Simplification�
To make this document easier to read, we are going to enable pretty printing.
>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> init_printing(use_unicode=True)
simplify
�
Now let�s jump in and do some interesting mathematics. One of the most useful
features of a symbolic manipulation system is the ability to simplify
mathematical expressions. SymPy has dozens of functions to perform various
kinds of simplification. There is also one general function called
simplify()
that attempts to apply all of these functions in an intelligent
way to arrive at the simplest form of an expression. Here are some examples
>>> simplify(sin(x)**2 + cos(x)**2)
1
>>> simplify((x**3 + x**2  x  1)/(x**2 + 2*x + 1))
x  1
>>> simplify(gamma(x)/gamma(x  2))
(x  2)?(x  1)
Here, gamma(x)
is \(\Gamma(x)\), the gamma function. We see that simplify()
is capable of handling a large class of expressions.
But simplify()
has a pitfall. It just applies all the major
simplification operations in SymPy, and uses heuristics to determine the
simplest result. But �simplest� is not a welldefined term. For example, say
we wanted to �simplify� \(x^2 + 2x + 1\) into \((x + 1)^2\):
>>> simplify(x**2 + 2*x + 1)
2
x + 2?x + 1
We did not get what we want. There is a function to perform this
simplification, called factor()
, which will be discussed below.
Another pitfall to simplify()
is that it can be unnecessarily slow, since
it tries many kinds of simplifications before picking the best one. If you
already know exactly what kind of simplification you are after, it is better
to apply the specific simplification function(s) that apply those
simplifications.
Applying specific simplification functions instead of simplify()
also has
the advantage that specific functions have certain guarantees about the form
of their output. These will be discussed with each function below. For
example, factor()
, when called on a polynomial with rational coefficients,
is guaranteed to factor the polynomial into irreducible factors.
simplify()
has no guarantees. It is entirely heuristical, and, as we saw
above, it may even miss a possible type of simplification that SymPy is
capable of doing.
simplify()
is best when used interactively, when you just want to whittle
down an expression to a simpler form. You may then choose to apply specific
functions once you see what simplify()
returns, to get a more precise
result. It is also useful when you have no idea what form an expression will
take, and you need a catchall function to simplify it.
Polynomial/Rational Function Simplification�
expand�
expand()
is one of the most common simplification functions in SymPy.
Although it has a lot of scopes, for now, we will consider its function in
expanding polynomial expressions. For example:
>>> expand((x + 1)**2)
2
x + 2?x + 1
>>> expand((x + 2)*(x  3))
2
x  x  6
Given a polynomial, expand()
will put it into a canonical form of a sum of
monomials.
expand()
may not sound like a simplification function. After all, by its
very name, it makes expressions bigger, not smaller. Usually this is the
case, but often an expression will become smaller upon calling expand()
on
it due to cancellation.
>>> expand((x + 1)*(x  2)  (x  1)*x)
2
factor�
factor()
takes a polynomial and factors it into irreducible factors over
the rational numbers. For example:
>>> factor(x**3  x**2 + x  1)
? 2 ?
(x  1)??x + 1?
>>> factor(x**2*z + 4*x*y*z + 4*y**2*z)
2
z?(x + 2?y)
For polynomials, factor()
is the opposite of expand()
. factor()
uses a complete multivariate factorization algorithm over the rational
numbers, which means that each of the factors returned by factor()
is
guaranteed to be irreducible.
If you are interested in the factors themselves, factor_list
returns a
more structured output.
>>> factor_list(x**2*z + 4*x*y*z + 4*y**2*z)
(1, [(z, 1), (x + 2?y, 2)])
Note that the input to factor
and expand
need not be polynomials in
the strict sense. They will intelligently factor or expand any kind of
expression (though note that the factors may not be irreducible if the input
is no longer a polynomial over the rationals).
>>> expand((cos(x) + sin(x))**2)
2 2
sin (x) + 2?sin(x)?cos(x) + cos (x)
>>> factor(cos(x)**2 + 2*cos(x)*sin(x) + sin(x)**2)
2
(sin(x) + cos(x))
collect�
collect()
collects common powers of a term in an expression. For example
>>> expr = x*y + x  3 + 2*x**2  z*x**2 + x**3
>>> expr
3 2 2
x  x ?z + 2?x + x?y + x  3
>>> collected_expr = collect(expr, x)
>>> collected_expr
3 2
x + x ?(z + 2) + x?(y + 1)  3
collect()
is particularly useful in conjunction with the .coeff()
method. expr.coeff(x, n)
gives the coefficient of x**n
in expr
:
>>> collected_expr.coeff(x, 2)
z + 2
cancel�
cancel()
will take any rational function and put it into the standard
canonical form, \(\frac{p}{q}\), where \(p\) and \(q\) are expanded polynomials with
no common factors, and the leading coefficients of \(p\) and \(q\) do not have
denominators (i.e., are integers).
>>> cancel((x**2 + 2*x + 1)/(x**2 + x))
x + 1

