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relation.py
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relation.py
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r"""
Symbolic Equations and Inequalities
Sage can solve symbolic equations and inequalities. For
example, we derive the quadratic formula as follows::
sage: a,b,c = var('a,b,c')
sage: qe = (a*x^2 + b*x + c == 0)
sage: qe
a*x^2 + b*x + c == 0
sage: print(solve(qe, x))
[
x == -1/2*(b + sqrt(b^2 - 4*a*c))/a,
x == -1/2*(b - sqrt(b^2 - 4*a*c))/a
]
The operator, left hand side, and right hand side
--------------------------------------------------
Operators::
sage: eqn = x^3 + 2/3 >= x - pi
sage: eqn.operator()
<built-in function ge>
sage: (x^3 + 2/3 < x - pi).operator()
<built-in function lt>
sage: (x^3 + 2/3 == x - pi).operator()
<built-in function eq>
Left hand side::
sage: eqn = x^3 + 2/3 >= x - pi
sage: eqn.lhs()
x^3 + 2/3
sage: eqn.left()
x^3 + 2/3
sage: eqn.left_hand_side()
x^3 + 2/3
Right hand side::
sage: (x + sqrt(2) >= sqrt(3) + 5/2).right()
sqrt(3) + 5/2
sage: (x + sqrt(2) >= sqrt(3) + 5/2).rhs()
sqrt(3) + 5/2
sage: (x + sqrt(2) >= sqrt(3) + 5/2).right_hand_side()
sqrt(3) + 5/2
Arithmetic
----------
Add two symbolic equations::
sage: var('a,b')
(a, b)
sage: m = 144 == -10 * a + b
sage: n = 136 == 10 * a + b
sage: m + n
280 == 2*b
sage: int(-144) + m
0 == -10*a + b - 144
Subtract two symbolic equations::
sage: var('a,b')
(a, b)
sage: m = 144 == 20 * a + b
sage: n = 136 == 10 * a + b
sage: m - n
8 == 10*a
sage: int(144) - m
0 == -20*a - b + 144
Multiply two symbolic equations::
sage: x = var('x')
sage: m = x == 5*x + 1
sage: n = sin(x) == sin(x+2*pi, hold=True)
sage: m * n
x*sin(x) == (5*x + 1)*sin(2*pi + x)
sage: m = 2*x == 3*x^2 - 5
sage: int(-1) * m
-2*x == -3*x^2 + 5
Divide two symbolic equations::
sage: x = var('x')
sage: m = x == 5*x + 1
sage: n = sin(x) == sin(x+2*pi, hold=True)
sage: m/n
x/sin(x) == (5*x + 1)/sin(2*pi + x)
sage: m = x != 5*x + 1
sage: n = sin(x) != sin(x+2*pi, hold=True)
sage: m/n
x/sin(x) != (5*x + 1)/sin(2*pi + x)
Substitution
------------
Substitution into relations::
sage: x, a = var('x, a')
sage: eq = (x^3 + a == sin(x/a)); eq
x^3 + a == sin(x/a)
sage: eq.substitute(x=5*x)
125*x^3 + a == sin(5*x/a)
sage: eq.substitute(a=1)
x^3 + 1 == sin(x)
sage: eq.substitute(a=x)
x^3 + x == sin(1)
sage: eq.substitute(a=x, x=1)
x + 1 == sin(1/x)
sage: eq.substitute({a:x, x:1})
x + 1 == sin(1/x)
You can even substitute multivariable and matrix
expressions::
sage: x,y = var('x, y')
sage: M = Matrix([[x+1,y],[x^2,y^3]]); M
[x + 1 y]
[ x^2 y^3]
sage: M.substitute({x:0,y:1})
[1 1]
[0 1]
Solving
-------
We can solve equations::
sage: x = var('x')
sage: S = solve(x^3 - 1 == 0, x)
sage: S
[x == 1/2*I*sqrt(3) - 1/2, x == -1/2*I*sqrt(3) - 1/2, x == 1]
sage: S[0]
x == 1/2*I*sqrt(3) - 1/2
sage: S[0].right()
1/2*I*sqrt(3) - 1/2
sage: S = solve(x^3 - 1 == 0, x, solution_dict=True)
sage: S
[{x: 1/2*I*sqrt(3) - 1/2}, {x: -1/2*I*sqrt(3) - 1/2}, {x: 1}]
sage: z = 5
sage: solve(z^2 == sqrt(3),z)
Traceback (most recent call last):
...
