164 On Expansions in Tetraclioric Functions 
Equating coefficients of 6^ we have 
^6s_2 = (s + l)6,.+i-s6s, 
i.e. = g ~ jA-sl (xii). 
By this difference formula successive 6's can be found very quickly if 60. ^i, K 
are known and we have already found these. 
Now Cs = {'^i^ycs 
or h^{^/pY+'Cs. ^ ■ 
Substituting in (xii) 
11,, , .... 
or c,,+j = ~ sc, + c.,_, (xm). 
(s + 1) 
This formula gives us very readily the coefficients uf c6 (— D) and thus the 
expansion is obtained. 
We had, Equation (xi), 
xP~^ e-?-/2p 
and all the c's are known since Cq = ^ , Cj = c, = 0. 
To find the area under the curve (xi) up to abscissa x, remembering that the 
left-hand side is zero from ^ = — 00 to — p, 
= "^P j (CoTi + Ci V2~! To + . . . + c,_i Vs ! T, + . ..) cZ^. 
Now I T.dz = - 
therefore 
a; g—x 
0 ro9^)"«^^ = ^/~' 
-<x. \/27r ' V2 ' VS "'" Vs 
1 
h (1 + «',)- {ciTi + C2 V2 ! T2+ ... +Cs_i v'(s- 1)! T,_i + ...j since Co = 