x
>>> expr = 1/x + (3*x/2  2)/(x  4)
>>> expr
3?x
  2
2 1
 + 
x  4 x
>>> cancel(expr)
2
3?x  2?x  8

2
2?x  8?x
>>> expr = (x*y**2  2*x*y*z + x*z**2 + y**2  2*y*z + z**2)/(x**2  1)
>>> expr
2 2 2 2
x?y  2?x?y?z + x?z + y  2?y?z + z

2
x  1
>>> cancel(expr)
2 2
y  2?y?z + z

x  1
Note that since factor()
will completely factorize both the numerator and
the denominator of an expression, it can also be used to do the same thing:
>>> factor(expr)
2
(y  z)

x  1
However, if you are only interested in making sure that the expression is in
canceled form, cancel()
is more efficient than factor()
.
apart�
apart()
performs a partial fraction decomposition on a rational
function.
>>> expr = (4*x**3 + 21*x**2 + 10*x + 12)/(x**4 + 5*x**3 + 5*x**2 + 4*x)
>>> expr
3 2
4?x + 21?x + 10?x + 12

4 3 2
x + 5?x + 5?x + 4?x
>>> apart(expr)
2?x  1 1 3
   + 
2 x + 4 x
x + x + 1
Trigonometric Simplification�
Note
SymPy follows Python�s naming conventions for inverse trigonometric
functions, which is to append an a
to the front of the function�s
name. For example, the inverse cosine, or arc cosine, is called acos()
.
>>> acos(x)
acos(x)
>>> cos(acos(x))
x
>>> asin(1)
?

2
trigsimp�
To simplify expressions using trigonometric identities, use trigsimp()
.
>>> trigsimp(sin(x)**2 + cos(x)**2)
1
>>> trigsimp(sin(x)**4  2*cos(x)**2*sin(x)**2 + cos(x)**4)
cos(4?x) 1
 + 
2 2
>>> trigsimp(sin(x)*tan(x)/sec(x))
2
sin (x)
trigsimp()
also works with hyperbolic trig functions.
>>> trigsimp(cosh(x)**2 + sinh(x)**2)
cosh(2?x)
>>> trigsimp(sinh(x)/tanh(x))
cosh(x)
Much like simplify()
, trigsimp()
applies various trigonometric identities to
the input expression, and then uses a heuristic to return the �best� one.
expand_trig�
To expand trigonometric functions, that is, apply the sum or double angle
identities, use expand_trig()
.
>>> expand_trig(sin(x + y))
sin(x)?cos(y) + sin(y)?cos(x)
>>> expand_trig(tan(2*x))
2?tan(x)

