J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
Y.-F. S. P
ÉTERMANN
J.-L. R
ÉMY
I. V
ARDI
On a functional-differential equation related to
Golomb’s self-described sequence
Journal de Théorie des Nombres de Bordeaux, tome
11, n
o1 (1999),
p. 211-230
<http://www.numdam.org/item?id=JTNB_1999__11_1_211_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On
afunctional-differential
equation
related
to
Golomb’s self-described
sequence
par Y.-F.S.
PÉTERMANN,
J.-L.RÉMY,
et I. VARDIRÉSUMÉ.
L’équation
différentielle fonctionnellef’(t)
a des liens étroits avec la suite auto-décrite F de
Golomb,
1, 2, 2, 3, 3, 4, 4, 4 5, 5, 5, 6, 6, 6, 6, ....
1, 2, 2, 3, 3, 4,
Nous décrivons les solutions croissantes de cette équation. Nous montrons
qu’une
telle solutionpossède
nécessairement un pointfixe non
négatif,
et que pourchaque
nombre p ~ 0 il y aex-actement une solution croissante ayant p pour point fixe. Nous
montrons
également qu’en
général
une condition initiale ne déter-mine pas une solution unique: les courbes représentatives de deuxsolutions croissantes distinctes se croisent en effet une infinité de
fois. En
fait,
nous conjecturons que la différence de deux solutionscroissantes se comporte de
façon
très similaire au terme d’erreurE(n)
dansl’expression
asymptotiqueF(n) =
~2-~n~-1
+E(n)
(où ~
est le nombred’or).
ABSTRACT. The functional-differential equation
f’(t)
=1/f(f(t))
is
closely
related to Golomb’s self-described sequence F, 1, 2, 2, 3, 3, 4, 4, 4 4 5, 5, 5,1, 2, 2, 3, 3, 4,
We describe the
increasing
solutions of this equation. We show that such a solution must have anonnegative
fixed point, andthat for every number p ~ 0 there is
exactly
oneincreasing
so-lution with p as a fixed point. We also show that in
general
aninitial condition doesn’t determine a unique solution: indeed the
graphs
of two distinct increasing solutions cross each otherinfin-itely
many times. In fact we conjecture that the difference of twoincreasing
solutions behaves verysimilarly
as the error termE(n)
in the asymptotic expression~2-~n~-1
+E(n) (where ~
is thegolden
number).
1. INTRODUCTION.
For a real number p we are
considering
the functional differentialequation
with initial condition
f (p)
= p. We came across thisequation
whilestudy-ing
a self-described sequence ofpositive
integers usually
referred to as"Golomb’s
sequence",
by
the name of S.W. Golomb who rediscovered itin 1966
[Go].
We say that a sequence("word")
W ofpositive
integers
isself-described
orself-generating
ifT(W)
=W,
whereT(W)
is the sequenceconsisting
of the numbers of consecutiveequal
entries of W. Golomb’s- . - - - - . "’.... - - -
-sequence P’
is the sequence,
w i m J m
the
only
nondecreasing
self-described sequencetaking
allpositive integral
values.
In[Go]
Golomb asks for aproof
of theasymptotic
equivalence
where 0
denotes thegolden
number. One suchproof
can be found in[Fi].
Another one, based on a heuristic
argument
of D. Marcus[Ma]
and latercompleted
in~Pe~,
establishes inparticular
the fact thatwhere
f
is anypositive
solution of(1).
In hisargument
Marcus mentions the solution of(1) f+(t) _
~2-~t~-1;
note that this function has two fixedpoints,
p = 0 andp =
0.
It seems natural to ask whether Marcus’ solution is the
only
one of(1)
with initial condition
f (0)
= 0. The answer to thisquestion
is yes, andproving
it isproposed
asproblem
10573 of the A.M.S.Monthly.
But italso seems natural then to ask whether other solutions of
(1)
than Marcus’exist,
and if so howthey
behave. Weeventually
realised that answers tothese
questions
were too voluminous for a"problems
section" such as in theA.M.S.
Monthly,
and that at least alarge
note was necessary to treat them.In this paper we describe all the
increasing
solutions of(1).
We provethat for each
p >
0 there isexactly
one(increasing)
solutionf
of(1)
withf (p)
=p
(Theorem 1),
and that there are no otherincreasing
solutions thanthose
having
(at least)
onenonnegative
fixedpoint
(Theorem 2).
Welong
thought
(and
tried toprove)
that solutions of(1)
satisfied a moregeneral
unicity condition,
i.e. that we could havef (p)
=q for at most one solution
of
(1),
for every real numbers p and q, and notonly for q
=p > 1. This
would have
implied
of course that two distinct solutions could never crosseach other. But in fact the
unicity
condition is not satified in a verystrong
sense: indeed each
pair
of distinctincreasing
solutions cross each otherinfinitely
often(Theorem 3).
But are there other solutions than
increasing
ones? We shall see below("Preliminary
remark3.3")
that a solution isnecessarily strictly
monotoneon an interval on which it is defined. We first
thought
there were node-creasing
solutions,
until we discoveredf _ (t)
=-~-~-1 (-t)-~;
note that this function has the fixedpoint
p= -1~~.
It is not our purpose to alsotreat
decreasing
solutions;
we shall nevertheless see that the existence ofmany other such solutions is not difficult to establish: see the last section
of the paper.
Note added in
May,
1999. Until well after the present paper was written wewere unaware of the existence of
[McK],
which wasbrought
to our attentionby
BertholdSchweizer,
of theUniversity
of Massachusetts at Amherst. In this work M.A. McKiernan establishes the existence of acomplex
solutionof
(1)
satisfying
f (p)
=p, for every
complex
number p0,
andannounces the
presentation,
in a future article(which
to ourknowledge
hasnot been
published),
of the extension of this result to all pwith Ipl
> 1. He doesn’t address theproblem
of theunicity
of a solution with agiven
initialcondition.
