Quod Erat Demonstrandum

2012/01/30

由 3 出發

Filed under: Fun — johnmayhk @ 3:52 下午

e.g. 1

$3$

$=\sqrt{1+8}$

$=\sqrt{1+2\times 4}$

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

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

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

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

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

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

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

$\dots$

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

e.g. 2

$3$

$=\sqrt{5+4}$

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

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

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

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

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

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

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

$\dots$

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

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 | 回應

• Not difficult at all. ^_^

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