the same height as the intended power : and the last term of the series will contain only the 2d part of the given root, when raised also to the same height of the intended power: but all the other or intermediate terms will contain the products of some powers of both the members of the root, in such sort, that the powers or indices of the 1st or leading member will always decrease by 1, while those of the 28 member always increase by 1. 2. To find the Co-efficients. The first co-efficient is always 1, and the second is the same as the index of the intended power; to find the 3d co-efficient, multiply that of the 2d term by the index of the leading letter in the same term, and divide the product by 2; and so on, that is, inultiply the coefficient of the term last found by the index of the leading quantity in that term, and divide the product by the number of terms to that place, and it will give the co-efficient of the term next following; which rule will find all the co-efficients, one after another. Note. The whole number of terms will be 1 more than the index of the given power : and when both terms of the root are +, all the terms of the power will be t; but if the ser cond term be –, all the odd terms will be +, and all the even terms –, which causes the terms to be to and alternately. Also the sum of the two indices, in each term, is always the same number, viz. the index of the required power : and, counting from the middle of the series, both ways, or towards the right and left, the indices of the two terms are the same figures at equal distances, but mutually changed places. Moreover, the co-efficients are the same numbers at equal distances from the middle of the series, towards the right and left; so by whatever numbers the increase to the middle, by the same in the reverse order they decrease to the end. EXAMPLES. 1. Let 8 + x be involved to the 5th power. The terms without the co-efficients, by the 1st rule, will be a', atx, a'r?, aʻr, er4, r", 5 X 4 10 X 3 10 x 2 5* ; 5 or, 1, 5, 10, 10, 5, 1; Therefore the 5th power altogether is But But it is best to set down both the co-efficients and the powers of the letters at once, in one line, without the intermediate lines in the above example, as in the example here below. 2. Let a- x be involved to the 6th power. The terms with the co-efficients will be a-6a.r + 15a'.r-200 r3 + 15a'.r-Gaxs + ... 2 3. Required the 4th power of a – x. Ans. at-4a2x + 6aRx? — 4ax3 + 24. And thus any other powers may be set down at once, in the same manner; which is the best way. EVOLUTION. EVOLUTION is the reverse of Involution, being the method of finding the square root, cube root, &c, of any given quantity, whether simple or compound. CASE I. To find the Roots of Simple Quantities. Extract the root of the co-efficient, for the numeral part; and divide the index of the letter or letters, by the index of the power, and it will give the root of the literal part; then annex this to the former, for the whole root sought* or : * Any even root of an affirmative quantity, may be either + hus the square root of + a2 is either + a, or – a; be. cause + axta = + a', and -ax - Q = + a2 also. But an odd root of any quantity will have the same sign as the quantity itself: thus the cube root of + a' is + a, and the cube root of , is a; for + ax + ax + a = + a, and - ax — а Х a3. Any even root of a negative quantity is impossible; for neither taxta, nor -ax – a can produce - az. Any root of a product, is equal to the like root of each of the factors multiplied together. And for the root of a fraction, take the root of the numerator, and the root of the denominator. a = EXAMPLES. 9c" 30 V 5. EXAMPLES. 1. The square root of 4a?, is 2a. 2. The cube root of 8a?, is 2a} or 2a. 5a2b2 5a2b2 ab 3. The square root of or 9c2 16a486 2ab2 4. The cube root of 5. To find the square root of 2a*b*. Ans. ab2. 6. To find the cube root of -64a3b6. Ans. -4ab. 8ab2 2 7. To find the square root of 8. To find the 4th root of 8 la46o. Ans. 3ab vb. 9. To find the 5th root of - 32a555. Ans. - 2ab Sb. 3c2a. Ans. 2ab V30 303 CASE II. To find the Square Root of a Compound Quantity. This is performed like as in numbers, thus : 1. Range the quantities according to the dimensions of one of the letters, and set the root of the first term in the quotient. 2. Subtract the square of the root, thus found, from the first term, and bring down the next two terms to the remainder for a dividend ; and take double the root for a divisor. 3. Divide the dividend by the divisor, and annex the result both to'the quotient and to the divisor. 4. Multiply the divisor, thus increased, by the term last set in the quotient, and subtract the product from the dividend. And so on, always the same, as in common arithmetic. EXAMPLES. 1. Extract the square root of at- 4a3b + 6aob? — 4ab}. + b*. a4 – 4a3b + 6a’b?— 4ab3 + 6+ ( a? - 2ab + b2 the root. 2. Find the root of + ta'b + 10a*b2 + 12ab + b*. at + tab + 10L + 12963 + 6+ ( a + 2ab + 3b. 20* + 2ab) 4a’b + 10.3b 3b + ab 2a + 4ab + 362 ) 6a 52 + 12ab3 + 6* 6a’b? + ] 2ab3 + 64 3. To find the square root of a* + 403 + 6a+ + 4a + 1. Ans. a + 2a +1. 4. Extract the square root of at – 2a + 2a + Ans. 2-3 5. It is required to find the square root of a’-ab. b 6 63 Ans. a &c. Sa 16a? CASE III. To find the Roots of any Powers in General. This is also done like the same roots in numbers, thus: Find the root of the first term, and set it in the quotient. -Subtract its power from that term, and bring down the second term for a dividend. Involve the root, last found, to the next lower power, and multiply it by the index of the given power, for a divisor.---Divide the dividend by the divisor, and set the quotient as the next term of the root. Involve now the whole root to the power to be extracted; then subtract the power thus arising from the given power, and divide the first term of the remainder by the divisor first found; and so on till the whole is finished *. EXAMPLES. * As this method, in high powers, may be thought too laborious, it will not be improper to observe, that the roots of compound quantities may sometimes be easily discovered, thus : Extract the roots of some of the most simple terms, and connect them together by the sign + or -, as may be judged most suitable for the purpose.--Involve the componnd root, thus found, to the proper power ; then, if this be the same with the given quantity, it is the root required.---But if it be found to differ only in me of the signs, change them from + to-, or from to + till its power agrees with the given one throughout. Thus, EXAMPLES. 1 1. To find the square root of a* - 2a’b+3ab? - 2ab3 + 6*. 24 - 2ab + 3ab2 - 2ab3 + 64 (a-ab + b2 44-2a3b + 3a2b2 - 2ab3 + 14 = (d-ab + b2)?. 2. Find the cube root of a' - 6as + 21a4 — 44a3 + 63a 54a + 27. ao -605 + 21a-- 44a3 + 63a2-54a + 37 ( a - 2a + 3. a° -6x5 + 2104-44a3+63a% - 54a +27 = (a*--20-3)*. 3. To find the square root of a? 2ab + 2ar + 62 2bx + x?. Ans. a-bot X. 4. Find the cube root of a 3as + 9a4 -- 13a3 + 18a? – 12a + 8. Ans. a’ – a + 2. 5. Find the 4th root of 8104 - 216a3b + 216ab2 96ab3 + 1664. Ans. 3a-2b. 6. Find the 5th root of as -- 10a4 + 40a? - 80a? + 800 - 32. Ans. a -2. 7. Required the square root of 1 - x?. 8. Required the cube root of 1 – 23. Thus, in the 5th example, the root 3a-2b, is the difference of the roots of the first and last terms; and in the 3d example, the root a-b + x, is the sum of the roots of the 1st, 4th, and 6th terms. The same may also be observed of the 6th example, where the root is found from the first and last terms. |