Jacques Sauloy
email: jacques.sauloy@gmail.com

Changgui Zhang
email: changgui.zhang@univ-lille.fr

On the vanishing of coefficients of the powers of a theta function

A result on the Galois theory of q-difference equations [3] leads to the following question: if q ∈ C * , |q| < 1 and if one sets θ q (z) := ∑ m∈Z q m(m-1)/2 z m , can some coefficients of the Laurent series expansion of θ k q (z), k ∈ N * , vanish ? We give a partial answer.

Introduction 1.Origin of the problem

Let q ∈ C such that |q| > 1. In [START_REF] Sauloy | Théorie analytique locale des équations aux q-différences de pentes arbitraires[END_REF], one uses the function 1 θ q (z) := ∑ m∈Z q -m(m+1)/2 z m , z ∈ C * . This is a holomorphic function over C * . For every k ∈ N * , one introduces the coefficients t (k) n , n ∈ Z, through the expansion into a Laurent series θ k q (z) = ∑t (k) n z n . These series converge normally over every compact subset of the domain C \ D(0, 1), whence the explicit formula:

t (k) n = ∑ m 1 ,...,m k ∈Z m 1 +•••+m k =n q -(m 1 (m 1 +1)+•••+m k (m k +1))/2 , which implies that each t (k)
n is,as a function of q, holomorphic over the domain C\D(0, 1). Since q > 0 ⇒ t (k) n (q) > 0, the function t (k) n is not identically 0, so the set of its zeroes is 1 discrete, whence denumerable. As a consequence, the set of all zeroes of all functions t (k) n is denumerable.

For all q such that no t (k) n (q) vanishes, the first author has described in [START_REF] Sauloy | Théorie analytique locale des équations aux q-différences de pentes arbitraires[END_REF] the "wild fundamental group" of analytic complex linear q-difference equations which are irregular at 0 (and with arbitrary slopes). This led to call "good values" of q such values. As we just saw, this property is generically satisfied. The goal of the present article is to study various cases of vanishing of t (k) n (q). However, to stick as much as possible to the classical litterature on q-special functions (above all Ramanujan's opus !), we shall assume in this work that |q| < 1, q = 0, and one shall begin with the following definition:

(1) θ q (x) := ∑ m∈Z q m(m-1)/2 x m , x ∈ C * , which again defines a holomorphic function over C * . For every k ∈ N * , we shall denote γ k,n , n ∈ Z the coefficients of the Laurent series expansion of θ k q :

(2)

θ k q (q) = ∑ n∈Z γ k,n q n =⇒ γ k,n (q) = γ k,n = ∑ m 1 ,...,m k ∈Z m 1 +•••+m k =n q (m 1 (m 1 -1)+•••+m k (m k -1))/2 ,
The relation to the formulas in [START_REF] Sauloy | Théorie analytique locale des équations aux q-différences de pentes arbitraires[END_REF] is clear: if |q| > 1 and if p := q -1 , then θ q (z) = θ p (x) with x := pz, and moreover t (k) n (q) = γ k,n (q -1 ); and the "good values" of q, |q| > 1, are those such that none of the γ k,n (q -1 ) vanishes.

In this work, we shall first study θ q , |q| < 1, q = 0, and the coefficients γ k,n . However, in 2.2 and then again in 3.2 we shall rather use (for simplicity of the computations) another of the classical theta functions 2 :

(3) ϑ(x) := ϑ q (x) := ∑ n∈Z q n 2 /2 x n .

We correspondingly introduce the coefficients of powers of ϑ:

(4) ϑ k q (z) = ∑ n∈Z c k,n z n =⇒ c k,n (q) = c k,n = ∑ (n 1 ,...,n k )∈Z k n 1 +•••+n k =n q (n 2 1 +•••+n 2 k )/2 .
From the obvious relations: ϑ(x) = θ q (q 1/2 x), whence c k,n (q) = q n/2 γ k,n , we see how both studies are easily related. Our main results consist in a series of sufficient conditions on k, n for the existence of a zero of γ k,n or, equivalently, of c k,n , in the punctured open disk

• D(0, 1) \ {0}. However we confess that we could not solve the main mystery: which are the bad values of q ?

First computations

Obviously γ 1,n (q) = q n(n-1)/2 never vanishes. The same is true for γ 2,n (q) thanks to the following calculation: γ 2,n (q) = ∑ l+m=n q (l(l-1)+m(m-1))/2 = ∑ l+m=n q n(n-1)/2 q -lm = q n(n-1)/2 ∑ m q m(m-n) = q n(n-1)/2 ∑ m (q 2 ) m(m-1)/2 (q -n+1 ) m = q n(n-1)/2 θ q 2 (q -n+1 ), which can only vanish for 0 < |q| < 1 by virtue of the triple product formula along with the impossibility of relation -q -n+1 ∈ q Z . Computing γ 2,n easily yields a closed formula for θ 2 q :

(5) θ 2 q (x) = θ q 2 (q)θ q 2 (x 2 ) + xθ q 2 (1)θ q 2 (qx 2 ).

Here is a direct argument (more general arguments will come later, see sections 2.1 and 2.2):

θ 2 q (x) = ∑ n∈Z γ 2,n (q)
x n = ∑ n∈Z q n(n-1)/2 θ q 2 (q -n+1 )x n = ∑ m∈Z q m(2m-1) θ q 2 (q -2m+1 )x 2m + ∑ m∈Z q m(2m+1) θ q 2 (q -2m )x 2m+1

= ∑ m∈Z q m(2m-1) q -m(m+1) q m θ q 2 (q)x 2m + ∑ m∈Z q m(2m+1) q -m(m+1) θ q 2 (1)x 2m+1 = θ q 2 (q) ∑ m∈Z q m(m-1) x 2m + θ q 2 (1) ∑ m∈Z q m 2 x 2m+1 = θ q 2 (q)θ q 2 (x 2 ) + xθ q 2 (1)θ q 2 (qx 2 ).

We used en passant the relation θ q (q -m x) = q -m(m+1)/2 x m θ q (x) (with q ← q 2 ).

It is difficult not to hope that such an identity can be extended to higher powers. According to Professor Bruce Berndt, powers of theta functions are tackled in volume III of the "Ramanujan Notebooks" [START_REF] Bruce | Ramanujan's notebooks. Part III[END_REF]; and indeed, he there gives the proof (Entry 29) of such a relation from which the above equality flows easily; and Entry 30 seems to give the clue to the calculation of θ 3 q (z) and maybe of θ 4 q (z), but it is not quite clear. At any rate, using Entries 18 to 30 does allow a first approach to our problem, see section 2.1.

Another kind of calculus, akin to the theory of divergent q-series [START_REF] Zhang | A discrete summation for linear q-difference equations with analytic coefficients: General theory and examples[END_REF] and of their transformations (q-Borel, q-Laplace . . . ), allows one to iteratively obtain such identities. We expound it in section 2.2. With a bit of real analysis one can then deduce for instance that γ 3,0 (q) vanishes for at least one negative value of q. Using some numerical/formal calculus with Maple, one can make the value more precise. Note that this section rather uses function ϑ in order to simplify the form of the identities; translating back the results is immediate.

It is possible to recognize in the series such as that giving γ 3,0 (q) the generating series of the number of representations by some quadratic form:

γ 3,0 (q) = ∑ a,b,c∈Z a+b+c=0 q (a 2 -a+b 2 -b+c 2 -c)/2 = ∑ a,b,c∈Z a+b+c=0 q -(ab+bc+ca) = ∑ a,b∈Z q a 2 +ab+b 2 = f (q),
where we have set f (x) := ∑ a,b∈Z x a 2 +ab+b 2 = ∑ n≥0 r(n)x n for |x| < 1; here r(n) denotes the number of pairs (a, b) ∈ Z × Z such that a 2 + ab + b 2 = n (the computation is wholly justified by the fact that this is a positive definite quadratic form). The integer:

R(n) := r(0) + • • • + r(n) = card{(a, b) ∈ Z × Z | a 2 + ab + b 2 ≤ n} is equivalent, when n → +∞, to Cn, where C is the area of the ellipse x 2 + xy + y 2 ≤ 1, i.e. 2π/ √ 3.
By standard abelian theorems we draw that, when x → 1 -:

f (x) 1 -x = ∑ n≥0 R(n)x n ∼ ∑ Cnx n = C x (1 -x) 2 =⇒ f (x) ∼ C 1 -x •
On the other hand, the exponent a 2 + ab + b 2 is even if, and only if a and b are. The even part of series f (x) is therefore the sub-sum on the such pairs, that is f (x 4 ), so that

f (-x) = 2 f (x 4 ) -f (x). Hence, still for x → 1 -: f (x 4 ) ∼ C 1 -x 4 ∼ C 4(1 -x) =⇒ f (-x) ∼ -C 2(1 -x) =⇒ lim x→-1 x>-1 f (x) = -∞.
Since f (0) = 1, the function f (x), thus also the function γ 3,0 (q), vanishes at least once on ]-1, 0[. One of our goals is to generalize this fact and the ideas which led to it.

A bit of theory 1.3.1 Linear relations among theta powers and theta functions

We shall write various sets of integers as:

N = {0, 1, . . .}, N * = N \ {0} = {1, 2, . . .}, Z = N ∪ (-N) = {0, 1, -1, 2, -2, . . .}.
Let k ∈ N * . The C-linear space of holomorphic functions over C * such that f (qx) = x -k f (x) has dimension k, as can be seen by expanding f into a Laurent series ∑ f m x m : the latter is uniquely and linearly determined by coefficients f 0 , . . ., f k-1 . The functions x i θ q k (q i x k ), i = 0, . . . , k -1, belong to that space and are linearly independent (their supports are pairwise disjoint). The function θ k q also belongs to that space, whence a relation:

(6) θ k q (x) = k-1 ∑ i=0 a k,i x i θ q k (q i x k ).
For instance identity (5) at the end of the calculation at the beginning of 1.2 says that a 2,0 = θ q 2 (q) and a 2,1 = θ q 2 (1).

Expanding the left hand side along with each term of the right hand side yields the relations:

(7) n = jk + i, 0 ≤ i ≤ k -1 =⇒ γ k,n = (q k ) j( j-1)/2 q i j a k,i .
In particular a k,i = γ k,i (i = 0, . . ., k -1) so our identity above boils down to:

θ k q (x) = k-1 ∑ i=0 γ k,i x i θ q k (q i x k ).
Note that the functional equation θ k q (qx) = x -k θ k q (x) immediately yields γ k,n+k = q n γ k,n which implies, by iteration, formula (7).

Thus, on the one hand vanishing of some γ k,n (q) only depends of the class of n modulo k; on the other hand, the k complex numbers a k,i , i = 0, . . . , k -1 contain all relevant information. We are therefore led to look for explicit identities (i.e. with known coefficients) of the above type.

Some easy facts about the coefficients

We collect here some facts about the coefficients γ k,n (and therefore also about the coefficients c k,n ) which follow directly from the explicit formulas ( 2) and ( 4).

First note that n(n -1) ≥ 0 for all n ∈ Z, and that it vanishes only for n = 0, 1. So γ k,n (q) involves only non negative powers of q and the power q 0 comes from terms with multiindex (m 1 , . . . , m k ) ∈ {0, 1} k such that m 1 + • • • + m k = n. There are k n such terms, whence:

(8) ∀k ∈ N * , ∀n ∈ Z , γ k,n (q) = k n + O(q).
Second, note that in the summation (2), changing (m 1 , . . ., m k ) to (1m 1 , . . . , 1m k ) does not change the general term q (•••) , permutes Z k and transforms the sum n into kn. Therefore:

(9) ∀k ∈ N * , ∀n ∈ Z , γ k,n = γ k,k-n .
Last, we give without proof an identity somehow related to the above symmetry; it can be checked by elementary manipulation of indexes:

∀k ∈ N * , ∀n ∈ {1, . . ., k -1} , γ k,n = ∑ m∈Z γ n,m γ k-n,m .

Contents of the paper

In order to have information on the coefficients, we first look for identities in section 2. Using Ramanujan's formulas in 2.1 gives some information on powers of θ q , unhappily not enough. So we develop in 2.2 a calculus inspired from the theory of transformations of divergent q-series [START_REF] Zhang | A discrete summation for linear q-difference equations with analytic coefficients: General theory and examples[END_REF]. However, this is more simply done using function ϑ instead of θ q (but the results are quite easy to translate). This calculus allows us to find some preliminary information on the vanishing of coefficients with some real analysis.

In 3 we generalize the real analysis argument through an asymptotic study of the coefficients of θ q k , resp. ϑ k , near q = -1. First, in 3.1 we relate said coefficients to generating series of the numbers of representations by quadratic forms. Usual considerations of geometry of numbers yield rather good estimations (theorem 1) and a first list of sufficient conditions for vanishing coefficients (corollary 2). Last, well known modular properties of the generating series allow for some complementary information (proposition 7). Last, in 3.2 we return to the formalism for ϑ introduced in 2.2 and use the classical modular transformation formula to obtain again asymptotic estimates (theorems 3 and 2) from which we deduce more sufficient conditions: actually, exactly what is needed to recover corollary 2 and its consequences.

The difference between the two approaches towards the same sufficient conditions is reflected in the nature of the asymptotic estimates: one is in terms of powers of t = log(e -πi q), the other is a transseries in the exponential scale.

Note that in the end one mysterious question remains unsolved: what can be said of the "bad values" of q ? Contrary to our first guesses, they are not all real negative . . . In the same spirit, if f : Z → C is any k-periodic function, one writes ∑ a (mod k) f (a) a sum taken over an arbitrary set of representatives a modulo k. Last, we shall write in underlined characters k-uples n = (n 1 , . . . , n k ), whether in Z k , R k . . .

