Monodromy , hypergeometric polynomials and

Monodromy , hypergeometric polynomials and diophantine approximations Marc Huttner October 11, 2007

UFR de Math´ematiques UMR AGAT CNRS Universit´e des Sciences et Technologies de Lille F- - 59665 Villeneuve d’Ascq Cedex ,France

Abstract The monodromy’s study of Fuchsian hypergeometric differential equation provides a natural framework for the explicit determination of rational approximations of polylogarithmic and hypergeometric functions .Thus , we can obtain almost without calculation explicit determination of many polynomials and hypergeometric power series related to their Pad´e approximations . From now on , using a classical way , one can study the arithmetic nature of numbers related to the values taken by these functions. R´ esum´ e L’´etude de la monodromie des equations diff´erentielles hyperg´eom`etriques Fuchsiennes fournit un cadre naturel pour la d´etermination explicite des approximations rationnelles des fonctions polylogarithmes et de certaines fonctions hyperg´eom`etriques .Ceci permet de donner presque sans calcul certains polynˆ omes et s´eries hyperg´eom`etriques intervenant dans la d´etermination des approximants de Pad´e li´es `a ces fonctions. Par un cheminement d´esormais classique , on peut alors ´etudier la nature arithm´etiques de certains nombres en relation avec les valeurs prises par ces fonctions en certains points.

Keywords Irrationality,linear independence over a field , generalized hypergeometric series, singularities, monodromy ,local behavior of solutions ,normal forms inverse problems (Riemann Hilbert inverse differential, etc) . Approximation by rational functions, Pad´e approximations.

1

AMS classification 11 J 72-33 C20-34 M 35-34 M

Contents 1

Introduction

The aim of this article is to devote to a better understanding of many constructions of effective rational approximations to solutions of particular linear hypergeometric functions from the perspective of the monodromy theory . The monodromy’s study of Fuchsian hypergeometric differential equation provides a natural framework for the explicit determination of rational approximations of polylogarithmic and hypergeometric functions . The main result of our study can be formulated as follows: We can effectively construct the system of Pad´e approximants at z = ∞ to system of functions g1 (z), g2 (z), · · · gq (z) where gq (z) =

∞ X n=1

(1/z)n (b1 + n)(b2 + n) · · · (bq + n)

gq−1 (z) = (θ − bq )(gq (z)), · · · , g1 (z) = (θ − b2 )(g2 (z)), d where as usual θ denotes the operator θz = z dz ,. gq (z) is a solution holomorphic at infinity of the following Fuchsian differential equation of order q + 1 which is easily related to an hypergeometric differential equation . This equation can also be written:

(θ − b1 )(θ − b2 ) · · · (θ − bq )(g(z)) =

(−1)q−1 1−z

This system of functions gives a very special contiguous system of local solutions at 0 of the previous Fuchsian differential equation. Indeed gq−1 (z), · · · , g1 (z) are solutions of fuchsian differentials equations of order q , q − 1, · · · 2. By effective construction ,we understand similtaneous solutions of the following problems: 1) Find explicit expression for differential equations that are satisfied by remainders functions and sometimes polynomials of Pad´e approximants. 2) Find some ”closed” formulae ,(i.e. integral expressions for remainders functions R∞ (z) as functions of n ) and explicit determination of polynomials ) . In general ,we can deduce the asymptotics of the remainders and for the 2

polynomials for fixed z and n → ∞. From now on , using a classical way , one can study the arithmetic nature of numbers related to the values taken by functions related to these hypergeometric power series where for 1 ≤ i ≤ q ,bi ∈ Q. To obtain , almost without calculation explicit determination of many polynomials and hypergeometric power series related to their Pad´e approximations, we use the method of [?] to solve in this particular case the realization problem: given in this particular case the subset, S = {0, 1, ∞} of P1 (C) and given local numerical data of rank q on Z =: P1 (C)\S we are asked to explicitly construct and analyze the behavior of Fuchsian differential equations that give rise to the given local system. When the local system has no accessory parameters, that is , when it is ”rigid” according to the terminology of Katz [?], the Fuchsian differential equation is unique and we obtain a Pad´e linear form as a special case of the contiguity relations. The rigidity of the local system or, equivalently the differential equation, has important arithmetic applications. In particular, the uniqueness of the equation corresponding, respectively, to the generalized hypergeometric functions , the polylogarithmic and the Lerch functions Lk (x, z) =

