本文整理汇总了Python中sympy.functions.binomial函数的典型用法代码示例。如果您正苦于以下问题：Python binomial函数的具体用法？Python binomial怎么用？Python 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binomial函数的8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_catalan
def test_catalan():
n = Symbol('n', integer=True)
m = Symbol('n', integer=True, positive=True)
catalans = [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786]
for i, c in enumerate(catalans):
assert catalan(i) == c
assert catalan(n).rewrite(factorial).subs(n, i) == c
assert catalan(n).rewrite(Product).subs(n, i).doit() == c
assert catalan(x) == catalan(x)
assert catalan(2*x).rewrite(binomial) == binomial(4*x, 2*x)/(2*x + 1)
assert catalan(Rational(1, 2)).rewrite(gamma) == 8/(3*pi)
assert catalan(Rational(1, 2)).rewrite(factorial).rewrite(gamma) ==\
8 / (3 * pi)
assert catalan(3*x).rewrite(gamma) == 4**(
3*x)*gamma(3*x + Rational(1, 2))/(sqrt(pi)*gamma(3*x + 2))
assert catalan(x).rewrite(hyper) == hyper((-x + 1, -x), (2,), 1)
assert catalan(n).rewrite(factorial) == factorial(2*n) / (factorial(n + 1)
* factorial(n))
assert isinstance(catalan(n).rewrite(Product), catalan)
assert isinstance(catalan(m).rewrite(Product), Product)
assert diff(catalan(x), x) == (polygamma(
0, x + Rational(1, 2)) - polygamma(0, x + 2) + log(4))*catalan(x)
assert catalan(x).evalf() == catalan(x)
c = catalan(S.Half).evalf()
assert str(c) == '0.848826363156775'
c = catalan(I).evalf(3)
assert str((re(c), im(c))) == '(0.398, -0.0209)'
开发者ID:A-turing-machine，项目名称:sympy，代码行数:32，代码来源:test_comb_numbers.py
示例2: test_catalan
def test_catalan():
assert catalan(1) == 1
assert catalan(2) == 2
assert catalan(3) == 5
assert catalan(4) == 14
assert catalan(x) == catalan(x)
assert catalan(2*x).rewrite(binomial) == binomial(4*x, 2*x)/(2*x + 1)
assert catalan(Rational(1, 2)).rewrite(gamma) == 8/(3*pi)
assert catalan(3*x).rewrite(gamma) == 4**(
3*x)*gamma(3*x + Rational(1, 2))/(sqrt(pi)*gamma(3*x + 2))
assert catalan(x).rewrite(hyper) == hyper((-x + 1, -x), (2,), 1)
assert diff(catalan(x), x) == (polygamma(
0, x + Rational(1, 2)) - polygamma(0, x + 2) + log(4))*catalan(x)
c = catalan(0.5).evalf()
assert str(c) == '0.848826363156775'
开发者ID:DVNSarma，项目名称:sympy，代码行数:18，代码来源:test_comb_numbers.py
示例3: rsolve_poly
def rsolve_poly(coeffs, f, n, **hints):
"""
Given linear recurrence operator `\operatorname{L}` of order
`k` with polynomial coefficients and inhomogeneous equation
`\operatorname{L} y = f`, where `f` is a polynomial, we seek for
all polynomial solutions over field `K` of characteristic zero.
The algorithm performs two basic steps:
(1) Compute degree `N` of the general polynomial solution.
(2) Find all polynomials of degree `N` or less
of `\operatorname{L} y = f`.
There are two methods for computing the polynomial solutions.
If the degree bound is relatively small, i.e. it's smaller than
or equal to the order of the recurrence, then naive method of
undetermined coefficients is being used. This gives system
of algebraic equations with `N+1` unknowns.
In the other case, the algorithm performs transformation of the
initial equation to an equivalent one, for which the system of
algebraic equations has only `r` indeterminates. This method is
quite sophisticated (in comparison with the naive one) and was
invented together by Abramov, Bronstein and Petkovsek.
It is possible to generalize the algorithm implemented here to
the case of linear q-difference and differential equations.
Lets say that we would like to compute `m`-th Bernoulli polynomial
up to a constant. For this we can use `b(n+1) - b(n) = m n^{m-1}`
recurrence, which has solution `b(n) = B_m + C`. For example:
>>> from sympy import Symbol, rsolve_poly
>>> n = Symbol('n', integer=True)
>>> rsolve_poly([-1, 1], 4*n**3, n)
C0 + n**4 - 2*n**3 + n**2
References
==========
.. [1] S. A. Abramov, M. Bronstein and M. Petkovsek, On polynomial
solutions of linear operator equations, in: T. Levelt, ed.,
Proc. ISSAC '95, ACM Press, New York, 1995, 290-296.
