division is negative ; and in both cases the co-efficient of the second term is formed from the first, ly multiplying the power of the first term made less by 1 by its exponent ; but here we must observe that the sign of the second term of the expansion is not entirely governed by the sign of the exponent in the first term, but depends also on the sign of the second term of the root, viz. when the expansion is the result of multiplication, the second term will be affirmative or negative according as the second term of the root is affirmative or negative; or when the expausion is the result of division, the second term will be negative or affirmative, according as the second term of the divisor is affirmative or negative, that is as the second term of the root of the divisor is affirmative or negative ; but when the second term of the root is affirmative, the second term of the expansion will always have the same sign as the exponent of the first term of the expansion ; that is to say, in the expansion the sign of the second term is entirely governed by the sign of the exponent in the first. (148.) To determine the nature of the reverse operation of extracting the mth root, we must attend to the particular cases of raising powers, and thence by inspection we will find in the result of evolution upon the quantity 3+x, that the root will always be of the form A+A'x+A"x2 +&c. Now we have already seen that Gm+mga-ty, are the first two terms of the expansion of B+x raised to the mth power, and that B-_MB-m-'x, are the first two terms of the exparsica which arises by dividing unity by the mth power of B+x; we find that if the exponent of the first term of the given power be divided by the number which indicates the root to be extracted, the result will be the first term of the root : again, if the exponent of the first term of the root thus found, the first term of the root itself having its exponent diminished by unity, and the constant power of x be multiplied together, the product will be the second terms of the root. Let the first and second terms of the given power whose mth root is required be am + mam-ix, then the first member of the root will be añ=a whose exponent is 1; whence the second term of the root is 1 xal-'xx=f. Again let the first and second terms of the given power whose mth root is required be a-m-ma kell, then a-m=the first term of the root; therefore the second member of the root is Ixal-ixr=x. Moreover let the first and second members of the given power be a+x, then am is the first term of the !Toot, and x XXE I is the second member of the root. (149.) The expansion arising by involving the series B+B's+ B":*+G"g) + &c. to the nth power, and extracting the mth root of the atb power, will still be of the form of the series proposed. 1 1 m m For the expansion of the nth power of any series, is another series of the same form, and the expansion of the mth root of any series, is a series of the same forn, and hence the expansion of (B+b'x + B".x2 +8"ro +&c.)" is a series of the form A+A'r+A"zo+A"'x+&c. and the first and second term will få + B8 x as is evident by pur 1 a suing the same modes of reasoning. Therefore if m be any number whole or fractional, affirmative or negative, it indicates a certain operation performed upon the binomial atx and the two first terms of this operation will be am + mam-it. If 1 n=-1, then a-'-la--'xor zzr is the first and second terms of the quotient arising by dividing 1 by a +r. (150.) To determine the developement of a function having the following property, 0(1+x) xP(1+y)=0(1+r+y+zy); and now, if we take ((1+x)=a+bx+cro+dx*+&c., then will P(1+y) be represented by a +by+cy + dy' + &c. and 01 ++y+xy) by a +6.(1+y+ry)+c.(r+y+xy) +d.(x+y+ry)' +e.(+y + xy)* +&c.; and expanding the latter expression, and multiplying the former, we shall find ad.rs асх? abr aby adys bcxoy +&c. +bPry + icry? box +&c. dixi CX 2 =P(1+3+ y+wy)=a+2+(+2c) xy + 2c +3d) xy? (2c+3d) ray cy? dy Comparing, now, the similar terms, we have a=1,b=1,6+20 6.6-1 1.6-1.6-2 =b?, or c= -- ; 2c +3d=bc, or d= ; ; and so on. 1.2 1.2.3 In the same manner we may determine all the subsequent coefficients ; but from those calculated above, the law of their formation is sufficiently apparent. It is evident, then, that the function 0(1+x) can be developed into an expression of the form Ն 6.6-1 1.1-1.6-2 6.6-1.6–2.643 1 x + -7° + -39 + &c.; and 1.2 1.2.3 1.2.3.4 it is also evident, that the quantity 1, being wholly independent, will generate as many different varieties as we assume different values for it. Let us, now, apply this reasoning to the binomial, (1+x)". It is plain that (1+x)" X (1+y)" is equivalent to (1+x+y+xy)"; and it therefore forms a particular case of the function 0(1+z); and of con -1.6–2x+ x2 + f(n).f(n)-1.f(n)-2 0 sequence the developement of (1+x)", as far as it depends on the variations of x, may be stated thus, b b.1-1 1.6-1.6--2 x*+&c. 1.2.3 Here, however, the quantity b is not indefinite, for, to whatever form the expression (1+x)* may be reduced, it cannot possibly contain more variable quantities than x and n. Hence, as v is completely independent of r, it must be some function of n; and if we denote this function by f(n), we shall have 1n .+ 1.2 1.2.3 &c, and changing n first into m, and then into m+n, we obtain f (m). f (m)-1 (1+x)*=l+fm) x+ .X + &c. 1.2 and (1+x)"+"=1+f (m+n).x+ f(m+n). f(m+n)-1. 