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Partial Fractions.

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Assume that you have an integral of the form \int_a^b \frac{f(x)}{g(x)} \; dx, where f, g \in \R[x] are polynomials. It is well-known that the best way to solve this is to do a partial fraction decomposition of \frac{f(x)}{g(x)}, which allows you to find an antiderivative of \frac{f(x)}{g(x)}; this can then be used to evaluate \int_a^b \frac{f(x)}{g(x)} \; dx. Partial fraction decomposition of r(x) = \frac{f(x)}{g(x)} allows you to write r(x) as a sum of a polynomial with a linear combination of terms of the form \frac{1}{(x - \lambda)^n}, where the \lambda \in \C range over the (complex) roots of g(x) and n lies between 1 and the multiplicity of the root \lambda of g(x). Now,

\frac{1}{(x - \lambda)^n} ={} & \frac{d}{dx} \left( -\frac{1}{n - 1} \cdot \frac{1}{(x - \lambda)^{n - 1}} \right) \text{ for } n > 1 \\ \text{and} \quad \frac{1}{x - \lambda} ={} & \frac{d}{dx} \log(x - \lambda),

whence finding an antiderivative of this is easy.

In case one has \lambda \in \C \setminus \R, one has that \overline{\lambda} is a root of g(x) as well, and one can combine the terms \frac{1}{(x - \lambda)^n} and \frac{1}{(x - \overline{\lambda})^n} to a term \frac{1}{x^2 - (2 \Re \lambda) x + \abs{\lambda}^2}, multiplied by a polynomial in \R[x] of degree < 2. This allows to give a partial fraction decomposition of \frac{f(x)}{g(x)} purely in term of real polynomials. Note that x^2 - (2 \Re \lambda) x + \abs{\lambda}^2 \in \R[x] is the minimal polynomial of \lambda over \R.

More generally, one can consider polynomials over an arbitrary field K. Now g \in K[x] will in general not split into linear (or at most quadratic) factors over K. But nonetheless, one can still do a partial fraction decomposition. Moreover, one can continue this to another level of abstractness by working with arbitrary principal ideal domains.

The main ingredient which we need is the Chinese Remainder Theorem:

Theorem (Chinese Remainder Theorem).

Let R be a principal ideal domain and let g \in R. Write g = \prod_{i=1}^n g_i, where g_1, \dots, g_n \in R are pairwise coprime elements. Then

R / (g) \to \prod_{i=1}^n R / (g_i), \qquad f \mapsto (f + (g_1), \dots, f + (g_n))

is an isomorphism. Moreover, there exist f_1, \dots, f_n \in R such that \prod_{j = 1 \atop j \neq i}^n g_j divides f_i, that the inverse of the above isomorphism is given by

(a_1 + (g_1), \dots, a_n + (g_n)) \mapsto \sum_{i=1}^n a_i f_i + (g).

In case (R, d) is Euclidean, one can choose the f_i such that d(f_i) < d(g_i).

Proof.

Consider \varphi : R \to \prod_{i=1}^n R / (g_i), f \mapsto (f + (g_1), \dots, f + (g_n)). We first show that \ker \varphi = (g). For that, note that

f \in \ker \varphi \Leftrightarrow \forall i : f \in (g_i) \Leftrightarrow \forall i : g_i \mid f.

Now the g_i are pairwise coprime, whence this is equivalent to g = \prod_{i=1}^n g_i dividing f. Hence, \ker \varphi = (f). Therefore, \varphi induces a monomorphism \hat{\varphi} : R / (f) \to prod_{i=1}^n R / (g_i). We have to show that \hat{\varphi} is surjective, or alternatively, that it admits an inverse.

