闲话 24.6.24

闲话

果然我还是喜欢阿育的罐头啊。

推歌:悲伤虚构反应 by 一般诗云p feat. 诗岸

多元拉反 \(1\)\(2\) 的推导

前传

x义x 博客里没有展开说如何暴力展开行列式,可能是 trivial。
我觉得不很 trivial 啊!来展开一下。

我们首先要处理的就是

\[A = \left\{[i = j] - \frac{x_i}{g_j(\bm x)}\frac{\partial g_j(\bm x)}{\partial x_i} \right\}_{m, m} \]

的行列式。

\([i = j]\) 会给每个对角线上元素贡献 \(1\),让每个 \(a_{i, i}\) 的值偏移通项 \(a_{i, j} = \dfrac{x_i}{g_j(\bm x)}\dfrac{\partial g_j(\bm x)}{\partial x_i}\)。这样我们可以先把对角线的括号展开,并枚举不选择 \(1\) 的位置 \(S\subseteq {1, 2, \dots, m}\),枚举 \(S\) 上的排列 \(\pi\)(其值域也为 \(S\),因此枚举 \(\pi_i, i \in S\) 时得到的还是 \(S\)),并选择每个 \(a_{i, \pi_i}, i \in S\)\([i = i], i \not\in S\)。这样选择得到的项的正负性取决于这整个排列的各个环。由于 \([i = i]\) 系数为 \(1\)\(a_{i, \pi_i}\) 系数为 \(-1\),这部分的贡献是 \((-1)^{\lvert S \rvert}\)。另一部分就是 \(\pi\) 的逆序数,它应当和 \(\pi\) 的偶数长度环的数量的奇偶性相同。由于 \(\lvert S \rvert\)\(\pi\) 中奇数长度环数量的奇偶性相同,令 \(c(\pi)\)\(\pi\) 中环的数量,我们知道

\[\det A = \sum_{S, \pi} (-1)^{c(\pi)} \prod_{i \in S} a_{i, \pi_i} \]

\(S = \varnothing\)\(\pi\) 也为空,最后方的 \(\prod\) 退化为 \(1\)\(c(\pi) = 0\)

将其带入多元拉反 \(1\),有

\[\begin{aligned} %%论文的 f 是我的 g,w 是我的 f,g 是我的 h%% & [\bm t^{\bm n}] h(\bm f(\bm t)) \\ = \ & [\bm x^{\bm n}] h(\bm x) \bm g(\bm x)^{\bm n} \det A \\ = \ & [\bm x^{\bm n}] h(\bm x) \bm g(\bm x)^{\bm n} \sum_{S, \pi} (-1)^{c(\pi)} \prod_{i \in S} a_{i, \pi_i} \\ = \ & [\bm x^{\bm n}] \sum_{S, \pi} (-1)^{c(\pi)} \left( h(\bm x) \bm g(\bm x)^{\bm n} \prod_{i \in S} \dfrac{x_i}{g_j(\bm x)}\dfrac{\partial g_j(\bm x)}{\partial x_i} \right) \\ = \ & [\bm x^{\bm n}] \sum_{S, \pi} (-1)^{c(\pi)} \left\{ \left(h(\bm x)\prod_{i\not \in S, i > 0} g_i(\bm x)^{n_i} \right) \left(\prod_{i \in S} g_i(\bm x)^{n_i} \dfrac{x_i}{g_{\pi_i}(\bm x)}\dfrac{\partial g_{\pi_i}(\bm x)}{\partial x_i} \right) \right\} \\ = \ & [\bm x^{\bm n}] \sum_{S, \pi} (-1)^{c(\pi)} \left\{ \left(h(\bm x)\prod_{i\not \in S, i > 0} g_i(\bm x)^{n_i} \right) \left(\prod_{i \in S} x_i g_{\pi_i}(\bm x)^{n_{\pi_i} - 1} \dfrac{\partial g_{\pi_i}(\bm x)}{\partial x_i} \right) \right\} \\ = \ & [\bm x^{\bm n - \bm 1}] \sum_{S, \pi} (-1)^{c(\pi)} \left\{ \left(\prod_{i\not \in S, i > 0} \frac{1}{x_i}\right)\left(h(\bm x)\prod_{i\not \in S, i > 0} g_i(\bm x)^{n_i} \right) \left(\prod_{i \in S} g_{\pi_i}(\bm x)^{n_{\pi_i} - 1} \dfrac{\partial g_{\pi_i}(\bm x)}{\partial x_i} \right) \right\} \end{aligned}\]