TypeError: 5 is not a valid variable.
We can also solve equations involving matrices. The following
example defines a multivariable function ``f(x,y)``, then solves
for where the partial derivatives with respect to ``x``
and ``y`` are zero. Then it substitutes one of the solutions
into the Hessian matrix ``H`` for ``f``::
sage: f(x,y) = x^2*y+y^2+y
sage: solutions = solve(list(f.diff()),[x,y],solution_dict=True)
sage: solutions == [{x: -I, y: 0}, {x: I, y: 0}, {x: 0, y: -1/2}]
True
sage: H = f.diff(2) # Hessian matrix
sage: H.subs(solutions[2])
[(x, y) |--> -1 (x, y) |--> 0]
[ (x, y) |--> 0 (x, y) |--> 2]
sage: H(x,y).subs(solutions[2])
[-1 0]
[ 0 2]
We illustrate finding multiplicities of solutions::
sage: f = (x-1)^5*(x^2+1)
sage: solve(f == 0, x)
[x == -I, x == I, x == 1]
sage: solve(f == 0, x, multiplicities=True)
([x == -I, x == I, x == 1], [1, 1, 5])
We can also solve many inequalities::
sage: solve(1/(x-1)<=8,x)
[[x < 1], [x >= (9/8)]]
We can numerically find roots of equations::
sage: (x == sin(x)).find_root(-2,2)
0.0
sage: (x^5 + 3*x + 2 == 0).find_root(-2,2,x)
-0.6328345202421523
sage: (cos(x) == sin(x)).find_root(10,20)
19.634954084936208
We illustrate some valid error conditions::
sage: (cos(x) != sin(x)).find_root(10,20)
Traceback (most recent call last):
...
ValueError: Symbolic equation must be an equality.
sage: (SR(3)==SR(2)).find_root(-1,1)
Traceback (most recent call last):
...
RuntimeError: no zero in the interval, since constant expression is not 0.
There must be at most one variable::
sage: x, y = var('x,y')
sage: (x == y).find_root(-2,2)
Traceback (most recent call last):
...
NotImplementedError: root finding currently only implemented in 1 dimension.
Assumptions
-----------
Forgetting assumptions::
sage: var('x,y')
(x, y)
sage: forget() #Clear assumptions
sage: assume(x>0, y < 2)
sage: assumptions()
[x > 0, y < 2]
sage: (y < 2).forget()
sage: assumptions()
[x > 0]
sage: forget()
sage: assumptions()
[]
Miscellaneous
-------------
Conversion to Maxima::
sage: x = var('x')
sage: eq = (x^(3/5) >= pi^2 + e^i)
sage: eq._maxima_init_()
'(_SAGE_VAR_x)^(3/5) >= ((%pi)^(2))+(exp(0+%i*1))'
sage: e1 = x^3 + x == sin(2*x)
sage: z = e1._maxima_()
sage: z.parent() is sage.calculus.calculus.maxima
True
sage: z = e1._maxima_(maxima)
sage: z.parent() is maxima
True
sage: z = maxima(e1)
sage: z.parent() is maxima
True
Conversion to Maple::
sage: x = var('x')
sage: eq = (x == 2)
sage: eq._maple_init_()
'x = 2'
Comparison::
sage: x = var('x')
sage: (x>0) == (x>0)
True
sage: (x>0) == (x>1)
False
sage: (x>0) != (x>1)
True
Variables appearing in the relation::
sage: var('x,y,z,w')
(x, y, z, w)
sage: f = (x+y+w) == (x^2 - y^2 - z^3); f
w + x + y == -z^3 + x^2 - y^2
sage: f.variables()
(w, x, y, z)
LaTeX output::
sage: latex(x^(3/5) >= pi)
x^{\frac{3}{5}} \geq \pi
When working with the symbolic complex number `I`, notice that comparisons do not
automatically simplify even in trivial situations::
sage: I^2 == -1
-1 == -1
sage: I^2 < 0
-1 < 0
sage: (I+1)^4 > 0
-4 > 0
Nevertheless, if you force the comparison, you get the right answer (:trac:`7160`)::
sage: bool(I^2 == -1)
True
sage: bool(I^2 < 0)
True
sage: bool((I+1)^4 > 0)
False
More Examples
-------------
::
sage: x,y,a = var('x,y,a')
sage: f = x^2 + y^2 == 1
sage: f.solve(x)
[x == -sqrt(-y^2 + 1), x == sqrt(-y^2 + 1)]
::
sage: f = x^5 + a
sage: solve(f==0,x)
