0
$\begingroup$

I want to find $a(m,n)$ with $~a(m,n)=\alpha * a(m-1,n) + \beta * a(m,n-1)$, where $\alpha$ and $\beta$ are constants. I expect an explicit $a(m,n)$ represented by $a(0,j),~j=1,2,\cdots,n$ and $a(i,0),~i=1,2,\cdots,m$.

But I could get nothing from the code below. How to make it work?

RSolve[a[m, n] == α a[m - 1, n] + β a[m, n - 1], a[m, n], {m, n}]
Improve this question
$\endgroup$
4
  • $\begingroup$ RSolve does not realize that you assume it is going to stop the recurrence at zero. Giving it a[m,0]==c, a[0,n]==d is sometimes required. But in this case that doesn't help. Maybe the unknowns alpha and beta are the problem, but even RSolve[{a[m, n] == 2 a[m-1, n] + 3 a[m, n-1], a[0, n] == 1, a[m, 0] == 1}, a[m, n], {m, n}] isn't solved. Perhaps there is no simple closed form solution to this even simpler example. $\endgroup$
    – Bill
    Apr 29, 2015 at 16:51
  • $\begingroup$ @Bill But what can I do to find out the answer? $\endgroup$
    – eccstartup
    Apr 30, 2015 at 14:27
  • 1
    $\begingroup$ If you evaluate this: a[0, 0] := o; a[0, n_] := p; a[m_, 0] := q; a[m_, n_] := \[Alpha] a[m-1, n] + \[Beta] a[m, n-1]; Simplify[Table[a[i, j], {i, 0, 4}, {j, 0, 4}]] and you neatly line up the rows and columns then it appears that the relationship between m and n and the entry in a particular location is not so complicated that you might never be able to see a general expression to write down. (I have used o, p, q instead of a(0,0) a(0,j) a(i,0) to stop the recursion). If this were all numbers then Mathematica might be able to find an expression, but with symbols I can't see how yet. $\endgroup$
    – Bill
    Apr 30, 2015 at 17:16
  • $\begingroup$ I think this is helpful. It verifies my proof. @Bill $\endgroup$
    – eccstartup
    May 1, 2015 at 0:07

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.