View source for Partial fraction decomposition

{{short description|Decomposition of a rational fraction into a sum of fractions with simple denominators}}
{{more footnotes|date=September 2012}}
In [[algebra]], the '''partial fraction decomposition''' or '''partial fraction expansion''' of a [[rational function]] (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&pg=PA662&dq=partial+fraction%27&hl=en&sa=X&ved=0ahUKEwi_86PhweXdAhWWWX0KHW_tAJ8Q6AEIRzAG#v=onepage&q=partial%20fraction'&f=false |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. "[ftp://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, [[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 |url= |location= |publisher=Kluwer Academic Publilshers |page=179 |isbn=978-90-481-5391-6 |author-link= }}</ref>

In symbols, one can use ''partial fraction expansion'' to change a rational fraction in the form

: <math> \frac{f(x)}{g(x)} </math>

where {{math|''f''}} and {{math|''g''}} are polynomials, into an expression of the form

: <math> \sum_j \frac{f_j(x)}{g_j(x)} </math>

where:

* the [[denominator]], {{math|''g''<sub>''j''</sub> (''x'')}}, of each fraction is a [[Exponentiation|power]] of an [[irreducible polynomial|irreducible]] (not factorable into polynomials of positive degree) [[polynomial]] and
* the [[numerator]] is a polynomial of smaller degree than this irreducible polynomial.

As [[factorization of polynomials]] may be difficult, a coarser decomposition is often preferred, which consists of replacing factorization by [[square-free factorization]]. This amounts to replace "irreducible" by "square-free" in the preceding description of the outcome.

== Basic principles ==

If a rational function <math>R(x) = \frac{f(x)}{g(x)}</math> in one [[Indeterminate (variable)|indeterminate]] {{math|''x''}} has a [[denominator]] that factors as

:<math> g(x) = P(x) \cdot Q(x) </math>

over a [[field (mathematics)|field]] {{math|''K''}} (we can take this to be [[real number]]s, or [[complex number]]s) and if in addition {{math|''P''}} and {{math|''Q''}} have no common factor, then by [[Polynomial greatest common divisor#Bézout's identity and extended GCD algorithm|Bézout's identity for polynomials]], there exist polynomials {{math|''C''(''x'')}} and {{math|''D''(''x'')}} such that <math>\deg(C) < \deg(Q)</math>, <math>\deg(D) < \deg(P)</math>, and

:<math>CP + DQ = 1</math>

Thus

:<math>\frac{1}{g(x)} = \frac{CP+DQ}{PQ} = \frac{C}{Q}+\frac{D}{P}</math>

and hence {{math|''R''}} may be written as

:<math> R = \frac{f(x)}{g(x)} = \frac{Df(x)}{P} + \frac{Cf(x)}{Q},</math>

where all numerators are polynomials.

Using this idea inductively the rational function {{math|''R''(''x'')}} can be written as a sum with denominators being powers of [[irreducible polynomial]]s. To take this further, if required, write:

:<math>\frac {G(x)}{F(x)^n}</math>

as a sum with denominators powers of {{math|''F''}} and [[numerator]]s of degree lower than the degree of {{math|''F''}}, plus a possible extra polynomial. This can be done by the [[Euclidean algorithm for polynomials]].  The result is the following [[theorem]]:

{{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>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>\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}}.}}

If {{math|''K''}} is 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>g=\prod_{i=1}^k p_i^{n_i}.</math>

There are (unique) polynomials ''b'' and ''c''<sub>''ij''</sub> with deg&nbsp;''c''<sub>''ij''</sub>&nbsp;<&nbsp;deg&nbsp;''p''<sub>''i''</sub> such that

: <math>\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. In fact, we have

:<math>\frac{c_{i1}}{p_i}=\sum_{\alpha_j:p_i(\alpha_j)=0}\frac{c_{i1}(\alpha_j)}{p'_i(\alpha_j)}\frac{1}{x-\alpha_j}.</math>

There are various methods to compute above decomposition. The one that is the simplest to describe is probably the so-called [[Charles Hermite|Hermite]]'s method. As the degree of ''c''<sub>''ij''</sub> is bounded by the degree of ''p''<sub>''i''</sub>, and the degree of ''b'' is the difference of the degrees of ''f'' and ''g'' (if this difference is non negative; otherwise, ''b''=0), one may write these unknowns polynomials as polynomials with unknown coefficients. Reducing the two members of above formula to the same denominator and writing that the coefficients of each power of ''x'' are the same in the two numerators, one gets a [[system of linear equations]] which can be solved to obtain the desired values for the unknowns 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>''i''</sub> are distinct constants and deg&nbsp;''P''&nbsp;<&nbsp;''n'', partial fractions are generally obtained by supposing that

: <math>\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]].)

A more direct computation, which is strongly related with [[Lagrange interpolation]] consists of writing

: <math>\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>.

