Partial fractions

also: apart

Rewriting a rational function as a polynomial plus one term per irreducible factor power of the denominator. Splitting across coprime denominators is the Bezout identity and nothing more; doing it over irreducible rather than square-free factors is what requires a factoriser.

Rewrite a rational function as a polynomial plus one term per irreducible factor power of the denominator:

\frac{n}{\prod _{i} d _{i}^{e_{i}}} = q + i \sum j = 1 \sum e_{i} \frac{c _{i, j}}{d _{i}^{j}}, de g c_{i, j} < de g d_{i}

Splitting across two coprime denominators is the Bezout identity and nothing more. From $s d_{1} + t d_{2} = 1$ , multiplying by $n$ and dividing by $d_{1} d_{2}$ gives the two pieces directly, and reducing each numerator modulo its denominator keeps the degrees where they belong.

Expanding a repeated factor is writing the numerator in base $d_{i}$ : divide by $d_{i}$ and the remainder is the coefficient of $1/ d_{i}^{e_{i}}$ , then recurse on the quotient.

Verbatim, including a repeated pole and an irreducible quadratic that stays whole.

> apart 1/((x - 1)^2*(x + 2)), x
-1/9*(-1 + x)^(-1) + 1/9*(2 + x)^(-1) + 1/3*(-1 + x)^(-2)
> apart 1/(x^3 + x), x
x^(-1) - x*(1 + x^2)^(-1)

Doing this over irreducible rather than square-free factors is what requires a factoriser, and it is why partial fractions sit near the top of a CAS rather than near the bottom. Integration wants the full decomposition, because each term integrates to a logarithm or a power and nothing else.