Generating Functions - Fibonacci sequence

Question

I'm new to generating functions, and my lecture showed how to obtain exact closed form expression for Fibonacci numbers.

The coefficients of the generating function F(x) is the Fibonacci sequence {f_n}. After some manipulation,

$$\begin{align} (1-x-x^2)F(x) &= x \tag{A} \\ \\ F(x) &= \frac{x}{1-x-x^2} \tag{B} \\ \\ F(x) &= \frac{\frac{A}{1-a_1x}+\frac{B}{1-a_2x}}{\sqrt 5} \tag{C} \\ \\ F(x) &= \sum_{n=0} f_nx^n \tag{D} \end{align}$$

After doing the partial fraction decomposition, F(x) can then be written as a sum of 2 geometric series the coefficients of the series can be read off to obtain the closed form expression.

The equality in the first line is between formal power series; and the equality in the following lines is in terms of the function. I struggling to understand how the second line follows the first line.

Can someone please explain the sense of the steps?

Answer 1

The second line follows the first line by division. You simply divide both sides by $(1-x-x^2)$

Answer 2

The solutions to $1-y-y^2=0$ are $A=(1+\sqrt 5)/2$ and $B=(1-\sqrt 5)/2$.Therefore for all $x$ we have $1-x-x^2=(1-x/A)(1-x/B)$. Then for $A \ne x \ne B$ we have $$\frac {1}{1-x-x^2}=$$ $$ \frac {1}{(1-x/A)(1-x/B)}=$$ $$\left( \frac {1/A}{1-x/A}- \frac {1/B}{1-x/B} \right) \frac{1}{1/A-1/B}$$ which simplifies a little because $1/A-1/B=(B-A)/AB$ and $B-A=-\sqrt 5$ and $AB=-1$