galgebra.atoms¶
Sympy primitives for representing atos ga expressions
Members¶

class
galgebra.atoms.
BasisVectorSymbol
[source]¶ A symbol representing a basis vector
Symbols are identified by name and assumptions:
>>> from sympy import Symbol >>> Symbol("x") == Symbol("x")
True >>> Symbol(“x”, real=True) == Symbol(“x”, real=False) False

apart
(x=None, **args)¶ See the apart function in sympy.polys

property
args
¶ Returns a tuple of arguments of ‘self’.
Examples
>>> from sympy import cot >>> from sympy.abc import x, y
>>> cot(x).args (x,)
>>> cot(x).args[0] x
>>> (x*y).args (x, y)
>>> (x*y).args[1] y
Notes
Never use self._args, always use self.args. Only use _args in __new__ when creating a new function. Don’t override .args() from Basic (so that it’s easy to change the interface in the future if needed).

args_cnc
(cset=False, warn=True, split_1=True)¶ Return [commutative factors, noncommutative factors] of self.
self is treated as a Mul and the ordering of the factors is maintained. If
cset
is True the commutative factors will be returned in a set. If there were repeated factors (as may happen with an unevaluated Mul) then an error will be raised unless it is explicitly suppressed by settingwarn
to False.Note: 1 is always separated from a Number unless split_1 is False.
>>> from sympy import symbols, oo >>> A, B = symbols('A B', commutative=0) >>> x, y = symbols('x y') >>> (2*x*y).args_cnc() [[1, 2, x, y], []] >>> (2.5*x).args_cnc() [[1, 2.5, x], []] >>> (2*x*A*B*y).args_cnc() [[1, 2, x, y], [A, B]] >>> (2*x*A*B*y).args_cnc(split_1=False) [[2, x, y], [A, B]] >>> (2*x*y).args_cnc(cset=True) [{1, 2, x, y}, []]
The arg is always treated as a Mul:
>>> (2 + x + A).args_cnc() [[], [x  2 + A]] >>> (oo).args_cnc() # oo is a singleton [[1, oo], []]

as_coeff_Add
(rational=False)¶ Efficiently extract the coefficient of a summation.

as_coeff_Mul
(rational=False)¶ Efficiently extract the coefficient of a product.

as_coeff_add
(*deps)¶ Return the tuple (c, args) where self is written as an Add,
a
.c should be a Rational added to any terms of the Add that are independent of deps.
args should be a tuple of all other terms of
a
; args is empty if self is a Number or if self is independent of deps (when given).This should be used when you don’t know if self is an Add or not but you want to treat self as an Add or if you want to process the individual arguments of the tail of self as an Add.
if you know self is an Add and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail.
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_add() (3, ()) >>> (3 + x).as_coeff_add() (3, (x,)) >>> (3 + x + y).as_coeff_add(x) (y + 3, (x,)) >>> (3 + y).as_coeff_add(x) (y + 3, ())

as_coeff_exponent
(x)¶ c*x**e > c,e
where x can be any symbolic expression.

as_coeff_mul
(*deps, **kwargs)¶ Return the tuple (c, args) where self is written as a Mul,
m
.c should be a Rational multiplied by any factors of the Mul that are independent of deps.
args should be a tuple of all other factors of m; args is empty if self is a Number or if self is independent of deps (when given).
This should be used when you don’t know if self is a Mul or not but you want to treat self as a Mul or if you want to process the individual arguments of the tail of self as a Mul.
if you know self is a Mul and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail;
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_mul() (3, ()) >>> (3*x*y).as_coeff_mul() (3, (x, y)) >>> (3*x*y).as_coeff_mul(x) (3*y, (x,)) >>> (3*y).as_coeff_mul(x) (3*y, ())

as_coefficient
(expr)¶ Extracts symbolic coefficient at the given expression. In other words, this functions separates ‘self’ into the product of ‘expr’ and ‘expr’free coefficient. If such separation is not possible it will return None.
Examples
>>> from sympy import E, pi, sin, I, Poly >>> from sympy.abc import x
>>> E.as_coefficient(E) 1 >>> (2*E).as_coefficient(E) 2 >>> (2*sin(E)*E).as_coefficient(E)
Two terms have E in them so a sum is returned. (If one were desiring the coefficient of the term exactly matching E then the constant from the returned expression could be selected. Or, for greater precision, a method of Poly can be used to indicate the desired term from which the coefficient is desired.)
>>> (2*E + x*E).as_coefficient(E) x + 2 >>> _.args[0] # just want the exact match 2 >>> p = Poly(2*E + x*E); p Poly(x*E + 2*E, x, E, domain='ZZ') >>> p.coeff_monomial(E) 2 >>> p.nth(0, 1) 2
Since the following cannot be written as a product containing E as a factor, None is returned. (If the coefficient
2*x
is desired then thecoeff
method should be used.)>>> (2*E*x + x).as_coefficient(E) >>> (2*E*x + x).coeff(E) 2*x
>>> (E*(x + 1) + x).as_coefficient(E)
>>> (2*pi*I).as_coefficient(pi*I) 2 >>> (2*I).as_coefficient(pi*I)
See also
coeff()
return sum of terms have a given factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used

as_coefficients_dict
()¶ Return a dictionary mapping terms to their Rational coefficient. Since the dictionary is a defaultdict, inquiries about terms which were not present will return a coefficient of 0. If an expression is not an Add it is considered to have a single term.
Examples
>>> from sympy.abc import a, x >>> (3*x + a*x + 4).as_coefficients_dict() {1: 4, x: 3, a*x: 1} >>> _[a] 0 >>> (3*a*x).as_coefficients_dict() {a*x: 3}

as_content_primitive
(radical=False, clear=True)¶ This method should recursively remove a Rational from all arguments and return that (content) and the new self (primitive). The content should always be positive and
Mul(*foo.as_content_primitive()) == foo
. The primitive need not be in canonical form and should try to preserve the underlying structure if possible (i.e. expand_mul should not be applied to self).Examples
>>> from sympy import sqrt >>> from sympy.abc import x, y, z
>>> eq = 2 + 2*x + 2*y*(3 + 3*y)
The as_content_primitive function is recursive and retains structure:
>>> eq.as_content_primitive() (2, x + 3*y*(y + 1) + 1)
Integer powers will have Rationals extracted from the base:
>>> ((2 + 6*x)**2).as_content_primitive() (4, (3*x + 1)**2) >>> ((2 + 6*x)**(2*y)).as_content_primitive() (1, (2*(3*x + 1))**(2*y))
Terms may end up joining once their as_content_primitives are added:
>>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (11, x*(y + 1)) >>> ((3*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (9, x*(y + 1)) >>> ((3*(z*(1 + y)) + 2.0*x*(3 + 3*y))).as_content_primitive() (1, 6.0*x*(y + 1) + 3*z*(y + 1)) >>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))**2).as_content_primitive() (121, x**2*(y + 1)**2) >>> ((x*(1 + y) + 0.4*x*(3 + 3*y))**2).as_content_primitive() (1, 4.84*x**2*(y + 1)**2)
Radical content can also be factored out of the primitive:
>>> (2*sqrt(2) + 4*sqrt(10)).as_content_primitive(radical=True) (2, sqrt(2)*(1 + 2*sqrt(5)))
If clear=False (default is True) then content will not be removed from an Add if it can be distributed to leave one or more terms with integer coefficients.
>>> (x/2 + y).as_content_primitive() (1/2, x + 2*y) >>> (x/2 + y).as_content_primitive(clear=False) (1, x/2 + y)

as_dummy
()¶ Return the expression with any objects having structurally bound symbols replaced with unique, canonical symbols within the object in which they appear and having only the default assumption for commutativity being True.
Examples
>>> from sympy import Integral, Symbol >>> from sympy.abc import x, y >>> r = Symbol('r', real=True) >>> Integral(r, (r, x)).as_dummy() Integral(_0, (_0, x)) >>> _.variables[0].is_real is None True
Notes
Any object that has structural dummy variables should have a property, bound_symbols that returns a list of structural dummy symbols of the object itself.
Lambda and Subs have bound symbols, but because of how they are cached, they already compare the same regardless of their bound symbols:
>>> from sympy import Lambda >>> Lambda(x, x + 1) == Lambda(y, y + 1) True

as_expr
(*gens)¶ Convert a polynomial to a SymPy expression.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y
>>> f = (x**2 + x*y).as_poly(x, y) >>> f.as_expr() x**2 + x*y
>>> sin(x).as_expr() sin(x)

as_independent
(*deps, **hint)¶ A mostly naive separation of a Mul or Add into arguments that are not are dependent on deps. To obtain as complete a separation of variables as possible, use a separation method first, e.g.:
separatevars() to change Mul, Add and Pow (including exp) into Mul
.expand(mul=True) to change Add or Mul into Add
.expand(log=True) to change log expr into an Add
The only nonnaive thing that is done here is to respect noncommutative ordering of variables and to always return (0, 0) for self of zero regardless of hints.
For nonzero self, the returned tuple (i, d) has the following interpretation:
i will has no variable that appears in deps
d will either have terms that contain variables that are in deps, or be equal to 0 (when self is an Add) or 1 (when self is a Mul)
if self is an Add then self = i + d
if self is a Mul then self = i*d
otherwise (self, S.One) or (S.One, self) is returned.
To force the expression to be treated as an Add, use the hint as_Add=True
Examples
– self is an Add
>>> from sympy import sin, cos, exp >>> from sympy.abc import x, y, z
>>> (x + x*y).as_independent(x) (0, x*y + x) >>> (x + x*y).as_independent(y) (x, x*y) >>> (2*x*sin(x) + y + x + z).as_independent(x) (y + z, 2*x*sin(x) + x) >>> (2*x*sin(x) + y + x + z).as_independent(x, y) (z, 2*x*sin(x) + x + y)
– self is a Mul
>>> (x*sin(x)*cos(y)).as_independent(x) (cos(y), x*sin(x))
noncommutative terms cannot always be separated out when self is a Mul
>>> from sympy import symbols >>> n1, n2, n3 = symbols('n1 n2 n3', commutative=False) >>> (n1 + n1*n2).as_independent(n2) (n1, n1*n2) >>> (n2*n1 + n1*n2).as_independent(n2) (0, n1*n2 + n2*n1) >>> (n1*n2*n3).as_independent(n1) (1, n1*n2*n3) >>> (n1*n2*n3).as_independent(n2) (n1, n2*n3) >>> ((xn1)*(xy)).as_independent(x) (1, (x  y)*(x  n1))
– self is anything else:
>>> (sin(x)).as_independent(x) (1, sin(x)) >>> (sin(x)).as_independent(y) (sin(x), 1) >>> exp(x+y).as_independent(x) (1, exp(x + y))
– force self to be treated as an Add:
>>> (3*x).as_independent(x, as_Add=True) (0, 3*x)
– force self to be treated as a Mul:
>>> (3+x).as_independent(x, as_Add=False) (1, x + 3) >>> (3+x).as_independent(x, as_Add=False) (1, x  3)
Note how the below differs from the above in making the constant on the dep term positive.
>>> (y*(3+x)).as_independent(x) (y, x  3)
 – use .as_independent() for true independence testing instead
of .has(). The former considers only symbols in the free symbols while the latter considers all symbols
>>> from sympy import Integral >>> I = Integral(x, (x, 1, 2)) >>> I.has(x) True >>> x in I.free_symbols False >>> I.as_independent(x) == (I, 1) True >>> (I + x).as_independent(x) == (I, x) True
Note: when trying to get independent terms, a separation method might need to be used first. In this case, it is important to keep track of what you send to this routine so you know how to interpret the returned values
>>> from sympy import separatevars, log >>> separatevars(exp(x+y)).as_independent(x) (exp(y), exp(x)) >>> (x + x*y).as_independent(y) (x, x*y) >>> separatevars(x + x*y).as_independent(y) (x, y + 1) >>> (x*(1 + y)).as_independent(y) (x, y + 1) >>> (x*(1 + y)).expand(mul=True).as_independent(y) (x, x*y) >>> a, b=symbols('a b', positive=True) >>> (log(a*b).expand(log=True)).as_independent(b) (log(a), log(b))
See also
separatevars()
,expand()
,sympy.core.add.Add.as_two_terms()
,sympy.core.mul.Mul.as_two_terms()
,as_coeff_add()
,as_coeff_mul()

as_leading_term
(*symbols)¶ Returns the leading (nonzero) term of the series expansion of self.
The _eval_as_leading_term routines are used to do this, and they must always return a nonzero value.
Examples
>>> from sympy.abc import x >>> (1 + x + x**2).as_leading_term(x) 1 >>> (1/x**2 + x + x**2).as_leading_term(x) x**(2)

as_numer_denom
()¶ expression > a/b > a, b
This is just a stub that should be defined by an object’s class methods to get anything else.
See also
normal()
return a/b instead of a, b

as_ordered_factors
(order=None)¶ Return list of ordered factors (if Mul) else [self].

as_ordered_terms
(order=None, data=False)¶ Transform an expression to an ordered list of terms.
Examples
>>> from sympy import sin, cos >>> from sympy.abc import x
>>> (sin(x)**2*cos(x) + sin(x)**2 + 1).as_ordered_terms() [sin(x)**2*cos(x), sin(x)**2, 1]

as_poly
(*gens, **args)¶ Converts
self
to a polynomial or returnsNone
.>>> from sympy import sin >>> from sympy.abc import x, y
>>> print((x**2 + x*y).as_poly()) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + x*y).as_poly(x, y)) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + sin(y)).as_poly(x, y)) None

as_powers_dict
()¶ Return self as a dictionary of factors with each factor being treated as a power. The keys are the bases of the factors and the values, the corresponding exponents. The resulting dictionary should be used with caution if the expression is a Mul and contains non commutative factors since the order that they appeared will be lost in the dictionary.
See also
as_ordered_factors()
An alternative for noncommutative applications, returning an ordered list of factors.
args_cnc()
Similar to as_ordered_factors, but guarantees separation of commutative and noncommutative factors.

as_real_imag
(deep=True, **hints)¶ Performs complex expansion on ‘self’ and returns a tuple containing collected both real and imaginary parts. This method can’t be confused with re() and im() functions, which does not perform complex expansion at evaluation.
However it is possible to expand both re() and im() functions and get exactly the same results as with a single call to this function.
>>> from sympy import symbols, I
>>> x, y = symbols('x,y', real=True)
>>> (x + y*I).as_real_imag() (x, y)
>>> from sympy.abc import z, w
>>> (z + w*I).as_real_imag() (re(z)  im(w), re(w) + im(z))

as_set
()¶ Rewrites Boolean expression in terms of real sets.
Examples
>>> from sympy import Symbol, Eq, Or, And >>> x = Symbol('x', real=True) >>> Eq(x, 0).as_set() FiniteSet(0) >>> (x > 0).as_set() Interval.open(0, oo) >>> And(2 < x, x < 2).as_set() Interval.open(2, 2) >>> Or(x < 2, 2 < x).as_set() Union(Interval.open(oo, 2), Interval.open(2, oo))

as_terms
()¶ Transform an expression to a list of terms.

aseries
(x=None, n=6, bound=0, hir=False)¶ Asymptotic Series expansion of self. This is equivalent to
self.series(x, oo, n)
. Parameters
self (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
n (Value) – The number of terms upto which the series is to be expanded.
hir (Boolean) – Set this parameter to be True to produce hierarchical series. It stops the recursion at an early level and may provide nicer and more useful results.
bound (Value, Integer) – Use the
bound
parameter to give limit on rewriting coefficients in its normalised form.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x, y
>>> e = sin(1/x + exp(x))  sin(1/x)
>>> e.aseries(x) (1/(24*x**4)  1/(2*x**2) + 1 + O(x**(6), (x, oo)))*exp(x)
>>> e.aseries(x, n=3, hir=True) exp(2*x)*sin(1/x)/2 + exp(x)*cos(1/x) + O(exp(3*x), (x, oo))
>>> e = exp(exp(x)/(1  1/x))
>>> e.aseries(x) exp(exp(x)/(1  1/x))
>>> e.aseries(x, bound=3) exp(exp(x)/x**2)*exp(exp(x)/x)*exp(exp(x) + exp(x)/(1  1/x)  exp(x)/x  exp(x)/x**2)*exp(exp(x))
 Returns
Asymptotic series expansion of the expression.
 Return type
Expr
Notes
This algorithm is directly induced from the limit computational algorithm provided by Gruntz. It majorly uses the mrv and rewrite subroutines. The overall idea of this algorithm is first to look for the most rapidly varying subexpression w of a given expression f and then expands f in a series in w. Then same thing is recursively done on the leading coefficient till we get constant coefficients.
If the most rapidly varying subexpression of a given expression f is f itself, the algorithm tries to find a normalised representation of the mrv set and rewrites f using this normalised representation.
If the expansion contains an order term, it will be either
O(x ** (n))
orO(w ** (n))
wherew
belongs to the most rapidly varying expression ofself
.References
 1
A New Algorithm for Computing Asymptotic Series  Dominik Gruntz
 2
Gruntz thesis  p90
 3
See also
Expr.aseries()
See the docstring of this function for complete details of this wrapper.

property
assumptions0
¶ Return object type assumptions.
For example:
Symbol(‘x’, real=True) Symbol(‘x’, integer=True)
are different objects. In other words, besides Python type (Symbol in this case), the initial assumptions are also forming their typeinfo.
Examples
>>> from sympy import Symbol >>> from sympy.abc import x >>> x.assumptions0 {'commutative': True} >>> x = Symbol("x", positive=True) >>> x.assumptions0 {'commutative': True, 'complex': True, 'extended_negative': False, 'extended_nonnegative': True, 'extended_nonpositive': False, 'extended_nonzero': True, 'extended_positive': True, 'extended_real': True, 'finite': True, 'hermitian': True, 'imaginary': False, 'infinite': False, 'negative': False, 'nonnegative': True, 'nonpositive': False, 'nonzero': True, 'positive': True, 'real': True, 'zero': False}

atoms
(*types)¶ Returns the atoms that form the current object.
By default, only objects that are truly atomic and can’t be divided into smaller pieces are returned: symbols, numbers, and number symbols like I and pi. It is possible to request atoms of any type, however, as demonstrated below.
Examples
>>> from sympy import I, pi, sin >>> from sympy.abc import x, y >>> (1 + x + 2*sin(y + I*pi)).atoms() {1, 2, I, pi, x, y}
If one or more types are given, the results will contain only those types of atoms.
>>> from sympy import Number, NumberSymbol, Symbol >>> (1 + x + 2*sin(y + I*pi)).atoms(Symbol) {x, y}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number) {1, 2}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol) {1, 2, pi}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol, I) {1, 2, I, pi}
Note that I (imaginary unit) and zoo (complex infinity) are special types of number symbols and are not part of the NumberSymbol class.
The type can be given implicitly, too:
>>> (1 + x + 2*sin(y + I*pi)).atoms(x) # x is a Symbol {x, y}
Be careful to check your assumptions when using the implicit option since
S(1).is_Integer = True
buttype(S(1))
isOne
, a special type of sympy atom, whiletype(S(2))
is typeInteger
and will find all integers in an expression:>>> from sympy import S >>> (1 + x + 2*sin(y + I*pi)).atoms(S(1)) {1}
>>> (1 + x + 2*sin(y + I*pi)).atoms(S(2)) {1, 2}
Finally, arguments to atoms() can select more than atomic atoms: any sympy type (loaded in core/__init__.py) can be listed as an argument and those types of “atoms” as found in scanning the arguments of the expression recursively:
>>> from sympy import Function, Mul >>> from sympy.core.function import AppliedUndef >>> f = Function('f') >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(Function) {f(x), sin(y + I*pi)} >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(AppliedUndef) {f(x)}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Mul) {I*pi, 2*sin(y + I*pi)}

property
binary_symbols
¶ Return from the atoms of self those which are free symbols.
For most expressions, all symbols are free symbols. For some classes this is not true. e.g. Integrals use Symbols for the dummy variables which are bound variables, so Integral has a method to return all symbols except those. Derivative keeps track of symbols with respect to which it will perform a derivative; those are bound variables, too, so it has its own free_symbols method.
Any other method that uses bound variables should implement a free_symbols method.

cancel
(*gens, **args)¶ See the cancel function in sympy.polys

property
canonical_variables
¶ Return a dictionary mapping any variable defined in
self.bound_symbols
to Symbols that do not clash with any existing symbol in the expression.Examples
>>> from sympy import Lambda >>> from sympy.abc import x >>> Lambda(x, 2*x).canonical_variables {x: _0}

classmethod
class_key
()¶ Nice order of classes.

coeff
(x, n=1, right=False)¶ Returns the coefficient from the term(s) containing
x**n
. Ifn
is zero then all terms independent ofx
will be returned.When
x
is noncommutative, the coefficient to the left (default) or right ofx
can be returned. The keyword ‘right’ is ignored whenx
is commutative.See also
as_coefficient()
separate the expression into a coefficient and factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used
Examples
>>> from sympy import symbols >>> from sympy.abc import x, y, z
You can select terms that have an explicit negative in front of them:
>>> (x + 2*y).coeff(1) x >>> (x  2*y).coeff(1) 2*y
You can select terms with no Rational coefficient:
>>> (x + 2*y).coeff(1) x >>> (3 + 2*x + 4*x**2).coeff(1) 0
You can select terms independent of x by making n=0; in this case expr.as_independent(x)[0] is returned (and 0 will be returned instead of None):
>>> (3 + 2*x + 4*x**2).coeff(x, 0) 3 >>> eq = ((x + 1)**3).expand() + 1 >>> eq x**3 + 3*x**2 + 3*x + 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 2] >>> eq = 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 0]
You can select terms that have a numerical term in front of them:
>>> (x  2*y).coeff(2) y >>> from sympy import sqrt >>> (x + sqrt(2)*x).coeff(sqrt(2)) x
The matching is exact:
>>> (3 + 2*x + 4*x**2).coeff(x) 2 >>> (3 + 2*x + 4*x**2).coeff(x**2) 4 >>> (3 + 2*x + 4*x**2).coeff(x**3) 0 >>> (z*(x + y)**2).coeff((x + y)**2) z >>> (z*(x + y)**2).coeff(x + y) 0
In addition, no factoring is done, so 1 + z*(1 + y) is not obtained from the following:
>>> (x + z*(x + x*y)).coeff(x) 1
If such factoring is desired, factor_terms can be used first:
>>> from sympy import factor_terms >>> factor_terms(x + z*(x + x*y)).coeff(x) z*(y + 1) + 1
>>> n, m, o = symbols('n m o', commutative=False) >>> n.coeff(n) 1 >>> (3*n).coeff(n) 3 >>> (n*m + m*n*m).coeff(n) # = (1 + m)*n*m 1 + m >>> (n*m + m*n*m).coeff(n, right=True) # = (1 + m)*n*m m
If there is more than one possible coefficient 0 is returned:
>>> (n*m + m*n).coeff(n) 0
If there is only one possible coefficient, it is returned:
>>> (n*m + x*m*n).coeff(m*n) x >>> (n*m + x*m*n).coeff(m*n, right=1) 1

collect
(syms, func=None, evaluate=True, exact=False, distribute_order_term=True)¶ See the collect function in sympy.simplify

combsimp
()¶ See the combsimp function in sympy.simplify

compare
(other)¶ Return 1, 0, 1 if the object is smaller, equal, or greater than other.
Not in the mathematical sense. If the object is of a different type from the “other” then their classes are ordered according to the sorted_classes list.
Examples
>>> from sympy.abc import x, y >>> x.compare(y) 1 >>> x.compare(x) 0 >>> y.compare(x) 1

compute_leading_term
(x, logx=None)¶ as_leading_term is only allowed for results of .series() This is a wrapper to compute a series first.

conjugate
()¶ Returns the complex conjugate of ‘self’.

could_extract_minus_sign
()¶ Return True if self is not in a canonical form with respect to its sign.
For most expressions, e, there will be a difference in e and e. When there is, True will be returned for one and False for the other; False will be returned if there is no difference.
Examples
>>> from sympy.abc import x, y >>> e = x  y >>> {i.could_extract_minus_sign() for i in (e, e)} {False, True}

count
(query)¶ Count the number of matching subexpressions.

count_ops
(visual=None)¶ wrapper for count_ops that returns the operation count.

doit
(**hints)¶ Evaluate objects that are not evaluated by default like limits, integrals, sums and products. All objects of this kind will be evaluated recursively, unless some species were excluded via ‘hints’ or unless the ‘deep’ hint was set to ‘False’.
>>> from sympy import Integral >>> from sympy.abc import x
>>> 2*Integral(x, x) 2*Integral(x, x)
>>> (2*Integral(x, x)).doit() x**2
>>> (2*Integral(x, x)).doit(deep=False) 2*Integral(x, x)

dummy_eq
(other, symbol=None)¶ Compare two expressions and handle dummy symbols.
Examples
>>> from sympy import Dummy >>> from sympy.abc import x, y
>>> u = Dummy('u')
>>> (u**2 + 1).dummy_eq(x**2 + 1) True >>> (u**2 + 1) == (x**2 + 1) False
>>> (u**2 + y).dummy_eq(x**2 + y, x) True >>> (u**2 + y).dummy_eq(x**2 + y, y) False

equals
(other, failing_expression=False)¶ Return True if self == other, False if it doesn’t, or None. If failing_expression is True then the expression which did not simplify to a 0 will be returned instead of None.
If
self
is a Number (or complex number) that is not zero, then the result is False.If
self
is a number and has not evaluated to zero, evalf will be used to test whether the expression evaluates to zero. If it does so and the result has significance (i.e. the precision is either 1, for a Rational result, or is greater than 1) then the evalf value will be used to return True or False.

