Editing Partial fraction decomposition

{{Short description|Rational fractions as sums of simple terms}}
{{more footnotes|date=September 2012}}
In [[algebra]], the '''partial fraction decomposition''' or '''partial fraction expansion''' of a [[rational fraction]] (that is, a [[fraction (mathematics)|fraction]] such that the numerator and the denominator are both [[polynomial]]s) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.<ref>{{cite book |last1=Larson |first1=Ron |title=Algebra & Trigonometry |date=2016 |publisher=Cengage Learning |isbn=9781337271172 |url=https://books.google.com/books?id=Ft-5DQAAQBAJ&q=partial+fraction%27&pg=PA662 |language=en}}</ref>

The importance of the partial fraction decomposition lies in the fact that it provides [[algorithm]]s for various computations with [[rational function]]s, including the explicit computation of [[antiderivative]]s,<ref>Horowitz, Ellis. "[https://ftp.cs.wisc.edu/pub/techreports/1970/TR91.pdf Algorithms for partial fraction decomposition and rational function integration]." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971.</ref> [[Taylor series| Taylor series expansions]], [[Z-transform|inverse Z-transform]]s, and [[Laplace transform|inverse Laplace transform]]s. The concept was discovered independently in 1702 by both [[Johann Bernoulli]] and [[Gottfried Leibniz]].<ref>{{cite book |last=Grosholz |first=Emily |date=2000 |title=The Growth of Mathematical Knowledge |publisher=Kluwer Academic Publilshers |page=179 |isbn=978-90-481-5391-6 }}</ref>

In symbols, the ''partial fraction decomposition'' of a rational fraction of the form <math display="inline"> \frac{f(x)}{g(x)}, </math> where {{math|''f''}} and {{math|''g''}} are polynomials, is the expression of the rational fraction as

<math display="block">\frac{f(x)}{g(x)}=p(x) + \sum_j \frac{f_j(x)}{g_j(x)} </math>

where
{{math|''p''(''x'')}} is a polynomial, and, for each {{mvar|j}},
the [[denominator]] {{math|''g''<sub>''j''</sub> (''x'')}} is a [[Exponentiation|power]] of an [[irreducible polynomial]] (i.e. not factorizable into polynomials of positive degrees), and
the [[numerator]] {{math|''f''<sub>''j''</sub> (''x'')}} is a polynomial of a smaller degree than the degree of this irreducible polynomial.

When explicit computation is involved, a coarser decomposition is often preferred, which consists of replacing "irreducible polynomial" by "[[square-free polynomial]]" in the description of the outcome. This allows replacing [[polynomial factorization]] by the much easier-to-compute [[square-free factorization]]. This is sufficient for most applications, and avoids introducing [[irrational number|irrational coefficients]] when the coefficients of the input polynomials are [[integer]]s or [[rational number]]s.

== Basic principles ==

Let 
<math display="block">R(x) = \frac FG</math> 
be a [[rational fraction]], where {{mvar|F}} and {{mvar|G}} are [[univariate polynomial]]s in the [[Indeterminate (variable)|indeterminate]] {{math|''x''}} over a field. The existence of the partial fraction can be proved by applying inductively the following reduction steps.

===Polynomial part===
There exist two polynomials {{mvar|E}} and {{math|''F''{{sub|1}}}} such that 
<math display="block">\frac FG=E+\frac{F_1}G,</math>
and
<math display="block">\deg F_1 <\deg G,</math>
where <math>\deg P</math> denotes the [[degree of a polynomial|degree]] of the polynomial {{mvar|P}}.

This results immediately from the [[Euclidean division of polynomials|Euclidean division]]  of {{mvar|F}} by {{mvar|G}}, which asserts the existence of {{mvar|E}} and {{math|''F''{{sub|1}}}} such that <math>F = EG + F_1</math> and <math>\deg F_1 < \deg G.</math>

This allows supposing in the next steps that <math>\deg F <\deg G.</math>

===Factors of the denominator===
If <math>\deg F < \deg G,</math> and 
<math display="block">G = G_1 G_2,</math>
where {{math|''G''{{sub|1}}}} and {{math|''G''{{sub|2}}}} are [[coprime polynomials]], then there exist polynomials <math>F_1</math> and <math>F_2</math> such that
<math display="block">\frac FG=\frac{F_1}{G_1}+\frac{F_2}{G_2},</math>
and
<math display="block">\deg F_1 < \deg G_1\quad\text{and}\quad\deg F_2 < \deg G_2.</math>

This can be proved as follows. [[Bézout's identity for polynomials|Bézout's identity]] asserts the existence of polynomials {{math|''C''}} and {{math|''D''}} such that 
<math display="block">CG_1 + DG_2 = 1</math>
(by hypothesis, {{math|1}} is a [[Polynomial greatest common divisor|greatest common divisor]] of {{math|''G''{{sub|1}}}} and {{math|''G''{{sub|2}}}}).

