def prde_no_cancel_b_large(b, Q, n, DE):
"""
Parametric Poly Risch Differential Equation - No cancellation: deg(b) large enough.
Given a derivation D on k[t], n in ZZ, and b, q1, ..., qm in k[t] with
b != 0 and either D == d/dt or deg(b) > max(0, deg(D) - 1), returns
h1, ..., hr in k[r] 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.
"""
db = b.degree(DE.t)
m = len(Q)
H = [Poly(0, DE.t)]*m
for N in range(n, -1, -1): # [n, ..., 0]
for i in range(m):
si = Q[i].nth(N + db)/b.LC()
sitn = Poly(si*DE.t**N, DE.t)
H[i] = H[i] + sitn
Q[i] = Q[i] - derivation(sitn, DE) - b*sitn
if all(qi.is_zero for qi in Q):
dc = -1
M = zeros(0, 2)
else:
dc = max([qi.degree(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
def limited_integrate(fa, fd, G, DE):
"""
Solves the limited integration problem: f = Dv + Sum(ci*wi, (i, 1, n))
"""
fa, fd = fa*Poly(1/fd.LC(), DE.t), fd.monic()
A, B, h, N, g, V = limited_integrate_reduce(fa, fd, G, DE)
V = [g] + V
g = A.gcd(B)
A, B, V = A.quo(g), B.quo(g), [via.cancel(vid*g, include=True) for
via, vid in V]
Q, M = prde_linear_constraints(A, B, V, DE)
M, _ = constant_system(M, zeros(M.rows, 1), DE)
l = M.nullspace()
if M == Matrix() or len(l) > 1:
# Continue with param_rischDE()
raise NotImplementedError("param_rischDE() is required to solve this "
"integral.")
elif len(l) == 0:
raise NonElementaryIntegralException
elif len(l) == 1:
# The c1 == 1. In this case, we can assume a normal Risch DE
if l[0][0].is_zero:
raise NonElementaryIntegralException
else:
l[0] *= 1/l[0][0]
C = sum(Poly(i, DE.t)*q for (i, q) in zip(l[0], Q))
# Custom version of rischDE() that uses the already computed
# denominator and degree bound from above.
B, C, m, alpha, beta = spde(A, B, C, N, DE)
y = solve_poly_rde(B, C, m, DE)
return (alpha*y + beta, h), list(l[0][1:])
else:
raise NotImplementedError
def no_cancel_b_large(b, c, n, DE):
"""
Poly Risch Differential Equation - No cancellation: deg(b) large enough.
Given a derivation D on k[t], n either an integer or +oo, and b, c
in k[t] with b != 0 and either D == d/dt or
deg(b) > max(0, 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.
"""
q = Poly(0, DE.t)
while not c.is_zero:
m = c.degree(DE.t) - b.degree(DE.t)
if not 0 <= m <= n: # n < 0 or m < 0 or m > n
raise NonElementaryIntegralException
p = Poly(c.as_poly(DE.t).LC() / b.as_poly(DE.t).LC() * DE.t**m,
DE.t,
expand=False)
q = q + p
n = m - 1
c = c - derivation(p, DE) - b * p
return q
def _sqrt_symbolic_denest(a, b, r):
"""Given an expression, sqrt(a + b*sqrt(b)), return the denested
expression or None.
Algorithm:
If r = ra + rb*sqrt(rr), try replacing sqrt(rr) in ``a`` with
(y**2 - ra)/rb, and if the result is a quadratic, ca*y**2 + cb*y + cc, and
(cb + b)**2 - 4*ca*cc is 0, then sqrt(a + b*sqrt(r)) can be rewritten as
sqrt(ca*(sqrt(r) + (cb + b)/(2*ca))**2).
