# How to solve the difference equation $W(x+h)=c(x)W(x)$?

Let $c(x)=\prod_{s=1}^{n}(x-z_s)$ be a polynomial, where $z_s$ are some complex numbers. How to solve the difference equation $W(x+h)=c(x)W(x)$ for $W(x)$? Thank you very much.

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Start with $n=1$.
Functional equation $W(x+h) = (x-a) W(x)$ is, after change of variables $x= h \cdot z$, becomes $g(z+1) = h (z - \frac{a}{h}) g(z)$, which is solved by $g(z) = \kappa \cdot h^z \Gamma\left(z - \frac{a}{h} \right)$ in a suitable class of functions, so $W(x) = \kappa \cdot h^\tfrac{x}{h} \Gamma\left( \frac{x-a}{h} \right)$.
When $n>1$, the solution is going to be a product of these: $$W(x) = \kappa \cdot h^{n x/h} \cdot \prod_{s=1}^n \Gamma\left( \frac{x-a_s}{h} \right)$$
thank you very much. But it seems that $g(z+1)=\kappa \cdot h^{z} \Gamma(z-\frac{a}{h})$. – LJR Nov 23 '11 at 16:10
@user9791 Well, $\Gamma(z + 1) = z \Gamma(z)$, thus $\Gamma\left(z + 1 - \frac{a}{h} \right) = \left(z - \frac{a}{h} \right) \Gamma\left(z - \frac{a}{h} \right)$. So I think I am correct. – Sasha Nov 23 '11 at 16:12
there is another problem. We can verify that $g(z)$ satisfies the equation. Is $g(z)$ the only solution? – LJR Nov 23 '11 at 17:12