本文整理汇总了Python中sympy.functions.elementary.exponential.exp_polar函数的典型用法代码示例。如果您正苦于以下问题:Python exp_polar函数的具体用法?Python exp_polar怎么用?Python exp_polar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exp_polar函数的4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: eval
def eval(cls, arg):
from sympy.functions.elementary.complexes import arg as argument
if arg.is_number:
ar = argument(arg)
# In general we want to affirm that something is known,
# e.g. `not ar.has(argument) and not ar.has(atan)`
# but for now we will just be more restrictive and
# see that it has evaluated to one of the known values.
if ar in (0, pi/2, -pi/2, pi):
return exp_polar(I*ar)*abs(arg)
if arg.is_Mul:
args = arg.args
else:
args = [arg]
included = []
excluded = []
positive = []
for arg in args:
if arg.is_polar:
included += [arg]
elif arg.is_positive:
positive += [arg]
else:
excluded += [arg]
if len(excluded) < len(args):
if excluded:
return Mul(*(included + positive))*polar_lift(Mul(*excluded))
elif included:
return Mul(*(included + positive))
else:
return Mul(*positive)*exp_polar(0)
开发者ID:asmeurer,项目名称:sympy,代码行数:32,代码来源:complexes.py
示例2: _eval_expand_func
def _eval_expand_func(self, **hints):
from sympy import exp, I, floor, Add, Poly, Dummy, exp_polar, unpolarify
z, s, a = self.args
if z == 1:
return zeta(s, a)
if s.is_Integer and s <= 0:
t = Dummy('t')
p = Poly((t + a)**(-s), t)
start = 1/(1 - t)
res = S(0)
for c in reversed(p.all_coeffs()):
res += c*start
start = t*start.diff(t)
return res.subs(t, z)
if a.is_Rational:
# See section 18 of
# Kelly B. Roach. Hypergeometric Function Representations.
# In: Proceedings of the 1997 International Symposium on Symbolic and
# Algebraic Computation, pages 205-211, New York, 1997. ACM.
# TODO should something be polarified here?
add = S(0)
mul = S(1)
# First reduce a to the interaval (0, 1]
if a > 1:
n = floor(a)
if n == a:
n -= 1
a -= n
mul = z**(-n)
add = Add(*[-z**(k - n)/(a + k)**s for k in range(n)])
elif a <= 0:
n = floor(-a) + 1
a += n
mul = z**n
add = Add(*[z**(n - 1 - k)/(a - k - 1)**s for k in range(n)])
m, n = S([a.p, a.q])
zet = exp_polar(2*pi*I/n)
root = z**(1/n)
return add + mul*n**(s - 1)*Add(
*[polylog(s, zet**k*root)._eval_expand_func(**hints)
/ (unpolarify(zet)**k*root)**m for k in range(n)])
# TODO use minpoly instead of ad-hoc methods when issue 5888 is fixed
if isinstance(z, exp) and (z.args[0]/(pi*I)).is_Rational or z in [-1, I, -I]:
# TODO reference?
if z == -1:
p, q = S([1, 2])
elif z == I:
p, q = S([1, 4])
elif z == -I:
p, q = S([-1, 4])
else:
arg = z.args[0]/(2*pi*I)
p, q = S([arg.p, arg.q])
return Add(*[exp(2*pi*I*k*p/q)/q**s*zeta(s, (k + a)/q)
for k in range(q)])
return lerchphi(z, s, a)
开发者ID:moorepants,项目名称:sympy,代码行数:60,代码来源:zeta_functions.py
示例3: unpolarify
def unpolarify(eq, subs={}, exponents_only=False):
"""
If p denotes the projection from the Riemann surface of the logarithm to
the complex line, return a simplified version eq' of `eq` such that
p(eq') == p(eq).
Also apply the substitution subs in the end. (This is a convenience, since
``unpolarify``, in a certain sense, undoes polarify.)
>>> from sympy import unpolarify, polar_lift, sin, I
>>> unpolarify(polar_lift(I + 2))
2 + I
>>> unpolarify(sin(polar_lift(I + 7)))
sin(7 + I)
"""
if isinstance(eq, bool):
return eq
eq = sympify(eq)
if subs != {}:
return unpolarify(eq.subs(subs))
changed = True
pause = False
if exponents_only:
pause = True
while changed:
changed = False
res = _unpolarify(eq, exponents_only, pause)
if res != eq:
changed = True
eq = res
if isinstance(res, bool):
return res
# Finally, replacing Exp(0) by 1 is always correct.
# So is polar_lift(0) -> 0.
return res.subs({exp_polar(0): 1, polar_lift(0): 0})
开发者ID:asmeurer,项目名称:sympy,代码行数:35,代码来源:complexes.py
示例4: test_sympy__functions__elementary__exponential__exp_polar
def test_sympy__functions__elementary__exponential__exp_polar():
from sympy.functions.elementary.exponential import exp_polar
assert _test_args(exp_polar(2))
开发者ID:Kimay,项目名称:sympy,代码行数:3,代码来源:test_args.py
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