You're growing desperate from the fight.
基本策略
- 已知系数的幂级数
首先是一些可以通过整体法得到封闭形式的幂级数，所谓整体法，即是通过将幂级数位移，用自己表示自己然后做差。
\[\begin{aligned} \left \langle 1,1,1,1,1, \dots \right \rangle &\rightarrow \frac{1}{1 - x} \\ \left \langle a^0,a^1,a^2,a^3,a^4, \dots \right \rangle &\rightarrow \frac{1}{1 - ax} \\ \sum_{i = 0}x^ik &= \frac{1}{1 - x^k} \\ \sum_{i = 0}c^ix^ik &= \frac{1}{1 - cx^k} \end{aligned} \]二项式的幂级数
\[\begin{aligned} \left\langle \binom{n}{0}, -\binom{n}{1}, \binom{n}{2}, -\binom{n}{3}, \dots, (-1)^n \binom{n}{n} \right\rangle &\rightarrow (1 - x)^n \\ \left\langle \binom{n}{0}, \binom{n + 1}{1}, \binom{n + 2}{2}, \dots, \binom{n + k}{k}, \dots \right\rangle &\rightarrow \frac{1}{(1 - x)^{n + 1}} \end{aligned} \]- 斐波那契数列
写成生产函数的形式 \(F(x) = \sum_{i = 0}^{\infty} fib_i x^i\)，其中 \(fbi_i\) 递推式为 \(fbi_0 = 0; fbi_1 = 1; fbi_i = fbi_{i - 1} + fbi_{i - 2} \ (i > 1)\)。
有下面的式子
\[\begin{matrix} F(x) = &fbi_0 x &+ &fbi_1 x &+ &fbi_2 x^2 &+ &fbi_3x^3 &+ \dots \\ xF(x) = & & &fbi_0 x &+ &fbi_1 x^2 &+ &fbi_2x^3 &+ \dots \\ x^2F(x) = & & & & &fbi_0 x^2 &+ &fbi_1x^3 &+ \dots \\ \end{matrix} \]于是得到
\[\begin{aligned} F(x) - xF(x) &- x^2F(x) = x \\ F(x) &= \frac{x}{1 - x - x^2} \end{aligned} \]- 构造幂级数
平移： \([x^n]x^tF(x) = f_{n - t}\)
标签：Function,dots,Generating,aligned,06,frac,binom,&+,fbi From： https://www.cnblogs.com/Iridescent41/p/17528182.html