Generalized Dirichlet series of n variables associated with automatic sequences

0.18 MB·

by  Shuo Li

Generalized Dirichlet series of n variables associated with automatic
sequences
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Shuo LI
1
0
2 CNRS, Institut de Math´ematiques de Jussieu-PRG
n Universit´e Pierre et Marie Curie, Case 247
a 4 Place Jussieu
J F-75252 Paris Cedex 05 (France)

0 [email protected]
3

]
O
R´esum´e
C
Thisarticle consists togiveanecessary and sufficientcondition of themeromorphiccontinuityofDirichlet
.
h series defined as ax , Where a is a q-automatic sequence of n parameters and P : Cn → C a
at polynomial, such Pthaxt∈NPndPo(exs)snot have zexros on Qn+. And some specific cases of n=1 will also be studied in
m thisarticleasexamplestoshowthepossibilitytohaveanholomorphiccontinuityonthewholecomplexplane.
Some equivalencesbetween infiniteproducts are also built as consequencesof these results.
[

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v1 Introduction
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0
An automatic sequence can be defined intuitively as a sequence of multi-index with values in a finite set, such
6
that for a given integer d≥2, each term of such sequence can be computed by using the base d expansion of the
8
its index. The phenomena of automates can be explained by using the language of sub-sequences , which will be
0
.given as definition in the next section, as well as the language of state machine. [7] gives an example to prove
1
the equivalence between the two definitions in the 2-index case, the statement used in the above article can be
0
7extended to an n-index case for an arbitrary entire number n. Such sequences show many interesting properties
1in a large span of researching fields, such as number theory, harmonic analysis, even in physics and theoretical
:computer science [1].
v
i TheDirichletsequencesoftheform ∞ an havebeenstudiedin[3],andourjobisanaturalgeneralizationof
X n=1 ns
the results in above article by using the same method of calculations. Remarking that the constantsequences are
r P nµ1nµ2...nµI
aa kind of particular automatic sequences, the Dirichlet sequences in the form (n1,n2...nI)∈NI+ p(1n1,n22,..nII)s have
been largely studied, where (µ ,µ ,...µ )∈NI . Mellin [5] firstly proved in 1900 that the above functions have a
1 2 I + P
meromorphic continuation to whole complex plan when µ = 0 for all indexes i, then Mahler [4] generalized the
i
resultto the casethatµ arearbitrarypositiveentirenumberswhenthe polynomialsatisfiesthe elliptic condition
i
in 1927. In 1987, Sargo [8] proved that the condition “lim|p(n ,n ...n )| → ∞ when |(n ,n ...n )| → ∞” is a
1 2 I 1 2 I
necessary and sufficient condition for the aboveDirichlet sequences to have a meromorphiccontinuationto whole
complex plan. In this article, we want to prove such a condition is also necessary and sufficient for the sequences
ax to have an meromorphic continuation on C.
x∈Nn P(x)s
P
2 notations, definitions and basic properties of automatic sequences
Herewedeclaresomenotationsusedinthis article.We denotebyxann-tuple (x ,x ...x )wesayx≥y(resp.
1 2 n
x > y) if and only if x−y ∈ Rn(reps. x−y ∈ Rn), and we have analogue definition for the symbol ≤ (resp.
+
<). We denote by xµ the n-tuple (xµ1,xµ2...xµn). For a constant c, we denote by c the tuple (c,c...c) and for two
1 2 n
tuples x and y, we denote by <x,y > the real number n x y
i=1 i i
P
1
Definition Let d ≥ 2 be an integer. A sequence (a ) with values in the set A is called d−automatic if and
n n≥0
only if its d−kernel N (a) is finite, where the d−kernel of the sequence(a ) is the set of subsequences defined
d x x≥0
by
N (a)= (m ,m ...m )7−→a ;k≥0,(0)≤l ≤(dk−1)
d 1 2 n (dkm1+l1,dkm2+l2,...,dkmn+ln)
n o
Remark A d−automatic sequence necessarily takes finitely many values. Hence we can assume that the set A is
finite.
Because of the definition of q-automatic with n variables, there are some basic properties.
