Editing Partial fraction decomposition (section)

== Procedure ==

Given two polynomials <math>P(x)</math> and <math>Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)</math>, where the ''α''<sub>''n''</sub> are distinct constants and {{math|deg ''P'' < ''n''}}, explicit expressions for partial fractions can be obtained by supposing that
<math display="block">\frac{P(x)}{Q(x)} = \frac{c_1}{x-\alpha_1} + \frac{c_2}{x-\alpha_2} + \cdots + \frac{c_n}{x-\alpha_n}</math>
and solving for the ''c''<sub>''i''</sub> constants, by substitution, by [[equating the coefficients]] of terms involving the powers of ''x'', or otherwise.  (This is a variant of the [[method of undetermined coefficients]]. After both sides of the equation are multiplied by Q(x), one side of the equation is a specific polynomial, and the other side is a polynomial with undetermined coefficients. The equality is possible only when the coefficients of like powers of ''x'' are equal. This yields n equations in n unknowns, the c<sub>k</sub>.)

A more direct computation, which is strongly related to [[Lagrange interpolation]], consists of writing
<math display="block">\frac{P(x)}{Q(x)} = \sum_{i=1}^n \frac{P(\alpha_i)}{Q'(\alpha_i)}\frac{1}{(x-\alpha_i)} </math>
where <math>Q'</math> is the derivative of the polynomial <math>Q</math>. The coefficients of <math>\tfrac{1}{x-\alpha_j}</math> are called the [[Residue (complex analysis)|residues]] of ''f/g''.

This approach does not account for several other cases, but can be modified accordingly:

* If <math>\deg P \geq \deg Q, </math> then it is necessary to perform the [[Polynomial#Divisibility|Euclidean division]] of ''P'' by ''Q'', using [[polynomial long division]], giving {{math|1=''P''(''x'') = ''E''(''x'') ''Q''(''x'') + ''R''(''x'')}} with {{math|deg ''R'' < ''n''}}. Dividing by ''Q''(''x'') this gives <math display="block">\frac{P(x)}{Q(x)} = E(x) + \frac{R(x)}{Q(x)},</math> and then seek partial fractions for the remainder fraction (which by definition satisfies {{math|deg ''R'' < deg ''Q''}}).
* If ''Q''(''x'') contains nonlinear factors which are irreducible over the given field, then the numerator ''N''(''x'') of each partial fraction with such a factor ''F''(''x'') in the denominator must be sought as a polynomial with {{math|deg ''N'' < deg ''F''}}, rather than as a constant. For example, take the following decomposition over '''R''': <math display="block">\frac{x^2 + 1}{(x+2)(x-1)\color{Blue}(x^2+x+1)} = \frac{a}{x+2} + \frac{b}{x-1} + \frac{\color{OliveGreen}cx + d}{\color{Blue}x^2 + x + 1}.</math>
* Suppose {{math|1=''Q''(''x'') = (''x'' − ''α'')<sup>''r''</sup> ''S''(''x'')}} and {{math|''S''(''α'') ≠ 0}}, that is {{math|''α''}} is a root of {{math|''Q''(''x'')}} of [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiplicity]] {{mvar|r}}. In the partial fraction decomposition, the {{mvar|r}} first powers of {{math|(''x'' − ''α'')}} will occur as denominators of the partial fractions (possibly with a zero numerator). For example, if {{math|1=''S''(''x'') = 1}} the partial fraction decomposition has the form <math display="block">\frac{P(x)}{Q(x)} = \frac{P(x)}{(x-\alpha)^r} = \frac{c_1}{x-\alpha} + \frac{c_2}{(x-\alpha)^2} + \cdots + \frac{c_r}{(x-\alpha)^r}.</math>

=== Illustration ===

In an example application of this procedure, {{math|(3''x'' + 5)/(1 − 2''x'')<sup>2</sup>}} can be decomposed in the form

<math display="block">\frac{3x + 5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}.</math>

[[Clearing denominators]] shows that {{math|1=3''x'' + 5 = ''A'' + ''B''(1 − 2''x'')}}. Expanding and equating the coefficients of powers of {{math|''x''}} gives
{{block indent | em = 1.5 | text = {{math|1=5 = ''A'' + ''B''}} and  {{math|1=3''x'' = −2''Bx''}}}}

Solving this [[system of linear equations]] for {{math|''A''}} and {{math|''B''}} yields {{math|1=''A'' = 13/2 and ''B'' = −3/2}}.  Hence,

<math display="block">\frac{3x + 5}{(1-2x)^2} = \frac{13/2}{(1-2x)^2} + \frac{-3/2}{(1-2x)}.</math>

=== Residue method ===
{{See also|Heaviside cover-up method}}
Over the complex numbers, suppose ''f''(''x'') is a rational proper fraction, and can be decomposed into

<math display="block">f(x) = \sum_i \left( \frac{a_{i1}}{x - x_i} + \frac{a_{i2}}{( x - x_i)^2} + \cdots + \frac{a_{i k_i}}{(x - x_i)^{k_i}} \right). </math>

Let
<math display="block"> g_{ij}(x) = (x - x_i)^{j-1}f(x),</math>
then according to the [[Laurent series#Uniqueness|uniqueness of Laurent series]], ''a''<sub>''ij''</sub> is the coefficient of the term {{math|(''x'' − ''x''<sub>''i''</sub>)<sup>−1</sup>}} in the Laurent expansion of ''g''<sub>''ij''</sub>(''x'') about the point ''x''<sub>''i''</sub>, i.e., its [[residue (complex analysis)|residue]]
<math display="block">a_{ij} = \operatorname{Res}(g_{ij},x_i).</math>

This is given directly by the formula
<math display="block">a_{ij} = \frac 1 {(k_i-j)!}\lim_{x\to x_i}\frac{d^{k_i-j}}{dx^{k_i-j}} \left((x-x_i)^{k_i} f(x)\right),</math>
or in the special case when ''x''<sub>''i''</sub> is a simple root,
<math display="block">a_{i1}=\frac{P(x_i)}{Q'(x_i)},</math>
when
<math display="block">f(x)=\frac{P(x)}{Q(x)}.</math>