468 MR T. MUIR ON NEW GENERAL FORMULA FOR 
and we are required to determine a, a,, a@,.... in terms of A,, Ay, A;,..., 
LBS BR DENS 5 5 
Writing C, for A, — B,, C, for A, — B,, &c., the left-hand member becomes 
Cie + Cy? + Cyw?+... 
a aE ee eh Rese ee op 
whence there results 
C.+Ce+Ce?+... 1 
Dee gee Ee SE oe ea 
aN nee cas 
Ag+... 
or 
(CtCw+ Catt...) (mtgy2_ 1+ Bet Batt Bat... 
from which, by equating the terms which are free of x, we have 
Cin 1s ait Zo, = 
Again, striking these terms out of the equation, and dividing both numbers 
by a, there results after transformation 
(4C2—B,) + (aC,— Bo + (a,C,— Ba? +... (Cr+ Cr+ Oya?-+ ...) 2 ON 
az 
tg +—— 

and now multiplying both sides by a, + an _,,, and equating the terms whieh 

are free of x, we have 
Ce 
(a,C,—B,)a,+ C,=9, and i A, —————— 
i By 
C, Cy 


Proceeding in this way with the equation (which increases rapidly in com- 
plexity), we derive equations for the determination of a;, aq,... . Writing 
K(m%, ad), K(a, a, ds), .. . for the continuants a,a,+1, a,a,4,+a,+43,...., that 
is,* for the successive denominators of the convergents of the continued frac- 
Ling Hage al 
ty +++... 
K (qa, @, @) with the annexation of C,; as a factor to the term of highest degree, 
and C, as a factor to the term of the next degree; and the equations for the 
determination of @3, @%, a; .... become 
K(q, G2, a)C3, C.) — K(a, a§B, Bs 
K(a, Ay, U3, a§C,, C;, C,) aa K(a, 3, a,\B;, B,) =9 
K(ai, G, %,°G, As)C;, Ca, C3) — K(a@, U3, U4, a\B,, B;, B,) = 0 
tion _> then K(a, a, a(C;, C,) will conveniently denote 
* See paper on “Continuants” in the Proc. R.S. E. for 1873-74, and pamphlet on “The 
Expression of a Quadratic Surd as a Continued Fraction,” Glasgow, 1874. 
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