Complex Fibonacci

Calculus Level 5

Binet's Formula is a closed form expression for Fibonacci numbers:

F t = φ t ( φ ) t 5 F_t=\dfrac{\varphi^t-(-\varphi)^{-t}}{\sqrt{5}}

It can be said that F t N F_t\in\mathbb{N} when t N t\in\mathbb{N} , however, F t C F_t\in\mathbb{C} when t R + t\in\mathbb{R}^+ .

If we plot F t F_t in the complex plane for t R + t\in\mathbb{R}^+ , we get an oscillating curve which intersects the real axis at the Fibonacci numbers, as shown in the picture above.

Let A n A_n denote the area under a segment between two consecutive Fibonacci numbers ( n th n^\text{th} and ( n + 1 ) th (n+1)^\text{th} ).

Find lim n A n \displaystyle \lim_{n\to\infty} A_n .

Note: φ \varphi denotes the golden ratio .

ln φ 2 π \dfrac{\ln{\varphi}}{2\pi} ln φ 3 π \dfrac{\ln{\varphi}}{3\pi} 4 ln φ 9 π \dfrac{4\ln{\varphi}}{9\pi} 2 ln φ 5 π \dfrac{2\ln{\varphi}}{5\pi}

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1 solution

Wesley Low
Jul 7, 2020

Binet's Formula can be rewritten as

φ n ( φ ) n 5 = φ n 5 φ n 5 cos n π + φ n i 5 sin n π \frac{{\varphi}^{n}-{\left(-\varphi\right)}^{-n}}{\sqrt{5}}=\frac{{\varphi}^n}{\sqrt{5}}-\frac{{\varphi}^{-n}}{\sqrt{5}}\cos{n\pi}+\frac{{\varphi}^{-n}i}{\sqrt{5}}\sin{n\pi}

Since the curve between two integers is either completely negative or positive, we can define A n A_n as

A n = n n + 1 ( φ t 5 sin t π ) d ( φ t 5 φ t 5 cos t π ) A_n=\left|\int_{n}^{n+1} \left(\frac{{\varphi}^{-t}}{\sqrt{5}}\sin{t\pi} \right)d\left(\frac{{\varphi}^t}{\sqrt{5}}-\frac{{\varphi}^{-t}}{\sqrt{5}}\cos{t\pi}\right) \right| = n n + 1 ( φ t 5 sin t π ) ( φ t ln φ 5 + φ t ln φ 5 cos t π + π φ t 5 sin t π ) d t =\left|\int_{n}^{n+1} \left(\frac{{\varphi}^{-t}}{\sqrt{5}}\sin{t\pi} \right)\left( \frac{{\varphi}^{t}\ln\varphi}{\sqrt{5}} +\frac{{\varphi}^{-t}\ln\varphi}{\sqrt{5}}\cos{t\pi}+\frac{\pi{\varphi}^{-t}}{\sqrt{5}}\sin{t\pi}\right)dt\right| = 1 5 n n + 1 ( ln φ sin t π + φ 2 t ln φ sin t π cos t π + φ 2 t π sin 2 ( t π ) ) d t = \frac{1}{5}\left|\int_{n}^{n+1} \left(\ln \varphi\sin{t\pi} +{\varphi}^{-2t}\ln\varphi\sin{t\pi}\cos{t\pi} +{\varphi}^{-2t}\pi\sin^{2}\left(t\pi\right)\right)dt \right| Consider 0 sin t π cos t π 1 0\leq\left|\sin{t\pi}\cos{t\pi}\right|\leq1 And a b f ( t ) d t a b f ( t ) d t \left|\int_{a}^{b}f\left(t\right)dt \right| \leq \int_{a}^{b}\left| f\left(t\right)\right| dt Applying both inequalities, we have φ 2 t sin t π cos t π φ 2 t {\varphi}^{-2t}\left|\sin{t\pi}\cos{t\pi}\right| \leq {\varphi}^{-2t} n n + 1 φ 2 t sin t π cos t π d t n n + 1 φ 2 t sin t π cos t π d t \left|\int_{n}^{n+1} {\varphi}^{-2t}\sin{t\pi}\cos{t\pi} dt \right|\leq \int_{n}^{n+1} \left|{\varphi}^{-2t}\sin{t\pi}\cos{t\pi} \right| dt n n + 1 φ 2 t d t \leq \int_{n}^{n+1} {\varphi}^{-2t}dt lim n n n + 1 φ 2 t sin t π cos t π d t lim n n n + 1 φ 2 t d t \left|\lim\limits_{n\to\infty}\int_{n}^{n+1} {\varphi}^{-2t}\sin{t\pi}\cos{t\pi} dt\right| \leq \lim\limits_{n\to\infty} \int_{n}^{n+1}{\varphi}^{-2t} dt = lim n 1 2 ( φ 2 ( n + 1 ) φ 2 n ) =\lim\limits_{n\to\infty}\frac{1}{-2}\left({\varphi}^{-2\left(n+1\right)}-{\varphi}^{-2n}\right) = 0 =0 By Squeeze Theorem, we get lim n n n + 1 φ 2 t sin t π cos t π d t = 0 \left|\lim\limits_{n\to\infty}\int_{n}^{n+1} {\varphi}^{-2t}\sin{t\pi}\cos{t\pi} dt\right|=0 lim n n n + 1 φ 2 t sin t π cos t π d t = 0 \lim\limits_{n\to\infty}\int_{n}^{n+1} {\varphi}^{-2t}\sin{t\pi}\cos{t\pi} dt=0 Using the same argument, we can show that lim n n n + 1 φ 2 t sin 2 ( t π ) d t = 0 \lim\limits_{n\to\infty}\int_{n}^{n+1}{\varphi}^{-2t}\sin^{2}\left(t\pi\right)dt=0

Hence lim n A n = lim n 1 5 n n + 1 ( ln φ sin t π + φ 2 t ln φ sin t π cos t π + φ 2 t π sin 2 ( t π ) ) d t \lim\limits_{n\to\infty}A_n=\lim\limits_{n\to\infty}\frac{1}{5}\left|\int_{n}^{n+1} \left(\ln \varphi\sin{t\pi} +{\varphi}^{-2t}\ln\varphi\sin{t\pi}\cos{t\pi} +{\varphi}^{-2t}\pi\sin^{2}\left(t\pi\right)\right)dt \right| = ln φ 5 lim n n n + 1 sin t π d t =\frac{\ln\varphi}{5}\lim\limits_{n\to\infty}\left|\int_{n}^{n+1} \sin{t\pi}dt\right| = ln φ 5 π lim n ( cos ( n + 1 ) π cos n π ) =\frac{-\ln\varphi}{5\pi}\lim\limits_{n\to\infty}\left|\left(\cos{\left(n+1\right)\pi}-\cos{n\pi}\right)\right| = 2 ln φ 5 π lim n cos n π =\frac{2\ln\varphi}{5\pi}\lim\limits_{n\to\infty}\left|\cos{n\pi}\right| For n Z 0 + n \in \Z_{0}^{+} , this gives lim n A n = 2 ln φ 5 π \lim\limits_{n\to\infty}A_n=\boxed{\frac{2\ln\varphi}{5\pi}}

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