递推式转和式
总结一下数学基础
求和因子法
用途
解形如 \(a_i = ca_{i - 1} + b\) 的线性递推式。
做法
寻找一个数 \(s_i\) 使得 \(cs_i = s_{i - 1}\) , 两边都乘上 \(s_i\)
那么 :
\[s_ia_i = cs_ia_{i - 1} + s_ib\\
s_ia_i = s_{i - 1}a_{i - 1} + s_ib\\
设 T_i = s_ia_i\\
T_i = T_{i - 1} + s_ib\\
即 T_n = T_0 + \sum_{i=1}^{n}s_ib\\
a_n = \frac{T_n}{s_n} = \frac{T_0 + \sum_{i=1}^{n}s_ib}{s_n}
\]
实践一下
解 \(T_0 = 0\) , \(T_i = 2T_{i - 1} + 1\)
取 \(s_i = \frac{1}{2^i}\)
那么 :
\[\frac{1}{2^i}T_i = \frac{1}{2^i}2T_{i - 1} + \frac{1}{2^i}\\
\frac{1}{2^i}T_i = \frac{1}{2^{i - 1}}T_{i - 1} + \frac{1}{2^i}\\
设 A_i = \frac{1}{2^i}T_i\\
A_i = A_{i - 1} + \frac{1}{2^i}\\
A_n = A_0 + \sum_{i=1}^{n} \frac{1}{2^i}\\
A_n = \sum_{i=1}^{n} \frac{1}{2^i}\\
A_n = 1 - \frac{1}{2^n}\\
T_n = 2^nA_n = 2^n(1 - \frac{1}{2^n}) = 2^n - 1
\]
等比数列法 (不知道叫什么,自己取的名)
用途
解形如 \(a_i = ca_{i - 1} + b\) 的线性递推式。
结论
\(a_n = c^n(a_0 + \frac{b}{c - 1}) - \frac{b}{c - 1}\)
证明
设存在一个 \(x\) 使得 \(a_i+x = c(a_{i - 1} + x)\)
尝试求出 \(x\)
\[\because a_i+x = c(a_{i - 1} + x)\\
\therefore a_i + x = ca_{i - 1} + cx\\
\because a_i = ca_{i - 1} + b\\
\therefore ca_{i - 1} + b + x = ca_{i - 1} + cx\\
\therefore b + x = cx\\
\therefore b = (c - 1)x\\
\therefore x = \frac{b}{c - 1}\\
\]
所以 \(a_i+\frac{b}{c - 1} = c(a_{i - 1} + \frac{b}{c - 1})\)
设 \(T_i\) 表示 \(a_i + \frac{b}{c - 1}\)
那么 \(T_i = cT_{i - 1}\)
\(T_n = c^nT_0\)
\(a_n + \frac{b}{c - 1} = c^n(a_0 + \frac{b}{c - 1})\)
\(a_n = c^n(a_0 + \frac{b}{c - 1}) - \frac{b}{c - 1}\)
实践一下
解 \(T_0 = 0\) , \(T_i = 2T_{i - 1} + 1\)
\[T_n = 2^n(T_0 + \frac{1}{2 - 1}) - \frac{1}{2 - 1}\\
T_n = 2^n(0 + 1) - 1\\
T_n = 2^n - 1
\]
剩下的先咕着