Linear differential and recurrence equations viewed as data-structures

Transcription

1 Linear differential and recurrence equations viewed as data-structures Bruno Salvy Inria & ENS de Lyon Leçons de Mathématiques et d'informatique d'aujourd'hui Bordeaux, Mai 2015

2 Computer Algebra Effective mathematics (what can we compute?) Their complexity (how fast?) Several million users. 30 years of algorithmic progress. Thesis in this presentation: linear differential and recurrence equations are a good data-structure. 2

3 Topics Fast computation with high precision; automatic proofs of identities; «computation» of expansions, of (multiple) integrals, of (multiple) sums. 3

4 The objects of study Def. A power series is called differentially finite (D-finite) when it is the solution of a linear differential equation with polynomial coefficients. Exs: sin, cos, exp, log, arcsin, arccos, arctan, arcsinh,hypergeometric series, Bessel functions, Def. A sequence is polynomially recursive (P-recursive) when it is the solution of a linear recurrence with polynomial coefficients. 1X Prop. f = f n z n D-finite P-recursive. n=0 f n 4

5 Example Coefficient of X in P(X)=(1+X) (1+X+X 2 ) 10000? Linear differential equation of order 1 linear recurrence of order 2 unroll (cleverly). > P := (1+x)^(2*N)*(1+x+x^2)^N: > deq := gfun:-holexprtodiffeq(p,y(x)): > rec := gfun:-diffeqtorec(%,y(x),u(k)): > p := gfun:-rectoproc(subs(n=10000,rec),u(k)): > p(20000); 23982[ digits...]33952 Total time: 0.5 sec 5

6 I. Fast computation at large precision From large integers to precise numerical values

7 Fast multiplication Fast Fourier Transform (Gauss, Cooley-Tuckey, Schönhage-Strassen). Two integers of n digits can be multiplied with O(n log(n) loglog(n)) bit operations. Direct consequence (by Newton iteration): inverses, square-roots, : same cost. 7

8 Binary Splitting for linear recurrences (70 s and 80 s) n! by divide-and-conquer: Cost: O(n log 3 n loglog n) using FFT linear recurrences of order 1 reduce to arbitrary order: same idea, same cost (matrix factorial): ex: e n := en e n 1 nx k=0 n! :=n bn/2c {z } size O(n log n) 1 k! = 1 n satisfies a 2nd order rec, computed via n n {z 0 } A(n) (bn/2c + 1) 1 {z } size O(n log n) p!(n) := (p(n) p(bn/2c)) (p(bn/2c + 1) p(1)) en 1 e n 2 = 1 1 n! A!(n). 0 8

9 Numerical evaluation of solutions of LDEs Principle: f(x) = NX n=0 a n x n {z } fast evaluation f solution of a LDE with coeffs in Q(x) (our data-structure!) 1. linear recurrence in N for the first sum (easy); 2. tight bounds on the tail (technical); + 3. no numerical roundoff errors. 1X n=n+1 a n x n {z } good bounds The technique used for fast evaluation of constants like 1 = 12 C 3/2 1X n=0 ( 1) n (6n)!(A + nb) (3n)!n! 3 C 3n with A= , B= , C=

10 Analytic continuation Compute f(x), f 0 (x),...,f (d 1) (x) as new initial conditions and handle error propagation: Ex: erf(π) with 15 digits: arctan(1+i) 0! 200 terms ! 18 terms ! 6terms Again: computation on integers. No roundoff errors. 10

11 II. Proofs of Identities

12 Confinement k+1 vectors in dimension k an identity LDE the function and all its derivatives are confined in a finite dimensional vector space the sum and product of solutions of LDEs satisfy LDEs same property for P-recursive sequences 12

13 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. Cauchy s theorem concludes. Proofs of non-linear identities by linear algebra! 13

14 Example: 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! 14

15 Dynamic Dictionary of Mathematical Functions User need Recent algorithmic progress Maths on the web 15

16 Demonstration 16

17 Guess & prove continued fractions 1. Differential equation produces first terms (easy): arctan x = Taylor x x x x Guess a formula (easy): a n = n2 4n Prove that the CF with these a n satisfies the differential equation. Continued fraction No human intervention needed. 17

18 Automatic Proof of the guessed CF arctan x =? Aim: RHS satisfies (x 2 +1)y -1=0; 1 + Convergents P n /Q n where P n and Q n satisfy the LRE u n =u n-1 +a n u n-2 (and Q n (0) 0); 1 + x n 2 4n 2 1 x2 1 + Define H n :=(Q n ) 2 ((x 2 +1)(P n /Q n ) -1); H n is a polynomial in P n,q n and their derivatives; therefore, it satisfies a LRE that can be computed; from it, H n =O(x n ) visible from there, (P n /Q n ) -1/(1+x 2 )=O(x n ) too; conclude P n /Q n arctan by integrating. 18

19 III. Ore Polynomials

20 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 Taylor morphism: D x (n+1)s n ; x S n -1 produces linear recurrence from LDE 20

