Пример #1
0
def test_nsolve():
    # onedimensional
    from sympy import Symbol, sin, pi
    x = Symbol('x')
    assert nsolve(sin(x), 2) - pi.evalf() < 1e-16
    assert nsolve(Eq(2 * x, 2), x, -10) == nsolve(2 * x - 2, -10)
    # multidimensional
    x1 = Symbol('x1')
    x2 = Symbol('x2')
    f1 = 3 * x1**2 - 2 * x2**2 - 1
    f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8
    f = Matrix((f1, f2)).T
    F = lambdify((x1, x2), f.T, modules='mpmath')
    for x0 in [(-1, 1), (1, -2), (4, 4), (-4, -4)]:
        x = nsolve(f, (x1, x2), x0, tol=1.e-8)
        assert mnorm(F(*x), 1) <= 1.e-10
    # The Chinese mathematician Zhu Shijie was the very first to solve this
    # nonlinear system 700 years ago (z was added to make it 3-dimensional)
    x = Symbol('x')
    y = Symbol('y')
    z = Symbol('z')
    f1 = -x + 2 * y
    f2 = (x**2 + x * (y**2 - 2) - 4 * y) / (x + 4)
    f3 = sqrt(x**2 + y**2) * z
    f = Matrix((f1, f2, f3)).T
    F = lambdify((x, y, z), f.T, modules='mpmath')

    def getroot(x0):
        root = nsolve((f1, f2, f3), (x, y, z), x0)
        assert mnorm(F(*root), 1) <= 1.e-8
        return root

    assert map(round, getroot((1, 1, 1))) == [2.0, 1.0, 0.0]
Пример #2
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def test_nsolve():
    # onedimensional
    from sympy import Symbol, sin, pi
    x = Symbol('x')
    assert nsolve(sin(x), 2) - pi.evalf() < 1e-16
    assert nsolve(Eq(2*x, 2), x, -10) == nsolve(2*x - 2, -10)
    # multidimensional
    x1 = Symbol('x1')
    x2 = Symbol('x2')
    f1 = 3 * x1**2 - 2 * x2**2 - 1
    f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8
    f = Matrix((f1, f2)).T
    F = lambdify((x1, x2), f.T, modules='mpmath')
    for x0 in [(-1, 1), (1, -2), (4, 4), (-4, -4)]:
        x = nsolve(f, (x1, x2), x0, tol=1.e-8)
        assert mnorm(F(*x),1) <= 1.e-10
    # The Chinese mathematician Zhu Shijie was the very first to solve this
    # nonlinear system 700 years ago (z was added to make it 3-dimensional)
    x = Symbol('x')
    y = Symbol('y')
    z = Symbol('z')
    f1 = -x + 2*y
    f2 = (x**2 + x*(y**2 - 2) - 4*y)  /  (x + 4)
    f3 = sqrt(x**2 + y**2)*z
    f = Matrix((f1, f2, f3)).T
    F = lambdify((x,  y,  z), f.T, modules='mpmath')
    def getroot(x0):
        root = nsolve((f1,  f2,  f3), (x,  y,  z), x0)
        assert mnorm(F(*root),1) <= 1.e-8
        return root
    assert map(round,  getroot((1,  1,  1))) == [2.0,  1.0,  0.0]
Пример #3
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def test_nsolve():
    # onedimensional
    from sympy import Symbol, sin, pi
    x = Symbol('x')
    assert nsolve(sin(x), 2) - pi.evalf() < 1e-16
    assert nsolve(Eq(2 * x, 2), x, -10) == nsolve(2 * x - 2, -10)
    # Testing checks on number of inputs
    raises(TypeError, "nsolve(Eq(2*x,2))")
    raises(TypeError, "nsolve(Eq(2*x,2),x,1,2)")
    # Issue 1730
    assert nsolve(x**2 / (1 - x) / (1 - 2 * x)**2 - 100, x, 0)  # doesn't fail
    # multidimensional
    x1 = Symbol('x1')
    x2 = Symbol('x2')
    f1 = 3 * x1**2 - 2 * x2**2 - 1
    f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8
    f = Matrix((f1, f2)).T
    F = lambdify((x1, x2), f.T, modules='mpmath')
    for x0 in [(-1, 1), (1, -2), (4, 4), (-4, -4)]:
        x = nsolve(f, (x1, x2), x0, tol=1.e-8)
        assert mnorm(F(*x), 1) <= 1.e-10
    # The Chinese mathematician Zhu Shijie was the very first to solve this
    # nonlinear system 700 years ago (z was added to make it 3-dimensional)
    x = Symbol('x')
    y = Symbol('y')
    z = Symbol('z')
    f1 = -x + 2 * y
    f2 = (x**2 + x * (y**2 - 2) - 4 * y) / (x + 4)
    f3 = sqrt(x**2 + y**2) * z
    f = Matrix((f1, f2, f3)).T
    F = lambdify((x, y, z), f.T, modules='mpmath')

