# Partial fraction decomposition

(Redirected from Partial fractions in integration)

In algebra, the partial fraction decomposition or partial fraction expansion of a rational function (that is, a fraction such that the numerator and the denominator are both polynomials) 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.[1]

The importance of the partial fraction decomposition lies in the fact that it provides algorithms for various computations with rational functions, including the explicit computation of antiderivatives,[2] Taylor series expansionsinverse Z-transformsinverse Laplace transforms. The concept was discovered in 1702 by both Johann Bernoulli and Gottfried Leibnizindependently.[3]

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

${displaystyle {frac {f(x)}{g(x)}}}$

where f and g are polynomials, into an expression of the form

${displaystyle sum _{j}{frac {f_{j}(x)}{g_{j}(x)}}}$

where:

• the denominatorgj (x), of each fraction is a power of an 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 ${displaystyle R(x)={frac {f(x)}{g(x)}}}$ in one indeterminate x has a denominator that factors as

${displaystyle g(x)=P(x)cdot Q(x)}$

over a field K (we can take this to be real numbers, or complex numbers) and if in addition P and Q have no common factor, then by Bézout's identity for polynomials, there exist polynomials C(x) and D(x) such that ${displaystyle deg(C)${displaystyle deg(D), and

${displaystyle CP+DQ=1}$

Thus

${displaystyle {frac {1}{g(x)}}={frac {CP+DQ}{PQ}}={frac {C}{Q}}+{frac {D}{P}}}$

and hence R may be written as

${displaystyle R={frac {f(x)}{g(x)}}={frac {Df(x)}{P}}+{frac {Cf(x)}{Q}},}$

where all numerators are polynomials.

Using this idea inductively the rational function R(x) can be written as a sum with denominators being powers of irreducible polynomials. To take this further, if required, write:

${displaystyle {frac {G(x)}{F(x)^{n}}}}$

as a sum with denominators powers of F and numerators of degree lower than the degree of F, plus a possible extra polynomial. This can be done by the Euclidean algorithm for polynomials. The result is the following theorem:

Theorem — Let f and g be nonzero polynomials over a field K. Write g as a product of powers of distinct irreducible polynomials :

${displaystyle g=prod _{i=1}^{k}p_{i}^{n_{i}}.}$

There are (unique) polynomials b and aij with deg aij < deg pi such that

${displaystyle {frac {f}{g}}=b+sum _{i=1}^{k}sum _{j=1}^{n_{i}}{frac {a_{ij}}{p_{i}^{j}}}.}$

If deg f < deg g, then b = 0.

If K is field of complex numbers, the fundamental theorem of algebra implies that all pi have degree one, and all numerators ${displaystyle a_{ij}}$ are constants. When K is the field of real numbers, some of the pi 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 pi may be the factors of the square-free factorization of g. When K is the field of rational numbers, as it is typically the case in computer algebra, this allows to replace factorization by 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

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:

${displaystyle g=prod _{i=1}^{k}p_{i}^{n_{i}}.}$

There are (unique) polynomials b and cij with deg cij < deg pi such that

${displaystyle {frac {f}{g}}=b+sum _{i=1}^{k}sum _{j=2}^{n_{i}}left({frac {c_{ij}}{p_{i}^{j-1}}} ight)'+sum _{i=1}^{k}{frac {c_{i1}}{p_{i}}}.}$

where ${displaystyle X'}$ denotes the derivative of ${displaystyle X.}$

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

${displaystyle {frac {c_{i1}}{p_{i}}}=sum _{alpha _{j}:p_{i}(alpha _{j})=0}{frac {c_{i1}(alpha _{j})}{p'_{i}(alpha _{j})}}{frac {1}{x-alpha _{j}}}.}$

There are various methods to compute above decomposition. The one that is the simplest to describe is probably the so-called Hermite's method. As the degree of cij is bounded by the degree of pi, 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 ${displaystyle P(x)}$ and ${displaystyle Q(x)=(x-alpha _{1})(x-alpha _{2})cdots (x-alpha _{n})}$, where the αi are distinct constants and deg P < n, partial fractions are generally obtained by supposing that

${displaystyle {frac {P(x)}{Q(x)}}={frac {c_{1}}{x-alpha _{1}}}+{frac {c_{2}}{x-alpha _{2}}}+cdots +{frac {c_{n}}{x-alpha _{n}}}}$

and solving for the ci 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

${displaystyle {frac {P(x)}{Q(x)}}=sum _{i=1}^{n}{frac {P(alpha _{i})}{Q'(alpha _{i})}}{frac {1}{(x-alpha _{i})}}}$

where ${displaystyle Q'}$ is the derivative of the polynomial ${displaystyle Q}$.