∞ X k=0

zk (n + x)k

[?], [?], [?] enables us to establish many new explicit formulae for Pad´e approximations. The polynomials discussed in this paper originate in the construction of simultaneous Pad´e approximation to these hypergeometric power series at 0. There are in particular hypergeometric or ’almost’ hypergeometric and often related to orthogonal polynomials. The arithmetic motivation of searching such effective rational approximations come for proving irrationality or transcendence of numbers arising as values of hypergeometric functions , such as Liq (1/p)) , ζ(2), ζ(3), · · · etc , P zn where : Liq (z) = ∞ n=1 nq denotes the polylogarithmic function. We recall that in 1978 R.Apery [?] proved that the number ∞ X 1 ζ(3) = n3 n=1

was irrational. Before Apery’s result , nothing was known about the arithmetic properties of the zeta- functions at odd points. As well known , Apery’s proof is based on his discovery of a sequence of good rational approximations of ζ(3). F.Beukers [?](1979) has given another interpretation of Apery’s rational

3

approximations to ζ(3) . He found the remarkable integral’s representation (Euler’s type integral) Z 0

1

− log(xy) Pn (x)Pn (y)dxdy = 2(Un ζ(3) − Vn ) 1 − xy

(1.1)

where

1 d n n ( ) (x (1 − x)n ) ∈ Z[x] n! dx are Legendre polynomials and Un , Vn are rational numbers where   −n, −n, n + 1, n + 1 Un = 4 F3 1 . 1, 1, 1 Pn (x) =

4 F3

denotes here the hypergeometric power series . L.A Gutnik (1979) [?] established that for any number q 6= 0 , there is an irrational number among the numbers 3ζ(3) + qζ(2) , ζ(2) + 2q. log 2 The polylogarithm Lik (z) =

∞ X zn , nk

n=1

the hypergeometric and the so-called Meijer G-functions play an important role in the proof.In fact Gutnik found another representation: Z 1 Γ4 (−t)Γ2 (n + 1 + t) dt = 2(Un ζ(3) − Vn ) 2iπ C Γ2 (n + 1 − t) Here Γ(t) is Euler gamma-function and the contour C is the vertical straight line that begins at −1/2 − i∞ and ends at −1/2 + i∞. By means of these functions , the following simultaneous approximation of 1, Li1 (1/z), Li2 (1/z), Li3 (1/z) were established R1 (z) = A3 (z)Li2 (1/z) + A2 (z)Li2 (1/z) + A0 (z) R2 (z) = 2A3 (z)Li3 (1/z) + A2 (z)Li2 (1/z) + A1 (z) Where A0 (z), A1 (z), A2 (z), A3 (z) are polynomials in z with rational coefficients of degree at most n satisfying the conditions ordz=∞ R1 (z) ≥ n + 1 ordz=∞ R2 (z) ≥ n + 1 This explicit expression implies that A2 (1) = 0 . Using the Gutnik’s approach Yu.V.Nesterenko [?] proposed a new proof of

4

the irrationality of ζ(3) an discovered new continued fractions for ζ(3) . In fact , F (z) = R1 (z) log z + R2 (z) satisfies the Fuchsian differential equation θ4 − z(θ − n)2 (θ + n + 1)2 F = 0 d where θ is the operator z dz . The Apery ’s constructions were generalized to the study of arithmetic of linear forms in odd zeta functions by Nesterenko, Ball-Rivoal and Zudilin in [?][?],[?],[?]. In order to make our exposition self-contained , we shall first review in the next subsections some relevant notations and background material ,omitting the proofs.