[x == 1/4*(-a)^(1/5)*(sqrt(5) + I*sqrt(2*sqrt(5) + 10) - 1), x == -1/4*(-a)^(1/5)*(sqrt(5) - I*sqrt(-2*sqrt(5) + 10) + 1), x == -1/4*(-a)^(1/5)*(sqrt(5) + I*sqrt(-2*sqrt(5) + 10) + 1), x == 1/4*(-a)^(1/5)*(sqrt(5) - I*sqrt(2*sqrt(5) + 10) - 1), x == (-a)^(1/5)]
You can also do arithmetic with inequalities, as illustrated
below::
sage: var('x y')
(x, y)
sage: f = x + 3 == y - 2
sage: f
x + 3 == y - 2
sage: g = f - 3; g
x == y - 5
sage: h = x^3 + sqrt(2) == x*y*sin(x)
sage: h
x^3 + sqrt(2) == x*y*sin(x)
sage: h - sqrt(2)
x^3 == x*y*sin(x) - sqrt(2)
sage: h + f
x^3 + x + sqrt(2) + 3 == x*y*sin(x) + y - 2
sage: f = x + 3 < y - 2
sage: g = 2 < x+10
sage: f - g
x + 1 < -x + y - 12
sage: f + g
x + 5 < x + y + 8
sage: f*(-1)
-x - 3 < -y + 2
TESTS:
We test serializing symbolic equations::
sage: eqn = x^3 + 2/3 >= x
sage: loads(dumps(eqn))
x^3 + 2/3 >= x
sage: loads(dumps(eqn)) == eqn
True
AUTHORS:
- Bobby Moretti: initial version (based on a trick that Robert
Bradshaw suggested).
- William Stein: second version
- William Stein (2007-07-16): added arithmetic with symbolic equations
"""
from __future__ import print_function
from six.moves import range
import operator
def test_relation_maxima(relation):
"""
Return True if this (in)equality is definitely true. Return False
if it is false or the algorithm for testing (in)equality is
inconclusive.
EXAMPLES::
sage: from sage.symbolic.relation import test_relation_maxima
sage: k = var('k')
sage: pol = 1/(k-1) - 1/k -1/k/(k-1);
sage: test_relation_maxima(pol == 0)
True
sage: f = sin(x)^2 + cos(x)^2 - 1
sage: test_relation_maxima(f == 0)
True
sage: test_relation_maxima( x == x )
True
sage: test_relation_maxima( x != x )
False
sage: test_relation_maxima( x > x )
False
sage: test_relation_maxima( x^2 > x )
False
sage: test_relation_maxima( x + 2 > x )
True
sage: test_relation_maxima( x - 2 > x )
False
Here are some examples involving assumptions::
sage: x, y, z = var('x, y, z')
sage: assume(x>=y,y>=z,z>=x)
sage: test_relation_maxima(x==z)
True
sage: test_relation_maxima(z<x)
False
sage: test_relation_maxima(z>y)
False
sage: test_relation_maxima(y==z)
True
sage: forget()
sage: assume(x>=1,x<=1)
sage: test_relation_maxima(x==1)
True
sage: test_relation_maxima(x>1)
False
sage: test_relation_maxima(x>=1)
True
sage: test_relation_maxima(x!=1)
False
sage: forget()
sage: assume(x>0)
sage: test_relation_maxima(x==0)
False
sage: test_relation_maxima(x>-1)
True
sage: test_relation_maxima(x!=0)
True
sage: test_relation_maxima(x!=1)
False
sage: forget()
TESTS:
Ensure that ``canonicalize_radical()`` and ``simplify_log`` are not
used inappropriately, :trac:`17389`. Either one would simplify ``f``
to zero below::
sage: x,y = SR.var('x,y')
sage: assume(y, 'complex')
sage: f = log(x*y) - (log(x) + log(y))
sage: f(x=-1, y=i)
-2*I*pi
sage: test_relation_maxima(f == 0)
False
sage: forget()
Ensure that the ``sqrt(x^2)`` -> ``abs(x)`` simplification is not
performed when testing equality::
sage: assume(x, 'complex')
sage: f = sqrt(x^2) - abs(x)
sage: test_relation_maxima(f == 0)
False
sage: forget()
If assumptions are made, ``simplify_rectform()`` is used::
sage: assume(x, 'real')
sage: f1 = ( e^(I*x) - e^(-I*x) ) / ( I*e^(I*x) + I*e^(-I*x) )
sage: f2 = sin(x)/cos(x)
sage: test_relation_maxima(f1 - f2 == 0)
True
sage: forget()
But not if ``x`` itself is complex::
sage: assume(x, 'complex')
sage: f1 = ( e^(I*x) - e^(-I*x) ) / ( I*e^(I*x) + I*e^(-I*x) )
sage: f2 = sin(x)/cos(x)
sage: test_relation_maxima(f1 - f2 == 0)
False
sage: forget()
