Theoretical Partial Fractions

We will follow the iterative procedure to partial fraction decomposition as described in this Wikipedia article.

Problem 1. Given polynomials $f, g$ , prove that there exist unique polynomials $q, r$ with $\deg(r) < \deg(g)$ such that

$\displaystyle \frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)},\quad g(x) \neq 0$

(Click for Solution)

Solution. By polynomial long division, there exist unique polynomials $q,r$ with $\deg(r) < \deg(g)$ such that

$\displaystyle f = q \cdot g + r.$

Hence, any $x \in \mathbb R$ such that $g(x) \neq 0$ ,

$\displaystyle \frac{f(x)}{g(x)} = \frac{q(x) \cdot g(x) + r(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}.$

Henceforth, we assume that $\deg(f) < \deg(g)$ .

Definition 1. For polynomials $f, g$ , denote $f \mid g$ if there exists a polynomial $d$ such that $g = d \cdot f$ . Denote $\gcd(f, g) = 1$ if for any polynomial $h$ , $h \mid f$ and $h \mid g$ implies that $h$ is a nonzero constant polynomial.

Problem 2. For non-constant polynomials $g_1, g_2$ such that $\gcd(g_1,g_2) = 1$ , prove that there exists unique polynomials $r_1, r_2$ with $\deg(r_i) < \deg(g_i)$ such that

$\displaystyle \frac{f}{g_1 \cdot g_2} = \frac{r_1}{g_1} + \frac{r_2}{g_2}.$

(Click for Solution)

Solution. The claim is equivalent to ensuring the polynomials $r_1,r_2$ satisfy the equation

$f = r_1 \cdot g_2 + r_2 \cdot g_1,$

which resembles Bézout’s lemma. In what follows, we adopt the proof therein. Consider the collection of polynomials

$\mathcal K := \{s_1 \cdot g_2 + s_2 \cdot g_1 : s_1,s_2 \in \mathbb R[x]\backslash \{0\}\},$

where we denote the set of all polynomials with real coefficients by $\mathbb R[x]$ , and the subset of $\mathbb N$ defined by

$K := \{\deg(s) : s \in \mathcal K\} \equiv \deg(\mathcal K).$

We claim that $\mathcal K$ is non-empty. Use polynomial long division to obtain suitable polynomials $q,r$ such that

$g_2 = q \cdot g_1 + r,\quad \deg(r) < \deg(g_1).$

If $r = 0$ , then $g_1 \mid g_2$ . Since $g_1 \mid g_1$ as well, we have $g_1$ is a nonzero constant, a contradiction. Therefore, $r \neq 0$ and $r = 1 \cdot g_2 + (-q) \cdot g_1 \in \mathcal K$ . Therefore, $\mathcal K$ is non-empty, so that $K \subseteq \mathbb N$ is non-empty. By the well-ordering principle, $K = \deg(\mathcal K)$ has a smallest element $d$ .

Find polynomials $s_1,s_2$ such that

$\deg(\underbrace{s_2 \cdot g_1 + s_1 \cdot g_2}_t) = d.$

We claim that $t \mid g_1$ . Use polynomial long division again to obtain suitable polynomials $\tilde q, \tilde r$ such that

$g_1 = \tilde q \cdot t + \tilde r,\quad \deg(\tilde r) < \deg(t) = d.$

If $\tilde r \neq 0$ , then

$\tilde r = g_1 - \tilde q \cdot t = (1 - s_2 \cdot t) \cdot g_1 + (- s_1 \cdot t) \cdot g_2 \in \mathcal K,$

and $d \leq \deg(\tilde r) < d$ , a contradiction. Therefore, $\tilde r = 0$ so that $t \mid g_1$ . Symmetrically, $t \mid g_2$ , so that $t$ is a nonzero constant. Without loss of generality, we can assume $t = 1$ , so that

$f = f \cdot 1 = f \cdot (s_2 \cdot g_1 + s_1 \cdot g_2) = (f \cdot s_2) \cdot g_1 + (f \cdot s_1) \cdot g_2.$

Dividing both sides by $g_1 \cdot g_2$ yields

$\displaystyle \frac{f}{g_1 \cdot g_2} = \frac{f \cdot s_1}{g_1} + \frac{f \cdot s_2}{g_2}.$

