In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is a fraction such that the numerator and the denominator are both polynomials) is the operation that consists in 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 an algorithm for computing the antiderivative of a rational function.
In symbols, one can use partial fraction expansion to change a rational fraction in the form
 $\backslash frac\{f(x)\}\{g(x)\}$
where ƒ and g are polynomials, into an expression of the form
 $\backslash sum\_j\; \backslash frac\{f\_j(x)\}\{g\_j(x)\}$
where g_{j} (x) are polynomials that are factors of g(x), and are in general of lower degree.
Thus, the partial fraction decomposition may be seen as the inverse procedure of the more elementary operation of addition of rational fractions, which produces a single rational fraction with a numerator and denominator usually of high degree.
The full decomposition pushes the reduction as far as it will go: in other words, the factorization of g is used as much as possible. Thus, the outcome of a full partial fraction expansion expresses that fraction as a sum of fractions, where:
 the denominator of each term is a power of an irreducible (not factorable) polynomial and
 the numerator is a polynomial of smaller degree than that irreducible polynomial. To decrease the degree of the numerator directly, the Euclidean division can be used, but in fact if ƒ already has lower degree than g this isn't helpful.
Basic principles
The basic principles involved are quite simple; it is the algorithmic aspects that require attention in particular cases. On the other hand, the existence of a decomposition of a certain kind is an assumption in practical cases, and the principles should explain which assumptions are justified.
Assume a rational function R(x) = ƒ(x)/g(x) in one indeterminate x has a denominator that factors as
 $g(x)\; =\; P(x)\; \backslash cdot\; Q(x)\; \backslash ,$
over a field K (we can take this to be real numbers, or complex numbers). If P and Q have no common factor, then R may be written as
 $\backslash frac\{A\}\{P\}\; +\; \backslash frac\{B\}\{Q\}$
for some polynomials A(x) and B(x) over K. The existence of such a decomposition is a consequence of the fact that the polynomial ring over K is a principal ideal domain, so that
 $CP\; +\; DQ\; =\; 1\; \backslash ,$
for some polynomials C(x) and D(x) (see Bézout's identity).
Using this idea inductively we can write R(x) as a sum with denominators powers of irreducible polynomials. To take this further, if required, write:
 $\backslash frac\; \{G(x)\}\{F(x)^n\}$
as a sum with denominators powers of F and numerators of degree less than F, plus a possible extra polynomial. This can be done by the Euclidean algorithm, polynomial case. The result is the following theorem:
Let ƒ and g be nonzero polynomials over a field K. Write g as a product of powers of distinct irreducible polynomials :
 $g=\backslash prod\_\{i=1\}^k\; p\_i^\{n\_i\}.$
There are (unique) polynomials b and a_{ij} with deg a_{ij} < deg p_{i} such that
 $\backslash frac\{f\}\{g\}=b+\backslash sum\_\{i=1\}^k\backslash sum\_\{j=1\}^\{n\_i\}\backslash frac\{a\_\{ij$
{p_i^j}.
If deg ƒ < deg g, then b = 0.}}
Therefore, when the field K is the complex numbers, we can assume that each p_{i} has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. When K is the real numbers, some of the p_{i} might be quadratic, so in the partial fraction decomposition a quotient of a linear polynomial by a power of a quadratic might occur.
In the preceding theorem, one may replace "distinct irreducible polynomials" by "pairwise coprime polynomials that are coprime with their derivative". For example, the p_{i} may be the factors of the squarefree factorization of g. When K is the field of the rational numbers, as it is typically the case in computer algebra, this allows to replace factorization by greatest common divisor to compute the partial fraction decomposition.
