7 ms·
Author here, in case anyone has any questions. :)
by akalin 9y ago
Author here, in case anyone has any questions. :)
- enriquto 9y agoWell, some quintics are solvable, aren't they? (like x^5-1=0). Maybe for each particular quintic there is a different formula that solves it. You just prove that there's no formula that works in all cases when you plug the coefficients at the same place. It would be more satisfying (and more general) to exhibit a specific unsolvable quintic.
- dullcrisp 9y agoAll quintics have solutions, but there is no general quintic formula.
- enriquto 9y agoOf course, but in this game "solution" means "a number that can be expressed as a composition of radicals and elementary operations".
- moomin 9y agoNot true, there's only no general quintic formula that employs a specific, restricted set of operations. In practice, there is a general quintic formula. It just needs one extra operation.
- popcorncolonel 9y agoWhat other operation? If we don't know what it is, how do you know it needs just one?
- vog 9y agoDoes that matter? If you are free to add any additional operation, this whole thing becomes meaningless. You can simply define your operation "NewOp(a0,a1,..,a4)" to something like "the smallest root of the quintic a0+a1x+...+a4x^4+x^5". (Here, "smallest" can be anything as long as it is a completely defined tie-breaker, such as: the value with the smallest real part, and among those the one with the smallest imaginary part.)
- moomin 9y agoI think it's something like p(x) = smallest positive solution of z^5 + z - x. Compare sqrt(x) = smallest positive solution of z^2 - x. But it's a _long_ time since I've done Galois theory and I can't find a decent math exchange answer for it right now, so don't treat this as gospel.
- akalin 9y agoYeah, I think you're thinking of a Bring radical: https://en.wikipedia.org/wiki/Bring_radical https://en.wikipedia.org/wiki/Bring_radical Edit: Oh, but that only lets you solve some quintics. https://news.ycombinator.com/item?id=14686886 https://news.ycombinator.com/item?id=14686886 describes the functions you need to solve all quintics.
- deleted 9y ago[deleted]
- akalin 9y agoYes, that's correct. That's the big disadvantage to the topological method I talk about (developed by Vladimir Arnold); it can only prove stuff about the general nth-degree formulas, whereas Galois theory can prove stuff about specific polynomials (e.g., that x^5 – x – 1 is unsolvable in radicals). On the other hand, the topological method can be easily extended to prove stuff about formulas including other continuous, single-valued operations, like exponential and trigonometric functions, and Galois theory has a harder time with this.
- wolfgke 9y ago> That's the big disadvantage to the topological method I talk about (developed by Vladimir Arnold); it can only prove stuff about the general nth-degree formulas, whereas Galois theory can prove stuff about specific polynomials (e.g., that x^5 – x – 1 is unsolvable in radicals). Can these topological methods be generalized to arbitrary fields?
- akalin 9y agoI'm a bit out of my depth here, but I think you can generalize it to differential fields, where you want to figure out e.g. which functions have a closed-form integral. A reference (the only reference?) would be Topological Galois theory: https://www.springer.com/us/book/9783642388705 https://www.springer.com/us/book/9783642388705
- TheRealPomax 9y agoWhich library/libraries did you use for the interactive graphs?
- akalin 9y agoJust JSXGraph: https://jsxgraph.uni-bayreuth.de/wp/index.html https://jsxgraph.uni-bayreuth.de/wp/index.html I'm not entirely happy with the framerate I get with the button-triggered animations (especially on mobile) but lack the front-end expertise to track down the problem.
- jacobolus 9y agoWould be nice if dragging the coefficients would also change the roots.
- akalin 9y agoYeah. I think I punted on that because, although it's easy for degrees <= 4 (since those have general formulas!), I'd have to implement some root-finding method for degrees >= 5, and somehow preserve the orderings of the roots. Not insurmountable problems -- I might go back and do that someday!