Use Problem 1 to obtain polynomials $q_1, r_1$ such that

$\displaystyle \frac{f}{g_1 \cdot g_2} = q_1 + \frac{r_1}{g_1} + \frac{f \cdot s_2}{g_2} = \frac{r_1}{g_1} + \frac{q_1 \cdot g_2 + f \cdot s_2}{g_2}.$

Defining $r_2 := q_1 \cdot g_2 + f \cdot s_2$ , it remains to prove that $\deg(r_2) < \deg(g_2)$ . Indeed, since

$f = r_1 \cdot g_2 + r_2 \cdot g_1 \quad \Rightarrow \quad r_2 \cdot g_1 = f - r_1 \cdot g_2,$

we have

$\begin{aligned}\deg(r_2) &= \deg(f - r_1 \cdot g_2) - \deg(g_1) \\ &\leq \max\{ \deg(f), \deg(r_1 \cdot g_2)\} - \deg(g_1) \\ &< \max\{ \deg(g_1 \cdot g_2), \deg(g_1 \cdot g_2)\} - \deg(g_1) \\ &= \deg(g_1 \cdot g_2) - \deg(g_1) \\ &= \deg(g_1) + \deg(g_2) - \deg(g_1) = \deg(g_2), \end{aligned}$

as required.

Problem 3. For non-constant polynomials $g_1, \dots, g_n$ such that $\gcd(g_i,g_j) = 1$ for any $i \neq j$ , prove that there exists unique polynomials $r_1, \dots, r_n$ with $\deg(r_i) < \deg(g_i)$ such that

$\displaystyle \frac{f}{g_1\cdot \dots \cdot g_n} = \sum_{i=1}^n \frac{r_i}{g_i}.$

(Click for Solution)

Solution. We prove by strong induction on $n$ . Problem 2 establishes the base case $n = 2$ . For the general case $n = k+1$ , observe that $\gcd(g_{k+1}, g_1 \cdot \dots \cdot g_k) = 1$ . Apply the $n = 2$ case and the $n = k$ case to obtain the unique decomposition

$\begin{aligned}\frac{f}{g_1\cdot \dots \cdot g_k g_{k+1}} &= \frac{r_0}{g_1\cdot \cdots \cdot g_k} + \frac{r_{k+1}}{g_{k+1}},\quad \deg(r_0) < \deg(g_1\cdot \cdots \cdot g_k), \\ &= \sum_{i=1}^k \frac{r_i}{g_i} + \frac{r_{k+1}}{g_{k+1}} = \sum_{i=1}^{k+1} \frac{r_i}{g_i}.\end{aligned}$

Definition 2. Call a polynomial $f$ irreducible if for polynomials $g,h \neq f$ , $g \cdot h = f$ implies that either $g$ or $h$ is constant.

Problem 4. Let $f, g$ be polynomials such that $\deg(g) = \deg(f)-1$ . Prove that if $f$ is irreducible, then $\gcd(f, g) = 1$ .

(Click for Solution)

Solution. Let $h$ be a polynomial such that $h \mid f$ and $h \mid g$ . Find polynomials $s, t$ such that $f = h \cdot s$ , $g = h \cdot t$ . If $h$ is non-constant, then $s \neq 0$ is, so that $g = \frac 1s \cdot f \cdot t$ , which implies

$\deg(g) = \deg(f) + \deg(t) \geq \deg(f) > \deg(g),$

a contradiction. Therefore, $h$ is constant.

Problem 5. Suppose $g$ is irreducible. Prove that there exists unique polynomials $s_1, \dots, s_n$ with $\deg(s_i) < \deg(g)$ such that

$\displaystyle \frac{f}{g^n} = \sum_{i=1}^n \frac{s_i}{g^i}.$

(Click for Solution)

Solution. For the polynomial $g(x) = \sum_{i=0}^n a_ix^i$ , define its formal derivative by $g'(x) = \sum_{i=1}^n i a_i x^{i-1}$ . Since $g$ is irreducible and $\deg(g') = \deg(g)-1$ , by Problem 4, $\gcd(g, g') = 1$ . By the proof in Problem 2, there exists polynomials $s, t$ such that $g \cdot s + g' \cdot t = 1$ . We prove the result by induction. The case $n = 1$ is obvious. For the general case $n = k+1$ ,

$\displaystyle \frac{f}{g^{k+1}} = \frac{f \cdot 1}{g^{k+1}} = \frac{f \cdot g \cdot s + f \cdot g' \cdot t}{g^{k+1}}.$

