1915-16.] On the Theory of Continued-Fractions. 
For we have by hypothesis 
(x + b 2 ) 2 + c 2 d 2 + c 3 d 3 c 3 {2x + b 2 + b z ) c 3 c 4 0 
d 3 (2x + b 2 + b 3 ) (x + b z ) 2 + c z d 3 + c 4 d 4 c 4 (2x + b 3 + b 4 ) c 4 c 5 
255 
dTj 
dx 
(x + by) 2 + c 2 d 2 
d 2 {2x + by + b 2 ) 
cJ2x + by + b 2 ) 
(x + b 2 ) 2 + c 2 d 2 + c 3 d 3 
c 3 {2x + b 2 +b 3 ) 
so we may write 
Cydy 0 0 0 
d 2 (2x + by + b 2 ) (x + b 2 f + c 2 d 2 + c 3 d 3 c 3 {2x + b 2 + b 3 ) c 3 c 4 
d 2 d 3 d 3 {2x-rb 2 + b 3 ) {x + b s ) 2 + c 3 d 3 + c 4 d 4 c 4 [2x + b 3 + b 4 ) . 
dT_ 
Ctl dx~ 
{x + by) 2 + c 2 d 2 
c 2 (2x + by + b 2 ) 
d 2 {2x + by + b 2 ) (x + b 2 ) 2 + c 2 d 2 + c 3 d 3 c 3 { 2x+b 2 + b 3 ) 
0 
c 3 c 4 
and therefore 
i dT 
1 _ a l T~ : 
dx 
(x + by) 2 + Cydy + c 2 d 2 c 2 {2x + by + b 2 ) c 2 c 3 
d 2 (2x + by + b 2 ) (x + b 2 ) 2 + c 2 d 2 + c 3 d 3 c 3 {2x + b 2 + b 3 ) 
{x + byf + c 2 d 2 c 2 (2x + by + b^ c 2 c 3 0 
d 2 {2x + by + b 2 ) {x + b 2 ) 2 + c 2 d 2 + c 3 d 3 c 3 {2x + b 2 + b 3 ) c 3 c 4 
But the equation S = ^ + J_ aT g ives - ^ = S 2 (l - af^). Substituting for 1 - the 
determinantal expression just found, and for S its expression as a quotient of two 
determinants given by Sylvester’s theorem, we obtain formula (19) for — , and the 
dx 
theorem is thus established Formula (18) may be derived by reversing the process by 
which (19) was derived from (18). 
{Issued separately January 19, 1917.) 
