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高中数学数列[原创]线性递归关系式求解

秦序>方法技巧>数列>正文 edit
秦序网
  2016-8-4 23:35:05

Abstract:

对于陈景润先生的《组合数学》 5.2节 线性递归关系式的解 给出的结论中的一种情形做一点讨论。

Keywords:

数列, 线性递归,通项公式,连加法  
陈景润先生在其《组合数学》在 5.2 线性递归关系式的解中介绍了形如
\[
a_n=c_1a_{n-1}+c_2a_{n-2}+\cdots+c_ma_{n-m}=\sum\limits_{    j=1}^mc_ja_{n-j} (n\ge m+1)
\]
递推公式决定的数列 ($a_1, a_2, \cdots, a_m$已知) 通项形为
\begin{equation}\label{generalsolution}
a_n=k_1\alpha_1^n+k_2\alpha_2^n+\cdots+k_m\alpha_m^n=\sum\limits_{j=1}^mk_j\alpha_j^n.
\end{equation}
其中$\alpha_1, \alpha_2, \cdots, \alpha_m$为方程$\alpha^m=c_1\alpha^{m-1}+c_2\alpha^{m-2}+\cdots+c_m$的$m$个根 (不论是否为复根,此处暂设无重根) .
在$\alpha_1, \alpha_2, \cdots, \alpha_m$若有重根,不妨设$\alpha_1$为$m_1$重,$\alpha_2$为$m_2$重,…,$\alpha_t$为$m_t$重,且$\sum\limits_{j=1}^tm_j=m$,则用
\[
\alpha_1^n, n\alpha_1^n, \cdots, n^{m_1-1}\alpha_1^n, \alpha_2^n, n\alpha_2^n,\cdots, n^{m_2-1}\alpha_2^n, \cdots, \alpha_t^n, n\alpha_t^n,\cdots, n^{m_t-1}\alpha_t^n
\]
分别对应替换 \eqref{generalsolution} 中的$\alpha_1^n, \alpha_2^n, \cdots, \alpha_m^n$即得.
以下证明$m=2$时的情形. (不当之处欢迎斧正)
**********************************
已知:$a_1, a_2, a_n=pa_{n-1}+qa_{n-2} (n\ge3)$,求$\{a_n\}$的通项公式.
解:令$a_n=\alpha^n (n\ge3)$,则由$a_n=pa_{n-1}+qa_{n-2}$得
\[
\alpha^2-p\alpha-q=0.
\]
解之得根$\alpha_1, \alpha_2$.
即$\alpha^2-(\alpha_1+\alpha_2)\alpha+\alpha_1\alpha_2=0$,得
\begin{equation}
a_n=(\alpha_+\alpha_2)a_{n-1}-\alpha_1\alpha_2.
\end{equation}
稍作调整即得
\begin{align*}
a_n-\alpha_1a_{n-1} & = \alpha_2a_{n-1}-\alpha_1\alpha_2a_{n-2}\\
& = \alpha_2(a_{n-1}-\alpha_1a_{n-2})\\
\dfrac{a_n-\alpha_1a_{n-1}}{a_{n-1}-\alpha_1a_{n-2}} & = \alpha_2.
\end{align*}
则$\{a_n-\alpha_1a_{n-1}\}$为以$a_2-\alpha_1a_1$为首项,以$\alpha_2$为等比的等比数列.
易得
\begin{equation}\label{ker}
a_n-\alpha_1a_{n-1}=\alpha_2^{n-2}(a_2-\alpha_1a_1).
\end{equation}
以下按照$\alpha_1=\alpha_2$和$\alpha_1\ne\alpha_2$分别讨论.

