J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
C. G. P
INNER
A. J. V
AN DER
P
OORTEN
N. S
ARADHA
Some infinite products with interesting continued
fraction expansions
Journal de Théorie des Nombres de Bordeaux, tome
5, n
o1 (1993),
p. 187-216
<http://www.numdam.org/item?id=JTNB_1993__5_1_187_0>
© Université Bordeaux 1, 1993, tous droits réservés.
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Some
infinite
products
with
interesting
continued fraction
expansions
by
C. G.PINNER1,
A. J. VAN DERPOORTEN2
AND N.SARADHA3
RÉSUMÉ. We display several infinite products with interesting continued fraction expansions. Specifically, for various small values of k -
necessarily excluding k = 3 since that case cannot occur, we display infinite products
in the field of formal power series whose truncations yield their every k-th convergent.
1. Introduction
Let ... be a sequence of rational
integers,
and letF(x)
be the infiniteproduct
with continued fraction
expansion
~ao(x~ , ai(z) , ... ] .
In[1]
it is shown that for n =0, 1, 2,
... each truncationis a
convergent
ofF( x ~
if,
andonly if,
1980 Mathematics Subject Classification (1985 Revision). 11A55. Mots-c]6s : continued fraction, infinite product.
Manuscrit reçu le 10 juin 1993.
1
This work was done in part at the Mathematical Sciences Research Institute, Berkeley. 2 Work
supported in part by grants from the Australian Research Council and by a
re-search agreement with Digital Equipment Corporation.
3
Honorary research fellow attached to the ceNTRe for Number Theory Research,
Moreover,
then thedegree
of thesubsequent partial
quotient
isgiven
by
Furthermore,
one sees that the truncationsprovide
precisely
every sec-ondconvergent
if,
andonly if,
for
n = 0, 1, 2, ....
Infact,
given
(5),
thepartial
quotients
ofF(x)
can
readily
be listedexplicitly
and are ofrapidly increasing degree.
Hence-forth,
weonly
considerproducts
for which(3)
is satisfied. So their everytruncation
yields
aconvergent.
Here we will be concerned with classes of infinite
products
whose contin-ued fractionexpansions
have theproperty
that thoseconvergents
arising
from the truncations of theproduct
areagain separated by
some fewcon-vergents.
In this context we say that two consecutive truncationsand of our
product
arek -apart if,
assuggested by
thenota-tion,
there areexactly k -
1 intermediateconvergents.
Our aim is to use the above mentioned2-apart
examples
to construct additionalproducts
with truncations a fixed bounded distanceapart.
We shallexplicitly give
examples
withregular
gaps of size4 , 5 , 6
and 8. We also illustrate howthe matrix
interpretation
of continued fractions can benicely employed
tocleanly
generate
thepartial
quotients
of theproducts
westudy.
2.
Terminology
We use the
terminology
of[1]
and[2].
Namely, given
a fieldK,
we let L =lI~((~-1 ))
denote the field of formal Laurent series inx-1
over K. Then each F E L is ofshape
and we say that the
degree
of F isdeg
F = d. Theintegral
part ofF,
in
K[x].
A series F E L has aunique
continued fractionexpansion
denotedby
where the
partial
quotients
ch arepolynomials
in x , and havedegree
atleast 1 for h > 1. From the
general theory
of continued fractions we have the fundamentalcorrespondence
whereby
for h =o, l, 2, ...
if,
andonly
if,
defining
theconvergents
ph/qh
by products
of certain matrices.Further,
on
setting
we observe that for c E
7~ ,
B /
Then we may write
formally
that,
, - , , /
whence if ~2013~ denotes the
correspondence
between matrixproducts
andcontinued fractions we have
The
correspondence
between continued fractions and the so called R-L sequences isparticularly
convenient instudying
themultiplication
or divi-sion of a continued fractionby
a rationalquantity.
This is illustrated indetail in
[1].
