294 
v a B—a(l —e) AA 
=2B| a—— (1—b!) — BET 
yp—b B 
— & (l—e) ze fs a jo (2a—l)a A + (82?—1) (1—0’) a]. 
v—b B 
When now for brevity C is written for 
NaA + AB = (2—2) a? —2e (l—e) aA + a (le?) (1-6) A? 
(according to (2) and (26)), then 
AC 
Cc 1—2) (2e—1) merker adt. ) a Ble = 
v ) A 
| = 2Ba — (Lb) — 2 C—w (l—a) — B la”? — ete, |, 
because a B—a(1—2«) AA is =C and (2—a’) C—a(1i—«) NA’? = BE 
Hence we have also: 
B: v q C 
3C—(1+a)— b"v=2aB es (1—6')— (le) - —| (2a@-—1)b'— Ne — ete.) 
N v—b Bv_b| 
For (2e—1)6’C + N (a? — etc.) may be written 5? (1 — 6’) — 
— A*(a’— os hence we have: 
(1+ #)B: pl TE En 
ore = 2aB— (1-8) «(1 “EB meh N si 
Before proceeding, we shall apply a control-calculation to this 
equation. When A is =O, then A becomes =a, B= (2—a)a, 
C = (2—a)a’, so that then (a) passes into 
(1 +) Em ", 
3 (2) a? —- N == 2 (24-2) a? DE (1—b') — | 
a(l—a) v (2—2«)' a? ef — a (w°— 4% + 2) 
ean N 
i.e. after division by (2—«#) «*: 
(la) (2-2) - (1-b')—(#?- 4a +2): (2-2)? 
—- — "y= 2 — b' l— ; 
N dende = (1- Je (1-2) — ar N | 
in which NV = (2—a) (14e) (1— 6’) —a (1a) = 2—(2— 2) (1+ a) 6’. 
This gives: 
bv 1 v x (1—.x) (1—’) 
3 — — — = 2(1—b' 
1—}' x (l—e) v—b | ( N * 
(2a) (1+) 0-5) 
w (1—a) ((2—a)?—2) | 
(2-20)? N 
or 
