The Fibonacci Numbers and Golden Ratio I

_{September 11, 2021.}

Let's talk some math today! In the morning, I read this interesting thing about Fibonacci numbers.

"n is a Fibonacci number if and only if either 5n

^{2}+4 or 5n^{2}-4 is a perfect square."

Let's try this out first. We all know the Fibonacci sequence looks like this, 0,1,1,2,3,5... where each number is the sum of its previous two. So, on applying the above theory,

5*0^{2}+4 = 9 = 3 * 3

5*1^{2}+4 = 9 = 3 * 3

5*2^{2}-4 = 16 = 4 * 4

5*3^{2}+4 = 49 = 7 * 7

5*5^{2}-4 = 121 = 11 * 11

we can conclude that this holds. But my fear is this isn't that practical. I mean, it doesn't give us a fast way to test the Fibonacci numbers. There are many ways to test the Fibonacci series. I thought I'd talk about the Fibonacci and their relationship with the Golden ratio (phi), then discuss a formula for testing out the Fibonacci in regards to both.

First, let's start small. The golden ratio (can call it phi), is to put it straightforwardly, is the positive solution to the equation x^{2}-x-1 = 0, which can be found using the formula: ɸ = 1+√5/2.

Like every other thing in math, there is a property of this golden ratio, that is, ɸ^{2} = ɸ+1. Now, let's see how this property has a thing for our Fibonacci.

Let us first try to increment the power for the above property and see what happens.

ɸ^{3} = ɸ.ɸ^{2} = ɸ.(ɸ+1) = ɸ^{2}+ɸ = 2ɸ+1

Good, let's try the same for the next one.

ɸ^{4} = ɸ.ɸ^{3} = ɸ.(2ɸ+1) = 2ɸ^{2}+ɸ = 3ɸ+2

Cool, just one more now.

ɸ^{5} = ɸ.ɸ^{4} = ɸ.(3ɸ+2) = 3ɸ^{2}+2ɸ = 5ɸ+3

Okay, enough of this, we are getting it! The Fibonacci series is involved here. So, we can state this generally as ɸ^{n} = F_{n}ɸ+F_{n-1}, where n>=2.

This is interesting, right? Well, can we prove it? Of course! Let's use Mathematical Induction for the proof. You can read about Mathematical Induction here, but for now, I'll keep it simple. If we're using MI, we first take a particular value of n (like n=2) and prove that our statement holds for this value. Then we prove that it also holds for n+1the value. So, if we say it is true for n=2, then we prove that it is for n=3 also, then n=4, and so on...

The statement we wrote is as follows: ɸ^{n} = F_{n}ɸ+F_{n-1}, where n>=2.

We know F1 = F2 = 1. So, now we take n as 2.

ɸ^{2} = F_{2}ɸ+F_{1}. So, as per MI, let's prove for n+1

ɸ^{n+1} = ɸ.ɸ^{n}

= ɸ.(F_{n}ɸ+F_{n-1})

= ɸ^{2}F_{n}+ɸF_{n-1}

= F_{n}(ɸ+1)+ɸF_{n-1}

= ɸ(F_{n}+F_{n-1})+F_{n}

= F_{n+1}ɸ+F_{n}

Hence, we can say that the statements hold for n=2 therefore, for n+1 too. So, this is nice but what's more interesting is the formula we can form with all this information. I'll be writing another post for that.