Simplification

From

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)

In Julia:

• " not ' are used for strings:
• pretty printing si enable by default
• we will deliberatately input extra functions from the sympy module such as powsimp, ...

julia> using SymPy

julia> import_from(sympy)

julia> @syms x, y, z
(x, y, z)

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)

In Julia:

• we need to load in SpecialFunctions to have access to gamma:

julia> simplify(sin(x)^2 + cos(x)^2)
1

julia> simplify((x^3 + x^2 - x - 1)/(x^2 + 2*x + 1))
x - 1

julia> using SpecialFunctions

julia> 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

In Julia:

julia> simplify(x^2 + 2*x + 1) |> string
"x^2 + 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

In Julia:

julia> expand((x + 1)^2) |> string
"x^2 + 2*x + 1"

julia> expand((x + 2)*(x - 3)) |> string
"x^2 - 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

In Julia:

julia> 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)

In Julia:

julia> factor(x^3 - x^2 + x - 1) |> string
"(x - 1)*(x^2 + 1)"
julia> factor(x^2*z + 4*x*y*z + 4*y^2*z) |>  string
"z*(x + 2*y)^2"

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)])

In Julia:

julia> factor_list(x^2*z + 4*x*y*z + 4*y^2*z)
(1, Tuple{Sym, Int64}[(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))

In Julia:

julia> expand((cos(x) + sin(x))^2) |> string
"sin(x)^2 + 2*sin(x)*cos(x) + cos(x)^2"

julia> factor(cos(x)^2 + 2*cos(x)*sin(x) + sin(x)^2) |> string
"(sin(x) + cos(x))^2"

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

In Julia:

julia> expr = x*y + x - 3 + 2*x^2 - z*x^2 + x^3
 3    2        2
x  - x ⋅z + 2⋅x  + x⋅y + x - 3

julia> collected_expr = collect(expr, x)
 3    2
x  + x ⋅(2 - z) + 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

In Julia:

julia> collected_expr.coeff(x, 2)
2 - z

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

In Julia:

julia> cancel((x^2 + 2*x + 1)/(x^2 + x))
x + 1
─────
  x  

julia> expr = 1/x + (3*x/2 - 2)/(x - 4)
3⋅x        
─── - 2    
 2        1
─────── + ─
 x - 4    x

julia> cancel(expr) |>  string
"(3*x^2 - 2*x - 8)/(2*x^2 - 8*x)"

julia> expr = (x*y^2 - 2*x*y*z + x*z^2 + y^2 - 2*y*z + z^2)/(x^2 - 1)
   2                2    2            2
x⋅y  - 2⋅x⋅y⋅z + x⋅z  + y  - 2⋅y⋅z + z 
───────────────────────────────────────
                  2                    
                 x  - 1     

julia> 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

In Julia:

julia> factor(expr)  |> string
"(y - z)^2/(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

In Julia:

julia> expr = (4*x^3 + 21*x^2 + 10*x + 12)/(x^4 + 5*x^3 + 5*x^2 + 4*x);  string(expr)
"(4*x^3 + 21*x^2 + 10*x + 12)/(x^4 + 5*x^3 + 5*x^2 + 4*x)"

julia> apart(expr)
 2⋅x - 1       1     3
────────── - ───── + ─
 2           x + 4   x
x  + x + 1            

Trigonometric Simplification

   >>> acos(x)
   acos(x)
   >>> cos(acos(x))
   x
   >>> asin(1)
   π
   ─
   2

In Julia:

julia> acos(x)
acos(x)

julia> cos(acos(x))
x

julia> asin(1)
1.5707963267948966

julia> sympy.asin(1)
π
─
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)

In Julia:

julia> trigsimp(sin(x)^2 + cos(x)^2)
1

julia> trigsimp(sin(x)^4 - 2*cos(x)^2*sin(x)^2 + cos(x)^4)
cos(4⋅x)   1
──────── + ─
   2       2

julia> trigsimp(sin(x)*tan(x)/sec(x)) |> string
"sin(x)^2"

trigsimp() also works with hyperbolic trig functions.

