1.1 Forms for fractions px + q An expression of the form (x + r) (x + s) quadratic expression which factorises) may be written as

Transcription

1 1 Partial Fractions x ny rational expression e.g. x (x 2 1) or x 4 x may be written () (x 3) as a sum of simpler fractions. This has uses in many areas e.g. integration or Laplace Transforms. The procedure for carrying out this process is as follows. 1.1 Forms for fractions px + q n expression of the form (x + r) (x + s) quadratic expression which factorises) may be written as (i.e. the denominator is a x + r + x + s. px 2 + qx + r n expression of the form (i.e. the denominator (x + s) (x + t) (x + u) is a cubic expression which factorises completely) may be written as x + s + x + t + C x + u. Similarly, if the denominator is an n th degree polynomial which factorises completely then the expression may be written as the sum of n fractions. n expression of the form px2 + qx + r (x + s) 2 (i.e. the denominator (x + t) consists of a simple factor and also a repeated factor) may be written as x + s + (x + s) 2 + C x + t. n expression of the form px3 + qx 2 + rx + s (x + t) 3 (i.e. the denominator (x + u) consists of a simple factor and also a repeated (cubed) factor) may be written as x + t + (x + t) 2 + C (x + t) 3 + D x + t. n expression of the form px 2 + qx + r (x + s) (x 2 + tx + u) quadratic factor which does not factorise) may be written as x + C x 2 + tx + u. Combinations of the above may occur e.g. may be written as (i.e. there is a x + s + px 4 + qx 3 + rx 2 + sx + t (x + u) (x + v) 2 (x 2 + gx + h). x + u + x + v + C (x + v) 2 + Dx + E x 2 + gx + h 1

2 One possible pitfall concerns fractions where the numerator is of equal or higher order than the denominator. Imagine that the numerator is of order m while the denominator is of order n. efore using one of the forms above, the fraction must be changed to a polynomial of degree m n added to a fraction where the numerator is of smaller order than px 3 + qx 2 + rx + s the denominator. For example, must be written in (x + t) (x + v) gx + h the form x + +. The numerator is of order 3 while (x + t) (x + v) the denominator is of order 2. the polynomial at the beginning is of order 1 i.e. x Finding Coefficients Once the form for a fraction is decided, the procedure for finding the coefficients, etc is as follows 1. Write down an equality. On the left-hand side will be the original expression; on the right-hand side will be the chosen form. 2. Multiply both sides by the denominator of the left-hand side. 3. The result should be a polynomial on the left-hand side and a sum of polynomials on the right-hand side. The equality should hold for all x. It is worth seeing if there are any values of x for which certain terms disappear giving a simple equation in terms of the coefficients, etc. 4. Equations for any remaining coefficients can be found by comparing coefficients of powers of x. The simplest equations can be found by considering the highest and lowest powers. 5. Solve the equations in parts 3 and 4 to find the coefficients. 1.3 Examples x 7 Example 1. in partial fraction form. (x + 3) (x 2) s the denominator consists of two distinct linear factors, the relevant form is x x 2. x 7 So (x + 3) (x 2) = x (Step 1) x 2 Multiply both sides by (x + 3) (x 2) to get x 7 = (x 2) + (x + 3) (Step 2) Setting x = 2 will allow one bracket to disappear i.e. 2 7 = (2 2) + (2 + 3) = 5 = 5. So = 1 2

3 Setting x = 3 will allow the other bracket to disappear i.e. 3 7 = ( 3 2) + ( 3 + 3) = 10 = 5. So = 2 Hence, = 2, = 1 and so x 7 (x + 3) (x 2) = 2 x x 2 Example 2 2x 2 6x 4 in partial fraction form. (x 4) () In this example, both the numerator and the denominator are of order 2., the first step is to express the numerator as a constant times the denominator plus a polynomial of degree less than the denominator i.e. Let 2x 2 6x 2 = P (x 4) () + (Qx + R) i.e. 2x 2 6x 2 = P ( x 2 2x 8 ) + (Qx + R) Comparing coefficients of x 2, 2 = P i.e. P = 2 Comparing coefficients of x, 6 = 2P +Q i.e. 6 = 4+Q = Q = 2 Comparing constants, 2 = 8P + R i.e. 2 = 16 + R = R = 14 2x 2 6x 4 2x + 14 So, = 2 + (x 4) () (x 4) () 2x + 14 The term, may be put into partial fraction form by (x 4) () methods similar to Example 1. gain, the form will be x 4 +, 2x + 14 (x 4) () = x 4 + Multiplying by (x 4) () gives 2x + 14 = () + (x 4) Setting, x = 2 gives = ( 2 + 2) + ( 2 4) = 18 = 6 so = 3 Setting x = 4 gives = (4 + 2) + (4 4) = 6 = 6 so = 1 Hence 2x + 14 (x 4) () = 1 x 4 3 and so 2x 2 6x 4 (x 4) () = x 4 3 Example 3 4x 2 + 9x 3 in partial fractions form. x (x + 1) (x 3) Here the term x can be treated as x + 0. s the denominator consists of three distinct linear terms, the correct partial fraction form is x + x C x 3 4x 2 + 9x 3 So, x (x + 1) (x 3) = x + x C x 3 Multiplying by x (x + 1) (x 3) gives 3

