# Quod Erat Demonstrandum

## 2018/03/14

### 黃金比某級數

Filed under: Fun,mathematics,NSS — johnmayhk @ 11:11 上午
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$\displaystyle \Phi=\frac{1}{\Phi}+\frac{1}{\Phi^2}+\frac{1}{\Phi^3}+\dots$

$\displaystyle \Phi=\frac{1+\sqrt{5}}{2}$

$\displaystyle \frac{\Phi-1}{1}=\frac{1}{\Phi}$

$\displaystyle \Phi-1=\frac{1}{\Phi}$ …………… (*)

$\displaystyle \frac{1}{\Phi}+\frac{1}{\Phi^2}=1$ ……………… (**)

$\displaystyle \frac{1}{\Phi^2}$

$\displaystyle =\frac{1}{\Phi^2}(\frac{1}{\Phi}+\frac{1}{\Phi^2})$

$\displaystyle =\frac{1}{\Phi^3}+\frac{1}{\Phi^4}$

$\displaystyle =\frac{1}{\Phi^3}+\frac{1}{\Phi^4}(\frac{1}{\Phi}+\frac{1}{\Phi^2})$

$\displaystyle =\frac{1}{\Phi^3}+\frac{1}{\Phi^5}+\frac{1}{\Phi^6}$

$\displaystyle =\frac{1}{\Phi^3}+\frac{1}{\Phi^5}+\frac{1}{\Phi^6}(\frac{1}{\Phi}+\frac{1}{\Phi^2})$

$\displaystyle =\frac{1}{\Phi^3}+\frac{1}{\Phi^5}+\frac{1}{\Phi^7}+\dots$

$\displaystyle \Phi=\frac{1}{\Phi}+\frac{1}{\Phi^2}+\frac{1}{\Phi^3}+\dots$

$\displaystyle \pi=\frac{5}{\Phi} \frac{2}{\sqrt{2+\sqrt{2+\Phi}}} \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+\Phi}}}} \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\Phi}}}}} \dots$