Replies: 6 comments 3 replies
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多项式牛顿迭代中的$g'(f_0(x))$怎么求啊 |
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简单地说,你就把 f 看成自变量就好了(相当于先求 g'(x),然后把里面的 x 都换成 f(x)) |
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讲得很棒 不过有点不明白为什么可以把多项式看作自变量 |
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严谨一点的写法应该是$g(f, x)$和$∂g/∂f(f, x)$。有解的条件是存在数值$a$使得$g(a, 0)=0$且$∂g/∂f(a, 0)$可逆。 比如多项式求逆,其实是$g(f,x) = 1/f - h(x)$, 偏导数自然就是$-f^{-2}$没有$h(x)$了。$f0(x) = 1/h(0)$。 |
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我能说多项式牛顿迭代实现「多项式求逆」和「多项式开方」和倍增法实现「多项式求逆」和「多项式开方」本质是一模一样的吗 |
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多项式求逆的例子,疑惑求解答,我感觉 |
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https://oi-wiki.org/math/poly/newton/
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