Rational approximations of values of the Gamma func- tion at - PowerPoint PPT Presentation

Rational approximations of values of the Gamma func- tion at rational points Tanguy Rivoal Institut Fourier, CNRS and Universit´ e Grenoble 1 Conference “Approximation et extrapolation des suites et des s´ eries convergentes et divergentes”, CIRM, 28/09–02/10/2009 1

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 u n lim = α v n n → + ∞ 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 n → + ∞ ( v n α − u n ) = 0 . lim If furthermore ∀ n v n α − u n � = 0, then α �∈ Q . 2

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 approx- imations are solutions of a linear recurrence of finite order with polynomial coefficients . 3

∀ 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

C 3 ( n, α, x ) = − ( n +3) 5 ( n +4) 2 (8 n 2 +4 αn − 3 xn +38 n − 6 x − αx +10 α +44) ( n + 2)(8 n 2 + 22 n + 4 αn − 3 xn + 6 α − 3 x − αx + 14) 2 (8 n 2 + 38 n + 4 αn − 3 xn + 10 α − 6 x − αx + 44) C 2 ( n, α, x ) = (24 n 5 +7 xn 4 +28 αn 4 +330 n 4 − 6 x 2 n 3 +91 xn 3 +1794 n 3 +310 αn 3 +7 αxn 3 +70 xαn 2 +13 xα 2 n 2 − 5 x 2 αn 2 − 4 α 3 n 2 +4824 n 2 − 6 α 2 n 2 − 45 x 2 n 2 +418 xn 2 +1272 αn 2 +576 x − 25 x 2 αn +218 αxn +79 xα 2 n − 26 α 3 n +5 α 3 xn − 4 x 2 α 2 n +816 xn +2296 αn +6420 n − 111 x 2 n − 30 α 2 n +3384+1540 α +576 x +216 αx − α 3 x 2 + 16 α 3 x − 10 x 2 α 2 − 31 x 2 α + 116 xα 2 − 40 α 3 − 90 x 2 − 36 α 2 ) ( n + 3) 4 ( n + 2)( − 8 n 2 αx − 4 αn − 38 n + 3 xn 6 x − 44 − 10 α ) ( − 8 n 2 − 22 n − 4 αn + 3 xn − 14 − 6 α + 3 x + αx ) 2 5

C 1 ( n, α, x ) = − (24 n 4 − 57 xn 3 +20 αn 3 +186 n 3 − 38 xαn 2 +518 n 2 +4 α 2 n 2 +26 x 2 n 2 − 315 xn 2 +120 αn 2 +13 x 2 αn − 148 αxn − 5 xα 2 n − 543 xn +232 αn +610 n +85 x 2 n +14 α 2 n − 3 x 3 n +254+144 α − 285 x − 138 αx − x 3 α + x 2 α 2 +24 x 2 α − 9 xα 2 +59 x 2 +10 α 2 − 3 x 3 )( n + 3) 2 (8 n 2 + 22 n + 4 αn − 3 xn + 14 + 6 α − 3 x − αx ) (8 n 2 + 4 αn + 54 n − 3 xn − αx + 14 α − 9 x + 90)( n − α + 2)( n + 2 + α ) 2 (8 n 2 − 3 xn + 38 αxn + 4 αn + 10 α − 6 x + 44)( n + 2) C 0 ( n, α, x ) = ( n − α +1)( n +1+ α ) 2 (8 n 2 − 3 xn +4 αn +38 nαx +10 α − 6 x +44) 2 ( n +3) 2 (8 n 2 +22 n +4 nα − 3 xn +14+6 α − 3 x − αx )( n − α +2)( n +2+ α ) 2 (8 n 2 − 3 xn + 54 n + 4 αn + 14 α − αx + 90 − 9 x ) . 6

( 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 +(6 α +2 α 2 − 4) x − α 3 − 4 − 9 α − 6 α 2 4 P 2 ( α, x ) = 1 � ( − 3 α + α 2 + 2) x 3 + ( − 3 α 3 + 30 − 9 α 2 ) x 2 36 � +( − 108+33 α 2 +24 α 3 +3 α 4 − 60 α ) x − 12 α 4 − α 5 − 24 − 90 α 2 − 49 α 3 − 76 α Q 1 ( α, x ) = 1 � � (1+ α 2 +2 α ) x 2 +( − 10 α − 6 α 2 − 4) x − 4+4 α 2 Q 0 ( α, x ) = x − 2 , 4 Q 2 ( α, x ) = 1 � ( α 4 +12 α +13 α 2 +4+6 α 3 ) x 3 +( − 60 α 3 − 12 α 4 − 96 α 2 − 48 α ) x 2 72 � + ( − 336 α − 216 + 84 α 3 + 30 α 4 − 66 α 2 ) x + 60 α 2 − 12 α 4 − 48 7