2

Pad´ e approximants at infinity

When we want to try to approximate simultaneously a system of functions S = {1, f1 (1/z), f2 (1/z), · · · , fq (1/z)}

(2.1)

near the point z = ∞ on the Riemann sphere P1 (C) where these functions have ”good” arithmetic expansions. There are two kinds of rational approximations than can be used. Definition 1 Approximations of the first kind Polynomials {A0 (z), A1 (z), · · · , Aq (z)}

(2.2)

such that Ai (z) 1 ≤ i ≤ q are of degrees at most di and such that expanding at z = ∞ we have : R(z) := A0 (z) + A1 (z)f1 (1/z) + · · · Aq (x)fq (1/z) where Ord∞ R(z) := σ∞ ≥

q X

(2.3)

(dk + 1)

k=0

is called Pad´e approximants of the first kind (of weights (d1 , d2 , · · · , dq )). Such polynomials exist since the condition (1,5) imposes on the coefficients of the polynomials σ − 1 linear equations while the number of undetermined coefficients of the polynomials is σ.By virtue of this homogeneous system of linear equations we have at least one non trivial solution. We note also that deg A0 (z) ≤ min1≤k≤q dk − 1 . The second one is called Pad´e approximants of the second kind . 5

Let r1 ≥ r2 ≥ · · · ≥ rq ≥ 0 be integers , and for eachPi = 1, 2, · · · q let ki the number of indices s for which bs = bi .We put N = qj=1 rj . Then the following inequality holds for some polynomials PN,j (z) ∈ C[z] of degrees at most N − 1 ,1 ≤ j ≤ q. Rj (z) = (QN (z)gj (z) − PN,j (z) ≥ rj + 1 and Ord∞ Rj (z) ≥ rj + 1. This Pad´e approximant is called Pad´e approximant of second kind at infinity. Like in the explicit continued fraction expansions,we are interested in cases of effective Pad´e approximations of the first case at infinity of the family S. In general all functions , f1 (1/z), f2 (1/z), · · · , fq (1/z) are explicitly known in terms of their power series expansions at z = ∞ and fk (1/z) =

∞ X

n a(k) n (1/z)

n=1

1≤k≤m. For instance and in this paper we shall take fk (1/z) = Lik (1/z) =

∞ X 1 (1/z)n nk

n=1

Here Lik is the polylogarihmic function.See also [?].

3

Differential hypergeometric equations

In the following if α ∈ C we put (α)0 = 1 and if n ≥ 1 , (α)n = α(α + 1)(α + 2) . . . (α + n − 1) (Pochhammer symbol )  F (a0 , . . . , aq , b1 , . . . bq |z) :=q+1 Fq =

∞ X n=0

Qq

Qj=0 q

a0 , a1 , · · · , aq b1 , b 2 , · · · , b q

 z

(3.1)

(aj )n z n

j=1 (bj )n

n!

denotes the hypergeometric power series . Provided that no denominator bi is a non positive integer , the series coefficients are finite and the series converge absolutly in an open disk |z| < 1 provided that q q+1 X X d = n . These analytic continuations permit us to give a ” natural way” to introduce the so called ”Levelt basis” [?] of global solutions of a linear Fuchsian differential equation. The Riemann scheme related to this equation is:   0 ∞ 1  σ0 σ∞ 0     b1 − 1 −n − b1 + 1 1     b2 − 1 −n − b2 + 1 2   |z  R  . . . . . .   . . .     .. .. ..   . . . bq − 1 −n − bq + 1 σ1 This Riemann scheme can also be written   ∞ 1 0  σ 0 + b1 − 1 σ ∞ + b1 − 1 0      b2 − b 1 −n − b + b 1 2 1    b3 − b 1 −n − b3 − +b1 2    .. .. .. |z  z b1 −1 R    . . .     .. ..   . −n − b + b . 1 q−1    bq − b1 −n + b1 − bq σ1  0 −n The quasi- analytic solution which belongs to the ”exponents be written (within a normalisation constant)  −n, −n + b1 , · · · , −n + bq b1 −1 z q+1 Fq 1 + σ ∞ + b1 , 1 + b 1 − b 2 · · · , 1 + b1 − b q The Riemann scheme of the remainder  0  σ 0   −σ∞ + b1 − n + 1   −σ∞ + b2 − n + 1 (1/z)σ∞ R   ..  .   ..  . −σ∞ + bq − n + 1 12

