多项式乘法逆
多项式乘法逆
给定\(F(x)\)
求\(G(x)\)满足
\[G(x)F(x)\equiv 1\ (mod\ x^n)\\
\]
假设已知
\[H(x)F(x)\equiv 1\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\
F(x)(G(x)-H(x))\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\
G(x)-H(x)\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\
(G(x)-H(x))^2\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\
G^2(x)+H^2(x)-2*G(x)H(x)\equiv 0\ (mod\ x^{\lceil\frac{n}{2}\rceil})\\
\]
两边同乘\(F(x)\),且\(F(x)G(x)\equiv 1\)
\[G(x)=2H(x)-H(x)^2F(x)\ (mod\ x^n)
\]
或硬上牛顿迭代
\[F(G(x))=\frac{1}{G_0(x)}-F(x)\equiv 0\ (mod\ x^n)\\
G(x)=G_0(x)-\frac{\frac{1}{G_0(x)}-F(x)}{-\frac{1}{G_0^2(x)}}\\
G(x)=2G_0(x)-F(x)G_0^2(x)
\]