Let i \in \{ 1, \dots, n \}. We have that \hat{g}_i := \prod_{j = 1 \atop j \neq i}^n g_j is coprime to g_i; hence, we have a Bézout identity, i.e. we can write 1 = h_{i,1} g_i + h_{i,2} \hat{g}_i with h_{i,1}, h_{i,2} \in R; in case (R, d) is Euclidean, we can also assume that d(h_{i,1}) < d(\hat{g}_i), d(h_{i,2}) < d(g_i) (see my post on the Extended Euclidean Algorithm). But this means that h_{i,2} \hat{g}_i is zero in R / (g_j) for all j \neq i, 1 in R / (g_i). Set f_i := h_{i,2}; therefore, \sum_{i=1}^n a_i f_i + (g_i) = a_i + (g_i), whence \varphi(\sum_{i=1}^n a_i f_i) = (a_1 + (g_1), \dots, a_n + (g_n)), as we had to show.

We need another simple lemma, which works for a certain class of Euclidean rings such as K[x] with d(f) = q^{\deg f} for some fixed q > 1, and \Z with d(x) = |x|. The two properties we need are the following:

Definition.

We will say that an Euclidean domain (R, d) is admissible if d : R \setminus \{ 0 \} \to \N_{>0} is multiplicative and for all f, g \in R, g \neq 0 one can write f = q g + r with q, r \in R such that d(r) < d(g) and d(f - r) \le d(f).

Lemma.

Let (R, d) be an admissible Euclidean domain and p \in R \setminus (R^* \cup \{ 0 \}). Any f \in R can be written in the form f = \sum_{i=0}^n f_i p^i with f_i \in R, d(f_i) < d(p), where n = \max\{ 0, \floor{\log_{d(p)} d(f)} \}.

Proof.

We show this by induction on \floor{\log_{d(p)} d(f)}. For \floor{\log_{d(p)} d(f)} \le 0, \log_{d(p)} d(f) < 1, whence d(f) < d(p). Hence, f = f p^0 is of the required form. Now, assume that \floor{\log_{d(p)} d(f)} > 0. Write f = g p + r with g, r \in R, d(r) < d(p) and d(f - r) \le d(f). Hence, d(g p) = d(f - r) \le d(f) and \log_{d(p)} d(f) \le \log_{d(p)} d(f) - 1, whence by induction, we can write g = \sum_{i=0}^{\floor{\log_{d(p)} d(f)} - 1} f_{i+1} p^i with suitable f_{i+1}, d(f_{i+1}) < d(p). If we set f_0 := r, we obtain f = \sum_{i=0}^n f_i p^i with n = \floor{\log_{d(p)} d(f)} = \max\{ 0, \floor{\log_{d(p)} d(f)} \} as required.

Now partial fraction decomposition is a direct corollary of these results:

Corollary (Partial Fraction Decomposition).

Let R be a principal ideal domain with field of fractions K, and let f, g \in R be two elements. Assume that g = \lambda \prod_{i=1}^n p_i^{e_i}, where \lambda \in R^*, p_1, \dots, p_n \in K[x] are pairwise coprime polynomials and e_i \in \N. Then there exist elements h, h_{ij} \in R such that

\frac{f}{g} = h + \sum_{i=1}^n \sum_{j=1}^{e_i} \frac{h_{ij}}{p_i^j}.

In case (R, d) is Euclidean with a multiplicative degree function, we can assume that d(h_{ij}) < d(p_i) for 1 \le j \le e_i, 1 \le i \le n.

As an example, we want to consider the Partial Fraction Decomposition for \frac{f}{g} = \frac{23}{12} \in \Q, i.e. for R = \Z, f = 23, g = 12. Now 12 = 2^2 \cdot 3 and 23 = 12 + 1 \cdot 3 + 2 \cdot 4, whence

\frac{23}{12} = 1 + \frac{1}{2^2} + \frac{0}{2^1} + \frac{2}{3^1}

is the partial fraction decomposition of \frac{23}{12} in \Q.

Proof.

We can assume that \lambda = 1 by replacing f by \lambda^{-1} f.

