In

_{''ij''}) < deg(''p''_{''i''}), so one may write each ''c_{ij}'' 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.

_{''i''} are distinct constants and deg ''P'' < ''n'', partial fractions are generally obtained by supposing that
: $\backslash frac\; =\; \backslash frac\; +\; \backslash frac\; +\; \backslash cdots\; +\; \backslash frac$
and solving for the ''c''_{''i''} 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 to Lagrange interpolation, consists of writing
: $\backslash frac\; =\; \backslash sum\_^n\; \backslash frac\backslash frac$
where $Q\text{'}$ is the derivative of the polynomial $Q$. The coefficients of $\backslash tfrac$ 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 $\backslash deg\; P\; \backslash geqslant\; \backslash deg\; Q,$ 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 ''R'' < ''n''. Dividing by ''Q''(''x'') this gives
:: $\backslash frac\; =\; E(x)\; +\; \backslash frac,$
:and then seek partial fractions for the remainder fraction (which by definition satisfies deg ''R'' < deg ''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 ''N'' < deg ''F'', rather than as a constant. For example, take the following decomposition over R:
:: $\backslash frac\; =\; \backslash frac\; +\; \backslash frac\; +\; \backslash frac.$
* Suppose and , that is is a root of of Multiplicity (mathematics)#Multiplicity of a root of a polynomial, multiplicity . In the partial fraction decomposition, the first powers of will occur as denominators of the partial fractions (possibly with a zero numerator). For example, if the partial fraction decomposition has the form
:: $\backslash frac\; =\; \backslash frac\; =\; \backslash frac\; +\; \backslash frac\; +\; \backslash cdots\; +\; \backslash frac.$

_{''ij''} is the coefficient of the term (''x'' − ''x''_{''i''})^{−1} in the Laurent expansion of ''g''_{''ij''}(''x'') about the point ''x''_{''i''}, i.e., its residue (complex analysis), residue
: $a\_\; =\; \backslash operatorname(g\_,x\_i).$
This is given directly by the formula
: $a\_\; =\; \backslash frac\; 1\; \backslash lim\_\backslash frac\; \backslash left((x-x\_i)^\; f(x)\backslash right),$
or in the special case when ''x''_{''i''} is a simple root,
: $a\_=\backslash frac,$
when
: $f(x)=\backslash frac.$

_{1},..., ''a''_{''m''}, ''b''_{1},..., ''b''_{''n''}, ''c''_{1},..., ''c''_{''n''} are real numbers with ''b''_{''i''}^{2} − 4''c''_{''i''} < 0, and ''j''_{1},..., ''j''_{''m''}, ''k''_{1},..., ''k''_{''n''} are positive integers. The terms (''x'' − ''a''_{''i''}) are the ''linear factors'' of ''q''(''x'') which correspond to real roots of ''q''(''x''), and the terms (''x''_{''i''}^{2} + ''b''_{''i''}''x'' + ''c''_{''i''}) 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:
: $f(x)\; =\; \backslash frac\; =\; P(x)\; +\; \backslash sum\_^m\backslash sum\_^\; \backslash frac\; +\; \backslash sum\_^n\backslash sum\_^\; \backslash frac$
Here, ''P''(''x'') is a (possibly zero) polynomial, and the ''A''_{''ir''}, ''B''_{''ir''}, and ''C''_{''ir''} 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''_{''ir''}, ''B''_{''ir''}, and ''C''_{''ir''}. 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).

