本文整理汇总了Python中sympy.polys.polyroots.roots_binomial函数的典型用法代码示例。如果您正苦于以下问题:Python roots_binomial函数的具体用法?Python roots_binomial怎么用?Python roots_binomial使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了roots_binomial函数的5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,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: test_roots_binomial
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]
_check(roots_binomial(Poly(x**8 - 2)))
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 cartes((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 8813
assert roots(Poly(2*x**3 - 16*y**3, x)) == {
2*y*(-S(1)/2 - sqrt(3)*I/2): 1,
2*y: 1,
2*y*(-S(1)/2 + sqrt(3)*I/2): 1}
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:31,代码来源:test_polyroots.py
示例2: _roots_trivial
def _roots_trivial(cls, poly, radicals):
"""Compute roots in linear, quadratic and binomial cases. """
if poly.degree() == 1:
return roots_linear(poly)
if not radicals:
return None
if poly.degree() == 2:
return roots_quadratic(poly)
elif poly.length() == 2 and poly.TC():
return roots_binomial(poly)
else:
return None
开发者ID:A-turing-machine,项目名称:sympy,代码行数:14,代码来源:rootoftools.py
示例3: roots_trivial
def roots_trivial(poly, radicals=True):
"""Compute roots in linear, quadratic and binomial cases. """
if poly.degree() == 1:
return roots_linear(poly)
else:
if not radicals:
return None
if poly in _rootof_trivial_cache:
roots = _rootof_trivial_cache[poly]
else:
if radicals and poly.degree() == 2:
roots = roots_quadratic(poly)
elif radicals and poly.length() == 2 and poly.TC():
roots = roots_binomial(poly)
else:
return None
_rootof_trivial_cache[poly] = roots
return roots
开发者ID:haz,项目名称:sympy,代码行数:21,代码来源:rootoftools.py
示例4: _roots_trivial
def _roots_trivial(cls, poly, radicals):
"""Compute roots in linear, quadratic and binomial cases. """
if poly.degree() == 1:
return roots_linear(poly)
if not radicals:
return None
free = len(poly.free_symbols)
sort = isinstance(poly, PurePoly) and free or free > 1
if poly.degree() == 2:
roots = roots_quadratic(poly, sort=sort)
elif poly.length() == 2 and poly.TC():
roots = roots_binomial(poly, sort=sort)
else:
return None
# put roots in same order as RootOf instances
if not sort:
key = [r.n(2) for r in roots]
key = [(1 if not r.is_real else 0, C.re(r), C.im(r))
for r in key]
_, roots = zip(*sorted(zip(key, roots)))
return roots
开发者ID:MCGallaspy,项目名称:sympy,代码行数:22,代码来源:rootoftools.py
示例5: test_roots_binomial
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 cartes((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)
roots = roots_binomial(p)
assert roots == _nsort(roots)
开发者ID:brajeshvit,项目名称:virtual,代码行数:23,代码来源:test_polyroots.py
注:本文中的sympy.polys.polyroots.roots_binomial函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。 |
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