Fast Computation of Power Series Solutions of Systems of Differential Equations

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Fast Computation of Power Series Solutions of Systems of Differential Equations Bruno.Salvy@inria.

1 Fast Computation of Power Series Solutions of Systems of Differential Equations Algorithms Project, Inria Joint work with A. Bostan, F. Chyzak, F. Ollivier, É. Schost, A. Sedoglavic

2 Series Expansions to High Order Motivations: combinatorics (generating functions of ordered structures); numerical analysis (e.g., Padé approximants); cryptography (e.g., morphisms between curves). Non-linear differential systems can be reduced to linear ones. Problem Input: linear differential equation/system of order r with power series coefficients, positive integer N r. Output: N first terms of one solution or of a basis of solutions. Complexity measure: number of arithmetic operations. Aim: quasi-linear wrt N, good wrt r. M(N) = O(N log N) polynomial multiplication [Schönhage-Strassen71]; MM(r, N) = O(r 2 M(N) + r ω N) polynomial matrix product [Bostan-Schost05].

3 Results input power series polynomial constant output /output coefficients coefficients coefficients size equation /basis O(r 2 N log N) O(dr 2 N) O(rN) rn equation /1 soln O(rN log 2 N) O(drN) O( log rn) N system /basis O(r 2 N log N) O(dr ω N) O(r 2 log rn) r 2 N system /1 soln O(r 2 N log 2 N) O(dr 2 N) O(r log rn) rn Previous results: O(r 2 N 2 ) undetermined coefficients; O(r r N log N) [Brent-Kung78]; O(r 2 N log 2 N) [van der Hoeven02].

4 Results input power series polynomial constant output /output coefficients coefficients coefficients size equation /basis O(r 2 N) O(dr 2 N) O(rN) rn equation /1 soln O(rN) O(drN) O(N) N system /basis O(r 2 N) O(dr ω N) O(r 2 N) r 2 N system /1 soln O(r 2 N) O(dr 2 N) O(rN) rn Previous results: O(r 2 N 2 ) undetermined coefficients; O(r r N) [Brent-Kung78]; O(r 2 N) [van der Hoeven02].

5 Results input power series polynomial constant output /output coefficients coefficients coefficients size equation /basis O(r 2 N) O(dr 2 N) O(rN) rn equation /1 soln O(rN) O(drN) O(N) N system /basis O(r 2 Nlog N) O(dr ω N) O(r 2 N) r 2 N system /1 soln O(r 2 N) O(dr 2 N) O(rN) rn Previous results: O(r 2 N 2 ) undetermined coefficients; O(r r N) [Brent-Kung78]; O(r 2 N) [van der Hoeven02].

6 Basic Idea: Newton Iteration has Good Complexity To solve φ(y) = 0, iterate y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Example 1: φ(y) = 3y 2 + y 1 y 0 = 0.5 y 1 = y 2 = y 3 = y 4 = The cost is dominated by the last iteration.

7 Basic Idea: Newton Iteration has Good Complexity To solve φ(y) = 0, iterate y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Example 2: φ(y) = ty 2 + y 1 y 0 = 0 y 1 = 1 y 2 = 1 t + 2t 2 4t 3 + 8t 4 16t t 6 64t y 3 = 1 t + 2t 2 5t t 4 42t t 6 428t The cost is dominated by the last iteration.

8 Newton for Power Series y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Reciprocal [Sieveking72, Kung74]: φ(y) = a(t) 1/y; y n+1 = y n u n+1, u n+1 = y n (y n a(t) 1). Same complexity O(M(N)) as polynomial multiplication.

9 Newton for Power Series y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Reciprocal [Sieveking72, Kung74]: φ(y) = a(t) 1/y; Logarithm [Brent75]: log a(t) = a (t)/a(t); Same complexity O(M(N)) as polynomial multiplication.

10 Newton for Power Series y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Reciprocal [Sieveking72, Kung74]: φ(y) = a(t) 1/y; Logarithm [Brent75]: log a(t) = a (t)/a(t); Exponential [Brent75]: φ(y) = a(t) log y. y n+1 = y n u n+1, u n+1 = y n (log y n a(t)). Same complexity O(M(N)) as polynomial multiplication.

