Re: Inductance calculations

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Original poster: "Godfrey Loudner by way of Terry Fritz <teslalist-at-qwest-dot-net>" <ggreen-at-gwtc-dot-net>

Hello Antonio

The following quadratically convergent algorithm for F(c) and E(c)
might be useful. More details can be found in the book: Pi and the
AGM, J. Borwein and P. Borwein, John Wiley & Sons. The AGM
for F and E is an iteration, a task at which computers excel.

Start with a(0) = 1
                 b(0) = (1 - c^2)^(1/2)
                 c(0) = c
where  c is in the interval [0 , 1).

The iteration is given by a(n+1) = [a(n) + b(n)] / 2
                                      b(n+1) = [a(n) b(n)]^(1/2)
                                      c(n+1) = [a(n) - b(n)] / 2

Now a(n) and b(n) converge to a common limit, denoted by
M(c).

Then F(c) = Pi / [2M(c)]  and
          E(c) = {1 - [c(0)^2 / 2 + c(1)^2 + 2c(2)^2 +

                       4c(3)^2 + 8c(4)^2 + ...]} F(c)

I use my hand calculator, which has a 10 significant figure display.
Here is an example for F(1/2) and E(1/2).

a(0) = 1
b(0) = 0.866025404
c(0) = 0.5

a(1) = 0.933012702
b(1) = 0.930604859
c(1) = 0.066987298

a(2) = 0.931808781
b(2) = 0.931808003
c(2) = 0.001203922

a(3) = 0.931808392
b(3) = 0.931808392
c(3) = 0.000000389

a(4) = 0.931808392
b(4) = 0.931808392
c(4) = 0

The iteration stops here, because we used only 10
significant figures. The iteration will continue if we
use more significant figures. Four iterations give
the accuracy below.

F(1/2) = Pi / (2 x 0.931808392) = 1.685750354

E(1/2) = {1 - [c(0)^2 / 2 + c(1)^2 + 2c(2)^2 + 4c(3)^2]} F(1/2)

            =  1.467462209

Mathematica gives F(1/2) = 1.6857503548125960429 and
                              E(1/2) = 1.4674622093394271555

F(1/100) = 1.570835598
E(1/100) = 1.570757055 with three iterations
F(1/100) = 1.5708355989121522360 (Mathematica)
E(1/100) = 1.5707570561503852873 (Mathematica)

F(99/100) = 3.356600519
E(99/100) = 1.028475808 with six iterations
F(99/100) = 3.356600523361192376
E(99/100) = 1.0284758090288040010

Higher order versions of the AGM are possible, which
converge even faster. But they become complicated.

Godfrey Loudner



----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Saturday, March 15, 2003 1:38 PM
Subject: Re: Inductance calculations


  Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz
<teslalist-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

 > (I had to improve the calculation of the elliptic functions

 > Antonio Carlos M. de Queiroz