Mathematical Methods for Physics and Engineering - Matematica.NET

1.4 PARTIAL FRACTIONSThus any one of the methods listed above shows that4x +2x 2 +3x +2 = −2x +1 + 6x +2 .The best method to use in any particular circumstance will depend on thecomplexity, in terms of the degrees of the polynomials and the multiplicities ofthe roots of the denominator, of the function being considered and, to someextent, on the individual inclinations of the student; some prefer lengthy butstraightforward solution of simultaneous equations, whilst others feel more athome carrying through shorter but more abstract calculations in their heads.1.4.1 Complications and special casesHaving established the basic method for partial fractions, we now show, throughfurther worked examples, how some complications are dealt with by extensionsto the procedure. These extensions are introduced one at a time, but of course inany practical application more than one may be involved.The degree of the numerator is greater than or equal to that of the denominatorAlthough we have not specifically mentioned the fact, it will be apparent fromtrying to apply method (i) of the previous subsection to such a case, that if thedegree of the numerator (m) is not less than that of the denominator (n) then theratio of two polynomials cannot be expressed in partial fractions.To get round this difficulty it is necessary to start by dividing the denominatorh(x) into the numerator g(x) to obtain a further polynomial, which we will denoteby s(x), together with a function t(x) thatis a ratio of two polynomials for whichthe degree of the numerator is less than that of the denominator. The functiont(x) can therefore be expanded in partial fractions. As a formula,f(x) = g(x)r(x)= s(x)+t(x) ≡ s(x)+h(x) h(x) . (1.45)It is apparent that the polynomial r(x) istheremainder obtained when g(x) isdivided by h(x), and, in general, will be a polynomial of degree n − 1. It is alsoclear that the polynomial s(x) will be of degree m − n. Again, the actual divisionprocess can be set out as an algebraic long division sum but is probably moreeasily handled by writing (1.45) in the formor, more explicitly, asg(x) =s(x)h(x)+r(x) (1.46)g(x) =(s m−n x m−n + s m−n−1 x m−n−1 + ···+ s 0 )h(x)+(r n−1 x n−1 + r n−2 x n−2 + ···+ r 0 )(1.47)and then equating coefficients.21