Proof that fibonacci goes to phi now!

Proof that fib(n+1)/fib(n) goes to phi (the golden ratio) as n gets large

The golden ratio is defined by AB/AC = AC/BC. Let AB=phi and AC=1. Then we get phi=1/(AB-AC) = 1/(phi-1), phi(phi-1)=1, phiˆ2-phi-1=0. One solution to this quadratic equation is the golden ratio, phi, while the other, called phi', is negative and doesn't interest us.

fib(n+1)/fib(n) = (fib(n)+fib(n-1))/fib(n) = 1 + fib(n-1)/fib(n) = 1 + 1/(fib(n)/fib(n-1))

As n gets very large, fib(n+1)/fib(n) and fib(n)/fib(n-1) become indistinguishable, so let's call those two by x.

x = 1 + 1/x, xˆ2 = x + 1, xˆ2 - x - 1 = 0.

As you can see, it is the same quadratic equation as the one mentioned before. Since fib only gives positive numbers, the ratio between two fibs which is x must be a positive number, which excludes the negative solution to that equation. So the solution must be phi (the golden ratio).

Therefore, fibonacci goes to phi.

Actually good post, I don't expect stuff like this when coming here. Thank you.

Wow... Bachelor level mathematics...

From the fact that phiˆ2 = phi+1 we get the following curiosity:
For a geometric progression of ratio phi, (a_0, a_0*phi, a_0*phiˆ2, a_0*phiˆ3, ...), i.e. such that a_n = a_{n-1}*phi for all n>=1, we also get the following rule:
a_n = a_{n-1} + a_{n-2}, for all n>=2.

Proof:
a_n = a_{n-1}*phi = a_{n-2}*phiˆ2 = a_{n-2}*(phi+1) = a_{n-2}*phi + a_{n-2} = a_{n-1} + a_{n-2}

This is the same sequence rule as that of the fibonacci sequence, except that fibonacci has two fixed initial numbers a_0=1 and a_1=1, so they aren't the same.

Still, as has already been proven, the ratio of a_{n+1}/a_n in the fibonacci sequence approaches phi!

What programming language is this?