ESSAYS 291 



If n is a positive integer, the series for <f> n ceases (owing to the exhaustion of 

 he binomial coefficients) at the term O m , if O m is the highest power in 0. 



Examples. 

 [O + cO*] n = + ncO i 4- 2« a c'0 3 4- « a (2« - 3)^0* + 2n 3 (^n - 4)<r 4 5 + etc. 

 [Iog(i+O)] n = 0-^0 2 + *«(3« + i)0 3 /3 !-J»(3«+i)(2»+ i)0 4 /4'- + i«(45« 8 + 



+65« s + 3o« + 4)o 5 /5 ! etc. 

 [t°-l] n = + \nO x + \n{^n - i)0 3 /3! +4»(3»- i)(2»- i)0*/4! + i«(45« 3 -65« a 4- 



+ 3o«-4)0 6 /5! + etc. 

 [Oe°] n = + nO , + «(2«+i)0 3 /2 ! +«(i2« 2 -i5« + 4)o 4 /3 ! 4- etc. 

 [ f °log(i + O)]" = O 4- InO 1 4- »(3» - i)0 3 /3 ! 4- 4»(» - 0(6» 4- n)0*/4 ! 4- »(45» 3 + 



+ 65« 2 -24o«+i84)0*/6 ! + etc. 

 [OcosO]" = 0-«0 3 /2 ! +«(9«-8)0 6 /4 ! -rc(i5>/- i6) 2 7 /6 ! + etc. 

 [log(i 4-0)] 12 (o-oi)=o-oo9433 . . . [e°- i] 12 (ooi) =0-010638 ... 3*6 



The iteration formulas for sinO, tanO, sin -1 0, tan _1 0, are already known 

 (Wolstenholme's Mathematical Problems, 3rd ed. 1891, Nos. 620, 621). 



4'0. If the absolute term of the proposed multinomial be wanting, but the 

 coefficient b of the first term O be not unity, we may proceed as before in the case 

 of 3'3, by equating coefficients in the values for 4><£ and <$<£, giving 



Bb = bB, CP + Bc^cW + bC, 



Bb* + 2Cbc+Bd=dB» + 2cBC+b£>, etc. 

 Hence C(b*-b) = c(B*- B), 



D(b* - b){b* -b) = 2c\B* - B)(B - b) + d{B* - B){b* - b), 



and so on. On commencing a few substitutions for the values of $, $ 2 , . . ., we 

 shall easily see that B = b n ; so that the sub-coefficients of c, d, e, . . . in the expan- 

 sion of </>"(= *) are composed of the sums of «-terms of geometric progressions of 

 various powers of b. Writing for brevity 



G r s (** - 1 W r - i) and K n m (G r - G s )l(b r ~ s - 1 ), 

 we have 



[bo 4- rt> 2 + dO s . . -] n = ^ n O 4- Gxc\b . b n O* + { G 2 djb + 2A* n c*/b i }b n Q* + { G z e\b + 

 4-(3A- 32 + 2A" 31 )^ 2 4-(4^3i + ^ 32 - *>K^)c*\&(b - i)}£»0 4 +{G i flb+(^K i3 4- 

 4-2A'«W3 2 + zKu&lb* 4- ((3* 4- u)K a + (bb + io)A" 4 i - (96 + 9K32 - 

 - 12^)^1 b\P - 1)+ (AKu + (2b + io)A\i-(4b + 4)A" 32 - {l*b+\2)Kn. + 

 4- (i4b+2)Kny*lb\b-i)(b i -i)}b n o i + elc. 4"i 



I am not aware that this series has been given before. It can be recast in 

 several ways, which I have no space for (see however io'o) ; but the following 

 theorems are useful for this purpose : 



G r =(B- l)(l+B+B* . . . B r ~ l )!{b- i)(i+b+b* . . . b r ~ l ) ; 



Krs=b s 4- (1 4- b'-^b* + (1 4- b r ~ s + b* r - s >)b Ss . . . (1 4- b r ~ s . . . Jfn-Wr-^a-l* 4 . 2 



Successive substitutions show that when n is positive, integral, and finite the 

 coefficients in the expansion of </>" must also be finite and integral. Hence, not 

 only must G r and K T$ be so (as just shown), but all the functions of K within 

 small brackets in 4-1 must be divisible by£-i, b s - 1, . . . where these denominators 

 occur. 



When n = o, then G>=o and $ =O. When #=l, G r =Gs and <f> n = </>. 



If b -> i, G T -> n ; and by taking the limits of the coefficients in 41, we shall 

 find that the whole series reduces to that of 3-3, as it should do. It is interesting 