Last remark on Golomb’s sequence. The sequence
F(n)
wasrecently
stud-ied in several papers, and a much more
precise asymptotic
estimate thanGolomb’s
(2)
is now known. I. Vardi[Va]
firstproved
that the error termassociated with
(2)
is «n~-1/
log
n. He alsoproposed
twoconjectures,
which were
proved respectively by
J.-L.Remy
andby
Y.-F.S. P6termannand J.-L.
Remy,
and which can be summarized as follows(putting
F(t)
=F([t])
if t is not aninteger).
Theorem 0
([R6]
and[P6R6]).
We havefor t >
2where the real
functions
h iscontinuous,
notidentically
zero, andsatisfies
h(x) = -h(x
+1)
for x
> 0.It is
interesting
to notethat,
while the main term is a solution of the functional differentialequation
(1),
the error term associated with(2)
(i.e.
the second and third terms on theright
of(3))
satisfies anapproximate
functional-integral
equation
As we
mentioned,
the functionf+(t) _ ~2-~t~-1,
which is the main termon the
right
of(3),
is anincreasing
solution of(1).
We also announcedsolutions of
(1)
cross each other aninfinity
of times. We think thesecross-ings
occur veryregularly,
in apattern
that can be describedsimilarly
asthe error term in
(3).
Conjecture
0.If f is
anincreasing
solutionof
(1)
which isdifferent
from
where the real
functions
1] iscontinuous,
satisfies
~(x) _ -q(x
+1),
and vanishesexactly
once on[0,
1).
Acknowledgements.
This work waspartly
done while the first author wasvisiting
the second andsupported by
the Université Henri Poincar6Nancy
1 andby
the INRIA Lorraine. The authors are verygrateful
to these insti-tutions for theirhelp.
2. STATEMENT OF THE RESULTS.
We
adopt
the convention thatf
is a maximal solution(MS
forshort)
of
(1)
above if thefollowing
conditions are satisfied.(a)
It is defined and continuous on some interval Icontaining
p, but notonly
p.(b) f(I)
C I.(c)
Equation
(1)
is satisfied forevery t
in I(where
by
convention we writef’(t)
=00if f ( f (t)) = 0).
(d)
f
cannot be extended to alarger
interval Jcontaining
I and whereConditions
(a), (b)
and(c)
are satisfied.The main result of this paper is
Theorem 1. For every real
number p >
0 there isexactly
one MSf of
(1)
with
f (p)
= p.The
proof
splits
up into three casesdepending
on whether p1,
p =1,
or p > 1. For p > 1 the
general
idea isfairly simple:
First one notes that the functional differentialequation
(1)
implies
that allhigher
derivativesf(k)(t)
off (t)
can beexpressed
as rational functions of iterates off (t).
Itthen follows that if p is a fixed
point
off (t),
then is auniquely
defined function of p. This then
implies
that theTaylor
series off (t)
at p isuniquely
defined in terms of(computable)
rational functions of p. To show that a solution exists and isunique
one then has to prove(2)
TheTaylor
series defined in this way converges in someneighbourhood
of p.
(3)
A solution of(1)
isanalytic
in thisneighbourhood.
Lemma 2 below shows that the case p 1 can be reduced to p > 1. For p = 1 we have not been able to prove
(2)
directly,
so another method isrequired.
It is not difficult toverify
(see
Preliminary
remark 3.3 nextsection)
that all the solutions of(1)
withp >
0 areincreasing.
But areall the
increasing
solutions of the functional differentialequation
(1)
the functions describedby
Theorem 1 and itsproof?
It is apriori
conceivablethat some other
increasing
solution(i.e.
with no fixedpoint)
might
exist. We show that this is not the case.Theorem 2.
If f
is a solutionsof
(1)
which isincreasing,
and rraaximal(in
the sense that its interval
of defcnition
Isatisfies
the conditions(b), (c),
and
(d)),
thenf
has a(nonnegative)
fired
point.
We shall see in the
proof
of Theorem 1 that if 1 qq’,
and iffq (t)
is the solution of
(1)
with p = q, withIq
its domain ofdefinition,
then for0 t q’
and t EIq
we havefq(t)
But this doesn’t remain truefor all t >
q’,
In fact thegraphs
of twoarbitrary
distinctincreasing
solutionscross each other
infinitely
many times.Theorem 3.
if f
and g areincreasing
maximal solutionsof
(1),
then thereis a sequence
of
numbers si -3 00(i
-~oo)
withf (si)
=g(si).
3. PRELIMINARY REMARKS.3.1. A maximal solution of
(1)
with p = 0 or withp =
~? where 0
denotes thegolden
number and I =[0,
+oo)
isA maximal solution of
(1)
with p= -1~~
and I =(-oo, 0)
isThe solution
f-
isclearly
maximal;
to see thatf+
ismaximal,
see Remark3.5.
3.2. If
f
is a MS of(1)
thenf’(t) ~
0 forevery t
in I.(Otherwise f ( f (t))
would not bedefined).
3.3. If
f
is a MS of(1)
thenf’
/ remains of the samesign
onI,
whencethe function
f
of
cannotchange
sign
on I. It also follows thatf
isstrictly
monotone on I.
Suppose
on thecontrary
thatf’
takespositive
andnegative
values onsuch that
f (tl) - f (t)
andf (t2) - f (t)
are of the samesign.
So forin-stance we have >
f (t)
andf (t2)
>f (t).
