本文整理汇总了Python中sympy.hypersimp函数的典型用法代码示例。如果您正苦于以下问题：Python hypersimp函数的具体用法？Python hypersimp怎么用？Python hypersimp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hypersimp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: hypsum
def hypsum(expr, n, start, prec):
"""
Sum a rapidly convergent infinite hypergeometric series with
given general term, e.g. e = hypsum(1/factorial(n), n). The
quotient between successive terms must be a quotient of integer
polynomials.
"""
from sympy import hypersimp, lambdify
if start:
expr = expr.subs(n, n + start)
hs = hypersimp(expr, n)
if hs is None:
raise NotImplementedError("a hypergeometric series is required")
num, den = hs.as_numer_denom()
func1 = lambdify(n, num)
func2 = lambdify(n, den)
h, g, p = check_convergence(num, den, n)
if h < 0:
raise ValueError("Sum diverges like (n!)^%i" % (-h))
# Direct summation if geometric or faster
if h > 0 or (h == 0 and abs(g) > 1):
term = expr.subs(n, 0)
term = (MPZ(term.p) << prec) // term.q
s = term
k = 1
while abs(term) > 5:
term *= MPZ(func1(k - 1))
term //= MPZ(func2(k - 1))
s += term
k += 1
return from_man_exp(s, -prec)
else:
alt = g < 0
if abs(g) < 1:
raise ValueError("Sum diverges like (%i)^n" % abs(1 / g))
if p < 1 or (p == 1 and not alt):
raise ValueError("Sum diverges like n^%i" % (-p))
# We have polynomial convergence: use Richardson extrapolation
# Need to use at least quad precision because a lot of cancellation
# might occur in the extrapolation process
prec2 = 4 * prec
term = expr.subs(n, 0)
term = (MPZ(term.p) << prec2) // term.q
def summand(k, _term=[term]):
if k:
k = int(k)
_term[0] *= MPZ(func1(k - 1))
_term[0] //= MPZ(func2(k - 1))
return make_mpf(from_man_exp(_term[0], -prec2))
with workprec(prec):
v = nsum(summand, [0, mpmath_inf], method="richardson")
return v._mpf_
开发者ID:brajeshvit，项目名称:virtual，代码行数:60，代码来源:evalf.py
示例2: test_hypersimp
def test_hypersimp():
n, k = symbols('nk', integer=True)
assert hypersimp(factorial(k), k) == k + 1
assert hypersimp(factorial(k**2), k) is None
assert hypersimp(1/factorial(k), k) == 1/(k + 1)
assert hypersimp(2**k/factorial(k)**2, k) == 2/(k**2+2*k+1)
assert hypersimp(binomial(n, k), k) == (n-k)/(k+1)
assert hypersimp(binomial(n+1, k), k) == (n-k+1)/(k+1)
term = (4*k+1)*factorial(k)/factorial(2*k+1)
assert hypersimp(term, k) == (4*k + 5)/(6 + 16*k**2 + 28*k)
term = 1/((2*k-1)*factorial(2*k+1))
assert hypersimp(term, k) == (2*k-1)/(6 + 22*k + 24*k**2 + 8*k**3)
term = binomial(n, k)*(-1)**k/factorial(k)
assert hypersimp(term, k) == (k - n)/(k**2+2*k+1)
开发者ID:KevinGoodsell，项目名称:sympy，代码行数:21，代码来源:test_simplify.py
示例3: hypsum
def hypsum(expr, n, start, prec):
"""
Sum a rapidly convergent infinite hypergeometric series with
given general term, e.g. e = hypsum(1/factorial(n), n). The
quotient between successive terms must be a quotient of integer
polynomials.
"""
from sympy import Float, hypersimp, lambdify
if prec == float('inf'):
raise NotImplementedError('does not support inf prec')
if start:
expr = expr.subs(n, n + start)
hs = hypersimp(expr, n)
if hs is None:
raise NotImplementedError("a hypergeometric series is required")
num, den = hs.as_numer_denom()
func1 = lambdify(n, num)
func2 = lambdify(n, den)
h, g, p = check_convergence(num, den, n)
if h < 0:
raise ValueError("Sum diverges like (n!)^%i" % (-h))
term = expr.subs(n, 0)
if not term.is_Rational:
raise NotImplementedError("Non rational term functionality is not implemented.")