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

• If ${displaystyle deg Pgeqslant deg Q,}$ then it is necessary to perform the Euclidean division of P by Q, using polynomial long division, giving P(x) = E(xQ(x) + R(x) with deg R < n. Dividing by Q(x) this gives
${displaystyle {frac {P(x)}{Q(x)}}=E(x)+{frac {R(x)}{Q(x)}},}$
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:
${displaystyle {frac {x^{2}+1}{(x+2)(x-1)color {Blue}(x^{2}+x+1)}}={frac {a}{x+2}}+{frac {b}{x-1}}+{frac {color {OliveGreen}cx+d}{color {Blue}x^{2}+x+1}}.}$
• Suppose Q(x) = (x − α)rS(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:
${displaystyle {frac {P(x)}{Q(x)}}={frac {P(x)}{(x-alpha )^{r}}}={frac {c_{1}}{x-alpha }}+{frac {c_{2}}{(x-alpha )^{2}}}+cdots +{frac {c_{r}}{(x-alpha )^{r}}}.}$

### Illustration

In an example application of this procedure, (3x + 5)/(1 − 2x)2 can be decomposed in the form

${displaystyle {frac {3x+5}{(1-2x)^{2}}}={frac {A}{(1-2x)^{2}}}+{frac {B}{(1-2x)}}.}$

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,

${displaystyle {frac {3x+5}{(1-2x)^{2}}}={frac {13/2}{(1-2x)^{2}}}+{frac {-3/2}{(1-2x)}}.}$

### Residue method

Over the complex numbers, suppose f(x) is a rational proper fraction, and can be decomposed into

${displaystyle f(x)=sum _{i}left({frac {a_{i1}}{x-x_{i}}}+{frac {a_{i2}}{(x-x_{i})^{2}}}+cdots +{frac {a_{ik_{i}}}{(x-x_{i})^{k_{i}}}} ight).}$

Let

${displaystyle g_{ij}(x)=(x-x_{i})^{j-1}f(x),}$

then according to the uniqueness of Laurent seriesaij is the coefficient of the term (x − xi)−1 in the Laurent expansion of gij(x) about the point xi, i.e., its residue

${displaystyle a_{ij}=operatorname {Res} (g_{ij},x_{i}).}$

This is given directly by the formula

${displaystyle a_{ij}={frac {1}{(k_{i}-j)!}}lim _{x o x_{i}}{frac {d^{k_{i}-j}}{dx^{k_{i}-j}}}left((x-x_{i})^{k_{i}}f(x) ight),}$

or in the special case when xi is a simple root,

${displaystyle a_{i1}={frac {P(x_{i})}{Q'(x_{i})}},}$

when

${displaystyle f(x)={frac {P(x)}{Q(x)}}.}$

## Over the reals

Partial fractions are used in real-variable integral calculus to find real-valued 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 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

${displaystyle f(x)={frac {p(x)}{q(x)}}}$

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

${displaystyle q(x)=(x-a_{1})^{j_{1}}cdots (x-a_{m})^{j_{m}}(x^{2}+b_{1}x+c_{1})^{k_{1}}cdots (x^{2}+b_{n}x+c_{n})^{k_{n}}}$

where a1,..., amb1,..., bnc1,..., cn are real numbers with bi2 − 4ci < 0, and j1,..., jmk1,..., kn are positive integers. The terms (x − ai) are the linear factors of q(x) which correspond to real roots of q(x), and the terms (xi2 + bix + ci) 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 f(x) is the following:

${displaystyle f(x)={frac {p(x)}{q(x)}}=P(x)+sum _{i=1}^{m}sum _{r=1}^{j_{i}}{frac {A_{ir}}{(x-a_{i})^{r}}}+sum _{i=1}^{n}sum _{r=1}^{k_{i}}{frac {B_{ir}x+C_{ir}}{(x^{2}+b_{i}x+c_{i})^{r}}}}$