evalf
(n=15, subs=None, maxn=100, chop=False, strict=False, quad=None, verbose=False)¶ Evaluate the given formula to an accuracy of n digits.
 Parameters
subs (dict, optional) – Substitute numerical values for symbols, e.g.
subs={x:3, y:1+pi}
. The substitutions must be given as a dictionary.maxn (int, optional) – Allow a maximum temporary working precision of maxn digits.
chop (bool or number, optional) –
Specifies how to replace tiny real or imaginary parts in subresults by exact zeros.
When
True
the chop value defaults to standard precision.Otherwise the chop value is used to determine the magnitude of “small” for purposes of chopping.
>>> from sympy import N >>> x = 1e4 >>> N(x, chop=True) 0.000100000000000000 >>> N(x, chop=1e5) 0.000100000000000000 >>> N(x, chop=1e4) 0
strict (bool, optional) – Raise
PrecisionExhausted
if any subresult fails to evaluate to full accuracy, given the available maxprec.quad (str, optional) – Choose algorithm for numerical quadrature. By default, tanhsinh quadrature is used. For oscillatory integrals on an infinite interval, try
quad='osc'
.verbose (bool, optional) – Print debug information.
Notes
When Floats are naively substituted into an expression, precision errors may adversely affect the result. For example, adding 1e16 (a Float) to 1 will truncate to 1e16; if 1e16 is then subtracted, the result will be 0. That is exactly what happens in the following:
>>> from sympy.abc import x, y, z >>> values = {x: 1e16, y: 1, z: 1e16} >>> (x + y  z).subs(values) 0
Using the subs argument for evalf is the accurate way to evaluate such an expression:
>>> (x + y  z).evalf(subs=values) 1.00000000000000

expand
(deep=True, modulus=None, power_base=True, power_exp=True, mul=True, log=True, multinomial=True, basic=True, **hints)¶ Expand an expression using hints.
See the docstring of the expand() function in sympy.core.function for more information.

property
expr_free_symbols
¶ Like
free_symbols
, but returns the free symbols only if they are contained in an expression node.Examples
>>> from sympy.abc import x, y >>> (x + y).expr_free_symbols {x, y}
If the expression is contained in a nonexpression object, don’t return the free symbols. Compare:
>>> from sympy import Tuple >>> t = Tuple(x + y) >>> t.expr_free_symbols set() >>> t.free_symbols {x, y}

extract_additively
(c)¶ Return self  c if it’s possible to subtract c from self and make all matching coefficients move towards zero, else return None.
Examples
>>> from sympy.abc import x, y >>> e = 2*x + 3 >>> e.extract_additively(x + 1) x + 2 >>> e.extract_additively(3*x) >>> e.extract_additively(4) >>> (y*(x + 1)).extract_additively(x + 1) >>> ((x + 1)*(x + 2*y + 1) + 3).extract_additively(x + 1) (x + 1)*(x + 2*y) + 3
Sometimes autoexpansion will return a less simplified result than desired; gcd_terms might be used in such cases:
>>> from sympy import gcd_terms >>> (4*x*(y + 1) + y).extract_additively(x) 4*x*(y + 1) + x*(4*y + 3)  x*(4*y + 4) + y >>> gcd_terms(_) x*(4*y + 3) + y
See also

extract_branch_factor
(allow_half=False)¶ Try to write self as
exp_polar(2*pi*I*n)*z
in a nice way. Return (z, n).>>> from sympy import exp_polar, I, pi >>> from sympy.abc import x, y >>> exp_polar(I*pi).extract_branch_factor() (exp_polar(I*pi), 0) >>> exp_polar(2*I*pi).extract_branch_factor() (1, 1) >>> exp_polar(pi*I).extract_branch_factor() (exp_polar(I*pi), 1) >>> exp_polar(3*pi*I + x).extract_branch_factor() (exp_polar(x + I*pi), 1) >>> (y*exp_polar(5*pi*I)*exp_polar(3*pi*I + 2*pi*x)).extract_branch_factor() (y*exp_polar(2*pi*x), 1) >>> exp_polar(I*pi/2).extract_branch_factor() (exp_polar(I*pi/2), 0)
If allow_half is True, also extract exp_polar(I*pi):
>>> exp_polar(I*pi).extract_branch_factor(allow_half=True) (1, 1/2) >>> exp_polar(2*I*pi).extract_branch_factor(allow_half=True) (1, 1) >>> exp_polar(3*I*pi).extract_branch_factor(allow_half=True) (1, 3/2) >>> exp_polar(I*pi).extract_branch_factor(allow_half=True) (1, 1/2)

extract_multiplicatively
(c)¶ Return None if it’s not possible to make self in the form c * something in a nice way, i.e. preserving the properties of arguments of self.
Examples
>>> from sympy import symbols, Rational
>>> x, y = symbols('x,y', real=True)
>>> ((x*y)**3).extract_multiplicatively(x**2 * y) x*y**2
>>> ((x*y)**3).extract_multiplicatively(x**4 * y)
>>> (2*x).extract_multiplicatively(2) x
>>> (2*x).extract_multiplicatively(3)
>>> (Rational(1, 2)*x).extract_multiplicatively(3) x/6

factor
(*gens, **args)¶ See the factor() function in sympy.polys.polytools

find
(query, group=False)¶ Find all subexpressions matching a query.

fourier_series
(limits=None)¶ Compute fourier sine/cosine series of self.
See the docstring of the
fourier_series()
in sympy.series.fourier for more information.

fps
(x=None, x0=0, dir=1, hyper=True, order=4, rational=True, full=False)¶ Compute formal power power series of self.
See the docstring of the
fps()
function in sympy.series.formal for more information.

property
free_symbols
¶ Return from the atoms of self those which are free symbols.
For most expressions, all symbols are free symbols. For some classes this is not true. e.g. Integrals use Symbols for the dummy variables which are bound variables, so Integral has a method to return all symbols except those. Derivative keeps track of symbols with respect to which it will perform a derivative; those are bound variables, too, so it has its own free_symbols method.
Any other method that uses bound variables should implement a free_symbols method.

classmethod
fromiter
(args, **assumptions)¶ Create a new object from an iterable.
This is a convenience function that allows one to create objects from any iterable, without having to convert to a list or tuple first.
Examples
>>> from sympy import Tuple >>> Tuple.fromiter(i for i in range(5)) (0, 1, 2, 3, 4)

property
func
¶ The toplevel function in an expression.
The following should hold for all objects:
>> x == x.func(*x.args)
Examples
>>> from sympy.abc import x >>> a = 2*x >>> a.func <class 'sympy.core.mul.Mul'> >>> a.args (2, x) >>> a.func(*a.args) 2*x >>> a == a.func(*a.args) True

gammasimp
()¶ See the gammasimp function in sympy.simplify

getO
()¶ Returns the additive O(..) symbol if there is one, else None.

getn
()¶ Returns the order of the expression.
The order is determined either from the O(…) term. If there is no O(…) term, it returns None.
Examples
>>> from sympy import O >>> from sympy.abc import x >>> (1 + x + O(x**2)).getn() 2 >>> (1 + x).getn()

has
(*patterns)¶ Test whether any subexpression matches any of the patterns.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y, z >>> (x**2 + sin(x*y)).has(z) False >>> (x**2 + sin(x*y)).has(x, y, z) True >>> x.has(x) True
Note
has
is a structural algorithm with no knowledge of mathematics. Consider the following halfopen interval:>>> from sympy.sets import Interval >>> i = Interval.Lopen(0, 5); i Interval.Lopen(0, 5) >>> i.args (0, 5, True, False) >>> i.has(4) # there is no "4" in the arguments False >>> i.has(0) # there *is* a "0" in the arguments True
Instead, use
contains
to determine whether a number is in the interval or not:>>> i.contains(4) True >>> i.contains(0) False
Note that
expr.has(*patterns)
is exactly equivalent toany(expr.has(p) for p in patterns)
. In particular,False
is returned when the list of patterns is empty.>>> x.has() False

integrate
(*args, **kwargs)¶ See the integrate function in sympy.integrals

invert
(g, *gens, **args)¶ Return the multiplicative inverse of
self
modg
whereself
(andg
) may be symbolic expressions).See also
sympy.core.numbers.mod_inverse()
,sympy.polys.polytools.invert()

is_algebraic_expr
(*syms)¶ This tests whether a given expression is algebraic or not, in the given symbols, syms. When syms is not given, all free symbols will be used. The rational function does not have to be in expanded or in any kind of canonical form.
This function returns False for expressions that are “algebraic expressions” with symbolic exponents. This is a simple extension to the is_rational_function, including rational exponentiation.
Examples
>>> from sympy import Symbol, sqrt >>> x = Symbol('x', real=True) >>> sqrt(1 + x).is_rational_function() False >>> sqrt(1 + x).is_algebraic_expr() True
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be an algebraic expression to become one.
>>> from sympy import exp, factor >>> a = sqrt(exp(x)**2 + 2*exp(x) + 1)/(exp(x) + 1) >>> a.is_algebraic_expr(x) False >>> factor(a).is_algebraic_expr() True
See also
References

is_constant
(*wrt, **flags)¶ Return True if self is constant, False if not, or None if the constancy could not be determined conclusively.
If an expression has no free symbols then it is a constant. If there are free symbols it is possible that the expression is a constant, perhaps (but not necessarily) zero. To test such expressions, a few strategies are tried:
1) numerical evaluation at two random points. If two such evaluations give two different values and the values have a precision greater than 1 then self is not constant. If the evaluations agree or could not be obtained with any precision, no decision is made. The numerical testing is done only if
wrt
is different than the free symbols.2) differentiation with respect to variables in ‘wrt’ (or all free symbols if omitted) to see if the expression is constant or not. This will not always lead to an expression that is zero even though an expression is constant (see added test in test_expr.py). If all derivatives are zero then self is constant with respect to the given symbols.
3) finding out zeros of denominator expression with free_symbols. It won’t be constant if there are zeros. It gives more negative answers for expression that are not constant.
If neither evaluation nor differentiation can prove the expression is constant, None is returned unless two numerical values happened to be the same and the flag
failing_number
is True – in that case the numerical value will be returned.If flag simplify=False is passed, self will not be simplified; the default is True since self should be simplified before testing.
Examples
>>> from sympy import cos, sin, Sum, S, pi >>> from sympy.abc import a, n, x, y >>> x.is_constant() False >>> S(2).is_constant() True >>> Sum(x, (x, 1, 10)).is_constant() True >>> Sum(x, (x, 1, n)).is_constant() False >>> Sum(x, (x, 1, n)).is_constant(y) True >>> Sum(x, (x, 1, n)).is_constant(n) False >>> Sum(x, (x, 1, n)).is_constant(x) True >>> eq = a*cos(x)**2 + a*sin(x)**2  a >>> eq.is_constant() True >>> eq.subs({x: pi, a: 2}) == eq.subs({x: pi, a: 3}) == 0 True
>>> (0**x).is_constant() False >>> x.is_constant() False >>> (x**x).is_constant() False >>> one = cos(x)**2 + sin(x)**2 >>> one.is_constant() True >>> ((one  1)**(x + 1)).is_constant() in (True, False) # could be 0 or 1 True

is_polynomial
(*syms)¶ Return True if self is a polynomial in syms and False otherwise.
This checks if self is an exact polynomial in syms. This function returns False for expressions that are “polynomials” with symbolic exponents. Thus, you should be able to apply polynomial algorithms to expressions for which this returns True, and Poly(expr, *syms) should work if and only if expr.is_polynomial(*syms) returns True. The polynomial does not have to be in expanded form. If no symbols are given, all free symbols in the expression will be used.
This is not part of the assumptions system. You cannot do Symbol(‘z’, polynomial=True).
Examples
>>> from sympy import Symbol >>> x = Symbol('x') >>> ((x**2 + 1)**4).is_polynomial(x) True >>> ((x**2 + 1)**4).is_polynomial() True >>> (2**x + 1).is_polynomial(x) False
>>> n = Symbol('n', nonnegative=True, integer=True) >>> (x**n + 1).is_polynomial(x) False
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be a polynomial to become one.
>>> from sympy import sqrt, factor, cancel >>> y = Symbol('y', positive=True) >>> a = sqrt(y**2 + 2*y + 1) >>> a.is_polynomial(y) False >>> factor(a) y + 1 >>> factor(a).is_polynomial(y) True
>>> b = (y**2 + 2*y + 1)/(y + 1) >>> b.is_polynomial(y) False >>> cancel(b) y + 1 >>> cancel(b).is_polynomial(y) True
See also .is_rational_function()

is_rational_function
(*syms)¶ Test whether function is a ratio of two polynomials in the given symbols, syms. When syms is not given, all free symbols will be used. The rational function does not have to be in expanded or in any kind of canonical form.
This function returns False for expressions that are “rational functions” with symbolic exponents. Thus, you should be able to call .as_numer_denom() and apply polynomial algorithms to the result for expressions for which this returns True.
This is not part of the assumptions system. You cannot do Symbol(‘z’, rational_function=True).
Examples
>>> from sympy import Symbol, sin >>> from sympy.abc import x, y
>>> (x/y).is_rational_function() True
>>> (x**2).is_rational_function() True
>>> (x/sin(y)).is_rational_function(y) False
>>> n = Symbol('n', integer=True) >>> (x**n + 1).is_rational_function(x) False
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be a rational function to become one.
>>> from sympy import sqrt, factor >>> y = Symbol('y', positive=True) >>> a = sqrt(y**2 + 2*y + 1)/y >>> a.is_rational_function(y) False >>> factor(a) (y + 1)/y >>> factor(a).is_rational_function(y) True
See also is_algebraic_expr().

leadterm
(x)¶ Returns the leading term a*x**b as a tuple (a, b).
Examples
>>> from sympy.abc import x >>> (1+x+x**2).leadterm(x) (1, 0) >>> (1/x**2+x+x**2).leadterm(x) (1, 2)

limit
(x, xlim, dir='+')¶ Compute limit x>xlim.

lseries
(x=None, x0=0, dir='+', logx=None)¶ Wrapper for series yielding an iterator of the terms of the series.
Note: an infinite series will yield an infinite iterator. The following, for exaxmple, will never terminate. It will just keep printing terms of the sin(x) series:
for term in sin(x).lseries(x): print term
The advantage of lseries() over nseries() is that many times you are just interested in the next term in the series (i.e. the first term for example), but you don’t know how many you should ask for in nseries() using the “n” parameter.
See also nseries().

match
(pattern, old=False)¶ Pattern matching.
Wild symbols match all.
Return
None
when expression (self) does not match with pattern. Otherwise return a dictionary such that:pattern.xreplace(self.match(pattern)) == self
Examples
>>> from sympy import Wild >>> from sympy.abc import x, y >>> p = Wild("p") >>> q = Wild("q") >>> r = Wild("r") >>> e = (x+y)**(x+y) >>> e.match(p**p) {p_: x + y} >>> e.match(p**q) {p_: x + y, q_: x + y} >>> e = (2*x)**2 >>> e.match(p*q**r) {p_: 4, q_: x, r_: 2} >>> (p*q**r).xreplace(e.match(p*q**r)) 4*x**2
The
old
flag will give the oldstyle pattern matching where expressions and patterns are essentially solved to give the match. Both of the following give None unlessold=True
:>>> (x  2).match(p  x, old=True) {p_: 2*x  2} >>> (2/x).match(p*x, old=True) {p_: 2/x**2}

matches
(expr, repl_dict={}, old=False)¶ Helper method for match() that looks for a match between Wild symbols in self and expressions in expr.
Examples
>>> from sympy import symbols, Wild, Basic >>> a, b, c = symbols('a b c') >>> x = Wild('x') >>> Basic(a + x, x).matches(Basic(a + b, c)) is None True >>> Basic(a + x, x).matches(Basic(a + b + c, b + c)) {x_: b + c}

n
(n=15, subs=None, maxn=100, chop=False, strict=False, quad=None, verbose=False)¶ Evaluate the given formula to an accuracy of n digits.
 Parameters
subs (dict, optional) – Substitute numerical values for symbols, e.g.
subs={x:3, y:1+pi}
. The substitutions must be given as a dictionary.maxn (int, optional) – Allow a maximum temporary working precision of maxn digits.
chop (bool or number, optional) –
Specifies how to replace tiny real or imaginary parts in subresults by exact zeros.
When
True
the chop value defaults to standard precision.Otherwise the chop value is used to determine the magnitude of “small” for purposes of chopping.
>>> from sympy import N >>> x = 1e4 >>> N(x, chop=True) 0.000100000000000000 >>> N(x, chop=1e5) 0.000100000000000000 >>> N(x, chop=1e4) 0
strict (bool, optional) – Raise
PrecisionExhausted
if any subresult fails to evaluate to full accuracy, given the available maxprec.quad (str, optional) – Choose algorithm for numerical quadrature. By default, tanhsinh quadrature is used. For oscillatory integrals on an infinite interval, try
quad='osc'
.verbose (bool, optional) – Print debug information.
Notes
When Floats are naively substituted into an expression, precision errors may adversely affect the result. For example, adding 1e16 (a Float) to 1 will truncate to 1e16; if 1e16 is then subtracted, the result will be 0. That is exactly what happens in the following:
>>> from sympy.abc import x, y, z >>> values = {x: 1e16, y: 1, z: 1e16} >>> (x + y  z).subs(values) 0
Using the subs argument for evalf is the accurate way to evaluate such an expression:
>>> (x + y  z).evalf(subs=values) 1.00000000000000

nseries
(x=None, x0=0, n=6, dir='+', logx=None)¶ Wrapper to _eval_nseries if assumptions allow, else to series.
If x is given, x0 is 0, dir=’+’, and self has x, then _eval_nseries is called. This calculates “n” terms in the innermost expressions and then builds up the final series just by “crossmultiplying” everything out.
The optional
logx
parameter can be used to replace any log(x) in the returned series with a symbolic value to avoid evaluating log(x) at 0. A symbol to use in place of log(x) should be provided.Advantage – it’s fast, because we don’t have to determine how many terms we need to calculate in advance.
Disadvantage – you may end up with less terms than you may have expected, but the O(x**n) term appended will always be correct and so the result, though perhaps shorter, will also be correct.
If any of those assumptions is not met, this is treated like a wrapper to series which will try harder to return the correct number of terms.
See also lseries().
Examples
>>> from sympy import sin, log, Symbol >>> from sympy.abc import x, y >>> sin(x).nseries(x, 0, 6) x  x**3/6 + x**5/120 + O(x**6) >>> log(x+1).nseries(x, 0, 5) x  x**2/2 + x**3/3  x**4/4 + O(x**5)
Handling of the
logx
parameter — in the following example the expansion fails sincesin
does not have an asymptotic expansion at oo (the limit of log(x) as x approaches 0):>>> e = sin(log(x)) >>> e.nseries(x, 0, 6) Traceback (most recent call last): ... PoleError: ... ... >>> logx = Symbol('logx') >>> e.nseries(x, 0, 6, logx=logx) sin(logx)
In the following example, the expansion works but gives only an Order term unless the
logx
parameter is used:>>> e = x**y >>> e.nseries(x, 0, 2) O(log(x)**2) >>> e.nseries(x, 0, 2, logx=logx) exp(logx*y)

nsimplify
(constants=[], tolerance=None, full=False)¶ See the nsimplify function in sympy.simplify

powsimp
(*args, **kwargs)¶ See the powsimp function in sympy.simplify

primitive
()¶ Return the positive Rational that can be extracted nonrecursively from every term of self (i.e., self is treated like an Add). This is like the as_coeff_Mul() method but primitive always extracts a positive Rational (never a negative or a Float).
Examples
>>> from sympy.abc import x >>> (3*(x + 1)**2).primitive() (3, (x + 1)**2) >>> a = (6*x + 2); a.primitive() (2, 3*x + 1) >>> b = (x/2 + 3); b.primitive() (1/2, x + 6) >>> (a*b).primitive() == (1, a*b) True

radsimp
(**kwargs)¶ See the radsimp function in sympy.simplify

ratsimp
()¶ See the ratsimp function in sympy.simplify

rcall
(*args)¶ Apply on the argument recursively through the expression tree.
This method is used to simulate a common abuse of notation for operators. For instance in SymPy the the following will not work:
(x+Lambda(y, 2*y))(z) == x+2*z
,however you can use
>>> from sympy import Lambda >>> from sympy.abc import x, y, z >>> (x + Lambda(y, 2*y)).rcall(z) x + 2*z

refine
(assumption=True)¶ See the refine function in sympy.assumptions

removeO
()¶ Removes the additive O(..) symbol if there is one

replace
(query, value, map=False, simultaneous=True, exact=None)¶ Replace matching subexpressions of
self
withvalue
.If
map = True
then also return the mapping {old: new} whereold
was a subexpression found with query andnew
is the replacement value for it. If the expression itself doesn’t match the query, then the returned value will beself.xreplace(map)
otherwise it should beself.subs(ordered(map.items()))
.Traverses an expression tree and performs replacement of matching subexpressions from the bottom to the top of the tree. The default approach is to do the replacement in a simultaneous fashion so changes made are targeted only once. If this is not desired or causes problems,
simultaneous
can be set to False.In addition, if an expression containing more than one Wild symbol is being used to match subexpressions and the
exact
flag is None it will be set to True so the match will only succeed if all nonzero values are received for each Wild that appears in the match pattern. Setting this to False accepts a match of 0; while setting it True accepts all matches that have a 0 in them. See example below for cautions.The list of possible combinations of queries and replacement values is listed below:
Examples
Initial setup
>>> from sympy import log, sin, cos, tan, Wild, Mul, Add >>> from sympy.abc import x, y >>> f = log(sin(x)) + tan(sin(x**2))
 1.1. type > type
obj.replace(type, newtype)
When object of type
type
is found, replace it with the result of passing its argument(s) tonewtype
.>>> f.replace(sin, cos) log(cos(x)) + tan(cos(x**2)) >>> sin(x).replace(sin, cos, map=True) (cos(x), {sin(x): cos(x)}) >>> (x*y).replace(Mul, Add) x + y
 1.2. type > func
obj.replace(type, func)
When object of type
type
is found, applyfunc
to its argument(s).func
must be written to handle the number of arguments oftype
.>>> f.replace(sin, lambda arg: sin(2*arg)) log(sin(2*x)) + tan(sin(2*x**2)) >>> (x*y).replace(Mul, lambda *args: sin(2*Mul(*args))) sin(2*x*y)
 2.1. pattern > expr
obj.replace(pattern(wild), expr(wild))
Replace subexpressions matching
pattern
with the expression written in terms of the Wild symbols inpattern
.>>> a, b = map(Wild, 'ab') >>> f.replace(sin(a), tan(a)) log(tan(x)) + tan(tan(x**2)) >>> f.replace(sin(a), tan(a/2)) log(tan(x/2)) + tan(tan(x**2/2)) >>> f.replace(sin(a), a) log(x) + tan(x**2) >>> (x*y).replace(a*x, a) y
Matching is exact by default when more than one Wild symbol is used: matching fails unless the match gives nonzero values for all Wild symbols:
>>> (2*x + y).replace(a*x + b, b  a) y  2 >>> (2*x).replace(a*x + b, b  a) 2*x
When set to False, the results may be nonintuitive:
>>> (2*x).replace(a*x + b, b  a, exact=False) 2/x
 2.2. pattern > func
obj.replace(pattern(wild), lambda wild: expr(wild))
All behavior is the same as in 2.1 but now a function in terms of pattern variables is used rather than an expression:
>>> f.replace(sin(a), lambda a: sin(2*a)) log(sin(2*x)) + tan(sin(2*x**2))
 3.1. func > func
obj.replace(filter, func)
Replace subexpression
e
withfunc(e)
iffilter(e)
is True.>>> g = 2*sin(x**3) >>> g.replace(lambda expr: expr.is_Number, lambda expr: expr**2) 4*sin(x**9)
The expression itself is also targeted by the query but is done in such a fashion that changes are not made twice.
>>> e = x*(x*y + 1) >>> e.replace(lambda x: x.is_Mul, lambda x: 2*x) 2*x*(2*x*y + 1)
When matching a single symbol, exact will default to True, but this may or may not be the behavior that is desired:
Here, we want exact=False:
>>> from sympy import Function >>> f = Function('f') >>> e = f(1) + f(0) >>> q = f(a), lambda a: f(a + 1) >>> e.replace(*q, exact=False) f(1) + f(2) >>> e.replace(*q, exact=True) f(0) + f(2)
But here, the nature of matching makes selecting the right setting tricky:
>>> e = x**(1 + y) >>> (x**(1 + y)).replace(x**(1 + a), lambda a: x**a, exact=False) 1 >>> (x**(1 + y)).replace(x**(1 + a), lambda a: x**a, exact=True) x**(x  y + 1) >>> (x**y).replace(x**(1 + a), lambda a: x**a, exact=False) 1 >>> (x**y).replace(x**(1 + a), lambda a: x**a, exact=True) x**(1  y)
It is probably better to use a different form of the query that describes the target expression more precisely:
>>> (1 + x**(1 + y)).replace( ... lambda x: x.is_Pow and x.exp.is_Add and x.exp.args[0] == 1, ... lambda x: x.base**(1  (x.exp  1))) ... x**(1  y) + 1
See also
subs()
substitution of subexpressions as defined by the objects themselves.
xreplace()
exact node replacement in expr tree; also capable of using matching rules