We assume the relevantterminology
here.We also recall from
[2]
that a rational functionp( x) / q( x)
is aconvergent
Phlqh
to Fif,
andonly if,
and that the
degree
of the nextpartial quotient
is thengiven by
3. The
Examples
RevealedAs mentioned
above,
wealready
know a class of infiniteproducts
such that the truncations arealways
2-apart;
namely
where A =
(Ao, al, ... ~
satisfies the conditionsfor all n > 0. It
transpires
that we can use such known cases to constructa slue of new
examples
with theproperty
we want. Inparticular
ifthen the truncations of G
(although
ingeneral
nolonger
2-apart)
will still be of bounded distanceapart:
THEOREM 1. Let
where
Then for a~I n =
0, 1, 2,
... , the successive truncations of
G(x)
are
convergents
of G at most(2r
+2)
-apart.
Proof.
By
criterion(3)
it follows that the stated truncationscertainly
areMoreover the condition
2Ài.1 I
Ài+1
leads to a, so tospeak, gratuitous
inter-mediate
convergent
with
as may be
readily
checked with anappeal
to(6).
Between them these twopartial quotients
soak up most of the availabledegree
and(as
theremaining
are of
degree
at leastone)
we obtainNote: we see at once the the truncations are
2-apart
if andonly
if,6;-- v
and that otherwise
(since
by
degree considerations j #
1 1~ -1 )
we must have
separation
k > 4. We shall see later that there are in factno
3-apart
examples.
Since one canenvisage
cases were all theremaining
quotients
are indeed linear the bound 2r + 2 seemsreasonably
precise;
we shall see cases where this bound issharp
when r =1,
2 or 3.Not
surprisingly,
thephenomenon
of oneproduct
inheriting
thebound-edness
property
from another is rather moregeneral:
THEOREM 2. If
F(x)
andG(x)
are infiniteproducts
(whose
truncationsare all
convergents)
which differonly by
a factorconsisting
of a rational function, , 1"/ ’B
then the truncations of
G(x)
are of bounded distanceapart
if,
andonly if,
the truncations of
F(x)
are of bounded distanceapart.
LEMMA 1.
Suppose
thatpilqi
and areconvergents
to a Laurent seriesF(x)
such thatare
convergents
toThen
Proof.
Theproof employs
a similardegree counting
argument:
We suppose that there are M + 1
convergents
(0
j
M)
to
G(x)
withFurther suppose that there are N
convergents
(1 j N)
to
F(x)
with 0n(j)
k anddeg
>deg fg.
Now sincewe see from criterion
(6)
thatall are
convergents
to G withthese
convergents
monopolise
most of the availabledegree. Hence,
as theremaining
convergents
are at worst linear we see thatThe bound follows at once since
trivially
N k and t M .Notice that we made no
prior
assumption
here about the nature ofF(x);
forexample
wemight
considerproducts
with
to ensure that the ’truncations’
are
convergents
(needless
to say we shall do no suchthing
here).
4. The Transduction Process Justified
We saw in Theorem 1 that
examples
with the desiredproperty
could beobtained
by
multiplying
aproduct
known to have bounded gaps(for
exam-ple
the2-apart
products)
by
a rational function. In this section we showhow the matrix formulation alluded to in
§2
can be used togenerate
thenew
quotients
from the old. We suppose thatso that in terms of the matrix
equivalence
Hence we may obtain the
b2’s
from theCi’8
by passing
themultiplying
matrixthrough
the and L~’s termby
term. At eachstage
this amounts tomultiplying
each RC or L’ on the leftby
a matrix of determinantf g
ofthe form
where
deg
Cdeg
A .By
applying
Euclid’salgorithm
to the rows of theresulting
matrix(that
is to sayby writing
where d = or d’ =
[0/6]
and soon)
weoutput
a successsion ofRd~’s
andLd’s until
we are left onceagain
with a matrixIn the
special
case that all theci’s
i > 1 havedeg
ci >deg fg
(as
will be the case in all of ourexamples)
theprocedure
can be describedfairly
concretely:
As
deg co
may well be zero the first transduction can beatypical.
