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发表于 2020-8-9 01:01
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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\(\{\frac{a_n}{b_n}\},\)
\((a_1,b_1)=(3,2),\;\;(a_{n+1},b_{n+1})=(a_n+2b_n,a_n+b_n)\) |
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