If assumptions are made, then ``simplify_factorial()`` is used::
sage: n,k = SR.var('n,k')
sage: assume(n, 'integer')
sage: assume(k, 'integer')
sage: f1 = factorial(n+1)/factorial(n)
sage: f2 = n + 1
sage: test_relation_maxima(f1 - f2 == 0)
True
sage: forget()
In case one of the solutions while solving an equation is a real number::
sage: var('K, d, R')
(K, d, R)
sage: assume(K>0)
sage: assume(K, 'noninteger')
sage: assume(R>0)
sage: assume(R<1)
sage: assume(d<R)
sage: assumptions()
[K > 0, K is noninteger, R > 0, R < 1, d < R]
"""
m = relation._maxima_()
#Handle some basic cases first
if repr(m) in ['0=0']:
return True
elif repr(m) in ['0#0', '1#1']:
return False
if relation.operator() == operator.eq: # operator is equality
try:
s = m.parent()._eval_line('is (equal(%s,%s))'%(repr(m.lhs()),repr(m.rhs())))
except TypeError:
raise ValueError("unable to evaluate the predicate '%s'" % repr(relation))
elif relation.operator() == operator.ne: # operator is not equal
try:
s = m.parent()._eval_line('is (notequal(%s,%s))'%(repr(m.lhs()),repr(m.rhs())))
except TypeError:
raise ValueError("unable to evaluate the predicate '%s'" % repr(relation))
else: # operator is < or > or <= or >=, which Maxima handles fine
try:
s = m.parent()._eval_line('is (%s)'%repr(m))
except TypeError:
raise ValueError("unable to evaluate the predicate '%s'" % repr(relation))
if s == 'true':
return True
elif s == 'false':
return False # if neither of these, s=='unknown' and we try a few other tricks
if relation.operator() != operator.eq:
return False
difference = relation.lhs() - relation.rhs()
if difference.is_trivial_zero():
return True
# Try to apply some simplifications to see if left - right == 0.
#
# TODO: If simplify_log() is ever removed from simplify_full(), we
# can replace all of these individual simplifications with a
# single call to simplify_full(). That would work in cases where
# two simplifications are needed consecutively; the current
# approach does not.
#
simp_list = [difference.simplify_factorial(),
difference.simplify_rational(),
difference.simplify_rectform(),
difference.simplify_trig()]
for f in simp_list:
try:
if f().is_trivial_zero():
return True
break
except Exception:
pass
return False
def string_to_list_of_solutions(s):
r"""
Used internally by the symbolic solve command to convert the output
of Maxima's solve command to a list of solutions in Sage's symbolic
package.
EXAMPLES:
We derive the (monic) quadratic formula::
sage: var('x,a,b')
(x, a, b)
sage: solve(x^2 + a*x + b == 0, x)
[x == -1/2*a - 1/2*sqrt(a^2 - 4*b), x == -1/2*a + 1/2*sqrt(a^2 - 4*b)]
Behind the scenes when the above is evaluated the function
:func:`string_to_list_of_solutions` is called with input the
string `s` below::
sage: s = '[x=-(sqrt(a^2-4*b)+a)/2,x=(sqrt(a^2-4*b)-a)/2]'
sage: sage.symbolic.relation.string_to_list_of_solutions(s)
[x == -1/2*a - 1/2*sqrt(a^2 - 4*b), x == -1/2*a + 1/2*sqrt(a^2 - 4*b)]
"""
from sage.categories.all import Objects
from sage.structure.sequence import Sequence
from sage.calculus.calculus import symbolic_expression_from_maxima_string
v = symbolic_expression_from_maxima_string(s, equals_sub=True)
return Sequence(v, universe=Objects(), cr_str=True)
###########
# Solving #
###########
def solve(f, *args, **kwds):
r"""
Algebraically solve an equation or system of equations (over the
complex numbers) for given variables. Inequalities and systems
of inequalities are also supported.
