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From: Bruno Luong on 26 May 2010 15:19
"Felipe G. Nievinski" <fgnievinski(a)gmail.com> wrote in message <htjlqp$e3q$1(a)news.acm.uiuc.edu>...

>
> I found this statement at <http://dlmf.nist.gov/19.25#v>:
> "[elliptic] integrals of the second kind, ... are not invertible
> in terms of single-valued functions"
> though I find the treatment given, in terms of the so-called elliptic
> symmetric integrals, a little hard to digest.
>

Are they talking about invertible in C or in R? To illustrate, y = x^3 is invertible in R but not in C. This is surely an "incomplete" statement. LOL

Your post is too long to read, and did not even spot any question. Is it just a discussion?

Bruno
From: John D'Errico on 26 May 2010 15:41
"Felipe G. Nievinski" <fgnievinski(a)gmail.com> wrote in message <htjlqp$e3q$1(a)news.acm.uiuc.edu>...
> Are you aware of an inverse of the incomplete elliptic integral of the
> second (not first) kind? I'm familiar with the Jacobi amplitude as the
> inverse of the incomplete elliptic integral of the first kind, but
> can't seem to find in the literature an inverse for the second kind.
> Alternatively, it'd help to know how to obtain the first kind given
> the second one (both incomplete).
>
> I found this statement at <http://dlmf.nist.gov/19.25#v>:
> "[elliptic] integrals of the second kind, ... are not invertible
> in terms of single-valued functions"
> though I find the treatment given, in terms of the so-called elliptic
> symmetric integrals, a little hard to digest.
>
> I'm trying to discretize the perimeter of an ellipse at regularly
> spaced points.Trivial discretizations end up causing an undesirable
> high/low density of points at maximum/minimum distance from the
> center. (A brute force approach would be to discretize at a
> sufficiently high density everywhere then discard excessive points
> where necessary.)

Better, and quite simple, is just to generate a set of
arbitrarily spaced points on the ellipse, and then use
my interparc function, to interpolate points equally
spaced on the ellipse.

http://www.mathworks.com/matlabcentral/fileexchange/27096-interparc

This will be reasonably fast, and quite accurate.

HTH,
John
From: John D'Errico on 26 May 2010 16:00
"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <htjtgg$gvj$1(a)fred.mathworks.com>...
> "Felipe G. Nievinski" <fgnievinski(a)gmail.com> wrote in message <htjlqp$e3q$1(a)news.acm.uiuc.edu>...
> > Are you aware of an inverse of the incomplete elliptic integral of the
> > second (not first) kind? I'm familiar with the Jacobi amplitude as the
> > inverse of the incomplete elliptic integral of the first kind, but
> > can't seem to find in the literature an inverse for the second kind.
> > Alternatively, it'd help to know how to obtain the first kind given
> > the second one (both incomplete).
> >
> > I found this statement at <http://dlmf.nist.gov/19.25#v>:
> > "[elliptic] integrals of the second kind, ... are not invertible
> > in terms of single-valued functions"
> > though I find the treatment given, in terms of the so-called elliptic
> > symmetric integrals, a little hard to digest.
> >
> > I'm trying to discretize the perimeter of an ellipse at regularly
> > spaced points.Trivial discretizations end up causing an undesirable
> > high/low density of points at maximum/minimum distance from the
> > center. (A brute force approach would be to discretize at a
> > sufficiently high density everywhere then discard excessive points
> > where necessary.)
>
> Better, and quite simple, is just to generate a set of
> arbitrarily spaced points on the ellipse, and then use
> my interparc function, to interpolate points equally
> spaced on the ellipse.
>
> http://www.mathworks.com/matlabcentral/fileexchange/27096-interparc
>
> This will be reasonably fast, and quite accurate.
>

Alternatively, you could also just use the same scheme I
use in that tool, here applying an odesolver (ode45) to
the ellipse, parameterized by angle around the center.

HTH,
John
From: Roger Stafford on 26 May 2010 22:56
"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <htjuko$1ug$1(a)fred.mathworks.com>...
> Alternatively, you could also just use the same scheme I
> use in that tool, here applying an odesolver (ode45) to
> the ellipse, parameterized by angle around the center.
>
> HTH,
> John

Applying John's idea, you could use arclength, s, as the independent variable in ode45 and solve the simultaneous equations

dx/ds = -a^2*y/sqrt(b^4*x^2+a^4*y^2)
dy/ds = +b^2*x/sqrt(b^4*x^2+a^4*y^2)

to travel counterclockwise around the ellipse. You need only go a quarter of the way around. You set the s steps returned and stopping point in accordance with the perimeter values you already know.

Roger Stafford
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