2D variable coefficient recurrence relation

  • Thread starter rsq_a
  • Start date
  • #1
rsq_a
107
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.
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
135
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.
 
  • #3
rsq_a
107
1
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 realise 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.
 

Suggested for: 2D variable coefficient recurrence relation

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
11
Views
924
Top