a 1 + The corresponding values 1 23 a And by a process similar to the above, it may be shown that XC x2 203 a XC x04 :1+-+ + + + +, &c. Where the law, by which either of these series may be continued at pleasure, is obvious.* 005 a a2 a 3 a4 a5 + a 3 * In this example, if x be less than a, the series is convergent, or the value of the terms continually diminish; but when x is greater than a, it is said to diverge. To explain this by numbers: suppose a 3, and x = 2. X x2 23 Then, 1++=+ &c. a a2 938 1 2 The corresponding values are, 2 4 1 + 3 + ğ t &c. 292 where the fractions or terms of the series grow less and less, and the farther they are extended, the more they converge or approximate to 0, which is supposed to be the last term or limit. But if a=2, and x=3, a2 Then 1++-+ X x2 a az are, 8 in which the terms become larger and larger. This is called a diverging series. 3 9 27 + 2 + 1 + 8 • If x=1, and a 1 in the preceding example: a X x2 203 Then, &c., will be a + x &c. Now, because, 1+1 infinitely continued, is; a singular conclusion, when it is perceived from the terms themselves, that their sum must necessarily be either 0 or +1, to whatever extent the division is supposed to be continued. The real question, however, results from the fractional parts, which 23 &c. 1, it has been said that 1-1+1—1, &c., 2 4. Let 2y3-1972 +26y 16 be 7. Let 4x4 8. Let x4 3. Let a3 + 5a2x + 5ax2 + x3 be divided by a + x. 11. Let ba by a + 26. EXAMPLES FOR PRACTICE. X. 2ax + x2 be divided by a 13. Let 1 and x — 1 by x B Ans. x4 6. Let 48x3 5. Let x5 +1 be divided by a + 1, x3 + x2 x+1, and x + x2 + x3 + x2 + x + 1. · 76ax2 — 64a3x + 105a3 be divided by 2x - 3a. Ans. 24x2 2αx-35a2 perkang 9x2+6x 1 be divided by 2x2+3x Ans. 2x2 a2x2 + 2α3x aa be divided by x2 Ans. a a. q-pa + a2 Ans. a4ax + x2. divided by y 8. Ans 2y2 — 3y + 2. M 9. Let 6x4 96 be divided by 3x-6, and a5 + x5 by a + x. Ans. 2x+4x2 + 8x+16, and aa — a3x + a2x2 3 ax3 + x4. 10. Let 32x5+ 243 be divided by 2x + 3, and xo— a by x Ans. 16x-24x3 +36x2 a. 5 - 54x+81, and x2+ ax1 + a2x3 + a3x2 + a2∞ + a5. y, and aa + 4a2b + 86a Ans. x2 2ax + a3 (by the division) is always when the when the sum is +1: consequently former case, and 1-in the other.-ED. y a + 26 12. Let x2 + px + q be divided by x + a, and x3 qx r by x- a. Ans. x+p−a+ OC +a2+q+ 5x+10x2-10x3 + 5x4 Ans. 1 3x + 3x2 14. Let a1 +46a be divided by a2 · 2ab + 2b2. 1 2x + x2. Cha 3x + 1. ax + a2. Ans. x2+ax a2. OC. and x2+(a-p)x-ap a x5 be divided by x23• Ans. a2+2ab + 2b2. 15. a5 by a2 5α*x + 10a3x2 10α2x2 + 5αx1 25 be divided 2ax +x2. Ans. a3 3a2x + 3ax2 23 16. Let a2 + b2 be divided by a2 + ab √2 +b2. Ans. a2 ab √2+ b2 sum of the terms is 0, and is the true quotient in the OF ALGEBRAIC FRACTIONS. ALGEBRAIC fractions have the same names and rules of operation as numeral fractions in common arithmetic; and the methods of reducing them, in either of these branches, to their most convenient forms, are as follows: CASE I. To find the greatest common measure of the terms of a fraction. RULE. 1. Arrange the two quantities according to the order of their powers, and divide that which is of the highest dimensions by the other, having first expunged any factor, that may be contained in all the terms of the divisor, without being common to those of the dividend. 2. Divide this divisor by the remainder, simplified, if necessary, as before; and so on, for each successive remainder and its preceding divisor, till nothing remains, when the divisor last used will be the greatest common measure required; and if such a divisor cannot be found, the terms of the fraction have no common measure.