Identities and coefficients

2.1 Some identities drawn from "Ramanujan Notebooks"

Our sole reference in this section is chapter 1 of volume III of the Ramanujan Notebooks [START_REF] Bruce | Ramanujan's notebooks. Part III[END_REF]. For non zero complex numbers a, b such that |ab| < 1, Ramanujan introduces the function:

f (a, b) := 1 + +∞ ∑ k=1 (ab) k(k-1)/2 (a k + b k ) = +∞ ∑ k=-∞ a k(k+1)/2 b k(k-1)/2 ,
in which can be recognized, in a very symmetric guise, our function θ q . Indeed, setting q := ab, one immediately sees that:

f (a, b) = θ q (a).
We shall translate in terms of θ q some identities of Ramanujan (among those for which this is not already done in [START_REF] Bruce | Ramanujan's notebooks. Part III[END_REF]) then draw consequences.

Translation of some identities from [1]

We give no proof of these translations, verifications are mechanical (even if the proof of the identities themselves are not !). Actually, Entry 22 is directly copied from [START_REF] Bruce | Ramanujan's notebooks. Part III[END_REF].

We omitted the important Entries 20 and 27, which express modularity, because we do not take advantage of that property in the present section (it will appear in section 3).

Entry (18).

1. θ q (x) = θ q (q/x) 2. θ q (q) = 2θ q 4 (q) = 2θ q 4 (q 3 ) 3. θ q (-1) = θ q (-q) = 0 4. θ q (x) = q n(n-1)/2 x n θ q (q n x)

The above equalities are elementary; the following is Jacobi's celebrated Triple Product Formula. Recall the Pochhammer symbols

(x; q) ∞ := ∏ n≥0 (1 -xq n ).
Entry (19). θ q (x) = (q; q) ∞ (-x; q) ∞ (-q/x; q) ∞ Entry (22).

1. φ(q) := θ q 2 (q) = ∑ k∈Z q k 2 = (-q; q 2 ) ∞ (q 2 ; q 2 ) ∞ (q; q 2 ) ∞ (-q 2 ; q 2 ) ∞ 2. ψ(q) := θ q 4 (q) = ∞ ∑ k=0 q k(k+1)/2 = (q 2 ; q 2 ) ∞ (q; q 2 ) ∞ 3. f (-q) := θ q 3 (-q) = ∑ k∈Z (-1) k q k(3k-1)/2 = (q; q) ∞ 4. χ(q) := (-q; q 2 ) ∞ (note that this item is purely a definition)

The following one is related to the distribution formula of Pochhammer symbols:

n-1 ∏ k=0 (q k x; q n ) ∞ = (x; q) ∞ .
Entry (28).

n-1

∏ k=0 θ q n (q k x) = (q n ; q n ) n

Consequences for the powers of θ q

One first proves anew identity (5):

Proposition 1. θ 2 q (x) = θ q 2 (q)θ q 2 (x 2 ) + xθ q 2 (1)θ q 2 (qx 2 ). Proof. The most direct path is to take the arithmetical mean of both items of Entry 29, with y = x. One can also take the arithmetical mean of the last two items of Entry 30, then replace φ(q) and ψ(q 2 ) according to the two first items of Entry 22, last apply the first two items of Entry 18.

As already noted, a 2,0 = θ q 2 (q) et a 2,1 = θ q 2 (1) (notations of 1.3).

Corollary 1.

a 4,0 = θ q 2 (q) 2 θ q 4 (q 2 ) + qθ q 2 (1) 2 θ q 4 (1), a 4,1 = 2θ q 2 (1)θ q 2 (q)θ q 4 (q 3 ), a 4,2 = θ q 2 (q) 2 θ q 4 (1) + θ q 2 (1) 2 θ q 4 (q 2 ), a 4,3 = 2θ q 2 (1)θ q 2 (q)θ q 4 (q). Proof. We square (5):

θ 4 q (x) = (θ q 2 (q)) 2 (θ q 2 (x 2 )) 2 + x 2 (θ q 2 (1)) 2 (θ q 2 (qx 2 )) 2 + 2xθ q 2 (1)θ q 2 (q)θ q 2 (x 2 )θ q 2 (qx 2 ).
First term, directly from (5) (replacing q by q 2 and x by x 2 ):

(θ q 2 (x 2 )) 2 = θ q 4 (q 2 )θ q 4 (x 4 ) + x 2 θ q 4 (1)θ q 4 (q 2 x 4 )

Second term, directly from (5) (replacing q by q 2 and x by qx 2 ):

(θ q 2 (qx 2 )) 2 = θ q 4 (q 2 )θ q 4 (q 2 x 4 )+qx 2 θ q 4 (1)θ q 4 (q 4 x 4 ) = θ q 4 (q 2 )θ q 4 (q 2 x 4 )+qx -2 θ q 4 (1)θ q 4 (x 4 )

Third term, appealling to Entry 29 with x ← x 2 and y ← qx 2 (mean of both items):

θ q 2 (x 2 )θ q 2 (qx 2 ) = θ q 4 (q 3
)θ q 4 (qx 4 ) + x 2 θ q 4 (q)θ q 4 (q 3 x 4 ).

Collecting:

θ 4 q (x) = (θ q 2 (q)) 2 θ q 4 (q 2 )θ q 4 (x 4 ) + x 2 θ q 4 (1)θ q 4 (q 2 x 4 ) + x 2 (θ q 2 (1)) 2 θ q 4 (q 2 )θ q 4 (q 2 x 4 ) + qx -2 θ q 4 (1)θ q 4 (x 4 )
+ 2xθ q 2 (1)θ q 2 (q) θ q 4 (q 3 )θ q 4 (qx 4 ) + x 2 θ q 4 (q)θ q 4 (q 3 x 4 ) = θ q 2 (q) 2 θ q 4 (q 2 ) + qθ q 2 (1) 2 θ q 4 (1) θ q 4 (x 4 ) + 2θ q 2 (1)θ q 2 (q)θ q 4 (q 3 )xθ q 4 (qx 4 ) + θ q 2 (q) 2 θ q 4 (1) + θ q 2 (1) 2 θ q 4 (q 2 ) x 2 θ q 4 (q 2 x 4 ) + 2θ q 2 (1)θ q 2 (q)θ q 4 (q)x 3 θ q 4 (q 3 x 4 ).

Note that, since θ q 4 (q 3 ) = θ q 4 (q), we have a 4,1 = a 4,3 as was to be expected by (9). Also note that various transformations are possible; for instance, using Entry 18 one can check that a 4,1 = a 4,3 = 1 2 θ q (1) 3 .

In all cases, these coefficients vanish for no value of q. However, corollary 2 in 3.1 predicts that a 4,0 and a 4,3 vanish for at least one q < 0 (but says nothing about a 4,1 and a 4,3 ).

The process can be (partially) generalized by squaring identity (6). By definition:

θ 2k q (x) = k-1 ∑ i=0 a k,i x i θ q k (q i x k ) 2 = k-1 ∑ i, j=0 a k,i a k, j x i+ j θ q k (q i x k )θ q k (q j x k ) = k-1 ∑ i, j=0
a k,i a k, j x i+ j θ q 2k (q i+ j x 2k )θ q 2k (q j-i+k ) + q i x k θ q 2k (q i+ j+k x 2k )θ q 2k (q j-i ) .

We used again consequence θ q (x)θ q (y) = θ q 2 (xy)θ q 2 (qy/x) + xθ q 2 (qxy)θ q 2 (y/x) of Entry 29. Replacing each term θ q 2k (q m y) such that m ≥ 2k by y -1 θ q 2k (q m-2k y) and collecting carefully, one finds, for l = 0, . . ., 2k -1:

a 2k,l = ∑ i+ j=l a k,i a k, j θ q 2k (q k+ j-i ) + ∑ i+ j+k=l a k,i a k, j q k-j θ q 2k (q j-i ) if l ≤ k -2, ∑ i+ j+k=l a k,i a k, j q i θ q 2k (q j-i ) if l ≥ k.
(There is no second term ∑ if l = k -1.) Taking k = 2 and using the values of a 2,0 , a 2,1 given in 1.3, one does recognize the a 4,i of corollary 1. Iterative application of these formulas would allow us to determine the a k,i when k is a power of 2; but even so, without giving a closed formula. The methods of 2.2 will prove more to be effective.

A calculus of coefficients of powers of theta 2.2.1 Notations and conventions for this subsection

We suppose chosen τ ∈ C such that q = e 2πiτ and Imτ > 0, which in particular gives an inambiguous meaning to non integral powers of q. In this subsection, in order to have simpler formulas, we rather consider the following theta function:

ϑ(x) := ϑ q (x) := ∑ n∈Z q n 2 /2 x n = θ q q 1/2 x .
The series ϑ plainly represents an analytic function all over the punctured complex plane. Let k ∈ N * and n ∈ Z. We write c k,n (q) the coefficient of x n in the Laurent series expansion of ϑ k (x); that is:

(10) c k,n (q) = ∑ (n 1 ,...,n k )∈Z k n 1 +•••+n k =n q (n 2 1 +•••+n 2 k )/2 .
We therefore have:

(11) c k,n = q n/2 γ k,n .
Setting n := 0 in formula (10), we get:

(12) c k,0 (q) = 1 + k(k -1)q + O(q 2 ),
where the constant term corresponds to (0, . . ., 0) and where the coefficient k(k -1) before q comes from the fact that k-uples

(n 1 , . . . , n k ) such that n 2 1 +• • • +n 2 k = 2, n 1 +• • • +n k = 0 are made of (k -2) zeroes
, one +1 and one -1. More generally, after (8) page 5:

(13) c k,n (q) = q n/2 k n + O(q) .
Actually, γ k,n (q) = q -n/2 c k,n (q) represents an analytic function over the unit disk, with value k n at 0 and its Taylor series at 0 has non negative integer coefficients.

Some notations.

1. If f = ∑ n∈Z a n x n ∈ C x, 1
x , one puts:

B f (x) := B q f (x) = ∑ n∈Z a n q n 2 /2 x n ∈ C x, 1
x .

2. If B q f represents a holomorphic function over C * , one puts:

T f (x) = T q f (x) = ∑ n∈Z B q f (q -n )x n ∈ C x, 1 x . 3. For all a ∈ C * and k ∈ N * , one denotes σ a f (x) := f (ax) and ρ k f (x) := f (x k ).
4. For (a 1 , ..., a m ) ∈ C m , one puts:

(a 1 , ..., a m ; q) ∞ := m ∏ n=1 (a n ; q) ∞ .
(Pochhammer symbols (a; q) ∞ were defined at the beginning of 2.1.1.)

Remark 1. We here recall some useful facts:

1. ϑ(1/x) = ϑ(x); ϑ(q n x) = q -n 2 /2 x -n ϑ(x) for n ∈ Z. 2. ϑ(x) = (q, - √ q x, - √ q x ; q) ∞ .
3. ϑ(x) = 0 if, and only if x ∈ -q Z+ 1 2 .

4.

k-1

∏ n=0 ϑ q k (xq n ) = (q k ; q k ) k ∞ (q; q) ∞ ϑ(xq (k-1)/2 ) for k ∈ N * . 5. B q k ϑ q m = ϑ q k+m for k, m ∈ N * . 6. B(x n f ) = q n 2 /2 x n B f for n ∈ Z. 7. σ a B = Bσ a for a ∈ C * . 8. B q ρ k = ρ k B q k 2 for k ∈ N * . Proposition 2. If B q f represents a holomorphic function over C * , one has: ϑ f = B(T f ).
Proof. Immediate calculation.

For f = ϑ, remark 1 (5) says that Bϑ = ϑ q 2 , which implies:

T ϑ(x) = ∑ n∈Z ϑ q 2 (q -n )x n = ∑ n∈Z ϑ q 2 (q n )x n .
With the help of proposition 2, one finds the following expression:

(14)

ϑ 2 (x) = ∑ n∈Z ϑ q 2 (q n )q n 2 /2 x n .
According to remark 1 (3), we know that ϑ q 2 (q n ) = 0 except if q n ∈ -q 2Z+1 , which is impossible since-1 / ∈ q Z . Therefore:

Fact. The series expanding ϑ 2 has no null coefficient.

In order to go on, let us introduce some more general notations. Let k ∈ N * . Replacing n by 1 and then ϑ by ϑ k in the second functional equation of remark 1 (1), we find that

σ q ϑ k (x) = q -k/2 x -k ϑ k (x). Moreover, if T k,n (x) = x n ϑ q k (x k q n ) with n ∈ Z, one has σ q T k,n (x) = q -k/2 x -k T k,n (x)
. This being said, the functions T k,n all satisfy the same linear equation as ϑ k (compare 1.3.1).

A recurrence relation

Proposition 3. For every k ∈ N * , we have:

(15) ϑ k (x) = k-1 ∑ n=0 c k,n (q) x n ϑ q k (x k q n ) .
Moreover, the coefficients c k,n (q) satisfy the following recurrence relation:

(16) c k+1,n (q) = k-1 ∑ n ′ =0 c k,n ′ (q) q (n-n ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)n ′ ) .
Proof. We shall apply proposition 2 to a function of the form

f (x) = ϑ q k (x k q n ), where k ∈ N * et n ∈ {0, ..., k -1}. Write ϑ q k (x k q n ) = σ q n/k ρ k ϑ q k (x)
and use remark 1. We have:

Bϑ q k (x k q n ) = Bσ q n/k ρ k ϑ q k (x) = • • • = ϑ q k 2 +k (x k q n ) .
We deduce that:

(17) ϑ(x) ϑ q k (x k q n ) = B(T f )(x) = ∑ m∈Z ϑ q k 2 +k (q -mk+n ) q m 2 /2 x n .
If m = (k+1)ν+r with ν ∈ Z and r ∈ {-n, 1-n, ..., k-n}, we write q -mk+n = q -(k+1)kν-kr+n and get: ϑ q k 2 +k (q -mk+n ) = q -k(k+1)ν 2 /2 q (-kr+n)ν ϑ q k 2 +k (q -kr+n ) .

