Solveset¶
This is the official documentation of the solveset
module in solvers.
It contains the frequently asked questions about our new module to solve
equations.
What’s wrong with solve():¶
SymPy already has a pretty powerful solve
function. But it has a lot of major
issues
It doesn’t have a consistent output for various types of solutions It needs to return a lot of types of solutions consistently:
Single solution : \(x = 1\)
Multiple solutions: \(x^2 = 1\)
No Solution: \(x^2 + 1 = 0 ; x \in \mathbb{R}\)
Interval of solution: \(\lfloor x \rfloor = 0\)
Infinitely many solutions: \(sin(x) = 0\)
Multivariate functions with point solutions: \(x^2 + y^2 = 0\)
Multivariate functions with nonpoint solution: \(x^2 + y^2 = 1\)
System of equations: \(x + y = 1\) and \(x  y = 0\)
Relational: \(x > 0\)
And the most important case: “We don’t Know”
The input API is also a mess, there are a lot of parameters. Many of them are not needed and they make it hard for the user and the developers to work on solvers.
There are cases like finding the maxima and minima of function using critical points where it is important to know if it has returned all the solutions.
solve
does not guarantee this.
Why Solveset?¶
solveset
has a cleaner input and output interface:solveset
returns a set object and a set object takes care of all types of output. For cases where it doesn’t “know” all the solutions aConditionSet
with a partial solution is returned. For input it only takes the equation, the variables to solve for and the optional argumentdomain
over which the equation is to be solved.solveset
can return infinitely many solutions. For example solving for \(\sin{(x)} = 0\) returns \(\{2 n \pi  n \in \mathbb{Z}\} \cup \{2 n \pi + \pi  n \in \mathbb{Z}\}\), whereassolve
only returns \([0, \pi]\).There is a clear code level and interface level separation between solvers for equations in the complex domain and the real domain. For example solving \(e^x = 1\) when \(x\) is to be solved in the complex domain, returns the set of all solutions, that is \(\{2 n i \pi  n \in \mathbb{Z}\}\), whereas if \(x\) is to be solved in the real domain then only \(\{0\}\) is returned.
Why do we use Sets as an output type?¶
SymPy has a well developed sets module, which can represent most of the set containers in Mathematics such as:
FiniteSet
Represents a finite set of discrete numbers.
Interval
Represents a real interval as a set.
ProductSet
Represents a Cartesian product of sets.
ImageSet
Represents the image of a set under a mathematical function
>>> from sympy import ImageSet, S, Lambda >>> from sympy.abc import x >>> squares = ImageSet(Lambda(x, x**2), S.Naturals) # {x**2 for x in N} >>> 4 in squares True
ComplexRegion
Represents the set of all complex numbers in a region in the Argand plane.
ConditionSet
Represents the set of elements, which satisfies a given condition.
Also, the predefined set classes such as:
Naturals
\(\mathbb{N}\)Represents the natural numbers (or counting numbers), which are all positive integers starting from 1.
Naturals0
\(\mathbb{N_0}\)Represents the whole numbers, which are all the nonnegative integers, inclusive of 0.
Integers
\(\mathbb{Z}\)Represents all integers: positive, negative and zero.
Reals
\(\mathbb{R}\)Represents the set of all real numbers.
Complexes
\(\mathbb{C}\)Represents the set of all complex numbers.
EmptySet
\(\phi\)Represents the empty set.
The above six sets are available as Singletons, like
S.Integers
.
It is capable of most of the set operations in mathematics:
Union
Intersection
Complement
SymmetricDifference
The main reason for using sets as output to solvers is that it can consistently represent many types of solutions. For the single variable case it can represent:
No solution (by the empty set).
Finitely many solutions (by
FiniteSet
).Infinitely many solutions, both countably and uncountably infinite solutions (using the
ImageSet
module).
Interval
There can also be bizarre solutions to equations like the set of rational numbers.
No other Python object (list, dictionary, generator, Python sets) provides the
flexibility of mathematical sets which our sets module tries to emulate. The
second reason to use sets is that they are close to the entities which
mathematicians deal with and it makes it easier to reason about them. Set
objects conform to Pythonic conventions when possible, i.e., x in A
and
for i in A
both work when they can be computed. Another advantage of using
objects closer to mathematical entities is that the user won’t have to “learn”
our representation and she can have her expectations transferred from her
mathematical experience.
For the multivariate case we represent solutions as a set of points in a
ndimensional space and a point is represented by a FiniteSet
of ordered
tuples, which is a point in \(\mathbb{R}^n\) or \(\mathbb{C}^n\).
Please note that, the general FiniteSet
is unordered, but a FiniteSet
with a tuple as its only argument becomes ordered, since a tuple is ordered.
So the order in the tuple is mapped to a predefined order of variables
while returning solutions.
For example:
>>> from sympy import FiniteSet
>>> FiniteSet(1, 2, 3) # Unordered
{1, 2, 3}
>>> FiniteSet((1, 2, 3)) # Ordered
{(1, 2, 3)}
Why not use dicts as output?
Dictionary are easy to deal with programmatically but mathematically they are not very precise and use of them can quickly lead to inconsistency and a lot of confusion. For example:
There are a lot of cases where we don’t know the complete solution and we may like to output a partial solution, consider the equation \(fg = 0\). The solution of this equation is the union of the solution of the following two equations: \(f = 0\), \(g = 0\). Let’s say that we are able to solve \(f = 0\) but solving \(g = 0\) isn’t supported yet. In this case we cannot represent partial solution of the given equation \(fg = 0\) using dicts. This problem is solved with sets using a
ConditionSet
object:\(sol_f \cup \{x  x ∊ \mathbb{R} ∧ g = 0\}\), where \(sol_f\) is the solution of the equation \(f = 0\).