^{2} − 4''x'' + 8 is irreducible over the reals, as its discriminant is negative. Thus the partial fraction decomposition over the reals has the shape
: $\backslash frac=\backslash frac+\backslash frac$
Multiplying through by ''x''^{3} − 4''x''^{2} + 8''x'', we have the polynomial identity
: $4x^2-8x+16\; =\; A(x^2-4x+8)+(Bx+C)x$
Taking ''x'' = 0, we see that 16 = 8''A'', so ''A'' = 2. Comparing the ''x''^{2} 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,
: $f(x)=1+2\backslash left(\backslash frac+\backslash frac\backslash right)$
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:
:$\backslash frac=\backslash frac+\backslash frac$
Multiplying through by the denominator gives:
:$x=D(x-(2-2i))+E(x-(2+2i))$
Equating the coefficients of and the constant (with respect to ) coefficients of both sides of this equation, one gets a system of two linear equations in and , whose solution is
:$D=\backslash frac=\backslash frac,\; \backslash qquad\; E=\backslash frac=\backslash frac.$
Thus we have a complete decomposition:
:$f(x)=\backslash frac=1+\backslash frac+\backslash frac+\backslash frac$
One may also compute directly and with the residue method (see also example 4 below).

^{6} and ''x''^{5} on both side and we have:
:$\backslash begin\; A+D=2\; \backslash \backslash \; -A-3D\; =\; -4\; \backslash end\; \backslash quad\; \backslash Rightarrow\; \backslash quad\; A=\; D\; =\; 1.$
Therefore:
:$2x^6-4x^5+5x^4-3x^3+x^2+3x\; =\; 2x^6\; -4x^5\; +\; (2B\; +\; 5)\; x^4\; +\; (-2B\; -\; 3)\; x^3\; +\; (2B\; +1)\; x^2\; +\; (-\; 2B\; +\; 3)\; x$
which gives us ''B'' = 0. Thus the partial fraction decomposition is given by:
: $f(x)=x^2+3x+4+\backslash frac\; +\; \backslash frac\; +\; \backslash frac+\backslash frac.$
Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at $x\; =\; 1,\; \backslash imath$ in the above polynomial identity. (To this end, recall that the derivative at ''x'' = ''a'' of (''x'' − ''a'')^{''m''}''p''(''x'') vanishes if ''m'' > 1 and is just ''p''(''a'') for ''m'' = 1.) For instance the first derivative at ''x'' = 1 gives
: $2\backslash cdot6-4\backslash cdot5+5\backslash cdot4-3\backslash cdot3+2+3\; =\; A\backslash cdot(0+0)\; +\; B\backslash cdot(\; 4+\; 0)\; +\; 8\; +\; D\backslash cdot0$
that is 8 = 4''B'' + 8 so ''B'' = 0.

Make partial fraction decompositions

with Scilab. {{Authority control Algebra Elementary algebra Partial fractions

algebra
Algebra (from ar, الجبر, lit=reunion of broken parts, bonesetting, translit=al-jabr) is one of the areas of mathematics, broad areas of mathematics, together with number theory, geometry and mathematical analysis, analysis. In its most ge ...

, the partial fraction decomposition or partial fraction expansion of a rational fraction
In algebra
Algebra (from ar, الجبر, lit=reunion of broken parts, bonesetting, translit=al-jabr) is one of the areas of mathematics, broad areas of mathematics, together with number theory, geometry and mathematical analysis, analysis. In ...

(that is, a fraction
A fraction (from Latin ', "broken") represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths ...

such that the numerator and the denominator are both polynomial
In mathematics, a polynomial is an expression (mathematics), expression consisting of indeterminate (variable), indeterminates (also called variable (mathematics), variables) and coefficients, that involves only the operations of addition, subtra ...

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.
The importance of the partial fraction decomposition lies in the fact that it provides algorithm
of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers ''a'' and ''b'' in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" ...

s for various computations with rational function
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...

s, including the explicit computation of antiderivative
In calculus
Calculus, originally called infinitesimal calculus or "the calculus of infinitesimal
In mathematics, infinitesimals or infinitesimal numbers are quantities that are closer to zero than any standard real number, but are not zer ...

s, Taylor series expansions, inverse Z-transforms, and inverse Laplace transform
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...

s. The concept was discovered independently in 1702 by both Johann Bernoulli
Johann Bernoulli (also known as Jean or John; – 1 January 1748) was a Swiss
Swiss may refer to:
* the adjectival form of Switzerland
*Swiss people
Places
*Swiss, Missouri
*Swiss, North Carolina
*Swiss, West Virginia
*Swiss, Wisconsin
Other ...

and Gottfried Leibniz
Gottfried Wilhelm (von) Leibniz ; see inscription of the engraving depicted in the "#1666–1676, 1666–1676" section. ( – 14 November 1716) was a German polymath active as a mathematician, philosopher, scientist, and diplomat. He is a promin ...

