Rational approximations of values of the Gamma function at rational points

Transcription

1 Rational approximations of values of the Gamma function at rational points Tanguy Rivoal Institut Fourier, CNRS and Université Grenoble 1 Conference Approximation et extrapolation des suites et des séries convergentes et divergentes, CIRM, 28/09 02/10/2009 1

2 Computation of a given a real number α. 1) Numerical point of view: find a sequence of rational numbers ( u n v n ) n 0 such that lim n + u n v n = α with fast convergence and u n /v n reasonably easy to compute. 2) Diophantine point of view: find two sequences of integers (u n ) n 0 and (v n ) n 0 such that lim (v nα u n ) = 0. n + If furthermore n v n α u n 0, then α Q. 2

3 Aim of this talk: to present rational approximations of the numbers Γ(a/b), where a/b Q and Γ denotes the usual Gamma function. A similar method enables us to get rational approximations of the numbers γ + log(x), where γ is Euler s constant and x > 0, x Q. None of these approximations are good enough to satisfy 2). From the point of view of 1), they are new. The rational approximations are solutions of a linear recurrence of finite order with polynomial coefficients. 3

4 x > 0 and α C, consider the linear recurrence of order 3: C 3 (n, α, x)u n+3 +C 2 (n, α, x)u n+2 +C 1 (n, α, x)u n+1 +C 0 (n, α, x)u n = 0 (1) where the coefficients C j (n, α, x), j = 0, 1, 2, 3, are polynomials in n, α, x, of degree 16 in n, whose expressions are 4

5 C 3 (n, α, x) = (n+3) 5 (n+4) 2 (8n 2 +4αn 3xn+38n 6x αx+10α+44) (n + 2)(8n n + 4αn 3xn + 6α 3x αx + 14) 2 (8n n + 4αn 3xn + 10α 6x αx + 44) C 2 (n, α, x) = (24n 5 +7xn 4 +28αn n 4 6x 2 n 3 +91xn n αn 3 +7αxn 3 +70xαn 2 +13xα 2 n 2 5x 2 αn 2 4α 3 n n 2 6α 2 n 2 45x 2 n xn αn x 25x 2 αn+218αxn+79xα 2 n 26α 3 n+5α 3 xn 4x 2 α 2 n +816xn+2296αn+6420n 111x 2 n 30α 2 n α+576x+216αx α 3 x α 3 x 10x 2 α 2 31x 2 α + 116xα 2 40α 3 90x 2 36α 2 ) (n + 3) 4 (n + 2)( 8n 2 αx 4αn 38n + 3xn6x 44 10α) ( 8n 2 22n 4αn + 3xn 14 6α + 3x + αx) 2 5

6 C 1 (n, α, x) = (24n 4 57xn 3 +20αn n 3 38xαn n 2 +4α 2 n 2 +26x 2 n 2 315xn αn 2 +13x 2 αn 148αxn 5xα 2 n 543xn+232αn+610n+85x 2 n+14α 2 n 3x 3 n α 285x 138αx x 3 α+x 2 α 2 +24x 2 α 9xα 2 +59x 2 +10α 2 3x 3 )(n + 3) 2 (8n n + 4αn 3xn α 3x αx) (8n 2 + 4αn + 54n 3xn αx + 14α 9x + 90)(n α + 2)(n α) 2 (8n 2 3xn + 38αxn + 4αn + 10α 6x + 44)(n + 2) C 0 (n, α, x) = (n α+1)(n+1+α) 2 (8n 2 3xn+4αn+38nαx+10α 6x+44) 2 (n+3) 2 (8n 2 +22n+4nα 3xn+14+6α 3x αx)(n α+2)(n+2+α) 2 (8n 2 3xn + 54n + 4αn + 14α αx x). 6

7 (P n (x, α)) n 0 and (Q n (x, α)) n 0 solutions of (1), with initial values: P 0 (α, x) = x α 2, P 1 (α, x) = 1 ) ((1 α)x 2 +(6α+2α 2 4)x α 3 4 9α 6α 2 4 P 2 (α, x) = 1 36 ( ( 3α + α 2 + 2)x 3 + ( 3α α 2 )x 2 +( α 2 +24α 3 +3α 4 60α)x 12α 4 α α 2 49α 3 76α ) Q 0 (α, x) = x 2, Q 1 (α, x) = 1 4 ((1+α 2 +2α)x 2 +( 10α 6α 2 4)x 4+4α 2 ) Q 2 (α, x) = 1 ( (α 4 +12α+13α α 3 )x 3 +( 60α 3 12α 4 96α 2 48α)x ( 336α α α 4 66α 2 )x + 60α 2 12α 4 48 ) 7