4 4x 2 + 9x 3 = (x + 1) (x 3) + x (x 3) + Cx (x + 1) Letting x = 0, 3 = (1) ( 3) + (0) ( 3) + C (0) (1) = 3 = 3 so = 1 Letting x = 1, 8 = (0) ( 4) + ( 1) ( 4) + C ( 1) (0) = 8 = 4 so = 2 Letting x = 3, 60 = (4) (0) + (3) (0) + C (3) (4) = 60 = 12C so C = 5, 4x 2 + 9x 3 x (x + 1) (x 3) = 1 x 2 x x 3 Example 4. 3x3 + 15x x + 9 (x 1) () 3 in partial fraction form. s the factor () is raised to the third power, there will be a term involving, a term involving () 2 and a term involving () 3. There will also be a term involving x 1. Hence the form is + () 2 + C () 3 + D x 1. 3x3 + 15x x + 9 (x 1) () 3 = + () 2 + C () 3 + D x 1 Multiplying by (x 1) () 3 gives 3x x x + 9 = () 2 (x 1) + () (x 1) + C (x 1) + D () 3 Letting x = 1, 54 = (9) (0) + (3) (0) + C (0) + D (27) = 54 = 27D i.e. D = 2 Letting x = 2, 9 = (0) ( 3) + (0) ( 3) + C ( 3) + D (0) = 9 = 3C i.e. C = 3 There are no more values of x for which brackets cancel, so instead coefficients will be compared. Comparing coefficients of x 3, 3 = + D so = 1 Comparing coefficients of x 0, 9 = 4 2 C + 8D i.e. 9 = so 2 = 0 i.e. = 0 3x3 + 15x x + 9 (x 1) () 3 = () x 1 Example 5. 5x x + 8 (x + 4) (x 2 in partial fraction form. + 3x + 4) s the factor x 2 + 3x + 4 is a quadratic which will not factorise, the x + relevant factor will be x 2. There will also be a term involving + 3x + 4 x + x + 4. Hence the form is x 2 + 3x C x x x + 8 (x + 4) (x 2 + 3x + 4) = x + x 2 + 3x C x + 4 4

5 Multiplying by (x + 4) ( x 2 + 3x + 4 ) gives 5x x + 8 = (x + ) (x + 4) + C ( x 2 + 3x + 4 ) Letting x = 4, 24 = (4 + ) (0) + C (8) = 24 = 8C so C = 3 There are no more values of x for which brackets cancel, so instead coefficients will be compared. Comparing coefficients of x 2, 5 = + C so = 2 Comparing coefficients of x 0, 8 = 4 + 4C i.e. 8 = so = 1 5x x + 8 (x + 4) (x 2 + 3x + 4) = 2x 1 x 2 + 3x x + 4 Example 6 :- nastier example showing the combination of the various patterns. x 6 17x x 4 384x x (x 5) (x 3) 2 in partial (x 2 + x + 8) fraction form. s the numerator is of degree 6 while the denominator is of degree 5, the numerator can be expressed as a polynomial of degree 6 5 = 1 multiplied by the denominator plus another polynomial of degree less than 5. x 6 17x x 4 384x x = (P x + Q) ( x 5 10x x 3 94x x 360 ) + ( Rx 4 + Sx 3 + T x 2 + Ux + V ) Comparing coefficients x 6 : 1 = P x 5 : 17 = 10P + Q = 17 = 10 + Q so Q = 7 x 4 : 111 = 36P 10Q + R = 111 = R so R = 5 x 3 : 384 = 94P + 36Q + S = 384 = S so S = 38 x 2 : 1079 = 267P 94Q + T = 1079 = T so T = 154 x 1 : 2615 = 360P + 267Q + U = 2615 = U so U = 386 x 0 : 2833 = 360Q + V = 2833 = V so V = 313 x 6 17x x 4 384x x ( Rx 4 + Sx 3 + T x 2 + Ux + V ) = (P x + Q) + (x 5 10x x 3 94x x 360) = (x 7) + 5x4 38x x 2 386x The task now becomes one of expressing 5x4 38x x 2 386x in partial fractions form. The single factor in x 5 will give a fraction of the form x 5 ; the factor (x 3)2 gives rise to fractions 5

6 x 3 + C (x 3) 2. Finally, the presence of the term x2 + x + 8 implies a term Dx + E x 2 + x x 4 38x x 2 386x = x 5 + x 3 + Mulitplying by (x 5) (x 3) 2 ( x 2 + x + 8 ) gives 5x 4 38x x 2 386x = (x 3) 2 ( x 2 + x + 8 ) C (x 3) 2 + Dx + E x 2 + x (x 3) (x 5) ( x 2 + x + 8 ) +C (x 5) ( x 2 + x + 8 ) + (Dx + E) (x 5) (x 3) 2 Letting x = 3 gives : 80 = (0) + (0) + C ( 2) (20) + (3D + E) (0) and so 80 = 40C giving C = 2 Letting x = 5 gives : 608 = ( 2 2) (38) + (0) + C (0) + (3D + E) (0) and so 608 = 152 giving = 4 Now, comparing coefficients : x 0 : 313 = C 45E = 105 = E (1) x 4 : 5 = + + D = + D = 1 (2) x 3 : 38 = C 11D + E = D E = 20 (3) From (2), D = 1 and substituting into (3) gives E = Substituting into (1) gives 105 = ( 4 = 9) and so = 1. From (2), D = 2 and from (1), E = 5 Hence, 2x 5 x 2 + x + 8 5x 4 38x x 2 386x and so = 4 x 5 1 x (x 3) 2 + x 6 17x x 4 384x x (x 5) (x 3) 2 = (x 2 + x + 8) x x 5 1 x (x 3) 2 + 2x 5 x 2 + x + 8 6