Examples
========
>>> from diofant.simplify.sqrtdenest import _sqrt_symbolic_denest, sqrtdenest
>>> from diofant import sqrt, Symbol
>>> from diofant.abc import x
>>> a, b, r = 16 - 2*sqrt(29), 2, -10*sqrt(29) + 55
>>> _sqrt_symbolic_denest(a, b, r)
sqrt(-2*sqrt(29) + 11) + sqrt(5)
If the expression is numeric, it will be simplified:
>>> w = sqrt(sqrt(sqrt(3) + 1) + 1) + 1 + sqrt(2)
>>> sqrtdenest(sqrt((w**2).expand()))
1 + sqrt(2) + sqrt(1 + sqrt(1 + sqrt(3)))
Otherwise, it will only be simplified if assumptions allow:
>>> w = w.subs(sqrt(3), sqrt(x + 3))
>>> sqrtdenest(sqrt((w**2).expand()))
sqrt((sqrt(sqrt(sqrt(x + 3) + 1) + 1) + 1 + sqrt(2))**2)
Notice that the argument of the sqrt is a square. If x is made positive
then the sqrt of the square is resolved:
>>> _.subs(x, Symbol('x', positive=True))
sqrt(sqrt(sqrt(x + 3) + 1) + 1) + 1 + sqrt(2)
"""
a, b, r = map(sympify, (a, b, r))
rval = _sqrt_match(r)
if not rval:
return
ra, rb, rr = rval
if rb:
y = Dummy('y', positive=True)
try:
newa = Poly(a.subs(sqrt(rr), (y**2 - ra) / rb), y)
except PolynomialError:
return
if newa.degree() == 2:
ca, cb, cc = newa.all_coeffs()
cb += b
if _mexpand(cb**2 - 4 * ca * cc).equals(0):
z = sqrt(ca * (sqrt(r) + cb / (2 * ca))**2)
if z.is_number:
z = _mexpand(Mul._from_args(z.as_content_primitive()))
return z
def test_sympyissue_8438():
p = Poly([1, y, -2, -3], x).as_expr()
roots = roots_cubic(Poly(p, x), x)
z = -Rational(3, 2) - 7*I/2 # this will fail in code given in commit msg
post = [r.subs({y: z}) for r in roots]
assert set(post) == set(roots_cubic(Poly(p.subs({y: z}), x)))
# /!\ if p is not made an expression, this is *very* slow
assert all(p.subs({y: z, x: i}).evalf(2, chop=True) == 0 for i in post)
def cancel_exp(b, c, n, DE):
"""
Poly Risch Differential Equation - Cancellation: Hyperexponential case.
Given a derivation D on k[t], n either an integer or +oo, b in k, and
c in k[t] with Dt/t 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 diofant.integrals.prde import parametric_log_deriv
eta = DE.d.quo(Poly(DE.t, DE.t)).as_expr()
with DecrementLevel(DE):
etaa, etad = frac_in(eta, DE.t)
ba, bd = frac_in(b, DE.t)
A = parametric_log_deriv(ba, bd, etaa, etad, DE)
if A is not None:
a, m, z = A
if a == 1:
raise NotImplementedError("is_deriv_in_field() is required to "
"solve this problem.")
# if c*z*t**m == Dp for p in k<t> and q = p/(z*t**m) in k[t] and
# deg(q) <= n:
# return q
# 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
# a1 = b + m*Dt/t
a1 = b.as_expr()
with DecrementLevel(DE):
# TODO: Write a dummy function that does this idiom
a1a, a1d = frac_in(a1, DE.t)
a1a = a1a * etad + etaa * a1d * Poly(m, DE.t)
a1d = a1d * etad
a2a, a2d = frac_in(c.LC(), DE.t)
sa, sd = rischDE(a1a, a1d, 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) # deg(c) becomes smaller
return q
def test_sympyissue_8289():
roots = (Poly(x**2 + 2) * Poly(x**4 + 2)).all_roots()
assert roots == _nsort(roots)
roots = Poly(x**6 + 3 * x**3 + 2, x).all_roots()
assert roots == _nsort(roots)
roots = Poly(x**6 - x + 1).all_roots()
assert roots == _nsort(roots)
# all imaginary roots
roots = Poly(x**4 + 4 * x**2 + 4, x).all_roots()
assert roots == _nsort(roots)
def test_roots_quartic():
assert roots_quartic(Poly(x**4, x)) == [0, 0, 0, 0]
