For an $n$th order linear differential equation, if you have $n-1$ independent solutions, the Wronskian lets you find the last one. I would like to know if there is a way of finding the last solution to a $n$th order difference equation where you have $n-1$ independent solutions.
I know the product rule doesn't work for difference equations, so it can't work exactly the same way as the Wronskian, but I'm wondering if there's some other systematic method which achieves the same end.
Since making this question, I have discovered something which seems work similarly. In the two dimensional case, if we have the recurrence relation
$$x''=ax'+bx$$
(where $a$ and $b$ may depend on $t$) (I am using $x'$ to denote $x_{t+1}$ and $x$ to denotes $x_n$ for compactness) and have two solutions, $f$ and $g$ to this relation, we can define $$W=fg' - f'g$$ Then we find $$W' = f'g'' - f''g'$$ $$= f'(ag'+bg)-(af'+bf)g'$$ $$=bf'g - bfg'$$ so $$W' = -bW$$ $$W_t= W_0 (-b)^t$$ which can be used to find a first order recurrence relation for $g$ in terms of $f$.
I have also found a derivation using the $\Delta$ operator instead. If $f, g$ are solutions to $$\Delta^2 x = a\Delta x + bx$$ let $W = f \Delta g -g \Delta f$ then using the 'product rule' for difference equations, $\Delta(xy) = x \Delta y + y' \Delta x$ we have $$\Delta W = \Delta f \Delta g + f' \Delta^2 g - \Delta f \Delta g - g' \Delta^2 f$$ $$= f'(a \Delta g + bg) - g'(a \Delta f + bf)$$ $$= f'(ag' + (b-a)g) - g'(af' + (b-a)f)$$ $$= (b-a)(f'g-fg') = (b-a)(g \Delta f - f \Delta g)$$ so $$\Delta W = (b-a)W$$
So my new questions are: is this derivation correct? is this derivation the standard approach? and does this derivation generalise to higher order recurrence relations / difference equations?