rewrite
(*args, **hints)¶ Rewrite functions in terms of other functions.
Rewrites expression containing applications of functions of one kind in terms of functions of different kind. For example you can rewrite trigonometric functions as complex exponentials or combinatorial functions as gamma function.
As a pattern this function accepts a list of functions to to rewrite (instances of DefinedFunction class). As rule you can use string or a destination function instance (in this case rewrite() will use the str() function).
There is also the possibility to pass hints on how to rewrite the given expressions. For now there is only one such hint defined called ‘deep’. When ‘deep’ is set to False it will forbid functions to rewrite their contents.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x
Unspecified pattern:
>>> sin(x).rewrite(exp) I*(exp(I*x)  exp(I*x))/2
Pattern as a single function:
>>> sin(x).rewrite(sin, exp) I*(exp(I*x)  exp(I*x))/2
Pattern as a list of functions:
>>> sin(x).rewrite([sin, ], exp) I*(exp(I*x)  exp(I*x))/2

round
(n=None)¶ Return x rounded to the given decimal place.
If a complex number would results, apply round to the real and imaginary components of the number.
Examples
>>> from sympy import pi, E, I, S, Add, Mul, Number >>> pi.round() 3 >>> pi.round(2) 3.14 >>> (2*pi + E*I).round() 6 + 3*I
The round method has a chopping effect:
>>> (2*pi + I/10).round() 6 >>> (pi/10 + 2*I).round() 2*I >>> (pi/10 + E*I).round(2) 0.31 + 2.72*I
Notes
The Python
round
function uses the SymPyround
method so it will always return a SymPy number (not a Python float or int):>>> isinstance(round(S(123), 2), Number) True

separate
(deep=False, force=False)¶ See the separate function in sympy.simplify

series
(x=None, x0=0, n=6, dir='+', logx=None)¶ Series expansion of “self” around
x = x0
yielding either terms of the series one by one (the lazy series given when n=None), else all the terms at once when n != None.Returns the series expansion of “self” around the point
x = x0
with respect tox
up toO((x  x0)**n, x, x0)
(default n is 6).If
x=None
andself
is univariate, the univariate symbol will be supplied, otherwise an error will be raised. Parameters
expr (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
x0 (Value) – The value around which
x
is calculated. Can be any value fromoo
tooo
.n (Value) – The number of terms upto which the series is to be expanded.
dir (String, optional) – The seriesexpansion can be bidirectional. If
dir="+"
, then (x>x0+). Ifdir="", then (x>x0). For infinite ``x0
(oo
oroo
), thedir
argument is determined from the direction of the infinity (i.e.,dir=""
foroo
).logx (optional) – It is used to replace any log(x) in the returned series with a symbolic value rather than evaluating the actual value.
Examples
>>> from sympy import cos, exp, tan, oo, series >>> from sympy.abc import x, y >>> cos(x).series() 1  x**2/2 + x**4/24 + O(x**6) >>> cos(x).series(n=4) 1  x**2/2 + O(x**4) >>> cos(x).series(x, x0=1, n=2) cos(1)  (x  1)*sin(1) + O((x  1)**2, (x, 1)) >>> e = cos(x + exp(y)) >>> e.series(y, n=2) cos(x + 1)  y*sin(x + 1) + O(y**2) >>> e.series(x, n=2) cos(exp(y))  x*sin(exp(y)) + O(x**2)
If
n=None
then a generator of the series terms will be returned.>>> term=cos(x).series(n=None) >>> [next(term) for i in range(2)] [1, x**2/2]
For
dir=+
(default) the series is calculated from the right and fordir=
the series from the left. For smooth functions this flag will not alter the results.>>> abs(x).series(dir="+") x >>> abs(x).series(dir="") x >>> f = tan(x) >>> f.series(x, 2, 6, "+") tan(2) + (1 + tan(2)**2)*(x  2) + (x  2)**2*(tan(2)**3 + tan(2)) + (x  2)**3*(1/3 + 4*tan(2)**2/3 + tan(2)**4) + (x  2)**4*(tan(2)**5 + 5*tan(2)**3/3 + 2*tan(2)/3) + (x  2)**5*(2/15 + 17*tan(2)**2/15 + 2*tan(2)**4 + tan(2)**6) + O((x  2)**6, (x, 2))
>>> f.series(x, 2, 3, "") tan(2) + (2  x)*(tan(2)**2  1) + (2  x)**2*(tan(2)**3 + tan(2)) + O((x  2)**3, (x, 2))
 Returns
Expr – Series expansion of the expression about x0
 Return type
Expression
 Raises
TypeError – If “n” and “x0” are infinity objects
PoleError – If “x0” is an infinity object

simplify
(**kwargs)¶ See the simplify function in sympy.simplify

sort_key
(order=None)¶ Return a sort key.
Examples
>>> from sympy.core import S, I
>>> sorted([S(1)/2, I, I], key=lambda x: x.sort_key()) [1/2, I, I]
>>> S("[x, 1/x, 1/x**2, x**2, x**(1/2), x**(1/4), x**(3/2)]") [x, 1/x, x**(2), x**2, sqrt(x), x**(1/4), x**(3/2)] >>> sorted(_, key=lambda x: x.sort_key()) [x**(2), 1/x, x**(1/4), sqrt(x), x, x**(3/2), x**2]

subs
(*args, **kwargs)¶ Substitutes old for new in an expression after sympifying args.
 args is either:
two arguments, e.g. foo.subs(old, new)
 one iterable argument, e.g. foo.subs(iterable). The iterable may be
 o an iterable container with (old, new) pairs. In this case the
replacements are processed in the order given with successive patterns possibly affecting replacements already made.
 o a dict or set whose key/value items correspond to old/new pairs.
In this case the old/new pairs will be sorted by op count and in case of a tie, by number of args and the default_sort_key. The resulting sorted list is then processed as an iterable container (see previous).
If the keyword
simultaneous
is True, the subexpressions will not be evaluated until all the substitutions have been made.Examples
>>> from sympy import pi, exp, limit, oo >>> from sympy.abc import x, y >>> (1 + x*y).subs(x, pi) pi*y + 1 >>> (1 + x*y).subs({x:pi, y:2}) 1 + 2*pi >>> (1 + x*y).subs([(x, pi), (y, 2)]) 1 + 2*pi >>> reps = [(y, x**2), (x, 2)] >>> (x + y).subs(reps) 6 >>> (x + y).subs(reversed(reps)) x**2 + 2
>>> (x**2 + x**4).subs(x**2, y) y**2 + y
To replace only the x**2 but not the x**4, use xreplace:
>>> (x**2 + x**4).xreplace({x**2: y}) x**4 + y
To delay evaluation until all substitutions have been made, set the keyword
simultaneous
to True:>>> (x/y).subs([(x, 0), (y, 0)]) 0 >>> (x/y).subs([(x, 0), (y, 0)], simultaneous=True) nan
This has the added feature of not allowing subsequent substitutions to affect those already made:
>>> ((x + y)/y).subs({x + y: y, y: x + y}) 1 >>> ((x + y)/y).subs({x + y: y, y: x + y}, simultaneous=True) y/(x + y)
In order to obtain a canonical result, unordered iterables are sorted by count_op length, number of arguments and by the default_sort_key to break any ties. All other iterables are left unsorted.
>>> from sympy import sqrt, sin, cos >>> from sympy.abc import a, b, c, d, e
>>> A = (sqrt(sin(2*x)), a) >>> B = (sin(2*x), b) >>> C = (cos(2*x), c) >>> D = (x, d) >>> E = (exp(x), e)
>>> expr = sqrt(sin(2*x))*sin(exp(x)*x)*cos(2*x) + sin(2*x)
>>> expr.subs(dict([A, B, C, D, E])) a*c*sin(d*e) + b
The resulting expression represents a literal replacement of the old arguments with the new arguments. This may not reflect the limiting behavior of the expression:
>>> (x**3  3*x).subs({x: oo}) nan
>>> limit(x**3  3*x, x, oo) oo
If the substitution will be followed by numerical evaluation, it is better to pass the substitution to evalf as
>>> (1/x).evalf(subs={x: 3.0}, n=21) 0.333333333333333333333
rather than
>>> (1/x).subs({x: 3.0}).evalf(21) 0.333333333333333314830
as the former will ensure that the desired level of precision is obtained.
See also
replace()
replacement capable of doing wildcardlike matching, parsing of match, and conditional replacements
xreplace()
exact node replacement in expr tree; also capable of using matching rules
sympy.core.evalf.EvalfMixin.evalf()
calculates the given formula to a desired level of precision

taylor_term
(n, x, *previous_terms)¶ General method for the taylor term.
This method is slow, because it differentiates ntimes. Subclasses can redefine it to make it faster by using the “previous_terms”.

together
(*args, **kwargs)¶ See the together function in sympy.polys

trigsimp
(**args)¶ See the trigsimp function in sympy.simplify

xreplace
(rule, hack2=False)¶ Replace occurrences of objects within the expression.
 Parameters
rule (dictlike) – Expresses a replacement rule
 Returns
xreplace
 Return type
the result of the replacement
Examples
>>> from sympy import symbols, pi, exp >>> x, y, z = symbols('x y z') >>> (1 + x*y).xreplace({x: pi}) pi*y + 1 >>> (1 + x*y).xreplace({x: pi, y: 2}) 1 + 2*pi
Replacements occur only if an entire node in the expression tree is matched:
>>> (x*y + z).xreplace({x*y: pi}) z + pi >>> (x*y*z).xreplace({x*y: pi}) x*y*z >>> (2*x).xreplace({2*x: y, x: z}) y >>> (2*2*x).xreplace({2*x: y, x: z}) 4*z >>> (x + y + 2).xreplace({x + y: 2}) x + y + 2 >>> (x + 2 + exp(x + 2)).xreplace({x + 2: y}) x + exp(y) + 2
xreplace doesn’t differentiate between free and bound symbols. In the following, subs(x, y) would not change x since it is a bound symbol, but xreplace does:
>>> from sympy import Integral >>> Integral(x, (x, 1, 2*x)).xreplace({x: y}) Integral(y, (y, 1, 2*y))
Trying to replace x with an expression raises an error:
>>> Integral(x, (x, 1, 2*x)).xreplace({x: 2*y}) ValueError: Invalid limits given: ((2*y, 1, 4*y),)


class
galgebra.atoms.
BasisBaseSymbol
[source]¶ A basis base in a nonorthogonal algebra, such as \(e_1 e_2\)

apart
(x=None, **args)¶ See the apart function in sympy.polys

property
args
¶ Returns a tuple of arguments of ‘self’.
Examples
>>> from sympy import cot >>> from sympy.abc import x, y
>>> cot(x).args (x,)
>>> cot(x).args[0] x
>>> (x*y).args (x, y)
>>> (x*y).args[1] y
Notes
Never use self._args, always use self.args. Only use _args in __new__ when creating a new function. Don’t override .args() from Basic (so that it’s easy to change the interface in the future if needed).

args_cnc
(cset=False, warn=True, split_1=True)¶ Return [commutative factors, noncommutative factors] of self.
self is treated as a Mul and the ordering of the factors is maintained. If
cset
is True the commutative factors will be returned in a set. If there were repeated factors (as may happen with an unevaluated Mul) then an error will be raised unless it is explicitly suppressed by settingwarn
to False.Note: 1 is always separated from a Number unless split_1 is False.
>>> from sympy import symbols, oo >>> A, B = symbols('A B', commutative=0) >>> x, y = symbols('x y') >>> (2*x*y).args_cnc() [[1, 2, x, y], []] >>> (2.5*x).args_cnc() [[1, 2.5, x], []] >>> (2*x*A*B*y).args_cnc() [[1, 2, x, y], [A, B]] >>> (2*x*A*B*y).args_cnc(split_1=False) [[2, x, y], [A, B]] >>> (2*x*y).args_cnc(cset=True) [{1, 2, x, y}, []]
The arg is always treated as a Mul:
>>> (2 + x + A).args_cnc() [[], [x  2 + A]] >>> (oo).args_cnc() # oo is a singleton [[1, oo], []]

as_coeff_Add
(rational=False)¶ Efficiently extract the coefficient of a summation.

as_coeff_Mul
(rational=False)¶ Efficiently extract the coefficient of a product.

as_coeff_add
(*deps)¶ Return the tuple (c, args) where self is written as an Add,
a
.c should be a Rational added to any terms of the Add that are independent of deps.
args should be a tuple of all other terms of
a
; args is empty if self is a Number or if self is independent of deps (when given).This should be used when you don’t know if self is an Add or not but you want to treat self as an Add or if you want to process the individual arguments of the tail of self as an Add.
if you know self is an Add and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail.
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_add() (3, ()) >>> (3 + x).as_coeff_add() (3, (x,)) >>> (3 + x + y).as_coeff_add(x) (y + 3, (x,)) >>> (3 + y).as_coeff_add(x) (y + 3, ())

as_coeff_exponent
(x)¶ c*x**e > c,e
where x can be any symbolic expression.

as_coeff_mul
(*deps, **kwargs)¶ Return the tuple (c, args) where self is written as a Mul,
m
.c should be a Rational multiplied by any factors of the Mul that are independent of deps.
args should be a tuple of all other factors of m; args is empty if self is a Number or if self is independent of deps (when given).
This should be used when you don’t know if self is a Mul or not but you want to treat self as a Mul or if you want to process the individual arguments of the tail of self as a Mul.
if you know self is a Mul and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail;
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_mul() (3, ()) >>> (3*x*y).as_coeff_mul() (3, (x, y)) >>> (3*x*y).as_coeff_mul(x) (3*y, (x,)) >>> (3*y).as_coeff_mul(x) (3*y, ())

as_coefficient
(expr)¶ Extracts symbolic coefficient at the given expression. In other words, this functions separates ‘self’ into the product of ‘expr’ and ‘expr’free coefficient. If such separation is not possible it will return None.
Examples
>>> from sympy import E, pi, sin, I, Poly >>> from sympy.abc import x
>>> E.as_coefficient(E) 1 >>> (2*E).as_coefficient(E) 2 >>> (2*sin(E)*E).as_coefficient(E)
Two terms have E in them so a sum is returned. (If one were desiring the coefficient of the term exactly matching E then the constant from the returned expression could be selected. Or, for greater precision, a method of Poly can be used to indicate the desired term from which the coefficient is desired.)
>>> (2*E + x*E).as_coefficient(E) x + 2 >>> _.args[0] # just want the exact match 2 >>> p = Poly(2*E + x*E); p Poly(x*E + 2*E, x, E, domain='ZZ') >>> p.coeff_monomial(E) 2 >>> p.nth(0, 1) 2
Since the following cannot be written as a product containing E as a factor, None is returned. (If the coefficient
2*x
is desired then thecoeff
method should be used.)>>> (2*E*x + x).as_coefficient(E) >>> (2*E*x + x).coeff(E) 2*x
>>> (E*(x + 1) + x).as_coefficient(E)
>>> (2*pi*I).as_coefficient(pi*I) 2 >>> (2*I).as_coefficient(pi*I)
See also
coeff()
return sum of terms have a given factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used

as_coefficients_dict
()¶ Return a dictionary mapping terms to their Rational coefficient. Since the dictionary is a defaultdict, inquiries about terms which were not present will return a coefficient of 0. If an expression is not an Add it is considered to have a single term.
Examples
>>> from sympy.abc import a, x >>> (3*x + a*x + 4).as_coefficients_dict() {1: 4, x: 3, a*x: 1} >>> _[a] 0 >>> (3*a*x).as_coefficients_dict() {a*x: 3}

as_content_primitive
(radical=False, clear=True)¶ This method should recursively remove a Rational from all arguments and return that (content) and the new self (primitive). The content should always be positive and
Mul(*foo.as_content_primitive()) == foo
. The primitive need not be in canonical form and should try to preserve the underlying structure if possible (i.e. expand_mul should not be applied to self).Examples
>>> from sympy import sqrt >>> from sympy.abc import x, y, z
>>> eq = 2 + 2*x + 2*y*(3 + 3*y)
The as_content_primitive function is recursive and retains structure:
>>> eq.as_content_primitive() (2, x + 3*y*(y + 1) + 1)
Integer powers will have Rationals extracted from the base:
>>> ((2 + 6*x)**2).as_content_primitive() (4, (3*x + 1)**2) >>> ((2 + 6*x)**(2*y)).as_content_primitive() (1, (2*(3*x + 1))**(2*y))
Terms may end up joining once their as_content_primitives are added:
>>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (11, x*(y + 1)) >>> ((3*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (9, x*(y + 1)) >>> ((3*(z*(1 + y)) + 2.0*x*(3 + 3*y))).as_content_primitive() (1, 6.0*x*(y + 1) + 3*z*(y + 1)) >>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))**2).as_content_primitive() (121, x**2*(y + 1)**2) >>> ((x*(1 + y) + 0.4*x*(3 + 3*y))**2).as_content_primitive() (1, 4.84*x**2*(y + 1)**2)
Radical content can also be factored out of the primitive:
>>> (2*sqrt(2) + 4*sqrt(10)).as_content_primitive(radical=True) (2, sqrt(2)*(1 + 2*sqrt(5)))
If clear=False (default is True) then content will not be removed from an Add if it can be distributed to leave one or more terms with integer coefficients.
>>> (x/2 + y).as_content_primitive() (1/2, x + 2*y) >>> (x/2 + y).as_content_primitive(clear=False) (1, x/2 + y)

as_dummy
()¶ Return the expression with any objects having structurally bound symbols replaced with unique, canonical symbols within the object in which they appear and having only the default assumption for commutativity being True.
Examples
>>> from sympy import Integral, Symbol >>> from sympy.abc import x, y >>> r = Symbol('r', real=True) >>> Integral(r, (r, x)).as_dummy() Integral(_0, (_0, x)) >>> _.variables[0].is_real is None True
Notes
Any object that has structural dummy variables should have a property, bound_symbols that returns a list of structural dummy symbols of the object itself.
Lambda and Subs have bound symbols, but because of how they are cached, they already compare the same regardless of their bound symbols:
>>> from sympy import Lambda >>> Lambda(x, x + 1) == Lambda(y, y + 1) True

as_expr
(*gens)¶ Convert a polynomial to a SymPy expression.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y
>>> f = (x**2 + x*y).as_poly(x, y) >>> f.as_expr() x**2 + x*y
>>> sin(x).as_expr() sin(x)

as_independent
(*deps, **hint)¶ A mostly naive separation of a Mul or Add into arguments that are not are dependent on deps. To obtain as complete a separation of variables as possible, use a separation method first, e.g.:
separatevars() to change Mul, Add and Pow (including exp) into Mul
.expand(mul=True) to change Add or Mul into Add
.expand(log=True) to change log expr into an Add
The only nonnaive thing that is done here is to respect noncommutative ordering of variables and to always return (0, 0) for self of zero regardless of hints.
For nonzero self, the returned tuple (i, d) has the following interpretation:
i will has no variable that appears in deps
d will either have terms that contain variables that are in deps, or be equal to 0 (when self is an Add) or 1 (when self is a Mul)
if self is an Add then self = i + d
if self is a Mul then self = i*d
otherwise (self, S.One) or (S.One, self) is returned.
To force the expression to be treated as an Add, use the hint as_Add=True
Examples
– self is an Add
>>> from sympy import sin, cos, exp >>> from sympy.abc import x, y, z
>>> (x + x*y).as_independent(x) (0, x*y + x) >>> (x + x*y).as_independent(y) (x, x*y) >>> (2*x*sin(x) + y + x + z).as_independent(x) (y + z, 2*x*sin(x) + x) >>> (2*x*sin(x) + y + x + z).as_independent(x, y) (z, 2*x*sin(x) + x + y)
– self is a Mul
>>> (x*sin(x)*cos(y)).as_independent(x) (cos(y), x*sin(x))
noncommutative terms cannot always be separated out when self is a Mul
>>> from sympy import symbols >>> n1, n2, n3 = symbols('n1 n2 n3', commutative=False) >>> (n1 + n1*n2).as_independent(n2) (n1, n1*n2) >>> (n2*n1 + n1*n2).as_independent(n2) (0, n1*n2 + n2*n1) >>> (n1*n2*n3).as_independent(n1) (1, n1*n2*n3) >>> (n1*n2*n3).as_independent(n2) (n1, n2*n3) >>> ((xn1)*(xy)).as_independent(x) (1, (x  y)*(x  n1))
– self is anything else:
>>> (sin(x)).as_independent(x) (1, sin(x)) >>> (sin(x)).as_independent(y) (sin(x), 1) >>> exp(x+y).as_independent(x) (1, exp(x + y))
– force self to be treated as an Add:
>>> (3*x).as_independent(x, as_Add=True) (0, 3*x)
– force self to be treated as a Mul:
>>> (3+x).as_independent(x, as_Add=False) (1, x + 3) >>> (3+x).as_independent(x, as_Add=False) (1, x  3)
Note how the below differs from the above in making the constant on the dep term positive.
>>> (y*(3+x)).as_independent(x) (y, x  3)
 – use .as_independent() for true independence testing instead
of .has(). The former considers only symbols in the free symbols while the latter considers all symbols
>>> from sympy import Integral >>> I = Integral(x, (x, 1, 2)) >>> I.has(x) True >>> x in I.free_symbols False >>> I.as_independent(x) == (I, 1) True >>> (I + x).as_independent(x) == (I, x) True
Note: when trying to get independent terms, a separation method might need to be used first. In this case, it is important to keep track of what you send to this routine so you know how to interpret the returned values
>>> from sympy import separatevars, log >>> separatevars(exp(x+y)).as_independent(x) (exp(y), exp(x)) >>> (x + x*y).as_independent(y) (x, x*y) >>> separatevars(x + x*y).as_independent(y) (x, y + 1) >>> (x*(1 + y)).as_independent(y) (x, y + 1) >>> (x*(1 + y)).expand(mul=True).as_independent(y) (x, x*y) >>> a, b=symbols('a b', positive=True) >>> (log(a*b).expand(log=True)).as_independent(b) (log(a), log(b))
See also
separatevars()
,expand()
,sympy.core.add.Add.as_two_terms()
,sympy.core.mul.Mul.as_two_terms()
,as_coeff_add()
,as_coeff_mul()

as_leading_term
(*symbols)¶ Returns the leading (nonzero) term of the series expansion of self.
The _eval_as_leading_term routines are used to do this, and they must always return a nonzero value.
Examples
>>> from sympy.abc import x >>> (1 + x + x**2).as_leading_term(x) 1 >>> (1/x**2 + x + x**2).as_leading_term(x) x**(2)

as_numer_denom
()¶ expression > a/b > a, b
This is just a stub that should be defined by an object’s class methods to get anything else.
See also
normal()
return a/b instead of a, b

as_ordered_factors
(order=None)¶ Return list of ordered factors (if Mul) else [self].

as_ordered_terms
(order=None, data=False)¶ Transform an expression to an ordered list of terms.
Examples
>>> from sympy import sin, cos >>> from sympy.abc import x
>>> (sin(x)**2*cos(x) + sin(x)**2 + 1).as_ordered_terms() [sin(x)**2*cos(x), sin(x)**2, 1]

as_poly
(*gens, **args)¶ Converts
self
to a polynomial or returnsNone
.>>> from sympy import sin >>> from sympy.abc import x, y
>>> print((x**2 + x*y).as_poly()) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + x*y).as_poly(x, y)) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + sin(y)).as_poly(x, y)) None

as_powers_dict
()¶ Return self as a dictionary of factors with each factor being treated as a power. The keys are the bases of the factors and the values, the corresponding exponents. The resulting dictionary should be used with caution if the expression is a Mul and contains non commutative factors since the order that they appeared will be lost in the dictionary.
See also
as_ordered_factors()
An alternative for noncommutative applications, returning an ordered list of factors.
args_cnc()
Similar to as_ordered_factors, but guarantees separation of commutative and noncommutative factors.

as_real_imag
(deep=True, **hints)¶ Performs complex expansion on ‘self’ and returns a tuple containing collected both real and imaginary parts. This method can’t be confused with re() and im() functions, which does not perform complex expansion at evaluation.
However it is possible to expand both re() and im() functions and get exactly the same results as with a single call to this function.
>>> from sympy import symbols, I
>>> x, y = symbols('x,y', real=True)
>>> (x + y*I).as_real_imag() (x, y)
>>> from sympy.abc import z, w
>>> (z + w*I).as_real_imag() (re(z)  im(w), re(w) + im(z))

as_terms
()¶ Transform an expression to a list of terms.

aseries
(x=None, n=6, bound=0, hir=False)¶ Asymptotic Series expansion of self. This is equivalent to
self.series(x, oo, n)
. Parameters
self (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
n (Value) – The number of terms upto which the series is to be expanded.
hir (Boolean) – Set this parameter to be True to produce hierarchical series. It stops the recursion at an early level and may provide nicer and more useful results.
bound (Value, Integer) – Use the
bound
parameter to give limit on rewriting coefficients in its normalised form.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x, y
>>> e = sin(1/x + exp(x))  sin(1/x)
>>> e.aseries(x) (1/(24*x**4)  1/(2*x**2) + 1 + O(x**(6), (x, oo)))*exp(x)
>>> e.aseries(x, n=3, hir=True) exp(2*x)*sin(1/x)/2 + exp(x)*cos(1/x) + O(exp(3*x), (x, oo))
>>> e = exp(exp(x)/(1  1/x))
>>> e.aseries(x) exp(exp(x)/(1  1/x))
>>> e.aseries(x, bound=3) exp(exp(x)/x**2)*exp(exp(x)/x)*exp(exp(x) + exp(x)/(1  1/x)  exp(x)/x  exp(x)/x**2)*exp(exp(x))
 Returns
Asymptotic series expansion of the expression.
 Return type
Expr
Notes
This algorithm is directly induced from the limit computational algorithm provided by Gruntz. It majorly uses the mrv and rewrite subroutines. The overall idea of this algorithm is first to look for the most rapidly varying subexpression w of a given expression f and then expands f in a series in w. Then same thing is recursively done on the leading coefficient till we get constant coefficients.
If the most rapidly varying subexpression of a given expression f is f itself, the algorithm tries to find a normalised representation of the mrv set and rewrites f using this normalised representation.
If the expansion contains an order term, it will be either
O(x ** (n))
orO(w ** (n))
wherew
belongs to the most rapidly varying expression ofself
.References
 1
A New Algorithm for Computing Asymptotic Series  Dominik Gruntz
 2
Gruntz thesis  p90
 3
See also
Expr.aseries()
See the docstring of this function for complete details of this wrapper.

