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}]
  • $\begingroup$ 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
    Commented Apr 29, 2015 at 16:51
  • $\begingroup$ @Bill But what can I do to find out the answer? $\endgroup$
    – eccstartup
    Commented 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
    Commented Apr 30, 2015 at 17:16
  • $\begingroup$ I think this is helpful. It verifies my proof. @Bill $\endgroup$
    – eccstartup
    Commented May 1, 2015 at 0:07


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