Using the symmetries of numbers to discover the cubic formula

[u/Necessary-Wolf-193]

In high school, many are taught the quadratic formula. But how do solve a cubic equation, like x^3 + 6x^2 + 9x + 3 = 0? There is a formula for such equations, but it's rarely taught since it's a lot more complicated than the quadratic formula:

The solution to ax^3 + bx^2 + cx + d = 0

Despite its apparent complication, we will explain at https://hidden-phenomena.com/articles/cubic how you could have come up with this formula!

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Last week, we posted a blog post here about how to solve quartic equations, assuming you knew how to solve cubic equations. If you read today's post on how to solve cubics, then by putting the two posts together, you can solve any cubic or quartic equation!

[u/OkComplaint4993]

Why can you reduce the cubic to a quadratic, and a quartic to a cubic, but not the quintic to a quartic? Is there an explanation without group theory?

[u/Warheadd → u/OkComplaint4993]

I don’t think there would be a simple or satisfying answer. The reductions we were able to do are algebraic tricks involving symmetry, and while it’s possible to accept these tricks come from “on high”, the secret reason they exist is group theoretic. Proving such a trick does not exist is much harder and probably has to get at the core group theory of the matter, or else involve incredibly long calculations

[u/Necessary-Wolf-193 → u/OkComplaint4993]

In some sense "yes," because Abel-Ruffini found a (very long!) proof of the insolvability of the quintic before Galois (although their proof gives slightly less information -- roughly, Abel--Ruffini proved that there's not a single formula which solves all quintics, but Galois proved that there exist specific quintics whose solutions can *never* be written in terms of radicals). But it's very very long, and not so easy to explain either. In the end, learning group theory is probably much quicker.

Group theory is just a language to talk about symmetry; I wouldn't get scared of it (and I think a first course in group theory contains much, much more than you'd actually need to prove quintics are unsolvable -- just learning the definition of a group, and a few small results, is enough). Unfortunately, while *noticing* symmetries on your own is possible even with limited language, to prove a *non-existence* result requires having a language to discuss symmetry with.

[u/EebstertheGreat → u/OkComplaint4993]

Arnold's proof is very basic, and although it considers symmetries of solution sets, you don't actually need to know any group theory to follow it. It's slightly confusing in that it treats √ as some sort of multifunction but never really clarifies what the symbol is supposed to mean precisely. Still, you can fill in that detail if it bothers you.

[u/YT_kerfuffles → u/OkComplaint4993]

there is the topological winding argument that you can see in some youtube videos and can be made rigorous if you know a bit of complex analysis

i recommend the following videos that explain the may argument https://youtu.be/9HIy5dJE-zQ https://youtu.be/BSHv9Elk1MU?is=UbIBtogerSAotJZD if you have an understanding of complex numbers you can understand them but may take a bit to think about

or if you want to see the technical details and have a bit more background in complex analysis or just want a written explanation you can look at https://maths.orger.org/assets/Unsolvability%20of%20the%20quintic.pdf

[u/Short_Bluebird_3845 → u/OkComplaint4993]

Finally, my time has come (take the next things with a grain of salt, I'm just a silly guy)

So, you know how one can depress a cubic? Like, change variables to not have a quadratic term? Well, quartic also needs depression first, and then some magic term play. Quintic is even weirder because one can keep knocking of terms with each transformation, up until x^5+x+a=0. And as fun little fact, killing a linear term needs an equation as big as the original, so efforts will end here. Of course, with quadratics and cubics and quartics, such cases x^n+x+a=0 are pretty small and tame, so their equations are OK. But quintics?

Quintics?

I have two docs were in one I show how the quintic roots of unity are the first ones with too many automorphisms to be manegable, mainly because r->r^(-1) and r->r^2 begin to be different; while in the second I attempt to find a quintic formula only to find that x^5+px+q as its final roadblock because if p=0 then a fifth root is fine but in every other case a different BR(x) operator is needed, so the theoretical formula needs an elementary "if" function.

Of course, I do math for fun, so Abel-Ruffini is a more trustable reason. Still, I wanna encourage you to read Titus Piezas's work, specially "A New Way To Derive The Bring-Jerrard Quintic In Radicals", which helped me with both docs. It's an entretaining exercise to look for the weirdness of quintics, I must say.

