Arithmetical applications of lagrangian interpolation Tanguy Rivoal Institut Fourier CNRS and Université de Grenoble Conference Diophantine and Analytic Problems in Number Theory, The 00th anniversary of Gel fond, Moscow, 29/0/2007 02/02/2007
Polynomial Interpolation Hermite s identity : For x, z, α,..., α n C, x z = n (z α )(z α 2 ) (z α j ) (x α )(x α 2 ) (x α j+ ) + (z α )(z α 2 ) (z α n ) (x α )(x α 2 ) (x α n ) x z, provided the denominators do not vanish. Newton interpolation formula : Let F be holomorphic in a simply connected domain Ω and α,..., α n Ω. Then F (z) = with A j = and n C A j (z α )(z α 2 ) (z α j )+R n (z) F (x) (x α )(x α 2 ) (x α j+ ) dx R n (z) = F (x) (z α )(z α 2 ) (z α n ) C (x z) (x α )(x α 2 ) (x α n ) dx, where the simple curve C Ω encloses z, α,..., α n. 2
Gel fond s theorem (929). The number e π is transcendental over Q. Sketch of Gel fond s own proof. Expand exp(πz) in a Newton interpolation series at the points of Z[i] ordered by increasing modulus and arguments z 0 = 0, z, z 2, etc. Fact. For all z C, exp(πz) = + j= A j (z z 0 )(z z ) (z z j ). = A n 0 for infinitely many n. Fact 2. A n = n k=0 0 j n j k e πz k (z k z j ) = n k=0 e πz k ω n,k = P n (e π ), with P n (X) Q(i)[X, X ] of degree n/π + o( n). 3
Fact 3. (Gel fond) Let Ω n = lcm{ω n,0,..., ω n,n } : ( ) Ω n exp n log(n) + 63n + o(n). 2 Fact 4. (Fukasawa) ω n,k = exp ( n log(n) + βn + o(n) 2 Fact 3 and Fact 4 = H(Ω n P n ) = exp(o(n)). Fact 5. From the integral expression for A n : P n (e π ) exp( n log(n) + O(n)). ). Transcendence criteria. Let θ C, Q n (X) Z[i][X, X ] and λ n + s.t. Then θ Q. 0 < Q n (θ) exp( λ n ), deg(q n ) + log H(Q n ) = o(λ n ). Conclusion with θ = e π, Q n (X) = Ω n P n (X), deg(q n ) n, log H(Q n ) n and λ n = n log(n)/2 + O(n). 4
Rational Interpolation René Lagrange s formula (935). Consider z, α,..., α n+, β,..., β n C, γ n = α n+ β n, where γ n (x β ) (x β n ) = (n = 0) (x β ) (x β n ) = (n = ). Then x z = n (x β ) (x β j ) (z α ) (z α j ) γ j (x α ) (x α j+ ) (z β ) (z β j ) + (x β ) (x β n ) (z α ) (z α n+ ) (x α ) (x α n+ ) (z β ) (z β n ) x z, provided the denominators do not vanish. 5
Lagrange s interpolation formula. Let F be a holomorphic function in a simply connected domain Ω and α,..., α n+ Ω. Then for any z Ω, z β,..., β n : F (z) = n A j (z α )(z α 2 ) (z α j ) (z β )(z β 2 ) (z β j ) + R n(z) where A j = γ j C F (x) (x β ) (x β j ) (x α ) (x α j+ ) dx and R n (z) = (z α ) (z α n+ ) (z β ) (z β n ) C F (x) x z (x β ) (x β n ) (x α ) (x α n+ ) dx, where the simple curve C Ω encloses z and α,..., α n+. 6
Arithmetical applications of Lagrange s formulae Rational interpolation is adaptated to functions with poles at β, β 2,..., β n, etc. The coefficients A n are linear forms in the numbers (F (j) (α n )) j 0,n. Example. Hurwitz zeta function ζ(s, z) = k= (k + z) s, where s N, s 2, z C \ {, 2,...}. a z aζ(s, n) = ( )a (s) a ( ζ(s + a) n k= ) k s+a Qζ(s + a) + Q. By definition : (x) 0 = and, for n, (x) n = x(x + ) (x + n ). 7
A new proof of Apéry s theorem. The irrationality of ζ(3) is a consequence of the following. Theorem (R, 2006). For all z C\{, 2,...}, we have ζ(2, z) = n=0 A 2n (z n + ) 2 n (z + ) 2 n + n=0 A 2n+ (z n + ) 2 n (z + ) 2 n z n z + n +, where A 0 = ζ(2) and, for all n 0, A 2n+ = 2n + C n (x + ) 2 n (x n) 2 ζ(2, x) dx n+ Qζ(3) + Q and similarly A 2n+2 Qζ(3) + Q. The simple curve C n encloses 0,,..., n but none of the poles of ζ(2, z). Sketch of the proof. We apply Lagrange s interpolation formulae with α 2n+ = α 2n+2 = n, β 2n+ = β 2n+2 = (n+). 8