Using a dict may lead to surprising results like:
solve(Eq(x**2, 1), x) != solve(Eq(y**2, 1), y)
Mathematically, this doesn’t make sense. Using
FiniteSet
here solves the problem.It also cannot represent solutions for equations like \(x < 1\), which is a disk of radius 1 in the Argand Plane. This problem is solved using complex sets implemented as
ComplexRegion
.
Input API of solveset
¶
solveset
has a cleaner input API, unlike solve
. It takes a maximum
of three arguments:
solveset(equation, variable=None, domain=S.Complexes)
Equation(s)
The equation(s) to solve.
Variable(s)
The variable(s) for which the equation is to be solved.
Domain
The domain in which the equation is to be solved.
solveset
removes theflags
argument ofsolve
, which had made the input API messy and output API inconsistent.
What is this domain argument about?¶
Solveset is designed to be independent of the assumptions on the variable being solved for and instead, uses the
domain
argument to decide the solver to dispatch the equation to, namelysolveset_real
orsolveset_complex
. It’s unlike the oldsolve
which considers the assumption on the variable.>>> from sympy import solveset, S >>> from sympy.abc import x >>> solveset(x**2 + 1, x) # domain=S.Complexes is default {I, I} >>> solveset(x**2 + 1, x, domain=S.Reals) EmptySet()
What are the general methods employed by solveset to solve an equation?¶
Solveset uses various methods to solve an equation, here is a brief overview of the methodology:
The
domain
argument is first considered to know the domain in which the user is interested to get the solution.If the given function is a relational (
>=
,<=
,>
,<
), and the domain is real, thensolve_univariate_inequality
and solutions are returned. Solving for complex solutions of inequalities, like \(x^2 < 0\) is not yet supported.Based on the
domain
, the equation is dispatched to one of the two functionssolveset_real
orsolveset_complex
, which solves the given equation in the complex or real domain, respectively.If the given expression is a product of two or more functions, like say \(gh = 0\), then the solution to the given equation is the Union of the solution of the equations \(g = 0\) and \(h = 0\), if and only if both \(g\) and \(h\) are finite for a finite input. So, the solution is built up recursively.
If the function is trigonometric or hyperbolic, the function
_solve_real_trig
is called, which solves it by converting it to complex exponential form.The function is now checked if there is any instance of a
Piecewise
expression, if it is, then it’s converted to explicit expression and set pairs and then solved recursively.The respective solver now tries to invert the equation using the routines
invert_real
andinvert_complex
. These routines are based on the concept of mathematical inverse (though not exactly). It reduces the real/complex valued equation \(f(x) = y\) to a set of equations: \(\{g(x) = h_1(y), g(x) = h_2(y), ..., g(x) = h_n(y) \}\) where \(g(x)\) is a simpler function than \(f(x)\). There is some work needed to be done in this to find invert of more complex expressions.After the invert, the equations are checked for radical or Abs (Modulus), then the method
_solve_radical
tries to simplify the radical, by removing it using techniques like squaring, cubing etc, and_solve_abs
solves nested Modulus by considering the positive and negative variants, iteratively.If none of the above method is successful, then methods of polynomial is used as follows:
The method to solve the rational function,
_solve_as_rational
, is called. Based on the domain, the respective poly solver_solve_as_poly_real
or_solve_as_poly_complex
is called to solvef
as a polynomial.The underlying method
_solve_as_poly
solves the equation using polynomial techniques if it’s already a polynomial equation or, with a change of variables, can be made so.The final solution set returned by
solveset
is the intersection of the set of solutions found above and the input domain.
How do we manipulate and return an infinite solution?¶
In the real domain, we use our
ImageSet
class in the sets module to return infinite solutions.ImageSet
is an image of a set under a mathematical function. For example, to represent the solution of the equation \(\sin{(x)} = 0\), we can use theImageSet
as:>>> from sympy import ImageSet, Lambda, pi, S, Dummy, pprint >>> n = Dummy('n') >>> pprint(ImageSet(Lambda(n, 2*pi*n), S.Integers), use_unicode=True) {2⋅n⋅π  n ∊ ℤ}Where
n
is a dummy variable. It is basically the image of the set of integers under the function \(2\pi n\).In the complex domain, we use complex sets, which are implemented as the
ComplexRegion
class in the sets module, to represent infinite solution in the Argand plane. For example to represent the solution of the equation \(z = 1\), which is a unit circle, we can use theComplexRegion
as:>>> from sympy import ComplexRegion, FiniteSet, Interval, pi, pprint >>> pprint(ComplexRegion(FiniteSet(1)*Interval(0, 2*pi), polar=True), use_unicode=True) {r⋅(ⅈ⋅sin(θ) + cos(θ))  r, θ ∊ {1} × [0, 2⋅π)}Where the
FiniteSet
in theProductSet
is the range of the value of \(r\), which is the radius of the circle and theInterval
is the range of \(\theta\), the angle from the \(x\) axis representing a unit circle in the Argand plane.Note: We also have nonpolar form notation for representing solution in rectangular form. For example, to represent first two quadrants in the Argand plane, we can write the
ComplexRegion
as:>>> from sympy import ComplexRegion, Interval, pi, oo, pprint >>> pprint(ComplexRegion(Interval(oo, oo)*Interval(0, oo)), use_unicode=True) {x + y⋅ⅈ  x, y ∊ (∞, ∞) × [0, ∞)}where the Intervals are the range of \(x\) and \(y\) for the set of complex numbers \(x + iy\).
How does solveset
ensure that it is not returning any wrong solution?¶
Solvers in a Computer Algebra System are based on heuristic algorithms, so it’s usually very hard to ensure 100% percent correctness, in every possible case. However there are still a lot of cases where we can ensure correctness. Solveset tries to verify correctness wherever it can. For example:
Consider the equation \(x = n\). A naive method to solve this equation would return
{n, n}
as its solution, which is not correct since{n, n}
can be its solution if and only ifn
is positive. Solveset returns this information as well to ensure correctness.>>> from sympy import symbols, S, pprint, solveset >>> x, n = symbols('x, n') >>> pprint(solveset(abs(x)  n, x, domain=S.Reals), use_unicode=True) {x  x ∊ {n, n} ∧ (n ∈ [0, ∞))}Though, there still a lot of work needs to be done in this regard.
