Partial Fraction Decomposition

When studying techniques of integration of rational functions in calculus, a typical problem we might run across is to find the partial fraction decomposition, dwhich is shown but not proven in calculus. The partial fraction decomposition of rational functions asserts that every rational function can be written as the sum of a polynomial and partial fractions.

Here we are interested in are understanding, why does setting things up the way we do in Step III always result in a system of linear equations that has a solution? It turns out that the answer lies in properties of polynomials that can be derived from the division algorithm. In order to prove this, we will once again examine and exploit the similarities between the positive integers and polynomials with coefficients in a field. We begin with a lemma about rational numbers which follows from the Euclidean Algorithm.

For example 7/15 can be decomposed as

where A, c₁, c₂ ∈ ℤ, 0 ≤ c₁ < 3, and 0 ≤ c₂ < 5.

Let’s see how to perform this decomposition. Since 5 and 3 are relatively prime, the division algorithm tells us that there exist b₁, b₁ ∈ ℤ such that

There are many possible choices for α₁, α₂. One such choice is α₁ = −1 and α₂ = 2, which gives us

Multiplying both sides by 7 yields

Then dividing both sides by 15 gives us

The division algorithm tells us that

Therefore we can replace −7/3 by −(2+1/3) and also replace 14/5 by 2+4/5. As as result, we now have

Thus A = 0, c₁ = −1 and c₂ = 4 yields the desired decomposition.

More generally, if the number of relatively prime factors exceeds two, we can prove that an appropriate decomposition exists by applying Mathematical Induction to the number of factors of the denominator.

Lemma 5.12.1. Let a/b a be a nonzero rational number, where b = d₁ d₂ ⋅·· d_n, such that each d_i is a positive integer with d_i and d_j relatively prime whenever i ≠ j. Then

where A, c₁, c₂ ∈ ℤ, 0 ≤ c_i < d_i, for all i.

Proof. We will proceed by Mathematical Induction and begin by letting

We must first show that 1 ∈ S. Therefore, we will examine the rational number a/b, where b = d₁. If use the division algorithm to divide a by d₁ and then let A denote the quotient and c₁ the remainder, we obtain

where A, c₁ ∈ ℤ, 0 ≤ c₁ < d₁. Dividing this equation by b = d₁ results in

Thus, we have decomposed a/b and so, 1 ∈ S.

To conclude the proof, it now suffices to show that if a natural number k belongs to S, then so does k + 1. We need to examine the rational number a/b, where b = d₁ d2 ··· d_k d_k+1, such that d_i and d_j are relative prime whenever i ≠ j. Let e = d₁ d2 ··· d_k; observe that e and d_k+1 are relative prime. Therefore, the Euclidean Algorithm implies that there exist α₁, α₂ ∈ ℤ such that

If we multiply this equation by a and let β₁ = aα₁ and β₂ = aα₂, we obtain

Since b = e · d_k+1, dividing this equation by b yields

Using the fact that k ∈ S, we can decompose β₁/e as

where A, c_k+1 ∈ ℤ, 0 ≤ c_k+1 < d_k+1. If we substitute both equation (5.12.2) and the previous equation back into equation (5.12.1) and then let A = A₁ + A₂, we obtain

Since A, c₁, c₂, ..., c_k+1 ∈ ℤ, 0 ≤ c_i < d_i, for all i, we have successfully decomposed a/b. Thus, k +1 ∈ S, thereby proving the inductive step.  □

The only tools used to prove this Lemma are the division algorithm, the Euclidean Algorithm, and Mathematical Induction. In light of this, it should come as no surprise that by using the division algorithm and Euclidean Algorithm for polynomials over a field, we can easily adapt the proof of Lemma 5.12.1 to prove an analogous result for decomposing rational functions. We will keep the notation in Lemma 5.12.2 as close as possible to the notation in Lemma 5.12.1 to make the similarities between positive integers and polynomials as transparent as possible.

Lemma 5.12.2. Let K be a field and let a(x), b(x) be nonzero elements of the polynomial ring K [x]. Suppose b(x) = d₁(x)d₂ (x) ··· d_n (x), where each d_i (x) ∈ K [x] has degree at least one and d_i (x), d_j (x) are relatively prime whenever i ≠ j. Then

where A(x), c₁(x), c₂(x), ..., c(x) ∈ K[x] and, for all i, either c_i(x) = 0 or has degree less than the degree of d_i (x).

Proof. All the key ideas of this proof are contained in the proof of Lemma 5.12.1. It will simply be a matter of making some minor modifications.
We will proceed by Mathematical Induction and begin by letting

We must first show that 1 ∈ S. Therefore, we begin with a(x)/ b(x), where b(x) = d₁(x). Using the division algorithm in K[x] to divide a(x) by d₁(x) and then letting A(x) denote the quotient and c₁(x) the remainder, we obtain

where A(x), c₁(x) ∈ K[x] and 0 ≤ deg c₁ (x) < d₁(x). Dividing this equation by b(x) = d₁(x) results in

Thus, we have decomposed a(x)/b(x) and so, 1 ∈ S.

It now suffices to show that if k belongs to S, then so does k + 1. We need to examine a(x) / b(x), where b(x) = d₁(x) d2(x) ··· d_k(x) d_k+1(x), where d_i (x), d_j (x) are relative prime whenever i ≠ j. Let e(x) = d₁(x)d₂(x) ··· d_k(x); observe that e(x) and d_k+1(x) are relative prime in K [x]. Therefore, the Euclidean Algorithm in K[x] implies that there exist α₁(x), α₂ (x) ∈ K[x] such that

If we multiply this equation by a(x) and let β₁(x) = a(x)α₁(x) and β₂(x) = aα₂(x), we obtain

Since b(x) = e(x)⋅d_{k +1}(x), dividing this equation by b(x) yields

However, k ∈ S, therefore we can decompose β₁(x)/e(x) as

where A₁(x), c₁, c₂, c_i(x) ∈ K[x] and for all i, 0 ≤ deg c_i(x) < d_i(x). Furthermore, since 1 ∈ S, we can rewrite β₂(x)/d_{k +1}(x) as

where A₂ (x), c_k+1 (x) ∈ K [x] and 0 ≤ deg c_k+1 (x) < deg d_k+1(x). Substituting both equation (13) and the previous equation back into equation (12) and then letting A(x) = A₁ (x) + A₂ (x), we obtain

Since A(x), c₁ (x), c₂ (x), ... , c_k+1 (x) ∈ K [x] and, for all i, 0 ≤ deg c_i(x) deg d_i (x), we have successfully decomposed a(x) b(x). Thus, k + 1 ∈ S, thereby concluding the proof.  □