Partial fraction expansion

Describes particle fraction expansion using the Heaviside method. Decompose rational functions of polynomials in Laplace or Z-transforms.

Oliver Heaviside (1850-1925), was an English electrical engineer, mathematician and physicist who among many things adapted complex numbers to the study of electrical circuits. He introduced a method to decompose rational function of polynomials as they occur when using the Laplace transform to solve differential equations.\(\)

Whenever the denominator of a rational function can be factored into distinct linear factors, the fraction can be expressed as the sum of partial fractions.

$$ \begin{align} f(x)=\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{1+a_1x+a_2x^2+a_3x^3+\ldots+a_Nx^N}\nonumber\\[10mu] \triangleq\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\label{eq:factors} \end{align} $$

A rational function \(\eqref{eq:factors}\) with \(N\) distinct poles \(r_k\), can be expressed as a summation of simple fractions

$$ \shaded{ \frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\triangleq\sum_{k=1}^N \frac{c_k}{x-r_k} } $$

Written out as

$$ \frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}=\frac{c_1}{x-r_1}+\frac{c_2}{x-r_2}+\cdots+\frac{c_{\small N}}{x-r_{\small N}}\label{eq:distinctpoles} $$

The constants \(c_k\) can be obtained by dividing out the \((x-r_k)\) factor in equation \(\eqref{eq:distinctpoles}\) and evaluating at \(x=r_k\)

$$ c_k = \left.\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\right|_{x=r_k} $$

Non-distinct poles

If a pole \(r_k\) is not distinct, the partial fraction expansion for that multiple pole becomes a bit more involved. When the pole occurs \(q\) times as in \((x-r_k)^q\), that pole expands to a summation of decreasing powers of \((x-r_k)\)

$$ \sum_{i=1}^q\frac{d_i}{(x-r_k)^i}\label{eq:nondistict} $$

where the constants \(d_i\) follow to satisfy

$$ d_i=\left.\frac{1}{(q-i)!}\ \left( \frac{\mathrm{d}^{q-i}}{\mathrm{d}x^{q-i}}\ (x-r_k)^q\ f(x) \right)\right|_{x=r_k} $$

In practice, it is easier to first find \(d_q\) by dividing out the highest power \((x-r_k)^q\) in equation \(\eqref{eq:factors}\) and evaluating at \(x=r_k\)

$$ d_q=\left.\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\right|_{x=r_k} $$

The remaining \(d_k\) constants are found by substituting known constants and matching like powers.

For example

his is one of those cases where examples might shed some light on the topic.

Example 1

Consider the rational polynomial

$$ f(x)=\frac{-8+24x}{1-2x+x^2}\label{eq:example} $$

The denominator is a second-order polynomial. The roots of any quadratic equation \(ax^2+bx+c=0\) are

$$ r_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \nonumber $$

The two roots of the denominator polynomial in equation \(\eqref{eq:example}\) are

$$ r_{1,2 } = \frac{2\pm\sqrt{4-4}}{2}\quad\Rightarrow\quad\shaded{r_{1,2}=r=1} $$

This gives equation \(\eqref{eq:example}\) with a factorized denominator

$$ f(x) = \frac{-8+24x}{\left(1-x\right)^2} $$

According to \(\eqref{eq:nondistict}\), this can be expressed as partial fractions

$$ f(x)=\frac{-8+24x}{(1-x)^2}\triangleq\frac{c_1}{(1-x)^2}+\frac{c_2}{1-x} \label{eq:heavyside2} $$

Find the constant \(c_1\) using using Heaviside’s cover up method: multiply all terms by \((1-x)^2\) and then evaluate for \(1-x=0\) so that only the term \(c_1\) is left on the right

$$ c_1=\left.\frac{-8+24x}{((1-x)^2}\right|_{x=1}=-8+24\Rightarrow \shaded{c_1=16} $$

Given the value of \(c_1\), constant \(c_2\) can be found by substituting any numerical value (other than \(0\) in equation \(\eqref{eq:heavyside2}\)

$$ \begin{align} \frac{-8+24x}{(1-x)^2} &= \frac{c_1}{(1-x)^2}+\frac{c_2}{1-x} \Big|_{x=2,\,c_1=16} \implies \nonumber \\ \frac{-8+48}{(1-2)^2} &= \frac{16}{(1-2)^2}+\frac{c_2}{1-2} \implies \nonumber \\ 40 &= 16-c_2\Rightarrow\shaded{c_2=-24} \end{align} $$