Search based solver and stepbystep solution¶
Note: This is under Development.
After the introduction of
ConditionSet
, the solving of equations can be seen as set transformations. Here is an abstract view of the things we can do to solve equations.
Apply various set transformations on the given set.
Define a metric of the usability of solutions, or a notion of some solutions being better than others.
Different transformations would be the nodes of a tree.
Suitable searching techniques could be applied to get the best solution.
ConditionSet
gives us the ability to represent unevaluated equations and inequalities in forms like \(\{xf(x)=0; x \in S\}\) and \(\{xf(x)>0; x \in S\}\) but a more powerful thing aboutConditionSet
is that it allows us to write the intermediate steps as set to set transformation. Some of the transformations are:
Composition: \(\{xf(g(x))=0;x \in S\} \Rightarrow \{xg(x)=y; x \in S, y \in \{zf(z)=0; z \in S\}\}\)
 Polynomial Solver: \(\{x  P(x) = 0;x \in S\} \Rightarrow \{x_1,x_2, ... ,x_n\} \cap S\),
where \(x_i\) are roots of \(P(x)\).
Invert solver: \(\{xf(x)=0;x \in S\} \Rightarrow \{g(0) \text{ all g such that } f(g(x)) = x\}\)
 logcombine: \(\{x \log(f(x)) + \log(g(x));x \in S\}\)
\(\Rightarrow \{x \log(f(x).g(x)); x \in S\} \text{ if } f(x) > 0 \text{ and } g(x) > 0\) \(\Rightarrow \{x \log(f(x)) + \log(g(x));x \in S\} \text{ otherwise}\)
 product solve: \(\{xf(x)g(x)=0; x \in S\}\)
\(\Rightarrow \{xf(x)=0; x \in S\} U \{xg(x)=0; x \in S\}\) \(\text{ given } f(x) \text{ and } g(x) \text{ are bounded.}\) \(\Rightarrow \{xf(x)g(x)=0; x \in S\}, \text{ otherwise}\)
Since the output type is same as the input type any composition of these transformations is also a valid transformation. And our aim is to find the right sequence of compositions (given the atoms) which transforms the given condition set to a set which is not a condition set i.e., FiniteSet, Interval, Set of Integers and their Union, Intersection, Complement or ImageSet. We can assign a cost function to each set, such that, the more desirable that form of set is to us, the less the value of the cost function. This way our problem is now reduced to finding the path from the initial ConditionSet to the lowest valued set on a graph where the atomic transformations forms the edges.
How do we deal with cases where only some of the solutions are known?¶
Creating a universal equation solver, which can solve each and every equation we encounter in mathematics is an ideal case for solvers in a Computer Algebra System. When cases which are not solved or can only be solved incompletely, a
ConditionSet
is used and acts as an unevaluated solveset object.Note that, mathematically, finding a complete set of solutions for an equation is undecidable. See Richardson’s theorem.
ConditionSet
is basically a Set of elements which satisfy a given condition. For example, to represent the solutions of the equation in the real domain:\[(x^2  4)(\sin(x) + x)\]We can represent it as:
\(\{2, 2\} ∪ \{x  x \in \mathbb{R} ∧ x + \sin(x) = 0\}\)
What will you do with the old solve?¶
There are still a few things
solveset
can’t do, which the oldsolve
can, such as solving non linear multivariate & LambertW type equations. Hence, it’s not yet a perfect replacement for oldsolve
. The ultimate goal is to:
Replace
solve
withsolveset
once solveset is at least as powerful assolve
, i.e.,solveset
does everything thatsolve
can do currently, andeventually rename
solveset
tosolve
.
How are symbolic parameters handled in solveset?¶
Solveset is in its initial phase of development, so the symbolic parameters aren’t handled well for all the cases, but some work has been done in this regard to depict our ideology towards symbolic parameters. As an example, consider the solving of \(x = n\) for real \(x\), where \(n\) is a symbolic parameter. Solveset returns the value of \(x\) considering the domain of the symbolic parameter \(n\) as well:
\[([0, \infty) \cap \{n\}) \cup ((\infty, 0] \cap \{n\}).\]This simply means \(n\) is the solution only when it belongs to the
Interval
\([0, \infty)\) and \(n\) is the solution only when \(n\) belongs to theInterval
\(( \infty, 0]\).There are other cases to address too, like solving \(2^x + (a  2)\) for \(x\) where \(a\) is a symbolic parameter. As of now, It returns the solution as an intersection with \(\mathbb{R}\), which is trivial, as it doesn’t reveal the domain of \(a\) in the solution.
Recently, we have also implemented a function to find the domain of the expression in a FiniteSet (Intersection with the interval) in which it is notempty. It is a useful addition for dealing with symbolic parameters. For example:
>>> from sympy import Symbol, FiniteSet, Interval, not_empty_in, sqrt, oo >>> from sympy.abc import x >>> not_empty_in(FiniteSet(x/2).intersect(Interval(0, 1)), x) Interval(0, 2) >>> not_empty_in(FiniteSet(x, x**2).intersect(Interval(1, 2)), x) Union(Interval(sqrt(2), 1), Interval(1, 2))
References¶
 1
 2
https://github.com/sympy/sympy/wiki/GSoC2014ApplicationHarshGupta:Solvers
 3
https://github.com/sympy/sympy/wiki/GSoC2015ApplicationAMiTKumar–Solvers:ExtendingSolveset
 5
 6
https://github.com/sympy/sympy/pull/2948 : Action Plan for improving solvers.