assert roots_quartic(Poly(x**4 + x**3, x)) in [
[-1, 0, 0, 0],
[0, -1, 0, 0],
[0, 0, -1, 0],
[0, 0, 0, -1]
]
assert roots_quartic(Poly(x**4 - x**3, x)) in [
[1, 0, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]
]
lhs = roots_quartic(Poly(x**4 + x, x))
rhs = [Rational(1, 2) + I*sqrt(3)/2, Rational(1, 2) - I*sqrt(3)/2, 0, -1]
assert sorted(lhs, key=hash) == sorted(rhs, key=hash)
# test of all branches of roots quartic
for i, (a, b, c, d) in enumerate([(1, 2, 3, 0),
(3, -7, -9, 9),
(1, 2, 3, 4),
(1, 2, 3, 4),
(-7, -3, 3, -6),
(-3, 5, -6, -4),
(6, -5, -10, -3)]):
if i == 2:
c = -a*(a**2/Integer(8) - b/Integer(2))
elif i == 3:
d = a*(a*(3*a**2/Integer(256) - b/Integer(16)) + c/Integer(4))
eq = x**4 + a*x**3 + b*x**2 + c*x + d
ans = roots_quartic(Poly(eq, x))
assert all(eq.subs({x: ai}).evalf(chop=True) == 0 for ai in ans)
# not all symbolic quartics are unresolvable
eq = Poly(q*x + q/4 + x**4 + x**3 + 2*x**2 - Rational(1, 3), x)
sol = roots_quartic(eq)
assert all(verify_numerically(eq.subs({x: i}), 0) for i in sol)
z = symbols('z', negative=True)
eq = x**4 + 2*x**3 + 3*x**2 + x*(z + 11) + 5
zans = roots_quartic(Poly(eq, x))
assert all(verify_numerically(eq.subs({x: i, z: -1}), 0) for i in zans)
# but some are (see also issue sympy/sympy#4989)
# it's ok if the solution is not Piecewise, but the tests below should pass
eq = Poly(y*x**4 + x**3 - x + z, x)
ans = roots_quartic(eq)
assert all(type(i) == Piecewise for i in ans)
reps = ({y: -Rational(1, 3), z: -Rational(1, 4)}, # 4 real
{y: -Rational(1, 3), z: -Rational(1, 2)}, # 2 real
{y: -Rational(1, 3), z: -2}) # 0 real
for rep in reps:
sol = roots_quartic(Poly(eq.subs(rep), x))
assert all(verify_numerically(w.subs(rep) - s, 0) for w, s in zip(ans, sol))
def test_roots_binomial():
assert roots_binomial(Poly(5 * x, x)) == [0]
assert roots_binomial(Poly(5 * x**4, x)) == [0, 0, 0, 0]
assert roots_binomial(Poly(5 * x + 2, x)) == [-Rational(2, 5)]
A = 10**Rational(3, 4) / 10
assert roots_binomial(Poly(5*x**4 + 2, x)) == \
[-A - A*I, -A + A*I, A - A*I, A + A*I]
a1 = Symbol('a1', nonnegative=True)
b1 = Symbol('b1', nonnegative=True)
r0 = roots_quadratic(Poly(a1 * x**2 + b1, x))
r1 = roots_binomial(Poly(a1 * x**2 + b1, x))
assert powsimp(r0[0]) == powsimp(r1[0])
assert powsimp(r0[1]) == powsimp(r1[1])
for a, b, s, n in itertools.product((1, 2), (1, 2), (-1, 1), (2, 3, 4, 5)):
if a == b and a != 1: # a == b == 1 is sufficient
continue
p = Poly(a * x**n + s * b)
ans = roots_binomial(p)
assert ans == _nsort(ans)
# issue sympy/sympy#8813
assert roots(Poly(2 * x**3 - 16 * y**3, x)) == {
2 * y * (-Rational(1, 2) - sqrt(3) * I / 2): 1,
2 * y: 1,
2 * y * (-Rational(1, 2) + sqrt(3) * I / 2): 1
}
def test_nroots2():
p = Poly(x**5 + 3 * x + 1, x)
roots = p.nroots(n=3)
# The order of roots matters. The roots are ordered by their real
# components (if they agree, then by their imaginary components),
# with real roots appearing first.
assert [str(r) for r in roots] == \
['-0.332', '-0.839 - 0.944*I', '-0.839 + 0.944*I',
'1.01 - 0.937*I', '1.01 + 0.937*I']
roots = p.nroots(n=5)
assert [str(r) for r in roots] == \
['-0.33199', '-0.83907 - 0.94385*I', '-0.83907 + 0.94385*I',
'1.0051 - 0.93726*I', '1.0051 + 0.93726*I']
def test_nroots2():
p = Poly(x**5 + 3*x + 1, x)
roots = p.nroots(n=3)
