
MR T. MUIR ON EISENSTEIN’S CONTINUED FRACTIONS, 141 


Hence, 
2 sin 5 
sim am =—,—- (A=t Bb) 
pt ke? 
1.€., 
. 2 ; : os 
sinamz = 4-{ sin a (A? + B’) 
ips 
as was to be proved. Ina perfectly similar manner (XII.)...(XVI.) may 
be established. For the case of (X VII.) we put z=g’, y=q, and p=q in 
(XX.), and there results after simplification 
(1 —q?) (1—q*) (1—9) .. yal ; 
(1—¢) (1—¢) (1-4)... 1--4~  #@ 


But the first member is known to be=1+q+@+q¢5+ g°+..., and 
therefore on writing w—* for g we have the identity required, in the case where 
m is infinite. Lastly, equating the reciprocals of the sides of (XX.), then 
putting «=0, p~'z for y and p~" for p, we obtain (XVIII) almost at 
once.* 
Having thus reduced the number of apparently independent results to jive, 
it now remains to show how these may be established. Consider, first, the 
most important of the five, viz., (XX.). Making use of the observed fact that 
the continued product 
(1—ax)(1—px)(1—p’a).... 
is divided by 1—z by writing in it px for x, we can readily transform 
(l—x)\(1—px)(1—p*x).... 
(—y)A—py)— py) -- +: 

* Equating the reciprocals of the sides of (XX.), putting y = 0 and writing z for x, we have the 
same continued fraction equal to 
1 
(l—z)1—pz)1—p%)... 

Hence the identity 
( = (é -aQ -z) y  ~ G=20 =e Mace 
the truth of which may also be seen from the fact that were we to try to find, by the method of 
undetermined coefficients, the expansion of the two expressions in ascending powers of z, we should 
in each case make use of the observation that the change of z intu pz is equivalent to multiplication by 
ml — 2, 


