partial fractions for polynomials
The theory is used for, for example, the method of partial fraction decomposition for integrating rational functions over the reals (http://planetmath.org/ALectureOnThePartialFractionDecompositionMethod).
The proofs involve fairly elementary algebra only. Although we refer to Euclidean domains in our proofs, the reader who is not familiar with abstract algebra may simply read that as “set of polynomials” (which is one particular Euclidean domain).
Also note that the proofs themselves furnish a method for actually computing the partial fraction decomposition, as a finite-time algorithm, provided the irreducible factorization of the denominator is known. It is not an efficient way to find the partial fraction decomposition; usually one uses instead the method of making substitutions into the polynomials, to derive linear constraints on the coefficients. But what is important is that the existence proofs here justify the substitution method. The uniqueness property proved here might also simplify some calculations: it shows that we never have to consider multiple solutions for the coefficients in the decomposition.
Let and be polynomials over a field, and be any positive integer. Then there exist unique polynomials such that
But according to the division algorithm for polynomials (also known as long division), the quotient and remainder polynomial after a division (by in this case) are unique. So must be uniquely determined. Then we can rearrange:
By uniqueness of division again (by ), is determined. Repeating this process, we see that all the polynomials and are uniquely determined. ∎
Let and be polynomials over a field. Let be the factorization of to irreducible factors (which is unique except for the ordering and constant factors). Then there exist unique polynomials such that
Existence has already been proven as a special case of partial fractions in Euclidean domains; we now prove uniqueness. Suppose equation (2) has been given. First, multiply the equation by :
The polynomial sum on the far right of this equation has degree , because each summand has degree . So the polynomial sum is the remainder of a division of by . Then the quotient polynomial is uniquely determined.
Now suppose and are polynomials of degree , such that
We claim that . Let and , and write
for some polynomials and . Rearranging, we get:
In particular, divides the left side. Since is relatively prime from , it must divide the factor . But , hence must be the zero polynomial. That is, .
So we can cancel the term on both sides of equation (3). And we could repeat the argument, and show that and are the same, and are the same, and so on. Therefore, we have shown that the polynomials in the following expression
are unique. In particular, is the following numerator that results when the fractions are put under a common denominator :
uniquely determines . (Note that the proof of Theorem 1 shows that , as .) ∎
|Title||partial fractions for polynomials|
|Date of creation||2013-03-22 15:40:22|
|Last modified on||2013-03-22 15:40:22|
|Last modified by||stevecheng (10074)|
|Synonym||partial fraction decomposition of rational functions|
|Synonym||partial fractions for rational functions|