Theorem 1 Let q ≥ 2 be an integer and (a ) be a sequence with values in A, the kernel of the automate.
x x≥0
Then, the following properties are equivalent :
(i) The sequence (a ) is q−automatic
x x≥0
(ii) There exists an integer t≥1 and a set of t sequences N′ = (a1) ,...,(at) ) such that
x x≥0 x x≥0
- the sequence (a1) is equal to the sequence (a ) n o
x x≥0 x x≥0
′
- the set N is closed under the maps (a ) 7−→(a ) for 0≤y ≤q−1
x x≥0 qx+y x≥0
(iii) There exist an integer t≥1 and a sequence(A ) with values in At, that we denote as a column vector,
x x≥0
as (A ,A ,A ...A ,A ...)t. There exist qn matrices of size t×t, say M ,M ...M
1,1...1 2,1...1 1,2...1 1,1...2 2,2...1 1,1...1 1,2...1 q,q..q
, with the property that each row of each M has exactly one entry equal to 1, and the other t−1 entries equal to
i
0, such that :
- the first component of the vector (A ) is the sequence (a )
x x≥0 x x≥0
- for each 0≤y ≤q−1, the equality A =M A holds
qx+y y x
′
Proof It is a natural consequence of the finiteness of the N , the 1-index case was firstly announced in [3].
Proposition 1 Let (a ) be a q-automatic sequence and (b ) be a periodic sequence of period c. Then the
x x≥0 x x≥0
sequence (a ×b ) is also q-automatic and its q-kernel can be completed in such a way that all transition
x x x≥0
matrices of the maps (a ×b ) 7−→ (a ×b ) on the new set are independent on the choice of the
x x x≥0 qx+y qx+y x≥0
values taken in the sequence (b ) .
x x≥0
Proof As (a ) is a q-automatic sequence, we denote by N : (a(1)) ,(a(2)) ,...,(a(l)) its q-kernel.
x x≥0 a x x≥0 x x≥0 x x≥0
The sequence (bx)x≥0 is a periodic sequence, by its nature, it isnalso an q-automatic sequence,owe denote by
N : (b(1)) ,(b(2)) ,...,(b(s)) the q-kernelof(b ) .As both ofthe q-kernelarefinite,we canconclude
b x x≥0 x x≥0 x x≥0 x x≥0
the snet of tensor product of these twooabove sets is finite :
N : (a(i)×b(j)) |0≤i≤l,0≤j ≤s
ab x x x≥0
n o
which is the q-kernel of the sequence (a ×b ) . We remark that there is a subjection from N‘ to N , where
x x x≥0 b b
N‘ is defined as
b
(m ,m ,...,m )−→(qkm +y ,qkm +y ,...,qkm +y ) mod c;k ≥0,(0)≤y ≤(dk−1) ,
1 2 n 1 1 2 2 n n
n o
such a subjection is defined as
N‘ →N :(m ) →(a ) .
b b x x≥0 mx x≥0
However, N‘ itself is the q-kernel of the c-periodic sequence b‘ :b‘(m ,m ,...,m )=(m ,m ,...,m ) mod c and
b 1 2 n 1 2 n
its cardinal is independent on choice of the values in the sequence (b ) .
x x≥0
For a specific mapping (a ×b ) 7−→ (a ×b ) , where 0 ≤ x < q, on the set N , it can be
x x x≥0 qx+y qx+y x≥0 ab
completed to (a ×b‘) 7−→ (a ×b‘ ) , where 0 ≤ x < q and this completion can be represented
x x x≥0 qx+y qx+y x≥0
uniquely by the tensor product of two mappings (a ) 7−→(a ) and(b‘) 7−→(b‘ ) , as each has
x x≥0 qx+y x≥0 x x≥0 qx+y x≥0
a unique transition matrix on their own q-kernel, the tensor product of them is also unique on N and all these
ab‘
transitionmatricesandcompletionofthe q-kernelareindependentonthe choiceofvaluesinthesequence(b )
x x≥0
2
Let us consider the Dirichlet series f(s) = ax , Where a is a q-automate, a necessary condition
x∈Nn/0 p(x)s x
of the convergence of such series is that |p(x)| → ∞ when |x| → ∞ (meaning p(x) → +∞ or p(x) → −∞ when
P
|x|→∞), here we want to show this is a sufficient condition.
Definition Let p(x) = a xα be an n-variable polynomial. We denote by supp(p) the set of multi-index α
α α
such that a is non zero. and E(p) the convex envelope of the subset of Rn
α P
α−β/α∈supp(p),β ∈Rn
We say a monomial xα is extremal if α is(cid:8)an extremal point of E(p). I(cid:9)f p does not have any extremal point, then
it has a biggest monomial such that p(x) is written as p(x)=a xα+ a xβ.
