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I am looking for a general rule to be able to decompose into partial fractions all expressions of the form:

$\frac{P(x)}{Q(x)}$

where $P(x)$ is a polynomial of degree $p$, and $Q(x)$ is a polynomial of degree $q$, such that $p<q$ $\forall$ $\{p, q\} \subset \mathbb{N}$

An attempt at doing this is already available but it is not a general rule, more a collection of cases.

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    $\begingroup$ Nice link btw, I like his notes. $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 11:38
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    $\begingroup$ Due to the large number of possible cases no single formula or rule would be able to cover it. Instead a general process could be given. Would that suffice? $\endgroup$ – Ian Miller Jan 22 '17 at 13:11
  • $\begingroup$ It would definitely suffice yes! $\endgroup$ – Psi Jan 22 '17 at 13:32
  • $\begingroup$ @IanMiller so did you post this general process somewhere? $\endgroup$ – Severo Raz Feb 27 '18 at 11:01
  • $\begingroup$ is the factorization of $Q(x)$ given for known ? $\endgroup$ – G Cab Feb 27 '18 at 16:17
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This may be done with residues (as per the Heaviside cover-up method):

$$\frac{P(x)}{Q(x)}=\sum_n\frac{\text{Res}_{x=n}f(x)}{x-n}$$

where $f(x)=\frac{P(x)}{Q(x)}$ and $Q(n)=0$, and $n\in\mathbb C$, provided the roots are distinct.

Beyond that, you may find useful information in Wikipedia.

A quick example of how it works, provided the fraction is in simplified form:

$$\frac{P(x)}{(x-a)(x-b)(x-c)}=\sum_{n\in\{a,b,c\}}\frac{\text{Res}_{x=n}f(x)}{x-n}$$

where

$\text{Res}_{x=a}f(x)=\frac{P(a)}{(a-b)(a-c)}$

$\text{Res}_{x=b}f(x)=\frac{P(b)}{(b-a)(b-c)}$

$\text{Res}_{x=c}f(x)=\frac{P(c)}{(c-a)(c-b)}$

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    $\begingroup$ I am afraid that this works only when $Q$ has distinct zeros. $\endgroup$ – Sangchul Lee Jan 22 '17 at 11:39
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    $\begingroup$ @SangchulLee Thanks for pointing that out. $\endgroup$ – Simply Beautiful Art Jan 22 '17 at 11:42
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There is no useful closed form for the general case. But there are simple algorithms to compute a partial fraction decomposition without requiring polynomial factorization (only gcds).

First we can use gcds to compute the squarefree fractorization of the denominator

$$ D = \prod B_k^k,\ \ B_k\ {\rm squarefree},\ \ \gcd(B_i,B_j) = 1\ {\rm if}\ i\neq j$$

Next we can split apart coprime factors $\,B_i\,$ in the denominator by using the extended Euclidean algorithm to compute the Bezout equation for their gcd follows

$$\begin{align} &\gcd(B,C) = 1\,\Rightarrow\, BD+CE = 1\\[0.5em] \Rightarrow\ \ &\dfrac{A}{BC}\, =\, \dfrac{A(BD+CE)}{BC} = \dfrac{AD}C + \dfrac{AE}{B}\end{align}$$

Finally the case where the denominator $= P^n$ is a power is handled in the usual way: iteratively apply the division algorithm to compute the $P$-adic expansion of the numerator, then divide by $P^n$

$$\begin{align} A\, &=\, a_0 + P q_0\\ &=\, a_0 + P(a_1 + P q_1)\\ &\ \ \vdots\\ &=\, a_0 + a_1 P + \cdots a_n P^n,\ \ \deg a_i < \deg P\\ \Rightarrow\, \dfrac{A}{P^n}\,&=\,\dfrac{a_0}{P^n}+\dfrac{a_1}{P^{n-1}}+\cdots+a_n \end{align}$$

These ideas go back to Hermite, who employed them to integrate rational functions. Nowadays we know much more efficient algorithms, which you can find in many expositions on integration algorithms (e.g. Manuel Bronstein's book).

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