多项式牛顿迭代
描述¶
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 \(f\left(x\right)\)。
将 \(g\left(f\left(x\right)\right)\) 在 \(f_{0}\left(x\right)\) 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：
\[ \sum_{i=0}^{+\infty}\frac{g^{\left(i\right)}\left(f_{0}\left(x\right)\right)}{i!}\left(f\left(x\right)-f_{0}\left(x\right)\right)^{i}\equiv g\left(f_{0}\left(x\right)\right)+g'\left(f_{0}\left(x\right)\right)\left(f\left(x\right)-f_{0}\left(x\right)\right)\equiv 0\pmod{x^{n}} \]
\[ f\left(x\right)\equiv f_{0}\left(x\right)-\frac{g\left(f_{0}\left(x\right)\right)}{g'\left(f_{0}\left(x\right)\right)}\pmod{x^{n}} \]
例题¶
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
\[ T\left(n\right)=T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log{n}\right) \]