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This page is meant as a comprehensive review of partial fraction expansion. Partial fraction expansion allows us to fit functions to the known ones given by the known Fourier Transform pairs table.

First, the denominator must be of a higher degree than the numerator. If this is not the case, then perform long division to make it such. Note: for the remainder of this guide it is assumed that the denominator is of a higher degree than the numerator. There are four cases that arise which one must consider:

Case 1 : Denominator is a product of distinct linear factors.

$\frac{(polynomial)}{(a_1x+b_1)(a_2x+b_2)\cdots(a_kx+b_k)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\cdots+\frac{A_k}{a_kx+b_k}$

Case 2 : Denominator is a product of linear factors, some of which are repeated.

$\frac{(polynomial)}{(a_1x+b_1)^r}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{(a_1x+b_1)^2}+\cdots+\frac{A_r}{(a_1x+b_1)^r}$

Case 3 : Denominator contains irreducible quadratic factors, none of which is repeated.

$\frac{(polynomial)}{ax^2+bx+c}=\frac{Ax+B}{ax^2+bx+c}$

Case 4 : Denominator contains a repeated irreducible quadratic factor.

$\frac{(polynomial)}{(ax^2+bx+c)^r}=\frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\cdots+\frac{A_rx+B_r}{(ax^2+bx+c)^r}$

Example encompassing all of the above cases:

$\frac {(polynomial degree < 10)}{x(x-1)(x^2+x+1)(x^2+1)^3} = \frac {A}{x} + \frac {B}{x^2+x+1} + \frac {Cx+D}{x^2+1} + \frac {Ex+F}{(x^2+1)^2} + \frac {Gx+H}{(x^2+1)^2} + \frac {Ix+J}{(x^2+1)^3}$

General method to find out what the Capital Letters equal:

The most general method is to multiply both sides by the left hand side's denominator. This clears all fractions. Then, expand and collect terms on the right hand side. Equate the right and left hand sides' coefficients. This leads to a system of linear equations which can be solved for obtaining the Capital Letters. Tricks to obtaining the Capital Letters quickly: Capital Letters whose denominator is the highest power of its kind can be found directly as follows:

First, multiply both sides by its denominator.

Second, find the value of x which would make the denominator equal 0.

Third, evaluate the equation for that value of x. This leaves you with the Capital Letter on the right and its value on the left. Example of finding Capital Letters:

$\frac{4x}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}$

The above trick can be used to find B and C but not A.

$B=\frac{4x}{x+1}\;(evaluated\;at\;x=1)=\frac{4}{2}=2$
$C=\frac{4x}{(x-1)^2}\;(evaluated\;at\;x=-1)=\frac{-4}{4}=-1$

A is then found using the general method given above.

\begin{align} 4x&=A(x-1)(x+1)+B(x+1)+C(x-1)^2 \\ &=A(x-1)(x+1)+2(x+1)-(x-1)^2 \\ &=(A-1)x^2+4x+(-A+1) \end{align}
$A-1=0 \rightarrow A=1$

Note: The four cases for finding the form of the partial fraction expansion as well as the general method of finding the capital letters were adapted from section 7.4 in Calculus Early Transcendentals, 5e. (Our Calc I, II, & III book.) The tricks to obtaining the capital letters quickly are from learning to do the Laplace Transform in ECE 202.