Simplify¶
simplify¶

sympy.simplify.simplify.
simplify
(expr, ratio=1.7, measure=<function count_ops>, fu=False)[source]¶ Simplifies the given expression.
Simplification is not a well defined term and the exact strategies this function tries can change in the future versions of SymPy. If your algorithm relies on “simplification” (whatever it is), try to determine what you need exactly  is it powsimp()?, radsimp()?, together()?, logcombine()?, or something else? And use this particular function directly, because those are well defined and thus your algorithm will be robust.
Nonetheless, especially for interactive use, or when you don’t know anything about the structure of the expression, simplify() tries to apply intelligent heuristics to make the input expression “simpler”. For example:
>>> from sympy import simplify, cos, sin >>> from sympy.abc import x, y >>> a = (x + x**2)/(x*sin(y)**2 + x*cos(y)**2) >>> a (x**2 + x)/(x*sin(y)**2 + x*cos(y)**2) >>> simplify(a) x + 1
Note that we could have obtained the same result by using specific simplification functions:
>>> from sympy import trigsimp, cancel >>> trigsimp(a) (x**2 + x)/x >>> cancel(_) x + 1
In some cases, applying
simplify()
may actually result in some more complicated expression. The defaultratio=1.7
prevents more extreme cases: if (result length)/(input length) > ratio, then input is returned unmodified. Themeasure
parameter lets you specify the function used to determine how complex an expression is. The function should take a single argument as an expression and return a number such that if expressiona
is more complex than expressionb
, thenmeasure(a) > measure(b)
. The default measure function iscount_ops()
, which returns the total number of operations in the expression.For example, if
ratio=1
,simplify
output can’t be longer than input.>>> from sympy import sqrt, simplify, count_ops, oo >>> root = 1/(sqrt(2)+3)
Since
simplify(root)
would result in a slightly longer expression, root is returned unchanged instead:>>> simplify(root, ratio=1) == root True
If
ratio=oo
, simplify will be applied anyway:>>> count_ops(simplify(root, ratio=oo)) > count_ops(root) True
Note that the shortest expression is not necessary the simplest, so setting
ratio
to 1 may not be a good idea. Heuristically, the default valueratio=1.7
seems like a reasonable choice.You can easily define your own measure function based on what you feel should represent the “size” or “complexity” of the input expression. Note that some choices, such as
lambda expr: len(str(expr))
may appear to be good metrics, but have other problems (in this case, the measure function may slow down simplify too much for very large expressions). If you don’t know what a good metric would be, the default,count_ops
, is a good one.For example:
>>> from sympy import symbols, log >>> a, b = symbols('a b', positive=True) >>> g = log(a) + log(b) + log(a)*log(1/b) >>> h = simplify(g) >>> h log(a*b**(log(a) + 1)) >>> count_ops(g) 8 >>> count_ops(h) 5
So you can see that
h
is simpler thang
using the count_ops metric. However, we may not like howsimplify
(in this case, usinglogcombine
) has created theb**(log(1/a) + 1)
term. A simple way to reduce this would be to give more weight to powers as operations incount_ops
. We can do this by using thevisual=True
option:>>> print(count_ops(g, visual=True)) 2*ADD + DIV + 4*LOG + MUL >>> print(count_ops(h, visual=True)) 2*LOG + MUL + POW + SUB
>>> from sympy import Symbol, S >>> def my_measure(expr): ... POW = Symbol('POW') ... # Discourage powers by giving POW a weight of 10 ... count = count_ops(expr, visual=True).subs(POW, 10) ... # Every other operation gets a weight of 1 (the default) ... count = count.replace(Symbol, type(S.One)) ... return count >>> my_measure(g) 8 >>> my_measure(h) 14 >>> 15./8 > 1.7 # 1.7 is the default ratio True >>> simplify(g, measure=my_measure) log(a)*log(b) + log(a) + log(b)
Note that because
simplify()
internally tries many different simplification strategies and then compares them using the measure function, we get a completely different result that is still different from the input expression by doing this.
separatevars¶

sympy.simplify.simplify.
separatevars
(expr, symbols=[], dict=False, force=False)[source]¶ Separates variables in an expression, if possible. By default, it separates with respect to all symbols in an expression and collects constant coefficients that are independent of symbols.
If dict=True then the separated terms will be returned in a dictionary keyed to their corresponding symbols. By default, all symbols in the expression will appear as keys; if symbols are provided, then all those symbols will be used as keys, and any terms in the expression containing other symbols or nonsymbols will be returned keyed to the string ‘coeff’. (Passing None for symbols will return the expression in a dictionary keyed to ‘coeff’.)
If force=True, then bases of powers will be separated regardless of assumptions on the symbols involved.
Notes
The order of the factors is determined by Mul, so that the separated expressions may not necessarily be grouped together.
Although factoring is necessary to separate variables in some expressions, it is not necessary in all cases, so one should not count on the returned factors being factored.
Examples
>>> from sympy.abc import x, y, z, alpha >>> from sympy import separatevars, sin >>> separatevars((x*y)**y) (x*y)**y >>> separatevars((x*y)**y, force=True) x**y*y**y
>>> e = 2*x**2*z*sin(y)+2*z*x**2 >>> separatevars(e) 2*x**2*z*(sin(y) + 1) >>> separatevars(e, symbols=(x, y), dict=True) {'coeff': 2*z, x: x**2, y: sin(y) + 1} >>> separatevars(e, [x, y, alpha], dict=True) {'coeff': 2*z, alpha: 1, x: x**2, y: sin(y) + 1}
If the expression is not really separable, or is only partially separable, separatevars will do the best it can to separate it by using factoring.
>>> separatevars(x + x*y  3*x**2) x*(3*x  y  1)
If the expression is not separable then expr is returned unchanged or (if dict=True) then None is returned.
>>> eq = 2*x + y*sin(x) >>> separatevars(eq) == eq True >>> separatevars(2*x + y*sin(x), symbols=(x, y), dict=True) == None True
nthroot¶

