普通型生成函数

普通型生成函数

一、定义

构造这么一个多项式函数 \(F(x)\)，使得 \(x\) 的 \(n\) 次方系数为 \(f(n)\)。

\[F[x] = \sum^\infty_{i = 0}f(i)\ x^i \]

二、格式声明

==为逻辑判断符，=为运算符号

[表达式] 表达式值为真返回\(\ 1\)，否则返回\(\ 0\)

<>中表示序列、f()、g()表示原函数、F()、G()表示生成函数

拿斐波那契数列举个例子：

\[<f_1,f_2,f_3,……> \]

\[f(n) = f(n-1) + f(n-2) + [n == 1] \]

\[F[x] = \sum^\infty_{i = 0}f(i)\ x^i \]

三、常见操作

h()表示操作后的函数，H()表示操作后的生成函数

1.数乘

\[<cf_0,cf_1,cf_2,…> \]

\[h(x) = cf(x) \]

\[H(x) = cF(x) \]

2.加减

\[<f_0+g_0,f_1+g_1,f_2+g_2,…> \]

\[h(x) = f(x) + g(x) \]

\[H(x) = F(x) + G(x) \]

3.右移(k位)

\[<0,0,0,0,f_0,f_1,f_2,f_3,…> \]

\[h(x) = f(x - k) \]

\[H(x)=x^kF(x) \]

4.求导

\[<f_1,2f_2,3f_3,…> \]

\[H(x) = F'(x) \]

5.卷积

\[<f_0g_0,f_1g_0+f_0f_1,f_0g_2+f_1g_1+f_2g_0,…,\sum_{i+j=n}f_ig_j,…> \]

\[h(x) =\sum_{i=0}^nf_ig_{n-i} \]

\[H(x)=F(x)\times G(x) \]

6.取项

\[<0,0,0,…，0,f(n),0,…,0,0> \]

\[g(x) = \begin{cases}h(x) & n = 1\\0 & otherwise\end{cases} \]

\[[x^n]F(x) = f(n) \]

四、全是例子

建议蒙住右边的生成函数一列，这样食用效果更佳

点击注脚查看证明，*为卷积

编号	记法	函数	数列	生成函数
\(T_1\)	无	\(f(x)\to f(x-1)\)	无	\(F(x)\to xF(x)\)
\(T_2\)	\(1\)	\(f(x) = 1\)	\(<1,1,1,…,1,1,1>\)	\(F(x)= 1 + x + x^2 + x^3 + …=\frac{1}{1-x}\)[1]
\(T_3\)	\([n == 1]\)	\(f(x) = \begin{cases}1&x==1\\0&otherwise\end{cases}\)	\(<0,1,0,…,0,0>\)	\(F(x) = 1\)
\(T_4\)	\([n\%3 == 0 ]\)	\(f(x) = \begin{cases}1&x\%3 == 0\\0&otherwise\end{cases}\)	\(<1,0,0,1,0,0,…>\)	\(F(x) = 1 + x^3 + x^6 + x^9 + … = \frac{1}{1-x^3}\)[2]
\(T_5\)	\([n\geq 3]\)	\(f(x) = \begin{cases}1&x\geq 3\\0&otherwise\end{cases}\)	\(<0,0,0,1,1,1,…>\)	\(F(x) = x^3 + x^4 + … = \frac{x^3}{1-x}\)[3]
\(T_6\)	\([n\leq3]\)	\(f(x) = \begin{cases}1 & x\leq3\\0&otherwise \end{cases}\)	\(<1,1,1,1,0,0,…>\)	\(F(x) = \frac{1}{1-x} - \frac{x^4}{1-x}\)[4]
\(T_7\)	\(2^n\)	\(f(x) = 2^x\)	\(<1,2,4,8,…>\)	\(F(x)=1 + 2x + 4x + … = \frac{1}{1-2x}\)[5]
\(T_8\)	\(n\)	\(f(x) = x\)	\(<0,1,2,3,…>\)	$F(x) = 1 + 2x + 3x^2+… =1*1 = \frac{1}{(1-x)^2} $（此处\(1*1\)的\(1\)为\(T_2\)的记法）

五、全是例题

Q1. 有1g、2g、3g、4g的砝码各一个，问对于一些重量，求方案数

\[\begin{cases}1g&<1,1,0,0,0,…>&F(x) = 1 + x\\ 2g&<1,0,1,0,0,…>&G(x) = 1 + x^2\\ 3g&<1,0,0,1,0,…>&H(x) = 1 + x^3\\ 4g&<1,0,0,0,1,0,…>&I(x) = 1 + x^4 \end{cases} \]

\[P(x) = F(x)\times G(x)\times H(x)\times I(x)\\=1+x+x^2+2x^3+2x^4+2x^5+2x^6+2x^7+x^8+x^9+x^{10} \]

答案数列为:\(<1,1,1,2,2,2,2,2,1,1,1>\)

Q2. n个盘子的汉诺塔问题

\[f(x) = 2\times f(x-1)+1 \]

\[F(x) = 2xF(x) + \frac{1}{1-x} \]

\[\Downarrow \]

\[F(x) = \frac{1}{(x-1)(2x-1)} \]

更详细推导过程详见^[6]

Q3.无限砝码1g,2g,3g,4g

\[\begin{cases}1g&<1,1,1,1,1,1,…>&F(x)=\frac{1}{1-x}\\2g&<1,0,1,0,1,0,…>&G(x)=\frac{1}{1-x^2}\\3g&<1,0,0,1,0,0,1,…>&H(x)=\frac{1}{1-x^3}\\4g&<1,0,0,0,1,0,0,0,1,…>&I(x) = \frac1{1-x^4}\end{cases} \]

\[P(x) = F(x)\times G(x)\times H(x)\times I(x) \]

Q4.斐波那契数列

\[f(n) = f(n-1)+f(n-2)+[n==1] \]

\[F(x) = xF(x) + x^2F(x)+x \]

\[(1-x-x^2)F(x) = x \]

\[F(x) = \frac x {1-x-x^2} \]

Q5.

\[e^x = 1 + x + \frac 1 2x^2 + \frac 1 {3!}x^3+… \]

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1. 证明：\(xF(x) + 1= 1 + x + x^2 + x^3 + … = \frac{1}{1-x} = F(x)\\ \Downarrow \\F(x) = \frac{1}{1-x}\)​ ↩︎

2. 证明：令\(t = x^3\)、套用\(T_2\)即可 ↩︎

3. 证明：\(\frac{F(x)}{x^3} = 1 + x + x^2 = \frac{x^3}{1-x}\) ↩︎

4. 证明：从\(0\)到\(\infty-\)从\(4\)到\(\infty=\)从\(0\)到\(3\) ↩︎

5. 证明：\(F(x) = 1 + (2x) + (2x)^2 + (2x)^3 + …\)再用 ↩︎

6. 推导：\(f(x-1)\)生成函数为\(G(x) = xF(x)\)(右移)，\(2f(x-1)\)生成函数为\(H(x) = 2G(x) = 2xF(x)\)(数乘)，\(2f(x-1)+1\)生成函数即为\(F(x) = H(x) + \frac{1}{1-x}=2xF(x) + \frac 1 {1-x}···(T_2)\) ↩︎