 7
https://github.com/sympy/sympy/issues/6659 :
solve()
is a giant mess 8
https://github.com/sympy/sympy/pull/7523 :
solveset
PR 9
 10
 11
Solveset Module Reference¶
Use solveset()
to solve equations or expressions (assumed to be equal to 0) for a single variable.
Solving an equation like \(x^2 == 1\) can be done as follows:
>>> from sympy import solveset
>>> from sympy import Symbol, Eq
>>> x = Symbol('x')
>>> solveset(Eq(x**2, 1), x)
{1, 1}
Or one may manually rewrite the equation as an expression equal to 0:
>>> solveset(x**2  1, x)
{1, 1}
The first argument for solveset()
is an expression (equal to zero) or an equation and the second argument
is the symbol that we want to solve the equation for.

sympy.solvers.solveset.
solveset
(f, symbol=None, domain=S.Complexes)[source]¶ Solves a given inequality or equation with set as output
 Parameters
f : Expr or a relational.
The target equation or inequality
symbol : Symbol
The variable for which the equation is solved
domain : Set
The domain over which the equation is solved
 Returns
Set
A set of values for \(symbol\) for which \(f\) is True or is equal to zero. An \(EmptySet\) is returned if \(f\) is False or nonzero. A \(ConditionSet\) is returned as unsolved object if algorithms to evaluate complete solution are not yet implemented.
\(solveset\) claims to be complete in the solution set that it returns.
 Raises
NotImplementedError
The algorithms to solve inequalities in complex domain are not yet implemented.
ValueError
The input is not valid.
RuntimeError
It is a bug, please report to the github issue tracker.
Notes
Python interprets 0 and 1 as False and True, respectively, but in this function they refer to solutions of an expression. So 0 and 1 return the Domain and EmptySet, respectively, while True and False return the opposite (as they are assumed to be solutions of relational expressions).
Examples
>>> from sympy import exp, sin, Symbol, pprint, S >>> from sympy.solvers.solveset import solveset, solveset_real
The default domain is complex. Not specifying a domain will lead to the solving of the equation in the complex domain (and this is not affected by the assumptions on the symbol):
>>> x = Symbol('x') >>> pprint(solveset(exp(x)  1, x), use_unicode=False) {2*n*I*pi  n in Integers}
>>> x = Symbol('x', real=True) >>> pprint(solveset(exp(x)  1, x), use_unicode=False) {2*n*I*pi  n in Integers}
If you want to use \(solveset\) to solve the equation in the real domain, provide a real domain. (Using
solveset_real
does this automatically.)
>>> R = S.Reals >>> x = Symbol('x') >>> solveset(exp(x)  1, x, R) {0} >>> solveset_real(exp(x)  1, x) {0}
The solution is mostly unaffected by assumptions on the symbol, but there may be some slight difference:
>>> pprint(solveset(sin(x)/x,x), use_unicode=False) ({2*n*pi  n in Integers} \ {0}) U ({2*n*pi + pi  n in Integers} \ {0})
>>> p = Symbol('p', positive=True) >>> pprint(solveset(sin(p)/p, p), use_unicode=False) {2*n*pi  n in Integers} U {2*n*pi + pi  n in Integers}
Inequalities can be solved over the real domain only. Use of a complex domain leads to a NotImplementedError.
>>> solveset(exp(x) > 1, x, R) Interval.open(0, oo)
See also
solveset_real
solver for real domain
solveset_complex
solver for complex domain

sympy.solvers.solveset.
invert_real
(f_x, y, x, domain=Reals)[source]¶ Inverts a realvalued function. Same as _invert, but sets the domain to
S.Reals
before inverting.

sympy.solvers.solveset.
invert_complex
(f_x, y, x, domain=S.Complexes)¶ Reduce the complex valued equation
f(x) = y
to a set of equations{g(x) = h_1(y), g(x) = h_2(y), ..., g(x) = h_n(y) }
whereg(x)
is a simpler function thanf(x)
. The return value is a tuple(g(x), set_h)
, whereg(x)
is a function ofx
andset_h
is the set of function{h_1(y), h_2(y), ..., h_n(y)}
. Here,y
is not necessarily a symbol.The
set_h
contains the functions, along with the information about the domain in which they are valid, through set operations. For instance, ify = Abs(x)  n
is inverted in the real domain, thenset_h
is not simply \({n, n}\) as the nature of \(n\) is unknown; rather, it is: \(Intersection([0, oo) {n}) U Intersection((oo, 0], {n})\)By default, the complex domain is used which means that inverting even seemingly simple functions like
exp(x)
will give very different results from those obtained in the real domain. (In the case ofexp(x)
, the inversion vialog
is multivalued in the complex domain, having infinitely many branches.)If you are working with real values only (or you are not sure which function to use) you should probably set the domain to
S.Reals
(or use \(invert\_real\) which does that automatically).Examples
>>> from sympy.solvers.solveset import invert_complex, invert_real >>> from sympy.abc import x, y >>> from sympy import exp, log
When does exp(x) == y?
>>> invert_complex(exp(x), y, x) (x, ImageSet(Lambda(_n, I*(2*_n*pi + arg(y)) + log(Abs(y))), Integers)) >>> invert_real(exp(x), y, x) (x, Intersection({log(y)}, Reals))
When does exp(x) == 1?
>>> invert_complex(exp(x), 1, x) (x, ImageSet(Lambda(_n, 2*_n*I*pi), Integers)) >>> invert_real(exp(x), 1, x) (x, {0})
See also

sympy.solvers.solveset.
domain_check
(f, symbol, p)[source]¶ Returns False if point p is infinite or any subexpression of f is infinite or becomes so after replacing symbol with p. If none of these conditions is met then True will be returned.
Examples
>>> from sympy import Mul, oo >>> from sympy.abc import x >>> from sympy.solvers.solveset import domain_check >>> g = 1/(1 + (1/(x + 1))**2) >>> domain_check(g, x, 1) False >>> domain_check(x**2, x, 0) True >>> domain_check(1/x, x, oo) False
The function relies on the assumption that the original form of the equation has not been changed by automatic simplification.