sympy.simplify.simplify.
nthroot
(expr, n, max_len=4, prec=15)[source]¶ compute a real nthroot of a sum of surds
Parameters: expr : sum of surds
n : integer
max_len : maximum number of surds passed as constants to
nsimplify
Examples
>>> from sympy.simplify.simplify import nthroot >>> from sympy import Rational, sqrt >>> nthroot(90 + 34*sqrt(7), 3) sqrt(7) + 3
Algorithm
First
nsimplify
is used to get a candidate root; if it is not a root the minimal polynomial is computed; the answer is one of its roots.
besselsimp¶

sympy.simplify.simplify.
besselsimp
(expr)[source]¶ Simplify besseltype functions.
This routine tries to simplify besseltype functions. Currently it only works on the Bessel J and I functions, however. It works by looking at all such functions in turn, and eliminating factors of “I” and “1” (actually their polar equivalents) in front of the argument. Then, functions of halfinteger order are rewritten using strigonometric functions and functions of integer order (> 1) are rewritten using functions of low order. Finally, if the expression was changed, compute factorization of the result with factor().
>>> from sympy import besselj, besseli, besselsimp, polar_lift, I, S >>> from sympy.abc import z, nu >>> besselsimp(besselj(nu, z*polar_lift(1))) exp(I*pi*nu)*besselj(nu, z) >>> besselsimp(besseli(nu, z*polar_lift(I))) exp(I*pi*nu/2)*besselj(nu, z) >>> besselsimp(besseli(S(1)/2, z)) sqrt(2)*cosh(z)/(sqrt(pi)*sqrt(z)) >>> besselsimp(z*besseli(0, z) + z*(besseli(2, z))/2 + besseli(1, z)) 3*z*besseli(0, z)/2
hypersimp¶

sympy.simplify.simplify.
hypersimp
(f, k)[source]¶ Given combinatorial term f(k) simplify its consecutive term ratio i.e. f(k+1)/f(k). The input term can be composed of functions and integer sequences which have equivalent representation in terms of gamma special function.
The algorithm performs three basic steps:
 Rewrite all functions in terms of gamma, if possible.
 Rewrite all occurrences of gamma in terms of products of gamma and rising factorial with integer, absolute constant exponent.
 Perform simplification of nested fractions, powers and if the resulting expression is a quotient of polynomials, reduce their total degree.
If f(k) is hypergeometric then as result we arrive with a quotient of polynomials of minimal degree. Otherwise None is returned.
For more information on the implemented algorithm refer to:
 W. Koepf, Algorithms for mfold Hypergeometric Summation, Journal of Symbolic Computation (1995) 20, 399417
hypersimilar¶

sympy.simplify.simplify.
hypersimilar
(f, g, k)[source]¶ Returns True if ‘f’ and ‘g’ are hypersimilar.
Similarity in hypergeometric sense means that a quotient of f(k) and g(k) is a rational function in k. This procedure is useful in solving recurrence relations.
For more information see hypersimp().
nsimplify¶

sympy.simplify.simplify.
nsimplify
(expr, constants=[], tolerance=None, full=False, rational=None)[source]¶ Find a simple representation for a number or, if there are free symbols or if rational=True, then replace Floats with their Rational equivalents. If no change is made and rational is not False then Floats will at least be converted to Rationals.
For numerical expressions, a simple formula that numerically matches the given numerical expression is sought (and the input should be possible to evalf to a precision of at least 30 digits).
Optionally, a list of (rationally independent) constants to include in the formula may be given.
A lower tolerance may be set to find less exact matches. If no tolerance is given then the least precise value will set the tolerance (e.g. Floats default to 15 digits of precision, so would be tolerance=10**15).
With full=True, a more extensive search is performed (this is useful to find simpler numbers when the tolerance is set low).
See also
Examples
>>> from sympy import nsimplify, sqrt, GoldenRatio, exp, I, exp, pi >>> nsimplify(4/(1+sqrt(5)), [GoldenRatio]) 2 + 2*GoldenRatio >>> nsimplify((1/(exp(3*pi*I/5)+1))) 1/2  I*sqrt(sqrt(5)/10 + 1/4) >>> nsimplify(I**I, [pi]) exp(pi/2) >>> nsimplify(pi, tolerance=0.01) 22/7
posify¶

sympy.simplify.simplify.
posify
(eq)[source]¶ Return eq (with generic symbols made positive) and a dictionary containing the mapping between the old and new symbols.
Any symbol that has positive=None will be replaced with a positive dummy symbol having the same name. This replacement will allow more symbolic processing of expressions, especially those involving powers and logarithms.
A dictionary that can be sent to subs to restore eq to its original symbols is also returned.
>>> from sympy import posify, Symbol, log, solve >>> from sympy.abc import x >>> posify(x + Symbol('p', positive=True) + Symbol('n', negative=True)) (_x + n + p, {_x: x})
>>> eq = 1/x >>> log(eq).expand() log(1/x) >>> log(posify(eq)[0]).expand() log(_x) >>> p, rep = posify(eq) >>> log(p).expand().subs(rep) log(x)
It is possible to apply the same transformations to an iterable of expressions:
>>> eq = x**2  4 >>> solve(eq, x) [2, 2] >>> eq_x, reps = posify([eq, x]); eq_x [_x**2  4, _x] >>> solve(*eq_x) [2]
logcombine¶

