I can solve any quintic equation!!
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- Опубліковано 25 кві 2024
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Quintics can't be solved by radicals, but Michael is a moderate so it's no problem.
He still had to solve it by bringing the radicals.
@@torlumnitor8230yes sir, i agree with you, that quintic never be solved by any formula.
(If the constant term is 0) I can formulaically solve any quintic equation.
I know how to solve a quintic equation where the constants in front of the x^5, x^4, x^3, and x^2 are all 0.
Bros like "I have a truly magnificent proof but I will put it behind a paywall"
@@RedRad1990 If the constant term is zero you can factor the polynomial into x times a fourth degree polynomial and luckily there is an extremely long formula to find the roots of a fourth degree polynomial
And how i mean can you explain in brief
@@gautamchoubey1275 if the constant is 0, then 0 solves it. Then use the quartic formula.
Wolfram Research once sold a fascinating poster explaining various approaches to solving the quintic equation. I had it on my wall for many years.
Sweet, i wish i have one of those when i was studying ciclytomic polynomial
@@EvidLekan How would it help you?
they still sell it. cool btw, especially if you like mathematica
_We spent so much time asking whether or not we could, we never stopped to ask whether or not we should_
I don't know if we should, but if Steve Mould could, Steve Mould would
@@Valenqueen how many molds could a steve mould mold if a steve mold could mould mould?
This is so correct. For all practical purposes, using numerical methods like NR is more than enough to get roots of polynomials with ANY degree 100 times easier
Galois rn: 😡
Galois was obviously someone of low competence and never should've been taken seriously, considering he died in a duel at the age of 20.
@@Gordy-io8sb man had very dramatic and eventful life unlike most mathematicians in his period
@@genres381 pretty sure the other commenter was joking
Edit: nvm they're a dumb moron
@@Gordy-io8sb I mean... Isn't he EPIC?!
@@NotBroihon No I wasn't. Don't try to speak on my behalf, it's rude. Didn't you ever learn manners, kid?
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).
can it be solved by trig sub for example?
@@cicik57 Do you mean the Bring-Jerrard form? I do not think so. At least I never saw this kind of solution (trigonometric substitution) for the general case (any values for p and q).
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.
When you read up on Bring radicals, you will see that you might just as well solve quintics numerically.
@@florisv559that is absolutely not the point or spirit of this question.
@@Luigiman-rc9fi You missed the irony then. Pity. This video doesn't even scratch the surface of what Bring radicals are about, it just mentions them. The difference between solving a quartic or quintic, or the difference in abstraction between "square root of three" and "Bring(a)" is like the difference between 3^3^3 and 3^3^3^3.
Now, the Newton-Raphson method defines a convergent process and the power series that define the trig and exponential functions are just that. So is the Bring function, but then at an entirely different level.
I can use my computer to solve equations too. It's also faster, but it never does backflips.
As someone who had a Galois Theory exam only 11½ hours ago, this is well-timed bliss
They teach Galois' awful work now & give you exams for it? What has this world come to?
@@Gordy-io8sbmine is coming up in a few weeks. Algebra PhD qualifying exams
@@Gordy-io8sb man, what a specific topic you chose to troll about. Hahahahahaha. It is original, I guess, I give you that.
@@samueldeandrade8535 I'm not a "troll", nit-wit. I am a mathematician.
@@samueldeandrade8535 You're an arrogant edgelord. You likely have no mathematical knowledge at all, aside from basic arithmetic probably maybe. Why should I have to prove anything to you? P.S. Stop watching Michael's videos until you learn some real math.
22:49
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
There is.Good old graphs.
@@samarthchohan106 Yeah, but if the solution is irrational number you don't have the closed form.
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
Sí es
Exactly, it is.
michael never fails to do hard math
Atleast he doesn't do math hard
@@XisonBlazin10 you never know
Took me a while to understand the long transformation process and solutions through hypergeometric functions in Bring-Jerrard form, I wish I had this resource when I was still learning it
Great that you brought that to out attention 🙂
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.
