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I am trying to solve the following system of recurrences with RSolve:

RSolve[
  {(1 - q^n) A[n] == 
    q^(3 n - 1) B[n - 1] + q^(2 n + 1) A[n - 1] + 
    q^(4 n - 2) A[n - 2] + q^(4 n - 2) B[n - 2],
   B[n] == 
     q^(n + 1) A[n - 1] - q^(5 n - 5) A[n - 3] - q^(5 n - 5) B[n - 3],
   A[0] == 1, A[1] == q^3/(1 - q), A[2] == q^6/((1 - q) (1 - q^2)), 
   B[0] == 0, B[1] == q^2, B[2] == q^6/(1 - q)},
  {A[n], B[n]}, n]

But the output is just the same as my input. I tried with initial conditions at for A and B at 1, 2, 3 (to avoid the singularity of 1 - q^0, but it also does not return anything.

Does this mean that Mathematica cannot solve this system?

m_goldberg
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Reimundo Heluani
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  • ONLY can solve for q=1: RSolve[{(1 - q^n) A[n] == q^(3 n - 1) B[n - 1] + q^(2 n + 1) A[n - 1] + q^(4 n - 2) A[n - 2] + q^(4 n - 2) B[n - 2], B[n] == q^(n + 1) A[n - 1] - q^(5 n - 5) A[n - 3] - q^(5 n - 5) B[n - 3]} /. q -> 1, {A[n], B[n]}, n] // FullSimplify – Mariusz Iwaniuk Feb 24 '20 at 18:40
  • at $q=1$ the system is not defined as the initial conditions blow up. In fact it is easy to see that $A[n]$ times $(q)n$ is a polynomial and $B[n]$ times $(q){n-1}$ is a polynomial. I tried writing the recurrence for these products (multiplying accordingly the system above) and also didn' t get a solution – Reimundo Heluani Feb 25 '20 at 11:33

0 Answers0