Let T be in the intervaland define I- .
f -1(T)
fl andI+
:=f -1 (T)
n(t,
t2l.
The sets I- andI+
are notempty,
asf
iscon-tinuous. And we can choose T with 0 since
f’
cannot vanishby
Remark 3.2. Now weput t’
:= sup I-and t2
=inf I+. Again
by
using
thecontinuity
off
wehave ti
E I-and t2
EI+.
This meansthat t’
tt2,
that =f (t2)
=T,
and thatf (s)
T forevery s in the interval
(~2)-And,
again
sincef’
cannotvanish,
we must havef’(t[ )
0 andf’(t2)
> 0. But this is in contradiction withf’(ti)
=f’(t2)
=1/ f (T),
sincef (T) ~
0.3.4.
If f
is a MSof (1)
andif f’(to)
= oo for some real number to inI,
thento is the
largest
lower bound or the smallest upper bound of the interval I.Indeed if a
neighbourhood
of to is included inI,
then with thehelp
ofRemark 3.3 we infer that a
neighbourhood
N ofTo
= is included inI,
thatf (To)
=0,
and thatf (T)
has a constantsign
in N. The relationf’(To)
=0,
being
excludedby
Remark3.2,
we now reach a contradiction verysimilarly
as in theproof
of Remark 3.3.3.5. A MS
f
of(1)
isindefinitely
differentiable in I.This will follow from
expression
(4)
in Section 4below,
whereby
conven-tion we write = oo
if f, (t) :
:= = 0 for some t with 2 ~ With thehelp
of Remarks 3.3 and 3.4 we see that this can in factpossibly
occuronly
at a tosatisfying
=0,
that is at to = inf I or to =sup 7.
3.6.If f
is a MS of(1)
andif f (a) >
0 andf’(a)
> 0 for some a, thenf (t)
is
positive
andstrictly increasing
for all t in Iexceeding
a, so thatf’(t)
ispositive
and(strictly)
decreasing
for all t in Iexceeding
a.3.7.
If f is
anincreasing
MS of(1)
and if I is unboundedabove,
then sois f (I).
Indeed the
assumption
0f (f (t))
A for all t leads to a directcontra-diction,
as thenf’ (t) 2:
1/A
for all t.3.8. Let a solution
f
of(I)
have a fixedpoint
p. Then we have thefollow-ing.
(i)
0 thenf
can have no fixedpoint
of the othersign.
Indeed the derivatives of
f
at fixedpoints
ofopposite signs
are ofopposite signs.
(ii)
If p
0,
thenf’(p)
=1/p
0,
f
isdecreasing
onI,
and thusf
canhave no other fixed
point.
(iii)
If p >
1,
thenf’(p)
=l/p
1,
f’(t)
isstrictly
decreasing
for t > p,(iv)
andif f
has alarger
fixedpoint
P2, then p2 > 1. Indeed sincef’(p)
=I/p
>1,
if wesuppose that p2 is the next fixed
point
off
we must havef (t)
> t for p t p2, whencef’(p2)
1 andp2 >
1. And p2 = 1 is ruled out since thenf’(1)
= 1 andby
Remark 3.6f’(t)
> 1 for p t1,
which is in contradiction withf(p)=pand f(1)=1.
(v)
If p = 0
thenf
can have nonegative
fixedpoint.
Note that
f’(0)
=oo, so that
by
Remark 3.4f (t)
is defined eitheron some
nonnegative,
or on somenonpositive,
values of t. If it ison
nonnegative
values(like
f+
of Remark3.1),
thenf
can have nonegative
fixedpoint.
If it is onnonpositive values,
thenby
Condition(b) f (t)
isalways
negative,
and so isf’(t) ( f’(t)
= oo(or "-oo")
being
allowed),
so thatf
isstrictly decreasing
andnegative.
But this is notcompatible
withf (0)
= 0.(vi)
From all this it follows thatif p
isnegative
orif p
=1,
thenf
hasno other fixed
point,
that if 0 p 1 thenf
can have at most oneother fixed
point
p2, which must belarger
than1,
and thatthen
f
can have at most one other fixedpoint
pl, which mustsatisfy
0pl1.
3.9. If
f
is a MS of the differentialequation
(1),
then(by
Remark3.3)
the inverse function
f -1
is defined on the intervalf (I).
So for t Ef (I)
the functionf
satisfies the indefiniteintegral
equation
To see
this,
note that ingeneral,
if a and bbelong
tof (I),
we havewhere
k(t)
is anintegrable
function.Equation
(1’)
followsby putting
k - 1. But also note that
(1’)
is notquite equivalent
to(1),
since ingeneral,
for a solutionf
of(1)
the inverse functionf -1
is notnecessarily
defined
everywhere
in I(i.e f (I)
can bestrictly
included inI).
4. PROOF OF THEOREM 1.
With the two first lemmas we
begin by
showing
that a MS of(1)
withsome p
satisfying
0 p 1 is also a MS of(1)
for some P2 > 1. Firstwe show that I contains no t p, which
by
Condition(a)
implies
that it contains some t > p.Lemma 1. If
f
is a MS of(1)
with 0p
1,
then I contains no numberProof.
If p = 0 this follows from Remark 3.8(see
theproof
of(v)).
If 0 p 1 and to pbelongs
toI,
thenf (to)
to sinceby
Remark 3.8f
has no fixedpoint
smaller than p. So if we note ao = to and =f (ai)
for i >0,
the sequence ai isdecreasing
andentirely
in I. If iteventually
becomesnegative,
then there is aTo >
0 withf (To)
=0,
and thus there is also aTl
withTo
T1
p,f (Tl )
=To,
andf’(Ti)
=oo, which is in
contradiction with Remark 3.4.