8 Theorem 1 (R, 2009). (i) P n (α, x), Q n (α, x) Q[α, x]. (ii) α = u/v Q and x = a/b Q, n! 2 (n+1)! 2 v 2n+1 b 4n 1 P n (α, x) Z, n! 2 (n+1)! 2 v 3n+2 b 4n 1 Q n (α, x) Z. (iii) x > 0 and α C\{0}, Re(α) > 1, s(α, x) 0 and q(α, x) 0 such that Q n (α, x)γ(α+1) P n (α, x)x α s(α, x) n 2 2Re(α)/3 exp ( 3 2 x1/3 n 2/ x2/3 n 1/3 and Q n (α, x) q(α, x) n 2 2α/3 exp ( 3x 1/3 n 2/3 x 2/3 n 1/3 ). ) 8

9 Example. Define (p n ) n 0 and (q n ) n 0 to be the solutions of 64(8n + 17)(n + 4) 2 (8n + 9)(n + 3) 3 (n + 2)U n+3 16(8n + 9)(n + 2)(24n n + 155)(2n + 7) 2 (n + 3) 2 U n+2 +4(2n+7)(n+2)(8n+25)(48n n n+32)(2n+5) 2 U n+1 (8n + 25)(8n + 17)(2n + 1)(2n + 7)(2n + 5) 2 (2n + 3) 2 U n = 0 with p 0 = 3 2, p 1 = 81 32, p 2 = , q 0 = 1, q 1 = 45 16, q 2 = 825 lim n + p n q n = Γ ( ) 3 2 = π Then 9

10 Idea of the proof. Euler s functions: For z [0, + ) and Re(β) > 1, set F β (z) := 0 t β e t z t dt k=1 Γ(β + k) z k. Laguerre type polynomials of degree 2n in x: A n,α (x) := 1 ( n! 2 ex x n α (e x x n+α ) (n)) (n) Q[α, x]. They are orthogonal on (0, + ) for the two weights e x and x α e x. (α 0) 10

11 Simultaneous type II Padé approximants at z = to F 0 (z) and F α (z). Lemma 1. For β {0, α}, α C\{0}, Re(α) > 1 and z C\[0, + ), R n,α,β (z) := 0 A n,α (t) z t t β e t dt = A n,α (z)f β (z) B n,α,β (z) k=1 (k β n) n (k α + β n) n n! 2 Γ(β + k) z k ( ) 1 = O z n+1. Here, 0 A n,α (z) A n,α (t) B n,α,β (z) := z t is of degree at most 2n 1 in z. t β e t dt Γ(1 + β)q[α, z] 11

12 Crucial fact: B n,α,0 (z) Q[α, z] and B n,α,α (z) = Γ(1 + α)c n,α (z) for some C n,α (z) Q[α, z]. For β C \ {0}, take z β to be given by its principal value whenever π < arg(z) < π. Set F (z) := z α F 0 (z) F α (z) = 0 z α t α z t e t dt, R n,α (z) := z α R n,α,0 (z) R n,α,α (z) = 0 z α t α z t A n,α (t)e t dt, which are both analytic on C \ (, 0]. 12

13 Obviously, R n,α (z) = A n,α (z)f (z) + Γ(1 + α)c n,α (z) z α B n,α,0 (z). To remove the term A n,α (z)f (z), set P n (α, z) = A n,α (z) A n+1,α (z) B n,α,0 (z) B n+1,α,0 (z) Q[α, z] Q n (α, z) = A n,α (z) A n+1,α (z) C n,α (z) C n+1,α (z) Q[α, z], S n (α, z) = A n,α (z) A n+1,α (z) R n,α (z) R n+1,α (z). 13

14 Lemma 2. (i) For any z C \ (, 0], we have S n (α, z) = Q n (α, z)γ(1 + α) z α P n (α, z) where n! 2 (n + 1)! 2 P n (α, z) Z[α, z], n! 2 (n + 1)! 2 Q n (α, z) Z[α, z]. (ii) P n (α, z), Q n (α, z) and S n (α, z) are solutions of the linear recurrence (1). (iii) The degrees of Q n and P n in z are at most 4n 1, those in α at most 3n + 2 and 2n + 1 respectively. 14

