"""Limits of sequences"""
from __future__ import print_function, division
from sympy.core.sympify import sympify
from sympy.core.singleton import S
from sympy.core.add import Add
from sympy.core.power import Pow
from sympy.core.symbol import Dummy
from sympy.core.function import PoleError
from sympy.series.limits import Limit
[docs]def difference_delta(expr, n=None, step=1):
"""Difference Operator.
Discrete analog of differential operator. Given a sequence x[n],
returns the sequence x[n + step] - x[n].
Examples
========
>>> from sympy import difference_delta as dd
>>> from sympy.abc import n
>>> dd(n*(n + 1), n)
2*n + 2
>>> dd(n*(n + 1), n, 2)
4*n + 6
References
==========
.. [1] https://reference.wolfram.com/language/ref/DifferenceDelta.html
"""
expr = sympify(expr)
if n is None:
f = expr.free_symbols
if len(f) == 1:
n = f.pop()
elif len(f) == 0:
return S.Zero
else:
raise ValueError("Since there is more than one variable in the"
" expression, a variable must be supplied to"
" take the difference of %s" % expr)
step = sympify(step)
if step.is_number is False or step.is_finite is False:
raise ValueError("Step should be a finite number.")
if hasattr(expr, '_eval_difference_delta'):
result = expr._eval_difference_delta(n, step)
if result:
return result
return expr.subs(n, n + step) - expr
[docs]def dominant(expr, n):
"""Finds the dominant term in a sum, that is a term that dominates
every other term.
If limit(a/b, n, oo) is oo then a dominates b.
If limit(a/b, n, oo) is 0 then b dominates a.
Otherwise, a and b are comparable.
If there is no unique dominant term, then returns ``None``.
Examples
========
>>> from sympy import Sum
>>> from sympy.series.limitseq import dominant
>>> from sympy.abc import n, k
>>> dominant(5*n**3 + 4*n**2 + n + 1, n)
5*n**3
>>> dominant(2**n + Sum(k, (k, 0, n)), n)
2**n
See Also
========
sympy.series.limitseq.dominant
"""
terms = Add.make_args(expr.expand(func=True))
term0 = terms[-1]
comp = [term0] # comparable terms
for t in terms[:-1]:
e = (term0 / t).gammasimp()
l = limit_seq(e, n)
if l is S.Zero:
term0 = t
comp = [term0]
elif l is None:
return None
elif l not in [S.Infinity, -S.Infinity]:
comp.append(t)
if len(comp) > 1:
return None
return term0
def _limit_inf(expr, n):
try:
return Limit(expr, n, S.Infinity).doit(deep=False, sequence=False)
except (NotImplementedError, PoleError):
return None
def _limit_seq(expr, n, trials):
from sympy.concrete.summations import Sum
for i in range(trials):
if not expr.has(Sum):
result = _limit_inf(expr, n)
if result is not None:
return result
num, den = expr.as_numer_denom()
if not den.has(n) or not num.has(n):
result = _limit_inf(expr.doit(), n)
if result is not None:
return result
return None
num, den = (difference_delta(t.expand(), n) for t in [num, den])
expr = (num / den).gammasimp()
if not expr.has(Sum):
result = _limit_inf(expr, n)
if result is not None:
return result
num, den = expr.as_numer_denom()
num = dominant(num, n)
if num is None:
return None
den = dominant(den, n)
if den is None:
return None
expr = (num / den).gammasimp()
[docs]def limit_seq(expr, n=None, trials=5):
"""Finds limits of terms having sequences at infinity.
Parameters
==========
expr : Expr
SymPy expression for the n-th term of the sequence
n : Symbol
The index of the sequence, an integer that tends to positive infinity.
trials: int, optional
The algorithm is highly recursive. ``trials`` is a safeguard from
infinite recursion in case the limit is not easily computed by the
algorithm. Try increasing ``trials`` if the algorithm returns ``None``.
Admissible Terms
================
The algorithm is designed for sequences built from rational functions,
indefinite sums, and indefinite products over an indeterminate n. Terms of
alternating sign are also allowed, but more complex oscillatory behavior is
not supported.
Examples
========
>>> from sympy import limit_seq, Sum, binomial
>>> from sympy.abc import n, k, m
>>> limit_seq((5*n**3 + 3*n**2 + 4) / (3*n**3 + 4*n - 5), n)
5/3
>>> limit_seq(binomial(2*n, n) / Sum(binomial(2*k, k), (k, 1, n)), n)
3/4
>>> limit_seq(Sum(k**2 * Sum(2**m/m, (m, 1, k)), (k, 1, n)) / (2**n*n), n)
4
See Also
========
sympy.series.limitseq.dominant
References
==========
.. [1] Computing Limits of Sequences - Manuel Kauers
"""
if n is None:
free = expr.free_symbols
if len(free) == 1:
n = free.pop()
elif not free:
return expr
else:
raise ValueError("expr %s has more than one variables. Please"
"specify a variable." % (expr))
elif n not in expr.free_symbols:
return expr
n_ = Dummy("n", integer=True, positive=True)
# If there is a negative term raised to a power involving n, consider
# even and odd n separately.
powers = (p.as_base_exp() for p in expr.atoms(Pow))
if any(b.is_negative and e.has(n) for b, e in powers):
L1 = _limit_seq(expr.xreplace({n: 2*n_}), n_, trials)
if L1 is not None:
L2 = _limit_seq(expr.xreplace({n: 2*n_ + 1}), n_, trials)
if L1 == L2:
return L1
else:
return _limit_seq(expr.xreplace({n: n_}), n_, trials)