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# Simplify¶

## simplify¶

sympy.simplify.simplify.simplify(expr, ratio=1.7)[source]

Naively 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.

In some cases, applying simplify() may actually result in some more complicated expression. By default ratio=1.7 prevents more extreme cases: if (result length)/(input length) > ratio, then input is returned unmodified (count_ops() is used to measure length).

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

>>> from sympy import S, simplify, count_ops, oo
>>> root = S("(5/2 + 21**(1/2)/2)**(1/3)*(1/2 - I*3**(1/2)/2)"
... "+ 1/((1/2 - I*3**(1/2)/2)*(5/2 + 21**(1/2)/2)**(1/3))")


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

>>> simplify(root, ratio=1) is 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, default value ratio=1.7 seems like a reasonable choice.

## collect¶

sympy.simplify.simplify.collect(expr, syms, evaluate=True, exact=False)[source]

Collect additive terms with respect to a list of symbols 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.

This function will not apply any redundant expanding to the input expression, so user is assumed to enter expression in final form. This makes ‘collect’ more predictable as there is no magic behind the scenes. However it is important to note, that powers of products are converted to products of powers using ‘separate’ function.

There are two possible types of output. First, if ‘evaluate’ flag is set, this function will return a single expression or else it will return a dictionary with separated symbols up to rational powers as keys and collected sub-expressions as values respectively.

>>> from sympy import collect, sympify, Wild
>>> from sympy.abc import a, b, c, x, y, z


This function can collect symbolic coefficients in polynomial 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[sympify(1)]
c


You can also work with multi-variate 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)
(a + b)*(x**2)**c


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,x)
(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


Note: arguments are expected to be in expanded form, so you might have to call expand() prior to calling this function.

## separate¶

sympy.simplify.simplify.separate(expr, deep=False, force=False)[source]

A wrapper to expand(power_base=True) which separates a power with a base that is a Mul into a product of powers, without performing any other expansions, provided that assumptions about the power’s base and exponent allow.

deep=True (default is False) will do separations inside functions.

force=True (default is False) will cause the expansion to ignore assumptions about the base and exponent. When False, the expansion will only happen if the base is non-negative or the exponent is an integer.

>>> from sympy.abc import x, y, z
>>> from sympy import separate, sin, cos, exp

>>> (x*y)**2
x**2*y**2

>>> (2*x)**y
(2*x)**y
>>> separate(_)
2**y*x**y

>>> separate((x*y)**z)
(x*y)**z
>>> separate((x*y)**z, force=True)
x**z*y**z
>>> separate(sin((x*y)**z))
sin((x*y)**z)
>>> separate(sin((x*y)**z), deep=True, force=True)
sin(x**z*y**z)

>>> separate((2*sin(x))**y + (2*cos(x))**y)
2**y*sin(x)**y + 2**y*cos(x)**y

>>> separate((2*exp(y))**x)
2**x*exp(x*y)

>>> separate((2*cos(x))**y)
2**y*cos(x)**y


Notice that summations are left untouched. If this is not the desired behavior, apply ‘expand’ to the expression:

>>> separate(((x+y)*z)**2)
z**2*(x + y)**2
>>> (((x+y)*z)**2).expand()
x**2*z**2 + 2*x*y*z**2 + y**2*z**2

>>> separate((2*y)**(1+z))
2**(z + 1)*y**(z + 1)
>>> ((2*y)**(1+z)).expand()
2*2**z*y*y**z


## radsimp¶

sympy.simplify.simplify.radsimp(expr)[source]

Rationalize the denominator.

Examples:
>>> from sympy import radsimp, sqrt, Symbol
>>> radsimp(1/(2+sqrt(2)))
-2**(1/2)/2 + 1
>>> x,y = map(Symbol, 'xy')
>>> e = ((2+2*sqrt(2))*x+(2+sqrt(8))*y)/(2+sqrt(2))
>>> radsimp(e)
2**(1/2)*x + 2**(1/2)*y


## ratsimp¶

sympy.simplify.simplify.ratsimp(expr)[source]

Put an expression over a common denominator, cancel and reduce.

== Examples ==
>>> from sympy import ratsimp
>>> from sympy.abc import x, y
>>> ratsimp(1/x + 1/y)
(x + y)/(x*y)


## fraction¶

sympy.simplify.simplify.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)


## trigsimp¶

sympy.simplify.simplify.trigsimp(expr, deep=False, recursive=False)[source]

== Usage ==

trigsimp(expr) -> reduces expression by using known trig identities

== Notes ==

deep: - Apply trigsimp inside functions

recursive: - Use common subexpression elimination (cse()) and apply trigsimp recursively (recursively==True is quite expensive operation if the expression is large)

== 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
>>> trigsimp(log(e))
log(2*sin(x)**2 + 2*cos(x)**2)
>>> trigsimp(log(e), deep=True)
log(2)


## powsimp¶

sympy.simplify.simplify.powsimp(expr, deep=False, combine='all', force=False)[source]
== Usage ==
powsimp(expr, deep) -> 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


## combsimp¶

sympy.simplify.simplify.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 result is be given in terms of binomials and factorials.

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 more familiar binomials and factorials.

All transformation rules can be found (or was derived from) here:

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)


## 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:

1. Rewrite all functions in terms of gamma, if possible.
2. Rewrite all occurrences of gamma in terms of products of gamma and rising factorial with integer, absolute constant exponent.
3. 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 m-fold Hypergeometric Summation,
Journal of Symbolic Computation (1995) 20, 399-417

## hypersimilar¶

sympy.simplify.simplify.hypersimilar(f, g, k)[source]

Returns True if ‘f’ and ‘g’ are hyper-similar.

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=False)[source]

Replace numbers with simple representations.

If rational=True then numbers are simply replaced with their rational equivalents.

If rational=False, a simple formula that numerically matches the given 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.

With full=True, a more extensive search is performed (this is useful to find simpler numbers when the tolerance is set low).

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*(5**(1/2)/10 + 1/4)**(1/2)
>>> nsimplify(I**I, [pi])
exp(-pi/2)
>>> nsimplify(pi, tolerance=0.01)
22/7


## Square Root Denest¶

### sqrtdenest¶

sympy.simplify.sqrtdenest.sqrtdenest(expr)[source]

Denests an expression that contains nested square roots. This algorithm is based on <http://www.almaden.ibm.com/cs/people/fagin/symb85.pdf>.

## Common Subexpresion Elimination¶

### cse¶

sympy.simplify.cse_main.cse(exprs, symbols=None, optimizations=None)[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, optional
The (preprocessor, postprocessor) pairs. If not provided, sympy.simplify.cse.cse_optimizations is used.

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.

## Hypergeometric Function Expansion¶

sympy.simplify.hyperexpand.hyperexpand(f, allow_hyper=False)[source]

Expand hypergeometric functions. If allow_hyper is True, allow partial simplification (that is a result different from input, but still containing hypergeometric functions).

Examples:

>>> from sympy.simplify.hyperexpand import hyperexpand
>>> from sympy.functions import hyper
>>> from sympy.abc import z
>>> hyperexpand(hyper([], [], z))
exp(z)


Non-hyperegeometric parts of the expression and hypergeometric expressions that are not recognised are left unchanged:

>>> hyperexpand(1 + hyper([1, 1, 1], [], z))
1 + hyper((1, 1, 1), (), z)


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