# Source code for sympy.solvers.recurr

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.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(0, 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(0, 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(0, N + 1): C.append(Symbol('C' + str(i))) y += C[i] * n**i for i in range(0, 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(0, A + 1): for k in range(0, 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(0, A): v[j] += B * V[i - k, j] denom = alpha[-A].subs(x, i) for j in range(0, 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(0, A): v[j] += B * V[i - k, j] g += B * G[i - k] denom = alpha[-A].subs(x, i) for j in range(0, 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(0, 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(0, 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(0, i + 1) ]) denoms = [ C.subs(n, n + i) for i in range(0, r + 1) ] for i in range(0, 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(0, 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*(-sqrt(5)/2 + 1/2)**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(0, 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(0, r + 1): a = Mul(*[ A.subs(n, n + j) for j in range(0, 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(0, 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 type(init) is 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