Application to symbolic integration
For the purpose of symbolic integration, the preceding result may be refined into
Let ƒ 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:
 $g=\backslash prod\_\{i=1\}^k\; p\_i^\{n\_i\}.$
There are (unique) polynomials b and c_{ij} with deg c_{ij} < deg p_{i} such that
 $\backslash frac\{f\}\{g\}=b+\backslash sum\_\{i=1\}^k\backslash sum\_\{j=2\}^\{n\_i\}\backslash left(\backslash frac\{c\_\{ij$
{p_i^{j1}}\right)' +
\sum_{i=1}^k \frac{c_{i1}}{p_i}.
where $X\text{'}$ denotes the derivative of $X.$}}
This reduces the computation of the antiderivative of a rational function to the integration of the last sum, with is called the logarithmic part, because its antiderivative is a linear combination of logarithms. In fact, we have
 $\backslash frac\{c\_\{i1\}\}\{p\_i\}=\backslash sum\_\{\backslash alpha\_j:p\_i(\backslash alpha\_j)=0\}\backslash frac\{c\_\{i1\}(\backslash alpha\_j)\}\{p\text{'}\_i(\backslash alpha\_j)\}\backslash frac\{1\}\{x\backslash alpha\_j\}.$
There are various methods to compute above decomposition. The one that is the simplest to describe is probably the socalled Hermite's method. As the degree of c_{ij} is bounded by the degree of p_{i}, 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 $P(x)$ and $Q(x)\; =\; (x\backslash alpha\_1)(x\backslash alpha\_2)\; \backslash cdots\; (x\backslash alpha\_n)$, where the α_{i} are distinct constants and deg P < n, partial fractions are generally obtained by supposing that
 $\backslash frac\{P(x)\}\{Q(x)\}\; =\; \backslash frac\{c\_1\}\{x\backslash alpha\_1\}\; +\; \backslash frac\{c\_2\}\{x\backslash alpha\_2\}\; +\; \backslash cdots\; +\; \backslash frac\{c\_n\}\{x\backslash alpha\_n\}$
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 with Lagrange interpolation consists in writing
 $\backslash frac\{P(x)\}\{Q(x)\}\; =\; \backslash sum\_\{i=1\}^n\; \backslash frac\{P(\backslash alpha\_i)\}\{Q\text{'}(\backslash alpha\_i)\}\backslash frac\{1\}\{(x\backslash alpha\_i)\}$
where $Q\text{'}$ is the derivative of the polynomial $Q$.
This approach does not account for several other cases, but can be modified accordingly:
 $\backslash frac\{P(x)\}\{Q(x)\}\; =\; E(x)\; +\; \backslash frac\{R(x)\}\{Q(x)\},$
 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:
 $\backslash frac\{x^2\; +\; 1\}\{(x+2)(x1)\backslash color\{Blue\}(x^2+x+1)\}\; =\; \backslash frac\{a\}\{x+2\}\; +\; \backslash frac\{b\}\{x1\}\; +\; \backslash frac\{\backslash color\{OliveGreen\}cx\; +\; d\}\{\backslash color\{Blue\}x^2\; +\; x\; +\; 1\}.$
 Suppose Q(x) = (x − α)^{r}S(x) and S(α) ≠ 0. Then Q(x) has a zero α of 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) = 1 to get the following decomposition:
 $\backslash frac\{P(x)\}\{Q(x)\}\; =\; \backslash frac\{P(x)\}\{(x\backslash alpha)^r\}\; =\; \backslash frac\{c\_1\}\{x\backslash alpha\}\; +\; \backslash frac\{c\_2\}\{(x\backslash alpha)^2\}\; +\; \backslash cdots\; +\; \backslash frac\{c\_r\}\{(x\backslash alpha)^r\}.$
Illustration
In an example application of this procedure, (3x + 5)/(1 − 2x)^{2} can be decomposed in the form
 $\backslash frac\{3x\; +\; 5\}\{(12x)^2\}\; =\; \backslash frac\{A\}\{(12x)^2\}\; +\; \backslash frac\{B\}\{(12x)\}.$
Clearing denominators shows that 3x + 5 = A + B(1 − 2x). Expanding and equating the coefficients of powers of x gives
 5 = A + B and 3x = −2Bx
Solving for A and B yields A = 13/2 and B = −3/2. Hence,
 $\backslash frac\{3x\; +\; 5\}\{(12x)^2\}\; =\; \backslash frac\{13/2\}\{(12x)^2\}\; +\; \backslash frac\{3/2\}\{(12x)\}.$
Residue method
Over the complex numbers, suppose ƒ(x) is a rational proper fraction, and can be decomposed into