sympy.simplify.simplify.
logcombine
(expr, force=False)[source]¶ Takes logarithms and combines them using the following rules:
 log(x) + log(y) == log(x*y) if both are not negative
 a*log(x) == log(x**a) if x is positive and a is real
If
force
is True then the assumptions above will be assumed to hold if there is no assumption already in place on a quantity. For example, ifa
is imaginary or the argument negative, force will not perform a combination but ifa
is a symbol with no assumptions the change will take place.See also
posify
 replace all symbols with symbols having positive assumptions
Examples
>>> from sympy import Symbol, symbols, log, logcombine, I >>> from sympy.abc import a, x, y, z >>> logcombine(a*log(x) + log(y)  log(z)) a*log(x) + log(y)  log(z) >>> logcombine(a*log(x) + log(y)  log(z), force=True) log(x**a*y/z) >>> x,y,z = symbols('x,y,z', positive=True) >>> a = Symbol('a', real=True) >>> logcombine(a*log(x) + log(y)  log(z)) log(x**a*y/z)
The transformation is limited to factors and/or terms that contain logs, so the result depends on the initial state of expansion:
>>> eq = (2 + 3*I)*log(x) >>> logcombine(eq, force=True) == eq True >>> logcombine(eq.expand(), force=True) log(x**2) + I*log(x**3)
Radsimp¶
radsimp¶

sympy.simplify.radsimp.
radsimp
(expr, symbolic=True, max_terms=4)[source]¶ Rationalize the denominator by removing square roots.
Note: the expression returned from radsimp must be used with caution since if the denominator contains symbols, it will be possible to make substitutions that violate the assumptions of the simplification process: that for a denominator matching a + b*sqrt(c), a != +/b*sqrt(c). (If there are no symbols, this assumptions is made valid by collecting terms of sqrt(c) so the match variable
a
does not containsqrt(c)
.) If you do not want the simplification to occur for symbolic denominators, setsymbolic
to False.If there are more than
max_terms
radical terms then the expression is returned unchanged.Examples
>>> from sympy import radsimp, sqrt, Symbol, denom, pprint, I >>> from sympy import factor_terms, fraction, signsimp >>> from sympy.simplify.radsimp import collect_sqrt >>> from sympy.abc import a, b, c
>>> radsimp(1/(I + 1)) (1  I)/2 >>> radsimp(1/(2 + sqrt(2))) (sqrt(2) + 2)/2 >>> x,y = map(Symbol, 'xy') >>> e = ((2 + 2*sqrt(2))*x + (2 + sqrt(8))*y)/(2 + sqrt(2)) >>> radsimp(e) sqrt(2)*(x + y)
No simplification beyond removal of the gcd is done. One might want to polish the result a little, however, by collecting square root terms:
>>> r2 = sqrt(2) >>> r5 = sqrt(5) >>> ans = radsimp(1/(y*r2 + x*r2 + a*r5 + b*r5)); pprint(ans) ___ ___ ___ ___ \/ 5 *a + \/ 5 *b  \/ 2 *x  \/ 2 *y  2 2 2 2 5*a + 10*a*b + 5*b  2*x  4*x*y  2*y
>>> n, d = fraction(ans) >>> pprint(factor_terms(signsimp(collect_sqrt(n))/d, radical=True)) ___ ___ \/ 5 *(a + b)  \/ 2 *(x + y)  2 2 2 2 5*a + 10*a*b + 5*b  2*x  4*x*y  2*y
If radicals in the denominator cannot be removed or there is no denominator, the original expression will be returned.
>>> radsimp(sqrt(2)*x + sqrt(2)) sqrt(2)*x + sqrt(2)
Results with symbols will not always be valid for all substitutions:
>>> eq = 1/(a + b*sqrt(c)) >>> eq.subs(a, b*sqrt(c)) 1/(2*b*sqrt(c)) >>> radsimp(eq).subs(a, b*sqrt(c)) nan
If symbolic=False, symbolic denominators will not be transformed (but numeric denominators will still be processed):
>>> radsimp(eq, symbolic=False) 1/(a + b*sqrt(c))
rad_rationalize¶

sympy.simplify.radsimp.
rad_rationalize
(num, den)[source]¶ Rationalize num/den by removing square roots in the denominator; num and den are sum of terms whose squares are rationals
Examples
>>> from sympy import sqrt >>> from sympy.simplify.radsimp import rad_rationalize >>> rad_rationalize(sqrt(3), 1 + sqrt(2)/3) (sqrt(3) + sqrt(6)/3, 7/9)
collect¶

