REDUCTION OF ERROR BY LINEAR COMPOUNDING. 
235 
(b) Next, suppose that g — f. Then 
= 1 /«/,/. j jj ... j — — exp — bP . d9 f d9 {) d0 1 
= 1/«/,/. ||j ••• \< f >/ exg — ^ P . d 0 f d 6 t ) d 0 1 ... ddi , 
by ( a). Hence 
N t = 1 /y/ ' a ft f . |jj ... | \ fs 2 exp — . exp —\Q . dp / d 0 0 d 9 x ... dO t 
. * doi 
= v 7 (2 Tr/a ftf ). ... exp—1-(). dO t) dO l ... dO 
Also 
B= l/\/«/,/ • j [J ... jexp— Tf\f/ . exp—^-(1 . ... 
= v 7 (2 Tr/ctf'f) . J exp—. d6 l) d0 1 ... c?0 z . 
Hence 
Ah/Z) = 1. 
.» y 
4. Hence the y s are related to the u’s in such a way that 
m.p.e. of y f and u g = 0 {g^f) or 1 (g = /); 
and therefore the r’s and the y s are conjugate sets ; which proves (i.). It follows 
that 
a fig — m.p.e. of y f and y g . 
This proves (iii.); and (iv.) is the similar result which we should have obtained by 
expressing P in terms of the (p's. Also 
P — a o,o^o 2 + 2 a Qtl 0p i + a lA 0^+cv tA 0i 
— #0 ( a 0,0$0 + ^0,l$l+ Cl! '0,2$2 + ... +«o ,fil) 
+ 0 1 (« lt O 0Q + a i, l$l + ^1,2^2 + ••• +<*1. A) 
4 " • • • 
+ (^, (A + A + d[, 2 0 2 + • . . + « 7 , A) 
= O 0 (p 0 + 6 1 cp 1 + 0 2 (p 2 + ... + 9 t (pi ; 
which proves (ii.). 
Appendix III. —-Improved Advancing Differences in terms of Sums. 
The expression for A f v 0 given by (136) differs from that given in (15) and §5 of 
Fitting,” in that it involves A"u 0 , 'Z" 2 u 1 , 2 //3 r 2 , ... , instead of A"u 0 , 'A" 2 u 0 , 2 //3 w 0 > ••• • 