.. [2] M. Petkovsek, Hypergeometric solutions of linear recurrences
with polynomial coefficients, J. Symbolic Computation,
14 (1992), 243-264.
.. [3] M. Petkovsek, H. S. Wilf, D. Zeilberger, A = B, 1996.
"""
f = sympify(f)
if not f.is_polynomial(n):
return None
homogeneous = f.is_zero
r = len(coeffs) - 1
coeffs = [ Poly(coeff, n) for coeff in coeffs ]
polys = [ Poly(0, n) ] * (r + 1)
terms = [ (S.Zero, S.NegativeInfinity) ] *(r + 1)
for i in xrange(0, r + 1):
for j in xrange(i, r + 1):
polys[i] += coeffs[j]*binomial(j, i)
if not polys[i].is_zero:
(exp,), coeff = polys[i].LT()
terms[i] = (coeff, exp)
d = b = terms[0][1]
for i in xrange(1, r + 1):
if terms[i][1] > d:
d = terms[i][1]
if terms[i][1] - i > b:
b = terms[i][1] - i
d, b = int(d), int(b)
x = Dummy('x')
degree_poly = S.Zero
for i in xrange(0, r + 1):
if terms[i][1] - i == b:
degree_poly += terms[i][0]*FallingFactorial(x, i)
nni_roots = roots(degree_poly, x, filter='Z',
predicate=lambda r: r >= 0).keys()
if nni_roots:
N = [max(nni_roots)]
else:
N = []
#.........这里部分代码省略.........
开发者ID:alhirzel，项目名称:sympy，代码行数:101，代码来源:recurr.py
示例4: test_euler_failing
def test_euler_failing():
# depends on dummy variables being implemented https://github.com/sympy/sympy/issues/5665
assert euler(2*n).rewrite(Sum) == I*Sum(Sum((-1)**_j*2**(-_k)*I**(-_k)*(-2*_j + _k)**(2*n + 1)*binomial(_k, _j)/_k, (_j, 0, _k)), (_k, 1, 2*n + 1))
开发者ID:DVNSarma，项目名称:sympy，代码行数:3，代码来源:test_comb_numbers.py
示例5: trigintegrate
def trigintegrate(f, x, conds='piecewise'):
"""Integrate f = Mul(trig) over x
>>> from sympy import Symbol, sin, cos, tan, sec, csc, cot
>>> from sympy.integrals.trigonometry import trigintegrate
>>> from sympy.abc import x
>>> trigintegrate(sin(x)*cos(x), x)
sin(x)**2/2
>>> trigintegrate(sin(x)**2, x)
x/2 - sin(x)*cos(x)/2
>>> trigintegrate(tan(x)*sec(x), x)
1/cos(x)
>>> trigintegrate(sin(x)*tan(x), x)
-log(sin(x) - 1)/2 + log(sin(x) + 1)/2 - sin(x)
http://en.wikibooks.org/wiki/Calculus/Integration_techniques
See Also
========
sympy.integrals.integrals.Integral.doit
sympy.integrals.integrals.Integral
"""
from sympy.integrals.integrals import integrate
pat, a, n, m = _pat_sincos(x)
f = f.rewrite('sincos')
M = f.match(pat)
if M is None:
return
n, m = M[n], M[m]
if n is S.Zero and m is S.Zero:
return x
zz = x if n is S.Zero else S.Zero
a = M[a]
if n.is_odd or m.is_odd:
u = _u
n_, m_ = n.is_odd, m.is_odd
# take smallest n or m -- to choose simplest substitution
if n_ and m_:
n_ = n_ and (n < m) # NB: careful here, one of the
m_ = m_ and not (n < m) # conditions *must* be true
# n m u=C (n-1)/2 m
# S(x) * C(x) dx --> -(1-u^2) * u du
if n_:
ff = -(1 - u**2)**((n - 1)/2) * u**m
uu = cos(a*x)
# n m u=S n (m-1)/2
# S(x) * C(x) dx --> u * (1-u^2) du
elif m_:
ff = u**n * (1 - u**2)**((m - 1)/2)
uu = sin(a*x)
fi = integrate(ff, u) # XXX cyclic deps
fx = fi.subs(u, uu)
if conds == 'piecewise':
return Piecewise((zz, Eq(a, 0)), (fx / a, True))
return fx / a
# n & m are both even
#
# 2k 2m 2l 2l
# we transform S (x) * C (x) into terms with only S (x) or C (x)
#
# example:
# 100 4 100 2 2 100 4 2
# S (x) * C (x) = S (x) * (1-S (x)) = S (x) * (1 + S (x) - 2*S (x))
#
# 104 102 100
# = S (x) - 2*S (x) + S (x)
# 2k
# then S is integrated with recursive formula
# take largest n or m -- to choose simplest substitution
n_ = (abs(n) > abs(m))
m_ = (abs(m) > abs(n))
res = S.Zero
if n_:
# 2k 2 k i 2i
# C = (1 - S ) = sum(i, (-) * B(k, i) * S )
if m > 0:
for i in range(0, m//2 + 1):
res += ((-1)**i * binomial(m//2, i) *
_sin_pow_integrate(n + 2*i, x))
elif m == 0:
res = _sin_pow_integrate(n, x)
else:
#.........这里部分代码省略.........
