def checksol(f, symbol, sol=None, **flags):
"""Checks whether sol is a solution of equation f == 0.
Input can be either a single symbol and corresponding value
or a dictionary of symbols and values. ``f`` can be a single
equation or an iterable of equations. A solution must satisfy
all equations in ``f`` to be considered valid; if a solution
does not satisfy any equation, False is returned; if one or
more checks are inconclusive (and none are False) then None
is returned.
Examples:
---------
>>> from sympy import symbols
>>> from sympy.solvers import checksol
>>> x, y = symbols('x,y')
>>> checksol(x**4-1, x, 1)
True
>>> checksol(x**4-1, x, 0)
False
>>> checksol(x**2 + y**2 - 5**2, {x:3, y: 4})
True
None is returned if checksol() could not conclude.
flags:
'numerical=True (default)'
do a fast numerical check if f has only one symbol.
'minimal=True (default is False)'
a very fast, minimal testing.
'warning=True (default is False)'
print a warning if checksol() could not conclude.
'simplify=True (default)'
simplify solution before substituting into function (which
will then be simplified regardless of the flag setting)
'force=True (default is False)'
make positive all symbols without assumptions regarding sign.
"""
if sol is not None:
sol = {symbol: sol}
elif isinstance(symbol, dict):
sol = symbol
else:
msg = 'Expecting sym, val or {sym: val}, None but got %s, %s'
raise ValueError(msg % (symbol, sol))
if is_sequence(f):
if not f:
raise ValueError('no functions to check')
rv = True
for fi in f:
check = checksol(fi, sol, **flags)
if check:
continue
if check is False:
return False
rv = None # don't return, wait to see if there's a False
return rv
if isinstance(f, Poly):
f = f.as_expr()
elif isinstance(f, Equality):
f = f.lhs - f.rhs
if not f:
return True
if not f.has(*sol.keys()):
return False
attempt = -1
numerical = flags.get('numerical', True)
while 1:
attempt += 1
if attempt == 0:
val = f.subs(sol)
elif attempt == 1:
if not val.atoms(Symbol) and numerical:
# val is a constant, so a fast numerical test may suffice
if val not in [S.Infinity, S.NegativeInfinity]:
# issue 2088 shows that +/-oo chops to 0
val = val.evalf(36).n(30, chop=True)
elif attempt == 2:
if flags.get('minimal', False):
return
# the flag 'simplify=False' is used in solve to avoid
# simplifying the solution. So if it is set to False there
# the simplification will not be attempted here, either. But
# if the simplification is done here then the flag should be
# set to False so it isn't done again there.
# FIXME: this can't work, since `flags` is not passed to
# `checksol()` as a dict, but as keywords.
# So, any modification to `flags` here will be lost when returning
# from `checksol()`.
for k in sol:
sol[k] = simplify(sympify(sol[k]))
val = simplify(f.subs(sol))
if flags.get('force', False):
val = posify(val)[0]
elif attempt == 3:
val = powsimp(val)
elif attempt == 4:
val = cancel(val)
elif attempt == 5:
val = val.expand()
elif attempt == 6:
val = together(val)
elif attempt == 7:
val = powsimp(val)
else:
break
if val.is_zero:
return True
elif attempt > 0 and numerical and val.is_nonzero:
return False
if flags.get('warning', False):
print("\n\tWarning: could not verify solution %s." % sol)
def checkodesol(ode, sol, func=None, order="auto", solve_for_func=True):
r"""
Substitutes ``sol`` into ``ode`` and checks that the result is ``0``.
This only works when ``func`` is one function, like `f(x)`. ``sol`` can
be a single solution or a list of solutions. Each solution may be an
:py:class:`~sympy.core.relational.Equality` that the solution satisfies,
e.g. ``Eq(f(x), C1), Eq(f(x) + C1, 0)``; or simply an
:py:class:`~sympy.core.expr.Expr`, e.g. ``f(x) - C1``. In most cases it
will not be necessary to explicitly identify the function, but if the
function cannot be inferred from the original equation it can be supplied
through the ``func`` argument.
If a sequence of solutions is passed, the same sort of container will be
used to return the result for each solution.
It tries the following methods, in order, until it finds zero equivalence:
1. Substitute the solution for `f` in the original equation. This only
works if ``ode`` is solved for `f`. It will attempt to solve it first
unless ``solve_for_func == False``.
2. Take `n` derivatives of the solution, where `n` is the order of
``ode``, and check to see if that is equal to the solution. This only
works on exact ODEs.
3. Take the 1st, 2nd, ..., `n`\th derivatives of the solution, each time
solving for the derivative of `f` of that order (this will always be
possible because `f` is a linear operator). Then back substitute each
derivative into ``ode`` in reverse order.
