Fibonacci sequence

The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, … starts with two 1s, and each later term is the sum of the two before it. The same rule $x_k = x_{k-1} + x_{k-2}$ can start from any two positive numbers $x_1 = a$ and $x_2 = b$; from 2, 1 it gives the Lucas numbers (2, 1, 3, 4, 7, 11, …). Here you can build such a sequence as squares added to a rectangle, follow the ratio of neighbouring terms $r_k = x_k / x_{k-1}$ as it closes in on the golden ratio φ, and see on a cobweb diagram why it gets there from any start.

The start is a rectangle a wide and b tall. Each new term is a square laid against a side as long as the newest term: a rectangle with sides $x_{k-1}$ and $x_{k-2}$ gains a square of side $x_{k-1}$ and becomes $x_k$ by $x_{k-1}$. So from the third term on, $r_k$ is the long side over the short side: the rectangle’s shape. The picture keeps the rectangle the same size, so you watch its shape settle rather than its numbers grow.

A square changes the ratio by a fixed rule, $r_{k+1} = (x_k + x_{k-1})/x_k = 1 + 1/r_k$. A shape that a square leaves unchanged needs $r = 1 + 1/r$, that is $r^2 = r + 1$, whose positive root is the golden ratio $\varphi = (1 + \sqrt5)/2 \approx 1.618$. That is the golden rectangle: add a square and you get the same shape, φ times as large. The cobweb diagram draws the rule. From a ratio r, go up to the curve $y = 1 + 1/r$ to read the next ratio, then across to the diagonal $y = r$ to start the next step from it. Curve and diagonal cross at (φ, φ).

Why does the ratio alternate, and why does it arrive from every positive start? Since $\varphi = 1 + 1/\varphi$, one step changes the gap to φ exactly by

The minus sign puts every ratio on the other side of φ from the one before, unless the start is exactly golden. From $r_3$ on every ratio is above 1, so each step multiplies the gap by less than $1/\varphi \approx 0.618$ and the gap shrinks to 0. Only the first step can move away: when $r_2 = b/a$ is below 1/φ, the gap grows. Near φ the factor approaches $-1/\varphi^2 \approx -0.382$, the slope of the curve where it crosses the diagonal, so in the end each step leaves the gap about 0.382 times as large.

The terms themselves grow almost geometrically: each comes ever closer to φ times the one before. For the Fibonacci numbers $F_1 = F_2 = 1$, Binet’s formula makes this exact:

Because $|\psi| < 1$, $F_n$ is the whole number nearest to $\varphi^n/\sqrt5$. On the eigenvector page the same two numbers appear as the eigenvalues of the matrix that takes one pair of neighbouring terms to the next.