.
In symbols, the ''partial fraction decomposition'' of a rational fraction of the form
$\backslash textstyle\; \backslash frac,$
where and are polynomials, is its expression as
: $\backslash frac=p(x)\; +\; \backslash sum\_j\; \backslash frac$
where
is a polynomial, and, for each ,
the denominator
A fraction (from Latin
Latin (, or , ) is a classical language belonging to the Italic languages, Italic branch of the Indo-European languages. Latin was originally spoken in the area around Rome, known as Latium. Through the power of the Rom ...

is a power
Power typically refers to:
* Power (physics)
In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, p ...

of an irreducible polynomial
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...

(that is not factorable into polynomials of positive degrees), and
the numerator
A fraction (from Latin
Latin (, or , ) is a classical language belonging to the Italic languages, Italic branch of the Indo-European languages. Latin was originally spoken in the area around Rome, known as Latium. Through the power of the Rom ...

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 mathematics, a square-free polynomial is a polynomial defined over a field (mathematics), field (or more generally, an integral domain) that does not have as a divisibility (ring theory), divisor any square of a constant polynomial, non-constant ...

" in the description of the outcome. This allows replacing polynomial factorization
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...

by the much easier to compute square-free factorization
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...

. This is sufficient for most applications, and avoids introducing irrational coefficients when the coefficients of the input polynomials are integer
An integer (from the Latin
Latin (, or , ) is a classical language belonging to the Italic languages, Italic branch of the Indo-European languages. Latin was originally spoken in the area around Rome, known as Latium. Through the power of t ...

s or rational number
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...

s.
Basic principles

Let :$R(x)\; =\; \backslash frac\; FG$ be arational fraction
In algebra
Algebra (from ar, الجبر, lit=reunion of broken parts, bonesetting, translit=al-jabr) is one of the areas of mathematics, broad areas of mathematics, together with number theory, geometry and mathematical analysis, analysis. In ...

, where and are univariate polynomial
In mathematics, a polynomial is an expression (mathematics), expression consisting of variable (mathematics), variables (also called indeterminate (variable), indeterminates) and coefficients, that involves only the operations of addition, subtra ...

s in the indeterminate 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 and such that :$\backslash frac\; FG=E+\backslash fracG,$ and :$\backslash deg\; F\_1\; <\backslash deg\; G,$ where $\backslash deg\; P$ denotes thedegree
Degree may refer to:
As a unit of measurement
* Degree symbol (°), a notation used in science, engineering, and mathematics
* Degree (angle), a unit of angle measurement
* Degree (temperature), any of various units of temperature measurement ...

of the polynomial .
This results immediately from the Euclidean division
In arithmetic, Euclidean division – or division with remainder – is the process of division (mathematics), dividing one integer (the dividend) by another (the divisor), in a way that produces a quotient and a remainder smaller than the divisor ...

of by , which asserts the existence of and such that $F=EG+F\_1$ and $\backslash deg\; F\_1\; <\backslash deg\; G.$
This allows supposing in the next steps that $\backslash deg\; F\; <\backslash deg\; G.$
Factors of the denominator

If $\backslash deg\; F\; <\backslash deg\; G,$ and :$G=G\_1G\_2,$ where and arecoprime polynomials
In algebra, the greatest common divisor (frequently abbreviated as GCD) of two polynomial
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, stru ...

