handling of rsolve coefficients

[j3parera]

The rsolve method does not always allow the use of coefficients with functions even if its result is constant or independent of the solution's function symbol. For instance, a classic second order system with two complex conjugate poles and expanded coefficients:

n = Symbol('n', integer=True)
y = Function('y')
r = Symbol('r', real=True, positive=True)
theta = Symbol('theta', real=True)
f = y(n) - 2*r*cos(theta)*y(n-1) + r**2*y(n-2)
sol = rsolve(f, y(n))

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-44-fbee907275eb> in <module>
      3 theta = Symbol('theta', real=True)
      4 f = y(n) - 2*r*cos(theta)*y(n-1) + r**2*y(n-2)
----> 5 sol = rsolve(f, y(n))

~/.local/lib/python3.7/site-packages/sympy/solvers/recurr.py in rsolve(f, y, init)
    739                 else:
    740                     raise ValueError(
--> 741                         "'%s' expected, got '%s'" % (y.func, h.func))
    742             else:
    743                 coeff *= h

ValueError: 'y' expected, got 'cos'

In this case cos(theta) is independent of n.
With some specific values of theta it doesn't work:

f = y(n) - 2*r*cos(S.Pi/16)*y(n-1) + r**2*y(n-2)
sol = rsolve(f, y(n))
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-45-0a4e9e18043f> in <module>
      1 f = y(n) - 2*r*cos(S.Pi/16)*y(n-1) + r**2*y(n-2)
----> 2 sol = rsolve(f, y(n))

~/.local/lib/python3.7/site-packages/sympy/solvers/recurr.py in rsolve(f, y, init)
    739                 else:
    740                     raise ValueError(
--> 741                         "'%s' expected, got '%s'" % (y.func, h.func))
    742             else:
    743                 coeff *= h

ValueError: 'y' expected, got 'cos'

But with some other values it does:

f = y(n) - 2*r*cos(S.Pi/4)*y(n-1) + r**2*y(n-2)
sol = rsolve(f, y(n))
print(sol)

with the independent symbol r in solution correctly handled:
C0*(sqrt(2)*r*(1 - I)/2)**n + C1*(sqrt(2)*r*(1 + I)/2)**n.

[j3parera]

rsolve also fails for quite simple recurrences:

y = Function('y')
n = Symbol('n', integer=True)
f = y(n) - 0.9 * y(n - 1) + 0.81 * y(n - 2)
sol = rsolve(f, y(n))

results in sol = None.
But, surprisingly, just changing the last coefficient from 0.81 to 0.8 it works!

[smichr]

rsolve also fails for quite simple recurrences:

my guess is that this is a Float vs Rational issue:

>>> y = Function('y')
... n = Symbol('n', integer=True)
... f = y(n) - S(9)/10 * y(n - 1) + Rational('0.81') * y(n - 2)
... sol = rsolve(f, y(n))

>>> sol
C0*(9/20 - 9*sqrt(3)*I/20)**n + C1*(9/20 + 9*sqrt(3)*I/20)**n