Examples of Generating Functions - Worked Example B: Fibonacci Numbers

Worked Example B: Fibonacci Numbers

Consider the problem of finding a closed formula for the Fibonacci numbers Fn defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. We form the ordinary generating function

f = \sum_{n \ge 0} F_n x^n

for this sequence. The generating function for the sequence (Fn−1) is xf and that of (Fn−2) is x2f. From the recurrence relation, we therefore see that the power series xf + x2f agrees with f except for the first two coefficients:

\begin{array}{rcrcrcrcrcrcr} f & = & F_0x^0 & + & F_1x^1 & + & F_2x^2 & + & \cdots & + & F_ix^i & + &\cdots\\ xf & = & & & F_0x^1 & + & F_1x^2 & + & \cdots & + &F_{i-1}x^i & + &\cdots\\ x^2f & = & & & & & F_0x^2 & + & \cdots & + &F_{i-2}x^i & +&\cdots\\ (x+x^2)f & = & & & F_0x^1 & + & (F_0+F_1)x^2 & + & \cdots & + & (F_{i-1}+F_{i-2})x^i & +&\cdots\\ & = & & & & & F_2x^2 & + & \cdots & + & F_ix^i & +& \cdots\\ \end{array}

Taking these into account, we find that

f = xf + x^2 f + x . \,\!

(This is the crucial step; recurrence relations can almost always be translated into equations for the generating functions.) Solving this equation for f, we get

f = \frac{x} {1 - x - x^2} .

The denominator can be factored using the golden ratio φ1 = (1 + √5)/2 and φ2 = (1 − √5)/2, and the technique of partial fraction decomposition yields

f = \frac{1}{\sqrt{5}} \left (\frac{1}{1-\varphi_1 x} - \frac{1} {1- \varphi_2 x} \right ) .

These two formal power series are known explicitly because they are geometric series; comparing coefficients, we find the explicit formula

F_n = \frac{1} {\sqrt{5}} (\varphi_1^n - \varphi_2^n).

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