With m 2 = (k + 1) 2 ν 2 + 2(k + 1)rν + r 2 , it follows that: ϑ q k 2 +k (q -mk+n ) q m 2 /2 = q (k+1)ν 2 /2 q (r+n)ν q r 2 /2 ϑ q k 2 +k (q -kr+n ) .

Hence expression (17) becomes:

(18) ϑ(x) ϑ q k (x k q n ) = k-n ∑ r=-n
q r 2 /2 ϑ q k 2 +k (q -kr+n ) x r ϑ q k+1 (xq r+n ) .

Using (15), we deduce that (18):

ϑ k+1 (x) = k-n ∑ r=-n q r 2 /2 k-1 ∑ n=0 c k,n (q) ϑ q k 2 +k (q -kr+n ) x r+n ϑ q k+1 (xq r+n ).
This completes the proof of the recurrence relation (16). In this way, we finish the proof of proposition 3.

A symmetry relation

As already noted in 1.3.2, equation (9), γ k,n = γ k,k-n . Therefore, iff 0 < n < k, one has the following symmetry:

(19) c k,n (q) = q n-k/2 c k,k-n (q) .
Let us explain how to use (19) so as to shorten the expression of recurrence (16). First assume that k is odd ≥ 1, and write k = 2k ′ + 1. Grouping the terms in the sum from (16) according to whether

n ′ = 0 or 1 ≤ n ′ ≤ k ′ or n ′ > k ′ , one has: c k+1,n (q) = c k,0 (q) q n 2 /2 ϑ q k 2 +k (q -kn ) + k ′ ∑ n ′ =1 c k,n ′ (q) q (n-n ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)n ′ ) + c k,k ′ +n ′ (q) q (n-k ′ -n ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)(k ′ +n ′ ) ) . (20) From (19), one has: c k,k ′ +n ′ (q) = q n ′ -1/2 c k,k ′ +1-n ′ . Writing k ′ + n ′ = k -1 -k ′ + n ′ , we draw (k + 1)(k ′ + n ′ ) = (k + 1)k + (k + 1)(n ′ -k ′ -1)
, which yields:

ϑ q k 2 +k (q -kn+(k+1)(k ′ +n ′ ) ) = ϑ q k 2 +k (q -kn+(k+1)(n ′ -k ′ -1)+(k+1)k ) = q kn-(k+1)(n ′ -1/2) ϑ q k 2 +k (q -kn+(k+1)(n ′ -k ′ -1) ) . Since kn -(k + 1)(n ′ -1/2) = (2k ′ + 1)(n -n ′ ) + k ′ + 1 -n ′ , one has: q n ′ -1/2 q (n-k ′ -n ′ ) 2 /2 q kn-(k+1)(n ′ -1/2) = q (n-n ′ +k ′ +1) 2 /2 .
As a consequence, if n ′′ = k ′ + 1n ′ , one finds:

c k,k ′ +n ′ (q) q (n-k ′ -n ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)(k ′ +n ′ ) ) = c k,n ′′ q (n+n ′′ ) 2 /2 ϑ q k 2 +k (q -kn-(k+1)n ′′ ) .
When n ′ goes from 1 to k ′ , n ′′ runs along the same values in descending order. Going back to formula (20), one is led to the following remark.

Fact. If k = 2k ′ + 1 and k ′ ∈ N, then: c k+1,n (q) =c k,0 (q) q n 2 /2 ϑ q k 2 +k (q kn ) + k ′ ∑ n ′ =1 c k,n ′ (q) × q (n ′ -n) 2 /2 ϑ q k 2 +k (q k(n ′ -n)+n ′ ) + q (n ′ +n) 2 /2 ϑ q k 2 +k (q k(n ′ +n)+n ′ ) . (21)
In particular, for n = 0, one has:

(22) c k+1,0 (q) = c k,0 (q) ϑ q k 2 +k (1) + 2 k ′ ∑ n=1 c k,n (q) q n 2 /2 ϑ q k 2 +k (q (k+1)n ) .
Consider now the case when k = 2k ′ with k ′ ∈ N * . Using relation ( 16), one can check that c k+1,n (q) = c k,0 (q) q n 2 /2 ϑ q k 2 +k (q -kn ) + c k,k ′ (q) q (n-k ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)k ′ )

+ k ′ -1 ∑ n ′ =1 c k,n ′ (q) q (n-n ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)n ′ ) + c k,k ′ +n ′ (q) q (n-k ′ -n ′ ) 2 /2 ϑ q k 2 +k (q -kn+(k+1)(k ′ +n ′ ) ) . (23)
A similar analysis would entail the following remark.

Fact. If k = 2k ′ et k ′ ∈ N * , then: c k+1,n (q) =c k,0 (q) q n 2 /2 ϑ q k 2 +k (q kn ) + c k,k ′ (q) q (n-k ′ ) 2 /2 ϑ q k 2 +k (q k(k ′ -n)+k ′ ) + k ′ -1 ∑ n ′ =1 c k,n ′ (q) q (n ′ -n) 2 /2 ϑ q k 2 +k (q k(n ′ -n)+n ′ ) + q (n ′ +n) 2 /2 ϑ q k 2 +k (q k(n ′ +n)+n ′ ) . ( 24
)
In particular, for n = 0, one has:

c k+1,0 (q) =c k,0 (q) ϑ q k 2 +k (1) + c k,k ′ (q) q k 2 /8 ϑ q k 2 +k (q (k+1)k/2 ) + 2 k ′ -1 ∑ n=1 c k,n (q) q n 2 /2 ϑ q k 2 +k (q (k+1)n ) .
(25) 2.2.4 About coefficients c k,n (q) for k = 2 or 3 Putting k := 1 in (16), since c 1,0 (q) = 1, we find:

(26) c 2,n (q) = ϑ q 2 (q n ) q n 2 /2 , n ∈ {0, 1} .
Moreover, with k = 2 and µ = -1 in (43), we have:

c 2,n (q) = q n 2 /2 x -n 2 ϑ q 2 (x 2 ) ϑ 2 (xq -n/2 ) + (-1) n ϑ 2 (-xq -n/2 ) .
Let us set x = q (n+1)/2 , and recall that ϑ(-√ q) = 0; one deduces the expression c 2,n (q) = q -n/2 ϑ 2 ( √ q) 2 ϑ q 2 (q n+1 ) . Comparing the latter with (26) for n = 0 gives the following identity:

(27) ϑ 2 ( √ q) = 2 ϑ q 2 (1) ϑ q 2 (q) .
Applying relation 1 (4) for k = 2 and x = 1, one has:

ϑ q 2 (1) ϑ q 2 (q) = (q 2 ; q 2 ) 2 ∞ (q; q) ∞ ϑ( √ q) = (q 2 ; q 2 ) 2 ∞ (q; q) ∞ ϑ( √ q) .
Identity (27) can then be read as follows:

(28) ϑ( √ q) = 2 (q 2 ; q 2 ) 2 ∞ (q; q) ∞ ,
where the factor 2 looks almost unexpected . . .

Let us compute c 3,n (q) for n = 0, 1, 2 with the help of (16): c 3,n (q) = c 2,0 (q) q n 2 /2 ϑ q 6 (q 2n ) + c 2,1 (q) q (n-1) 2 /2 ϑ q 6 (q 2n-3 ) .

With (26), one finds:

(29) c 3,n (q) = q n 2 /2 ϑ q 2 (1) ϑ q 6 (q 2n ) + q (1+(n-1) 2 )/2 ϑ q 2 (q) ϑ q 6 (q 2n-3 ) .

In particular, one notes that: c 3,0 (q) = ϑ q 2 (1) ϑ q 6 (1) + q ϑ q 2 (q) ϑ q 6 (q 3 ) .

If one puts f (q) = ϑ q 2 (1) and g(q) = ϑ q 2 (q), one has:

(30) c 3,0 (q) = f (q) f (q 3 ) + q g(q) g(q 3 ) .

The functions f and g may be written in the following form:

f (q) = 1 + 2 ∑ n≥1 q n 2 , g(q) = 2 + 2 ∑ n≥1 q n(n+1) ,
where the power series all have the unit circle as convergence boundary. If q → -1 -, one sees that | f (q)| < 1 but g(q) → +∞. Considering expression (30), one draws that c 3,0 (q) → -∞ for q → -1 -. Since c 3,0 (0) = 1, one infers:

Fact. c 3,0 (q) admits (at least) a zero over ] -1, 0[. Using Maple, by chosing values of n up to N = 600, one more or less gets the graph of q → c 3,0 (q) over interval ] -0.5, 0.5[. Depending on the look of the graph, one can ask Maple to evaluate c 3,0 (q) for a = -0.163034 and b = -0.163033. Herebelow some of the numerical values.

For N = 400: Remark 2. It seems that c 3,0 (q) vanishes somewhere between a = -0.163034 and b = -0.163033.

Put n = 1 in (29); then:

(31) c 3,1 (q) = q 1/2 ϑ q 2 (1) ϑ q 6 (q 2 ) + ϑ q 2 (q) ϑ q 6 (q) .

We find likewise: c 3,2 (q) = q q ϑ q 2 (1) ϑ q 6 (q 4 ) + ϑ q 2 (q) ϑ q 6 (q) .

Noting that ϑ q 6 (q 4 ) = ϑ q 6 (q -4 ) = q -1 ϑ q 6 (q 2 ), we get:

(32) c 3,2 (q) = q 1/2 c 3,1 (q) .
This is consistent with the symmetry stated in (19).

About coefficients c 4,n (q)

Putting k ′ = 1 and k = 3 in (21), we obtain that: c 4,n (q) =c 3,0 (q) q n 2 /2 ϑ q 12 (q kn ) + c 3,1 (q) × q (n-1) 2 /2 ϑ q 12 (q -3n+4 ) + q (n+1) 2 /2 ϑ q 12 (q 3n+4 ) .

For n = 0, formula (22) implies: c 4,0 (q) = c 3,0 (q) ϑ q 12 (1) + 2 c 3,1 (q) q 1/2 ϑ q 12 (q 4 ) .

In account of (29), one can check that: c 4,0 (q) =ϑ q 2 (1) ϑ q 6 (1) ϑ q 12 (1) + q ϑ q 2 (q) ϑ q 6 (q 3 ) ϑ q 12 (1)

+ 2 ϑ q 2 (1) ϑ q 6 (q 2 ) + ϑ q 2 (q) ϑ q 6 (q) ϑ q 12 (q 4 ) .

(

) 34
It may be noted that both (33) and (34) are much more complicated that the formulas found in Corollary 1.

3 Asymptotics of the powers of theta and vanishing of the coefficients 3.1 Some sufficient conditions for the vanishing of γ k,n (q)

Main statement and reduction to an asymptotic estimation

Recall that, for k ≥ 1, underlined characters denote k-uples: x = (x 1 , . . . , x k ).

We first introduce a linear and a quadratic form on R k :

S(x) := ∑ 1≤i≤k x i , Q(x) := ∑ 1≤i≤ j≤k x i x j = 1 2 ∑ 1≤i≤k x 2 i + S(x) 2 .
The last equality shows that Q is a positive definite form.

Lemma 1. Let k ∈ N * and n ∈ Z. Put, for x ∈ C, |x| < 1:

f k,n (x) := ∑ m∈Z k
x Q(m)-nS(m) .

Then:

(35) ∀q ∈ C \ D(0, 1) , γ k+1,n (q) = q n(n-1)/2 f k,n (q).
The function f k,n is holomorphic over the punctured open disk

• D(0, 1) \ {0} with a pole of order n(n -1)/2 at 0 (meaning that if n ∈ {0, 1} then f k,n is actually holomorphic at 0).

Proof. From formula (2) one draws the following calculation:

γ k+1,n (q) = ∑ m 0 ,...,m k ∈Z m 0 +•••+m k =n q m 2 0 +•••+m 2 k -m 0 -•••-m k 2 = ∑ m 1 ,...,m k ∈Z q (m 1 +•••+m k -n) 2 +m 2 1 +•••+m 2 k -n 2 = ∑ m 1 ,...,m k ∈Z q (m 1 +•••+m k ) 2 -2n(m 1 +•••+m k )+n 2 +m 2 1 +•••+m 2 k -n 2 = q n(n-1)/2 ∑ m 1 ,...,m k ∈Z q (m 1 +•••+m k ) 2 -2n(m 1 +•••+m k )+m 2 1 +•••+m 2 k 2 = q n(n-1)/2 ∑ m∈Z k q Q(m)-nS(m) . Theorem 1. Let k ∈ N * and n ∈ Z. Writing ε := (-1) n , one has, when x → 1 -: (36) f k,n (-x) = π k 2 √ 2 √ k + 1 cos((k + ε)π/4) (1 -x) k 2 + O 1 (1 -x) k-1 2
.

From 3.1.2 on, most of the rest3 of subsection 3.1 is devoted to the proof of theorem 1.

