Wikipedia defines a generating function as

a formal power series in one indeterminate, whose coefficients encode information about a sequence of numbers an that is indexed by the natural numbers.

Generating functions are useful tools with many applications to discrete mathematics. In this post, we’ll show how they can be used to find a closed form expression for certain recurrence relations by proving that

\[ \begin{align*} F_n & = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right) \end{align*} \]

where \(F_n\) is the \(n\)-th Fibonacci number. The derivation of this formula is quite accessible to anyone comfortable with algebra and geometric series. For a much broader introduction to many of the uses of generating functions, refer to Prof. Herbert Wilf’s excellent book generatingfunctionology, the second edition of which is available as a free download.

Recall that the Fibonacci numbers are defined by the recurrence relation \[ \begin{align*} F_n & = F_{n - 1} + F_{n - 2} \end{align*} \]

for \(n \geq 2\), with \(F_0 = 0\) and \(F_1 = 1\).

We begin by defining the generating function for the Fibonacci numbers as the formal power series whose coefficients are the Fibonacci numbers themselves,

\[ \begin{align*} F(x) & = \sum_{n = 0}^\infty F_n x^n = \sum_{n = 1}^\infty F_n x^n, \end{align*} \]

since \(F_0 = 0\).

We then separate the two initial terms from the sum and subsitute the recurrence relation for \(F_n\) into the coefficients of the sum.

\[ \begin{align*} F(x) & = x + \sum_{n = 2}^\infty F_n x^n \\ & = x + \sum_{n = 2}^\infty (F_{n - 1} + F_{n - 2}) x^n \\ & = x + \sum_{n = 2}^\infty F_{n - 1} x^n + \sum_{n = 2}^\infty F_{n - 2} x^n \end{align*} \]

We now focus on rewriting each of these two sums in terms of the generating function. For the first sum, we have

\[ \begin{align*} \sum_{n = 2}^\infty F_{n - 1} x^n & = x \sum_{n = 2}^\infty F_{n - 1} x^{n - 1} = x \sum_{n = 1}^\infty F_n x^n = x F(x). \end{align*} \]

Similarly, for the second sum, we have \[ \begin{align*} \sum_{n = 2}^\infty F_{n - 2} x^n & = x^2 \sum_{n = 2}^\infty F_{n - 2} x^{n - 2} = x^2 \sum_{n = 0}^\infty F_n x^n = x^2 F(x). \end{align*} \]

We therefore have that

\[ \begin{align*} F(x) & = x + x F(x) + x^2 F(x). \end{align*} \]

Sovling for the generating function, we get

\[ \begin{align*} F(x) & = \frac{x}{1 - x - x^2}. \end{align*} \]

Now that we have found a closed form for the generating function, all that remains is to express this function as a power series. After doing so, we may match its coefficients term-by-term with the corresponding Fibonacci numbers. The roots of the polynomial \(1 - x - x^2\) are \(-\phi\) and \(-\psi\), where

\[ \begin{align*} \phi & = \frac{1 + \sqrt{5}}{2} \end{align*} \]

and

\[ \begin{align*} \psi & = \frac{1 - \sqrt{5}}{2}, \end{align*} \]

so the polynomial factors as \(1 - x - x^2 = - (x + \phi) (x + \psi)\).

In order to express the generating function as a power series, we will use the partial fraction decomposition to express it in the form

\[ \begin{align*} F(x) & = -\frac{x}{(x + \phi) (x + \psi)} = \frac{A}{x + \phi} + \frac{B}{x + \psi}, \end{align*} \]

which is equivalent to

\[ \begin{align*} -x & = A (x + \psi) + B (x + \phi). \end{align*} \]

Letting \(x = -\phi\), we find that \(A = -\frac{\phi}{\sqrt{5}}\). Similarly, letting \(x = -\psi\), we get that \(B = \frac{\psi}{\sqrt{5}}\). Therefore

\[ \begin{align*} F(x) & = \frac{1}{\sqrt{5}} \left( \frac{\psi}{x + \psi} - \frac{\phi}{x + \phi} \right). \end{align*} \]

We now wish to express each of these two terms as the sum of a geometric series. Recall that the sum of a geometric series is given by

\[ \begin{align*} \frac{1}{1 - x} & = \sum_{n = 0}^\infty x^n. \end{align*} \]

Note that this infinite sum converges if and only if \(|x| < 1\). However, considered as a formal power series, this identity always holds. We use this identity, and the fact that \(\phi = -\frac{1}{\psi}\), to rewrite the first term of the generating function as

\[ \begin{align*} \frac{\psi}{x + \psi} & = \frac{1}{1 + \frac{x}{\psi}} \\ & = \frac{1}{1 - \phi x} \\ & = \sum_{n = 0}^\infty \phi^n x^n. \end{align*} \]

Similarly,

\[ \begin{align*} \frac{\phi}{x + \phi} & = \sum_{n = 0}^\infty \psi^n x^n, \end{align*} \]

so

\[ \begin{align*} F(x) & = \frac{1}{\sqrt{5}} \left( \frac{\psi}{x + \psi} - \frac{\phi}{x + \phi} \right) \\ & = \frac{1}{\sqrt{5}} \left( \sum_{n = 0}^\infty \phi^n x^n - \sum_{n = 0}^\infty \psi^n x^n \right) \\ & = \sum_{n = 0}^\infty \frac{1}{\sqrt{5}} \left( \phi^n - \psi^n \right) x^n. \end{align*} \]

Since the definition of \(F(x)\) was

\[ \begin{align*} F(x) & = \sum_{n = 0}^\infty F_n x^n, \end{align*} \]

we match the coefficients on corresponding powers of \(x\) in these two expressions for \(F(x)\) to finally arrive at the desired closed form for the \(n\)-th Fibonacci number,

\[ \begin{align*} F_n & = \frac{1}{\sqrt{5}} \left( \phi^n - \psi^n \right). \end{align*} \]

Deriving this identity gives an excellent glimpse of the power of generating functions. Again, for a much more thorough treatment of their many applications, consult generatingfunctionology.

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