def _solve_as_rational(f, symbol, domain):
    """ solve rational functions"""
    f = together(f, deep=True)
    g, h = fraction(f)
    if not h.has(symbol):
        return _solve_as_poly(g, symbol, domain)
    else:
        valid_solns = _solveset(g, symbol, domain)
        invalid_solns = _solveset(h, symbol, domain)
        return valid_solns - invalid_solns

def _solve_as_rational(f, symbol, solveset_solver, as_poly_solver):
    """ solve rational functions"""
    f = together(f, deep=True)
    g, h = fraction(f)
    if not h.has(symbol):
        return as_poly_solver(g, symbol)
    else:
        valid_solns = solveset_solver(g, symbol)
        invalid_solns = solveset_solver(h, symbol)
        return valid_solns - invalid_solns

def _solve_real_trig(f, symbol):
    """ Helper to solve trigonometric equations """
    f = trigsimp(f)
    f = f.rewrite(exp)
    f = together(f)
    g, h = fraction(f)
    y = Dummy('y')
    g, h = g.expand(), h.expand()
    g, h = g.subs(exp(I*symbol), y), h.subs(exp(I*symbol), y)
    if g.has(symbol) or h.has(symbol):
        raise NotImplementedError

    solns = solveset_complex(g, y) - solveset_complex(h, y)

    if isinstance(solns, FiniteSet):
        return Union(*[invert_complex(exp(I*symbol), s, symbol)[1]
                       for s in solns])
    elif solns is S.EmptySet:
        return S.EmptySet
    else:
        raise NotImplementedError

def _solve_trig(f, symbol, domain):
    """ Helper to solve trigonometric equations """
    f = trigsimp(f)
    f_original = f
    f = f.rewrite(exp)
    f = together(f)
    g, h = fraction(f)
    y = Dummy("y")
    g, h = g.expand(), h.expand()
    g, h = g.subs(exp(I * symbol), y), h.subs(exp(I * symbol), y)
    if g.has(symbol) or h.has(symbol):
        return ConditionSet(symbol, Eq(f, 0), S.Reals)

    solns = solveset_complex(g, y) - solveset_complex(h, y)

    if isinstance(solns, FiniteSet):
        result = Union(*[invert_complex(exp(I * symbol), s, symbol)[1] for s in solns])
        return Intersection(result, domain)
    elif solns is S.EmptySet:
        return S.EmptySet
    else:
        return ConditionSet(symbol, Eq(f_original, 0), S.Reals)

#.........这里部分代码省略.........

       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.
           'simplified=True (default)'
               solution should be simplified before substituting into function
               and function should be simplified after making substitution.
           '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 hasattr(f, '__iter__') and hasattr(f, '__len__'):
        if not f:
            raise ValueError('no functions to check')
        rv = set()
        for fi in f:
            check = checksol(fi, sol, **flags)
            if check is False:
                return False
            rv.add(check)
        if None in rv: # rv might contain True and/or None
            return None
        assert len(rv) == 1 # True
        return True

    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 'simplified=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()`.
            if flags.get('simplified', True):
                for k in sol:
                    sol[k] = simplify(sympify(sol[k]))
                flags['simplified'] = False
                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 _solveset(f, symbol, domain, _check=False):
    """Helper for solveset to return a result from an expression
    that has already been sympify'ed and is known to contain the
    given symbol."""
    # _check controls whether the answer is checked or not

    from sympy.simplify.simplify import signsimp
    orig_f = f
    f = together(f)
    if f.is_Mul:
        _, f = f.as_independent(symbol, as_Add=False)
    if f.is_Add:
        a, h = f.as_independent(symbol)
        m, h = h.as_independent(symbol, as_Add=False)
        f = a/m + h  # XXX condition `m != 0` should be added to soln
    f = piecewise_fold(f)

