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What Is a Holonomic Function? What is a holonomic function? Christoph Koutschan Johann Radon Institute for Computational and Applied Mathematics (RICAM) Austrian Academy of Sciences December 16, 2013 SFB Status Seminar, Strobl Algorithmic and Enumerative Combinatorics What is a holonomic function? Notation: Let K be a computable field of characteristic zero. Preliminaries: A finitely generated left module over the Weyl algebra As(K) is holonomic if it is zero or if it has Bernstein dimension s, i.e., the dimension of the graded associated module with respect to the Bernstein filtration (which filters according to total degree). Definition 1: A function f(x1; : : : ; xs) is holonomic if the left K K As( )-module As( )= AnnAs(K)(f) is holonomic. In other words, if it is the solution of a maximally overdetermined system of differential equations. Definition 2: A function f is holonomic if the following K(x1; : : : ; xs)-vector space is finite-dimensional: O O dimK(x1;:::;xs)( = AnnO(f)) < 1, where denotes the Ore algebra K(x ; : : : ; x )[D ; 1; d ] ··· [D ; 1; d ]. 1 s x1 dx1 xs dxs 1 / 38 A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients: pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0; p0; : : : ; pd 2 K[n] not all zero. −! In both cases, only finitely many initial conditions! −! Also called holonomic function resp. holonomic sequence. D-finite and P-recursive A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients: (d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0; p0; : : : ; pd 2 K[x] not all zero. 2 / 38 −! In both cases, only finitely many initial conditions! −! Also called holonomic function resp. holonomic sequence. D-finite and P-recursive A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients: (d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0; p0; : : : ; pd 2 K[x] not all zero. A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients: pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0; p0; : : : ; pd 2 K[n] not all zero. 2 / 38 −! Also called holonomic function resp. holonomic sequence. D-finite and P-recursive A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients: (d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0; p0; : : : ; pd 2 K[x] not all zero. A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients: pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0; p0; : : : ; pd 2 K[n] not all zero. −! In both cases, only finitely many initial conditions! 2 / 38 D-finite and P-recursive A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients: (d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0; p0; : : : ; pd 2 K[x] not all zero. A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients: pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0; p0; : : : ; pd 2 K[n] not all zero. −! In both cases, only finitely many initial conditions! −! Also called holonomic function resp. holonomic sequence. 2 / 38 Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1. 3 / 38 2n+7 n+3 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1. Can express any shift as K(n)-linear combination of Hn and Hn+1: Hn+4 3 / 38 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1. Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 3 / 38 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1. Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3 / 38 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1. Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3 / 38 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1. Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 3 / 38 Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.
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