property
assumptions0
¶ Return object type assumptions.
For example:
Symbol(‘x’, real=True) Symbol(‘x’, integer=True)
are different objects. In other words, besides Python type (Symbol in this case), the initial assumptions are also forming their typeinfo.
Examples
>>> from sympy import Symbol >>> from sympy.abc import x >>> x.assumptions0 {'commutative': True} >>> x = Symbol("x", positive=True) >>> x.assumptions0 {'commutative': True, 'complex': True, 'extended_negative': False, 'extended_nonnegative': True, 'extended_nonpositive': False, 'extended_nonzero': True, 'extended_positive': True, 'extended_real': True, 'finite': True, 'hermitian': True, 'imaginary': False, 'infinite': False, 'negative': False, 'nonnegative': True, 'nonpositive': False, 'nonzero': True, 'positive': True, 'real': True, 'zero': False}

atoms
(*types)¶ Returns the atoms that form the current object.
By default, only objects that are truly atomic and can’t be divided into smaller pieces are returned: symbols, numbers, and number symbols like I and pi. It is possible to request atoms of any type, however, as demonstrated below.
Examples
>>> from sympy import I, pi, sin >>> from sympy.abc import x, y >>> (1 + x + 2*sin(y + I*pi)).atoms() {1, 2, I, pi, x, y}
If one or more types are given, the results will contain only those types of atoms.
>>> from sympy import Number, NumberSymbol, Symbol >>> (1 + x + 2*sin(y + I*pi)).atoms(Symbol) {x, y}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number) {1, 2}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol) {1, 2, pi}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol, I) {1, 2, I, pi}
Note that I (imaginary unit) and zoo (complex infinity) are special types of number symbols and are not part of the NumberSymbol class.
The type can be given implicitly, too:
>>> (1 + x + 2*sin(y + I*pi)).atoms(x) # x is a Symbol {x, y}
Be careful to check your assumptions when using the implicit option since
S(1).is_Integer = True
buttype(S(1))
isOne
, a special type of sympy atom, whiletype(S(2))
is typeInteger
and will find all integers in an expression:>>> from sympy import S >>> (1 + x + 2*sin(y + I*pi)).atoms(S(1)) {1}
>>> (1 + x + 2*sin(y + I*pi)).atoms(S(2)) {1, 2}
Finally, arguments to atoms() can select more than atomic atoms: any sympy type (loaded in core/__init__.py) can be listed as an argument and those types of “atoms” as found in scanning the arguments of the expression recursively:
>>> from sympy import Function, Mul >>> from sympy.core.function import AppliedUndef >>> f = Function('f') >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(Function) {f(x), sin(y + I*pi)} >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(AppliedUndef) {f(x)}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Mul) {I*pi, 2*sin(y + I*pi)}

cancel
(*gens, **args)¶ See the cancel function in sympy.polys

property
canonical_variables
¶ Return a dictionary mapping any variable defined in
self.bound_symbols
to Symbols that do not clash with any existing symbol in the expression.Examples
>>> from sympy import Lambda >>> from sympy.abc import x >>> Lambda(x, 2*x).canonical_variables {x: _0}

classmethod
class_key
()¶ Nice order of classes.

coeff
(x, n=1, right=False)¶ Returns the coefficient from the term(s) containing
x**n
. Ifn
is zero then all terms independent ofx
will be returned.When
x
is noncommutative, the coefficient to the left (default) or right ofx
can be returned. The keyword ‘right’ is ignored whenx
is commutative.See also
as_coefficient()
separate the expression into a coefficient and factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used
Examples
>>> from sympy import symbols >>> from sympy.abc import x, y, z
You can select terms that have an explicit negative in front of them:
>>> (x + 2*y).coeff(1) x >>> (x  2*y).coeff(1) 2*y
You can select terms with no Rational coefficient:
>>> (x + 2*y).coeff(1) x >>> (3 + 2*x + 4*x**2).coeff(1) 0
You can select terms independent of x by making n=0; in this case expr.as_independent(x)[0] is returned (and 0 will be returned instead of None):
>>> (3 + 2*x + 4*x**2).coeff(x, 0) 3 >>> eq = ((x + 1)**3).expand() + 1 >>> eq x**3 + 3*x**2 + 3*x + 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 2] >>> eq = 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 0]
You can select terms that have a numerical term in front of them:
>>> (x  2*y).coeff(2) y >>> from sympy import sqrt >>> (x + sqrt(2)*x).coeff(sqrt(2)) x
The matching is exact:
>>> (3 + 2*x + 4*x**2).coeff(x) 2 >>> (3 + 2*x + 4*x**2).coeff(x**2) 4 >>> (3 + 2*x + 4*x**2).coeff(x**3) 0 >>> (z*(x + y)**2).coeff((x + y)**2) z >>> (z*(x + y)**2).coeff(x + y) 0
In addition, no factoring is done, so 1 + z*(1 + y) is not obtained from the following:
>>> (x + z*(x + x*y)).coeff(x) 1
If such factoring is desired, factor_terms can be used first:
>>> from sympy import factor_terms >>> factor_terms(x + z*(x + x*y)).coeff(x) z*(y + 1) + 1
>>> n, m, o = symbols('n m o', commutative=False) >>> n.coeff(n) 1 >>> (3*n).coeff(n) 3 >>> (n*m + m*n*m).coeff(n) # = (1 + m)*n*m 1 + m >>> (n*m + m*n*m).coeff(n, right=True) # = (1 + m)*n*m m
If there is more than one possible coefficient 0 is returned:
>>> (n*m + m*n).coeff(n) 0
If there is only one possible coefficient, it is returned:
>>> (n*m + x*m*n).coeff(m*n) x >>> (n*m + x*m*n).coeff(m*n, right=1) 1

collect
(syms, func=None, evaluate=True, exact=False, distribute_order_term=True)¶ See the collect function in sympy.simplify

combsimp
()¶ See the combsimp function in sympy.simplify

compare
(other)¶ Return 1, 0, 1 if the object is smaller, equal, or greater than other.
Not in the mathematical sense. If the object is of a different type from the “other” then their classes are ordered according to the sorted_classes list.
Examples
>>> from sympy.abc import x, y >>> x.compare(y) 1 >>> x.compare(x) 0 >>> y.compare(x) 1

compute_leading_term
(x, logx=None)¶ as_leading_term is only allowed for results of .series() This is a wrapper to compute a series first.

conjugate
()¶ Returns the complex conjugate of ‘self’.

could_extract_minus_sign
()¶ Return True if self is not in a canonical form with respect to its sign.
For most expressions, e, there will be a difference in e and e. When there is, True will be returned for one and False for the other; False will be returned if there is no difference.
Examples
>>> from sympy.abc import x, y >>> e = x  y >>> {i.could_extract_minus_sign() for i in (e, e)} {False, True}

count
(query)¶ Count the number of matching subexpressions.

count_ops
(visual=None)¶ wrapper for count_ops that returns the operation count.

doit
(**hints)¶ Evaluate objects that are not evaluated by default like limits, integrals, sums and products. All objects of this kind will be evaluated recursively, unless some species were excluded via ‘hints’ or unless the ‘deep’ hint was set to ‘False’.
>>> from sympy import Integral >>> from sympy.abc import x
>>> 2*Integral(x, x) 2*Integral(x, x)
>>> (2*Integral(x, x)).doit() x**2
>>> (2*Integral(x, x)).doit(deep=False) 2*Integral(x, x)

dummy_eq
(other, symbol=None)¶ Compare two expressions and handle dummy symbols.
Examples
>>> from sympy import Dummy >>> from sympy.abc import x, y
>>> u = Dummy('u')
>>> (u**2 + 1).dummy_eq(x**2 + 1) True >>> (u**2 + 1) == (x**2 + 1) False
>>> (u**2 + y).dummy_eq(x**2 + y, x) True >>> (u**2 + y).dummy_eq(x**2 + y, y) False

equals
(other, failing_expression=False)¶ Return True if self == other, False if it doesn’t, or None. If failing_expression is True then the expression which did not simplify to a 0 will be returned instead of None.
If
self
is a Number (or complex number) that is not zero, then the result is False.If
self
is a number and has not evaluated to zero, evalf will be used to test whether the expression evaluates to zero. If it does so and the result has significance (i.e. the precision is either 1, for a Rational result, or is greater than 1) then the evalf value will be used to return True or False.

evalf
(n=15, subs=None, maxn=100, chop=False, strict=False, quad=None, verbose=False)¶ Evaluate the given formula to an accuracy of n digits.
 Parameters
subs (dict, optional) – Substitute numerical values for symbols, e.g.
subs={x:3, y:1+pi}
. The substitutions must be given as a dictionary.maxn (int, optional) – Allow a maximum temporary working precision of maxn digits.
chop (bool or number, optional) –
Specifies how to replace tiny real or imaginary parts in subresults by exact zeros.
When
True
the chop value defaults to standard precision.Otherwise the chop value is used to determine the magnitude of “small” for purposes of chopping.
>>> from sympy import N >>> x = 1e4 >>> N(x, chop=True) 0.000100000000000000 >>> N(x, chop=1e5) 0.000100000000000000 >>> N(x, chop=1e4) 0
strict (bool, optional) – Raise
PrecisionExhausted
if any subresult fails to evaluate to full accuracy, given the available maxprec.quad (str, optional) – Choose algorithm for numerical quadrature. By default, tanhsinh quadrature is used. For oscillatory integrals on an infinite interval, try
quad='osc'
.verbose (bool, optional) – Print debug information.
Notes
When Floats are naively substituted into an expression, precision errors may adversely affect the result. For example, adding 1e16 (a Float) to 1 will truncate to 1e16; if 1e16 is then subtracted, the result will be 0. That is exactly what happens in the following:
>>> from sympy.abc import x, y, z >>> values = {x: 1e16, y: 1, z: 1e16} >>> (x + y  z).subs(values) 0
Using the subs argument for evalf is the accurate way to evaluate such an expression:
>>> (x + y  z).evalf(subs=values) 1.00000000000000

expand
(deep=True, modulus=None, power_base=True, power_exp=True, mul=True, log=True, multinomial=True, basic=True, **hints)¶ Expand an expression using hints.
See the docstring of the expand() function in sympy.core.function for more information.

property
expr_free_symbols
¶ Like
free_symbols
, but returns the free symbols only if they are contained in an expression node.Examples
>>> from sympy.abc import x, y >>> (x + y).expr_free_symbols {x, y}
If the expression is contained in a nonexpression object, don’t return the free symbols. Compare:
>>> from sympy import Tuple >>> t = Tuple(x + y) >>> t.expr_free_symbols set() >>> t.free_symbols {x, y}

extract_additively
(c)¶ Return self  c if it’s possible to subtract c from self and make all matching coefficients move towards zero, else return None.
Examples
>>> from sympy.abc import x, y >>> e = 2*x + 3 >>> e.extract_additively(x + 1) x + 2 >>> e.extract_additively(3*x) >>> e.extract_additively(4) >>> (y*(x + 1)).extract_additively(x + 1) >>> ((x + 1)*(x + 2*y + 1) + 3).extract_additively(x + 1) (x + 1)*(x + 2*y) + 3
Sometimes autoexpansion will return a less simplified result than desired; gcd_terms might be used in such cases:
>>> from sympy import gcd_terms >>> (4*x*(y + 1) + y).extract_additively(x) 4*x*(y + 1) + x*(4*y + 3)  x*(4*y + 4) + y >>> gcd_terms(_) x*(4*y + 3) + y
See also

extract_branch_factor
(allow_half=False)¶ Try to write self as
exp_polar(2*pi*I*n)*z
in a nice way. Return (z, n).>>> from sympy import exp_polar, I, pi >>> from sympy.abc import x, y >>> exp_polar(I*pi).extract_branch_factor() (exp_polar(I*pi), 0) >>> exp_polar(2*I*pi).extract_branch_factor() (1, 1) >>> exp_polar(pi*I).extract_branch_factor() (exp_polar(I*pi), 1) >>> exp_polar(3*pi*I + x).extract_branch_factor() (exp_polar(x + I*pi), 1) >>> (y*exp_polar(5*pi*I)*exp_polar(3*pi*I + 2*pi*x)).extract_branch_factor() (y*exp_polar(2*pi*x), 1) >>> exp_polar(I*pi/2).extract_branch_factor() (exp_polar(I*pi/2), 0)
If allow_half is True, also extract exp_polar(I*pi):
>>> exp_polar(I*pi).extract_branch_factor(allow_half=True) (1, 1/2) >>> exp_polar(2*I*pi).extract_branch_factor(allow_half=True) (1, 1) >>> exp_polar(3*I*pi).extract_branch_factor(allow_half=True) (1, 3/2) >>> exp_polar(I*pi).extract_branch_factor(allow_half=True) (1, 1/2)

extract_multiplicatively
(c)¶ Return None if it’s not possible to make self in the form c * something in a nice way, i.e. preserving the properties of arguments of self.
Examples
>>> from sympy import symbols, Rational
>>> x, y = symbols('x,y', real=True)
>>> ((x*y)**3).extract_multiplicatively(x**2 * y) x*y**2
>>> ((x*y)**3).extract_multiplicatively(x**4 * y)
>>> (2*x).extract_multiplicatively(2) x
>>> (2*x).extract_multiplicatively(3)
>>> (Rational(1, 2)*x).extract_multiplicatively(3) x/6

factor
(*gens, **args)¶ See the factor() function in sympy.polys.polytools

find
(query, group=False)¶ Find all subexpressions matching a query.

fourier_series
(limits=None)¶ Compute fourier sine/cosine series of self.
See the docstring of the
fourier_series()
in sympy.series.fourier for more information.

fps
(x=None, x0=0, dir=1, hyper=True, order=4, rational=True, full=False)¶ Compute formal power power series of self.
See the docstring of the
fps()
function in sympy.series.formal for more information.

property
free_symbols
¶ Return from the atoms of self those which are free symbols.
For most expressions, all symbols are free symbols. For some classes this is not true. e.g. Integrals use Symbols for the dummy variables which are bound variables, so Integral has a method to return all symbols except those. Derivative keeps track of symbols with respect to which it will perform a derivative; those are bound variables, too, so it has its own free_symbols method.
Any other method that uses bound variables should implement a free_symbols method.

classmethod
fromiter
(args, **assumptions)¶ Create a new object from an iterable.
This is a convenience function that allows one to create objects from any iterable, without having to convert to a list or tuple first.
Examples
>>> from sympy import Tuple >>> Tuple.fromiter(i for i in range(5)) (0, 1, 2, 3, 4)

property
func
¶ The toplevel function in an expression.
The following should hold for all objects:
>> x == x.func(*x.args)
Examples
>>> from sympy.abc import x >>> a = 2*x >>> a.func <class 'sympy.core.mul.Mul'> >>> a.args (2, x) >>> a.func(*a.args) 2*x >>> a == a.func(*a.args) True

gammasimp
()¶ See the gammasimp function in sympy.simplify

getO
()¶ Returns the additive O(..) symbol if there is one, else None.

getn
()¶ Returns the order of the expression.
The order is determined either from the O(…) term. If there is no O(…) term, it returns None.
Examples
>>> from sympy import O >>> from sympy.abc import x >>> (1 + x + O(x**2)).getn() 2 >>> (1 + x).getn()

has
(*patterns)¶ Test whether any subexpression matches any of the patterns.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y, z >>> (x**2 + sin(x*y)).has(z) False >>> (x**2 + sin(x*y)).has(x, y, z) True >>> x.has(x) True
Note
has
is a structural algorithm with no knowledge of mathematics. Consider the following halfopen interval:>>> from sympy.sets import Interval >>> i = Interval.Lopen(0, 5); i Interval.Lopen(0, 5) >>> i.args (0, 5, True, False) >>> i.has(4) # there is no "4" in the arguments False >>> i.has(0) # there *is* a "0" in the arguments True
Instead, use
contains
to determine whether a number is in the interval or not:>>> i.contains(4) True >>> i.contains(0) False
Note that
expr.has(*patterns)
is exactly equivalent toany(expr.has(p) for p in patterns)
. In particular,False
is returned when the list of patterns is empty.>>> x.has() False

integrate
(*args, **kwargs)¶ See the integrate function in sympy.integrals

invert
(g, *gens, **args)¶ Return the multiplicative inverse of
self
modg
whereself
(andg
) may be symbolic expressions).See also
sympy.core.numbers.mod_inverse()
,sympy.polys.polytools.invert()

is_algebraic_expr
(*syms)¶ This tests whether a given expression is algebraic or not, in the given symbols, syms. When syms is not given, all free symbols will be used. The rational function does not have to be in expanded or in any kind of canonical form.
This function returns False for expressions that are “algebraic expressions” with symbolic exponents. This is a simple extension to the is_rational_function, including rational exponentiation.
Examples
>>> from sympy import Symbol, sqrt >>> x = Symbol('x', real=True) >>> sqrt(1 + x).is_rational_function() False >>> sqrt(1 + x).is_algebraic_expr() True
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be an algebraic expression to become one.
>>> from sympy import exp, factor >>> a = sqrt(exp(x)**2 + 2*exp(x) + 1)/(exp(x) + 1) >>> a.is_algebraic_expr(x) False >>> factor(a).is_algebraic_expr() True
See also
References

property
is_comparable
¶ Return True if self can be computed to a real number (or already is a real number) with precision, else False.
Examples
>>> from sympy import exp_polar, pi, I >>> (I*exp_polar(I*pi/2)).is_comparable True >>> (I*exp_polar(I*pi*2)).is_comparable False
A False result does not mean that self cannot be rewritten into a form that would be comparable. For example, the difference computed below is zero but without simplification it does not evaluate to a zero with precision:
>>> e = 2**pi*(1 + 2**pi) >>> dif = e  e.expand() >>> dif.is_comparable False >>> dif.n(2)._prec 1

is_constant
(*wrt, **flags)¶ Return True if self is constant, False if not, or None if the constancy could not be determined conclusively.
If an expression has no free symbols then it is a constant. If there are free symbols it is possible that the expression is a constant, perhaps (but not necessarily) zero. To test such expressions, a few strategies are tried:
1) numerical evaluation at two random points. If two such evaluations give two different values and the values have a precision greater than 1 then self is not constant. If the evaluations agree or could not be obtained with any precision, no decision is made. The numerical testing is done only if
wrt
is different than the free symbols.2) differentiation with respect to variables in ‘wrt’ (or all free symbols if omitted) to see if the expression is constant or not. This will not always lead to an expression that is zero even though an expression is constant (see added test in test_expr.py). If all derivatives are zero then self is constant with respect to the given symbols.
3) finding out zeros of denominator expression with free_symbols. It won’t be constant if there are zeros. It gives more negative answers for expression that are not constant.
If neither evaluation nor differentiation can prove the expression is constant, None is returned unless two numerical values happened to be the same and the flag
failing_number
is True – in that case the numerical value will be returned.If flag simplify=False is passed, self will not be simplified; the default is True since self should be simplified before testing.
Examples
>>> from sympy import cos, sin, Sum, S, pi >>> from sympy.abc import a, n, x, y >>> x.is_constant() False >>> S(2).is_constant() True >>> Sum(x, (x, 1, 10)).is_constant() True >>> Sum(x, (x, 1, n)).is_constant() False >>> Sum(x, (x, 1, n)).is_constant(y) True >>> Sum(x, (x, 1, n)).is_constant(n) False >>> Sum(x, (x, 1, n)).is_constant(x) True >>> eq = a*cos(x)**2 + a*sin(x)**2  a >>> eq.is_constant() True >>> eq.subs({x: pi, a: 2}) == eq.subs({x: pi, a: 3}) == 0 True
>>> (0**x).is_constant() False >>> x.is_constant() False >>> (x**x).is_constant() False >>> one = cos(x)**2 + sin(x)**2 >>> one.is_constant() True >>> ((one  1)**(x + 1)).is_constant() in (True, False) # could be 0 or 1 True

is_polynomial
(*syms)¶ Return True if self is a polynomial in syms and False otherwise.
This checks if self is an exact polynomial in syms. This function returns False for expressions that are “polynomials” with symbolic exponents. Thus, you should be able to apply polynomial algorithms to expressions for which this returns True, and Poly(expr, *syms) should work if and only if expr.is_polynomial(*syms) returns True. The polynomial does not have to be in expanded form. If no symbols are given, all free symbols in the expression will be used.
This is not part of the assumptions system. You cannot do Symbol(‘z’, polynomial=True).
Examples
>>> from sympy import Symbol >>> x = Symbol('x') >>> ((x**2 + 1)**4).is_polynomial(x) True >>> ((x**2 + 1)**4).is_polynomial() True >>> (2**x + 1).is_polynomial(x) False
>>> n = Symbol('n', nonnegative=True, integer=True) >>> (x**n + 1).is_polynomial(x) False
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be a polynomial to become one.
>>> from sympy import sqrt, factor, cancel >>> y = Symbol('y', positive=True) >>> a = sqrt(y**2 + 2*y + 1) >>> a.is_polynomial(y) False >>> factor(a) y + 1 >>> factor(a).is_polynomial(y) True
>>> b = (y**2 + 2*y + 1)/(y + 1) >>> b.is_polynomial(y) False >>> cancel(b) y + 1 >>> cancel(b).is_polynomial(y) True
See also .is_rational_function()

is_rational_function
(*syms)¶ Test whether function is a ratio of two polynomials in the given symbols, syms. When syms is not given, all free symbols will be used. The rational function does not have to be in expanded or in any kind of canonical form.
This function returns False for expressions that are “rational functions” with symbolic exponents. Thus, you should be able to call .as_numer_denom() and apply polynomial algorithms to the result for expressions for which this returns True.
This is not part of the assumptions system. You cannot do Symbol(‘z’, rational_function=True).
Examples
>>> from sympy import Symbol, sin >>> from sympy.abc import x, y
>>> (x/y).is_rational_function() True
>>> (x**2).is_rational_function() True
>>> (x/sin(y)).is_rational_function(y) False
>>> n = Symbol('n', integer=True) >>> (x**n + 1).is_rational_function(x) False
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be a rational function to become one.
>>> from sympy import sqrt, factor >>> y = Symbol('y', positive=True) >>> a = sqrt(y**2 + 2*y + 1)/y >>> a.is_rational_function(y) False >>> factor(a) (y + 1)/y >>> factor(a).is_rational_function(y) True
See also is_algebraic_expr().

leadterm
(x)¶ Returns the leading term a*x**b as a tuple (a, b).
Examples
>>> from sympy.abc import x >>> (1+x+x**2).leadterm(x) (1, 0) >>> (1/x**2+x+x**2).leadterm(x) (1, 2)

limit
(x, xlim, dir='+')¶ Compute limit x>xlim.

lseries
(x=None, x0=0, dir='+', logx=None)¶ Wrapper for series yielding an iterator of the terms of the series.
Note: an infinite series will yield an infinite iterator. The following, for exaxmple, will never terminate. It will just keep printing terms of the sin(x) series:
for term in sin(x).lseries(x): print term
The advantage of lseries() over nseries() is that many times you are just interested in the next term in the series (i.e. the first term for example), but you don’t know how many you should ask for in nseries() using the “n” parameter.
See also nseries().