    >>> trigsimp(cosh(x)**2 + sinh(x)**2)
    cosh(2⋅x)
    >>> trigsimp(sinh(x)/tanh(x))
    cosh(x)

In Julia:

julia> trigsimp(cosh(x)^2 + sinh(x)^2)
cosh(2⋅x)

julia> 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

In Julia:

julia> expand_trig(sin(x + y))
sin(x)⋅cos(y) + sin(y)⋅cos(x)

julia> expand_trig(tan(2*x))
  2⋅tan(x) 
───────────
       2   
1 - tan (x)

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)

In Julia:

julia> 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

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 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

1. This: $x^ax^b = x^{a + b}$ is always true

2. This: $x^ay^a = (xy)^a$ is true when $x, y \geq 0$ and $a \in \mathbb{R}$; but note $(-1)^{1/2}(-1)^{1/2} \neq (-1\cdot-1)^{1/2}$ and $\sqrt{x}\sqrt{y} \neq \sqrt{xy}$ in general

3. This: $(x^a)^b = x^{ab}$ when $b \in \mathbb{Z}$; but note ${\left((-1)^2\right )}^{1/2} \neq (-1)^{2\cdot1/2}$ and $\sqrt{x^2}\neq x$ 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')

In Julia:

The same notation can be used:

julia> x, y = symbols("x y", positive=true)
(x, y)

julia> a, b = symbols("a b", real=true)
(a, b)

julia> z, t, c = symbols("z t c")
(z, t, c)

However, the recommended way is to use @syms for symbol construction with assumptions:

julia> @syms x::positive, y::positive
(x, y)

julia> @syms a::real, b::real
(a, b)

julia> @syms z, t, c
(z, t, c)



----

!!! note "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.

python >>> sqrt(x) == x**Rational(1, 2) True

##### In `Julia`:

* we can construction rational numbers with `//`

jldoctest simplification julia> sqrt(x) == x^(1//2) true

----

powsimp
-------

`powsimp()` applies identities 1 and 2 from above, from left to right.

python

powsimp(xa*xb)

 a + b
x

powsimp(xa*ya)

    a

(x⋅y)

##### In `Julia`:

jldoctest simplification julia> powsimp(x^a*x^b) a + b x

julia> powsimp(x^ay^a) |> string "(xy)^a"

----

Notice that `powsimp()` refuses to do the simplification if it is not valid.

python >>> powsimp(tc*zc) c c t ⋅z

##### In `Julia`:

jldoctest simplification julia> 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.

python >>> powsimp(tc*zc, force=True) c (t⋅z)

##### In `Julia`:

jldoctest simplification julia> powsimp(t^cz^c, force=true) |> string "(tz)^c"

----

Note that in some instances, in particular, when the exponents are integers or
rational numbers, and identity 2 holds, it will be applied automatically.

python

(z*t)**2

 2  2
t ⋅z

sqrt(x*y)

√x⋅√y

##### In `Julia`:

jldoctest simplification julia> (z*t)^2 2 2 t ⋅z

julia> 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.

python

powsimp(z2*t2)

 2  2
t ⋅z

powsimp(sqrt(x)*sqrt(y))

√x⋅√y

##### In `Julia`:

jldoctest simplification julia> powsimp(z^2*t^2) 2 2 t ⋅z

julia> 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.

python >>> expandpowerexp(x**(a + b)) a b x ⋅x

>>> expand_power_base((x*y)**a)
 a  a
x ⋅y

##### In `Julia`:

jldoctest simplification julia> expandpowerexp(x^(a + b)) a b x ⋅x

julia> expandpowerbase((x*y)^a) a a x ⋅y

----

As with `powsimp()`, identity 2 is not applied if it is not valid.

python >>> expandpowerbase((z*t)**c) c (t⋅z)

##### In `Julia`:

jldoctest simplification julia> expandpowerbase((zt)^c) |> string "(tz)^c"