727&c. 1.2 Now to ascertain the nature of the operations denoted by f, we multiply the two first of the three preceding equations together ; and as (1+r)" X (1+x)=(1+x)"+", the product on the right hand side should exactly correspord with the developement of (1+x)"+". Performing these operations, which are two simple and obvious to require that they should be detailed at length, and then comparing the resulting terms, we shall find from the whole of them only one conclusion, namely, that the function f(n) added to the function f (m), is equal to the function f (m+n). Assuming, therefore, that f (n) can be interpreted by the series of terms a+bn+cno+dno+&c., the function f (m) will have the form a+bm +cmø + dm' + &c.; and f (m+n) will become a+b. (m+n) +c. (m® +2mn+ n°)+d. (m'+ 3mon +3mn? +13)+&c.; and adding the two first series together, and comparing with the latter, we shall find a=o, b=b, and c, d, &c. all =o. Hence we discover that the value of f(n) is bn, and quently, that the form of the developement depending on the joint variations of 10 and X, is bn.bn-1 bn.bn-1.bn-2 (1+5)=1+b+ + **+&c. 1.2 1.2.3 The quantity b, however, remains still undetermined; but what. ever its value may be, is not affected by the values of n or x; and therefore, if we substitute i forn, the right hand side of the equation should be reduced to 1+x. Now this cannot take place for any other value of b, than b=1, and, consequently, we obtain this final result, viz. that n.n-1.1(1+x)=1+ foto conse n.nl x + &c. 1.2 1.2.3 most 2at by ezt eytt&c ba by (151.) If we have to determine the developement of a function designated by the following property, P(1+x) +011+y)=P(1+x+y+zy), we assume p(1+x)=a+bx+cx + dx' +er*+&c.; and adding to this the equation P(1+y)=a+by+cy + dys + ey*+&c. it is evident that the sum of the two, viz. br , cx dx + should be equivalent to dx' cri (2c +3d) zhy (1+x+y+xy)= a+|+6+2c) xy + + &c. dys and comparing the terms, we shall find -6 b a=o, b=l, c= 2 &c. 31 and hence we have x 0 (1+x)=bx(x + 2 4 In this developement the quantity b is understood, but as it is not subject to alteration from the variations of x, it merely indicates that the function possesses a variety of systems. Questions to exercise the foregoing Rules. 1. A Smuggler had a quantity of Brandy which he expected would raise €9 18; after he sold 10 gallons, a Revenue officer seized one third of the remainder, in consequence of which he makes only €8 2. Required the number of gallons he had, and the price per gallon. Ans. 22 gallons, at gs. per gallon. 2. Three guineas were to be raised on two estates, to be charged proportionably to their values. Of this sum, A.'s estate which was 4 acres more than B.'s, but worse by two shillings an acre, paid £1 15; but had A. possessed 6 acres more, and had B.'s land been worth three shillings an acre less, it would have paid £2 5. Required the values of the estates, Ans. A.'s £6, B.'s £4 16. 3. A coach set out from Cambridge to London with a certain number of passengers, four more being on the outside than within. Seven outside passengers could travel at 2s. less expence than four inside. The fare of the whole amounted to £9, but at the end of half the journey it took up three more outside, and one more inside passengers ; in consequence of which the fare of the whole became increased in the proportion of 17 to 15. Required the number of passengers, and the fare of the inside and outside. Ans. There were 5 inside, and 9 outside passengers, and the fares were 18 and 10 shillings, respectively. 4. Some persons agreed to give sixpence each to a waterman for carrying them from London to Gravesend ; but with this condition that for every other person taken in by the way, three-pence should be abated in their joint fare. Now the waterman took in three more than a fourth part of the number of the first passengers, in consideration of which, he took of them but 5 pence each. How many persons were there at first ? Ans. 36 passengers. 5. The hold of a ship contained 442 gallons of water. This was emptied out by two buckets, the greater of wbich, holding twice as much as the other, was emptied twice in three minutes, but the less three times in two minutes; and the whole time of emptying was twelve minutes. Required the size of each. Ans. the least held 13, and the greater 26 gallons. 6. If 10 apples cost a penny, and 25 pears cost two-pence, and I buy 100 apples and pears for nine-pence halfpenny, how many of each shall I have ? Ans. 75 Apples, and 25 Pears. 7. A man at a party of cards, betted three shillings to two upon every deal; after twenty deals he won five shillings. How many deals did he win? Ans. 13. 8. Being sent to market to buy a certain quantity of meat, I found that if I bought beef, which was then four-pence a pound, I should lay out all the money I was entrusted with; but that if I bought mutton, which was then three-pence halfpenny a pound, I should have two shillings left. How much meat was sent for ? Ans. 48lbs. 9. A General having lost a battle, found that he had only half his army, +3600 men left, fit for action ; one-eighth of his men +600 being wounded, and the rest, which were one-fifth of the whole army, either slain, taken prisoners, or missing. Of how many men did his army consist ? Ans. 24,000. 10. Suppose that for every ten sheep a farme Aept, he should plough an acre of land, and be allowed one acre of pasture for every four sheep. How many sheep may that person keep who farms 700 acres ? Ans, 2000. 11. A and B began to trade with equal sums of money. In the first year A gained 40 pounds, B lost 40; but in the second, A lost one-third of what he then had, and B gained a sum less by 40 pounds than twice the sum that A had lost; when it appeared that B had twice as much money as A. What money did each begin with ? Ans. £320. 12. Upon measuring the corn produced by a field, it appeared that it yielded only one-third part more than was sown. How much was that ? Ans. 36 quarters. 13. A certain sum of money is divided every week among the resident members of a corporation. It happened one week that the number resident was the square root of the number of pounds to be divided. Two men, however, coming into residence the week after diminished the dividend of each of the former individuals £i 6s, ed. What was the sum to be divided > tus. £16. |