Mathematical Methods for Physics and Engineering - Matematica.NET

1.4 PARTIAL FRACTIONSA i but linear functions of x, i.e.oftheformB i x + C i . Thus, in the expansion,linear terms (first-degree polynomials) in the denominator have constants (zerodegreepolynomials) in their numerators, whilst quadratic terms (second-degreepolynomials) in the denominator have linear terms (first-degree polynomials) intheir numerators. As a symbolic formula, the partial fraction expansion ofg(x)(x − α 1 )(x − α 2 ) ···(x − α p )(x 2 + a 2 1 )(x2 + a 2 2 ) ···(x2 + a 2 q )should take the formA 1x − α 1+ A 2x − α 2+ ···+A p+ B 1x + C 1x − α p x 2 + a 2 + B 2x + C 21x 2 + a 2 2+ ···+ B qx + C qx 2 + a 2 .qOf course, the degree of g(x) must be less than p +2q; if it is not, an initialdivision must be carried out as demonstrated earlier.Repeated factors in the denominatorConsider trying (incorrectly) to expandf(x) =x − 4(x +1)(x − 2) 2in partial fraction form as follows:x − 4(x +1)(x − 2) 2 = A 1x +1 + A 2(x − 2) 2 .Multiplying both sides of this supposed equality by (x +1)(x − 2) 2 produces anequation whose LHS is linear in x, whilst its RHS is quadratic. This is clearlywrong and so an expansion in the above form cannot be valid. The correction wemust make is very similar to that needed in the previous subsection, namely thatsince (x − 2) 2 is a quadratic polynomial the numerator of the term containing itmust be a first-degree polynomial, and not simply a constant.The correct form for the part of the expansion containing the doubly repeatedroot is therefore (Bx+ C)/(x − 2) 2 . Using this form and either of methods (i) and(ii) for determining the constants gives the full partial fraction expansion asx − 4(x +1)(x − 2) 2 = − 5 5x − 16+9(x +1) 9(x − 2) 2 ,as the reader may verify.Since any term of the form (Bx + C)/(x − α) 2 can be written asB(x − α)+C + Bα(x − α) 2 = Bx − α + C + Bα(x − α) 2 ,and similarly for multiply repeated roots, an alternative form for the part of thepartial fraction expansion containing a repeated root α isD 1x − α + D 2(x − α) 2 + ···+ D p(x − α) p . (1.48)23