# Direct summation if geometric or faster
if h > 0 or (h == 0 and abs(g) > 1):
term = (MPZ(term.p) << prec) // term.q
s = term
k = 1
while abs(term) > 5:
term *= MPZ(func1(k - 1))
term //= MPZ(func2(k - 1))
s += term
k += 1
return from_man_exp(s, -prec)
else:
alt = g < 0
if abs(g) < 1:
raise ValueError("Sum diverges like (%i)^n" % abs(1/g))
if p < 1 or (p == 1 and not alt):
raise ValueError("Sum diverges like n^%i" % (-p))
# We have polynomial convergence: use Richardson extrapolation
vold = None
ndig = prec_to_dps(prec)
while True:
# Need to use at least quad precision because a lot of cancellation
# might occur in the extrapolation process; we check the answer to
# make sure that the desired precision has been reached, too.
prec2 = 4*prec
term0 = (MPZ(term.p) << prec2) // term.q
def summand(k, _term=[term0]):
if k:
k = int(k)
_term[0] *= MPZ(func1(k - 1))
_term[0] //= MPZ(func2(k - 1))
return make_mpf(from_man_exp(_term[0], -prec2))
with workprec(prec):
v = nsum(summand, [0, mpmath_inf], method='richardson')
vf = Float(v, ndig)
if vold is not None and vold == vf:
break
prec += prec # double precision each time
vold = vf
return v._mpf_
开发者ID:arghdos，项目名称:sympy，代码行数:74，代码来源:evalf.py
示例4: hypsum
def hypsum(expr, n, start, prec):
"""
Sum a rapidly convergent infinite hypergeometric series with
given general term, e.g. e = hypsum(1/factorial(n), n). The
quotient between successive terms must be a quotient of integer
polynomials.
"""
from sympy import hypersimp, lambdify
if start:
expr = expr.subs(n, n+start)
hs = hypersimp(expr, n)
if hs is None:
raise NotImplementedError("a hypergeometric series is required")
num, den = hs.as_numer_denom()
func1 = lambdify(n, num)
func2 = lambdify(n, den)
h, g, p = check_convergence(num, den, n)
if h < 0:
raise ValueError("Sum diverges like (n!)^%i" % (-h))
# Direct summation if geometric or faster
if h > 0 or (h == 0 and abs(g) > 1):
one = MP_BASE(1) << prec
term = expr.subs(n, 0)
term = (MP_BASE(term.p) << prec) // term.q
s = term
k = 1
while abs(term) > 5:
term *= MP_BASE(func1(k-1))
term //= MP_BASE(func2(k-1))
s += term
k += 1
return from_man_exp(s, -prec)
else:
alt = g < 0
if abs(g) < 1:
raise ValueError("Sum diverges like (%i)^n" % abs(1/g))
if p < 1 or (p == 1 and not alt):
raise ValueError("Sum diverges like n^%i" % (-p))
# We have polynomial convergence:
# Use Shanks extrapolation for alternating series,
# Richardson extrapolation for nonalternating series
if alt:
# XXX: better parameters for Shanks transformation
# This tends to get bad somewhere > 50 digits
N = 5 + int(prec*0.36)
M = 2 + N//3
NTERMS = M + N + 2
else:
N = 3 + int(prec*0.15)
M = 2*N
NTERMS = M + N + 2
# Need to use at least double precision because a lot of cancellation
# might occur in the extrapolation process
prec2 = 2*prec
one = MP_BASE(1) << prec2
term = expr.subs(n, 0)
term = (MP_BASE(term.p) << prec2) // term.q
s = term
table = [make_mpf(from_man_exp(s, -prec2))]
for k in xrange(1, NTERMS):
term *= MP_BASE(func1(k-1))
term //= MP_BASE(func2(k-1))
s += term
table.append(make_mpf(from_man_exp(s, -prec2)))
k += 1
orig = mp.prec
try:
mp.prec = prec
if alt:
v = shanks_extrapolation(table, N, M)
else:
v = richardson_extrapolation(table, N, M)
finally:
mp.prec = orig
return v._mpf_
开发者ID:jcockayne，项目名称:sympy-rkern，代码行数:79，代码来源:evalf.py
注：本文中的sympy.hypersimp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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