Fibonacci's Rabbits

The Fibonacci sequence first appears in the book Liber Abaci (1202) by Leonardo Bonacci, otherwise known as Fibonacci. He first used it to calculate the rate of growth of rabbit populations.

Fibonacci considers the growth of a hypothetical rabbit population, assuming that: a newly born breeding pair of rabbits are put in a field; each breeding pair mates at the age of one month, and at the end of their second month they always produce another pair of rabbits (one male, one female); and rabbits never die, but continue breeding forever. Fibonacci posed the puzzle: how many pairs will there be in one year?

Let the function $f(n)$ represent the number of pairs of rabbits in the month $n$.
If $f(0)=0$ and $f(1)=1$, we then have,
\begin{equation} f(n) = f(n-1)+f(n-2) \end{equation} otherwise known as Fibonacci's sequence:
\begin{equation} 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597,... \end{equation} But how does this population grow? Does its rate of growth increase as the population increases? If so, in what manner? To answer this, we examine the ratio of one population $f(n)$ and the population preceding it $f(n-1)$. Substituting Fibonacci's relationship above, $f(n) = f(n-1)+f(n-2)$ and some light algebra:

\begin{equation} \frac{f(n)}{f(n-1)} = \frac{f(n-1)+f(n-2)}{f(n-1)} = 1 + \frac{f(n-2)}{f(n-1)} \end{equation}
We also know by the properties of limits, as $n \rightarrow \infty$,

\begin{equation} \frac{f(n)}{f(n-1)} = \frac{f(n-1)}{f(n-2)} \end{equation}
Combinining the two statements above for large $n$, we have

\begin{equation} \frac{f(n-1)}{f(n-2)} = 1 + \frac{f(n-2)}{f(n-1)} \end{equation}
For simplicity of notation, let's define $\phi = \frac{f(n-1)}{f(n-2)}$. In other words, $\phi$ is a measure of how much larger one population is than the previous. Then the above relationship becomes

\begin{equation} \phi = 1 + \frac{1}{\phi} \end{equation}
We can write this as $\phi^2 - \phi -1 = 0$, and solve with the quadratic formula

\begin{equation} \phi = \frac{1\pm\sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2} \end{equation}
Since we're dealing with populations, the number can only be positive, so we're left with

\begin{equation} \phi = \frac{1+\sqrt{5}}{2} \end{equation}
Therefore, as rabbit populations grow large, the size of each next generation is a constant $\frac{1+\sqrt{5}}{2}$ times larger than the last.

Now, this should look very familiar as it's Fibonacci's famous number, otherwise known as the most important number to architecture and design: The Golden Mean!
Stay tuned for a future Exploration on how this number has been used for thousands of years to create everything from temples to the paintings of Salvador Dali.