α β<α β
P
3 proof of meromorphic continuation
In this section we prove the main result read as below :
Theorem 2 Let p be a polynomial of n variables that |p(x)| → ∞ when |x| → ∞, then for a given n-tuple µ, if
there exists an l ∈ N such that ni=1xµii ∈ E(pl) then (x)∈Nn/(0) axQp(nix=)1sxµii admits an abscissa of convergence
σ such that it converge absolutely on the semi plan R (s) > σ and has a meromorphic continuation on whole
Q Pe
complex plan. What is more, the number of poles of this function (if any) is finite and located on a finite set.
This result will be declared by proving several lemmas successively :
Lemma 1 Let a be an q-automate, and p(x) = m xα be a n-variable homogeneous polynomial,of degree d,
x α α
with positive coefficients such that |p(x)| → ∞ when |x| → ∞, let µ ∈ Nn be a multi-index and l ∈ N such that
P +
n xµi ∈ E(pl), for any 0 ≤ β ≤ q, define p (x) = q−n(p(qx+β)−p(qx)), then for any k ∈ N, the function
i=1 i β
fQk,β,µ : s −→ (x)∈Nn/0 axpβ(px()xk)Qs+nik=1xµii admits an abscissa of convergence σk,µ such that Fk,β,µ converge to an
holomorphic function on the right half plan R (s)>σ .
P e k,µ
Proof We firstly prove that f (s) converge when R (s)>n.
0,0,0 e
a |a | |a |
|f (s)|=| x |≤ x ≤ x
0,0,0 (x)∈XNn/0p(x)s (x)∈XNn/0p(x)Re(s) (x)∈XNn/0(min(mα)<x,1>)Re(s)
≤ max(|ax|) ( 1 + 1 )
(min(mα)Re(s) <x,1>Re(s) <x,1>Re(s)
<xX,1><n <xX,1>≥n
≤ max(|ax|) ( 1 + mn+−n1−1 ) (1)
(min(mα)Re(s) <x,1>Re(s) (cid:0)mRe(s)(cid:1)
<xX,1><n mX≥n
≤ max(|ax|) ( 1 + mn−1 )
(min(mα)Re(s) <x,1>Re(s) mRe(s)
<xX,1><n mX≥n
max(|a |) 1 1
≤ x ( + )
(min(mα)Re(s) <x,1>Re(s) mRe(s)+1−n
<xX,1><n mX≥n
the summation 1 exists and is bounded when R (s)>n.
m≥n mRe(s)+1−n e
For any 0 ≤ β <Pp, and any multi-index k ∈ supp(p), we remark that ni=0(xi+βi)ki = l≤kCl ni=0(xi)ki
whereC <pn,whichshowsallmonomialsofthepolynomialp ,ofdegreed−1,canbemajoredbyoneofp’s.This
l β
Q P Q
effect shows that, pβ(x) → 0 when |x| → ∞. There is an C ∈ R+ such that for all < x,1 >≥ C ,|pβ(x)| < 1/2.
p(x) 0 0 p(x)
For any k >0, take C =max(n,C )
1 0
|axpkβ(x) ni=1xµii|≤ ax pkβ(x) ≤ ax +(1/2)k ax
p(x)Qs+k pRe(s)−l(x)pk(x) pRe(s)−l(x) pRe(s)−l(x)
(x)∈XNn/0 (x)∈XNn/0 <x,X1><C1 <x,X1>≥C1
3
With K a constant in R+, the above function converge to an holomorphic function on the half plan R (s)>
e
n+l. what is more, for all b>k, | axpbβ(x) | is bounded on this half plan.
<x,1>≥x1 ps−l+b(x)
Lemma 2 With the same notatioPns and hypotheses as above, The function F : s −→ ax admit a
(x)∈Nn/0 p(x)s
meromorphic continuation on the whole complex plan.
P
Proof In this proof, we consider the q-automatic sequence (a ) as itself multiplied by an constant sequence
x x≥0
(b ) = 1, which is a q-automatic sequence. Because of the proposition 1, the q-kernel of this sequence admits
x x≥0
a completion, we can define a sequence of vector (A ) and the matrices of transitionon this completion as we
x x≥0
do in the theory 1.
For any µ∈Nn+, considering the function Fµ defined as Fµ(s)= (x)∈Nn/(0) AzQp(nix=)1sxµii, where Ax is defined
as above, and an constant N ∈N such that C <N np, where C is defined as in the previous lemma.