INPUT:
- ``f`` - equation or system of equations (given by a
list or tuple)
- ``*args`` - variables to solve for.
- ``solution_dict`` - bool (default: False); if True or non-zero,
return a list of dictionaries containing the solutions. If there
are no solutions, return an empty list (rather than a list containing
an empty dictionary). Likewise, if there's only a single solution,
return a list containing one dictionary with that solution.
There are a few optional keywords if you are trying to solve a single
equation. They may only be used in that context.
- ``multiplicities`` - bool (default: False); if True,
return corresponding multiplicities. This keyword is
incompatible with ``to_poly_solve=True`` and does not make
any sense when solving inequalities.
- ``explicit_solutions`` - bool (default: False); require that
all roots be explicit rather than implicit. Not used
when solving inequalities.
- ``to_poly_solve`` - bool (default: False) or string; use
Maxima's ``to_poly_solver`` package to search for more possible
solutions, but possibly encounter approximate solutions.
This keyword is incompatible with ``multiplicities=True``
and is not used when solving inequalities. Setting ``to_poly_solve``
to 'force' (string) omits Maxima's solve command (useful when
some solutions of trigonometric equations are lost).
- ``algorithm`` - string (default: 'maxima'); to use SymPy's
solvers set this to 'sympy'. Note that SymPy is always used
for diophantine equations.
- ``domain`` - string (default: 'complex'); setting this to 'real'
changes the way SymPy solves single equations; inequalities
are always solvedin the real domain.
EXAMPLES::
sage: x, y = var('x, y')
sage: solve([x+y==6, x-y==4], x, y)
[[x == 5, y == 1]]
sage: solve([x^2+y^2 == 1, y^2 == x^3 + x + 1], x, y)
[[x == -1/2*I*sqrt(3) - 1/2, y == -sqrt(-1/2*I*sqrt(3) + 3/2)],
[x == -1/2*I*sqrt(3) - 1/2, y == sqrt(-1/2*I*sqrt(3) + 3/2)],
[x == 1/2*I*sqrt(3) - 1/2, y == -sqrt(1/2*I*sqrt(3) + 3/2)],
[x == 1/2*I*sqrt(3) - 1/2, y == sqrt(1/2*I*sqrt(3) + 3/2)],
[x == 0, y == -1],
[x == 0, y == 1]]
sage: solve([sqrt(x) + sqrt(y) == 5, x + y == 10], x, y)
[[x == -5/2*I*sqrt(5) + 5, y == 5/2*I*sqrt(5) + 5], [x == 5/2*I*sqrt(5) + 5, y == -5/2*I*sqrt(5) + 5]]
sage: solutions=solve([x^2+y^2 == 1, y^2 == x^3 + x + 1], x, y, solution_dict=True)
sage: for solution in solutions: print("{} , {}".format(solution[x].n(digits=3), solution[y].n(digits=3)))
-0.500 - 0.866*I , -1.27 + 0.341*I
-0.500 - 0.866*I , 1.27 - 0.341*I
-0.500 + 0.866*I , -1.27 - 0.341*I
-0.500 + 0.866*I , 1.27 + 0.341*I
0.000 , -1.00
0.000 , 1.00
Whenever possible, answers will be symbolic, but with systems of
equations, at times approximations will be given by Maxima, due to the
underlying algorithm::
sage: sols = solve([x^3==y,y^2==x], [x,y]); sols[-1], sols[0]
([x == 0, y == 0],
[x == (0.3090169943749475 + 0.9510565162951535*I),
y == (-0.8090169943749475 - 0.5877852522924731*I)])
sage: sols[0][0].rhs().pyobject().parent()
Complex Double Field
sage: solve([y^6==y],y)
[y == 1/4*sqrt(5) + 1/4*I*sqrt(2*sqrt(5) + 10) - 1/4,
y == -1/4*sqrt(5) + 1/4*I*sqrt(-2*sqrt(5) + 10) - 1/4,
y == -1/4*sqrt(5) - 1/4*I*sqrt(-2*sqrt(5) + 10) - 1/4,
y == 1/4*sqrt(5) - 1/4*I*sqrt(2*sqrt(5) + 10) - 1/4,
y == 1,
y == 0]
sage: solve( [y^6 == y], y)==solve( y^6 == y, y)
