discussion it should be distinctly understood that assumption (6) holds when and only when these equations are solvable. In any particular case we can find out immediately whether the equations are solvable by attempting to solve them. If the numbers A, B,..., N do not exist, the fact will appear by our inability to solve the linear equations. As a matter of fact, one and only one solution always exists under the assumption of this section. N stand for expres

If in (6) we assume that several of the symbols A, B, sions linear in x, as, for instance, ax + b, we should then have a larger number of variables to determine than there are equations. Under these circumstances there is an infinite number of solutions of the equations. Thus if we should seek to f(x) express as the sum of partial fractions where the numerators are not con$ (x) stants but functions of x, we could get any number of such developments.

We have the following

RULE. Factor (x) into linear factors, as

Multiply both sides of the expression by $(x), equate coefficients of like powers of x, and solve the resulting linear equations for A, B,

N.

Replace A, B,..., N by these values and check by substituting for x some number distinct from a, ß, ν.

x22A(x-2)x + B(x-1)x + C(x-1) (x-2),

=

x22 (A+B+C) x2 - (2 A+B+3 C) x + 2 C.

204. Development when (x) = 0 has imaginary roots. In the preceding section no mention has been made of any distinction between real and imaginary values of a, ß,

V. In fact

the method given is valid whether they are real or imaginary. It is, however, desirable to obtain a development in which only real numbers appear.

where let us suppose that μ and v are the only pair of conjugate

imaginary roots of (x), m and n being conjugate complex numbers.

Then adding the corresponding terms of (1), we obtain

Since μ and v are the only imaginary roots of (x) = 0, the last term of (2) is real, as is also the entire right-hand member (§ 152). Hence, letting the numerator

x (m + n) — a (m + n) + ib (m − n) = Mx + N,

we have the development

By complete induction we can establish this form of numerator where there is any number of pairs of imaginary roots of (x) = 0. We have proved the form (3) where there is one pair of imaginary roots. Assuming the form where there are k pairs, we can prove it similarly where there are k+ 1 pairs. Hence we have the

THEOREM. If (x) is factorable into distinct linear and quadratic factors, but the quadratic factors are not further reducible

into real* factors, then (a) is separable into partial fractions

f(x)
$(x)

where x2 + μx + v is an irreducible quadratic factor of $(x).

This theorem is of course true only under the condition that the linear equations obtained in the process of determining the constants are solvable. It turns out, however, that in this case as in § 203 the linear equations obtained always have one and only one solution provided that the roots are all distinct.

*A real factor is one whose coefficients are all real numbers.

or

Multiply by (x − 1) (x2 + x + 1),

x2 + 1 = A (x2 + x + 1) + (Bx + C) (x 1).

205. Development when (x) = (x − a)". In this case the method given in the previous sections fails, as the equations for determining the values of the numerators are incompatible. If we let + (n-1,

f'(x)
(x)

2-1

=

2

(x — α)n

we can separate into partial fractions as follows.

(1)

Let x α = =y, that is, x = = y +a, and substitute in (1). We obtain after collecting powers of y

where the A's are constants. Replacing y by x

following development: