遞推式轉和式

總結一下數學基礎

求和因子法

用途

解形如 \(a_i = ca_{i - 1} + b\) 的線性遞推式。

作法

尋找一個數 \(s_i\) 使得 \(cs_i = s_{i - 1}\) , 兩邊都乘上 \(s_i\)

那麼 :

 

\[s_ia_i = cs_ia_{i - 1} + s_ib\\ s_ia_i = s_{i - 1}a_{i - 1} + s_ib\\ 設 T_i = s_ia_i\\ T_i = T_{i - 1} + s_ib\\ 即 T_n = T_0 + \sum_{i=1}^{n}s_ib\\ a_n = \frac{T_n}{s_n} = \frac{T_0 + \sum_{i=1}^{n}s_ib}{s_n} \]

 

實踐一下

解 \(T_0 = 0\) , \(T_i = 2T_{i - 1} + 1\)

取 \(s_i = \frac{1}{2^i}\)

那麼 :

 

\[\frac{1}{2^i}T_i = \frac{1}{2^i}2T_{i - 1} + \frac{1}{2^i}\\ \frac{1}{2^i}T_i = \frac{1}{2^{i - 1}}T_{i - 1} + \frac{1}{2^i}\\ 設 A_i = \frac{1}{2^i}T_i\\ A_i = A_{i - 1} + \frac{1}{2^i}\\ A_n = A_0 + \sum_{i=1}^{n} \frac{1}{2^i}\\ A_n = \sum_{i=1}^{n} \frac{1}{2^i}\\ A_n = 1 - \frac{1}{2^n}\\ T_n = 2^nA_n = 2^n(1 - \frac{1}{2^n}) = 2^n - 1 \]

 

等比數列法 （不知道叫什麼，本身取的名）

用途

解形如 \(a_i = ca_{i - 1} + b\) 的線性遞推式。

結論

\(a_n = c^n(a_0 + \frac{b}{c - 1}) - \frac{b}{c - 1}\)

證實

設存在一個 \(x\) 使得 \(a_i+x = c(a_{i - 1} + x)\)

嘗試求出 \(x\)

 

\[\because a_i+x = c(a_{i - 1} + x)\\ \therefore a_i + x = ca_{i - 1} + cx\\ \because a_i = ca_{i - 1} + b\\ \therefore ca_{i - 1} + b + x = ca_{i - 1} + cx\\ \therefore b + x = cx\\ \therefore b = (c - 1)x\\ \therefore x = \frac{b}{c - 1}\\ \]

 

因此 \(a_i+\frac{b}{c - 1} = c(a_{i - 1} + \frac{b}{c - 1})\)

設 \(T_i\) 表示 \(a_i + \frac{b}{c - 1}\)

那麼 \(T_i = cT_{i - 1}\)

\(T_n = c^nT_0\)

\(a_n + \frac{b}{c - 1} = c^n(a_0 + \frac{b}{c - 1})\)

\(a_n = c^n(a_0 + \frac{b}{c - 1}) - \frac{b}{c - 1}\)

實踐一下

解 \(T_0 = 0\) , \(T_i = 2T_{i - 1} + 1\)

 

\[T_n = 2^n(T_0 + \frac{1}{2 - 1}) - \frac{1}{2 - 1}\\ T_n = 2^n(0 + 1) - 1\\ T_n = 2^n - 1 \]

 

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