proceedings of the
american mathematical
Volume 97. Number
2. June 1986
society
LONG PERIODIC ORBITS OF THE TRIANGLE MAP
MANNY SCAROWSKY AND ABRAHAM BOYARSKY1
Abstract. Let r: [0,1] - [0,1] be defined by t(x) = 2x on [0,1/2] and r(x) =
2(1 - x) on [1/2,1]. We consider t restricted to the domain DN = {2a/pN, N >
1,0 « la < pN,(a,p) - 1} where p is any odd prime. Let k > 1 be the minimum
integer such that pN \2k + 1. Then there are ((p - 1) ■pN~1)/2k periodic orbits of
t| _ , having equal length, and there are k points in each orbit. Furthermore, the
proportion of points in any of these periodic orbits which lie in an interval (c, d)
approaches d - c as pN~] -» 00. An application to the irreducibility of certain
nonnegative matrices is given.
1. Introduction. Let t: [0,1] -» [0,1] denote the triangle map, defined by
T(X)
(2x,
0<x<
1/2,
\2(l-x),
1/2 < jc < 1.
This map, together with its topological conjugate map g(x) = 4x(l - x), have in
recent years played an important role in understanding the dynamical behaviour of
one-dimensional models of certain nonlinear phenomena [1, 2]. These maps have
periodic orbits of all periods and aperiodic orbits which approach neither an
equilibrium nor a periodic orbit. "From a practical viewpoint it is probably
impossible to distinguish an aperiodic orbit from one with a very long period" [2, p.
107]. The purpose of this note is to give mathematical substance to this conjecture,
by proving the existence of long periodic orbits which "exhibit" Lebesque measure,
the unique absolutely continuous measure invariant under t. We say that a periodic
orbit "exhibits" a measure p on [0,1] if the proportion of points in the orbit that lie
in an interval (c,d) approaches fi([c, d]).
As an application of the long periodic orbits that we shall display, we shall be able
to prove the reducibility or irreducibility of certain large, sparse, nonnegative
matrices.
2. Main result.
= 1}, where p is
that x and p are
D -» D, and that
Consider t with the domain D = D n = {0 < 2x/pN < l,(x, p)
an odd prime, N is an integer > 1, and (x, p) = 1 denotes the fact
relatively prime. Let / = t|0: D -* [0,1]. It is easy to show that /:
\D\, the number of points in D, is equal to ((p - l)/2) •pN~1.
Received by the editors September 21, 1984.
1980 Mathematics Subject Classification. Primary 28D05; Secondary 15A36.
'Research supported by NSERC Grant #A9072 and an FCAC grant from the Quebec Department of
Education.
©1986 American
Mathematical
Society
0002-9939/86 $1.00 + $.25 per page
247
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248
MANNY SCAROWSKY AND ABRAHAM BOYARSKY
Theorem 1. The map f: D -» D is an isomorphism and induces a partition of D into
disjoint periodic orbits of equal length. Let k = k(pN) be the minimum integer > 1
such that pN 12k ± 1. Then the number of points in each orbit is k, and the number of
periodic orbits is ((p — 1) ■pN~x)/2k.
Proof. Using the facts that
f(2x/pN)
= 2(even integer)//?",
f(2x/pN)
= 2(odd integer)/p",
if 2x/pN < \,
and
if 2x/pN > \,
it is easy to see that / is an isomorphism.
Now, if 2x/pN has period /, then A ± 2'(2x/pN) = 2x/pN, A g Z. This implies
that pN 121+ 1, and it follows that / must be a multiple of k. Note that
k = k(p») =
X{pN)/2
if A(/?N)iseven,
X(pN)
if A(/?")isodd,
where X(pN) is the order of 2 (modp"). This follows from the fact that X(pN) is
the minimum integer ^ 1 such that pN \2X(-P' — 1: since pN\22k - 1, we have that
k < X(pN) < 2k, i.e., 1 < 2k/X(pN) < 2. Since 2k/X(pN) is an integer, X(/?Ar)=
k,2k. Thus /c = X(pN), X(pN)/2. Clearly, if X(pN) is odd the first case occurs,
and if X(pN) is even, the second case occurs.