Consequences of theorem 1. We first discuss here consequences of this theorem of interest for our problem (possibility of vanishing of the coefficients γ k,n ). Assume k ≥ 2 and 0 ≤ n ≤ k, so that γ k,n (0) = k n > 0 by formula (8) page 5. On the other hand, the sign of γ k,n (q) as q → -1 while staying within ]-1, 0[ is (after lemma 35) (-1) n(n-1)/2 × cos k ′ π/4, where k ′ := k -1 + (-1) n ; so this sign depends on k (mod 8) ∈ {0, . . . , 7} and on n (mod 4) ∈ {0, . . ., 3} (see our notational conventions at the very end of section 1). More precisely:

(-1) n(n-1)/2 = +1 if n (mod 4) ∈ {0, 1}, -1 if n (mod 4) ∈ {2, 3},
and

sign cos k ′ π/4 =      +1 if k ′ (mod 8) ∈ {7, 0, 1}, 0 if k ′ (mod 8) ∈ {2, 6}, -1 if k ′ (mod 8) ∈ {3, 4, 5}. Now, k ′ = k if n is
even and k -2 if n is odd, so that we list the cases when γ k,n (q) < 0 as q → -1 , we know that in all these cases it must vanish somewhere.

Corollary 2. Assume k ≥ 2 and 0 ≤ n ≤ k. In all the following cases, γ k,n (q) vanishes for at least one value of q ∈ ]-1, 0[:

• n (mod 4) = 0 and k (mod 8) ∈ {3, 4, 5},
• n (mod 4) = 1 and k (mod 8) ∈ {5, 6, 7},

• n (mod 4) = 2 and k (mod 8) ∈ {7, 0, 1},

• n (mod 4) = 3 and k (mod 8) ∈ {1, 2, 3}, All these conditions (i.e. their logical disjunction) can be summarized in a unique one: k -2n (mod 8) ∈ {3, 4, 5}.

Note that the above conditions exclude a priori the case k = 2 (because for n = 0, 1, 2 one does not have k -2n(mod 8) ∈ {3, 4, 5}), so in what follows we assume that k ≥ 3 and write:

Z k := {n ∈ Z | γ k,n (q) vanishes somewhere in ]-1, 0[}.
By the remark at the end of 1.3.1, Z k is k-periodic i.e. n ∈ Z k ⇒ n + kZ ⊂ Z k . In order to use that fact, we introduce some more notations:

X k := {n ∈ {0, . . . , k} | k -2n (mod 8) ∈ {3, 4, 5}}, Y k := X k + kZ.
Note that X k is invariant under the symmetry n ↔ k -n, and therefore so is Y k : we shall not have to exploit further the symmetry expressed by equation (9), it is already built-in. With the previous notations, we thus have:

Corollary 3. Let k ≥ 3. Then Y k ⊂ Z k .
To make this more useful, we describe more precisely X k , and, in some cases, Y k .

Case k even: k = 8a + 2b, b ∈ {0, 1, 2, 3}. Then in condition k -2n(mod 8) ∈ {3, 4, 5} only the remainder 4 is possible:

X k = {n ∈ {0, . . ., k} | k ≡ 2n + 4 (mod 8)} = {n ∈ {0, . . ., k} | n ≡ b + 2 (mod 4)}.
For instance, one easily checks the following cases:

• If k = 8a, then X k = 4 {0, . . ., 2a} + 2 and Y k = 4Z + 2. • If k = 8a + 4, then X k = 4 {0, . . ., 2a} and Y k = 4Z.
Corollary 4. All γ 8a,4c+2 (q) and all γ 8a+4,4c (q) vanish for some q ∈] -1, 0[. Cases k = 8a + 2 and k = 8a + 6 can be similarly described (but less simply).

Case k odd: k = 8a + 2b + 1, b ∈ {0, 1, 2, 3}. Then in condition k -2n(mod 8) ∈ {3, 4, 5} only cases 3 and 5 ≡ -3 are possible:

X k = {n ∈ {0, . . ., k} | k ≡ 2n±3 (mod 8)} = {n ∈ {0, . . ., k} | n ≡ b+2 or b-1 (mod 4)}.
Again these cases can be similarly but less simply described. We just give an example: if k = 3, we find that X 3 = {0, 2} and Y 3 = 3Z ± 1.

Two auxiliary propositions and an application to a generating series

E M := {y ∈ R k | F(y) + L(y) ≤ M} and Λ M := E M ∩ Z k .
Since F is positive definite, E M is a compact ellipsoid and Λ M is a finite set.

The volume of E M is easily computed as follows. Let V k the volume of the unit ball of R k and let D F the discriminant of F. The matrix of F writes t SS, where S is invertible, and D F = (det S) 2 . Therefore F(y) = (Sy) 2 := Sy, Sy . In the same way, L(y) = By for some line matrix B. If y 0 := 1 2 t (BS -1 ), one sees that Sy+y 0 , Sy+y 0 2 = F(y)+L(y)+y 2 0 . Thus, the mapping y → Sy+y 0 transforms E M into a ball with radius M + y 2 0 , the volume of which is V k (M + y 2 0 ) k/2 . On the other hand, this affine linear mapping multiplies all volumes by |det S| = D 1/2 F . One concludes:

µ(E M ) = V k D 1/2 F (M + y 2 0 ) k/2 .
Proposition 4. Let V k the volume of the unit ball of R k and let D F the discriminant of F. Then, when M → +∞:

(37) card Λ M = V k D 1/2 F M k/2 + O M (k-1)/2 . Proof. Let E ′ M := Λ M + [-1/2, 1/2] k , the volume µ(E ′ M ) of which is card Λ M . We shall first determine M ′ , M ′′ > 0 such that E M ′ ⊂ E ′
M ⊂ E M ′′ , which will provide us uper and lower bound for card Λ M = µ(E ′ M ). Let y ∈ Λ M and u ∈ [-1/2, 1/2] k and let us write φ := F + L. Triangle inequalities applied to the norm √ F imply:

F(y) -a ≤ F(y) -F(u) ≤ F(y + u) ≤ F(y) + F(u) ≤ F(y) + a,
where a := 1 2 ∑ 1≤i≤ j≤k a i, j , because F(u) ≤ a 2 for every u ∈ [-1/2, 1/2] k . One deduces:

F(y) -2a F(y) + a 2 ≤ + u) ≤ F(y) + 2a F(y) + a 2 .
In the same way, L(y

) -b ≤ L(y + u) = L(y) + L(u) ≤ L(y) + b, where b := 1 2 ∑ 1≤i≤k |b i |.
One draws:

φ(y) + a 2 -b -2a F(y) ≤ φ(y + u) ≤ φ(y) + a 2 + b + 2a F(y).
On the other hand, there exists a constant c > 0 such that |L| ≤ c √ F (continuity of the linear form L with respect to the topology defined by the norm √ F, all this in finite dimension). One deduces:

φ(y) ≥ F(y) -c F(y) = ( F(y) -c/2) 2 -c 2 /4 =⇒ F(y) ≤ c/2 + φ(y) + c 2 /4.
In the end, we find, for d ′ := a 2bac and d ′′ := a 2 + b + ac, the bounds:

φ(y) -2a φ(y) + c 2 /4 + d ′ ≤ φ(y + u) ≤ φ(y) + 2a φ(y) + c 2 /4 + d ′′ valid for all y ∈ Λ M and u ∈ [-1/2, 1/2] k .
It is then obvious that putting:

M ′ := M -2a M + c 2 /4 + d ′ and M ′′ := M + 2a M + c 2 /4 + d ′′ one has indeed: E M ′ ⊂ E ′ M ⊂ E M ′′ =⇒ µ(E M ′ ) ≤ card Λ M ≤ µ(E M ′′ ),
said otherwise:

V k D 1/2 F (M -2a M + c 2 /4+d ′ +y 2 0 ) k/2 ≤ card Λ M ≤ V k D 1/2 F (M +2a M + c 2 /4+d ′′ +y 2 0 ) k/2 .
Since, for all α, β, γ ∈ R, one has:

(M + α M + β + γ) k 2 = M k 2 1 + O M -1 2 k 2 = M k 2 + O M k-1 2
, the desired estimation follows.

Second auxiliary proposition.

Proposition 5. For any α > 0 and x ∈ C, |x| < 1, let S α (x) := ∑ n≥0 n α x n . Then, when

x → 1 -:

(38) S α (x) = Γ(α + 1) (1 -x) α+1 + O 1 (1 -x) α .
Proof. Write, for 0 < x < 1, a :=ln x > 0 and, for t ≥ 0, h a (t) := t α x t = t α e -at . We thus have:

I := +∞ 0 h a (t) dt = Γ(α + 1) a α+1 •
The function h a grows over 0, α a and decreases (with limit 0 + ) over α a , +∞ ; its max-

imum is h a α a = α ae α .
If α = a/m, m ∈ N * , whence α a = m, one deduces the bounds:

h a (0) + • • • + h a (m -1) ≤ m 0 h a (t) dt ≤ h a (1) + • • • + h a (m) h a (m + 1) + • • • + h a (n) + • • • ≤ +∞ m h a (t) dt ≤ h a (m) + • • • + h a (n) + • • • , whence ∑ n≥0 h a (n) ∈ [I -h a (m), I + h a (m)]. Therefore: S α (x) = Γ(α + 1) (-ln x) α+1 + O 1 (-ln x) α = Γ(α + 1) (1 -x) α+1 + O 1 (1 -x) α
for ln x ∼ x -1 when x → 1. The estimation is therefore right for all a := α/m, but since S α (x) is monotonous as a function of a, it stays valid without restriction when a → 0 + and x → 1 -.

Application to a generating series. Let F and L as defined at the beginning of 3.1.2 and define the generating series:

(39) Proposition 6. For x → 1 -, one has the asymptotic estimation:

Φ F,L (x) := ∑ m∈Z k x F(m)+L(m) = ∑ n∈ Z r F,L (n)x n , where r F,L (n) := card {m ∈ Z k | F(m) + L(m) = n}.
(40) Φ F,L (x) = 1 D 1/2 F π k 2 (1 -x) k 2 + O 1 (1 -x) k-1 2
.

Proof. We abreviate

C := V k D 1/2 F
•. Then:

Φ F,L (x) 1 -x = ∑ n∈Z r F,L (n)x n ∑ n∈Z x n = ∑ n∈Z R F,L (n)x n , or R F,L (n) := ∑ p≤n r F,L (p) = card Λ n
with the same notations as before. From proposition 4, there is a constant C ′ > 0 such that:

∀n > 0 , R F,L (n) -Cn k 2 < C ′ n k-1 2 . Let Ψ(x) := ∑ n>0 R F,L (n)
x n , so that, when x → 1 -:

Φ F,L (x) 1 -x = Ψ(x) + O(1).
We immediately get, for 0 < x < 1:

Ψ(x) -CS k 2 (x) ≤ C ′ S k-1 2 (x).
But, after proposition 5:

S k 2 (x) = Γ( k 2 + 1) (1 -x) k 2 +1 + O 1 (1 -x) k 2 and S k-1 2 (x) = Γ( k-1 2 + 1) (1 -x) k-1 2 +1 + O 1 (1 -x) k-1 2
, whence:

Φ F,L (x) 1 -x = C Γ( k 2 + 1) (1 -x) k 2 +1 + O 1 (1 -x) k+1 2
.

Since we know that

V k = π k 2 Γ( k 2 + 1)
, the desired estimation follows.

End of the proof of theorem 1

Odd and even parts of f k,n . By partitioning Z k into 2 k classes modulo the subgroup (2Z) k , one decomposes the generating series f k,n :

f k,n (x) = ∑ m∈Z k x Q(m)-nS(m) = ∑ ε∈{0,1} k ∑ m∈Z k x Q(2m+ε)-nS(2m+ε) = ∑ ε∈{0,1} k x Q(ε)-nS(ε) ∑ m∈Z k x 4Q(m)+4B(m,ε)-2nS(m) ,
where B denotes the symmetric bilinear form such that B(m, m) = Q(m). Each sum ∑ m∈Z k

x 4Q(m)+4B(m,ε)-2nS(m) is even as a function of m, because of the form of the exponents:

4Q(m) + 4B(m, ε) -2nS(m) = 4 ∑ 1≤i≤ j≤k m i m j + 2 ∑ 1≤i≤ j≤k (m i ε j + m j ε i ) -2n ∑ 1≤i≤k m i .
Therefore:

f k,n (-x) = ∑ ε∈{0,1} k (-1) Q(ε)-nS(ε) x Q(ε)-nS(ε) ∑ m∈Z k x 4Q(m)+4B(m,ε)-2nS(m) = ∑ ε∈{0,1} k (-1) Q(ε)-nS(ε) x Q(ε)-nS(ε) Φ Q,L ε (x 4 ),
where we use again the generating series Φ F,L of (39), here with the quadratic form F := Q and the linear form L := L ε defined by:

L ε (m) := B(m, ε) -(n/2)S(m).
Lemma 2. When x → 1 -, we have:

(41) f (-x) =   ∑ ε∈{0,1} k (-1) Q(ε)-nS(ε)   1 2 k D 1/2 Q π k 2 (1 -x) k 2 + O 1 (1 -x) k-1 2
.