True
Here we demonstrate very basic use of the optional keywords::
sage: ((x^2-1)^2).solve(x)
[x == -1, x == 1]
sage: ((x^2-1)^2).solve(x,multiplicities=True)
([x == -1, x == 1], [2, 2])
sage: solve(sin(x)==x,x)
[x == sin(x)]
sage: solve(sin(x)==x,x,explicit_solutions=True)
[]
sage: solve(abs(1-abs(1-x)) == 10, x)
[abs(abs(x - 1) - 1) == 10]
sage: solve(abs(1-abs(1-x)) == 10, x, to_poly_solve=True)
[x == -10, x == 12]
sage: from sage.symbolic.expression import Expression
sage: Expression.solve(x^2==1,x)
[x == -1, x == 1]
We must solve with respect to actual variables::
sage: z = 5
sage: solve([8*z + y == 3, -z +7*y == 0],y,z)
Traceback (most recent call last):
...
TypeError: 5 is not a valid variable.
If we ask for dictionaries containing the solutions, we get them::
sage: solve([x^2-1],x,solution_dict=True)
[{x: -1}, {x: 1}]
sage: solve([x^2-4*x+4],x,solution_dict=True)
[{x: 2}]
sage: res = solve([x^2 == y, y == 4],x,y,solution_dict=True)
sage: for soln in res: print("x: %s, y: %s" % (soln[x], soln[y]))
x: 2, y: 4
x: -2, y: 4
If there is a parameter in the answer, that will show up as
a new variable. In the following example, ``r1`` is a real free
variable (because of the ``r``)::
sage: forget()
sage: x, y = var('x,y')
sage: solve([x+y == 3, 2*x+2*y == 6],x,y)
[[x == -r1 + 3, y == r1]]
sage: var('b, c')
(b, c)
sage: solve((b-1)*(c-1), [b,c])
[[b == 1, c == r...], [b == r..., c == 1]]
Especially with trigonometric functions, the dummy variable may
be implicitly an integer (hence the ``z``)::
sage: solve( sin(x)==cos(x), x, to_poly_solve=True)
[x == 1/4*pi + pi*z...]
sage: solve([cos(x)*sin(x) == 1/2, x+y == 0],x,y)
[[x == 1/4*pi + pi*z..., y == -1/4*pi - pi*z...]]
Expressions which are not equations are assumed to be set equal
to zero, as with `x` in the following example::
sage: solve([x, y == 2],x,y)
[[x == 0, y == 2]]
If ``True`` appears in the list of equations it is
ignored, and if ``False`` appears in the list then no
solutions are returned. E.g., note that the first
``3==3`` evaluates to ``True``, not to a
symbolic equation.
::
sage: solve([3==3, 1.00000000000000*x^3 == 0], x)
[x == 0]
sage: solve([1.00000000000000*x^3 == 0], x)
[x == 0]
Here, the first equation evaluates to ``False``, so
there are no solutions::
sage: solve([1==3, 1.00000000000000*x^3 == 0], x)
[]
Completely symbolic solutions are supported::
sage: var('s,j,b,m,g')
(s, j, b, m, g)
sage: sys = [ m*(1-s) - b*s*j, b*s*j-g*j ];
sage: solve(sys,s,j)
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: solve(sys,(s,j))
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: solve(sys,[s,j])
[[s == 1, j == 0], [s == g/b, j == (b - g)*m/(b*g)]]
sage: z = var('z')
sage: solve((x-z)^2==2, x)
[x == z - sqrt(2), x == z + sqrt(2)]
Inequalities can be also solved::
sage: solve(x^2>8,x)
[[x < -2*sqrt(2)], [x > 2*sqrt(2)]]
sage: x,y=var('x,y'); (ln(x)-ln(y)>0).solve(x)
[[log(x) - log(y) > 0]]
sage: x,y=var('x,y'); (ln(x)>ln(y)).solve(x) # random
[[0 < y, y < x, 0 < x]]
[[y < x, 0 < y]]
A simple example to show the use of the keyword
``multiplicities``::
sage: ((x^2-1)^2).solve(x)
[x == -1, x == 1]
sage: ((x^2-1)^2).solve(x,multiplicities=True)
([x == -1, x == 1], [2, 2])
sage: ((x^2-1)^2).solve(x,multiplicities=True,to_poly_solve=True)
Traceback (most recent call last):
...