So ai
> 0 for alli,
and a :=0 exists. If a is in I then
f (a)
=limi-+oo
f (ai)
=a, and a is a fixed
point
smaller than p, a contradiction. If a is not in I we definef(a) =
limi-+oo
f (ai),
and on the one hand we have a =f (a)
=f ( f (a)),
and on the otherhand,
sincef’
isdecreasing
on(a, oo),
f’(a)
exists and isequal
to
f’(ai).
It follows that we can extend the definition off
at a insuch a way that Conditions
(a), (b)
and(c)
be satisfied on J := I U acontradiction to Condition
(d).
DWe are now in
position
to prove thefollowing.
Lemma 2.
If f
is a MSof
(l~
with 0 p 1 there isexactly
one otherfixed
point
P2of f ,
which islarger
than 1.Proof. By
Remark 3.8 above it is sufficient to show thatf
has another fixedpoint.
By
Lemma 1 there is someto
> p such thatf
is defined on[p, to].
We
put
as before ao = to and =f(ai)
for i > 0. Then I containsIP, ai]
for every i. If no such interval contains a fixedpoint
larger
than p,then ai+1 =
f (ai)
> ai forevery i. This
implies
that the sequenceof ai
is bounded(if
itweren’t,
thenf’(ai)
would tend to0,
andf (ai)/ai
couldnot remain
larger
than1).
Thus a :=limi,,,.
ai exists. If a is inI,
thenf (a)
= a and we are done. And theassumption
that a is not in I leads toa contradiction as in Lemma 1: we
defines
f (a)
=ai and we infer
that
f
satisfies(1)
at t = a. This concludes theproof.
0Now we prove Theorem
1,
considering separately
the three cases p >1,
0
Case 1: p > 1. In the next three lemmas we first establish the existence of
a solution and describe its domain of definition I.
Lemma 3.
If p
> 1 then there is afunction
f (t)
defined
in the interval(p -
P2(p -
1)~2, p
+ p2(p -
1)/2)
andsatisfying
(1)
there.Proof.
Letf (t)
satisfy
theequation
(1)
and definef1(t)
=f (t)
andfk+1(t)
= 1.
Also,
one hasand
similarly
forany k
> 1where the sum runs over a finite collection
Sk
of sets ofexponents
a2, ~ ~ ~ , ak+1, and where the
C(a2, ... , ak+1)
arepositive
constants.Now,
provided t
is such thatfk (t)
> 0 forevery k >
2(it
follows from Remark 3.5 that this will be the case iff2(t)
>0),
we haveNow note that when a term under the sum of
(4)
isdifferentiated,
theexponent
bi
off i (t)
appearing
in one of theresulting
termsalways
satisfiesai
b2
ai + 2: it follows that max ai 2k. Hence if weput t
= p in theexpression
above we obtainthat is
So the
Taylor
series forf
has a radius of convergence at leastP2(p - 1)/2.
Now considerthe formal power series solution of +
s)
= +s)),
withh(p)
(p)
=p. Then
h(p)
(p
+s)
is also its own formalTaylor
series at s = 0. Note that the abovecomputation
is valid at t =p if we
replace
f by
h~p~.
Itfollows that this
Taylor
series has a radius of convergence at leastp2(p-1)/2,
and thusrepresents
a solution of(1)
within that radius.Lemma 4.
If p
> 1 then there is afunction
f (t)
defined
in the intervalProof.
Consider the functionf
of Lemma2,
defined andsatisfying
(1)
onKo
:=~p, Po~,
wherePo
p-I- p2(p -
1)/2,
andput
for t inKo
Since
f
isstrictly increasing
its inverse functionf -1
is well defined onf (Ko).
So for t inf (Ko)
we have.. 1 ;, - ., ...
The inverse
function g-1
is thus well defined onKo,
=f (v) .
there. But sincef (t)
> 1 for t > p we see from(6)
that in fact the inversefunction
g-1
is well defined ong(Ko)
=f-1(Ko)
=:Kl,
whereKl
is an intervalcontaining Ko
andlarger
thanKo. Moreover,
since for t inKo
weinfer from
(6)
thatg’(v)
=f (v),
we have for t inK,
Thus the
function g-1
coincides withf
onKo,
and satisfies(1)
onKl.
Wewrite
f(t)
:= for t inKi.
Continuing
in this way we obtain a sequence of intervalsKi
=~p, Pi]
(i
=0,1,2,... ),
withPi+1
>Pi,
on each of which the functionf
of Lemma3 can be extended and still
satisfy
(1).
And we havef (Pi+1)
=Pi.
To conclude theproof
of the lemma we have to ensure that the sequence ofPi
is unbounded. If it is bounded we can do as in Lemmas 3 and4,
thatis put P =
limi-+oo
Pi
anddefine
f (P)
=f (Pi)
=limi-+oo Pi
= P. We theneasily
see thatf
still satisfies(1)
at t = P. Butf (P)
=P,
acontradiction to Remark 3.8. 0
Lemma 5.
If p
> 1 then there is afunction
f (t)
defined
in an interval[Q, p]
andsatisfying
(1)
there,
where eitherQ
is afixed
point
of f
withsatisfies
f(Q)
= Z andf ( f (Q))
=f(Z)
= 0for
some Z withQ
Z 0.Proof.
This time we use the fact that the functionf
of Lemma 3 is definedand satisfies
(1)
onJo
:=(Qo, p~,
whereQo
>p - p2(p -
1)/2.
As abovewe define
g(t)
by
(6)
for t inJo.
Similarly
as before we obtainrecursively
asequence of intervals
Ji
=[Qi, p]
(i
=0,1,2,...
