Source code for sympy.series.residues

"""
This module implements the Residue function and related tools for working
with residues.
"""

from __future__ import print_function, division

from sympy import sympify
from sympy.utilities.timeutils import timethis


[docs]@timethis('residue') def residue(expr, x, x0): """ Finds the residue of ``expr`` at the point x=x0. The residue is defined as the coefficient of 1/(x-x0) in the power series expansion about x=x0. Examples ======== >>> from sympy import Symbol, residue, sin >>> x = Symbol("x") >>> residue(1/x, x, 0) 1 >>> residue(1/x**2, x, 0) 0 >>> residue(2/sin(x), x, 0) 2 This function is essential for the Residue Theorem [1]. References ========== 1. http://en.wikipedia.org/wiki/Residue_theorem """ # The current implementation uses series expansion to # calculate it. A more general implementation is explained in # the section 5.6 of the Bronstein's book {M. Bronstein: # Symbolic Integration I, Springer Verlag (2005)}. For purely # rational functions, the algorithm is much easier. See # sections 2.4, 2.5, and 2.7 (this section actually gives an # algorithm for computing any Laurent series coefficient for # a rational function). The theory in section 2.4 will help to # understand why the resultant works in the general algorithm. # For the definition of a resultant, see section 1.4 (and any # previous sections for more review). from sympy import collect, Mul, Order, S expr = sympify(expr) if x0 != 0: expr = expr.subs(x, x + x0) for n in [0, 1, 2, 4, 8, 16, 32]: if n == 0: s = expr.series(x, n=0) else: s = expr.nseries(x, n=n) if s.has(Order) and s.removeO() == 0: # bug in nseries continue if not s.has(Order) or s.getn() >= 0: break if s.has(Order) and s.getn() < 0: raise NotImplementedError('Bug in nseries?') s = collect(s.removeO(), x) if s.is_Add: args = s.args else: args = [s] res = S(0) for arg in args: c, m = arg.as_coeff_mul(x) m = Mul(*m) if not (m == 1 or m == x or (m.is_Pow and m.exp.is_Integer)): raise NotImplementedError('term of unexpected form: %s' % m) if m == 1/x: res += c return res