NotImplementedError: to_poly_solve does not return multiplicities
Here is how the ``explicit_solutions`` keyword functions::
sage: solve(sin(x)==x,x)
[x == sin(x)]
sage: solve(sin(x)==x,x,explicit_solutions=True)
[]
sage: solve(x*sin(x)==x^2,x)
[x == 0, x == sin(x)]
sage: solve(x*sin(x)==x^2,x,explicit_solutions=True)
[x == 0]
The following examples show the use of the keyword ``to_poly_solve``::
sage: solve(abs(1-abs(1-x)) == 10, x)
[abs(abs(x - 1) - 1) == 10]
sage: solve(abs(1-abs(1-x)) == 10, x, to_poly_solve=True)
[x == -10, x == 12]
sage: var('Q')
Q
sage: solve(Q*sqrt(Q^2 + 2) - 1, Q)
[Q == 1/sqrt(Q^2 + 2)]
The following example is a regression in Maxima 5.39.0.
It used to be possible to get one more solution here,
namely ``1/sqrt(sqrt(2) + 1)``, see
https://sourceforge.net/p/maxima/bugs/3276/::
sage: solve(Q*sqrt(Q^2 + 2) - 1, Q, to_poly_solve=True)
[Q == -sqrt(-sqrt(2) - 1)]
An effort is made to only return solutions that satisfy
the current assumptions::
sage: solve(x^2==4, x)
[x == -2, x == 2]
sage: assume(x<0)
sage: solve(x^2==4, x)
[x == -2]
sage: solve((x^2-4)^2 == 0, x, multiplicities=True)
([x == -2], [2])
sage: solve(x^2==2, x)
[x == -sqrt(2)]
sage: z = var('z')
sage: solve(x^2==2-z, x)
[x == -sqrt(-z + 2)]
sage: assume(x, 'rational')
sage: solve(x^2 == 2, x)
[]
In some cases it may be worthwhile to directly use ``to_poly_solve``
if one suspects some answers are being missed::
sage: forget()
sage: solve(cos(x)==0, x)
[x == 1/2*pi]
sage: solve(cos(x)==0, x, to_poly_solve=True)
[x == 1/2*pi]
sage: solve(cos(x)==0, x, to_poly_solve='force')
[x == 1/2*pi + pi*z...]
The same may also apply if a returned unsolved expression has a
denominator, but the original one did not::
sage: solve(cos(x) * sin(x) == 1/2, x, to_poly_solve=True)
[sin(x) == 1/2/cos(x)]
sage: solve(cos(x) * sin(x) == 1/2, x, to_poly_solve=True, explicit_solutions=True)
[x == 1/4*pi + pi*z...]
sage: solve(cos(x) * sin(x) == 1/2, x, to_poly_solve='force')
[x == 1/4*pi + pi*z...]
