# Partial Fraction Expansion

$$\require{AMSsymbols}\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \require{cancel} \newcommand\ccancel[black] {\color{#1}{\cancel{\color{black}{#2}}}}$$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

This 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
\begin{align} f(x)&=\frac{-8+24x}{\left(1-x\right)^2} \end{align}

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 $$\eqref{eq:heavyside2}$$
\left.\frac{-8+24x}{(1-x)^2}=\frac{c_1}{(1-x)^2}+\frac{c_2}{1-x}\right|_{x=2,\,c_1=16}\Rightarrow\nonumber\\ \begin{align} \frac{-8+48}{(1-2)^2}&=\frac{16}{(1-2)^2}+\frac{c_2}{1-2}\Rightarrow\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
$$\shaded{f(x)=-\frac{1}{2}+11\frac{x}{(x-2)^3}+8\frac{x}{(x-2)^2}+\frac{5}{2}\frac{x}{x-2}}$$

References
[MIT-cu]

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