# The order of roots matters. The roots are ordered by their real
# components (if they agree, then by their imaginary components),
# with real roots appearing first.
assert [str(r) for r in roots] == \
['-0.332', '-0.839 - 0.944*I', '-0.839 + 0.944*I',
'1.01 - 0.937*I', '1.01 + 0.937*I']
roots = p.nroots(n=5)
assert [str(r) for r in roots] == \
['-0.33199', '-0.83907 - 0.94385*I', '-0.83907 + 0.94385*I',
'1.0051 - 0.93726*I', '1.0051 + 0.93726*I']
def test_roots_cubic():
assert roots_cubic(Poly(2 * x**3, x)) == [0, 0, 0]
assert roots_cubic(Poly(x**3 - 3 * x**2 + 3 * x - 1, x)) == [1, 1, 1]
assert roots_cubic(Poly(x**3 + 1, x)) == \
[-1, Rational(1, 2) - I*sqrt(3)/2, Rational(1, 2) + I*sqrt(3)/2]
assert roots_cubic(Poly(2*x**3 - 3*x**2 - 3*x - 1, x))[0] == \
Rational(1, 2) + cbrt(3)/2 + 3**Rational(2, 3)/2
eq = -x**3 + 2 * x**2 + 3 * x - 2
assert roots(eq, trig=True, multiple=True) == \
roots_cubic(Poly(eq, x), trig=True) == [
Rational(2, 3) + 2*sqrt(13)*cos(acos(8*sqrt(13)/169)/3)/3,
-2*sqrt(13)*sin(-acos(8*sqrt(13)/169)/3 + pi/6)/3 + Rational(2, 3),
-2*sqrt(13)*cos(-acos(8*sqrt(13)/169)/3 + pi/3)/3 + Rational(2, 3),
]
def prde_normal_denom(fa, fd, G, DE):
"""
Parametric Risch Differential Equation - Normal part of the denominator.
Given a derivation D on k[t] and f, g1, ..., gm in k(t) with f weakly
normalized with respect to t, return the tuple (a, b, G, h) such that
a, h in k[t], b in k<t>, G = [g1, ..., gm] in k(t)^m, and for any solution
c1, ..., cm in Const(k) and y in k(t) of Dy + f*y == Sum(ci*gi, (i, 1, m)),
q == y*h in k<t> satisfies a*Dq + b*q == Sum(ci*Gi, (i, 1, m)).
"""
dn, ds = splitfactor(fd, DE)
Gas, Gds = list(zip(*G))
gd = reduce(lambda i, j: i.lcm(j), Gds, Poly(1, DE.t))
en, es = splitfactor(gd, DE)
p = dn.gcd(en)
h = en.gcd(en.diff(DE.t)).quo(p.gcd(p.diff(DE.t)))
a = dn*h
c = a*h
ba = a*fa - dn*derivation(h, DE)*fd
ba, bd = ba.cancel(fd, include=True)
G = [(c*A).cancel(D, include=True) for A, D in G]
return a, (ba, bd), 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)),
(c1, ..., cm) is a solution of Mx == 0, and p and the ci satisfy
a*Dp + b*p == Sum(ci*qi, (i, 1, m)).
Because M has entries in k(t), and because Matrix doesn't play well with
Poly, M will be a Matrix of Basic expressions.
"""
m = len(G)
Gns, Gds = list(zip(*G))
d = reduce(lambda i, j: i.lcm(j), Gds)
d = Poly(d, field=True)
Q = [(ga*(d).quo(gd)).div(d) for ga, gd in G]
if not all([ri.is_zero for _, ri in Q]):
N = max([ri.degree(DE.t) for _, ri in Q])
M = Matrix(N + 1, m, lambda i, j: Q[j][1].nth(i))
else:
M = Matrix() # No constraints, return the empty matrix.
qs, _ = list(zip(*Q))
return qs, M
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 diofant.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 test_roots_quadratic():
assert roots_quadratic(Poly(2*x**2, x)) == [0, 0]
assert roots_quadratic(Poly(2*x**2 + 3*x, x)) == [-Rational(3, 2), 0]
assert roots_quadratic(Poly(2*x**2 + 3, x)) == [-I*sqrt(6)/2, I*sqrt(6)/2]
assert roots_quadratic(Poly(2*x**2 + 4*x + 3, x)) == [-1 - I*sqrt(2)/2, -1 + I*sqrt(2)/2]
f = x**2 + (2*a*e + 2*c*e)/(a - c)*x + (d - b + a*e**2 - c*e**2)/(a - c)
assert (roots_quadratic(Poly(f, x)) ==
[-e*(a + c)/(a - c) - sqrt((a*b + 4*a*c*e**2 -
a*d - b*c + c*d)/(a - c)**2),
-e*(a + c)/(a - c) + sqrt((a*b + 4*a*c*e**2 -
a*d - b*c + c*d)/(a - c)**2)])
# check for simplification
f = Poly(y*x**2 - 2*x - 2*y, x)
assert roots_quadratic(f) == [-sqrt((2*y**2 + 1)/y**2) + 1/y,
sqrt((2*y**2 + 1)/y**2) + 1/y]
f = Poly(x**2 + (-y**2 - 2)*x + y**2 + 1, x)
assert roots_quadratic(f) == [y**2/2 - sqrt(y**4)/2 + 1,
y**2/2 + sqrt(y**4)/2 + 1]
f = Poly(sqrt(2)*x**2 - 1, x)
r = roots_quadratic(f)
assert r == _nsort(r)
# issue sympy/sympy#8255
f = Poly(-24*x**2 - 180*x + 264)
assert [w.evalf(2) for w in f.all_roots(radicals=True)] == \
[w.evalf(2) for w in f.all_roots(radicals=False)]
for _a, _b, _c in itertools.product((-2, 2), (-2, 2), (0, -1)):
f = Poly(_a*x**2 + _b*x + _c)
roots = roots_quadratic(f)
assert roots == _nsort(roots)
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 spde(a, b, c, n, DE):
"""
Rothstein's Special Polynomial Differential Equation algorithm.
