Linear Differential Equations as a Data-Structure

Transcription

1 Linear Differential Equations as a Data-Structure Bruno Salvy Inria & ENS de Lyon FoCM, July 14, 2017

2 Computer Algebra Effective mathematics: what can we compute exactly? And complexity: how fast? (also, how big is the result?) Systems with several million users 50+ years of algorithmic progress in computational mathematics! 1/26

3 Sources of Linear Differential Equations Generating functions in combinatorics M. Bousquet-Mélou Periods Classical elementary and special functions (small order) A. Bostan P. Lairez 2/26

4 LDEs as a Data-Structure Polynomial equations Numerical evaluation Diagonals Definite sums and integrals Linear Differential Equations Conversions Local and asymptotic expansions Proofs of identities Closed forms M. Singer Solutions called differentially finite (abbrev. D-finite) 3/26

5 A. Using Linear Differential Equations Exactly

6 A. Using Linear Differential Equations Exactly I. Numerical Values

7 Fast Computation with Linear Recurrences (70 s and 80 s) 1. Multiplication of integers is fast (Fast Fourier Transform): 2. n! millions of digits 1sec. 3. Linear recurrence: convert into 1st order recurrence on vectors and apply the same idea. Ex: in complexity O(n) by divide-and-conquer n! :=n dn/2e {z } size O(n log n) e n := nx k=0 1 k! satisfies a 2nd order rec, computed via en e n 1 = 1 n (dn/2e 1) 1 {z } size O(n log n) n n {z 0 } A(n) en 1 e n 2 Conclusion: Nth element in O(N) ops. = 1 n! A!(n) 1 0 Notation: O(n) means O(n log k n) for some k. 4/26

8 Numerical evaluation of solutions of LDEs Principle: f(x) = NX n=0 a n x n {z } fast evaluation + 1X n=n+1 a n x n {z } good bounds f solution of a LDE with coeffs in Q(x) 1. linear recurrence in N for the first sum (easy); 2. tight bounds on the tail (technical); 3. extend to C by analytic continuation. Computation on integers. No roundoff errors. arctan(1+i) Conclusion: value anywhere with N digits in Õ(N) ops. Sage code available M. Mezzarobba [Chudnovsky-Chudnovsky87;van der Hoeven99;Mezzarobba-S.10;Mezzarobba16] 5/26

9 A. Using Linear Differential Equations Exactly II. Local and Asymptotic Expansions

10 Dynamic Dictionary of Mathematical Functions User need Recent algorithmic progress Maths on the web [Benoit-Chyzak-Darrasse-Gerhold-Mezzarobba-S.2010] 6/26

11 A. Using Linear Differential Equations Exactly III. Proofs of Identities

12 Proof technique > series(sin(x)^2+cos(x)^2-1,x,4); f satisfies a LDE f,f,f, live in a finite-dim. vector space O(x 4 ) Why is this a proof? 1. sin and cos satisfy a 2nd order LDE: y +y=0; 2. their squares and their sum satisfy a 3rd order LDE; 3. the constant -1 satisfies y =0; 4. thus sin 2 +cos 2-1 satisfies a LDE of order at most 4; 5. the Cauchy-Lipschitz theorem concludes. Proofs of non-linear identities by linear algebra! 7/26

13 Mehler s identity for Hermite polynomials 1X n=0 H n (x)h n (y) un n! = exp 4u(xy u(x 2 +y 2 )) 1 4u 2 p 1 4u 2 1. Definition of Hermite polynomials: recurrence of order 2; 2. Product by linear algebra: H n+k (x)h n+k (y)/(n+k)!, k N generated over H n (x)h n (y) n! Q (x,n) by, H n+1(x)h n (y) n!, H n(x)h n+1 (y) n! recurrence of order at most 4; 3. Translate into differential equation., H n+1(x)h n+1 (y) n! 8/26

14 Guess & Prove Continued Fractions 1. Taylor expansion produces first terms (easy): arctan x = x 1 3 x x x Guess a formula (easy): a n = n2 4n Prove that the CF with these a n converges to arctan: show that H n := Q 2 n (x 2 + 1)(P n /Q n ) 0 1 = O(x n ) where P n /Q n is the nth convergent. Algo compute a LRE for H n and simplify it. gfun[contfrac] No human intervention needed. 9/26

15 It Works! This method has been applied to all explicit C-fractions in Cuyt et alii, starting from either: a Riccati equation: a q-riccati equation: y 0 = A(z)+B(z)y + C(z)y 2 y(qz) =A(z)+B(z)y(z)+C(z)y(z)y(qz) a difference Riccati equation: y(s + 1) = A(s) + B(s)y(s) + C(s)y(s)y(s + 1) It works in all cases, including Gauss s CF, Heine s q- analogue and Brouncker s CF for Gamma. In all cases, H n satisfies a recurrence of small order. In progress: 1. explain why this method works so well, [Maulat-S. 15,17] 2. classify the formulas it yields. 10/26