2
 tan (x) + 1
Because expand_trig()
tends to make trigonometric expressions larger, and
trigsimp()
tends to make them smaller, these identities can be applied in
reverse using trigsimp()
>>> trigsimp(sin(x)*cos(y) + sin(y)*cos(x))
sin(x + y)
Powers�
Before we introduce the power simplification functions, a mathematical discussion on the identities held by powers is in order. There are three kinds of identities satisfied by exponents
 \(x^ax^b = x^{a + b}\)
 \(x^ay^a = (xy)^a\)
 \((x^a)^b = x^{ab}\)
Identity 1 is always true.
Identity 2 is not always true. For example, if \(x = y = 1\) and \(a = \frac{1}{2}\), then \(x^ay^a = \sqrt{1}\sqrt{1} = i\cdot i = 1\), whereas \((xy)^a = \sqrt{1\cdot1} = \sqrt{1} = 1\). However, identity 2 is true at least if \(x\) and \(y\) are nonnegative and \(a\) is real (it may also be true under other conditions as well). A common consequence of the failure of identity 2 is that \(\sqrt{x}\sqrt{y} \neq \sqrt{xy}\).
Identity 3 is not always true. For example, if \(x = 1\), \(a = 2\), and \(b = \frac{1}{2}\), then \((x^a)^b = {\left ((1)^2\right )}^{1/2} = \sqrt{1} = 1\) and \(x^{ab} = (1)^{2\cdot1/2} = (1)^1 = 1\). However, identity 3 is true when \(b\) is an integer (again, it may also hold in other cases as well). Two common consequences of the failure of identity 3 are that \(\sqrt{x^2}\neq x\) and that \(\sqrt{\frac{1}{x}} \neq \frac{1}{\sqrt{x}}\).
To summarize
Identity  Sufficient conditions to hold  Counterexample when conditions are not met  Important consequences 


Always true  None  None 

\(x, y \geq 0\) and \(a \in \mathbb{R}\)  \((1)^{1/2}(1)^{1/2} \neq (1\cdot1)^{1/2}\)  \(\sqrt{x}\sqrt{y} \neq \sqrt{xy}\) in general 