- duetosymmetry 9y agoI was just pondering about this. I imagine that moving a coefficient a small amount (and so long as the roots are not too close) that Newton-Raphson, initialized with a previously-known root, would converge very quickly to the new root while preserving the order. If there truly are roots of multiplicity >= 2, then preserving the order of the multiple roots doesn't matter. Another issue is that Newton-Raphson might fail if the roots are too close.
- TheRealPomax 9y agoAh, interesting, I had not heard of that one.
- 19eightyfour 9y agoThanks for the post. What are the special icosahedron related functions one commenter said could be used to solve quintics?
- ocfnash 9y agoI am the other commenter! The special icosahedron functions can be expressed in many ways but perhaps the most elementary _explicit_ representation is as particular Gaussian hypergeometric functions. See this repo for example: https://github.com/ocfnash/icosahedral_quintic https://github.com/ocfnash/icosahedral_quintic The special functions are used here: https://github.com/ocfnash/icosahedral_quintic/blob/master/quintic_icosahedral_solution_demo.py#L87 https://github.com/ocfnash/icosahedral_quintic/blob/master/q... Geometrically these functions locally invert the (branched) covering represented by the diagram at the top of this page: http://olivernash.org/2012/02/05/on-kleins-icosahedral-solution-of-the-quintic/index.html http://olivernash.org/2012/02/05/on-kleins-icosahedral-solut...
- 19eightyfour 9y agoThanks then.
- pierrebai 9y agoThis page is an example why I gave up and still give up on advance mathematics. Maths is just the extreme example of science, where to make communication possible between its practitioners, new words encoding known facts are constantly created. Then another layer of new words with definitions based on the first layers are defined... and so on. Rapidly, we end-up with total gibberish for the non-initiated. For example on that page, we start with quintic and radicals (I can grok that). Then icosahedral functions, then hypergeometric functions, finite monodromy, 60-fold branched covering of the complex projective line (lost!).
- akalin 9y agoHey, this is really cool!
- IshKebab 9y agoThis sentence is very confusing: > Now this formula looks right, since x_1x and x_2x are at the same coordinates as r_1r and r_2r. I assume you mean "since before they have been moved...", but by the time the reader gets to that point you've already told them to move r1 and r2 and x1 and x2 obviously aren't at the same coordinates.
- akalin 9y agoThanks, I see why it's confusing. I'll fix it!
- alecbenzer 9y agoI've seen this argument several times and always get lost at around the same point: > But that means that our candidate solution cannot be the quadratic formula! If it were, then x_1 and x_2 would have ended up swapped, too. Why?
- jacksnipe 9y agoBecause by definition, if we have found the right solution, then x_1 = r_1 and x_2 = r_2.
- alecbenzer 9y agoI don't see why that's true by definition. We're looking for a function f st. f(a,b,c) = {r1, r2} iff 0 = a·r1^2 + b·r1 + c = a·r2^2 + b·r2 + c. Right? So as long as the correct two roots are produced, why is the order that they are produced in relevant?
- akalin 9y agoBecause we start off with the candidate solution matching the real roots, and we continuously move the roots to swap them. At every point of the movement, the candidate solution match the real roots (including order) by assumption, and so if at the end of the movement the candidate solution returns to its initial state, i.e. it's equal to the swapped roots, then that would contradict continuity. Does that make sense?
- HurrdurrHodor 9y agoIt doesn't work with Firefox, right?
- akalin 9y agoI just tried with Firefox 54.0.1 on macOS, and it seems to work. What version are you using?
- alanbernstein 9y agoThanks for this, this is a topic I've been curious about for a long time, but never knew how to learn about it without diving into a textbook. I'm reading through it now, and I'm confused by this statement: "Second, and more surprisingly, if you swap r_1 and r_2r, x_1 and x_2 also exchange places, seemingly contradicting Theorem 1!" Why is there any connection to theorem 1 here? We're talking about a radical, and theorem 1 is concerned with rational expressions.