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Fibonacci Numbers and the Pascal Triangle |
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Golden`s Section Formulas
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The connection between the Fibonacci and Lucas numbers and the Golden Section is expressed by the well-known mathematical formulas, so-called Binet's formulas. Jacques Philippe Marie Binet was born on February 2, 1776 in Renje and died on May 12, 1856 in Paris.Binet graduated from the Polytechnic School in Paris and after its graduation in 1806 he worked at the Bridges and Roads Department of the French government.
He became a teacher of the Polytechnic school in 1807 and in one year became an assistant-professor of the applied analysis and descriptive geometry.
Binet investigated foundations of matrix theory and his works in this direction were continued then by other researchers. In 1812 he discovered the rule of matrix multiplication and this discovery glorified his name more, than other his works.
 Most interesting is the fact that Binet studied the linear difference equations; the Fibonacci recurrent equation is their particular case. His famous Binet's formulas connect Fibonacci and Lucas numbers with the Golden Section.
Binet's Fibonacci Number Formula was derived by Binet in 1843, although the result was known to Euler and to Daniel Bernoulli more than a century ago.
To get Binet's formulas let us take the expression of connecting the powers of the golden ratio :
phi2 + phi = 1
or:
(-1)phi2 + F(1)(2phi + 1)phi = 1
By multiplying the previous identity by -phi2 we get the next identity:
(-1)2phi4 + 1*(2phi + 1)phi2 = 1
or:
(-1)2phi4 + F(2)(2phi + 1)phi2 = 1
Also by multiplying the previous identity by -phi2 we get the next identity:
(-1)3phi6 + 2*(2phi + 1)phi3 = 1
or:
(-1)3phi6 + F(3)(2phi + 1)phi3 = 1
and so on:
(-1)4phi8 + 3*(2phi + 1)phi4 = 1
or:
(-1)4phi8 + F(4)(2phi + 1)phi4 = 1
and:
(-1)5phi10 + 5*(2phi + 1)phi5 = 1
or:
(-1)5phi10 + F(5)(2phi + 1)phi5 = 1
In general case we have:
(-1)nphi2n + F(n)(2phi + 1)phin = 1
Note that the equation is valid for any integer n. Using the previous equation it is also possible to exppress Fibonacci number F(n) throught the Golden Section:
 or:
 Since phi is less than one in size, its powers decrease rapidly. We can use this to derive the following simpler formula for nth Fibonacci number F(n):
 It is interesting that A de Moivre (1667-1754) had written about Binet`s Formula, in 1730, and had indeed found a method for finding formula for any general series of numbers formed in a similar way to the Fibonacci series.
Let`s remember again that the degrees of the Golden Section are connected by the following identity:
phi2 + phi = 1
or:
(-1)2phi2 + L(1)phi = 1
By multiplying the previous identity by -phi2 we get the next identity:
(-1)3phi4 + 3*phi2 = 1
(-1)3phi4 + L(2)phi2 = 1
Also by multiplying the previous identity by -phi2 we get the next identity:
(-1)4phi6 + 4* phi3 = 1
or:
(-1)4phi6 + L(3)phi3 = 1
and so on:
(-1)5phi8 + 7*phi4 = 1
or:
(-1)5phi8 + L(4)phi4 = 1
and
(-1)6phi10 + 11*phi5 = 1
or:
(-1)6phi10 + L(5)phi5 = 1
In general case we have:
(-1)n+1phi2n + L(n)phin = 1
Note that the equation is valid for any integer n. Using the previous equation it is also possible to exppress Lucas number L(n) throught the Golden Section:
L(n) = Phin + (-1)nphin
or:
 We only used this formula for positive whole values on n and it only gives integer results. Well, perheps it was not really so surprising since the formula is supposed to define the Lucas numbers which are integers; but it is surprising in that this formula involves Phi and phi which are both irrational numbers.
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© 2001-2003 Radoslav Jovanovic translated: D.Filipovic created: January 2003. | |