A
LFRED
J. V
AN DER
P
OORTEN
Symmetry and folding of continued fractions
Journal de Théorie des Nombres de Bordeaux, tome
14, n
o2 (2002),
p. 603-611
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Symmetry
and
folding
of continued
fractions
par ALFRED J. VAN DER POORTEN
To Michel France on his 65th birthday
RÉSUMÉ. Le ’lemme de
pliage’
de Michel Mendès Francefour-nit une nouvelle
justification
de lasymétrie
dudéveloppement
enfraction continue d’un irrationnel
quadratique.
ABSTRACT. Michel Mendès France’s
’Folding
Lemma’ forcontin-ued fraction
expansions provides
an unusualexplanation
for thewell known symmetry in the
expansion
of aquadratic
irrationalinteger.
1. Continued fractions
A continued
frdction
expansion
is
conveniently represented by
[ago ,
al ,a2 , ... .
Set[ ao ,
al , ... ,ah] =
Xh/Yh
for h =0, 1, 2,
.... Then the definition( ao ,
al ,... ,
ah
]
=ao
+ 1 / [ al ,
... , ahI
immediately implies by
induction on h that there is acorrespondence
between certain
products
of twoby
two matrices and continued fractions. It followsby transposition
of matrices thatManuscrit requ le 5 f6vrier 2001.
and
by
taking
determinants thatProposition
2. One then sees thatIt follows that if a :
Specifically,
if thepartial
quotients
ai, a2, ... arepositive integers,
as isthe case for the
simple
continued fractionexpansions
we discussbelow,
anda is
irrational,
thenillustrating
thequality
ofapproximation
of theconvergents
xh/yh
of a.Continued fractions are well studied
objects,
so newelementary
resultsconcerning
their structure are asurprise.
Proposition
3(Folding
Lemma[2]).
We haveHere we have set Wh =
al, a2 ... , ah;
accordingly
the notation-w
de-notes -ah, -ah-17- -- , -al. It will save space to write a for -a.Proof.
Here E--~ denotes the’correspondence’
between twoby
two matrices and continued fractions. We haveas
alleged.
Moreover,
( c -
== c ,
]
by Proposition
1. 0The
adjective
’folding’
isappropriate
because iteration of theperturbed
symmetry w -
w, c,-iii
yields
apattern
ofsigns
on the word wcorre-sponding
to thepattern
of creases in a sheet of paperrepeatedly
folded inhalf;
see[1].
Incidentally,
it is often convenient todrop subscripts
and to use theconvention
whereby if,
say, y denotes yh, theny’
denotes theprevious
y;that
is,
y‘
= yh-i .The
folding
lemma makes it easy,given
theexpansion
of apartial
sumof a
series,
toadjust
a continued fractionexpansion
for anappended
termif the
intervening ’gap’
is wideenough.
Ourexample
is from the function field case, where’degree’
in Xreplaces
thelogarithm
of the absolute value.of
degree -2n
or less. Of course if thedegree
of theappended
term is much~ less,
that introduces apartial
quotient
ofcorrespondingly high degree.
As an
application
of the lemma: the continued fractionexpansion
of the sumis
given
sequentially
by
Jwhere the
addition of each term is done
by
a ’fold’ with c =-X ;
see~4~ * .
2.
Symmetry
The
computation
shows how to make
negative partial quotients
positive.
Specifically
Here we use
having
recalled thatThus
folding
yields
asymmetric
continued fractionexpansion, perturbed
(cf
[3])
in thegiven example by
the word ah , c -1,1,
ah - 1.This raises the
question
as to whether every such’symmetric’
expansion
arises fromfolding.
Mind you, it should not raise any suchthing.
If thegiven
symmetric
expansion
yields
theconvergent
X/Y
and its ’first half’ isx / y
thennecessarily
is of the for someappropriate
c.So,
a betterquestion
is to ask what onemight
learn fromviewing
asym-metric
expansion
as a foldedexpansion.
The best known case of asym-metric
expansion
is theperiod
of a realquadratic
irrationalinteger.