 $f(x)\; =\; \backslash sum\_i\; \backslash left(\; \backslash frac\{a\_\{i1\}\}\{x\; \; x\_i\}\; +\; \backslash frac\{a\_\{i2\}\}\{(\; x\; \; x\_i)^2\}\; +\; \backslash cdots\; +\; \backslash frac\{a\_\{i\; k\_i\}\}\{(x\; \; x\_i)^\{k\_i\}\}\; \backslash right).$
Let
 $g\_\{ij\}(x)=(xx\_\{i\})^\{j1\}f(x),$
then according to the uniqueness of Laurent series, a_{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
 $a\_\{ij\}\; =\; \backslash operatorname\{Res\}(g\_\{ij\},x\_i).$
This is given directly by the formula
 $a\_\{ij\}=\backslash frac\{1\}\{(k\_\{i\}j)!\}\backslash lim\_\{x\backslash to\; x\_i\}\backslash frac\{d^\{k\_\{i\}j\}\}\{dx^\{k\_\{i\}j\}\}\backslash left((xx\_\{i\})^\{k\_\{i\}\}f(x)\backslash right),$
or in the special case when x_{i} is a simple root,
 $a\_\{i1\}=\backslash frac\{P(x\_\{i\})\}\{Q\text{'}(x\_\{i\})\},$
when
 $f(x)=\backslash frac\{P(x)\}\{Q(x)\}.$
Note that P(x) and Q(x) may or may not be polynomials.
Over the reals
Partial fractions are used in realvariable integral calculus to find realvalued antiderivatives of rational functions. Partial fraction decomposition of real rational functions is also used to find their Inverse Laplace transforms. For applications of partial fraction decomposition over the reals, see
General result
Let ƒ(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
 $f(x)\; =\; \backslash frac\{p(x)\}\{q(x)\}$
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. By the fundamental theorem of algebra, we can write
 $q(x)\; =\; (xa\_1)^\{j\_1\}\backslash cdots(xa\_m)^\{j\_m\}(x^2+b\_1x+c\_1)^\{k\_1\}\backslash cdots(x^2+b\_nx+c\_n)^\{k\_n\}$
where a_{1},..., a_{m}, b_{1},..., b_{n}, c_{1},..., c_{n} are real numbers with b_{i}^{2} − 4c_{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 conjugate roots of q(x).
Then the partial fraction decomposition of ƒ(x) is the following:
 $f(x)\; =\; \backslash frac\{p(x)\}\{q(x)\}\; =\; P(x)\; +\; \backslash sum\_\{i=1\}^m\backslash sum\_\{r=1\}^\{j\_i\}\; \backslash frac\{A\_\{ir\}\}\{(xa\_i)^r\}\; +\; \backslash sum\_\{i=1\}^n\backslash sum\_\{r=1\}^\{k\_i\}\; \backslash frac\{B\_\{ir\}x+C\_\{ir\}\}\{(x^2+b\_ix+c\_i)^r\}$
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 lefthand side is simply p(x) and whose righthand 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 limits (see Example 5).
Examples
Example 1
 $f(x)=\backslash frac\{1\}\{x^2+2x3\}$
Here, the denominator splits into two distinct linear factors:
 $q(x)=x^2+2x3=(x+3)(x1)$
so we have the partial fraction decomposition
 $f(x)=\backslash frac\{1\}\{x^2+2x3\}\; =\backslash frac\{A\}\{x+3\}+\backslash frac\{B\}\{x1\}$
Multiplying through by x^{2} + 2x − 3, we have the polynomial identity
 $1=A(x1)+B(x+3)$
Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that
 $f(x)\; =\backslash frac\{1\}\{x^2+2x3\}\; =\backslash frac\{1\}\{4\}\backslash left(\backslash frac\{1\}\{x+3\}+\backslash frac\{1\}\{x1\}\backslash right)$
Example 2
 $f(x)=\backslash frac\{x^3+16\}\{x^34x^2+8x\}$
After longdivision, we have
 $f(x)=1+\backslash frac\{4x^28x+16\}\{x^34x^2+8x\}=1+\backslash frac\{4x^28x+16\}\{x(x^24x+8)\}$
Since (−4)^{2} − 4(8) = −16 < 0, x^{2} − 4x + 8 is irreducible, and so
 $\backslash frac\{4x^28x+16\}\{x(x^24x+8)\}=\backslash frac\{A\}\{x\}+\backslash frac\{Bx+C\}\{x^24x+8\}$
Multiplying through by x^{3} − 4x^{2} + 8x, we have the polynomial identity
 $4x^28x+16\; =\; A(x^24x+8)+(Bx+C)x$
Taking x = 0, we see that 16 = 8A, 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 = −4A + C = −8 + C, so C = 0. Altogether,
 $f(x)=1+2\backslash left(\backslash frac\{1\}\{x\}+\backslash frac\{x\}\{x^24x+8\}\backslash right)$
The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.
