r"""
This module is intended for solving recurrences or, in other words,
difference equations. Currently supported are linear, inhomogeneous
equations with polynomial or rational coefficients.
The solutions are obtained among polynomials, rational functions,
hypergeometric terms, or combinations of hypergeometric term which
are pairwise dissimilar.
``rsolve_X`` functions were meant as a low level interface
for ``rsolve`` which would use Mathematica's syntax.
Given a recurrence relation:
.. math:: a_{k}(n) y(n+k) + a_{k-1}(n) y(n+k-1) +
... + a_{0}(n) y(n) = f(n)
where `k > 0` and `a_{i}(n)` are polynomials in `n`. To use
``rsolve_X`` we need to put all coefficients in to a list ``L`` of
`k+1` elements the following way:
``L = [a_{0}(n), ..., a_{k-1}(n), a_{k}(n)]``
where ``L[i]``, for `i=0, \ldots, k`, maps to
`a_{i}(n) y(n+i)` (`y(n+i)` is implicit).
For example if we would like to compute `m`-th Bernoulli polynomial
up to a constant (example was taken from rsolve_poly docstring),
then we would use `b(n+1) - b(n) = m n^{m-1}` recurrence, which
has solution `b(n) = B_m + C`.
Then ``L = [-1, 1]`` and `f(n) = m n^(m-1)` and finally for `m=4`:
>>> from sympy import Symbol, bernoulli, rsolve_poly
>>> n = Symbol('n', integer=True)
>>> rsolve_poly([-1, 1], 4*n**3, n)
C0 + n**4 - 2*n**3 + n**2
>>> bernoulli(4, n)
n**4 - 2*n**3 + n**2 - 1/30
For the sake of completeness, `f(n)` can be:
[1] a polynomial -> rsolve_poly
[2] a rational function -> rsolve_ratio
[3] a hypergeometric function -> rsolve_hyper
"""
from __future__ import print_function, division
from collections import defaultdict
from sympy.core.singleton import S
from sympy.core.numbers import Rational, I
from sympy.core.symbol import Symbol, Wild, Dummy
from sympy.core.relational import Equality
from sympy.core.add import Add
from sympy.core.mul import Mul
from sympy.core import sympify
from sympy.simplify import simplify, hypersimp, hypersimilar
from sympy.solvers import solve, solve_undetermined_coeffs
from sympy.polys import Poly, quo, gcd, lcm, roots, resultant
from sympy.functions import binomial, factorial, FallingFactorial, RisingFactorial
from sympy.matrices import Matrix, casoratian
from sympy.concrete import product
from sympy.core.compatibility import default_sort_key, range
from sympy.utilities.iterables import numbered_symbols
[docs]def rsolve_poly(coeffs, f, n, **hints):
r"""
Given linear recurrence operator `\operatorname{L}` of order
`k` with polynomial coefficients and inhomogeneous equation
`\operatorname{L} y = f`, where `f` is a polynomial, we seek for
all polynomial solutions over field `K` of characteristic zero.
The algorithm performs two basic steps:
(1) Compute degree `N` of the general polynomial solution.
(2) Find all polynomials of degree `N` or less
of `\operatorname{L} y = f`.
There are two methods for computing the polynomial solutions.
If the degree bound is relatively small, i.e. it's smaller than
or equal to the order of the recurrence, then naive method of
undetermined coefficients is being used. This gives system
of algebraic equations with `N+1` unknowns.
In the other case, the algorithm performs transformation of the
initial equation to an equivalent one, for which the system of
algebraic equations has only `r` indeterminates. This method is
quite sophisticated (in comparison with the naive one) and was
invented together by Abramov, Bronstein and Petkovsek.
It is possible to generalize the algorithm implemented here to
the case of linear q-difference and differential equations.