* NOTE. If any of the divisors, in the course of the operation, become negative, they may have their signs changed, or be taken affirmatively, without altering the truth of the result; and if the first term of a divisor should not be exactly contained in the first term of the dividend, the several terms of the latter may be multiplied by any number or quantity, that will render the division complete.f * If, by proceeding in this manner, no compound divisor can be found, that is, if the last remainder be only a simple quantity, we may conclude the case proposed does not admit of any, but is already in its lowest terms. Thus, for instance, if the fraction proposed were to be a3+2a2x+3αx2+4x3. a2 + ax + x2 it is plain by inspection, that it is not reducible by any simple divisor; but to know whether it may not, by a compound one, I proceed as above, and find the last remainder to be the simple quantity 7x2: whence I conclude that the fraction is already in its lowest terms. † In finding the greatest common measure of two quantities, either of them may be multiplied, or divided, by any quantity, which is not a divisor of the other, or that contains no factor which is common to them both, without in any respect changing the result. It may here also be farther added, that the common measure, or 1. Required the greatest common measure of the fraction 0C4 2,5 EXAMPLES. x2+2bx+b2° x2 + 2bx+b2) x3 — b3x (x 4a3 2a3-3a+1 — x2 1 Whence x2+1 is the greatest common measure required. 2. Required the greatest common measure of the fraction x3- b2x * 2bx2 - 2b2x or x + b * x2+2bx+b2(x + b bx+b2 Where xb is the greatest common measure required. 3. Required the greatest common measure of the fraction 3a2-2a-1 divisor, of any number of quantities, may be determined in a similar manner to that given above, by first finding the common measure of two of them, and then of that common measure and a third; and so on to the last. *Here, I divide the remainder · 2x2 — 2b2x by — 2xb, (its greatest simple divisor) and the quotient is x+b; and then I divide the last divisor by xb, &c.--ED. S Where, since a last divisor a a3 a2x x2 + ax3 4 12a3 4. It is required to find the greatest common measure of 113 a3 x4 a1 3a+1 2a2 - 5à + 3) 3a2 2 5. Required the greatest common measure of the fraction 204 Ans. a2. x2. 2a - 1 ax2+x3° 6. Required the greatest common measure of the fraction x2 + a2x2 + aa 4 Ans. x2+ax'+ a2. 11a-11 or a 1 J + ∙1) 2a2 — 5a † 3 (3a — 3, it follows that the 1 is the common measure required. a4° 7. Required the greatest common measure of the fraction 7a2 - 23ab +6b2 Ans. x Ans. a 3b. 503 18a2b+11ab2 — 6b3° 8. *Required the greatest common measure of the fraction x2 + ax2 + bx2 - 2a2x + bax — 2ba2 Ans. x+2a. x2- bx+2ax - 2ab *This fraction can be reduced by Simpson's rule (page 48) thus:Fractions that have in them more than two different letters, and one of the letters rises only to a single dimension, either in the numerator or denominator, it will be best to divide the said numerator or denominator (whichever it is) into two parts, so that the said letter may be found in every term of the one part, and be totally excluded out of the other; this being done, let the greatest common divisor of these two parts be found, which will evidently be a divisor to the whole, and by which the division of the quantity is to be tried; as in the following example, where the fraction given is 23+ ax2 + bx22a2x+bax-2 baz 22 ·bx+2ax-2ab Here the denominator being the least compounded, and b rising therein |
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