----

And as with `powsimp()`, you can force the expansion to happen without
fiddling with assumptions by using `force=True`.

python

expandpowerbase((z*t)**c, force=True)

 c  c
t ⋅z

##### In `Julia`:

jldoctest simplification julia> expandpowerbase((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()`.

python

x2*x3

 5
x

expandpowerexp(x**5)

 5
x

##### In `Julia`:

jldoctest simplification julia> x^2*x^3 |> string "x^5"

julia> expandpowerexp(x^5) |> string "x^5"

----

### powdenest


`powdenest()` applies identity 3, from left to right.

python >>> powdenest((xa)b) a⋅b x

##### In `Julia`:

jldoctest simplification julia> powdenest((x^a)^b) a⋅b x

----

As before, the identity is not applied if it is not true under the given
assumptions.

python >>> powdenest((za)b) b ⎛ a⎞ ⎝z ⎠

##### In `Julia`:

jldoctest simplification julia> powdenest((z^a)^b) |> string "(z^a)^b"

----

And as before, this can be manually overridden with `force=True`.

python >>> powdenest((za)b, force=True) a⋅b z

##### In `Julia`:

jldoctest simplification julia> 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.

python >>> ln(x) log(x)

##### In `Julia`:

* `ln` is exported

jldoctest simplification julia> 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.

python >>> x, y = symbols('x y', positive=True) >>> n = symbols('n', real=True)

##### In `Julia`:

jldoctest simplification julia> @syms x::positive, y::positive (x, y)

julia> @syms n::real n

----

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.

python >>> expandlog(log(x*y)) log(x) + log(y) >>> expandlog(log(x/y)) log(x) - log(y) >>> expandlog(log(x**2)) 2⋅log(x) >>> expandlog(log(x**n)) n⋅log(x) >>> expand_log(log(z*t)) log(t⋅z)

##### In `Julia`:

jldoctest simplification julia> expand_log(log(x*y)) log(x) + log(y)

julia> expand_log(log(x/y)) log(x) - log(y)

julia> expand_log(log(x^2)) 2⋅log(x)

julia> expand_log(log(x^n)) n⋅log(x)

julia> 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.

python >>> expandlog(log(z**2)) ⎛ 2⎞ log⎝z ⎠ >>> expandlog(log(z**2), force=True) 2⋅log(z)

##### In `Julia`:

jldoctest simplification julia> expand_log(log(z^2)) ⎛ 2⎞ log⎝z ⎠

jldoctest simplification julia> expand_log(log(z^2), force=true) 2⋅log(z)

----

logcombine
----------

To apply identities 1 and 2 from right to left, use `logcombine()`.

python >>> logcombine(log(x) + log(y)) log(x⋅y) >>> logcombine(nlog(x)) ⎛ n⎞ log⎝x ⎠ >>> logcombine(nlog(z)) n⋅log(z)

##### In `Julia`:

jldoctest simplification julia> logcombine(log(x) + log(y)) log(x⋅y)

julia> logcombine(n*log(x)) ⎛ n⎞ log⎝x ⎠

julia> logcombine(n*log(z)) n⋅log(z)

----

`logcombine()` also has a `force` option that can be used to ignore
assumptions.

python >>> logcombine(n*log(z), force=True) ⎛ n⎞ log⎝z ⎠

##### In `Julia`:

jldoctest simplification julia> 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 :ref:`Functions Module <functions-contents>` 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`.

python >>> x, y, z = symbols('x y z') >>> k, m, n = symbols('k m n')

##### In `Julia`:

jldoctest simplification julia> @syms x, y, z (x, y, z)

julia> @syms k, m, n (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.

python >>> factorial(n) n!

