And equating the corresponding coefficients, m n m m =A, A = 2B+A, B = 3 C + 2 B, n n and hence the law of composition of the coefficients is evident. Let N be the coefficient of the (r+1)th term, and M that of the rth, then it is evident that (491.) The coefficients being now found, the developement is known. By making n = 1 in the preceding investigation, the theorem will be established for an integral exponent, namely, m; and it may be remarked, that the preceding demonstration is independent of induction. For the case of a fractional negative exponent, as m let m be changed into - m, in the demonstration, as far as equation [2] inclusive, then the first member of this equation becomes an equation which differs from [5] only in the sign of m, and therefore the values of the coefficients will be found from the preceding values, by merely changing the sign of m. Hence, whatever be the value or the sign of m, the formula (492.) If the coefficients of the successive terms, beginning with the second, be respectively A, B, C, D, ... then And the coefficients for these four cases are in order, m A = n = "B=1" = ", C = Bm—2n D C 2n 3n (493.) As the series of coefficients in the second and fourth lines are already adapted to the negative exponents, the values of m and n in them must be taken with the positive sign. 5 = 1 4 — — 3, B = 3 × 3 + 1 = 6, C =—6 × 3 =— 10, 3+1 1 By [III] m = 1, n = 5, m = {}, Y = a2, X = − x2, n a2 3. Find the developement of † (a2 + x2)2 or a3 (a2+x2) ̄ŝ X x2 a2, Here by [IV], m = 2, n = 3, X=x2, Y=a2, Y and Y32 2; and in order that Y might be an of the binomial, and therefore 4 is the second term. The coefficients are the same as in example 2; hence integer, 32 was assumed as the first term -- THE EXPONENTIAL THEOREM. (495.) 'An exponential function is a quantity with a variable exponent.' Thus, in the equation axb, x is a variable exponent, and a is an exponential function of x. (496.) To develope a, assume a=1+b, then (491) a* = (1+b)* = = 1+xb+* (x−1)f2+ 2 x (x−1) (x—2) 63+... [1] 2x3 Hence it is evident that ar may be developed in a series containing only integral and positive powers of x, and also that the first term is 1, and the coefficients are functions of hence let b or a; ax = 1+kx+ Ax2 + Bx3 + CÃa + .. ... [2] By inspection of the series [1], it is evident that the coefficients of x in the different terms are b2 1,3 74 b5 k-b- + + 2 3 4 5 [3] in which b or a— 1 is known, and hence k is known. If in [2] be assumed = z, then a2 = 1+kz+ Az2 + Bz3 + ... hence, subtracting the equation [2] from this, - ... a2 — a2 = k(≈ — x) + A (z2 — x2) + B (≈3 — 2-3) + ... If now be assumed =x+h, the first member of the last equation becomes axth — ax = ax (ah 1); and if in [2] h be substituted for x, then a2 — 1 — kh + Ah2 + and ax (ar-1)= ax (kh + Ah2 + Bh3 + ...) Now, the first members of the two last equations are equal, for a= arth; and hence the second members are also equal, and are divisible by ≈-x=h; hence dividing ax(k+Ah+Bh2+...)=k+A (≈+x)+B(x2+zx+x2)+... and as h is arbitrary, let it =0, and then z=x, and the |