match
(pattern, old=False)¶ Pattern matching.
Wild symbols match all.
Return
None
when expression (self) does not match with pattern. Otherwise return a dictionary such that:pattern.xreplace(self.match(pattern)) == self
Examples
>>> from sympy import Wild >>> from sympy.abc import x, y >>> p = Wild("p") >>> q = Wild("q") >>> r = Wild("r") >>> e = (x+y)**(x+y) >>> e.match(p**p) {p_: x + y} >>> e.match(p**q) {p_: x + y, q_: x + y} >>> e = (2*x)**2 >>> e.match(p*q**r) {p_: 4, q_: x, r_: 2} >>> (p*q**r).xreplace(e.match(p*q**r)) 4*x**2
The
old
flag will give the oldstyle pattern matching where expressions and patterns are essentially solved to give the match. Both of the following give None unlessold=True
:>>> (x  2).match(p  x, old=True) {p_: 2*x  2} >>> (2/x).match(p*x, old=True) {p_: 2/x**2}

matches
(expr, repl_dict={}, old=False)¶ Helper method for match() that looks for a match between Wild symbols in self and expressions in expr.
Examples
>>> from sympy import symbols, Wild, Basic >>> a, b, c = symbols('a b c') >>> x = Wild('x') >>> Basic(a + x, x).matches(Basic(a + b, c)) is None True >>> Basic(a + x, x).matches(Basic(a + b + c, b + c)) {x_: b + c}

n
(n=15, subs=None, maxn=100, chop=False, strict=False, quad=None, verbose=False)¶ Evaluate the given formula to an accuracy of n digits.
 Parameters
subs (dict, optional) – Substitute numerical values for symbols, e.g.
subs={x:3, y:1+pi}
. The substitutions must be given as a dictionary.maxn (int, optional) – Allow a maximum temporary working precision of maxn digits.
chop (bool or number, optional) –
Specifies how to replace tiny real or imaginary parts in subresults by exact zeros.
When
True
the chop value defaults to standard precision.Otherwise the chop value is used to determine the magnitude of “small” for purposes of chopping.
>>> from sympy import N >>> x = 1e4 >>> N(x, chop=True) 0.000100000000000000 >>> N(x, chop=1e5) 0.000100000000000000 >>> N(x, chop=1e4) 0
strict (bool, optional) – Raise
PrecisionExhausted
if any subresult fails to evaluate to full accuracy, given the available maxprec.quad (str, optional) – Choose algorithm for numerical quadrature. By default, tanhsinh quadrature is used. For oscillatory integrals on an infinite interval, try
quad='osc'
.verbose (bool, optional) – Print debug information.
Notes
When Floats are naively substituted into an expression, precision errors may adversely affect the result. For example, adding 1e16 (a Float) to 1 will truncate to 1e16; if 1e16 is then subtracted, the result will be 0. That is exactly what happens in the following:
>>> from sympy.abc import x, y, z >>> values = {x: 1e16, y: 1, z: 1e16} >>> (x + y  z).subs(values) 0
Using the subs argument for evalf is the accurate way to evaluate such an expression:
>>> (x + y  z).evalf(subs=values) 1.00000000000000

nseries
(x=None, x0=0, n=6, dir='+', logx=None)¶ Wrapper to _eval_nseries if assumptions allow, else to series.
If x is given, x0 is 0, dir=’+’, and self has x, then _eval_nseries is called. This calculates “n” terms in the innermost expressions and then builds up the final series just by “crossmultiplying” everything out.
The optional
logx
parameter can be used to replace any log(x) in the returned series with a symbolic value to avoid evaluating log(x) at 0. A symbol to use in place of log(x) should be provided.Advantage – it’s fast, because we don’t have to determine how many terms we need to calculate in advance.
Disadvantage – you may end up with less terms than you may have expected, but the O(x**n) term appended will always be correct and so the result, though perhaps shorter, will also be correct.
If any of those assumptions is not met, this is treated like a wrapper to series which will try harder to return the correct number of terms.
See also lseries().
Examples
>>> from sympy import sin, log, Symbol >>> from sympy.abc import x, y >>> sin(x).nseries(x, 0, 6) x  x**3/6 + x**5/120 + O(x**6) >>> log(x+1).nseries(x, 0, 5) x  x**2/2 + x**3/3  x**4/4 + O(x**5)
Handling of the
logx
parameter — in the following example the expansion fails sincesin
does not have an asymptotic expansion at oo (the limit of log(x) as x approaches 0):>>> e = sin(log(x)) >>> e.nseries(x, 0, 6) Traceback (most recent call last): ... PoleError: ... ... >>> logx = Symbol('logx') >>> e.nseries(x, 0, 6, logx=logx) sin(logx)
In the following example, the expansion works but gives only an Order term unless the
logx
parameter is used:>>> e = x**y >>> e.nseries(x, 0, 2) O(log(x)**2) >>> e.nseries(x, 0, 2, logx=logx) exp(logx*y)

nsimplify
(constants=[], tolerance=None, full=False)¶ See the nsimplify function in sympy.simplify

powsimp
(*args, **kwargs)¶ See the powsimp function in sympy.simplify

primitive
()¶ Return the positive Rational that can be extracted nonrecursively from every term of self (i.e., self is treated like an Add). This is like the as_coeff_Mul() method but primitive always extracts a positive Rational (never a negative or a Float).
Examples
>>> from sympy.abc import x >>> (3*(x + 1)**2).primitive() (3, (x + 1)**2) >>> a = (6*x + 2); a.primitive() (2, 3*x + 1) >>> b = (x/2 + 3); b.primitive() (1/2, x + 6) >>> (a*b).primitive() == (1, a*b) True

radsimp
(**kwargs)¶ See the radsimp function in sympy.simplify

ratsimp
()¶ See the ratsimp function in sympy.simplify

rcall
(*args)¶ Apply on the argument recursively through the expression tree.
This method is used to simulate a common abuse of notation for operators. For instance in SymPy the the following will not work:
(x+Lambda(y, 2*y))(z) == x+2*z
,however you can use
>>> from sympy import Lambda >>> from sympy.abc import x, y, z >>> (x + Lambda(y, 2*y)).rcall(z) x + 2*z

refine
(assumption=True)¶ See the refine function in sympy.assumptions

removeO
()¶ Removes the additive O(..) symbol if there is one

replace
(query, value, map=False, simultaneous=True, exact=None)¶ Replace matching subexpressions of
self
withvalue
.If
map = True
then also return the mapping {old: new} whereold
was a subexpression found with query andnew
is the replacement value for it. If the expression itself doesn’t match the query, then the returned value will beself.xreplace(map)
otherwise it should beself.subs(ordered(map.items()))
.Traverses an expression tree and performs replacement of matching subexpressions from the bottom to the top of the tree. The default approach is to do the replacement in a simultaneous fashion so changes made are targeted only once. If this is not desired or causes problems,
simultaneous
can be set to False.In addition, if an expression containing more than one Wild symbol is being used to match subexpressions and the
exact
flag is None it will be set to True so the match will only succeed if all nonzero values are received for each Wild that appears in the match pattern. Setting this to False accepts a match of 0; while setting it True accepts all matches that have a 0 in them. See example below for cautions.The list of possible combinations of queries and replacement values is listed below:
Examples
Initial setup
>>> from sympy import log, sin, cos, tan, Wild, Mul, Add >>> from sympy.abc import x, y >>> f = log(sin(x)) + tan(sin(x**2))
 1.1. type > type
obj.replace(type, newtype)
When object of type
type
is found, replace it with the result of passing its argument(s) tonewtype
.>>> f.replace(sin, cos) log(cos(x)) + tan(cos(x**2)) >>> sin(x).replace(sin, cos, map=True) (cos(x), {sin(x): cos(x)}) >>> (x*y).replace(Mul, Add) x + y
 1.2. type > func
obj.replace(type, func)
When object of type
type
is found, applyfunc
to its argument(s).func
must be written to handle the number of arguments oftype
.>>> f.replace(sin, lambda arg: sin(2*arg)) log(sin(2*x)) + tan(sin(2*x**2)) >>> (x*y).replace(Mul, lambda *args: sin(2*Mul(*args))) sin(2*x*y)
 2.1. pattern > expr
obj.replace(pattern(wild), expr(wild))
Replace subexpressions matching
pattern
with the expression written in terms of the Wild symbols inpattern
.>>> a, b = map(Wild, 'ab') >>> f.replace(sin(a), tan(a)) log(tan(x)) + tan(tan(x**2)) >>> f.replace(sin(a), tan(a/2)) log(tan(x/2)) + tan(tan(x**2/2)) >>> f.replace(sin(a), a) log(x) + tan(x**2) >>> (x*y).replace(a*x, a) y
Matching is exact by default when more than one Wild symbol is used: matching fails unless the match gives nonzero values for all Wild symbols:
>>> (2*x + y).replace(a*x + b, b  a) y  2 >>> (2*x).replace(a*x + b, b  a) 2*x
When set to False, the results may be nonintuitive:
>>> (2*x).replace(a*x + b, b  a, exact=False) 2/x
 2.2. pattern > func
obj.replace(pattern(wild), lambda wild: expr(wild))
All behavior is the same as in 2.1 but now a function in terms of pattern variables is used rather than an expression:
>>> f.replace(sin(a), lambda a: sin(2*a)) log(sin(2*x)) + tan(sin(2*x**2))
 3.1. func > func
obj.replace(filter, func)
Replace subexpression
e
withfunc(e)
iffilter(e)
is True.>>> g = 2*sin(x**3) >>> g.replace(lambda expr: expr.is_Number, lambda expr: expr**2) 4*sin(x**9)
The expression itself is also targeted by the query but is done in such a fashion that changes are not made twice.
>>> e = x*(x*y + 1) >>> e.replace(lambda x: x.is_Mul, lambda x: 2*x) 2*x*(2*x*y + 1)
When matching a single symbol, exact will default to True, but this may or may not be the behavior that is desired:
Here, we want exact=False:
>>> from sympy import Function >>> f = Function('f') >>> e = f(1) + f(0) >>> q = f(a), lambda a: f(a + 1) >>> e.replace(*q, exact=False) f(1) + f(2) >>> e.replace(*q, exact=True) f(0) + f(2)
But here, the nature of matching makes selecting the right setting tricky:
>>> e = x**(1 + y) >>> (x**(1 + y)).replace(x**(1 + a), lambda a: x**a, exact=False) 1 >>> (x**(1 + y)).replace(x**(1 + a), lambda a: x**a, exact=True) x**(x  y + 1) >>> (x**y).replace(x**(1 + a), lambda a: x**a, exact=False) 1 >>> (x**y).replace(x**(1 + a), lambda a: x**a, exact=True) x**(1  y)
It is probably better to use a different form of the query that describes the target expression more precisely:
>>> (1 + x**(1 + y)).replace( ... lambda x: x.is_Pow and x.exp.is_Add and x.exp.args[0] == 1, ... lambda x: x.base**(1  (x.exp  1))) ... x**(1  y) + 1
See also
subs()
substitution of subexpressions as defined by the objects themselves.
xreplace()
exact node replacement in expr tree; also capable of using matching rules

rewrite
(*args, **hints)¶ Rewrite functions in terms of other functions.
Rewrites expression containing applications of functions of one kind in terms of functions of different kind. For example you can rewrite trigonometric functions as complex exponentials or combinatorial functions as gamma function.
As a pattern this function accepts a list of functions to to rewrite (instances of DefinedFunction class). As rule you can use string or a destination function instance (in this case rewrite() will use the str() function).
There is also the possibility to pass hints on how to rewrite the given expressions. For now there is only one such hint defined called ‘deep’. When ‘deep’ is set to False it will forbid functions to rewrite their contents.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x
Unspecified pattern:
>>> sin(x).rewrite(exp) I*(exp(I*x)  exp(I*x))/2
Pattern as a single function:
>>> sin(x).rewrite(sin, exp) I*(exp(I*x)  exp(I*x))/2
Pattern as a list of functions:
>>> sin(x).rewrite([sin, ], exp) I*(exp(I*x)  exp(I*x))/2

round
(n=None)¶ Return x rounded to the given decimal place.
If a complex number would results, apply round to the real and imaginary components of the number.
Examples
>>> from sympy import pi, E, I, S, Add, Mul, Number >>> pi.round() 3 >>> pi.round(2) 3.14 >>> (2*pi + E*I).round() 6 + 3*I
The round method has a chopping effect:
>>> (2*pi + I/10).round() 6 >>> (pi/10 + 2*I).round() 2*I >>> (pi/10 + E*I).round(2) 0.31 + 2.72*I
Notes
The Python
round
function uses the SymPyround
method so it will always return a SymPy number (not a Python float or int):>>> isinstance(round(S(123), 2), Number) True

separate
(deep=False, force=False)¶ See the separate function in sympy.simplify

series
(x=None, x0=0, n=6, dir='+', logx=None)¶ Series expansion of “self” around
x = x0
yielding either terms of the series one by one (the lazy series given when n=None), else all the terms at once when n != None.Returns the series expansion of “self” around the point
x = x0
with respect tox
up toO((x  x0)**n, x, x0)
(default n is 6).If
x=None
andself
is univariate, the univariate symbol will be supplied, otherwise an error will be raised. Parameters
expr (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
x0 (Value) – The value around which
x
is calculated. Can be any value fromoo
tooo
.n (Value) – The number of terms upto which the series is to be expanded.
dir (String, optional) – The seriesexpansion can be bidirectional. If
dir="+"
, then (x>x0+). Ifdir="", then (x>x0). For infinite ``x0
(oo
oroo
), thedir
argument is determined from the direction of the infinity (i.e.,dir=""
foroo
).logx (optional) – It is used to replace any log(x) in the returned series with a symbolic value rather than evaluating the actual value.
Examples
>>> from sympy import cos, exp, tan, oo, series >>> from sympy.abc import x, y >>> cos(x).series() 1  x**2/2 + x**4/24 + O(x**6) >>> cos(x).series(n=4) 1  x**2/2 + O(x**4) >>> cos(x).series(x, x0=1, n=2) cos(1)  (x  1)*sin(1) + O((x  1)**2, (x, 1)) >>> e = cos(x + exp(y)) >>> e.series(y, n=2) cos(x + 1)  y*sin(x + 1) + O(y**2) >>> e.series(x, n=2) cos(exp(y))  x*sin(exp(y)) + O(x**2)
If
n=None
then a generator of the series terms will be returned.>>> term=cos(x).series(n=None) >>> [next(term) for i in range(2)] [1, x**2/2]
For
dir=+
(default) the series is calculated from the right and fordir=
the series from the left. For smooth functions this flag will not alter the results.>>> abs(x).series(dir="+") x >>> abs(x).series(dir="") x >>> f = tan(x) >>> f.series(x, 2, 6, "+") tan(2) + (1 + tan(2)**2)*(x  2) + (x  2)**2*(tan(2)**3 + tan(2)) + (x  2)**3*(1/3 + 4*tan(2)**2/3 + tan(2)**4) + (x  2)**4*(tan(2)**5 + 5*tan(2)**3/3 + 2*tan(2)/3) + (x  2)**5*(2/15 + 17*tan(2)**2/15 + 2*tan(2)**4 + tan(2)**6) + O((x  2)**6, (x, 2))
>>> f.series(x, 2, 3, "") tan(2) + (2  x)*(tan(2)**2  1) + (2  x)**2*(tan(2)**3 + tan(2)) + O((x  2)**3, (x, 2))
 Returns
Expr – Series expansion of the expression about x0
 Return type
Expression
 Raises
TypeError – If “n” and “x0” are infinity objects
PoleError – If “x0” is an infinity object

simplify
(**kwargs)¶ See the simplify function in sympy.simplify

sort_key
(order=None)¶ Return a sort key.
Examples
>>> from sympy.core import S, I
>>> sorted([S(1)/2, I, I], key=lambda x: x.sort_key()) [1/2, I, I]
>>> S("[x, 1/x, 1/x**2, x**2, x**(1/2), x**(1/4), x**(3/2)]") [x, 1/x, x**(2), x**2, sqrt(x), x**(1/4), x**(3/2)] >>> sorted(_, key=lambda x: x.sort_key()) [x**(2), 1/x, x**(1/4), sqrt(x), x, x**(3/2), x**2]

subs
(*args, **kwargs)¶ Substitutes old for new in an expression after sympifying args.
 args is either:
two arguments, e.g. foo.subs(old, new)
 one iterable argument, e.g. foo.subs(iterable). The iterable may be
 o an iterable container with (old, new) pairs. In this case the
replacements are processed in the order given with successive patterns possibly affecting replacements already made.
 o a dict or set whose key/value items correspond to old/new pairs.
In this case the old/new pairs will be sorted by op count and in case of a tie, by number of args and the default_sort_key. The resulting sorted list is then processed as an iterable container (see previous).
If the keyword
simultaneous
is True, the subexpressions will not be evaluated until all the substitutions have been made.Examples
>>> from sympy import pi, exp, limit, oo >>> from sympy.abc import x, y >>> (1 + x*y).subs(x, pi) pi*y + 1 >>> (1 + x*y).subs({x:pi, y:2}) 1 + 2*pi >>> (1 + x*y).subs([(x, pi), (y, 2)]) 1 + 2*pi >>> reps = [(y, x**2), (x, 2)] >>> (x + y).subs(reps) 6 >>> (x + y).subs(reversed(reps)) x**2 + 2
>>> (x**2 + x**4).subs(x**2, y) y**2 + y
To replace only the x**2 but not the x**4, use xreplace:
>>> (x**2 + x**4).xreplace({x**2: y}) x**4 + y
To delay evaluation until all substitutions have been made, set the keyword
simultaneous
to True:>>> (x/y).subs([(x, 0), (y, 0)]) 0 >>> (x/y).subs([(x, 0), (y, 0)], simultaneous=True) nan
This has the added feature of not allowing subsequent substitutions to affect those already made:
>>> ((x + y)/y).subs({x + y: y, y: x + y}) 1 >>> ((x + y)/y).subs({x + y: y, y: x + y}, simultaneous=True) y/(x + y)
In order to obtain a canonical result, unordered iterables are sorted by count_op length, number of arguments and by the default_sort_key to break any ties. All other iterables are left unsorted.
>>> from sympy import sqrt, sin, cos >>> from sympy.abc import a, b, c, d, e
>>> A = (sqrt(sin(2*x)), a) >>> B = (sin(2*x), b) >>> C = (cos(2*x), c) >>> D = (x, d) >>> E = (exp(x), e)
>>> expr = sqrt(sin(2*x))*sin(exp(x)*x)*cos(2*x) + sin(2*x)
>>> expr.subs(dict([A, B, C, D, E])) a*c*sin(d*e) + b
The resulting expression represents a literal replacement of the old arguments with the new arguments. This may not reflect the limiting behavior of the expression:
>>> (x**3  3*x).subs({x: oo}) nan
>>> limit(x**3  3*x, x, oo) oo
If the substitution will be followed by numerical evaluation, it is better to pass the substitution to evalf as
>>> (1/x).evalf(subs={x: 3.0}, n=21) 0.333333333333333333333
rather than
>>> (1/x).subs({x: 3.0}).evalf(21) 0.333333333333333314830
as the former will ensure that the desired level of precision is obtained.
See also
replace()
replacement capable of doing wildcardlike matching, parsing of match, and conditional replacements
xreplace()
exact node replacement in expr tree; also capable of using matching rules
sympy.core.evalf.EvalfMixin.evalf()
calculates the given formula to a desired level of precision

taylor_term
(n, x, *previous_terms)¶ General method for the taylor term.
This method is slow, because it differentiates ntimes. Subclasses can redefine it to make it faster by using the “previous_terms”.

together
(*args, **kwargs)¶ See the together function in sympy.polys

trigsimp
(**args)¶ See the trigsimp function in sympy.simplify

xreplace
(rule, hack2=False)¶ Replace occurrences of objects within the expression.
 Parameters
rule (dictlike) – Expresses a replacement rule
 Returns
xreplace
 Return type
the result of the replacement
Examples
>>> from sympy import symbols, pi, exp >>> x, y, z = symbols('x y z') >>> (1 + x*y).xreplace({x: pi}) pi*y + 1 >>> (1 + x*y).xreplace({x: pi, y: 2}) 1 + 2*pi
Replacements occur only if an entire node in the expression tree is matched:
>>> (x*y + z).xreplace({x*y: pi}) z + pi >>> (x*y*z).xreplace({x*y: pi}) x*y*z >>> (2*x).xreplace({2*x: y, x: z}) y >>> (2*2*x).xreplace({2*x: y, x: z}) 4*z >>> (x + y + 2).xreplace({x + y: 2}) x + y + 2 >>> (x + 2 + exp(x + 2)).xreplace({x + 2: y}) x + exp(y) + 2
xreplace doesn’t differentiate between free and bound symbols. In the following, subs(x, y) would not change x since it is a bound symbol, but xreplace does:
>>> from sympy import Integral >>> Integral(x, (x, 1, 2*x)).xreplace({x: y}) Integral(y, (y, 1, 2*y))
Trying to replace x with an expression raises an error:
>>> Integral(x, (x, 1, 2*x)).xreplace({x: 2*y}) ValueError: Invalid limits given: ((2*y, 1, 4*y),)


class
galgebra.atoms.
BasisBladeSymbol
[source]¶ A basis blade such as \(e_1 \wedge e_2\)

apart
(x=None, **args)¶ See the apart function in sympy.polys

property
args
¶ Returns a tuple of arguments of ‘self’.
Examples
>>> from sympy import cot >>> from sympy.abc import x, y
>>> cot(x).args (x,)
>>> cot(x).args[0] x
>>> (x*y).args (x, y)
>>> (x*y).args[1] y
Notes
Never use self._args, always use self.args. Only use _args in __new__ when creating a new function. Don’t override .args() from Basic (so that it’s easy to change the interface in the future if needed).

args_cnc
(cset=False, warn=True, split_1=True)¶ Return [commutative factors, noncommutative factors] of self.
self is treated as a Mul and the ordering of the factors is maintained. If
cset
is True the commutative factors will be returned in a set. If there were repeated factors (as may happen with an unevaluated Mul) then an error will be raised unless it is explicitly suppressed by settingwarn
to False.Note: 1 is always separated from a Number unless split_1 is False.
>>> from sympy import symbols, oo >>> A, B = symbols('A B', commutative=0) >>> x, y = symbols('x y') >>> (2*x*y).args_cnc() [[1, 2, x, y], []] >>> (2.5*x).args_cnc() [[1, 2.5, x], []] >>> (2*x*A*B*y).args_cnc() [[1, 2, x, y], [A, B]] >>> (2*x*A*B*y).args_cnc(split_1=False) [[2, x, y], [A, B]] >>> (2*x*y).args_cnc(cset=True) [{1, 2, x, y}, []]
The arg is always treated as a Mul:
>>> (2 + x + A).args_cnc() [[], [x  2 + A]] >>> (oo).args_cnc() # oo is a singleton [[1, oo], []]

as_coeff_Add
(rational=False)¶ Efficiently extract the coefficient of a summation.

as_coeff_Mul
(rational=False)¶ Efficiently extract the coefficient of a product.

as_coeff_add
(*deps)¶ Return the tuple (c, args) where self is written as an Add,
a
.c should be a Rational added to any terms of the Add that are independent of deps.
args should be a tuple of all other terms of
a
; args is empty if self is a Number or if self is independent of deps (when given).This should be used when you don’t know if self is an Add or not but you want to treat self as an Add or if you want to process the individual arguments of the tail of self as an Add.
if you know self is an Add and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail.
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_add() (3, ()) >>> (3 + x).as_coeff_add() (3, (x,)) >>> (3 + x + y).as_coeff_add(x) (y + 3, (x,)) >>> (3 + y).as_coeff_add(x) (y + 3, ())

as_coeff_exponent
(x)¶ c*x**e > c,e
where x can be any symbolic expression.

