def test_RootSum():
f = x**5 + 2 * x - 1
assert str(RootSum(f, Lambda(z, z),
auto=False)) == "RootSum(x**5 + 2*x - 1)"
assert str(RootSum(f, Lambda(
z, z**2), auto=False)) == "RootSum(x**5 + 2*x - 1, Lambda(z, z**2))"
def test_RootSum():
r = RootSum(x**3 + x + 3, Lambda(y, log(y*z)))
assert mcode(r) == ("RootSum[Function[{x}, x^3 + x + 3], "
"Function[{y}, Log[y*z]]]")
def assemble_partfrac_list(partial_list):
r"""Reassemble a full partial fraction decomposition
from a structured result obtained by the function ``apart_list``.
Examples
========
This example is taken from Bronstein's original paper:
>>> from diofant.abc import x, y
>>> f = 36 / (x**5 - 2*x**4 - 2*x**3 + 4*x**2 + x - 2)
>>> pfd = apart_list(f)
>>> pfd
(1,
Poly(0, x, domain='ZZ'),
[(Poly(_w - 2, _w, domain='ZZ'), Lambda(_a, 4), Lambda(_a, x - _a), 1),
(Poly(_w**2 - 1, _w, domain='ZZ'), Lambda(_a, -3*_a - 6), Lambda(_a, x - _a), 2),
(Poly(_w + 1, _w, domain='ZZ'), Lambda(_a, -4), Lambda(_a, x - _a), 1)])
>>> assemble_partfrac_list(pfd)
-4/(x + 1) - 3/(x + 1)**2 - 9/(x - 1)**2 + 4/(x - 2)
If we happen to know some roots we can provide them easily inside the structure:
>>> pfd = apart_list(2/(x**2-2))
>>> pfd
(1,
Poly(0, x, domain='ZZ'),
[(Poly(_w**2 - 2, _w, domain='ZZ'),
Lambda(_a, _a/2), Lambda(_a, x - _a),
1)])
>>> pfda = assemble_partfrac_list(pfd)
>>> pfda
RootSum(_w**2 - 2, Lambda(_a, _a/(x - _a)))/2
>>> pfda.doit()
-sqrt(2)/(2*(x + sqrt(2))) + sqrt(2)/(2*(x - sqrt(2)))
>>> from diofant import Dummy, Poly, Lambda, sqrt
>>> a = Dummy("a")
>>> pfd = (1, Poly(0, x, domain='ZZ'), [([sqrt(2),-sqrt(2)], Lambda(a, a/2), Lambda(a, -a + x), 1)])
>>> assemble_partfrac_list(pfd)
-sqrt(2)/(2*(x + sqrt(2))) + sqrt(2)/(2*(x - sqrt(2)))
See also
========
apart, apart_list
"""
# Common factor
common = partial_list[0]
# Polynomial part
polypart = partial_list[1]
pfd = polypart.as_expr()
# Rational parts
for r, nf, df, ex in partial_list[2]:
if isinstance(r, Poly):
# Assemble in case the roots are given implicitly by a polynomials
an, nu = nf.variables, nf.expr
ad, de = df.variables, df.expr
# Hack to make dummies equal because Lambda created new Dummies
de = de.subs(ad[0], an[0])
func = Lambda(an, nu / de**ex)
pfd += RootSum(r, func, auto=False, quadratic=False)
else:
# Assemble in case the roots are given explicitely by a list of algebraic numbers
for root in r:
pfd += nf(root) / df(root)**ex
return common * pfd