Demonstrating the coefficients of the power series of $\frac{1}{1-z-z^2}$ satisfies a recurrence relation.

Question

I have the power series $$\frac{1}{1-z-z^2} = \sum_{n=0}^{\infty} c_nz^n$$ and I'd like to show that the coefficients of this power series satisfy $c_n=c_{n-1}+c_{n-2}$. I thought the most obvious way to go about this would be induction, but it is giving me some problems.

The base case is easy. So let's assume that $c_n=c_{n-1}+c_{n-2}$ holds as our inductive hypothesis. The goal is to show that $c_{n+1}=c_n+c_{n-1}$.

Now, if we consider $$c_{n+1} = \frac{f^{(n+1)}(0)}{(n+1)!}=\frac{[f^{(n)}(0)]'}{(n+1)!}$$ Then we can substitute in via our inductive hypothesis, and get the result

$$\frac{[nf^{(n-1)}(0) + n(n-1)f^{(n-2)}(0)]'}{(n+1)!}$$

which does not get me any closer to the desired result.

I also tried determining the nth derivative of $\frac{1}{1-z-z^2}$, but that did not turn out to be an easy task. Am I going about this the wrong way? Is there an easier method to consider?

Thanks very much, your thoughts are greatly appreciated. ~Dom

Answer 1

Form the generating function

$$g(z) = \sum_{k=0}^{\infty} c_n z^n$$

Noting that $c_n-c_{n-1}-c_{n-2}=0$, multiply by $z^n$ and sum this relation from $n=2$ to $\infty$:

$$\sum_{k=2}^{\infty} c_n z^n - \sum_{k=2}^{\infty} c_{n-1} z^n - \sum_{k=2}^{\infty} c_{n-2} z^n = 0$$

Express each sum in terms of $g(z)$ and the initial conditions; you will get an equation for $g(z)$ which you can solve.