def test_diofantissue_453(): x = Symbol('x', real=True) assert isolve(abs((x - 1) / (x - 5)) <= Rational(1, 3), x) == And(Integer(-1) <= x, x <= 2) assert solve(abs((x - 1) / (x - 5)) - Rational(1, 3), x) == [{ x: -1 }, { x: 2 }]
def test_sympyissue_8974(): assert isolve(-oo < x, x) == And(-oo < x, x < oo) assert isolve(oo > x, x) == And(-oo < x, x < oo)
def test_solve_univariate_inequality(): assert isolve(x**2 >= 4, x, relational=False) == Union(Interval(-oo, -2, True), Interval(2, oo, False, True)) assert isolve(x**2 >= 4, x) == Or(And(Le(2, x), Lt(x, oo)), And(Le(x, -2), Lt(-oo, x))) assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x, relational=False) == \ Union(Interval(1, 2), Interval(3, oo, False, True)) assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x) == \ Or(And(Le(1, x), Le(x, 2)), And(Le(3, x), Lt(x, oo))) # issue sympy/sympy#2785: assert isolve(x**3 - 2*x - 1 > 0, x, relational=False) == \ Union(Interval(-1, -sqrt(5)/2 + Rational(1, 2), True, True), Interval(Rational(1, 2) + sqrt(5)/2, oo, True, True)) # issue sympy/sympy#2794: assert isolve(x**3 - x**2 + x - 1 > 0, x, relational=False) == \ Interval(1, oo, True, True) # XXX should be limited in domain, e.g. between 0 and 2*pi assert isolve(sin(x) < S.Half, x) == \ Or(And(-oo < x, x < pi/6), And(5*pi/6 < x, x < oo)) assert isolve(sin(x) > S.Half, x) == And(pi/6 < x, x < 5*pi/6) # numerical testing in valid() is needed assert isolve(x**7 - x - 2 > 0, x) == \ And(RootOf(x**7 - x - 2, 0) < x, x < oo) # handle numerator and denominator; although these would be handled as # rational inequalities, these test confirm that the right thing is done # when the domain is EX (e.g. when 2 is replaced with sqrt(2)) assert isolve(1/(x - 2) > 0, x) == And(Integer(2) < x, x < oo) den = ((x - 1)*(x - 2)).expand() assert isolve((x - 1)/den <= 0, x) == \ Or(And(-oo < x, x < 1), And(Integer(1) < x, x < 2)) assert isolve(x > oo, x) is S.false
def test_solve_univariate_inequality(): assert isolve(x**2 >= 4, x, relational=False) == Union(Interval(-oo, -2), Interval(2, oo)) assert isolve(x**2 >= 4, x) == (Integer(2) <= x) | (x <= -2) assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x, relational=False) == \ Union(Interval(1, 2), Interval(3, oo)) assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x) == \ ((Integer(1) <= x) & (x <= 2)) | (Integer(3) <= x) # issue sympy/sympy#2785: assert isolve(x**3 - 2*x - 1 > 0, x, relational=False) == \ Union(Interval(-1, -sqrt(5)/2 + Rational(1, 2), True, True), Interval(Rational(1, 2) + sqrt(5)/2, oo, True)) # issue sympy/sympy#2794: assert isolve(x**3 - x**2 + x - 1 > 0, x, relational=False) == \ Interval(1, oo, True) # XXX should be limited in domain, e.g. between 0 and 2*pi assert isolve(sin(x) < Rational(1, 2), x) == (x < pi / 6) | (5 * pi / 6 < x) assert isolve(sin(x) > Rational(1, 2), x) == (pi / 6 < x) & (x < 5 * pi / 6) # numerical testing in valid() is needed assert isolve(x**7 - x - 2 > 0, x) == (RootOf(x**7 - x - 2, 0) < x) # handle numerator and denominator; although these would be handled as # rational inequalities, these test confirm that the right thing is done # when the domain is EX (e.g. when 2 is replaced with sqrt(2)) assert isolve(1 / (x - 2) > 0, x) == (Integer(2) < x) den = ((x - 1) * (x - 2)).expand() assert isolve((x - 1) / den <= 0, x) == (x < 1) | ((Integer(1) < x) & (x < 2)) assert isolve(x > oo, x) is false # issue sympy/sympy#10268 assert reduce_inequalities(log(x) < 300) == (-oo < x) & (x < E**300)
def test_diofantissue_453(): x = Symbol('x', real=True) assert isolve(abs((x - 1)/(x - 5)) <= Rational(1, 3), x) == And(Integer(-1) <= x, x <= 2) assert solve(abs((x - 1)/(x - 5)) - Rational(1, 3), x) == [{x: -1}, {x: 2}]
def test_solve_univariate_inequality(): assert isolve(x**2 >= 4, x, relational=False) == Union(Interval(-oo, -2, True), Interval(2, oo, False, True)) assert isolve(x**2 >= 4, x) == Or(And(Le(2, x), Lt(x, oo)), And(Le(x, -2), Lt(-oo, x))) assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x, relational=False) == \ Union(Interval(1, 2), Interval(3, oo, False, True)) assert isolve((x - 1)*(x - 2)*(x - 3) >= 0, x) == \ Or(And(Le(1, x), Le(x, 2)), And(Le(3, x), Lt(x, oo))) # issue sympy/sympy#2785: assert isolve(x**3 - 2*x - 1 > 0, x, relational=False) == \ Union(Interval(-1, -sqrt(5)/2 + Rational(1, 2), True, True), Interval(Rational(1, 2) + sqrt(5)/2, oo, True, True)) # issue sympy/sympy#2794: assert isolve(x**3 - x**2 + x - 1 > 0, x, relational=False) == \ Interval(1, oo, True, True) # XXX should be limited in domain, e.g. between 0 and 2*pi assert isolve(sin(x) < Rational(1, 2), x) == \ Or(And(-oo < x, x < pi/6), And(5*pi/6 < x, x < oo)) assert isolve(sin(x) > Rational(1, 2), x) == And(pi/6 < x, x < 5*pi/6) # numerical testing in valid() is needed assert isolve(x**7 - x - 2 > 0, x) == \ And(RootOf(x**7 - x - 2, 0) < x, x < oo) # handle numerator and denominator; although these would be handled as # rational inequalities, these test confirm that the right thing is done # when the domain is EX (e.g. when 2 is replaced with sqrt(2)) assert isolve(1/(x - 2) > 0, x) == And(Integer(2) < x, x < oo) den = ((x - 1)*(x - 2)).expand() assert isolve((x - 1)/den <= 0, x) == \ Or(And(-oo < x, x < 1), And(Integer(1) < x, x < 2)) assert isolve(x > oo, x) is false