The sum of partial fractions follows as

$$ \shaded{ f(x)=\frac{-24}{1-x}+\frac{16}{\left(1-x\right)^{2}} } $$

Example 2

Consider the rational polynomial where the order of the numerator is the same as that of the denominator

$$ f(x) = \frac{2x^3+x^2-x+4}{(x-2)^3} $$

Make the degree of the numerator \(1\) less than that of the denominator by dividing both sides by \(x\). This makes \(\frac{f(x)}{z}\) is a strictly proper rational function. According to Heaviside, this can be expressed as partial fractions

$$ \frac{f(x)}{x} = \frac{2x^3+x^2-x+4}{x(x-2)^3}\triangleq\frac{c_1}{x}+\frac{c_2}{(x-2)^3}+\frac{c_3}{(x-2)^2}+\frac{c_4}{x-2}\label{eq:example2a} $$

Multiply both sides of equation \(\eqref{eq:example2a}\) by \(x\) so that only \(c_0\) is left on the right; then evaluate for \(x=0\)

$$ c_1 = \left.\frac{2x^3+x^2-x+4}{(x-2)^3}\right|_{x=0}=\frac{4}{(-2)^3}\Rightarrow \shaded{c_1=-\frac{1}{2}} $$

Use Heaviside’s cover up method to find the constant \(c_3\). (multiply both sides with \((x-2)^3\) so that only \(c_1\) is left on the right; then evaluate for \(x=2\)

$$ c_2 = \left.\frac{2x^3+x^2-x+4}{x}\right|_{x=2}=\frac{16+4-2+4}{2}\Rightarrow \shaded{c_2=11} $$

Bring all the terms into a common denominator

$$ \begin{align} \frac{2x^3+x^2-x+4}{x(x-2)^3}&=\frac{c_1}{x}+\frac{c_2}{(x-2)^3}+\frac{c_3}{(x-2)^2}+\frac{c_4}{x-2}\Rightarrow\nonumber\\[10mu] \frac{2x^3+x^2-x+4}{x(x-2)^3}&=\frac{c_1(x-2)^3}{x(x-2)^3}+\frac{c_2x}{x(x-2)^3}+\frac{c_3x(x-2)}{x(x-2)(x-2)^2}+\frac{c_4x(x-2)^2}{x(x-2)^2(x-2)}\Rightarrow\nonumber\\[10mu] 2x^3+x^2-x+4&=c_1(x-2)^3+c_2x+c_3x(x-2)+c_4x(x-2)^2 \end{align} $$

Substitute the known values \(c_1\) and \(c_2\), expand and regroup

$$ \begin{align} 2x^3+x^2-x+4&=-\frac{1}{2}(x^3-6x^2+12x-8)+11x+c_3x(x-2)+c_4x(x-2)^2\quad\Rightarrow\nonumber\\ 2x^3+x^2-x\cancel{+4}&=-\frac{1}{2}x^3+3x^2-6x\cancel{+4}+11x+c_3x^2-2c_3x+c_4x^3-4c_4x^2+4c_4x\ \Rightarrow\nonumber\\ 2x^3+x^2-x&=(c_4-\frac{1}{2})x^3+(3-4c_4+c_3)x^2+(-6+11+4c_4-2c_3)x \end{align} $$

Match corresponding powers of \(x\)

$$ \left. \begin{align} 2\cancel{x^3}&=(c_4-\frac{1}{2})\cancel{x^3}\Rightarrow 2=c_4-\frac{1}{2}\Rightarrow \shaded{c_4=\frac{5}{2}}\nonumber\\ \cancelto{1}{x^2}&=(3-4c_4+c_3)\cancel{x^2}\Rightarrow 1=3-4c_4+c_3\nonumber\\ -\cancelto{1}{x}&=(-6+11+4c_4-2c_3)\cancel{x}\Rightarrow -1=-6+11+4c_4-2c_3\nonumber\\ \end{align}\right\}\Rightarrow\nonumber $$

Given \(c_4\), we only need one equation to solve for \(c_3\)

$$ c_3 = 4c_4-2=10-2\Rightarrow \shaded{c_3=8} $$

Substitute the values of \(c_{1\ldots4}\) in \(\eqref{eq:example2a}\)

$$ \frac{f(x)}{x}=-\frac{\frac{1}{2}}{x}+\frac{11}{(x-2)^3}+\frac{8}{(x-2)^2}+\frac{\frac{5}{2}}{x-2} $$

Multiplying both sides by \(x\) gives the sum of partial fractions

$$ \shaded{f(x)=-\frac{1}{2}+11\frac{x}{(x-2)^3}+8\frac{x}{(x-2)^2}+\frac{5}{2}\frac{x}{x-2}} $$

Reference: MIT-cu