The remainder has two complicated different expressions, depending on the parity of n : R 2n (z) = (z n + )2 n (z n) (z + ) 2 n C ζ(2, x) x z (x + ) 2 n (x n + ) 2 dx. n (x n) Fact. The function ζ(2, x)(x + ) 2 n poles at x =, 2,..., n. has no Fact 2. For all x C \ Z, ζ(2, x) + ζ(2, x) = π2 sin(πx) 2. Consequence. For R(z) > 0 and any κ ]0, [, we have R 2n (z) = (z n + )2 n (z n) (z + ) 2 n n2 κ i κ+i t + tn + z Γ(nt) 4 Γ(n nt + ) 2 Γ(n + nt + ) 2 dt. What is the best value of κ? 9
From Stirling s formula : R 2n (z) min κ ]0,[ κ2κ ( κ) κ ( + κ) +κ ( 2 ) 4n+o(n). ) 2n+o(n) Same bound for R 2n+ (z), A 2n and A 2n+2. We deduce the normal convergence on any compact of R(z) > 0 of both series in the theorem and even on any compact at positive distance from, 2, etc. Hence, the identity holds on C \ {, 2,...} by analytic continuation. Proof of Apéry s theorem. We have A 2n+2 2n + 2 = 2 n ( n j ) 2 ( n + j n )2 ζ(3) + v n where d 3 n v n Z and d n = lcm{, 2,..., n}. Similarly, A 2n+ 2n + = ũ nζ(3) + ṽ n where ũ n Z and d 3 nṽ n Z. 0
For all n : d 3 n A n = U n ζ(3) + V n Zζ(3) + Z. Since ζ(2, z) is not a rational function, its lagrangian expansion implies that d 3 n A n 0 for infinitely many n. Finally, since lim sup n + d 3 n A n /n e 3 ( 2 ) 4 <, the irrationality of ζ(3) follows. A similar proof of the irrationality of log(2) follows from the expansion where k= B n+ 2n + = ( ) k k + z = n n= B n (z n + 2) n (z + ) n, ( ) n j( n )( n + j j j ) k= ( ) k k + j Q log(2) + Q. What about ζ(2)?
A new type of polynomial-rational interpolation Proof of Hermite s identity. Start with x z = x α + z α x αx z. H(α) In H(α ), replace the final /(x z) by the RHS of H(α 2 ), replace the final /(x z) by the RHS of H(α 3 ) and so on. Proof of Lagrange s identity. Start with x z = α β (x α)(z β) + x β z α. L(α, β) x αz αx z In H(α ), replace the final /(x z) by the RHS of L(α 2, β ), replace the final /(x z) by the RHS of L(α 3, β 2 ), replace the final /(x z) by the RHS of L(α 4, β 2 ) and so on. 2
We can obtain infinitely many new infinite expansions of /(x z) by mixing H(α) and L(δ, β). The sequence H(α ), L(α 2, β ), H(α 3 ), L(α 4, β 2 ), H(α 5 ), L(α 6, β 3 ), H(α 6 ), L(α 8, β 4 ), etc, leads to the following. Proposition (R, 2006). For an integer n 0, let n = n or n. We have x z = n n (x β ) (x β j ) (x α ) (x α 2j+ ) + (z α ) (z α 2j ) (z β ) (z β j ) (x β ) (x β j ) (z α ) (z α 2j+ ) ω j (x α ) (x α 2j+2 ) (z β ) (z β j+ ) with ω j = α 2j+2 β j+, and + S n,n, S n,n = (x β ) (x β n ) (z α ) (z α 2n+ ) (x α ) (x α 2n+ ) (z β ) (z β n ) x z, S n,n = (x β ) (x β n+ ) (z α ) (z α 2n+2 ) (x α ) (x α 2n+2 ) (z β ) (z β n+ ) x z. 3
Arithmetical application. The irrationality of ζ(2) is a corollary of the following result. Theorem (R, 2006). For all z C\{, 2,...}, we have n= ( n ) n + z = n=0 A n (z n + ) 2 n (z + ) n + n=0 B n (z n + ) 2 n (z + ) n z n z + n +, where A 0 = B 0 = 0 and, for all n, A n = C n (x + ) n (x n) (x n) 2 n+ ζ(, x) dx Qζ(2) + Q and B n = 2n C n (x + ) n (x n) 2 n+ ζ(, x) dx Qζ(2) + Q. The simple curve C n encloses 0,,..., n but none of, 2, etc. 4