sympy.simplify.radsimp.
collect
(expr, syms, func=None, evaluate=None, exact=False, distribute_order_term=True)[source]¶ Collect additive terms of an expression.
This function collects additive terms of an expression with respect to a list of expression up to powers with rational exponents. By the term symbol here are meant arbitrary expressions, which can contain powers, products, sums etc. In other words symbol is a pattern which will be searched for in the expression’s terms.
The input expression is not expanded by
collect()
, so user is expected to provide an expression is an appropriate form. This makescollect()
more predictable as there is no magic happening behind the scenes. However, it is important to note, that powers of products are converted to products of powers using theexpand_power_base()
function.There are two possible types of output. First, if
evaluate
flag is set, this function will return an expression with collected terms or else it will return a dictionary with expressions up to rational powers as keys and collected coefficients as values.See also
Examples
>>> from sympy import S, collect, expand, factor, Wild >>> from sympy.abc import a, b, c, x, y, z
This function can collect symbolic coefficients in polynomials or rational expressions. It will manage to find all integer or rational powers of collection variable:
>>> collect(a*x**2 + b*x**2 + a*x  b*x + c, x) c + x**2*(a + b) + x*(a  b)
The same result can be achieved in dictionary form:
>>> d = collect(a*x**2 + b*x**2 + a*x  b*x + c, x, evaluate=False) >>> d[x**2] a + b >>> d[x] a  b >>> d[S.One] c
You can also work with multivariate polynomials. However, remember that this function is greedy so it will care only about a single symbol at time, in specification order:
>>> collect(x**2 + y*x**2 + x*y + y + a*y, [x, y]) x**2*(y + 1) + x*y + y*(a + 1)
Also more complicated expressions can be used as patterns:
>>> from sympy import sin, log >>> collect(a*sin(2*x) + b*sin(2*x), sin(2*x)) (a + b)*sin(2*x) >>> collect(a*x*log(x) + b*(x*log(x)), x*log(x)) x*(a + b)*log(x)
You can use wildcards in the pattern:
>>> w = Wild('w1') >>> collect(a*x**y  b*x**y, w**y) x**y*(a  b)
It is also possible to work with symbolic powers, although it has more complicated behavior, because in this case power’s base and symbolic part of the exponent are treated as a single symbol:
>>> collect(a*x**c + b*x**c, x) a*x**c + b*x**c >>> collect(a*x**c + b*x**c, x**c) x**c*(a + b)
However if you incorporate rationals to the exponents, then you will get well known behavior:
>>> collect(a*x**(2*c) + b*x**(2*c), x**c) x**(2*c)*(a + b)
Note also that all previously stated facts about
collect()
function apply to the exponential function, so you can get:>>> from sympy import exp >>> collect(a*exp(2*x) + b*exp(2*x), exp(x)) (a + b)*exp(2*x)
If you are interested only in collecting specific powers of some symbols then set
exact
flag in arguments:>>> collect(a*x**7 + b*x**7, x, exact=True) a*x**7 + b*x**7 >>> collect(a*x**7 + b*x**7, x**7, exact=True) x**7*(a + b)
You can also apply this function to differential equations, where derivatives of arbitrary order can be collected. Note that if you collect with respect to a function or a derivative of a function, all derivatives of that function will also be collected. Use
exact=True
to prevent this from happening:>>> from sympy import Derivative as D, collect, Function >>> f = Function('f') (x) >>> collect(a*D(f,x) + b*D(f,x), D(f,x)) (a + b)*Derivative(f(x), x) >>> collect(a*D(D(f,x),x) + b*D(D(f,x),x), f) (a + b)*Derivative(f(x), x, x) >>> collect(a*D(D(f,x),x) + b*D(D(f,x),x), D(f,x), exact=True) a*Derivative(f(x), x, x) + b*Derivative(f(x), x, x) >>> collect(a*D(f,x) + b*D(f,x) + a*f + b*f, f) (a + b)*f(x) + (a + b)*Derivative(f(x), x)
Or you can even match both derivative order and exponent at the same time:
>>> collect(a*D(D(f,x),x)**2 + b*D(D(f,x),x)**2, D(f,x)) (a + b)*Derivative(f(x), x, x)**2
Finally, you can apply a function to each of the collected coefficients. For example you can factorize symbolic coefficients of polynomial:
>>> f = expand((x + a + 1)**3) >>> collect(f, x, factor) x**3 + 3*x**2*(a + 1) + 3*x*(a + 1)**2 + (a + 1)**3
Note
Arguments are expected to be in expanded form, so you might have to call
expand()
prior to calling this function.
collect_sqrt¶

sympy.simplify.radsimp.
collect_sqrt
(expr, evaluate=None)[source]¶ Return expr with terms having common square roots collected together. If
evaluate
is False a count indicating the number of sqrtcontaining terms will be returned and, if nonzero, the terms of the Add will be returned, else the expression itself will be returned as a single term. Ifevaluate
is True, the expression with any collected terms will be returned.Note: since I = sqrt(1), it is collected, too.
See also
Examples
>>> from sympy import sqrt >>> from sympy.simplify.radsimp import collect_sqrt >>> from sympy.abc import a, b
>>> r2, r3, r5 = [sqrt(i) for i in [2, 3, 5]] >>> collect_sqrt(a*r2 + b*r2) sqrt(2)*(a + b) >>> collect_sqrt(a*r2 + b*r2 + a*r3 + b*r3) sqrt(2)*(a + b) + sqrt(3)*(a + b) >>> collect_sqrt(a*r2 + b*r2 + a*r3 + b*r5) sqrt(3)*a + sqrt(5)*b + sqrt(2)*(a + b)
If evaluate is False then the arguments will be sorted and returned as a list and a count of the number of sqrtcontaining terms will be returned:
>>> collect_sqrt(a*r2 + b*r2 + a*r3 + b*r5, evaluate=False) ((sqrt(3)*a, sqrt(5)*b, sqrt(2)*(a + b)), 3) >>> collect_sqrt(a*sqrt(2) + b, evaluate=False) ((b, sqrt(2)*a), 1) >>> collect_sqrt(a + b, evaluate=False) ((a + b,), 0)
collect_const¶

