SciPy_SymPy/tutorial_sphinx/src/tutorial/simplification.rst

.. _tutorial-simplify:

================
 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
<http://en.wikipedia.org/wiki/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 well-defined 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``, which 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

.. TODO: Discuss coeff method in more detail in some other section (maybe
   basic expression manipulation tools)

``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
<http://en.wikipedia.org/wiki/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)
   π
   ─
   2

.. TODO: Can we actually do anything with inverse trig functions,
   simplification wise?

``trigsimp``
------------

To simplify expressions using trigonometric identities, use ``trigsimp``.

    >>> trigsimp(sin(x)**2 + cos(x)**2)
    1

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(cos(2*x))
         2
    2⋅cos (x) - 1

.. TODO: It would be much better to teach individual trig rewriting functions
   here, but they don't exist yet.  See
   https://code.google.com/p/sympy/issues/detail?id=357.

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

1. `x^ax^b = x^{a + b}`
2. `x^ay^a = (xy)^a`
3. `(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\cdot-1} = \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 at
least if `x` is nonnegative or `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                                                       |
+=======================+====================================+====================================================+=============================================================================+
|1. `x^ax^b = x^{a + b}`|Always true                         |None                                                |None                                                                         |
+-----------------------+------------------------------------+----------------------------------------------------+-----------------------------------------------------------------------------+
|2. `x^ay^a = (xy)^a`   |`x, y \geq 0` and `a \in \mathbb{R}`|`(-1)^{1/2}(-1)^{1/2} \neq (-1\cdot-1)^{1/2}`       |`\sqrt{x}\sqrt{y} \neq \sqrt{xy}` in general                                 |
+-----------------------+------------------------------------+----------------------------------------------------+-----------------------------------------------------------------------------+
|3. `(x^a)^b = x^{ab}`  |`x \geq 0` or `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 that ``x``
  and ``y`` are positive, and that ``a`` and ``b`` are real.  We will leave
  ``z``, ``t``, and ``c`` 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')

  .. TODO: Rewrite this using the new assumptions

.. 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)
      ___   ___
    ╲╱ x ⋅╲╱ y

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))
      ___   ___
    ╲╱ x ⋅╲╱ y

``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

1. `\log{(xy)} = \log{(x)} + \log{(y)}`
2. `\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)
    >>> z, t = symbols('z t')

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`` 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 in the SymPy documentation.

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 <http://en.wikipedia.org/wiki/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
<http://en.wikipedia.org/wiki/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
<http://en.wikipedia.org/wiki/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)
    Γ(z)

The `generalized hypergeometric function
<http://en.wikipedia.org/wiki/Generalized_hypergeometric_function>`_ is
``hyper``.  ``hyper([a_1, ..., a_p], [b_1, ..., b_q], z)`` represents
`{}_pF_q\left(\begin{matrix} a_1, \dots, a_p \\ b_1, \dots, b_q \end{matrix}
\middle| z \right)`.  The most common case is `{}_2F_1`, which is often
referred to as the `ordinary hypergeometric function
<http://en.wikipedia.org/wiki/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)
    Γ(x + 1)

For some tips on applying more targeted rewriting, see the
:ref:`tutorial-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)⋅Γ(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 G-function (see
:py:meth:`its documentation <sympy.functions.special.hyper.meijerg>` for more information).

    >>> expr = meijerg([[1],[1]], [[1],[]], -z)
    >>> expr
    ╭─╮1, 1 ⎛1  1 │   ⎞
    │╶┐     ⎜     │ -z⎟
    ╰─╯2, 1 ⎝1    │   ⎠
    >>> hyperexpand(expr)
     1
     ─
     z
    ℯ

``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))
       π
    ────────
    sin(π⋅x)

Example: Continued Fractions
============================

Let's use SymPy to explore continued fractions.  A `continued fraction
<http://en.wikipedia.org/wiki/Continued_fraction>`_ is an expression of the
form

.. math::

   a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{ \ddots + \cfrac{1}{a_n}
   }}}

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
    (a₀, a₁, a₂, a₃, a₄)
    >>> a0, a1, a2, a3, a4 = syms
    >>> frac = list_to_frac(syms)
    >>> frac
                 1
    a₀ + ─────────────────
                   1
         a₁ + ────────────
                      1
              a₂ + ───────
                        1
                   a₃ + ──
                        a₄

This form is useful for understanding continued fractions, but lets put it
into standard rational function form using ``cancel``.

    >>> frac = cancel(frac)
    >>> frac
    a₀⋅a₁⋅a₂⋅a₃⋅a₄ + a₀⋅a₁⋅a₂ + a₀⋅a₁⋅a₄ + a₀⋅a₃⋅a₄ + a₀ + a₂⋅a₃⋅a₄ + a₂ + a₄
    ─────────────────────────────────────────────────────────────────────────
                     a₁⋅a₂⋅a₃⋅a₄ + a₁⋅a₂ + a₁⋅a₄ + a₃⋅a₄ + 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
                    a₂⋅a₃⋅a₄ + a₂ + a₄
    a₀ + ───────────────────────────────────────
         a₁⋅a₂⋅a₃⋅a₄ + a₁⋅a₂ + a₁⋅a₄ + a₃⋅a₄ + 1
    >>> l.append(a0)
    >>> frac = 1/(frac - a0)
    >>> frac
    a₁⋅a₂⋅a₃⋅a₄ + a₁⋅a₂ + a₁⋅a₄ + a₃⋅a₄ + 1
    ───────────────────────────────────────
               a₂⋅a₃⋅a₄ + a₂ + a₄

Now we repeat this process

    >>> frac = apart(frac, a1)
    >>> frac
             a₃⋅a₄ + 1
    a₁ + ──────────────────
         a₂⋅a₃⋅a₄ + a₂ + a₄
    >>> l.append(a1)
    >>> frac = 1/(frac - a1)
    >>> frac = apart(frac, a2)
    >>> frac
             a₄
    a₂ + ─────────
         a₃⋅a₄ + 1
    >>> l.append(a2)
    >>> frac = 1/(frac - a2)
    >>> frac = apart(frac, a3)
    >>> frac
         1
    a₃ + ──
         a₄
    >>> l.append(a3)
    >>> frac = 1/(frac - a3)
    >>> frac = apart(frac, a4)
    >>> frac
    a₄
    >>> l.append(a4)
    >>> list_to_frac(l)
                 1
    a₀ + ─────────────────
                   1
         a₁ + ────────────
                      1
              a₂ + ───────
                        1
                   a₃ + ──
                        a₄


.. sidebar:: Quick Tip

   You can execute multiple lines at once in SymPy Live.  Typing
   ``Shift-Enter`` instead of ``Enter`` will enter a newline instead of
   executing.

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).

.. Answer: a0 is the only symbol that does not appear in the denominator