21 Framework: Ore polynomials (fg) 0 = f 0 g+fg 0, S n (f n g n )=f n+1 S n (g n ), n(f n g n )=f n+1 n (g n )+ n (f n )g n, and many more (e.g., q-analogues) are captured by A (A integral domain) with = (a)@ + (a) σ a ring morphism, δ a σ-derivation (δ(ab)=σ(a)δ(b)+δ(a)b). Main property: A,B in A, then deg AB=deg A+deg B. Consequence 1: (non-commutative) Euclidean division Consequence 2: (non-commutative) Euclidean algorithm 21

22 GCRD & LCLM greatest common right divisor & least common left multiple GCRD(A,B): maximal operator whose solutions are common to A and B. LCLM(A,B): minimal operator having the solutions of A and B for solutions. Example: closure by sum. Computation: Euclidean algorithm or linear algebra. 22

23 Reduction of order Input: a (large) linear recurrence equation + ini. cond Output: a factor annihilating this solution Step 1: use the recurrence and its initial conditions to compute a large number of terms; Step 2: guess a linear recurrence equation annihilating this sequence (linear algebra); Step 3: take the gcrd of this operator and the initial one; Step 3: prove that this factor annihilates the solution by checking sufficiently many initial conditions. 23

24 Example from a continued fraction expansion P k = a k x 2 P k 2 + P k 1, a k = ( 2k (2k+1)(2k+3), 2(k+2) (2k+1)(2k+3), k even, k odd. Aim: a recurrence for all k. Step 1: use both recurrences to find a relation between P k,p k+2,p k+4 for even k and one for odd k; Step 2: compute their LCLM (order 8); Step 3: use the initial conditions to reduce (order 4). 24

25 Chebyshev expansions arctan Taylor Chebyshev z 1 3 z z5 + 2( p T1 (x) 2 + 1) (2 p 2 + 3) T 3 (x) 3(2 p 2 + 3) + T 5 (x) 2 5(2 p 2 + 3)

26 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). 26

27 Application: Chebyshev expansions Extend Taylor morphism to 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+S -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-s -1 )/2. Prop. If y is a solution of L(x,d/dx), then its Chebyshev coefficients annihilate the numerator of L(X,D). 27

28 IV. Systems of equations

29 Example: Contiguity of Hypergeometric Series F(a, b; c; z) = 1X n=0 (a) n (b) n (c) n n! {z } u a,n z n, (x) n := x(x + 1) (x + n 1). u a,n+1 (a + n)(b + n) = u a,n (c + n)(n + 1)! z(1 z)f00 +(c (a + b + 1)z)F 0 abf = 0, u a+1,n = n u a,n a + 1! S af := F(a + 1, b; c; z) = z a F0 + F. S a Gauss 1812: contiguity relation. dim=2 ) S 2 af, S a F, F linearly dependent (coordinates in Q(a,b,c,z)) (a + 1)(z 1)S 2 af +((b a 1)z + 2 c + 2a)S a F +(c a 1)F = 0. z 29

30 Ore Algebras O:=K(x 1,,x r ) 1,, r :=K(x 1,,x r ) 1 r, with commuting i s and for i j, δ i ( j )=0 and σ i ( j )= j. Def. LM (leading monomial) on next slide. Main property: A,B in O, then LM(AB)=LM(A)LM(B). Consequence: (non-commutative) Gröbner bases Gröbner bases as a data-structure to encode special functions 30

31 Gröbner Bases 1. Monomial ordering: order on N k, compatible with +, 0 minimal. 2. Leading monomial of a polynomial: the largest one. 3. Gröbner basis of a (left) ideal I: corners of stairs. 4. Quotient mod I: basis below the stairs (Vect{ α f}). 5. Reduction of P: Rewrite P mod I on this basis. 6. Dimension: «size» of the quotient. 7. D-finiteness: dimension 0. An access to (finite-dimensional) vector spaces. 31

32 Closure Properties + Proposition. dim ann(f + g) apple max(dim ann f, dim ann g), dim ann(fg) apple dim ann f + dim ann g, dim ann(@f) apple dim ann f. Algorithms by linear algebra simple definitions data-structures for more complicated functions 32

33 V. Sums and Integrals

34 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! nx k=0 q k2 (q; q) k (q; q) n k = nx k= n ( 1) k q (5k2 k)/2 (q; q) n k (q; q) n+k Aims: 1. Prove them automatically 2. Find the rhs given the lhs Note: at least one free variable 34

35 I(x) = Example: u(n, k) = S(n + 1) = X k Creative telescoping Z Input: equations (differential for f or recurrence for u). n def. by k n + 1 f(x, t) dt =? or U(n) = X k k n + 1 = n + 1 n n, k n + 1 k k k + 1 = X n + 1 n + 1 n + k k + 1 k + 1 k {z } telesc. Output: equations for the sum or the integral. {z } telesc. IF one knows A(n,S n ) and B(n,k,S n,s k ) such that u(n, k) =? = n k n k + 1 k n n +2 k k Pascal = 2S(n). (A(n,S n )+Δ k B(n,k,S n,s k )) u(n,k)=0, certificate then the sum telescopes, leading to A(n,S n ) U(n)=0. 35