\(b \in \mathbb{Z}\)  \({\left((1)^2\right )}^{1/2} \neq (1)^{2\cdot1/2}\)  \(\sqrt{x^2}\neq x\) and \(\sqrt{\frac{1}{x}}\neq\frac{1}{\sqrt{x}}\) in general 
This is important to remember, because by default, SymPy will not perform simplifications if they are not true in general.
In order to make SymPy perform simplifications involving identities that are only true under certain assumptions, we need to put assumptions on our Symbols. We will undertake a full discussion of the assumptions system later, but for now, all we need to know are the following.
By default, SymPy Symbols are assumed to be complex (elements of \(\mathbb{C}\)). That is, a simplification will not be applied to an expression with a given Symbol unless it holds for all complex numbers.
Symbols can be given different assumptions by passing the assumption to
symbols()
. For the rest of this section, we will be assuming thatx
andy
are positive, and thata
andb
are real. We will leavez
,t
, andc
as arbitrary complex Symbols to demonstrate what happens in that case.>>> x, y = symbols('x y', positive=True) >>> a, b = symbols('a b', real=True) >>> z, t, c = symbols('z t c')
Note
In SymPy, sqrt(x)
is just a shortcut to x**Rational(1, 2)
. They
are exactly the same object.
>>> sqrt(x) == x**Rational(1, 2)
True
powsimp�
powsimp()
applies identities 1 and 2 from above, from left to right.
>>> powsimp(x**a*x**b)
a + b
x
>>> powsimp(x**a*y**a)
a
(x?y)
Notice that powsimp()
refuses to do the simplification if it is not valid.
>>> powsimp(t**c*z**c)
c c
t ?z
If you know that you want to apply this simplification, but you don�t want to
mess with assumptions, you can pass the force=True
flag. This will force
the simplification to take place, regardless of assumptions.
>>> powsimp(t**c*z**c, force=True)
c
(t?z)
Note that in some instances, in particular, when the exponents are integers or rational numbers, and identity 2 holds, it will be applied automatically.
>>> (z*t)**2
2 2
t ?z
>>> sqrt(x*y)
vx?vy
This means that it will be impossible to undo this identity with
powsimp()
, because even if powsimp()
were to put the bases together,
they would be automatically split apart again.
>>> powsimp(z**2*t**2)
2 2
t ?z
>>> powsimp(sqrt(x)*sqrt(y))
vx?vy
expand_power_exp / expand_power_base�
expand_power_exp()
and expand_power_base()
apply identities 1 and 2
from right to left, respectively.
>>> expand_power_exp(x**(a + b))
a b
x ?x
>>> expand_power_base((x*y)**a)
a a
x ?y
As with powsimp()
, identity 2 is not applied if it is not valid.
>>> expand_power_base((z*t)**c)
c
(t?z)
And as with powsimp()
, you can force the expansion to happen without
fiddling with assumptions by using force=True
.
>>> expand_power_base((z*t)**c, force=True)
c c
t ?z
As with identity 2, identity 1 is applied automatically if the power is a
number, and hence cannot be undone with expand_power_exp()
.
>>> x**2*x**3
5
x
>>> expand_power_exp(x**5)
5
x
powdenest�
powdenest()
applies identity 3, from left to right.
>>> powdenest((x**a)**b)
a?b
x
As before, the identity is not applied if it is not true under the given assumptions.
>>> powdenest((z**a)**b)
b
? a?
?z ?
And as before, this can be manually overridden with force=True
.
>>> powdenest((z**a)**b, force=True)
a?b
z
Exponentials and logarithms�
Note
In SymPy, as in Python and most programming languages, log
is the
natural logarithm, also known as ln
. SymPy automatically provides an
alias ln = log
in case you forget this.
>>> ln(x)
log(x)
Logarithms have similar issues as powers. There are two main identities
 \(\log{(xy)} = \log{(x)} + \log{(y)}\)
 \(\log{(x^n)} = n\log{(x)}\)
Neither identity is true for arbitrary complex \(x\) and \(y\), due to the branch cut in the complex plane for the complex logarithm. However, sufficient conditions for the identities to hold are if \(x\) and \(y\) are positive and \(n\) is real.
>>> x, y = symbols('x y', positive=True)
>>> n = symbols('n', real=True)
As before, z
and t
will be Symbols with no additional assumptions.
Note that the identity \(\log{\left (\frac{x}{y}\right )} = \log(x)  \log(y)\) is a special case of identities 1 and 2 by \(\log{\left (\frac{x}{y}\right )} =\) \(\log{\left (x\cdot\frac{1}{y}\right )} =\) \(\log(x) + \log{\left( y^{1}\right )} =\) \(\log(x)  \log(y)\), and thus it also holds if \(x\) and \(y\) are positive, but may not hold in general.
We also see that \(\log{\left( e^x \right)} = x\) comes from \(\log{\left ( e^x \right)} = x\log(e) = x\), and thus holds when \(x\) is real (and it can be verified that it does not hold in general for arbitrary complex \(x\), for example, \(\log{\left (e^{x + 2\pi i}\right)} = \log{\left (e^x\right )} = x \neq x + 2\pi i\)).
expand_log�
To apply identities 1 and 2 from left to right, use expand_log()
. As
always, the identities will not be applied unless they are valid.
>>> expand_log(log(x*y))
log(x) + log(y)
>>> expand_log(log(x/y))
log(x)  log(y)
>>> expand_log(log(x**2))
2?log(x)
>>> expand_log(log(x**n))
n?log(x)
>>> expand_log(log(z*t))
log(t?z)
As with powsimp()
and powdenest()
, expand_log()
has a force
option that can be used to ignore assumptions.
>>> expand_log(log(z**2))
? 2?
log?z ?
>>> expand_log(log(z**2), force=True)
2?log(z)
logcombine�
To apply identities 1 and 2 from right to left, use logcombine()
.
>>> logcombine(log(x) + log(y))
log(x?y)
>>> logcombine(n*log(x))
? n?
log?x ?
>>> logcombine(n*log(z))
n?log(z)
logcombine()
also has a force
option that can be used to ignore
assumptions.
>>> logcombine(n*log(z), force=True)
? n?
log?z ?
Special Functions�
SymPy implements dozens of special functions, ranging from functions in combinatorics to mathematical physics.
An extensive list of the special functions included with SymPy and their documentation is at the Functions Module page.
For the purposes of this tutorial, let�s introduce a few special functions in SymPy.
Let�s define x
, y
, and z
as regular, complex Symbols, removing any
assumptions we put on them in the previous section. We will also define k
,
m
, and n
.
>>> x, y, z = symbols('x y z')
>>> k, m, n = symbols('k m n')
The factorial function is
factorial
. factorial(n)
represents \(n!= 1\cdot2\cdots(n  1)\cdot
n\). \(n!\) represents the number of permutations of \(n\) distinct items.
>>> factorial(n)
n!
The binomial coefficient function is
binomial
. binomial(n, k)
represents \(\binom{n}{k}\), the number of
ways to choose \(k\) items from a set of \(n\) distinct items. It is also often
written as \(nCk\), and is pronounced �\(n\) choose \(k\)�.
>>> binomial(n, k)
?n?
? ?
?k?
The factorial function is closely related to the gamma function, gamma
. gamma(z)
represents \(\Gamma(z) = \int_0^\infty t^{z  1}e^{t}\,dt\), which for positive integer
\(z\) is the same as \((z  1)!\).
>>> gamma(z)
?(z)
The generalized hypergeometric function is
hyper
. hyper([a_1, ..., a_p], [b_1, ..., b_q], z)
represents
\({}_pF_q\left(\begin{matrix} a_1, \cdots, a_p \\ b_1, \cdots, b_q \end{matrix}
\middle z \right)\). The most common case is \({}_2F_1\), which is often
referred to as the ordinary hypergeometric function.
>>> hyper([1, 2], [3], z)
 ?1, 2 � ?
+ ? � z?
2? 1 ? 3 � ?
rewrite�
A common way to deal with special functions is to rewrite them in terms of one
another. This works for any function in SymPy, not just special functions.
To rewrite an expression in terms of a function, use
expr.rewrite(function)
. For example,
>>> tan(x).rewrite(sin)
2
2?sin (x)

