1915-16.] On the Theory of Continued-Fractions. 
253 
Y 3 , . . . Y n are replaced by zero. That is to say, the value of is 
equal to the value of yj Y 0 derived from the equations 
Y 0 = ip + x ) z o "t c i z i 
0 = dyZ Q + ( by + x)z x + c 2 z 2 
0 = d 2 Zy + (b 2 + x)z 2 + c 3 z 3 
< 0 = d n z n _ 1 + (b n + x)z n 
O = (b + x)y 0 + c 1 y 1 -z 0 
0 = d-^y 0 + {by + x)y-y + c 2 y 2 — Zy 
0 = d n y n _ i + (b n + x)y n - z n . 
But these equations give at once 
i 
Kj 
o 
o 
0 
0 . . . 
0 
0 
b + x 
C 1 
0 . . , 
. 0 
0 
0 
0 
0 . . . 
0 
0 
d. 
by+x 
C 2 • • ■ 
, 0 
0 
0 
0 
0 . . . 
0 
0 
0 
0 
0 . . . 
■ dn 
b n + x 
b + x 
C 1 
0 . . . 
0 
0 
- 1 
0 
0 . . , 
. 0 
0 
d i 
b 
i+z 
c 2 . . . 
0 
0 
0 
- 1 
0 . . . 
. 0 
0 
0 
0 
0 . . . 
dn l 
i n + x 
0 
0 
0 . . . 
, 0 
- 1 
and therefore 
we 
have finally 
0 
0 
0 . . 
. 0 
0 
dy 
by + x 
6*2 . . 
. 0 
0 
0 
0 
0 . . 
. 0 
0 
0 
d 2 b 2 
+ x . . 
. 0 
0 
0 
0 
0 . . 
. 0 
0 
0 
0 
0 . . 
• d n 
b n + x 
C 1 
0 
0 . . 
. 0 
0 
-1 
0 
0 . . 
. 0 
0 
Oy “I - X 
0 . . 
. 0 
0 
0 
-1 
0 . . 
. 0 
0 
dS 
0 
0 
0 . . 
. d n 
b n + x 
0 
0 
0 . . 
. 0 
- 1 
dx 
0 
0 
0 
0 . . 
. 0 
0 
b + x 
c i 
0 . . 
. 0 
0 
0 
0 
0 
0 . . 
. 0 
0 
d x 
by+X 
e 2 . . 
. 0 
0 
0 
0 
0 
0 . . 
. 0 
0 
0 
d, b, 
2 + X . . 
. 0 
0 
0 
0 
0 
0 . . 
. 0 
0 
0 
0 
0 . . 
. dn \ 
b n + x 
b + x 
'o 
0 
0 . . 
. 0 
0 
-1 
0 
0 . . 
. 0 
0 
dy b 
y+X 
C 2 
0 . . 
. 0 
0 
0 
-1 
0 . . 
. 0 
0 
0 
rf 2 
h 
, + x c s . . 
. 0 
0 
0 
0 - 
-1 . . 
. 0 
0 
0 
0 
0 
0 . . 
. d n 
b n + x 
0 
0 
0 . . 
. 0 
-1 