, then there exist polynomials $F\_1$ and $F\_2$ such that
:$\backslash frac\; FG=\backslash frac+\backslash frac,$
and
:$\backslash deg\; F\_1\; <\; \backslash deg\; G\_1\backslash quad\backslash text\backslash quad\backslash deg\; F\_2\; <\; \backslash deg\; G\_2.$
This can be proved as follows. Bézout's identity
In elementary number theory, Bézout's identity (also called Bézout's lemma) is the following theorem
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical ...

asserts the existence of polynomials and such that
:$CG\_1\; +\; DG\_2\; =\; 1$
(by hypothesis, is a Polynomial greatest common divisor, greatest common divisor of and ).
Let $DF=G\_1Q+F\_1$ with $\backslash deg\; F\_1\; <\; \backslash deg\; G\_1$ be the Euclidean division
In arithmetic, Euclidean division – or division with remainder – is the process of division (mathematics), dividing one integer (the dividend) by another (the divisor), in a way that produces a quotient and a remainder smaller than the divisor ...

of by $G\_1.$ Setting $F\_2=CF+QG\_2,$ one gets
:$\backslash begin\; \backslash frac\; FG\&=\backslash frac\; =\backslash frac+\backslash frac\backslash \backslash \; \&=\backslash frac+\backslash frac\backslash \backslash \; \&=\backslash frac+\backslash frac.\; \backslash end$
It remains to show that $\backslash deg\; F\_2\; <\; \backslash deg\; G\_2.$ By reducing the last sum of fractions to a common denominator, one gets
$F=F\_2G\_1+F\_1G\_2,$
and thus
:$\backslash begin\; \backslash deg\; F\_2\; \&=\backslash deg(F-F\_1G\_2)-\backslash deg\; G\_1\; \backslash le\; \backslash max(\backslash deg\; F,\backslash deg\; (F\_1G\_2))-\backslash deg\; G\_1\backslash \backslash \; \&<\; \backslash max(\backslash deg\; G,\backslash deg(G\_1G\_2))-\backslash deg\; G\_1=\; \backslash deg\; G\_2\; \backslash end$
Powers in the denominator

Using the preceding decomposition inductively one gets fractions of the form $\backslash frac\; F\; ,$ with $\backslash deg\; F\; <\; \backslash deg\; G^k=\; k\backslash deg\; G,$ where is anirreducible polynomial
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...

. If , one can decompose further, by using that an irreducible polynomial is a square-free polynomial
In mathematics, a square-free polynomial is a polynomial defined over a field (mathematics), field (or more generally, an integral domain) that does not have as a divisibility (ring theory), divisor any square of a constant polynomial, non-constant ...

, that is, $1$ is a Polynomial greatest common divisor, greatest common divisor of the polynomial and its derivative. If $G\text{'}$ is the derivative of , Bézout's identity
In elementary number theory, Bézout's identity (also called Bézout's lemma) is the following theorem
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical ...

provides polynomials and such that $CG+DG\text{'}=1$ and thus $F=FCG+FDG\text{'}.$ Euclidean division of $FDG\text{'}$ by $G$ gives polynomials $H\_k$ and $Q$ such that $FDG\text{'}=QG+H\_k$ and $\backslash deg\; H\_k\; <\backslash deg\; G.$ Setting $F\_=FC+Q,$ one gets
:$\backslash frac\; F\; =\backslash frac+\backslash frac,$
with $\backslash deg\; H\_k\; <\backslash deg\; G.$
Iterating this process with $\backslash frac$ in place of $\backslash frac\; F$ leads eventually to the following theorem.
Statement

The uniqueness can be proved as follows. Let . All together, and the have coefficients. The shape of the decomposition defines a linear map from coefficient vectors to polynomials of degree less than . The existence proof means that this map is surjective. As the two vector spaces 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 is field of complex numbers, the fundamental theorem of algebra implies that all have degree one, and all numerators $a\_$ are constants. When is the field of real numbers, some of the 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 may be the factors of thesquare-free factorization
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...

of . When is the field of rational number
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...

