def cancel_primitive(b, c, n, DE):
"""
Poly Risch Differential Equation - Cancellation: Primitive case.
Given a derivation D on k[t], n either an integer or +oo, b in k, and
c in k[t] with Dt in k and b != 0, either raise
NonElementaryIntegralException, in which case the equation Dq + b*q == c
has no solution of degree at most n in k[t], or a solution q in k[t] of
this equation with deg(q) <= n.
"""
from sympy.integrals.prde import is_log_deriv_k_t_radical_in_field
with DecrementLevel(DE):
ba, bd = frac_in(b, DE.t)
A = is_log_deriv_k_t_radical_in_field(ba, bd, DE)
if A is not None:
n, z = A
if n == 1: # b == Dz/z
raise NotImplementedError("is_deriv_in_field() is required to "
" solve this problem.")
# if z*c == Dp for p in k[t] and deg(p) <= n:
# return p/z
# else:
# raise NonElementaryIntegralException
if c.is_zero:
return c # return 0
if n < c.degree(DE.t):
raise NonElementaryIntegralException
q = Poly(0, DE.t)
while not c.is_zero:
m = c.degree(DE.t)
if n < m:
raise NonElementaryIntegralException
with DecrementLevel(DE):
a2a, a2d = frac_in(c.LC(), DE.t)
sa, sd = rischDE(ba, bd, a2a, a2d, DE)
stm = Poly(sa.as_expr() / sd.as_expr() * DE.t**m, DE.t, expand=False)
q += stm
n = m - 1
c -= b * stm + derivation(stm, DE)
return q
def no_cancel_b_small(b, c, n, DE):
"""
Poly Risch Differential Equation - No cancellation: deg(b) small enough.
Given a derivation D on k[t], n either an integer or +oo, and b, c
in k[t] with deg(b) < deg(D) - 1 and either D == d/dt or
deg(D) >= 2, either raise NonElementaryIntegralException, in which case the
equation Dq + b*q == c has no solution of degree at most n in k[t],
or a solution q in k[t] of this equation with deg(q) <= n, or the
tuple (h, b0, c0) such that h in k[t], b0, c0, in k, and for any
solution q in k[t] of degree at most n of Dq + bq == c, y == q - h
is a solution in k of Dy + b0*y == c0.
"""
q = Poly(0, DE.t)
while not c.is_zero:
if n == 0:
m = 0
else:
m = c.degree(DE.t) - DE.d.degree(DE.t) + 1
if not 0 <= m <= n: # n < 0 or m < 0 or m > n
raise NonElementaryIntegralException
if m > 0:
p = Poly(
c.as_poly(DE.t).LC() / (m * DE.d.as_poly(DE.t).LC()) * DE.t**m,
DE.t,
expand=False,
)
else:
if b.degree(DE.t) != c.degree(DE.t):
raise NonElementaryIntegralException
if b.degree(DE.t) == 0:
return (q, b.as_poly(DE.T[DE.level - 1]),
c.as_poly(DE.T[DE.level - 1]))
p = Poly(c.as_poly(DE.t).LC() / b.as_poly(DE.t).LC(),
DE.t,
expand=False)
q = q + p
n = m - 1
c = c - derivation(p, DE) - b * p
return q
def no_cancel_equal(b, c, n, DE):
"""
Poly Risch Differential Equation - No cancellation: deg(b) == deg(D) - 1
Explanation
===========
Given a derivation D on k[t] with deg(D) >= 2, n either an integer
or +oo, and b, c in k[t] with deg(b) == deg(D) - 1, either raise
NonElementaryIntegralException, in which case the equation Dq + b*q == c has
no solution of degree at most n in k[t], or a solution q in k[t] of
this equation with deg(q) <= n, or the tuple (h, m, C) such that h
in k[t], m in ZZ, and C in k[t], and for any solution q in k[t] of
degree at most n of Dq + b*q == c, y == q - h is a solution in k[t]
of degree at most m of Dy + b*y == C.
