## Tuesday, December 04, 2007

### 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 Pi is less than deg Si 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 Qm, then:

P/Qm = P1/Q + P2/Q2 + ... + Pm/Qm

where deg Pi 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 = P1Qm-1 + R1 where deg R1 is less than deg Qm-1

(2) deg P1 is less than deg Q since:

(a) Assume deg P1 ≥ deg Q

(b) Then P1Qm-1 is greater than deg Qm

(c) But this is impossible since deg P is less than deg Qm and since P1Qm-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 Qm-2 and so on.

(4) Eventually, we get to:

P = P1Qm-1 + P2Qm-2 + ... + Pm-1Q + Pm

(5) If we now divide both sides by Qm, we get:

P/Qm = P1/Q + P2/Q2 + ... + Pm-1/Qm-1 + Pm/Qm

QED

References
• Jean-Pierre Tignol, , World Scientific, 2001