Proof. One calls upon proposition 6 and the fact that 1

(1 -x 4 ) k 2 ∼ 1 2 k (1 -x) k 2 when x → 1.
We are left with the task of determining the value (and the sign !) of the factor

∑ ε∈{0,1} k (-1) Q(ε)-nS(ε) .
Calculation of the factor ∑ ε∈{0,1} k (-1) Q(ε)-nS(ε) and conclusion. Let

A := card{ε ∈ {0, 1} k | Q(ε) ≡ nS(ε) (mod 2)} and B := card{ε ∈ {0, 1} k | Q(ε) ≡ nS(ε) (mod 2)}.
One thus has:

A -B = ∑ ε∈{0,1} k (-1) Q(ε)-nS(ε) and A + B = 2 k .
If l among the k components ε j of ε have value 1 (so that kl have value 0), then Q(ε) = l(l + 1)/2 and S(ε) = l, whence:

A = ∑ 0≤l≤k l(l+1) 2 -nl even k l and B = ∑ 0≤l≤k l(l+1) 2 -nl odd k l
We thus must discuss the parity of l(l + 1)/2nl. If n is even, it is the same as that of l(l + 1)/2, which is even if, and only if, the remainder of l modulo 4 is 0 or 3; if n is odd, it is the same as that of l(l -1)/2, which is even if, and only if the remainder of l modulo 4 is 0 or 1. We are led to introduce:

∀ j = 0, 1, 2, 3 , s j := ∑ 0≤l≤k l≡ j (mod 4) k l .
We then have:

(A, B) = (s 0 + s 3 , s 1 + s 2 ) if n is even, (s 0 + s 1 , s 2 + s 3 ) if n is odd.
The integers s j can be deduced from formulas (given here for k > 0):

s 0 + s 1 + s 2 + s 3 = (1 + 1) k = 2 k , s 0 + is 1 -s 2 -is 3 = (1 + i) k = 2 k/2 (cos kπ/4 + i sin kπ/4), s 0 -s 1 + s 2 -s 3 = (1 -1) k = 0, s 0 -is 1 -s 2 + is 3 = (1 -i) k = 2 k/2 (cos kπ/4 -i sin kπ/4).
One draws:

s 0 = 1 4 2 k + 2 k 2 +1 cos kπ/4 , s 1 = 1 4 2 k + 2 k 2 +1 sin kπ/4 , s 2 = 1 4 2 k -2 k 2 +1 cos kπ/4 , s 3 = 1 4 2 k -2 k 2 +1 sin kπ/4 , 25
whence, if n is even:

A -B = s 0 -s 1 -s 2 + s 3 = 1 4 2 k 2 +1 (2 cos kπ/4 -2 sinkπ/4) = 2 k+1 2 cos(k + 1)π/4,
and, if n is odd:

A -B = s 0 + s 1 -s 2 -s 3 = 1 4 2 k 2 +1 (2 cos kπ/4 + 2 sinkπ/4) = 2 k+1 2 cos(k -1)π/4.
Combining these values with lemma 2, one finds:

f k,n (-x) = π k 2 2 k-1 2 D 1 2 Q cos((k + ε)π/4) (1 -x) k 2 + O 1 (1 -x) k-1 2
.

This formula can be made more precise by noting that D Q = (k +1)2 -k . (Quick argument: a ∈ C being fixed, let D n (x) the determinant of the n × n matrix having x on the diagonal and a elsewhere. One checks that D ′ n (x) = nD n-1 (x) and D n (a) = 0, from which one deduces by induction that D n (x) = (xa) n-1 (x + (n -1)a). One then sets x := 1 and a := 1/2.) Replacing D Q by (k + 1)2 -k in the above formula:

f k,n (-x) = π k 2 √ 2 √ k + 1 cos((k + ε)π/4) (1 -x) k 2 + O 1 (1 -x) k-1 2
. This is indeed formula (36) of theorem 1.

Complement: a first attempt at exploiting modularity

The generating series of the numbers of representations by positive definite quadratic form are known to have modular properties, it has been, since Jacobi, one of the clues of the appearance of theta functions in number theory. We shall here try (and only partially succeed) to exploit this fact to improve the above result.

Use of Poisson's summation formula. We use Poisson's formula in the form quoted in [START_REF] Serre | A course in arithmetic. Translation of "Cours d'arithmetique[END_REF]§VII.6]. Let f a function over R k , of Schwartz class, meaning that f is C ∞ and all its partial derivatives D f at all orders have fast decay, i.e. they are such that ∀N , |D f (x)| = O |x| -N when |x| → +∞. We write µ the Lebesgue measure and x.y the canonical inner product on R k . The Fourier transform f , defined as:

f (y) := R k e -2iπx.y f (x) dµ(x)
also is of Schwartz class; and Poisson's summation formula can be stated as follows:

∑ m∈Z k f (m) = ∑ m∈Z k f (m).
Now let F(x) := ∑ a i, j x i x j a positive definite quadratic form and L(x) := ∑ b i x i a linear form over R k . The function:

φ F,L (x) := e -(F(x)+L(x))
is of Schwartz class. We shall compute its Fourier transform. We write X ,Y,U,V, B, . . . the column vectors associated with x, y, u, v, b . . . and A := (a i, j ) the (real symmetric) matrix of F, so that F(x) = t X AX . Likewise L(x) = t BX , where B is the column vector associated with b := (b 1 , . . . , b k ). Let S a real square invertible matrix such that A = t SS (such S is known to exist) and let T := S -1 . Then:

φ F,L (y) = R k e -2iπx.y-(F(x)+L(x)) dµ(x) = R k e -t XAX-t BX-2iπ t Y X dµ(x) = 1 |det S| R k e -t UU-t BTU-2iπ t Y TU dµ(u) by change of variable X = SU = 1 |det S| R k e -u 2 -vu dµ(u) where one put t V := t BT + 2iπ t Y T = e v 2 /4 |det S| R k e -(u+v/2) 2 dµ(u) = e v 2 /4 |det S| R k e -u 2 dµ(u) = π k/2 e v 2 /4 |det S| • (Because R k e -u 2 dµ(u)
is the gaussian integral over R k .) At last, we compute:

v 2 = ( t BT + 2iπ t Y T )( t T B + 2iπ t TY ) = ( t B + 2iπ t Y )A -1 (B + 2iπY ) = F(b + 2iπy),
where was introduced the quadratic form F with matrix A -1 (this form also is positive definite). And we eventually reaches the result:

φ F,L (y) = π k/2 e F(b/2) √ det A e -π 2 ( F(y)+ L(y)) .
Poisson's summation formula then yields:

(42)

∑ m∈Z k e -(F(m)+L(m)) = π k/2 e F(b/2) √ det A ∑ m∈Z k e -π 2 ( F(m)+ L(m)) .
Application to a generating series. Recall that f k,n (x) = ∑ m∈Z k

x Q(m)-nS(m) and that we are looking for conditions ensuring the vanishing of this function. We have:

f k,n (x) = ∑ m∈Z k φ F,L (m)
, where F := tQ, L := -tnS, with x = e -t .

Note that |x| < 1 ⇔ ℜ(t) > 0. In order to apply formula 42, we thus take:

A := t      1 1/2 . . . 1
     1 1 . . . 1      .
One easily checks that:

A -1 = 2 (k + 1)t      k -1/2 . . . -1 -1 k . . . -1 . . . . . . . . . . . . -1 -1 . . . k      ,
whence:

F(y) = 2 (k + 1)t Q(y), where Q(y) := k ∑ 1≤i≤k y 2 i -2 ∑
1≤i< j≤k y i y j , and L(y) = -2n (k + 1)iπ S(y).

Setting y := e -2π 2 (k+1)t , we then see that f k,n (x) equals, up to a non zero constant factor:

g(y) := ∑ m∈Z k e riπS(m) y Q(m) ,
where one put r := -2n k + 1 ∈ Q. The function g(y) is well defined for |y| < 1, which is consistent with the above definition of y and the fact that |x| < 1 ⇔ ℜ(t) > 0 ⇔ ℜ(1/t) > 0. Note that, since Q is positive definite, g(0) = 1. But the presence of a "phase" riS(m) on the one hand, the arithmetical properties of Q on the other hand, complicate slightly the use of real analysis here.

We follow, as far as possible, the same method as before:

g(y) = ∑ ε∈{0,1} k ∑ m∈Z k e riπS(2m+ε) y Q(2m+ε) = ∑ ε∈{0,1} k e riπS(ε) y Q(ε) ∑ m∈Z k e 2riπS(m) (y 4 ) Q(m)+ B(m,ε) , where B(u, v) := k ∑ u i v i -2 ∑ i< j
(u i v j +u j v i ) denotes the symmetric bilinear form such that B(m, m) = Q(m). One also finds out that, if ε has l coefficients with value 1, then S(ε) = l and Q(ε) = kll(l -1).

If r is an integer. In that case, e 2riπS(m) = 1, which somehow gets rid of phase problems. One want to make y 4 → 1 -but in such a way that g(y) stays real. We will take y = i p z, 0 ≤ z < 1, p ∈ {0, 1, 2, 3} being chosen such that pk be even: pk = 2s. Then:

y Q(ε) = i pkl-pl(l-1) z Q(ε) = (-1) sl-pl(l-1)/2 z Q(ε) ,
and all terms are indeed real. Moreover, we have the following asymptotic estimate when z → 1 -:

g(y) = k ∑ l=0 k l (-1) rl+sl-pl(l-1)/2 C (1 -z) k/2 + O 1 (1 -z) (k-1)/2 ,
where C > 0 is computed as before using the discriminant of Q.

If the left factor of the main term is strictly negative, the limit of g(y) is -∞ and one deduces again the existence of a zero of g(y). To compute said factor, one notes that l(l -1)/2 is even if, and only if l ≡ 0 or l ≡ 1 (mod 4). One then has, with the same notations as before:

k ∑ l=0 k l
(-1) rl+sl-pl(l-1)/2 = s 0 + (-1) s+r s 1 + (-1) p (s 2 + (-1) s+r s 3 ) = (s 0 + (-1) p s 2 ) + (-1) s+r (s 1 + (-1) p s 3 ).

If p is even, since s 0 + s 2 = s 1 + s 3 = 2 k-1 , one finds 0 or 2 k and therefore certainly not a strictly negative number. So we suppose that p is odd, i.e. p = 1 or p = 3. Either case amounts to the same, so we take p = 1 and therefore k = pk = 2s. Since r = -2n/(k + 1) has been assumed integral and since the denominator is odd, we see that n is a multiple of k + 1 and that r is even.

Let us resume the calculation:

k ∑ l=0 k l (-1
) rl+sl-pl(l-1)/2 = (s 0s 2 ) + (-1) s+r (s 1s 3 ) = 2 k/2 (cos kπ/4 + (-1) s+r sin kπ/4) = 2 s (cos sπ/2 + (-1) s+r sin sπ/2).

This number is strictly negative if, and only if, s ≡ 1 or 2 (mod 4). So we can conclude this study:

Proposition 7. If k ≡ 2 or 4 (mod 8) and if k + 1|n, then γ k,n (q) vanishes for at least one value of q.

It is not easy here to characterize that value: one would require that the argument of

y := e -2π 2
(k+1)t be π/2, whence a condition on t which we shall not try to write down.

If r ′ := r -1/2 is an integer. In that case, one checks that k is necessarily odd: k = 2s -1. An analysis similar to the previous one leads us to take y = iz, whence again y 4 = z 4 . With the same notations, one then gets:

e riπS(ε) y Q(ε) = (-1) r ′ +sl-l(l-1)/2 z l(2s-l)

and

∑ m∈Z k e 2riπS(m) (y 4 ) Q(m)+ B(m,ε) = ∑ m∈Z k (-1) S(m) (z 4 ) Q(m)+ B(m,ε) .
It is the latter sum that causes problems: one can indeed check that the terms with S(m) even, resp. odd, mutually compensate in such a way that the main term of the asymptotic estimation disappears. Our method does not allow to conclude in this case.

3.2 Another approach to asymptotics when q → -1

Here again and until the end of the paper, we shall use the function ϑ and the coefficients c k,n (q) introduced in 2.2. We begin with a kind of "distribution formula".

Recall that c 1,0 (q) = 1. From now on, we shall assume that k ≥ 2; also, from equation (15):

ϑ k (x) = k-1 ∑ n=0 c k,n (q) x n ϑ q k (x k q n ) .
Last, recall the notation ∑ j (mod k) f ( j) which was introduced at the very end of section 1.

Proposition 8. If µ is a primitive k th root of unity, one has:

(43) c k,n (q) = q n 2 /k x -n k ϑ q k (x k ) ∑ j (mod k) µ -n j ϑ k (xq -n/k µ j ) .
In particular:

(44) c k,n (q) = q n 2 /k k ϑ q k (1) ∑ j (mod k)
e 2n jπi/k ϑ k (q n/k e 2 jπi/k ) .

Proof. For r ∈ {0, . . ., k -1} and j ∈ Z, divide by x r both members of (15) and then replace x by xµ j ; it follows that:

x -r µ -r j ϑ k (xµ j ) = k-1 ∑ n=0 c k,n (q) x n-r µ (n-r) j ϑ q k (x k q n ) .
With r = n and index j running along a complete system of integers modulo k, we find that coefficients c k,n (q) satisfy the following identities:

x -n ∑ j (mod k) µ -n j ϑ k (xµ j ) = k c k,n (q) ϑ q k (x k q n ) .
Substituting xq -n/k for x above, we deduce (43). Putting x = 1 and µ = e -2πi/k in (43), we immediately obtain (44).

Note that, since ϑ(q -n/k µ j ) = ϑ(q n/k µ -j ) = q 1/2-n/k µ j ϑ(q (n-k)/k µ -j ), one can put (44) under the following form:

c k,n (q) = q ((k-n) 2 +n 2 )/(2k) k ϑ q k (1) k-1 ∑ j=0 µ (k-n) j ϑ k (q (n-k)/k µ -j ) .
Comparing it with formula (44) where one has replaced n by kn, one recovers the symmetry property of c k,n already encountered.

