In partial fraction decomposition, do the terms in the denominator have to be irreducible polynomials?

Question

Somewhere I read that in PFD the polynomials in the denominator should be irreducible. For curiosity I tried it with a reducible polynomial and got an answer out of it. I think I might have not understood what they meant.

This the the fraction and the result I got.

$$ \frac{1}{(x^2+x)(x+2)}=-\frac{0.5x-0.5}{x^2+x}+\frac{0.5}{x+2} $$

Answer 1

First of all, at present you have a confusing typo in your displayed equation: presumably the first factor of the denominator of the left hand side should be $x^2+x$.

It is part of the definition of the PFD that you should restrict to powers of irreducible polynomials, yes. This is a natural condition that is helpful for many applications: for instance in finding antiderivatives of rational functions (over $\mathbb{R}$) and finding Laurent series expansions (over $\mathbb{C}$).

So I would say that your identity above is not a PFD. If you're asking whether it's a true identity, yes it is, assuming the typo mentioned above is fixed. If you're asking for the possibility of establishing identities like this in some generality: sure, it's pretty clear that there will always be an identity of the form

$\frac{A}{x-c_1} + \frac{B}{x-c_2} = \frac{Cx+D}{(x-c_1)(x-c_2)}$,

right? Just do the addition on the left hand side.

Answer 2

I'd like to know where you read that, since it's wrong. If the denominator of the fraction you are working with has repeated factors, then (in general) you must allow those repeated factors in the denominators of the solution.

Answer 3

The answer depends on what you plan to do with the partial fraction decomposition. To compute such decompositions requires only the the factors in the denominator be coprime (vs. powers of primes = irreducibles). For example, if $\rm\,(P,Q) = 1,\:$ i.e. $\rm\,P,Q\,$ are coprime, then

$$\rm \dfrac{A}{PQ} \,=\ \dfrac{B}P\, +\, \dfrac{C}Q \ \iff\ A\, =\, BQ + CP $$

Since $\rm\,(P,Q)=1,\,$ the latter equation can be solved for $\rm\,B,C\,$ using the extended gcd algorithm to compute the Bezout identity $\rm\ 1 = bQ + cP\,\ $ then multiplying it through by $\rm\,A.\:$ Then, if desired, one can decompose this further if the denominators decompose further. For example, if $\rm\:P = p^k\:$ is a prime power, and $\rm\:Q\:$ is a product of prime powers, then one can apply the above recusively, decomposing $\rm\:C/Q = C/P'Q',\:$ etc, till all the denominators are all prime powers. But many applications do not need such full decompositions into prime power denominators.

For example, let's consider Hermite's algorithm for integrating rational functions. It works as follows. By squarefree decomposing the denominator and partial fraction expanding, we reduce to integrating $\rm\:A/D^k\in \mathbb Q(x)\:,\:$ where $\rm\:\deg\:A < \deg\:D^k,\:$ and where $\rm\:D\:$ is squarefree, so $\rm\:\gcd(D,D') = 1\:.\:$ Thus by Bezout (extended Euclidean algorithm) there are $\rm\:B,C\in \mathbb Q[x]\:$ such that $\rm\ B\ D' + C\ D\ =\ A/(1-k)\:.\:$ Then a little algebra shows that

$$\rm\int \frac{A}{D^k}\ =\ \frac{B}{D^{k-1}}\ +\ \int \frac{(1-k)\ C - B'}{D^{k-1}} $$

Iterating the above rule we eventually reduce to the case $\rm\:k=1\:,\:$ i.e. squarefree denominator $\rm\:D\:.\:$ Thus using the above "quotient rule" and nothing deeper than Euclid's algorithm for polynomials (without requiring any factorization) one can mechanically compute the "rational part" of the integral of a rational function, i.e. the part of the integral not involving logarithms. This Hermite reduction rule is the basis of an algorithm due to Hermite (1872). It plays a fundamental role in the transcendental case of some integration algorithms.