I can solve any quintic equation!!

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[disgruntledtoons]

Quintics can't be solved by radicals, but Michael is a moderate so it's no problem.

[xinpingdonohoe]

(If the constant term is 0) I can formulaically solve any quintic equation.

[zh]

Wolfram Research once sold a fascinating poster explaining various approaches to solving the quintic equation. I had it on my wall for many years.

[rubensramos]

Great! One can use two Tschirnhaus transformations to transform any general quintic to the Bring-Jerrard quintic (x^5+px+q = 0) (As Michael said, it is a long procedure). Then, one can easily solve this quintic using the Lambert-Tsallis Wq function, as it was done in this paper: "Analytical solutions of cubic and quintic polynomials in micro and nanoelectronics using the Lambert-Tsallis Wq function", Journal of Computational Electronics, Volume 21, pages 396–400, (2022).

[titush.]

_We spent so much time asking whether or not we could, we never stopped to ask whether or not we should_

[Luigiman-rcfi]

I appreciate this video a lot, because I always knew about it not being generally solvable by radicals, but then that always gave me the question of "well, can't we just do something else for the other ones that aren't solvable by radicals?" It's good to have a clear answer for that now, I like this.

[jcubed]

As someone who had a Galois Theory exam only 11½ hours ago, this is well-timed bliss

[aidenmcdonald]

can’t the bring-radical constants also be expressed by hypergeometric functions? it might not be in terms of radicals, but honestly it’s a pretty good “quintic formula”

[marc-andredesrosiers]

Great that you brought that to out attention 🙂

[tomholroyd]

I can use my computer to solve equations too. It's also faster, but it never does backflips.

[josepherhardt]

To the tune of "Climb Every Mountain" :
SOLVE! EV'RY! QUINTIC!
FIND! EV'RY! ROOT!
They say it's IM-POS-SI-BLE
but now the question's MOOT!

[Chalisque]

My final year undergraduate project involved at one point digging up the Bring-Jerrard reduction to x^5+x+C for some C, and the solution via elliptic functions. Expressing that C directly in terms if the original coefficients of a general monic quintic is complicated.

[johnsavard]

They have no solution by radicals, but that doesn't mean they have no solutions. You can use Elliptic Integrals, and you can use Bring radicals. But I don't recall seeing an explanation of how to do it, or even a definition of the Bring radicals in terms of elliptic integrals (as opposed to the definition in terms of which quintics they're the solutions of, which is reproduced often).

[OmnipotentEntity]

In your general example you set x = t+2, and all of the coefficients in your original equation are positive, so I don't see how the substitution could have cancelled everything. Should the substitution have been x = t-2 or similar?

[stephenhamer]

Read about the Bring Radical here:
For real a, the map a -> BR(a) - where BR(a) is the unique real solution of x^5 + x + a - is odd, monotonically decreasing, and unbounded, with asymptotic behavior BR(a) -> a^(1/5) for large a
I take it that BR(a) is not itself expressible by radicals, at least for general a

[Happy_Abe]

With radicals like the Bring radicals, is there some way to formulate why being expressable as normal radicals is so much nicer, since radicals and the Bring radicals are kind of defined similarly as being a root of specific polynomial. Is it that being expressable as radicals gives some nice algebraic properties? Roots from the Bring radical that are not expressable as radicals are still algebraic elements over Q so they should still be nice enough so what’s so special about expressability via radicals. Maybe it’s similar to constructible numbers being expressable using square roots and is more of an aesthetic preference for these nice operations(though constructible has a geometric origin to it), but I’m curious if there’s a way to formulate this.

Also, for arbitrary degree n polynomials. Is there some collection of functions B_n(x) that will give roots to simpler degree n polynomials that can always be used to solve the general degree n polynomial? If so we can define B1, B2, B3, and B4 to be the general 1st, 2nd, 3rd, and 4th roots since those are needed to solve the first degree 4 cases. And then B5 can be the Bring radical and so on for the ultraradical needed to solve the degree n case. It would be very interesting if this idea generalized, but I’m doubtful it does, but I’m curious nonetheless.

[WK-]

Is there a connection between the Bring story and Watson's theorem, or are these just two different methods to find solutions to particular classes of quintics?

[alikaperdue]

I now know that there is no way to solve quintics. According to my limitations

[dogbiscuituk]

I like to make the substitution h=-a/5, x=y+h right at the outset, so when you calculate p, q, r, s, all the denominators vanish:
p = b − 10h^2
q = c + 3bh − 20h^3
r = d + 2ch + 3bh^2 − 15h^4
s = e + dh + ch^2 + bh^3 − 4h^5

[yqisq]

Reminds me of Lambert W function. You can't write it out in simple terms, but you can use it to solve a whole family of transcendental equations.