>>> domain_check(x/x, x, 0) # x/x is automatically simplified to 1 True
To deal with automatic evaluations use evaluate=False:
>>> domain_check(Mul(x, 1/x, evaluate=False), x, 0) False
linear_eq_to_matrix¶

sympy.solvers.solveset.
linear_eq_to_matrix
(equations, *symbols)[source]¶ Converts a given System of Equations into Matrix form. Here \(equations\) must be a linear system of equations in \(symbols\). Element M[i, j] corresponds to the coefficient of the jth symbol in the ith equation.
The Matrix form corresponds to the augmented matrix form. For example:
\[4x + 2y + 3z = 1\]\[3x + y + z = 6\]\[2x + 4y + 9z = 2\]This system would return \(A\) & \(b\) as given below:
[ 4 2 3 ] [ 1 ] A = [ 3 1 1 ] b = [6 ] [ 2 4 9 ] [ 2 ]
The only simplification performed is to convert \(Eq(a, b) > a  b\).
 Raises
ValueError
The equations contain a nonlinear term. The symbols are not given or are not unique.
Examples
>>> from sympy import linear_eq_to_matrix, symbols >>> c, x, y, z = symbols('c, x, y, z')
The coefficients (numerical or symbolic) of the symbols will be returned as matrices:
>>> eqns = [c*x + z  1  c, y + z, x  y] >>> A, b = linear_eq_to_matrix(eqns, [x, y, z]) >>> A Matrix([ [c, 0, 1], [0, 1, 1], [1, 1, 0]]) >>> b Matrix([ [c + 1], [ 0], [ 0]])
This routine does not simplify expressions and will raise an error if nonlinearity is encountered:
>>> eqns = [ ... (x**2  3*x)/(x  3)  3, ... y**2  3*y  y*(y  4) + x  4] >>> linear_eq_to_matrix(eqns, [x, y]) Traceback (most recent call last): ... ValueError: The term (x**2  3*x)/(x  3) is nonlinear in {x, y}
Simplifying these equations will discard the removable singularity in the first, reveal the linear structure of the second:
>>> [e.simplify() for e in eqns] [x  3, x + y  4]
Any such simplification needed to eliminate nonlinear terms must be done before calling this routine.
linsolve¶

sympy.solvers.solveset.
linsolve
(system, *symbols)[source]¶ Solve system of N linear equations with M variables; both underdetermined and overdetermined systems are supported. The possible number of solutions is zero, one or infinite. Zero solutions throws a ValueError, whereas infinite solutions are represented parametrically in terms of the given symbols. For unique solution a FiniteSet of ordered tuples is returned.
All Standard input formats are supported: For the given set of Equations, the respective input types are given below:
\[3x + 2y  z = 1\]\[2x  2y + 4z = 2\]\[2x  y + 2z = 0\]Augmented Matrix Form, \(system\) given below:
[3 2 1 1] system = [2 2 4 2] [2 1 2 0]
List Of Equations Form
\(system = [3x + 2y  z  1, 2x  2y + 4z + 2, 2x  y + 2z]\)
Input A & b Matrix Form (from Ax = b) are given as below:
[3 2 1 ] [ 1 ] A = [2 2 4 ] b = [ 2 ] [2 1 2 ] [ 0 ]
\(system = (A, b)\)
Symbols can always be passed but are actually only needed when 1) a system of equations is being passed and 2) the system is passed as an underdetermined matrix and one wants to control the name of the free variables in the result. An error is raised if no symbols are used for case 1, but if no symbols are provided for case 2, internally generated symbols will be provided. When providing symbols for case 2, there should be at least as many symbols are there are columns in matrix A.
The algorithm used here is GaussJordan elimination, which results, after elimination, in a row echelon form matrix.
 Returns
A FiniteSet containing an ordered tuple of values for the
unknowns for which the \(system\) has a solution. (Wrapping
the tuple in FiniteSet is used to maintain a consistent
output format throughout solveset.)
Returns EmptySet(), if the linear system is inconsistent.
 Raises
ValueError
The input is not valid. The symbols are not given.
Examples
>>> from sympy import Matrix, S, linsolve, symbols >>> x, y, z = symbols("x, y, z") >>> A = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 10]]) >>> b = Matrix([3, 6, 9]) >>> A Matrix([ [1, 2, 3], [4, 5, 6], [7, 8, 10]]) >>> b Matrix([ [3], [6], [9]]) >>> linsolve((A, b), [x, y, z]) {(1, 2, 0)}
Parametric Solution: In case the system is underdetermined, the function will return a parametric solution in terms of the given symbols. Those that are free will be returned unchanged. e.g. in the system below, \(z\) is returned as the solution for variable z; it can take on any value.
>>> A = Matrix([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) >>> b = Matrix([3, 6, 9]) >>> linsolve((A, b), x, y, z) {(z  1, 2  2*z, z)}
If no symbols are given, internally generated symbols will be used. The \(tau0\) in the 3rd position indicates (as before) that the 3rd variable – whatever it’s named – can take on any value:
>>> linsolve((A, b)) {(tau0  1, 2  2*tau0, tau0)}
List of Equations as input
>>> Eqns = [3*x + 2*y  z  1, 2*x  2*y + 4*z + 2,  x + y/2  z] >>> linsolve(Eqns, x, y, z) {(1, 2, 2)}
Augmented Matrix as input
>>> aug = Matrix([[2, 1, 3, 1], [2, 6, 8, 3], [6, 8, 18, 5]]) >>> aug Matrix([ [2, 1, 3, 1], [2, 6, 8, 3], [6, 8, 18, 5]]) >>> linsolve(aug, x, y, z) {(3/10, 2/5, 0)}
Solve for symbolic coefficients
>>> a, b, c, d, e, f = symbols('a, b, c, d, e, f') >>> eqns = [a*x + b*y  c, d*x + e*y  f] >>> linsolve(eqns, x, y) {((b*f + c*e)/(a*e  b*d), (a*f  c*d)/(a*e  b*d))}
A degenerate system returns solution as set of given symbols.
