PartialFractions - the expressions become more complicated....

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Partial Fractions by Michael Pejic 10/4/10 For any n th degree polynomial p ( x ) with distinct roots { x 1 ,...,x n } , it is possible to write 1 p ( x ) = A 1 x - r 1 + A 2 x - r 2 + ··· + A n x - r n for some constants { A 1 ,...,A n } . The usual way to calculate these is to put the right hand side over a common denominator, multiply out the resulting terms, and then equate the coefficients of the various powers of x , giving a system of n linear equations in the n unknowns { A 1 ,...,A n } . This is a lot of unnecessary work. Instead, to find A k , multiply through by x - r k , giving x - r k p ( x ) = A 1 ( x - r k ) x - r 1 + A 2 ( x - r k ) x - r 2 + ··· + A k + ··· + A n ( x - r k ) x - r n Taking the limit of each side as x r k then gives 1 p 0 ( r k ) = A k where L’Hospital’s rule has been used to evaluate the left-hand side. This technique can be extended to polynomials with multiple roots, but
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Unformatted text preview: the expressions become more complicated. For example, if x 1 is a root of multiplicity two, then for 1 p ( x ) = B x-r 1 + C ( x-r 1 ) 2 + A 2 x-r 2 + + A n-1 x-r n-1 A k is still given by 1 p ( r k ) , but C is given by lim x r 1 ( x-r 1 ) 2 p ( x ) = 2 p 00 ( r 1 ) where LHospitals rule has been used twice, and B is given by lim x r 1 x-r 1 p ( x )-C x-r 1 =-2 p 000 ( r 1 ) 3 p 00 ( r 1 ) 2 where LHospitals rule has been used three times and the value of C has been substituted in. 1...
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This note was uploaded on 02/05/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at University of California, Berkeley.

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