多项式牛顿迭代

多项式牛顿迭代

设

\[f(g(x))\equiv 0 (mod\ x^n) \]

求出此模意义下的\(g(x)\)

当\(n=1\)时，单独求出\([x^0]f(g(x))\)

假设已经得到了模\(x^{\lceil\frac{n}{2}\rceil}\)意义下的解\(g_0(x)\)，要求模\(x^{n}\)意义下的解\(g(x)\)

将\(f(g(x))\)在\(g_0(x)\)处泰勒展开得到

\[\sum_{n}\frac{f^{(n)}(g_0(x))}{n!}(g(x)-g_0(x))^n\equiv 0\ (mod\ x^n) \]

因为\(g(x)-g_0(x)\)的最低非零次项的次数为\(\lceil\frac{n}{2}\rceil\)，故有

\[\forall n\geq 2,(g(x)-g_0(x))^n\equiv 0\ (mod\ x^n) \]

则

\[\sum_{n}\frac{f^{(n)}(g_0(x))}{n!}(g(x)-g_0(x))^n\equiv f(g_0(x))+f'(g_0(x))(g(x)-g_0(x))\ (mod\ x^n)\\ g(x)\equiv g_0(x)-\frac{f(g_0(x))}{f'(g_0(x))}\ (mod\ x^n)\\ \]