We now show that if 2xx/pN g D, then fk(2xx/pN) = (2xx/pN). It is easy to
verify that
(1)
*(*)-
2mx - /,
x g [i/2m,(i
+ l)/2m],
/ even,
-2mjc + i + 1,
* g [i/2m,(i
+ l)/2m],
/odd.
This follows from the continuity of rm, which has slope ±2m, and from the fact that
Tm(0) = rm(l) = 0. Let x = 2xx/pN. Then i is determined by i/2m < 2xx/pN <
(i + Y)/2m,
i.e.,
ipN < 2m+1xx < (i + \)pN,
2m+lx1/pN - i < 1. Therefore,
0 < 2m+lxx - ipN < pN,
i = [2m+1xx/pN],
Using
T*(2x1/piV):
(2)
(3)
2xx
)* + L
if
2*+1^
2xx
is even,
+ 1 if
L P'
is odd.
L P
But
2k+lxx/pN
= (2*+1 ± 2)xx/pN
+ 2*!//»^.
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or
0 <
(1), we can determine
249
LONG PERIODIC ORBITS OF THE TRIANGLE MAP
If p"|2* + 1, then [2k + 1xx/pN] = (2* + 1 + 2)xx/pN - 1 is odd, and using (3), we
obtain rk(2xx/pN) = 2xx/pN. Similarly, if pN\2k - 1, [2k + 1xx/pN] is even and
from (2) we obtain rk(2xx/pN) = 2xx/pN. Thus, the period of 2xx/pN is exactly k
and the number of periodic orbits is \D\/k = ((/? - 1) • pN'l)/(2
■k).
Q.E.D.
Remark. If for some positive integer m', pm'\\2k(p) ± 1, i.e., the division is exact,
then k(p) = k(p2)=
■■■ = k(pm).
In Table 1, we display for N = 1, the period k and the number of the periodic
orbits for 3 < p < 100. Note that /c = k(p) satisfies 1 « it < (p - l)/2 and
k\(p - l)/2. Since p 12*± 1, 2* >/> - 1, and k > log(p - 1). Therefore, /c(p)->
oo as p -» oo.
Table 1
period of
orbit
prime
number
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
(f-l)/2
1
2
3
5
6
8
9
11
14
15
18
20
21
23
26
29
30
33
35
36
39
41
44
48
number of
periodic orbits
k
)/2k
(P
1
2
3
5
6
4
9
11
14
5
18
10
7
23
26
29
30
33
35
9
39
41
4
11
24
2
Lemma 1. Let p\2X{p) — 1, where X(p) > 1 is minimal and assumep'
'\\2Mp)-l.
Then for any N > m', pN\2X' - 1 iff X' is a multiple of X(p)pN~ ", and the
minimum value of X' > 0 such thatpN \2X — 1 equals X(p)pNm
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.
MANNY SCAROWSKY AND ABRAHAM BOYARSKY
250
Proof. Clearly the result is true for N = m'. Assume the result is true for
N = / > m'. Then p/+1 |2V - 1 implies p'|2v - 1 implies X = a- X(p)pl~m', and
pi+^\2aMp) "'""' - 1. But
2«*/>V-'
- 1 = (1 + 8pm')ap'~m - 1 = aSp1 + multiple of p'+1,
where (8, p) = 1. Thus p \ a and the lemma is proved.
Q.E.D.
Remark. For p < 500,000, m' > 1 only for p = 1093 and p = 3511, in which
case m' = 2.
Theorem
2. The proportion of points in a fixed period orbit which lies in an interval
(c, d) approaches
Proof.
d — c aspN~m
-» oo.
Let 2m + 1xx = qmpN + rm, where qm, rm G Z, 0 < rm < pN and 2xx/pN
G
D. For technical reasons we let m vary from 1 to 2k. Then 2m+ ixx/pN = qm + rm/pN
and
2x
rm/PN>
rm even,
(/»"-O/P*.
''«odd-
Let # { } denote the number of points in the set { }. Then
#{m:
(4)
Tm(2xx/pN)
= #{m:
g (c,d),
rm/pN <=(c,d),
1 < m < 2k)
rm wtn;{pN
- rm)/pN
^(c,d),rm
odd)
= # {m: pNc < rm < pNd, rm even; pNc < pN - rm< pNd, rm odd}.