开发者ID:AALEKH，项目名称:sympy，代码行数:101，代码来源:trigonometry.py
示例6: trigintegrate
#.........这里部分代码省略.........
# S(x) * C(x) dx --> u * (1-u^2) du
elif m_:
ff = u**n * (1-u**2)**((m-1)/2)
uu = sin(a*x)
fi= sympy.integrals.integrate(ff, u) # XXX cyclic deps
fx= fi.subs(u, uu)
return fx / a
# n & m are even
else:
# 2k 2m 2l 2l
# we transform S (x) * C (x) into terms with only S (x) or C (x)
#
# example:
# 100 4 100 2 2 100 4 2
# S (x) * C (x) = S (x) * (1-S (x)) = S (x) * (1 + S (x) - 2*S (x))
#
# 104 102 100
# = S (x) - 2*S (x) + S (x)
# 2k
# then S is integrated with recursive formula
# take largest n or m -- to choose simplest substitution
n_ = (abs(n) > abs(m))
m_ = (abs(m) > abs(n))
res = S.Zero
if n_:
# 2k 2 k i 2i
# C = (1-S ) = sum(i, (-) * B(k,i) * S )
if m > 0 :
for i in range(0,m/2+1):
res += (-1)**i * binomial(m/2,i) * _sin_pow_integrate(n+2*i, x)
elif m == 0:
res=_sin_pow_integrate(n,x)
else:
# m < 0 , |n| > |m|
# / /
# | |
# | m n -1 m+1 n-1 n - 1 | m+2 n-2
# | cos (x) sin (x) dx = ________ cos (x) sin (x) + _______ | cos (x) sin (x) dx
# | |
# | m + 1 m + 1 |
#/ /
#
#
res=Rational(-1,m+1)*cos(x)**(m+1)*sin(x)**(n-1) + Rational(n-1,m+1)*trigintegrate(cos(x)**(m+2)*sin(x)**(n-2),x)
elif m_:
# 2k 2 k i 2i
# S = (1 -C ) = sum(i, (-) * B(k,i) * C )
if n > 0:
# / /
# | |
# | m n | -m n
# | cos (x)*sin (x) dx or | cos (x) * sin (x) dx
# | |
# / /
#
# |m| > |n| ; m,n >0 ; m,n belong to Z - {0}
# n 2
# sin (x) term is expanded here interms of cos (x), and then integrated.
for i in range(0,n/2+1):
开发者ID:hector1618，项目名称:sympy，代码行数:67，代码来源:trigonometry.py
示例7: trigintegrate
def trigintegrate(f, x):
"""Integrate f = Mul(trig) over x
>>> from sympy import Symbol, sin, cos
>>> from sympy.integrals.trigonometry import trigintegrate
>>> from sympy.abc import x
>>> trigintegrate(sin(x)*cos(x), x)
sin(x)**2/2
>>> trigintegrate(sin(x)**2, x)
x/2 - cos(x)*sin(x)/2
http://en.wikibooks.org/wiki/Calculus/Further_integration_techniques
"""
pat, a,n,m = _pat_sincos(x)
##m - cos
##n - sin
M = f.match(pat)
if M is None:
return
n, m = M[n], M[m] # should always be there
if n is S.Zero and m is S.Zero:
return x
a = M[a]
if n.is_integer and m.is_integer:
if n.is_odd or m.is_odd:
u = _u
n_, m_ = n.is_odd, m.is_odd
# take smallest n or m -- to choose simplest substitution
if n_ and m_:
n_ = n_ and (n < m) # NB: careful here, one of the
m_ = m_ and not (n < m) # conditions *must* be true
# n m u=C (n-1)/2 m
# S(x) * C(x) dx --> -(1-u^2) * u du
if n_:
ff = -(1-u**2)**((n-1)/2) * u**m
uu = cos(a*x)
# n m u=S n (m-1)/2
# S(x) * C(x) dx --> u * (1-u^2) du
elif m_:
ff = u**n * (1-u**2)**((m-1)/2)
uu = sin(a*x)
fi= sympy.integrals.integrate(ff, u) # XXX cyclic deps
fx= fi.subs(u, uu)
return fx / a
# n & m are even
else:
