Source code for sympy.series.limitseq

"""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.function import PoleError
from sympy.series.limits import Limit

[docs]def difference_delta(expr, n=None, step=1): """Difference Operator. Discrete analogous to differential operator. Examples ======== >>> from sympy import difference_delta as dd >>> from import n >>> dd(n*(n + 1), n) 2*n + 2 >>> dd(n*(n + 1), n, 2) 4*n + 6 References ========== .. [1] """ 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: raise ValueError("Step should be a number.") elif step in [S.Infinity, -S.Infinity]: raise ValueError("Step should be bounded.") 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 most dominating term in an expression. if limit(a/b, n, oo) is oo then a dominates b. if limit(a/b, n, oo) is 0 then b dominates a. else a and b are comparable. returns the most dominant term. If no unique domiant term, then returns ``None``. Examples ======== >>> from sympy import Sum >>> from sympy.series.limitseq import dominant >>> from 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).combsimp() 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
[docs]def limit_seq(expr, n=None, trials=5): """Finds limits of terms having sequences at infinity. Parameters ========== expr : Expr SymPy expression that is admissible (see section below). n : Symbol Find the limit wrt to n at infinity. trials: int, optional The algorithm is highly recursive. ``trials`` is a safeguard from infinite recursion incase limit is not easily computed by the algorithm. Try increasing ``trials`` if the algorithm returns ``None``. Admissible Terms ================ The terms should be built from rational functions, indefinite sums, and indefinite products over an indeterminate n. A term is admissible if the scope of all product quantifiers are asymptotically positive. Every admissible term is asymptoticically monotonous. Examples ======== >>> from sympy import limit_seq, Sum, binomial >>> from 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 """ from sympy.concrete.summations import Sum 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 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).combsimp() 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).combsimp()