a(1)=3，b(1)=2，a(n+1)=a(n)+2b(n)，b(n+1)=a(n)+b(n)，则 lim(n→∞)a(n)/b(n)=√2

elim

题: 设\(\;(a_1,b_1)=(3,2),\;(a_{n+1},b_{n+1})=(a_n+2b_n,a_n+b_n).\)
\(\quad\)试证\(\{\frac{a_n}{b_n}\}\)收敛,\(\;\displaystyle\lim_{n\to\infty}{\small\frac{a_n}{b_n}}=\sqrt{2}\)
\(\quad\)试证\(\,{\small\sqrt{2}=1+}\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}}\)

luyuanhong

elim

谢谢陆老师的解. 很漂亮!
令\(\,\sqrt{2}=1+y,\,\text{i.e. }y(y+2)=1,\,\boxed{y=\small\frac{1}{2+y}}\)
\(\therefore\quad{\small\sqrt{2}=1+y=1+}\frac{1}{2+y}{\small=1+}\frac{1}{2+\frac{1}{2+y}}{\small=\cdots=1+}\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}}}\)
\(\therefore\;\;\sqrt{2}\,\)的渐近分数是\(\,{\small 1+}\frac{1}{2},\,{\small 1+}\frac{1}{2+\frac{1}{2}},\,{\small 1+}\frac{1}{2+\frac{1}{2+\frac{1}{2}}},\ldots\)
即\(\quad\frac{3}{2},\,\frac{7}{5},\,\frac{17}{12},\,\frac{41}{29},\ldots\)
所以\({\small\,\sqrt{2}}\,\)的渐近分数序列是\(\{\frac{a_n}{b_n}\},\)
\((a_1,b_1)=(3,2),\;\;(a_{n+1},b_{n+1})=(a_n+2b_n,a_n+b_n)\)

elim

\(\because\quad\underset{\,}{\frac{a_{n+1}}{b_{n+1}}}=\frac{a_n+2b_n}{a_n+b_n}\implies\frac{a_{n+2}}{b_{n+2}}=\frac{3a_n+4b_n}{2a_n+3b_n}\)
\(\therefore\underset{\,}{\quad\frac{a_{n+2}}{b_{n+2}}}-\frac{a_n}{b_n}=\frac{2b_n}{2a_n+3b_n}\big({\small 2-}(\frac{a_n}{b_n})^2\big)\)
\(\qquad{\small 2-}\big(\frac{3a_n+4b_n}{2a_n+3b_n}\big)^2=\big(\frac{b_n}{2a_n+3b_n}\big)^2\big({\small 2-}\big(\frac{a_n}{b_n}\big)^2\big)\)
\(\therefore\quad\sqrt{2}-\frac{a_{n+2}}{b_{n+2}},\;\sqrt{2}-\frac{a_n}{b_n},\,\overset{\,}{\frac{a_{n+2}}{b_{n+2}}}-\frac{a_n}{b_n}\,\)同号. 进而得
\(\qquad\overset{\,}{\frac{a_2}{b_2}}< \cdots< \frac{a_{2n}}{b_{2n}}< \sqrt{2}< \frac{a_{2n-1}}{b_{2n-1}}< \cdots< \frac{a_1}{b_1}.\;\)
\(\qquad\)据此易见\(\,\displaystyle\lim_{n\to\infty}{\small\frac{a_n}{b_n}}=\sqrt{2}.\)

另外从递归关系得\(\,a_{n+1}=b_{n+1}+b_n=b_{n+2}-b_{n+1}.\,\)于是
\(b_{n+2}-2b_{n+1}-b_n=0,\;\; b_n=\frac{1}{2\sqrt{2}}\big({\small(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}}\big)\)
进而得\(\,a_n=\frac{1}{2}\big({\small(1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1}}\big)\)
注意\(\,\small(1+\sqrt{2})^n\,(\color{blue}{(1-\sqrt{2})^n}))\to\infty\,(\color{blue}{0})\;{\small(n\to\infty),}\;\,\displaystyle\lim_{n\to\infty}{\small\frac{a_n}{b_n}}=\sqrt{2}.\)
再解:\(\small\begin{bmatrix}1& 2\\1& 1\end{bmatrix}\,\)的特征值,单位特征向量由下式给出:
\(\qquad\;\begin{bmatrix}1& 2\\1& 1\end{bmatrix}\begin{bmatrix}(2/3)^{1/2}\\ \pm 3^{-1/2}\end{bmatrix}=(1\pm\sqrt{2})\begin{bmatrix}(2/3)^{1/2}\\ \pm 3^{-1/2}\end{bmatrix}\)
于是\(\,\underset{\,}{\begin{bmatrix}a_n\\ b_n\end{bmatrix}}=\begin{bmatrix}1& 2\\1& 1\end{bmatrix}^{n-1}\begin{bmatrix}a_1\\ b_1\end{bmatrix}\)
\(\small=\scriptsize\begin{bmatrix}(2/3)^{1/2}&(2/3)^{1/2}\\ 3^{-1/2}&-3^{-1/2}\end{bmatrix}\begin{bmatrix}(1+\sqrt{2})^{n-1}& 0\\ 0&(1-\sqrt{2})^{n-1}\end{bmatrix}\begin{bmatrix}(2/3)^{1/2}&(2/3)^{1/2}\\ 3^{-1/2}&-3^{-1/2}\end{bmatrix}^{-1}\begin{bmatrix}3\\2\end{bmatrix}\)
\(\small=\begin{bmatrix}\frac{1}{2}((1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1})\\ \frac{1}{2\sqrt{2}}((1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}) \end{bmatrix}.\quad\displaystyle\lim_{n\to\infty}{\small\frac{a_n}{b_n}}=\sqrt{2}\)

luyuanhong

楼上 elim 的解答也很好！已收藏。