# 2k 2m 2l 2l
# we transform S (x) * C (x) into terms with only S (x) or C (x)
#
# example:
# 100 4 100 2 2 100 4 2
# S (x) * C (x) = S (x) * (1-S (x)) = S (x) * (1 + S (x) - 2*S (x))
#
# 104 102 100
# = S (x) - 2*S (x) + S (x)
# 2k
# then S is integrated with recursive formula
# take largest n or m -- to choose simplest substitution
n_ = (abs(n) > abs(m))
m_ = (abs(m) > abs(n))
res = S.Zero
if n_:
# 2k 2 k i 2i
# C = (1-S ) = sum(i, (-) * B(k,i) * S )
if m > 0 :
for i in range(0,m/2+1):
res += (-1)**i * binomial(m/2,i) * sin_pow_integrate(n+2*i, x)
elif m == 0:
res=sin_pow_integrate(n,x)
else:
# m < 0 , |n| > |m|
# / /
# | |
# | m n -1 m+1 n-1 n - 1 | m+2 n-2
# | cos (x) sin (x) dx = ________ cos (x) sin (x) + _______ | cos (x) sin (x) dx
# | |
# | m + 1 m + 1 |
#/ /
#
#
res=Rational(-1,m+1)*cos(x)**(m+1)*sin(x)**(n-1) + Rational(n-1,m+1)*trigintegrate(cos(x)**(m+2)*sin(x)**(n-2),x)
#.........这里部分代码省略.........
开发者ID:Praveen-Ramanujam，项目名称:MobRAVE，代码行数:101，代码来源:trigonometry.py
示例8: trigintegrate
def trigintegrate(f, x):
"""Integrate f = Mul(trig) over x
>>> from sympy import Symbol, sin, cos
>>> from sympy.integrals.trigonometry import trigintegrate
>>> x = Symbol('x')
>>> trigintegrate(sin(x)*cos(x), x)
1/2*sin(x)**2
>>> trigintegrate(sin(x)**2, x)
x/2 - cos(x)*sin(x)/2
http://en.wikibooks.org/wiki/Calculus/Further_integration_techniques
"""
pat, a,n,m = _pat_sincos(x)
M = f.match(pat)
if M is None:
return
n, m = M[n], M[m] # should always be there
if n is S.Zero and m is S.Zero:
return x
a = M[a]
if n.is_integer and n.is_integer:
if n.is_odd or m.is_odd:
u = _u
n_, m_ = n.is_odd, m.is_odd
# take smallest n or m -- to choose simplest substitution
if n_ and m_:
n_ = n_ and (n < m) # NB: careful here, one of the
m_ = m_ and not (n < m) # conditions *must* be true
# n m u=C (n-1)/2 m
# S(x) * C(x) dx --> -(1-u^2) * u du
if n_:
ff = -(1-u**2)**((n-1)/2) * u**m
uu = cos(a*x)
# n m u=S n (m-1)/2
# S(x) * C(x) dx --> u * (1-u^2) du
elif m_:
ff = u**n * (1-u**2)**((m-1)/2)
uu = sin(a*x)
fi= sympy.integrals.integrate(ff, u) # XXX cyclic deps
fx= fi.subs(u, uu)
return fx / a
# n & m are even
else:
# 2k 2m 2l 2l
# we transform S (x) * C (x) into terms with only S (x) or C (x)
#
# example:
# 100 4 100 2 2 100 4 2
# S (x) * C (x) = S (x) * (1-S (x)) = S (x) * (1 + S (x) - 2*S (x))
#
# 104 102 100
# = S (x) - 2*S (x) + S (x)
# 2k
# then S is integrated with recursive formula
# take largest n or m -- to choose simplest substitution
n_ = (n > m) # NB: careful here, one of the
m_ = not (n > m) # conditions *must* be true
res = S.Zero
if n_:
# 2k 2 k i 2i
# C = (1-S ) = sum(i, (-) * B(k,i) * S )
for i in range(0,m/2+1):
res += (-1)**i * binomial(m/2,i) * Sin_2k_integrate(n/2+i, x)
elif m_:
# 2k 2 k i 2i
# S = (1 -C ) = sum(i, (-) * B(k,i) * C )
for i in range(0,n/2+1):
res += (-1)**i * binomial(n/2,i) * Cos_2k_integrate(m/2+i, x)
return res.subs(x, a*x) / a
开发者ID:jcockayne，项目名称:sympy-rkern，代码行数:88，代码来源:trigonometry.py
注：本文中的sympy.functions.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。 |
客服电话
APP下载
官方微信
请发表评论