Let <math>DF=G_1Q+F_1</math> with <math>\deg F_1 < \deg G_1</math> be the [[Euclidean division of polynomials|Euclidean division]] of {{mvar|DF}} by <math>G_1.</math> Setting  <math>F_2=CF+QG_2,</math> one gets
<math display="block">\begin{align}
\frac FG&=\frac{F(CG_1 + DG_2)}{G_1G_2}
=\frac{D F}{G_1}+\frac{CF}{G_2}\\
&=\frac{F_1+G_1Q}{G_1}+\frac{F_2-G_2Q}{G_2}\\
&=\frac{F_1}{G_1} + Q + \frac{F_2}{G_2} - Q\\
&=\frac{F_1}{G_1}+\frac{F_2}{G_2}.
\end{align}</math>
It remains to show that <math>\deg F_2 < \deg G_2.</math> By reducing the last sum of fractions to a common denominator, one gets
<math>F=F_2G_1+F_1G_2,</math>
and thus
<math display="block">\begin{align}
\deg F_2 &=\deg(F-F_1G_2)-\deg G_1 \le \max(\deg F,\deg (F_1G_2))-\deg G_1\\
&< \max(\deg G,\deg(G_1G_2))-\deg G_1= \deg G_2 
\end{align}</math>

===Powers in the denominator===
Using the preceding decomposition inductively one gets fractions of the form <math>\frac F {G^k},</math> with <math>\deg F < \deg G^k= k\deg G,</math> where {{mvar|G}} is an [[irreducible polynomial]]. If {{math|''k'' > 1}}, one can decompose further, by using that an irreducible polynomial is a [[square-free polynomial]], that is, <math>1</math> is a [[Polynomial greatest common divisor|greatest common divisor]] of the polynomial and its [[derivative]]. If <math>G'</math> is the derivative of {{mvar|G}}, [[Polynomial greatest common divisor#Bézout's identity and extended GCD algorithm|Bézout's identity]] provides polynomials {{mvar|C}} and {{mvar|D}} such that <math>CG + DG' = 1</math> and thus <math>F=FCG+FDG'.</math> Euclidean division of <math>FDG'</math> by <math>G</math>  gives polynomials <math>H_k</math> and <math>Q</math> such that <math>FDG' = QG + H_k</math> and <math>\deg H_k < \deg G.</math> Setting <math>F_{k-1}=FC+Q,</math> one gets
<math display="block">\frac F {G^k} = \frac{H_k}{G^k}+\frac{F_{k-1}}{G^{k-1}},</math>
with <math>\deg H_k <\deg G.</math>

Iterating this process with <math>\frac{F_{k-1}}{G^{k-1}}</math> in place of <math>\frac F{G^k}</math> leads eventually to the following theorem.

===Statement===
{{math_theorem|name=Theorem|Let {{math|''f''}} and {{math|''g''}} be nonzero polynomials over a field {{math|''K''}}. Write {{math|''g''}} as a product of powers of distinct irreducible polynomials :
<math display="block">g=\prod_{i=1}^k p_i^{n_i}.</math>

There are (unique) polynomials {{math|''b''}} and {{math|''a''<sub>''ij''</sub>}} with {{math|deg ''a''<sub>''ij''</sub> < deg ''p''<sub>''i''</sub>}} such that
<math display="block">\frac{f}{g}=b+\sum_{i=1}^k\sum_{j=1}^{n_i}\frac{a_{ij}}{p_i^j}.</math>

If {{math|deg ''f'' < deg ''g''}}, then {{math|''b'' {{=}} 0}}.}}

The uniqueness can be proved as follows. Let {{math|1=''d'' = max(1 + deg ''f'', deg ''g'')}}. All together, {{math|''b''}} and the {{math|''a''<sub>''ij''</sub>}} have {{mvar|d}} coefficients. The shape of the decomposition defines a [[linear map]] from coefficient vectors to polynomials {{mvar|f}} of degree less than {{mvar|d}}. The existence proof means that this map is [[surjective]]. As the two [[vector space]]s have the same dimension, the map is also [[injective]], which means uniqueness of the decomposition. By the way, this proof induces an algorithm for computing the decomposition through [[linear algebra]].