    # assign the solvers to use
    solver = lambda f, x, domain=domain: _solveset(f, x, domain)
    if domain.is_subset(S.Reals):
        inverter_func = invert_real
    else:
        inverter_func = invert_complex
    inverter = lambda f, rhs, symbol: inverter_func(f, rhs, symbol, domain)

    result = EmptySet()

    if f.expand().is_zero:
        return domain
    elif not f.has(symbol):
        return EmptySet()
    elif f.is_Mul and all(_is_finite_with_finite_vars(m, domain)
            for m in f.args):
        # if f(x) and g(x) are both finite we can say that the solution of
        # f(x)*g(x) == 0 is same as Union(f(x) == 0, g(x) == 0) is not true in
        # general. g(x) can grow to infinitely large for the values where
        # f(x) == 0. To be sure that we are not silently allowing any
        # wrong solutions we are using this technique only if both f and g are
        # finite for a finite input.
        result = Union(*[solver(m, symbol) for m in f.args])
    elif _is_function_class_equation(TrigonometricFunction, f, symbol) or \
            _is_function_class_equation(HyperbolicFunction, f, symbol):
        result = _solve_trig(f, symbol, domain)
    elif f.is_Piecewise:
        dom = domain
        result = EmptySet()
        expr_set_pairs = f.as_expr_set_pairs()
        for (expr, in_set) in expr_set_pairs:
            if in_set.is_Relational:
                in_set = in_set.as_set()
            if in_set.is_Interval:
                dom -= in_set
            solns = solver(expr, symbol, in_set)
            result += solns
    else:
        lhs, rhs_s = inverter(f, 0, symbol)
        if lhs == symbol:
            # do some very minimal simplification since
            # repeated inversion may have left the result
            # in a state that other solvers (e.g. poly)
            # would have simplified; this is done here
            # rather than in the inverter since here it
            # is only done once whereas there it would
            # be repeated for each step of the inversion
            if isinstance(rhs_s, FiniteSet):
                rhs_s = FiniteSet(*[Mul(*
                    signsimp(i).as_content_primitive())
                    for i in rhs_s])
            result = rhs_s
        elif isinstance(rhs_s, FiniteSet):
            for equation in [lhs - rhs for rhs in rhs_s]:
                if equation == f:
                    if any(_has_rational_power(g, symbol)[0]
                           for g in equation.args) or _has_rational_power(
                           equation, symbol)[0]:
                        result += _solve_radical(equation,
                                                 symbol,
                                                 solver)
                    elif equation.has(Abs):
                        result += _solve_abs(f, symbol, domain)
                    else:
                        result += _solve_as_rational(equation, symbol, domain)
                else:
                    result += solver(equation, symbol)
        else:
            result = ConditionSet(symbol, Eq(f, 0), domain)

    if _check:
        if isinstance(result, ConditionSet):
            # it wasn't solved or has enumerated all conditions
            # -- leave it alone
            return result

        # whittle away all but the symbol-containing core
        # to use this for testing
        fx = orig_f.as_independent(symbol, as_Add=True)[1]
        fx = fx.as_independent(symbol, as_Add=False)[1]

        if isinstance(result, FiniteSet):
            # check the result for invalid solutions
#.........这里部分代码省略.........

def simplify(expr, ratio=1.7, measure=count_ops, fu=False):
    """
    Simplifies the given expression.

    Simplification is not a well defined term and the exact strategies
    this function tries can change in the future versions of SymPy. If
    your algorithm relies on "simplification" (whatever it is), try to
    determine what you need exactly  -  is it powsimp()?, radsimp()?,
    together()?, logcombine()?, or something else? And use this particular
    function directly, because those are well defined and thus your algorithm
    will be robust.