This function returns a tuple. The first item in the tuple is ``True`` if
the substitution results in ``0``, and ``False`` otherwise. The second
item in the tuple is what the substitution results in. It should always
be ``0`` if the first item is ``True``. Sometimes this function will
return ``False`` even when an expression is identically equal to ``0``.
This happens when :py:meth:`~sympy.simplify.simplify.simplify` does not
reduce the expression to ``0``. If an expression returned by this
function vanishes identically, then ``sol`` really is a solution to
the ``ode``.
If this function seems to hang, it is probably because of a hard
simplification.
To use this function to test, test the first item of the tuple.
Examples
========
>>> from sympy import Eq, Function, checkodesol, symbols
>>> x, C1 = symbols('x,C1')
>>> f = Function('f')
>>> checkodesol(f(x).diff(x), Eq(f(x), C1))
(True, 0)
>>> assert checkodesol(f(x).diff(x), C1)[0]
>>> assert not checkodesol(f(x).diff(x), x)[0]
>>> checkodesol(f(x).diff(x, 2), x**2)
(False, 2)
"""
if not isinstance(ode, Equality):
ode = Eq(ode, 0)
if func is None:
try:
_, func = _preprocess(ode.lhs)
except ValueError:
funcs = [
s.atoms(AppliedUndef)
for s in (sol if is_sequence(sol, set) else [sol])
]
funcs = set().union(*funcs)
if len(funcs) != 1:
raise ValueError(
"must pass func arg to checkodesol for this case.")
func = funcs.pop()
if not isinstance(func, AppliedUndef) or len(func.args) != 1:
raise ValueError("func must be a function of one variable, not %s" %
func)
if is_sequence(sol, set):
return type(sol)([
checkodesol(ode, i, order=order, solve_for_func=solve_for_func)
for i in sol
])
if not isinstance(sol, Equality):
sol = Eq(func, sol)
elif sol.rhs == func:
sol = sol.reversed
if order == "auto":
order = ode_order(ode, func)
solved = sol.lhs == func and not sol.rhs.has(func)
if solve_for_func and not solved:
rhs = solve(sol, func)
if rhs:
eqs = [Eq(func, t) for t in rhs]
if len(rhs) == 1:
eqs = eqs[0]
return checkodesol(ode, eqs, order=order, solve_for_func=False)
x = func.args[0]
# Handle series solutions here
if sol.has(Order):
assert sol.lhs == func
Oterm = sol.rhs.getO()
solrhs = sol.rhs.removeO()
Oexpr = Oterm.expr
assert isinstance(Oexpr, Pow)
sorder = Oexpr.exp
assert Oterm == Order(x**sorder)
odesubs = (ode.lhs - ode.rhs).subs(func, solrhs).doit().expand()
neworder = Order(x**(sorder - order))
odesubs = odesubs + neworder
assert odesubs.getO() == neworder
residual = odesubs.removeO()
return (residual == 0, residual)
s = True
testnum = 0
while s:
if testnum == 0:
# First pass, try substituting a solved solution directly into the
# ODE. This has the highest chance of succeeding.
ode_diff = ode.lhs - ode.rhs
if sol.lhs == func:
s = sub_func_doit(ode_diff, func, sol.rhs)
s = besselsimp(s)
else:
testnum += 1
continue
ss = simplify(s.rewrite(exp))
if ss:
# with the new numer_denom in power.py, if we do a simple
# expansion then testnum == 0 verifies all solutions.
s = ss.expand(force=True)
else:
s = 0
testnum += 1
elif testnum == 1:
# Second pass. If we cannot substitute f, try seeing if the nth
# derivative is equal, this will only work for odes that are exact,
# by definition.
s = simplify(
trigsimp(diff(sol.lhs, x, order) - diff(sol.rhs, x, order)) -
trigsimp(ode.lhs) + trigsimp(ode.rhs))
# s2 = simplify(
# diff(sol.lhs, x, order) - diff(sol.rhs, x, order) - \
# ode.lhs + ode.rhs)
testnum += 1
elif testnum == 2:
# Third pass. Try solving for df/dx and substituting that into the
# ODE. Thanks to Chris Smith for suggesting this method. Many of
# the comments below are his, too.
# The method:
# - Take each of 1..n derivatives of the solution.
# - Solve each nth derivative for d^(n)f/dx^(n)
# (the differential of that order)
# - Back substitute into the ODE in decreasing order
# (i.e., n, n-1, ...)
# - Check the result for zero equivalence
if sol.lhs == func and not sol.rhs.has(func):
diffsols = {0: sol.rhs}
elif sol.rhs == func and not sol.lhs.has(func):
diffsols = {0: sol.lhs}
else:
diffsols = {}
sol = sol.lhs - sol.rhs
for i in range(1, order + 1):
