REDUCTION OF ERROR BY LINEAR COMPOUNDING. 
221 
and so on. There is a formula connecting these, which makes it unnecessary to deal 
with more than the first quantity in each row ; or, if we deal with them all, the 
formula can be used for checking the results. (An example is given at the end of 
§ 15 of “Reduction.”) 
We have 
1 = terms in S j+1 , § j+2 , .... 
By (XV.), this is the improved value of S f — Of using os after <5j; and therefore, 
by (XIII.), 
1 = ( e f)j — Qf,j( e j)i — (e/)j —.(76) 
Re-arranging, and replacing j by j— 1, j— 2, ...,/+ 1, and remembering that, by (75), 
Of f = 1, we have 
i € f)j ~i c f)j-i = °f.j 
( e /)/+i — ( € f)f — fy,/+i E /+i> 
( e f)f — fy,f E /- 
Hence, by addition, 
( €f )j = 0 M E f + d/, /+ iE /+1 + 0 /i/+2 E /+2 +... + OfjEj .(77) 
19. Mean Products of Error (Alternative Formula).—{ i.) By (77) and (38), 
f f , g )j = m.p.e. of S g and (ey) ; - 
= m.p.e. of S g and e f.f E ‘f +0 ff+ i E /+i+ ••• + 
= 6/,/(\,/)/+ 0/./+1 (\,/+l)/+l + ••• + 
. = t t)t- 
The summation has to be made from t =f to t = j. But, if (/>/, we see from (37) 
that it is sufficient to make the summation from t = g. Hence, using “ t =f, g” to 
denote summation from t — f or from t — g, according as f or g is the greater, 
we have 
iff,g)j = X Qf,t{\,t) t . (78) 
t=f, <7 
But, by putting g — j in (78) (or taking the m.p.e. of Sj and each member of (76)) 
and then replacing f and j by g and t, 
Hence, substituting in (78), 
ifg, t)t @g,t Aj 
iff , g)j — X 0f,t0g,t Aj 
■ t=f<9 
(79) 
(80) 