    Nonetheless, especially for interactive use, or when you don't know
    anything about the structure of the expression, simplify() tries to apply
    intelligent heuristics to make the input expression "simpler".  For
    example:

    >>> from sympy import simplify, cos, sin
    >>> from sympy.abc import x, y
    >>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2)
    >>> a
    (x**2 + x)/(x*sin(y)**2 + x*cos(y)**2)
    >>> simplify(a)
    x + 1

    Note that we could have obtained the same result by using specific
    simplification functions:

    >>> from sympy import trigsimp, cancel
    >>> trigsimp(a)
    (x**2 + x)/x
    >>> cancel(_)
    x + 1

    In some cases, applying :func:`simplify` may actually result in some more
    complicated expression. The default ``ratio=1.7`` prevents more extreme
    cases: if (result length)/(input length) > ratio, then input is returned
    unmodified.  The ``measure`` parameter lets you specify the function used
    to determine how complex an expression is.  The function should take a
    single argument as an expression and return a number such that if
    expression ``a`` is more complex than expression ``b``, then
    ``measure(a) > measure(b)``.  The default measure function is
    :func:`count_ops`, which returns the total number of operations in the
    expression.

    For example, if ``ratio=1``, ``simplify`` output can't be longer
    than input.

    ::

        >>> from sympy import sqrt, simplify, count_ops, oo
        >>> root = 1/(sqrt(2)+3)

    Since ``simplify(root)`` would result in a slightly longer expression,
    root is returned unchanged instead::

       >>> simplify(root, ratio=1) == root
       True

    If ``ratio=oo``, simplify will be applied anyway::

        >>> count_ops(simplify(root, ratio=oo)) > count_ops(root)
        True

    Note that the shortest expression is not necessary the simplest, so
    setting ``ratio`` to 1 may not be a good idea.
    Heuristically, the default value ``ratio=1.7`` seems like a reasonable
    choice.

    You can easily define your own measure function based on what you feel
    should represent the "size" or "complexity" of the input expression.  Note
    that some choices, such as ``lambda expr: len(str(expr))`` may appear to be
    good metrics, but have other problems (in this case, the measure function
    may slow down simplify too much for very large expressions).  If you don't
    know what a good metric would be, the default, ``count_ops``, is a good
    one.

    For example:

    >>> from sympy import symbols, log
    >>> a, b = symbols('a b', positive=True)
    >>> g = log(a) + log(b) + log(a)*log(1/b)
    >>> h = simplify(g)
    >>> h
    log(a*b**(-log(a) + 1))
    >>> count_ops(g)
    8
    >>> count_ops(h)
    5

    So you can see that ``h`` is simpler than ``g`` using the count_ops metric.
    However, we may not like how ``simplify`` (in this case, using
    ``logcombine``) has created the ``b**(log(1/a) + 1)`` term.  A simple way
    to reduce this would be to give more weight to powers as operations in
    ``count_ops``.  We can do this by using the ``visual=True`` option:

    >>> print(count_ops(g, visual=True))
    2*ADD + DIV + 4*LOG + MUL
    >>> print(count_ops(h, visual=True))
    2*LOG + MUL + POW + SUB
#.........这里部分代码省略.........

def solveset_complex(f, symbol):
    """ Solve a complex valued equation.

    Parameters
    ==========

    f : Expr
        The target equation
    symbol : Symbol
        The variable for which the equation is solved

    Returns
    =======

    Set
        A set of values for `symbol` for which `f` equal to
        zero. An `EmptySet` is returned if no solution is found.

    `solveset_complex` claims to be complete in the solution set that
    it returns.

    Raises
    ======

    NotImplementedError
        The algorithms for to find the solution of the given equation are
        not yet implemented.
    ValueError
        The input is not valid.
    RuntimeError
        It is a bug, please report to the github issue tracker.

    See Also
    ========

    solveset_real: solver for real domain

    Examples
    ========

    >>> from sympy import Symbol, exp
    >>> from sympy.solvers.solveset import solveset_complex
    >>> from sympy.abc import x, a, b, c
    >>> solveset_complex(a*x**2 + b*x +c, x)
    {-b/(2*a) - sqrt(-4*a*c + b**2)/(2*a), -b/(2*a) + sqrt(-4*a*c + b**2)/(2*a)}

    Due to the fact that complex extension of my real valued functions are
    multivariate even some simple equations can have infinitely many solution.