0 1 0 1 P
F (s)= Ax ni=1xµii = Ax ni=1xµii + Aqz+y ni=1(qzi+yi)µi
µ p(x)s p(x)s ps(qz+y)
(x)∈XNn/(0) Q (x)<X(N0q) Q (yX)<(q)(z)∈NnX/{t<N0} Q
= Ax ni=1xµii + Aqz+y ni=1(qzi)µi + Aqz+yCψ,y ni=1(qzi+yi)ψi
p(x)s ps(qz+y) ps(qz+y)
(x)<X(N0q) Q (yX)<(q)(z)∈NnX/{t<N0} Q (ψX)<(µ)(yX)<(q)(z)∈NnX/{t<N0} Q
(2)
Where C is a periodic coefficient in function of x, we denote by Res (s) the therm
ψ,y µ
Aqz+yCψ,y ni=1(qzi+yi)ψi
Res (s)= .
µ ps(qz+y)
Q
(ψX)<(µ)(yX)<(q)(z)∈NnX/{t<N0}
We remark that all sequences a C are in the form of a production of a specific q-automatic sequence by
qz+y ψ,y
a q-periodic one, because of the proposition1, such sequences admit a unique completion,the same one as a has,
x
and the transition matrices on this completion is invariant on function of the q-periodic sequences C .
ψ,y
F (s)= Ax ni=1xµii = Ax ni=1xiµi + Aqz+y ni=1(qzi+yi)µi
µ p(x)s p(x)s ps(qz+y)
(x)∈XNn/(0) Q (x)<X(N0q) Q (yX)<(q)(z)∈NnX/{t<N0} Q
= Ax ni=1xµii + Aqz+y ni=1(qzi)µi +Res (s)
p(x)s ps(qz+y) µ
(x)<X(N0q) Q (yX)<(q)(z)∈NnX/{t<N0} Q
= Ax ni=1xµii + M Aqz ni=1(qzi)µi 1 +Res (s)
(x)<X(N0q) Qp(x)s (yX)<(q) y(z)∈NnX/{t<N0} Qps(qz) (1+ ppy((zz)))s µ
= Ax ni=1xµii + M Az ni=1(qzi)µi −s+k (−py(z))k+Res (s)
(x)<X(N0q) Qp(x)s (yX)<(q) y(z)∈NnX/{t<N0} Qps(qz) kX≥0 k ! p(z) µ
= Ax ni=1xµii +q<ζ,1>−ns M −s+k Az ni=1(zi)µi(−py(z))k +Res (s)
(x)<X(N0q) Qp(x)s (yX)<(q) ykX≥0 k !(z)∈NnX/{t<N0} Q (p(z))s+k µ
(3)
The above equation gives :
(Id−q<µ,1>−ns M )F (s)=q<µ,1>−ns M −s+k Az ni=1(zi)µi(−py(z))k
(yX)<(q) y µ (yX)<(q) ykX≥1 k !(z)∈NnX/{t<N0} Q (p(z))s+k (4)
A n xµi
+ x i=1 i +Res (s)
p(x)s µ
(x)<X(N0q) Q
4
By multiplying comt(Id−q<µ,1>−ns M ) on both side, we have :
(y)<(q) y
P
A n xµi
det(Id−q<µ,1>−ns M )F (s)=comt(Id−q<µ,1>−ns M )( x i=1 i +Res (s)
y µ y p(x)s µ
(yX)<(q) (yX)<(q) (x)<X(N0q) Q
+q<µ,1>−ns M −s+k Az ni=1(zi)µi(−py(z))k)
y k (p(z))s+k
(yX)<(q) kX≥1(cid:18) (cid:19)(z)∈NnX/{t<N0} Q
(5)
The above formula shows that F has a meromorphiccontinuation onthe half plan on R (s)>n−1. Thanks
0 e
to the previous lemma, we have for any µ ∈ Nn, such that n xµi ∈ E(p), the function F (s) converges and
+ i=1 i µ
is bounded when R (s) > n + 1. The equation 5 shows that, all of these functions F have a meromorphic
e Q µ
continuation on R (s) > n. In particular, for any 0 ≤ β < q, the functions s −→ Axpβ(x) have a
e (x)∈Nn/(0) p(x)s
meromorphic continuation on R (s) > n, this effect proves that F has a meromorphic continuation on the half
e 0 P
plan on R (s) > n−2. We do this successively to prove the meromorphic continuation on R (s) > n−3... To
e e
conclusion, The function F has a meromorphic continuation on whole complex plan.