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).
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
I've been waiting for this one 😁
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?
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.
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?
Yes, I tried the substitution x=t-2 with the symbolic toolbox of Matlab and it gives the correct answer.
@@Rudi_F_Vienna However, I did the math on paper, and I found out that the constant term of the original quintic has to be 215, not 67, to get a Bring quintic constant term of 3, ***but then*** I found out that I was wrong, since I had forgotten to multiply the constant terms of the (t-2)^2, (t-2)^3, and (t-2)^4 by the coefficients of the x^2, x^3, and x^4 terms, so I had gotten the wrong answer for the constant term of the Bring quintic, instead of 3, when I used 67 as the original quintic's constant term.
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.
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.
I’d like to see polynomial division (by hand) with the found solution and then solving the resulting fourth degree equation :)
@MichaelPennMath I have always been curious. how does this change for higher order polynomials?
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
This is called the quartic order Tschinhaus transformation, for each the Bring-Jerrad z^5+a*z+b=0 are obtained.
The substitution requires 120 coeficients.
The arguments inside the Bring radical function can multiplied by exp(2*pi*I*k/5) which is the five unitary quintic complex Roots.
Basically this conundrum bring all five roots, but are too complicated comparable to numerical methods.
A sextic equation can be always solved using the Kampe de Feriet function which is a two variable hypergeometric function 😂.
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.
Interesting! Is there always a transformation that removes the x^4 and x^3 term?
How can we numerically approximate Br(a)?
There are both taylor series and hypergeometric representations for the Bring radical.
Yes, there is always such a transformation, as described in the Wikipedia article on the Bring radical.
Since I'm sadistic tonight. Suppose 2024^p - 2023^p is divisible by 2027 for some prime p. Deduce the value of p.
glad we have computers
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.
I don't think all solutions to x^100=1 are radicals, but the rest are trigonometric solutions.
15:50 Shouldn't we have sqrt(-4) and sqrt(-16) instead?
P.S. Finally, someone brought Bring radicals into UA-cam!
No, since those are 5th roots, ***not*** square roots.
Happens to not be the first one but yea it is crazy how scarce it is to find content on this
does the bring radical have a power series expansion?
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 😉
That's one solution to the arbitrary quintic. What of the other four? Are they always complex?
If you can solve for one of the roots, you can use synthetic division and solve for the remaining quartic, which has a formula
If you know one solution, you can factor it out to get a degree four equation, which is solvable by radicals.
the bring radical has 5 roots. The other four may or may not be complex
@@user-mw6tq2nq2e Other series reversion techniques and things like differential resolvents or elliptic functions can get the other 4 solns for the Bring-Jerrard quintic in a cleaner form than factoring and solving the remaining quartic, but def harder to understand
Bring Radical is a function with five branches, thus Will return five values.
This video breaks the Galois theory.
..very usefull !!
"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.
I remember one of my college professors saying “when it comes to quintics, you take what you can get.” Actually I just made that up. But it sounds like something an old professor might say, right??
Huh? The video _did_ tell how you to approach all quintic polynomials. (Although unfortunately Michael did not specify the transformation which is needed near the end.)
There's a series expansion of Br(a), so while it's not a finite expression, you can get an approximate solution (within the radius of convergence) if you sum enough terms... You know, better than a poke in the eye...
@@Kapomafioso And, to be fair, it's not like we have anything better for radicals.
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.
No, he’s right, it’s not at the core or ridiculous to gloss over
x=t-2 ?
Yes, it should be x=t-2. I tried x=t+2 in WolframAlpha and it didn't work, wherein I suddenly realized there would be only positive terms and so no cancellation.
This doesn't seem much harder than applying the quadratic formula. /s
Rational Root theorem can’t help me now 😭😭😭
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.
Nice!
This is the best clickbait title yet
Suggestion : use timestamps to indicate sections that can be used as homework, that is non theory workings.
Abel-Ruffini teorem states that this is impossible (of course if constant term is non-zero!)
