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19 Elliptic IntegralsLegendre’s Integrals

§19.8 Quadratic Transformations

Contents
  1. §19.8(i) Gauss’s Arithmetic-Geometric Mean (AGM)
  2. §19.8(ii) Landen Transformations
  3. §19.8(iii) Gauss Transformation

§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 =an⁢gn,
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 a0≥g0 and define

19.8.2 cn=an2−gn2.

Then

19.8.3 cn+1=an−gn2=cn24⁢an+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)=2π⁢∫0π/2dθa02⁢cos2⁡θ+g02⁢sin2⁡θ=1π⁢∫0∞dtt⁢(t+a02)⁢(t+g02).

The first of these shows that

The AGM appears in

and in

where a0=1, g0=k′, p02=1−α2, Q0=1, and

19.8.8 pn+1 =pn2+an⁢gn2⁢pn,
εn =pn2−an⁢gnpn2+an⁢gn,
Qn+1 =12⁢Qn⁢εn,
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

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−k′1+k′,
ϕ1 =ϕ+arctan⁡(k′⁢tan⁡ϕ)=arcsin⁡((1+k′)⁢sin⁡ϕ⁢cos⁡ϕ1−k2⁢sin2⁡ϕ).

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

19.8.13 F⁡(ϕ,k) =12⁢(1+k1)⁢F⁡(ϕ1,k1),
E⁡(ϕ,k) =12⁢(1+k′)⁢E⁡(ϕ1,k1)−k′⁢F⁡(ϕ,k)+12⁢(1−k′)⁢sin⁡ϕ1.

where

19.8.15 ω2 =k2−α21−α2,
α12 =α2⁢ω2(1+k′)2,
c1 =csc2⁡ϕ1.

Ascending Landen Transformation

Let

19.8.16 k2 =2⁢k/(1+k),
2⁢ϕ2 =ϕ+arcsin⁡(k⁢sin⁡ϕ).

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

§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=sin⁡ϕ=1. Let

19.8.18 k1 =(1−k′)/(1+k′),
sin⁡ψ1 =(1+k′)⁢sin⁡ϕ1+Δ,
Δ =1−k2⁢sin2⁡ϕ.

(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

where

19.8.21 ρ =1−(k2/α2),
α12 =α2⁢(1+ρ)2/(1+k′)2,
c =csc2⁡ϕ.

If 0<α2<k2, then ρ is pure imaginary.