def _eval_nseries(self, x, n, logx):
# NOTE Please see the comment at the beginning of this file, labelled
# IMPORTANT.
from sympy import cancel
if not logx:
logx = log(x)
if self.args[0] == x:
return logx
arg = self.args[0]
k, l = Wild("k"), Wild("l")
r = arg.match(k * x**l)
if r is not None:
#k = r.get(r, S.One)
#l = r.get(l, S.Zero)
k, l = r[k], r[l]
if l != 0 and not l.has(x) and not k.has(x):
r = log(k) + l * logx # XXX true regardless of assumptions?
return r
# TODO new and probably slow
s = self.args[0].nseries(x, n=n, logx=logx)
while s.is_Order:
n += 1
s = self.args[0].nseries(x, n=n, logx=logx)
a, b = s.leadterm(x)
p = cancel(s / (a * x**b) - 1)
g = None
l = []
for i in xrange(n + 2):
g = log.taylor_term(i, p, g)
g = g.nseries(x, n=n, logx=logx)
l.append(g)
return log(a) + b * logx + Add(*l) + C.Order(p**n, x)
def _eval_as_leading_term(self, x):
arg = self.args[0].as_leading_term(x)
if x in arg.free_symbols and C.Order(1, x).contains(arg):
return 2 * x / sqrt(pi)
else:
return self.func(arg)
def _eval_as_leading_term(self, x):
arg = self.args[0].as_leading_term(x)
if x in arg.free_symbols and C.Order(1, x).contains(arg):
return S.ImaginaryUnit * S.Pi / 2
else:
return self.func(arg)
def _eval_as_leading_term(self, x):
arg = self.args[0].as_leading_term(x)
if C.Order(1, x).contains(arg):
return arg
else:
return self.func(arg)
def _eval_as_leading_term(self, x):
n, z = [a.as_leading_term(x) for a in self.args]
o = C.Order(z, x)
if n == 0 and o.contains(1 / x):
return o.getn() * log(x)
else:
return self.func(n, z)
def _taylor(self, x, n):
l = []
g = None
for i in xrange(n):
g = self.taylor_term(i, self.args[0], g)
g = g.nseries(x, n=n)
l.append(g)
return Add(*l) + C.Order(x**n, x)
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
def _eval_as_leading_term(self, x):
arg = self.args[0]
if arg.is_Add:
return Mul(*[exp(f).as_leading_term(x) for f in arg.args])
arg = self.args[0].as_leading_term(x)
if C.Order(1, x).contains(arg):
return S.One
return exp(arg)
def doit(self, **hints):
"""Evaluates limit"""
e, z, z0, dir = self.args
if hints.get('deep', True):
e = e.doit(**hints)
z = z.doit(**hints)
z0 = z0.doit(**hints)
if e == z:
return z0
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.is_Mul:
if abs(z0) is S.Infinity:
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and
any(a.is_polynomial(z) or
any(z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a))
for a in Add.make_args(w)))
if all(ok(w) for w in e.as_numer_denom()):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = e.subs(z, 1/u)
r = limit(inve.as_leading_term(u), u,
S.Zero, "+" if z0 is S.Infinity else "-")
if isinstance(r, Limit):
return self
else:
return r
if e.is_Order:
return C.Order(limit(e.expr, z, z0), *e.args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
if r is None:
return self
return r
def _eval_aseries(self, n, args0, x, logx):
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 = C.Order(1 / z, x)
else:
m = C.ceiling((n + 1) / 2)
l = [
bernoulli(2 * k) / (2 * k * z**(2 * k))
for k in range(1, m)
]
r -= Add(*l)
o = C.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 = C.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 = C.Order(1 / z**(2 * m), x)
if n == 0:
o = C.Order(1 / z, x)
elif n == 1:
o = C.Order(1 / z**2, x)
r = e0._eval_nseries(z, n, logx) + o
return -1 * (-1 / z)**N * r
def _eval_aseries(self, n, args0, x, logx):
if args0[0] != S.Infinity:
return super(_erfs, self)._eval_aseries(n, args0, x, logx)
z = self.args[0]
l = [ 1/sqrt(S.Pi) * C.factorial(2*k)*(-S(
4))**(-k)/C.factorial(k) * (1/z)**(2*k + 1) for k in xrange(0, n) ]
o = C.Order(1/z**(2*n + 1), x)
# It is very inefficient to first add the order and then do the nseries
return (Add(*l))._eval_nseries(x, n, logx) + o
def _eval_nseries(self, x, n, logx):
