ESSAYS 293 



If in this we substitute the values of Ca, c% ...,di, a% . . . , e 9 , e t ... obtained in 

 5-1, we return to the expansion of <£» given in 3-3. If we suppress the subscripts, 

 we return to 5*1. The same groups of letters occur in both expansions. 



Another method for establishing both theorems is by use of the primitive 

 identity often employed in the Finite Calculus. If a , a h a 2 , a s . . . are any 

 quantities, functions, or operations capable of addition and,, subtraction, then 



a\ = <2o + {a\ — # 0) ; 



«a = a + i(a\ - a ) + (#2 - 2a i + a o) ', 



a 3 = a + 3(«i -a ) + 3(^2 - 2«i + «o) + («s - 3«2 + 3<h - a <>) ; 



a n = a + n{a x - a ) + n%{a 9 -2a + a ) + n 3 (a 3 - 3«2 + 3«i - a ) + etc. 



Hence if </>•( = o), j>, <£*, . • . <£" be the given series, 



cf>* = + n(4> -0) + n 2 ((f> t -2<l>+0) + n3((f> 3 -3<t> i + 3<f> ~0) + etc - 5*3 



The values of the expressions within brackets containing <j>», <f>*, ...are 

 worked out by substitutions or by the use of the formula 2*4, and the coefficients 

 of the successive powers of O are collected. The result is the formula 3-3. Or 

 we may let the series stand as it is in successive values of « ls » a , n 3 , . . . If 

 <f)»=[(a + bx + cx 2 . ..)0] n , the series gives the expansion of (a + bx+cx* . . . )» 

 that is, the algebraic multinomial theorem. See end of yo. 



6'0. In 3'2 it was found impossible to obtain the iteration-formula for a + bO + 

 cO* + . . . , that is, for an operation of a degree higher than one with an absolute 

 term, by means of successive substitutions. We now obtain it by the means of 

 the elementary operative theorem, first used, I believe, by Babbage in his work on 

 periodic functions, namely that if # = £££" x > then 



and * n =fl"C" 1 . 6>I 



First let (=p + 0, so that £- 1 =0-p-* Then if 



Z=[o-pl<j>lp + o]=<i>(P + o)-p, 



and ^=[p + 0\^>{p J rO)-p} n {0-p\ 6-2 



as an identity. Now let p be, if possible, a number m such that <pm = m. Then 

 if <f> be any operation which can be expanded in ascending powers of O, the 

 nuclear operation £, that is <£(»* + 0)-<£w, will contain no absolute term, and 

 may therefore be iterated by 2*1, 3:3, or 4"i ; so that 



[a + bO + CO* + . . .lT=m + [<f>'m . O + W m • O 2 + W' m • O s + . . .]"(0 - m). 6'3 



Suppose, for a simple example, that we require to iterate i2-6^r + ^ 3 upon 

 itself 7z- 1 times. Then m must be one of the roots of i2-6x + x 2 =x, that is, 

 must be either 3 or 4. Thus 



^ = 3 + [0 2 Kr-3) = 3 + [0 1 "](r- 3 ) = 3 + (ar-3)»"; (by 2*1) 



or (j> n x = 4 + [2 O + 0*] n (x - 4) 



= 4 + [2 n O + i2"(2"-l)0 2 + £2"(2"-l)(2"-2)0 3 +. . .Jx ~ 4) 0>y 4'0 



= 3 + [(i + 0) 2 "](-r-4) = 3 + (^-3) 2 ", 

 as before. 



a + bO 

 * To invert simple operations proceed as follows. Suppose </>=,. Oper- 

 ate with both sides on <6 -1 . Then = ^-j— ,. Now solve algebraically and we 



c+a<p~ l 



get <£-! = — - — — . Or we may first solve algebraically and then operate on <£ _1 . 

 b — do 



