506 XII Elliptic Functions
In this case
ψ(t 1 )=q 1 ±p 1 [(p 12 −a 12 )(p 12 −b^21 )]^1 /^2 /(p^21 −t 12 ),where
p 1 =( 1 / 2 )(p+ab/p),
q 1 =(p^21 −a^21 )^1 /^2 =[(p^2 −a^2 )(p^2 −b^2 )]^1 /^2 / 2 p,and the+or−sign is taken according asp>aor 0<p<b.Sincep 1 >a 1 in either
event, without loss of generalitywe now assume that p>a.Thenalsop 1 <pand
P(a,b,p)=q 1 K(a,b)+P(a 1 ,b 1 ,p 1 ).Define the sequence{pn}inductively by
p 0 =p, pn+ 1 =( 1 / 2 )(pn+anbn/pn)(n= 0 , 1 ,...),and put
qn+ 1 =(p^2 n+ 1 −an^2 + 1 )^1 /^2 =[(p^2 n−an^2 )(pn^2 −bn^2 )]^1 /^2 / 2 pn.Thenpn→v≥M(a,b)asn→∞,sincean<pn<pn− 1. In factv=M(a,b),
as one sees by lettingn→∞in the recurrence relation defining the sequence{pn}.
Moreover
δn:=(a^2 n−b^2 n)/(p^2 n−an^2 )→0asn→∞,since
δn+ 1 =δn(
pn^2
4 an^2 + 1)(
a^2 n−b^2 n
p^2 n−bn^2)
<δnpn^2 / 4 a^2 n+ 1.Hence
(p^2 n−b^2 n)/(p^2 n−a^2 n)= 1 +δn→ 1.SinceP(an,bn,pn)
=pn[(p^2 n−an^2 )(p^2 n−b^2 n)]^1 /^2∫π/ 20(an^2 sin^2 θ+b^2 ncos^2 θ)−^1 /^2 dθ
(p^2 n−a^2 n)sin^2 θ+(p^2 n−b^2 n)cos^2 θ,
it follows thatP(an,bn,pn)→π/2asn→∞. Hence
P(a,b,p)=(q 1 +q 2 +···)K(a,b)+π/ 2. (3)To avoid taking differences of nearly equal quantities, the constantsqnmay be calcu-
lated by means of the recurrence relations
δn+ 1 =δn^2 pn^2 / 4 a^2 n+ 1 ( 1 +δn), qn+ 1 =( 1 +δn)^1 /^2 qn^2 / 2 pn(n= 1 , 2 ,...).