I can solve any quintic equation!! 

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

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

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

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

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).

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.

Galois rn: 😡

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

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

22:49

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.

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”

21:24 I think it's x=t-2

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!

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

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.

michael never fails to do hard math

Great that you brought that to out attention 🙂

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).

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

Hi there. You got the two quadratics in 10:55 from my answer to the question "How to solve a quintic polynomial equation?" from MathStackExchange, didn't you? They do not appear anywhere else in the literature, not even in Watson's or Berndt's paper.

15:50 Shouldn't we have sqrt(-4) and sqrt(-16) instead? P.S. Finally, someone brought Bring radicals into RU-vid!

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?

Read about the Bring Radical here: en.wikipedia.org/wiki/Bring_radical 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

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?

I’d like to see polynomial division (by hand) with the found solution and then solving the resulting fourth degree equation :)

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.

You really didn't have to write down any of p,q,r,s in terms of A-E; once it's established that we can eliminate the quartic term with the x->y change of variables, the p-q-r-s form becomes *the* most general form of quintic. Any particular equation with a quartic term can be reduced to it, and then the specific p,q,r,s coefficients picked up.

What is the general order of the substitution required to transform any quintic into t^5 + t + p form? I think that with the linear substitution like x = t+2 you can only get rid of the coefficient of x^4 in general

Interesting! Is there always a transformation that removes the x^4 and x^3 term? How can we numerically approximate Br(a)?

Galois´s Theorem: a irredutible quintic (or irredutible polynomial with prime degree ) is solvable by radicals if only if all roots are the form p(a_1,a_2) where a_1,a_2 are roots of polynomial, Corollary, A solvable irredutible quintic (or solvable irredutible polynomial of prime degree) has two real roots or all roots are reals. Lacroix and Poisson dissappointed us.

I would imagine that any polynomial equation of degree n is solvable by a function of the form (x^n)+x+a=0 (I’ll call it the Bring Radical sub n or Brn(a)). For example the Br5(3) is (x^5)+x+3=0, and Br6(11) is (x^6)+x+11.

@MichaelPennMath I have always been curious. how does this change for higher order polynomials?

I can solve any quintic equation of this type: 😂 x^5 + 5px^4 + 10p^2x^3 + 10p^3x^2 + 5p^4^x + p^5 , where p is a constant 😉

Since I'm sadistic tonight. Suppose 2024^p - 2023^p is divisible by 2027 for some prime p. Deduce the value of p.

"I can solve some quintic equations under such and such condition." Fair nuff. I was hoping there was some esoteric way of approaching the rest in closed form, such as with logarithms or other non radical functions. But has Galois outfoxed us after all? Pardon me as an engineering-level math duffer.

What's the "very very complicated substitution" in 20:20 like? It seems to be the core of the things here, so saying that it's "definitely not worth looking at it" is ridiculous.

Before we go to the general quintic equation, we must seriously verify the true existence of the alleged real root for the simplest case say like (x^5 = 2), where I only claim that alleged real root never exist (except only in human minds), However, the non-existence of such an alleged real root was proven non-existing in very simple proofs that were publically published in my many posts on sci.math, Quora, MS, etc Noting that understanding such simple proofs is requiring only a mid-school level FOR SURE Regards Bassam Karzeddin 🔊

This doesn't seem much harder than applying the quadratic formula. /s

That's one solution to the arbitrary quintic. What of the other four? Are they always complex?

This is the best clickbait title yet

glad we have computers

The solvable quintic in the video can be written as the sum of two powers of 5 multiplied by scalars. In this case it's equivalent to solving 2(x−1)^5+(x+2)^5 = 0. If someone tells you a quintic is solvable by radicals chances it reduces to a similar form.

It can not be solved by radicals, this thesis does not say that it could not be solved general by some expression of primitive functions.

When a b c d e values are known, Newton Iteration method is a better choice.

I don't think all solutions to x^100=1 are radicals, but the rest are trigonometric solutions.

Abel-Ruffini teorem states that this is impossible (of course if constant term is non-zero!)

Suggestion : use timestamps to indicate sections that can be used as homework, that is non theory workings.

x=t-2 ?

This video breaks the Galois theory.

I've been waiting for this one 😁

Как выпускник мехмата, могу заявить, что медали продаются на втором этаже магазина напротив.

You showed how to eliminate x^4 term with linear substitution, but what's the idea how to eliminate x^3 and x^2 terms? It must be linear too so degree stays, but there is no general way to reduce ANY quintic equation to x^5+px+q form, or I'm getting something wrong

Rational Root theorem can’t help me now 😭😭😭

Isn't math amazing? Normal person: Find X such that so and so. Math: That's an impossible problem. Normal person: Oh no! Math: But there's hope! Normal person: Oh yeah? Math: Yeah! If you solved this other impossible problem in terms of Y, I can tell you what X is. Normal person: Math: Normal person: Math: what?

You sometimes send out content trying to gauge your viewer's preferences for future topics. This is an exceptionally high quality channel already with the topics you have been presenting the last 4 years. I don't come here near as frequently as I used to because of your refusal to respond to ANY comment. I understand you are busy as hell with family and full time physical classes you have to manage. You expect us to subscribe, become Patrons and buy Michael Penn "products", ask us to make comments which you don't respond to.... You can disable the comment sections if you don't have time to respond to them. But wait, that could cut into your ad revenue.....

@16:10 why are the negative signs moved outside the 5th roots? 😮

does the bring radical have a power series expansion?

Step 1: Guess one of the roots. Step 2: Solve the rest 😏

How does x=t+2 cancel everything if all the coefficients are positive?!

I can I can I can solve quintic, I use comptuer 🙂

Let’s just use newtons method

Suppose that, I want to draw the image from the screen, copy. I don't want to print the image. I want to copy using pen and paper. Because I only have a blue print and a ruler. It's not exact science. still, it's close.

Meh. Just give me some special function to represent the roots of an arbitrary polynomial of arbitrary degree.

No way Galois had skill issue

How can Degree 5 polynomial has only 4 roots??

..very usefull !!

why not horner?

They stopped reaching it for The last 500 years

Hypergeometric function ?

Using fixed-point or Newton-Raphson iteration and complex programming, one can evaluate all roots of most polynomials up to the maximum accuracy allowed by the computational processes. No need to employ complicated solution techniques.

This isnt Malfatti solver?😅

I'll just use Newton's method Lol

Is it clickbate???

This was way too complicated I can follow a lot of the videos but this was out of my league

but that is like cheating: we turn a given quintic into a simpler one and then go: "well, we can't really solve this so let's invent this function that gives us a solution in terms of a hidden-inside-it quintic polynomial"

Lame.. Give the solution in terms of A, B, C, D and E, not in some other function

why would you hate me.