The specific form in question is basically sqrt(any quartic). Seems like almost all of these will be able to be expressed in terms of elliptic integrals and that's it. Outer post summarizes this just fine.
We're talking about Bezier curves in the context of CAD and graphics. In this case, I believe there is no reason to assume that these curves will have a more special form than sqrt(arbitrary quartic). Do you think that they do have a more special form, and that this form will simplify nicely? Or are you suggesting that sqrt(abrbitrary quartic) might simplify more?
If I may, your confusion seems to come from the meaning of "elliptic curves in general have no closed form".
You seem to interpret that as "given any elliptic integral A, there is no closed form for the solution of A". This is false (there are many counterexamples).
What it actually means is "there is no single closed-form that produces the solution of any given elliptic integral". The quantifiers are the other way around.
This is why I brought up the halting problem. There's a similar confusion that often comes up, where people think that it means there's no way to determine if any given program halts. But this is false - the program "let X=5*6" trivially halts, for instance.
What it actually means is that there's no single program that can uniformly determine whether any given input program halts. It's exactly the same situation.
No, I'm not getting this confused at all. I understand the point. What I'm saying is that because the general elliptic integral corresponding to sqrt(quartic) doesn't have a closed form, and because this (or maybe a slightly more specific form) is what's of interest in this context (CAD), saying something about the closed form of specific sqrt(quartic) elliptic integrals isn't very interesting as far as I can see.
the general elliptic integral corresponding to sqrt(quartic) doesn't have a closed form
I see, so you’re claiming that there is no closed form for the integral of sqrt(quartic). That is a different statement that neither directly implies nor is implied by the statement in the Wikipedia article. Maybe it is true and has been proven though, I don’t know!
You're getting hung up on a logical point which, while correct, isn't particularly salient to the context in which this conversation is occurring (CAD and graphics). The original article, while mathematically interesting, is also not particularly salient within that context.
Take whatever drugs you like. You just seem a little bewildered why people aren't more fired up about what you have to say about the halting problem. Just trying to suggest that it may be because this isn't the right venue for it. I wish you many happy mathematical returns. :-)
The issue is that "sqrt(arbitrary quartic)" is already more specialized than "elliptic integral". So we can't just talk about elliptic integrals in general, we need proof that this specialization doesn't give rise to closed forms.
Yes, if they had then those particular elliptic integrals would have a closed form. Some degenerate classes of elliptic integrals do have closed forms, in precisely the same way that you can determine whether certain subsets of programs halt despite the halting problem in general being insoluble! I think you misunderstand what is meant by the statement "elliptic integrals in general have no closed form".
No, I don't think I misunderstand, and I don't think my reduction is invalid. I take a Bezier curve, and reduce the expression of its arc length to an elliptic integral. If this particular elliptic integral has no closed form, that means that this particular Bezier curve doesn't have closed form arc length parametrization and also that Bezier curves in general don't admit such parametrization. Because that's what "X in general is not Y" means: that there is at least one X which is not Y.
I agree with everything you've written in this reply, but originally you were appealing to the fact that elliptic curves in general don't have closed forms ("they don't, though") to prove your point, which isn't valid because we're talking about a strict subset of them here.