sympy.simplify.radsimp.
collect_const
(expr, *vars, **kwargs)[source]¶ A nongreedy collection of terms with similar number coefficients in an Add expr. If
vars
is given then only those constants will be targeted. Although any Number can also be targeted, if this is not desired setNumbers=False
and no Float or Rational will be collected.See also
Examples
>>> from sympy import sqrt >>> from sympy.abc import a, s, x, y, z >>> from sympy.simplify.radsimp import collect_const >>> collect_const(sqrt(3) + sqrt(3)*(1 + sqrt(2))) sqrt(3)*(sqrt(2) + 2) >>> collect_const(sqrt(3)*s + sqrt(7)*s + sqrt(3) + sqrt(7)) (sqrt(3) + sqrt(7))*(s + 1) >>> s = sqrt(2) + 2 >>> collect_const(sqrt(3)*s + sqrt(3) + sqrt(7)*s + sqrt(7)) (sqrt(2) + 3)*(sqrt(3) + sqrt(7)) >>> collect_const(sqrt(3)*s + sqrt(3) + sqrt(7)*s + sqrt(7), sqrt(3)) sqrt(7) + sqrt(3)*(sqrt(2) + 3) + sqrt(7)*(sqrt(2) + 2)
The collection is signsensitive, giving higher precedence to the unsigned values:
>>> collect_const(x  y  z) x  (y + z) >>> collect_const(y  z) (y + z) >>> collect_const(2*x  2*y  2*z, 2) 2*(x  y  z) >>> collect_const(2*x  2*y  2*z, 2) 2*x  2*(y + z)
fraction¶

sympy.simplify.radsimp.
fraction
(expr, exact=False)[source]¶ Returns a pair with expression’s numerator and denominator. If the given expression is not a fraction then this function will return the tuple (expr, 1).
This function will not make any attempt to simplify nested fractions or to do any term rewriting at all.
If only one of the numerator/denominator pair is needed then use numer(expr) or denom(expr) functions respectively.
>>> from sympy import fraction, Rational, Symbol >>> from sympy.abc import x, y
>>> fraction(x/y) (x, y) >>> fraction(x) (x, 1)
>>> fraction(1/y**2) (1, y**2)
>>> fraction(x*y/2) (x*y, 2) >>> fraction(Rational(1, 2)) (1, 2)
This function will also work fine with assumptions:
>>> k = Symbol('k', negative=True) >>> fraction(x * y**k) (x, y**(k))
If we know nothing about sign of some exponent and ‘exact’ flag is unset, then structure this exponent’s structure will be analyzed and pretty fraction will be returned:
>>> from sympy import exp >>> fraction(2*x**(y)) (2, x**y)
>>> fraction(exp(x)) (1, exp(x))
>>> fraction(exp(x), exact=True) (exp(x), 1)
Ratsimp¶
Trigonometric simplification¶
trigsimp¶

sympy.simplify.trigsimp.
trigsimp
(expr, **opts)[source]¶ reduces expression by using known trig identities
Notes
method:  Determine the method to use. Valid choices are ‘matching’ (default), ‘groebner’, ‘combined’, and ‘fu’. If ‘matching’, simplify the expression recursively by targeting common patterns. If ‘groebner’, apply an experimental groebner basis algorithm. In this case further options are forwarded to
trigsimp_groebner
, please refer to its docstring. If ‘combined’, first run the groebner basis algorithm with small default parameters, then run the ‘matching’ algorithm. ‘fu’ runs the collection of trigonometric transformations described by Fu, et al. (see the \(fu\) docstring).Examples
>>> from sympy import trigsimp, sin, cos, log >>> from sympy.abc import x, y >>> e = 2*sin(x)**2 + 2*cos(x)**2 >>> trigsimp(e) 2
Simplification occurs wherever trigonometric functions are located.
>>> trigsimp(log(e)) log(2)
Using \(method="groebner"\) (or \("combined"\)) might lead to greater simplification.
The old trigsimp routine can be accessed as with method ‘old’.
>>> from sympy import coth, tanh >>> t = 3*tanh(x)**7  2/coth(x)**7 >>> trigsimp(t, method='old') == t True >>> trigsimp(t) tanh(x)**7
Power simplification¶
powsimp¶

