Expanding $\frac{1}{1-z-z^2}$ to a power series. How would you expand the analytic function $$\frac{1}{1-z-z^2}$$ to a series of the form $$\sum_{k=0}^\infty a_k z^k \, \, ?$$
 A: Write $1-z-z^2=(a-z)(z+b)$ and and using this, write the partial fraction of
$$
\frac 1{1-z-z^2}=\frac 1{a+b}\left(\frac 1 {a-z} +\frac 1 {b+z} \right)
$$
A: Three ways:


*

*Write as partial fractions:
$$
\frac{1}{1 - z - z^2}
  = \frac{1}{(1 - \tau z) (1 - \overline{\tau} z)}
  = \frac{\tau}{\sqrt{5} (1 - \tau z)} 
      - \frac{\overline{\tau}}{\sqrt{5} (1 -\overline{\tau} z)}
$$
Here $\tau$ is the positive root of $r^2 - r - 1 = 0$, $\overline{\tau}$ the negative one ($\tau$s are zeros of the denominator $1 - z - z^2$). That is:
\begin{align}
\tau            &= \frac{-1 + \sqrt{5}}{2} \\
\overline{\tau} &= \frac{-1 - \sqrt{5}}{2}
\end{align}
This is a pair of geometric series:
$$
[z^n] \frac{1}{1 - z - z^2}
  = \frac{\tau^{n + 1} - \overline{\tau}^{n + 1}}{\sqrt{5}}
$$

*Expand:
\begin{align}
\frac{1}{1 - z(1 + z)}
  &= \sum_{r \ge 0} z^r (1 + z)^r \\
  &= \sum_{r \ge 0} z^r \sum_{0 \le s \le r} \binom{r}{s} z^s \\
[z^n] \frac{1}{1 - z - z^2}
  &= \sum_{r + s = n} \binom{r}{s} \\
  &= \sum_{0 \le k \le n} \binom{k}{n - k}
\end{align}

*Recognize the generating function of the Fibonacci numbers:
$$
F_0 = 0, F_1 = 1, F_{n + 2} = F_{n + 1} + F_n
$$
gives:
$$
F(z) = \sum_{n \ge 0} F_n z^n = \frac{z}{1 - z - z^2}
$$
so that:
$$
\frac{1}{1 - z - z^2}
  = \frac{F(z) - F_0}{z}
  = \sum_{n \ge 0} F_{n + 1} z^n
$$

A: Hint. Consider $$(1-z-z^2)\left(\sum a_kz^k\right)=1$$ as a formal polynomial equation, where the RHS is the polynomial $1=1+0z+0z^2+\dotsb$.  Multiplying out the LHS and equating with the RHS term by term, we find that $$\eqalign{a_0\,&=&\;1\cr -a_0+a_1\,&=&\;0\cr -a_0 - a_1 + a_2\,&=&\;0 \cr -a_1 - a_2 + a_3 \,&=&\;0\cr -a_2 - a_3 + a_4 \,&=&\;0 \cr \vdots\;\;\; &&\;\, \vdots\cr}$$
which we can rearrange to
$$\eqalign{a_0\,&=&\;1\\a_1\,&=&\;a_0\\a_2\,&=&\;a_0 + a_1\\ a_3\,&=&\;a_1 + a_2\\ a_4\,&=&\;a_2 + a_3 \\  \vdots \;\;&&\;\;\vdots}$$
Does this sequence look familiar$\ldots$?
