w nence if the coefficients of all the factors are the same, the coefficients of all their second terms will be the same. Now, suppose the coefficient of the second term of every factor to be b', and the coefficient of the second term of their product will be ; whence the first and second terms of their product will be 1 nd mix. If the series 6+ B′x+B′′x2 + B′′x3+ &c. be raised to any power m, the expansion will be of the form (3+ß′x+ß′′x2+ß'''x3+&c.)"=ß*(1+b1x+b′′x2 + b'x3 +&c.)m =G"(1+mor+Ar+A" r® +&e =Çm+mßm l'x+A”B” x2 + A11ßm x3 +&c. =6m+mG"-1 B'x+B′′x2 + B'"'x3 + &c. That is, the first and second terms of the expansion will be 6" and mem-16'x and the remaining terms of the form B"x2+B""x3+&c. If B+ had been raised to the mth power instead of the series B+ß'x+ ß′′x2+ ß'"+&c. the first and second terms of the expansion would have been, 3m and m Gm-1x, and the remaining terms of Bm the form B"x2+B""x+&c.; that is, the whole expansion of the form B+B'x+B"x2+B""x" +&c. (144.) If the series p°+p'x+p"x2+p+&c. be divided by the series 1 — b'x—b"x" — b'"'-&c. the quotient will be of the (form q+q'x+ 9°x2+9"x3+&c. For let the operation be performed, so that q may be substituted for p, q' for the sum of the parts which forms the coefficient of the first term of the first remainder, q" for the sum of the parts which forms the coefficient of the first term of the second remainder, and so on, as follows +qx+&c. \—Vx—b"x2—L'!'! x3—&c.)q+p'x +p′′x2 +p!!!x3 +&c.(q+q'x+q′′x® q —q b' x — q b " x 2 —qb11x3 — &c. Then the co-efficients of the terms of the quotient may be thus derived, viz. Therefore the co-efficients q, q', q", q'" &c. of the terms of the quotient are functions of the co-efficients, of the terms of the divisor and dividend, and the quotient is of the form asserted. Hence, also, the quotient arising by dividing the dividend by the product of any number of series must always be of the same form. (145.) If 1 be divided by any given power m of the series 3+ B'x+ "2+6x+&c. the quotient will be of the form C+C'x+C"x2+ C+&c. and the first two terms B and B'x will be ß-, —m ß—m―1ß'x. For the first two terms of the mth power of the series ß+ß' x+ Bx+x+&c. are 6m+mßßx, and the remaining terms are of the form Br+B+&c.; whence the quotient arising by dividing 1 by the series B+B'x+B"2+B'"+ &c. will be of the form C+ C'x+C"x2+C'"+ &c. and with respect to the absolute values of the first two terms of the form C,+C'x, it will only be necessary to divide 1 by ßm+m ßm-1ß'x, the first two terms of the series B+B′x+B"x+ &c. thus mth If the series B+B'x+ B'x2+ &c. had been the expansion of the power of the binomial ẞ+x, the first two terms would have been B" + 6m-1x. 146.) Therefore whether the operation be that of involving a binomial or a whole series to any power m, or of dividing unity by the mth power of the binomial or mth power of the whole series, the expansion will always be of the form A+A'x+A"x2 +All+&c. of which the two first terms will be Gm Gm-13'x or ß-m‡ mß—=—1ß'x, according as the expansion arises from multiplication or division, and if the root be a binomial, then = 1; and consequently 'will disappear in the second term. (147.) In each case the co-efficient of the first term is the same. Dt the sign of the exponent, which in the expansion arising by division is negative; and in both cases the co-efficient of the second term is formed from the first, by 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 expansion 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+, that the root will always be of the form A+A'x+Ax2+&c. Now we have already seen that G +m fm−1 x, are the first two terms of the expansion of ẞ+ raised to the mth power, and that 3-"-m ß--x, are the first two terms of the expansion which arises by dividing unity by the mth power of ẞ+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 term of the root. Let the first and second terms of the given power whose mth root is required be am+mamix, then the first member of the root will be a"=a whose exponent is 1; whence the second term of the root is 1 xa11×x=x. Again let the first and second terms of the given power whose mth root is required be a-"-ma then aa the first term of the root; therefore the second member of the root is 1×a1-1×x=x. Moreover let the first and second members of the given power be a+x, then am is the first term of the 1 Xx= -a r is the second member of the m (149.) The expansion arising by involving the series 3+ B'x+ 6"+""+&c. to the nth power, and extracting the mth root of the nth power, will still be of the form of the series proposed. For the expansion of the nth power of any series, is another teries of the same form, and the expansion of the mth root of any series, is a series of the same form, and hence the expansion of (3+3'x + n ("x2+6""'x3+&c.) is a series of the form A+A'x+A"x2+A'''x3 +&c. and the first and second term will β r as is evident by pur suing the same modes of reasoning. m Therefore if m be any number whole or fractional, affirmative or negative, it indicates a certain operation performed upon the binomial a+r and the two first terms of this operation will be am+mam-x. If the quotient arising by dividing 1 by a+x. (150.) To determine the developement of a function having the following property, (1+x) × (1+y)=4(1+x+y+xy); and now, if we take ¢(1+x)=a+bx+cx2+dx3+&c., then will (1+y) be represented by a+by+cy2+ dy3 + &c. and Q(1+x+y+xy) by a+b.(x+y+xy)+c.(x+y+xy)2+d.(x+y+xy)3 + e.(x + y + xy) * +&c.; and expanding the latter expression, and multiplying the former, we shall find Comparing, now, the similar terms, we have a=1, b=b, b+2c ī x+ -x2+ 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 (1+x) can be developed into an expression of the form ს b.b-1 b.b-1.b-2 b.b-1.b-2.6-3 1 + -x+&c.; and it is also evident, that the quantity b, 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+)" × (1+y)" is equivalent to (1+x+y+xy)"; and it therefore forms a particular case of the function (1+); and of con 1.2 1.2.3 1.2.3.4 sequence the developement of (1+x)", as far as it depends on the variations of x, may be stated thus, 1+1 b.b-1 b. b-1.b-2x+&c. x+ ·x2+· 1.2 1.2.3 Here, however, the quantity is not indefinite, for, to whatever form the expression (1+r) may be reduced, it cannot possibly con- · tain more variable quantities than r and n. Hence, as b is completely independent of x, it must be some function of n; and if we denote this function by ƒ (n), we shall have &c, and changing n first into m, and then into m+n, we obtain 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 + x)* × (1+x)”=(1+x)"+", the product on the right hand side should exactly correspond 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+cn2+dn2+&c., the function f(m) will have the form a+b+cm2 + dm3 +&c.: and f(m+n) will become a+b. (m+n)+c. (m2+2mn+n3)+d. (m3 + 3m2n+3mn2+n3)+&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 =0. Hence we discover that the value of ƒ (n) is bn, and consequently, that the form of the developement depending on the joint variations of 1 and x, is (1+x)=1+b7x+ bn.bn-1 1.2 bn.bn-1.bn-2 -x3+ &c. The quantity b, however, remains still undetermined; but whatever its value may be, is not affected by the values of n or x; and therefore, if we substitute 1 for n, 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 |