m(m − 1) (b + x)m−272 + etc.; consequently, if in this we 1. 2 x= put = 0, and change h into x, it will become (b + x)m = m(m −1)(m −2)bm-33 bm + mbm-1x + etc., as required: m(m 1)zm-2x2 + 1 2 1 2 3 + It may be added, that if any function of x, when expanded by MacLaurin's Theorem, contains infinite terms, it will clearly show that the Theorem is not applicable to the expansion. 3. If we subtract X from the two members of (a), we shall get X'-X = X1.h + X, h2 + the variation of X, arising from changing x into x + h in X, and then subtracting X from the result. If we represent X'-X by writing the Greek letter A (delta) before or to the left of X, we shall have X' - X = AX; and in a similar manner, if we represent x + h by x', we may express h = x'x by Ax; consequently, if we use AX and Ax for X' - X and h, (d) may be written in the form AX=X. Δ. + (e); which is (sometimes) called an equation of finite differences, because AX and Ax are supposed to have definite. values, and is said to be the characteristic of finite differences. Again, if we represent the first term of the right member of (d) by writing d before or to the left of X, we shall have the equation dX = X1. h, (ƒ), which is called the differential of X, since it is only a part of the difference in the right member of (d). If, for uniformity of notation, we represent h in (f) by da, and then divide both members of the equation by h = dX dx, we shall get = X1, (g); in which is called the dX dx first differential coefficient, since de multiplied by it gives dX, the differential of X; and it is clear, from (g), that the value of the first differential coefficient is expressed by the first derived function of X. If we divide the two members of (d) by h, it will be re and if in this we put h = 0, the right member of the equation will be reduced to X1, the first derived function of X, 0 since the factor which member of (h) is reduced to the form is common to the numerator and denominator is put equal to 0. Hence, if we actually divide the variation or difference X' - X of X by h, the corresponding variation or difference of the independent variable x, and put h = 0 in the quotient, the result will be the first derived function of X, or it will be dX the value of the first differential coefficient dx Thus, we get the first differential coefficient of 2 by changing x into x + h, which gives (x + h)2 = x2 + 2xh + h2 or (x + h)2 — x2 = 2xh+h2; and by indicating the division by h, in the first member of this equation, and performing it in the second member, we get (x + h)2x2 h =2x+h; conse quently, putting h0 in the quotient, we have 2x for the Because X, is the first derived function of X1, it follows,. from what has been done, that we shall have X2 = since X1 dX d Because the dx dX quotient is actually independent of de, we may clearly da (da) second differential coefficient of X; and in the same way we may express the third differential coefficient by and so (da) dX .h dx consequently, by substituting the differential coefficients for X1, X2, X, etc., in (a), it will be reduced to X' = X + + + d2X h2 d3X h3 + etc., (a'). Because X' is supposed to be the same function of x + h that X is of x, if we represent X by f(x) = a function of x, we must express X' by f(x + h) : = a function of x+h. Hence, putting f(x) and f(x + h) for X and X', (a') becomes f(x + h) = f(x) + df(x) d2f(x) h2 dx .h + (dx)2 1.2 + etc., (a''); which agrees with the form in which Taylor's Theorem is usually given. (50.) Still supposing X to be a function of x, and that X, X'1, X'2, X'3, . . . . X'n are the values of X, X'n are the values of X, which correspond to x, x + h, x + 2h, x + 3h,....x + nh, severally; we propose to find X'n. 1. We shall use AX, AX1, AX', etc., to represent the variations of X, X'1, X', etc., which result from changing x into h; and in a similar way we shall express the variations of AX, ▲X'1, ▲X'2, etc. (which result from changing x into xh), by ▲2X, ▲2X'1, AX', etc., and the variations of these will be expressed by ▲3X, ▲3X'1, ▲3X′2, etc., and so on. Hence, if we change x in X into x + h, we shall get X + AX = X'1, since X+ AX and X', are different expressions for the value of X, which corresponds to x + h. And if in X + get X+2AX + changed into X + 1 1 X = X', we change x into x + h, we shall ^2X = X'1 + AX1 = X', since X is Δ ΔΙ Δ Χ and Δ Χ into Δ Χ + Δ X, and that X', corresponds to x + 2h. Similarly, from X + 2^X + ▲2X = X'2, we shall get X + 3^X+3 ^3X + ^3X = X'2 + ▲ X'2 = X's, and so on to any extent. Hence, it is clear that we shall have X'n = X +n^X+ n(n − 1)2X + n(n - 1)(n-2) A3X+...