0
What is more, the poles of such function can only locate at the zeros of the function s −→ det(Id −
q<µ,1>−ns M ) for an arbitrary µ ∈ Nn, so we conclude that all poles of function F(s) are located
(y)<(q) y +
in the set
P 1 logλ 2ikπ
s= ( + −l)
n logq logq
with λ any eigenvalue of the matrix M ,k∈Z,l∈Z and log is defined as complex logarithm.
(y)<(q) y
P
Remark The previous lemma proves indeed that for all monomial n xµi ∈E(pl) for some l ∈N the function
i=1 i
s−→ (x)∈Nn/(0) AxQp(nix=)1sxµii admits a meromorphic continuation oQn whole complex plan.
P
Lemma 3 Let p be an n-variable polynomial, of degree d , such that |p(x)|→∞ when |x|→∞, and all its terms
are extreme, let a be an automate, then G:s−→ ax is meromorphic.
x (x)∈Nn/(0) ps(x)
P
Proof The hypotheses |p(x)| → ∞ when |x| → ∞ shows that all coefficients of this polynomial are positive or
negative at the same time. Without loss of generality, we suppose that all its coefficient are positive. We build
another sequence b of n+1-variable in such a way that b = 0 if y 6= 1 and b = a if y = 1. we
y y n+1 y y1,y2...yn n+1
remark that the sequence b is also automatic because its q-kernel is finite. And we complete the function p to
y
an n+1-variable polynomial g of degree d in such a way that we multiply x to all terms whose degrees are
n+1
smaller then d until they all have the same degree d. so we have
a b
x x
G(s)= =
ps(x) gs(x)
(x)∈XNn/(0) (y)∈NXn+1/(0)
The previous lemma proves that this function has a meromorphic continuation to whole complex plan.
Remark We have the same remark as before : for any monomial n xµi ∈ E(gl) for some l ∈ N the function
i=1 i
s−→ (x)∈Nn/(0) BxQp(nix=)1sxµii admits a meromorphic continuation oQn whole complex plan.
P
proof of the theorem 2 Letus writethe polynomialp(x) inthe formp(x)=p (x)+Res(x)where p (x)is the
0 0
summationofallextreme monomialsofp(x),itis easytosee Res(x) =O(x−1). Sothere existsx ∈N suchthat
p0(x) 0 +
forall|x|>x ,|Res(x)|<1.Foranygivensemiplan{s|R (s)>m,m∈R},takeanintegers >2−dm+<µ,1>
0 p0(x) e 0
whereddenotedthedegreeofthepolynomial.wecanconcludethat AxQni=1xµii ∞ −s+k Resk(x) =
|x|>x0 (p0(x))s k=s0+1 k pk0(x)
O(|s|s0). So we have the equivalence as below :
P P (cid:0) (cid:1)
5
A n xµi A n xµi A n xµi 1
x i=1 i = x i=1 i + x i=1 i
(x)∈XNn+/(0) Qp(x)s |xX|≤x0 Qp(x)s |xX|>x0 (pQ0(x))s (1+ Rpe0s((xx)))s (6)
A n xµi A n xµi s0 −s+k Resk(x)
= x i=1 i + x i=1 i +O(|s|s0)
p(x)s (p (x))s k pk(x)
|xX|≤x0 Q |xX|>x0 Q0 Xk=0(cid:18) (cid:19) 0
For each k∈N,
A n xµi Resk(x) A C n
x(p (ix=)1)si pk(x) = (p (xx))is+k xjii (7)
|xX|>x0 Q0 0 jX≤kµ|xX|>x0 0 iY=1
Where C are constants to x, and s−→ AxCi n xji are meromorphic functions because of the
i |x|>x0 (p0(x))s+k i=1 i
previous lemma, in effect, if we complete the polynomial p to an n+1-variable polynomial g, then all term of
the form n xµi are in the convexset E(glP) for some l, an0d wQe use the remark of lemma 3. As there are finitely
i=1 i
many of meromorphic function in (6), we can conclude that for every k >0, s−→ AxQni=1xµiiResk(x) is
Q |x|>x0 (p0(x))s
meromorphic. This proves that for an arbitrary s ∈ R the function f(s) is a finite summation of meromorphic
0 P
continuous functions on the half plan R (s) > s , so itself is meromorphic continuous on this half plan. To
e 0
conclusion, the function s−→ (x)∈Nn/(0) axQp(nix=)1sxµii is meromorphic on whole complex plane.