# Differentiation is a linear operator, so there should always
# be 1 solution. Nonetheless, we test just to make sure.
# We only need to solve once. After that, we automatically
# have the solution to the differential in the order we want.
if i == 1:
ds = sol.diff(x)
try:
sdf = solve(ds, func.diff(x, i))
if not sdf:
raise NotImplementedError
except NotImplementedError:
testnum += 1
break
else:
diffsols[i] = sdf[0]
else:
# This is what the solution says df/dx should be.
diffsols[i] = diffsols[i - 1].diff(x)
# Make sure the above didn't fail.
if testnum > 2:
continue
else:
# Substitute it into ODE to check for self consistency.
lhs, rhs = ode.lhs, ode.rhs
for i in range(order, -1, -1):
if i == 0 and 0 not in diffsols:
# We can only substitute f(x) if the solution was
# solved for f(x).
break
lhs = sub_func_doit(lhs, func.diff(x, i), diffsols[i])
rhs = sub_func_doit(rhs, func.diff(x, i), diffsols[i])
ode_or_bool = Eq(lhs, rhs)
ode_or_bool = simplify(ode_or_bool)
if isinstance(ode_or_bool, (bool, BooleanAtom)):
if ode_or_bool:
lhs = rhs = S.Zero
else:
lhs = ode_or_bool.lhs
rhs = ode_or_bool.rhs
# No sense in overworking simplify -- just prove that the
# numerator goes to zero
num = trigsimp((lhs - rhs).as_numer_denom()[0])
# since solutions are obtained using force=True we test
# using the same level of assumptions
## replace function with dummy so assumptions will work
_func = Dummy("func")
num = num.subs(func, _func)
## posify the expression
num, reps = posify(num)
s = simplify(num).xreplace(reps).xreplace({_func: func})
testnum += 1
else:
break
if not s:
return (True, s)
elif s is True: # The code above never was able to change s
raise NotImplementedError("Unable to test if " + str(sol) +
" is a solution to " + str(ode) + ".")
else:
return (False, s)
def _tsolve(eq, sym, **flags):
"""
Helper for _solve that solves a transcendental equation with respect
to the given symbol. Various equations containing powers and logarithms,
can be solved.
Only a single solution is returned. This solution is generally
not unique. In some cases, a complex solution may be returned
even though a real solution exists.
>>> from sympy import log
>>> from sympy.solvers.solvers import _tsolve as tsolve
>>> from sympy.abc import x
>>> tsolve(3**(2*x+5)-4, x)
[(-5*log(3) + 2*log(2))/(2*log(3))]
>>> tsolve(log(x) + 2*x, x)
[LambertW(2)/2]
"""
if _patterns is None:
_generate_patterns()
eq2 = eq.subs(sym, _x)
for p, sol in _patterns:
m = eq2.match(p)
if m:
soln = sol.subs(m).subs(_x, sym)
if not(soln is S.NaN or
soln.has(S.Infinity) or
soln.has(S.NegativeInfinity) or
sym in soln.free_symbols):
return [soln]
# let's also try to invert the equation
lhs = eq
rhs = S.Zero
while True:
indep, dep = lhs.as_independent(sym)
# dep + indep == rhs
if lhs.is_Add:
# this indicates we have done it all
if indep is S.Zero:
break
lhs = dep
rhs-= indep
# dep * indep == rhs
else:
# this indicates we have done it all
if indep is S.One:
break
lhs = dep
rhs/= indep
# -1
# f(x) = g -> x = f (g)
if lhs.is_Function and lhs.nargs==1 and hasattr(lhs, 'inverse'):
rhs = lhs.inverse() (rhs)
lhs = lhs.args[0]
sol = solve(lhs-rhs, sym)
return sol
elif lhs.is_Add:
# just a simple case - we do variable substitution for first function,
# and if it removes all functions - let's call solve.
# x -x -1
# UC: e + e = y -> t + t = y
t = Dummy('t')
terms = lhs.args
# find first term which is a Function
for f1 in lhs.args:
if f1.is_Function:
ok = True
break
else:
ok = False # didn't find a function
if ok:
# perform the substitution
lhs_ = lhs.subs(f1, t)
# if no Functions left, we can proceed with usual solve
if not (lhs_.is_Function or
any(term.is_Function for term in lhs_.args)):
cv_sols = solve(lhs_ - rhs, t)
for sol in cv_sols:
if sol.has(sym):
# there is more than one function
break
else:
cv_inv = solve( t - f1, sym )[0]
sols = list()
for sol in cv_sols:
sols.append(cv_inv.subs(t, sol))
return sols
if flags.pop('posify', True):
flags['posify'] = False
pos, reps = posify(lhs)
u = sym
for u, s in reps.iteritems():
if s == sym:
break
try:
soln = _solve(pos - rhs, u, **flags)
except NotImplementedError:
return
return [s.subs(reps) for s in soln]