Given a derivation D on k[t], an integer n and a, b, c in k[t] with
a != 0, either raise NonElementaryIntegralException, in which case the
equation a*Dq + b*q == c has no solution of degree at most n in
k[t], or return the tuple (B, C, m, alpha, beta) such that B, C,
alpha, beta in k[t], m in ZZ, and any solution q in k[t] of degree
at most n of a*Dq + b*q == c must be of the form
q == alpha*h + beta, where h in k[t], deg(h) <= m, and Dh + B*h == C.
This constitutes step 4 of the outline given in the rde.py docstring.
"""
zero = Poly(0, DE.t)
alpha = Poly(1, DE.t)
beta = Poly(0, DE.t)
while True:
if c.is_zero:
return zero, zero, 0, zero, beta # -1 is more to the point
if (n < 0) is True:
raise NonElementaryIntegralException
g = a.gcd(b)
if not c.rem(g).is_zero: # g does not divide c
raise NonElementaryIntegralException
a, b, c = a.quo(g), b.quo(g), c.quo(g)
if a.degree(DE.t) == 0:
b = b.to_field().quo(a)
c = c.to_field().quo(a)
return b, c, n, alpha, beta
r, z = gcdex_diophantine(b, a, c)
b += derivation(a, DE)
c = z - derivation(r, DE)
n -= a.degree(DE.t)
beta += alpha * r
alpha *= a
def test_root_factors():
assert root_factors(Poly(1, x)) == [Poly(1, x)]
assert root_factors(Poly(x, x)) == [Poly(x, x)]
assert root_factors(x**2 - 1, x) == [x + 1, x - 1]
assert root_factors(x**2 - y, x) == [x - sqrt(y), x + sqrt(y)]
assert root_factors((x**4 - 1)**2) == \
[x + 1, x + 1, x - 1, x - 1, x - I, x - I, x + I, x + I]
assert root_factors(Poly(x**4 - 1, x), filter='Z') == \
[Poly(x + 1, x), Poly(x - 1, x), Poly(x**2 + 1, x)]
assert root_factors(8*x**2 + 12*x**4 + 6*x**6 + x**8, x, filter='Q') == \
[x, x, x**6 + 6*x**4 + 12*x**2 + 8]
pytest.raises(ValueError, lambda: root_factors(Poly(x * y)))
def no_cancel_equal(b, c, n, DE):
"""
Poly Risch Differential Equation - No cancellation: deg(b) == deg(D) - 1
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_roots_cubic():
assert roots_cubic(Poly(2 * x**3, x)) == [0, 0, 0]
assert roots_cubic(Poly(x**3 - 3 * x**2 + 3 * x - 1, x)) == [1, 1, 1]
assert roots_cubic(Poly(x**3 + 1, x)) == \
[-1, Rational(1, 2) - I*sqrt(3)/2, Rational(1, 2) + I*sqrt(3)/2]
assert roots_cubic(Poly(2*x**3 - 3*x**2 - 3*x - 1, x))[0] == \
Rational(1, 2) + cbrt(3)/2 + 3**Rational(2, 3)/2
eq = -x**3 + 2 * x**2 + 3 * x - 2
assert roots(eq, trig=True, multiple=True) == \
roots_cubic(Poly(eq, x), trig=True) == [
Rational(2, 3) + 2*sqrt(13)*cos(acos(8*sqrt(13)/169)/3)/3,
-2*sqrt(13)*sin(-acos(8*sqrt(13)/169)/3 + pi/6)/3 + Rational(2, 3),
-2*sqrt(13)*cos(-acos(8*sqrt(13)/169)/3 + pi/3)/3 + Rational(2, 3),
]
res = roots_cubic(Poly(x**3 + 2 * a / 27, x))
assert res == [
-root(a + sqrt(a**2), 3) / 3,
Mul(Rational(-1, 3),
Rational(-1, 2) + sqrt(3) * I / 2,
root(a + sqrt(a**2), 3),
evaluate=False),
Mul(Rational(-1, 3),
Rational(-1, 2) - sqrt(3) * I / 2,
root(a + sqrt(a**2), 3),
evaluate=False)
]
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:
# TODO: implement this (requires recursive param_rischDE() call)
raise NotImplementedError
def order_at(a, p, t):
"""
Computes the order of a at p, with respect to t.