sympy.simplify.powsimp.
powsimp
(expr, deep=False, combine='all', force=False, measure=<function count_ops>)[source]¶ reduces expression by combining powers with similar bases and exponents.
Notes
If deep is True then powsimp() will also simplify arguments of functions. By default deep is set to False.
If force is True then bases will be combined without checking for assumptions, e.g. sqrt(x)*sqrt(y) > sqrt(x*y) which is not true if x and y are both negative.
You can make powsimp() only combine bases or only combine exponents by changing combine=’base’ or combine=’exp’. By default, combine=’all’, which does both. combine=’base’ will only combine:
a a a 2x x x * y => (x*y) as well as things like 2 => 4
and combine=’exp’ will only combine
a b (a + b) x * x => x
combine=’exp’ will strictly only combine exponents in the way that used to be automatic. Also use deep=True if you need the old behavior.
When combine=’all’, ‘exp’ is evaluated first. Consider the first example below for when there could be an ambiguity relating to this. This is done so things like the second example can be completely combined. If you want ‘base’ combined first, do something like powsimp(powsimp(expr, combine=’base’), combine=’exp’).
Examples
>>> from sympy import powsimp, exp, log, symbols >>> from sympy.abc import x, y, z, n >>> powsimp(x**y*x**z*y**z, combine='all') x**(y + z)*y**z >>> powsimp(x**y*x**z*y**z, combine='exp') x**(y + z)*y**z >>> powsimp(x**y*x**z*y**z, combine='base', force=True) x**y*(x*y)**z
>>> powsimp(x**z*x**y*n**z*n**y, combine='all', force=True) (n*x)**(y + z) >>> powsimp(x**z*x**y*n**z*n**y, combine='exp') n**(y + z)*x**(y + z) >>> powsimp(x**z*x**y*n**z*n**y, combine='base', force=True) (n*x)**y*(n*x)**z
>>> x, y = symbols('x y', positive=True) >>> powsimp(log(exp(x)*exp(y))) log(exp(x)*exp(y)) >>> powsimp(log(exp(x)*exp(y)), deep=True) x + y
Radicals with Mul bases will be combined if combine=’exp’
>>> from sympy import sqrt, Mul >>> x, y = symbols('x y')
Two radicals are automatically joined through Mul:
>>> a=sqrt(x*sqrt(y)) >>> a*a**3 == a**4 True
But if an integer power of that radical has been autoexpanded then Mul does not join the resulting factors:
>>> a**4 # auto expands to a Mul, no longer a Pow x**2*y >>> _*a # so Mul doesn't combine them x**2*y*sqrt(x*sqrt(y)) >>> powsimp(_) # but powsimp will (x*sqrt(y))**(5/2) >>> powsimp(x*y*a) # but won't when doing so would violate assumptions x*y*sqrt(x*sqrt(y))
powdenest¶

sympy.simplify.powsimp.
powdenest
(eq, force=False, polar=False)[source]¶ Collect exponents on powers as assumptions allow.
 Given
(bb**be)**e
, this can be simplified as follows:  if
bb
is positive, or e
is an integer, orbe < 1
then this simplifies tobb**(be*e)
 if
Given a product of powers raised to a power,
(bb1**be1 * bb2**be2...)**e
, simplification can be done as follows: if e is positive, the gcd of all bei can be joined with e;
 all nonnegative bb can be separated from those that are negative and their gcd can be joined with e; autosimplification already handles this separation.
 integer factors from powers that have integers in the denominator of the exponent can be removed from any term and the gcd of such integers can be joined with e
Setting
force
to True will make symbols that are not explicitly negative behave as though they are positive, resulting in more denesting.Setting
polar
to True will do simplifications on the Riemann surface of the logarithm, also resulting in more denestings.When there are sums of logs in exp() then a product of powers may be obtained e.g.
exp(3*(log(a) + 2*log(b)))
 >a**3*b**6
.Examples
>>> from sympy.abc import a, b, x, y, z >>> from sympy import Symbol, exp, log, sqrt, symbols, powdenest
>>> powdenest((x**(2*a/3))**(3*x)) (x**(2*a/3))**(3*x) >>> powdenest(exp(3*x*log(2))) 2**(3*x)
Assumptions may prevent expansion:
>>> powdenest(sqrt(x**2)) sqrt(x**2)
>>> p = symbols('p', positive=True) >>> powdenest(sqrt(p**2)) p
No other expansion is done.
>>> i, j = symbols('i,j', integer=True) >>> powdenest((x**x)**(i + j)) # X> (x**x)**i*(x**x)**j x**(x*(i + j))
But exp() will be denested by moving all nonlog terms outside of the function; this may result in the collapsing of the exp to a power with a different base:
>>> powdenest(exp(3*y*log(x))) x**(3*y) >>> powdenest(exp(y*(log(a) + log(b)))) (a*b)**y >>> powdenest(exp(3*(log(a) + log(b)))) a**3*b**3
If assumptions allow, symbols can also be moved to the outermost exponent:
>>> i = Symbol('i', integer=True) >>> powdenest(((x**(2*i))**(3*y))**x) ((x**(2*i))**(3*y))**x >>> powdenest(((x**(2*i))**(3*y))**x, force=True) x**(6*i*x*y)
>>> powdenest(((x**(2*a/3))**(3*y/i))**x) ((x**(2*a/3))**(3*y/i))**x >>> powdenest((x**(2*i)*y**(4*i))**z, force=True) (x*y**2)**(2*i*z)
>>> n = Symbol('n', negative=True)
>>> powdenest((x**i)**y, force=True) x**(i*y) >>> powdenest((n**i)**x, force=True) (n**i)**x
 Given
Combinatrial simplification¶
combsimp¶

sympy.simplify.combsimp.
combsimp
(expr)[source]¶ Simplify combinatorial expressions.
This function takes as input an expression containing factorials, binomials, Pochhammer symbol and other “combinatorial” functions, and tries to minimize the number of those functions and reduce the size of their arguments.
The algorithm works by rewriting all combinatorial functions as expressions involving rising factorials (Pochhammer symbols) and applies recurrence relations and other transformations applicable to rising factorials, to reduce their arguments, possibly letting the resulting rising factorial to cancel. Rising factorials with the second argument being an integer are expanded into polynomial forms and finally all other rising factorial are rewritten in terms of more familiar functions. If the initial expression consisted of gamma functions alone, the result is expressed in terms of gamma functions. If the initial expression consists of gamma function with some other combinatorial, the result is expressed in terms of gamma functions.
If the result is expressed using gamma functions, the following three additional steps are performed:
 Reduce the number of gammas by applying the reflection theorem gamma(x)*gamma(1x) == pi/sin(pi*x).
 Reduce the number of gammas by applying the multiplication theorem gamma(x)*gamma(x+1/n)*...*gamma(x+(n1)/n) == C*gamma(n*x).
 Reduce the number of prefactors by absorbing them into gammas, where possible.
All transformation rules can be found (or was derived from) here:
 http://functions.wolfram.com/GammaBetaErf/Pochhammer/17/01/02/
 http://functions.wolfram.com/GammaBetaErf/Pochhammer/27/01/0005/
Examples
>>> from sympy.simplify import combsimp >>> from sympy import factorial, binomial >>> from sympy.abc import n, k
>>> combsimp(factorial(n)/factorial(n  3)) n*(n  2)*(n  1) >>> combsimp(binomial(n+1, k+1)/binomial(n, k)) (n + 1)/(k + 1)
Square Root Denest¶
sqrtdenest¶

