302.S12B: Quintic Impossible 2 - An Insolvable Quintic

  Рет қаралды 8,113

Matthew Salomone

Matthew Salomone

Күн бұрын

Пікірлер: 48
@M4THandMUS1C
@M4THandMUS1C 10 жыл бұрын
You are one of a kind. Don't think I've ever come across an instructor with your level of knowledge, clarity and enthusiasm! Higher mathematics education could use more people like you.
@nickruffmath
@nickruffmath 3 жыл бұрын
Thank you so much for making & sharing these videos. You make a complex & abstract subject so accessible and exciting!
@NgocAnhNguyen-si5rq
@NgocAnhNguyen-si5rq Жыл бұрын
You are a genius teacher. Your videos are so funny and understandable at the same time.
@rhke6789
@rhke6789 3 жыл бұрын
Bravo..the finale! Your logic, enthusiasm, and great articulation are great assets to my learning. To do it over again, I would watch this video first...then, knowing what I need to learn, and then start methodically with some perspective that what is being taught and how it will be used. I have search KZbin high and low looking for a good understanding of Galois' s proof. This is clearly the best without smoke and mirrors. Congrats to you for your teaching ability and congrats to me for finding your lessons and patiently learning from them.
@BillShillito
@BillShillito 10 жыл бұрын
That. Felt. Good. Satisfying QED indeed. And the music helped. XD Great series of lectures - you have a style that's actually energizing to listen to. The quintic was the one topic from my Abstract Algebra book we never went over, and it always felt like a crime we never did so. Now I understand it, and it was so worth it. Thanks so much!
@Trunks47r786
@Trunks47r786 9 жыл бұрын
Bill Shillito Are you going to make new videos??
@speedbird7587
@speedbird7587 3 жыл бұрын
Very energetic ! Your explanation is brilliant!
@heisenbergmuzik4948
@heisenbergmuzik4948 4 жыл бұрын
Eureka!!! Man love your energy and style... Hats off!!!
@NoNTr1v1aL
@NoNTr1v1aL 4 жыл бұрын
Thank you very much for these videos!
@imaneferradj9606
@imaneferradj9606 6 жыл бұрын
the best video I've ever watched !!!!
@duckymomo7935
@duckymomo7935 7 жыл бұрын
I love his enthusiasm
@Jim-vr2lx
@Jim-vr2lx 6 ай бұрын
Bravo sir! Bravo.
@Hythloday71
@Hythloday71 10 жыл бұрын
AWESOME ! thanks for sharing all this. I did this in group theory and it was tough stuff. What do you think of this as a lay summary ? ROOTS can be like permutations of objects, permutations are a really core fundamental type of group that contain all other groups (non-abelian), some groups play nice, they all have formulas, some groups don't play nice, if a polynomial type was to be associated with a group that didn't play nice it wouldn't have a formula.
@PunmasterSTP
@PunmasterSTP 5 ай бұрын
14:36 Peak math education 👌
@mathador4467
@mathador4467 7 жыл бұрын
This is a great introduction to Galois' Theory, but without good definitions and real proofs, you can't truly say you proved that fifth degree equations are unsolvable by radicals.
@stapleman007
@stapleman007 2 жыл бұрын
14:33 Exactly how Galois felt about 5th degree polynomial roots.
@RubenHogenhout
@RubenHogenhout 6 жыл бұрын
If you allow also the matrix transformation sometimes you can solve more then the eye can tell. For example. If I have got the X^5 + 5*X^3 + 5*X +2 = 0 and I transform it with the quadratic X^2 + X + 2 + Z = 0 then the determinant of the matrix will give you the quintic Z^5 + 15*Z + 44 = 0 . And the X in this case is equal too ( -1 + (2)^(1/2) )^(1/5) + ( -1 + -(2)^(1/2) )^(1/5) is about equal to -0.354378367 now you can just fill it in the the quadratic to calculate the Z . Z is equal about -1.77120573 and this is the same zero of the Quintic in Z. Now in this case I wonder. What is the solable group of this quintic in Z?
@RubenHogenhout
@RubenHogenhout 6 жыл бұрын
Nice , can you show too that for example X^5 + 15X + 12 = 0 has a solvable group?
@ravenmitchell820
@ravenmitchell820 9 жыл бұрын
It can be solvable if use i=(√-1). You can also use the sum or difference of 2 radicals in a root or radical. That would make it seem unsolvable but it can work if you try. And you can use substitution, ex. p=a+b, or you can combine theories to make the algebraic roots of the equation and have answers of the polynomial. Ex. x=(-b±(√(b^2-4ac))/2a
@nguyen2003
@nguyen2003 9 жыл бұрын
+Damien Mitchell That's not how it works. It won't work for "all" quintic polynomials, trust me. If the solution to all quintic polynomials existed in closed form, I would know it. You are confusing this with whether or not a "case-by-case" scenario is solvable.
@ravenmitchell820
@ravenmitchell820 9 жыл бұрын
+Peter Nguyen what about 3rd power equation and 4th power equation theory
@MatthewSalomone
@MatthewSalomone 9 жыл бұрын