),
withQi+1
Qi,
on each of which the functionf
of Lemma 2 can be extended and stillsatisfy
(1),
and with
f (Qi+1 )
=Qi ~
But here the constructionmight
end after a finite number ofsteps:
indeed if for some i we obtainf (Qi)
0 theexpression
(6)
does notproduce
anymore a well defined function onJi.
So we considerI. At each
step
we havef (Qi)
>0,
so that the construction never ends. Sincef’(p)
=1 j p
1 wemay suppose that
f (Qo)
>Qo.
Then if for some i > 0f (Qi)
>Qi,
we havef-1(Qi) _
Qi+1
Qi
whenceI(Qi+1) =
Qi
>Qi+,.
Thusf (Qi)
>Qi
for each i.(Note
that this will also be true for each iio
in the case where the constructionends at
Jio).
Now sincef"(t) =
-(f3(t)f23(t))-’
isnegative
at least for all t for whichf (t)
>0,
that is inparticular
on every.li,
f’
isdecreasing
on everyJi,
and thus the sequence ofQi
cannotdiverge.
So
put
Q
:=limi-4oo Qi.
Again
we definef (Q)
=limi_
f(Qi) .
We theneasily
see thatf (Q)
=Q
and thatf
still satisfies(1)
at t= Q.
And
by
Remark 3.8 we must haveQ >
0.II. At the i-th
step
of the construction we reach aQi
withf (Qi)
0. So there is a Z >Qi
withf (Z)
= 0. Note that Z isnegative
sinceQi Z
Qi-l
= 0. Sincef
is well defined andnonnegative
on
~Z, p~,
expression
(6)
produces
afunction g
well defined on~Z, p~,
and whose inverse
coinciding
withf
on[Z,p],
is well defined on[Q
:=g(Z),p]
and satisfies(1)
there. So we may extend the definitionof
f
on(Q, p~
by putting
f
= there. And we havef (Q)
=Z,
as claimed.0 This concludes the
proof
of Lemma 5. So if p > 1 we established theexistence of a
function f satisfying
(1)
in an interval~Q, oo)
whereQ
is either a fixedpoint
forf
with 0Q
1,
or is such thatf’(Q)
= oo. With Lemma 1 and Remark 3.4 this shows thatf
is a MS of(1).
To prove the
unicity,
we first note that theproof
of Lemma 2 shows that any solutionh(t)
of(1)
must have at t =p the same
Taylor
series asf .
If h is
analytic
in some interval[Qo, Po]
withQo
pPo
then h =f
there. In order toverify
this,
note in addition that there is some po withp > p - po >
1,
suchthat if p > t > p - po
thenh(t)
> t > 0 andh,’(t)
> 0: this follows from the facts that 0h’(p)
=1/p
1,
that h’is
continuous,
and that h has no other fixedpoint
p’ >
1. It follows that ifhk-1(t)
> t> 0 for p >
t > p - po thenhk(t)
>h(t)
> t > 0 for p > t > p - po.Thus,
for some po >0,
the functionshk
allsatisfy
hk (t)
> t if p > t > p - po. It follows in turn that estimate(5)
in theproof
of Lemma 3 remains true if wereplace f by h
and pby
some to withThus
Also note that from estimate
(4)
in the sameproof
ispositive
anddecreasing
if k isodd,
andnegative
andincreasing
if k is even, whencemodulus of the
Taylor
expansion
of order k ofh(t)
at p does not exceedAnd estimate
(7)
shows that if t is closeenough
to p the lastexpression
tends to 0 as k - oo. Thus h isanalytic
in some[Qo, Po]
as claimed.Now we show that h is
f ,
with I =[Q,
oo)
its domain of definition.Suppose
it is not the case and let I’ =[Q~)
or I’ =(Q’, w~
be thedo-main of definition of h.
Suppose
first that h =f
on I fl I’. Then I = I’would contradict the
hypothesis,
I’ C I orQ’
Q
would contradict themaximality
off ,
and I C I’ oo would contradict themaximality
ofh. To see this recall that
Io
CI’,
and recall theproofs
of Lemmas 4 and 5. So there must be some t E I rl I’ withf (t) ~ h(t).
Nowby
Remark 3.8 and Lemma1,
forevery t
in I which is not a fixedpoint
off
we have0
( f (t) - p)/(t - p)
1,
and forevery t
in I’ which is not a fixedpoint
of h we have 0
(h(t) -
p)/(t - p)
1’Suppose
for instance there issome t p with
f (t)
#
h,(t) (the
other case is treatedsimilarly).
If weput
to :=
sup~t
p, f (t)
~
h(t)},
we have h -f
on[to, p]
and to Notethat to cannot be a fixed
point
off
orh,
as it then would be the infimumof I or I’
(by
Remark3.8),
which would contradict themaximality
off
orof
h, f
and hbeing
defined(and distinct)
on some t to. Thush(to) >
to andf (to)
> to. Nowby hypothesis
f (t)
andh(t)
are defined for values oft
to;
moreoverf
and h arestrictly increasing.
So the inverse functionsf-1(t)
andh-1(t)
are defined for values of th (to)
=f (to).
It follows that there is some so withto
soh(to)
=f (to),
and such thatf-1
andh-1
are both defined on
~so, p~.
Thusby
Remark 3.9 wehave,
for t > so,and
f -1 - h-1
on(so, p~.
The strictmonotonicity
off
and h nowimply
that
f -
h on[ti,p]
for some t1 to(satisfying f (tl)
=h(tl)
=so),
contradicting
the definition of to. This concludes theproof
of Theorem 1 incase p > 1.
Case 2: 0 p 1. We first prove the
unicity.