Below*
Jeffrey Shallit recently noticed that the evaluation just given is foreshadowed in remarks of Walter Leighton and W. T. Scott, ’A general continued fraction expansion’, Bull. Amer. Math. Soc. 45 (1939) 596-605.
3. Periodic continued fractions
Suppose
the realquadratic
irrationalinteger
6 has norm n and trace t.That
is, b2 -
t6 -I- n = 0 withintegers n and t, whe_re t2 -
4n > 0 is not asquare in Z. To fix
6,
we suppose that 6 >6,
where 6 denotes theconjugate
of 6.Let
ao ,
al ,... , ah ~ - ~ ao , ]
be an initialsegment
of thecontinued fraction
expansion
of 6. We willplay
with the matricesnoticing
first that thedecomposition
defines sequences
(Ph)h>o
and(Qh)h>o
ofintegers.
Here,
thedeterminant x 2 -
txy
+ny2
ofNh .
AlsoBecause
x/y
is aconvergent
of 6 we have1/y,
and soQ
is atmost 6 - 3. It follows from several
applications
of the boxprinciple
that there arepositive
integers
X and Yproviding
a solution to ’Pell’sequation’
Proposition
4. tXY +nY2
= ::1:1 andX/Y
=( ao ,
... , then
By
the matrixcorrespondencet
this isNotice moreover that
Thus the word ao, al, ... , ar + t must be a
palindrome~.
t There is a subtle point here at which I tacitly suppose that X 2 - tXY + ny2 = ( -1 )’’ .
However, that loses no generality, because if X/Y = [... ar also X/Y = [...
ar - 1 , 1 ). t Anubis: Dog as a devil deified lived as a god.
4.
Halfway
in theperiod
Suppose, first,
that theperiod
of 6 is of evenlength
r = 2h + 2. ThenN =
Nh
corresponds
to half theperiod
in the sense thatN2
= (~
x Y
provides
X/Y
yielding
theperiod
as inProposition
4.Thus we consider
whence
By
(2)
we haveand so,
indeed,
theFolding
Lemma entails thatProposition
5.Suppose
that thequadratic
irrationalinteger
6satisfies
6 >3
and has a continuedfraction
expansion
period of
evenlength
2h + 2.Then
if
andonly if Q
divides 2P + t.Proof.
Thenecessity
of the condition is clear since otherwise thegiven
expansion
is not admissible. For itssufficiency
we need to confirm thatX/Y
doesyield
a solution to ’Pell’sequation’.
That is not obvious:plainly
X2 -
tXY +ny2
=Q2.
However,
it turns out that bothQIX
andQI Y.
To seethat,
considerthe conditions
(2)
moduloQ,
thatis,
Px - -ny andPy - x -
ty.
Recall we suppose 2P+t - 0. Then X
= x2 -ny2 ==
= 0 and Y =2xy-ty2
= =(Py+(P+t)y)y -
0,
assuggested.
0Suppose,
second,
that theperiod
of 6 is of oddlength
r = 2h + 1. Then Ncorresponds
to half theperiod
in the sense thatprovides
whence
By
(2)
and so
provided
thatQ
=Q’;
here w’ denotes the wordaI, a2,... , ah-1.
Proposition
6.Suppose
that thequadratic
irrationalinteger J satisfies
6 >3
and has a continuedf raction
expansion
period of
oddlength
2h + 1.Then
if
andonly if Q
=Q’.
Remark. We will see below that
Pk
+ + t =akqk
for k =0, 1,
... , so
when
Q’
=Q
therequirement
+ P +t)
isautomatically
satisfied.Proof.
Thenecessity
of the condition is clear since otherwise thegiven
expansion
is notsymmetric;
inparticular
we would not have ah =(P’
+P +
t) /Q.
For itssufficiency
we need to confirm that doesyield
asolution to ’Pell’s
equation’.
That’s not obvious:plainly
=_QIQ = -Q2.
However,
bothQ I X
andQ I Y.