Expression as a finite sum of theta quotients

We shall now consider, when q → -1, the asymptotic behaviour of c k,n (q). We shall for that use the modular transformation of ϑ. So we borrow from [2, p. 166, (76.1)] the notation ϑ 3 (v|τ):

ϑ 3 (v|τ) := ∑ n∈Z e πi(n 2
τ+2nv) , whence ϑ q (x) = ϑ 3 (v|τ) for x = e 2πiv and q = e 2πiτ .

Let

M := -1 0 2 -1 , τ ′ := Mτ = -τ 2τ -1 and q ′ := e 2πiτ ′ . If a = -1, b = 0, c = 2 and d = -1, note that ϑ 1-a-c,1-b-d = ϑ 0,2 = ϑ 3 , where ϑ µ,ν is defined in [2, p. 181, (81.2)].
From that, the modular formula given for ϑ 3 in [2, p. 182, Theorem] implies that:

ϑ q (x) = i ε 2 e -2πivv ′ i 2τ -1 ϑ q ′ (x ′ ) ,
where

ε 2 := e πi/4 2 -1 , x := e 2πiv , v ′ = v 2τ -1 and x ′ = e 2πiv ′ . Since 2 -1 = 2 1 =
1, one deduces that:

(45) ϑ q (x) = e 2πi(1/8-vv ′ ) i 2τ -1 ϑ q ′ (x ′ ) .
Recall that γ k,n (q) = c k,n (q) q -n/2 . Proposition 9. The following formula holds:

(46) γ k,n (q) = e πi(k-2n)/4 (qe -πi ) n(n-k)/(2k) k ϑ q k (1) i 2τ -1 k/2 S k,n (q) , where (47) S k,n (q) := ∑ j (mod k) q ′ e πi (2 j+n) 2 /(2k) ϑ k q ′ q ′(2 j+n)/k e 2πi j/k . 2. k = 4k ′ + 2, n = 2n ′ + 1; 3. k = 4k ′ + 1, n ∈ {2n ′ , 2n ′ + 1}; 4. k = 4k ′ -1, n ∈ {2n ′ , 2n ′ -1}.
Proof of theorem 2 comes through a series of lemmas and remarks. We now give a direct consequence on zeroes of γ k,n (q). Assume that k ≥ 3 and n lies between 0 and k -1. Applying theorem 2, we find that γ k,n (q) → -∞ for q → -1 only in the following cases: (i) k ≡ 0 or 4 (mod 8) and n ≡ 2 or 0 (mod 4) respectively; (ii) k ≡ 2 or 6 (mod 8) and n ≡ 3 or 1 (mod 4) respectively; (iii) k ≡ 1 or 5 (mod 8) and n ≡ 2, 3 or 0, 1 (mod 4) respectively; (iv) k ≡ 3 or 7 (mod 8) and n ≡ 0, 3 or 2, 1 (mod 4) respectively.

In all these cases γ k,n (q) admits a zero somewhere over ]-1, 0[. From that point on, we easily recover corollary 2 and its consequences.

Application of modular forms to ϑ q k (1)

Suppose that τ → 1 2 and q → -1. More precisely, write τ = 1 2 + it with t > 0; then:

(49) q = e πi-2πt , τ ′ = - 1 2 + i 4t , q ′ = e -πi-π/2t .
It follows that q ′ will tend to 0 exponentally fast as t → 0 + . For the sake of simplicity shall denote ∅ any function f defined over ]0, +∞[ and such that f (t) = O(e -κ/t ) for t → 0 for some κ > 0. This being said, taking x := 1 in (45), one gets:

(50) ϑ q (1) = e πi/4 √ 2t (1 + ∅) .

This will be generalized as follows.

Lemma 3. Let k ∈ N * . If q = e πi-2πt with t → 0 + , we have:

(51) ϑ q k (1) =                                    1 √ kt (1 + ∅) k ≡ 0 (mod 4); e πi/4 √ 2kt (1 + ∅) k ≡ 1 (mod 4); 2 √ kt e -1/(4kt) (1 + ∅) k ≡ 2 (mod 4); e -πi/4 √ 2kt (1 + ∅) k ≡ 3 (mod 4).
Proof. We first write ϑ q k (1) = ϑ 3 (0|kτ) = ϑ 3 (0| k 2 + kti). Taking in account formula [2, p.181, (81.12)], we obtain the following expressions: Last, if k ≡ 3 (mod 4), we shall use the last relation of (52). Applying the last modular formula from [START_REF] Rademacher | Topics in analytic number theory[END_REF]p. 182,Theorem] to the quadruple (a, b, c, d) = (-1, 0, 2, -1), we see that, for τ ′ = -τ 2τ -1 : So let us make up a "fundamental" system with k elements as follows:

ϑ 4 (v ′ | τ ′ ) = i 2 ε 2 2τ -1 i e 2πivv ′
Λ k,n ≡ 2 j + n k (mod 2) : j ∈ Z .

More precisely, we will choose Λ k,n in the following way:

(58) When α runs in Λ k,n , the terms s α defined in (55) give rise to the following identity, coming from (47):

Λ k,n =                                    -1 + 2
(59) S k,n (q) = ∑ α∈Λ k,n s k α (q) .

Successively taking α := 0 and 1 in expression (57), one finds that:

s 0 (q) = 1 + 2 ∞ ∑ m=1
cos(mnπ/k) e -πim 2 /2-πm 2 /(4t) and:

s 1 (q) = e πi(n/k-1/2) 1 + 2 ∞ ∑ m=1 (-1) m cos(mnπ/k) e -πim 2 /2-πm 2 /(4t) .

One deduces that:

(60) s k 0 (q) = 1 -2ki cos(nπ/k) e -π/(4t) + O(e -π/(2t) ) and (61) s k 1 (q) = e πi(n-k/2) 1 + 2ki cos(nπ/k) e -π/(4t) + O(e -π/(2t) ) .

Let α ∈]0, 1[ and put α ′ := min((α -2) 2 , (α + 1) 2 ) > 1; we shall assume that k ≥ 3. Using expression (57), we obtain that s α (q) = e -πα 2 /(4t)i e πin/k-π(1-α) 2 /(4t) + O(e -πα ′ /(4t) ) Theorem 3. Let k ≥ 3 and 0 ≤ n < k and let Γ k,n the function defined in (53). We have the following asymptotic estimations for γ k,n (q) when q = e πi-2πt and t → 0 + : 1. If k is even and k -2n ≡ 2 (mod 4), then γ k,n (q) = ∅.

2. If k = 4k ′ and n = 2n ′ or if k = 4k ′ + 2 and n = 2n ′ + 1, where k ′ , n ′ ∈ Z, then:

(69) γ k,n (q) = (-1) k ′ -n ′ 2 Γ k,n (t) (1 + ∅) .

3. If k = 4k ′ + 1 and n ∈ {2n ′ , 2n ′ + 1} or if k = 4k ′ -1 and n ∈ {2n ′ , 2n ′ -1}, where k ′ , n ′ ∈ Z, then:

(70) γ k,n (q) = (-1) k ′ -n ′ √ 2 Γ k,n (t) (1 + ∅) .

Proof. Suppose first that k ≡ 0 (mod 4) and n ≡ 1 (mod 2). As k ≥ 4, relation (66) shows that S k,n (q) = O(e -π min(k-2,4)/(4kt) ) = O(e -π/(2kt) ); one deduces with the help of first and third relations of (54), that γ k,n (q) = O(Γ k,n (q) e -π/(4kt) ) = ∅. Moreover, if k ≡ 2 (mod 4) and n ≡ 0 (mod 2), relation (65) combined with the third formula of (54) implies that γ k,n = O(Γ k,n (q) e -π/(4kt) ) = ∅. Thus the first statement of the theorem is obtained.

If k = 4k ′ and n = 2n ′ with k ′ , n ′ ∈ Z, one simultaneously considers the first relation of (54) and formula (65) and one finds that

γ k,n (q) = 2Γ k,n cos(n ′ -k ′ )π + ∅,
which is equivalent to (69). The case of k = 4k ′ + 2 and n = 2n ′ + 1 may be treated in a similar way, using the third relation of (54) and formula (66).

In the case k = 4k ′ + 1 and n ∈ {2n ′ , 2n ′ + 1} one concludes directly by combining the second relation of (54) with (67) or (68).

As for the remaining cases, if k = 4k ′ -1 and n = 2n ′ , relation (67) and the last formula of (54) imply (70). Last, if k = 4k ′ -1 and n = 2n ′ -1, relation (68) and the last formula of (54) also imply (70).

Proof of theorem 2. It follows directly from theorem 3.

c 3 ,

 3 0 (a) = -2.96590 * 10 -6 , c 3,0 (b) = 3.41022854 * 10 -6 . For N = 500: c 3,0 (a) = -2.96589725 * 10 -6 , c 3,0 (b) = 3.41023 * 10 -6 . For N = 600: c 3,0 (a) = -2.96589725 * 10 -6 , c 3,0 (b) = 3.41022854 * 10 -6 .

First

  auxiliary proposition. Let L(y) := ∑ 1≤i≤k b i y i a linear form and F(y) := ∑ 1≤i≤ j≤k a i, j y i y j a positive definite quadratic form on R k . Write, for M > 0:

  The series Φ F,L (x) only has a finite number of terms with negative exponent and it converges normally over every compact subset of the punctured open disk • D(0, 1) \ {0}.

  One has: F(b + 2iπy) = F(b) -4π 2 F(y) + L(y) , where L(y)

2 + 2 +

 22 kti) k ≡ 1 (mod 4); ϑ 4 (0 | kti) k ≡ 2 (mod 4);ϑ 4 (0 | 1 kti) k ≡ 3 (mod 4).If k ≡ 0 (mod 4), looking at the first relation of (52) and the second modular relation from [2, p. 177, (79.9)] implies thatϑ 3 (0 | kti) 2 /(kt) ,whence the first asymptotic estimation stated in (51).If k ≡ 1 (mod 4), one obtains the corresponding asymptotic estimate (51) from replacing t by kt in (50).If k ≡ 2 (mod 4), consider the third relation in (52) and the first modular relation in [2, p. 177, (79.9)]: this implies thatϑ 4 (0 | kti) (n+1/2) 2 /(kt) ,where ϑ 2 is defined in[2, p.166, (76.1)]. One deduces:ϑ 4 (0 | kti) = 1 √ kt 2e -π/(4kt) + 2e -9π/(4kt) + • • • = 2 √ kt e -π/(4kt) (1 + ∅) ,which yields the third formula in (51).

ϑ 2 , 1

 21 (v | τ) ,where, as before, ε 2 = e πi/4 and v, v ′ are the same as in (45). Modifying the first relation from [2, p. 181, (81.31)] into the formϑ µ+2,ν (v | τ) = (-1) ν ϑ µ,ν (v | τ), we find that ϑ 2,1 (v | τ) = -ϑ 0,1 (v | τ) = -ϑ 4 (v | τ). Putting v = 0 and τ =

  k , . . . , 0, . . ., 1 (k, n) ≡ (0, 0) (mod 2);

	-1 +	1 k	, . . . ,	1 k	, . . . , 1 -	1 k	(k, n) ≡ (0, 1) (mod 2);
	-1 +	1 k	, . . . , 0, . . ., 1 -	1 k	(k, n) ≡ (1, 0) (mod 2);
	-1 +	2 k	, . . . ,	1 k	, . . . , 1

(k, n) ≡ (1, 1) (mod 2).

This definition requires an unambiguous determination of q 1/2 , the necessary conventions are explained in 2.2.1.

At any rate, it seems difficult to find in[START_REF] Bruce | Ramanujan's notebooks. Part III[END_REF] equalities involving products of functions θ q with different bases q, which explains in part the limited range of the following consequences with respect to our goals.

Note however that in 3.1.4 we apply modular properties of generating series of number of representations by quadratic forms to obtain some complementary informations (proposition 7).

Acknowledgements

The first author is indebted to Professor Berndt for suggesting to look at Ramanujan's Notebooks and positive comments on first attempts.

The second author was supported by Labex CEMPI (Centre Européen pour les Mathématiques, la Physique et leurs Interaction).

Some more notations

We already introduced N, N * , Z at the beginning of 1.3.1. For every k ∈ N * and a ∈ Z, we write a (mod k) either for the remainder of the euclidean division of a by k (thus a (mod k) ∈ {0, . . ., k -1}) or for the congruence class of a modulo k (thus a (mod k) ∈ Z/kZ). The context should hopefully lift any ambiguity.

Proof. Note that τ = τ ′ 2τ ′ + 1 and (2τ -1)(2τ ′ + 1) = -1. For x := q n/k e 2 jπi/k , write v := nτ + j k so that:

Applying (45) to x = q n/k e 2 jπi/k and considering relation ϑ q ′ (x ′ ) = ϑ q ′ (1/x ′ ), one gets:

(48)

where C j := e 2πi(k/8-(n 2 τ-2 j 2 )/(2k)) . Since

the relations (44) and (48) imply:

q ′ e πi (2 j+n) 2 /(2k) ϑ k q ′ q ′(2 j+n)/k e 2πi j/k , which is plainly equivalent to (46).

Sufficient conditions for vanishing at real negative values

The main goal of all the rest of this section is the proof of the following result.

Theorem 2. Let k an integer ≥ 3 and n an integer lying between 0 and k -1, and suppose that q tends to -1 within the real interval ] -1, 0[.

• One has lim

q→-1 γ k,n (q) = 0 if, and only if, k is even and k -2n ≡ 2 (mod 4).