##### In `Julia`:

jldoctest simplification julia> 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`".

python >>> binomial(n, k) ⎛n⎞ ⎜ ⎟ ⎝k⎠

##### In `Julia`:

jldoctest simplification julia> binomial(n, k) ⎛n⎞ ⎜ ⎟ ⎝k⎠

----

The factorial function is closely related to the [gamma  function](http://en.wikipedia.org/wiki/Gamma_function). 
$\Gamma(z) = \int_0^\infty t^{z - 1}e^{-t}dt$ is implemented in  `gamma(z)`, which for positive integer has:
`z` is the same as `(z - 1)!`.

python >>> gamma(z) Γ(z)

##### In `Julia`:

* recall, we need to load `SpecialFunctions` for `gamma` to be available

jldoctest simplification julia> 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
${}_pF_q\left(\begin{matrix} a_1, \cdots, a_p \\ b_1, \cdots, b_q \end{matrix}
\middle| z \right)$.  The most common case is ${}_2F_1$, which is often
referred to as the `ordinary hypergeometric function
<http://en.wikipedia.org/wiki/Hypergeometric_function>`.

python >>> hyper([1, 2], [3], z) ┌─ ⎛1, 2 │ ⎞ ├─ ⎜ │ z⎟ 2╵ 1 ⎝ 3 │ ⎠

##### In `Julia`:

* as `[1,2]` is not symbolic, we qualify `hyper`

jldoctest simplification julia> sympy.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,

python >>> tan(x).rewrite(sin) 2 2⋅sin (x) ───────── sin(2⋅x) >>> factorial(x).rewrite(gamma) Γ(x + 1)

##### In `Julia`:

jldoctest simplification julia> tan(x).rewrite(sin) |> string "2sin(x)^2/sin(2x)"

jldoctest simplification julia> 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

python >>> expand_func(gamma(x + 3)) x⋅(x + 1)⋅(x + 2)⋅Γ(x)

##### In `Julia`:

jldoctest simplification julia> expand_func(gamma(x + 3)) x⋅(x + 1)⋅(x + 2)⋅Γ(x)

----

hyperexpand
-----------

To rewrite `hyper` in terms of more standard functions, use
`hyperexpand()`.

python >>> hyperexpand(hyper([1, 1], [2], z)) -log(-z + 1) ───────────── z

##### In `Julia`:

* As `[1,1]` is not symbolic, we qualify `hyperexpand`:

jldoctest simplification julia> sympy.hyperexpand(sympy.hyper([1, 1], [2], z)) -log(1 - z) ──────────── z

----

`hyperexpand()` also works on the more general Meijer G-function (see
:py:meth:`its documentation <sympy.functions.special.hyper.meijerg>` for more
information).

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

##### In `Julia`:

* again, we qualify `meijerg`

jldoctest simplification julia> expr = sympy.meijerg([[1],[1]], [[1],[]], -z) ╭─╮1, 1 ⎛1 1 │ ⎞ │╶┐ ⎜ │ -z⎟ ╰─╯2, 1 ⎝1 │ ⎠ julia> hyperexpand(expr) |> string "exp(1/z)"

----

### combsimp


To simplify combinatorial expressions, use `combsimp()`.

python >>> n, k = symbols('n k', integer = True) >>> combsimp(factorial(n)/factorial(n - 3)) n⋅(n - 2)⋅(n - 1) >>> combsimp(binomial(n+1, k+1)/binomial(n, k)) n + 1 ───── k + 1

##### In `Julia`:

jldoctest simplification julia> @syms n::integer, k::integer (n, k)

julia> combsimp(factorial(n)/factorial(n - 3)) n⋅(n - 2)⋅(n - 1)

julia> combsimp(binomial(n+1, k+1)/binomial(n, k)) n + 1 ───── k + 1

----

### gammasimp


To simplify expressions with gamma functions or combinatorial functions with
non-integer argument, use `gammasimp()`.

python >>> gammasimp(gamma(x)*gamma(1 - x)) π ──────── sin(π⋅x)

##### In `Julia`:

jldoctest simplification julia> gammasimp(gamma(x)gamma(1 - x)) |> string "pi/sin(pix)"

----

### 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

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

where $a_0, \ldots, a_n$ are integers, and $a_1, \ldots, a_n$ are positive. Acontinued 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.

