Quod Erat Demonstrandum

2012/01/30

由 3 出發

Filed under: Fun — johnmayhk @ 3:52 下午

由 3 出發,建立「有趣」關係。

e.g. 1

3

=\sqrt{1+8}

=\sqrt{1+2\times 4}

=\sqrt{1+2\sqrt{16}}

=\sqrt{1+2\sqrt{1+15}}

=\sqrt{1+2\sqrt{1+3\times 5}}

=\sqrt{1+2\sqrt{1+3\sqrt{25}}}

=\sqrt{1+2\sqrt{1+3\sqrt{1+24}}}

=\sqrt{1+2\sqrt{1+3\sqrt{1+4\times 6}}}

=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{36}}}}

\dots

再做一些分析,得

\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots}}}}=3

e.g. 2

3

=\sqrt{5+4}

=\sqrt{5+\sqrt{16}}

=\sqrt{5+\sqrt{6+2\times 5}}

=\sqrt{5+\sqrt{6+2\sqrt{25}}}

=\sqrt{5+\sqrt{6+2\sqrt{7+3\times 6}}}

=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{36}}}}

=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{8+4\times 7}}}}

=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{8+4\sqrt{49}}}}}

\dots

再做一些分析,得

\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{8+4\sqrt{9+\dots}}}}}=3

2 則迴響 »

  1. Is it difficult to prove the convergence? (or is it converge)
    (I havent attempt yet :P)

    迴響 由 Justin — 2012/01/30 @ 10:45 下午 | 回覆

    • Not difficult at all. ^_^

      迴響 由 johnmayhk — 2012/01/31 @ 1:41 下午 | 回覆


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