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19 Elliptic IntegralsLegendre’s Integrals

§19.8 Quadratic Transformations

Contents

§19.8(i) Gauss’s Arithmetic-Geometric Mean (AGM)

When a0 and g0 are positive numbers, define

19.8.1 an+1 =an+gn2,
gn+1 =angn,
n=0,1,2,.

As n, an and gn converge to a common limit M(a0,g0) called the AGM (Arithmetic-Geometric Mean) of a0 and g0. By symmetry in a0 and g0 we may assume a0g0 and define

19.8.2 cn=an2-gn2.

Then

19.8.3 cn+1=an-gn2=cn24an+1,

showing that the convergence of cn to 0 and of an and gn to M(a0,g0) is quadratic in each case.

The AGM has the integral representations

19.8.4 1M(a0,g0)=2π0π/2dθa02cos2θ+g02sin2θ=1π0dtt(t+a02)(t+g02).

The first of these shows that

19.8.5 K(k)=π2M(1,k),
-<k2<1.

The AGM appears in

19.8.6 E(k)=π2M(1,k)(a02-n=02n-1cn2)=K(k)(a12-n=22n-1cn2),
-<k2<1, a0=1, g0=k,

and in

19.8.7 Π(α2,k)=π4M(1,k)(2+α21-α2n=0Qn),
-<k2<1, -<α2<1,

where a0=1, g0=k, p02=1-α2, Q0=1, and

19.8.8 pn+1 =pn2+angn2pn,
εn =pn2-angnpn2+angn,
Qn+1 =12Qnεn,
n=0,1,.

Again, pn and εn converge quadratically to M(a0,g0) and 0, respectively, and Qn converges to 0 faster than quadratically. If α2>1, then the Cauchy principal value is

19.8.9 Π(α2,k)=π4M(1,k)k2k2-α2n=0Qn,
-<k2<1, 1<α2<,

where (19.8.8) still applies, but with

19.8.10 p02=1-(k2/α2).

§19.8(ii) Landen Transformations

Descending Landen Transformation

Let

19.8.11 k1 =1-k1+k,
ϕ1 =ϕ+arctan(ktanϕ)
=arcsin((1+k)sinϕcosϕ1-k2sin2ϕ).

(Note that 0<k<1 and 0<ϕ<π/2 imply k1<k and ϕ<ϕ1<2ϕ, and also that ϕ=π/2 implies ϕ1=π.) Then

19.8.12 K(k) =(1+k1)K(k1),
E(k) =(1+k)E(k1)-kK(k).
19.8.13 F(ϕ,k) =12(1+k1)F(ϕ1,k1),
E(ϕ,k) =12(1+k)E(ϕ1,k1)-kF(ϕ,k)+12(1-k)sinϕ1.
19.8.14 2(k2-α2)Π(ϕ,α2,k)=ω2-α21+kΠ(ϕ1,α12,k1)+k2F(ϕ,k)-(1+k)α12RC(c1,c1-α12),

where

19.8.15 ω2 =k2-α21-α2,
α12 =α2ω2(1+k)2,
c1 =csc2ϕ1.

Ascending Landen Transformation

Let

19.8.16 k2 =2k/(1+k),
2ϕ2 =ϕ+arcsin(ksinϕ).

(Note that 0<k<1 and 0<ϕπ/2 imply k<k2<1 and ϕ2<ϕ.) Then

19.8.17 F(ϕ,k) =21+kF(ϕ2,k2),
E(ϕ,k) =(1+k)E(ϕ2,k2)+(1-k)F(ϕ2,k2)-ksinϕ.

§19.8(iii) Gauss Transformation

We consider only the descending Gauss transformation because its (ascending) inverse moves F(ϕ,k) closer to the singularity at k=sinϕ=1. Let

19.8.18 k1 =(1-k)/(1+k),
sinψ1 =(1+k)sinϕ1+Δ,
Δ =1-k2sin2ϕ.

(Note that 0<k<1 and 0<ϕ<π/2 imply k1<k and ψ1<ϕ, and also that ϕ=π/2 implies ψ1=π/2, thus preserving completeness.) Then

19.8.19 F(ϕ,k) =(1+k1)F(ψ1,k1),
E(ϕ,k) =(1+k)E(ψ1,k1)-kF(ϕ,k)+(1-Δ)cotϕ,
19.8.20 ρΠ(ϕ,α2,k)=41+kΠ(ψ1,α12,k1)+(ρ-1)F(ϕ,k)-RC(c-1,c-α2),

where

19.8.21 ρ =1-(k2/α2),
α12 =α2(1+ρ)2/(1+k)2,
c =csc2ϕ.

If 0<α2<k2, then ρ is pure imaginary.