sympy.simplify.sqrtdenest.
sqrtdenest
(expr, max_iter=3)[source]¶ Denests sqrts in an expression that contain other square roots if possible, otherwise returns the expr unchanged. This is based on the algorithms of [1].
See also
sympy.solvers.solvers.unrad
References
[1] http://researcher.watson.ibm.com/researcher/files/usfagin/symb85.pdf
[2] D. J. Jeffrey and A. D. Rich, ‘Symplifying Square Roots of Square Roots by Denesting’ (available at http://www.cybertester.com/data/denest.pdf)
Examples
>>> from sympy.simplify.sqrtdenest import sqrtdenest >>> from sympy import sqrt >>> sqrtdenest(sqrt(5 + 2 * sqrt(6))) sqrt(2) + sqrt(3)
Common Subexpresion Elimination¶
cse¶

sympy.simplify.cse_main.
cse
(exprs, symbols=None, optimizations=None, postprocess=None, order='canonical', ignore=())[source]¶ Perform common subexpression elimination on an expression.
Parameters: exprs : list of sympy expressions, or a single sympy expression
The expressions to reduce.
symbols : infinite iterator yielding unique Symbols
The symbols used to label the common subexpressions which are pulled out. The
numbered_symbols
generator is useful. The default is a stream of symbols of the form “x0”, “x1”, etc. This must be an infinite iterator.optimizations : list of (callable, callable) pairs
The (preprocessor, postprocessor) pairs of external optimization functions. Optionally ‘basic’ can be passed for a set of predefined basic optimizations. Such ‘basic’ optimizations were used by default in old implementation, however they can be really slow on larger expressions. Now, no pre or post optimizations are made by default.
postprocess : a function which accepts the two return values of cse and
returns the desired form of output from cse, e.g. if you want the replacements reversed the function might be the following lambda: lambda r, e: return reversed(r), e
order : string, ‘none’ or ‘canonical’
The order by which Mul and Add arguments are processed. If set to ‘canonical’, arguments will be canonically ordered. If set to ‘none’, ordering will be faster but dependent on expressions hashes, thus machine dependent and variable. For large expressions where speed is a concern, use the setting order=’none’.
ignore : iterable of Symbols
Substitutions containing any Symbol from
ignore
will be ignored.Returns: replacements : list of (Symbol, expression) pairs
All of the common subexpressions that were replaced. Subexpressions earlier in this list might show up in subexpressions later in this list.
reduced_exprs : list of sympy expressions
The reduced expressions with all of the replacements above.
Examples
>>> from sympy import cse, SparseMatrix >>> from sympy.abc import x, y, z, w >>> cse(((w + x + y + z)*(w + y + z))/(w + x)**3) ([(x0, w + y + z)], [x0*(x + x0)/(w + x)**3])
Note that currently, y + z will not get substituted if y  z is used.
>>> cse(((w + x + y + z)*(w  y  z))/(w + x)**3) ([(x0, w + x)], [(w  y  z)*(x0 + y + z)/x0**3])
List of expressions with recursive substitutions:
>>> m = SparseMatrix([x + y, x + y + z]) >>> cse([(x+y)**2, x + y + z, y + z, x + z + y, m]) ([(x0, x + y), (x1, x0 + z)], [x0**2, x1, y + z, x1, Matrix([ [x0], [x1]])])
Note: the type and mutability of input matrices is retained.
>>> isinstance(_[1][1], SparseMatrix) True
The user may disallow substitutions containing certain symbols: >>> cse([y**2*(x + 1), 3*y**2*(x + 1)], ignore=(y,)) ([(x0, x + 1)], [x0*y**2, 3*x0*y**2])
opt_cse¶

sympy.simplify.cse_main.
opt_cse
(exprs, order='canonical', verbose=False)[source]¶ Find optimization opportunities in Adds, Muls, Pows and negative coefficient Muls
Parameters: exprs : list of sympy expressions
The expressions to optimize.
order : string, ‘none’ or ‘canonical’
The order by which Mul and Add arguments are processed. For large expressions where speed is a concern, use the setting order=’none’.
verbose : bool
Print debug information (default=False)
Returns: opt_subs : dictionary of expression substitutions
The expression substitutions which can be useful to optimize CSE.
Examples
>>> from sympy.simplify.cse_main import opt_cse >>> from sympy.abc import x >>> opt_subs = opt_cse([x**2]) >>> print(opt_subs) {x**(2): 1/(x**2)}
tree_cse¶

