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that case the term containing it would reduce to zero when x =

= 0, (367, 1), and we should have A = 0, which is contrary to the implied conditions of the proposition.

2d. The exponent, a'-a, can not be a negative quantity; for in that case the term containing it would reduce to infinity when x=0, (367,2), and we should have A = 0, which is also contrary to the implied conditions of the proposition. Now since a'-a can neither be a positive nor a negative quantity, it must be nothing or zero; that is,

a'-a = 0, It follows also that each of the other exponents, b--a, c'-a, etc., in equation (3), is positive, being algebraically greater than zero; hence all the terms after the first in the second member of this equation, must disappear when x = 0, (367, 1), and we shall have,

A = A'x' = A'.
Now since A and A' are independent of x, we shall have

Axa A'xa', whatever be the value of x. We may therefore suppress these terms in equation (1). There will result

B+Cao+.... = B'zo+Cao+ whence, by reasoning as before, we shall find that

b = b', c=c', etc. B = B

eto. That is, equation (1) is an identical equation, the two members having the same form. Hence, the given equation is also identical, and the proposition is proved.

It is obvious that the preceding demonstration will apply if one or more of the exponents, a, b, c, .. are negative; or if a = = 0, in which case each member will contain an absolute term.

III. In every equation which is satisfied for any value whatever of the unknown quantity, and which involves like powers of this quantity in the two members, the coefficients of the corresponding powers will be equal, each to each. Let us assume the equation, Ax+BX+Cx+....= A'x"+B'X+0+....

+, the number of terms being either limited or unlimited.

....2

C=C',

of x.

Now if this equation is capable of being satisfied for any value of ir, then according to the preceding demonstration, not only must the exponents of x in the two members be equal respectively, but the coefficients also must be equal, each to each ; that is,

A = A', B= B', C = C', etc. Every such equation is obviously identical, though it is not necessary that A, B, C, etc. should be of the same form respectively, as A', B', C", etc.

IV. In every equation which is satisfied for any value whatever of the unknown quantity, and which has zero for one of its members, the coefficients of the different powers of the unknown quantity are separately equal to zero, Let

Ax* + Bx + Cxo+Dad to... = 0, (1) represent the equation, arranged according to the ascending powers

The coefficients, A, B, C, D, etc., are supposed to be independent of x, and consequently the same for all values of x. Divide every term in this equation by *; we shall have

A+B.c6. +Ccm +Dach +....= 0. (2) In this equation make x = 0; then since the exponents, ba, C-a, d-a, etc., are all positive, every term after the first will reduce to zero, (367, 1), and we shall have

A=0. Suppressing A.x in equation (1), and then dividing through by 20, we obtain

B+Cocond+Doodto... = 0. (3) In this equation make x = 0, and we have

B=0. In like manner we may prove that each of the other coefficients is equal to zero.

It is important to observe in this connection that the coefficients, A, B, C, D, etc., must be supposed to represent polynomial expressions, which reduce to zero in consequence of having positive and regrtive parts that are respectively equal to each other,

DECOMPOSITION OF RATIONAL FRACTIONS.

369. By means of the properties of identical equations, a fraction may often be separated into two or more partial fractions, whose denominators shall be simpler than the given denominator. In every such case, the given fraction is the sum of the partial fractions; hence its denominator will be a common multiple of the denominators of the partial fractions.

8x-31 1. Separate

into partial fractions.

x-7x+10 By inspection, we perceive that

2—7x+10 = (2-5)(x-2). Now assume

8x-31

A B
+

(1)
(x5)(x-2)

-2 Since the first member is simply the sum of the two fractions in the second member, this is obviously an identical equation. Clearing of fractions and uniting terms, we have

8x—31 =(A+B)x—(24+5B), (2) in which 31 in the first member, and (2A+5B) in the second, may be considered as coefficients of x'. Now according to (368, III), the coefficients of the like powers of x in the two members must be equal ; and we have, therefore,

A+B=8,

2A+5B= 31. From these equations we readily obtain

A=3, and B=5; whence from equation (1), we have 8x_31 3 5

Ans. x*—7x+10 X-5 It should be observed that equations (3) and (4) are, the equations of condition, which must exist in order that equation (1) shall be true for all values of x.

(3)

+

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7x* +3

7x+ 2. Separate

into partial fractions.
(x+1)(2x-1)
Suppose, if possible,

А B
(x+1)(2x-1) 2+1 +2.-1;

(1) clearing of fractions, we obtain

700* + x = (2A+B)x+(B-A); transposing all the terms to the first member, we have

7xo+(1-2A-B)x+(A-B) = 0. (2) If this equation be possible, it must be an identical equation; and as one member is zero, the coefficients of the different powers

of x must be separately equal to zero (368, IV); and we shall have

7=0, which is absurd. Hence, we infer that the fraction can not be separated into partial fractions, having numerators independent of x. Again, assume 7x +30 Ах Вх

+ i

(1) (x+1)(2x-1)x+1

2x-1 clearing of fractions and collecting terms,

7xo+x = (2A + B)x+(B-A)x; ; (2) equating the coefficients of like powers of x,

2A+B=7,

B-A=1; whence we obtain A=2, B = 3; and by substitution in equation (1), 7x* +

2

320 +

Ans. (x+1)(2x-1)

2x-1' From this example we learn that if we assume an impossible form for the partial fractions, the fact will be made apparent by some absurdity in the equations of condition.

NOTE.—If the given denominator consists of three or more factors, there will be three or more partial fractions. But there will always be as inany equations of condition as there are numerators to be determined,

+1

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+

20x+2 2. Resolve

into partial fractions. 2x+3x—20

8

6 Ans.

2x-5 æ+4 6.x— 22x + 18 3. Resolve

into partial fractions. (x - 1)(x - 5x+6)

1

2 3 Ans. + +

-2 30 2+2 4. Resolve into partial fractions. 2c

1

3 2 Ans.

+

xx1
10
5. Resolve
x*_13x? +36

into partial fractions.
1
1
1

1
Ans.

2(x+2) 23-2) 3(x+3) ):

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+3(2-3

THE RESIDUAL FORMULA.

370. It has been shown in (89, 4) that x*—y" is exactly divisible by -y, if m is a positive whole number. The form of the quotient is as follows:

m

2

= "+y+"y"+y+...+y Xy the number of terms in the quotient being equal to m.

Now suppose x = =y; then each term will become am, and since there are m terms, we have the formula,

mc-.

(A) XY The subscript equation, y=x, is used to indicate the condition under which the first member of (A) will be equal to the second.

m

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