15 (i) and (iii) are easy. (ii) is consequence of the facts that the sequences A n,α, B n,α,0, B n,α,α, R n,α,0, R n,α,α (hence R n,α ) are all solutions of the same linear recurrence of order 3, say R, obtained explicitly using Zeilberger s algorithm EKHAD. that if a n and b n are solutions of a recurrence of order 3: U n+3 = p n U n+2 + q n U n+1 + r n U n, then the sequence of determinants a n b n+1 a n+1 b n is solution of the linear recurrence also of order 3: U n+3 = q n+1 U n+2 p n r n+1 U n+1 + r n r n+1 U n. 15

16 Final steps. Lemma 3. (i) x > 0, the modulus of A n,α, B n,α,0, B n,α,α are bounded by u(x, α) n 1 Re(α)/3 exp(3/2 x1/3 n 2/3 1/2 x 2/3 n 1/3 ), where u(x, α) depends on the sequence. (ii) x > 0, r(x, α) 0 s.t. R n,α (x) r(x, α) n 1 α/3 exp( 3x1/3 n 2/3 + x 2/3 n 1/3 ). 16

17 (i) follows from Birkhoff-Trjitzinsky theory applied to the recurrence R. (ii) is a consequence of the identity R n,α (x) = αx α (1 α) n(1 + α) n n! u n t n+α e t (1 + u) n+1 α (x + ut) n+1+α dtdu which implies that R n,α (x) 0 when n 0, and then we apply Birkhoff-Trjitzinsky theory again to R. 17

18 Lemma 3 (both (i) and (ii)) implies that S n (α, x) 0 when n +. Apply again Birkhoff-Trjitzinsky theory to recurrence (1) to get the bounds in Theorem 1, ie., x > 0, Q n (α, x)γ(α+1) P n (α, x)x α s(α, x) n 2 2Re(α)/3 exp ( 3 2 x1/3 n 2/ x2/3 n 1/3 ), Q n (α, x) q(α, x) n 2 2α/3 exp ( 3x 1/3 n 2/3 x 2/3 n 1/3 ). 18

19 Results related to Euler s constant γ. Theorem 2 (R, 2008, Transactions AMS). There exists two polynomial solutions P n (z) and Q n (z) of a linear recurrence of order 3, of degree at most n + 1 such that (i) n! 2 P n (z) and n! 2 Q n (z) belong to Z[z]. (ii) x > 0, s(x) 0 and q(x) 0 s.t. Q n (x) ( ln(x) + γ ) P n (x) s(x) n 2 exp( 3 2 x1/3 n 2/ x2/3 n 1/3 ), Q n (x) q(x) n 2 exp(3x1/3 n 2/3 x 2/3 n 1/3 ). 19

20 Examples The recurrence (n + 3) 2 (8n + 11)(8n + 19)U n+3 = (24n n + 215)(8n + 11)U n+2 (24n n n + 25)(8n + 27)U n+1 + (n + 2) 2 (8n + 19)(8n + 27)U n provides two sequences of rational numbers (p n ) n 0 and (q n ) n 0 with p 0 = 1, p 1 = 4, p 2 = 77/4 and q 0 = 1, q 1 = 7, q 2 = 65/2 such that p n q n γ. The first such recurrence for γ was obtained by Aptekarev in

21 The recurrence (n + 1)(n + 2)(n + 3)U n+3 = (3n n + 29)(n + 1)U n+2 (3n 3 + 6n 2 7n 13)U n+1 + (n + 2) 3 U n provides two sequences of rational numbers (p n ) n 0 and (q n ) n 0 with p 0 = 1, p 1 = 11, p 2 = 71 and q 0 = 0, q 1 = 8, q 2 = 56 such that p n q n ln(2) + γ. 21

22 The construction is based on Euler type functions 0 log(t) s e t z t dt k=1 Γ (s) (k) z k. Laguerre type polynomials with α = 0: A n,0 (x) = 1 ( x n (e x x n ) (n)) (n) Q[x], n! 2 ex orthogonal on (0, + ) for the weigths e x and log(x)e x. (Remember that the case α = 0 was excluded.) This construction can be viewed as a limiting case of our results for Γ(1 + α) because lim α 0 Γ(1 + α) 1 α = Γ (1) = γ. 22