For a, p in k[t], the order of a at p is defined as nu_p(a) = max({n
in Z+ such that p**n|a}), where a != 0. If a == 0, nu_p(a) = +oo.
To compute the order at a rational function, a/b, use the fact that
nu_p(a/b) == nu_p(a) - nu_p(b).
"""
if a.is_zero:
return oo
if p == Poly(t, t):
return a.as_poly(t).ET()[0][0]
# Uses binary search for calculating the power. power_list collects the tuples
# (p^k,k) where each k is some power of 2. After deciding the largest k
# such that k is power of 2 and p^k|a the loop iteratively calculates
# the actual power.
power_list = []
p1 = p
r = a.rem(p1)
tracks_power = 1
while r.is_zero:
power_list.append((p1, tracks_power))
p1 = p1 * p1
tracks_power *= 2
r = a.rem(p1)
n = 0
product = Poly(1, t)
while len(power_list) != 0:
final = power_list.pop()
productf = product * final[0]
r = a.rem(productf)
if r.is_zero:
n += final[1]
product = productf
return n
def test_printing():
f, g = [dmp_normal([], 0, EX)] * 2
e = PolynomialDivisionFailed(f, g, EX)
assert str(e)[:57] == ("couldn't reduce degree in a polynomial "
"division algorithm")
assert str(e)[-140:][:57] == ("You may want to use a different "
"simplification algorithm.")
f, g = [dmp_normal([], 0, RR)] * 2
e = PolynomialDivisionFailed(f, g, RR)
assert str(e)[-139:][:74] == ("Your working precision or tolerance of "
"computations may be set improperly.")
f, g = [dmp_normal([], 0, ZZ)] * 2
e = PolynomialDivisionFailed(f, g, ZZ)
assert str(e)[-168:][:80] == (
"Zero detection is guaranteed in this "
"coefficient domain. This may indicate a bug")
e = OperationNotSupported(Poly(x), 'spam')
assert str(e).find('spam') >= 0
assert str(e).find('operation not supported') >= 0
exc = PolificationFailed(1, x, x**2)
assert str(exc).find("can't construct a polynomial from x") >= 0
exc = PolificationFailed(1, [x], [x**2], True)
assert str(exc).find("can't construct polynomials from x") >= 0
e = ComputationFailed('LT', 1, exc)
assert str(e).find('failed without generators') >= 0
assert str(e).find('x**2') >= 0
e = ExactQuotientFailed(Poly(x), Poly(x**2))
assert str(e).find('does not divide') >= 0
assert str(e).find('x**2') >= 0
assert str(e).find('in ZZ') < 0
e = ExactQuotientFailed(Poly(x), Poly(x**2), ZZ)
assert str(e).find('in ZZ') >= 0
def test_sympyissue_8438():
p = Poly([1, y, -2, -3], x).as_expr()
roots = roots_cubic(Poly(p, x), x)
z = -Rational(3,
2) - 7 * I / 2 # this will fail in code given in commit msg
post = [r.subs({y: z}) for r in roots]
assert set(post) == set(roots_cubic(Poly(p.subs({y: z}), x)))
# /!\ if p is not made an expression, this is *very* slow
assert all(p.subs({y: z, x: i}).evalf(2, chop=True) == 0 for i in post)
def weak_normalizer(a, d, DE, z=None):
"""
Weak normalization.
Given a derivation D on k[t] and f == a/d in k(t), return q in k[t]
such that f - Dq/q is weakly normalized with respect to t.
f in k(t) is said to be "weakly normalized" with respect to t if
residue_p(f) is not a positive integer for any normal irreducible p
in k[t] such that f is in R_p (Definition 6.1.1). If f has an
elementary integral, this is equivalent to no logarithm of
integral(f) whose argument depends on t has a positive integer
coefficient, where the arguments of the logarithms not in k(t) are
in k[t].
Returns (q, f - Dq/q)
"""
z = z or Dummy('z')
dn, ds = splitfactor(d, DE)
# Compute d1, where dn == d1*d2**2*...*dn**n is a square-free
# factorization of d.
g = gcd(dn, dn.diff(DE.t))
d_sqf_part = dn.quo(g)
d1 = d_sqf_part.quo(gcd(d_sqf_part, g))
a1, b = gcdex_diophantine(
d.quo(d1).as_poly(DE.t), d1.as_poly(DE.t), a.as_poly(DE.t))
r = (a - Poly(z, DE.t) * derivation(d1, DE)).as_poly(DE.t).resultant(
d1.as_poly(DE.t))
r = Poly(r, z)
if not r.has(z):
return Poly(1, DE.t), (a, d)
N = [i for i in r.real_roots() if i in ZZ and i > 0]
q = reduce(mul,
[gcd(a - Poly(n, DE.t) * derivation(d1, DE), d1) for n in N],
Poly(1, DE.t))
dq = derivation(q, DE)
sn = q * a - d * dq
sd = q * d
sn, sd = sn.cancel(sd, include=True)
return q, (sn, sd)
def solve_biquadratic(f, g, opt):
"""Solve a system of two bivariate quadratic polynomial equations.
