REDUCTION OF ERROR BY LINEAR COMPOUNDING. 
229 
(v \ M .. ( \s-n {2&— 1, 2k— 1} ( £m, 2.<?—1~1 (2s + 2k-i, s + k- 2) 
1 ' liM - 2_ s r/ ' {2k-l, 2s-l}(±m,2k-l] 2s + 2k-2 
= (4& —4, 2k—2)f(^k—2). F {—n + k, n + k, 2k —§ ; 1, 2/:, 2k — 
= (4&—4, 2k —2)/{m, 4& —3]. 
(130) 
It would, of course, in view of the identity of the E’s as stated in (iv.) above, have 
been sufficient to obtain (126) and deduce (127)-(130) from it. 
(vii.) The A’s having been found, the A’s, i.e., the m.pp.e. of the.improved values of 
A% 0 and A% 0 , etc., are obtained by (80). The 6’ s and <//s being as in (ii.) above, 
(A) 
'V'? 
2 0/,*0** (2«, 0/( m > 2£+l],. 
<=/, g 
• (131) 
(B) 
'W, 2 ? = 
2 ffi/,2* 2t (^h 2t)/(m, 4t+ l],. 
<=/, <7 
■ (132) 
(C) 
M2/-l,2 ? -l = 
t = k 
2 02/-l,2«-l 02j-],2«-l (d^ 2, 2t l)/(m, 4t l], 
t =/, <7 
. (133) 
(D) 
^2/-l,2</-l ~ 
2 ffi/_i,2t-i ffi ? -i,2s-i (4t — 2, 2t l)/(?n, 4t l], . 
t =/. <7 
■ (134) 
(E) 
f x 2f-2,2g-2 = 
t = k 
2 <p2f-2,2t-2 ( f ) 2g-2,2t-2 4, 2 1 2 )/(w+ 4£ 3]. 
. (135) 
To find the m.s.e. of the improved value of any l.c. of the differences, or the m.p.e. of 
two such l.cc., we apply (II.) of § 6, as in § 19 (ii-)- Thus, for m = 2n + 1, 
m.p.e. of b 0 v 0 + b 1 fiSv 0 + b 2 ^v 0 + ... + b 2k S 2k v D and c 0 v 0 + + c 2 <f u :) + ... + c 2k 8 2k v a 
[g = t 
— 2 | ^2 ^ 2 /^ 2 /,2«jj 2 C2ff@2ff,2tj' (4i, 24£ + 1 ] 
t = k (J =t "I (g = t 
+ 2,2 &2/-102/-l,2i-l n ^ (: 2fl-102,7 —1,2*-1 
t = 1 L/= 1 
J l.'7 = 1 
•2, 2£-l)/(m, U-l]. 
(viii.) The As having been found, the e’s are then given in terms of the <r’s by (40). 
Using the expressions for the o-’s given in (26) and (31)—(35), we get the following :— 
(A) 
A%> - "2 A, „2 "° +l u n 
<7 = 0 
V’9‘ 
9 = 3 
2 
g = 0 
L 2 {-y\ 
(B) 
(C) 
= 
f!7 = fc 
L\ 2 X 2/ ,^/eo- J - 9+ ffi 0 ; <m, 
_ ' (7 — 0 
f/ = k 2 
A *1 2 fJ-2f-l,2g-l cr ?7 A) > 
S' = 1 
t = m 
t =0 
£ = 
. 1 
£ = — 5 m 
t — 2 111 
t = — fm 
.... (136) 
;.(136a) 
. (137) 
; . . . . (138) 