16 B. Conversions (LDE LDE)

17 From equations to operators D x d/dx x mult by x product composition D x x=xd x +1 S n (n n+1) n mult by n product composition S n n=(n+1)s n Ex. (erf): Taylor morphism: D x (n+1)s n ; x S n -1 produces linear recurrence from LDE D 2 x +2xD x 7! (n + 1)S n (n + 1)S n +2S 1 n (n + 1)S n =(n + 1)(n + 2)S 2 n +2n 11/26

18 Chebyshev expansions arctan Taylor Chebyshev z 1 3 z z5 + 2( p T1 (x) 2 + 1) (2 p 2 + 3) c k = 2 Z 1 T 3 (x) 3(2 p 2 + 3) + T 5 (x) 2 5(2 p 2 + 3) f(x)t k (x) p 1 x 2 dx 12/26

19 Ore fractions Generalize commutative case: R=Q -1 P with P & Q operators. B -1 A=D -1 C when ba=dc with bb=dd=lclm(b,d). Algorithms for sum and product: B -1 A+D -1 C=LCLM(B,D) -1 (ba+dc), with bb=dd=lclm(b,d) B -1 AD -1 C=(aB) -1 dc, with aa=dd=lclm(a,d). [Ore1933] 13/26

20 Application: Chebyshev expansions Taylor x n+1 =x x n x X:=S -1 (x n ) =nx n-1 d/dx D:=(n+1)S Chebyshev 2xT n (x)=t n+1 (x)+t n-1 (x) x X:=(S n +S n -1 )/2 2(1-x 2 )T n (x)=-nt n+1 (x)+nt n-1 (x) d/dx D:=(1-X 2 ) -1 n(s n -S n -1 )/2. Prop. If y is a solution of L(x,d/dx), then its Chebyshev coefficients annihilate the numerator of L(X,D). > deqarctan:=(x^2+1)*diff(y(x),x)-1: > diffeqtogfsrec(deqarctan,y(x),u(n),functions=chebyshevt(n,x)); nu(n) + 6(n + 2)u(n + 2) + (n + 4)u(n + 4) Applications to Validated Numerical Approximation [Benoit-S.09;Benoit12;BenoitJoldesMezzarobba17] M. Joldes 14/26

21 C. Computing Linear Differential Equations (Efficiently)

22 C. Computing Linear Differential Equations (Efficiently) I. Algebraic Series and Questions of Size

23 Algebraic Series can be Computed Fast P (X, Y (X)) = 0 P irreducible Wanted: the first N Taylor coefficients of Y. P x (X, Y (X)) + P y (X, Y (X)) Y 0 (X) =0 Y 0 (X) =( P x P 1 y mod P )(X, Y (X)) a polynomial Note: F sol LDE F(Y(X)) sol LDE (same argument) Y (X),Y 0 (X),Y 00 (X),...in Vect Q(X) (1,Y,Y 2,...) finite dimension (deg P) a LDE by linear algebra [Abel1827;Cockle1861;Harley1862;Tannery1875] 15/26

24 degree O(D^3) Order-Degree Curve The cost of minimality minimal LDE nice recurrence minimal recurrence degree D=deg P O(D^2) O(D^2) O(D^2) O(D) O(D) O(D) O(D) O(D^2) order O(D^2) O(D^2) O(D^3) order differential equations corresponding recurrences [Bostan-Chyzak-Lecerf-S.-Schost07;Chen-Kauers12;Chen-Jaroschek-Kauers-Singer13] 16/26

25 C. Computing Linear Differential Equations (Efficiently) II. Creative Telescoping

26 Examples X j,k nx 2 2 n n + k nx kx 3 n n + k k = k k k k j k=0 k=0 j=0 j + k r n s + n j k n + r s r ( 1) j+k =( 1) l k + l j k m j n + l m n l Z i xj 1 (ax)i 1 (ax)y 0 (x)k 0 (x) dx = ln(1 a4 ) 2 a 2 I (1 + 2xy + 4y 2 )exp 4x 2 y 2 1+4y 2 dy = H n(x) y n+1 (1 + 4y 2 ) 3 2 bn/2c! Aims: nx k=0 q k2 (q; q) k (q; q) n k = nx k= n ( 1) k q (5k2 k)/2 (q; q) n 1. Prove them automatically 2. Find the rhs given the lhs k (q; q) n+k Note: at least one free variable First: find a LDE (or LRE) 17/26