• Otherwise, lim

) can be expressed by means of a pair of integers (k ′ , n ′ ) in one of the following manners:

Expanding ϑ 4 (0 | τ) with the help of [2, p.166, (76.1)], we have:

Substituting kt for t in the last relation, we eventually obtain the complete asymptotic estimation (51).

Combining (51) with (46), we deduce the following remark.

If q = e πi-2πt with t → 0 + , one has:

Asymptotic study of theta functions with modular variables

We now are left to study the asymptotic behaviour of S k,n (q) for q → -1. In order to do so, let us consider expression (47), and put:

(55) s α (q) = s k,n,α (q) = q ′ e πi α 2 /2 ϑ q ′ (q ′ e πi ) α e -πin/k for every α ∈ R. Since ϑ q (x) = 0 for x ∈ -q 1 2 +Z , we draw the following remark:

Remark 4. If n/k = 1/2 and α = -1/2, one has:

Expanding ϑ q ′ in series with q ′ = e -πi-π/(2t) (see ( 49)), we have: (57) s α (q) = ∑ m∈Z e -πim(m/2+n/k)-π(α+m) 2 /(4t) , which implies that s α+2 (q) = s α (q). and s -α (q) = e -πα 2 /(4t)i e -πin/k-π(1-α) 2 /(4t) + O(e -πα ′ /(4t) ) .

In particular, if α = 1/k, then α ′ = (k +1) 2 /k 2 ; taking in account relations (1-α) 2 -α 2 = (k -2)/k and α ′ -α 2 = (k + 2)/k entails: s 1/k (q) = e -π/(4k 2 t) 1i e πin/k-π(k-2)/(4kt) + O(e -π(k+2)/(4kt) ) and s -1/k (q) = e -π/(4k 2 t) 1i e -πin/k-π(k-2)/(4kt) + O(e -π(k+2)/(4kt) ) .

If k * = min(2k -4, k + 2), one gets that: (62)

In a similar way, one will find that:

Proposition 10. Let k ≥ 3 and 0 ≤ n < k. The following asymptotic estimates hold S k,n (q) whenever q = e πi-2πt and t → 0 + :

(68) S k,n (q) = i k + 2 e -π/(4kt) + O(e -π min(k-1,4)/(4kt) ) .

Proof. We shall use the expression of S k,n (q) given in (59) with the various forms of Λ k,n shown in (58).

(1) Suppose that (k, n) ≡ (0, 0) (mod 2); one has k ≥ 4. Because of the first relation of (58), one can write (59) in the following form:

Consider ( 60), (61) together with (64), and note that 1 2 ≥ 2 k ; then: S k,n (q) = 1 + e πi(n-k/2) + O(e -π/(kt) ) , which takes us to relation (65).

(2) Suppose that (k, n) ≡ (0, 1) (mod 2); again, k ≥ 4. Thanks to the second relation of (58), expression (59) can be put in the following form:

Since (-1) k+n i k = -i k , one obtains directly (66), taking in account relations (62), (63) together with (64).

(3) Suppose that (k, n) ≡ (1, 0) (mod 2). By virtue of the third relation of (58), one transforms (59) into:

Relation (67) flows immediately from the estimations contained in (60), (63) combined with (64). (4) Last, suppose that (k, n) ≡ (1, 1) (mod 2). By virtue of the last relation of (58), one expresses (59) in the following form:

Meanwhile, also note that e πi(n-k/2) = (-1) n-k i k = i k . By way of consequence, relation (68) directly follows from formulas (61), (62) with (64).

End of the proof of theorem 2

Considering remark 3 together with proposition 10, we obtain the following result.


Hua Chen
email: chenhua@whu.edu.cn

Zhuangchu Luo

Changgui Zhang
email: zhang@math.univ-lille1.fr

On the summability of divergent power series satisfying singular PDEs

.

Résumé

.

Version française abrégée L'équation différentielle x 2 y -y + x = 0 admet pour solution formelle la série dite d'Euler Ê(x) = .

n≥0 n!x n+1 . Borel-sommable dans toutes les directions sauf l'axe réel positif [0, +∞), celle-ci jour un rôle de première importance dans la théorie analytique des équations différentielles à singularités irrégulières [START_REF] Balser | Formal power series and linear systems of meromorphic ordinary differential equations[END_REF][START_REF] Ramis | Gevrey asymptotics and applications to holomorphic ordinary differential equations[END_REF]. L'équation aux dérivée partielles (2) ci-dessous, variante bidimensionnelle de la précédente équation, est satisfaite terme à terme par la série entière F (x 1 , x 2 ) = n1,n2≥0 (n 1 + n 2 )! x n1+1 1 x n2+1 2 , qui donne lieu à la série d'Euler en considérant Fx1 (0, x 2 ) ou Fx2 (x 1 , 0). Voir aussi [START_REF] Costin | Existence and uniqueness for a class of nonlinear higher-order partial differential equations in the complex plane[END_REF] pour une expression de F au moyen du taux d'accroisements de Ê évalué en deux points distincts. L'étude de l'équation (2) avec sa solutoin formelle F sera développée dans §3. On verra, en effet, que cette dernière est Borel-sommable au sens de Definition 2.1 dans toutes les directions d'argument (d 1 , d 2 ) ∈ [0, 2π] 2 tel que |d 1 -d 2 | < π. Les multiplicateurs de Stokes ainsi associés seront décrits dans Theorem 3.1. On pourrait s'étonner du fait que les directions singulières (d 1 , d 2 ) ∈ [0, 2π] 2 , avec |d 1 -d 2 | ≥ π, occupent un quart du tore R 2 /(2πZ) 2 des directions issues de l'origine du plan C 2 . Ce phénomène est pourtant inévitable, vu l'expression de la somme de Borel de F fournie dans Theorem 3.2.

Par ailleurs, l'équation (2) fait partie d'une famille d'équations linéaires comme indiquée dans [START_REF] Luo | Exponential-type Nagumo norms and summability of formal solutions of singular partial differential equations[END_REF], lesquelles pourront ensuite être rendues non linéaires sous la forme générale [START_REF] Luo | On a family of symmetric hypergeometric functions of several variables and their Euler type integral representation[END_REF] ; voir §4.1-2. Nous nous interesserons enfin, dans §4.3, à l'étude du problème de Cauchy pour (1), avec n = 2 et u(0, x) = 0, sous certaines hypothèses de généricité relatives à la singularité en (x 1 , x 2 ) = (0, 0) ∈ C 2 ; voir [START_REF] Majima | Asymptotic analysis for integrable connections with irregular singular points[END_REF].

Nous verrons que les solutions formelles des équations évoquées ci-dessus sont toutes Borel-sommables dans les directions convenablement choisies ; voir Theorems 4.1, 4.2 and 4.3. Pour ce faire, nous allons convertir chacune des ces équations aux dérivées partielles en une équation mélangeant les produits de convolution et les différentielles partiels dans le plan de Borel. Le problème de sommabilité sera alors ammené à une analyse d'existence de solutions à croissance au plus exponentielle à l'infini dans des secteurs dans le plan complexe bidimensionnel. Une telle analyse sera rendue possible grâce à une version adoptée au cas de deux variables de la norme Nagumo initialement introduite dans [START_REF] Luo | Exponential-type Nagumo norms and summability of formal solutions of singular partial differential equations[END_REF]Definition 4.2].

Enfin, notons que la notion de sommabilité à la Borel d'une série de deux ou plusieurs variables a été largement étudiée dans [START_REF] Sanz | Summability in a direction of formal power series in several variables[END_REF], [START_REF] Mozo-Fernández | Topological tensor products and asymptotic developments[END_REF], [START_REF] Zurro | A new Taylor type formula and C ∞ extensions for asymptotically developable functions[END_REF] et bien d'autres travaux. Dans [START_REF] Chen | On the Borel-Laplace summation of power series of two variables[END_REF], nous avons étudié plusieurs conditions équivalentes pour décrire une série Borel-summable dans une direction donnée en cas de deux variables. Comme le montrera Theorem 2.1, nous allons priviléger le point de vue des fonctions ayant un développement asymptotique au sens de H. Majima [START_REF] Majima | Asymptotic analysis for integrable connections with irregular singular points[END_REF]. On verra que, sur certains aspects, notre approche diffère de celle entreprise par [START_REF] Sanz | Summability in a direction of formal power series in several variables[END_REF] ; voir les discussions développées au §2.2.

Les résultats annoncés dans la présente Note sont extraits de notre article [START_REF] Chen | Summability of Formal Solutions of Singular PDEs by means of two-dimensional Borel-Laplace method[END_REF].

Introduction

Let us consider the following non-linear singular PDE of the form

t∂ t u = F (t, x, u, ∇ x u), (t, x) ∈ C t × C n x , (1)
where u = u(t, x) is an unknown function, ∇ x is the usual derivative operator (∂ x1 , • • • , ∂ xn ) and F is a function of (t, x, u, v). Such equations can be found already in [START_REF] Baouendi | Cauchy problems with characteristic initial hypersurface[END_REF][START_REF] Tahara | Fuchsian type equations and Fuchsian hyperbolic equations[END_REF]; see also [START_REF] Chen | On the holomorphic solution of non-linear totally characteristic equations[END_REF][START_REF] Costin | Existence and uniqueness for a class of nonlinear higher-order partial differential equations in the complex plane[END_REF][START_REF] Gérard | Singular nonliear partial differential equations[END_REF][START_REF] Luo | On a family of symmetric hypergeometric functions of several variables and their Euler type integral representation[END_REF]. In [START_REF] Chen | Formal solutions for first-order nonlinear PDEs with irregular singularity[END_REF][START_REF] Luo | Exponential-type Nagumo norms and summability of formal solutions of singular partial differential equations[END_REF], we dealt with the case of n = 1 under the condition that the equation in (1) admits x = 0 as a generic irregular singular point. It has been show that the formal solution of (1) is k-summable for x and convergent for t. In this paper, we will study the case of n = 2; the main idea is to make use of the 2-dimensional Borel-Laplace summation method considered in our previous paper [START_REF] Chen | On the Borel-Laplace summation of power series of two variables[END_REF]. Under some suitable conditions, it will be shown that the formal solution is Borel-summable for the variable (x 1 , x 2 ) while the variable t remains near zero; see Theorem 4.3.

As t = 0 can be viewed as a Fuchsian type singular point in (1), we will start with some PDEs without the variable t. The first differential equation that we will study is the following one:

x 2 1 ∂ x1 u + x 2 2 ∂ x2 u -u + x 1 x 2 = 0 , (2)
which is a PDE version of the ODE satisfied by the Euler series; see §2. After having obtained the directions along which the formal solution of ( 2) is Borel-summable, we will consider the Stokes phenomenon, for what a link with the Euler series will be established. Section 4 will be essentially devoted to two families of nonlinear PDEs, one of which is a special case of (1). In the below, we will start with some general statements about the Borel-summability of power series of two variables.

Borel-summability of power series of two variables

Let C to denote the Riemann surface of the Logarithm, which will be identified with R >0 ×R via the map (r, ϕ)

→ re iϕ . Given (d, θ, ρ) ∈ R × R 2 >0 , one writes S d (θ; ρ) = {x ∈ C : |x| < ρ, | arg(x) -d| < θ/2}; by extension, S d (θ) = S d (θ; ∞)
, which is an unbounded sector. Unless explicitly stated, we usually suppose that θ < 2π, in such a way that each of S d (θ; ρ) and S d (θ) can be viewed as one part of the complex plane C. As usual, let D(0; R) ( D(0; R), resp.) be the open (closed, resp.) disc centered at 0 with radius R > 0. Furthermore, one defines S d R (θ) = D(0; R) ∪ S d (θ). In the case of two variables, we will write

d = (d 1 , d 2 ), θ = (θ 1 , θ 2 ), ρ = (ρ 1 , ρ 2 ) and R = (R 1 , R 2 )
. This permits to define the domains S d (θ; ρ), S d (θ) and S d R (θ) in an evident way. Furthermore, we will use the partial order < in R 2 given by the relation (a

1 , a 2 ) < (b 1 , b 2 ) ⇔ a 1 < b 1 and a 2 < b 2 .
By substituting ≤ to < here, we obtain also the relation ≤ in R 2 . Similarly we can define the partial orders > and ≥ in R 2 respectively. Lastly, we will write

x n = x n1 1 x n2 2 for all x = (x 1 , x 2 ) ∈ C 2 and n = (n 1 , n 2 ) ∈ Z 2 ≥0 .
2.1 -On the one hand, given S = S d (θ; ρ), we will denote by A 1 (S) the space of analytic functions admitting a strong asymptotic expansion as x = (x 1 , x 2 ) → (0, 0) in S in the sense of H. Majima [START_REF] Majima | Asymptotic analysis for integrable connections with irregular singular points[END_REF].

On the other hand, if S = S d R (θ), we will denote by E 1 (S) the set of analytic functions φ in S with an exponential growth of order at most one at infinity. This is to say, for any (R , θ

) ∈ R 2 >0 × R 2 >0 with R < R and θ < θ, one can find (C, ν) ∈ R 2 >0 such that |φ(ξ)| ≤ C exp(ν |ξ|) for all ξ = (ξ 1 , ξ 2 ) ∈ S d R (θ ), where |ξ| = |ξ 1 | + |ξ 2 |. Thus, to each d ∈ R 2 ,
we can associate the inductive limit spaces A d

1 and E d 1 in the following way:

A d 1 = ∪ θ>π,ρ>0 A 1 (S d (θ; ρ)) , E d 1 = ∪ θ>0,R>0 E 1 (S d R (θ)) ,
where π = (π, π) and 0 = (0, 0). Furthermore, we introduce the following generalized space Ẽd

1 : Ẽd 1 = C δ 0 ⊕ E d1 1 δ 0;1 ⊕ E d2 1 δ 0;2 ⊕ E d 1 .
Here δ 0 , δ 0;1 and δ 0;2 are the Dirac operators related to 0 = (0, 0) ∈ C 2 = C ξ1 × C ξ2 , 0 ∈ C ξ1 and 0 ∈ C ξ2 , respectively; E d1

1 and E d2 1 denote the spaces of analytic functions with an exponential growth of order at most one at infinity in some sector containing the directions d 1 and d 2 , respectively.

