Let $∑_{n=0}^∞c_n z^n $ be a representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ Let $∑_{n=0}^∞c_n z^n $ be a power series representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ and radius of convergence of the series.
Clearly this is a power series with center $z_0=0$, and $f(z)=\frac{1}{1-z-z^2 }$ is analytic, because it's represented by a power series. I also know that
$$c_n =\frac{1}{n!} f^{(n)}(0)$$
but this doesn't get me anywhere, I also try the special case of Taylor series, but nothing look like this. I wonder if any one would give  me a hint please.
 A: Using the factorization $1 - z - z^2 = -(z + \varphi)(z + \bar{\varphi})$, where $\varphi = (1 + \sqrt{5})/2$ and $\bar{\varphi} = (1 - \sqrt{5})/2$, we write
\begin{align}\frac{1}{1 - z - z^2} &= \frac{1}{\sqrt{5}}\left(\frac{1}{z + \varphi} - \frac{1}{z + \bar{\varphi}}\right) \\
& = \frac{1}{\sqrt{5}}\left(\frac{1/\varphi}{(z/\varphi) + 1} - \frac{1/\bar{\varphi}}{(z/\bar{\varphi}) + 1}\right)\\
& = \frac{1}{\sqrt{5}}\left(\frac{-\bar{\varphi}}{1 - \bar{\varphi}z} + \frac{\varphi}{1 - \varphi z}\right)\\
& = \frac{1}{\sqrt{5}}\sum_{n = 0}^\infty [-\bar{\varphi}(\bar{\varphi} z)^n +  \varphi(\varphi z)^n]\\
& = \sum_{n = 0}^\infty \left(\frac{\varphi^{n+1} - \bar{\varphi}^{n+1}}{\sqrt{5}}\right)z^n.
\end{align}
So $c_n = (\varphi^{n+1} - \bar{\varphi}^{n+1})/\sqrt{5}$ (which is equal to $F_{n+1}$, the $(n+1)$st Fibonacci number) and the radius of convergence is $1/\varphi$.
A: Besides decomposing in elementary fractions, you can start from
$$f(z)(1-z-z^2)=1,$$
and by identification $$c_nz^n-c_{n-1}z^{n-1}z-c_{n-2}z^{n-2}z^2=(c_n-c_{n-1}-c_{n-2})z^n=0$$
for $n\ge2$.
A: Taylor series of function is
$$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+21z^7+....$$
the coefficients are Fibonacci numbers
$$F(n)=F(n-1)+F(n-2)$$
hence
$C_n=F(n)$
A: Represent it as a geometric series:
$$\frac{1}{1 - z - z^2} = \sum_{k=0}^{\infty} z^k(1 + z)^k$$
The $c_n$ coefficient is effected by the terms $k = \lceil n/2\rceil$ to $k = n$. In particular,
$$c_n = \sum_{k = \lceil n/2 \rceil}^{n} {k  \choose n - k}$$
Can you show that these coefficients are the Fibonacci numbers? (Hint: The fibonacci sequence are the shallow diagonals on Pascal's triangle)