    >>> solveset_complex(exp(x) - 1, x)
    ImageSet(Lambda(_n, 2*_n*I*pi), Integers())

    """
    if not symbol.is_Symbol:
        raise ValueError(" %s is not a symbol" % (symbol))

    f = sympify(f)
    original_eq = f
    if not isinstance(f, (Expr, Number)):
        raise ValueError(" %s is not a valid sympy expression" % (f))

    f = together(f)
    # Without this equations like a + 4*x**2 - E keep oscillating
    # into form  a/4 + x**2 - E/4 and (a + 4*x**2 - E)/4
    if not fraction(f)[1].has(symbol):
        f = expand(f)

    if f.is_zero:
        raise NotImplementedError("S.Complex set is not yet implemented")
    elif not f.has(symbol):
        result = EmptySet()
    elif f.is_Mul and all([_is_finite_with_finite_vars(m) for m in f.args]):
        result = Union(*[solveset_complex(m, symbol) for m in f.args])
    else:
        lhs, rhs_s = invert_complex(f, 0, symbol)
        if lhs == symbol:
            result = rhs_s
        elif isinstance(rhs_s, FiniteSet):
            equations = [lhs - rhs for rhs in rhs_s]
            result = EmptySet()
            for equation in equations:
                if equation == f:
                    result += _solve_as_rational(equation, symbol,
                                                 solveset_solver=solveset_complex,
                                                 as_poly_solver=_solve_as_poly_complex)
                else:
                    result += solveset_complex(equation, symbol)
        else:
            raise NotImplementedError

    if isinstance(result, FiniteSet):
        result = [s for s in result
                  if isinstance(s, RootOf)
                  or domain_check(original_eq, symbol, s)]
        return FiniteSet(*result)
    else:
        return result

def solveset_real(f, symbol):
    """ Solves a real valued equation.

    Parameters
    ==========

    f : Expr
        The target equation
    symbol : Symbol
        The variable for which the equation is solved

    Returns
    =======

    Set
        A set of values for `symbol` for which `f` is equal to
        zero. An `EmptySet` is returned if no solution is found.

    `solveset_real` claims to be complete in the set of the solution it
    returns.

    Raises
    ======

    NotImplementedError
        The algorithms for to find the solution of the given equation are
        not yet implemented.
    ValueError
        The input is not valid.
    RuntimeError
        It is a bug, please report to the github issue tracker.


    See Also
    =======

    solveset_complex : solver for complex domain

    Examples
    ========

    >>> from sympy import Symbol, exp, sin, sqrt, I
    >>> from sympy.solvers.solveset import solveset_real
    >>> x = Symbol('x', real=True)
    >>> a = Symbol('a', real=True, finite=True, positive=True)
    >>> solveset_real(x**2 - 1, x)
    {-1, 1}
    >>> solveset_real(sqrt(5*x + 6) - 2 - x, x)
    {-1, 2}
    >>> solveset_real(x - I, x)
    EmptySet()
    >>> solveset_real(x - a, x)
    {a}
    >>> solveset_real(exp(x) - a, x)
    {log(a)}

    In case the equation has infinitely many solutions an infinitely indexed
    `ImageSet` is returned.

    >>> solveset_real(sin(x) - 1, x)
    ImageSet(Lambda(_n, 2*_n*pi + pi/2), Integers())

    If the equation is true for any arbitrary value of the symbol a `S.Reals`
    set is returned.

    >>> solveset_real(x - x, x)
    (-oo, oo)