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 2 n +1 b 4 n − 1 P n ( α, x ) ∈ Z , n ! 2 ( n +1)! 2 v 3 n +2 b 4 n − 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 s ( α, x ) − 3 2 x 1 / 3 n 2 / 3 +1 � � Q n ( α, x )Γ( α +1) − P n ( α, x ) x α � � 2 x 2 / 3 n 1 / 3 � � ≤ n 2 − 2Re( α ) / 3 exp � � and Q n ( α, x ) ∼ q ( α, x ) 3 x 1 / 3 n 2 / 3 − x 2 / 3 n 1 / 3 � � n 2 − 2 α/ 3 exp . 8

Example. Define ( p n ) n ≥ 0 and ( q n ) n ≥ 0 to be the solutions of 64(8 n + 17)( n + 4) 2 (8 n + 9)( n + 3) 3 ( n + 2) U n +3 − 16(8 n + 9)( n + 2)(24 n 2 + 123 n + 155)(2 n + 7) 2 ( n + 3) 2 U n +2 +4(2 n +7)( n +2)(8 n +25)(48 n 3 +158 n 2 +147 n +32)(2 n +5) 2 U n +1 − (8 n + 25)(8 n + 17)(2 n + 1)(2 n + 7)(2 n + 5) 2 (2 n + 3) 2 U n = 0 with p 0 = 3 2 , p 1 = 81 32 , p 2 = 2185 384 , q 0 = 1, q 1 = 45 16 , q 2 = 825 128 . Then √ π p n � 3 � lim = Γ = 2 . 2 q n n → + ∞ 9

Idea of the proof. • Euler’s functions: For z �∈ [0 , + ∞ ) and Re( β ) > − 1, set � ∞ ∞ t β e − t Γ( β + k ) � F β ( z ) := z − t d t ∼ . z k 0 k =1 • Laguerre type polynomials of degree 2 n in x : 1 x n − α ( e − x x n + α ) ( n ) � ( n ) ∈ Q [ α, x ] . n ! 2 e x � A n,α ( x ) := They are orthogonal on (0 , + ∞ ) for the two weights e − x and x α e − x . ( α � = 0) 10

e approximants at z = ∞ to F 0 ( z ) and Simultaneous type II Pad´ F α ( z ) . Lemma 1. For β ∈ { 0 , α } , α ∈ C \{ 0 } , Re( α ) > − 1 and z ∈ C \ [0 , + ∞ ) , � ∞ A n,α ( t ) t β e − t d t = A n,α ( z ) F β ( z ) − B n,α,β ( z ) R n,α,β ( z ) := z − t 0 ∞ ( k − β − n ) n ( k − α + β − n ) n · Γ( β + k ) 1 � � � ∼ = O . n ! 2 z k z n +1 k =1 Here, � ∞ A n,α ( z ) − A n,α ( t ) t β e − t d t ∈ Γ(1 + β ) Q [ α, z ] B n,α,β ( z ) := z − t 0 is of degree at most 2 n − 1 in z . 11

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 � ∞ z α − t α e − t d t, F ( z ) := z α F 0 ( z ) − F α ( z ) = z − t 0 � ∞ z α − t α A n,α ( t ) e − t d t, R n,α ( z ) := z α R n,α, 0 ( z ) − R n,α,α ( z ) = z − t 0 which are both analytic on C \ ( −∞ , 0]. 12

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 � � A n,α ( z ) B n,α, 0 ( z ) � � P n ( α, z ) = � ∈ Q [ α, z ] � � A n +1 ,α ( z ) B n +1 ,α, 0 ( z ) � � � � � A n,α ( z ) C n,α ( z ) � � Q n ( α, z ) = � ∈ Q [ α, z ] , � � A n +1 ,α ( z ) C n +1 ,α ( z ) � � � � � A n,α ( z ) R n,α ( z ) � � S n ( α, z ) = � . � � A n +1 ,α ( z ) R n +1 ,α ( z ) � � � 13

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 recur- rence (1) . ( iii ) The degrees of Q n and P n in z are at most 4 n − 1 , those in α at most 3 n + 2 and 2 n + 1 respectively. 14

( 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

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 · x 1 / 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( − 3 x 1 / 3 n 2 / 3 + x 2 / 3 n 1 / 3 ) . 16

( i ) follows from Birkhoff-Trjitzinsky theory applied to the recurrence R . ( ii ) is a consequence of the identity R n,α ( x ) = � ∞ � ∞ u n t n + α e − t αx α (1 − α ) n (1 + α ) n (1 + u ) n +1 − α ( x + ut ) n +1+ α d t d u n ! 2 0 0 which implies that R n,α ( x ) → 0 when n → 0, and then we apply Birkhoff-Trjitzinsky theory again to R . 17

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, s ( α, x ) − 3 2 x 1 / 3 n 2 / 3 +1 � � Q n ( α, x )Γ( α +1) − P n ( α, x ) x α � � 2 x 2 / 3 n 1 / 3 � n 2 − 2Re( α ) / 3 exp � ≤ , � � Q n ( α, x ) ∼ q ( α, x ) 3 x 1 / 3 n 2 / 3 − x 2 / 3 n 1 / 3 � � n 2 − 2 α/ 3 exp . 18