be considered, so far as may be necessary, in subsequent chapters. Ex. 1. To develope un in a series consisting of u and its successive differences (Ex. of Art. 7, resumed). Ex. 2. To express ▲"u, in terms of u, and its successive values. Since u x = Eux - ux, we have ▲ux = Ux+1 and as, the operations being performed, each side remains a function of x, Of particular applications of this theorem those are the most important which result from supposing u= x”. Now let the notation A"0" be adopted to express what the first member of the above equation becomes when x=0; then + n (n-1) (n − 2)m_ n (n − 1) (n − 2) (n − 3)m 1.2 1.2.3 + &c......(5). The systems of numbers expressed by A"0" are of frequent occurrence in the theory of series*. From (2) Art. 1, we have ▲"0" 1.2... n, and, equating this with the corresponding value given by (5), we have 1.2... n=n" -n (n − 1)" + n (n - 1) (n − 2)" — &c....(6)†. Ex. 3. To obtain developed expressions for the nth difference of the product of two functions u, and v. where E applies to u alone, and E' to v, alone, we have ▲u ̧vã = (EE' — 1) UxVx, and generally ▲"u ̧vx = (EE' − 1)" uxVx .(7). It now only remains to transform, if needful, and to develope the operative function in the second member according to the nature of the expansion required. Thus if it be required to express Aruv, in ascending differ * A very simple method of calculating their values will be given in Ex. 8 of this chapter. + This formula is of use in demonstrating Wilson's Theorem, that 1+|n· [n-1 is divisible by n when n is a prime number. ences of v, we must change E' into A'+1, regarding A' as operating only on v. We then have Remembering then that A and E operate only on u and ▲′ only on v, and that the accent on the latter symbol may be dropped when that symbol only precedes v,, we have ▲ˆuxvx= ▲”ux • Vx + nA”-1ux+1 · Avx As a particular illustration, suppose u2 = a*. Then, since Again, if the expansion is to be ordered according to successive values of v, it is necessary to expand the untransformed operative function in the second member of (7) in ascending powers of E' and develope the result. We find Lastly, if the expansion is to involve only the differences of u, and v2, then, changing E into 1 +▲, and E' into 1+A', we have Ux ▲"uxvx = (A + A' + ▲▲')" už v 2.........(11), and the symbolic trinomial in the second member is now to be developed and the result interpreted. Ex. 4. To express A"u, in terms of the differential coefficients of ux• Now t being a symbol of quantity, we have (12). &c.)...... (13), + + &c. 1.2.3 1. = t” + A ̧tn+1 +  ̧†¤+2 + &c., on expansion, A1, A,, being numerical coefficients. Hence d n d n+1 d n+2 (e* — 1)* = ( 2 ) " + 4, ( 2 )TM + 4, (2)** + &c., dx dx Ux dx The coefficients A,, A,,... &c. may be determined in various ways, the simplest in principle being perhaps to develope the right-hand member of (13) by the polynomial theorem, and then seek the aggregate coefficients of the successive powers of t. But the expansion may also be effected with complete determination of the constants by a remarkable secondary form of Maclaurin's theorem, which we shall proceed to demonstrate. Secondary form of Maclaurin's Theorem. PROP. The development of (t) in positive and integral powers of t, when such development is possible, may be expressed in the form First, we shall shew that if (x) and (x) are any two functions of admitting of development in the form a + bx + cx2 + &c., provided that x be made equal to 0, after the implied operations are performed. For, developing all the functions, each member of the above equation is resolved into a series of terms of the form A () m x", while in corresponding terms of the two members the order of the indices m and n will be reversed. Now dym dx " is equal to 0 if m is greater than n, to 1.2...n if m is equal to n, and again to 0 if m is less than n and at the same time x equal to 0; for in this case x-TM is a factor. Hence if x = 0, n-m dx and therefore under the same condition the equation (15) is true, or, adopting the notation above explained, d d • ( 1 ) + (0) = ↓ ( 2 ) + (0.0).... Now by Maclaurin's theorem in its known form d $ (t) = $ (0) + do d2 $(0).t+ 12 ..... (16). do2 $ (0) · 1. z + &c. ... ... (17). Hence, applying the above theorem of reciprocity, (中 $ (t) = $ (0) + $ (~) t2 0 . t + $ (2) 02. — — 2 + &c. ... (18), '1.2 the secondary form in question. The two forms of Maclaurin's theorem (17), (18) may with propriety be termed conjugate. |