sin(2?x)
>>> factorial(x).rewrite(gamma)
?(x + 1)
For some tips on applying more targeted rewriting, see the Advanced Expression Manipulation section.
expand_func�
To expand special functions in terms of some identities, use
expand_func()
. For example
>>> expand_func(gamma(x + 3))
x?(x + 1)?(x + 2)??(x)
hyperexpand�
To rewrite hyper
in terms of more standard functions, use
hyperexpand()
.
>>> hyperexpand(hyper([1, 1], [2], z))
log(z + 1)

z
hyperexpand()
also works on the more general Meijer Gfunction (see
its documentation
for more
information).
>>> expr = meijerg([[1],[1]], [[1],[]], z)
>>> expr
??1, 1 ?1 1 � ?
�?� ? � z?
??2, 1 ?1 � ?
>>> hyperexpand(expr)
1

z
?
combsimp�
To simplify combinatorial expressions, use combsimp()
.
>>> combsimp(factorial(n)/factorial(n  3))
n?(n  2)?(n  1)
>>> combsimp(binomial(n+1, k+1)/binomial(n, k))
n + 1

k + 1
combsimp()
also simplifies expressions with gamma
.
>>> combsimp(gamma(x)*gamma(1  x))
?

sin(??x)
Example: Continued Fractions�
Let�s use SymPy to explore continued fractions. A continued fraction is an expression of the form
where \(a_0, \ldots, a_n\) are integers, and \(a_1, \ldots, a_n\) are positive. A continued fraction can also be infinite, but infinite objects are more difficult to represent in computers, so we will only examine the finite case here.
A continued fraction of the above form is often represented as a list \([a_0; a_1, \ldots, a_n]\). Let�s write a simple function that converts such a list to its continued fraction form. The easiest way to construct a continued fraction from a list is to work backwards. Note that despite the apparent symmetry of the definition, the first element, \(a_0\), must usually be handled differently from the rest.
>>> def list_to_frac(l):
... expr = Integer(0)
... for i in reversed(l[1:]):
... expr += i
... expr = 1/expr
... return l[0] + expr
>>> list_to_frac([x, y, z])
1
x + 
1
y + 
z
We use Integer(0)
in list_to_frac
so that the result will always be a
SymPy object, even if we only pass in Python ints.
>>> list_to_frac([1, 2, 3, 4])
43