+Peter Nguyen Indeed, the point of this video (and of the whole video series in general) is to prove that no formula exists for the quintic in general. The substitutions and radical tricks you describe would work for some quintics (in particular, if their Galois group is solvable, e.g. S4) but there is no hope for a one-size-fits-all formula for all quintic polynomials.
@ravenmitchell820
@ravenmitchell820 9 жыл бұрын
+Matthew Salomone It could be be possible to start with Ax^5+Bx+C=0 then divide A on both sides and then make a perfect quint is and go from there.
@ravenmitchell820
@ravenmitchell820 9 жыл бұрын
+Damien Mitchell i=(-1)^(1/2) would just be part of it
@ravenmitchell820
@ravenmitchell820 9 жыл бұрын
The quintic equation can have 5 solutions but you need to try roots in roots or try to factor out the roots you know from the polynomials.
@nguyen2003
@nguyen2003 9 жыл бұрын
+Damien Mitchell Actually, it could have the same root more than once, making it appear as though it has less than 5 roots. If you are interested in math, try math.stackexchange.com
@RubenHogenhout
@RubenHogenhout 6 жыл бұрын
The quintic is this form ( Bring Gerard ) can have a maximum of three real roots. Three or one this is right.
@utashiori141
@utashiori141 10 жыл бұрын
Hello. Thanks for this. I subscribed to you. Could you kindly put up a playlist of the videos from the beginning of group theory up to this video of quintic insolvability. Thanks.
@MatthewSalomone
@MatthewSalomone 10 жыл бұрын
Start here: Exploring Abstract Algebra II: Contents Playlist: kzbin.info/aero/PLL0ATV5XYF8DTGAPKRPtYa4E8rOLcw88y
@utashiori141
@utashiori141 10 жыл бұрын
Matthew Salomone Hello. I'm really poor in abstract algebra right now so I hope you can answer my question in a non-sophisticated manner (and hopefully a video). What is the direct connection of the solvability of a group & also the abelian part of the group tower in the solvability of the quintic ? Does my question make sense ?
@nguyen2003
@nguyen2003 9 жыл бұрын
+Uta Shiori My recommendation for questions like these is to ask them in a math forum, particularly math.stackexchange.com
@ckclasses9835
@ckclasses9835 7 жыл бұрын
Sir chandrakant here from India sir ple teach us how to find onto ,into, homomorphism.
@robkim55
@robkim55 9 жыл бұрын
Bravo ......wonderful videos........now the question is how do you generally find a solution to a 5th degree equation??? without using extensions Numerically?
@nathanjue2838
@nathanjue2838 7 жыл бұрын
I believe numerical approximations are the only way to find "solutions" to the general 5th degree equation.
@RubenHogenhout
@RubenHogenhout 6 жыл бұрын
There exists serie expansions.
@duckymomo7935
@duckymomo7935 6 жыл бұрын
Elliptic curves And some other methods If approximations are acceptable, use a computer
@matthewcory4733
@matthewcory4733 7 жыл бұрын
Except if you're using a 10-line bisection algorithm...
@FrancisCWolfe
@FrancisCWolfe 10 жыл бұрын
Do be pedantic, you don't actually *have* to find one whose Galois group is all of S5, you just have to prove that one exists. I don't know how you would do that without constructing one, but I bet you could.
@RubenHogenhout
@RubenHogenhout 4 жыл бұрын
Nice prove. This quintic is indeed not solvable in radicals. But what you can do is two make the last coeeficienten equal like this. X^5 + -6*X + 3 = 0 X = k*Y -6/k^4 = 3/k^5 (-1/96)*Y^5 + Y + 1 = 0 u= -1/96 k = -1/2 Y = -1.011002460811 Y = 3.341870528962 Y = -2.803283758532 Y = -1 + u + -5*u^2 + 35*u^3 + -285*u^4 + 2555*u^5 Y = -1 + -1/96 + -5/9216 + -35/884736 + -285/84934656 + -2555/8153726976 = = -1.01100243008 X = -1/2 * -1.01100243008 = 0.5050122 the quintic has three real zeros. The others are. X = -1.670935264 X = 1.4016418
@goboy6882
@goboy6882 3 жыл бұрын
Check it off of my "Bucket List".
302.S13A: Fundamental Theorem of Algebra -- Statements
8:17
Matthew Salomone
Рет қаралды 3,3 М.
302.S9A: Galois Groups and "Stubborn" Polynomials
13:49
Matthew Salomone
Рет қаралды 6 М.
Chain Game Strong ⛓️
00:21
Anwar Jibawi
Рет қаралды 41 МЛН
Правильный подход к детям
00:18
Beatrise
Рет қаралды 11 МЛН
We Attempted The Impossible 😱
00:54
Topper Guild
Рет қаралды 56 МЛН
302.S9B: The Galois Correspondence
18:03
Matthew Salomone
Рет қаралды 13 М.
Why There's 'No' Quintic Formula (proof without Galois theory)
45:04
not all wrong
Рет қаралды 556 М.
302.S10B: Radical Extensions & Solvable Groups
12:56
Matthew Salomone
Рет қаралды 8 М.
I Tried to Put the Bible to the Test
8:24
Cold-Case Christianity - J. Warner & Jimmy Wallace
Рет қаралды 178
Lutheran Pastor Becomes Orthodox (Fr. John Fenton) | FULL INTERVIEW
1:46:44
Roots of Orthodoxy
Рет қаралды 1,2 М.
Galois Theory Explained Simply
14:45
Math Visualized
Рет қаралды 490 М.
Chain Game Strong ⛓️
00:21
Anwar Jibawi
Рет қаралды 41 МЛН