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partial-fractions

# partial-fractions - Partial Fractions Review Introduction...

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— Partial Fractions Review — Introduction. The goal of partial fraction decomposition is to rewrite a rational function in a sim- pler form: split it into a sum of smaller rational functions where each denominator is an irreducible polynomial. Here, irreducible means that it cannot be factored as a product of smaller polynomials. Definition 1. A rational function is a fraction where both the numerator and denominator are (real) polynomials. If the numerator is a polynomial P ( s ) and the denominator is a polynomial Q ( s ), then R ( s ) defined by R ( s ) = P ( s ) Q ( s ) = p 0 + p 1 s + p 2 s 2 + · · · + p m s m q 0 + q 1 s + q 2 s 2 + · · · + q s , p i , q i R , is a rational function. Theorem 2 (Fundamental Theorem of Algebra) . Suppose P ( s ) = p 0 + p 1 s + p 2 s 2 + · · · + p n s n is a polynomial of degree n with p i R . There exists a unique set of numbers { a 0 , a 1 , . . . , a k } ⊆ C such that P ( s ) factors into a product of linear terms P ( s ) = a 0 k Y i = 1 ( s - a i ) ε i = a 0 ( s - a 1 ) ε 1 ( s - a 2 ) ε 2 . . . ( s - a k ) ε k where the degrees match: ε 1 + ε 2 + · · · + ε k = n, ε i > 0 . In this case, each a i is a pole of P ( s ) . However, if you are unwilling to deal with complex numbers, then you may be stuck with quadratic factors that are not reducible in terms of reals (i.e., that are “irreducible over R ”). For example, you can factor s 2 + 1 = ( s - )( s + ) over C , but you cannot write s 2 + 1 in terms of linear factors with real coe ffi cients. More generally, you have the following: Corollary 3. Suppose P ( s ) = p 0 + p 1 s + p 2 s 2 + · · · + p n s n is a polynomial of degree n with p i R . There exists a unique set of real numbers { a 0 , a 1 , . . . , a j , c 1 , c 2 , . . . , c k } such that P ( s ) factors into a product of linear and irreducible quadratic terms P ( s ) = a 0 h ( s - a 1 ) ε 1 ( s - a 2 ) ε 2 . . . ( s - a j ) ε j i h ( s 2 + β 1 s + c 1 ) δ 1 ( s 2 + β 2 s + c 2 ) δ 2 . . . ( s 2 + β m s + c m ) δ m i where the degrees match: ε 1 + ε 2 + · · · + ε j + 2 δ 1 + 2 δ 2 + · · · + 2 δ m = n , ε i , δ i > 0 . Here, each a i is a pole of P ( s ) , and each quadratic factor corresponds to a pair of complex conjugate poles of P ( s ) . Cor. 3 makes partial fractions a bit more obnoxious whenever there are irreducible quadratic factors. Since we want to factor the denominator Q ( s ), let’s fix the notation and assume we have Q ( s ) factored like so: Q ( s ) = a 0 j Y i = 1 ( s - a i ) ε i · m Y k = 1 (( s - a k ) 2 + b 2 k ) δ k (1) Recall that one can always write the quadratic terms like this by completing the square: s 2 + β i s + c i = s 2 + β i s + β i 2 2 - β i 2 2 !

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