"""
q = Poly(0, DE.t)
lc = cancel(-b.as_poly(DE.t).LC() / DE.d.as_poly(DE.t).LC())
if lc.is_Integer and lc.is_positive:
M = lc
else:
M = -1
while not c.is_zero:
m = max(M, c.degree(DE.t) - DE.d.degree(DE.t) + 1)
if not 0 <= m <= n: # n < 0 or m < 0 or m > n
raise NonElementaryIntegralException
u = cancel(m * DE.d.as_poly(DE.t).LC() + b.as_poly(DE.t).LC())
if u.is_zero:
return (q, m, c)
if m > 0:
p = Poly(c.as_poly(DE.t).LC() / u * DE.t**m, DE.t, expand=False)
else:
if c.degree(DE.t) != DE.d.degree(DE.t) - 1:
raise NonElementaryIntegralException
else:
p = c.as_poly(DE.t).LC() / b.as_poly(DE.t).LC()
q = q + p
n = m - 1
c = c - derivation(p, DE) - b * p
return q
def test_param_poly_rischDE():
DE = DifferentialExtension(extension={"D": [Poly(1, x)]})
a = Poly(x ** 2 - x, x, field=True)
b = Poly(1, x, field=True)
q = [Poly(x, x, field=True), Poly(x ** 2, x, field=True)]
h, A = param_poly_rischDE(a, b, q, 3, DE)
assert A.nullspace() == [Matrix([0, 1, 1, 1])] # c1, c2, d1, d2
# Solution of a*Dp + b*p = c1*q1 + c2*q2 = q2 = x**2
# is d1*h1 + d2*h2 = h1 + h2 = x.
assert h[0] + h[1] == Poly(x, x)
# a*Dp + b*p = q1 = x has no solution.
a = Poly(x ** 2 - x, x, field=True)
b = Poly(x ** 2 - 5 * x + 3, x, field=True)
q = [Poly(1, x, field=True), Poly(x, x, field=True), Poly(x ** 2, x, field=True)]
h, A = param_poly_rischDE(a, b, q, 3, DE)
assert A.nullspace() == [Matrix([3, -5, 1, -5, 1, 1])]
p = -5 * h[0] + h[1] + h[2] # Poly(1, x)
assert a * derivation(p, DE) + b * p == Poly(x ** 2 - 5 * x + 3, x)
etaa, etad = frac_in(dcoeff, DE.t)
if recognize_log_derivative(2 * betaa, betad, DE):
A = parametric_log_deriv(
alphaa * sqrt(-1) * betad + alphad * betaa,
alphad * betad, etaa, etad, DE)
if A is not None:
a, m, z = A
if a == 1:
n = min(n, m)
N = max(0, -nb, n - nc)
pN = p**N
pn = p**-n
A = a * pN
B = ba * pN.quo(bd) + Poly(n, DE.t) * a * derivation(p, DE).quo(p) * pN
C = (ca * pN * pn).quo(cd)
h = pn
# (a*p**N, (b + n*a*Dp/p)*p**N, c*p**(N - n), p**-n)
return (A, B, C, h)
def bound_degree(a, b, cQ, DE, case='auto', parametric=False):
"""
Bound on polynomial solutions.
Given a derivation D on k[t] and a, b, c in k[t] with a != 0, return
n in ZZ such that deg(q) <= n for any solution q in k[t] of
a*Dq + b*q == c, when parametric=False, or deg(q) <= n for any solution
c1, ..., cm in Const(k) and q in k[t] of a*Dq + b*q == Sum(ci*gi, (i, 1, m))
def is_log_deriv_k_t_radical(fa, fd, DE, Df=True):
"""
Checks if Df is the logarithmic derivative of a k(t)-radical.
b in k(t) can be written as the logarithmic derivative of a k(t) radical if
there exist n in ZZ and u in k(t) with n, u != 0 such that n*b == Du/u.
Either returns (ans, u, n, const) or None, which means that Df cannot be
written as the logarithmic derivative of a k(t)-radical. ans is a list of
tuples such that Mul(*[i**j for i, j in ans]) == u. This is useful for
seeing exactly what elements of k(t) produce u.
This function uses the structure theorem approach, which says that for any
f in K, Df is the logarithmic derivative of a K-radical if and only if there
are ri in QQ such that::
--- --- Dt
\ r * Dt + \ r * i
/ i i / i --- = Df.
