4 ms·

Second paragraph of the Wikipedia article on elliptic integrals: https://en.wikipedia.org/wiki/Elliptic_integral https://en.wikipedia.org/wiki/Elliptic_integral

kevinventullo 2y ago
The point is that the general non-reducibility of elliptic integrals to closed form does not preclude the possibility of reducing to closed form some particular elliptic integral or combination thereof.
- sfpotter 2y ago
  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.
  - deleted 2y ago
    [deleted]
  - bubblyworld 2y ago
    They're just pointing out that strictly speaking this is not a valid proof that these integrals have no closed form. Compare: halting problem being uncomputable tells you nothing about whether you can solve it for a subset of valid programs.
    - deleted 2y ago
      [deleted]
    - sfpotter 2y ago
      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?
      - Dylan16807 2y ago
        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.
      - bubblyworld 2y ago
        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.
        sfpotter 2y ago
        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.
        bubblyworld 2y ago
        Okay, I'm not really sure what you're talking about, apologies. I don't have anything else to add.
        sfpotter 2y ago
        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.
    - Joker_vD 2y ago
      If they had then the elliptic integrals that they're reducible to would also have closed form. They don't though.
      - bubblyworld 2y ago
        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".
        Joker_vD 2y ago
        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.
        bubblyworld 2y ago
        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.
- eigenket 2y ago
  This is obvious because a straight line is a (degenerate) example of a quadratic or cubic Bezier curve.
dataflow 2y ago
"every elliptic integral can be brought into a form that [...]" Yes? That is a reduction in the other direction. It starts off with an elliptical integral, then reduces it to something else. Not the other way around.

kevinventullo 2y ago

The point is that the general non-reducibility of elliptic integrals to closed form does not preclude the possibility of reducing to closed form some particular elliptic integral or combination thereof.

sfpotter 2y ago
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.
- deleted 2y ago
  [deleted]
- bubblyworld 2y ago
  They're just pointing out that strictly speaking this is not a valid proof that these integrals have no closed form. Compare: halting problem being uncomputable tells you nothing about whether you can solve it for a subset of valid programs.
  - deleted 2y ago
    [deleted]
  - sfpotter 2y ago
    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?
    - Dylan16807 2y ago
      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.
    - bubblyworld 2y ago
      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.
      - sfpotter 2y ago
        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.
        bubblyworld 2y ago
        Okay, I'm not really sure what you're talking about, apologies. I don't have anything else to add.
        sfpotter 2y ago
        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.
  - Joker_vD 2y ago
    If they had then the elliptic integrals that they're reducible to would also have closed form. They don't though.
    - bubblyworld 2y ago
      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".
      - Joker_vD 2y ago
        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.
        bubblyworld 2y ago
        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.
eigenket 2y ago
This is obvious because a straight line is a (degenerate) example of a quadratic or cubic Bezier curve.

dataflow 2y ago

"every elliptic integral can be brought into a form that [...]" Yes? That is a reduction in the other direction. It starts off with an elliptical integral, then reduces it to something else. Not the other way around.