( 215 ) 
AS «>(«) : 
.£_(«”)_£# (f) we f mc i by integration of (14) 
gXas) 2 g'*(a) 
1 pg{*)+ 1 , P9i x ) + 1 , „ 
/w=+ ° = r % ^wTi + 
In order to satisfy {A) we dispose in such a way of the constants 
X , p, </ and a, that 
;+l 
i i T 
. _1, ff/ao +i, 
+ a ~;i%<K*)+l 
To this end we must put 
fa— <n 4- t'fa—d)’+40y _(a—d)—t/(a-0)’+4(?y 
* = --5F- ’ q - - 28 ’ 
X =z log 
|«4-<f+t/(W) 2 +W _ 
= % s 
4(«<f—0y) 
whilst we choose a = 0. 
We then find for /fir) 
1 b /^)+ 1 
/( ) i 
where p, q and X are determined by (16). 
Let us still put 
«_±d 
2 2 v 
■=S)| 
(17) 
» = 
(a+r)"-(«_-r)" _ Q(n) = («+»)"+ (»-»)*, ' 
we then find on account of (a) for <p n {%) 
_ now + («-<o ^W} fa) + wwi (d) 
,fn (X) - '-'L Zymfa) + |«n)-(a-d)iln)! J 
By putting <j(x) = S and therefore x = g-\ (S)=G( 5) we can make 
the standard form (D) to pass into 
-*»• >=°(S 9 .^ 
or in a more symmetrical form, 
•=<gS) • - m 
When thus we succeed in finding such a parameter 5 that y = <p{x) 
can be brought to the form (&) or (D"), then for such a function <p 
the iterationproblem is solved. 
15 * 
