# Partial fraction decomposition¶

The partial fraction decomposition of a univariate rational function:

$f(x) = \frac{p(x)}{q(x)}$

where $$p$$ and $$q$$ are co-prime and $$\deg(p) < \deg(q)$$, is an expression of the form:

$\sum_{i=1}^k \sum_{j=1}^{n_i} \frac{a_{ij}(x)}{q_i^j(x)}$

where $$q_i$$ for $$i=1 \ldots k$$ are factors (e.g. over rationals or Gaussian rationals) of $$q$$:

$q(x) = \prod_{i=1}^k q_i^{n_i}$

If $$p$$ and $$q$$ aren’t co-prime, we can use cancel() to remove common factors and if $$\deg(p) >= \deg(q)$$, then div() can be used to extract the polynomial part of $$f$$ and reduce the degree of $$p$$.

Suppose we would like to compute partial fraction decomposition of:

>>> f = 1/(x**2*(x**2 + 1))
>>> f
1
───────────
2 ⎛ 2    ⎞
x ⋅⎝x  + 1⎠


This can be achieved with SymPy’s built-in function apart():

>>> apart(f)
1      1
- ────── + ──
2        2
x  + 1   x


We can use together() to verify this result:

>>> together(_)
1
───────────
2 ⎛ 2    ⎞
x ⋅⎝x  + 1⎠


Now we would like to compute this decomposition step-by-step. The rational function $$f$$ is already in factored form and has two factors $$x^2$$ and $$x^2 + 1$$. If $$f$$ was in expanded from, we could use factor() to obtain the desired factorization:

>>> numer(f)/expand(denom(f))
1
───────
4    2
x  + x

>>> factor(_)
1
───────────
2 ⎛ 2    ⎞
x ⋅⎝x  + 1⎠


Based on the definition, the partial fraction expansion of $$f$$ will be of the following form:

$\frac{A}{x} + \frac{B}{x^2} + \frac{C x + D}{x^2 + 1}$

Let’s do this with SymPy. We will use undetermined coefficients method to solve this problem. Let’s start by defining some symbols:

>>> var('A:D')
(A, B, C, D)


We use here the lexicographic syntax of var(). Next we can define three rational functions:

>>> p1 = A/x
>>> p2 = B/x**2
>>> p3 = (C*x + D)/(x**2 + 1)

>>> p1, p2, p3
⎛A  B   C⋅x + D⎞
⎜─, ──, ───────⎟
⎜x   2    2    ⎟
⎝   x    x  + 1⎠


Let’s add them together to get the desired form:

>>> h = sum(_)
>>> h
A   B    C⋅x + D
─ + ── + ───────
x    2     2
x     x  + 1


The next step is to rewrite this expression as rational function in $$x$$:

>>> together(h)
⎛ 2    ⎞     ⎛ 2    ⎞    2
A⋅x⋅⎝x  + 1⎠ + B⋅⎝x  + 1⎠ + x ⋅(C⋅x + D)
────────────────────────────────────────
2 ⎛ 2    ⎞
x ⋅⎝x  + 1⎠

>>> factor(_, x)
3            2
A⋅x + B + x ⋅(A + C) + x ⋅(B + D)
─────────────────────────────────
2 ⎛ 2    ⎞
x ⋅⎝x  + 1⎠


Let’s now visually compare the last expression with $$f$$:

>>> Eq(_, f)
3            2
a⋅x + b + x ⋅(a + c) + x ⋅(b + d)        1
───────────────────────────────── = ───────────
2 ⎛ 2    ⎞               2 ⎛ 2    ⎞
x ⋅⎝x  + 1⎠              x ⋅⎝x  + 1⎠


Our task boils down to finding $$A$$, $$B$$, $$C$$ and $$D$$. We notice that denominators are equal so we will proceed only with numerators:

>>> eq = Eq(numer(_.lhs), numer(_.rhs))
>>> eq
3            2
a⋅x + b + x ⋅(a + c) + x ⋅(b + d) = 1


To solve this equation, we use solve_undetermined_coeffs():

>>> solve_undetermined_coeffs(eq, [A, B, C, D], x)
{A: 0, B: 1, C: 0, D: -1}


This gave us values for our parameters, which now can be put into the initial expression:

>>> h.subs(_)
1      1
- ────── + ──
2        2
x  + 1   x


This result is identical to the result we got from apart(f). Suppose however, we would like to see how undetermined coefficients method works. First we have to extract coefficients of $$x$$ of both sides of the equation:

>>> lhs, rhs = Poly(eq.lhs, x), Poly(eq.rhs, x)

>>> lhs
Poly((A + C)*x**3 + (B + D)*x**2 + A*x + B, x, domain='ZZ[A,B,C,D]')
>>> rhs
Poly(1, x, domain='ZZ')


Now we can use Poly.nth() to obtain coefficients of $$x$$:

>>> [ Eq(lhs.nth(i), rhs.nth(i)) for i in xrange(4) ]
[b = 1, a = 0, b + d = 0, a + c = 0]


Solving this system of linear equations gives the same solution set as previously:

>>> solve(_)
{a: 0, b: 1, c: 0, d: -1}

>>> f.subs(_)
1      1
- ────── + ──
2        2
x  + 1   x


There are several other ways we can approach undetermined coefficients method. For example we could use collect() for this:

>>> collect(eq.lhs - eq.rhs, x, evaluate=False)
⎧                 2          3       ⎫
⎨1: B - 1, x: A, x : B + D, x : A + C⎬
⎩                                    ⎭

>>> solve(_.values())
{A: 0, B: 1, C: 0, D: -1}


Notice that even though the expressions were not Eq()‘s, this still worked. This is because SymPy assumes by default that expressions are identically equal to 0, so solve(Eq(expr, 0)) is the same as solve(expr).

This approach is even simpler than using Poly.nth(). Finally we use a little trick with Symbol and visually present solution to partial fraction decomposition of $$f$$:

>>> Eq(Symbol('apart')(f), f.subs(_))
⎛     1     ⎞       1      1
apart⎜───────────⎟ = - ────── + ──
⎜ 2 ⎛ 2    ⎞⎟      2        2
⎝x ⋅⎝x  + 1⎠⎠     x  + 1   x


1. Compute partial fraction decomposition of:
• $$\frac{3 x + 5}{(2 x + 1)^2}$$
• $$\frac{3 x + 5}{(u x + v)^2}$$
• $$\frac{(3 x + 5)^2}{(2 x + 1)^2}$$

(solution)

1. Can you use Expr.coeff() in place of Poly.nth()?

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