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Unformatted text preview: 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 + p 1 s + p 2 s 2 + + p m s m q + 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 + 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 , a 1 , . . . , a k } C such that P ( s ) factors into a product of linear terms P ( s ) = a k Y i = 1 ( s a i ) i = a ( s a 1 ) 1 ( s a 2 ) 2 . . . ( s a k ) k where the degrees match: 1 + 2 + + k = n, i > . 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 i )( s + i ) 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 + 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 , 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 h ( s a 1 ) 1 ( s a 2 ) 2 . . . ( s a j ) j ih ( s 2 + 1 s + c 1 ) 1 ( s 2 + 2 s + c 2 ) 2 . . . ( s 2 + m s + c m ) m i where the degrees match: 1 + 2 + + j + 2 1 + 2 2 + + 2 m = n , i , i > . 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 ), lets fix the notation and assume we have Q ( s ) factored like so: Q ( s ) = a j Y i = 1 ( s a i ) i m Y k = 1 (( s a k ) 2 + b 2 k ) k (1) Recall that one can always write the quadratic terms like this by completing the square: s 2 + i s + c i = s 2 + i s + i 2 2 i 2 2 !...
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This note was uploaded on 09/26/2011 for the course MATH 3113 taught by Professor Dickey during the Fall '08 term at The University of Oklahoma.
 Fall '08
 DICKEY
 Fractions, Rational Functions

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