def test_idiff():
# the use of idiff in ellipse also provides coverage
circ = x**2 + y**2 - 4
ans = -3*x*(x**2 + y**2)/y**5
assert ans == idiff(circ, y, x, 3).simplify()
assert ans == idiff(circ, [y], x, 3).simplify()
assert idiff(circ, y, x, 3).simplify() == ans
explicit = 12*x/sqrt(-x**2 + 4)**5
assert ans.subs(y, solve(circ, y)[0]).equals(explicit)
assert True in [sol.diff(x, 3).equals(explicit) for sol in solve(circ, y)]
assert idiff(x + t + y, [y, t], x) == -Derivative(t, x) - 1
def test_issue_8235():
assert reduce_inequalities(x**2 - 1 < 0) == \
And(Integer(-1) < x, x < Integer(1))
assert reduce_inequalities(x**2 - 1 <= 0) == \
And(Integer(-1) <= x, x <= 1)
assert reduce_inequalities(x**2 - 1 > 0) == \
Or(And(-oo < x, x < -1), And(x < oo, Integer(1) < x))
assert reduce_inequalities(x**2 - 1 >= 0) == \
Or(And(-oo < x, x <= Integer(-1)), And(Integer(1) <= x, x < oo))
eq = x**8 + x - 9 # we want RootOf solns here
sol = solve(eq >= 0)
tru = Or(And(RootOf(eq, 1) <= x, x < oo), And(-oo < x, x <= RootOf(eq, 0)))
assert sol == tru
# recast vanilla as real
assert solve(sqrt((-x + 1)**2) < 1) == And(Integer(0) < x, x < 2)
def test_diofantissue_453():
x = Symbol('x', real=True)
assert isolve(abs((x - 1) / (x - 5)) <= Rational(1, 3),
x) == And(Integer(-1) <= x, x <= 2)
assert solve(abs((x - 1) / (x - 5)) - Rational(1, 3), x) == [{
x: -1
}, {
x: 2
}]
def solve_univariate_inequality(expr, gen, relational=True):
"""
Solves a real univariate inequality.
Examples
========
>>> from diofant.solvers.inequalities import solve_univariate_inequality
>>> from diofant.core.symbol import Symbol
>>> x = Symbol('x', real=True)
>>> solve_univariate_inequality(x**2 >= 4, x)
Or(2 <= x, x <= -2)
>>> solve_univariate_inequality(x**2 >= 4, x, relational=False)
(-oo, -2] U [2, oo)
"""
from diofant.simplify.simplify import simplify
from diofant.solvers.solvers import solve, denoms
e = expr.lhs - expr.rhs
parts = n, d = e.as_numer_denom()
if all(i.is_polynomial(gen) for i in parts):
solns = solve(n, gen, check=False)
singularities = solve(d, gen, check=False)
else:
solns = solve(e, gen, check=False)
singularities = []
for d in denoms(e):
singularities.extend(solve(d, gen))
include_x = expr.func(0, 0)
def valid(x):
v = e.subs(gen, x)
try:
r = expr.func(v, 0)
except TypeError:
r = S.false
r = simplify(r)
if r in (S.true, S.false):
return r
if v.is_extended_real is False:
return S.false
else:
if v.is_comparable:
v = v.n(2)
if v._prec > 1:
return expr.func(v, 0)
elif v.is_comparable is False:
return False
raise NotImplementedError
start = S.NegativeInfinity
sol_sets = [S.EmptySet]
try:
reals = _nsort(set(solns + singularities), separated=True)[0]
except NotImplementedError:
raise NotImplementedError('sorting of these roots is not supported')
for x in reals:
end = x
if end in [S.NegativeInfinity, S.Infinity]:
if valid(Integer(0)):
sol_sets.append(Interval(start, S.Infinity, True, True))
break
if valid((start + end) / 2 if start != S.NegativeInfinity else end -
1):
sol_sets.append(Interval(start, end, True, True))
if x in singularities:
singularities.remove(x)
elif include_x:
sol_sets.append(FiniteSet(x))
start = end
end = S.Infinity
if valid(start + 1):
sol_sets.append(Interval(start, end, True, True))
rv = Union(*sol_sets)
return rv if not relational else rv.as_relational(gen)
def transform(self, x, u):
r"""
Performs a change of variables from `x` to `u` using the relationship
given by `x` and `u` which will define the transformations `f` and `F`
(which are inverses of each other) as follows:
1) If `x` is a Symbol (which is a variable of integration) then `u`
will be interpreted as some function, f(u), with inverse F(u).
This, in effect, just makes the substitution of x with f(x).
2) If `u` is a Symbol then `x` will be interpreted as some function,
F(x), with inverse f(u). This is commonly referred to as
u-substitution.
