I am required to solve the below system of recurrence relations:$$\begin{cases}a_n-a_{n-4}=t_n-t_{n-3}\\a_n-a_{n-1}=25t_{n-1}-t_{n-3}\end{cases}$$As you can see, I can't isolate $a_i$ or $t_i$. I would like to know if a non-constant closed-form solution exists and if yes, how to obtain it.
Solving a system of coupled recurrence relations
discrete mathematicsrecurrence-relationssystems of equations
Related Solutions
Although it is possible to solve selected non-linear recurrence relations if you happen to be lucky, in general all sorts of peculiar and difficult-to-characterize things can happen.
One example is found in chaotic systems. These are hypersensitive to initial conditions, meaning that the behavior after many iterations is extremely sensitive to tiny variations in the initial conditions, and thus any formula expressing the relationship will grow impossibly large. These recurrence equations can be amazingly simple, with xn+1 = 4xn(1-xn) with x0 between 0 and 1 as one of the classic simple examples (i.e. merely quadratic; this is the logistic map).
User @Did has already given the Mandelbrot set example--similarly simple to express, and similarly difficult to characterize analytically (e.g. by giving a finite closed-form solution).
Finally, note that to solve every non-linear recurrence relation would imply that one could solve the Halting problem, since one could encode a program as initial states and the workings of the Turing machine as the recurrence relations. So it is certainly hopeless in the most general case. (Which highly restricted cases admit solutions is still an interesting question.)
You did well. You just stopped too soon! Now you solve $a_{n+1}=4a_{n-1}$, which is easy: $a_{2n}=0,a_{2n+1}=4^na_1$.
So from the first given equation $b_2=-a_2-a_1=-a_1$ and hence From the second given equation $b_2=b_1-3a_0=b_1=1$, so $a_1=-1$.
Now the first equation gives $b_{2n}=-a_{2n-1}=4^{n-1}=2^{2n-2}$ and $b_{2n+1}=-a_{2n+1}=4^n=2^{2n}$.
Check: $-a_{2n-1}-a_{2n}=-a_{2n-1}=2^{2n-2}=b_{2n-1}$
$-a_{2n}-a_{2n+1}=-a_{2n+1}=2^{2n}=b_{2n+1}$
$b_{2n}+3a_{2n-1}=2^{2n-2}+3\cdot2^{2n-2}=2^{2n}=b_{2n+1}$
$b_{2n+1}+3a_{2n}=2^{2n}=b_{2n+2}$
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On your first question, I have no idea where this particular problem came from, but recurrence relations often appear when trying to solve combinatorial problems. Sometimes you get two different kinds of configuration mixed up, so it is convenient to denote their numbers as $a_n,b_n$ and derive relations like those in the Q.
Best Answer
Hint:
$$a_n - a_{n-4}=(a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + (a_{n-2} - a_{n-3}) + (a_{n-3} - a_{n-4})$$
Here, the LHS and each summand on the RHS can be expressed via $t$ using one of your two equations, giving a recurrence relation in just $t$.