python >>> def listtofrac(l): ... expr = Integer(0) ... for i in reversed(l[1:]): ... expr += i ... expr = 1/expr ... return l[0] + expr >>> listtofrac([x, y, z]) 1 x + ───── 1 y + ─ z

##### In `Julia`:

jldoctest simplification julia> function listtofrac(l) expr = Sym(0) for i in reverse(l[2:end]) expr += i expr = 1/expr end l[1] + expr end listtofrac (generic function with 1 method)

julia> listtofrac([x, y, z]) |> string "x + 1/(y + 1/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.

python >>> listtofrac([1, 2, 3, 4]) 43 ── 30

##### In `Julia`:

jldoctest simplification julia> listtofrac([1, 2, 3, 4]) |> N 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`,
..., `a4`.

python >>> syms = symbols('a0:5') >>> syms (a₀, a₁, a₂, a₃, a₄) >>> a0, a1, a2, a3, a4 = syms >>> frac = listtofrac(syms) >>> frac 1 a₀ + ───────────────── 1 a₁ + ──────────── 1 a₂ + ─────── 1 a₃ + ── a₄

##### In `Julia`:

We can pass tensor-like notation to `@syms` to create containers of indexed variables

jldoctest simplification julia> @syms a0:4

julia> frac = listtofrac(a); string(frac)

"a₀ + 1/(a₁ + 1/(a₂ + 1/(a₃ + 1/a₄)))"

----

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

python >>> 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

##### In `Julia`:

jldoctest simplification julia> frac = cancel(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.

python >>> 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

##### In `Julia`:

jldoctest simplification julia> l = Sym[] Sym[]

julia> a0,a1,a2,a3,a4,a5 = a; # destructure

julia> frac = apart(frac, a0); string(frac) "a0 + (a2a3a4 + a2 + a4)/(a1a2a3a4 + a1a2 + a1a4 + a3a4 + 1)"

julia> push!(l, a0) 1-element Array{Sym,1}: a₀

julia> frac = 1/(frac - a0); string(frac) "(a1a2a3a4 + a1a2 + a1a4 + a3a4 + 1)/(a2a3a4 + a2 + a4)"

----

Now we repeat this process

python >>> 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) >>> listtofrac(l) 1 a₀ + ───────────────── 1 a₁ + ──────────── 1 a₂ + ─────── 1 a₃ + ── a₄

##### In `Julia`:

jldoctest simplification julia> frac = apart(frac, a1);

julia> push!(l, a1);

julia> frac = 1/(frac - a1);

julia> frac = apart(frac, a2);

julia> push!(l, a2);

julia> frac = 1/(frac - a2);

julia> frac = apart(frac, a3);

julia> push!(l, a3);

julia> frac = 1/(frac - a3);

julia> frac = apart(frac, a4);

julia> push!(l, a4);

julia> listtofrac(l) |> string "a0 + 1/(a1 + 1/(a2 + 1/(a3 + 1/a4)))"

----




!!! note "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

python >>> import random >>> l = list(symbols('a0:5')) >>> random.shuffle(l) >>> origfrac = frac = cancel(listto_frac(l)) >>> del l

##### In `Julia`:

* `shuffle` from Python is `randperm` in the `Random` module

julia julia> using Random

julia> @syms a0:4

julia> a = a[randperm(length(a))] 5-element Vector{Sym}: a₄ a₀ a₃ a₂ a₁

julia> origfrac = frac = cancel(listto_frac(a)) a₀⋅a₁⋅a₂⋅a₃⋅a₄ + a₀⋅a₁⋅a₄ + a₀⋅a₃⋅a₄ + a₁⋅a₂⋅a₃ + a₁⋅a₂⋅a₄ + a₁ + a₃ + a₄ ───────────────────────────────────────────────────────────────────────── a₀⋅a₁⋅a₂⋅a₃ + a₀⋅a₁ + a₀⋅a₃ + a₁⋅a₂ + 1 ```

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).