Here, P(x) is a (possibly zero) polynomial, and the AirBir, and Cir 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 AirBir, and Cir. 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

${displaystyle f(x)={frac {1}{x^{2}+2x-3}}}$

Here, the denominator splits into two distinct linear factors:

${displaystyle q(x)=x^{2}+2x-3=(x+3)(x-1)}$

so we have the partial fraction decomposition

${displaystyle f(x)={frac {1}{x^{2}+2x-3}}={frac {A}{x+3}}+{frac {B}{x-1}}}$

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

${displaystyle 1=A(x-1)+B(x+3)}$

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

${displaystyle f(x)={frac {1}{x^{2}+2x-3}}={frac {1}{4}}left({frac {-1}{x+3}}+{frac {1}{x-1}} ight)}$

### Example 2

${displaystyle f(x)={frac {x^{3}+16}{x^{3}-4x^{2}+8x}}}$

After long-division, we have

${displaystyle f(x)=1+{frac {4x^{2}-8x+16}{x^{3}-4x^{2}+8x}}=1+{frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}}$

The factor x2 − 4x + 8 is irreducible over the reals, as its discriminant (−4)2 − 4×8 = − 16 is negative. Thus the partial fraction decomposition over the reals has the shape

${displaystyle {frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}={frac {A}{x}}+{frac {Bx+C}{x^{2}-4x+8}}}$

Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity

${displaystyle 4x^{2}-8x+16=A(x^{2}-4x+8)+(Bx+C)x}$

Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 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,

${displaystyle f(x)=1+2left({frac {1}{x}}+{frac {x}{x^{2}-4x+8}} ight)}$

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:

${displaystyle {frac {x}{x^{2}-4x+8}}={frac {D}{x-(2+2i)}}+{frac {E}{x-(2-2i)}}}$

Multiplying through by the denominator gives:

${displaystyle x=D(x-(2-2i))+E(x-(2+2i))}$

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

${displaystyle D={frac {1+i}{2i}}={frac {1-i}{2}},qquad E={frac {1-i}{-2i}}={frac {1+i}{2}}.}$

Thus we have a complete decomposition:

${displaystyle f(x)={frac {x^{3}+16}{x^{3}-4x^{2}+8x}}=1+{frac {2}{x}}+{frac {1-i}{x-(2+2i)}}+{frac {1+i}{x-(2-2i)}}}$

One may also compute directly AD and 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.

${displaystyle f(x)={frac {x^{9}-2x^{6}+2x^{5}-7x^{4}+13x^{3}-11x^{2}+12x-4}{x^{7}-3x^{6}+5x^{5}-7x^{4}+7x^{3}-5x^{2}+3x-1}}}$

After long-division and factoring the denominator, we have

${displaystyle f(x)=x^{2}+3x+4+{frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}}$

The partial fraction decomposition takes the form

${displaystyle {frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}={frac {A}{x-1}}+{frac {B}{(x-1)^{2}}}+{frac {C}{(x-1)^{3}}}+{frac {Dx+E}{x^{2}+1}}+{frac {Fx+G}{(x^{2}+1)^{2}}}.}$

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

{displaystyle {egin{aligned}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{aligned}}}

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

${displaystyle {egin{cases}4=4C&x=1\2+2i=(Fi+G)(2+2i)&x=i\0=A-B+C-E-G&x=0end{cases}}}$

Solving this we have:

${displaystyle {egin{cases}C=1\F=0,G=1\E=A-Bend{cases}}}$

Using these values we can write:

{displaystyle {egin{aligned}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)xend{aligned}}}

We compare the coefficients of x6 and x5 on both side and we have:

${displaystyle {egin{cases}A+D=2\-A-3D=-4end{cases}}quad Rightarrow quad A=D=1.}$

Therefore:

${displaystyle 2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x=2x^{6}-4x^{5}+(2B+5)x^{4}+(-2B-3)x^{3}+(2B+1)x^{2}+(-2B+3)x}$

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

${displaystyle f(x)=x^{2}+3x+4+{frac {1}{(x-1)}}+{frac {1}{(x-1)^{3}}}+{frac {x+1}{x^{2}+1}}+{frac {1}{(x^{2}+1)^{2}}}.}$

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

${displaystyle 2cdot 6-4cdot 5+5cdot 4-3cdot 3+2+3=Acdot (0+0)+Bcdot (4+0)+8+Dcdot 0}$

that is 8 = 4B + 8 so B = 0.