# NOTE Please see the comment at the beginning of this file, labelled
# IMPORTANT.
from sympy import limit, oo, powsimp
arg = self.args[0]
arg_series = arg._eval_nseries(x, n=n, logx=logx)
if arg_series.is_Order:
return 1 + arg_series
arg0 = limit(arg_series.removeO(), x, 0)
if arg0 in [-oo, oo]:
return self
t = Dummy("t")
exp_series = exp(t)._taylor(t, n)
o = exp_series.getO()
exp_series = exp_series.removeO()
r = exp(arg0)*exp_series.subs(t, arg_series - arg0)
r += C.Order(o.expr.subs(t, (arg_series - arg0)), x)
r = r.expand()
return powsimp(r, deep=True, combine='exp')
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.is_Mul:
if abs(z0) is S.Infinity:
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and any(
a.is_polynomial(z) or any(
z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a)) for a in Add.make_args(w)))
if all(ok(w) for w in e.as_numer_denom()):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = e.subs(z, 1 / u)
return limit(inve.as_leading_term(u), u, S.Zero,
"+" if z0 is S.Infinity else "-")
if e.is_Order:
return C.Order(limit(e.expr, z, z0), *e.args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi/2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign*e.args[0], z, z0)/disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign*e.args[0], z, z0)/disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2*i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base != 0 and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex)*S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
if abs(z0) is S.Infinity:
n, d = e.as_numer_denom()
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and
any(a.is_polynomial(z) or
any(z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a))
for a in Add.make_args(w)))
if all(ok(w) for w in (n, d)):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = (n/d).subs(z, 1/u)
return limit(inve.as_leading_term(u), u,
S.Zero, "+" if z0 is S.Infinity else "-")
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i*limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo*i
else:
if p >= 0:
return S.Zero
else:
return -oo*i
if e.is_Add:
if e.is_polynomial():
if not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
elif e.is_rational_function(z):
rval = Add(*[limit(term, z, z0, dir) for term in e.args])
if rval != S.NaN:
return rval
if not any([a.is_unbounded for a in e.args]):
e = e.normal() # workaround for issue 3744
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
def _eval_nseries(self, x, x0, n):
from sympy import powsimp
arg = self.args[0]
k, l = Wild("k"), Wild("l")
r = arg.match(k * x**l)
if r is not None:
k, l = r[k], r[l]
if l != 0 and not l.has(x) and not k.has(x):
r = log(k) + l * log(x)
return r
order = C.Order(x**n, x)
arg = self.args[0]
x = order.symbols[0]
ln = C.log
use_lt = not C.Order(1, x).contains(arg)
if not use_lt:
arg0 = arg.limit(x, 0)
use_lt = (arg0 is S.Zero)
if use_lt: # singularity, #example: self = log(sin(x))
# arg = (arg / lt) * lt
lt = arg.as_leading_term(x) # arg = sin(x); lt = x
a = powsimp((arg / lt).expand(), deep=True,
combine='exp') # a = sin(x)/x
# the idea is to recursively call ln(a).series(), but one needs to
# make sure that ln(sin(x)/x) doesn't get "simplified" to
# -log(x)+ln(sin(x)) and an infinite recursion occurs, see also the
# issue 252.
obj = ln(lt) + ln(a)._eval_nseries(x, x0, n)
else:
# arg -> arg0 + (arg - arg0) -> arg0 * (1 + (arg/arg0 - 1))
z = (arg / arg0 - 1)
x = order.symbols[0]
ln = C.log
o = C.Order(z, x)
if o is S.Zero:
return ln(1 + z) + ln(arg0)
if o.expr.is_number:
e = ln(order.expr * x) / ln(x)
else:
e = ln(order.expr) / ln(o.expr)
n = e.limit(x, 0) + 1
if n.is_unbounded:
# requested accuracy gives infinite series,
# order is probably nonpolynomial e.g. O(exp(-1/x), x).
return ln(1 + z) + ln(arg0)
try:
n = int(n)
except TypeError:
#well, the n is something more complicated (like 1+log(2))
n = int(n.evalf()) + 1
assert n >= 0, ` n `
l = []
g = None
for i in xrange(n + 2):
g = ln.taylor_term(i, z, g)
g = g.nseries(x, x0, n)
l.append(g)
obj = C.Add(*l) + ln(arg0)
obj2 = expand_log(powsimp(obj, deep=True, combine='exp'))
if obj2 != obj:
r = obj2.nseries(x, x0, n)
else:
r = obj
if r == self:
return self
return r + order
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi / 2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2 * i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_bounded and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex) * S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i * limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo * i
else:
if p >= 0:
return S.Zero
else:
return -oo * i
if e.is_Add:
if e.is_polynomial() and not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []
unbounded_result = []
finite = []
unknown = []
ok = True
for term in e.args:
if not term.has(z) and not term.is_unbounded:
finite.append(term)
continue
result = term.subs(z, z0)
bounded = result.is_bounded
if bounded is False or result is S.NaN:
if unknown:
ok = False
break
unbounded.append(term)
if result != S.NaN:
# take result from direction given
result = limit(term, z, z0, dir)
unbounded_result.append(result)
elif bounded:
finite.append(result)
else:
if unbounded:
ok = False
break
unknown.append(result)
if not ok:
# we won't be able to resolve this with unbounded
# terms, e.g. Sum(1/k, (k, 1, n)) - log(n) as n -> oo:
# since the Sum is unevaluated it's boundedness is
# unknown and the log(n) is oo so you get Sum - oo
# which is unsatisfactory.
raise NotImplementedError('unknown boundedness for %s' %
(unknown or result))
u = Add(*unknown)
if unbounded:
inf_limit = Add(*unbounded_result)
if inf_limit is not S.NaN:
return inf_limit + u
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir) + u
else:
return Add(*finite) + u
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except PoleError:
r = heuristics(e, z, z0, dir)
return r
def limit(e, z, z0, dir="+"):
"""
Compute the limit of e(z) at the point z0.
z0 can be any expression, including oo and -oo.
For dir="+" (default) it calculates the limit from the right
(z->z0+) and for dir="-" the limit from the left (z->z0-). For infinite z0
(oo or -oo), the dir argument doesn't matter.
Examples
========
>>> from sympy import limit, sin, Symbol, oo
>>> from sympy.abc import x
>>> limit(sin(x)/x, x, 0)
1
>>> limit(1/x, x, 0, dir="+")
oo
>>> limit(1/x, x, 0, dir="-")
-oo
>>> limit(1/x, x, oo)
0
Notes
=====
First we try some heuristics for easy and frequent cases like "x", "1/x",
"x**2" and similar, so that it's fast. For all other cases, we use the
Gruntz algorithm (see the gruntz() function).
"""
from sympy import Wild, log
e = sympify(e)
z = sympify(z)
z0 = sympify(z0)
if e == z:
return z0
if e.is_Rational:
return e
if not e.has(z):
return e
# gruntz fails on factorials but works with the gamma function
# If no factorial term is present, e should remain unchanged.
# factorial is defined to be zero for negative inputs (which
# differs from gamma) so only rewrite for positive z0.
if z0.is_positive:
e = e.rewrite(factorial, gamma)
if e.func is tan:
# discontinuity at odd multiples of pi/2; 0 at even
disc = S.Pi / 2
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if i.is_even:
return S.Zero
elif i.is_odd:
if dir == '+':
return S.NegativeInfinity
else:
return S.Infinity
if e.func is cot:
# discontinuity at multiples of pi; 0 at odd pi/2 multiples
disc = S.Pi
sign = 1
if dir == '-':
sign *= -1
i = limit(sign * e.args[0], z, z0) / disc
if i.is_integer:
if dir == '-':
return S.NegativeInfinity
else:
return S.Infinity
elif (2 * i).is_integer:
return S.Zero
if e.is_Pow:
b, ex = e.args
c = None # records sign of b if b is +/-z or has a bounded value
if b.is_Mul:
c, b = b.as_two_terms()
if c is S.NegativeOne and b == z:
c = '-'
elif b == z:
c = '+'
if ex.is_number:
if c is None:
base = b.subs(z, z0)
if base.is_finite and (ex.is_bounded or base is not S.One):
return base**ex
else:
if z0 == 0 and ex < 0:
if dir != c:
# integer
if ex.is_even:
return S.Infinity
elif ex.is_odd:
return S.NegativeInfinity
# rational
elif ex.is_Rational:
return (S.NegativeOne**ex) * S.Infinity