Proposition 2 Let f(s)= (Px)∈Nn/(0) axQp(nix=)1sxµii be the function defined as in theorem 2. Let s0 be its first pole
on the axis of real number counting from plus infinity to minus infinity. Then the function H(s) has a simple pole
P
on this point.
Proof Werecallaclassicalresultofmatrices(see[6]):LetB beamatrixofsizet×toveranycommutativefield,
p (X)be itscharacteristicpolynomial,andπ (X)be its monic minimalpolynomial.Let∆(X)be the monicgcd
B B
of the entries of (the transpose of ) the comatrix of the matrix (B−XI), then :
p (X)=(−1)tπ (X)∆(X)
B B
Let us denote by B the matrix (nq)−1 M and by T its size. By devising ∆(qn(s−1)+<µ,1>) on both
(y)<(q) y
sides of the formula 5, we get :
P
π (qn(s−1)+<µ,1>)H(s)= comt(qn(s−1)+<µ,1>Id− (y)<(q)My)( Ax ni=1xµii +Res (s)
B ∆(qn(s−1)+<µ,1P>) (x)<X(N0q) Qp(x)s µ (8)
+q<µ,1>−ns M −s+k Az ni=1(zi)µi(−py(z))k)
y k (p(z))s+k
(yX)<(q) kX≥1(cid:18) (cid:19)(z)∈NnX/{t<N0} Q
The right hand side of above function is holomorphic when Re(s) > s . As s is the first pole of H(s) on
0 0
real axis counting from plus infinity, it is a zero point of function π (qn(s−1)+<µ,1>) associated to the eigenvalue
B
1 of the matrix B. On the other hand, as B is a stochastic matrix, it has a simple root on 1, so the function
π (qn(s−1)+<µ,1>) has a simple root on s which conclude the proposition.
B 0
4 Infinite products
LetP(x)= d a xi be apolynomialwhichdoesnothavezerosonQandP˜(x) be anothersuchthatP˜(x)=
i=0 i
d−1−aixn−i, by definition, we have P(x)=a xd−a xdP(1). Let us define c = ai for all i=0,1,...,d−1.
i=0 ad P d d x i ad
In this section we consider two Dirichlet series generated by 1-index automatic sequences :
P
∞ (ζ)Sq(n)
f(s)=
(P(n+1))s
n=0
X
6
∞ (ζ)Sq(n)
g(s)=
(P(n))s
n=1
X
Where q and r are two integers satisfying 2 ≤ r ≤ q and r divides q, ζ is a r-th root of unity, such that ζ 6= 1,
s (n)isthesumofdigitsofnintheq-aryexpansionsatisfyings (0)=0ands (qn+a)=s (n)+afor0≤a≤q−1.
q q q q
Let us denote by :
∞ (ζ)Sq(n)
φ(s)=
(n+1)s
n=0
X
∞ (ζ)Sq(n)
ψ(s)=
(n)s
n=1
X
Itisprovedin[2]thatφandψ haveholomorphiccontinuationstothewholecomplexplane,andψ(s)(qs−1)=
φ(s)(ζqs−1) for all s∈C
Proposition 3 f and g also have holomorphic continuations to the whole complex plane if all their coefficients
c satisfy max|c |< 1.
i i d
Proof We firstly remark that the hypothesis of max|c | < 1 induces the fact |P˜( 1 )| < 1 for any n ∈ N .
i d n+1 +
Indeed, |P˜( 1 )|=| d−1−ai(n+1)d−i|≤ d−1|ai|<1.
n+1 i=0 ad i=0 ad
P ∞ (ζ)SqP(n)
f(s)=
(P(n+1))s
n=0
X
∞ (ζ)Sq(n)
=a−s
d (n+1)ds(1−P˜( 1 ))s
n=0 n+1
X
∞ (ζ)Sq(n) ∞ s+k−1 1
=a−s P˜( ))k
d (n+1)ds k n+1
n=0 k=0(cid:18) (cid:19)
X X
∞ (ζ)Sq(n) ∞ s+k−1 dk
=a−ds (n+1)ds k mk,l(n+1)−l (9)
n=0 k=0(cid:18) (cid:19)l=k
X X X
∞ s+k−1 dk ∞ (ζ)Sq(n)
=a−s m
d k k,l (n+1)ds+l
k=0(cid:18) (cid:19)l=k n=0
X X X
∞ dk
s+k−1
=a−s m φ(ds+l)
d k k,l
k=0(cid:18) (cid:19)l=k
X X
∞ dk
s+k−1
=a−sφ(ds)+as m φ(ds+l)
d d k k,l
k=1(cid:18) (cid:19)l=k
X X
Wherem = c andM aresetsofk elementsincludedin{c |1≤i≤n−1}
k,l Mk,l∈P({ci|1≤i≤n−1}) ci∈Mk,l i k,l i
and the sum of indices of its elements equals l. The hypothesis m = max|ci| < 1 shows that | dk−km | ≤
P Q d l=k k,l
(md)k < 1, so the left part of (9) converge because φ(s) is bounded for large |s|, which proves the holomorphic
P
continuation of f on the whole complex plane.
