Basic functions: Rational functions
Partial fraction decomposition: distinct linear factors for the denominator
Addition of two rational functions with distinct denominators is done by reducing the functions to a common denominator and the result is a fraction with a compound denominator.
Sometimes it is more convenient to go the opposite way and rewrite a rational function with a compound denominator as a sum of simpler fractions. This reverse process is called partial fraction decomposition. Let us first have a look at two simple examples, which we can generalise.
Example 1 \[\begin{aligned} \frac{1}{x}+\frac{1}{1-x} &= \frac{1-x}{x(1-x)}+\frac{x}{x(1-x)} & \blue{\text{reduction to a common denominator}}\\[0.25cm]
&= \frac{1-x+x}{x(1-x)}& \blue{\text{addition of numerators}}\\[0.25cm]
&= \frac{1}{x(1-x)}& \blue{\text{fraction with a compound denominator}}\end{aligned}\]
But how do you get from \(\frac{1}{x(1-x)}\) to the sum of \(\frac{1}{x}+\frac{1}{1-x}\)?
The basic idea is to write \(\frac{1}{x(1-x)}\) as a sum of multiples of \(\frac{1}{x}\) and \(\frac{1}{1-x}\).
In other words, we assume that \[\frac{1}{x(1-x)}=\frac{a}{x}+\frac{b}{1-x}\] for certain, yet to be determined, constants \(a\) and \(b\). For the right-hand side we have \[\begin{aligned}\frac{a}{x}+\frac{b}{1-x} &= \frac{a(1-x)}{x(1-x)}+\frac{bx}{x(1-x)} & \blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(1-x)+bx}{x(1-x)} & \blue{\text{addition of numerators}} \\[0.25cm] &=\frac{(b-a)x+a}{x(1-x)} & \blue{\text{collection of terms}}\end{aligned}\] Then we have \((b-a)x+a=1\) for all \(x\). This is only possible if \(b-a=0\) and \(a=1\).
It follows that \(a=1\) and \(b=1\) (you also get these results by \(x=0\) and \(x=1\) in \((b-a)x+a=1\) ).
In this way we have found the sum that we started with at the beginning of this example.
Example 2 Assume that \[\frac{4x-3}{(x-2)(x-1)}=\frac{a}{x-2}+\frac{b}{x-1}\] for certain numbers \(a\) and \(b\), the values of which must be determined. For the right-hand side we have \[\begin{aligned}\frac{a}{x-2}+\frac{b}{x-1} &= \frac{a(x-1)}{(x-2)(x-1)}+\frac{b(x-2)}{(x-2)(x-1)} & \blue{\text{reduction to a common denominator}} \\[0.25cm]&=\frac{a(x-1)+b(x-2)}{(x-2)(x-1)} & \blue{\text{addition of numerators}} \\[0.25cm] &=\frac{(a+b)x-(a+2b)}{(x-2)(x-1)} & \blue{\text{collection of terms}}\end{aligned}\] Then we have \((a+b)x-(a+2b)=4x-3\) for all \(x\).
This is only possible if \(a+b=4\) and \(a+2b=3\).
This is a system of two linear equations in two unknowns, which can easily be solved (do this!):
\(a=5\) and \(b=-1\). So: \[\frac{4x-3}{(x-2)(x-1)}=\frac{5}{x-2}-\frac{1}{x-1}\]
Note that the scalar \(5\) of \(\frac{1}{x-2}\) is equal to the value at substitution of \(x=2\) into \(\frac{4x-3}{(x-1)}\); similarly, the scalar \(-1\) of \(\frac{1}{x-1}\) is equal to the value at substitution of \(x=1\) into \(\frac{4x-3}{(x-2)}\).
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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 \(q(x)\) only consists of distinct linear factors (so without repetition).
Partial fraction decomposition for a denominator that can be factorised into distinct linear factors 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 \(n\) deistinct linear factors, say, \[q(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)\] where \(a_i\neq\alpha_j\) for \(i\neq j\) and \(1\le i,j,\le n\). Then the partial fraction decomposition of \(\frac{p(x)}{q(x)}\) has the following form: \[\frac{p(x)}{q(x)}=\frac{a_1}{x-\alpha_1}+\frac{a_2}{x-\alpha_2}+\cdots +\frac{a_n}{x-\alpha_n}\] The coefficients can be calculated by setting up and solving a system of \(n\) linear equations with \(n\) unknowns.
But there is also a direct method. If we multiply the above partial fraction decomposition on the left-hand and right-hand side by \(x-\alpha_j\) for \(1\le j\le n\), we get \[(x-\alpha_j)\cdot \frac{p(x)}{q(x)}=a_1\frac{x-\alpha_j}{x-\alpha_1}+\cdots +a_{j-1}\frac{x-\alpha_j}{x-\alpha_{j-1}}+a_j+a_{j+1}\frac{x-\alpha_j}{x-\alpha_{j+1}}+\cdots +a_n\frac{x-a_j}{x-\alpha_n}\] So: \[\begin{aligned}a_j &=\lim_{x\to\alpha_j}(x-\alpha_j)\cdot\frac{p(x)}{q(x)}\\[0.25cm] &=\frac{p(\alpha_j)}{(\alpha_j-\alpha_1)\cdots(\alpha_j-\alpha_{j-1})\cdot (\alpha_{j}-\alpha_{j+1})\cdots (\alpha_{j}-\alpha_n)} \end{aligned}\] In practice, you calculate \(a_j\) by cancelling the factor \(x-\alpha_j\) in the denominator of \(\frac{p(x)}{q(x)}\) followed by substituting \(x=\alpha_j\) in the obtained rational function.