Once f and F have been identified, the transformation is made as
follows:
.. math:: \int_a^b x \mathrm{d}x \rightarrow \int_{F(a)}^{F(b)} f(x)
\frac{\mathrm{d}}{\mathrm{d}x}
where `F(x)` is the inverse of `f(x)` and the limits and integrand have
been corrected so as to retain the same value after integration.
Notes
=====
The mappings, F(x) or f(u), must lead to a unique integral. Linear
or rational linear expression, `2*x`, `1/x` and `sqrt(x)`, will
always work; quadratic expressions like `x**2 - 1` are acceptable
as long as the resulting integrand does not depend on the sign of
the solutions (see examples).
The integral will be returned unchanged if `x` is not a variable of
integration.
`x` must be (or contain) only one of of the integration variables. If
`u` has more than one free symbol then it should be sent as a tuple
(`u`, `uvar`) where `uvar` identifies which variable is replacing
the integration variable.
XXX can it contain another integration variable?
Examples
========
>>> from diofant.abc import a, b, c, d, x, u, y
>>> from diofant import Integral, S, cos, sqrt
>>> i = Integral(x*cos(x**2 - 1), (x, 0, 1))
transform can change the variable of integration
>>> i.transform(x, u)
Integral(u*cos(u**2 - 1), (u, 0, 1))
transform can perform u-substitution as long as a unique
integrand is obtained:
>>> i.transform(x**2 - 1, u)
Integral(cos(u)/2, (u, -1, 0))
This attempt fails because x = +/-sqrt(u + 1) and the
sign does not cancel out of the integrand:
>>> Integral(cos(x**2 - 1), (x, 0, 1)).transform(x**2 - 1, u)
Traceback (most recent call last):
...
ValueError:
The mapping between F(x) and f(u) did not give a unique integrand.
transform can do a substitution. Here, the previous
result is transformed back into the original expression
using "u-substitution":
>>> ui = _
>>> _.transform(sqrt(u + 1), x) == i
True
We can accomplish the same with a regular substitution:
>>> ui.transform(u, x**2 - 1) == i
True
If the `x` does not contain a symbol of integration then
the integral will be returned unchanged. Integral `i` does
not have an integration variable `a` so no change is made:
>>> i.transform(a, x) == i
True
When `u` has more than one free symbol the symbol that is
replacing `x` must be identified by passing `u` as a tuple:
>>> Integral(x, (x, 0, 1)).transform(x, (u + a, u))
Integral(a + u, (u, -a, -a + 1))
>>> Integral(x, (x, 0, 1)).transform(x, (u + a, a))
Integral(a + u, (a, -u, -u + 1))
See Also
========
diofant.concrete.expr_with_limits.ExprWithLimits.variables : Lists the integration variables
diofant.concrete.expr_with_limits.ExprWithLimits.as_dummy : Replace integration variables with dummy ones
"""
from diofant.solvers.solvers import solve, posify
d = Dummy('d')
xfree = x.free_symbols.intersection(self.variables)
if len(xfree) > 1:
raise ValueError('F(x) can only contain one of: %s' %
self.variables)
xvar = xfree.pop() if xfree else d
if xvar not in self.variables:
return self
u = sympify(u)
if isinstance(u, Expr):
ufree = u.free_symbols
if len(ufree) != 1:
raise ValueError(
filldedent('''
When f(u) has more than one free symbol, the one replacing x
must be identified: pass f(u) as (f(u), u)'''))
uvar = ufree.pop()
else:
u, uvar = u
if uvar not in u.free_symbols:
raise ValueError(
filldedent('''
Expecting a tuple (expr, symbol) where symbol identified
a free symbol in expr, but symbol is not in expr's free
symbols.'''))
if not isinstance(uvar, (Dummy, Symbol)):
raise ValueError(
filldedent('''
Expecting a tuple (expr, symbol) but didn't get
a symbol; got %s''' % uvar))
if x.is_Symbol and u.is_Symbol:
return self.xreplace({x: u})
if not x.is_Symbol and not u.is_Symbol:
raise ValueError('either x or u must be a symbol')
if uvar == xvar:
return self.transform(x, (u.subs(uvar, d), d)).xreplace({d: uvar})
if uvar in self.limits:
raise ValueError(
filldedent('''
u must contain the same variable as in x
or a variable that is not already an integration variable'''))
if not x.is_Symbol:
F = [x.subs(xvar, d)]
soln = solve(u - x, xvar, check=False)
if not soln:
raise ValueError('no solution for solve(F(x) - f(u), x)')
f = [fi.subs(uvar, d) for fi in soln]
else:
f = [u.subs(uvar, d)]
pdiff, reps = posify(u - x)
puvar = uvar.subs([(v, k) for k, v in reps.items()])
soln = [s.subs(reps) for s in solve(pdiff, puvar)]
if not soln:
raise ValueError('no solution for solve(F(x) - f(u), u)')
F = [fi.subs(xvar, d) for fi in soln]
newfuncs = {(self.function.subs(xvar, fi) * fi.diff(d)).subs(d, uvar)
for fi in f}
if len(newfuncs) > 1:
raise ValueError(
filldedent('''
The mapping between F(x) and f(u) did not give
a unique integrand.'''))
