Suppose I have the following recurrence equation:

$$ F_{i+1}P_{i+1}+F_iP_i-P_{i+1}+P_i=2J\,, \quad\textrm{with}\quad F_{i+L}=F_i,P_{i+L}=P_i\,. $$

$J$ is constant, $\{P_i\}$ are probabilities with normalization $\sum_{i=1}^LP_i=1$. Now I want to use RSolve to solve for $\{P_i\}$ (or is it possible?), but I don't know how to set RSolve up for such periodic relations.


1 Answer 1


This should probably be a comment but it's too large. Let's look at the system for a particular case, say $L=3$. This means $f_1=f_4=f_7...$ and $f_2=f_5=f_8...$ and similarly for the $p_i$s. Then the relationship is really three relationships:

$f_2 p_2 + f_1 p_1 -p_2+p_1=2J$

$f_3 p_3 + f_2 p_2 -p_3+p_2=2J$

$f_1 p_1 + f_3 p_3 -p_1+p_3=2J$

which uses the fact that $f_4=f_1$ and $p_4=p_1$. You also have $p_1+p_2+p_3=1$.

So there is no real recurrence here. Instead, this is a collection of 4 equations and 6 unknowns. Hence there is no way to solve this.

  • $\begingroup$ I see your point. Actually I want to point out that $f_i$ is given, we don't need to solve for $f_i$. $\endgroup$
    – wdg
    Commented Jun 13, 2013 at 14:49
  • $\begingroup$ Then you have a system with three parameters (the $p_i$s) and four equations. There may or may not be a solution. But the point remains that it is a linear system and not a recurrence relation and so should be easy to solve (or to verify that there is no solution). $\endgroup$
    – bill s
    Commented Jun 13, 2013 at 14:54

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