本文整理汇总了Python中sympy.functions.combinatorial.numbers.bernoulli函数的典型用法代码示例。如果您正苦于以下问题:Python bernoulli函数的具体用法?Python bernoulli怎么用?Python bernoulli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bernoulli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_issue_14177
def test_issue_14177():
n = Symbol('n', positive=True, integer=True)
assert zeta(2*n) == (-1)**(n + 1)*2**(2*n - 1)*pi**(2*n)*bernoulli(2*n)/factorial(2*n)
assert zeta(-n) == (-1)**(-n)*bernoulli(n + 1)/(n + 1)
n = Symbol('n')
assert zeta(2*n) == zeta(2*n) # As sign of z (= 2*n) is not determined
开发者ID:KonstantinTogoi,项目名称:sympy,代码行数:9,代码来源:test_zeta_functions.py
示例2: _eval_aseries
def _eval_aseries(self, n, args0, x, logx):
from sympy import Order
if args0[1] != oo or not \
(self.args[0].is_Integer and self.args[0].is_nonnegative):
return super(polygamma, self)._eval_aseries(n, args0, x, logx)
z = self.args[1]
N = self.args[0]
if N == 0:
# digamma function series
# Abramowitz & Stegun, p. 259, 6.3.18
r = log(z) - 1/(2*z)
o = None
if n < 2:
o = Order(1/z, x)
else:
m = ceiling((n + 1)//2)
l = [bernoulli(2*k) / (2*k*z**(2*k)) for k in range(1, m)]
r -= Add(*l)
o = Order(1/z**(2*m), x)
return r._eval_nseries(x, n, logx) + o
else:
# proper polygamma function
# Abramowitz & Stegun, p. 260, 6.4.10
# We return terms to order higher than O(x**n) on purpose
# -- otherwise we would not be able to return any terms for
# quite a long time!
fac = gamma(N)
e0 = fac + N*fac/(2*z)
m = ceiling((n + 1)//2)
for k in range(1, m):
fac = fac*(2*k + N - 1)*(2*k + N - 2) / ((2*k)*(2*k - 1))
e0 += bernoulli(2*k)*fac/z**(2*k)
o = Order(1/z**(2*m), x)
if n == 0:
o = Order(1/z, x)
elif n == 1:
o = Order(1/z**2, x)
r = e0._eval_nseries(z, n, logx) + o
return (-1 * (-1/z)**N * r)._eval_nseries(x, n, logx)
开发者ID:SungSingSong,项目名称:sympy,代码行数:40,代码来源:gamma_functions.py
示例3: eval
def eval(cls, z, a_=None):
if a_ is None:
z, a = list(map(sympify, (z, 1)))
else:
z, a = list(map(sympify, (z, a_)))
if a.is_Number:
if a is S.NaN:
return S.NaN
elif a is S.One and a_ is not None:
return cls(z)
# TODO Should a == 0 return S.NaN as well?
if z.is_Number:
if z is S.NaN:
return S.NaN
elif z is S.Infinity:
return S.One
elif z is S.Zero:
if a.is_negative:
return S.Half - a - 1
else:
return S.Half - a
elif z is S.One:
return S.ComplexInfinity
elif z.is_Integer:
if a.is_Integer:
if z.is_negative:
zeta = (-1)**z * bernoulli(-z + 1)/(-z + 1)
elif z.is_even:
B, F = bernoulli(z), factorial(z)
zeta = 2**(z - 1) * abs(B) * pi**z / F
else:
return
if a.is_negative:
return zeta + harmonic(abs(a), z)
else:
return zeta - harmonic(a - 1, z)
开发者ID:amitsaha,项目名称:sympy,代码行数:39,代码来源:zeta_functions.py
示例4: _eval_aseries
def _eval_aseries(self, n, args0, x, logx):
if args0[0] != oo:
return super(loggamma, self)._eval_aseries(n, args0, x, logx)
z = self.args[0]
m = min(n, C.ceiling((n + S(1)) / 2))
r = log(z) * (z - S(1) / 2) - z + log(2 * pi) / 2
l = [bernoulli(2 * k) / (2 * k * (2 * k - 1) * z ** (2 * k - 1)) for k in range(1, m)]
o = None
if m == 0:
o = C.Order(1, x)
else:
o = C.Order(1 / z ** (2 * m - 1), x)
# It is very inefficient to first add the order and then do the nseries
return (r + Add(*l))._eval_nseries(x, n, logx) + o
开发者ID:Krastanov,项目名称:sympy,代码行数:14,代码来源:gamma_functions.py
示例5: test_sympy__functions__combinatorial__numbers__bernoulli
def test_sympy__functions__combinatorial__numbers__bernoulli():
from sympy.functions.combinatorial.numbers import bernoulli
assert _test_args(bernoulli(x))
开发者ID:101man,项目名称:sympy,代码行数:3,代码来源:test_args.py
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