math-ai/BlueMO

1{2    "source_file": "./raw_volume-zh/volume6/chapter1-7.tex",3    "problem_type": "calculation",4    "problem": "例2. 数列 $\\left\\{a_n\\right\\}$ 满足 $a_1==2$, 对 $n=1,2, \\cdots$ 有\n$$\na_{n+1}=\\frac{a_n}{2}+\\frac{1}{a_n} . \\label{eq1}\n$$\n求该数列的通项公式.",5    "solution": "解:这里介绍不动点处理的方法 (它源于函数迭代的思想), 先求方程\n$$\n\\lambda=\\frac{\\lambda}{2}+\\frac{1}{\\lambda} . \\label{eq2}\n$$\n的解, 得 $\\lambda_{1,2}= \\pm \\sqrt{2}$.\n注意到 $a_1=2$, 结合式\\ref{eq1}可知, $\\left\\{a_n\\right\\}$ 的每一项都是正有理数, 现在用式\\ref{eq1}-\\ref{eq2}, 可知对 $\\lambda= \\pm \\sqrt{2}$ 都有\n$$\na_{n+1}-\\lambda=\\frac{a_n-\\lambda}{2}+\\left(\\frac{1}{a_n}-\\frac{1}{\\lambda}\\right),\n$$\n变形为\n$$\n\\frac{a_{n+1}-\\lambda}{a_n-\\lambda}=\\frac{1}{2}-\\frac{1}{\\lambda a_n}=\\frac{\\lambda a_n-2}{2 \\lambda a_n} . \\label{eq3}\n$$\n在式\\ref{eq3}中分别取 $\\lambda=\\sqrt{2} 、-\\sqrt{2}$ 所得两式作商, 得\n$$\n\\frac{a_{n+1}-\\sqrt{2}}{a_{n+1}+\\sqrt{2}}=\\left(\\frac{a_n-\\sqrt{2}}{a_n+\\sqrt{2}}\\right)^2 \\text {. }\n$$\n于是, 我们有\n$$\n\\begin{aligned}\n\\frac{a_{n+1}-\\sqrt{2}}{a_{n+1}+\\sqrt{2}} & =\\left(\\frac{a_n-\\sqrt{2}}{a_n+\\sqrt{2}}\\right)^2=\\left(\\frac{a_{n-1}-\\sqrt{2}}{a_{n-1}+\\sqrt{2}}\\right)^{2^2} \\\\\n& =\\cdots=\\left(\\frac{a_1-\\sqrt{2}}{a_1+\\sqrt{2}}\\right)^{2^n}=(3-2 \\sqrt{2})^{2^n} \\\\\n& =(\\sqrt{2}-1)^{2^{n+1}} .\n\\end{aligned}\n$$\n所以 $\\frac{a_n-\\sqrt{2}}{a_n+\\sqrt{2}}=(\\sqrt{2}-1)^{2^n}$, 解得 $a_n=\\frac{\\sqrt{2}\\left(1+(\\sqrt{2}-1)^{2^n}\\right)}{1-(\\sqrt{2}-1)^{2^n}}$.",6    "remark": "",7    "figures": []8}