”b1 ” can also  1/z

is:  ∞ 1  σ∞ + σ0 0   σ ∞ + b1 + 1 1   σ ∞ + b2 + 1 2 |1/z   .. ..  . .   .. ..  . . σ ∞ + bq + 1 σ 1

(4.9)

then within a multiplicative constant , one finds: 

σ∞ + b1 − 1, · · · , σ∞ + b2 − 1, σ∞ + bq − 1, σ∞ + σ0 R∞ (z) = (1/z) q+1 Fq σ∞ + b1 + n, · · · , σ∞ + n + bq (4.10) If we choose the normalisation’s constant of the remainder Qq (Γ(σ∞ + bk − 1))q Γ(σ∞ + σ0 ) C∞ (n) = k=1 , Γ(σ∞ + bk + n)q (n!)q σ∞

we obtain the following integral formula : Z 1 Qq Z 1 σ∞ +bk −1 (1 − tk )n k=1 tk R∞ (z) = ··· dt1 · · · dtq (z − t1 t2 · · · tq )σ∞ +σ0 0 0 The analytic continuation of the linear form i.e. monodromy around the singular point ”1” and ”0” given by the block of the matrix Λ(x) R∞ (z) = A0 (z) +

q−1 X

Ak (x)gk (z).

k=0

gives the other branches. Now ,we use analytic continuation given by the matrix Λ(z) (see also [?] ) multiplied by the matrix   Aq (z) Aq−1 (z)    ...  A0 (z) which is invariant by use of monodromy . For instance:  −n, −n + b1 − b2 , · · · , −n + b1 − bq , σ∞ + b1 − 1 Aq (z) =q+1 Fq 1 + b 1 − b 2 , · · · , 1 + b1 − b q , 1 + b 1 − σ 0

 z (4.11) The identification of these two expressions gives the polynomials and ends the proof of the theorem.

5

Approximations of Lerch functions

Now,using the previous section in the particular case b1 = b2 = · · · bq = x , one obtains results about the Lerch function. (See also [?]) . For an integer q ≥ 1 ,x and z are complex numbers such |z| ≤ 1 ,x is not a negative integer ,we put   x, x · · · , x, 1 z = 1 Φq (x, z) Lq (z) =q+1 Fq (5.1) 1 + x1 , 1 + x, · · · , 1 + x xq 13

 1/z

Using the same method as [?] , we can find the Pad´e approximations of the family S and S =: C(z){g1 (z), · · · , gq z} is of rank q + 1 over C(z). We use of analytic continuation of gq (z), · · · , g1 z along loops γ1 and γ0 based in a vicinity of 1 resp 0. We use the following loops : γ1 , γ0 ◦ γ1 · · · , γ0m−2 ◦ γ1 , γ0q−1 ◦ γ1 . to obtain the following matrix of ”periods” whose the first block related to the root z is given by: We find:     R∞ (z) A0 (z) 1 Φ1 (1/z) · · · Φq (1/z)  R1 (z)    .. (q−1) x x  (1/z)/(q − 1)!(1/z)      0 2iπ(1/z) 2iπ log .    ..   .    .. .   Ak (z)  =     .. .. m−1 m−1  0 0 (2iπ) (2iπ) log (1/z)     . . q x 0 ··· 0 (2iπ) (1/z) Rq (z) Aq (x) (5.2) If σ∞ = q(n + 1) − 1 , we are in the Pad´e’s case. If 1 − n = 1 − σ0 + x + σ∞ + x , i.e , we obtain the same result as [?]. Using the same method as [?] , we can obtain the well-poised and the very well poised case . For instance , we obtain the same result as [?] with the polynomial Aq (z). The first quasi-polynomial Aq (z)(1/z)x satisfies the same hypergeometric differential equation as R∞ (z). We thus obtain: Corollar 1  Aq (z) =q+1 Fq

−n, −n, · · · , −n, σ∞ + x − 1 1, 1, · · · , 1, x − σ0

 z

(5.3)