as_coeff_mul
(*deps, **kwargs)¶ Return the tuple (c, args) where self is written as a Mul,
m
.c should be a Rational multiplied by any factors of the Mul that are independent of deps.
args should be a tuple of all other factors of m; args is empty if self is a Number or if self is independent of deps (when given).
This should be used when you don’t know if self is a Mul or not but you want to treat self as a Mul or if you want to process the individual arguments of the tail of self as a Mul.
if you know self is a Mul and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail;
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_mul() (3, ()) >>> (3*x*y).as_coeff_mul() (3, (x, y)) >>> (3*x*y).as_coeff_mul(x) (3*y, (x,)) >>> (3*y).as_coeff_mul(x) (3*y, ())

as_coefficient
(expr)¶ Extracts symbolic coefficient at the given expression. In other words, this functions separates ‘self’ into the product of ‘expr’ and ‘expr’free coefficient. If such separation is not possible it will return None.
Examples
>>> from sympy import E, pi, sin, I, Poly >>> from sympy.abc import x
>>> E.as_coefficient(E) 1 >>> (2*E).as_coefficient(E) 2 >>> (2*sin(E)*E).as_coefficient(E)
Two terms have E in them so a sum is returned. (If one were desiring the coefficient of the term exactly matching E then the constant from the returned expression could be selected. Or, for greater precision, a method of Poly can be used to indicate the desired term from which the coefficient is desired.)
>>> (2*E + x*E).as_coefficient(E) x + 2 >>> _.args[0] # just want the exact match 2 >>> p = Poly(2*E + x*E); p Poly(x*E + 2*E, x, E, domain='ZZ') >>> p.coeff_monomial(E) 2 >>> p.nth(0, 1) 2
Since the following cannot be written as a product containing E as a factor, None is returned. (If the coefficient
2*x
is desired then thecoeff
method should be used.)>>> (2*E*x + x).as_coefficient(E) >>> (2*E*x + x).coeff(E) 2*x
>>> (E*(x + 1) + x).as_coefficient(E)
>>> (2*pi*I).as_coefficient(pi*I) 2 >>> (2*I).as_coefficient(pi*I)
See also
coeff()
return sum of terms have a given factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used

as_coefficients_dict
()¶ Return a dictionary mapping terms to their Rational coefficient. Since the dictionary is a defaultdict, inquiries about terms which were not present will return a coefficient of 0. If an expression is not an Add it is considered to have a single term.
Examples
>>> from sympy.abc import a, x >>> (3*x + a*x + 4).as_coefficients_dict() {1: 4, x: 3, a*x: 1} >>> _[a] 0 >>> (3*a*x).as_coefficients_dict() {a*x: 3}

as_content_primitive
(radical=False, clear=True)¶ This method should recursively remove a Rational from all arguments and return that (content) and the new self (primitive). The content should always be positive and
Mul(*foo.as_content_primitive()) == foo
. The primitive need not be in canonical form and should try to preserve the underlying structure if possible (i.e. expand_mul should not be applied to self).Examples
>>> from sympy import sqrt >>> from sympy.abc import x, y, z
>>> eq = 2 + 2*x + 2*y*(3 + 3*y)
The as_content_primitive function is recursive and retains structure:
>>> eq.as_content_primitive() (2, x + 3*y*(y + 1) + 1)
Integer powers will have Rationals extracted from the base:
>>> ((2 + 6*x)**2).as_content_primitive() (4, (3*x + 1)**2) >>> ((2 + 6*x)**(2*y)).as_content_primitive() (1, (2*(3*x + 1))**(2*y))
Terms may end up joining once their as_content_primitives are added:
>>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (11, x*(y + 1)) >>> ((3*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (9, x*(y + 1)) >>> ((3*(z*(1 + y)) + 2.0*x*(3 + 3*y))).as_content_primitive() (1, 6.0*x*(y + 1) + 3*z*(y + 1)) >>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))**2).as_content_primitive() (121, x**2*(y + 1)**2) >>> ((x*(1 + y) + 0.4*x*(3 + 3*y))**2).as_content_primitive() (1, 4.84*x**2*(y + 1)**2)
Radical content can also be factored out of the primitive:
>>> (2*sqrt(2) + 4*sqrt(10)).as_content_primitive(radical=True) (2, sqrt(2)*(1 + 2*sqrt(5)))
If clear=False (default is True) then content will not be removed from an Add if it can be distributed to leave one or more terms with integer coefficients.
>>> (x/2 + y).as_content_primitive() (1/2, x + 2*y) >>> (x/2 + y).as_content_primitive(clear=False) (1, x/2 + y)

as_dummy
()¶ Return the expression with any objects having structurally bound symbols replaced with unique, canonical symbols within the object in which they appear and having only the default assumption for commutativity being True.
Examples
>>> from sympy import Integral, Symbol >>> from sympy.abc import x, y >>> r = Symbol('r', real=True) >>> Integral(r, (r, x)).as_dummy() Integral(_0, (_0, x)) >>> _.variables[0].is_real is None True
Notes
Any object that has structural dummy variables should have a property, bound_symbols that returns a list of structural dummy symbols of the object itself.
Lambda and Subs have bound symbols, but because of how they are cached, they already compare the same regardless of their bound symbols:
>>> from sympy import Lambda >>> Lambda(x, x + 1) == Lambda(y, y + 1) True

as_expr
(*gens)¶ Convert a polynomial to a SymPy expression.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y
>>> f = (x**2 + x*y).as_poly(x, y) >>> f.as_expr() x**2 + x*y
>>> sin(x).as_expr() sin(x)

as_independent
(*deps, **hint)¶ A mostly naive separation of a Mul or Add into arguments that are not are dependent on deps. To obtain as complete a separation of variables as possible, use a separation method first, e.g.:
separatevars() to change Mul, Add and Pow (including exp) into Mul
.expand(mul=True) to change Add or Mul into Add
.expand(log=True) to change log expr into an Add
The only nonnaive thing that is done here is to respect noncommutative ordering of variables and to always return (0, 0) for self of zero regardless of hints.
For nonzero self, the returned tuple (i, d) has the following interpretation:
i will has no variable that appears in deps
d will either have terms that contain variables that are in deps, or be equal to 0 (when self is an Add) or 1 (when self is a Mul)
if self is an Add then self = i + d
if self is a Mul then self = i*d
otherwise (self, S.One) or (S.One, self) is returned.
To force the expression to be treated as an Add, use the hint as_Add=True
Examples
– self is an Add
>>> from sympy import sin, cos, exp >>> from sympy.abc import x, y, z
>>> (x + x*y).as_independent(x) (0, x*y + x) >>> (x + x*y).as_independent(y) (x, x*y) >>> (2*x*sin(x) + y + x + z).as_independent(x) (y + z, 2*x*sin(x) + x) >>> (2*x*sin(x) + y + x + z).as_independent(x, y) (z, 2*x*sin(x) + x + y)
– self is a Mul
>>> (x*sin(x)*cos(y)).as_independent(x) (cos(y), x*sin(x))
noncommutative terms cannot always be separated out when self is a Mul
>>> from sympy import symbols >>> n1, n2, n3 = symbols('n1 n2 n3', commutative=False) >>> (n1 + n1*n2).as_independent(n2) (n1, n1*n2) >>> (n2*n1 + n1*n2).as_independent(n2) (0, n1*n2 + n2*n1) >>> (n1*n2*n3).as_independent(n1) (1, n1*n2*n3) >>> (n1*n2*n3).as_independent(n2) (n1, n2*n3) >>> ((xn1)*(xy)).as_independent(x) (1, (x  y)*(x  n1))
– self is anything else:
>>> (sin(x)).as_independent(x) (1, sin(x)) >>> (sin(x)).as_independent(y) (sin(x), 1) >>> exp(x+y).as_independent(x) (1, exp(x + y))
– force self to be treated as an Add:
>>> (3*x).as_independent(x, as_Add=True) (0, 3*x)
– force self to be treated as a Mul:
>>> (3+x).as_independent(x, as_Add=False) (1, x + 3) >>> (3+x).as_independent(x, as_Add=False) (1, x  3)
Note how the below differs from the above in making the constant on the dep term positive.
>>> (y*(3+x)).as_independent(x) (y, x  3)
 – use .as_independent() for true independence testing instead
of .has(). The former considers only symbols in the free symbols while the latter considers all symbols
>>> from sympy import Integral >>> I = Integral(x, (x, 1, 2)) >>> I.has(x) True >>> x in I.free_symbols False >>> I.as_independent(x) == (I, 1) True >>> (I + x).as_independent(x) == (I, x) True
Note: when trying to get independent terms, a separation method might need to be used first. In this case, it is important to keep track of what you send to this routine so you know how to interpret the returned values
>>> from sympy import separatevars, log >>> separatevars(exp(x+y)).as_independent(x) (exp(y), exp(x)) >>> (x + x*y).as_independent(y) (x, x*y) >>> separatevars(x + x*y).as_independent(y) (x, y + 1) >>> (x*(1 + y)).as_independent(y) (x, y + 1) >>> (x*(1 + y)).expand(mul=True).as_independent(y) (x, x*y) >>> a, b=symbols('a b', positive=True) >>> (log(a*b).expand(log=True)).as_independent(b) (log(a), log(b))
See also
separatevars()
,expand()
,sympy.core.add.Add.as_two_terms()
,sympy.core.mul.Mul.as_two_terms()
,as_coeff_add()
,as_coeff_mul()

as_leading_term
(*symbols)¶ Returns the leading (nonzero) term of the series expansion of self.
The _eval_as_leading_term routines are used to do this, and they must always return a nonzero value.
Examples
>>> from sympy.abc import x >>> (1 + x + x**2).as_leading_term(x) 1 >>> (1/x**2 + x + x**2).as_leading_term(x) x**(2)

as_numer_denom
()¶ expression > a/b > a, b
This is just a stub that should be defined by an object’s class methods to get anything else.
See also
normal()
return a/b instead of a, b

as_ordered_factors
(order=None)¶ Return list of ordered factors (if Mul) else [self].

as_ordered_terms
(order=None, data=False)¶ Transform an expression to an ordered list of terms.
Examples
>>> from sympy import sin, cos >>> from sympy.abc import x
>>> (sin(x)**2*cos(x) + sin(x)**2 + 1).as_ordered_terms() [sin(x)**2*cos(x), sin(x)**2, 1]

as_poly
(*gens, **args)¶ Converts
self
to a polynomial or returnsNone
.>>> from sympy import sin >>> from sympy.abc import x, y
>>> print((x**2 + x*y).as_poly()) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + x*y).as_poly(x, y)) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + sin(y)).as_poly(x, y)) None

as_powers_dict
()¶ Return self as a dictionary of factors with each factor being treated as a power. The keys are the bases of the factors and the values, the corresponding exponents. The resulting dictionary should be used with caution if the expression is a Mul and contains non commutative factors since the order that they appeared will be lost in the dictionary.
See also
as_ordered_factors()
An alternative for noncommutative applications, returning an ordered list of factors.
args_cnc()
Similar to as_ordered_factors, but guarantees separation of commutative and noncommutative factors.

as_real_imag
(deep=True, **hints)¶ Performs complex expansion on ‘self’ and returns a tuple containing collected both real and imaginary parts. This method can’t be confused with re() and im() functions, which does not perform complex expansion at evaluation.
However it is possible to expand both re() and im() functions and get exactly the same results as with a single call to this function.
>>> from sympy import symbols, I
>>> x, y = symbols('x,y', real=True)
>>> (x + y*I).as_real_imag() (x, y)
>>> from sympy.abc import z, w
>>> (z + w*I).as_real_imag() (re(z)  im(w), re(w) + im(z))

as_terms
()¶ Transform an expression to a list of terms.

aseries
(x=None, n=6, bound=0, hir=False)¶ Asymptotic Series expansion of self. This is equivalent to
self.series(x, oo, n)
. Parameters
self (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
n (Value) – The number of terms upto which the series is to be expanded.
hir (Boolean) – Set this parameter to be True to produce hierarchical series. It stops the recursion at an early level and may provide nicer and more useful results.
bound (Value, Integer) – Use the
bound
parameter to give limit on rewriting coefficients in its normalised form.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x, y
>>> e = sin(1/x + exp(x))  sin(1/x)
>>> e.aseries(x) (1/(24*x**4)  1/(2*x**2) + 1 + O(x**(6), (x, oo)))*exp(x)
>>> e.aseries(x, n=3, hir=True) exp(2*x)*sin(1/x)/2 + exp(x)*cos(1/x) + O(exp(3*x), (x, oo))
>>> e = exp(exp(x)/(1  1/x))
>>> e.aseries(x) exp(exp(x)/(1  1/x))
>>> e.aseries(x, bound=3) exp(exp(x)/x**2)*exp(exp(x)/x)*exp(exp(x) + exp(x)/(1  1/x)  exp(x)/x  exp(x)/x**2)*exp(exp(x))
 Returns
Asymptotic series expansion of the expression.
 Return type
Expr
Notes
This algorithm is directly induced from the limit computational algorithm provided by Gruntz. It majorly uses the mrv and rewrite subroutines. The overall idea of this algorithm is first to look for the most rapidly varying subexpression w of a given expression f and then expands f in a series in w. Then same thing is recursively done on the leading coefficient till we get constant coefficients.
If the most rapidly varying subexpression of a given expression f is f itself, the algorithm tries to find a normalised representation of the mrv set and rewrites f using this normalised representation.
If the expansion contains an order term, it will be either
O(x ** (n))
orO(w ** (n))
wherew
belongs to the most rapidly varying expression ofself
.References
 1
A New Algorithm for Computing Asymptotic Series  Dominik Gruntz
 2
Gruntz thesis  p90
 3
See also
Expr.aseries()
See the docstring of this function for complete details of this wrapper.

property
assumptions0
¶ Return object type assumptions.
For example:
Symbol(‘x’, real=True) Symbol(‘x’, integer=True)
are different objects. In other words, besides Python type (Symbol in this case), the initial assumptions are also forming their typeinfo.
Examples
>>> from sympy import Symbol >>> from sympy.abc import x >>> x.assumptions0 {'commutative': True} >>> x = Symbol("x", positive=True) >>> x.assumptions0 {'commutative': True, 'complex': True, 'extended_negative': False, 'extended_nonnegative': True, 'extended_nonpositive': False, 'extended_nonzero': True, 'extended_positive': True, 'extended_real': True, 'finite': True, 'hermitian': True, 'imaginary': False, 'infinite': False, 'negative': False, 'nonnegative': True, 'nonpositive': False, 'nonzero': True, 'positive': True, 'real': True, 'zero': False}

atoms
(*types)¶ Returns the atoms that form the current object.
By default, only objects that are truly atomic and can’t be divided into smaller pieces are returned: symbols, numbers, and number symbols like I and pi. It is possible to request atoms of any type, however, as demonstrated below.
Examples
>>> from sympy import I, pi, sin >>> from sympy.abc import x, y >>> (1 + x + 2*sin(y + I*pi)).atoms() {1, 2, I, pi, x, y}
If one or more types are given, the results will contain only those types of atoms.
>>> from sympy import Number, NumberSymbol, Symbol >>> (1 + x + 2*sin(y + I*pi)).atoms(Symbol) {x, y}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number) {1, 2}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol) {1, 2, pi}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol, I) {1, 2, I, pi}
Note that I (imaginary unit) and zoo (complex infinity) are special types of number symbols and are not part of the NumberSymbol class.
The type can be given implicitly, too:
>>> (1 + x + 2*sin(y + I*pi)).atoms(x) # x is a Symbol {x, y}
Be careful to check your assumptions when using the implicit option since
S(1).is_Integer = True
buttype(S(1))
isOne
, a special type of sympy atom, whiletype(S(2))
is typeInteger
and will find all integers in an expression:>>> from sympy import S >>> (1 + x + 2*sin(y + I*pi)).atoms(S(1)) {1}
>>> (1 + x + 2*sin(y + I*pi)).atoms(S(2)) {1, 2}
Finally, arguments to atoms() can select more than atomic atoms: any sympy type (loaded in core/__init__.py) can be listed as an argument and those types of “atoms” as found in scanning the arguments of the expression recursively:
>>> from sympy import Function, Mul >>> from sympy.core.function import AppliedUndef >>> f = Function('f') >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(Function) {f(x), sin(y + I*pi)} >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(AppliedUndef) {f(x)}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Mul) {I*pi, 2*sin(y + I*pi)}

cancel
(*gens, **args)¶ See the cancel function in sympy.polys

property
canonical_variables
¶ Return a dictionary mapping any variable defined in
self.bound_symbols
to Symbols that do not clash with any existing symbol in the expression.Examples
>>> from sympy import Lambda >>> from sympy.abc import x >>> Lambda(x, 2*x).canonical_variables {x: _0}

classmethod
class_key
()¶ Nice order of classes.

coeff
(x, n=1, right=False)¶ Returns the coefficient from the term(s) containing
x**n
. Ifn
is zero then all terms independent ofx
will be returned.When
x
is noncommutative, the coefficient to the left (default) or right ofx
can be returned. The keyword ‘right’ is ignored whenx
is commutative.See also
as_coefficient()
separate the expression into a coefficient and factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used
Examples
>>> from sympy import symbols >>> from sympy.abc import x, y, z
You can select terms that have an explicit negative in front of them:
>>> (x + 2*y).coeff(1) x >>> (x  2*y).coeff(1) 2*y
You can select terms with no Rational coefficient:
>>> (x + 2*y).coeff(1) x >>> (3 + 2*x + 4*x**2).coeff(1) 0
You can select terms independent of x by making n=0; in this case expr.as_independent(x)[0] is returned (and 0 will be returned instead of None):
>>> (3 + 2*x + 4*x**2).coeff(x, 0) 3 >>> eq = ((x + 1)**3).expand() + 1 >>> eq x**3 + 3*x**2 + 3*x + 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 2] >>> eq = 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 0]
You can select terms that have a numerical term in front of them:
>>> (x  2*y).coeff(2) y >>> from sympy import sqrt >>> (x + sqrt(2)*x).coeff(sqrt(2)) x
The matching is exact:
>>> (3 + 2*x + 4*x**2).coeff(x) 2 >>> (3 + 2*x + 4*x**2).coeff(x**2) 4 >>> (3 + 2*x + 4*x**2).coeff(x**3) 0 >>> (z*(x + y)**2).coeff((x + y)**2) z >>> (z*(x + y)**2).coeff(x + y) 0
In addition, no factoring is done, so 1 + z*(1 + y) is not obtained from the following:
>>> (x + z*(x + x*y)).coeff(x) 1
If such factoring is desired, factor_terms can be used first:
>>> from sympy import factor_terms >>> factor_terms(x + z*(x + x*y)).coeff(x) z*(y + 1) + 1
>>> n, m, o = symbols('n m o', commutative=False) >>> n.coeff(n) 1 >>> (3*n).coeff(n) 3 >>> (n*m + m*n*m).coeff(n) # = (1 + m)*n*m 1 + m >>> (n*m + m*n*m).coeff(n, right=True) # = (1 + m)*n*m m
If there is more than one possible coefficient 0 is returned:
>>> (n*m + m*n).coeff(n) 0
If there is only one possible coefficient, it is returned:
>>> (n*m + x*m*n).coeff(m*n) x >>> (n*m + x*m*n).coeff(m*n, right=1) 1

collect
(syms, func=None, evaluate=True, exact=False, distribute_order_term=True)¶ See the collect function in sympy.simplify

combsimp
()¶ See the combsimp function in sympy.simplify

compare
(other)¶ Return 1, 0, 1 if the object is smaller, equal, or greater than other.
Not in the mathematical sense. If the object is of a different type from the “other” then their classes are ordered according to the sorted_classes list.
Examples
>>> from sympy.abc import x, y >>> x.compare(y) 1 >>> x.compare(x) 0 >>> y.compare(x) 1

compute_leading_term
(x, logx=None)¶ as_leading_term is only allowed for results of .series() This is a wrapper to compute a series first.

conjugate
()¶ Returns the complex conjugate of ‘self’.

could_extract_minus_sign
()¶ Return True if self is not in a canonical form with respect to its sign.
For most expressions, e, there will be a difference in e and e. When there is, True will be returned for one and False for the other; False will be returned if there is no difference.
Examples
>>> from sympy.abc import x, y >>> e = x  y >>> {i.could_extract_minus_sign() for i in (e, e)} {False, True}

count
(query)¶ Count the number of matching subexpressions.

count_ops
(visual=None)¶ wrapper for count_ops that returns the operation count.

doit
(**hints)¶ Evaluate objects that are not evaluated by default like limits, integrals, sums and products. All objects of this kind will be evaluated recursively, unless some species were excluded via ‘hints’ or unless the ‘deep’ hint was set to ‘False’.
>>> from sympy import Integral >>> from sympy.abc import x
>>> 2*Integral(x, x) 2*Integral(x, x)
>>> (2*Integral(x, x)).doit() x**2
>>> (2*Integral(x, x)).doit(deep=False) 2*Integral(x, x)

dummy_eq
(other, symbol=None)¶ Compare two expressions and handle dummy symbols.
Examples
>>> from sympy import Dummy >>> from sympy.abc import x, y
>>> u = Dummy('u')
>>> (u**2 + 1).dummy_eq(x**2 + 1) True >>> (u**2 + 1) == (x**2 + 1) False
>>> (u**2 + y).dummy_eq(x**2 + y, x) True >>> (u**2 + y).dummy_eq(x**2 + y, y) False

equals
(other, failing_expression=False)¶ Return True if self == other, False if it doesn’t, or None. If failing_expression is True then the expression which did not simplify to a 0 will be returned instead of None.
If
self
is a Number (or complex number) that is not zero, then the result is False.If
self
is a number and has not evaluated to zero, evalf will be used to test whether the expression evaluates to zero. If it does so and the result has significance (i.e. the precision is either 1, for a Rational result, or is greater than 1) then the evalf value will be used to return True or False.

evalf
(n=15, subs=None, maxn=100, chop=False, strict=False, quad=None, verbose=False)¶ Evaluate the given formula to an accuracy of n digits.
 Parameters
subs (dict, optional) – Substitute numerical values for symbols, e.g.
subs={x:3, y:1+pi}
. The substitutions must be given as a dictionary.maxn (int, optional) – Allow a maximum temporary working precision of maxn digits.
chop (bool or number, optional) –
Specifies how to replace tiny real or imaginary parts in subresults by exact zeros.
When
True
the chop value defaults to standard precision.Otherwise the chop value is used to determine the magnitude of “small” for purposes of chopping.
>>> from sympy import N >>> x = 1e4 >>> N(x, chop=True) 0.000100000000000000 >>> N(x, chop=1e5) 0.000100000000000000 >>> N(x, chop=1e4) 0
strict (bool, optional) – Raise
PrecisionExhausted
if any subresult fails to evaluate to full accuracy, given the available maxprec.quad (str, optional) – Choose algorithm for numerical quadrature. By default, tanhsinh quadrature is used. For oscillatory integrals on an infinite interval, try
quad='osc'
.verbose (bool, optional) – Print debug information.
Notes
When Floats are naively substituted into an expression, precision errors may adversely affect the result. For example, adding 1e16 (a Float) to 1 will truncate to 1e16; if 1e16 is then subtracted, the result will be 0. That is exactly what happens in the following:
>>> from sympy.abc import x, y, z >>> values = {x: 1e16, y: 1, z: 1e16} >>> (x + y  z).subs(values) 0
Using the subs argument for evalf is the accurate way to evaluate such an expression:
>>> (x + y  z).evalf(subs=values) 1.00000000000000

expand
(deep=True, modulus=None, power_base=True, power_exp=True, mul=True, log=True, multinomial=True, basic=True, **hints)¶ Expand an expression using hints.
See the docstring of the expand() function in sympy.core.function for more information.

property
expr_free_symbols
¶ Like
free_symbols
, but returns the free symbols only if they are contained in an expression node.Examples
>>> from sympy.abc import x, y >>> (x + y).expr_free_symbols {x, y}
If the expression is contained in a nonexpression object, don’t return the free symbols. Compare:
>>> from sympy import Tuple >>> t = Tuple(x + y) >>> t.expr_free_symbols set() >>> t.free_symbols {x, y}

extract_additively
(c)¶ Return self  c if it’s possible to subtract c from self and make all matching coefficients move towards zero, else return None.
Examples
>>> from sympy.abc import x, y >>> e = 2*x + 3 >>> e.extract_additively(x + 1) x + 2 >>> e.extract_additively(3*x) >>> e.extract_additively(4) >>> (y*(x + 1)).extract_additively(x + 1) >>> ((x + 1)*(x + 2*y + 1) + 3).extract_additively(x + 1) (x + 1)*(x + 2*y) + 3
Sometimes autoexpansion will return a less simplified result than desired; gcd_terms might be used in such cases:
>>> from sympy import gcd_terms >>> (4*x*(y + 1) + y).extract_additively(x) 4*x*(y + 1) + x*(4*y + 3)  x*(4*y + 4) + y >>> gcd_terms(_) x*(4*y + 3) + y
See also

extract_branch_factor
(allow_half=False)¶ Try to write self as
exp_polar(2*pi*I*n)*z
in a nice way. Return (z, n).>>> from sympy import exp_polar, I, pi >>> from sympy.abc import x, y >>> exp_polar(I*pi).extract_branch_factor() (exp_polar(I*pi), 0) >>> exp_polar(2*I*pi).extract_branch_factor() (1, 1) >>> exp_polar(pi*I).extract_branch_factor() (exp_polar(I*pi), 1) >>> exp_polar(3*pi*I + x).extract_branch_factor() (exp_polar(x + I*pi), 1) >>> (y*exp_polar(5*pi*I)*exp_polar(3*pi*I + 2*pi*x)).extract_branch_factor() (y*exp_polar(2*pi*x), 1) >>> exp_polar(I*pi/2).extract_branch_factor() (exp_polar(I*pi/2), 0)
If allow_half is True, also extract exp_polar(I*pi):
>>> exp_polar(I*pi).extract_branch_factor(allow_half=True) (1, 1/2) >>> exp_polar(2*I*pi).extract_branch_factor(allow_half=True) (1, 1) >>> exp_polar(3*I*pi).extract_branch_factor(allow_half=True) (1, 3/2) >>> exp_polar(I*pi).extract_branch_factor(allow_half=True) (1, 1/2)

extract_multiplicatively
(c)¶ Return None if it’s not possible to make self in the form c * something in a nice way, i.e. preserving the properties of arguments of self.
Examples
>>> from sympy import symbols, Rational
>>> x, y = symbols('x,y', real=True)
>>> ((x*y)**3).extract_multiplicatively(x**2 * y) x*y**2
>>> ((x*y)**3).extract_multiplicatively(x**4 * y)
>>> (2*x).extract_multiplicatively(2) x
>>> (2*x).extract_multiplicatively(3)
>>> (Rational(1, 2)*x).extract_multiplicatively(3) x/6

factor
(*gens, **args)¶ See the factor() function in sympy.polys.polytools

find
(query, group=False)¶ Find all subexpressions matching a query.

fourier_series
(limits=None)¶ Compute fourier sine/cosine series of self.
See the docstring of the
fourier_series()
in sympy.series.fourier for more information.

fps
(x=None, x0=0, dir=1, hyper=True, order=4, rational=True, full=False)¶ Compute formal power power series of self.
See the docstring of the
fps()
function in sympy.series.formal for more information.

property
free_symbols
¶ Return from the atoms of self those which are free symbols.
For most expressions, all symbols are free symbols. For some classes this is not true. e.g. Integrals use Symbols for the dummy variables which are bound variables, so Integral has a method to return all symbols except those. Derivative keeps track of symbols with respect to which it will perform a derivative; those are bound variables, too, so it has its own free_symbols method.
Any other method that uses bound variables should implement a free_symbols method.

classmethod
fromiter
(args, **assumptions)¶ Create a new object from an iterable.
This is a convenience function that allows one to create objects from any iterable, without having to convert to a list or tuple first.
Examples
>>> from sympy import Tuple >>> Tuple.fromiter(i for i in range(5)) (0, 1, 2, 3, 4)

property
func
¶ The toplevel function in an expression.
The following should hold for all objects:
>> x == x.func(*x.args)
Examples
>>> from sympy.abc import x >>> a = 2*x >>> a.func <class 'sympy.core.mul.Mul'> >>> a.args (2, x) >>> a.func(*a.args) 2*x >>> a == a.func(*a.args) True

gammasimp
()¶ See the gammasimp function in sympy.simplify

getO
()¶ Returns the additive O(..) symbol if there is one, else None.

