Instead of functions and zeroes and infinity, or a geometric construction, let's simplify things and talk about addition in this part. We're going to work with a spreadsheet on the following problem.

Start with two non-negative integers, f(0) and f(1). Compute f(2) = f(0) + f(1). Compute f(3) = f(1) + f(2).

Continue, creating f(n) = f(n-1) + f(n-2), where each new number is the sum of the prior two numbers in the sequence.

This problem has been studied for a long time. If we set f(0) = 0 and f(1) = 1, we have a series of numbers called a Fibonacci sequence, after the Italian Leonardo Pisano Bigollo Fibonacci. He proposed this sequence his book published in 1202, the Book of Calculation. If you understand Latin you can read Liber Abaci online. The book covers many topics, including Egyptian fractions, interest computation, and square root approximations. Even more important than the problem we'll be discussing is the number system. Fibonacci promoted the Hindi-Arabic system of numbers, our modern system, over the Roman I, V, X, and such. While he wasn't the first, his book made a big difference in European usage of 0 through 9.

Leaving the history of zero for another time, we'll look at Fibonacci's rabbit mating problem. Say rabbits are able to mate at the age of one month, and pregnancy takes one month. Thus, at the end of its second month, a female can produce another pair of rabbits. Rabbits never die, and a mating pair always produces one new pair (one male, one female) every month from the second month on. The puzzle that Fibonacci posed was: if we start with a new pair from birth, how many pairs will there be in one year?

The circumstances and restrictions are not realistic, though rabbit pregnancies actually do last roughly one month and female rabbits can become pregnant again within a few days after giving birth. Birth litters are not always exactly 2, though, as they're usually more. Also, rabbits eventually die. Still, this isn't THAT unrealistic a situation in the short term.

To solve Fibonacci's problem, we'll let f(n) be the number of pairs during month n. By convention, f(0) = 0. f(1) = 1 for our new first pair. f(2) = 1 as well, as conception just occurred. The new pair is born at the end of month 2, so during month 3, f(3) = 2. Only the initial pair produces offspring in month 3, so f(4) = 3. In month 4, the initial pair and the month 2 pair breed, so f(5) = 5. We can proceed this way, presenting the results in a table. At the end of a year, Fibonacci has 144 pairs of rabbits.

Month | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Pairs | 0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | 89 | 144 |

While it's nice addition practice, we don't want to do this for too many steps. Additionally, we might want to examine ratios or other factors. A spreadsheet like Microsoft Excel will assist this process. I will show output from Excel; for a zero-cost option, try Google Documents.

The Excel file has places to enter starting values f(0) and f(1). The columns keep track of the month n and the number of rabbit pairs. Things get large in a hurry. At the end of 1 year, we have 144 pairs, enough for a couple classrooms. At the end of 3 years, we have over 14 Million pairs, more than enough to give one pair to every person in Georgia. It would take us only 49 months to have over 7 Billion pairs, one pair for every living human. Maybe it's a good thing Fibonacci's rabbits aren't real.

To further explore the sequence, let's add a new column to our spreadsheet, the ratio of successive terms. For month n, the ratio will equal \frac{ f(n) }{ f(n-1) } . We'll start at month 2 to avoid possibly dividing by zero. The ratio reaches a limit pretty quickly, about 1.61803.

Is this ratio specific to our starting numbers? No. As long as we don't start with f(0) = 0, f(1) = 0, the post-apocalyptic case which will produce zero rabbits forever, we quickly reach the ratio 1.61803. For another example, I display f(0) = 2 and f(1) = 13. If you want to try it yourself, the Excel spreadsheet is fibratio.xlsx. You might try the Lucas sequence with f(0) = 2, f(1) = 1, or some other pairings.

To prove that the ratio becomes constant, we'll need to return to more complicated mathematics and work through some algebra. Let's look at the ratio for terms fairly far into the sequence, eliminating the effects of the initial start point. Symbolically, we are looking for the limit of the ratio, \lim_{n \rightarrow \infty} \frac{ f(n) }{ f(n-1) } . We won't proceed extremely formally. Because our Excel spreadsheets suggest that there is a limit, we will hypothesize that for large n, \frac{ f(n-1) }{ f(n-2) } = \frac{ f(n) }{ f(n-1) } . Substituting the Fibonacci relation f(n) = f(n-1) + f(n-2) gives us

\frac{ f(n-1) }{ f(n-2) } = \frac{ f(n-1) + f(n-2) }{ f(n-1) }

\frac{ f(n-1) }{ f(n-2) } = 1 + \frac{ f(n-2) }{ f(n-1) }

Let r be the limiting ratio r = \frac{ f(n-1) }{ f(n-2) } . Then we have r = 1 + (1/r) which becomes the quadratic equation r^2 = r + 1 . We can solve this via the quadratic formula, from its standard form r^2 - r - 1 = 0 .

r = \frac{ -(-1) \pm \sqrt{(-1)^2 - 4 (1) (-1) } } { 2 (1) } = \frac{ 1 \pm \sqrt{ 5 } }{ 2 }

The negative root would lead to a negative r, which makes no sense since we only have non-negative numbers. The proper limit for the ratio is \frac{ 1 + \sqrt{ 5 } }{ 2 } , approximately 1.618, as we've seen on the spreadsheets.

Not coincidentally, that number \frac{ 1 + \sqrt{ 5 } }{ 2 } should look familiar. We saw it in part 2 of the final, for the ratio of the side of the square to the additional part:

\frac{ S }{ a } = \frac{ 1 + \sqrt{5} }{ 2 }

The number \frac{ 1 + \sqrt{ 5 } }{ 2 } has a special name, the golden ratio, and is usually denoted by the Greek letter "phi", \phi . Mathematicians like it a lot. It has a Wikipedia page, a University of Surrey website, and even was featured on two BBC Radio programs, one in 2002 and another in 2007. The Golden Ratio appears in nature, like the spirals in the Nautilus sea-shell below. Some flowers, like buttercups, only have Fibonacci numbers for petal count. Pine cones usually have 5, 8, or 13 outside spirals.

There's a lot more about the golden ratio. If you're interested, according to the BBC's 5 numbers quiz, this means "You have a tendency to show up where least expected bringing balance and order to an otherwise chaotic world." Like rabbit ratios and geometric constructions, that's not a bad place to be.

This additional activity comes from Assignment 12; I chose it because it ties in very well to Final Part 2.

The Nautilus sea shell picture comes from allposters.com.