跳转至

多项式牛顿迭代

Description

给定多项式 g\left(x\right),已知有 f\left(x\right)满足:

g\left(f\left(x\right)\right)\equiv 0\pmod{x^{n}}

求出模 x^{n}意义下的 f\left(x\right).

Newton's Method

考虑倍增.

首先当 n=1时,\left[x^{0}\right]g\left(f\left(x\right)\right)=0的解需要单独求出.

假设现在已经得到了模 x^{\left\lceil\frac{n}{2}\right\rceil}意义下的解 f_{0}\left(x\right),要求模 x^{n}意义下的解 f\left(x\right).

g\left(f\left(x\right)\right)f_{0}\left(x\right)处进行泰勒展开,有:

\sum_{i=0}^{+\infty}\frac{g^{\left(i\right)}\left(f_{0}\left(x\right)\right)}{i!}\left(f\left(x\right)-f_{0}\left(x\right)\right)^{i}\equiv 0\pmod{x^{n}}

因为 f\left(x\right)-f_{0}\left(x\right)的最低非零项次数最低为 \left\lceil\frac{n}{2}\right\rceil,故有:

\forall 2\leqslant i:\left(f\left(x\right)-f_{0}\left(x\right)\right)^{i}\equiv 0\pmod{x^{n}}

则:

\sum_{i=0}^{+\infty}\frac{g^{\left(i\right)}\left(f_{0}\left(x\right)\right)}{i!}\left(f\left(x\right)-f_{0}\left(x\right)\right)^{i}\equiv g\left(f_{0}\left(x\right)\right)+g'\left(f_{0}\left(x\right)\right)\left(f\left(x\right)-f_{0}\left(x\right)\right)\equiv 0\pmod{x^{n}}
f\left(x\right)\equiv f_{0}\left(x\right)-\frac{g\left(f_{0}\left(x\right)\right)}{g'\left(f_{0}\left(x\right)\right)}\pmod{x^{n}}

Examples

多项式求逆

设给定函数为 h\left(x\right),有方程:

g\left(f\left(x\right)\right)=\frac{1}{f\left(x\right)}-h\left(x\right)\equiv 0\pmod{x^{n}}

应用 Newton's Method 可得:

\begin{aligned} f\left(x\right)&\equiv f_{0}\left(x\right)-\frac{\frac{1}{f_{0}\left(x\right)}-h\left(x\right)}{-\frac{1}{f_{0}^{2}\left(x\right)}}&\pmod{x^{n}}\\ &\equiv 2f_{0}\left(x\right)-f_{0}^{2}\left(x\right)h\left(x\right)&\pmod{x^{n}} \end{aligned}

时间复杂度

T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right)

多项式开方

设给定函数为 h\left(x\right),有方程:

g\left(f\left(x\right)\right)=f^{2}\left(x\right)-h\left(x\right)\equiv 0\pmod{x^{n}}

应用 Newton's Method 可得:

\begin{aligned} f\left(x\right)&\equiv f_{0}\left(x\right)-\frac{f_{0}^{2}\left(x\right)-h\left(x\right)}{2f_{0}\left(x\right)}&\pmod{x^{n}}\\ &\equiv\frac{f_{0}^{2}\left(x\right)+h\left(x\right)}{2f_{0}\left(x\right)}&\pmod{x^{n}} \end{aligned}

时间复杂度

T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right)

多项式 exp

设给定函数为 h\left(x\right),有方程:

g\left(f\left(x\right)\right)=\ln{f\left(x\right)}-h\left(x\right)\pmod{x^{n}}

应用 Newton's Method 可得:

\begin{aligned} f\left(x\right)&\equiv f_{0}\left(x\right)-\frac{\ln{f_{0}\left(x\right)}-h\left(x\right)}{\frac{1}{f_{0}\left(x\right)}}&\pmod{x^{n}}\\ &\equiv f_{0}\left(x\right)\left(1-\ln{f_{0}\left(x\right)+h\left(x\right)}\right)&\pmod{x^{n}} \end{aligned}

时间复杂度

T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right)

评论