36 Creative Telescoping Z I(x) = f(x, t) dt =? IF one knows A(x, x ) and B(x,t, x, t ) such that (A(x, x ) + t B(x,t, x, t )) f(x,t)=0, then the integral «telescopes», leading to A(x, x ) I(x)=0. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals. Richard P. Feynman 1985 Method: integration (summation) by parts and differentiation (difference) under the integral (sum) sign 36

37 Telescoping Ideal T t (f) := Ann f t Q(x,t)h@ x,@ t i {z } int. by parts \ Q(x)h@ x i. {z } di. under R hypergeometric summation: dim=1 + param. Gosper. [Zeilberger] holonomy: restrict int. by parts to Q(x)h@ x,@ t i and Gröbner bases. [Wilf-Zeilberger, also Sister Celine] finite dim, Ore algebras & GB [Chyzak] infinite dim & GB rational f and restrict to Q(x)[t, 1/den f]h@ x,@ t i in very good complexity

38 Chyzak s Algorithm T t (f) := Ann f t Q(x,t)h@ t i {z } int. by parts \ Q(x)h@ x i. {z } di. under R Input: a Gröbner basis G for Ann f in A= Q(x, t)h@ t i Output: P in Q(x)h@ x i and Q in A, reduced wrt G and such that (P+ t Q)f=0. For r=1,2,3, 1. use indeterminate coefficients to define Q = X q i,j (x, t)@ i x@ j t, P = X (i,j) below stairs appler p (x)@ x. 2. reduce P+ t Q using G, leading to a 1st order system for q i,j (x,t) and p α (x); 3. stop if a rational solution is found. 38

39 Examples of applications Hypergeometric: binomial sums, hypergeometric series; 2nX k=0 ( 1) k 2n k Higher dimension: classical orthogonal polynomials, special functions like Bessel, Airy, Struve, Weber, Anger, hypergeometric and generalized hypergeometric, J 0 (z) = 2 3 =( Z 1 Infinite dimension: Bernoulli, Stirling or Eulerian numbers, incomplete Gamma function, Z ) n (3n)! n! 3 cos(zt) p dt 1 t 2 exp( xy) (n, x) dx = (n) y 1 1 (y + 1) n 39

40 VI. Faster Creative Telescoping

41 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 Next, in T t (f) := Ann f t Q(x,t)h@ t i {z } int. by parts \ Q(x)h@ x i. {z } di. under R ' 12 pages we restrict to rational f t Q(x)[t, 1/ den f]h@ t i 41

42 Bivariate integrals by Hermite reduction I(t) = I P(t, x) Q m (t, x) dx Q square-free Int. over a cycle where Q 0. If m=1, Euclidean division: P=aQ+r, deg x r<deg x Q apple P P Def. Reduced form: Q = r Z Q x a Q If m>1, Bézout identity and integration by parts P = uq + v@ x Q! P u Q m = m 1 v/(1 m) Q m 1 +@ x Q m 1 {z } A m 1 Algorithm: R 0 :=[P/Q m ] for i=1,2, do R i :=[ t R i-1 ] when there is a relation c 0 (t)r 0 + +c i (t)r i =0 return c 0 + +c i t i apple P := r Q Q m := [A m 1] 42

43 More variables: Griffiths-Dwork reduction I(t) = I P(t, x) Q m (t, x) dx 1. Control degrees by homogenizing (x 1,,x n ) (x 0,,x n ) 2. If m=1, [P/Q]:=P/Q 3. If m>1, reduce modulo Jacobian ideal J := h@ 0 Q,...,@ n Qi P= r + v 0 Q + + v n Q P Q m = r 1 v 0 Q 0 m 1 Q m apple P Q m := r Q m +[A m 1] 1 + n v n Q m m 1 Thm. [Griffiths] In the regular case ( Q(t)[x]/J Q square-free Int. over a cycle where Q 0 v 0 + n v n Q m 1 {z } A m 1 finite dim), if R=P/Q m H hom of degree -n-1,[r] =0, Rdx = 0. SAME ALGORITHM. 43

44 Size and complexity I(t) = I P(t, x) Q m (t, x) {z } 2Q(t,x) dx no regularity assumed N := deg x Q, d t := max(deg t Q, deg t P) deg x P not too big Thm. A linear differential equation for I(t) can be computed in O(e 3n N 8n d t ) operations in Q. It has order N n and degree O(e n N 3n d t ). tight Note: generically, the certificate has at least N n2 /2 monomials. This has consequences for multiple binomial sums. 44

45 Conclusion Linear differential equations and recurrences are a great data-structure; Numerous algorithms have been developed in computer algebra; Efficient code is available; More is to be found (certificate-free algorithms, diagonals, )