30
Every finite continued fraction is a rational number, but we are interested in
symbolics here, so let�s create a symbolic continued fraction. The
symbols()
function that we have been using has a shortcut to create
numbered symbols. symbols('a0:5')
will create the symbols a0
, a1
,
�, a5
.
>>> syms = symbols('a0:5')
>>> syms
(a?, a?, a?, a?, a?)
>>> a0, a1, a2, a3, a4 = syms
>>> frac = list_to_frac(syms)
>>> frac
1
a? + 
1
a? + 
1
a? + 
1
a? + 
a?
This form is useful for understanding continued fractions, but lets put it
into standard rational function form using cancel()
.
>>> frac = cancel(frac)
>>> frac
a??a??a??a??a? + a??a??a? + a??a??a? + a??a??a? + a? + a??a??a? + a? + a?

a??a??a??a? + a??a? + a??a? + a??a? + 1
Now suppose we were given frac
in the above canceled form. In fact, we
might be given the fraction in any form, but we can always put it into the
above canonical form with cancel()
. Suppose that we knew that it could be
rewritten as a continued fraction. How could we do this with SymPy? A
continued fraction is recursively \(c + \frac{1}{f}\), where \(c\) is an integer
and \(f\) is a (smaller) continued fraction. If we could write the expression
in this form, we could pull out each \(c\) recursively and add it to a list. We
could then get a continued fraction with our list_to_frac()
function.
The key observation here is that we can convert an expression to the form \(c +
\frac{1}{f}\) by doing a partial fraction decomposition with respect to
\(c\). This is because \(f\) does not contain \(c\). This means we need to use the
apart()
function. We use apart()
to pull the term out, then subtract
it from the expression, and take the reciprocal to get the \(f\) part.
>>> l = []
>>> frac = apart(frac, a0)
>>> frac
a??a??a? + a? + a?
a? + 
a??a??a??a? + a??a? + a??a? + a??a? + 1
>>> l.append(a0)
>>> frac = 1/(frac  a0)
>>> frac
a??a??a??a? + a??a? + a??a? + a??a? + 1

a??a??a? + a? + a?
Now we repeat this process
>>> frac = apart(frac, a1)
>>> frac
a??a? + 1
a? + 
a??a??a? + a? + a?
>>> l.append(a1)
>>> frac = 1/(frac  a1)
>>> frac = apart(frac, a2)
>>> frac
a?
a? + 
a??a? + 1
>>> l.append(a2)
>>> frac = 1/(frac  a2)
>>> frac = apart(frac, a3)
>>> frac
1
a? + 
a?
>>> l.append(a3)
>>> frac = 1/(frac  a3)
>>> frac = apart(frac, a4)
>>> frac
a?
>>> l.append(a4)
>>> list_to_frac(l)
1
a? + 
1
a? + 
1
a? + 
1
a? + 
a?
Of course, this exercise seems pointless, because we already know that our
frac
is list_to_frac([a0, a1, a2, a3, a4])
. So try the following
exercise. Take a list of symbols and randomize them, and create the canceled
continued fraction, and see if you can reproduce the original list. For
example
>>> import random
>>> l = list(symbols('a0:5'))
>>> random.shuffle(l)
>>> orig_frac = frac = cancel(list_to_frac(l))
>>> del l
Click on �Run code block in SymPy Live� on the definition of list_to_frac()
above, and then on the above example, and try to reproduce l
from
frac
. I have deleted l
at the end to remove the temptation for
peeking (you can check your answer at the end by calling
cancel(list_to_frac(l))
on the list that you generate at the end, and
comparing it to orig_frac
.
See if you can think of a way to figure out what symbol to pass to apart()
at each stage (hint: think of what happens to \(a_0\) in the formula \(a_0 +
\frac{1}{a_1 + \cdots}\) when it is canceled).