Partial Fractions.

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 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 and lies between 1 and the multiplicity of the root $\lambda$ of . 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 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 . 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 . Now $g \in K[x]$ will in general not split into linear (or at most quadratic) factors over . 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 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 , 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 is Euclidean, one can choose the such that .

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 are pairwise coprime, whence this is equivalent to $g = \prod_{i=1}^n g_i$ dividing . 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 ; 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 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 for all $j \neq i$ , in . 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 with $d(f) = q^{\deg f}$ for some fixed , and $\Z$ with . The two properties we need are the following:

Definition.

We will say that an Euclidean domain 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 with $q, r \in R$ such that and $d(f - r) \le d(f)$ .

Lemma.

Let 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 $f_i \in R$ , , 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 . Hence, is of the required form. Now, assume that $\floor{\log_{d(p)} d(f)} > 0$ . Write with $g, r \in R$ , 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 , 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 be a principal ideal domain with field of fractions , 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 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 $R = \Z$ , , . 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 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 is Euclidean, 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 is not Euclidean.

In case 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 to , we can therefore assume that 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 . 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 is prime, this can be considered an element in the localization $K[x]_{p_i}$ of 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 .

In the special case that $p_i = x - \lambda$ , we see that $K[x] / (p_i) \cong K$ . Hence, since 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 , 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 respectively (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.

Felix' Math Place

Focussed on, but not limited to Computational Number Theory

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Felix' Math Place

Focussed on, but not limited to Computational Number Theory

Partial Fractions.

Comments.

About Me.

About This Blog.

Overview Pages.

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Recent Posts.

Archives.

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