    """
    if not symbol.is_Symbol:
        raise ValueError(" %s is not a symbol" % (symbol))

    f = sympify(f)
    if not isinstance(f, (Expr, Number)):
        raise ValueError(" %s is not a valid sympy expression" % (f))

    original_eq = f
    f = together(f)

    if f.has(Piecewise):
        f = piecewise_fold(f)
    result = EmptySet()

    if f.expand().is_zero:
        return S.Reals
    elif not f.has(symbol):
        return EmptySet()
    elif f.is_Mul and all([_is_finite_with_finite_vars(m) for m in f.args]):
        # if f(x) and g(x) are both finite we can say that the solution of
        # f(x)*g(x) == 0 is same as Union(f(x) == 0, g(x) == 0) is not true in
        # general. g(x) can grow to infinitely large for the values where
        # f(x) == 0. To be sure that we not are silently allowing any
        # wrong solutions we are using this technique only if both f and g and
        # finite for a finite input.
        result = Union(*[solveset_real(m, symbol) for m in f.args])
    elif _is_function_class_equation(C.TrigonometricFunction, f, symbol) or \
            _is_function_class_equation(C.HyperbolicFunction, f, symbol):
        result = _solve_real_trig(f, symbol)
    elif f.is_Piecewise:
        result = EmptySet()
#.........这里部分代码省略.........

def eval_sum_symbolic(f, limits):
    from sympy.functions import harmonic, bernoulli

    f_orig = f
    (i, a, b) = limits
    if not f.has(i):
        return f*(b - a + 1)

    # Linearity
    if f.is_Mul:
        L, R = f.as_two_terms()

        if not L.has(i):
            sR = eval_sum_symbolic(R, (i, a, b))
            if sR:
                return L*sR

        if not R.has(i):
            sL = eval_sum_symbolic(L, (i, a, b))
            if sL:
                return R*sL

        try:
            f = apart(f, i)  # see if it becomes an Add
        except PolynomialError:
            pass

    if f.is_Add:
        L, R = f.as_two_terms()
        lrsum = telescopic(L, R, (i, a, b))

        if lrsum:
            return lrsum

        lsum = eval_sum_symbolic(L, (i, a, b))
        rsum = eval_sum_symbolic(R, (i, a, b))

        if None not in (lsum, rsum):
            r = lsum + rsum
            if not r is S.NaN:
                return r

    # Polynomial terms with Faulhaber's formula
    n = Wild('n')
    result = f.match(i**n)

    if result is not None:
        n = result[n]

        if n.is_Integer:
            if n >= 0:
                if (b is S.Infinity and not a is S.NegativeInfinity) or \
                   (a is S.NegativeInfinity and not b is S.Infinity):
                    return S.Infinity
                return ((bernoulli(n + 1, b + 1) - bernoulli(n + 1, a))/(n + 1)).expand()
            elif a.is_Integer and a >= 1:
                if n == -1:
                    return harmonic(b) - harmonic(a - 1)
                else:
                    return harmonic(b, abs(n)) - harmonic(a - 1, abs(n))

    if not (a.has(S.Infinity, S.NegativeInfinity) or
            b.has(S.Infinity, S.NegativeInfinity)):
        # Geometric terms
        c1 = Wild('c1', exclude=[i])
        c2 = Wild('c2', exclude=[i])
        c3 = Wild('c3', exclude=[i])
        wexp = Wild('wexp')

        # Here we first attempt powsimp on f for easier matching with the
        # exponential pattern, and attempt expansion on the exponent for easier
        # matching with the linear pattern.
        e = f.powsimp().match(c1 ** wexp)
        if e is not None:
            e_exp = e.pop(wexp).expand().match(c2*i + c3)
            if e_exp is not None:
                e.update(e_exp)

        if e is not None:
            p = (c1**c3).subs(e)
            q = (c1**c2).subs(e)

            r = p*(q**a - q**(b + 1))/(1 - q)
            l = p*(b - a + 1)

            return Piecewise((l, Eq(q, S.One)), (r, True))

        r = gosper_sum(f, (i, a, b))

        if isinstance(r, (Mul,Add)):
            from sympy import ordered, Tuple
            non_limit = r.free_symbols - Tuple(*limits[1:]).free_symbols
            den = denom(together(r))
            den_sym = non_limit & den.free_symbols
            args = []
            for v in ordered(den_sym):
                try:
                    s = solve(den, v)
                    m = Eq(v, s[0]) if s else S.false
                    if m != False:
#.........这里部分代码省略.........