James Henderson 
171 
Obviously 
ci = - pliv + q) + pliv + ?) = f*. 
1 1 pq 
Co — 
+ 
pi 
2l{p + qf 2 ip + qfip + q + l) 2 I {p + q) {p + q + 1) {p + q} 
P 
p + l 
P 
2 {p + q\ Kp + q + l) p + q {p + q){p + q+l) 
2 {p + q\ \{p + q){p + q + i) + + 2 + 
= 0. 
Similarly the other c's can be determined but the work becomes more and more 
laborious as we go on. 
Unfortunately, as far as the numerical work is concerned, we have failed after 
many attempts to find a relation connecting successive c's, similar to that found 
in the case of the Incomplete F-function. At first it was thought that the 
following treatment would facilitate the numerical calculation of these coefficients. 
Let e P+1 =K + h,0 + h,d-+ ...+b,0' +..., 
then - pKp + q)d- ^a'd'' = log, [b, + b^d + h^O- + ... + b.d" +...]. 
Diff'erentiate this and then equate coefficients of powers of 6 : 
{bo + b,d + bS' + ...+ b^e^ +...) i-pKp + 2) - a'd) 
= b, + 2b,d + ^b,&' +... +sb,d'-' + .... 
Equate coefficients of 6^~^ : 
sb, = - p/{p + q) - o-' bs-.; 
therefore 
s 
P 
P+q 
.(xxiii). 
This formula enables us to calculate the b's very rapidly on the machine when 
p/(p + q) and have been determined. 
From equation (xxii) 
Co + c,0 + c.,6- + ... +0,0' + ... =(6o+ b,0 + bn0"- + . . .) 
1 + 
P 
+ 
0' p(p + l) 
p + q 21 {p + q){p + q + l) 
+ ... + ... 
Equate coefficients of 0': 
c^ = b^l. p{p + 'i-)---{p + s--l) 
s \ {p + q){p + q + 1} ... ip + q + s - I) 
I p(p + l)...{p + s-2) 
+ k 
+ ■■ +h^, 
I.e. 
(s ~ 1)1 {p + q){p+q + l)(p + q + s- 2) 
. i ^^._J p{p + l)...{p+s-r-l) 
I P 
II p + q 
r=o {s-r)\ {p + q){p + q + l),,,(p + q + s-r-l) 