Consider two MS of(1)
with 0 p 1f
and h. Thenby
Theorem 1f
has another fixedpoint
p2 >
1,
and h has another fixedpoint
p3 > 1.Suppose
they
areunequal,
say p2 P3. Let c be the
largest
number notexceeding
p3 such thatf (c)
=h(c).
This number exists sincef (p)
=h(p)
and sincef
and h are continuous. Note that c p2 and thatf (t)
h(t)
for t E(c,P3~,
and recall thatf (t)
> t andh(t)
> t for t E(p,P2).
It follows thatf’(c) h’(c).
It also follows that if c > p thenf ( f (c))
h( f (c))
=h(h(c)),
a contradiction.So c =
p. But then
So p2 =
P3, and
by
Case 1 h =f
on[p,
oo).
Now we prove the existence. For q =p2 > 1 denote
by
fq
the MS of(1)
andby
Iq
=[Qq,
oo)
its domain of definition. Recall that the fixedpoints
offp
are
and 0. So from Remark 3.1 and theargument
used to provethe
unicity
we infer that for every q with 1 q0
we havefq (t)
for t E
~max{0,
0],
and that for every q> 0
we havefq (t)
> for t E[max{0,
Q9~, q~.
It follows that eachfq
with 1 q0
has a fixedpoint
p =
p(q)
with 0 p 1(and
that nofq
with q > 0
has such a fixedpoint).
Moreover the same
argument
shows that if 1 qq’
0,
thenp(q’)
p(q)
and
fq(t)
for t E[p(q), q] -
So the function p :(1,~~
-(0,1)
is
decreasing.
What we need to show is that it iscontinuous,
and thatp
To prove the
continuity
we consider some q = p2 with 1 q~
and letq’
--> q with 1q’
0.
Without loss ofgenerality
we may suppose thatq’ -
q remains of the samesign. Suppose
for instance thatq’
increases to q;the other case is treated
similarly.
Weput f
for the MS of(1)
with fixedpoint
q, and h for the MS of(1)
with fixedpoint
q’.
Let s be a real number.For some s’ and s" between 0 and s we have
If,
for somenegative
real number so we haveso s
-so, and(this
will hold for instance if q:= (q +
1)/2
andq - q’ (q -
1)/2)
we seewith the
help
of estimate(7)
that the last term does not exceed in modulusprovided
so+q > 1. The lastexpression
tends to 0 as K ~ oo if -so is smallenough.
Now from(2)
it is clear that for everyk,
h,~~~ (q’) -
-a 0 asq’
~ q. So the sum tends to 0 asq’
2013~ andh(q’+s)
convergesuniformly
tof (q
+s)
in[so, -so~.
Iteasily
follows thath(t)
convergesuniformly
tof (t)
in
[Qo, q]
for someQo
q. Now we shallessentially
follow the constructionof Lemma
5,
after thefollowing.
Remark. Let p =
p(q)
andp’
=p(q’)
be the other fixedpoints
off
andh,
and let c > 0. Note thatp’
decreases ifq’
increases. There is aj3
_such that the derivatives of
f (t)
in[p
+E, q~
and of the functionsh(t)
in(p’ + 6,g]
are all within the interval~1/~, 1/~].
Now from
for t E
(Qo, q),
we see convergesuniformly
tof -1 (t)
for t E[Qo, q~.
For E > 0 as in the Remarkput
p, =inffpl
+ E. Then if 6 >0,
h(t)
convergesuniformly
tof (t)
for t E(f -1(Qo)
+ 6 :=provided
Q1
> p~. And we can as well fix 9 = E.Recursively
we see that ifh(t)
convergesuniformly
tof (t)
for t E[Qi,
q],
then it does also for t E + E:= QZ+i, Q~~
provided
Qi+1
> Pf.So for t E
[Q’,q],
and for everyQ’
>max{liminfi-tooQi
=:Q; pE},
h(t)
converges
uniformly
tof (t).
We havef (Q)
=limsuPi-too
+ E)
Q + e /p
Q+6.
Soh(t)
convergesuniformly
tof (t)
for t Eq~,
where
[p, Qf]
is thelargest
intervalbeginning
at p where 0f (t) -
t E.It now
easily
follows thatinf p’
=p.
Finally,
if p2 = q = 1 +,E where E1/9,
we show thatp(q) >
q - 9E. Ifnot it is easy to
verify
thatf’(t)
> q - 2E if tq
and thatf’(t) >
q + cif t
q -
3E,
whencef (q - 9E)
q -9E,
a contradiction. This shows thatp(q)
~ 1 as q ~ 1 and concludes theproof
of Theorem 1 in Case 2. Case 3: p = 1. We firstprove the existence of a solution. Let
and be the two sequences of functions
[1, oo)
-3[I, cxJ)
defined asfollows.
We see
recursively
thatthey
are well defined.Indeed,
is well definedwith 1 for
every t > 1,
then isstrictly
increasing
andunbounded,
whencef n
is well defined withf n (t) >
1 forevery t
> 1. Nowwe
have,
forevery t
>1,
And if
h+
and h- are twoincreasing
and unbounded functions(1, oo)
-with
h+(1)
=h_(1)
= 1 andh+(t) > h- (t)
forevery t
>1,
then wehave
-+ -I-lOt
whence
k-’(t)
forevery t >
1. Thus an inductionargument
starting
with(8)
shows that the sequences andt >
1),
and that in additionf2k(t) >
forevery k >
0. Now weput,
for t >1,
F(t)
= andG(t)
= and we shall prove that F - G and that F is a(maximal)
solution of(1)
for p = 1. First weverify
that F and G arestrictly increasing
andunbounded,
andthat their inverse functions
satisfy
J 1 J 1
whence in
particular
they
areindefinitely
differentiable functions. We provethe first
equation
in(9);
the second one is treatedsimilarly.
