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# Partial Fraction Decomposition

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Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions. To decompose a fraction, you first factor the denominator. Let's work backwards from the example above. The denominator is x2 + x, which factors as x(x + 1). Then you write the fractions with one of the factors for each of the denominators. Of course, you don't know what the numerators are yet, so you assign variables (usually capital letters) for these unknown values: There is another method for solving for the values of A and B. Since the equation "3x + 2 = A(x + 1) + B(x)" is supposed to be true for any value of x, we can pick useful values of x, plug- n-chug, and find the values for A and B. Looking at the equation "3x + 2 = A(x + 1) + B(x)", you can see that, if x = 0, then we quickly find that 2 = A: I've never seen this second method in textbooks, but it can often save you a whole lot of time over the "equate the coefficients and solve the system of equations" method that they usually teach.
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Partial Fraction Decomposition
Partial Fraction Decomposition
Partial-fraction decomposition is the process of starting with the simplified answer and taking it
back apart, of "decomposing" the final expression into its initial polynomial fractions.
To decompose a fraction, you first factor the denominator. Let's work backwards from the
example above. The denominator is x2 + x, which factors as x(x + 1).
Then you write the fractions with one of the factors for each of the denominators. Of course,
you don't know what the numerators are yet, so you assign variables (usual y capital letters)
for these unknown values:
There is another method for solving for the values of A and B. Since the equation "3x + 2 =
A(x + 1) + B(x)" is supposed to be true for any value of x, we can pick useful values of x, plug-
n-chug, and find the values for A and B. Looking at the equation "3x + 2 = A(x + 1) + B(x)",
you can see that, if x = 0, then we quickly find that 2 = A:
I've never seen this second method in textbooks, but it can often save you a whole lot of time
over the "equate the coefficients and solve the system of equations" method that they usual y
teach.
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where gj (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 algebraic fractions, that produces a single rational function
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 ful partial fraction
expansion expresses that function 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 algorithm can be used, but in fact if already
has lower degree than g this isn't helpful.
The main motivation to decompose a rational function into a sum of simpler fractions is that it
makes it simpler to perform linear operations on it. Therefore the problem of computing
derivatives, antiderivatives, integrals, power series expansions.
Fourier series, residues, and linear functional transformations of rational functions can be
reduced, via partial fraction decomposition, to making the computation on each single element
used in the decomposition.
See e.g. partial fractions in integration for an account of the use of the partial fractions in
finding antiderivatives.
Just which polynomials are irreducible depends on which field of scalars one adopts. Thus if
one al ows only real numbers, then irreducible polynomials are of degree either 1 or 2.
If complex numbers are al owed, only 1st-degree polynomials can be irreducible. If one allows
only rational numbers, or a finite field, then some higher-degree polynomials are irreducible.

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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 Air, Bir, 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.

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Partial Fraction Decomposition

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