<s>[INST] Generating functions and a closed form for the Fibonacci sequence - the big picture I have spent the last few hours trying to understand one way of deriving a closed form for the Fibonacci sequence. As part of improving my mathematical maturity, I am trying to learn to see the "big picture" of what I'm doing as opposed to myopically following the details of a derivation or proof and then forgetting most of what I've done within a few weeks or months.
Below I will post my derivation so that it can be checked for errors and general clarity, but what I would like also ask what the proper intuition or high-level perspective is on what I've done.
As I look over my work, the key takeaways I see are that we put the Fibonacci sequence into the form of a generating function, and in particular we managed to put the generating function into a compact rational form. Once this was done, the rest of the work essentially involved analyzing the rational form of the generating function and extracting information from it until we were able to write the generating function (in its formal power series form) in two different ways, compare coefficients and thus derive a closed form for the Fibonacci sequence.
If these observations are apt, then perhaps the big takeaway is that generating functions can be useful when they make it possible to package a sequence into a form that allows algebraic and/or analytic techniques to be brought to bear on the sequence, when it is represented as a generating function.
I would appreciate if anyone can tell me whether my perspective is reasonable or not.
My exposition:
The Fibonacci numbers are a sequence $1, 1, 2, 3, 5, 8, 13, \dots$. The first two numbers are 1, and then every subsequent number is the sum of the prior two. Let $(a_n)$ be the sequence of Fibonacci numbers and $f(x) = a_0 + a_1 x + a_2 x^2 + \dots$ be the generating function. Consider that
\begin{align*}
x^2 f(x) + x f(x) &= a_0 x^2 + a_1 x^3 + a_2 x^4 + \dots + a_0 x + a_1 x^2 + a_2 x^3 + \dots\\
&= a_0 x + (a_0 + a_1) x^2 + (a_1 + a_2) x^3 + \dots\\
&= a_0 x + a_2 x^2 + a_3 x^3 + \dots\\
&= f(x) - a_0 - a_1 x + a_0 x\\
&= f(x) - 1 - x + x \text{ (using the known values for } a_0 \text{ and } a_1)\\
&= f(x) - 1.
\end{align*}
Thus $f(x) = \frac{1}{1 - x - x^2}$. Now factor $1 - x - x^2$ as $(1 - \alpha x)(1 - \beta x)$, so that $- \alpha - \beta = -1$ (or more naturally, $\alpha + \beta = 1$) and $\alpha \beta = -1$. If you solve this system of equations you will end up with $\alpha = \frac{1 + \sqrt 5}{2}$ and $\beta = \frac{1 - \sqrt 5}{2}$, or the reverse. Now consider the partial fraction decomposition
\begin{align*}
f(x) &= a_0 + a_1 x + a_2 x^2 + \dots\\
&= \frac{1}{1 - x - x^2}\\
&= \frac{1}{(1 - \alpha x)(1 - \beta x)}\\
&= \frac{a}{1 - \alpha x} + \frac{b}{1 - \beta x}\\
\Rightarrow 1 &= a(1 - \beta x) + b(1 - \alpha x).
\end{align*}
This gives us another system of equations such that $a + b = 1$ and $-a \beta - \alpha b = 0$ (or more naturally, $a \beta + \alpha b = 0$.) If you solve \textit{this} system of equations you get $a = \frac{\sqrt 5 + 1}{2 \sqrt 5}$ and $b = \frac{\sqrt 5 - 1}{2 \sqrt 5}$.\
Thus,
\begin{align*}
f(x) &= a_0 + a_1 x + a_2 x^2 + \dots\\
&= \frac{1}{1 - x - x^2}\\
&= \frac{1}{(1 - \alpha x)(1 - \beta x)}\\
&= \frac{a}{1 - \alpha x} + \frac{b}{1 - \beta x}\\
&= a(1 + \alpha x + \alpha^2 x^2 + \dots) + b(1 + \beta x + \beta ^2 x^2 + \dots)\\
&= (a + b) + (a \alpha + b \beta)x + (a \alpha ^2 + b \beta^2)x^2 + \dots
\end{align*}
which explicitly means that $a_k = (\frac{\sqrt 5 + 1}{2 \sqrt 5})(\frac{1 + \sqrt 5}{2})^k + (\frac{\sqrt 5 - 1}{2 \sqrt 5})(\frac{1 - \sqrt 5}{2})^k$, as desired.
 [/INST] It's a good approach. One thing that can be simplified a little bit is:
$$
f(x)=\frac{1}{(1-\alpha x)(1-\beta x)} = \frac{1}{\alpha - \beta} \cdot \frac{\alpha (1-\beta x)  - \beta(1-\alpha x)}{(1-\alpha x)(1-\beta x)} = \frac{\alpha/(\alpha - \beta)}{1-\alpha x} - \frac{\beta/(\alpha - \beta)}{1-\beta x}.
$$
(And this is not hindsight.)
</s>