a/(ax + c) - b/(bx + d) = (ad - bc)/[(ax + c)(bx + d)]. This one identity suffices to underpin the entire theory of partial fractions. Note that a(-d/b) + c = (-1/b)(ad - bc) b(-c/a) + d = (+1/a)(ad - bc) These truths are neither accidental nor insignificant.
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Oh, wait -- strictly speaking, the theory of partial fractions also requires this identity: (ax + b)/(cx + d) = a/c - (ad - bc)/[c(cx + d)], which comes into play when a the rational function to be decomposed has a numerator (with degree less than that of the denominator.)
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Also, perhaps I ought to have noted that 1/[(ax + c)(bx + d)] = 1/[(a(-d/b) + c)(bx + d)] + 1/[(ax + c)(b(-c/a) + d)]. From the above it then follows that (mx + n)/[(ax + c)(bx + d)] = (m(-d/b) + n)/[(a(-d/b) + c)(bx + d)] + (m(-c/a) + n)/[(ax + c)(b(-c/a) + d)]
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So here's the general story: P(x)/Q(x) = [(mx + n) ... (px + q)]/[(ax + c) ... (bx + d)] = [(m(-d/b) + n) ... (p(-d/b) + q)]/[(a(-d/b) + c) ... (bx + d)] + ... [(m(-c/a) + n) ... (p(-c/a) + q)]/[(ax + c) ... (b(-c/a) + d)], provided deg(P) < deg(Q).
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Are these secretly facts about SL(2), the group of fractional linear transformations? (I'm just groping around here.)
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Yes, indeed they are. At the bottom of all this is the Euclidean algorithm, which is our natural means of navigation through what John Horton Conway calls the "topograph" of the rational numbers. The theory of partial fractions is inseparable from this object and its structure.
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