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 This reduces the computation of theantiderivative
In calculus
Calculus, originally called infinitesimal calculus or "the calculus of infinitesimal
In mathematics, infinitesimals or infinitesimal numbers are quantities that are closer to zero than any standard real number, but are not zer ...

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''Procedure

Given two polynomials $P(x)$ and $Q(x)\; =\; (x-\backslash alpha\_1)(x-\backslash alpha\_2)\; \backslash cdots\; (x-\backslash alpha\_n)$, where the ''α''Illustration

In an example application of this procedure, can be decomposed in the form : $\backslash frac\; =\; \backslash frac\; +\; \backslash frac.$ Clearing denominators shows that . Expanding and equating the coefficients of powers of gives : and Solving this system of linear equations for and yields . Hence, : $\backslash frac\; =\; \backslash frac\; +\; \backslash frac.$Residue method

Over the complex numbers, suppose ''f''(''x'') is a rational proper fraction, and can be decomposed into : $f(x)\; =\; \backslash sum\_i\; \backslash left(\; \backslash frac\; +\; \backslash frac\; +\; \backslash cdots\; +\; \backslash frac\; \backslash right).$ Let : $g\_(x)=(x-x\_i)^f(x),$ then according to the Laurent series#Uniqueness, uniqueness of Laurent series, ''a''Over the reals

Partial fractions are used in real number, real-variable integral calculus to find real-valuedantiderivative
In calculus
Calculus, originally called infinitesimal calculus or "the calculus of infinitesimal
In mathematics, infinitesimals or infinitesimal numbers are quantities that are closer to zero than any standard real number, but are not zer ...

s of rational function
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...

s. Partial fraction decomposition of real rational function
In mathematics
Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...

s is also used to find their Inverse Laplace transforms. 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 numbers. In other words, suppose there exist real polynomials functions ''p''(''x'') and ''q''(''x'')≠ 0, such that : $f(x)\; =\; \backslash frac$ 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 : $q(x)\; =\; (x-a\_1)^\backslash cdots(x-a\_m)^(x^2+b\_1x+c\_1)^\backslash cdots(x^2+b\_nx+c\_n)^$ where ''a''Examples

Example 1

: $f(x)=\backslash frac$ Here, the denominator splits into two distinct linear factors: : $q(x)=x^2+2x-3=(x+3)(x-1)$ so we have the partial fraction decomposition : $f(x)=\backslash frac\; =\backslash frac+\backslash frac$ Multiplying through by the denominator on the left-hand side gives us the polynomial identity : $1=A(x-1)+B(x+3)$ Substituting ''x'' = −3 into this equation gives ''A'' = −1/4, and substituting ''x'' = 1 gives ''B'' = 1/4, so that : $f(x)\; =\backslash frac\; =\backslash frac\backslash left(\backslash frac+\backslash frac\backslash right)$Example 2

: $f(x)=\backslash frac$ After Polynomial long division, long division, we have : $f(x)=1+\backslash frac=1+\backslash frac$ The factor ''x''Example 3

