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Math!

Golden Fibonacci

You've heard of it, but do you know it?

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Preface

I had quite an odd past few days. I’ve been doing more math than sleeping, and more sleeping than math homework. In short—I’m Gödel. Anyways, let’s get started before I get even more sidetracked.

Getting Started

First, I’ll “restate my assumptions”:

F0=1F1=1Fn=Fn−1+Fn−2lim⁡n→∞FnFn−1=φ\begin{aligned} & F_0=1 \\ & F_1=1 \\ & F_n=F_{n-1}+F_{n-2} \\ & \lim\limits_{n \to \infty } \frac{F_n}{F_{n-1}} = \varphi \end{aligned}

Okay that’s the definition of the Fibonacci sequence.

Some Algebra

First, we’re just going to do some substitution:

lim⁡n→∞Fn−1+Fn−2Fn−1=φ\lim\limits_{n \to \infty } \frac{F_{n-1} + F_{n-2}}{F_{n-1}} = \varphi

And then some algebra:

lim⁡n→∞Fn−1Fn−1+lim⁡n→∞Fn−2Fn−1=φ\lim\limits_{n \to \infty } \frac{F_{n-1}}{F_{n-1}} + \lim\limits_{n \to \infty } \frac{F_{n-2}}{F_{n-1}} = \varphi

Doesn’t that first term look just like 1? It is!

1+lim⁡n→∞Fn−2Fn−1=φ1 + \lim\limits_{n \to \infty } \frac{F_{n-2}}{F_{n-1}} = \varphi

And that second limit? Doesn’t it look similar to the limit in our assumptions? Well, the indices are n−1n-1 and n−2n-2 rather than nn and n−1n-1, but considering we are taking the limit to infinity, they should mean the same thing. The only difference is that it’s upside-down right? Well luckily we know that, as a great mathematician once told me, “in the same cadence as a Buddhist proverb: the inverse of a limit is the limit of the inverse.”

1+(lim⁡n→∞Fn−1Fn−2)−1=φ1 + (\lim\limits_{n \to \infty } \frac{F_{n-1}}{F_{n-2}})^{-1} = \varphi

1+(lim⁡n→∞FnFn−1)−1=φ1 + (\lim\limits_{n \to \infty } \frac{F_{n}}{F_{n-1}})^{-1} = \varphi

1+(φ)−1=φ1 + (\varphi)^{-1} = \varphi

Wait what? How are we going to use φ\varphi to define itself? Well luckily algebra has an answer. Let’s multiply everything by φ\varphi:

φ+1=φ2\varphi + 1 = \varphi^2

1=φ2−φ1 = \varphi^2 - \varphi

And to finish up, let’s complete the square:

1+14=φ2−φ+141 + \frac{1}{4} = \varphi^2 - \varphi + \frac{1}{4}

54=(φ−12)2\frac{5}{4} = (\varphi - \frac{1}{2})^2

54=φ−12\sqrt{\frac{5}{4}} = \varphi - \frac{1}{2}

1+52=φ\frac{1 + \sqrt{5}}{2} = \varphi

That’s it! The golden ratio! For reference, here is the definition of the golden ratio:

φ=ab\varphi = \frac{a}{b}

And yes! The two are related, but I’ll leave that as an exercise to the reader.

Have a golden day,
Ilan Bernstein