--- --- t
i in L i in E i
K/C(x) K/C(x)
Where C = Const(K), L_K/C(x) = { i in {1, ..., n} such that t_i is
transcendental over C(x)(t_1, ..., t_i-1) and Dt_i = Da_i/a_i, for some a_i
in C(x)(t_1, ..., t_i-1)* } (i.e., the set of all indices of logarithmic
monomials of K over C(x)), and E_K/C(x) = { i in {1, ..., n} such that t_i
is transcendental over C(x)(t_1, ..., t_i-1) and Dt_i/t_i = Da_i, for some
a_i in C(x)(t_1, ..., t_i-1) } (i.e., the set of all indices of
hyperexponential monomials of K over C(x)). If K is an elementary extension
over C(x), then the cardinality of L_K/C(x) U E_K/C(x) is exactly the
transcendence degree of K over C(x). Furthermore, because Const_D(K) ==
Const_D(C(x)) == C, deg(Dt_i) == 1 when t_i is in E_K/C(x) and
deg(Dt_i) == 0 when t_i is in L_K/C(x), implying in particular that E_K/C(x)
and L_K/C(x) are disjoint.
The sets L_K/C(x) and E_K/C(x) must, by their nature, be computed
recursively using this same function. Therefore, it is required to pass
them as indices to D (or T). L_args are the arguments of the logarithms
indexed by L_K (i.e., if i is in L_K, then T[i] == log(L_args[i])). This is
needed to compute the final answer u such that n*f == Du/u.
exp(f) will be the same as u up to a multiplicative constant. This is
because they will both behave the same as monomials. For example, both
exp(x) and exp(x + 1) == E*exp(x) satisfy Dt == t. Therefore, the term const
is returned. const is such that exp(const)*f == u. This is calculated by
subtracting the arguments of one exponential from the other. Therefore, it
is necessary to pass the arguments of the exponential terms in E_args.
To handle the case where we are given Df, not f, use
is_log_deriv_k_t_radical_in_field().
"""
H = []
if Df:
dfa, dfd = (fd * derivation(fa, DE) - fa * derivation(fd, DE)).cancel(
fd**2, include=True)
else:
dfa, dfd = fa, fd
# Our assumption here is that each monomial is recursively transcendental
if len(DE.L_K) + len(DE.E_K) != len(DE.D) - 1:
if [i for i in DE.cases if i == 'tan'] or \
set([i for i in DE.cases if i == 'primitive']) - set(DE.L_K):
raise NotImplementedError(
"Real version of the structure "
"theorems with hypertangent support is not yet implemented.")
# TODO: What should really be done in this case?
raise NotImplementedError("Nonelementary extensions not supported "
"in the structure theorems.")
E_part = [DE.D[i].quo(Poly(DE.T[i], DE.T[i])).as_expr() for i in DE.E_K]
L_part = [DE.D[i].as_expr() for i in DE.L_K]
lhs = Matrix([E_part + L_part])
rhs = Matrix([dfa.as_expr() / dfd.as_expr()])
A, u = constant_system(lhs, rhs, DE)
if not all(derivation(i, DE, basic=True).is_zero for i in u) or not A:
# If the elements of u are not all constant
# Note: See comment in constant_system
# Also note: derivation(basic=True) calls cancel()
return None
else:
if not all(i.is_Rational for i in u):
# TODO: But maybe we can tell if they're not rational, like
# log(2)/log(3). Also, there should be an option to continue
# anyway, even if the result might potentially be wrong.
raise NotImplementedError("Cannot work with non-rational "
"coefficients in this case.")
else:
n = reduce(ilcm, [i.as_numer_denom()[1] for i in u])
u *= n
terms = [DE.T[i] for i in DE.E_K] + DE.L_args
ans = list(zip(terms, u))
result = Mul(*[Pow(i, j) for i, j in ans])
# exp(f) will be the same as result up to a multiplicative
# constant. We now find the log of that constant.
argterms = DE.E_args + [DE.T[i] for i in DE.L_K]
const = cancel(fa.as_expr() / fd.as_expr() -
Add(*[Mul(i, j / n) for i, j in zip(argterms, u)]))
return (ans, result, n, const)
def is_deriv_k(fa, fd, DE):
"""
Checks if Df/f is the derivative of an element of k(t).
a in k(t) is the derivative of an element of k(t) if there exists b in k(t)
such that a = Db. Either returns (ans, u), such that Df/f == Du, or None,
which means that Df/f is not the derivative of an element of k(t). ans is
a list of tuples such that Add(*[i*j for i, j in ans]) == u. This is useful
for seeing exactly which elements of k(t) produce u.