Sincef2~(t) >
1 and isdecreasing,
F isintegrable
andThus we see that the inverse function
H-1
exists. And we know that thefunction g2~ is
increasing
forevery k >
0,
and that the sequenceis
decreasing
forevery t
> 1. Now if E > 0let 77
=H(y) - H(y - E).
Then for klarge enough
we havewhence t :=
H(y)
e(g2k(y - 6)~2A;(~/))
and(y - e, y).
It follows thatg2~ (t)
converges toH-1 (t).
But also converges toG(t),
and the firstequality
of(9)
isproved.
Now we show that F * G.
Lemma 6. Let F and G be two
increasing
andunbounded functions
e
indefinitely differentiable,
andsatisfying
Then
if
theequations (9)
hold we have P’ =- G.Proof.
First note that F and G must bestrictly
increasing,
unboundedfunctions,
and that their inverse functionsF-1
andG-1
are well-defined.From _ , , _ , ,
we see that
Now let
H(u)
:=maxIF(G(u)) -
G(F(u)), 0~.
Then from the aboveex-pression
and sinceF(v) >
1 andG(v) >
1 for all v > 1 me may writeSince on the other hand
F(F(u)) >
F(G(u))
and . we haveF(F(u)) - G(F(u))
>H(u)
whenceThere is some
(necessarily nonnegative)
constant C such thatF(t) -G(t)
C for
every t
with 1 t3/2.
Let C be the smallest such constant. Then sinceG(t~ F(t~ t
we haveif 1 t
3/2.
Thus C = 0 and F - G on(1, 3/2~.
In order to see that F = G on(1, oo)
werecursively
use theexpression
which we
just proved
holds when tbelongs
to the interval~1, 3/2~.
If weput to
=3/2
and ti
=F-1(ti-1)
(i
>0),
then the sequenceItil
isstrictly
increasing
sinceF-1(t)
> t forevery t
> 1by hypothesis.
And if(10)
holds for tti,
thenF-1(t)
=G-1(t)
for tti,
whenceF(t)
=G(t)
for tti+1
and
(10)
holds for tti+1.
So weonly
need to ensure that the sequence{ti~
cannot converge. If itdid,
to some T oo, say, then we would haveT =
limi-too
ti =F-1(ti)
=F-1(T),
in contradiction to one of thehypotheses.
This concludes theproof
of the Lemma. 0 Nowby differentiating
(10)
andputting v
=F(t)
we see that F is a solution of(1)
for p = 1(and
it is a MSby
Lemma1).
Finally
we show that the solution isunique.
First note thatby
Lemma 1a MS
f
of(1)
for p = 1 is defined on an interval I whose lower bound is1,
thatby
Remark 3.3f
isstrictly
increasing,
and that we thus havef (u) >
1,
whence
f’(u)
1,
for every u in I. Inparticular
we also have forevery u in I. Also note that
similarly
as weproved
above that F - G on[1, oo),
we can put ti =f -1(ti-1)
(i
2:1)
where to > 1 is some number inI,
and see that in fact I =
[1, oo).
If F is the solution we constructedabove,
then we haveNow there is some
nonnegative
constant C such thatIF(t) - f (t)~
C forevery t
in~l, 5/4~.
Let C be minimal with thisproperty.
Then for somefunction v =
v(v,)
wherev(v,)
is betweenf (u)
andF(u),
we have" .J.
Thus C = 0 and F -
f
on[1,5/4].
And in fact F -f’
on[1,
oo):
again
thiscan be seen
similarly
as in theproof
of F - G above. This concludes theproof
of Theorem 1.5. PROOFS OF THEOREMS 2 AND 3.
We now show that there are no other
increasing
MS of(1)
than thosedescribed
by
Theorem 1.Theorem 2.
If f
is a solutionsof
f’(t)
=1/f (f (t))
which isincreasing,
and maximal(in
the sense that its intervalof definition
Isatisfies
the conditions(b), (c),
and(d~),
thenf
has a(nonnegative)
fixed
point.
Proof.
Let A be thelargest
lower bound of I. If A >0,
thenconsidering
limits we see thatf (A)
exists and A EI;
and we must havef (A) >
A. If A 0 then sincef
takespositive
values and I is aninterval,
f
is defined ont = 0.
Suppose
f (0)
0. Then sincef
isincreasing,
f (t)
0 if t0,
and thusf’(0)
=1/ f ( f (0))
0,
which is not true. So we must havef (0)
> 0. Hence there is ato >
0 withf (to) >
to. Iff (to)
= to the theorem isproved.
So supposef (to)
> to. Thenf
is defined at least on the interval[to,
f (ta)
=:tl].
Now define tn =f(tn-1) (n
=1, 2, ~ ~ ~ ).
Then either for every n, or for some n. In the second casethere is some t > 0 with
f(t)
= t, andagain
the theorem isproved.
In the first case, we first note that theincreasing
sequence tn cannotdiverge:
if it did then we would have
f’(tn)
e0,
which isincompatible
with theassumption
that for every n.Thus tn
converges to some numberT as n e oo.
Again
considering
limits we see thatf(T)
exists and T E I.We have
f (T)
=limn-too
= =T,
andagain
the theorem isproved.
DRemark. It is conceivable that some solution
f
of(1)
withf (t)
> 0 ifderivatives of
opposite signs
forarguments
ofopposite signs,
and could thusnot be continuous on an interval
containing
0. In otherwords,
it could notbe an MS on any interval I.
Now we prove that the
graphs
of twoarbitrary
distinctincreasing
solu-tions cross each otherinfinitely
many times.Theorem 3.