It is easy to check f(0)=0, so we have :
7
∞ dk
f′(0)=dφ′(0)+a−s lim 1 s+k−1 m φ(ds+l)
d s→0s k k,l
k=1 (cid:18) (cid:19)l=k
X X
∞ dk
=−dlogq/(ζ−1)+ k−1 m φ(l)
k,l
k=1 l=k
X X
∞ dk ∞ (ζ)Sq(n)
=−dlogq/(ζ−1)+ k−1 m
k,l (n+1)l
kX=1 Xl=k nX=0 (10)
∞ ∞ dk
1
=−dlogq/(ζ−1)+ (ζ)Sq(n) k−1 m
k,l(n+1)l
n=0 k=1 l=k
X X X
∞ ∞
1
=−dlogq/(ζ−1)+ (ζ)Sq(n) k−1P˜k( )
n+1
n=0 k=1
X X
∞
1
=−dlogq/(ζ−1)+ (ζ)Sq(n)log(1−P˜( ))
n+1
n=0
X
on the other hand, one has for all s, ψ(s)(qs−1)=φ(s)(ζqs−1)
∞ dk
f′(0)=dφ′(0)+a−s lim 1 s+k−1 m φ(ds+l)
d s→0 s k k,l
k=1 (cid:18) (cid:19)l=k
X X
∞ dk
=−dlogq/(ζ−1)+ k−1 m ψ(l)(ql−1)/(ζql−1) (11)
k,l
k=1 l=k
X X
∞ dk ∞ dk
=−dlogq/(ζ−1)+ζ−1 k−1 m ψ(l)+(ζ−1−1) k−1 m ψ(l)/(ζql−1)
k,l k,l
k=1 l=k k=1 l=k
X X X X
by the same method as above, we can deduce
∞ dk ∞
1
k−1 m ψ(l)= (ζ)Sq(n)log(1−P˜( )) (12)
k,l
n
k=1 l=k n=1
X X X
and
∞ dk ∞ dk ∞ (ζ)Sq(n) ∞
k−1 m ψ(l)/(ζql−1)= k−1 m (ζql)−r
k,l k,l (n)l
k=1 l=k k=1 l=k n=1 r=1
X X X X X X
∞ ∞ ∞ dk
1
= (ζ)−r (ζ)Sq(n) k−1 m (13)
k,l(nqr)l
r=1 n=1 k=1 l=k
X X X X
∞ ∞
1
= (ζ)−r (ζ)Sq(n)log(1−P˜( ))
nqr
r=1 n=1
X X
As a consequence,
∞ ∞ ∞ ∞
1 1 1
(ζ)Sq(n)log(1−P˜( ))=ζ−1 (ζ)Sq(n)log(1−P˜( ))+(ζ−1−1) (ζ)Sq(n)−rlog(1−P˜( )) (14)
n+1 n nqr
n=0 n=1 r=1n=1
X X XX
Proposition 4 We have the equality
∞ ∞ ∞ ∞
(1−P˜( 1 ))ζSqn × (1−P˜(1))−ζSqn−1 ×( (1−P˜( 1 ))ζSqn−r)1−ζ−1 =1 (15)
n+1 n nqr
n=0 n=1 r=1n=1
Y Y Y Y
8
Let us consider an example as follows : for q = 2, then ζ = −1, if we take p(x) = 5x2 − x − 1 then
p˜(x)=0.2x2+0.2x.Wecomputeseparatelythevaluesoftheseequations:A(N)= N (1−P˜( 1 ))ζSqn,B(N)=
n=0 n+1
N (1−P˜(1))−ζSqn−1,C(R,N)= R N (1−P˜( 1 ))ζSqn−r andD(R,N)=A(N)B(N)C(R,N)2 forgiven
n=1 n r=1 n=1 nqr Q
N and R. It shows that, A(N) convergesto 0.73, B(N) convergesto 1.80, and for all R>50, C(R,N) converges
Q Q Q
to 0.87, as a consequence, D(R,N) converges to 1. This example can be considered as an evidence to show that
the formula (15) holds.
