Fibonacci numbers and the Golden Ratio

The ancient Greek geometers believed that a rectangle with sides in the ratio $\tau=(1+\sqrt{5})/2=1.618033989\ldots$ was the most pleasing to the eye. They called any such rectangle a Golden Rectangle, and so $\tau$ became known as the Golden Ratio. Naturally, the Greeks defined the Golden Ratio geometrically instead of writing it as a number.

Definition 792

  A Golden Rectangle is a rectangle R having the following property. If a square on one of the shorter sides of R is removed from R, then the remaining rectangle is similar to R.

In other words, if a square on one side of a Golden Rectangle is removed from the rectangle, the rectangle that remains is also golden!

If u and u+v are, respectively, the short and the long side of the Golden Rectangle above, then clearly  

$$\tau = \frac{u+v}{u} = \frac{u}{v}\,,$$ (2)

which shows that $\tau=1+1/\tau$ or $\tau^2=\tau+1$. Solving, we find $\tau=(1+\sqrt{5})/2$, demonstrating that the numerical and geometric definitions are the same.

The default window in which 2-dimensional graphics are drawn in the Mathematica software package is a Golden Rectangle.

There is a remarkable connection between $\tau$ and Fibonnaci numbers. The reason is that the powers $T_n=\tau^{n-1}$ satisfy the same basic recurrence as do the F_n. Indeed, since $\tau^2=\tau+1$, we have

 

T_n=T_n-1+T_n-2

for all $n\ge2$; but the initial values of the recurrence (4) are $T_0=1/\tau$ and T₁=1, so the T_n are not the Fibonacci numbers. Nevertheless, we have the following proposition, which provides the basic idea for estimating the efficiency of the Euclidean algorithm.

Proposition 810

  For all $n\ge0$, $F_n\ge T_n=\tau^{n-1}$.

Proof.Since $F_0=1\ge 1/\tau=T_0$ and F₁=1=T₁, the proposition is valid when n=0 and n=1. Now we appeal to mathematical induction: $F_{n+1}=F_n+F_{n-1}\ge T_n+T_{n-1}=T_{n+1}$ for all $n\ge1$. $\hfill \rule{2mm}{3.2mm} \vspace{4mm}$