He was the greatest European mathematician of the Middle Ages. He lived in Pisa, at a time—about 1170 to about 1250—when Europeans were beginning to use Arabic numerals. His book Liber Abaci was a widely used introduction to this exciting new technology.
What to call him can be a confusing matter. His first name is easy: Leonardo, or Leonard, or Leonardus. But Italian usage of last names was very flexible, then and for a long time afterward. Here are some versions that were official enough to appear on manuscripts of his books.
The Fibonacci sequence (F_{n}, n= 0,1, ... ) satisfies F_{0} = 0, F_{1} = 1, and
F_{n+2} = F_{n+1} + F_{n} .
Here is a brief table of values of F_{n} .
n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
F_{n} | 0 | 1 | 1 | 2 | 3 | 5 | 8 | 13 | 21 | 34 | 55 | 89 |
The Fibonacci numbers are related to the golden ratio, which will be denoted here by τ.
The recurrence relation A_{n+2} = A_{n+1} + A_{n} is satisfied if we let A_{n} = F_{n}; it is also satisfied if we let A_{n} = τ^{n}. This is because τ is a solution of
x^{2} = x + 1 .
Now that is a quadratic equation, so it has a second solution, which turns out to be -ρ = -τ^{-1} . Therefore the recurrence relation is also satisfied if we let A_{n} = (-τ)^{-n}.
That recurrence relation is a linear equation, so two solutions to it can be combined, with arbitrary constant factors. If we put together the two solutions we just found in this way:
u_{n} = (τ^{n} - (-τ)^{-n})/ √5 ,
then we find that u_{0} = 0 and u_{1} = 1; from this it follows that F_{n} = u_{n}.
Because τ > 1, positive powers of τ dominate negative powers, and so that expression for F_{n} implies that for large n, F_{n} is close to (τ^{n})/√5. Therefore F_{n+1}/F_{n} is approximately τ.
Although τ is an irrational number, computation with it is not difficult, and expressions involving τ can often be simplified. For instance, consider the powers of τ. We already have used the fact that τ^{n+2} = τ^{n+1} + τ^{n} ; we can apply it to reduce powers of τ thus:
τ^{0} | = | 1 | ||
τ^{1} | = | τ | ||
τ^{2} | = | τ | + | 1 |
τ^{3} | = | 2τ | + | 1 |
τ^{4} | = | 3τ | + | 2 |
τ^{5} | = | 5τ | + | 3 |
τ^{6} | = | 8τ | + | 5 |
τ^{7} | = | 13τ | + | 8 |
and so on; in general,
τ^{n} = F_{n}τ + F_{n-1} .
The negative powers of τ are found similarly, using the recurrence relation in the form τ^{−n} = −τ^{−(n−1)} + τ^{−(n−2)}. We find:
τ^{0} | = | 1 | ||
τ^{−1} | = | τ | − | 1 |
τ^{−2} | = | −τ | + | 2 |
τ^{−3} | = | 2τ | − | 3 |
τ^{−4} | = | −3τ | + | 5 |
τ^{−5} | = | 5τ | − | 8 |
τ^{−6} | = | −8τ | + | 13 |
τ^{−7} | = | 13τ | − | 21 |
and so on; in general,
τ^{−n} = (−1)^{n}F_{n}τ + (−1)^{n+1}F_{n+1} .
From these expressions for powers of τ, we can get a better idea of how good the approximate values of τ are. For example, we see that 13τ − 21 = τ^{−7}. This implies
τ = 21/13 + 1/(13τ^{7}).
How big is the error term? Well, we have an expression for τ^{7}, namely 13τ+8; and we know that 13τ is close to 21. Therefore the error term is close to
1/(13(21+8)) = 1/377.
Hmm, 377 = F_{14}. Coincidence? I think not...
Conway, John Horton and Guy,Richard K., The Book of Numbers, New York, Copernicus Books, 1997; pp. 113-126.
Gardner, Martin, The 2nd Scientific American Book of Mathematical Puzzles and Diversions, New York, Simon & Schuster, 1961; chapter 8.
Gies, Joseph and Gies, Frances, Leonard of Pisa, Gainesville, Georgia, Elliott Wave International, Inc., 1983.
Graham, Ronald L., Knuth, Donald E., and Patashnik, Oren , Concrete Mathematics, second ed., Reading, Massachusetts, Addison-Wesley Longmans, 1997; section 1.2.8.
Knuth, Donald E., The Art of Computer Programming, vol. 1, Reading, Massachusetts, Addison-Wesley Longmans, 1997; section 1.2.8.
Sigler, Laurence, Fibonacci's Liber Abaci: A Translation into Modern English of Leonardo Pisano's Book of Calculation, New York, Springer Verlag, 2004.
Stewart, Ian, Letters to a Young Mathematician, New York, Basic Books, 2006.