>>> system = Matrix(([0, 0, 0], [0, 0, 0], [0, 0, 0])) >>> linsolve(system, x, y) {(x, y)}
For an empty system linsolve returns empty set
>>> linsolve([], x) EmptySet()
An error is raised if, after expansion, any nonlinearity is detected:
>>> linsolve([x*(1/x  1), (y  1)**2  y**2 + 1], x, y) {(1, 1)} >>> linsolve([x**2  1], x) Traceback (most recent call last): ... ValueError: The term x**2 is nonlinear in {x}
nonlinsolve¶

sympy.solvers.solveset.
nonlinsolve
(system, *symbols)[source]¶ Solve system of N non linear equations with M variables, which means both under and overdetermined systems are supported. Positive dimensional system is also supported (A system with infinitely many solutions is said to be positivedimensional). In Positive dimensional system solution will be dependent on at least one symbol. Returns both real solution and complex solution(If system have). The possible number of solutions is zero, one or infinite.
 Parameters
system : list of equations
The target system of equations
symbols : list of Symbols
symbols should be given as a sequence eg. list
 Returns
A FiniteSet of ordered tuple of values of \(symbols\) for which the \(system\)
has solution. Order of values in the tuple is same as symbols present in
the parameter \(symbols\).
Please note that general FiniteSet is unordered, the solution returned
here is not simply a FiniteSet of solutions, rather it is a FiniteSet of
ordered tuple, i.e. the first & only argument to FiniteSet is a tuple of
solutions, which is ordered, & hence the returned solution is ordered.
Also note that solution could also have been returned as an ordered tuple,
FiniteSet is just a wrapper \({}\) around the tuple. It has no other
significance except for the fact it is just used to maintain a consistent
output format throughout the solveset.
For the given set of Equations, the respective input types
are given below:
\[x*y  1 = 0\]\[4*x**2 + y**2  5 = 0\]\(system = [x*y  1, 4*x**2 + y**2  5]\)
\(symbols = [x, y]\)
 Raises
ValueError
The input is not valid. The symbols are not given.
AttributeError
The input symbols are not \(Symbol\) type.
Examples
>>> from sympy.core.symbol import symbols >>> from sympy.solvers.solveset import nonlinsolve >>> x, y, z = symbols('x, y, z', real=True) >>> nonlinsolve([x*y  1, 4*x**2 + y**2  5], [x, y]) {(1, 1), (1/2, 2), (1/2, 2), (1, 1)}
Positive dimensional system and complements:
>>> from sympy import pprint >>> from sympy.polys.polytools import is_zero_dimensional >>> a, b, c, d = symbols('a, b, c, d', extended_real=True) >>> eq1 = a + b + c + d >>> eq2 = a*b + b*c + c*d + d*a >>> eq3 = a*b*c + b*c*d + c*d*a + d*a*b >>> eq4 = a*b*c*d  1 >>> system = [eq1, eq2, eq3, eq4] >>> is_zero_dimensional(system) False >>> pprint(nonlinsolve(system, [a, b, c, d]), use_unicode=False) 1 1 1 1 {(, d, , {d} \ {0}), (, d, , {d} \ {0})} d d d d >>> nonlinsolve([(x+y)**2  4, x + y  2], [x, y]) {(2  y, y)}
2. If some of the equations are nonpolynomial then \(nonlinsolve\) will call the \(substitution\) function and return real and complex solutions, if present.
>>> from sympy import exp, sin >>> nonlinsolve([exp(x)  sin(y), y**2  4], [x, y]) {(ImageSet(Lambda(_n, I*(2*_n*pi + pi) + log(sin(2))), Integers), 2), (ImageSet(Lambda(_n, 2*_n*I*pi + log(sin(2))), Integers), 2)}
3. If system is nonlinear polynomial and zerodimensional then it returns both solution (real and complex solutions, if present) using \(solve_poly_system\):
>>> from sympy import sqrt >>> nonlinsolve([x**2  2*y**2 2, x*y  2], [x, y]) {(2, 1), (2, 1), (sqrt(2)*I, sqrt(2)*I), (sqrt(2)*I, sqrt(2)*I)}
4. \(nonlinsolve\) can solve some linear (zero or positive dimensional) system (because it uses the \(groebner\) function to get the groebner basis and then uses the \(substitution\) function basis as the new \(system\)). But it is not recommended to solve linear system using \(nonlinsolve\), because \(linsolve\) is better for general linear systems.
>>> nonlinsolve([x + 2*y z  3, x  y  4*z + 9 , y + z  4], [x, y, z]) {(3*z  5, 4  z, z)}
5. System having polynomial equations and only real solution is solved using \(solve_poly_system\):
>>> e1 = sqrt(x**2 + y**2)  10 >>> e2 = sqrt(y**2 + (x + 10)**2)  3 >>> nonlinsolve((e1, e2), (x, y)) {(191/20, 3*sqrt(391)/20), (191/20, 3*sqrt(391)/20)} >>> nonlinsolve([x**2 + 2/y  2, x + y  3], [x, y]) {(1, 2), (1  sqrt(5), 2 + sqrt(5)), (1 + sqrt(5), 2  sqrt(5))} >>> nonlinsolve([x**2 + 2/y  2, x + y  3], [y, x]) {(2, 1), (2  sqrt(5), 1 + sqrt(5)), (2 + sqrt(5), 1  sqrt(5))}
6. It is better to use symbols instead of Trigonometric Function or Function (e.g. replace \(sin(x)\) with symbol, replace \(f(x)\) with symbol and so on. Get soln from \(nonlinsolve\) and then using \(solveset\) get the value of \(x\))
How Nonlinsolve Is Better Than Old Solver \(_solve_system\) :
1. A positive dimensional system solver : nonlinsolve can return solution for positive dimensional system. It finds the Groebner Basis of the positive dimensional system(calling it as basis) then we can start solving equation(having least number of variable first in the basis) using solveset and substituting that solved solutions into other equation(of basis) to get solution in terms of minimum variables. Here the important thing is how we are substituting the known values and in which equations.
