Breakdown of rational fractions as sums of partial fractions

Lemma 1: Rational fraction as sum of relatively prime polynomials

If P,Q,S₁,S₂ are polynomials such that:

Q = S₁S₂

S₁,S₂ are relatively prime polynomials

Then, there exists polynomials P₀, P₁, P₂ such that:

P/Q = P₀ + P₁/S₁ + P₂/S₂

deg P_i is less than deg S_i for both i=1 and i=2

Proof:

(1) Since GCD(S₁,S₂) = 1, there exists polynomials T₁, T₂ such that [See Corollary 3.1, here]:

1 = S₁T₁ + S₂T₂

(2) Multiply each side by P/Q to get:

P/Q = (PS₁T₁)/Q + (PS₂T₂)/Q

(3) Since Q=S₁S₂, we now have:

P/Q = (PS₁T₁)/(S₁S₂) + (PS₂T₂)/(S₁S₂) = (PT₁)/S₂ + (PT₂)/S₁

(4) By the Euclidean Division Algorithm for Polynomials (see Theorem, here), there exists U₁, U₂, R₁, R₂ such that:

PT₁ = S₂U₂ + R₂

where deg R₂ is less than deg S₂

PT₂ = S₁U₁ + R₁

where deg R₁ is less than deg S₁

(5) Replacing PT₁ and PT₂ in step #3 gives us:

P/Q = (S₂U₂ + R₂)/S₂ + (S₁U₁ + R₁)/S₁ =

= (S₂U₂)/S₂ + R₂/S₂ + (S₁U₁)/S₁ + R₁/S₁ =

= (U₂ + U₁) + R₁/S₁ + R₂/S₂

QED

Lemma 2:

For polynomials Q,P:

If Q is irreducible and deg P is less than deg Q^m, then:

P/Q^m = P₁/Q + P₂/Q² + ... + P_m/Q^m

where deg P_i is less than deg Q for i = 1, ..., m

Proof:

(1) Using Euclidean Division for Polynomials (see Theorem, here), there exists P₁,R₁ such that:

P = P₁Q^m-1 + R₁ where deg R₁ is less than deg Q^m-1

(2) deg P₁ is less than deg Q since:

(a) Assume deg P₁ ≥ deg Q

(b) Then P₁Q^m-1 is greater than deg Q^m

(c) But this is impossible since deg P is less than deg Q^m and since P₁Q^m-1 is less than P.

(d) So we reject our assumption at step #2a.

(3) We can now repeat this same step with P₂ as a quotient for Q^m-2 and so on.

(4) Eventually, we get to:

P = P₁Q^m-1 + P₂Q^m-2 + ... + P_m-1Q + P_m

(5) If we now divide both sides by Q^m, we get:

P/Q^m = P₁/Q + P₂/Q² + ... + P^m-1/Q^m-1 + P^m/Q^m

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001