De-noting
by
pi/qi
theconvergents
tocoflg
=[do ?
... ,
dr]
whereco f
=a( x )Pr
and g
=a(z)qr
we see thatThe
general
transitionthrough
an Lc isparticularly
simple:
, , ,,/
with
deg C’
=deg
(Bci
+ C - Adeg A .
The
general
passagethrough
an Rc is a little morecomplicated
and breaks down mostnaturally
into two parts:We first let
pilqi
denote theconvergents
toA/C
=[eo ,
el ,... ,
et]
where A =
ti(x)pt
and C = and observe thatwhere u = +
(-l)’Bpt-,
and v =(-1)t-lBpt;
of course there is theconvention that
/-. -B
/ B / -. B
so that if C = 0 we and do
nothing.
Hence if we let denote theconvergents
tou/v
=[ho ,
... , with u =
A(x)p’ s
and v = we obtainwhere
deg
C’deg
A’ .So as we pass
through
successive R’s and L’s the R’s of theoutput
arealways
followedby
L’s and vice versa. Moreover due to the conditiondeg
ci >deg f g
=deg
AB we see that(with
thepossible exception
of thevery first
output
do )
all ouroutputs
are ofdegree
greater
then orequal
to one. Hence in this case we are indeed
obtaining
(term
by
term)
thelegitimate
expansion bo, bl,
... ofG(x) .
Since ouroutput
is never less then ourinput
we can(by
feeding
theoutput
back in asinput)
use thisapproach
onF(x)
satisfying
a functionalequation
of the form:We shall
begin
the next section withjust
such anexample.
Of course the described process still functions if we allow the
possibility
ofinput quotients
withdeg
ei =deg f g .
In that case theoutput
is still sustainable but may well now contain rogue constant terms -although
techniques
do exist forindividually removing
suchillegalities relying
on the observations thatInput
containing
terms ei withdeg ei
deg( f g)
is moreproblematical;
not
only
is the above program nolonger
quite
suitable but theregenuinely
seems no way ofpreventing
thepossibility
of R ’s followed R’s or L’sby
L’s so that(with
thedanger
of wholescaleback-tracking through
cancellation)
we nolonger
have anyright
to assume that theoutput
weproduce
from afinite number of transductions bears much relation to the final
expansion.
Note that with theexception
of the twopossibly large
quotients
l BCA+C J
and
ho
= theremaining quotients
produced
inpassing
through
an are(by
considering
thedegree
of the denominators C andv )
oflimited
degree;
aphenomenon
that we havealready
encountered in the case described in Theorem 1. In fact since the sum of theirdegrees
is at most2 deg f g
we see that each L’i RCi+lproduces
at most 2deg f g
+ 2 ofoutput.
More carefulanalysis
of common factors could no doubt be usedto recover the bound of Theorem 1 in this way. When we are interested in truncations we can
(by
considering
thepotentially
large degree
of thesubsequent
partial
quotient)
use these observations to locate theirposition
in theoutput;
if the truncation is(as indicated)
the(n
+l)-st
convergent
toF(z)
then when n is even theexpansion
for the truncationconvergent
fp,,Igq,,
toG(x)
corresponds simply
to theoutput
generated
by passing
through
When n is odd we need in addition the eo, el, ... , etproduced
(when
C ~
0)
inpartially passing through
Of course the former situation is
preferable
and indeed whenF(x)
is ourfamiliar
2-apart
product
we can arrange for the truncations to beexactly
the ’evenconvergents’
(at
the expense of atolerably illegal
and in someways
preferable expansion
as we shall seebelow).