27 I(x) = Z Creative telescoping f(x, t) dt =? or U(n) = X k u(n, k) =? Input: equations (differential for f or recurrence for u). Output: equations for the sum or the integral. Ex.: U n := X k U n+1 2U n = X k n k n +1 k n 2 k = X k n +1 n +1 k k +1 {z } telescopes + n n k +1 k {z } telescopes Aim: find A(n,S n ) and B(n,k,S n,s k ) such that (A(n,S n )+Δ k B(n,k,S n,s k )) u(n,k)=0, Def: k: =S k -1. certificate then the sum telescopes, leading to A(n,S n ) U(n)=0. Integrals: differentiate under the sign and integrate by parts. 18/26

28 Telescoping Ideal T t (f) := Ann f t Q(x,t)h@ x,@ t i {z } int. by parts (certificate) \ Q(x)h@ x i. {z } di. under R First generation of algorithms relying on holonomy Restrict int. by parts to Q(x)h@ x,@ t i and use elimination. Second generation: faster using better certificates & algorithms Hypergeometric summation: dim=1 + param. Gosper. Undetermined coefficients in finite dim, Ore algebras & GB. Idem in infinite dim. X X c k (x)@ x t a i,j (x, t)@ x@ i j t 2 Ann f k i,j2s t z z F. Chyzak [Zeilberger et alii 90,91,92;Chyzak00;Chyzak-Kauers-S.09] 28 x x x y y 19/26

29 C. Computing Linear Differential Equations (Efficiently) III. 3rd Generation Creative Telescoping

30 Certificates are big n n + s n + r 2n r s C n := X r,s n ( 1) n+r+s r s s r n {z } f n,r,s (n + 2) 3 C n+2 2(2n + 3)(3n 2 + 9n + 7)C n+1 (4n + 3)(4n + 4)(4n + 5)C n = 180 kb ' 2 pages I(z) = I (1 + t 3 ) 2 dt 1 dt 2 dt 3 t 1 t 2 t 3 (1 + t 3 (1 + t 1 ))(1 + t 3 (1 + t 2 )) + z(1 + t 1 )(1 + t 2 )(1 + t 3 ) 4 z 2 (4z + 1)(16z 1)I 000 (z)+3z(128z z 1)I 00 (z)+(444z z 1)I 0 (z)+2(30z + 1)I(z) =1 080 kb ' 12 pages 3rd-generation algorithms: avoid computing the certificate 20/26

31 Periods I(t) = N := degx Q, I P(t, x) dx m Q (t, x) {z } 2Q(t,x) Q square-free Int. over a cycle where Q 0. x = (x1,..., xn ) dt := max(degt Q, degt P) degxp not too big Thm. A linear differential equation for I(t) can be computed in O(e3nN8ndt) operations in ℚ. It has order Nn and degree O(enN3ndt). tight Note: generically, the certificate has at least N n2/2 monomials. Applications to diagonals & to multiple binomial sums. [Picard1902;Dwork62;Griffiths69;Christol85;Bostan-Lairez-Salvy13;Lairez16] 21/26

32 Diagonals in this talk If F (z) = G(z) H(z) its diagonal is 1 k +1 2k k 2k k Apéry s a k : : : is a multivariate rational function with Taylor expansion F (z) = X i2n n c i z i, F (t) = X k2n c k,k,...,k t k. 1 1 x y =1+x + y +2xy + x2 + y x 2 y x (1 x y)(1 x) =1+y+1xy x2 +y x 2 y t(1 + x)(1 + y)(1 + z)(1 + y + z + yz + xyz) =1+ +5xyzt + Christol s conjecture: All differentially finite power series with integer coefficients and radius of convergence in (0, ) are diagonals. 22/26

33 Diagonals are Differentially Finite [Christol84,Lipshitz88] F (z 1,...,z d )= d 1 2 i 1 I F t dz2,z 2,...,z d z 2 z d z 2 dz d z d Thm. If F has degree d in n variables, ΔF satisfies a LDE with order d n, coeffs of degree d O(n). D-finite diag. alg. rat. Univariate power series + algo in ops. Õ(d 8n ) [Bostan-Lairez-S.13;Lairez16] 23/26

34 Multiple Binomial Sums Ex. S n = X r 0 X n ( 1) n+r+s r s 0 n s n + s n + r 2n r s s r n Thm. Diagonals binomial sums with 1 free index. > BinomSums[sumtores](S,u): ( ) defined properly 1 1 t(1 + u 1 )(1 + u 2 )(1 u 1 u 3 )(1 u 2 u 3 ) has for diagonal the generating function of S n LDE LRE [Bostan-Lairez-S.17] 24/26

35 (Non-)Commercial Algorithmes E caces en Calcul Formel z Alin Bostan Frédéric Chyzak Marc Giusti Romain Lebreton Grégoire Lecerf Bruno Salvy Éric Schost New book ( 700p.), based on our course. Freely available from our web pages, forever. Paper version before the end of y x 25/26

36 Conclusion Combinatorics Approximation Linear Differential Equations + Computer Algebra Special Functions The End Your Work? 26/26