I'm trying to solve a recurrence relation for a function of three variables. It works with 2 variables but not with 3. A simple example is:
RSolve[a[i + 1, j + 1, l + 1] == a[i, j, l] + 3, a[i, j, l], {i, j, l}]
Do you have any idea how to solve this?
Edit: The actual recurrence relation i want to solve is the following:
$$ C\left(h_1+1,h_2,h_3\right)=\frac{\left(h_1+h_2-h_3\right) C\left(h_1,h_2,h_3\right)}{2 h_1+2 \text{j}_1-k-2}\\ C\left(h_1,h_2+1,h_3\right)=\frac{\left(h_1+h_2-h_3\right) C\left(h_1,h_2,h_3\right)}{2 h_2+2 \text{j}_2-k-2} \\ C\left(h_1,h_2,h_3-1\right)=\frac{\left(h_1+h_2-h_3\right) C\left(h_1,h_2,h_3\right)}{2 h_3-2 \text{j}_3-3 (k+2)} $$ It can be brought into the form of my original question of course. However I know the solution to the problem already, it is a combination of Gamma-functions. But I am interested on how to solve it with Mathematica so I can use it on more complicated relations that will appear in the future.
C[h1, h2, h3 + 1]
rather thanC[h1, h2, h3 - 1]
. I know how to answer your question but wish to be sure that the equations are correct before doing so. $\endgroup$C[h1,h2,h3-1]
is correct as it stands. $\endgroup$C[h1 + 1, h2 +1, h3 + 1] == f[I, j, l] C[h1, h2, h3
.] Or, do you wish to solve the set of three equations explicitly? The former can be solved without difficulty in terms ofGamma
functions. The latter is overdetermined, unless specific constraints are placed on the initial conditions. $\endgroup$