I am going through MIT OCW course, and I suppose I've missed some point in the explanation.
So, we have the system's functional. The poles are clearly $\phi$ and $-1/\phi$ (where $\phi = 1.618\ldots$ is a golden ratio). So, we get $$H = \frac 1 {(1-\phi R)(1+\frac 1 \phi R)}.$$ Usually I would proceed with partial fractions to get the sum of two fractions. Something similar is done here, but I don't really understand where the numerators provided in the explanation are coming from.
Any help would be appreciated. Thanks.
$$ \frac 1 {(1-\varphi R)(1+\frac 1 \varphi R)} = \frac A {1-\varphi R} + \frac B {1+\frac 1 \varphi R} $$ Multiplying both sides by the denominator that appears on the left side, we get $$ 1 = A\left( 1 + \frac 1 \varphi R \right) + B\left( 1 - \varphi R \right). \tag 1 $$ This should hold if $R$ is any number at all. So in particular, it should hold if $R=-\varphi,$ which makes the first term on the right vanish: $$ 1 = A\cdot 0 + B(1 - \varphi (-\varphi)), $$ so $$ 1 = B(1 + \varphi^2) $$ and so \begin{align} B & = \frac 1 {1 + \varphi^2} \\[10pt] & = \frac 1 {1 + (\varphi + 1)} \text{ (Here we used the identity } \varphi^2 = \varphi + 1.) \\[10pt] & = \frac 1 {\varphi\sqrt 5} \text{ (by the identity } 2 + \varphi = \varphi\sqrt 5. \end{align} Just as $R = -\varphi$ makes the first term in $(1)$ vanish, so setting $R=1/\varphi$ makes the second term vanish, and then you can proceed similarly.
So you need to use the particular nature of the number $\varphi$ to do some of the arithmetic: $$\varphi^2 = \varphi + 1 \text{ and } \varphi\sqrt 5 = \varphi + 2.$$