This function uses the structure theorem approach, which says that for any
f in K, Df/f is the derivative of a element of K if and only if there are ri
in QQ such that::
--- --- Dt
\ r * Dt + \ r * i Df
/ i i / i --- = --.
--- --- t f
i in L i in E i
K/C(x) K/C(x)
Where C = Const(K), L_K/C(x) = { i in {1, ..., n} such that t_i is
transcendental over C(x)(t_1, ..., t_i-1) and Dt_i = Da_i/a_i, for some a_i
in C(x)(t_1, ..., t_i-1)* } (i.e., the set of all indices of logarithmic
monomials of K over C(x)), and E_K/C(x) = { i in {1, ..., n} such that t_i
is transcendental over C(x)(t_1, ..., t_i-1) and Dt_i/t_i = Da_i, for some
a_i in C(x)(t_1, ..., t_i-1) } (i.e., the set of all indices of
hyperexponential monomials of K over C(x)). If K is an elementary extension
over C(x), then the cardinality of L_K/C(x) U E_K/C(x) is exactly the
transcendence degree of K over C(x). Furthermore, because Const_D(K) ==
Const_D(C(x)) == C, deg(Dt_i) == 1 when t_i is in E_K/C(x) and
deg(Dt_i) == 0 when t_i is in L_K/C(x), implying in particular that E_K/C(x)
and L_K/C(x) are disjoint.
The sets L_K/C(x) and E_K/C(x) must, by their nature, be computed
recursively using this same function. Therefore, it is required to pass
them as indices to D (or T). E_args are the arguments of the
hyperexponentials indexed by E_K (i.e., if i is in E_K, then T[i] ==
exp(E_args[i])). This is needed to compute the final answer u such that
Df/f == Du.
log(f) will be the same as u up to a additive constant. This is because
they will both behave the same as monomials. For example, both log(x) and
log(2*x) == log(x) + log(2) satisfy Dt == 1/x, because log(2) is constant.
Therefore, the term const is returned. const is such that
log(const) + f == u. This is calculated by dividing the arguments of one
logarithm from the other. Therefore, it is necessary to pass the arguments
of the logarithmic terms in L_args.
To handle the case where we are given Df/f, not f, use is_deriv_k_in_field().
"""
# Compute Df/f
dfa, dfd = fd * (fd * derivation(fa, DE) -
fa * derivation(fd, DE)), fd**2 * fa
dfa, dfd = dfa.cancel(dfd, include=True)
# Our assumption here is that each monomial is recursively transcendental
if len(DE.L_K) + len(DE.E_K) != len(DE.D) - 1:
if [i for i in DE.cases if i == 'tan'] or \
set([i for i in DE.cases if i == 'primitive']) - set(DE.L_K):
raise NotImplementedError(
"Real version of the structure "
"theorems with hypertangent support is not yet implemented.")
# TODO: What should really be done in this case?
raise NotImplementedError("Nonelementary extensions not supported "
"in the structure theorems.")
E_part = [DE.D[i].quo(Poly(DE.T[i], DE.T[i])).as_expr() for i in DE.E_K]
L_part = [DE.D[i].as_expr() for i in DE.L_K]
lhs = Matrix([E_part + L_part])
rhs = Matrix([dfa.as_expr() / dfd.as_expr()])
A, u = constant_system(lhs, rhs, DE)
if not all(derivation(i, DE, basic=True).is_zero for i in u) or not A:
# If the elements of u are not all constant
# Note: See comment in constant_system
# Also note: derivation(basic=True) calls cancel()
return None
else:
if not all(i.is_Rational for i in u):
raise NotImplementedError("Cannot work with non-rational "
"coefficients in this case.")
