Calculus: Analytic Geometry and Calculus, with Vectors

516 Exponential and logarithmic functions

Remark: Determination of constants 11, B, C, D for which

1 Bx+B Cx+D

(x2+1)(x2+4) x2+1 + x2+4

or

(2) 1 = (Ax+B)(x2+4)+(Cx+D)(x2+1)

can be made in various ways. Particularly efficient work can be done by those

fortunate persons who know about the algebra of complex numbers, includ-

ing the "imaginary unit" i for which i2 = -1. Putting x = i in (2) gives

1 = (Ai + B)(3), so 11 = 0 and B =. Putting x = 2i in (2) gives 1 =

(2Ci + D)(-3), so C = 0 and D = -. Hence

1 1 1 1 1

(x2+1)(x2+4) =3x2+1 -3x2+4'

a result that is easily checked.

8 Obtain the partial fraction expansion of the left member of the formula

x3 _[(x+p) -p]3_(x + p)3 - 3p(x + p)2 + ...

(x+P)5 (x+p)s (x+p)b

by taking advantage of the broad hint in the formula.

9 We should not be too busy to see how Euler determined the constants in

x2 B C Dx+E

(1) (1 - x)3(1 + x2) (1 - x)3+ (1- x)2+ 1 __X+ 1+ x2

in his great textbook "Introductio in Analysin Infinitorum," Lausannae, 1748,

volume 1, page 31. Clear of fractions to obtain

(2) x2 =11(1 + x2) + B(1 - x)(1 + x2)

+C(1 -x)2(1+x2)+(Dx+E)(1 -x)3.

Put x = I to obtain A = -. Subtract (1 + x2) from both members of (2)

and divide by (1 - x) to obtain

(3) -fix- = B(1 + x2) + C(1 - x)(1 + x2) + (Dx + E)(1 - x)2.

Put x = 1 to obtain B = -_1. Add (1 + x2) and divide by (1 - x) to obtain

(4) -Tx = C(1 + x2) + (Dx + E)(1 - x).

Put x = 1 to obtain C = --. Add I(l + x2) and divide by (1 - x) to obtain

(5)wx=Dx+E,

and hence D -;,E_

10 One who wishes to think about a partial fraction problem in arithmetic

(or theory of numbers) may seek integers 11 and B for which

11 11 B

15= 3 +5