Examples
========
>>> from diofant.polys import Options, Poly
>>> from diofant.abc import x, y
>>> from diofant.solvers.polysys import solve_biquadratic
>>> NewOption = Options((x, y), {'domain': 'ZZ'})
>>> a = Poly(y**2 - 4 + x, y, x, domain='ZZ')
>>> b = Poly(y*2 + 3*x - 7, y, x, domain='ZZ')
>>> solve_biquadratic(a, b, NewOption)
[(1/3, 3), (41/27, 11/9)]
>>> a = Poly(y + x**2 - 3, y, x, domain='ZZ')
>>> b = Poly(-y + x - 4, y, x, domain='ZZ')
>>> solve_biquadratic(a, b, NewOption)
[(-sqrt(29)/2 + 7/2, -sqrt(29)/2 - 1/2), (sqrt(29)/2 + 7/2, -1/2 + \
sqrt(29)/2)]
"""
G = groebner([f, g])
if len(G) == 1 and G[0].is_ground:
return
if len(G) != 2:
raise SolveFailed
p, q = G
x, y = opt.gens
p = Poly(p, x, expand=False)
q = q.ltrim(-1)
p_roots = [ rcollect(expr, y) for expr in roots(p).keys() ]
q_roots = list(roots(q).keys())
solutions = []
for q_root in q_roots:
for p_root in p_roots:
solution = (p_root.subs(y, q_root), q_root)
solutions.append(solution)
return sorted(solutions, key=default_sort_key)
def test_sympyissue_8285():
roots = (Poly(4 * x**8 - 1, x) * Poly(x**2 + 1)).all_roots()
assert roots == _nsort(roots)
f = Poly(x**4 + 5 * x**2 + 6, x)
ro = [RootOf(f, i) for i in range(4)]
roots = Poly(x**4 + 5 * x**2 + 6, x).all_roots()
assert roots == ro
assert roots == _nsort(roots)
# more than 2 complex roots from which to identify the
# imaginary ones
roots = Poly(2 * x**8 - 1).all_roots()
assert roots == _nsort(roots)
assert len(Poly(2 * x**10 - 1).all_roots()) == 10 # doesn't fail
def test_roots1():
assert roots(1) == {}
assert roots(1, multiple=True) == []
q = Symbol('q', real=True)
assert roots(x**3 - q, x) == {
cbrt(q): 1,
-cbrt(q) / 2 - sqrt(3) * I * cbrt(q) / 2: 1,
-cbrt(q) / 2 + sqrt(3) * I * cbrt(q) / 2: 1
}
assert roots_cubic(Poly(x**3 - 1)) == [
1,
Rational(-1, 2) + sqrt(3) * I / 2,
Rational(-1, 2) - sqrt(3) * I / 2
]
assert roots([1, x, y]) == {
-x / 2 - sqrt(x**2 - 4 * y) / 2: 1,
-x / 2 + sqrt(x**2 - 4 * y) / 2: 1
}
pytest.raises(ValueError, lambda: roots([1, x, y], z))
def apart_undetermined_coeffs(P, Q):
"""Partial fractions via method of undetermined coefficients. """
X = numbered_symbols(cls=Dummy)
partial, symbols = [], []
_, factors = Q.factor_list()
for f, k in factors:
n, q = f.degree(), Q
for i in range(1, k + 1):
coeffs, q = take(X, n), q.quo(f)
partial.append((coeffs, q, f, i))
symbols.extend(coeffs)
dom = Q.get_domain().inject(*symbols)
F = Poly(0, Q.gen, domain=dom)
for i, (coeffs, q, f, k) in enumerate(partial):
h = Poly(coeffs, Q.gen, domain=dom)
partial[i] = (h, f, k)
q = q.set_domain(dom)
F += h * q
system, result = [], Integer(0)
for (k, ), coeff in F.terms():
system.append(coeff - P.nth(k))
from diofant.solvers import solve
solution = solve(system, symbols)
for h, f, k in partial:
h = h.as_expr().subs(solution)
result += h / f.as_expr()**k
return result
def dispersionset(p, q=None, *gens, **args):
r"""Compute the *dispersion set* of two polynomials.