If {{math|''K''}} is the field of [[complex number]]s, the [[fundamental theorem of algebra]] implies that all  {{math|''p''<sub>''i''</sub>}} have degree one, and all numerators <math>a_{ij}</math> are constants. When {{math|''K''}} is the field of [[real number]]s, some of the {{math|''p''<sub>''i''</sub>}} may be quadratic, so, in the partial fraction decomposition, quotients of linear polynomials by powers of quadratic polynomials may also occur.

In the preceding theorem, one may replace "distinct irreducible polynomials" by "[[pairwise coprime]] polynomials that are coprime with their derivative". For example, the {{math|''p''<sub>''i''</sub>}} may be the factors of the [[square-free factorization]] of {{math|''g''}}. When {{math|''K''}} is the field of [[rational number]]s, as it is typically the case in [[computer algebra]], this allows to replace factorization by [[polynomial greatest common divisor|greatest common divisor]] computation for computing a partial fraction decomposition.

==Application to symbolic integration==

For the purpose of [[symbolic integration]], the preceding result may be refined into

{{math_theorem|name=Theorem|Let ''f'' and ''g'' be nonzero polynomials over a field ''K''. Write ''g'' as a product of powers of pairwise coprime polynomials which have no multiple root in an algebraically closed field:

<math display="block">g=\prod_{i=1}^k p_i^{n_i}.</math>

There are (unique) polynomials ''b'' and ''c''<sub>''ij''</sub> with {{math|deg ''c''<sub>''ij''</sub> < deg ''p''<sub>''i''</sub>}} such that
<math display="block">\frac{f}{g} = b+\sum_{i=1}^k\sum_{j=2}^{n_i}\left(\frac{c_{ij}}{p_i^{j-1}}\right)' + \sum_{i=1}^k \frac{c_{i1}}{p_i}.</math>
where <math> X'</math> denotes the derivative of <math>X.</math>}}

This reduces the computation of the [[antiderivative]] of a rational function to the integration of the last sum, which is called the ''logarithmic part'', because its antiderivative is a linear combination of logarithms. 

There are various methods to compute decomposition in the Theorem. One simple way is called [[Charles Hermite|Hermite]]'s method. First, ''b'' is immediately computed by Euclidean division of ''f'' by ''g'', reducing to the case where deg(''f'') < deg(''g''). Next, one knows deg(''c''<sub>''ij''</sub>) < deg(''p''<sub>''i''</sub>), so one may write each ''c<sub>ij</sub>'' as a polynomial with unknown coefficients. Reducing the sum of fractions in the Theorem to a common denominator, and equating the coefficients of each power of ''x'' in the two numerators, one gets a [[system of linear equations]] which can be solved to obtain the desired (unique) values for the unknown coefficients.

== 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>

== Over the reals ==

Partial fractions are used in [[real number|real-variable]] [[integral calculus]] to find real-valued [[antiderivative]]s of [[rational function]]s. Partial fraction decomposition of real [[rational function]]s is also used to find their [[Inverse Laplace transform]]s. For applications of '''partial fraction decomposition over the reals''', see

* [[#Application to symbolic integration|Application to symbolic integration]], above
* [[Partial fractions in Laplace transforms]]

=== General result ===

Let <math>f(x)</math> be any rational function over the [[real number]]s.  In other words, suppose there exist real polynomials functions <math>p(x)</math> and <math>q(x) \neq 0</math>, such that
<math display="block">f(x) = \frac{p(x)}{q(x)}</math>

By dividing both the numerator and the denominator by the leading coefficient of <math>q(x)</math>, we may assume [[without loss of generality]] that <math>q(x)</math> is [[monic polynomial|monic]]. By the [[fundamental theorem of algebra]], we can write

<math display="block">q(x) = (x-a_1)^{j_1}\cdots(x-a_m)^{j_m}(x^2+b_1x+c_1)^{k_1}\cdots(x^2 + b_n x + c_n)^{k_n}</math>

where <math>a_1, \dots, a_m</math>, <math>b_1, \dots, b_n</math>, <math>c_1, \dots, c_n</math> are real numbers with <math>b_{i}^{2} -4c_{i} < 0</math>, and <math>j_1, \dots, j_m </math>, <math>k_1, \dots, k_n </math> are positive integers.  The terms <math> (x -a_i) </math> are the ''linear factors'' of <math>q(x)</math> which correspond to real roots of <math>q(x)</math>, and the terms <math> ( x_i^2 + b_ix + c_i ) </math> are the ''irreducible quadratic factors'' of <math>q(x)</math> which correspond to pairs of [[complex number|complex]] conjugate roots of  <math>q(x)</math>.