else:
return S.ComplexInfinity
return S.Infinity
return z0**ex
if e.is_Mul or not z0 and e.is_Pow and b.func is log:
if e.is_Mul:
if abs(z0) is S.Infinity:
n, d = e.as_numer_denom()
# XXX todo: this should probably be stated in the
# negative -- i.e. to exclude expressions that should
# not be handled this way but I'm not sure what that
# condition is; when ok is True it means that the leading
# term approach is going to succeed (hopefully)
ok = lambda w: (z in w.free_symbols and any(
a.is_polynomial(z) or any(
z in m.free_symbols and m.is_polynomial(z)
for m in Mul.make_args(a)) for a in Add.make_args(w)))
if all(ok(w) for w in (n, d)):
u = C.Dummy(positive=(z0 is S.Infinity))
inve = (n / d).subs(z, 1 / u)
return limit(inve.as_leading_term(u), u, S.Zero,
"+" if z0 is S.Infinity else "-")
# weed out the z-independent terms
i, d = e.as_independent(z)
if i is not S.One and i.is_bounded:
return i * limit(d, z, z0, dir)
else:
i, d = S.One, e
if not z0:
# look for log(z)**q or z**p*log(z)**q
p, q = Wild("p"), Wild("q")
r = d.match(z**p * log(z)**q)
if r:
p, q = [r.get(w, w) for w in [p, q]]
if q and q.is_number and p.is_number:
if q > 0:
if p > 0:
return S.Zero
else:
return -oo * i
else:
if p >= 0:
return S.Zero
else:
return -oo * i
if e.is_Add:
if e.is_polynomial() and not z0.is_unbounded:
return Add(*[limit(term, z, z0, dir) for term in e.args])
# this is a case like limit(x*y+x*z, z, 2) == x*y+2*x
# but we need to make sure, that the general gruntz() algorithm is
# executed for a case like "limit(sqrt(x+1)-sqrt(x),x,oo)==0"
unbounded = []
unbounded_result = []
unbounded_const = []
unknown = []
unknown_result = []
finite = []
zero = []
def _sift(term):
if z not in term.free_symbols:
if term.is_unbounded:
unbounded_const.append(term)
else:
finite.append(term)
else:
result = term.subs(z, z0)
bounded = result.is_bounded
if bounded is False or result is S.NaN:
unbounded.append(term)
if result != S.NaN:
# take result from direction given
result = limit(term, z, z0, dir)
unbounded_result.append(result)
elif bounded:
if result:
finite.append(result)
else:
zero.append(term)
else:
unknown.append(term)
unknown_result.append(result)
for term in e.args:
_sift(term)
bad = bool(unknown and unbounded)
if bad or len(unknown) > 1 or len(unbounded) > 1 and not zero:
uu = unknown + unbounded
# we won't be able to resolve this with unbounded
# terms, e.g. Sum(1/k, (k, 1, n)) - log(n) as n -> oo:
# since the Sum is unevaluated it's boundedness is
# unknown and the log(n) is oo so you get Sum - oo
# which is unsatisfactory. BUT...if there are both
# unknown and unbounded terms (condition 'bad') or
# there are multiple terms that are unknown, or
# there are multiple symbolic unbounded terms they may
# respond better if they are made into a rational
# function, so give them a chance to do so before
# reporting failure.
u = Add(*uu)
f = u.normal()
if f != u:
unknown = []
unbounded = []
unbounded_result = []
unknown_result = []
_sift(limit(f, z, z0, dir))
# We came in with a) unknown and unbounded terms or b) had multiple
# unknown terms
# At this point we've done one of 3 things.
# (1) We did nothing with f so we now report the error
# showing the troublesome terms which are now in uu. OR
# (2) We did something with f but the result came back as unknown.
# Normally this wouldn't be a problem,
# but we had either multiple terms that were troublesome (unk and
# unbounded or multiple unknown terms) so if we
# weren't able to resolve the boundedness by now, that indicates a
# problem so we report the error showing the troublesome terms which are
# now in uu.
if unknown:
if bad:
msg = 'unknown and unbounded terms present in %s'
elif unknown:
msg = 'multiple terms with unknown boundedness in %s'
raise NotImplementedError(msg % uu)
# OR
# (3) the troublesome terms have been identified as finite or unbounded
# and we proceed with the non-error code since the lists have been updated.
u = Add(*unknown_result)
if unbounded_result or unbounded_const:
unbounded.extend(zero)
inf_limit = Add(*(unbounded_result + unbounded_const))
if inf_limit is not S.NaN:
return inf_limit + u
if finite:
return Add(*finite) + limit(Add(*unbounded), z, z0, dir) + u
else:
return Add(*finite) + u
if e.is_Order:
args = e.args
return C.Order(limit(args[0], z, z0), *args[1:])
try:
r = gruntz(e, z, z0, dir)
if r is S.NaN:
raise PoleError()
except (PoleError, ValueError):
r = heuristics(e, z, z0, dir)
return r
