Let's consider the following simple recurrence relation
Can I use sympy to symbolically compute the resulting expression at index n recursively? I'm looking for a generic way to do this, i.e. I do not want to have an explicit solution to the above equation (as it may in general be to difficult for sympy to solve).
Based on http://gerin.perso.math.cnrs.fr/Enseignements/Symbolic3_Solutions.pdf I was able to write a solution
from sympy import *
from IPython.display import display
init_printing()
a, b = symbols('a b')
y = [1,1] # store sequence in list, start with initial conditions
N = 5
for n in range(2, N+1):
y += [simplify(a*y[-1]+b*y[-2])]
display(y[-1])
This gives the output
+
(+)+
((+)+)+(+)
(((+)+)+(+))+((+)+)
You can use rsolve:
In [10]: from sympy import *
In [11]: y = Function('y')
In [12]: a, b, n = symbols('a, b, n')
In [13]: eq = Eq(y(n), a*y(n-1) + b*y(n-2))
In [14]: eq
Out[14]: y(n) = a⋅y(n - 1) + b⋅y(n - 2)
In [15]: rsolve(eq, y(n))
Out[15]:
n n
⎛ __________⎞ ⎛ __________⎞
⎜ ╱ 2 ⎟ ⎜ ╱ 2 ⎟
⎜a ╲╱ a + 4⋅b ⎟ ⎜a ╲╱ a + 4⋅b ⎟
C₀⋅⎜─ - ─────────────⎟ + C₁⋅⎜─ + ─────────────⎟
⎝2 2 ⎠ ⎝2 2 ⎠
https://docs.sympy.org/latest/modules/solvers/solvers.html#sympy.solvers.recurr.rsolve