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Transcript
partial fractions∗ pahio† 2013-03-21 17:18:37 Every fractional number, i. e. such a rational number m n that the integer m is not divisible by the integer n, can be decomposed to a sum of partial fractions as follows: m m1 m2 mt = ν1 + ν2 + · · · + νt n p1 p2 pt Here, the pi ’s are distinct positive prime numbers, the νi ’s positive integers and the mi ’s some integers. Cf. the partial fractions of expressions. Examples: 6 6 = 289 172 3 1 1 = − 3+ 1 − 24 2 3 1 1 32 24 = − 3+ 2 − 1 504 2 3 7 How to get the numerators mi for decomposing a fractional number n1 to partial fractions? First one can take the highest power pν of a prime p which divides the denominator n. Then n = pν u, where gcd (u, pν ) = 1. Euclid’s algorithm gives some integers x and y such that 1 = xu+ypν . Dividing this equation by pν u gives the decomposition 1 1 x y = ν = ν+ . n p u p u If u has more than one distinct prime factors, a similar procedure can be made for the fraction uy , and so on. ∗ hPartialFractionsi created: h2013-03-21i by: hpahioi version: h35761i Privacy setting: h1i hDefinitioni h11A41i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Note. The numerators m1 , m2 , . . . , mt in the decomposition are not unique. E. g., we have also − 4 1 11 = − 3 + 1. 24 2 3 Cf. the programme “Murto” (in Finnish) or “Murd” (in Estonian) or “Bruch” (in German) or “Bråk” (in Swedish) or “Fraction”(in French) here. 2