[Math] the connection and the difference between the Golden Ratio and Fibonacci Sequence

fibonacci-numbersgolden ratiotrigonometry

The Golden Ratio, i.e.

$\varphi = \frac{1+\sqrt{5}}{2}$

and Fibonacci sequence, i.e.

$F_n=F_{n-1}+F_{n-2}$ with the initial conditions $F_0=0$ and $F_1=1$

are clearly connected, but not perfectly so, and I'm seeking to understand this. I've been trying to read up a bit to understand the similarities and differences between these 2.

A Live Science article, for instance, says:

Around 1200, mathematician Leonardo Fibonacci discovered the unique
properties of the Fibonacci sequence. This sequence ties directly into
the Golden ratio because if you take any two successive Fibonacci
numbers, their ratio is very close to the Golden ratio. As the numbers
get higher, the ratio becomes even closer to 1.618. For example, the
ratio of 3 to 5 is 1.666. But the ratio of 13 to 21 is 1.625. Getting
even higher, the ratio of 144 to 233 is 1.618. These numbers are all
successive numbers in the Fibonacci sequence.

Or, as another source put it:

The quotient of any Fibonacci number and it's predecessor approaches
Phi, represented as ϕ (1.618), the Golden ratio.

Based on these descriptions, it sounds like the the ratio of consecutive Fibonacci numbers and the Golden Ratio converge asymptotically but are not identical (especially with the initial ratios). I would like to understand this a little bit better.

Asides from Live Science and Google, I did some preliminary research on Mathematics SE and Cross Validated. The closest question I could find was an unanswered one which primarily focused on the relationship between Arctangents and the Fibonacci sequence.

Best Answer

If we set $\varphi = \frac{1+\sqrt{5}}{2}$ and $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$, then the Binet formula for Fibonacci numbers says:

$F_n=\frac{\varphi^n-(\bar\varphi)^{ n}}{\sqrt{5}}$.

Because $\mid \bar\varphi \mid < 1$, the term $(\bar\varphi)^n$ decays fairly quickly toward $0$. This points to why successive ratios of Fibonacci numbers get close to $\varphi$.


Note: Here is a reference for the Binet formula: http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html

Related Question