getn
()¶ Returns the order of the expression.
The order is determined either from the O(…) term. If there is no O(…) term, it returns None.
Examples
>>> from sympy import O >>> from sympy.abc import x >>> (1 + x + O(x**2)).getn() 2 >>> (1 + x).getn()

has
(*patterns)¶ Test whether any subexpression matches any of the patterns.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y, z >>> (x**2 + sin(x*y)).has(z) False >>> (x**2 + sin(x*y)).has(x, y, z) True >>> x.has(x) True
Note
has
is a structural algorithm with no knowledge of mathematics. Consider the following halfopen interval:>>> from sympy.sets import Interval >>> i = Interval.Lopen(0, 5); i Interval.Lopen(0, 5) >>> i.args (0, 5, True, False) >>> i.has(4) # there is no "4" in the arguments False >>> i.has(0) # there *is* a "0" in the arguments True
Instead, use
contains
to determine whether a number is in the interval or not:>>> i.contains(4) True >>> i.contains(0) False
Note that
expr.has(*patterns)
is exactly equivalent toany(expr.has(p) for p in patterns)
. In particular,False
is returned when the list of patterns is empty.>>> x.has() False

integrate
(*args, **kwargs)¶ See the integrate function in sympy.integrals

invert
(g, *gens, **args)¶ Return the multiplicative inverse of
self
modg
whereself
(andg
) may be symbolic expressions).See also
sympy.core.numbers.mod_inverse()
,sympy.polys.polytools.invert()

is_algebraic_expr
(*syms)¶ This tests whether a given expression is algebraic or not, in the given symbols, syms. When syms is not given, all free symbols will be used. The rational function does not have to be in expanded or in any kind of canonical form.
This function returns False for expressions that are “algebraic expressions” with symbolic exponents. This is a simple extension to the is_rational_function, including rational exponentiation.
Examples
>>> from sympy import Symbol, sqrt >>> x = Symbol('x', real=True) >>> sqrt(1 + x).is_rational_function() False >>> sqrt(1 + x).is_algebraic_expr() True
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be an algebraic expression to become one.
>>> from sympy import exp, factor >>> a = sqrt(exp(x)**2 + 2*exp(x) + 1)/(exp(x) + 1) >>> a.is_algebraic_expr(x) False >>> factor(a).is_algebraic_expr() True
See also
References

property
is_comparable
¶ Return True if self can be computed to a real number (or already is a real number) with precision, else False.
Examples
>>> from sympy import exp_polar, pi, I >>> (I*exp_polar(I*pi/2)).is_comparable True >>> (I*exp_polar(I*pi*2)).is_comparable False
A False result does not mean that self cannot be rewritten into a form that would be comparable. For example, the difference computed below is zero but without simplification it does not evaluate to a zero with precision:
>>> e = 2**pi*(1 + 2**pi) >>> dif = e  e.expand() >>> dif.is_comparable False >>> dif.n(2)._prec 1

is_constant
(*wrt, **flags)¶ Return True if self is constant, False if not, or None if the constancy could not be determined conclusively.
If an expression has no free symbols then it is a constant. If there are free symbols it is possible that the expression is a constant, perhaps (but not necessarily) zero. To test such expressions, a few strategies are tried:
1) numerical evaluation at two random points. If two such evaluations give two different values and the values have a precision greater than 1 then self is not constant. If the evaluations agree or could not be obtained with any precision, no decision is made. The numerical testing is done only if
wrt
is different than the free symbols.2) differentiation with respect to variables in ‘wrt’ (or all free symbols if omitted) to see if the expression is constant or not. This will not always lead to an expression that is zero even though an expression is constant (see added test in test_expr.py). If all derivatives are zero then self is constant with respect to the given symbols.
3) finding out zeros of denominator expression with free_symbols. It won’t be constant if there are zeros. It gives more negative answers for expression that are not constant.
If neither evaluation nor differentiation can prove the expression is constant, None is returned unless two numerical values happened to be the same and the flag
failing_number
is True – in that case the numerical value will be returned.If flag simplify=False is passed, self will not be simplified; the default is True since self should be simplified before testing.
Examples
>>> from sympy import cos, sin, Sum, S, pi >>> from sympy.abc import a, n, x, y >>> x.is_constant() False >>> S(2).is_constant() True >>> Sum(x, (x, 1, 10)).is_constant() True >>> Sum(x, (x, 1, n)).is_constant() False >>> Sum(x, (x, 1, n)).is_constant(y) True >>> Sum(x, (x, 1, n)).is_constant(n) False >>> Sum(x, (x, 1, n)).is_constant(x) True >>> eq = a*cos(x)**2 + a*sin(x)**2  a >>> eq.is_constant() True >>> eq.subs({x: pi, a: 2}) == eq.subs({x: pi, a: 3}) == 0 True
>>> (0**x).is_constant() False >>> x.is_constant() False >>> (x**x).is_constant() False >>> one = cos(x)**2 + sin(x)**2 >>> one.is_constant() True >>> ((one  1)**(x + 1)).is_constant() in (True, False) # could be 0 or 1 True

is_polynomial
(*syms)¶ Return True if self is a polynomial in syms and False otherwise.
This checks if self is an exact polynomial in syms. This function returns False for expressions that are “polynomials” with symbolic exponents. Thus, you should be able to apply polynomial algorithms to expressions for which this returns True, and Poly(expr, *syms) should work if and only if expr.is_polynomial(*syms) returns True. The polynomial does not have to be in expanded form. If no symbols are given, all free symbols in the expression will be used.
This is not part of the assumptions system. You cannot do Symbol(‘z’, polynomial=True).
Examples
>>> from sympy import Symbol >>> x = Symbol('x') >>> ((x**2 + 1)**4).is_polynomial(x) True >>> ((x**2 + 1)**4).is_polynomial() True >>> (2**x + 1).is_polynomial(x) False
>>> n = Symbol('n', nonnegative=True, integer=True) >>> (x**n + 1).is_polynomial(x) False
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be a polynomial to become one.
>>> from sympy import sqrt, factor, cancel >>> y = Symbol('y', positive=True) >>> a = sqrt(y**2 + 2*y + 1) >>> a.is_polynomial(y) False >>> factor(a) y + 1 >>> factor(a).is_polynomial(y) True
>>> b = (y**2 + 2*y + 1)/(y + 1) >>> b.is_polynomial(y) False >>> cancel(b) y + 1 >>> cancel(b).is_polynomial(y) True
See also .is_rational_function()

is_rational_function
(*syms)¶ Test whether function is a ratio of two polynomials in the given symbols, syms. When syms is not given, all free symbols will be used. The rational function does not have to be in expanded or in any kind of canonical form.
This function returns False for expressions that are “rational functions” with symbolic exponents. Thus, you should be able to call .as_numer_denom() and apply polynomial algorithms to the result for expressions for which this returns True.
This is not part of the assumptions system. You cannot do Symbol(‘z’, rational_function=True).
Examples
>>> from sympy import Symbol, sin >>> from sympy.abc import x, y
>>> (x/y).is_rational_function() True
>>> (x**2).is_rational_function() True
>>> (x/sin(y)).is_rational_function(y) False
>>> n = Symbol('n', integer=True) >>> (x**n + 1).is_rational_function(x) False
This function does not attempt any nontrivial simplifications that may result in an expression that does not appear to be a rational function to become one.
>>> from sympy import sqrt, factor >>> y = Symbol('y', positive=True) >>> a = sqrt(y**2 + 2*y + 1)/y >>> a.is_rational_function(y) False >>> factor(a) (y + 1)/y >>> factor(a).is_rational_function(y) True
See also is_algebraic_expr().

leadterm
(x)¶ Returns the leading term a*x**b as a tuple (a, b).
Examples
>>> from sympy.abc import x >>> (1+x+x**2).leadterm(x) (1, 0) >>> (1/x**2+x+x**2).leadterm(x) (1, 2)

limit
(x, xlim, dir='+')¶ Compute limit x>xlim.

lseries
(x=None, x0=0, dir='+', logx=None)¶ Wrapper for series yielding an iterator of the terms of the series.
Note: an infinite series will yield an infinite iterator. The following, for exaxmple, will never terminate. It will just keep printing terms of the sin(x) series:
for term in sin(x).lseries(x): print term
The advantage of lseries() over nseries() is that many times you are just interested in the next term in the series (i.e. the first term for example), but you don’t know how many you should ask for in nseries() using the “n” parameter.
See also nseries().

match
(pattern, old=False)¶ Pattern matching.
Wild symbols match all.
Return
None
when expression (self) does not match with pattern. Otherwise return a dictionary such that:pattern.xreplace(self.match(pattern)) == self
Examples
>>> from sympy import Wild >>> from sympy.abc import x, y >>> p = Wild("p") >>> q = Wild("q") >>> r = Wild("r") >>> e = (x+y)**(x+y) >>> e.match(p**p) {p_: x + y} >>> e.match(p**q) {p_: x + y, q_: x + y} >>> e = (2*x)**2 >>> e.match(p*q**r) {p_: 4, q_: x, r_: 2} >>> (p*q**r).xreplace(e.match(p*q**r)) 4*x**2
The
old
flag will give the oldstyle pattern matching where expressions and patterns are essentially solved to give the match. Both of the following give None unlessold=True
:>>> (x  2).match(p  x, old=True) {p_: 2*x  2} >>> (2/x).match(p*x, old=True) {p_: 2/x**2}

matches
(expr, repl_dict={}, old=False)¶ Helper method for match() that looks for a match between Wild symbols in self and expressions in expr.
Examples
>>> from sympy import symbols, Wild, Basic >>> a, b, c = symbols('a b c') >>> x = Wild('x') >>> Basic(a + x, x).matches(Basic(a + b, c)) is None True >>> Basic(a + x, x).matches(Basic(a + b + c, b + c)) {x_: b + c}

n
(n=15, subs=None, maxn=100, chop=False, strict=False, quad=None, verbose=False)¶ Evaluate the given formula to an accuracy of n digits.
 Parameters
subs (dict, optional) – Substitute numerical values for symbols, e.g.
subs={x:3, y:1+pi}
. The substitutions must be given as a dictionary.maxn (int, optional) – Allow a maximum temporary working precision of maxn digits.
chop (bool or number, optional) –
Specifies how to replace tiny real or imaginary parts in subresults by exact zeros.
When
True
the chop value defaults to standard precision.Otherwise the chop value is used to determine the magnitude of “small” for purposes of chopping.
>>> from sympy import N >>> x = 1e4 >>> N(x, chop=True) 0.000100000000000000 >>> N(x, chop=1e5) 0.000100000000000000 >>> N(x, chop=1e4) 0
strict (bool, optional) – Raise
PrecisionExhausted
if any subresult fails to evaluate to full accuracy, given the available maxprec.quad (str, optional) – Choose algorithm for numerical quadrature. By default, tanhsinh quadrature is used. For oscillatory integrals on an infinite interval, try
quad='osc'
.verbose (bool, optional) – Print debug information.
Notes
When Floats are naively substituted into an expression, precision errors may adversely affect the result. For example, adding 1e16 (a Float) to 1 will truncate to 1e16; if 1e16 is then subtracted, the result will be 0. That is exactly what happens in the following:
>>> from sympy.abc import x, y, z >>> values = {x: 1e16, y: 1, z: 1e16} >>> (x + y  z).subs(values) 0
Using the subs argument for evalf is the accurate way to evaluate such an expression:
>>> (x + y  z).evalf(subs=values) 1.00000000000000

nseries
(x=None, x0=0, n=6, dir='+', logx=None)¶ Wrapper to _eval_nseries if assumptions allow, else to series.
If x is given, x0 is 0, dir=’+’, and self has x, then _eval_nseries is called. This calculates “n” terms in the innermost expressions and then builds up the final series just by “crossmultiplying” everything out.
The optional
logx
parameter can be used to replace any log(x) in the returned series with a symbolic value to avoid evaluating log(x) at 0. A symbol to use in place of log(x) should be provided.Advantage – it’s fast, because we don’t have to determine how many terms we need to calculate in advance.
Disadvantage – you may end up with less terms than you may have expected, but the O(x**n) term appended will always be correct and so the result, though perhaps shorter, will also be correct.
If any of those assumptions is not met, this is treated like a wrapper to series which will try harder to return the correct number of terms.
See also lseries().
Examples
>>> from sympy import sin, log, Symbol >>> from sympy.abc import x, y >>> sin(x).nseries(x, 0, 6) x  x**3/6 + x**5/120 + O(x**6) >>> log(x+1).nseries(x, 0, 5) x  x**2/2 + x**3/3  x**4/4 + O(x**5)
Handling of the
logx
parameter — in the following example the expansion fails sincesin
does not have an asymptotic expansion at oo (the limit of log(x) as x approaches 0):>>> e = sin(log(x)) >>> e.nseries(x, 0, 6) Traceback (most recent call last): ... PoleError: ... ... >>> logx = Symbol('logx') >>> e.nseries(x, 0, 6, logx=logx) sin(logx)
In the following example, the expansion works but gives only an Order term unless the
logx
parameter is used:>>> e = x**y >>> e.nseries(x, 0, 2) O(log(x)**2) >>> e.nseries(x, 0, 2, logx=logx) exp(logx*y)

nsimplify
(constants=[], tolerance=None, full=False)¶ See the nsimplify function in sympy.simplify

powsimp
(*args, **kwargs)¶ See the powsimp function in sympy.simplify

primitive
()¶ Return the positive Rational that can be extracted nonrecursively from every term of self (i.e., self is treated like an Add). This is like the as_coeff_Mul() method but primitive always extracts a positive Rational (never a negative or a Float).
Examples
>>> from sympy.abc import x >>> (3*(x + 1)**2).primitive() (3, (x + 1)**2) >>> a = (6*x + 2); a.primitive() (2, 3*x + 1) >>> b = (x/2 + 3); b.primitive() (1/2, x + 6) >>> (a*b).primitive() == (1, a*b) True

radsimp
(**kwargs)¶ See the radsimp function in sympy.simplify

ratsimp
()¶ See the ratsimp function in sympy.simplify

rcall
(*args)¶ Apply on the argument recursively through the expression tree.
This method is used to simulate a common abuse of notation for operators. For instance in SymPy the the following will not work:
(x+Lambda(y, 2*y))(z) == x+2*z
,however you can use
>>> from sympy import Lambda >>> from sympy.abc import x, y, z >>> (x + Lambda(y, 2*y)).rcall(z) x + 2*z

refine
(assumption=True)¶ See the refine function in sympy.assumptions

removeO
()¶ Removes the additive O(..) symbol if there is one

replace
(query, value, map=False, simultaneous=True, exact=None)¶ Replace matching subexpressions of
self
withvalue
.If
map = True
then also return the mapping {old: new} whereold
was a subexpression found with query andnew
is the replacement value for it. If the expression itself doesn’t match the query, then the returned value will beself.xreplace(map)
otherwise it should beself.subs(ordered(map.items()))
.Traverses an expression tree and performs replacement of matching subexpressions from the bottom to the top of the tree. The default approach is to do the replacement in a simultaneous fashion so changes made are targeted only once. If this is not desired or causes problems,
simultaneous
can be set to False.In addition, if an expression containing more than one Wild symbol is being used to match subexpressions and the
exact
flag is None it will be set to True so the match will only succeed if all nonzero values are received for each Wild that appears in the match pattern. Setting this to False accepts a match of 0; while setting it True accepts all matches that have a 0 in them. See example below for cautions.The list of possible combinations of queries and replacement values is listed below:
Examples
Initial setup
>>> from sympy import log, sin, cos, tan, Wild, Mul, Add >>> from sympy.abc import x, y >>> f = log(sin(x)) + tan(sin(x**2))
 1.1. type > type
obj.replace(type, newtype)
When object of type
type
is found, replace it with the result of passing its argument(s) tonewtype
.>>> f.replace(sin, cos) log(cos(x)) + tan(cos(x**2)) >>> sin(x).replace(sin, cos, map=True) (cos(x), {sin(x): cos(x)}) >>> (x*y).replace(Mul, Add) x + y
 1.2. type > func
obj.replace(type, func)
When object of type
type
is found, applyfunc
to its argument(s).func
must be written to handle the number of arguments oftype
.>>> f.replace(sin, lambda arg: sin(2*arg)) log(sin(2*x)) + tan(sin(2*x**2)) >>> (x*y).replace(Mul, lambda *args: sin(2*Mul(*args))) sin(2*x*y)
 2.1. pattern > expr
obj.replace(pattern(wild), expr(wild))
Replace subexpressions matching
pattern
with the expression written in terms of the Wild symbols inpattern
.>>> a, b = map(Wild, 'ab') >>> f.replace(sin(a), tan(a)) log(tan(x)) + tan(tan(x**2)) >>> f.replace(sin(a), tan(a/2)) log(tan(x/2)) + tan(tan(x**2/2)) >>> f.replace(sin(a), a) log(x) + tan(x**2) >>> (x*y).replace(a*x, a) y
Matching is exact by default when more than one Wild symbol is used: matching fails unless the match gives nonzero values for all Wild symbols:
>>> (2*x + y).replace(a*x + b, b  a) y  2 >>> (2*x).replace(a*x + b, b  a) 2*x
When set to False, the results may be nonintuitive:
>>> (2*x).replace(a*x + b, b  a, exact=False) 2/x
 2.2. pattern > func
obj.replace(pattern(wild), lambda wild: expr(wild))
All behavior is the same as in 2.1 but now a function in terms of pattern variables is used rather than an expression:
>>> f.replace(sin(a), lambda a: sin(2*a)) log(sin(2*x)) + tan(sin(2*x**2))
 3.1. func > func
obj.replace(filter, func)
Replace subexpression
e
withfunc(e)
iffilter(e)
is True.>>> g = 2*sin(x**3) >>> g.replace(lambda expr: expr.is_Number, lambda expr: expr**2) 4*sin(x**9)
The expression itself is also targeted by the query but is done in such a fashion that changes are not made twice.
>>> e = x*(x*y + 1) >>> e.replace(lambda x: x.is_Mul, lambda x: 2*x) 2*x*(2*x*y + 1)
When matching a single symbol, exact will default to True, but this may or may not be the behavior that is desired:
Here, we want exact=False:
>>> from sympy import Function >>> f = Function('f') >>> e = f(1) + f(0) >>> q = f(a), lambda a: f(a + 1) >>> e.replace(*q, exact=False) f(1) + f(2) >>> e.replace(*q, exact=True) f(0) + f(2)
But here, the nature of matching makes selecting the right setting tricky:
>>> e = x**(1 + y) >>> (x**(1 + y)).replace(x**(1 + a), lambda a: x**a, exact=False) 1 >>> (x**(1 + y)).replace(x**(1 + a), lambda a: x**a, exact=True) x**(x  y + 1) >>> (x**y).replace(x**(1 + a), lambda a: x**a, exact=False) 1 >>> (x**y).replace(x**(1 + a), lambda a: x**a, exact=True) x**(1  y)
It is probably better to use a different form of the query that describes the target expression more precisely:
>>> (1 + x**(1 + y)).replace( ... lambda x: x.is_Pow and x.exp.is_Add and x.exp.args[0] == 1, ... lambda x: x.base**(1  (x.exp  1))) ... x**(1  y) + 1
See also
subs()
substitution of subexpressions as defined by the objects themselves.
xreplace()
exact node replacement in expr tree; also capable of using matching rules

rewrite
(*args, **hints)¶ Rewrite functions in terms of other functions.
Rewrites expression containing applications of functions of one kind in terms of functions of different kind. For example you can rewrite trigonometric functions as complex exponentials or combinatorial functions as gamma function.
As a pattern this function accepts a list of functions to to rewrite (instances of DefinedFunction class). As rule you can use string or a destination function instance (in this case rewrite() will use the str() function).
There is also the possibility to pass hints on how to rewrite the given expressions. For now there is only one such hint defined called ‘deep’. When ‘deep’ is set to False it will forbid functions to rewrite their contents.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x
Unspecified pattern:
>>> sin(x).rewrite(exp) I*(exp(I*x)  exp(I*x))/2
Pattern as a single function:
>>> sin(x).rewrite(sin, exp) I*(exp(I*x)  exp(I*x))/2
Pattern as a list of functions:
>>> sin(x).rewrite([sin, ], exp) I*(exp(I*x)  exp(I*x))/2

round
(n=None)¶ Return x rounded to the given decimal place.
If a complex number would results, apply round to the real and imaginary components of the number.
Examples
>>> from sympy import pi, E, I, S, Add, Mul, Number >>> pi.round() 3 >>> pi.round(2) 3.14 >>> (2*pi + E*I).round() 6 + 3*I
The round method has a chopping effect:
>>> (2*pi + I/10).round() 6 >>> (pi/10 + 2*I).round() 2*I >>> (pi/10 + E*I).round(2) 0.31 + 2.72*I
Notes
The Python
round
function uses the SymPyround
method so it will always return a SymPy number (not a Python float or int):>>> isinstance(round(S(123), 2), Number) True

separate
(deep=False, force=False)¶ See the separate function in sympy.simplify

series
(x=None, x0=0, n=6, dir='+', logx=None)¶ Series expansion of “self” around
x = x0
yielding either terms of the series one by one (the lazy series given when n=None), else all the terms at once when n != None.Returns the series expansion of “self” around the point
x = x0
with respect tox
up toO((x  x0)**n, x, x0)
(default n is 6).If
x=None
andself
is univariate, the univariate symbol will be supplied, otherwise an error will be raised. Parameters
expr (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
x0 (Value) – The value around which
x
is calculated. Can be any value fromoo
tooo
.n (Value) – The number of terms upto which the series is to be expanded.
dir (String, optional) – The seriesexpansion can be bidirectional. If
dir="+"
, then (x>x0+). Ifdir="", then (x>x0). For infinite ``x0
(oo
oroo
), thedir
argument is determined from the direction of the infinity (i.e.,dir=""
foroo
).logx (optional) – It is used to replace any log(x) in the returned series with a symbolic value rather than evaluating the actual value.
Examples
>>> from sympy import cos, exp, tan, oo, series >>> from sympy.abc import x, y >>> cos(x).series() 1  x**2/2 + x**4/24 + O(x**6) >>> cos(x).series(n=4) 1  x**2/2 + O(x**4) >>> cos(x).series(x, x0=1, n=2) cos(1)  (x  1)*sin(1) + O((x  1)**2, (x, 1)) >>> e = cos(x + exp(y)) >>> e.series(y, n=2) cos(x + 1)  y*sin(x + 1) + O(y**2) >>> e.series(x, n=2) cos(exp(y))  x*sin(exp(y)) + O(x**2)
If
n=None
then a generator of the series terms will be returned.>>> term=cos(x).series(n=None) >>> [next(term) for i in range(2)] [1, x**2/2]
For
dir=+
(default) the series is calculated from the right and fordir=
the series from the left. For smooth functions this flag will not alter the results.>>> abs(x).series(dir="+") x >>> abs(x).series(dir="") x >>> f = tan(x) >>> f.series(x, 2, 6, "+") tan(2) + (1 + tan(2)**2)*(x  2) + (x  2)**2*(tan(2)**3 + tan(2)) + (x  2)**3*(1/3 + 4*tan(2)**2/3 + tan(2)**4) + (x  2)**4*(tan(2)**5 + 5*tan(2)**3/3 + 2*tan(2)/3) + (x  2)**5*(2/15 + 17*tan(2)**2/15 + 2*tan(2)**4 + tan(2)**6) + O((x  2)**6, (x, 2))
>>> f.series(x, 2, 3, "") tan(2) + (2  x)*(tan(2)**2  1) + (2  x)**2*(tan(2)**3 + tan(2)) + O((x  2)**3, (x, 2))
 Returns
Expr – Series expansion of the expression about x0
 Return type
Expression
 Raises
TypeError – If “n” and “x0” are infinity objects
PoleError – If “x0” is an infinity object

simplify
(**kwargs)¶ See the simplify function in sympy.simplify

sort_key
(order=None)¶ Return a sort key.
Examples
>>> from sympy.core import S, I
>>> sorted([S(1)/2, I, I], key=lambda x: x.sort_key()) [1/2, I, I]
>>> S("[x, 1/x, 1/x**2, x**2, x**(1/2), x**(1/4), x**(3/2)]") [x, 1/x, x**(2), x**2, sqrt(x), x**(1/4), x**(3/2)] >>> sorted(_, key=lambda x: x.sort_key()) [x**(2), 1/x, x**(1/4), sqrt(x), x, x**(3/2), x**2]

subs
(*args, **kwargs)¶ Substitutes old for new in an expression after sympifying args.
 args is either:
two arguments, e.g. foo.subs(old, new)
 one iterable argument, e.g. foo.subs(iterable). The iterable may be
 o an iterable container with (old, new) pairs. In this case the
replacements are processed in the order given with successive patterns possibly affecting replacements already made.
 o a dict or set whose key/value items correspond to old/new pairs.
In this case the old/new pairs will be sorted by op count and in case of a tie, by number of args and the default_sort_key. The resulting sorted list is then processed as an iterable container (see previous).
If the keyword
simultaneous
is True, the subexpressions will not be evaluated until all the substitutions have been made.Examples
>>> from sympy import pi, exp, limit, oo >>> from sympy.abc import x, y >>> (1 + x*y).subs(x, pi) pi*y + 1 >>> (1 + x*y).subs({x:pi, y:2}) 1 + 2*pi >>> (1 + x*y).subs([(x, pi), (y, 2)]) 1 + 2*pi >>> reps = [(y, x**2), (x, 2)] >>> (x + y).subs(reps) 6 >>> (x + y).subs(reversed(reps)) x**2 + 2
>>> (x**2 + x**4).subs(x**2, y) y**2 + y
To replace only the x**2 but not the x**4, use xreplace:
>>> (x**2 + x**4).xreplace({x**2: y}) x**4 + y
To delay evaluation until all substitutions have been made, set the keyword
simultaneous
to True:>>> (x/y).subs([(x, 0), (y, 0)]) 0 >>> (x/y).subs([(x, 0), (y, 0)], simultaneous=True) nan
This has the added feature of not allowing subsequent substitutions to affect those already made:
>>> ((x + y)/y).subs({x + y: y, y: x + y}) 1 >>> ((x + y)/y).subs({x + y: y, y: x + y}, simultaneous=True) y/(x + y)
In order to obtain a canonical result, unordered iterables are sorted by count_op length, number of arguments and by the default_sort_key to break any ties. All other iterables are left unsorted.
>>> from sympy import sqrt, sin, cos >>> from sympy.abc import a, b, c, d, e
>>> A = (sqrt(sin(2*x)), a) >>> B = (sin(2*x), b) >>> C = (cos(2*x), c) >>> D = (x, d) >>> E = (exp(x), e)
>>> expr = sqrt(sin(2*x))*sin(exp(x)*x)*cos(2*x) + sin(2*x)
>>> expr.subs(dict([A, B, C, D, E])) a*c*sin(d*e) + b
The resulting expression represents a literal replacement of the old arguments with the new arguments. This may not reflect the limiting behavior of the expression:
>>> (x**3  3*x).subs({x: oo}) nan
>>> limit(x**3  3*x, x, oo) oo
If the substitution will be followed by numerical evaluation, it is better to pass the substitution to evalf as
>>> (1/x).evalf(subs={x: 3.0}, n=21) 0.333333333333333333333
rather than
>>> (1/x).subs({x: 3.0}).evalf(21) 0.333333333333333314830
as the former will ensure that the desired level of precision is obtained.
See also
replace()
replacement capable of doing wildcardlike matching, parsing of match, and conditional replacements
xreplace()
exact node replacement in expr tree; also capable of using matching rules
sympy.core.evalf.EvalfMixin.evalf()
calculates the given formula to a desired level of precision

taylor_term
(n, x, *previous_terms)¶ General method for the taylor term.
This method is slow, because it differentiates ntimes. Subclasses can redefine it to make it faster by using the “previous_terms”.

