拉格朗日反演

· · 算法·理论

不会证,但是记录一下公式:

以下记 G (x)F (x) 的复合逆,即 G (F (x)) = x

  1. \displaystyle [x ^ n] F (x) = \frac 1 n [x ^ {n - 1}] \left(\frac x {G (x)}\right) ^ n
  2. \displaystyle [x ^ n] F ^ k (x) = \frac k n [x ^ {n - k}] \left(\frac x {G (x)}\right) ^ n
  3. \displaystyle [x ^ n] H (F (x)) = \frac 1 n [x ^ {n - 1}] H ^ \prime (x) \left(\frac x {G (x)}\right) ^ n