else:
terms = DE.E_args + [DE.T[i] for i in DE.L_K]
ans = list(zip(terms, u))
result = Add(*[Mul(i, j) for i, j in ans])
argterms = [DE.T[i] for i in DE.E_K] + DE.L_args
l = []
for i, j in zip(argterms, u):
# We need to get around things like sqrt(x**2) != x
# and also sqrt(x**2 + 2*x + 1) != x + 1
icoeff, iterms = sqf_list(i)
l.append(
Mul(*([Pow(icoeff, j)] +
[Pow(b, e * j) for b, e in iterms])))
const = cancel(fa.as_expr() / fd.as_expr() / Mul(*l))
return (ans, result, const)
def parametric_log_deriv_heu(fa, fd, wa, wd, DE, c1=None):
"""
Parametric logarithmic derivative heuristic.
Given a derivation D on k[t], f in k(t), and a hyperexponential monomial
theta over k(t), raises either NotImplementedError, in which case the
heuristic failed, or returns None, in which case it has proven that no
solution exists, or returns a solution (n, m, v) of the equation
n*f == Dv/v + m*Dtheta/theta, with v in k(t)* and n, m in ZZ with n != 0.
If this heuristic fails, the structure theorem approach will need to be
used.
The argument w == Dtheta/theta
"""
# TODO: finish writing this and write tests
c1 = c1 or Dummy('c1')
p, a = fa.div(fd)
q, b = wa.div(wd)
B = max(0, derivation(DE.t, DE).degree(DE.t) - 1)
C = max(p.degree(DE.t), q.degree(DE.t))
if q.degree(DE.t) > B:
eqs = [p.nth(i) - c1 * q.nth(i) for i in range(B + 1, C + 1)]
s = solve(eqs, c1)
if not s or not s[c1].is_Rational:
# deg(q) > B, no solution for c.
return None
N, M = s[c1].as_numer_denom() # N and M are integers
N, M = Poly(N, DE.t), Poly(M, DE.t)
nfmwa = N * fa * wd - M * wa * fd
nfmwd = fd * wd
Qv = is_log_deriv_k_t_radical_in_field(N * fa * wd - M * wa * fd,
fd * wd, DE, 'auto')
if Qv is None:
# (N*f - M*w) is not the logarithmic derivative of a k(t)-radical.
return None
Q, e, v = Qv
if e != 1:
return None
if Q.is_zero or v.is_zero:
# Q == 0 or v == 0.
return None
return (Q * N, Q * M, v)
if p.degree(DE.t) > B:
return None
c = lcm(fd.as_poly(DE.t).LC(), wd.as_poly(DE.t).LC())
l = fd.monic().lcm(wd.monic()) * Poly(c, DE.t)
ln, ls = splitfactor(l, DE)
z = ls * ln.gcd(ln.diff(DE.t))
if not z.has(DE.t):
raise NotImplementedError("parametric_log_deriv_heu() "
"heuristic failed: z in k.")
u1, r1 = (fa * l.quo(fd)).div(z) # (l*f).div(z)
u2, r2 = (wa * l.quo(wd)).div(z) # (l*w).div(z)
eqs = [r1.nth(i) - c1 * r2.nth(i) for i in range(z.degree(DE.t))]
s = solve(eqs, c1)
if not s or not s[c1].is_Rational:
# deg(q) <= B, no solution for c.
return None
M, N = s[c1].as_numer_denom()
nfmwa = N.as_poly(DE.t) * fa * wd - M.as_poly(DE.t) * wa * fd
nfmwd = fd * wd
Qv = is_log_deriv_k_t_radical_in_field(nfmwa, nfmwd, DE)
if Qv is None:
# (N*f - M*w) is not the logarithmic derivative of a k(t)-radical.
return None
Q, v = Qv
if Q.is_zero or v.is_zero:
# Q == 0 or v == 0.
return None
return (Q * N, Q * M, v)
def constant_system(A, u, DE):
"""
Generate a system for the constant solutions.
Given a differential field (K, D) with constant field C = Const(K), a Matrix
A, and a vector (Matrix) u with coefficients in K, returns the tuple
(B, v, s), where B is a Matrix with coefficients in C and v is a vector
(Matrix) such that either v has coefficients in C, in which case s is True
and the solutions in C of Ax == u are exactly all the solutions of Bx == v,
or v has a non-constant coefficient, in which case s is False Ax == u has no
constant solution.