Example 3
 $f(x)=\backslash frac\{x^92x^6+2x^57x^4+13x^311x^2+12x4\}\{x^73x^6+5x^57x^4+7x^35x^2+3x1\}$
After longdivision and factoring the denominator, we have
 $f(x)=x^2+3x+4+\backslash frac\{2x^64x^5+5x^43x^3+x^2+3x\}\{(x1)^3(x^2+1)^2\}$
The partial fraction decomposition takes the form
 $\backslash frac\{2x^64x^5+5x^43x^3+x^2+3x\}\{(x1)^3(x^2+1)^2\}=\backslash frac\{A\}\{x1\}+\backslash frac\{B\}\{(x1)^2\}+\backslash frac\{C\}\{(x1)^3\}+\backslash frac\{Dx+E\}\{x^2+1\}+\backslash frac\{Fx+G\}\{(x^2+1)^2\}$
Multiplying through by (x − 1)^{3}(x^{2} + 1)^{2} we have the polynomial identity
 $$
\begin{align}
& {} \quad 2x^64x^5+5x^43x^3+x^2+3x \\
& =A(x1)^2(x^2+1)^2+B(x1)(x^2+1)^2+C(x^2+1)^2+(Dx+E)(x1)^3(x^2+1)+(Fx+G)(x1)^3
\end{align}
Taking x = 1 gives 4 = 4C, so C = 1. Similarly, taking x = i gives 2 + 2i = (Fi + G)(2 + 2i), so Fi + G = 1, so F = 0 and G = 1 by equating real and imaginary parts. With C = G = 1 and F = 0, taking x = 0 we get A − B + 1 − E − 1 = 0, thus E = A − B.
We now have the identity
 $$
\begin{align}
& {} 2x^64x^5+5x^43x^3+x^2+3x \\
& = A(x1)^2(x^2+1)^2+B(x1)(x^2+1)^2+(x^2+1)^2+(Dx+(AB))(x1)^3(x^2+1)+(x1)^3 \\
& = A((x1)^2(x^2+1)^2 + (x1)^3(x^2+1)) + B((x1)(x^2+1)  (x1)^3(x^2+1)) + (x^2+1)^2 + Dx(x1)^3(x^2+1)+(x1)^3
\end{align}
Expanding and sorting by exponents of x we get
 $$
\begin{align}
& {} 2 x^6 4 x^5 +5 x^4 3 x^3 + x^2 +3 x \\
& = (A + D) x^6 + (A  3D) x^5 + (2B + 4D + 1) x^4 + (2B  4D + 1) x^3 + (A + 2B + 3D  1) x^2 + (A  2B  D + 3) x
\end{align}
We can now compare the coefficients and see that
 $$
\begin{align}
A + D &=& 2 \\
A  3D &=& 4 \\
2B + 4D + 1 &=& 5 \\
2B  4D + 1 &=& 3 \\
A + 2B + 3D  1 &=& 1 \\
A  2B  D + 3 &=& 3 ,
\end{align}
with A = 2 − D and −A −3 D =−4 we get A = D = 1 and so B = 0, furthermore is C = 1, E = A − B = 1, F = 0 and G = 1.
The partial fraction decomposition of ƒ(x) is thus
 $f(x)=x^2+3x+4+\backslash frac\{1\}\{(x1)\}\; +\; \backslash frac\{1\}\{(x\; \; 1)^3\}\; +\; \backslash frac\{x\; +\; 1\}\{x^2+1\}+\backslash frac\{1\}\{(x^2+1)^2\}.$
Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at x=1 and at x=i 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 it is just p(a) if m=1.)