11 Newton for Power Series y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Reciprocal [Sieveking72, Kung74]: φ(y) = a(t) 1/y; Logarithm [Brent75]: log a(t) = a (t)/a(t); Exponential [Brent75]: φ(y) = a(t) log y. 1st order LODE [Brent & Kung 78]: y (t) a(t)y(t) = b(t) Homogeneous Case (b = 0): y(t) = y 0 (t) := exp(a(t)); Non-homogeneous Case: variation of constants, y(t) = y 0 (t) b(t)/y 0 (t). Same complexity O(M(N)) as polynomial multiplication.

12 Newton for Power Series y n+1 = y n u n+1, φ (y n )u n+1 = φ(y n ). Reciprocal [Sieveking72, Kung74]: φ(y) = a(t) 1/y; Logarithm [Brent75]: log a(t) = a (t)/a(t); Exponential [Brent75]: φ(y) = a(t) log y. 1st order LODE [Brent & Kung 78]: y (t) a(t)y(t) = b(t) Homogeneous Case (b = 0): y(t) = y 0 (t) := exp(a(t)); Non-homogeneous Case: variation of constants, y(t) = y 0 (t) b(t)/y 0 (t). Natural idea: LDE of order r equivalent to 1st order matrix equation: Y (t) A(t)Y (t) = B(t), but exp A(t) is not a solution of Y (t) = A(t)Y (t).

13 Newton Iteration for Operators Newton iteration for φ(y ) = 0: Y k+1 = Y k U k+1, Dφ Yk U k+1 = φ(y k ). }{{} differential Special cases:

14 Newton Iteration for Operators Newton iteration for φ(y ) = 0: Y k+1 = Y k U k+1, Dφ Yk U k+1 = φ(y k ). }{{} differential Special cases: Matrix inversion [Schulz33]: φ(y ) = AY I, Dφ U = AU. equation AU k+1 = AY k I ; U k+1 = Y k (AY k I ). Complexity: MM(r, N) = O(r 2 M(N) + r ω N).

15 Newton Iteration for Operators Newton iteration for φ(y ) = 0: Y k+1 = Y k U k+1, Dφ Yk U k+1 = φ(y k ). }{{} differential Special cases: Matrix inversion [Schulz33]: φ(y ) = AY I, Dφ U = AU. equation AU k+1 = AY k I ; U k+1 = Y k (AY k I ). Complexity: MM(r, N) = O(r 2 M(N) + r ω N). Our problem: φ(y ) = Y AY. Note that Dφ = φ!

16 New Algorithm and its Complexity To iterate Y k+1 = Y k U k+1, we need to solve Variation of constants: U k+1 A(t)U k+1 = Y k A(t)Y k. U k+1 = Y k Z k+1, Y k Z k+1 + Y k Z k+1 A(t)Y k Z }{{ k+1 = Y k } A(t)Y k, 0 Y k+1 = Y k Y k Quadratic convergence; Y 1 k (Y k A(t)Y k) mod t 2k+1, det Y 0 0 a whole basis of solutions is computed; its inverse is computed simultaneously; complexity O(r 2 M(N) + r ω N); good for r = 1 ([Hanrot-Zimmermann02] for exp(a(t))).

17 Experimental results "MatMul.dat" "Newton.dat" time (in seconds) time (in seconds) precision order precision order 9 10 Polynomial matrix multiplication vs. solving Y = AY by our method.

18 Non-Linear Differential Equations by Linearization Newton again: y k+1 = y k u k+1, Dφ yk }{{} differential u k+1 = φ(y k ). A 1st order example from number theory (related to ): Differential: φ(y) = (x 3 + ax + b)y 2 y 3 Ay B = 0. φ(y k +u k+1 ) φ(y k ) = (x 3 + ax + b)2y k u k+1 (3y k 2 A)u k+1 +. }{{} Dφ yk u k+1 Linear differential equation with power series coefficients fast algorithm for isogenies between elliptic curves [BoMoSaSc06]. Higher order a linear differential system at each step.

19 Final Comments Have a look at the paper for constant coefficients (including nice and fast algorithm for exp(ta)); a divide-and-conquer algorithm for only one solution (instead of a basis) in the general case. Still to be saved: a factor r for (1 equation, a basis of solutions); a factor log N for only one solution.

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