2 The Arithmetic-Geometric Mean 505and
bn≤(an^2 sin^2 θ+b^2 ncos^2 θ)^1 /^2 ≤an.Consequently, by lettingn→∞we obtain
K(a,b)=π/ 2 M(a,b). (1)Now takeφ(t)=a^2 −t^2 and putE(a,b):=∫ab[(a^2 −t^2 )/(t^2 −b^2 )]^1 /^2 dt.In this case
ψ(t 1 )=(a^2 −b^2 )/ 2 + 2 (a 12 −t 12 )and hence
E(a,b)=(a^2 −b^2 )K(a,b)/ 2 + 2 E(a 1 ,b 1 ).If we write
en= 2 n(a^2 n−b^2 n)then, sinceK(a,b)=K(an,bn), by repeating the process we obtain
E(a,b)/K(a,b)=(e 0 +e 1 +···+en− 1 )/ 2 + 2 nE(an,bn)/K(an,bn).But
2 nE(an,bn)=en∫π/ 20cos^2 θ(a^2 nsin^2 θ+bn^2 cos^2 θ)−^1 /^2 dθanden→0 (rapidly) asn→∞,since
en= 2 n(an− 1 −bn− 1 )^2 / 4 =en− 1 (an− 1 −bn− 1 )/ 4 an.Hence
E(a,b)/K(a,b)=(e 0 +e 1 +e 2 +···)/ 2. (2)To avoid taking differences of nearly equal quantities, the constantsenmay be calcu-
lated by means of the recurrence relations
en=e^2 n− 1 / 2 n+^2 a^2 n(n= 1 , 2 ,...).Next takeφ(t)=p[(p^2 −a^2 )(p^2 −b^2 )]^1 /^2 /(p^2 −t^2 ),where eitherp>aor 0<p<b, and put
P(a,b,p):=∫abp[(p^2 −a^2 )(p^2 −b^2 )]^1 /^2 dt/(p^2 −t^2 )[(a^2 −t^2 )(t^2 −b^2 )]^1 /^2.