Changing recurrence to matrix

Question

$$F(x) = aF(x-k+1) + bF(x-k+2) + cF(x-k+4) + dF(x-k+7)$$

where $F(x) = 1$ if $x<k$.

$a,b,c,d,k$ are known (and positive) and $x$ is chosen.

Can anyone show how to set this up in matrix form using $k=4$?

Answer 1

If $k=4$ the relationship is anticipatory because of $d F(x-4+7)$ term and if $d \neq 0$ then you may have to solve for $F(x+3)$. Also, $F(x-k+4)$ would cause problems if $c=1$.

I will show how to do it for $k>7$ say $k=8$.

Then

$$F(x) = a F(x-7) + b F(x-6)+cF(x-4)+d F(x-1)$$

So we have to go all the way back to $F(x-7)$. So let $$G(x) = \pmatrix{F(x)\\F(x-1)\\F(x-2)\\F(x-3)\\F(x-4)\\F(x-5) \\ F(x-6)} $$ Then $$G(x-1) = \pmatrix{F(x-1)\\F(x-2)\\F(x-3)\\F(x-4)\\F(x-5) \\ F(x-6)\\F(x-7)} $$ Hence $$ G(x) = \pmatrix{ d & 0&0&c&0&b&a\\ 1 & 0&0&0&0&0&0\\ 0&1 & 0&0&0&0&0\\ 0&0&1 & 0&0&0&0\\ 0&0&0&1 & 0&0&0\\ 0&0&0&0&1 & 0&0\\ 0&0&0&0&0 & 1&0\\ } G(x-1)$$

Added in response to OP's comment

Suppose $k=5$. Then the original equation is

$$ F(x) = a F(x-4)+bF(x-3)+cF(x-1)+dF(x+2)$$ replace $x$ by $x-2$ to get $$ F(x-2)=a F(x-6)+bF(x-5)+c F(x-3)+d F(x)$$ Solving we get $$ F(x) = \frac{1}{d} F(x-2)-\frac a d F(x-6) -\frac b d F(x-5) -\frac c d F(x-3) $$

Now proceed as before.

Additional comments Note that $F(x)$ depends on past values of $F$. $G$ is just a vector with $F$ and its past values. How long should $G$ be? At least as long as to be able to write $F(x)$ in terms of $G(x-1)$. Make it longer does not hurt except we need to keep more of the past history.