Use polynomial long division to obtain polynomials $q, s_{k+1}$ such that

$f \cdot g' \cdot t = q \cdot g + s_{k+1},\quad \deg(s_{k+1}) < \deg(g).$

Substituting,

$\displaystyle \frac{f}{g^{k+1}} = \frac{f \cdot g \cdot s + q \cdot g + s_{k+1}}{g^{k+1}} = \frac{f \cdot s + q}{g^k} + \frac{s_{k+1}}{g^{k+1}}.$

All that remains is to verify that $\deg(f \cdot s + q) < k \cdot \deg(g)$ :

$\begin{aligned} \deg(f \cdot s + q) &\leq \max \{\deg(f ), \deg(s_{k+1})\}-\deg(g) \\ &< (k+1) \cdot \deg(g) - \deg(g) = k \cdot \deg(g). \end{aligned}$

By the induction hypothesis,

$\displaystyle \frac{f}{g^{k+1}} = \frac{f \cdot s + q}{g^k} + \frac{s_{k+1}}{g^{k+1}} = \sum_{i=1}^k \frac{s_i}{g^i} + \frac{s_{k+1}}{g^{k+1}} = \sum_{i=1}^{k+1} \frac{s_i}{g^i}.$

Problem 6. Given irreducible polynomials $g_1,\dots, g_n$ such that $\gcd(g_i,g_j) = 1$ for any $i \neq j$ , prove that there exist unique polynomials $s_{k,\ell}$ with $\deg(s_{k,\ell}) < \deg(g_k)$ such that

$\displaystyle \frac{f}{\prod_{k=1}^n g_k^{\alpha_k}} = \sum_{k=1}^n \sum_{\ell=1}^{\alpha_k} \frac{s_{k,\ell}}{g_k^\ell},\quad \deg(s_{k,\ell}) < \deg(g_k).$

(Click for Solution)

Solution. We note that for any $i \neq j$ , $\gcd(g_i^{\alpha_i}, g_j^{\alpha_j}) = 1$ . By Problem 3, there exist unique polynomials $r_1,\dots, r_n$ such that

$\displaystyle \frac{f}{\prod_{k=1}^n g_k^{\alpha_k}} = \sum_{k=1}^n \frac{r_k}{g_k^{\alpha_k}},\quad \deg(r_k) < \deg(g_k^{\alpha_k}).$

By Problem 5, for each $k$ , there exist unique polynomials $s_{k,\ell}$ such that

$\displaystyle \frac{f}{\prod_{k=1}^n g_k^{\alpha_k}} = \sum_{k=1}^n \frac{r_k}{g_k^{\alpha_k}} = \sum_{k=1}^n \sum_{\ell=1}^{\alpha_k} \frac{s_{k,\ell}}{g_k^{\ell}},\quad \deg(s_{k,\ell}) < \deg(g_k).$

Problem 7. Given distinct real numbers $\alpha_i$ and irreducible quadratic polynomials $q_j$ that have no common factors, determine the partial fraction decomposition of

$\displaystyle \frac{f(x)}{\prod_{i=1}^m (x-\alpha_i)^{\beta_i} \cdot \prod_{j=1}^n q_j^{\gamma_j}(x)} ,\quad \deg(f) < \sum_{i=1}^m \beta_i + 2 \cdot \sum_{j=1}^n \gamma_j.$

(Click for Solution)

Solution. By Problem 6,

$\displaystyle \frac{f(x)}{\prod_{i=1}^m (x-\alpha_i)^{\beta_i} \cdot \prod_{j=1}^n q_j^{\gamma_j}(x)} = \sum_{i=1}^m \sum_{k=1}^{\beta_i} \frac{A_{i,k}}{(x-\alpha_i)^{k}} + \sum_{j=1}^n \sum_{\ell=1}^{\alpha_j} \frac{B_{j,\ell} \cdot x + C_{j,\ell}}{q_j^{\ell}(x)}.$

Finally, by the fundamental theorem of algebra, every polynomial can be factorised into a product of linear and quadratic factors. By Problem 7, therefore, any ratio of polynomials can be written as a sum of partial fractions.

—Joel Kindiak, 16 Sept 25, 1713H

KindiakMath

Theoretical Partial Fractions

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Theoretical Partial Fractions

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