Then the partial fraction decomposition of <math>f(x)</math>  is the following:

<math display="block">f(x) = \frac{p(x)}{q(x)} = P(x) + \sum_{i=1}^m\sum_{r=1}^{j_i} \frac{A_{ir}}{(x-a_i)^r} + \sum_{i=1}^n\sum_{r=1}^{k_i} \frac{B_{ir}x+C_{ir}}{(x^2+b_ix+c_i)^r}</math>

Here, ''P''(''x'') is a (possibly zero) polynomial, and the ''A''<sub>''ir''</sub>, ''B''<sub>''ir''</sub>, and ''C''<sub>''ir''</sub> are real constants.  There are a number of ways the constants can be found.

The most straightforward method is to multiply through by the common denominator ''q''(''x'').  We then obtain an equation of polynomials whose left-hand side is simply ''p''(''x'') and whose right-hand side has coefficients which are linear expressions of the constants ''A''<sub>''ir''</sub>, ''B''<sub>''ir''</sub>, and ''C''<sub>''ir''</sub>.  Since two polynomials are equal if and only if their corresponding coefficients are equal, we can equate the coefficients of like terms.  In this way, a system of linear equations is obtained which ''always'' has a unique solution.  This solution can be found using any of the standard methods of [[linear algebra]]. It can also be found with [[limit (mathematics)|limits]] (see [[#Example 5 (limit method)|Example 5]]).


== Examples ==

=== Example 1 ===

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

Here, the denominator splits into two distinct linear factors:

<math display="block">q(x)=x^2+2x-3=(x+3)(x-1)</math>

so we have the partial fraction decomposition

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

Multiplying through by the denominator on the left-hand side gives us the polynomial identity

<math display="block">1=A(x-1)+B(x+3)</math>

Substituting ''x''  = −3 into this equation gives ''A'' = −1/4, and substituting ''x''  = 1 gives ''B'' = 1/4, so that

<math display="block">f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)</math>

=== Example 2 ===

<math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}</math>

After [[Polynomial long division|long division]], we have

<math display="block">f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}</math>

The factor ''x''<sup>2</sup> − 4''x'' + 8 is irreducible over the reals, as its [[discriminant]] {{math|1=(−4)<sup>2</sup> − 4×8 = −16}} is negative. Thus the partial fraction decomposition over the reals has the shape

<math display="block">\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}</math>

Multiplying through by ''x''<sup>3</sup> − 4''x''<sup>2</sup> + 8''x'', we have the polynomial identity

<math display="block">4x^2-8x+16 = A \left(x^2-4x+8\right) + \left(Bx+C\right)x</math>

Taking ''x'' = 0, we see that 16 = 8''A'', so ''A'' = 2.  Comparing the ''x''<sup>2</sup> coefficients, we see that 4 = ''A'' + ''B'' = 2 + ''B'', so ''B'' = 2.  Comparing linear coefficients, we see that −8 = −4''A'' + ''C'' = −8 + ''C'', so ''C'' = 0.  Altogether,

<math display="block">f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)</math>

The fraction can be completely decomposed using [[complex numbers]]. According to the [[fundamental theorem of algebra]] every complex polynomial of degree ''n'' has ''n'' (complex) roots (some of which can be repeated). The second fraction can be decomposed to:

<math display="block">\frac{x}{x^2-4x+8}=\frac{D}{x-(2+2i)}+\frac{E}{x-(2-2i)}</math>

Multiplying through by the denominator gives:

<math display="block">x=D(x-(2-2i))+E(x-(2+2i)) </math>

Equating the coefficients of {{math|''x''}} and the constant (with respect to {{math|''x''}}) coefficients of both sides of this equation, one gets a system of two linear equations in {{math|''D''}} and {{math|''E''}}, whose solution is

<math display="block">D=\frac{1+i}{2i}=\frac{1-i}{2}, \qquad E=\frac{1-i}{-2i}=\frac{1+i}{2}.</math>

Thus we have a complete decomposition:

<math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}=1+\frac{2}{x}+\frac{1-i}{x-(2+2i)}+\frac{1+i}{x-(2-2i)}</math>

One may also compute directly {{math|''A'', ''D''}} and {{math|''E''}} with the residue method (see also example 4 below).