This algorithm is used both in solving parametric problems and in
determining if an element a of K is a derivative of an element of K or the
logarithmic derivative of a K-radical using the structure theorem approach.
Because Poly does not play well with Matrix yet, this algorithm assumes that
all matrix entries are Basic expressions.
"""
if not A:
return A, u
Au = A.row_join(u)
Au = Au.rref(simplify=cancel)[0]
# Warning: This will NOT return correct results if cancel() cannot reduce
# an identically zero expression to 0. The danger is that we might
# incorrectly prove that an integral is nonelementary (such as
# risch_integrate(exp((sin(x)**2 + cos(x)**2 - 1)*x**2), x).
# But this is a limitation in computer algebra in general, and implicit
# in the correctness of the Risch Algorithm is the computability of the
# constant field (actually, this same correctness problem exists in any
# algorithm that uses rref()).
#
# We therefore limit ourselves to constant fields that are computable
# via the cancel() function, in order to prevent a speed bottleneck from
# calling some more complex simplification function (rational function
# coefficients will fall into this class). Furthermore, (I believe) this
# problem will only crop up if the integral explicitly contains an
# expression in the constant field that is identically zero, but cannot
# be reduced to such by cancel(). Therefore, a careful user can avoid this
# problem entirely by being careful with the sorts of expressions that
# appear in his integrand in the variables other than the integration
# variable (the structure theorems should be able to completely decide these
# problems in the integration variable).
Au = Au.applyfunc(cancel)
A, u = Au[:, :-1], Au[:, -1]
for j in range(A.cols):
for i in range(A.rows):
if A[i, j].has(*DE.T):
# This assumes that const(F(t0, ..., tn) == const(K) == F
Ri = A[i, :]
# Rm+1; m = A.rows
Rm1 = Ri.applyfunc(lambda x: derivation(x, DE, basic=True) /
derivation(A[i, j], DE, basic=True))
Rm1 = Rm1.applyfunc(cancel)
um1 = cancel(
derivation(u[i], DE, basic=True) /
derivation(A[i, j], DE, basic=True))
for s in range(A.rows):
# A[s, :] = A[s, :] - A[s, i]*A[:, m+1]
Asj = A[s, j]
A.row_op(s, lambda r, jj: cancel(r - Asj * Rm1[jj]))
# u[s] = u[s] - A[s, j]*u[m+1
u.row_op(s, lambda r, jj: cancel(r - Asj * um1))
A = A.col_join(Rm1)
u = u.col_join(Matrix([um1]))
return (A, u)
betad = alphad
etaa, etad = frac_in(dcoeff, DE.t)
if recognize_log_derivative(2 * betaa, betad, DE):
A = parametric_log_deriv(alphaa, alphad, etaa, etad, DE)
B = parametric_log_deriv(betaa, betad, etaa, etad, DE)
if A is not None and B is not None:
a, s, z = A
if a == 1:
n = min(n, s / 2)
N = max(0, -nb)
pN = p**N
pn = p**-n # This is 1/h
A = a * pN
B = ba * pN.quo(bd) + Poly(n, DE.t) * a * derivation(p, DE).quo(p) * pN
G = [(Ga * pN * pn).cancel(Gd, include=True) for Ga, Gd in G]
h = pn
# (a*p**N, (b + n*a*Dp/p)*p**N, g1*p**(N - n), ..., gm*p**(N - n), p**-n)
return (A, B, G, h)
def prde_linear_constraints(a, b, G, DE):
"""
Parametric Risch Differential Equation - Generate linear constraints on the constants.
Given a derivation D on k[t], a, b, in k[t] with gcd(a, b) == 1, and
G = [g1, ..., gm] in k(t)^m, return Q = [q1, ..., qm] in k[t]^m and a
matrix M with entries in k(t) such that for any solution c1, ..., cm in
Const(k) and p in k[t] of a*Dp + b*p == Sum(ci*gi, (i, 1, m)),
def prde_no_cancel_b_small(b, Q, n, DE):
"""
Parametric Poly Risch Differential Equation - No cancellation: deg(b) small enough.