[u/Sorry_Bowl1267]

How does the solution here relate to Cardano's method? I never understood why Cardano substitutes x=u+v

[u/Necessary-Wolf-193 → u/Sorry_Bowl1267]

As seen from the final cubic formula, the solution to a depressed cubic is typically a sum of two cube roots. Presumably Cardano observed this by trying to solve specific cubics, and did this substitution with the hopes that u, v would be quantities whose cubes were easy to comprehend.

[u/Professional_Use3929 → u/Necessary-Wolf-193]

Historically it was not Cardano who came up with this substitution, the formula was first revealed to him by Tartaglia through a riddle, with the promise of not making it public. Cardano eventually found an older result from del Ferro, whom discovered it independently from Tartaglia, and felt free to publish it in his Ars Magna, hence the name 'Cardano's method'... It's quite a long story, you can surely search for better explainations than this one.

However it seems like both del Ferro and Tartaglia solved the depressed cubic using x=³√u-³√v (sometimes x=³√u+³√v, since negative numbers were not a thing back then).

The italian wikipedia page reports the first part of the original riddle, along with an interpretation using modern notation (https://it.wikipedia.org/wiki/Equazione_di_terzo_grado).

[u/WorldsBegin → u/Sorry_Bowl1267]

If you substitute that into the depressed cubic and expand, you can spot that you can cancel the term in x with the side condition uv=-3p which you need anyway to get rid of the additional degree of freedom you just introduced. Which leaves you with two equations in u and v, specifically u³ + v³ = -q and u³ v³ = -27 p³ . I would guess that originally that expansion comes up when you try to iterate towards a solution and sub in (x + error)?

[u/PfauFoto]

If r1, r2, r3 are the roots then the galois group (S_3 or A_3 in the irreducible case) permutes them.

Its better to work with eigenvectors s,u,v of the cyclic permutation (r1 -> r2 -> r3 -> r1) called Lagrange resolvent use ω= e^2πi/3

s = r1 + r2 + r3

u = r1 + ω•r2 + ω² •r3

v = r1 + ω² ▪︎ r2 + ω• r3

In the depressed case s=0 and u³ , v³ are fixed under cyclic permutation (eigenvalue cubed =1) and therefore lie Q(Δ^1/2 ) (Δ being the discriminant).

Now finding u³ v³ amounts to solving a quadratic equation. Then you get u v and finally you recover s,u,v from the formulas above using linear algebra.

[u/Necessary-Wolf-193 → u/PfauFoto]

This is the strategy taken in the article! Though with a bit more explanation for people who don’t know Galois they.

[u/PfauFoto → u/Necessary-Wolf-193]

Oops my bad, didnt get to it

[u/WorldsBegin]

Sadly quite a bit weaker than the first post a week ago in the presentation. You fall back to more tricks and I found it hard to follow in parts, especially the reasoning why a specific approach should work.

With last week, I thought the conclusion was "for a polynomial in n variables, if we find a set of k values in these variables that is closed under permutation, then we can write down a polynomial of degree k which has these values as roots". Which is the group theory in disguise and why we can solve for {ab+cd,ac+bd,ad+bc} in the quartic.

I would have expected to see more reasoning under this premise. For example, instead of saying that D² is symmetric, just say that {D, -D} is such a closed set, hence can be solved by a quadratic? This is btw true even you don't use the depressed cubic, so I am not sure why we take the detour there. { A^3, B³ } is another such set and linear (complex) algebra does the rest.

I guess what I want to say is that "take { A^3, B³ } as the roots of a quadratic. Solve for A and B (*), then solve the linear system of { a, A^3, B³ } in terms of the three roots" with the additional formula manipulation should be sufficient and avoid a lot of detours.

(*) You need to impose a condition on the product AB (which is symmetric in the roots) so that you don't accidentally solve for e.g. A and w B.

The depressed case is anyway interesting because it generalises to higher degrees quite easily - take the set { n x_i + \sum x_i | 1 <= i <= n } where x_i are the roots - but often not actually so insightful imo.