=== Example 3 ===

This example illustrates almost all the "tricks" we might need to use, short of consulting a [[computer algebra system]].

<math display="block">f(x)=\frac{x^9-2x^6+2x^5-7x^4+13x^3-11x^2+12x-4}{x^7-3x^6+5x^5-7x^4+7x^3-5x^2+3x-1}</math>

After [[Polynomial long division|long division]] and [[polynomial factorization|factoring]] the denominator, we have

<math display="block">f(x)=x^2+3x+4+\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2}</math>

The partial fraction decomposition takes the form

<math display="block">\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}.</math>

Multiplying through by the denominator on the left-hand side we have the polynomial identity

<math display="block">\begin{align}
&2x^6 - 4x^5 + 5x^4 - 3x^3 + x^2 + 3x \\[4pt] 
={}&A\left(x-1\right)^2 \left(x^2+1\right)^2+B\left(x-1\right)\left(x^2+1\right)^2 +C\left(x^2+1\right)^2 + \left(Dx+E\right)\left(x-1\right)^3\left(x^2+1\right)+\left(Fx+G\right)\left(x-1\right)^3
\end{align}</math>

Now we use different values of ''x'' to compute the coefficients:

<math display="block">\begin{cases} 4 = 4C & x =1 \\ 2 + 2i = (Fi + G) (2+ 2i) & x = i \\ 0 = A- B +C - E - G & x = 0 \end{cases}</math>

Solving this we have:

<math display="block">\begin{cases} C = 1 \\ F =0, G =1 \\ E = A-B\end{cases}</math>

Using these values we can write:

<math display="block">\begin{align}
&2x^6-4x^5+5x^4-3x^3+x^2+3x \\[4pt]
={}& A\left(x-1\right)^2 \left(x^2+1\right)^2 + B\left(x-1\right)\left(x^2+1\right)^2 + \left(x^2 + 1\right)^2 + \left(Dx + \left(A-B\right)\right)\left(x-1\right)^3 \left(x^2+1\right) + \left(x-1\right)^3 \\[4pt]
={}& \left(A + D\right) x^6 + \left(-A - 3D\right) x^5 + \left(2B + 4D + 1\right) x^4 + \left(-2B - 4D + 1\right) x^3 + \left(-A + 2B + 3D - 1\right) x^2 + \left(A - 2B - D + 3\right) x
\end{align}</math>

We compare the coefficients of ''x''<sup>6</sup> and ''x''<sup>5</sup> on both side and we have:

<math display="block">\begin{cases} A+D=2 \\ -A-3D = -4 \end{cases} \quad \Rightarrow \quad A= D = 1.</math>

Therefore:

<math display="block">2x^6-4x^5+5x^4-3x^3+x^2+3x = 2x^6 -4x^5 + (2B + 5) x^4 + (-2B - 3) x^3 + (2B +1) x^2 + (- 2B + 3) x</math>

which gives us ''B'' = 0. Thus the partial fraction decomposition is given by:

<math display="block">f(x)=x^2+3x+4+\frac{1}{(x-1)} + \frac{1}{(x - 1)^3} + \frac{x + 1}{x^2+1}+\frac{1}{(x^2+1)^2}.</math>

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at <math>x = 1, \imath</math> in the above polynomial identity. (To this end, recall that the derivative at ''x'' = ''a'' of (''x'' − ''a'')<sup>''m''</sup>''p''(''x'') vanishes if ''m'' > 1 and is just ''p''(''a'') for ''m'' = 1.) For instance the first derivative at ''x'' = 1 gives

<math display="block"> 2\cdot6-4\cdot5+5\cdot4-3\cdot3+2+3   = A\cdot(0+0) + B\cdot( 4+ 0) + 8 + D\cdot0 </math>

that is 8 = 4''B'' + 8 so ''B'' = 0.

