### Basic functions: Rational functions

### Partial fraction decomposition: repeated linear factors for the denominator

Let us start with three simple examples that we can generalise.

Example 1 \[\begin{aligned}\frac{3x+2}{x^2+2x+1} &= \frac{3x+2}{(x+1)^2}\\[0.25cm] &= \frac{3(x+1)-1}{(x+1)^2}\\[0.25cm]&= \frac{3}{x+1}-\frac{1}{(x+1)^2}\end{aligned}\]

Example 2 \[\begin{aligned}\frac{4x^2+3x+2}{(x+1)^3} &= \frac{4(x^2+2x+1)-5x-2}{(x+1)^3} &\blue{\text{note that }(x+1)^2=x^2+2x+1}\\[0.25cm]&=\frac{4(x+1)^2-5(x+1)+3}{(x+1)^3}&\blue{\text{create } x+1\text{ terms}}\\[0.25cm]&=\frac{3}{(x+1)^3}-\frac{5}{(x+1)^2}+\frac{4}{x+1}&\blue{\text{decompose the fraction}}\end{aligned}\] \(\phantom{abc}\)

Rewriting the numerator as a sum of powers of \(x+1\) is a bit of a puzzle; those who don't like this much can use the direct method below.

Assume that \[\frac{4x^2+3x+2}{(x+1)^3}=\frac{a}{x+1}+\frac{b}{(x+1)^2}+\frac{c}{(x+1)^3}\] for certain numbers \(a\), \(b\) and \(c\), the values of which must be determined. For the right-hand side we have \[\begin{aligned}\frac{a}{x+1}+\frac{b}{(x+1)^2}+\frac{c}{(x+1)^3} &= \frac{a(x+1)^2}{(x+1)^3}+\frac{b(x+1)}{(x+1)^3}+\frac{c}{(x+1)^3}\\[0.25cm] &\phantom{abcdefg}\blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(x+1)^2+b(x+1)+c}{(x+1)^3} \\[0.25cm] &\phantom{abcdefg}\blue{\text{addition of numerators}} \\[0.25cm]&=\frac{a(x^2+2x+1)+b(x+1)+c}{(x+1)^3}\\[0.25cm] &\phantom{abcdefg}\blue{\text{expansion of power}}\\[0.25cm] &=\frac{ax^2+(b+2a)x+(a+b+c)}{(x+1)^3}\\[0.25cm]&\phantom{abcdefg}\blue{\text{collection of terms}}\end{aligned}\] Then we have \(ax^2+(b+2a)x+(a+b+c)=4x^2+3x+2\) for all \(x\).

But polynomials are equal only when their coefficients are equal.

This is here only possible if \(a=4\), \(b+2a=3\) and \(a+b+c=2\).

It follows that \(a=4\), \(b=-5\) and \(c=3\).

In this way we have found the sum that we started with at the beginning of this example.

Example 3 Assume that \[\frac{x^2+2x+3}{x(x-1)^2}=\frac{a}{x}+\frac{b}{x-1}+\frac{c}{(x-1)^2}\] for certain numbers \(a\), \(b\) and \(c\), the values of which must be determined. For the right-hand side we have \[\begin{aligned}\frac{a}{x}+\frac{b}{x-1}+\frac{c}{(x-1)^2} &= \frac{a(x-1)^2}{x(x-1)^2} +\frac{bx(x-1)}{x(x-1)^2}+\frac{cx}{x(x-1)^2}\\[0.25cm] &\phantom{abcdefg}\blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(x-1)^2+bx(x-1)+cx}{x(x-1)^2} \\[0.25cm] &\phantom{abcdefg}\blue{\text{addition of numerators}} \\[0.25cm]&=\frac{a(x^2-2x+1)+b(x^2-x)+cx}{x(x-1)^2}\\[0.25cm] &\phantom{abcdefg}\blue{\text{expansion of terms at }a\text{ and }b}\\[0.25cm] &=\frac{(a+b)x^2+(c-b-2a)x+c}{x(x-1)^2}\\[0.25cm]&\phantom{abcdefg}\blue{\text{collection of terms}}\end{aligned}\] Then we have \((a+b)x^2+(c-b-2a)x+a=x^2+2x+3\) for all \(x\).

But polynomials are equal only when their coefficients are equal.

This is here only possible if \(a+b=1\), \(c-b-2a=2\) and \(a=3\). It follows that \(a=3\), \(b=-2\) and \(c=6\).

In this way we have determined the partial fraction decomposition: \[\frac{x^2+2x+3}{x(x-1)^2}=\frac{3}{x}-\frac{2}{x-1}+\frac{6}{(x-1)^2}\]

\(\phantom{abc}\)

The above examples illustrate how partial fraction decomposition can be done for a rational function \(\frac{p(x)}{q(x)}\) with the degree of \(p(x)\) less than the degree of \(q(x)\) and where the factorisation of the denominator \(q(x)\) still only consists of linear factors, but with repetitions. Every distinct factor may therefore be raised to a certain power with positive natural exponent in the factorisation of the denominator.

Partial fraction decomposition for a denominator that can be factorised into linear factors, but possibly with repetitions We consider a rational function \(\frac{p(x)}{q(x)}\) with the degree of \(p(x)\) less than the degree of \(q(x)\) and with \(q(x)\) a polynomial that can be written as \(n\) disticnt linear factors, say, \[q(x)=(x-\alpha_1)^{m_1}(x-\alpha_2)^{m_1}\cdots(x-\alpha_n)^{m_n}\] with \(a_i\neq\alpha_j\) for \(i\neq j, 1\le i,j,\le n\) and \(m_i\in\{1,2,3,\ldots\}\) for \(1\le i\le n\). Then, each power of a factor, say \((x-\alpha)^m\), in the partial fraction decomposition of \(\frac{p(x)}{q(x)}\) corresponds with the following sum of fractions: \[\frac{a_1}{x-\alpha}+\frac{a_2}{(x-\alpha)^2}+\cdots +\frac{a_m}{(x-\alpha)^m}\] All coefficients can be computed by adding up all fractions in the decomposition and equating the coefficients of like powers of \(x\) in the numerator with those in \(p(x)\), followed by solving the obtained system of equations.