Although
the cases are somewhat limited in form we are faced withpotentially infinitely
many differentpossibilities
for the transition matrixWe would
hope
that the transduction processproduces,
if not aCc
B)’
finite set, then at least
only
matrices of a similarenough
shape
that we can avoidsimply
performing
transitions ad infinitum. In the next section wesee this
technique
in action.5. The Matrices in Action
As mentioned
earlier,
it was shown in[2]
that any infiniteproduct
of theform
where A _
(Ao, Ail, ... )
satisfieshas truncations
2-apart
withexpansion
and
We
begin
this sectionby using
the method of transduction described aboveto
give
an alternativeproof
of this fact: From the functionalequation
we see that
The trivial observation
bo(x; A’)
= 1 entails that the initial transitionyields
Now since
Aj )
I
we see thatF(x; A’)
is apolynomial
inXÀl
and sopossesses
partial
quotients
withÀl
>2Ao
for all i > 1.Hence from our
previous
comments we are allowed to read off thebi(x; A)
directly
from ouroutput
as it isgenerated
termby
term. Thus we obtainbo(x;
A)
=1,
b1(x;
A)
= and weproceed
to the next transition armedwith the
knowledge
thatA’)
=XÀl
and theassumption
that2AO
I
À1:
Hence
b3(x; A)
andb4(x; A)
are of the stated form.From this
point
onwards the transition matrix movesthrough
the R’s and L’sunchanged.
Inparticular
under theassumption
(as
we havejust
seen for N =
1)
that1)
dividesb2N.+.2(x; A’)
we obtain at theN-th
subsequent
stage
and
A
simple
inductionargument
leads at once to the relationsand the stated formulae for
b2N+l(x; A)
andA).
Ourprevious
comments or
simple degree considerations
show us that the truncations arealways
2-apart.
From now on we will assume such an
expansion
which(in
the manner of[2])
it will prove convenient to write in theslightly illegal
form, - - I
-with
where
with
From the functional
equation
we obtain the matrix
correspondence
where the ci all are
polynomials
in and hence ofdegree deg ci
>Ao
>2r . In
particular
(from
our earlieranalysis)
it is clear that(once
we havesafely passed through
the initial we are allowed to read off the ai from theoutput,
transitionby
transition. Of course ingeneral
wecannot
expect
our transition matrix to remain constant as occurred in the aboveexample;
however we canhope
that itchanges
in afairly
predictable
manner.
THEOREM 3. Let
with
Ai
satisfying
(12).
Then the truncations of G areeventually
4-apart
for n > 1. The first 5
irregular partial quotients
aregiven implicitly
below.Proof. We first note the
general
matrix transition formulaePassing
themultiplier
through
the first three termsgives
us the first fivepartial quotients
ao , ... , a4 :Now
applying
the above formulae withA = 4 ,
D= 4 ,
Bo - - 2
we seeB that passage
through
the n-th block LC2n+l RC2n+2 takes the formwith
Bn
=Bn-1 -
tC2nH( -1) = -
Hence thealleged
partial
quotients.
Moreoverby
previous
comments orby simple degree
considerations we see that the truncations coincideexactly
with the finite continued fractionexpansions
produced
by
these transitions and hence are all(bar
thefirst)
four-apart.
THEOREM 4. Let
with
Ai
satisfying
(12).
Then(i)
if 4I
Àl
then the truncations areeventually
2-apart
(initial
gap5)
with
partial quotients
(ii)
then the truncations areeventually
6-apart
(initial
gap5~
withpartial
quotients
for n > 1 where
The first 8 or 13
partial
quotients
can be discovered below. -, Proof. In both cases the initial transitionsproduce
the first 7quotients
after which the transition matrix remains
unchanged
withfor n > 1. The stated
quotients
are then immediate. When4 t Ai
werequire
two furtherirregular
transitionsbefore the process settles down with
for n >
2,
whereThe
given partial
quotients
then followfairly easily.
with
Ai
satisfying
(12).