Thus, for instance the first derivative at x=1 gives
 $2\backslash cdot64\backslash cdot5+5\backslash cdot43\backslash cdot3+2+3\; =\; A\backslash cdot(0+0)\; +\; B\backslash cdot(\; 2+\; 0)\; +\; 8\; +\; D\backslash cdot0$
that is 8 = 2B + 8 so B=0.
Example 4 (residue method)
 $f(z)=\backslash frac\{z^\{2\}5\}\{(z^21)(z^2+1)\}=\backslash frac\{z^\{2\}5\}\{(z+1)(z1)(z+i)(zi)\}$
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
 $\backslash frac\{P(z\_i)\}\{Q\text{'}(z\_i)\}\; =\; \backslash frac\{z\_i^2\; \; 5\}\{4z\_i^3\}$,
are
 $1,\; 1,\; \backslash tfrac\{3i\}\{2\},\; \backslash tfrac\{3i\}\{2\}$,
respectively, and
 $f(z)=\backslash frac\{1\}\{z+1\}\backslash frac\{1\}\{z1\}+\backslash frac\{3i\}\{2\}\backslash frac\{1\}\{z+i\}\backslash frac\{3i\}\{2\}\backslash frac\{1\}\{zi\}$.
Example 5 (limit method)
Limits can be used to find a partial fraction decomposition.^{[1]}
 $f(x)\; =\; \backslash frac\{1\}\{x^3\; \; 1\}$
First, factor the denominator:
 $f(x)\; =\; \backslash frac\{1\}\{(x\; \; 1)(x^2\; +\; x\; +\; 1)\}$
The decomposition takes the form of
 $\backslash frac\{1\}\{(x1)(x^2+x+1)\}\; =\; \backslash frac\{A\}\{x\; \; 1\}\; +\; \backslash frac\{Bx\; +\; C\}\{x^2\; +\; x\; +\; 1\}$
As $x\; \backslash to\; 1$, the A term dominates, so the righthand side approaches $\backslash frac\{A\}\{x\; \; 1\}$. Thus, we have
 $\backslash frac\{1\}\{(x\; \; 1)(x^2\; +\; x\; +\; 1)\}\; =\; \backslash frac\{A\}\{x\; \; 1\}$
 $A\; =\; \backslash lim\_\{x\; \backslash to\; 1\}\{\backslash frac\{1\}\{x^2\; +\; x\; +\; 1\}\}\; =\; \backslash frac\{1\}\{3\}$
As $x\; \backslash to\; \backslash infty$, the righthand side is
 $\backslash lim\_\{x\; \backslash to\; \backslash infty\}\{\backslash frac\{A\}\{x\; \; 1\}\; +\; \backslash frac\{Bx\; +\; C\}\{x^2\; +\; x\; +\; 1\}\}\; =\; \backslash frac\{A\}\{x\}\; +\; \backslash frac\{Bx\}\{x^2\}\; =\; \backslash frac\{A\; +\; B\}\{x\}.$
 $\backslash frac\{A\; +\; B\}\{x\}\; =\; \backslash lim\_\{x\; \backslash to\; \backslash infty\}\{\backslash frac\{1\}\{x^3\; \; 1\}\}\; =\; 0$
Thus, $B\; =\; \backslash frac\{1\}\{3\}$.
At $x=0$, $1\; =\; A\; +\; C$. Therefore, $C\; =\; \backslash frac\{2\}\{3\}$.
The decomposition is thus $\backslash frac\{\backslash frac\{1\}\{3\}\}\{x\; \; 1\}\; +\; \backslash frac\{\backslash frac\{1\}\{3\}x\; \; \backslash frac\{2\}\{3\}\}\{x^2\; +\; x\; +\; 1\}$.