Impossible in terms of elementary functions**
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
Yes, there _is_ such a general way. Look at the Wikipedia article about the Bring radical.
There is a quadratic transformation that can eliminate the x⁴ and x³ terms, and then a quartic transformation is required to eliminate the x² term without elevating the polynomial degree. But, the P and A you get may very well be complex numbers.
In fact, linear substitution may very well be useless here. A quadratic substitution won't elevate the degree nor would the following quartic one.
IE the first substitution y= x² + mx + n and then there is a way to get rid of the x terms using collection (I don't quite get it but eh) leaving you with y⁵ + Ay² + By + C, and then the other transformation Z= y⁴ + Py³ + Qy² + Ry + S, and then collect the terms and solve the necessary equations to finally get something in the form z⁵+tz+u=0, and I'd recommend googling this since there is a more in depth explanation on the internet. But you just run a system of quadratic equations to figure out what the coefficients of the x⁵+Ax²+Bx+C would be equal in respect to the original equation, and then what the T and U terms will be.
@16:10 why are the negative signs moved outside the 5th roots? 😮
Obviously e. g. (-2)^5 = -32, so you could say that -2 is (a) fifth root of -32. It's not really mathematically rigorous, since roots are usually defined to be non-negative, but obviously it works here.
Step 1: Guess one of the roots. Step 2: Solve the rest 😏
When a b c d e values are known, Newton Iteration method is a better choice.
No way Galois had skill issue
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?
How does x=t+2 cancel everything if all the coefficients are positive?!
I think it was a (t-2) substitution instead
Как выпускник мехмата, могу заявить, что медали продаются на втором этаже магазина напротив.
Let’s just use newtons method
why not horner?
I can I can I can solve quintic, I use comptuer 🙂
How can Degree 5 polynomial has only 4 roots??
One of them is repeated.
@@carultchSo, a degree 5 always have a repeated root??
@@amulyasrivastava2985 No. Only special cases of quntics, or any degree polynomial in general, will have a repeated root.
For a polynomial to have a repeated root, a stationary point must coincide with one of its x-intercepts. A thrice-repeated root will have a stationary inflection point coinciding with an x-intercept.
The video doesn't show a 5th degree polynomial with only 4 roots, so what do you mean?
@@bjornfeuerbacher5514 Bro, the point of the video was to solve the degree 5 polynomial. He showed us only 4 roots. So , I am wondering about that 5th root
Meh. Just give me some special function to represent the roots of an arbitrary polynomial of arbitrary degree.
Honestly that’s kind of what Thomae’s formula feels like lol, but apparently it is grounded by geometry of hyperelliptic curves
@@MichaelMaths_ Interesting
Hypergeometric function ?
Where do you see a hypergeometric function here?
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.
They stopped reaching it for The last 500 years
#teaching
This isnt Malfatti solver?😅
I'll just use Newton's method
Lol
... which doesn't actually give the solutions.
Is it clickbate???
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.....
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 isn’t supposed to be an efficient way to approximate a solution to a quintic, rather he’s demonstrating that all quintic solutions are in the abstract of the form of a closed radical formula plus some other terms that are Bring radical solutions. He’s saying that once you introduce this new type of function into the allowable mix within a closed form then all quintics have closed forms available. (It’s not too dissimilar to having differential equation solutions that include the Lambert W function in them which is another function that isn’t expressible using radicals but provides useful information about the solutions in the abstract.)
Pov you completely missed the point of the video
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"
Isn't a radical also "cheating" by the same token? Both are defined to be the solution of a very simple polynomial
I mean this is how most common functions are defined. We don't really have a closed way of writing exp, ln, sin, cos etc., same with any root really. The only difference is that you're more familiar with these other functions and their usage is pretty "normalized"
The fact that that special "radical" is enough to solve any quintic is the actual interesting part of it. A lot of algebra is adding things to solve equations minimally.
Lame.. Give the solution in terms of A, B, C, D and E, not in some other function
XD
You would need an astronomically large board/paper to do that lol
why would you hate me.