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

* If <math>\deg P \geqslant \deg Q, </math> then it is necessary to perform the [[Polynomial#Divisibility|Euclidean division]] of ''P'' by ''Q'', using [[polynomial long division]], giving ''P''(''x'') = ''E''(''x'') ''Q''(''x'') + ''R''(''x'') with deg&nbsp;''R''&nbsp;<&nbsp;''n''. Dividing by ''Q''(''x'') this gives
:: <math>\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 deg&nbsp;''R''&nbsp;<&nbsp;deg&nbsp;''Q'').
* If ''Q''(''x'') contains 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 deg&nbsp;''N''&nbsp;<&nbsp;deg&nbsp;''F'', rather than as a constant. For example, take the following decomposition over '''R''':
:: <math>\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 ''Q''(''x'') = (''x'' − ''α'')<sup>''r''</sup>''S''(''x'') and ''S''(''α'')&nbsp;≠&nbsp;0. Then ''Q''(''x'') has a zero ''α'' of [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiplicity]] ''r'', and in the partial fraction decomposition, ''r'' of the partial fractions will involve the powers of (''x'' − ''α''). For illustration, take ''S''(''x'') =&nbsp;1 to get the following decomposition:
:: <math>\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, {{nowrap|(3''x'' + 5)/(1 &minus; 2''x'')<sup>2</sup>}} can be decomposed in the form

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

[[Clearing denominators]] shows that {{nowrap|1=3''x'' + 5 = ''A'' + ''B''(1&nbsp;&minus;&nbsp;2''x'')}}. Expanding and equating the coefficients of powers of ''x'' gives

: 5 = ''A'' + ''B'' and  3''x'' = &minus;2''Bx''

Solving for ''A'' and ''B'' yields ''A'' = 13/2 and ''B'' = &minus;3/2.  Hence,

: <math>\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>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> 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 (''x''&nbsp;−&nbsp;''x''<sub>''i''</sub>)<sup>&minus;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>a_{ij} = \operatorname{Res}(g_{ij},x_i).</math>

This is given directly by the formula

: <math>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>a_{i1}=\frac{P(x_i)}{Q'(x_i)},</math>

when

: <math>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 ''f''(''x'') be any rational function over the [[real number]]s.  In other words, suppose there exist real polynomials functions ''p''(''x'') and ''q''(''x'')≠ 0, such that

: <math>f(x) = \frac{p(x)}{q(x)}</math>

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

: <math>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_nx+c_n)^{k_n}</math>

where ''a''<sub>1</sub>,..., ''a''<sub>''m''</sub>, ''b''<sub>1</sub>,..., ''b''<sub>''n''</sub>, ''c''<sub>1</sub>,..., ''c''<sub>''n''</sub> are real numbers with ''b''<sub>''i''</sub><sup>2</sup> − 4''c''<sub>''i''</sub> < 0, and ''j''<sub>1</sub>,..., ''j''<sub>''m''</sub>, ''k''<sub>1</sub>,..., ''k''<sub>''n''</sub> are positive integers.  The terms (''x'' − ''a''<sub>''i''</sub>) are the ''linear factors'' of ''q''(''x'') which correspond to real roots of ''q''(''x''), and the terms (''x''<sub>''i''</sub><sup>2</sup> + ''b''<sub>''i''</sub>''x'' + ''c''<sub>''i''</sub>) are the ''irreducible quadratic factors'' of ''q''(''x'') which correspond to pairs of [[complex number|complex]] conjugate roots of ''q''(''x'').

Then the partial fraction decomposition of ''f''(''x'') is the following:

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

Here, the denominator splits into two distinct linear factors:

: <math>q(x)=x^2+2x-3=(x+3)(x-1)</math>

so we have the partial fraction decomposition

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

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

: <math>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>&nbsp;−&nbsp;4×8 = − 16}} is negative. Thus the partial fraction decomposition over the reals has the shape

: <math>\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>4x^2-8x+16 = A(x^2-4x+8)+(Bx+C)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>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>\frac{x}{x^2-4x+8}=\frac{D}{x-(2+2i)}+\frac{E}{x-(2-2i)}</math>

Multiplying through by the denominator gives:

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

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

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

Using these values we can write:

: <math>\begin{align}
2x^6-4x^5&+5x^4-3x^3+x^2+3x = \\[4pt]
&= A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+(x^2+1)^2+(Dx+(A-B))(x-1)^3(x^2+1)+(x-1)^3 \\[4pt]
&= (A + D) x^6 + (-A - 3D) x^5 + (2B + 4D + 1) x^4 + (-2B - 4D + 1) x^3 + (-A + 2B + 3D - 1) x^2 + (A - 2B - D + 3) 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>\begin{cases} A+D=2 \\ -A-3D = -4 \end{cases} \quad \Rightarrow \quad A= D = 1.</math>

Therefore:

:<math>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>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> 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> 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>\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3},</math>
are 
:<math> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2},</math>
respectively, and

:<math> 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> \frac{1}{x^3 - 1}</math>

First, factor the denominator which determines the decomposition:

:<math> \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>\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>\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>A = \frac{1}{3}.</math>

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

:<math>\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>\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>\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>\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>\int x^2+3+ \frac{-3x+7}{(x+2)(x-1)} \,dx</math>

Upon this, we may now perform partial fraction decomposition.

:<math>\int x^2+3+ \frac{-3x+7}{(x+2)(x-1)} \,dx = \int x^2+3+ \frac{A}{(x+2)}+\frac{B}{(x-1)} \,dx</math>
so:
:<math>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>A=\frac{-13}{3} \ , B=\frac{4}{3} </math>

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

:<math>\int x^2+3+ \frac{-13/3}{(x+2)}+\frac{4/3}{(x-1)} \,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>P(x), Q(x), A_1(x),\ldots, A_r(x)</math>

be real or complex polynomials 
assume that

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

satisfies
:<math>\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>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>\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>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>\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>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>  P- \sum_{j=1}^{r}Q_jA_j</math>

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

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

we then have

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

== Notes ==
{{Reflist}}

== References ==
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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]].

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