Title: Note on fast division algorithm for polynomials using Newton iteration
ArXiv ID: 1112.4014
Date: 2011-12-20
Authors: Zhengjun Cao and Hanyue Cao
📝 Abstract
The classical division algorithm for polynomials requires $O(n^2)$ operations for inputs of size $n$. Using reversal technique and Newton iteration, it can be improved to $O({M}(n))$, where ${M}$ is a multiplication time. But the method requires that the degree of the modulo, $x^l$, should be the power of 2. If $l$ is not a power of 2 and $f(0)=1$, Gathen and Gerhard suggest to compute the inverse,$f^{-1}$, modulo $x^{\lceil l/2^r\rceil}, x^{\lceil l/2^{r-1}\rceil},..., x^{\lceil l/2\rceil}, x^l$, separately. But they did not specify the iterative step. In this note, we show that the original Newton iteration formula can be directly used to compute $f^{-1}\,{mod}\,x^{l}$ without any additional cost, when $l$ is not a power of 2.
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arXiv:1112.4014v1 [cs.SC] 17 Dec 2011
Note on fast division algorithm for polynomials
using Newton iteration
Zhengjun Cao ∗,
Hanyue Cao
Department of Mathematics, Shanghai University, Shanghai, China.
∗caozhj@shu.edu.cn
Abstract
The classical division algorithm for polynomials requires O(n2) operations for inputs
of size n. Using reversal technique and Newton iteration, it can be improved to O(M(n)),
where M is a multiplication time. But the method requires that the degree of the modulo,
xl, should be the power of 2. If l is not a power of 2 and f(0) = 1, Gathen and Gerhard
suggest to compute the inverse, f −1, modulo x⌈l/2r⌉, x⌈l/2r−1⌉, · · · , x⌈l/2⌉, xl, separately.
But they did not specify the iterative step. In this note, we show that the original Newton
iteration formula can be directly used to compute f −1 mod xl without any additional cost,
when l is not a power of 2.
Keywords: Newton iteration, revisal, multiplication time
1
Introduction
Polynomials over a field form a Euclidean domain. This means that for all a, b with b ̸= 0
there exist unique q, r such that a = qb+r where deg r