Lets say that we would like to compute `m`-th Bernoulli polynomial
up to a constant. For this we can use `b(n+1) - b(n) = m n^{m-1}`
recurrence, which has solution `b(n) = B_m + C`. For example:
>>> from sympy import Symbol, rsolve_poly
>>> n = Symbol('n', integer=True)
>>> rsolve_poly([-1, 1], 4*n**3, n)
C0 + n**4 - 2*n**3 + n**2
References
==========
.. [1] S. A. Abramov, M. Bronstein and M. Petkovsek, On polynomial
solutions of linear operator equations, in: T. Levelt, ed.,
Proc. ISSAC '95, ACM Press, New York, 1995, 290-296.
.. [2] M. Petkovsek, Hypergeometric solutions of linear recurrences
with polynomial coefficients, J. Symbolic Computation,
14 (1992), 243-264.
.. [3] M. Petkovsek, H. S. Wilf, D. Zeilberger, A = B, 1996.
"""
f = sympify(f)
if not f.is_polynomial(n):
return None
homogeneous = f.is_zero
r = len(coeffs) - 1
coeffs = [Poly(coeff, n) for coeff in coeffs]
polys = [Poly(0, n)]*(r + 1)
terms = [(S.Zero, S.NegativeInfinity)]*(r + 1)
for i in range(r + 1):
for j in range(i, r + 1):
polys[i] += coeffs[j]*binomial(j, i)
if not polys[i].is_zero:
(exp,), coeff = polys[i].LT()
terms[i] = (coeff, exp)
d = b = terms[0][1]
for i in range(1, r + 1):
if terms[i][1] > d:
d = terms[i][1]
if terms[i][1] - i > b:
b = terms[i][1] - i
d, b = int(d), int(b)
x = Dummy('x')
degree_poly = S.Zero
for i in range(r + 1):
if terms[i][1] - i == b:
degree_poly += terms[i][0]*FallingFactorial(x, i)
nni_roots = list(roots(degree_poly, x, filter='Z',
predicate=lambda r: r >= 0).keys())
if nni_roots:
N = [max(nni_roots)]
else:
N = []
if homogeneous:
N += [-b - 1]
else:
N += [f.as_poly(n).degree() - b, -b - 1]
N = int(max(N))
if N < 0:
if homogeneous:
if hints.get('symbols', False):
return (S.Zero, [])
else:
return S.Zero
else:
return None
if N <= r:
C = []
y = E = S.Zero
for i in range(N + 1):
C.append(Symbol('C' + str(i)))
y += C[i] * n**i
for i in range(r + 1):
E += coeffs[i].as_expr()*y.subs(n, n + i)
solutions = solve_undetermined_coeffs(E - f, C, n)
if solutions is not None:
C = [c for c in C if (c not in solutions)]
result = y.subs(solutions)
else:
return None # TBD
else:
A = r
U = N + A + b + 1
nni_roots = list(roots(polys[r], filter='Z',
predicate=lambda r: r >= 0).keys())
if nni_roots != []:
a = max(nni_roots) + 1
else:
a = S.Zero
def _zero_vector(k):
return [S.Zero] * k
def _one_vector(k):
return [S.One] * k
def _delta(p, k):
B = S.One
D = p.subs(n, a + k)
for i in range(1, k + 1):
B *= -Rational(k - i + 1, i)
D += B * p.subs(n, a + k - i)
return D
alpha = {}
for i in range(-A, d + 1):
I = _one_vector(d + 1)
for k in range(1, d + 1):
I[k] = I[k - 1] * (x + i - k + 1)/k
alpha[i] = S.Zero
for j in range(A + 1):
for k in range(d + 1):
B = binomial(k, i + j)
D = _delta(polys[j].as_expr(), k)
alpha[i] += I[k]*B*D
V = Matrix(U, A, lambda i, j: int(i == j))