The other polynomials are given by Ak (z) =

n X dk (ck (n + t))

dtk

k=0

|t=0 z k

The remainder σ∞

R∞ (z) = C∞ (n)(1/z)

 q+1 Fq

σ∞ + σ0 , σ∞ + x − 1, σ∞ + x − 1 σ∞ + n + x, · · · , σ∞ + n + x

Where the normalisation’s constant is C∞ (n) =

(Γ(σ∞ + x − 1)q . Γ(σ∞ + x + n)q (n!)q 14

 1/z (5.4)

       

If 1 − n = 1 − σ0 + x + σ∞ + x , i.e , we obtain the same result as [?]. Using the same method as [?] , we can obtain the well-poised and the very well poised case . For instance , we obtain the same result as [?] with the polynomial Aq (z).

5.1

Pad´ e approximation of the second kind

Let r1 ≥ r2 ≥ · · · ≥ rq ≥ 0 be integers , and for eachPi = 1, 2, · · · q let ki the number of indices s for which bs = bi .We put N = qj=1 rj . Let b1 , b2 , · · · , bq be arbitrary nonnegative complex numbers , we give the Pad´e approximant of second kind at infinity for g1 (z), · · · , gq (z) . Theorem 2 QN (z) =  q+1 Fq

(b1 + r1 )r1 · · · (bq + rq )rq

(5.5)

r1 !r2 ! · · · rq !

−N, r1 + b1 , · · · , r2 + b2 , rq + bq b1 , b 2 , · · · , b q

 z

which can be written

QN (z) =

N X k=0

     N r1 + k + b 1 r 1 + k + bq k (−1) ··· z k r1 rq k

(5.6)

Proof Using monodromy around ’1’ ,we see that QN (z), (1/z)−b1 R1 (z), , (1/z)−b2 R2 (z), · · · (1/z)−bq Rq (z) satisfy the same differential equation .The Riemann Symbol related to QN (z) is   0 ∞ 1  0 −N 0     −b1 + 1 r1 + b1 1     −b2 + 1 r2 + b2 2  R |z  (5.7)  .. .. ..    . . .    .. .. ..   . . .  −bq + 1 rq + bq 0 whose analytic solution is the polynomial QN (z). Now one use normalisation to find Hata’s formulas [?]. The polynomials Pj are given by Z 1 Qn (x) − QN (t) −bi 1 Pj (z) = t (log(1/t))ki −1 dt (rj − 1)! 0 z−t 15

and the remainders : 1 Ri (z) = (rj − 1)!

1

Z 0

QN (t) −bi t (log(1/t))ki −1 dt z−t

We can also write : q

Q(z) =

Y 1 [ z 1−bk Drk z bk +rk −1 ](1 − z)N ) r1 !r2 ! · · · rq !

(5.8)

k=1

If r1 ≥ r2 ≥ · · · ≥ rq ,and for bi = 1 , i = 1 · · · q ,one obtains simultan´eous rational approximations of 1, L1 (1/z), · · · Liq (1/z) (see [?]) and [?]) where we found  −N, r1 + 1, · · · , r2 + 1, rq + 1 q+1 Fq 1, 1, · · · , 1 =

 n   X N r1 + k k=0

k

k

 ···

 z

 rq + k (−1)k z k k

d If we consider the derivation’s operator D = dz , we can write this polynomial : Dk xk = (θ + 1) · · · (θ + k)

yields the relation q

Q(z) =

Y 1 { (θ +1)(θ +2) · · · (θ +rk ) q+1 Fq r1 !r2 ! · · · rq !



k=1

 −N, 1, 1, · · · 1 z . 1, 1, · · · , 1, 1 (5.9)

and the relation  1 F0

−N z −



 =q+1 Fq

−N, 1, 1, · · · 1 z 1, 1, · · · , 1, 1



gives for this hypergeometric polynomial the formula : q

Y 1 Q(z) = [ Drk z rk ](1 − z)N ) r1 !r2 ! · · · rq !