2. Real and Complex both solutions : nonlinsolve returns both real and complex solution. If all the equations in the system are polynomial then using \(solve_poly_system\) both real and complex solution is returned. If all the equations in the system are not polynomial equation then goes to \(substitution\) method with this polynomial and non polynomial equation(s), to solve for unsolved variables. Here to solve for particular variable solveset_real and solveset_complex is used. For both real and complex solution function \(_solve_using_know_values\) is used inside \(substitution\) function.(\(substitution\) function will be called when there is any non polynomial equation(s) is present). When solution is valid then add its general solution in the final result.
3. Complement and Intersection will be added if any : nonlinsolve maintains dict for complements and Intersections. If solveset find complements or/and Intersection with any Interval or set during the execution of \(substitution\) function ,then complement or/and Intersection for that variable is added before returning final solution.
transolve¶

sympy.solvers.solveset.
_transolve
(f, symbol, domain)[source]¶ Function to solve transcendental equations. It is a helper to
solveset
and should be used internally._transolve
currently supports the following class of equations:Exponential equations
Logarithmic equations
 Parameters
f : Any transcendental equation that needs to be solved.
This needs to be an expression, which is assumed to be equal to
0
.symbol : The variable for which the equation is solved.
This needs to be of class
Symbol
.domain : A set over which the equation is solved.
This needs to be of class
Set
. Returns
Set
A set of values for
symbol
for whichf
is equal to zero. AnEmptySet
is returned iff
does not have solutions in respective domain. AConditionSet
is returned as unsolved object if algorithms to evaluate complete solution are not yet implemented.
How To Use
_transolve
_transolve
should not be used as an independent function, because it assumes that the equation (f
) and thesymbol
comes fromsolveset
and might have undergone a few modification(s). To use_transolve
as an independent function the equation (f
) and thesymbol
should be passed as they would have been bysolveset
.Examples
>>> from sympy.solvers.solveset import _transolve as transolve >>> from sympy.solvers.solvers import _tsolve as tsolve >>> from sympy import symbols, S, pprint >>> x = symbols('x', real=True) # assumption added >>> transolve(5**(x  3)  3**(2*x + 1), x, S.Reals) {(log(3) + 3*log(5))/(log(5) + 2*log(3))}
How
_transolve
Works_transolve
uses two types of helper functions to solve equations of a particular class:Identifying helpers: To determine whether a given equation belongs to a certain class of equation or not. Returns either
True
orFalse
.Solving helpers: Once an equation is identified, a corresponding helper either solves the equation or returns a form of the equation that
solveset
might better be able to handle.Philosophy behind the module
The purpose of
_transolve
is to take equations which are not already polynomial in their generator(s) and to either recast them as such through a valid transformation or to solve them outright. A pair of helper functions for each class of supported transcendental functions are employed for this purpose. One identifies the transcendental form of an equation and the other either solves it or recasts it into a tractable form that can be solved bysolveset
. For example, an equation in the form \(ab^{f(x)}  cd^{g(x)} = 0\) can be transformed to \(\log(a) + f(x)\log(b)  \log(c)  g(x)\log(d) = 0\) (under certain assumptions) and this can be solved withsolveset
if \(f(x)\) and \(g(x)\) are in polynomial form.How
_transolve
Is Better Than_tsolve
Better output
_transolve
provides expressions in a more simplified form.Consider a simple exponential equation
>>> f = 3**(2*x)  2**(x + 3) >>> pprint(transolve(f, x, S.Reals), use_unicode=False) 3*log(2) {} 2*log(3) + log(2) >>> pprint(tsolve(f, x), use_unicode=False) / 3 \    log(2/9) [log\2 /]
Extensible
The API of
_transolve
is designed such that it is easily extensible, i.e. the code that solves a given class of equations is encapsulated in a helper and not mixed in with the code of_transolve
itself.Modular
_transolve
is designed to be modular i.e, for every class of equation a separate helper for identification and solving is implemented. This makes it easy to change or modify any of the method implemented directly in the helpers without interfering with the actual structure of the API.Faster Computation
Solving equation via
_transolve
is much faster as compared to_tsolve
. Insolve
, attempts are made computing every possibility to get the solutions. This series of attempts makes solving a bit slow. In_transolve
, computation begins only after a particular type of equation is identified.How To Add New Class Of Equations
Adding a new class of equation solver is a threestep procedure:
Identify the type of the equations
Determine the type of the class of equations to which they belong: it could be of
Add
,Pow
, etc. types. Separate internal functions are used for each type. Write identification and solving helpers and use them from within the routine for the given type of equation (after adding it, if necessary). Something like:def add_type(lhs, rhs, x): .... if _is_exponential(lhs, x): new_eq = _solve_exponential(lhs, rhs, x) .... rhs, lhs = eq.as_independent(x) if lhs.is_Add: result = add_type(lhs, rhs, x)
Define the identification helper.
Define the solving helper.
Apart from this, a few other things needs to be taken care while adding an equation solver:
Naming conventions: Name of the identification helper should be as
_is_class
where class will be the name or abbreviation of the class of equation. The solving helper will be named as_solve_class
. For example: for exponential equations it becomes_is_exponential
and_solve_expo
.The identifying helpers should take two input parameters, the equation to be checked and the variable for which a solution is being sought, while solving helpers would require an additional domain parameter.
Be sure to consider corner cases.
