Computing the N-th term of a q-holonomic sequence

Abstract

In 1977, Strassen invented a famous baby-step / giant-step algorithm that computes the factorial N ! in arithmetic complexity quasi-linear in √ N. In 1988, the Chudnovsky brothers generalized Strassen's algorithm to the computation of the N-th term of any holonomic sequence in the same arithmetic complexity. We design q-analogues of these algorithms. We first extend Strassen's algorithm to the computation of the q-factorial of N , then Chudnovskys' algorithm to the computation of the N-th term of any q-holonomic sequence. Both algorithms work in arithmetic complexity quasi-linear in √ N. We describe various algorithmic consequences, including the acceleration of polynomial and rational solving of linear q-differential equations, and the fast evaluation of large classes of polynomials, including a family recently considered by Nogneng and Schost. CCS CONCEPTS • Computing methodologies → Algebraic algorithms.

A Simple and Fast Algorithm for Computing the N-th Term of a Linearly Recurrent Sequence

“…The simplest algorithm is to apply the inverse DFT of length n to obtain A, and then to apply the DFT of length 2n to A. This costs E(n) + E(2n) arithmetic operations, that is 9 2 n log n + 3n operations in K. This algorithm can be improved using the following formulas…”

“…(Q), the best known algorithms are presented in [16,12], resp. in [9]. In the algebraic model, they rely on an algorithmic technique called baby-step/giant-step, allowing to compute u N in a number of operations in R that is almost linear in √ N , up to logarithmic factors.…”

A Simple and Fast Algorithm for Computing the $N$-th Term of a Linearly Recurrent Sequence

“…(Q), the best known algorithms are presented in [16,12], resp. in [9]. In the algebraic model, they rely on an algorithmic technique called baby-step / giant-step, which allows to compute u N using a number of operations in R that is almost linear in √ N , up to logarithmic factors.…”

“…The simplest algorithm is to apply the inverse DFT of length n to obtain A, and then to apply the DFT of length 2n to A. This costs E(n)+E(2n) arithmetic operations, that is 9 2 n log n + 3n operations in K.…”