Two-level Recurrence in Matlab

Hi All,

I want to solve a (system of) recurrence equation in Matlab's symbolic toolbox. Formally, I would like to have a formula for the value of a(n) if a(n) is defines, as follows:

a(n+1) = 0.5*b(n)+0.5*a(n)

b(n+1) = 0.2 + 0.8*b(n)

For one-level recurrence the task is easy and the solution by Matlab is

eq1 := rec(b(n+1) = 0.2 + 0.8*b(n),b(n),b(0)=b0)

rec(b(n + 1) - 0.8*b(n) - 0.2, b(n), {b(0) = b0})

solve(eq1)

{0.8^n*(1.0*b0 - 1.0) + 1.0}

That is, Matlab determined the close for solution for b(n). Do you have an idea how I can achieve the same for a(n)?

eq2 := rec(a(n+1) = 0.5*b(n) + 0.5*a(n),a(n),a(0)=a0)

rec(a(n + 1) - 0.5*a(n) - 0.5*b(n), a(n), {a(0) = a0})

But solving the system of equations

solve({eq1,eq2})

gives an error of Error: Recursive definition: the maximal depth for nested procedure calls is reached. Evaluating: numeric::rationalize@rec_rat

despite a closed form solution exists.

Any help is appreciated very much.

Thanks! Gerzson

Answer 1

[Torsten]

You could try to insert the solution expression for b(n) in the calculation for a(n).

Another way:

https://www.wolframalpha.com/input/?i=Rsolve%5B%7Ba%5Bn%2B1%5D+%3D%3D+0.5*b%5Bn%5D%2B0.5*a%5Bn%5D,+b%5Bn%2B1%5D+%3D%3D+0.2+%2B+0.8*b%5Bn%5D,+a%5B0%5D%3D%3Da0,+b%5B0%5D%3D%3Db0%7D,%7Ba%5Bn%5D,b%5Bn%5D%7D,n%5D

Or this approach:

https://de.mathworks.com/help/symbolic/compute-z-transforms-and-inverse-z-transforms.html

Best wishes

Torsten.