Add tests for each helper.
Add a docstring to your helper that describes the method implemented. The documentation of the helpers should identify:
the purpose of the helper,
the method used to identify and solve the equation,
a proof of correctness
the return values of the helpers

sympy.solvers.solveset.
_is_exponential
(f, symbol)[source]¶ Return
True
if one or more terms containsymbol
only in exponents, elseFalse
. Parameters
f : Expr
The equation to be checked
symbol : Symbol
The variable in which the equation is checked
Examples
>>> from sympy import symbols, cos, exp >>> from sympy.solvers.solveset import _is_exponential as check >>> x, y = symbols('x y') >>> check(y, y) False >>> check(x**y  1, y) True >>> check(x**y*2**y  1, y) True >>> check(exp(x + 3) + 3**x, x) True >>> check(cos(2**x), x) False
Philosophy behind the helper
The function extracts each term of the equation and checks if it is of exponential form w.r.t
symbol
.

sympy.solvers.solveset.
_solve_exponential
(lhs, rhs, symbol, domain)[source]¶ Helper function for solving (supported) exponential equations.
Exponential equations are the sum of (currently) at most two terms with one or both of them having a power with a symboldependent exponent.
For example
\[5^{2x + 3}  5^{3x  1}\]\[4^{5  9x}  e^{2  x}\] Parameters
lhs, rhs : Expr
The exponential equation to be solved, \(lhs = rhs\)
symbol : Symbol
The variable in which the equation is solved
domain : Set
A set over which the equation is solved.
 Returns
A set of solutions satisfying the given equation.
A
ConditionSet
if the equation is unsolvable orif the assumptions are not properly defined, in that case
a different style of
ConditionSet
is returned having thesolution(s) of the equation with the desired assumptions.
Examples
>>> from sympy.solvers.solveset import _solve_exponential as solve_expo >>> from sympy import symbols, S >>> x = symbols('x', real=True) >>> a, b = symbols('a b') >>> solve_expo(2**x + 3**x  5**x, 0, x, S.Reals) # not solvable ConditionSet(x, Eq(2**x + 3**x  5**x, 0), Reals) >>> solve_expo(a**x  b**x, 0, x, S.Reals) # solvable but incorrect assumptions ConditionSet(x, (a > 0) & (b > 0), {0}) >>> solve_expo(3**(2*x)  2**(x + 3), 0, x, S.Reals) {3*log(2)/(2*log(3) + log(2))} >>> solve_expo(2**x  4**x, 0, x, S.Reals) {0}
Proof of correctness of the method
The logarithm function is the inverse of the exponential function. The defining relation between exponentiation and logarithm is:
\[{\log_b x} = y \enspace if \enspace b^y = x\]Therefore if we are given an equation with exponent terms, we can convert every term to its corresponding logarithmic form. This is achieved by taking logarithms and expanding the equation using logarithmic identities so that it can easily be handled by
solveset
.For example:
\[3^{2x} = 2^{x + 3}\]Taking log both sides will reduce the equation to
\[(2x)\log(3) = (x + 3)\log(2)\]This form can be easily handed by
solveset
.

sympy.solvers.solveset.
_solve_logarithm
(lhs, rhs, symbol, domain)[source]¶ Helper to solve logarithmic equations which are reducible to a single instance of \(\log\).
Logarithmic equations are (currently) the equations that contains \(\log\) terms which can be reduced to a single \(\log\) term or a constant using various logarithmic identities.
For example:
\[\log(x) + \log(x  4)\]can be reduced to:
\[\log(x(x  4))\] Parameters
lhs, rhs : Expr
The logarithmic equation to be solved, \(lhs = rhs\)
symbol : Symbol
The variable in which the equation is solved
domain : Set
A set over which the equation is solved.
 Returns
A set of solutions satisfying the given equation.
A
ConditionSet
if the equation is unsolvable.
Examples
>>> from sympy import symbols, log, S >>> from sympy.solvers.solveset import _solve_logarithm as solve_log >>> x = symbols('x') >>> f = log(x  3) + log(x + 3) >>> solve_log(f, 0, x, S.Reals) {sqrt(10), sqrt(10)}
Proof of correctness
A logarithm is another way to write exponent and is defined by
\[{\log_b x} = y \enspace if \enspace b^y = x\]When one side of the equation contains a single logarithm, the equation can be solved by rewriting the equation as an equivalent exponential equation as defined above. But if one side contains more than one logarithm, we need to use the properties of logarithm to condense it into a single logarithm.
Take for example
\[\log(2x)  15 = 0\]contains single logarithm, therefore we can directly rewrite it to exponential form as
\[x = \frac{e^{15}}{2}\]But if the equation has more than one logarithm as
\[\log(x  3) + \log(x + 3) = 0\]we use logarithmic identities to convert it into a reduced form
Using,
\[\log(a) + \log(b) = \log(ab)\]the equation becomes,
\[\log((x  3)(x + 3))\]This equation contains one logarithm and can be solved by rewriting to exponents.

sympy.solvers.solveset.
_is_logarithmic
(f, symbol)[source]¶ Return
True
if the equation is in the form \(a\log(f(x)) + b\log(g(x)) + ... + c\) elseFalse
. Parameters
f : Expr
The equation to be checked
symbol : Symbol
The variable in which the equation is checked
 Returns
True
if the equation is logarithmic otherwiseFalse
.
Examples
>>> from sympy import symbols, tan, log >>> from sympy.solvers.solveset import _is_logarithmic as check >>> x, y = symbols('x y') >>> check(log(x + 2)  log(x + 3), x) True >>> check(tan(log(2*x)), x) False >>> check(x*log(x), x) False >>> check(x + log(x), x) False >>> check(y + log(x), x) True
Philosophy behind the helper
The function extracts each term and checks whether it is logarithmic w.r.t
symbol
.
Diophantine Equations (DEs)¶
See Diophantine