    def getroot(x0):
        root = nsolve((f1, f2, f3), (x, y, z), x0)
        assert mnorm(F(*root), 1) <= 1.e-8
        return root

    assert map(round, getroot((1, 1, 1))) == [2.0, 1.0, 0.0]
    assert nsolve([Eq(f1), Eq(f2), Eq(f3)], [x, y, z],
                  (1, 1, 1))  # just see that it works
    a = Symbol('a')
    assert nsolve(1 / (0.001 + a)**3 - 6 / (0.9 - a)**3, a,
                  0.3).ae(mpf('0.31883011387318591'))
Пример #4
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def test_nsolve():
    # onedimensional
    x = Symbol('x')
    assert nsolve(sin(x), 2) - pi.evalf() < 1e-15
    assert nsolve(Eq(2*x, 2), x, -10) == nsolve(2*x - 2, -10)
    # Testing checks on number of inputs
    raises(TypeError, lambda: nsolve(Eq(2*x, 2)))
    raises(TypeError, lambda: nsolve(Eq(2*x, 2), x, 1, 2))
    # issue 4829
    assert nsolve(x**2/(1 - x)/(1 - 2*x)**2 - 100, x, 0)  # doesn't fail
    # multidimensional
    x1 = Symbol('x1')
    x2 = Symbol('x2')
    f1 = 3 * x1**2 - 2 * x2**2 - 1
    f2 = x1**2 - 2 * x1 + x2**2 + 2 * x2 - 8
    f = Matrix((f1, f2)).T
    F = lambdify((x1, x2), f.T, modules='mpmath')
    for x0 in [(-1, 1), (1, -2), (4, 4), (-4, -4)]:
        x = nsolve(f, (x1, x2), x0, tol=1.e-8)
        assert mnorm(F(*x), 1) <= 1.e-10
    # The Chinese mathematician Zhu Shijie was the very first to solve this
    # nonlinear system 700 years ago (z was added to make it 3-dimensional)
    x = Symbol('x')
    y = Symbol('y')
    z = Symbol('z')
    f1 = -x + 2*y
    f2 = (x**2 + x*(y**2 - 2) - 4*y) / (x + 4)
    f3 = sqrt(x**2 + y**2)*z
    f = Matrix((f1, f2, f3)).T
    F = lambdify((x, y, z), f.T, modules='mpmath')

    def getroot(x0):
        root = nsolve(f, (x, y, z), x0)
        assert mnorm(F(*root), 1) <= 1.e-8
        return root
    assert list(map(round, getroot((1, 1, 1)))) == [2.0, 1.0, 0.0]
    assert nsolve([Eq(
        f1), Eq(f2), Eq(f3)], [x, y, z], (1, 1, 1))  # just see that it works
    a = Symbol('a')
    assert nsolve(1/(0.001 + a)**3 - 6/(0.9 - a)**3, a, 0.3).ae(
        mpf('0.31883011387318591'))
Пример #5
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 def getroot(x0):
     root = nsolve(f, (x, y, z), x0)
     assert mnorm(F(*root), 1) <= 1.e-8
     return root
Пример #6
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 def getroot(x0):
     root = nsolve(f, (x, y, z), x0)
     assert mnorm(F(*root), 1) <= 1.e-8
     return root
Пример #7
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 def getroot(x0):
     root = nsolve((f1, f2, f3), (x, y, z), x0)
     assert mnorm(F(*root), 1) <= 1.0e-8
     return root