if homogeneous:
for i in range(A, U):
v = _zero_vector(A)
for k in range(1, A + b + 1):
if i - k < 0:
break
B = alpha[k - A].subs(x, i - k)
for j in range(A):
v[j] += B * V[i - k, j]
denom = alpha[-A].subs(x, i)
for j in range(A):
V[i, j] = -v[j] / denom
else:
G = _zero_vector(U)
for i in range(A, U):
v = _zero_vector(A)
g = S.Zero
for k in range(1, A + b + 1):
if i - k < 0:
break
B = alpha[k - A].subs(x, i - k)
for j in range(A):
v[j] += B * V[i - k, j]
g += B * G[i - k]
denom = alpha[-A].subs(x, i)
for j in range(A):
V[i, j] = -v[j] / denom
G[i] = (_delta(f, i - A) - g) / denom
P, Q = _one_vector(U), _zero_vector(A)
for i in range(1, U):
P[i] = (P[i - 1] * (n - a - i + 1)/i).expand()
for i in range(A):
Q[i] = Add(*[(v*p).expand() for v, p in zip(V[:, i], P)])
if not homogeneous:
h = Add(*[(g*p).expand() for g, p in zip(G, P)])
C = [Symbol('C' + str(i)) for i in range(A)]
g = lambda i: Add(*[c*_delta(q, i) for c, q in zip(C, Q)])
if homogeneous:
E = [g(i) for i in range(N + 1, U)]
else:
E = [g(i) + _delta(h, i) for i in range(N + 1, U)]
if E != []:
solutions = solve(E, *C)
if not solutions:
if homogeneous:
if hints.get('symbols', False):
return (S.Zero, [])
else:
return S.Zero
else:
return None
else:
solutions = {}
if homogeneous:
result = S.Zero
else:
result = h
for c, q in list(zip(C, Q)):
if c in solutions:
s = solutions[c]*q
C.remove(c)
else:
s = c*q
result += s.expand()
if hints.get('symbols', False):
return (result, C)
else:
return result
[docs]def rsolve_ratio(coeffs, f, n, **hints):
r"""
Given linear recurrence operator `\operatorname{L}` of order `k`
with polynomial coefficients and inhomogeneous equation
`\operatorname{L} y = f`, where `f` is a polynomial, we seek
for all rational solutions over field `K` of characteristic zero.
This procedure accepts only polynomials, however if you are
interested in solving recurrence with rational coefficients
then use ``rsolve`` which will pre-process the given equation
and run this procedure with polynomial arguments.
The algorithm performs two basic steps:
(1) Compute polynomial `v(n)` which can be used as universal
denominator of any rational solution of equation
`\operatorname{L} y = f`.
(2) Construct new linear difference equation by substitution
`y(n) = u(n)/v(n)` and solve it for `u(n)` finding all its
polynomial solutions. Return ``None`` if none were found.
Algorithm implemented here is a revised version of the original
Abramov's algorithm, developed in 1989. The new approach is much
simpler to implement and has better overall efficiency. This
method can be easily adapted to q-difference equations case.
Besides finding rational solutions alone, this functions is
an important part of Hyper algorithm were it is used to find
particular solution of inhomogeneous part of a recurrence.