More over, if we suppose, for any j
r−1 if s (m)=j (mod r)
x (m)= r q (16)
j (−r1 if sq(m)6=j (mod r)
We clearly have
x (m)=0 (17)
j
j modr
X
Furthermore,
x (m)ζj =ζSq(m) (18)
j
j modr
X
The Formula (14) can be reformulated as
∞ ∞ ∞
1 1 1
ζj x (n)(log(1−P˜( ))+ ζ−rlog(1−P˜( )))= ζj−1 x (n)(log(1−P˜( ))
j n+1 nqr j n
j modr n=0 r=1 j modr n=1
X X X X X
∞
1
+ ζ−rlog(1−P˜( )))
nqr
r=1
X
(19)
Let now η be a primitive root of unity, we can apply relation (19) successively to ζ =ηa for a=1,2,...,r−1.
Because of (17), we also have
∞ ∞
1 1
x (n)(log(1−P˜( ))+ ζ−rlog(1−P˜( )))=
j n+1 nqr
j modrn=0 r=1
X X X (20)
∞ ∞
1 1
x (n)(log(1−P˜( ))+ ζ−rlog(1−P˜( )))=0
j n nqr
j modrn=1 r=1
X X X
Definethematrices:Mat tobeMat =(ηij)andMat tobeMat =(ηij−i),i=0,1,...,r−1;j =0,1,...,r−1,
1 1 2 2
define λ and β by
∞ ∞
1 1
λ(j)= x (n)(log(1−P˜( ))+ ζ−rlog(1−P˜( )))
j n+1 nqr
n=0 r=1
X X
∞ ∞
1 1
β(j)= x (n)(log(1−P˜( ))+ ζ−rlog(1−P˜( )))
j n nqr
n=1 r=1
X X
Let
λ(0)
λ(1)
A= 
...
 λ(r−1) 
 
 
9
β(0)
β(1)
B = 
...
 β(r−1) 
 
 
Then we have
Mat A=Mat B
1 2
On the other hand, A is invertible and Mat =Mat ×Mat with
1 2 3
0 0 ... 0 1
1 0 ... 0 0
Mat3 = 0 1 ... 0 0 
 0 0 ... 1 0 
 
 
So we have
A=Mat ×B (21)
3
Proposition 5 We have the equality λ(i)=β(i−1) for i=1,2,...,r−1 and λ(0)=β(r−1), which lead to, for
i=1,2,...,r−1,
∞ ∞ ∞ ∞
1 1 1 1
((1−P˜( )× (1−P˜( )))xj(n) = ((1−P˜( )× (1−P˜( )))xj−1(n)
n+1 nqr n nqr
n=0 r=1 n=1 r=1
Y Y Y Y
and for i=0
∞ ∞ ∞ ∞
1 1 1 1
((1−P˜( )× (1−P˜( )))x0(n) = ((1−P˜( )× (1−P˜( )))xr−1(n)
n+1 nqr n nqr
n=0 r=1 n=1 r=1
Y Y Y Y
5 Reference
1. J.-P. Allouche, Automates finis et s´eries de Dirichlet, S´eminaire de Th´eorie des Nombres de Bordeaux
1984-1985(1985), Expos´e 8, p. 10.
2.J.-P.AlloucheandH.Cohen,Dirichletseriesandcuriousinfiniteproducts,Bull.Lond.Math.Soc.17(1985),
531–538.
3. J.-P. Allouche, M. Mend`es France, J. Peyri`ere, Automatic Dirichlet series, J. Number Theory 81, (2000)
359–373.
4. K. Mahler, U¨ber einen Satz von Mellin, Math. Ann, 100 (1928), 384–395
5.H.Mellin,EineFormelFu¨rdenLogarithmustranscendenterFunktionenvonendlichenGeschlechtActaSoc.
Sci. Fennicae, 29 (1900), pp. 3–49
6. H. Minc, “Nonnegative matrices,” Wiley, 1988.
7.O. Salon,Suites automatiques `a multi-indices et alg´ebricit´e,C. R. Acad. Sci. ParisS´er.I Math. 305(1987),
no. 12, 501–504.MR 916320
8.P.Sargos,Prolongementm´eromorphedess´eriesdeDirichletassoci´ees`adesfractionsrationnelles`aplusieurs
variables Annales de l’Institut Fourier, Fasc. 3 (1984)
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