Recall we haveQ’
=Q
and so P’ +5. Remarks and
explanations
5.1. Some identities. The
equations
Px +Qx’ =
-ny andPy
+Qy’ =
x -
ty
together
entail thatBrute
computation
now shows thatWe will avoid the need for such
brutality by showing
that our alsotaking
the
equations
(2)
in the formP’x’ -f- Q’x" - -ny’
andP’y’ +Q’y"
=x’ - ty’
provides
a base from which one can prove the facts we wantby
induction. Here we note that x =ax’
+ x"
and y
=ay’ + y",
where a = ah. Thenand
Multiplying
theseequations
respectively by x
andby
y, andsubtracting,
yields
That
is,
aspromised,
As
just
noted,
we therefore haveand
Multiplying
theseequations
respectively by
n andby P,
andadding, yields
is
step h
in the continued fractionexpansion
of J. Just so,(4)
fixes thereciprocal
of the remainder in(5)
above and thusQh+l
and the nextcom-plete quotient
(6
~-Moreover,
because theQ’s
arepositive
itfollows from
(4)
alone that -a-3
for h =0, 1,
.... Mind you,we also know that the
complete quotients satisfy
(6
+P)/Q
> 1 and the remainders -1(~ + P)/Q’
0.Given the rules
(3)
and(4)
defining
the continued fractionexpansion
it is aninteresting
exercise to retrieve the conditions(2)
and so thedecompo-sitions
(1)
of the matricesNh.
The trick is to useinduction,
showing
that+
-ny’
andP’y’
+Q’y" = x’ - ty’
entails Px +Qx’ =
-nyand
Py
+Qy’
= x -ty.
Thepoint
of this all is toemphasise
that thetwo
approaches
areequivalent,
confirming
that the matricesNh
detail thecontinued fraction
expansion
of 6.5.3.
Binary
quadratic
forms. The matrixNh
determines aquadratic
form
(2P
+t)uv
+(-1)hQ’v2
of discriminant4p2
+ 4Pt +t2
+4QQ’
= t2 -
4n= (6 -
a)2.
That is, it provides
P andQ,
and thediscriminant,
and thereforeQ’.
It
might
take us a little too far afield toprovide details,
so I leavecon-firmation of the
following
to the interested reader. It turns out thatmul-tiplication
of apair
andNj-l is
composition
of thecorresponding
quadratic
forms.Indeed,
this is a natural way to definecomposition.
Thereare
provisos.
Whereas the forms determinedby
theNh
arereduced,
thecomposite,
which apriori
hasleading
coefficients is ingeneral
notreduced.
Moreover,
it may not beprimitive.
Its coefficients will have thecommon factor
G2,
where G =gcd(Qt,
Q~,
+t).
One must thereforealways
consider the’primitive’
matrixproduct
G-1Ni-lNj-l.
5.4. The
’results’,
specifically Propositions
5 and6,
given
here are of course well known. I doubt however whetherthey
have heretofore beenproved by
theFolding
Lemma.My
remarks on the matricesNh
containsome
previously unpublished
observations. One motivation forviewing
continued fractions of
quadratic
irrationals as astudy
of theNh
is thatdoing
itsuggests
an immediategeneralisation
toapproximation
algorithms
for
higher degree algebraic
irrationals. References[1] F. M. DEKKING, M. MENDÈS FRANCE, A. J. VAN DER POORTEN, FOLDS!. Math. Intelligencer
4 (1982), 130-138; II: Symmetry disturbed. ibid. 173-181; III: More morphisms. ibid.
[2] M. MENDÈS FRANCE, Sur les fractions continues limitées. Acta Arith. 23 (1973), 207-215. [3] M. MENDÈS FRANCE, Principe de la symétrie perturbée. Seminar on Number Theory, Paris
1979-80, 77-98, Progr. Math. 12, Birkhäuser, Boston, Mass., 1981. [MR 83a:10089]
[4] A. J. VAN DER POORTEN, J. SHALLIT, Folded continued fractions. J. Number Theory 40
(1992), 237-250.
Alf VAN DER POORTEN
ceNTR,e for Number Theory Research
Macquarie University
Sydney, NSW 2109
Australia