sympy.simplify.cse_main.
tree_cse
(exprs, symbols, opt_subs=None, order='canonical', ignore=())[source]¶ Perform raw CSE on expression tree, taking opt_subs into account.
Parameters: exprs : list of sympy expressions
The expressions to reduce.
symbols : infinite iterator yielding unique Symbols
The symbols used to label the common subexpressions which are pulled out.
opt_subs : dictionary of expression substitutions
The expressions to be substituted before any CSE action is performed.
order : string, ‘none’ or ‘canonical’
The order by which Mul and Add arguments are processed. For large expressions where speed is a concern, use the setting order=’none’.
ignore : iterable of Symbols
Substitutions containing any Symbol from
ignore
will be ignored.
Hypergeometric Function Expansion¶
hyperexpand¶

sympy.simplify.hyperexpand.
hyperexpand
(f, allow_hyper=False, rewrite='default', place=None)[source]¶ Expand hypergeometric functions. If allow_hyper is True, allow partial simplification (that is a result different from input, but still containing hypergeometric functions).
If a Gfunction has expansions both at zero and at infinity,
place
can be set to0
orzoo
to indicate the preferred choice.Examples
>>> from sympy.simplify.hyperexpand import hyperexpand >>> from sympy.functions import hyper >>> from sympy.abc import z >>> hyperexpand(hyper([], [], z)) exp(z)
Nonhyperegeometric parts of the expression and hypergeometric expressions that are not recognised are left unchanged:
>>> hyperexpand(1 + hyper([1, 1, 1], [], z)) hyper((1, 1, 1), (), z) + 1
Traversal Tools¶
use¶

sympy.simplify.traversaltools.
use
(expr, func, level=0, args=(), kwargs={})[source]¶ Use
func
to transformexpr
at the given level.Examples
>>> from sympy import use, expand >>> from sympy.abc import x, y
>>> f = (x + y)**2*x + 1
>>> use(f, expand, level=2) x*(x**2 + 2*x*y + y**2) + 1 >>> expand(f) x**3 + 2*x**2*y + x*y**2 + 1
EPath Tools¶
EPath class¶

class
sympy.simplify.epathtools.
EPath
[source]¶ Manipulate expressions using paths.
EPath grammar in EBNF notation:
literal ::= /[AZaz_][AZaz_09]*/ number ::= /?\d+/ type ::= literal attribute ::= literal "?" all ::= "*" slice ::= "[" number? (":" number? (":" number?)?)? "]" range ::= all  slice query ::= (type  attribute) ("" (type  attribute))* selector ::= range  query range? path ::= "/" selector ("/" selector)*
See the docstring of the epath() function.

apply
(expr, func, args=None, kwargs=None)[source]¶ Modify parts of an expression selected by a path.
Examples
>>> from sympy.simplify.epathtools import EPath >>> from sympy import sin, cos, E >>> from sympy.abc import x, y, z, t
>>> path = EPath("/*/[0]/Symbol") >>> expr = [((x, 1), 2), ((3, y), z)]
>>> path.apply(expr, lambda expr: expr**2) [((x**2, 1), 2), ((3, y**2), z)]
>>> path = EPath("/*/*/Symbol") >>> expr = t + sin(x + 1) + cos(x + y + E)
>>> path.apply(expr, lambda expr: 2*expr) t + sin(2*x + 1) + cos(2*x + 2*y + E)

select
(expr)[source]¶ Retrieve parts of an expression selected by a path.
Examples
>>> from sympy.simplify.epathtools import EPath >>> from sympy import sin, cos, E >>> from sympy.abc import x, y, z, t
>>> path = EPath("/*/[0]/Symbol") >>> expr = [((x, 1), 2), ((3, y), z)]
>>> path.select(expr) [x, y]
>>> path = EPath("/*/*/Symbol") >>> expr = t + sin(x + 1) + cos(x + y + E)
>>> path.select(expr) [x, x, y]

epath¶

sympy.simplify.epathtools.
epath
(path, expr=None, func=None, args=None, kwargs=None)[source]¶ Manipulate parts of an expression selected by a path.
This function allows to manipulate large nested expressions in single line of code, utilizing techniques to those applied in XML processing standards (e.g. XPath).
If
func
isNone
,epath()
retrieves elements selected by thepath
. Otherwise it appliesfunc
to each matching element.Note that it is more efficient to create an EPath object and use the select and apply methods of that object, since this will compile the path string only once. This function should only be used as a convenient shortcut for interactive use.
This is the supported syntax:
 select all:
/*
Equivalent of
for arg in args:
.
 select all:
 select slice:
/[0]
or/[1:5]
or/[1:5:2]
Supports standard Python’s slice syntax.
 select slice:
 select by type:
/list
or/listtuple
Emulates
isinstance()
.
 select by type:
 select by attribute:
/__iter__?
Emulates
hasattr()
.
 select by attribute:
Parameters: path : str  EPath
A path as a string or a compiled EPath.
expr : Basic  iterable
An expression or a container of expressions.
func : callable (optional)
A callable that will be applied to matching parts.
args : tuple (optional)
Additional positional arguments to
func
.kwargs : dict (optional)
Additional keyword arguments to
func
.Examples
>>> from sympy.simplify.epathtools import epath >>> from sympy import sin, cos, E >>> from sympy.abc import x, y, z, t
>>> path = "/*/[0]/Symbol" >>> expr = [((x, 1), 2), ((3, y), z)]
>>> epath(path, expr) [x, y] >>> epath(path, expr, lambda expr: expr**2) [((x**2, 1), 2), ((3, y**2), z)]
>>> path = "/*/*/Symbol" >>> expr = t + sin(x + 1) + cos(x + y + E)
>>> epath(path, expr) [x, x, y] >>> epath(path, expr, lambda expr: 2*expr) t + sin(2*x + 1) + cos(2*x + 2*y + E)