两侧乘入 \(\prod n_i\),并令 \(h_0 = h, h_1 = g_1^{n_1}, \dots, h_m = g_m^{n_2}\) 换元。由于

\[\dfrac{\partial}{\partial x_i}h_k(\bm x) = \dfrac{\partial}{\partial x_i} g_k^{n_k}(\bm x) = n_k g_k^{n_k - 1} (\bm x) \dfrac{\partial}{\partial x_i} g_k(\bm x) \ , \]

\[\begin{aligned} & \left( \prod_{i = 1}^m n_i\right)[\bm t^{\bm n}] h(\bm f(\bm t)) \\ = \ & [\bm x^{\bm n - \bm 1}] \sum_{S, \pi} (-1)^{c(\pi)} \left\{ \left(\prod_{i\not \in S, i > 0} \frac{n_i}{x_i}\right)\left(\prod_{i\not \in S} h_i(\bm x) \right) \left(\prod_{i \in S}\frac{\partial h_{\pi_i}(\bm x)}{\partial x_i} \right) \right\} \end{aligned}\]

注意到

\[[x^{n - 1}] g'(x) = n[x^n] g(x) = [x^{n - 1}] \frac{n}{x} g(x) \]

因此在提取 \(x^{n - 1}\) 意义下 \(\dfrac{\partial}{\partial x} = \dfrac{n}{x}\)。因此原式

\[\begin{aligned} \\ = \ & [\bm x^{\bm n - \bm 1}] \sum_{S, \pi} (-1)^{c(\pi)} \left\{ \left(\prod_{i\not \in S, i > 0} \frac{\partial}{\partial x_i}\right)\left(\prod_{i\not \in S} h_i(\bm x) \times \prod_{i \in S}\frac{\partial h_{\pi_i}(\bm x)}{\partial x_i} \right) \right\} \end{aligned}\]

展开某一由 \(S,\pi\) 确定的项,考察大括号内部的部分。可以发现任意 \(h_i(\bm x)\) 关于某个变元最多只会被求一阶偏导,并且每个偏导 \(\dfrac{\partial}{\partial x_i}, i > 0\) 只会出现恰好一次,我们不妨使用图结构刻画其关系。

取点集 \(V = \{0,1,\dots, m\}\),边集 \(E\subseteq V^2\)。若存在 \(h_v\)\(x_u\) 求偏导,则 \(u\to v\) 连接有向边。这样就构造出了一张图 \(G = (V, E)\)。注意到除 \(0\) 外每个点的出度都为 \(1\),而 \(0\) 的出度为 \(0\),因此这张图是由一棵包含点 \(0\) 的内向树和一系列环组成的,并且这些环和 \(\pi\) 对应的环相同。这些边同样具有性质,初始时我们只限制了 \(i\to \pi_i, i\in S\) 有边,这是由于上式的最后一项,但对剩余的部分(即 \(\partial/\partial x_i, i \not \in S\))没有限制对象,因此这些边的终点是任意的。

因此令 \(\bm h = (h_0, h_1, \dots, h_m)\),我们知道

\[\left(\prod_{i\not \in S, i > 0} \frac{\partial}{\partial x_i}\right)\left(\prod_{i\not \in S} h_i(\bm x) \times \prod_{i \in S}\frac{\partial h_{\pi_i}(\bm x)}{\partial x_i} \right) = \sum_{G} \frac{\partial \bm h}{\partial G} \]

这里的 \(G\) 需要满足上述条件,即

  1. 点集 \(V = \{0,1,\dots, m\}\)
  2. \(0\) 外每个点的出度都为 \(1\),而 \(0\) 的出度为 \(0\)
  3. \(G\) 中的环包含 \(\pi\)\(\pi \subset G\))。

最终有

\[\begin{aligned} \left( \prod_{i = 1}^m n_i\right) [\bm t^{\bm n}] h(\bm f(\bm t)) = \ & [\bm x^{\bm n - \bm 1}] \sum_{S, \pi} (-1)^{c(\pi)} \sum_{G \supseteq \pi} \frac{\partial \bm h}{\partial G} \\ = \ & [\bm x^{\bm n - \bm 1}] \sum_{G} \frac{\partial \bm h}{\partial G}\sum_{\pi\subseteq G} (-1)^{c(\pi)} \end{aligned}\]

\(G\) 中存在环时,由于 \(\pi\) 要么完全包含环,要么完全不包含,由容斥知道 \(\sum_{\pi\subseteq G} (-1)^{c(\pi)} = 0\)。反之,\(\sum_{\pi\subseteq G} (-1)^{c(\pi)} = 1\),因为只有空排列满足该要求。整理得到

\[[\bm t^{\bm n}] h(\bm f(\bm t)) = \frac{1}{\prod_{i = 1}^m n_i} [\bm x^{\bm n - \bm 1}] \sum_{\mathcal T} \frac{\partial \bm h}{\partial \mathcal T} \]

其中 \(\mathcal T\) 为以 \(0\) 为根的内向树。这就是多元拉反 \(2\)

Reference:

[1] Edward A. Bender et al., A Multivariate Lagrange Inversion Formula for Asymptotic Calculations.

posted @ 2024-06-24 11:14  joke3579  阅读(39)  评论(0编辑  收藏  举报