之前讀研究所的作業。 😰
題目
Let $F_n$ denote the $n$-th fibonacci number. Let $\alpha=\frac{1+\sqrt{5}}{2}.$ Prove that
$$ F_n\geq\frac{\alpha^{n-\frac{1}{n}}}{\sqrt{5}}. $$
解答
首先我們知道幾件事
- $F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}, \forall n>0,\text{ where }\beta=\frac{1-\sqrt{5}}{2}$
- $\beta^n \leq \alpha^{-n}$
- $\alpha^2 - \alpha - 1 = 0$
所以這題其實只要證下面這件事情就好。
$$ \alpha^n-\beta^n\geq\alpha^{n-\frac{1}{n}} $$
為了容易閱讀，先令 $\gamma=\frac{1}{\sqrt[n]{\alpha}}$。我們知道 $\alpha>1>\gamma$，接著就開始 推倒 推導吧！
$$
\begin{aligned}
&\quad &\sum\limits_{i=0}^{n-1} {n-1 \choose i}\alpha^{i} &\geq \sum\limits_{i=0}^{n-1} \gamma^{i}\\
&\implies &(\alpha+1)^{n-1} &\geq \frac{1-\gamma^n}{1-\gamma}\\
&\implies &(\alpha+1)^{n-1}(1-\gamma) &\geq 1-\gamma^n\\
&\implies &(\alpha^2)^{n-1}(1-\frac{1}{\sqrt[n]{\alpha}}) &\geq 1-\frac{1}{\alpha}\\
&\implies &\alpha^{2n-2}-\alpha^{2n-2-\frac{1}{n}} &\geq 1-\frac{1}{\alpha}\\
&\implies &\alpha^{2n}-\alpha^{2n-\frac{1}{n}} &\geq \alpha^2-\alpha =1\\
&\implies &\alpha^n-\alpha^{n-\frac{1}{n}} &\geq \alpha^{-n}\\
&\implies &\alpha^n-\alpha^{-n} &\geq \alpha^{n-\frac{1}{n}}
\end{aligned}
$$
所以由上可得
$$ \alpha^n-\beta^n \geq \alpha^n-\alpha^{-n} \geq \alpha^{n-\frac{1}{n}} $$