本文整理汇总了Python中sympy.holonomic.HolonomicFunction.to_sequence方法的典型用法代码示例。如果您正苦于以下问题：Python HolonomicFunction.to_sequence方法的具体用法？Python HolonomicFunction.to_sequence怎么用？Python HolonomicFunction.to_sequence使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sympy.holonomic.HolonomicFunction的用法示例。
在下文中一共展示了HolonomicFunction.to_sequence方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: test_to_Sequence_Initial_Coniditons
# 需要导入模块: from sympy.holonomic import HolonomicFunction [as 别名]
# 或者: from sympy.holonomic.HolonomicFunction import to_sequence [as 别名]
def test_to_Sequence_Initial_Coniditons():
x = symbols('x')
R, Dx = DifferentialOperators(QQ.old_poly_ring(x), 'Dx')
n = symbols('n', integer=True)
_, Sn = RecurrenceOperators(QQ.old_poly_ring(n), 'Sn')
p = HolonomicFunction(Dx - 1, x, 0, [1]).to_sequence()
q = [(HolonomicSequence(-1 + (n + 1)*Sn, 1), 0)]
assert p == q
p = HolonomicFunction(Dx**2 + 1, x, 0, [0, 1]).to_sequence()
q = [(HolonomicSequence(1 + (n**2 + 3*n + 2)*Sn**2, [0, 1]), 0)]
assert p == q
p = HolonomicFunction(Dx**2 + 1 + x**3*Dx, x, 0, [2, 3]).to_sequence()
q = [(HolonomicSequence(n + Sn**2 + (n**2 + 7*n + 12)*Sn**4, [2, 3, -1, -1/2, 1/12]), 1)]
assert p == q
p = HolonomicFunction(x**3*Dx**5 + 1 + Dx, x).to_sequence()
q = [(HolonomicSequence(1 + (n + 1)*Sn + (n**5 - 5*n**3 + 4*n)*Sn**2), 0, 3)]
assert p == q
C_0, C_1, C_2, C_3 = symbols('C_0, C_1, C_2, C_3')
p = expr_to_holonomic(log(1+x**2))
q = [(HolonomicSequence(n**2 + (n**2 + 2*n)*Sn**2, [0, 0, C_2]), 0, 1)]
assert p.to_sequence() == q
p = p.diff()
q = [(HolonomicSequence((n + 2) + (n + 2)*Sn**2, [C_0, 0]), 1, 0)]
assert p.to_sequence() == q
p = expr_to_holonomic(erf(x) + x).to_sequence()
q = [(HolonomicSequence((2*n**2 - 2*n) + (n**3 + 2*n**2 - n - 2)*Sn**2, [0, 1 + 2/sqrt(pi), 0, C_3]), 0, 2)]
assert p == q