+n^"-1X+ 1 2 1.2 3 AX, (a); for it is plain, from the law of the coefficients in the successive changed values of X, that the coefficients of ▲X, ▲oX, ▲3X, etc., must be the same as the coefficients of the second, third, fourth, etc., terms of a binomial when raised to the nth power; while the numbers to be applied to A in the successive terms are the same as the indices of the second term of the binomial in the expansion. It is clear that in order to get X', from (a), n, X, ▲X, AX, etc., must be known. 2. Supposing n, X, X'1, X'2 . . . . X', to be given, we will now show how to find AX, 4X, etc., which are called the first terms of the first, second, etc., orders of differences. From the equations X + ^X = X'1, X'1 + ^ X'1 = X'2, X'2+ ^X'2 = X's, and so on, we easily get X = X'1- X AX'1 = X'2 - X'1, ▲ X'2 = X's - X'2, and so 2 2 2 1 2 1 on. — (X'1 — X) = X', And from AX = ^X'1 — ^X, ^'X'1 = ^X'2 — ^ X'1, and so on, we get 'X = X', — X'1 2X'1+ X 4'X'1 = X's — 2X'2 + X'1, +X', and so on; similarly, we get 3X', - X, ▲3X'1 = X', — 3X's + 3X'1⁄2 — 1 1 2 3 2 ▲2X22 = X'1 — 2X', X = X'1⁄2 − 3X'1⁄2 + X1, etc.; the law of continuation of the successive differences being evident. Hence, it is clear that the first difference of the nth order 1 will be expressed by "X = X'n — nX'n-1 + 1x'. n n(n − 1) (n − 2) X',- + etc., (b); observing that the coef ficients are the same and have the same signs as in the corresponding terms of (X' — 1)" when expanded according to the descending powers of X', while the numbers n, n-1, 2, etc., joined to X' (below) are the same as the exponents of X' in the corresponding terms of the expansion; noticing that the last term is X if n is an even number, and -X when n is an odd number. If we write the right member of (b) in a contrary order, it will evidently become "X n(n-1)x',n(n − 1) (n − 2)x', = ± [X−nX+ X's + etc.], (c); observing that + must be used for ± when n is an even number, and that must be used for when n is an odd number; and it is clear that the coefficients and their signs are the same as those which correspond to them in the expansion of (-1+ X')", according to the ascending powers of X', while the numbers 1, 2, 3, 4, etc., which are joined to X (below) are the same as the exponents of X' in the corresponding terms of the expansion. 3. To show the use of the formulæ (a), (b), and (c), take the following EXAMPLES. Ex. 1. To find the (n + 1) term of the series 1, 3, 6, 10, 15, 21, etc. Here, X, X1, X2, X's, etc., are severally represented by 1, 3, 6, 10, etc.; consequently, we have AX = X', - X= ▲ X'2 = X's — ΔΧ, order of differ 3 −1 = 2, ^X1' = X', - X'1 = 6-3 = 3, X', 10-64, and so on, for the first = ences; and X = X'2— X', — (X'1— X) = X'2 − 2X'1 + X =3-2=1, 'X'1 = X',- X'2- (X'2- X'1) = 4 — 3 = 1, (X'2— and so on, are the differences of the second order; and the differences of the third, fourth, etc., orders are 0. Hence, putting 1, 2, 1, 0, 0, etc., for X, 4X, 4X, ▲3X, ▲1X, etc., severally, (a) becomes X' = 1 + 2n + n n(n − 1) 1 2 the term whose num ber, reckoned from the first term 1 (of the series), is n, or it is the (n+1)th term of the series; and because it expresses the terms of the series in a general form, it is said to be its general term; thus, if for n we put 0, 1, 2, 3, 4, etc., we get (from (n + 2). (n + 1)) the series. 1 2 1)), 1, 3, 6, 10, 15, 21, etc., the terms of Ex. 2. To find the sum of n terms of the preceding series. Here, we form the new series 0, 1, (1 + 3), (1 + 3 + 6), (1+3+6+10), (1 + 3 + 6 + 10 + 15), etc., whose differences of the first order are 1, 3, 6, 10, 15, etc., which are the terms of the given series; and 2, 3, 4, etc., are the differences of the second order; also, 1, 1, 1, 1, etc., are the differences of the third order; and the differences of the following orders equal 0. Hence, putting 0, 1, 2. 1, 0, 0, etc., for X, AX, 4X, ▲3X, etc., in (a), we have X', 0+ n + n(n − ΔΧ, Δ'Χ, Thus, if we put 5 for n, we get 1 + 3 + 6 + 10 + 15 = |