We use ``use_grobner`` in Maxima if no solution is obtained from
Maxima's ``to_poly_solve``::
sage: x,y=var('x y'); c1(x,y)=(x-5)^2+y^2-16; c2(x,y)=(y-3)^2+x^2-9
sage: solve([c1(x,y),c2(x,y)],[x,y])
[[x == -9/68*sqrt(55) + 135/68, y == -15/68*sqrt(55) + 123/68],
[x == 9/68*sqrt(55) + 135/68, y == 15/68*sqrt(55) + 123/68]]
We use SymPy for Diophantine equations, see
``Expression.solve_diophantine``::
sage: assume(x, 'integer')
sage: assume(z, 'integer')
sage: solve((x-z)^2==2, x)
[]
sage: forget()
The following shows some more of SymPy's capabilities that cannot be
handled by Maxima::
sage: _ = var('t')
sage: r = solve([x^2 - y^2/exp(x), y-1], x, y, algorithm='sympy')
sage: (r[0][x], r[0][y])
(2*lambert_w(1/2), 1)
sage: solve(-2*x**3 + 4*x**2 - 2*x + 6 > 0, x, algorithm='sympy')
[x < 1/3*(1/2)^(1/3)*(9*sqrt(77) + 79)^(1/3) + 2/3*(1/2)^(2/3)/(9*sqrt(77) + 79)^(1/3) + 2/3]
sage: solve(sqrt(2*x^2 - 7) - (3 - x),x,algorithm='sympy')
[x == -8, x == 2]
sage: solve(sqrt(2*x + 9) - sqrt(x + 1) - sqrt(x + 4),x,algorithm='sympy')
[x == 0]
sage: r = solve([x + y + z + t, -z - t], x, y, z, t, algorithm='sympy')
sage: (r[0][x], r[0][z])
(-y, -t)
sage: r = solve([x^2+y+z, y+x^2+z, x+y+z^2], x, y,z, algorithm='sympy')
sage: (r[0][x], r[0][y])
(z, -(z + 1)*z)
sage: (r[1][x], r[1][y])
(-z + 1, -z^2 + z - 1)
sage: solve(abs(x + 3) - 2*abs(x - 3),x,algorithm='sympy',domain='real')
[x == 1, x == 9]
We cannot translate all results from SymPy but we can at least
print them::
sage: solve(sinh(x) - 2*cosh(x),x,algorithm='sympy')
ConditionSet(x, Eq((-exp(2*x) - 3)*exp(-x)/2, 0), S.Reals)
sage: solve(2*sin(x) - 2*sin(2*x), x,algorithm='sympy')
...
[ImageSet(Lambda(_n, 2*_n*pi), S.Integers),
ImageSet(Lambda(_n, 2*_n*pi + pi), S.Integers),
ImageSet(Lambda(_n, 2*_n*pi + 5*pi/3), S.Integers),
ImageSet(Lambda(_n, 2*_n*pi + pi/3), S.Integers)]
sage: solve(x^5 + 3*x^3 + 7, x, algorithm='sympy')[0] # known bug
complex_root_of(x^5 + 3*x^3 + 7, 0)
TESTS::
sage: solve([sin(x)==x,y^2==x],x,y)
[sin(x) == x, y^2 == x]
sage: solve(0==1,x)
Traceback (most recent call last):
...
TypeError: The first argument must be a symbolic expression or a list of symbolic expressions.
Test if the empty list is returned, too, when (a list of)
dictionaries (is) are requested (:trac:`8553`)::
sage: solve([SR(0)==1],x)
[]
sage: solve([SR(0)==1],x,solution_dict=True)
[]
sage: solve([x==1,x==-1],x)
[]
sage: solve([x==1,x==-1],x,solution_dict=True)
[]
sage: solve((x==1,x==-1),x,solution_dict=0)
[]
Relaxed form, suggested by Mike Hansen (:trac:`8553`)::
sage: solve([x^2-1],x,solution_dict=-1)
[{x: -1}, {x: 1}]
sage: solve([x^2-1],x,solution_dict=1)
[{x: -1}, {x: 1}]
sage: solve((x==1,x==-1),x,solution_dict=-1)
[]
sage: solve((x==1,x==-1),x,solution_dict=1)
[]
This inequality holds for any real ``x`` (:trac:`8078`)::
sage: solve(x^4+2>0,x)
[x < +Infinity]
Test for user friendly input handling :trac:`13645`::
sage: poly.<a,b> = PolynomialRing(RR)
sage: solve([a+b+a*b == 1], a)
Traceback (most recent call last):
...
TypeError: a is not a valid variable.
sage: a,b = var('a,b')
sage: solve([a+b+a*b == 1], a)
[a == -(b - 1)/(b + 1)]
sage: solve([a, b], (1, a))
Traceback (most recent call last):
...
TypeError: 1 is not a valid variable.
sage: solve([x == 1], (1, a))
Traceback (most recent call last):
...
TypeError: 1 is not a valid variable.
sage: x.solve((1,2))
Traceback (most recent call last):
...
TypeError: 1 is not a valid variable.
Test that the original version of a system in the French Sage book
now works (:trac:`14306`)::
sage: var('y,z')
(y, z)
sage: solve([x^2 * y * z == 18, x * y^3 * z == 24, x * y * z^4 == 6], x, y, z)
[[x == 3, y == 2, z == 1],
[x == (1.337215067... - 2.685489874...*I),
y == (-1.700434271... + 1.052864325...*I),