The role of the Taylor polynomial
The partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let
 $P(x),\; Q(x),\; A\_1(x),\backslash dots,\; A\_r(x)$
be real or complex polynomials; assume that
 $\backslash textstyle\; Q=\backslash prod\_\{j=1\}^\{r\}(x\backslash lambda\_j)^\{\backslash nu\_j\},$
that
 $\backslash textstyle\backslash deg(P)<\backslash deg(Q)=\backslash sum\_\{j=1\}^\{r\}\backslash nu\_j,$
and that
 $\backslash textstyle\backslash deg\; A\_j<\backslash nu\_j\backslash text\{\; for\; \}j=1,\backslash dots,r.$
Define also
 $\backslash textstyle\; Q\_i=\backslash prod\_\{j\backslash neq\; i\}(x\backslash lambda\_j)^\{\backslash nu\_j\}=\backslash frac\{Q\}\{(x\backslash lambda\_i)^\{\backslash nu\_i\}\}\; \backslash text\{\; for\; \}i=1,\backslash dots,r.$
Then we have
 $\backslash frac\{P\}\{Q\}=\backslash sum\_\{j=1\}^\{r\}\backslash frac\{A\_j\}\{(x\backslash lambda\_j)^\{\backslash nu\_j\}\}$
if, and only if, for each $\backslash textstyle\; i$ the polynomial $\backslash textstyle\; A\_i(x)$ is the Taylor polynomial of $\backslash textstyle\backslash frac\{P\}\{Q\_i\}$ of order $\backslash textstyle\backslash nu\_i1$ at the point $\backslash textstyle\backslash lambda\_i$:
 $A\_i(x):=\backslash sum\_\{k=0\}^\{\backslash nu\_i1\}\; \backslash frac\{1\}\{k!\}\backslash left(\backslash frac\{P\}\{Q\_i\}\backslash right)^\{(k)\}(\backslash lambda\_i)\backslash \; (x\backslash lambda\_i)^k.$
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
 $\backslash frac\{P\}\{Q\_i\}=A\_i\; +\; O((x\backslash lambda\_i)^\{\backslash nu\_i\})\backslash qquad$, as $x\backslash to\backslash lambda\_i;$
so $\backslash textstyle\; A\_i$ is the Taylor polynomial of $\backslash textstyle\backslash frac\{P\}\{Q\_i\}$, because of the unicity of the polynomial expansion of order $\backslash textstyle\backslash nu\_i1$, and by assumption $\backslash textstyle\backslash deg\; A\_i<\backslash nu\_i$.
Conversely, if the $\backslash textstyle\; A\_i$ are the Taylor polynomials, the above expansions at each $\backslash textstyle\backslash lambda\_i$ hold, therefore we also have
 $PQ\_i\; A\_i\; =\; O((x\backslash lambda\_i)^\{\backslash nu\_i\})\backslash qquad$, as $x\backslash to\backslash lambda\_i,$
which implies that the polynomial $\backslash textstyle\; PQ\_iA\_i$ is divisible by $\backslash textstyle\; (x\backslash lambda\_i)^\{\backslash nu\_i\}.$
For $\backslash textstyle\; j\backslash neq\; i$ also $\backslash textstyle\; Q\_jA\_j$ is divisible by $\backslash textstyle\; (x\backslash lambda\_i)^\{\backslash nu\_i\}$, so we have in turn that $\backslash textstyle\; P\; \backslash sum\_\{j=1\}^\{r\}Q\_jA\_j$ is divisible by $\backslash textstyle\; Q$. Since $\backslash textstyle\backslash deg\backslash left(\; P\; \backslash sum\_\{j=1\}^\{r\}Q\_jA\_j\; \backslash right)\; <\; \backslash deg(Q)$ we then have
$\backslash textstyle\; P\; \backslash sum\_\{j=1\}^\{r\}Q\_jA\_j=0$, and we find the partial fraction decomposition dividing by $\backslash textstyle\; Q$.
Fractions of integers
The idea of partial fractions can be generalized to other integral domains,
say the ring of integers where prime numbers take the role of irreducible denominators.
For example:
 $\backslash frac\{1\}\{18\}\; =\; \backslash frac\{1\}\{2\}\; \; \backslash frac\{1\}\{3\}\; \; \backslash frac\{1\}\{3^2\}.$
Notes
References







 Template:Springer

 Xin, Guoce. "A fast algorithm for partical fraction decompositions". math/0408189.



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