Examples
========
>>> from sympy.abc import x
>>> from sympy.solvers.recurr import rsolve_ratio
>>> rsolve_ratio([-2*x**3 + x**2 + 2*x - 1, 2*x**3 + x**2 - 6*x,
... - 2*x**3 - 11*x**2 - 18*x - 9, 2*x**3 + 13*x**2 + 22*x + 8], 0, x)
C2*(2*x - 3)/(2*(x**2 - 1))
References
==========
.. [1] S. A. Abramov, Rational solutions of linear difference
and q-difference equations with polynomial coefficients,
in: T. Levelt, ed., Proc. ISSAC '95, ACM Press, New York,
1995, 285-289
See Also
========
rsolve_hyper
"""
f = sympify(f)
if not f.is_polynomial(n):
return None
coeffs = list(map(sympify, coeffs))
r = len(coeffs) - 1
A, B = coeffs[r], coeffs[0]
A = A.subs(n, n - r).expand()
h = Dummy('h')
res = resultant(A, B.subs(n, n + h), n)
if not res.is_polynomial(h):
p, q = res.as_numer_denom()
res = quo(p, q, h)
nni_roots = list(roots(res, h, filter='Z',
predicate=lambda r: r >= 0).keys())
if not nni_roots:
return rsolve_poly(coeffs, f, n, **hints)
else:
C, numers = S.One, [S.Zero]*(r + 1)
for i in range(int(max(nni_roots)), -1, -1):
d = gcd(A, B.subs(n, n + i), n)
A = quo(A, d, n)
B = quo(B, d.subs(n, n - i), n)
C *= Mul(*[d.subs(n, n - j) for j in range(i + 1)])
denoms = [C.subs(n, n + i) for i in range(r + 1)]
for i in range(r + 1):
g = gcd(coeffs[i], denoms[i], n)
numers[i] = quo(coeffs[i], g, n)
denoms[i] = quo(denoms[i], g, n)
for i in range(r + 1):
numers[i] *= Mul(*(denoms[:i] + denoms[i + 1:]))
result = rsolve_poly(numers, f * Mul(*denoms), n, **hints)
if result is not None:
if hints.get('symbols', False):
return (simplify(result[0] / C), result[1])
else:
return simplify(result / C)
else:
return None
[docs]def rsolve_hyper(coeffs, f, n, **hints):
r"""
Given linear recurrence operator `\operatorname{L}` of order `k`
with polynomial coefficients and inhomogeneous equation
`\operatorname{L} y = f` we seek for all hypergeometric solutions
over field `K` of characteristic zero.
The inhomogeneous part can be either hypergeometric or a sum
of a fixed number of pairwise dissimilar hypergeometric terms.
The algorithm performs three basic steps:
(1) Group together similar hypergeometric terms in the
inhomogeneous part of `\operatorname{L} y = f`, and find
particular solution using Abramov's algorithm.
(2) Compute generating set of `\operatorname{L}` and find basis
in it, so that all solutions are linearly independent.
(3) Form final solution with the number of arbitrary
constants equal to dimension of basis of `\operatorname{L}`.
Term `a(n)` is hypergeometric if it is annihilated by first order
linear difference equations with polynomial coefficients or, in
simpler words, if consecutive term ratio is a rational function.
The output of this procedure is a linear combination of fixed
number of hypergeometric terms. However the underlying method
can generate larger class of solutions - D'Alembertian terms.
Note also that this method not only computes the kernel of the
inhomogeneous equation, but also reduces in to a basis so that
solutions generated by this procedure are linearly independent
Examples
========
>>> from sympy.solvers import rsolve_hyper
>>> from sympy.abc import x
>>> rsolve_hyper([-1, -1, 1], 0, x)
C0*(1/2 - sqrt(5)/2)**x + C1*(1/2 + sqrt(5)/2)**x
>>> rsolve_hyper([-1, 1], 1 + x, x)
C0 + x*(x + 1)/2
References
==========
.. [1] M. Petkovsek, Hypergeometric solutions of linear recurrences
with polynomial coefficients, J. Symbolic Computation,
14 (1992), 243-264.
.. [2] M. Petkovsek, H. S. Wilf, D. Zeilberger, A = B, 1996.