This example illustrates almost all the "tricks" we might need to use, short of consulting a computer algebra system. : $f(x)=\backslash frac$ After Polynomial long division, long division and polynomial factorization, factoring the denominator, we have : $f(x)=x^2+3x+4+\backslash frac$ The partial fraction decomposition takes the form :$\backslash frac\; =\; \backslash frac+\backslash frac+\backslash frac+\backslash frac+\backslash frac.$ Multiplying through by the denominator on the left-hand side we have the polynomial identity :$\backslash begin\; 2x^6-4x^5\&+5x^4-3x^3+x^2+3x\; =\; \backslash \backslash [4pt]\; \&=A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2\; +C(x^2+1)^2\; +(Dx+E)(x-1)^3(x^2+1)+(Fx+G)(x-1)^3\; \backslash end$ Now we use different values of ''x'' to compute the coefficients: :$\backslash begin\; 4\; =\; 4C\; \&\; x\; =1\; \backslash \backslash \; 2\; +\; 2i\; =\; (Fi\; +\; G)\; (2+\; 2i)\; \&\; x\; =\; i\; \backslash \backslash \; 0\; =\; A-\; B\; +C\; -\; E\; -\; G\; \&\; x\; =\; 0\; \backslash end$ Solving this we have: :$\backslash begin\; C\; =\; 1\; \backslash \backslash \; F\; =0,\; G\; =1\; \backslash \backslash \; E\; =\; A-B\backslash end$ Using these values we can write: : $\backslash begin\; 2x^6-4x^5\&+5x^4-3x^3+x^2+3x\; =\; \backslash \backslash [4pt]\; \&=\; A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+(x^2+1)^2+(Dx+(A-B))(x-1)^3(x^2+1)+(x-1)^3\; \backslash \backslash [4pt]\; \&=\; (A\; +\; D)\; x^6\; +\; (-A\; -\; 3D)\; x^5\; +\; (2B\; +\; 4D\; +\; 1)\; x^4\; +\; (-2B\; -\; 4D\; +\; 1)\; x^3\; +\; (-A\; +\; 2B\; +\; 3D\; -\; 1)\; x^2\; +\; (A\; -\; 2B\; -\; D\; +\; 3)\; x\; \backslash end$ We compare the coefficients of ''x''Example 4 (residue method)

:$f(z)=\backslash frac=\backslash frac$ 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 :$\backslash frac\; =\; \backslash frac,$ are :$1,\; -1,\; \backslash tfrac,\; -\backslash tfrac,$ respectively, and :$f(z)=\backslash frac-\backslash frac+\backslash frac\backslash frac-\backslash frac\backslash frac.$Example 5 (limit method)

Limit (mathematics), Limits can be used to find a partial fraction decomposition. Consider the following example: :$\backslash frac$ First, factor the denominator which determines the decomposition: :$\backslash frac\; =\; \backslash frac\; =\; \backslash frac\; +\; \backslash frac.$ Multiplying everything by $x-1$, and taking the limit when $x\; \backslash to\; 1$, we get :$\backslash lim\_\; \backslash left((x-1)\backslash left\; (\; \backslash frac\; +\; \backslash frac\; \backslash right\; )\backslash right)\; =\; \backslash lim\_\; A\; +\; \backslash lim\_\backslash frac\; =A.$ On the other hand, :$\backslash lim\_\; \backslash frac\; =\; \backslash lim\_\backslash frac\; =\; \backslash frac13,$ and thus: :$A\; =\; \backslash frac.$ Multiplying by and taking the limit when $x\; \backslash to\; \backslash infty$, we have :$\backslash lim\_\; x\backslash left(\; \backslash frac\; +\; \backslash frac\; \backslash right\; )=\; \backslash lim\_\; \backslash frac\; +\; \backslash lim\_\; \backslash frac=\; A+B,$ and :$\backslash lim\_\; \backslash frac\; =0.$ This implies and so $B\; =\; -\backslash frac$. For , we get $-1\; =\; -A\; +\; C,$ and thus $C\; =\; -\backslash tfrac$. Putting everything together, we get the decomposition :$\backslash frac\; =\; \backslash frac\; \backslash left(\; \backslash frac\; +\; \backslash frac\; \backslash right\; ).$Example 6 (integral)