By the Chinese Remainder Theorem, the map R / (g) \to \prod_{i=1}^n R / (p_i^{e_i}), a \mapsto (a + (p_1^{e_1}), \dots, a + (p_n^{e_n})) is an isomorphism. The inverse is given by (a_1 + (p_1^{e_1}), \dots, a_n + (p_n^{e_n})) \mapsto \sum_{i=1}^n a_i b_i \prod_{j=1 \atop j \neq i}^n p_j^{e_j} + (g) for some b_i \in R; in case (R, d) is Euclidean, b_i can be chosen such that d(b_i) < d(p_i)^{e_i}. Therefore, there exist a_i \in R (where in case (R, \deg) is Euclidean, d(a_i) < d(p_i)^{e_i}) and h \in R such that

f = \sum_{i=1}^n a_i b_i \prod_{j=1 \atop j \neq i}^n p_j^{e_j} + h g.

Dividing by g = \prod_{i=1}^n p_i^{e_i}, we get

\frac{f}{g} = h + \sum_{i=1}^n \frac{a_i b_i}{p_i^{e_i}}.

This shows the claim in case R is not Euclidean.

In case (R, d) is Euclidean, we can write a_i b_i = h_i p_i^{e_i} + r_i with d(r_i) < d(p_i)^{e_i}; by adding h_i to h, we can therefore assume that a_i b_i = r_i satisfies d(r_i) < d(p_i)^{e_i}. By the previous lemma, we can now write r_i = \sum_{j=0}^{n_i} h_{ij} p_i^j with d(h_{ij}) < d(p_i) and n_i \le \floor{\log_{d(p_i)} d(r_i)}; as d(r_i) < d(p_i)^{e_i}, we get n_i < e_i. Therefore, we obtain r_i = \sum_{i=0}^{e_i - 1} h_{ij} p_i^j with d(h_{ij}) < d(p_i), which gives

\frac{f}{g} = h + \sum_{i=1}^n \frac{\sum_{j=0}^{e_j - 1} h_{ij} p_i^j}{p_i^{e_i}} = h + \sum_{i=1}^n \sum_{j=0}^{e_j - 1} \frac{h_{ij}}{p_i^{e_i - j}},

and by reindexing the second sum we obtain the formula from the statement of the theorem.

Now assume that we have a rational function r \in K(x) written as

r = h + \sum_{i=1}^n \sum_{j=1}^{e_i} \frac{h_{ij}}{p_i^{e_j}},

where p_1, \dots, p_n \in K[x] \setminus K are pairwise coprime, h, h_{ij} \in K[x] with \deg h_{ij} < \deg p_i. Then

p_i^{e_i} r = \left( h + \sum_{j=1 \atop j\neq i}^n \sum_{k=1}^{e_j} \frac{h_{jk}}{p_j^{e_k}} \right) p_i^{e_i} + \sum_{j=1}^{e_i} h_{ij} p_i^{e_i - j}.

In case p_i is prime, this can be considered an element in the localization K[x]_{p_i} of K at the prime ideal \frakp_i = p_i K[x]. Let \widehat{K[x]_{p_i}} \supseteq K[x]_{p_i} be the completion of K[x]_{p_i}. Note that both K[x]_{p_i} and \widehat{K[x]_{p_i}} have the residue field K[x] / (p_i).

In the special case that p_i = x - \lambda, we see that K[x] / (p_i) \cong K. Hence, since p_i is a prime element in K[x]_{p_i}, we obtain an isomorphism K[[t]] \cong \widehat{K[x]_{p_i}} with t \mapsto p_i = x - \lambda. In K[[t]], the element p_i^{e_i} r \in \widehat{K[x]_{p_i}} corresponds to

\underbrace{(\ldots)}_{\in K[[t]]} t^{e_i} + \sum_{j=1}^{e_i} h_{ij} t^{e_i - j}.

In particular, we can use the Hasse derivative D_R^{(k)} of R = K[x] respectively R = K[[t]] (the definition can be translated to power series rings as well) to obtain

h_{ij} = \left( D_{K[[t]]}^{(e_i - j)} \bigl( (x - \lambda)^{e_i} r \bigr) \right)(0) = \left( D_{K[x]}^{(e_i - j)} \bigl( (x - \lambda)^{e_i} r \bigr) \right)(\lambda) \in K.

In the case K \in \{ \R, \C \}, this is Heaviside's formula.

Categories: Algebra Analysis
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