Ordinary generating function is very similar in concept to the $Z$ transform. For linear recurrences with constant coefficients we get rational functions. Suppose that we are limited to real partial fractions. We have rational function$$\frac{P(x)}{Q(x)}.$$We have following cases:
- Denominator has distinct linear factors (we do not consider complex numbers). Here it is easy to calculate $n$th derivative or simply get geometric series.
- Denominator has distinct irreducible quadratic factors. Here we know that\begin{align*}\operatorname{OGF}\left (\sin \left (\frac{\pi}{2}n\right )\right ) & =\frac{x}{1+x^2}, \\ \operatorname{OGF}\left (\cos \left (\frac{\pi}{2}n\right )\right ) & =\frac{1}{1+x^2}. \end{align*} But in this case we have following fractions:$$\frac{Ax+B}{x^2+px+q}.$$Can we conclude something if we complete the square in the denominator? Is there shift property like in Laplace transform?
- Denominator has repeated factors (linear of irreducible quadratics). Here we have that$$\operatorname{OGF}(f*g)=F(x)G(x),$$but convolution is given by sum instead of an integral$$(f*g)(n)=\sum \limits _{k=0}^nf(k)g(n-k).$$
Maybe this question help me.
What is $\operatorname{OGF}(a^n\cos (\omega n+\theta ))$ or $\operatorname{OGF}(a^n\sin (\omega n+\theta ))$?
I think that one of these is enough.
Is it necessary to play with complex numbers to get this $\operatorname{OGF}$?
If I don't make mistake in my calculations$$\operatorname{OGF}(a^n\cos (\omega n+\theta ))=\frac{\cos (\theta )-a\cos (\omega -\theta )x}{1-2a\cos (\omega )x+a^2x^2}.$$