Suppose we have the indefinite integral: :$\backslash int\; \backslash frac\; \backslash ,dx$ Before performing decomposition, it is obvious we must perform polynomial long division and Factorization, factor the denominator. Doing this would result in: :$\backslash int\; x^2+3+\; \backslash frac\; \backslash ,dx$ Upon this, we may now perform partial fraction decomposition. :$\backslash int\; x^2+3+\; \backslash frac\; \backslash ,dx\; =\; \backslash int\; x^2+3+\; \backslash frac+\backslash frac\; \backslash ,dx$ so: :$A(x-1)+B(x+2)=-3x+7$. 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: :$A=\backslash frac\; \backslash \; ,\; B=\backslash frac$ Plugging all of this back into our integral allows us to find the answer: :$\backslash int\; x^2+3+\; \backslash frac+\backslash frac\; \backslash ,dx\; =\; \backslash frac\; \backslash \; +3x-\backslash frac\; \backslash ln(,\; x+2,\; )+\backslash frac\; \backslash ln(,\; x-1,\; )+C$The role of the Taylor polynomial

The partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let :$P(x),\; Q(x),\; A\_1(x),\backslash ldots,\; A\_r(x)$ be real or complex polynomials assume that :$Q=\backslash prod\_^(x-\backslash lambda\_j)^,$ satisfies :$\backslash deg\; A\_1<\backslash nu\_1,\; \backslash ldots,\; \backslash deg\; A\_r<\backslash nu\_r,\; \backslash quad\; \backslash text\; \backslash quad\; \backslash deg(P)<\backslash deg(Q)=\backslash sum\_^\backslash nu\_j.$ Also define :$Q\_i=\backslash prod\_(x-\backslash lambda\_j)^=\backslash frac,\; \backslash qquad\; 1\; \backslash leqslant\; i\; \backslash leqslant\; r.$ Then we have :$\backslash frac=\backslash sum\_^\backslash frac$ if, and only if, each polynomial $A\_i(x)$ is the Taylor polynomial of $\backslash tfrac$ of order $\backslash nu\_i-1$ at the point $\backslash lambda\_i$: :$A\_i(x):=\backslash sum\_^\; \backslash frac\backslash left(\backslash frac\backslash right)^(\backslash lambda\_i)\backslash \; (x-\backslash lambda\_i)^k.$ 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 ≤ ''i'' ≤ ''r'', a polynomial expansion :$\backslash frac=A\_i\; +\; O((x-\backslash lambda\_i)^),\; \backslash qquad\; \backslash text\; x\backslash to\backslash lambda\_i,$ so $A\_i$ is the Taylor polynomial of $\backslash tfrac$, because of the unicity of the polynomial expansion of order $\backslash nu\_i-1$, and by assumption $\backslash deg\; A\_i<\backslash nu\_i$. Conversely, if the $A\_i$ are the Taylor polynomials, the above expansions at each $\backslash lambda\_i$ hold, therefore we also have :$P-Q\_i\; A\_i\; =\; O((x-\backslash lambda\_i)^),\; \backslash qquad\; \backslash text\; x\backslash to\backslash lambda\_i,$ which implies that the polynomial $P-Q\_iA\_i$ is divisible by $(x-\backslash lambda\_i)^.$ For $j\backslash neq\; i,\; Q\_jA\_j$ is also divisible by $(x-\backslash lambda\_i)^$, so :$P-\; \backslash sum\_^Q\_jA\_j$ is divisible by $Q$. Since :$\backslash deg\backslash left(\; P-\; \backslash sum\_^Q\_jA\_j\; \backslash right)\; <\; \backslash deg(Q)$ we then have :$P-\; \backslash sum\_^Q\_jA\_j=0,$ and we find the partial fraction decomposition dividing by $Q$.Fractions of integers

The idea of partial fractions can be generalized to other integral domains, say the ring ofinteger
An integer (from the Latin
Latin (, or , ) is a classical language belonging to the Italic languages, Italic branch of the Indo-European languages. Latin was originally spoken in the area around Rome, known as Latium. Through the power of t ...

s where prime numbers take the role of irreducible denominators. For example:
: $\backslash frac\; =\; \backslash frac\; -\; \backslash frac\; -\; \backslash frac.$
Notes

References

* * * * * * * * * * * * *External links

* *Make partial fraction decompositions

with Scilab. {{Authority control Algebra Elementary algebra Partial fractions