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My title - Departamento de Matemática da Universidade do Minho

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2 ITERATION/RECURSION 13<br />

Nice pictures. Em baixo, está uma imagem que nos tempos <strong>de</strong> Julia e Fatou apenas era possível<br />

ver com uns olhos matemáticos bem afina<strong>do</strong>s (um applet Java que produz a figura está no meu<br />

bestiario). O laço <strong>de</strong> corações vermelhos à esquer<strong>da</strong>, chama<strong>do</strong> Man<strong>de</strong>lbrot set, consiste nos valores<br />

<strong>do</strong> parâmetro complexo c tais que a órbita <strong>do</strong> ponto crítico z 0 = 0 permanece limita<strong>da</strong>. A região<br />

cinzenta à direita, chama<strong>da</strong> filled-in Julia set, consiste no conjunto <strong>da</strong>s condições iniciais z 0 cuja<br />

órbita é limita<strong>da</strong>. As outras côres (que permitem ver os conjuntos “invisíveis” <strong>de</strong> Cantor) são<br />

escolhi<strong>da</strong>s <strong>de</strong>pen<strong>de</strong>n<strong>do</strong> <strong>da</strong> veloci<strong>da</strong><strong>de</strong> com que as trajectórias z n fogem para o infinito.<br />

Conjunto <strong>de</strong> Man<strong>de</strong>lbrot (esquer<strong>da</strong>) e conjunto <strong>de</strong> Julia <strong>do</strong> polinómio z 2 + c com c ≃ −0.7645 − i · 0.1595 (direita).<br />

(from http://w3.math.uminho.pt/~scosentino/bestiario/julia.html)<br />

2.5 Finite difference equations<br />

Fibonacci mo<strong>de</strong>l is the prototype of<br />

Recursive linear equations. A recursive linear equation (or “finite difference linear equation”)<br />

is a law<br />

a p x n+p + a p−1 x n+p−1 + · · · + a 1 x n+1 + a 0 x n = 0<br />

which <strong>de</strong>fines a sequence (x n ) given a set of “initial conditions” x 0 , x 1 , . . . , x p−1 . Above, a 0 ≠<br />

0, a 1 , . . . , a p−1 , a p ≠ 0 are real or complex parameters. It is a discrete version of a linear ordinary<br />

differential equation of <strong>de</strong>gree p with constant coefficients.<br />

Eigenfunctions. The general recipe is: “linear equations have exponential solutions”. The<br />

conjecture x n = z n solves the recursive equation if z is a root of the characteristic polynomial<br />

P (z) = a p z p + a p−1 z p−1 + · · · + a 1 z + a 0<br />

In particular, if P has p distinct roots (in C), say z 1 , z 2 , . . . , z p , then the general solution of the<br />

recursive equation is a linear combination<br />

x n = c 1 z1 n + c 2 z2 n + · · · + c p zp<br />

n<br />

where the c 1 , c 2 , . . . , c p are constants which <strong>de</strong>pend on the initial conditions x 0 , x 1 , . . . x p−1 .<br />

Exercícios.<br />

• Find an explicit formula for the Fibonacci numbers f n ’s (which is known as Binet’s formula).

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