For two polynomials `f(x)` and `g(x)` with `\deg f > 0`
and `\deg g > 0` the dispersion set `\operatorname{J}(f, g)` is defined as:
.. math::
\operatorname{J}(f, g)
& := \{a \in \mathbb{N}_0 | \gcd(f(x), g(x+a)) \neq 1\} \\
& = \{a \in \mathbb{N}_0 | \deg \gcd(f(x), g(x+a)) \geq 1\}
For a single polynomial one defines `\operatorname{J}(f) := \operatorname{J}(f, f)`.
Examples
========
>>> from diofant import poly
>>> from diofant.abc import x
Dispersion set and dispersion of a simple polynomial:
>>> fp = poly((x - 3)*(x + 3), x)
>>> sorted(dispersionset(fp))
[0, 6]
>>> dispersion(fp)
6
Note that the definition of the dispersion is not symmetric:
>>> fp = poly(x**4 - 3*x**2 + 1, x)
>>> gp = fp.shift(-3)
>>> sorted(dispersionset(fp, gp))
[2, 3, 4]
>>> dispersion(fp, gp)
4
>>> sorted(dispersionset(gp, fp))
[]
>>> dispersion(gp, fp)
-oo
Computing the dispersion also works over field extensions:
>>> from diofant import sqrt
>>> fp = poly(x**2 + sqrt(5)*x - 1, x, domain='QQ<sqrt(5)>')
>>> gp = poly(x**2 + (2 + sqrt(5))*x + sqrt(5), x, domain='QQ<sqrt(5)>')
>>> sorted(dispersionset(fp, gp))
[2]
>>> sorted(dispersionset(gp, fp))
[1, 4]
We can even perform the computations for polynomials
having symbolic coefficients:
>>> from diofant.abc import a
>>> fp = poly(4*x**4 + (4*a + 8)*x**3 + (a**2 + 6*a + 4)*x**2 + (a**2 + 2*a)*x, x)
>>> sorted(dispersionset(fp))
[0, 1]
See Also
========
diofant.polys.dispersion.dispersion
References
==========
.. [1] [ManWright94]_
.. [2] [Koepf98]_
.. [3] [Abramov71]_
.. [4] [Man93]_
"""
# Check for valid input
same = False if q is not None else True
if same:
q = p
p = Poly(p, *gens, **args)
q = Poly(q, *gens, **args)
if not p.is_univariate or not q.is_univariate:
raise ValueError("Polynomials need to be univariate")
# The generator
if not p.gen == q.gen:
raise ValueError("Polynomials must have the same generator")
gen = p.gen
# We define the dispersion of constant polynomials to be zero
if p.degree() < 1 or q.degree() < 1:
return {0}
# Factor p and q over the rationals
fp = p.factor_list()
fq = q.factor_list() if not same else fp
# Iterate over all pairs of factors
J = set()
for s, unused in fp[1]:
for t, unused in fq[1]:
m = s.degree()
n = t.degree()
if n != m:
continue
an = s.LC()
bn = t.LC()
if not (an - bn).is_zero:
continue
# Note that the roles of `s` and `t` below are switched
# w.r.t. the original paper. This is for consistency
# with the description in the book of W. Koepf.
anm1 = s.coeff_monomial(gen**(m - 1))
bnm1 = t.coeff_monomial(gen**(n - 1))
alpha = (anm1 - bnm1) / (n * bn)
if not alpha.is_Integer:
continue
if alpha < 0 or alpha in J:
continue
if n > 1 and not (s - t.shift(alpha)).is_zero:
continue
J.add(alpha)
return J
def test_RootOf():
p = Poly(x**3 + y*x + 1, x)
assert mcode(RootOf(p, 0)) == 'Root[#^3 + #*y + 1 &, 1]'
def test_sympyissue_14291():
p = Poly(((x - 1)**2 + 1) * ((x - 1)**2 + 2) * (x - 1))
assert set(
p.all_roots()) == {1, 1 - I, 1 + I, 1 - I * sqrt(2), 1 + sqrt(2) * I}
def test_sympyissue_7724():
e = x**4 * I + x**2 + I
r1, r2 = roots(e, x), Poly(e, x).all_roots()
assert len(r1) == 4
assert {_.evalf() for _ in r1} == {_.evalf() for _ in r2}
def test_sympyissue_14291():
p = Poly(((x - 1)**2 + 1)*((x - 1)**2 + 2)*(x - 1))
assert set(p.all_roots()) == {1, 1 - I, 1 + I, 1 - I*sqrt(2), 1 + sqrt(2)*I}