# Quod Erat Demonstrandum

## 2018/11/20

### 費氏講

Filed under: Additional / Applied Mathematics,Fun,Junior Form Mathematics — johnmayhk @ 6:36 下午
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N 年前往中一班代堂，必談「64 = 65」謎題：

(圖片來源：https://i.stack.imgur.com/fWdMd.jpg)

(圖片來源：https://i.stack.imgur.com/nWWOQ.png)

$(a+b)c+1=b(b+c)$

$\Rightarrow b^2=ac+1$ …………………………. (*)

$F_1=1$
$F_2=1$
$F_3=F_2+F_1=2$
$F_4=F_3+F_2=3$
$F_5=F_4+F_3=5$
$F_6=F_5+F_4=8$

… …

$F_{n-1}F_{n+1}-F_{n}^2=(-1)^n$

$\left(\begin{array}{rcl}F_{n-1} &F_n \\F_n &F_{n+1} \\\end{array}\right)=\left(\begin{array}{rcl}0 &1 \\1 &1 \\\end{array}\right)^n$

$F_{n-1}F_{n+1}-F_{n}^2=\det\left(\begin{array}{rcl}F_{n-1} &F_n \\F_n &F_{n+1} \\\end{array}\right)=\det\left(\begin{array}{rcl}0 &1 \\1 &1 \\\end{array}\right)^n=(\det\left(\begin{array}{rcl}0 &1 \\1 &1 \\\end{array}\right))^n=(-1)^n$

$\cot^{-1}F_{2n-2}=\cot^{-1}F_{2n-1}+\cot^{-1}F_{2n}$

$\alpha=\cot^{-1}F_{2n-2}$
$\beta=\cot^{-1}F_{2n-1}$
$\gamma=\cot^{-1}F_{2n}$

$\cot\alpha=F_{2n-2}$
$\cot\beta=F_{2n-1}$
$\cot\gamma=F_{2n}$

$\tan\alpha=1/F_{2n-2}$
$\tan\beta=1/F_{2n-1}$
$\tan\gamma=1/F_{2n}$

$\displaystyle \tan(\beta+\gamma)=\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}$

$\displaystyle =\frac{1/F_{2n-1}+1/F_{2n}}{1-1/(F_{2n-1}F_{2n})}$

$\displaystyle =\frac{F_{2n-1}+F_{2n}}{F_{2n-1}F_{2n}-1}$

$\displaystyle =\frac{F_{2n-1}+F_{2n}}{F_{2n-1}(F_{2n-1}+F_{2n-2})-1}$

$\displaystyle =\frac{F_{2n-1}+F_{2n}}{F_{2n-1}^2+F_{2n-1}F_{2n-2}-1}$

$\displaystyle =\frac{F_{2n-1}+F_{2n}}{F_{2n-2}F_{2n}+F_{2n-1}F_{2n-2}}$　　（這裡用了卡西尼公式）

$\displaystyle =\frac{1}{F_{2n-2}}$

$\displaystyle =\tan\alpha$

$\alpha=\beta+\gamma$

$\cot^{-1}F_{2n-2}=\cot^{-1}F_{2n-1}+\cot^{-1}F_{2n}$

$\cot^{-1}F_2=\cot^{-1}F_3+\cot^{-1}F_4=\cot^{-1}F_3+\cot^{-1}F_5+\cot^{-1}F_6=\cot^{-1}F_3+\cot^{-1}F_5+\cot^{-1}F_7+\dots$

$\displaystyle \frac{4}{\pi}=\cot^{-1}2+\cot^{-1}5+\cot^{-1}13+\cot^{-1}34+\dots$

$\displaystyle \pi=4\sum^\infty_{i=1}\tan^{-1}\frac{1}{F_{2i+1}}$

$\displaystyle \tan^{-1}\frac{1}{xF_{2n-1}+F_{2n}}=\tan^{-1}\frac{1}{xF_{2n-1}+F_{2n+1}}+\tan^{-1}\frac{1}{x^2F_{2n-1}+xF_{2n+2}+F_{2n+2}}$

http://www.m-hikari.com/ijcms/ijcms-2015/5-8-2015/p/frontczakIJCMS5-8-2015.pdf

$b^2=ac$

$\displaystyle \frac{b}{a}$ 必然是黃金比（golden ratio），（中四同學試證之）這也和費氏數到有著連繫，即 $\displaystyle \frac{b}{a}=\lim_{n\rightarrow \infty}\frac{F_{n+1}}{F_n}$