Given a derivation D on k[t], n in ZZ, and b, q1, ..., qm in k[t] with
deg(b) < deg(D) - 1 and either D == d/dt or deg(D) >= 2, returns
h1, ..., hr in k[t] and a matrix A with coefficients in Const(k) such that
if c1, ..., cm in Const(k) and q in k[t] satisfy deg(q) <= n and
Dq + b*q == Sum(ci*qi, (i, 1, m)) then q = Sum(dj*hj, (j, 1, r)) where
d1, ..., dr in Const(k) and A*Matrix([[c1, ..., cm, d1, ..., dr]]).T == 0.
"""
m = len(Q)
H = [Poly(0, DE.t)]*m
for N in range(n, 0, -1): # [n, ..., 1]
for i in range(m):
si = Q[i].nth(N + DE.d.degree(DE.t) - 1)/(N*DE.d.LC())
sitn = Poly(si*DE.t**N, DE.t)
H[i] = H[i] + sitn
Q[i] = Q[i] - derivation(sitn, DE) - b*sitn
if b.degree(DE.t) > 0:
for i in range(m):
si = Poly(Q[i].nth(b.degree(DE.t))/b.LC(), DE.t)
H[i] = H[i] + si
Q[i] = Q[i] - derivation(si, DE) - b*si
if all(qi.is_zero for qi in Q):
dc = -1
M = Matrix()
else:
dc = max([qi.degree(DE.t) for qi in Q])
M = Matrix(dc + 1, m, lambda i, j: Q[j].nth(i))
A, u = constant_system(M, zeros(dc + 1, 1), DE)
c = eye(m)
A = A.row_join(zeros(A.rows, m)).col_join(c.row_join(-c))
return (H, A)
# else: b is in k, deg(qi) < deg(Dt)
t = DE.t
if DE.case != 'base':
with DecrementLevel(DE):
t0 = DE.t # k = k0(t0)
ba, bd = frac_in(b, t0, field=True)
Q0 = [frac_in(qi.TC(), t0, field=True) for qi in Q]
f, B = param_rischDE(ba, bd, Q0, DE)
# f = [f1, ..., fr] in k^r and B is a matrix with
# m + r columns and entries in Const(k) = Const(k0)
# such that Dy0 + b*y0 = Sum(ci*qi, (i, 1, m)) has
# a solution y0 in k with c1, ..., cm in Const(k)
# if and only y0 = Sum(dj*fj, (j, 1, r)) where
# d1, ..., dr ar in Const(k) and
# B*Matrix([c1, ..., cm, d1, ..., dr]) == 0.
# Transform fractions (fa, fd) in f into constant
# polynomials fa/fd in k[t].
# (Is there a better way?)
f = [Poly(fa.as_expr()/fd.as_expr(), t, field=True)
for fa, fd in f]
else:
# Base case. Dy == 0 for all y in k and b == 0.
# Dy + b*y = Sum(ci*qi) is solvable if and only if
# Sum(ci*qi) == 0 in which case the solutions are
# y = d1*f1 for f1 = 1 and any d1 in Const(k) = k.
f = [Poly(1, t, field=True)] # r = 1
B = Matrix([[qi.TC() for qi in Q] + [S(0)]])
# The condition for solvability is
# B*Matrix([c1, ..., cm, d1]) == 0
# There are no constraints on d1.
# Coefficients of t^j (j > 0) in Sum(ci*qi) must be zero.
d = max([qi.degree(DE.t) for qi in Q])
if d > 0:
M = Matrix(d, m, lambda i, j: Q[j].nth(i + 1))
A, _ = constant_system(M, zeros(d, 1), DE)
else:
# No constraints on the hj.
A = Matrix(0, m, [])
# Solutions of the original equation are
# y = Sum(dj*fj, (j, 1, r) + Sum(ei*hi, (i, 1, m)),
# where ei == ci (i = 1, ..., m), when
# A*Matrix([c1, ..., cm]) == 0 and
# B*Matrix([c1, ..., cm, d1, ..., dr]) == 0
# Build combined constraint matrix with m + r + m columns.
r = len(f)
I = eye(m)
A = A.row_join(zeros(A.rows, r + m))
B = B.row_join(zeros(B.rows, m))
C = I.row_join(zeros(m, r)).row_join(-I)
return f + H, A.col_join(B).col_join(C)