"""
coeffs = list(map(sympify, coeffs))
f = sympify(f)
r, kernel, symbols = len(coeffs) - 1, [], set()
if not f.is_zero:
if f.is_Add:
similar = {}
for g in f.expand().args:
if not g.is_hypergeometric(n):
return None
for h in similar.keys():
if hypersimilar(g, h, n):
similar[h] += g
break
else:
similar[g] = S.Zero
inhomogeneous = []
for g, h in similar.items():
inhomogeneous.append(g + h)
elif f.is_hypergeometric(n):
inhomogeneous = [f]
else:
return None
for i, g in enumerate(inhomogeneous):
coeff, polys = S.One, coeffs[:]
denoms = [S.One]*(r + 1)
s = hypersimp(g, n)
for j in range(1, r + 1):
coeff *= s.subs(n, n + j - 1)
p, q = coeff.as_numer_denom()
polys[j] *= p
denoms[j] = q
for j in range(r + 1):
polys[j] *= Mul(*(denoms[:j] + denoms[j + 1:]))
R = rsolve_poly(polys, Mul(*denoms), n)
if not (R is None or R is S.Zero):
inhomogeneous[i] *= R
else:
return None
result = Add(*inhomogeneous)
else:
result = S.Zero
Z = Dummy('Z')
p, q = coeffs[0], coeffs[r].subs(n, n - r + 1)
p_factors = [z for z in roots(p, n).keys()]
q_factors = [z for z in roots(q, n).keys()]
factors = [(S.One, S.One)]
for p in p_factors:
for q in q_factors:
if p.is_integer and q.is_integer and p <= q:
continue
else:
factors += [(n - p, n - q)]
p = [(n - p, S.One) for p in p_factors]
q = [(S.One, n - q) for q in q_factors]
factors = p + factors + q
for A, B in factors:
polys, degrees = [], []
D = A*B.subs(n, n + r - 1)
for i in range(r + 1):
a = Mul(*[A.subs(n, n + j) for j in range(i)])
b = Mul(*[B.subs(n, n + j) for j in range(i, r)])
poly = quo(coeffs[i]*a*b, D, n)
polys.append(poly.as_poly(n))
if not poly.is_zero:
degrees.append(polys[i].degree())
if degrees:
d, poly = max(degrees), S.Zero
else:
return None
for i in range(r + 1):
coeff = polys[i].nth(d)
if coeff is not S.Zero:
poly += coeff * Z**i
for z in roots(poly, Z).keys():
if z.is_zero:
continue
(C, s) = rsolve_poly([polys[i]*z**i for i in range(r + 1)], 0, n, symbols=True)
if C is not None and C is not S.Zero:
symbols |= set(s)
ratio = z * A * C.subs(n, n + 1) / B / C
ratio = simplify(ratio)
# If there is a nonnegative root in the denominator of the ratio,
# this indicates that the term y(n_root) is zero, and one should
# start the product with the term y(n_root + 1).
n0 = 0
for n_root in roots(ratio.as_numer_denom()[1], n).keys():
if n_root.has(I):
return None
elif (n0 < (n_root + 1)) == True:
n0 = n_root + 1
K = product(ratio, (n, n0, n - 1))
if K.has(factorial, FallingFactorial, RisingFactorial):
K = simplify(K)
if casoratian(kernel + [K], n, zero=False) != 0:
kernel.append(K)
kernel.sort(key=default_sort_key)
sk = list(zip(numbered_symbols('C'), kernel))
if sk:
for C, ker in sk:
result += C * ker
else:
return None
if hints.get('symbols', False):
symbols |= {s for s, k in sk}
return (result, list(symbols))
else:
return result
[docs]def rsolve(f, y, init=None):
r"""
Solve univariate recurrence with rational coefficients.
Given `k`-th order linear recurrence `\operatorname{L} y = f`,
or equivalently:
.. math:: a_{k}(n) y(n+k) + a_{k-1}(n) y(n+k-1) +
\cdots + a_{0}(n) y(n) = f(n)
where `a_{i}(n)`, for `i=0, \ldots, k`, are polynomials or rational
functions in `n`, and `f` is a hypergeometric function or a sum
of a fixed number of pairwise dissimilar hypergeometric terms in
`n`, finds all solutions or returns ``None``, if none were found.
