多项式操作的 n^2 递推

求逆

\[AB=1\\ \sum_{i=0}^nA_iB_{n-i}=[n=0]\\ A_0B_n+\sum_{i=1}^nA_iB_{n-i}=[n=0]\\ B_n=-\frac{1}{A_0}\sum_{i=1}^nA_iB_{n-i} \]

其中 \(B_0=\frac{1}{A_0}\)

exp

\[B=\exp(A)\\ B'=A'\exp(A)\\ B'=A'B\\ B_{n+1}(n+1)=\sum_{i=0}^{n}A_{i+1}(i+1)B_{n-i}\\ B_nn=\sum_{i=1}^nA_iB_{n-i}i\\ B_n=\frac{1}{n}\sum_{i=1}^nA_iB_{n-i}i \]

其中 \(B_0=1\)

ln

\[B=ln(A)\\ B'=\frac{A'}{A}\\ B'A=A'\\ \sum_{i=0}^nB_{i+1}(i+1)A_{n-i}=A_{n+1}(n+1)\\ B_{n+1}(n+1)=-\sum_{i=1}^nB_iiA_n-i+1\\ B_nn=nA_n-\sum_{i=1}^{n-1}B_iiA_{n-i} B_n=A_n-\frac{1}{n}\sum_{i=1}^{n-1}B_iiA_{n-i} \]

其中 \(B_0=0\)