296 SCIENCE PROGRESS 



a straight line parallel to the axis of x and at a distance m from it. For a some- 

 what different proof, let xo, X\ t xi, . . . denote the iterants x, <£*, <£*.r, .... Then 

 Dcf ) i xo = [cji'cf> . <f>']xo = <f>'xi . <fjxo, and D<prx ^<f>'x . <f>'xi, <f>'x* . . . <f>'x H „i. Hence 

 if | <f>'x | < I for the whole region containing these iterants, or even at the point 

 of condensation m, then Dcp^xo vanishes, that is, the slope of the curve <£ M is zero 

 within that region. The same thing may occur at all the midaxial roots where 

 the necessary conditions hold — in which case <£°° becomes a series of straight 

 lines parallel to x (see 5, p. 411). A good example is <£ = + 0(0 - i)(0-2) . . . 

 (0-r+i)/r. 



100 This concludes most of the matter proposed for this paper ; but some 

 further examples of iteration will be useful. 



(1) The following method gives the result of 4'i as a Dirichlet series : 

 Let 4>"s=[>0 + £-0* + ^0«. • .] n =Pb n +Qb*» + Rb in + etc., 



where P, Q, R, . . . contain O but not n. Then 



0»+i= Pbb n + QPb* n +Rb 3 b 3 » + etc. 

 But also <p + 1 = 4>$ i =b{Pb«+Qb' 1 * . . .) + c(Pb"+ Qb*~ . . .) + etc. 



= bPb* 4- {bQ + cP*)b*» + (bR + 2cPQ + dP 3 )b 3 " + etc. 

 By equating coefficients, 



c_ 2c 2 +(b-i)bd 



Q- fr-b F k ~ (p - b) (b* - b P * etc " 



Hence Q, R, S, . . . may be written qP*, rP 3 , sP*, . . . where q, r, s, . . . are com- 

 posed of b, c, d, . . . but contain neither P, O, nor n. Now to find P, 



0n = Pb* + qPW + rP 3 b 3 » + etc. = [O + qO* + rO* . . .]b n P ; 

 <£ =[O + ?O 2 + rO 3 . . .]P=0 ; 

 and P =[0 + ?0' + r0 3 .. .]-*; 



the invert being given by 3*5. Substituting these values of P, q, r, s, . . . we 

 obtain the required expansion (connecting Dirichlet's series with iteration). io'l 



(2) We may now write cf> n also in the form 



<£ B = [0 + ?0* + r0 3 . . .][bO]"[0-qO*-(r-2q i )0 3 -etc], 



which is simply the t£t _1 form given in 6"i. Or we may assume this form from 

 the first. Expanding by substitutions or by 2'4, and first putting n= 1, we have 

 ^bO + c-Ot + dO 3 . . . =bO + (b i -b)q0 2 +{(b 3 -b)r-2(b*-b)q*\0* + 



+ {(b* - b)s - {3b* -3b + 2b a - 2b)qr + ${b* - b)q 3 } O 4 + etc. io"2 



By equating coefficients (or by the previous theorem) we find q, r, s, . . . , and 

 place them in the series without disturbing the functions of b already there. Then 

 in order to find <£" we have only to put b n for b in these last-named functions. 



(3) An elegant way to put a power series in the form ££t -1 is to assume the 

 form of ££ as a power series, and then to divide it operatively by f as shown 

 in my papers (1, 4, 5). Suppose we wish to iterate bo + c -f dO~ x . . . 

 Assume that it = [O + p + qO' 1 . . .Jbojo + p + qO~ l . . .]-\ Then since 

 f |f -1 = ££//£ we divide operatively as follows : 



+ p +qO- 1 ...']{>0+p + q?>- 1 0- 1 + rb-*0-' s . . . [bO+p(i -b^+qib- 1 -b)0' 1 + 



bO+pt> + q&0- 1 + rbO-*... + {r(b*-b)+pq{b~ l -b)}0-* 



^(i-^)+^- 1 -^)0- 1 + r(^ _ g)o -* . . . + etc. 



q{b-*-b)0- A -q{b- l -b)pO-*.. . 



{r(b*-b)+pq(b- 1 -b}0-*+ . , , 



