2D variable coefficient recurrence relation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 3K views
rsq_a
Messages
103
Reaction score
1
Consider a 2D variable coefficient linear recurrence relation. An example might be:

[tex]b_{n,j+1} (j+1)(2n-1)(2n-2) = (2n-2+j)(2n-1+j)b_{n-1,j}[/tex]

which has the solution
[tex]b_{n,j} = \frac{(2n-1+j)!}{(2n-1)!j!}[/tex]

Is there any algorithm that can be used to derive this result? I have a recurrence relation which is a bit more complex than this one.
 
Mathematics news on Phys.org
Difference relations tend to be UGLY with respect to finding nice, explicit formulae.
The continuous analogue, diff. eqs, tend to be easier to generalize about.
 
arildno said:
Difference relations tend to be UGLY with respect to finding nice, explicit formulae.
The continuous analogue, diff. eqs, tend to be easier to generalize about.

I realize that. I know that for constant coefficient 2D linear recurrence relations, it seems that Z-Transforms can be applied and are mostly successful. I was wondering whether anybody had seen an analogous technique for non-constant coefficient ones.