Like as in the case of one variable, the spaces Ẽd 1 and A d 1 may be viewed as differential algebras.

Theorem 2.1 Given d = (d 1 , d 2 ) ∈ R 2 , the following Laplace integral L d 1 φ(x) = ∞e id 1 0 ∞e id 2 0 φ(ξ 1 , ξ 2 ) e -ξ1/x1-ξ2/x2 dξ 2 dξ 1
gives an isomorphism of differential algebras from Ẽd

1 onto A d 1 . If f (x) = n≥0 a n x n ∈ C[[x]], one defines its (formal) Borel-transform as follows (ξ = (ξ 1 , ξ 2 )): B1 f (ξ) = a 0 δ 0 + n1≥0 a n1+1,0 n 1 ! ξ n1 1 δ 0;1 + n2≥0 a 0,n2+1 n 2 ! ξ n2 2 δ 0;2 + n≥0 a n+1 n! ξ n . ( 3
) Definition 2.1 Let d = (d 1 , d 2 ) ∈ R 2 . A given power series f (x) ∈ C[[x]] will be called Borel-summable in the direction of argument d if B1 f ∈ Ẽd 1 .
In this case, we write f ∈ C{x} d 1 .

In view of Theorem 2.1, one finds that the composite map L d • B1 gives an isomorphism of differential algebras from C{x} d

1 onto A d 1 . So, L d • B1 f can be called Borel-sum of f ∈ C{x} d 1 in d.

-

The above-introduced spaces A 1 (S d (θ; ρ)) are already used in [START_REF] Sanz | Summability in a direction of formal power series in several variables[END_REF] with regard to the Borelsummability of power series of two or more variables; see also [START_REF] Carrillo | An extension of Borel-Laplace methods and monomial summability[END_REF][START_REF] Mozo-Fernández | Topological tensor products and asymptotic developments[END_REF] and [START_REF] Zurro | A new Taylor type formula and C ∞ extensions for asymptotically developable functions[END_REF]. However, instead of our spaces E 1 (S d R (θ)), one finds the spaces of type E 1 (S d 0 (θ)) in [START_REF] Sanz | Summability in a direction of formal power series in several variables[END_REF], where S d 0 (θ) = S d (θ) denotes an unbounded sector containing d. This difference may be significant for some situations, as it will be shown in the below.

If one considers the partial differential operator

L = x 2 2 ∂ x2 (x 2 1 ∂ x1 + 1) + 1, then the following series f (x 1 , x 2 ) = n2≥n1≥0 (-1) n2 (n 2 !) 2 (n 2 -n 1 )! x n1+1 1 x n2+1 2 (4)
will satisfy L u = x 1 x 2 . Letting φ = B f , one finds that φ(ξ) = 1 1+ξ2(1+ξ1) , so φ is analytic at 0 ∈ C 2 and can be analytically continued in any sector S d 0 (θ) such that |d j ± θ j /2| < π/2 for j = 1 and 2. Thus, the series f given by ( 4) could be Borel-summable in such directions d in the sense of [START_REF] Sanz | Summability in a direction of formal power series in several variables[END_REF]. What may be surprising is that φ can not be analytically continued in any domain of the form

S d R (θ) with R ∈ R 2 >0 , this means f ∈ C{x} d 1 for all d ∈ R 2
, by considering Definition 2.1 here. In our previous work [START_REF] Chen | On the Borel-Laplace summation of power series of two variables[END_REF], we introduced intermediate spaces situated between the inductive spaces E d 1 and ∪ θ>0 E 1 (S d 0 (θ)), and this defines what we call (1, 2) or (2, 1)-iteratively Borel summable series in a given direction d. In this setting, the above series f is (2, 1)-iteratively Borel summable but not (1, 2)-iteratively Borel summable in the direction (0, 0), contrarily to the statement of [START_REF] Sanz | Summability in a direction of formal power series in several variables[END_REF]Prop. 5.4]. S;µ,n ), with µ ∈ R ≥0 and n ∈ Z ≥0 . Let m ∈ N, D = D(0; ρ 1 ) × ... × D(0; ρ m ) with ρ j > 0, and let Φ be an analytic function defined in S × D. By using the results in [3, §3], one can prove that Φ ∈ Ẽ1;µ,n (S){z} ρ , with ρ = (ρ 1 , ..., ρ m ) if the map z → Φ(., z) is an analytic mapping from D to Ẽ1;µ,n (S). Now, we define Ẽd 1;µ,n {z} = ∪ Ẽ1;µ,n (S){z} ρ by taking the union for all S = S d R (θ) with θ > 0, R > 0 and for all ρ 1 > 0, ..., and ρ m > 0, where (ρ 1 , ..., ρ m ) = ρ. By Theorem 2. 1,A d 1 can be found from Ẽd 1 via the Laplace integral L d 1 . Thus, we define

-Let

A d 1;µ,n {z} = L d 1 Φ(., z) : Φ(., z) ∈ Ẽd 1;µ,n {z} , A d 1 {z} = ∪ µ≥0 A d 1;µ,n {z} (n ≥ 0) . (5)

2-dimensional Euler equation

Let Ê(x) = n≥0 n! x n+1 be the so-called Euler's series that satisfies x 2 y -y + x = 0. One can see that (2) is satisfied by the 2-dimensional Euler's series F 2 , the quotient of the additive group R 2 by (2πZ) 2 , and set

(x) = n1,n2≥0 (n 1 + n 2 )! x n1+1 1 x n2+1 2 . 3.1 -Let T 2 = R 2 /(2πZ)
D = {(d 1 , d 2 ) ∈ (0, 2π) 2 : |d 1 -d 2 | < π} , S = {d ∈ T 2 : d ∈ [0, 2π] 2 \ D} . (6)
Given d ∈ D and λ > 0, there exist θ ∈ R 2 >0 and δ > 0 such that |1 -

ξ 1 -ξ 2 | ≥ δ (λ + |ξ|) for all ξ = (ξ 1 , ξ 2 ) ∈ S d (θ). Consequently, F ∈ C{x} d 1 if, and only if, d ∈ D. Let {j, k} = {1, 2}, d = (d 1 , d 2 ) ∈ D, and define α j = max(0, d k -π), β j = min(2π, d k + π). Since β j > α j + π and d k ∈ (α j , α j + π], one can choose 0 in such a way that, letting d - j = α j + and d + j = β j -, one has d k ∈ (d - j , d - j + π) ⊂ (d - j , d + j ). Thus, if U = (d 1 -, d 1 + ) × (d - 2 , d + 2 ) ∪ (d - 1 , d + 1 ) × (d 2 -, d 2 + ) ,
then U ⊂ D and F ∈ C{x} δ 1 for δ ∈ U . Let F be the Borel-sum of F associated with a given δ ∈ U . It follows that F ∈ A 1 (S) with S = S 1 × S 2 or S = S 1 × S 2 , where

S j = {x j ∈ C * : arg(x j ) ∈ (d - j - π 2 , d + j + π 2 )} , S k = {x k ∈ C * : arg(x k ) ∈ (d k -- π 2 , d k + + π 2 } . (7)
Theorem 3.1 Let T j = {z ∈ S j : ze 2πi ∈ S j } for j = 1 and 2. The following relations hold respectively for x ∈ S 1 × T 2 and x ∈ T 1 × S 2 :

F (x 1 , x 2 e 2πi ) -F (x 1 , x 2 ) = 2πi x 1 x 2 x 1 -x 2 e -1/x2 , F (x 1 e 2πi , x 2 ) -F (x 1 , x 2 ) = 2πi x 1 x 2 x 2 -x 1 e -1/x1 . ( 8
)
3.2 -The above series F can be also expressed as follows:

F (x) = n≥0 n! n k=0 x k+1 1 x n-k+1 2 = x 1 x 2 x 1 -x 2 n≥0 n! (x n+1 1 -x n+1 2 ) = x 1 x 2 x 1 -x 2 Ê(x 1 ) -Ê(x 2 ) . (9)
In other words, if we denote by ∆ the operator defined by the following relation:

∆f (x 1 , x 2 ) = f (x 1 ) -f (x 2 ) x 1 -x 2 if x 1 = x 2 , ∆f (x 1 , x 2 ) = f (x 1 ) if x 1 = x 2 , (10)
then F (x) = x 1 x 2 ∆f (x). One remembers that the Euler series Ê is Borel-summable in all direction excepted in R >0 , and its Borel-sum in R -can be continued to be analytic in S π (3π); see [START_REF] Ramis | Gevrey asymptotics and applications to holomorphic ordinary differential equations[END_REF].

Theorem 3.2 Let d = (d 1 , d 2 ) ∈ T 2 \ S,
conserve S j and S k as in [START_REF] Chen | Summability of Formal Solutions of Singular PDEs by means of two-dimensional Borel-Laplace method[END_REF], and consider the Borel-sum F of

F in d. Then F (x) = x 1 x 2 ∆E(x) for all x = (x 1 , x 2 ) ∈ (S 1 × S 2 ) ∪ (S 1 × S 2 )
, where ∆E is obtained from (10) by replacing f with the Borel-sum E of Ê in S π (3π).

Summability of formal solutions of linear or non-linear PDEs

Let D be as in [START_REF] Chen | On the Borel-Laplace summation of power series of two variables[END_REF], and choose arg t ∈ [0, 2π) for any t ∈ C * ; given (β, γ 1 , γ 2 ) ∈ C * 3 , define

D β;γ1,γ2 = d ∈ R 2 : d + (arg γ 1 β , arg γ 2 β ) ∈ T 2 \ S .
3.1 -In order to generalize the study of (2), consider the following linear PDE:

c 1 (x) x 2 1 ∂ x1 u + c 2 (x) x 2 2 ∂ x2 u -b(x) u + a(x) = 0 , (11)
where a, b, c 1 and c 2 are some given functions of the variable x = (x 1 , x 2 ).

b(0) -c 1 (0) ξ 1 -c 2 (0) ξ 2 ) U = A -B * U + C 1 * (ξ 1 U ) + C 2 * (ξ 2 U ) ,
in which A, B, C 1 and C 2 are linked with the Borel-transforms of a, b, c 1 and c 2 . Let A d 1 {z} be as in [START_REF] Chen | Formal solutions for first-order nonlinear PDEs with irregular singularity[END_REF], and consider the following Cauchy problem:

-

c 1 (x) x 2 1 ∂ x1 u + c 2 (x) x 2 2 ∂ x2 u -G(x, u) = 0, u(0) = 0 . (12)
In what follows, we will assume the following hypothesis for ( 12

t∂ t u = c 1 (x)x 2 1 ∂ x1 u + c 2 (x)
x 2 2 ∂ x2 u -G(x, t, u, δ x u), u(x, 0) = 0, [START_REF] Majima | Asymptotic analysis for integrable connections with irregular singular points[END_REF] which is a special case of (1). We will conserve Hypothesis (H1) made in §4.2 and replace (H2) and (H3) with the following ones:

  S = S d R (θ), and consider Ẽ1 (S). By generalizing the Nagumo norm considered in [11, Definition 4.2], one can get a family of Banach spaces ( Ẽ1;µ,n (S),

Theorem 4 . 1

 41 Let d ∈ R 2 , and consider (11) with a, b, c 1 and c 2 ∈ A d 1 . If b(0) c 1 (0) c 2 (0) = 0 and d ∈ D b(0);c1(0),c2(0) , then (11) has a unique solution u ∈ A d 1 . To prove Theorem 4.1, one can apply Borel transform to rewrite (11) into a convolution equation: (

  ): (H1) d = (d 1 , d 2 ) ∈ R 2 , c j ∈ A d 1 with c j (0) = γ j = 0 for j = 1 and 2. (H2) G ∈ A d 1 {u}, G(0, 0) = 0 and ∂ u G(0, 0) = β = 0. (H3) The relation d ∈ D β;γ1,γ2 holds.

Theorem 4 . 2 3 . 3 -

 4233 Under (H1), (H2) and (H3), the Cauchy problem in (12) admits a unique solution u ∈ A d 1 . Let δ x = (x 1 ∂ x1 , x 2 ∂ x2 ) be a Euler vector fields, and consider the following Cauchy problem:

(Theorem 4 . 3

 43 H2a) G ∈ A d 1 {t, u, v 1 , v 2 }, G(x, 0 4 ) = ∂ v1 G(x, 0 4 ) = ∂ v2 G(x, 0 4 ) = 0 and ∂ u G(0 6 ) = β / ∈ Z <0 . (H3a) The relation d ∈ D 1;c1(0),c2(0) ∩ ∩ ∈N * D β+ ;c1(0),c2(0) holds. Under (H1), (H2a) and (H3a), the Cauchy problem in (13) admits a unique solution u ∈ A d 1 {t}, that is to say, there exist a unique sequence of functions u ∈ A d 1 such that the power seriesu(x, t) = ∞ =1 u (x) t(14)converges into a solution of (13) in a domain of the form S d R (θ) × D(0; T ), where θ > (π, π).

The research supported partially by the NSFC Grants (No 11171261 and 11371282)