裴礼文 一元函数极限 78页 例1.5.7 解答

castelu

例1.5.7：

　　设数列$\left\{p_n\right\}$，$\left\{q_n\right\}$满足
$$p_{n+1}=p_n+2q_n，q_{n+1}=p_n+q_n，p_1=q_1=1$$
　　求
$$\lim\limits_{n \to \infty}\frac{p_n}{q_n}$$

---

解法1：
　　令
$$a_n=\frac{p_n}{q_n}$$
　　则
$$\begin{eqnarray*}
\left|a_{n+1}-a_n \right|&=&\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n} \right|\\
&=&\left|\frac{p_n+2q_n}{p_n+q_n}-\frac{p_{n-1}+2q_{n-1}}{p_{n-1}+q_{n-1}} \right|\\
&=&\left|\frac{p_{n-1}q_n-p_nq_{n-1}}{\left(p_n+q_n\right)\left(p_{n-1}+q_{n-1}\right)} \right|\\
&=&\left|\frac{q_nq_{n-1}}{\left(p_n+q_n\right)\left(p_{n-1}+q_{n-1}\right)} \right|\left|a_n-a_{n-1} \right|\\
&=&\left|\frac{1}{\left(\frac{p_n}{q_n}+1\right)\left(\frac{p_{n-1}}{q_{n-1}}+1\right)} \right|\left|a_n-a_{n-1} \right|\\
&<&\frac{1}{4}\left|a_n-a_{n-1} \right|\\
\end{eqnarray*}$$
　　根据压缩映像原理，数列$\left\{a_n\right\}$收敛
　　注意到
$$a_{n+1}=\frac{p_{n+1}}{q_{n+1}}=\frac{p_n+2q_n}{p_n+q_n}=1+\frac{q_n}{p_n+q_n}=1+\frac{1}{1+a_n}$$
　　不妨设
$$\lim\limits_{n \to \infty}a_n=a$$
　　对递推式两边求极限
$$a=1+\frac{1}{1+a}$$
　　由于
$$a_n \ge 1，a=\sqrt{2}$$
　　即
$$\lim\limits_{n \to \infty}\frac{p_n}{q_n}=\sqrt{2}$$

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解法2：
　　将双变量递推改写成矩阵形式
$$\left( {\begin{array}{*{20}{c}}
{p_n}\\
{q_n}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
1&2\\
1&1
\end{array}} \right)
\left( {\begin{array}{*{20}{c}}
{p_{n-1}}\\
{q_{n-1}}
\end{array}} \right)$$
　　由递推公式得到
$$\left( {\begin{array}{*{20}{c}}
{p_n}\\
{q_n}
\end{array}} \right)=\left( {\begin{array}{*{20}{c}}
1&2\\
1&1
\end{array}} \right)^{n-1}
\left( {\begin{array}{*{20}{c}}
{p_1}\\
{q_1}
\end{array}} \right)$$
　　令
$$A = \left( {\begin{array}{*{20}{c}}
1&2\\
1&1
\end{array}} \right)$$
　　可知$A$的特征多项式
$$\left|\lambda E - A\right|=\lambda^2-2\lambda-1$$
　　可知$A$的特征值
$$\lambda_1=1+\sqrt{2}，\lambda_2=1-\sqrt{2}$$
　　经过计算
$$\exists P=
\left( {\begin{array}{*{20}{c}}
\sqrt{2}&-\sqrt{2}\\
1&1
\end{array}} \right)，
P^{-1}AP=
\left( {\begin{array}{*{20}{c}}
{1+\sqrt{2}}&0\\
0&{1-\sqrt{2}}
\end{array}} \right)$$
　　于是
$$\begin{eqnarray*}
A^{n-1}&=&P
\left( {\begin{array}{*{20}{c}}
\left(1+\sqrt{2}\right)^{n-1}&0\\
0&\left(1-\sqrt{2}\right)^{n-1}
\end{array}} \right)P^{-1}\\&=&
\frac{1}{2\sqrt{2}}\left( {\begin{array}{*{20}{c}}
{\sqrt{2} \left( \left( 1+\sqrt{2} \right)^{n-1}+\left( 1-\sqrt{2} \right)^{n-1} \right)}&{2 \left( \left( 1+\sqrt{2} \right)^{n-1}-\left( 1-\sqrt{2} \right)^{n-1} \right)}\\
{\left( 1+\sqrt{2} \right)^{n-1}-\left( 1-\sqrt{2} \right)^{n-1}}&{\sqrt{2} \left( \left( 1+\sqrt{2} \right)^{n-1}+\left( 1-\sqrt{2} \right)^{n-1} \right)}
\end{array}} \right)\\
\end{eqnarray*}$$
　　所以
$$\frac{p_n}{q_n}=\frac{\left( 2+\sqrt{2} \right) + \left( -2+\sqrt{2} \right)\left( 2\sqrt{2} - 3 \right)^{n-1}}{\left( 1+\sqrt{2} \right) + \left( -1+\sqrt{2} \right)\left( 2\sqrt{2} - 3 \right)^{n - 1}}$$
　　令
$$n \to \infty$$
　　得到
$$\lim\limits_{n \to \infty}\frac{p_n}{q_n}=\sqrt{2}$$