1904-5.] 
Dr Muir on the Theory of Continuants. 
665 
- C£ 0 d 2 a B 1 
a x a 2 a 3 
. a* 
% a l 
and therefore by the preceding 
for the two lengthiest elements, 
— a,Q a 2 
a 1 af - a 3 a Y 
a 0 a 2 a 1 - a 3 a 0 U 
a 2 a 0 Y 
and consequently 
<h 
a 2 
a 3 
1 
a 0 
a Y 
a 2 
a 3 
a o 
a x 
a 2 
. 
«o 
a 2 
transformation, if we write U, Y 
is equal to 
a i a 2 
a 0 a Y af - a 3 a 2 
a Q a 2 a 1 - a 3 a 0 U 
a 2 a 0 V , 
$0(^2 
a 0 a. 
- «3«l) 
-a 3 a 0 
a 2 a 0 U 
This does not agree with Gunther’s result, an awkward error 
appearing in the last column of both his determinants, and thus 
also in his continued fraction.* 
* With the correct values for U and V as here given, it is possible to take 
a step farther. For it will be found that 
a 2 
a 3 
1 
<h 
a 3 
1 
U — a 3 . 
a i 
a. 2 
a 3 
’ll 
> 
O'O 
a 2 
Ob 
a 0 
a i 
a 2 % 
, 
a-L 
a 2 
and consequently that the continued fraction is 
I Cbo I 
a 2 a 3 1 
a n 
1 a i 
a x a 2 a 3 
I a l a 3 \ 
a 0 a x a 2 
<x 0 a 2 1 
a x a 3 1 
a 0 a 2 a 3 
a x a 2 
where the three-line determinants are the complementary minors of the 
elements (4, 1), (4, 2) of either of the original determinants, and where the 
two-line determinants are the cofactors of the last elements in the three-line 
determinants, and so on. 