Now, by Lemma 1, jfc- k(pN) - k' pN~m', where k' = k(p) and JV> w'. Let
/■(,...,r'k„ be the residue classes of the ri (modp), i.e., the residue classes when
N = 1. Note that k" = k' or 2A:'. Then it is easy to see that rm = pm'i + rj,
0 < / < pN~m\ 1 <j < 2k', for N > m'. Hence, the right-hand side of (4) equals
# {m : pNc < pm'i + rj < pNd, i + rj even;
pNc <pN
- pm'i - rj < pNd, i + rj odd)
2k'
(5)
= L # {'■ PNC- rj < pm'i < pNc - rj, i + rj even}
2*'
+ E # {'': />"c + /•/ - pN < -pm'i < pNd + rj - pN, i + rj odd}.
7= 1
Since the number of even (odd) integers in an interval (e, f) is ((/ - e)/2) + 0(1),
the right-hand side of (5) sums to
g(£^^+0(l))+|(/^pl+o(1))
= 2k'pN-m'(d-
c) + O(k').
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251
LONG PERIODIC ORBITS OF THE TRIANGLE MAP
Dividing through by2k = 2k'pN~m',N^m',weget
#{m:
rm(2xx/pN)
G (c,d),l
< m < k}/k
= (d - c) + 0(\/pN-m).
The right-hand side approaches d - c as pN~m' -» oo. Q.E.D.
Remark. For p fixed, we need N -» oo. For p increasing it is enough that
N = N(p) > m', as p -* oo.
Corollary
1. Le/ Aka
bounded, continuous function on [0,1]. TÄe«
lim lÍh(Ac))=fh(x)dx.
¡,"--'-,00
« ,= 1
•'O
Proof. Given any e > 0, Theorem 2 implies that
1 k
(6)
/"i
K i-i
< e
'o
if pN~m' is sufficiently large, uniformly in (c, d) and x g Z), where x,4 is the
characteristic function of the set A. The same is true if (c, d) is replaced by any
closed or half-open interval. Thus (6) is true if Xte,d) is replaced by X/ where / is
any interval, and also for step functions hn = La,X/. where the summation is over a
finite number of terms. It follows that if sup\hn(x) - h(x)\ -» 0, where hn is a step
function, then (6) is true with X(C,d) replaced by h = lim hn, i.e., if x g D, and h g
the space of regular functions which is the closure of the space of step-functions
under the uniform norm, and includes the space of functions of BV. See [4, p. 139].
1 *
lim
- Y,h(Ti(x))=
pN-m'-> oo k k = x
jefl
i
( h(x)dx.
Q.E.D.
Jq
3. Application to the irreducibility of certain nonnegative matrices. Consider the
following class si of n X n matrices for various values of n. If n is even then
A „ g s/ has the form
(0
0\
1
0
1
(7)
0
A„ =
0
0
U
1
1
0 0
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0
0
0
1
1
0
MANNY SCAROWSKY AND ABRAHAM BOYARSKY
252
and if n is odd, then An G s/ has the form
CO 1 0
0
.
...
10..
0
1
0
(8)
0
0
0
0 1
0 0
10
0
1
0
1
U o .
It is not hard to verify that A2 and Ab are irreducible, but that As is reducible
into two irreducible blocks. Also, A3, A5, Ag, Axx are irreducible, but AX5 is
reducible into 3 irreducible blocks. The problem is to determine the dependence of
the irreducibility of An on n. We shall present a partial solution to this problem
using the results of §2.
Let us associate with the map /: D -* D a 0-1 matrix A where the Z/'th entry is
equal to 1 iff f(2i/pN) = 2j/pN, 1 < /, j < pN - 1, and (/, p) = (j, p) = 1 (i.e.,
we omit the /th row or yth column where p divides / or j); otherwise the yth entry
is 0. A is an n X n matrix where n = ((p — l)/2) • pN~l. If (p - l)/2 is even, and
N = 1, then A has the form (7), where n = (p - l)/2.If (/? — l)/2 is odd, and
N — 1, then A has the form (8), where n = (p - l)/2.
Let N = 1. Then n = (p - l)/2 is the dimension of the square matrix An. Using
the results in Table 1, we see from the fourth column how many periodic orbits
correspond to a given p. Clearly, when there is only 1 such orbit, An is irreducible.