(i)
If(3, Aj )
= 1 for all i then the truncations of G all are5-apart
withpartial quotients
for n > 1. The first seven
atypical partial
quotients
can be found below.(ii)
If 3I
AN
for some N > 1(where
we take N to beminimal)
then the truncations ofG(x)
areeventually
(after
the N’thtruncation)
2 -apart
withpartial
quotients
for n > 1. The first
(5N
-~-2)
quotients
are as in case(i) ,
and a5N+2 can be found below.After the initial transitions
we have the
following
identities: If n is oddand
and
So each time we pass
through
a block we output 5partial
quotients
and if n is even(respectively
odd)
either an A(respectively C)
ifAn *
4(mod 6)
or a B(respectively
D )
if 2(mod 6).
The statedquotients
can be read off and thendegree
considerations confirm that theNow if 3
1
AN
for some N > 1(assumed minimal)
then the process remains the same until we hit L C2N +1 . Inparticular
the ao, a1, ... , a5N+1 aregenerated
as above.Writing
we see that
after which the transition matrix remains
unchanged
withfor n > N . The stated
partial
quotients
are then evident anddegree
considerations or otherwise show that the truncations are indeed2-apart
as claimed.THEOREM 6. Let
with
Ai
satisfying
(12).
The the truncations of G areeventually
4-apart
The first nine
partial quotients
ao, ... , a8 can be foundProof. If
Ao
is odd and 6 !I
A,
then we have thefollowing
transition relations:Here we have to endure some rather
painful
initial transitions:if
Ao T
1(mod 6)
andif
Ao m
5(mod 6),
where!3o
= - 3 (~1 ~~0~ .
Hence after
passing through
the k -th block weoutput
4partial quotients
and a transition matrix of the formwhere
f3k == f3k-l -
3 czk+s(-1) _
-Àk+1/2k3Ào.
This makes the theorem evident.Notice that we have covered here all cases of
(11)
with r = 1 or 2 anda number of cases when r = 3. The
remaining
cases with r = 3 becomerather too
unwieldy
to handleby
locating
thequotients
in this way(when
2 r
Ao
and3 ~
notonly
do we have to forcethrough
multiplication
by
the twopolynomials
(x
+1)
and(x2 - x
+1)
but,
as we have seenbefore,
themultiple
factor(x-~-1)2
in(1-~x-3)(1-f-x-~°)
spawnsunwieldy
constants).
Thesemissing
cases will begiven
in§6
and are foundby
considering
convergents,
asopposed
toevaluating
partial quotients.
6. The
Convergents
ConsideredGiven the
convergents
andpn+h/qn+h
and the intermediatepartial
quotients
an+1, ... , an+h it isfairly
easy to retrieve themissing
convergents
Pn+h-1 /qn+h-1
in thefollowing
manner:We define
and observe that
s(r, m)
can bestraightforwardly
written in terms of aT+2,... ,
by
means of theeasily proved
iteration relationwhere,
as is wellknown,
It is not hard to then show
(by
induction onr )
that the intermediatepn+r ’s
and
qn+r’s,
1 r t~ - 1 aregiven by;
If the reader should
object
thatknowing
pn /qn
andPn+h/qn+h
canonly
determine pn, qn and pn+h, qn+ h up to a constant we observe that to obtain we need
only
determine the ratio of theleading
coefficients ofand Pn and this is
simply
theproduct
of theleading
coeflicients ofIn fact not
only
must themissing
convergents
Pn+r/qn+r
be of aspecial
form butconversely given polynomials
A and B withdeg
ABdeg
S(n,
h)
such that.&...
satisfies
then
p/q
is an intermediateconvergent
for some 0 i h This fact is an easy consequence of criterion(6)
sincewhere
clearly
Hence in order to find the
(n
+i)-th
missing
convergent
it is necessary and sufficient to look forpolynomials A, B
(sharing
nosuperfluous
commonfactors)
withand
giving
with
The limited form of the intermediate
convergents
allowedby
(14)
enablesTH.EOREM 7. Let
with the
Ai
E I~satisfying
Then the truncation
convergents
are
(i)
2-apart
if andonly
ifwith intermediate
convergent
(ii)
3-apart
only
in thespecial
casewhere X =
We
immediately
obtain thefollowing:
COROLLARY 1. The infiniteproduct
has truncations which are
eventually
2-apart
(that
is,
from the truncationonwards)
if andonly
ifThere are no infinite
products
of the form(15)
whose truncations(except
possibly
thefirst)
are everthree-apart.
