
4
305
44
305
44
305
44
305
44
305
44
x5 −
x4 −
x3 −
x2 −
x −
≡ 5x − 13 (mod x2 − 3x + 1),
327
305
305
305
520
x5 −
x4 −
x3 −
x2 −
x − 7 ≡ 5x − 13 (mod x2 − 3x + 1),
77
44
44
44
77
371
305
305
553
x5 −
x4 −
x4 −
x3 −
x3 −
x2 − 7x − 7 ≡ 5x − 13 (mod x2 − 3x + 1),
x2 − 7x − 7 ≡ 5x − 13 (mod x2 − 3x + 1),
88
44
44
88
305
44
229
44
305
44
4x5 −
4x5 − 7x4 − 5x3 − 7x2 − 7x − 7 ≡ 5x − 13 (mod x2 − 3x + 1).
Thus we have found a height 7 integer polynomial p(x) where l7(q) = 5q−13 = p(q).
4. Non-unit quadratic Pisot numbers
It is worth noting that Theorem 1 does not work for all quadratic Pisot numbers.
The problem is that Lemma 6 does not work for all Pisot polynomials x2 − ax − b.
For example, if q is the Pisot root of x2 − 2x − 2 (of approximately 2.732 050 808)
then we see that
7
x8 − 2x7 + 3x6 − 3x5 + 3x4 − 3x3 + 3x2 − 3x + 3 ≡ 8 − 3x (mod x2 − 2x − 2)
16
ˆ
ˆ
is in 3R[x] but it is not in Z[x] ∩ 3R[x]. Worse, we have |8 − 3q| = 0.196 152 424 <
0.267 949 192 = l3(q).
5. The existence of an infinite family of Pisot numbers where l(q) = a(q)
It is easy to give an infinite set of Pisot numbers q where l(q) = a(q). We know
from [5] that if qn − qn−1 − . . . − 1 = 0 then l(q) = qn−1 − qn−2 − . . . − 1. It is clear
in this case that l(q) = a(q). This answers [1, question 1] in the negative, as it gives
an infinite family of Pisot numbers where l(q) = a(q).
6. Further research
The counterexample of Section 4 shows that Theorem 1 does not work for all
quadratic Pisot numbers. Despite this, the spirit of Theorem 1 appears to be true.

54
peter borwein and kevin g. hare
Let {Ck/Dk} be the convergents of a quadratic Pisot number q, with conjugate r, not
necessarily a unit. If Theorem 1 could be extended to this case, then lm(q) = |Dkq−Ck|
where k is maximal such that |Dkr − Ck| 6 m/(1 − |r|). Computationally, it appears
that lm(q) = |Dkq − Ck| or |Dk−1q − Ck−1|, but no proof of this is known. It would
be interesting to know if this is indeed the case.
Secondly it would be of interest if a lemma similar to Lemma 6 could be found
that would work for all polynomials p where p(0) = ±1, regardless of the degree
of p(x). If something like this could be found then this could be used to prove, for
q ∈ (1, 2), that l(q) > 0 if and only if q is Pisot. This is believed to be true by a
number of people, see for example [1, 12]. The second part of this lemma easily
extends to arbitrary degree, but it is not clear that there is an algorithm that forces
all but d consecutive terms to be integers (where d is the degree of p(x)).
While searching for patterns among various Pisot numbers, it appears that a nice
description exists for the Pisot roots of x3 − x − 1 and x3 − x2 − 1. Some work has
been done on this [2], but the question is not fully answered as yet.
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Peter Borwein
Department of Mathematics
and Statistics
Kevin G. Hare
Department of Mathematics
and Statistics
Simon Fraser University
Burnaby
Simon Fraser University
Burnaby
British Columbia
Canada V5A 1S6
British Columbia
Canada V5A 1S6
pborwein@cecm.math.sfu.ca
kghare@cecm.math.sfu.ca