Hence for n = 1,2,3,5,6,9,...,
An is irreducible. But for n = 12,20,21,36,...,
An is reducible, with the number of irreducible blocks given in the fourth column of
Table 1. Thus, An can be permuted into a matrix with d = (p - l)/(2 • k)
irreducible blocks (if A^> 1, then d = ((p
l)/(2 • k)) ■pN~1). That is, there exists
a permutation matrix P such that
Ai
PAP'1
where all the A'n's are irreducible matrices having the same dimension
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k given in
253
LONG PERIODIC ORBITS OF THE TRIANGLE MAP
column 3 of Table 1. Each block A'n corresponds to a period of /, say ax —>a2 ■■■ -» a,.. We can label the rows and columns of A'„ so that each block looks like
0
1
0
/0
0
0
0
1
BL=
1
0
0
0
1
0,
Now this is a circulant matrix and its eigenvalues are the klh roots of unity [3, p.
309]. Thus B'n has jc* — 1 as its characteristic polynomial. Since xk - 1 has no
repeated roots, xk — 1 is also the minimal polynomial. It follows that the characteristic polynomial of An is (xk - \)d, and its minimal polynomial is xk - 1.
4. Concluding remarks, (i) All the foregoing results carry over to maps g topologically conjugate to t, i.e., when there exists a homeomorphism h: [0,1] -* [0,1] such
that g = h o t o h'1. An important example is the logistic map g(x) = 4.x(l — x) [2].
(ii) It is easy to show from the law of quadratic reciprocity that the number of
periodic orbits is even iff p = 1 + 8r, where r is any positive integer.
(iii) For TV= 1 the uniform distribution result of Theorem 2 does not necessarily
hold as p -* oo. Assuming, for instance, that there are an infinite number of primes
of the form p = 2' ± 1 (the primes of the form 2' - 1 are known as the Mersenne
primes), one sees easily that
a = #{m:
rm(2/p)
g (c,d),l
< m < k' = 1} = log(d/c)
+ 0(\),
c> 0.
Dividing through by /, we get a/1 -» 0 as p -» oo. We conjecture that Theorem 2
and Corollary 1 still hold if we restrict p to a class of primes such that X(p), the
order of 2 (mod p) is greater than c ■p, for some constant c.
(iv) Theorem 1 is true for Dn = {0 < 2x/n < 1, n odd, (x, n) = 1}, where
n = [1*!^'.
We conjecture that Theorem 2 is true if IïfLiPii'~m' -* oo. Also,
since fractions of the form x/2an, a > 0, (x, 2) = 1 are clearly eventually periodic
but not periodic, it follows that we have characterized the periodic behaviour of all
rational points under the map t.
(v) If we take /: [0,1] -» [0,1] to be the continuous map with /(0) = 0, f(\/n ) = 1,
f(2/n) = 0,...
so that / has slopes n, -n, alternately, and D = DN =
[a/pN,(a, p) = 1, a even}, where p is an odd prime and (p,n) = 1, then Theorem
1 still holds. It follows from well-known results about primtive roots; see e.g. [5, p.
50], that for any odd prime p there exists an n such that « is a primitive root of pN
for all N and thus for this n, DN will consist exactly of one orbit (N > 1). More
interestingly, it follows from a recent result of Gupta and Murty [6] that there is an
explicitly given set of 13 numbers—in fact, many such sets—such that one of those
numbers,—call it n—is a primitive root of infinitely many primes. So for these
primes, /: Dx -» Dx consists of exactly one orbit.
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254
MANNY SCAROWSKY AND ABRAHAM BOYARSKY
References
1. R. May, Stability and complexity in model ecosystems, 2nd ed., Princeton
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N.J., 1978.
2. J. Guckenheimer,
G. Oster and A. Ipaktchi, The dynamics of density dependent population models, J.
Math. Bio. 4 (1977), 101-147.
3. B. Noble, Applied linear algebra, Prentice-Hall, Englewood Cliffs, N.J., 1969.
4. J. Dieudonné,
Foundations of modern analysis, vol. 10, Academic Press, New York, 1960.
5. A. A. Gioia, 77ie theory of numbers, Markham, Chicago, 111.,1970.
6. R. Gupta and M. Murty, A remark on Artin's conjecture. Invent. Math. 78 (1984), 127-130.
Department of Mathematics,
Canada H4B 1R6
Loyola
Campus, Concordia
University,
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Montreal,
Quebec,