Proof. We suppose
that p
and p are constants such thatso that
Now if h = 2 we see from
(14)
that the middleconvergents
arenecessarily
given by
For the first of these to be a
polynomial
werequire
thatjib,,
=p;
sub-stituting
this in the secondgives
the stated criterion.Conversely
if such a condition is satisfied thenby
appealing
to ourprevious
comments or tocriterion
(6)
we see that thegiven
fraction is an intermediateconvergent
and moreoverby
degree
considerations that it is theonly
one.Now if h = 3 the
equations
(13)
and(14)
produce
From the
equations
for pm+2 we see thatfor some
polynomials
A(x)
andB(~).
However sincewe see that
A, B
are both in fact non-zero constants. Hencesubstituting
theseexpressions
for am+2 and back into(16)
we obtain anexpres-sion of the form
In view of the
rapid
increase of theÀi
> 2~2.- o ~~ ,
comparison
of the number of terms on theright
shows that such a relation isonly possible
if n = 1.Finally
substituting
foram+3 in q?.",.+1
gives
from which one
easily
deduces thatao
I
Ai .
An obviouschange
of variable reduces this to the case( 1
+X-I )(1
+cX -~‘)
which as we have seen is2-apart
and whenc ~
(- 1)~
iseasily
checked to be3-apart
with
expansion:
where b =
(-I) A c.
Our
examples
have featured constant gaps between thetruncations;
lest the reader be misled intobelieving
that this isgenerally
so wegive
theTHEOREM 8. Let
with the
Ai
satisfying
(12)
andThen the truncation
convergents
are
5-apart
when m - 3(mod
4),
6-apart
when rra - 2(mod 4),
7 -apart
when m - 0(mod 4)
and8-apart
when m - 1(mod 4)
with intermediateconvergents:
with
u(x)
=1 - x -~- x2 - x3 ,
and 1 - x .Proof. The above
expressions
may appear formidable but the process forconvergent
pN+k/qN+k .
Hence to find theconvergents
between andPN+k/qN+k
it isenough
to findpolynomials
Ai, Bi
withdeg AjBj
3such that the stated denominators are
polynomials.
Thusstarting
withA1
= 1 we find successive withdeg Aj
= 4 -deg Bj -1
until wehave filled in the gap
(that
is,
E(4 -
degAjBj)
=4).
Similarly
for theOf course we have made
things
heresimple by
choosing
theAi
such that for aroot
of we haveNote that it is not hard to see that a
judicious replacement
of asubsequence
of theAi
by Am m 6 ~ 2"~+2
(mod
10)
can eveninject
an appearance of utter randomness in these differentspacings.
As
promised,
we conclude with theremaining
r = 3 cases:THEOREM 9. Let
with
Ai
satisfying
(12).
Then the truncation
convergents
are
where - and k are the constants
Proof. As
before,
theproof
amounts totediously
finding
Ai,
BZ
such that the denominators arepolynomial
and so that the variousdegrees
add up.REFERENCES
[1]
J.-P. Allouche, M. Mendès France and A. J. van der Poorten, An infinite product[2]
M. Mendès France and A. J. van der Poorten, Some explicit continued fraction expansions, Mathematika, 38 (1991), 1-9.Christopher G. Pinner Alfred J. van der Poorten
Natarajan Saradha
Centre for Number Theory Research
Macquarie University
NSW 2109 Australia
pinner~mpce.mq.edu.au alftmpce.mq.edu.au saradha~mpce.mq.edu.au