(5.10)

k=1

The differential hypergeometric equation satisfied by Q(z) and the remainders Rk (z) are θq+1 − z(θ − N )(θ + r1 + 1) · · · (θ + rq + 1) = 0

16

This differential equation is unipotent ! Let us recall the sketch of the method to find the solutions of this differential equation. We use inverse-Mellin transform i.e..To find the polynomial’ (given by analytic solution at 0 of this equation . One set Z 1 f (z) = g(t)(−z)t dt. 2iπ C We must find a condition on the function g(t) under which this inverse Mellin transform converges and gives a solution of this differential equation (.Here C denotes a vertical line with possible deviation to avoid singularities of the integrand. We find that g(t) is a solution of the difference equation : g(t + 1) = −

(t − N )(t + r1 + 1) · · · ((t + rq + 1) (t + 1)q+1

We take a path of integration C as mentionned above.Then if lim (t2 g(t)(−z)t ) = 0

τ →∞

,(t = σ + iτ ) holds uniformly as t → ∞ in any finite vertical trip , then the integral converges uniformly with respect to z and the above calculation can be legitimated. In the following , we put r1 = r2 = · · · rq = n , N = qn and we consider the solutions of this differential equation belonging to the exponent n + 1 at infinity. We find since u = 1/z gives the transformation θu = −θz (with the same notation ). (θ + qn)(θ − n − 1)q − uθq+1 = 0 g(t) satisfies g(t + 1) = −g(t) and we can take g(t) =

tq+1 (t + qn)(t − n − 1)q

Γ(t)q+1 Γ(t + qn + 1)Γ(t − n)q

We simplify g(t) and we find the rational function R(t) =

((t − 1)(t − 2) · · · (t − n))q t(t + 1) · · · (t + qn)

We now set t = s + n + 1 to find the integral form of the solutions for the previous differential equation F (z) =

Γ(s + n + 1)q+1 . Γ(s + 1)q Γ(s + (q + 1)n + 1)q! 17

1 2iπ

Z

C+i∞

(R(s + n + 1)( C−i∞

π )q+1 (−1/z)s+n+1 ds sin(πt)

C is an arbitrary constant in the interval 0 < C < n + 1. Using the residue’s theorem we find : F (z) = R1 (z) + R2 (z) log(1/z) + · · · Rq (z) log(1/z)q This gives a sense to the substitution z = 1. In particular ,we obtain for q = 2 and for z = −1 simultaneous rational 2 approximations of log 2, − π12 ) and for z = 1 , q = 3 simultaneous approximations for ζ(2) and ζ(3). Remark 2 The method gives also the simultaneous approximations of L2 (1), Li3 (1), · · · Liq (1) see the construction given in[?] and [?] which uses the method of perturbing power series and construction of new differential equations.

6

The method of perturbing power series and construction of new differential equations

In [?] , Zudilin gives simultaneous approximations of Li1 (1/z), Li2 (1/z), Li3 (1/z) P (z)

but where the numerator Pj (z) are replaced by (1−z)j (q−1)n . Using our study ,we can find and generalize his result. We put ∞ X z n+t Lip (z, t) = (n + t)p n=1

and for k ≥ 1 , Lkp (z) =

1 ∂ k (Lp ( z1 , t)) |t=0 k! ∂tk

R(z) =

1 ∂ q (R(z, t)) |t=0 q! ∂tq

We consider

for R(z, t) = Q(z)L1 (1/z, t) −

P (z, t)( z1 )t (1 − z)(q−1)n+1

where P (z, t) ∈ C(t)[z]. We can deduce that : R(z) and Q(z) have the following Riemann scheme 18

      R     

 ∞ 1 0  0 −qn 0   0 n+1 1   0 n+1 2 |z   .. .. ..  . . .   .. .. ..  . . . 0 n + 1 −(q − 1)n

(6.1)

(We must remark here that the exponent −(q − 1)n is related to a no logarithmic exponent .) We then find   −qn, n + 1, n + 1, · · · n + 1 Q(z) =q+1 Fq z . 1, 1, · · · , 1, 1 We can use the same method to find the rational function given in [?] R( t) =