If f and g
areincreasing
maximal solutionsof
(1),
then thereis a sequences
of
numbers si
~ oo(i
eoo)
withf (si)
=g(si).
Proof.
First note that anyincreasing
solution off’(t)
= satisfies theasymptotic
equivalence
(t
2013~oo):
this follows fromProposition
2 in[P6]
(The
conditionf :
[0, oo)
~[0,
oo)
of thisgeneral
result can
clearly
be removed for the solutions of(1)
weconsider).
Let
[r, oo)
be the interval ofnonnegative
arguments
on which bothfunc-tions
f
and g
are defined. Now first assume thatf and g
never cross eachother,
for instance thatf (t)
>g(t)
for all t > T. It follows thatf’(t)
g’(t)
for all
these t,
and that if weput
f (to) - g(to)
=: C for some to > T to befixed later on, then
f (t) - g(t) >
C for T t G to. Now for T t to lett1 =
tl (t)
=f -1 (t).
Then wehave,
if to islarge enough,
where C’ =
C~~ ~/2.
Note that we have to = and choose tolarge
enough
to ensure thatf (tp) >
T. Thenby integrating
weget
from(11)
But
(13)
holds inparticular
forti (to)
=f-1(to) ~
0’-"to 0
as to e oo, and it follows thatl(t1(t))
cannot remainnegative
for all t tp.So there is at least one
crossing.
Now suppose the set S ofcrossings
is bounded. Then s =sup S
is itself acrossing.
(Note
inpassing
that theassumption
f - g
on an interval leads to a contradiction: i.e.crossings
arecrossings
in the strictestsense).
Suppose
that for instancef (t)
>g(t)
forall t > s =:
To.
LetTl
be such thatg(T1)
=To;
then for t >T1
we havef ( f (t))
>f (g(t))
>g(g(t)),
whenceg’(t)
>f’(t).
So if weput,
for someto >
Tl,
>f (tp) - g(to)
_:C,
thenf (t) - g(t)
> C forTl
t to.Now for
Tl t
to let ti = =f -1 (t).
Wemay suppose that
T1
islarge enough
to ensure that t1 >t;
and also that to islarge enough
to ensurethat
f (to)
belarger
thanTl.
Then the estimates(11), (12)
and(13)
hold if to islarge
enough,
and it follows as before thatg(t1(t)) - f (tl(t))
cannotremain
negative
for all t to. This concludes theproof
of the theorem. 0 6. LAST REMARK ON DECREASING SOLUTIONS.Although
it is not our purpose to also treatdecreasing
solutions of(1)
(like
f-(t)
=2013~’"~"~(2013~)~),
we note that it is nowfairly
easy to establish the existence of
infinitely
many such functions.Indeed,
for every p -1 theargument
in theproof
of Case 1 of Theorem1
helps
verify
that there isexactly
one MSfp
of(1)
withfP(p)
= p. Consider the formal power series solutionh(p) (t)
of(1)
with fixedpoint
p. From(4)
it is clear that at t = p this series has a radius of convergence at least aslarge
as that of the power seriesexpansion
of the solution of(1)
with fixedpoint
at t =lp 1,
which ispositive by
Lemma 3. Thus represents within that radius a solution of(1),
which we may callfp(t) .
We can thenextend
fP
to a MS of(1)
by
using
if the fixed
point
p is stable(as
in Case 1 of Theorem1),
respectively
unstable.(In
view of the functionf-
it seemslikely
that p isunstable).
Unicity
is obtainedsimilarly
as in Case 1 of Theorem1,
andby using
theanalyticity
offlpl(t).
We mention the fact
that,
by
comparing
the coefficients of the(formal)
solutionh(p)(t)
written as a(formal)
Taylor
series at t =p with those of
the solution
f_1~~(t)
:=f- (t) (also
at t =p),
one can in fact establish theexistence of a solution for every
p -1/~.
We have no
opinion
to offerregarding
the existence of solutions when-1/~ p 0.
REFERENCES
[Fi] N.J. Fine. Solution to problem 5407. Amer. Math. Monthly 74 (1967), 740-743.
[Go] S.W. Golomb. Problem 5407. Amer. Math. Monthly 73 (1966), 674.
[Ma] Daniel Marcus. Solution to problem 5407. Amer. Math. Monthly 74 (1967), 740.
[McK] M.A. McKiernan. The functional differential equation D f = Proc. Amer. Math.
Soc. 8 (1957), 230-233.
[Pé] Y.-F.S. Pétermann. On Golomb’s self describing sequence II. Arch. Math. 67 (1996), 473-477.
[PéRé] Y.-F.S. Pétermann and Jean-Luc Rémy. Golomb’s self-described sequence and functional differential equations. Illinois J. Math. 42 (1998), 420-440.
[Ré] Jean-Luc Rémy. Sur la suite autoconstruite de Golomb. J. Number Theory 66 (1997), 1-28.
[Va] Ilan Vardi. The error term in Golomb’s sequence. J. Number Theory 40 (1992), 1-11. Y.-F.S. PGTERMANN Université de Geneve Section de Math6matiques 2-4, rue du Lievre, C.P. 240 CH-1211 Geneve 24 SUISSE
E-mail: petermanos c2a . unige . ch
Jean-Luc REMY
Centre de Recherche en Informatique de Nancy
CNRS et INRIA Lorraine
B.P. 239
F-54506 Vandoeuvre-les-Nancy Cedex FRANCE
E-maid : Jean-Luc . Remyoloria . fr Ilan VARDI IHES 35 route de Chartres F-91440 Bures-sur-Yvette FRANCE E-mail : ilaneihes . fr