Initial conditions can be given as a dictionary in two forms:
(1) ``{ n_0 : v_0, n_1 : v_1, ..., n_m : v_m}``
(2) ``{y(n_0) : v_0, y(n_1) : v_1, ..., y(n_m) : v_m}``
or as a list ``L`` of values:
``L = [v_0, v_1, ..., v_m]``
where ``L[i] = v_i``, for `i=0, \ldots, m`, maps to `y(n_i)`.
Examples
========
Lets consider the following recurrence:
.. math:: (n - 1) y(n + 2) - (n^2 + 3 n - 2) y(n + 1) +
2 n (n + 1) y(n) = 0
>>> from sympy import Function, rsolve
>>> from sympy.abc import n
>>> y = Function('y')
>>> f = (n - 1)*y(n + 2) - (n**2 + 3*n - 2)*y(n + 1) + 2*n*(n + 1)*y(n)
>>> rsolve(f, y(n))
2**n*C0 + C1*factorial(n)
>>> rsolve(f, y(n), {y(0):0, y(1):3})
3*2**n - 3*factorial(n)
See Also
========
rsolve_poly, rsolve_ratio, rsolve_hyper
"""
if isinstance(f, Equality):
f = f.lhs - f.rhs
n = y.args[0]
k = Wild('k', exclude=(n,))
# Preprocess user input to allow things like
# y(n) + a*(y(n + 1) + y(n - 1))/2
f = f.expand().collect(y.func(Wild('m', integer=True)))
h_part = defaultdict(lambda: S.Zero)
i_part = S.Zero
for g in Add.make_args(f):
coeff = S.One
kspec = None
for h in Mul.make_args(g):
if h.is_Function:
if h.func == y.func:
result = h.args[0].match(n + k)
if result is not None:
kspec = int(result[k])
else:
raise ValueError(
"'%s(%s + k)' expected, got '%s'" % (y.func, n, h))
else:
raise ValueError(
"'%s' expected, got '%s'" % (y.func, h.func))
else:
coeff *= h
if kspec is not None:
h_part[kspec] += coeff
else:
i_part += coeff
for k, coeff in h_part.items():
h_part[k] = simplify(coeff)
common = S.One
for coeff in h_part.values():
if coeff.is_rational_function(n):
if not coeff.is_polynomial(n):
common = lcm(common, coeff.as_numer_denom()[1], n)
else:
raise ValueError(
"Polynomial or rational function expected, got '%s'" % coeff)
i_numer, i_denom = i_part.as_numer_denom()
if i_denom.is_polynomial(n):
common = lcm(common, i_denom, n)
if common is not S.One:
for k, coeff in h_part.items():
numer, denom = coeff.as_numer_denom()
h_part[k] = numer*quo(common, denom, n)
i_part = i_numer*quo(common, i_denom, n)
K_min = min(h_part.keys())
if K_min < 0:
K = abs(K_min)
H_part = defaultdict(lambda: S.Zero)
i_part = i_part.subs(n, n + K).expand()
common = common.subs(n, n + K).expand()
for k, coeff in h_part.items():
H_part[k + K] = coeff.subs(n, n + K).expand()
else:
H_part = h_part
K_max = max(H_part.keys())
coeffs = [H_part[i] for i in range(K_max + 1)]
result = rsolve_hyper(coeffs, -i_part, n, symbols=True)
if result is None:
return None
solution, symbols = result
if init == {} or init == []:
init = None
if symbols and init is not None:
if isinstance(init, list):
init = {i: init[i] for i in range(len(init))}
equations = []
for k, v in init.items():
try:
i = int(k)
except TypeError:
if k.is_Function and k.func == y.func:
i = int(k.args[0])
else:
raise ValueError("Integer or term expected, got '%s'" % k)
try:
eq = solution.limit(n, i) - v
except NotImplementedError:
eq = solution.subs(n, i) - v
equations.append(eq)
result = solve(equations, *symbols)
if not result:
return None
else:
solution = solution.subs(result)
return solution