若$\alpha_1=\alpha_2$,则 \eqref{ker} 为
\begin{equation}\label{eqker}
a_n-\alpha_1a_{n-1}=\alpha_1^{n-2}(a_2-\alpha_1a_1).
\end{equation}
稍作化简即得
\[
\dfrac{a_n}{\alpha_1^n}-\dfrac{a_{n-1}}{\alpha_1^{n-1}}=\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}.
\]
即$\{\dfrac{a_n}{\alpha_1^n}\}$为以$\dfrac{a_1}{\alpha_1}$为首项,以$\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}$为等差的等差数列. 得
\begin{align*}
\dfrac{a_n}{\alpha_1^n} & = \dfrac{a_1}{\alpha_1}+(n-1)\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}\\
& = \dfrac{a_1}{\alpha_1}-\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}+\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}\cdot n\\
& = \dfrac{2\alpha_1a_1-a_2}{\alpha_1^2}+\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}\cdot n.
\end{align*}

\begin{align}
a_n & = \dfrac{2\alpha_1a_1-a_2}{\alpha_1^2}\cdot\alpha_1^n+\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}\cdot n\alpha_1^n\\
& = k_1\alpha_1^n+k_2n\alpha_1^n,
\end{align}
其中$k_1=\dfrac{2\alpha_1a_1-a_2}{\alpha_1^2}, k_2=\dfrac{a_2-\alpha_1a_1}{\alpha_1^2}$.

若$\alpha_1\ne\alpha_2$,则对 \eqref{ker} 等式两边同除以$\alpha_1^n$得
\begin{equation}\label{neker}
\dfrac{a_n}{\alpha_1^n}-\dfrac{a_{n-1}}{\alpha_1^{n-1}}=\dfrac{\alpha_2^{n-2}}{\alpha_1^n}(a_2-\alpha_1a_1)=\dfrac{a_2-\alpha_1a_1}{\alpha_2^2}(\dfrac{\alpha_2}{\alpha_1})^n.
\end{equation}
利用连加法由 \eqref{neker} 可得
\begin{align*}
\dfrac{a_n}{\alpha_1^n}-\dfrac{a_1}{\alpha_1} & = \dfrac{a_2-\alpha_1a_1}{\alpha_2^2}\sum\limits_{j=2}^n(\dfrac{\alpha_2}{\alpha_1})^j\\
& = \dfrac{a_2-\alpha_1a_1}{\alpha_2^2}\cdot(\dfrac{\alpha_2}{\alpha_1})^2\cdot\dfrac{1-(\dfrac{\alpha_2}{\alpha_1})^{n-1}}{1-\dfrac{\alpha_2}{\alpha_1}}\\
& = \dfrac{a_2-\alpha_1a_1}{\alpha_1^2}\cdot\dfrac{\alpha_1-\dfrac{\alpha_2^{n-1}}{\alpha_1^{n-2}}}{\alpha_1-\alpha_2}\\
& = \dfrac{a_2-\alpha_1a_1}{\alpha_1^2(\alpha_1-\alpha_2)}\cdot(\alpha_1-\dfrac{\alpha_2^{n-1}}{\alpha_1^{n-2}}).
\end{align*}
再作化简得
\begin{align*}
\dfrac{a_n}{\alpha_1^n} & = \dfrac{a_1}{\alpha_1}+\dfrac{a_2-\alpha_1a_1}{\alpha_1(\alpha_1-\alpha_2)}-\dfrac{a_2-\alpha_1a_1}{\alpha_2(\alpha_1-\alpha_2)}\cdot\dfrac{\alpha_2^n}{\alpha_1^n}\\
& = \dfrac{a_2-\alpha_2a_1}{\alpha_1(\alpha_1-\alpha_2)}+\dfrac{\alpha_1a_1-a_2}{\alpha_2(\alpha_1-\alpha_2)}\cdot\dfrac{\alpha_2^n}{\alpha_1^n}.
\end{align*}
即得
\begin{align}
a_n & = \dfrac{a_2-\alpha_2a_1}{\alpha_1(\alpha_1-\alpha_2)}\cdot\alpha_1^n+\dfrac{\alpha_1a_1-a_2}{\alpha_2(\alpha_1-\alpha_2)}\cdot\alpha_2^n\\
& = k_1\alpha_1^n+k_2\alpha_2^n,
\end{align}
其中$k_1=\dfrac{a_2-\alpha_2a_1}{\alpha_1(\alpha_1-\alpha_2)}, k_2=\dfrac{\alpha_1a_1-a_2}{\alpha_2(\alpha_1-\alpha_2)}$.

至此,即证明了 $m=2$ 时的情形.

**************************

从以上过程很容易发现不论是结果,还是推导过程均比较繁琐,故采用一开始的提到的待定系数法求解$k_j$.

Reference:

陈景润著《组合数学》
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