newfunc = newfuncs.pop()
def _calc_limit_1(F, a, b):
"""
replace d with a, using subs if possible, otherwise limit
where sign of b is considered
"""
wok = F.subs(d, a)
if wok is S.NaN or wok.is_finite is False and a.is_finite:
return limit(sign(b) * F, d, a)
return wok
def _calc_limit(a, b):
"""
replace d with a, using subs if possible, otherwise limit
where sign of b is considered
"""
avals = list({_calc_limit_1(Fi, a, b) for Fi in F})
if len(avals) > 1:
raise ValueError(
filldedent('''
The mapping between F(x) and f(u) did not
give a unique limit.'''))
return avals[0]
newlimits = []
for xab in self.limits:
sym = xab[0]
if sym == xvar:
if len(xab) == 3:
a, b = xab[1:]
a, b = _calc_limit(a, b), _calc_limit(b, a)
if a - b > 0:
a, b = b, a
newfunc = -newfunc
newlimits.append((uvar, a, b))
elif len(xab) == 2:
a = _calc_limit(xab[1], 1)
newlimits.append((uvar, a))
else:
newlimits.append(uvar)
else:
newlimits.append(xab)
return self.func(newfunc, *newlimits)
def test_diofantissue_453():
x = Symbol('x', real=True)
assert isolve(abs((x - 1)/(x - 5)) <= Rational(1, 3),
x) == And(Integer(-1) <= x, x <= 2)
assert solve(abs((x - 1)/(x - 5)) - Rational(1, 3), x) == [{x: -1}, {x: 2}]
def heurisch_wrapper(f,
x,
rewrite=False,
hints=None,
mappings=None,
retries=3,
degree_offset=0,
unnecessary_permutations=None):
"""
A wrapper around the heurisch integration algorithm.
This method takes the result from heurisch and checks for poles in the
denominator. For each of these poles, the integral is reevaluated, and
the final integration result is given in terms of a Piecewise.
Examples
========
>>> from diofant.core import symbols
>>> from diofant.functions import cos
>>> from diofant.integrals.heurisch import heurisch, heurisch_wrapper
>>> n, x = symbols('n x')
>>> heurisch(cos(n*x), x)
sin(n*x)/n
>>> heurisch_wrapper(cos(n*x), x)
Piecewise((x, Eq(n, 0)), (sin(n*x)/n, true))
See Also
========
diofant.integrals.heurisch.heurisch
"""
from diofant.solvers.solvers import solve, denoms
f = sympify(f)
if x not in f.free_symbols:
return f * x
res = heurisch(f, x, rewrite, hints, mappings, retries, degree_offset,
unnecessary_permutations)
if not isinstance(res, Basic):
return res
# We consider each denominator in the expression, and try to find
# cases where one or more symbolic denominator might be zero. The
# conditions for these cases are stored in the list slns.
slns = []
for d in denoms(res):
try:
slns += solve(d, dict=True, exclude=(x, ))
except NotImplementedError:
pass
if not slns:
return res
slns = list(uniq(slns))
# Remove the solutions corresponding to poles in the original expression.
slns0 = []
for d in denoms(f):
try:
slns0 += solve(d, dict=True, exclude=(x, ))
except NotImplementedError:
pass
slns = [s for s in slns if s not in slns0]
if not slns:
return res
if len(slns) > 1:
eqs = []
for sub_dict in slns:
eqs.extend([Eq(key, value) for key, value in sub_dict.items()])
slns = solve(eqs, dict=True, exclude=(x, )) + slns
# For each case listed in the list slns, we reevaluate the integral.
pairs = []
for sub_dict in slns:
expr = heurisch(f.subs(sub_dict), x, rewrite, hints, mappings, retries,
degree_offset, unnecessary_permutations)
cond = And(*[Eq(key, value) for key, value in sub_dict.items()])
pairs.append((expr, cond))
pairs.append((heurisch(f, x, rewrite, hints, mappings, retries,
degree_offset, unnecessary_permutations), True))
return Piecewise(*pairs)
def test_slow_general_univariate():
r = RootOf(x**5 - x**2 + 1, 0)
assert solve(sqrt(x) + 1/root(x, 3) > 1) == \
Or(And(Integer(0) < x, x < r**6), And(r**6 < x, x < oo))