===Example 4 (residue method)===

<math display="block"> f(z)=\frac{z^{2}-5}{(z^2-1)(z^2+1)}=\frac{z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}</math>

Thus, ''f''(''z'') can be decomposed into rational functions whose denominators are ''z''+1, ''z''−1, ''z''+i, ''z''−i. Since each term is of power one, −1, 1, −''i'' and ''i'' are simple poles.

Hence, the residues associated with each pole, given by
<math display="block">\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3},</math>
are 
<math display="block"> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2},</math>
respectively, and

<math display="block"> f(z)=\frac{1}{z+1}-\frac{1}{z-1}+\frac{3i}{2}\frac{1}{z+i}-\frac{3i}{2}\frac{1}{z-i}. </math>

===Example 5 (limit method)===

[[Limit (mathematics)|Limits]] can be used to find a partial fraction decomposition.<ref>{{cite book|last=Bluman|first=George W.| title=Problem Book for First Year Calculus|year=1984|publisher=Springer-Verlag|location=New York|pages=250–251}}</ref> Consider the following example:

<math display="block"> \frac{1}{x^3 - 1}</math>

First, factor the denominator which determines the decomposition:

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

Multiplying everything by <math>x-1</math>, and taking the limit when <math>x \to 1</math>, we get

<math display="block">\lim_{x \to 1} \left((x-1)\left ( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )\right) = \lim_{x \to 1} A + \lim_{x \to 1}\frac{(x-1)(Bx + C)}{x^2 + x + 1} =A.</math>

On the other hand,

<math display="block">\lim_{x \to 1} \frac{(x-1)}{(x - 1)(x^2 + x + 1)} =  \lim_{x \to 1}\frac{1}{x^2 + x + 1} = \frac13,</math>

and thus:

<math display="block">A = \frac{1}{3}.</math>

Multiplying by {{math|''x''}} and taking the limit when <math>x \to \infty</math>, we have

<math display="block">\lim_{x \to \infty} x\left( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )= \lim_{x \to \infty} \frac{Ax}{x-1} + \lim_{x \to \infty} \frac{Bx^2+Cx}{x^2+x+1}= A+B,</math>

and

<math display="block">\lim_{x \to \infty} \frac{x}{(x - 1)(x^2 + x + 1)} =0.</math>

This implies {{math|1=''A'' + ''B'' = 0}} and so <math>B = -\frac{1}{3}</math>.

For {{math|1=''x'' = 0}}, we get  <math>-1 = -A + C,</math> and thus  <math>C = -\tfrac{2}{3}</math>.

Putting everything together, we get the decomposition

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

===Example 6 (integral)===
Suppose we have the indefinite [[integral]]:

<math display="block">\int \frac{x^4+x^3+x^2+1}{x^2+x-2} \,dx</math>

Before performing decomposition, it is obvious we must perform polynomial long division and [[Factorization|factor]] the denominator. Doing this would result in:

<math display="block">\int \left(x^2 + 3 + \frac{-3x+7}{(x+2)(x-1)}\right) dx</math>

Upon this, we may now perform partial fraction decomposition.

<math display="block">\int \left(x^2+3+ \frac{-3x+7}{(x+2)(x-1)}\right) dx = \int \left(x^2+3+ \frac{A}{(x+2)}+\frac{B}{(x-1)}\right) dx</math>
so:
<math display="block">A(x-1)+B(x+2)=-3x+7</math>.
Upon substituting our values, in this case, where x=1 to solve for B and x=-2 to solve for A, we will result in:

<math display="block">A=\frac{-13}{3} \ , B=\frac{4}{3} </math>

Plugging all of this back into our integral allows us to find the answer:

<math display="block">\int \left(x^2+3+ \frac{-13/3}{(x+2)}+\frac{4/3}{(x-1)}\right) \,dx = \frac{x^3}{3} \ + 3x-\frac{13}{3} \ln(|x+2|)+\frac{4}{3} \ln(|x-1|)+C </math>

== The role of the Taylor polynomial ==

The partial fraction decomposition of a rational function can be related to [[Taylor's theorem]] as follows. Let

<math display="block">P(x), Q(x), A_1(x),\ldots, A_r(x)</math>

be real or complex polynomials 
assume that

<math display="block">Q=\prod_{j=1}^{r}(x-\lambda_j)^{\nu_j},</math>

satisfies
<math display="block">\deg A_1<\nu_1, \ldots, \deg A_r<\nu_r,  \quad \text{and} \quad \deg(P)<\deg(Q)=\sum_{j=1}^{r}\nu_j.</math>

