3G 
PROFESSOR K. PEARSON ON MATHEMATICAL CONTRIBUTIONS 
Thence e = C 1 = ‘619,068 from (xxi.). 
jn -±ilv 
Calculating out the coefficients of the series in r in (xix.) we find 
•619,068 = r + T88,345r 3 + ‘064,0814r 3 + T07,8220r 4 + ‘005,9986r 3 + ‘067,2682r 6 
+ &c. 
Neglecting powers of r above the second, we find by solving the quadratic and 
taking the positive root 
r = ‘5600. 
Solving by two approximations the sextic we finally determine 
r = *5422, 
correct, I think, to four places of figures. 
Turning now to the probable error as given by Equation (1.), I find 
and from (xlix.) 
Further : 
h 2 + k 2 - 2 rhk - ‘348,924, 
log Xo = F170,0947. 
k - rh 
- = *275,642 , 
x/1 - /- 
Hence from (xlvii.) and (xlviii.) we find 
^ f'393,078 
li — rk 
% = ‘393,078. 
*= M 
-mz 
* - M 
1 r -275,642 
and by means of the probability integral table 
Vq = -108,884, = *152,865. 
By substituting in (1.), we find 
probable error of r = ‘0288. 
From (xxxiv.) and (xxxv.) we find 
p.e. of h = ‘0282. p.e. of k = ‘0278. 
Thus, finally, we may sum up our results 
h = ‘6463 ± ‘0282, k — ‘5828 ± ‘0278, 
r = *5422 ± -0288. 
The probable error of this r, if we had been able to find it from the product 
moment, would have been ‘0147 s or only about one-half its present value. 
