نضع
$\displaystyle{\displaylines{ \varphi = \frac{1+\sqrt{5}}{2}}}$نُعرف
متتالية فيبوناشي Suite de Fibonacci كالآتي :
$\displaystyle{\displaylines{\left\{\begin{matrix}u_0 = u_1 = 1\\ u_{n+2} = u_{n+1} + u_{n}\end{matrix}\right.}}$1) بين أن $\displaystyle{\displaylines{\varphi^2 = \varphi+ 1}}$ و $\displaystyle{\displaylines{ \frac{1}{ \varphi } = \varphi - 1}}$
2) بين أن $\displaystyle{\displaylines{\forall n \in \mathbb{N} \, : \, u_n = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+1} + \frac{(-1)^{n}}{\varphi^{n+1}}\right)}}$
3) بين أن : $\displaystyle{\displaylines{ \lim_{n\rightarrow + \infty} \frac{u_{n+1}}{u_n} = \varphi }}$
نعتبر المعادلة التالية في $\displaystyle{\displaylines{\mathbb{R}}}$ : $\displaystyle{\displaylines{x^2 = x + 1}}$
لدينا : $\displaystyle{\displaylines{ \Delta = \sqrt{(-1)^2 - 4 \times (-1)} = 5 }}$
إذن حلول المعادلة هما : $\displaystyle{\displaylines{x_1 = \frac{1+\sqrt{5}}{2}}}$ و $\displaystyle{\displaylines{x_2 = \frac{1-\sqrt{5}}{2}}}$
لدينا : $\displaystyle{\displaylines{ \varphi = x_1 }}$
إذن $\displaystyle{\displaylines{ \varphi^2 = \varphi+ 1 }}$
لدينا $\displaystyle{\displaylines{ 1 = \varphi^2 - \varphi }}$
إذن $\displaystyle{\displaylines{ \frac{1}{ \varphi } = \varphi - 1}}$
البرهان بالترجع من الدرجة 2, démonstration par récurrence d’ordre 2 :
من أجل $\displaystyle{\displaylines{ n = 0}}$ لدينا : $\displaystyle{\displaylines{u_0 = \frac{1}{\sqrt{5}} \, \left(\varphi^{0+1} + \frac{(-1)^{0}}{\varphi^{0+1}}\right) = 1}}$
من أجل $\displaystyle{\displaylines{ n = 1}}$ لدينا : $\displaystyle{\displaylines{u_1 \, = \frac{1}{\sqrt{5}} \, \left(\varphi^{1+1} + \frac{(-1)^{1}}{\varphi^{1+1}}\right) = 1}}$
لدينا الخاصية صحيحة من أجل $\displaystyle{\displaylines{ n = 0 , 1}}$, نفترض أن $\displaystyle{\displaylines{u_n = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+1} + \frac{(-1)^{n}}{\varphi^{n+1}}\right)}}$ و $\displaystyle{\displaylines{u_{n+1} = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+2} + \frac{(-1)^{n+1}}{\varphi^{n+2}}\right) }}$
لنبين أن : $\displaystyle{\displaylines{u_{n+2} = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+3} + \frac{(-1)^{n+2}}{\varphi^{n+3}}\right)}}$
لدينا : $\displaystyle{\displaylines{u_{n+2} = u_{n+1} + u_{n}}}$
إذن :
$\displaystyle{\displaylines{u_{n+2} = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+1} \, (\varphi + 1) + \frac{(-1)^{n+2}}{\varphi^{n+1}} \, \left(1 - \frac{1}{\varphi}\right)\right)}}$
لدينا حسب السؤال السابق : $\displaystyle{\displaylines{\varphi + 1 = \varphi^2}}$ و $\displaystyle{\displaylines{1 - \frac{1}{\varphi} = \frac{1}{\varphi^2}}}$
إذن :
$\displaystyle{\displaylines{u_{n+2} = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+3} + \frac{(-1)^{n+2}}{\varphi^{n+3}}\right)}}$
إذن حسب البرهان بالترجع من الدرجة 2 لدينا :
$\displaystyle{\displaylines{\forall n \in \mathbb{N} \ : \ u_n = \frac{1}{\sqrt{5}} \, \left(\varphi^{n+1} + \frac{(-1)^{n}}{\varphi^{n+1}}\right)}}$
لدينا :
$\displaystyle{\displaylines{\forall n \in \mathbb{N} \ : \\frac{u_{n+1}}{u_n} = \frac{\varphi^{n+2} + \frac{(-1)^{n+1}}{\varphi^{n+2}}}{\varphi^{n+1} + \frac{(-1)^{n}}{\varphi^{n+1}}}}}$
إذن :
$\displaystyle{\displaylines{\frac{u_{n+1}}{u_n} = \frac{\varphi + \frac{(-1)^{n+1}}{\varphi^{2 n+3}}}{1 + \frac{(-1)^{n}}{\varphi^{2 n+2}}}}}$
لدينا $\displaystyle{\displaylines{\varphi = \frac{1+\sqrt{5}}{2} > 1}}$ إذن : $\displaystyle{\displaylines{ \lim_{N \rightarrow + \infty} \frac{1}{\varphi^N} = 0}}$
إذن :
$\displaystyle{\displaylines{\lim_{n \rightarrow + \infty} \frac{(-1)^{n+1}}{\varphi^{2 n+3}} = \lim_{n \rightarrow + \infty} \frac{(-1)^{n}}{\varphi^{2 n+2}} = 0}}$
وبالتالي :
$\displaystyle{\displaylines{\lim_{n\rightarrow + \infty} \frac{u_{n+1}}{u_n} = \varphi}}$
