Raymond Toy wrote:
>
>>
> FWIW, I tried implementing the reduction algorithm in A&S for converting
> R(x,q) to elliptic integals of the first, second, and third kind. Parts
> of it work, but many parts do not. In particular, I never got it to
> recognize that sqrt((1-x^2)/(1-m*x^2)) is an elliptic integral.
>
> Ray
Hi,
thank you for your comments, Raymond and Richard. The example above is the
occasion to show how to reduce such things to canonical form.
First one writes sqrt((1-x^2)/(1-m*x^2)) = 1/(1-mx^2) sqrt((1-x^2)(1-mx^2))
which allows to reduce to rational function times square root of a
polynomial of degree 4. This polynomial has roots +1, -1,
+sqrt(m) , -sqrt(m). Sending one root to infinity by e.g. x -> 1/(x-1)
reduces to a polynomial of degree 3. Since i am now at my office, i have
checked in Whittaker and Watson (a course of modern analysis) it is in
section 20.6 chapter 20, page 452-455.
By the way, i think one of the problems when trying to compute elliptic
integrals is trying to coerce everything to old Jacobian elliptic
notations, cn, sn etc. which are in fact extremely unnatural from the point
of view of Riemann surface theory, and are rather inspired by an improper
analogy with trigonometric functions. Things are natural and simple when
expressed through the Weierstrass p function, which uniformizes the
genus 1 Riemann surface.
It is true that CAS tools seem to favor these elliptic functions, for
reasons i don't know.
As an answer to the question q^2=P(x) with P a polynomial of degree larger
than 4, this is an hyperelliptic Riemann surface, and as such the integral
is guaranteed to be impossible to express in terms of elliptic functions.
Hence this is a completely void generalization. Of course, as always, there
is an exception, if P has a double root, e.g. (x-a)^2, then you can bring
(x-a) out of the square root, and the degree of P falls by 2 units. So these
assertions are for a polynomial with simple roots.
What would be a legitimate generalization is an algebraic relation of the
form P(q,x)=0 which in fact describes a curve of genus 1 while it is hidden
in this explicit form. Then one needs to resort to algorithms (which exist,
i know a package exists for Maple) to change variables to good
variables q and w such that q^2=P(x). But i think this more complicated
case, which can occur in theory, is less likely to occur in practice than
the straightforward integration of a square root of a polynomial of degree
4 multiplied by a rational function.
--
Michel Talon