=

By successive repetitions of this process we find a, b, abg, and in general, a, b,

This is the principle of indeterminate coefficients as applied to infinite series.

APPLICATION TO INTEGRAL FUNCTIONS.

331. The principle of indeterminate coefficients was applied, without formal notice of the fact, to Examples 3 and 4 of Art. 153. The following are other examples of this class:

Ex. 1. Determine the values of c and d that will make

Ex. 2. Determine under what conditions ax3 + bx2 + cx + d is divisible by px2 + qx + r without remainder.

If the division be performed, the quotient will consist of two terms; namely, (a/p)x, and a second term independent of x. Hence, the quotient is of the form (a/p)x + k, where k is a quantity to be determined by the condition that the division shall be exact. We have, therefore,

α

ax3 + bx2 + cx + d = ( x + k) (рx2 + qx + r)

ε)

aq

(a + pk) x2

qk) x

+ qkx + rk,

and since this is to be true for all values of x, the coefficients of like powers of x in the two members of this equation must be equal [Art. 328]; whence

Replacing k in the first two of these equations by its value d/r derived from the third, we have

bp = aq + dp2/r, cp = ar + dpq/r.

These are the conditions necessary and sufficient in order that the division may be exact.

Ex. 3. Transform x2 + pxy + qy2 into the sum of two squares. Assume u = · (x +√qy)h, v = (x − √qy)k, and determine h and k by the condition that

u2 + v2 = x2 + pxy + qy2.

For this purpose we make the coefficients of xy and y2 in

that is, in

(x+√qy)2h2+(x −√qy)2k2,

(x2 + qy2) (h2 + k2) + 2√qxy (h2 — k2)

equal to p and q respectively. We thus obtain

or

h

1

h2 + k2 = 1, 2√¶(h2 — k2) = p,

kc

√ √(1 + 2 D1), * = 1/2 √(1 − 229) ·

Hence, h and k having these values,

x2 + pxy + qy2={(x + √qy)h}2 + {(x − √qy)k}2.

Ex. 4. Find the factors of

(b2c2 — a1) (bc) + (c2a2 — b4) (c − a) + (a2b2 — c1) (a - b).

This expression vanishes if c = b, or if a = c, or if b = a, and therefore bc, c — a, and a − b are factors. [Art. 148.]

Also, since the expression is symmetrical and is of five dimen

sions, there must be a fourth symmetrical factor of two dimensions, which must therefore have the form

L (a2 + b2 + c2) + M (bc + ca + ab).

The given expression is therefore identically equal to

(b −c) (c − a) (a − b){ L (a2 + b2 + c2) + M (bc + ca + ab)}, in which L and M are to be determined by equating coefficients in accordance with the principle of indeterminate coefficients.

The coefficients of a in the two expressions are respectively - (bc) and L (bc), and the coefficients of a3 are b2 — c2 and b2 — c2 – M (b2 - c2); hence L= 1 and M=0, and the required factors are (b −c), (c − a), (a − b), (a2 + b2 + c2).

EXAMPLES LXXXV.

1. Determine the value of k that will make x5 x2 + 2x2 + k divisible by x3 + x + 1 without remainder.

2. Determine the values of p, q, and r that will make x5 5 px3 + 5 qx2 − r divisible by (x — c)3 without remainder.

3. If ax2 + bx + c and a'x2 + b'x + c' have a common factor of the first degree, this factor, and also the remaining (second) factor, in each case, will be rational in all the letters. Prove this by determining the three distinct factors. [Compare Ex. 3, Art. 173.] 4. Prove that if

ax2+2hxy + by2 + 2 gx + 2fy + c

be expressible as the square of a rational integral function of x and y of the first degree, then

5. Determine the relation between a and b that will make (x+ay)" and x2 + bxy + y2 have a common factor of the first degree in x and y.

6. Determine k such that x2 y2-x-3y+k may be the product of two rational factors of the first degree in x and y.

7. What value of k will make the three equations 2x 3 = 0, x + y − 1 = 0, and x + ky + 1 = 0 simultaneous in x and y ?

8. Supposing a and b to be given numbers, determine a binomial expression (containing an arbitrary factor) which, when substituted for x, will make ax + b2 a perfect square (for all values of the arbitrary factor).

9. Prove that if ax3 + bx2 + cx + d is divisible by x2 - k2, then ad= bc.

10. Find the factors of

4 abc (a+b+c) + bc (b2 + c2) + ca (c2 + a2) + ab(a2 + b2).

11. Find the factors of

b2c2 (b − c) + c2a2 (c − a)+a2b2 (a — b).

12. Find the factors of

a1 (b − c) + b1 (c − a) + c1 ( a −b).

13. Find the cube root of

1+ 3x+6x2 + 7 x3 + 6 x2 + 3x5 + x6.

APPLICATION TO PARTIAL FRACTIONS.

332. In Art. 165, the process of obtaining a single fraction as the result of adding together any number of given fractions was explained. It is sometimes important to be able to perform the converse operation of decomposing a complex fraction into the sum of a series of simpler partial fractions, and for this purpose the principle of indeterminate coefficients is employed.

The sum of the two fractions A/(x-1), B/(x 2) will produce a fraction whose denominator is (x − 1)(x − 2). Let it therefore be proposed to determine A and B such that

1 x- 2 (x − 1)(x-2)

For this purpose it is obviously sufficient that

2x-5= (A + B)x − (2 A + B).

Since the left member of this identity contains no power of x higher than the first, both A and B are assumed to be constants, and therefore, by the principle of indeterminate coefficients,

partial fractions to a common denominator.

nators being omitted, we have

and reduce the

Then, the denomi

x2+1=A(x − 1)(x − 2) + Bx(x − 2) + Сx(x − 1).

We might now expand and equate coefficients, but it is simpler to proceed as follows:

Since the identity is true for all values of x, put x equal to 0, 1, and 2 in succession. From these substitutions the following results are at once obtained:

The process of decomposition cannot be applied directly to this fraction, because the numerator is not of lower degree than the denominator. But, by division, any such fraction can be replaced