Take a recurrence relation, like the way the Fibonacci sequence is defined:

$a_n = a_{n-1} + a_{n-2}$, with $a_0 = 0$ and $a_1 = 1$.

I’m going to make up an infinite degree polynomial that looks like this:

$A(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_k x^k + \dots$

… and figure out another way to write that polynomial.

If I multiply $A(x)$ by $x$, I get:

$xA(x) = a_0 x + a_1 x^2 + \dots + a_{k-1} x^{k} + \dots$

Same again:

$x^2A(x) = a_0 x^2 + a_1 x^3 + \dots + a_{k-2} x^{k} + \dots$.

Comparing coefficients, we can make most of the terms cancel out by combining these polynomials – the $x^k$ terms of $A(x) - xA(x) - x^2A(x)$ would all be zero (except near the very beginning) because of how the Fibonacci sequence is defined.

In fact, $(1 - x - x^2)A(x) = a_0 + (a_1 - a_0)x$ exactly, so $A(x) = \frac{a_0 + (a_1 - a_0)x}{1 - x - x^2}$. In our case, $a_0 = 0$ and $a_1 = 1$, so it’s $A(x) = \frac{x}{1-x-x^2}$

(This sort of trick can be done for any recurrence relation that relies on linear combinations of finitely many previous terms, and probably others as well.)

### So what?

Now, the bottom of that definition doesn’t factorise.

Hold up, let’s be a bit more precise: it doesn’t factorise over the rationals. It factorises perfectly happily over the reals ((in general, the complex numbers might be needed)) as $(1 - \phi x)\br{1 + \frac{x}{\phi}}$.

Which means we can write it in partial fractions. It’s a bit ugly, but it works out fine enough: $\frac{x}{\br{1-\phi x}\br{1+\frac{x}{\phi}}} \equiv \frac{1}{\sqrt{5}}\br{\frac{1}{1+ \frac{x}{\phi}} - \frac{1}{1-\phi x}}$.

### Still so what?

Hold on, I’m getting to it.

There’s a simple binomial expansion for $\frac{1}{1-kx}$ (it’s $1 + kx + k^2x^2 + \dots$). Here, we’ve got two of those, one with $k = -\frac{1}{\phi}$ and one with $k = \phi$.

So, the coefficient of $x^k$ in $A(x)$ is $\frac{1}{\sqrt{5}} \br{ \phi^k - \br{\frac{1}{\phi}}^k }$, which is Binet’s formula! (Remember, the coefficients by definition are the terms of the Fibonacci sequence.)

### I suppose there’s a clincher here, too

But of course!

If you have any finite linear recurrence relation, you can come up with a generating function in a similar way: if you’ve got $P a_n + Q a_{n-1} + R a_{n-2} + \dots + Z a_{n-k}= 0$, then your generating function is $A(x) = \frac{a_0 + x a_1 + \dots x^{k-1} a_{k-1}}{P + Qx + \dots + Zx^{k}}$.

If you can find the zeros of the denominator – which you definitely can if $k \le 4$ – you can split it up into partial fractions (possibly requiring a little finesse with repeated roots) and come up with an explicit form for each of the terms – which are the elements of the sequence you defined recurrently!