Also define

<math display="block">Q_i=\prod_{j\neq i}(x-\lambda_j)^{\nu_j}=\frac{Q}{(x-\lambda_i)^{\nu_i}}, \qquad 1 \leqslant i \leqslant r.</math>

Then we have

<math display="block">\frac{P}{Q}=\sum_{j=1}^{r}\frac{A_j}{(x-\lambda_j)^{\nu_j}}</math>

if, and only if, each polynomial <math>A_i(x)</math> is the Taylor polynomial of <math>\tfrac{P}{Q_i}</math> of order <math>\nu_i-1</math> at the point <math>\lambda_i</math>:

<math display="block">A_i(x):=\sum_{k=0}^{\nu_i-1} \frac{1}{k!}\left(\frac{P}{Q_i}\right)^{(k)}(\lambda_i)\ (x-\lambda_i)^k. </math>

Taylor's theorem (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients.

===Sketch of the proof===

The above partial fraction decomposition implies, for each 1&nbsp;≤&nbsp;''i''&nbsp;≤&nbsp;''r'', a polynomial expansion

<math display="block">\frac{P}{Q_i}=A_i + O((x-\lambda_i)^{\nu_i}), \qquad \text{for } x\to\lambda_i,</math>

so <math>A_i</math> is the Taylor polynomial of <math>\tfrac{P}{Q_i}</math>, because of the unicity of the polynomial expansion of order <math>\nu_i-1</math>, and by assumption <math>\deg A_i<\nu_i</math>.

Conversely, if the <math>A_i</math> are the Taylor polynomials, the above expansions at each <math>\lambda_i</math> hold, therefore we also have

<math display="block">P-Q_i A_i = O((x-\lambda_i)^{\nu_i}), \qquad \text{for } x\to\lambda_i,</math>

which implies that the polynomial <math> P-Q_iA_i</math> is divisible by <math>  (x-\lambda_i)^{\nu_i}.</math>

For <math> j\neq i, Q_jA_j</math> is also divisible by <math>(x-\lambda_i)^{\nu_i}</math>, so

<math display="block">  P- \sum_{j=1}^{r}Q_jA_j</math>

is divisible by <math>Q</math>. Since

<math display="block"> \deg\left( P- \sum_{j=1}^{r}Q_jA_j \right) < \deg(Q)</math>

we then have

<math display="block">  P- \sum_{j=1}^{r}Q_jA_j=0,</math>

and we find the partial fraction decomposition dividing by <math>  Q</math>.

== Fractions of integers ==

The idea of partial fractions can be generalized to other [[integral domain]]s, say the ring of [[integer]]s where [[prime numbers]] take the role of irreducible denominators. For example:

<math display="block">\frac{1}{18} = \frac{1}{2} - \frac{1}{3} - \frac{1}{3^2}. </math>

== Notes ==
{{Reflist}}

== References ==
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*{{Cite journal | last1 = Kung | first1 = H. T. | last2 = Tong | first2 = D. M. | doi = 10.1137/0206042 | title = Fast Algorithms for Partial Fraction Decomposition | journal = SIAM Journal on Computing | volume = 6 | issue = 3 | pages = 582 | year = 1977 | s2cid = 5857432 | url = https://figshare.com/articles/journal_contribution/6605561 }}
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*{{springer|id=u/u095160|title=Undetermined coefficients, method of|first=L. D.|last=Kudryavtsev}}
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*{{cite journal | first2=Damian | last2=Slota | first1=Roman | last1=Witula|year=2008 | journal=Appl. Math. Comput. |title=Partial fractions decompositions of some rational functions|pages=328–336 | volume=197 | doi=10.1016/j.amc.2007.07.048 | mr=2396331
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== External links ==
* {{MathWorld |urlname=PartialFractionDecomposition |title=Partial Fraction Decomposition}}
* {{cite web|first1= Sam | last1=Blake
|url=http://calc101.com/webMathematica/partial-fractions.jsp
|title=Step-by-Step Partial Fractions}}
* [http://cajael.com/eng/control/LaplaceT/LaplaceT-1_Example_2_6_OGATA_4editio.php Make partial fraction decompositions] with [[Scilab]].

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[[Category:Algebra]]
[[Category:Elementary algebra]]
[[Category:Partial fractions]]