((t − 1)(t − 2) · · · (t − n))q n!(t + 1)(t + 2) · · · (t + qn)

The Barnes integral that we consider is Z C+i∞ π 1 R(t + n + 1)( )q+1 (−1/z)t+n+1 dt 2iπ C−i∞ sin(πt) C is an arbitrary constant in the interval 0 < C < n + 1 (We can take C = −1/2 ). This result coincides with Zudilin’result for q = 2 and q = 3. We use the same construction to find the Apery’s differential equation :

6.1

Differential equations and simultaneous approximations of ζ(2) and ζ(3)

For p = 2, k = 1 , we have the Apery’s case.We use The linear form R(z) =

∂ 1 ((A2 )(z)Li2 (1/z, t) + A1 (z)Li1 (1/z, t) + A0 (z, t)( t )|t=0 ∂t z

where A0 (z, t) ∈ C(t)[z] . The differential operator related to this function is (for more details see [?])). In this case , using the previous study it is not difficult to show that the Riemann scheme is given by: see also [?]   0 ∞ 1  0 −n 0     (6.2) R  0 −n 1 |z  .  0 n+1 2  0 n+1 1 19

It is related to the hypergeometric differential equation θ4 − z(θ − n)2 (θ + n + 1)2 = 0

(6.3)

the Ap´ery’s polynomial [?] is given by   −n, −n, n + 1, n + 1 A3 (z) = 4 F3 z . 1, 1, 1

(6.4)

This differential equation is also unipotent.The previous riemann sheme can be written :   0 ∞ 1   0 n+1 0   1  0 n + 1 1 |1/z  (6.5) R . n+1 z   −2n − 1 n + 1 2 −2n − 1 n + 1 1 The nonlogarithmic solution is thus given by  1 n + 1, n + 1, n + 1, n + 1 F 4 3 1, 2n + 2, 2n + 2 z n+1

1 z

 (6.6)

(within a multiplicative constant). (n!4 1 If one puts ,R1 (z) = ( ((2n+1)!) 2 z n+1 )r1 (z) and denoting by r1 (z) =

∞ X

cn (1/z)n )

n=0 ∂ P∞ The logarithmic function is given by r2 (z) = ∂t ( k=0 cn+t (1/z)n ) . Let us recall the sketch of the method given in [?] and [?] to find the solutions of this differential equation . They use inverse-Mellin transform i.e..To find the ’Apery’s polynomial’ (given by analytic solution at 0 of (6,3) , one set Z 1 f (z) = g(t)(−z)t dt 2iπ C

and we find a condition on the function g(t) under which this inverse Mellin transform converges and gives a solution of this differential equation . Here C denotes a vertical line with possible deviation to avoid singularities of the integrand. We find that g(t) is a solution of the difference equation g(t + 1) = −g(t)

(t − n)2 (t + n + 1)2 (t + 1)4

We take a path of integration C as mentionned above.Then if lim (t2 g(t)(−z)t ) = 0

τ →∞

20

,(t = σ + iτ ) holds uniformly as t → ∞ in any finite vertical trip , then the integral converges uniformly with respect to z. For instance with Γ(−t)Γ(1 + n + t)2 g(t) = (Γ(1 + t)3 Γ(1 + n − t)2 after simplification, we find: 1 f (z) = 2iπ R(t) = [

Z

R(t)(−z)t dt

C

(t − 1)(t − 2) · · · (t − n) 2 ] (t)(t + 1) · · · (t + n)

We must find the remainder (i.e.) the solution belonging to the exponent n + 1 at ∞. We set u = 1/z , the new equation becomes uθu4 − (θu + n)2 (θu − n − 1)2 = 0 and g(u) =

Γ(u)4 Γ(u)4 Γ(1 + n − u)2 π = ( )2 2 2 2 Γ(u + n + 1) Γ(u − n) Γ(u + n + 1) sin(πu)

We can write f (u) as Z

1 f (u) = 2iπ or

1 f (u) = 2iπ

g(t)(−u)t dt

C

Z R(t)( C

π )2 sin(πt)

C is the vertical line