### Example 4 (residue method)

${displaystyle f(z)={frac {z^{2}-5}{(z^{2}-1)(z^{2}+1)}}={frac {z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}}}$

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

${displaystyle {frac {P(z_{i})}{Q'(z_{i})}}={frac {z_{i}^{2}-5}{4z_{i}^{3}}},}$

are

${displaystyle 1,-1,{ frac {3i}{2}},-{ frac {3i}{2}},}$

respectively, and

${displaystyle f(z)={frac {1}{z+1}}-{frac {1}{z-1}}+{frac {3i}{2}}{frac {1}{z+i}}-{frac {3i}{2}}{frac {1}{z-i}}.}$

### Example 5 (limit method)

Limits can be used to find a partial fraction decomposition.[4] Consider the following example:

${displaystyle {frac {1}{x^{3}-1}}}$

First, factor the denominator which determines the decomposition:

${displaystyle {frac {1}{x^{3}-1}}={frac {1}{(x-1)(x^{2}+x+1)}}={frac {A}{x-1}}+{frac {Bx+C}{x^{2}+x+1}}.}$

Multiplying everything by ${displaystyle x-1}$, and taking the limit when ${displaystyle x o 1}$, we get

${displaystyle lim _{x o 1}left((x-1)left({frac {A}{x-1}}+{frac {Bx+C}{x^{2}+x+1}} ight) ight)=lim _{x o 1}A+lim _{x o 1}{frac {(x-1)(Bx+C)}{x^{2}+x+1}}=A.}$

On the other hand

${displaystyle lim _{x o 1}{frac {(x-1)}{(x-1)(x^{2}+x+1)}}=lim _{x o 1}{frac {1}{x^{2}+x+1}}={frac {1}{3}},}$

and thus:

${displaystyle A={frac {1}{3}}.}$

Multiplying by x and taking the limit when ${displaystyle x o infty }$, we have

${displaystyle lim _{x o infty }xleft({frac {A}{x-1}}+{frac {Bx+C}{x^{2}+x+1}} ight)=lim _{x o infty }{frac {Ax}{x-1}}+lim _{x o infty }{frac {Bx^{2}+Cx}{x^{2}+x+1}}=A+B,}$

and

${displaystyle lim _{x o infty }{frac {x}{(x-1)(x^{2}+x+1)}}=0.}$

This implies A + B = 0 and so ${displaystyle B=-{frac {1}{3}}}$.

For x = 0, we get ${displaystyle -1=-A+C,}$ and thus ${displaystyle C=-{ frac {2}{3}}}$.

Putting everything together, we get the decomposition

${displaystyle {frac {1}{x^{3}-1}}={frac {1}{3}}left({frac {1}{x-1}}+{frac {-x-2}{x^{2}+x+1}} ight).}$

### Example 6 (integral)

Suppose we have the indefinite integral:

${displaystyle int {frac {x^{4}+x^{3}+x^{2}+1}{x^{2}+x-2}},dx}$

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

${displaystyle int x^{2}+3+{frac {-3x+7}{(x+2)(x-1)}},dx}$

Upon this, we may now perform partial fraction decomposition.

${displaystyle int x^{2}+3+{frac {-3x+7}{(x+2)(x-1)}},dx=int x^{2}+3+{frac {A}{(x+2)}}+{frac {B}{(x-1)}},dx}$

so:

${displaystyle A(x-1)+B(x+2)=-3x+7}$.

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:

${displaystyle A={frac {-13}{3}} ,B={frac {4}{3}}}$

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

${displaystyle int x^{2}+3+{frac {-13/3}{(x+2)}}+{frac {4/3}{(x-1)}},dx={frac {x^{3}}{3}} +3x-{frac {13}{3}}ln(|x+2|)+{frac {4}{3}}ln(|x-1|)+C}$

## The role of the Taylor polynomial

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

${displaystyle P(x),Q(x),A_{1}(x),ldots ,A_{r}(x)}$

be real or complex polynomials assume that

${displaystyle Q=prod _{j=1}^{r}(x-lambda _{j})^{ u _{j}},}$

satisfies

${displaystyle deg A_{1}< u _{1},ldots ,deg A_{r}< u _{r},quad { ext{and}}quad deg(P)