together
(*args, **kwargs)¶ See the together function in sympy.polys

trigsimp
(**args)¶ See the trigsimp function in sympy.simplify

xreplace
(rule, hack2=False)¶ Replace occurrences of objects within the expression.
 Parameters
rule (dictlike) – Expresses a replacement rule
 Returns
xreplace
 Return type
the result of the replacement
Examples
>>> from sympy import symbols, pi, exp >>> x, y, z = symbols('x y z') >>> (1 + x*y).xreplace({x: pi}) pi*y + 1 >>> (1 + x*y).xreplace({x: pi, y: 2}) 1 + 2*pi
Replacements occur only if an entire node in the expression tree is matched:
>>> (x*y + z).xreplace({x*y: pi}) z + pi >>> (x*y*z).xreplace({x*y: pi}) x*y*z >>> (2*x).xreplace({2*x: y, x: z}) y >>> (2*2*x).xreplace({2*x: y, x: z}) 4*z >>> (x + y + 2).xreplace({x + y: 2}) x + y + 2 >>> (x + 2 + exp(x + 2)).xreplace({x + 2: y}) x + exp(y) + 2
xreplace doesn’t differentiate between free and bound symbols. In the following, subs(x, y) would not change x since it is a bound symbol, but xreplace does:
>>> from sympy import Integral >>> Integral(x, (x, 1, 2*x)).xreplace({x: y}) Integral(y, (y, 1, 2*y))
Trying to replace x with an expression raises an error:
>>> Integral(x, (x, 1, 2*x)).xreplace({x: 2*y}) ValueError: Invalid limits given: ((2*y, 1, 4*y),)


class
galgebra.atoms.
BasisBladeNoWedgeSymbol
[source]¶ A basis blade with shortened rendering such as \(e_{12}\)

apart
(x=None, **args)¶ See the apart function in sympy.polys

property
args
¶ Returns a tuple of arguments of ‘self’.
Examples
>>> from sympy import cot >>> from sympy.abc import x, y
>>> cot(x).args (x,)
>>> cot(x).args[0] x
>>> (x*y).args (x, y)
>>> (x*y).args[1] y
Notes
Never use self._args, always use self.args. Only use _args in __new__ when creating a new function. Don’t override .args() from Basic (so that it’s easy to change the interface in the future if needed).

args_cnc
(cset=False, warn=True, split_1=True)¶ Return [commutative factors, noncommutative factors] of self.
self is treated as a Mul and the ordering of the factors is maintained. If
cset
is True the commutative factors will be returned in a set. If there were repeated factors (as may happen with an unevaluated Mul) then an error will be raised unless it is explicitly suppressed by settingwarn
to False.Note: 1 is always separated from a Number unless split_1 is False.
>>> from sympy import symbols, oo >>> A, B = symbols('A B', commutative=0) >>> x, y = symbols('x y') >>> (2*x*y).args_cnc() [[1, 2, x, y], []] >>> (2.5*x).args_cnc() [[1, 2.5, x], []] >>> (2*x*A*B*y).args_cnc() [[1, 2, x, y], [A, B]] >>> (2*x*A*B*y).args_cnc(split_1=False) [[2, x, y], [A, B]] >>> (2*x*y).args_cnc(cset=True) [{1, 2, x, y}, []]
The arg is always treated as a Mul:
>>> (2 + x + A).args_cnc() [[], [x  2 + A]] >>> (oo).args_cnc() # oo is a singleton [[1, oo], []]

as_coeff_Add
(rational=False)¶ Efficiently extract the coefficient of a summation.

as_coeff_Mul
(rational=False)¶ Efficiently extract the coefficient of a product.

as_coeff_add
(*deps)¶ Return the tuple (c, args) where self is written as an Add,
a
.c should be a Rational added to any terms of the Add that are independent of deps.
args should be a tuple of all other terms of
a
; args is empty if self is a Number or if self is independent of deps (when given).This should be used when you don’t know if self is an Add or not but you want to treat self as an Add or if you want to process the individual arguments of the tail of self as an Add.
if you know self is an Add and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail.
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_add() (3, ()) >>> (3 + x).as_coeff_add() (3, (x,)) >>> (3 + x + y).as_coeff_add(x) (y + 3, (x,)) >>> (3 + y).as_coeff_add(x) (y + 3, ())

as_coeff_exponent
(x)¶ c*x**e > c,e
where x can be any symbolic expression.

as_coeff_mul
(*deps, **kwargs)¶ Return the tuple (c, args) where self is written as a Mul,
m
.c should be a Rational multiplied by any factors of the Mul that are independent of deps.
args should be a tuple of all other factors of m; args is empty if self is a Number or if self is independent of deps (when given).
This should be used when you don’t know if self is a Mul or not but you want to treat self as a Mul or if you want to process the individual arguments of the tail of self as a Mul.
if you know self is a Mul and want only the head, use self.args[0];
if you don’t want to process the arguments of the tail but need the tail then use self.as_two_terms() which gives the head and tail;
if you want to split self into an independent and dependent parts use
self.as_independent(*deps)
>>> from sympy import S >>> from sympy.abc import x, y >>> (S(3)).as_coeff_mul() (3, ()) >>> (3*x*y).as_coeff_mul() (3, (x, y)) >>> (3*x*y).as_coeff_mul(x) (3*y, (x,)) >>> (3*y).as_coeff_mul(x) (3*y, ())

as_coefficient
(expr)¶ Extracts symbolic coefficient at the given expression. In other words, this functions separates ‘self’ into the product of ‘expr’ and ‘expr’free coefficient. If such separation is not possible it will return None.
Examples
>>> from sympy import E, pi, sin, I, Poly >>> from sympy.abc import x
>>> E.as_coefficient(E) 1 >>> (2*E).as_coefficient(E) 2 >>> (2*sin(E)*E).as_coefficient(E)
Two terms have E in them so a sum is returned. (If one were desiring the coefficient of the term exactly matching E then the constant from the returned expression could be selected. Or, for greater precision, a method of Poly can be used to indicate the desired term from which the coefficient is desired.)
>>> (2*E + x*E).as_coefficient(E) x + 2 >>> _.args[0] # just want the exact match 2 >>> p = Poly(2*E + x*E); p Poly(x*E + 2*E, x, E, domain='ZZ') >>> p.coeff_monomial(E) 2 >>> p.nth(0, 1) 2
Since the following cannot be written as a product containing E as a factor, None is returned. (If the coefficient
2*x
is desired then thecoeff
method should be used.)>>> (2*E*x + x).as_coefficient(E) >>> (2*E*x + x).coeff(E) 2*x
>>> (E*(x + 1) + x).as_coefficient(E)
>>> (2*pi*I).as_coefficient(pi*I) 2 >>> (2*I).as_coefficient(pi*I)
See also
coeff()
return sum of terms have a given factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used

as_coefficients_dict
()¶ Return a dictionary mapping terms to their Rational coefficient. Since the dictionary is a defaultdict, inquiries about terms which were not present will return a coefficient of 0. If an expression is not an Add it is considered to have a single term.
Examples
>>> from sympy.abc import a, x >>> (3*x + a*x + 4).as_coefficients_dict() {1: 4, x: 3, a*x: 1} >>> _[a] 0 >>> (3*a*x).as_coefficients_dict() {a*x: 3}

as_content_primitive
(radical=False, clear=True)¶ This method should recursively remove a Rational from all arguments and return that (content) and the new self (primitive). The content should always be positive and
Mul(*foo.as_content_primitive()) == foo
. The primitive need not be in canonical form and should try to preserve the underlying structure if possible (i.e. expand_mul should not be applied to self).Examples
>>> from sympy import sqrt >>> from sympy.abc import x, y, z
>>> eq = 2 + 2*x + 2*y*(3 + 3*y)
The as_content_primitive function is recursive and retains structure:
>>> eq.as_content_primitive() (2, x + 3*y*(y + 1) + 1)
Integer powers will have Rationals extracted from the base:
>>> ((2 + 6*x)**2).as_content_primitive() (4, (3*x + 1)**2) >>> ((2 + 6*x)**(2*y)).as_content_primitive() (1, (2*(3*x + 1))**(2*y))
Terms may end up joining once their as_content_primitives are added:
>>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (11, x*(y + 1)) >>> ((3*(x*(1 + y)) + 2*x*(3 + 3*y))).as_content_primitive() (9, x*(y + 1)) >>> ((3*(z*(1 + y)) + 2.0*x*(3 + 3*y))).as_content_primitive() (1, 6.0*x*(y + 1) + 3*z*(y + 1)) >>> ((5*(x*(1 + y)) + 2*x*(3 + 3*y))**2).as_content_primitive() (121, x**2*(y + 1)**2) >>> ((x*(1 + y) + 0.4*x*(3 + 3*y))**2).as_content_primitive() (1, 4.84*x**2*(y + 1)**2)
Radical content can also be factored out of the primitive:
>>> (2*sqrt(2) + 4*sqrt(10)).as_content_primitive(radical=True) (2, sqrt(2)*(1 + 2*sqrt(5)))
If clear=False (default is True) then content will not be removed from an Add if it can be distributed to leave one or more terms with integer coefficients.
>>> (x/2 + y).as_content_primitive() (1/2, x + 2*y) >>> (x/2 + y).as_content_primitive(clear=False) (1, x/2 + y)

as_dummy
()¶ Return the expression with any objects having structurally bound symbols replaced with unique, canonical symbols within the object in which they appear and having only the default assumption for commutativity being True.
Examples
>>> from sympy import Integral, Symbol >>> from sympy.abc import x, y >>> r = Symbol('r', real=True) >>> Integral(r, (r, x)).as_dummy() Integral(_0, (_0, x)) >>> _.variables[0].is_real is None True
Notes
Any object that has structural dummy variables should have a property, bound_symbols that returns a list of structural dummy symbols of the object itself.
Lambda and Subs have bound symbols, but because of how they are cached, they already compare the same regardless of their bound symbols:
>>> from sympy import Lambda >>> Lambda(x, x + 1) == Lambda(y, y + 1) True

as_expr
(*gens)¶ Convert a polynomial to a SymPy expression.
Examples
>>> from sympy import sin >>> from sympy.abc import x, y
>>> f = (x**2 + x*y).as_poly(x, y) >>> f.as_expr() x**2 + x*y
>>> sin(x).as_expr() sin(x)

as_independent
(*deps, **hint)¶ A mostly naive separation of a Mul or Add into arguments that are not are dependent on deps. To obtain as complete a separation of variables as possible, use a separation method first, e.g.:
separatevars() to change Mul, Add and Pow (including exp) into Mul
.expand(mul=True) to change Add or Mul into Add
.expand(log=True) to change log expr into an Add
The only nonnaive thing that is done here is to respect noncommutative ordering of variables and to always return (0, 0) for self of zero regardless of hints.
For nonzero self, the returned tuple (i, d) has the following interpretation:
i will has no variable that appears in deps
d will either have terms that contain variables that are in deps, or be equal to 0 (when self is an Add) or 1 (when self is a Mul)
if self is an Add then self = i + d
if self is a Mul then self = i*d
otherwise (self, S.One) or (S.One, self) is returned.
To force the expression to be treated as an Add, use the hint as_Add=True
Examples
– self is an Add
>>> from sympy import sin, cos, exp >>> from sympy.abc import x, y, z
>>> (x + x*y).as_independent(x) (0, x*y + x) >>> (x + x*y).as_independent(y) (x, x*y) >>> (2*x*sin(x) + y + x + z).as_independent(x) (y + z, 2*x*sin(x) + x) >>> (2*x*sin(x) + y + x + z).as_independent(x, y) (z, 2*x*sin(x) + x + y)
– self is a Mul
>>> (x*sin(x)*cos(y)).as_independent(x) (cos(y), x*sin(x))
noncommutative terms cannot always be separated out when self is a Mul
>>> from sympy import symbols >>> n1, n2, n3 = symbols('n1 n2 n3', commutative=False) >>> (n1 + n1*n2).as_independent(n2) (n1, n1*n2) >>> (n2*n1 + n1*n2).as_independent(n2) (0, n1*n2 + n2*n1) >>> (n1*n2*n3).as_independent(n1) (1, n1*n2*n3) >>> (n1*n2*n3).as_independent(n2) (n1, n2*n3) >>> ((xn1)*(xy)).as_independent(x) (1, (x  y)*(x  n1))
– self is anything else:
>>> (sin(x)).as_independent(x) (1, sin(x)) >>> (sin(x)).as_independent(y) (sin(x), 1) >>> exp(x+y).as_independent(x) (1, exp(x + y))
– force self to be treated as an Add:
>>> (3*x).as_independent(x, as_Add=True) (0, 3*x)
– force self to be treated as a Mul:
>>> (3+x).as_independent(x, as_Add=False) (1, x + 3) >>> (3+x).as_independent(x, as_Add=False) (1, x  3)
Note how the below differs from the above in making the constant on the dep term positive.
>>> (y*(3+x)).as_independent(x) (y, x  3)
 – use .as_independent() for true independence testing instead
of .has(). The former considers only symbols in the free symbols while the latter considers all symbols
>>> from sympy import Integral >>> I = Integral(x, (x, 1, 2)) >>> I.has(x) True >>> x in I.free_symbols False >>> I.as_independent(x) == (I, 1) True >>> (I + x).as_independent(x) == (I, x) True
Note: when trying to get independent terms, a separation method might need to be used first. In this case, it is important to keep track of what you send to this routine so you know how to interpret the returned values
>>> from sympy import separatevars, log >>> separatevars(exp(x+y)).as_independent(x) (exp(y), exp(x)) >>> (x + x*y).as_independent(y) (x, x*y) >>> separatevars(x + x*y).as_independent(y) (x, y + 1) >>> (x*(1 + y)).as_independent(y) (x, y + 1) >>> (x*(1 + y)).expand(mul=True).as_independent(y) (x, x*y) >>> a, b=symbols('a b', positive=True) >>> (log(a*b).expand(log=True)).as_independent(b) (log(a), log(b))
See also
separatevars()
,expand()
,sympy.core.add.Add.as_two_terms()
,sympy.core.mul.Mul.as_two_terms()
,as_coeff_add()
,as_coeff_mul()

as_leading_term
(*symbols)¶ Returns the leading (nonzero) term of the series expansion of self.
The _eval_as_leading_term routines are used to do this, and they must always return a nonzero value.
Examples
>>> from sympy.abc import x >>> (1 + x + x**2).as_leading_term(x) 1 >>> (1/x**2 + x + x**2).as_leading_term(x) x**(2)

as_numer_denom
()¶ expression > a/b > a, b
This is just a stub that should be defined by an object’s class methods to get anything else.
See also
normal()
return a/b instead of a, b

as_ordered_factors
(order=None)¶ Return list of ordered factors (if Mul) else [self].

as_ordered_terms
(order=None, data=False)¶ Transform an expression to an ordered list of terms.
Examples
>>> from sympy import sin, cos >>> from sympy.abc import x
>>> (sin(x)**2*cos(x) + sin(x)**2 + 1).as_ordered_terms() [sin(x)**2*cos(x), sin(x)**2, 1]

as_poly
(*gens, **args)¶ Converts
self
to a polynomial or returnsNone
.>>> from sympy import sin >>> from sympy.abc import x, y
>>> print((x**2 + x*y).as_poly()) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + x*y).as_poly(x, y)) Poly(x**2 + x*y, x, y, domain='ZZ')
>>> print((x**2 + sin(y)).as_poly(x, y)) None

as_powers_dict
()¶ Return self as a dictionary of factors with each factor being treated as a power. The keys are the bases of the factors and the values, the corresponding exponents. The resulting dictionary should be used with caution if the expression is a Mul and contains non commutative factors since the order that they appeared will be lost in the dictionary.
See also
as_ordered_factors()
An alternative for noncommutative applications, returning an ordered list of factors.
args_cnc()
Similar to as_ordered_factors, but guarantees separation of commutative and noncommutative factors.

as_real_imag
(deep=True, **hints)¶ Performs complex expansion on ‘self’ and returns a tuple containing collected both real and imaginary parts. This method can’t be confused with re() and im() functions, which does not perform complex expansion at evaluation.
However it is possible to expand both re() and im() functions and get exactly the same results as with a single call to this function.
>>> from sympy import symbols, I
>>> x, y = symbols('x,y', real=True)
>>> (x + y*I).as_real_imag() (x, y)
>>> from sympy.abc import z, w
>>> (z + w*I).as_real_imag() (re(z)  im(w), re(w) + im(z))

as_terms
()¶ Transform an expression to a list of terms.

aseries
(x=None, n=6, bound=0, hir=False)¶ Asymptotic Series expansion of self. This is equivalent to
self.series(x, oo, n)
. Parameters
self (Expression) – The expression whose series is to be expanded.
x (Symbol) – It is the variable of the expression to be calculated.
n (Value) – The number of terms upto which the series is to be expanded.
hir (Boolean) – Set this parameter to be True to produce hierarchical series. It stops the recursion at an early level and may provide nicer and more useful results.
bound (Value, Integer) – Use the
bound
parameter to give limit on rewriting coefficients in its normalised form.
Examples
>>> from sympy import sin, exp >>> from sympy.abc import x, y
>>> e = sin(1/x + exp(x))  sin(1/x)
>>> e.aseries(x) (1/(24*x**4)  1/(2*x**2) + 1 + O(x**(6), (x, oo)))*exp(x)
>>> e.aseries(x, n=3, hir=True) exp(2*x)*sin(1/x)/2 + exp(x)*cos(1/x) + O(exp(3*x), (x, oo))
>>> e = exp(exp(x)/(1  1/x))
>>> e.aseries(x) exp(exp(x)/(1  1/x))
>>> e.aseries(x, bound=3) exp(exp(x)/x**2)*exp(exp(x)/x)*exp(exp(x) + exp(x)/(1  1/x)  exp(x)/x  exp(x)/x**2)*exp(exp(x))
 Returns
Asymptotic series expansion of the expression.
 Return type
Expr
Notes
This algorithm is directly induced from the limit computational algorithm provided by Gruntz. It majorly uses the mrv and rewrite subroutines. The overall idea of this algorithm is first to look for the most rapidly varying subexpression w of a given expression f and then expands f in a series in w. Then same thing is recursively done on the leading coefficient till we get constant coefficients.
If the most rapidly varying subexpression of a given expression f is f itself, the algorithm tries to find a normalised representation of the mrv set and rewrites f using this normalised representation.
If the expansion contains an order term, it will be either
O(x ** (n))
orO(w ** (n))
wherew
belongs to the most rapidly varying expression ofself
.References
 1
A New Algorithm for Computing Asymptotic Series  Dominik Gruntz
 2
Gruntz thesis  p90
 3
See also
Expr.aseries()
See the docstring of this function for complete details of this wrapper.

property
assumptions0
¶ Return object type assumptions.
For example:
Symbol(‘x’, real=True) Symbol(‘x’, integer=True)
are different objects. In other words, besides Python type (Symbol in this case), the initial assumptions are also forming their typeinfo.
Examples
>>> from sympy import Symbol >>> from sympy.abc import x >>> x.assumptions0 {'commutative': True} >>> x = Symbol("x", positive=True) >>> x.assumptions0 {'commutative': True, 'complex': True, 'extended_negative': False, 'extended_nonnegative': True, 'extended_nonpositive': False, 'extended_nonzero': True, 'extended_positive': True, 'extended_real': True, 'finite': True, 'hermitian': True, 'imaginary': False, 'infinite': False, 'negative': False, 'nonnegative': True, 'nonpositive': False, 'nonzero': True, 'positive': True, 'real': True, 'zero': False}

atoms
(*types)¶ Returns the atoms that form the current object.
By default, only objects that are truly atomic and can’t be divided into smaller pieces are returned: symbols, numbers, and number symbols like I and pi. It is possible to request atoms of any type, however, as demonstrated below.
Examples
>>> from sympy import I, pi, sin >>> from sympy.abc import x, y >>> (1 + x + 2*sin(y + I*pi)).atoms() {1, 2, I, pi, x, y}
If one or more types are given, the results will contain only those types of atoms.
>>> from sympy import Number, NumberSymbol, Symbol >>> (1 + x + 2*sin(y + I*pi)).atoms(Symbol) {x, y}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number) {1, 2}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol) {1, 2, pi}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Number, NumberSymbol, I) {1, 2, I, pi}
Note that I (imaginary unit) and zoo (complex infinity) are special types of number symbols and are not part of the NumberSymbol class.
The type can be given implicitly, too:
>>> (1 + x + 2*sin(y + I*pi)).atoms(x) # x is a Symbol {x, y}
Be careful to check your assumptions when using the implicit option since
S(1).is_Integer = True
buttype(S(1))
isOne
, a special type of sympy atom, whiletype(S(2))
is typeInteger
and will find all integers in an expression:>>> from sympy import S >>> (1 + x + 2*sin(y + I*pi)).atoms(S(1)) {1}
>>> (1 + x + 2*sin(y + I*pi)).atoms(S(2)) {1, 2}
Finally, arguments to atoms() can select more than atomic atoms: any sympy type (loaded in core/__init__.py) can be listed as an argument and those types of “atoms” as found in scanning the arguments of the expression recursively:
>>> from sympy import Function, Mul >>> from sympy.core.function import AppliedUndef >>> f = Function('f') >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(Function) {f(x), sin(y + I*pi)} >>> (1 + f(x) + 2*sin(y + I*pi)).atoms(AppliedUndef) {f(x)}
>>> (1 + x + 2*sin(y + I*pi)).atoms(Mul) {I*pi, 2*sin(y + I*pi)}

cancel
(*gens, **args)¶ See the cancel function in sympy.polys

property
canonical_variables
¶ Return a dictionary mapping any variable defined in
self.bound_symbols
to Symbols that do not clash with any existing symbol in the expression.Examples
>>> from sympy import Lambda >>> from sympy.abc import x >>> Lambda(x, 2*x).canonical_variables {x: _0}

classmethod
class_key
()¶ Nice order of classes.

coeff
(x, n=1, right=False)¶ Returns the coefficient from the term(s) containing
x**n
. Ifn
is zero then all terms independent ofx
will be returned.When
x
is noncommutative, the coefficient to the left (default) or right ofx
can be returned. The keyword ‘right’ is ignored whenx
is commutative.See also
as_coefficient()
separate the expression into a coefficient and factor
as_coeff_Add()
separate the additive constant from an expression
as_coeff_Mul()
separate the multiplicative constant from an expression
as_independent()
separate xdependent terms/factors from others
sympy.polys.polytools.Poly.coeff_monomial()
efficiently find the single coefficient of a monomial in Poly
sympy.polys.polytools.Poly.nth()
like coeff_monomial but powers of monomial terms are used
Examples
>>> from sympy import symbols >>> from sympy.abc import x, y, z
You can select terms that have an explicit negative in front of them:
>>> (x + 2*y).coeff(1) x >>> (x  2*y).coeff(1) 2*y
You can select terms with no Rational coefficient:
>>> (x + 2*y).coeff(1) x >>> (3 + 2*x + 4*x**2).coeff(1) 0
You can select terms independent of x by making n=0; in this case expr.as_independent(x)[0] is returned (and 0 will be returned instead of None):
>>> (3 + 2*x + 4*x**2).coeff(x, 0) 3 >>> eq = ((x + 1)**3).expand() + 1 >>> eq x**3 + 3*x**2 + 3*x + 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 2] >>> eq = 2 >>> [eq.coeff(x, i) for i in reversed(range(4))] [1, 3, 3, 0]
You can select terms that have a numerical term in front of them:
>>> (x  2*y).coeff(2) y >>> from sympy import sqrt >>> (x + sqrt(2)*x).coeff(sqrt(2)) x
The matching is exact:
>>> (3 + 2*x + 4*x**2).coeff(x) 2 >>> (3 + 2*x + 4*x**2).coeff(x**2) 4 >>> (3 + 2*x + 4*x**2).coeff(x**3) 0 >>> (z*(x + y)**2).coeff((x + y)**2) z >>> (z*(x + y)**2).coeff(x + y) 0
In addition, no factoring is done, so 1 + z*(1 + y) is not obtained from the following:
>>> (x + z*(x + x*y)).coeff(x) 1
If such factoring is desired, factor_terms can be used first:
>>> from sympy import factor_terms >>> factor_terms(x + z*(x + x*y)).coeff(x) z*(y + 1) + 1
>>> n, m, o = symbols('n m o', commutative=False) <
