Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. 9 The Mathematics of Spiral Growth 9.1Fibonacci’s Rabbits 9.2Fibonacci Numbers 9.3 The Golden Ratio 9.4Gnomons 9.5Spiral Growth in Nature

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. In 1202 a young Italian named Leonardo Fibonacci published a book titled Liber Abaci (which roughly translated from Latin means “The Book of Calculation”). Although not an immediate success, Liber Abaci turned out to be one of the most important books in the history of Western civilization. Leonardo Fibonacci

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. Liber Abaci was a remarkable book full of wonderful ideas and problems, but our story in this chapter focuses on just one of those problems–a purely hypothetical question about the growth of a very special family of rabbits. Here is the question, presented in Fibonacci’s own (translated) words: “The Book of Calculation”

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. A man puts one pair of rabbits in a certain place entirely surrounded by a wall. How many pairs of rabbits can be produced from that pair in a year if the nature of these rabbits is such that every month each pair bears a new pair which from the second month on becomes productive? Fibonacci’s Rabbits

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. We will call P 1 the number of pairs of rabbits in the first month, P 2 the number of pairs of rabbits in the second month, P 3 the number of pairs of rabbits in the third month, and so on. With this notation the question asked by Fibonacci (...how many pairs of rabbits can be produced from [the original] pair in a year?) is answered by the value P 12 (the number of pairs of rabbits in month 12). For good measure we will add one more value, P 0, representing the original pair of rabbits introduced by “the man” at the start. Fibonacci’s Rabbits

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. Let’s see now how the number of pairs of rabbits grows month by month. We start with the original pair, which we will assume is a pair of young rabbits. In the first month we still have just the original pair (for convenience, let’s call them Pair A), so P 1 = 1. By the second month the original pair matures, becomes “productive,” and generates a new pair of young rabbits. Thus, by the second month we have the original mature Pair A plus the new young pair we will call Pair B, so P 2 = 2. Fibonacci’s Rabbits

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. By the third month Pair B is still too young to breed, but Pair A generates another new young pair, Pair C, so P 3 = 3. By the fourth month Pair C is still young, but both Pair A and Pair B are mature and generate a new pair each (Pairs D and E). It follows that P 4 = 5. We could continue this way, but our analysis can be greatly simplified by the following two observations: Fibonacci’s Rabbits

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. 1.In any given month (call it month N) the number of pairs of rabbits equals the total number of pairs in the previous month (i.e., in month N – 1 ) plus the number of mature pairs of rabbits in month N (these are the pairs that produce offspring–one new pair for each mature pair). 2.The number of mature rabbits in month N equals the total number of rabbits in month N – 2 (it takes two months for newborn rabbits to become mature). Fibonacci’s Rabbits

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. Observations 1 and 2 can be combined and simplified into a single mathematical formula: Fibonacci’s Rabbits P N = P N – 1 + P N – 2 The above formula reads as follows: The number of pairs of rabbits in any given month (P N ) equals the number of pairs of rabbits the previous month (P N – 1 ) plus the number of pairs of rabbits two months back (P N – 2 ).

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. It follows, in order, that Fibonacci’s Rabbits P 5 = P 4 + P 3 = = 8 P 6 = P 5 + P 4 = = 13 P 7 = P 6 + P 5 = = 21 P 8 = P 7 + P 6 = = 34 P 9 = P 8 + P 7 = = 55 P 10 = P 9 + P 8 = = 89 P 11 = P 10 + P 9 = = 144 P 12 = P 11 + P 10 = = 233

Excursions in Modern Mathematics, 7e: Copyright © 2010 Pearson Education, Inc. So there is the answer to Fibonacci’s question: In one year the man will have raised 233 pairs of rabbits. This is the end of the story about Fibonacci’s rabbits and also the beginning of a much more interesting story about a truly remarkable sequence of numbers called Fibonacci numbers. Fibonacci’s Rabbits