Also define

${displaystyle Q_{i}=prod _{j eq i}(x-lambda _{j})^{ u _{j}}={frac {Q}{(x-lambda _{i})^{ u _{i}}}},qquad 1leqslant ileqslant r.}$

Then we have

${displaystyle {frac {P}{Q}}=sum _{j=1}^{r}{frac {A_{j}}{(x-lambda _{j})^{ u _{j}}}}}$

if, and only if, each polynomial ${displaystyle A_{i}(x)}$ is the Taylor polynomial of ${displaystyle { frac {P}{Q_{i}}}}$ of order ${displaystyle u _{i}-1}$ at the point ${displaystyle lambda _{i}}$:

${displaystyle A_{i}(x):=sum _{k=0}^{ u _{i}-1}{frac {1}{k!}}left({frac {P}{Q_{i}}} ight)^{(k)}(lambda _{i}) (x-lambda _{i})^{k}.}$

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

${displaystyle {frac {P}{Q_{i}}}=A_{i}+O((x-lambda _{i})^{ u _{i}}),qquad { ext{for }}x o lambda _{i},}$

so ${displaystyle A_{i}}$ is the Taylor polynomial of ${displaystyle { frac {P}{Q_{i}}}}$, because of the unicity of the polynomial expansion of order ${displaystyle u _{i}-1}$, and by assumption ${displaystyle deg A_{i}< u _{i}}$.

Conversely, if the ${displaystyle A_{i}}$ are the Taylor polynomials, the above expansions at each ${displaystyle lambda _{i}}$ hold, therefore we also have

${displaystyle P-Q_{i}A_{i}=O((x-lambda _{i})^{ u _{i}}),qquad { ext{for }}x o lambda _{i},}$

which implies that the polynomial ${displaystyle P-Q_{i}A_{i}}$ is divisible by ${displaystyle (x-lambda _{i})^{ u _{i}}.}$

For ${displaystyle j eq i,Q_{j}A_{j}}$ is also divisible by ${displaystyle (x-lambda _{i})^{ u _{i}}}$, so

${displaystyle P-sum _{j=1}^{r}Q_{j}A_{j}}$

is divisible by ${displaystyle Q}$. Since

${displaystyle deg left(P-sum _{j=1}^{r}Q_{j}A_{j} ight)

we then have

${displaystyle P-sum _{j=1}^{r}Q_{j}A_{j}=0,}$

and we find the partial fraction decomposition dividing by ${displaystyle Q}$.

## 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:

${displaystyle {frac {1}{18}}={frac {1}{2}}-{frac {1}{3}}-{frac {1}{3^{2}}}.}$

## Notes

1. ^ Larson, Ron (2016). Algebra & Trigonometry. Cengage Learning. ISBN 9781337271172.
2. ^ Horowitz, Ellis. "Algorithms for partial fraction decomposition and rational function integration." Proceedings of the second ACM symposium on Symbolic and algebraic manipulation. ACM, 1971.
3. ^ Grosholz, Emily (2000). The Growth of Mathematical Knowledge. Kluwer Academic Publilshers. p. 179. ISBN 978-90-481-5391-6.
4. ^ Bluman, George W. (1984). Problem Book for First Year Calculus. New York: Springer-Verlag. pp. 250–251.

## References

• Rao, K. R.; Ahmed, N. (1968). "Recursive techniques for obtaining the partial fraction expansion of a rational function". IEEE Trans. Educ11 (2). pp. 152–154. doi:10.1109/TE.1968.4320370.
• Henrici, Peter (1971). "An algorithm for the incomplete decomposition of a rational function into partial fractions". Z. Angew. Math. Phys22 (4). pp. 751–755. doi:10.1007/BF01587772.
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## قسم الرياضيات

### التوقيت والتقويم

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### إعلانات

1- الاختبار الفصلى الثانى لمقرر التحليل العددى (يوم الاحد الموافق 3 / 7/ 1440 هـ)

2- الاختبار الفصلى الثانى لمقررحساب المتجهات (يوم الثلاثاء الموافق 5 / 7 / 1440 هـ)

الأثنين: 10 - 12

الثلاثاء: 8 - 10

الأربعاء: 8 - 10

### آلة حاسبة

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### إحصائية الموقع

عدد الصفحات: 256

البحوث والمحاضرات: 155

الزيارات: 140439