Fundamental theorem of Algebra - A Simple but Beautiful Geometric proof

Рет қаралды 13,093

Күн бұрын

Proof that any polynomial with degree n has exactly n roots.
Website source : www.cantorsparadise.com/an-el...
Time stamps :
0:00 Terms and Definitions
3:25 Complex Number Multiplication
8:34 Proof

Пікірлер: 49

@artherladett442 2 жыл бұрын

Wonderful. Thank you.

@momeme2925 2 жыл бұрын

the pleasure's mine

@deeppurrple Жыл бұрын

Math teachers, you can plagiarise. Best explanation in this genre.

@worldnotworld 9 ай бұрын

error at 5:10: vector b shoud be 1+2i

2 жыл бұрын

Muito bom! Great!

@momeme2925 2 жыл бұрын

@denysvlasenko1865 Жыл бұрын

You do not need to prove there are n roots. Existence of just one root R means that (if n>1) polynomial has a factor: (x-R)(polynomial_n-1). Then by induction it's trivial that polynomial_n-1 also has another root.

@worldnotworld 9 ай бұрын

You've just provided an informal proof!

@aanjaneymishra9722 4 ай бұрын

@@worldnotworldwhy is it not formal?

@FishSticker Жыл бұрын

6:43 small correction: you got 2root(2) by rating the root of 8, but then when you square it, you say that it is equal to 4root(2) when you should have said it was equal to eight

@momeme2925 Жыл бұрын

yes I definitely made a mistake there. Thanks for pointing it out :)

@mgmartin51 Жыл бұрын

Excuse me. What do you mean by “Gauss chad”?

@amadeodante Жыл бұрын

Chad is a piece of internet slang generally used to describe someone who’s admirable, great, amazing or any general good trait

@michaeledwards1432 Жыл бұрын

2root2 squared ?….8, no?

@charlievane Жыл бұрын

quite nice

@pausesmaths3086 Жыл бұрын

10:22 " It might dance around this circle maybe a various a little bit up a little bit down but generally it will be something like this " But at 11:11 , we understand that a circle has to pass through the origin But as you said, those are not perfect circles, maybe the origin 0 is somehow avoided from a circle to another one. For me, your proof suffers from a lack of an argument of continuity along the line joining C to O ... which is for me the key of the proof. Tell me if I'm wrong and why ... Good work though. It illimunates this proof.

@ingiford175 Жыл бұрын

Two things most these proofs lack are the following two elements: You need to show that each of the 'almost circles' in the target space is 'dense' and has no gaps in the output 'almost circles' as you go around the input circle You need to show that going from the 'near point' 'almost circle' to the 'far almost circle' that the family of 'near circles' going from one extreme to the other is a continuous change First one shows there is no gaps, and second one shows that you go from near C (which does not have 0+0i within it) to a large circle which has the point 0+0i within it and somewhere 'in between' one of them must cross the 0+0i point

@pausesmaths3086 Жыл бұрын

@@ingiford175 Agree. First : what looks tricky is that we go to the infinity to have what looks most like a circle (because ax^n dominates), but then we come back next to C, where the circle is willing to be deformed/distorted. Second : we now know that this proof has to be analytic. algebra or geometry aren't sufficient to demonstrate it. But no word about continuity here. Third : if it was so simple, Euler, Gauss, Lagrange ... would have guessed it faster than we understand it. But they failed to prove it. Let's call this video a good illustration of the idea of the proof, a good guideline to the demonstration, but not a "geometrical proof".

@markhughes7927 Жыл бұрын

@@pausesmaths3086 as a non-mathematician is there any value in the thought that vectors having a definition of direction and amplitude cannot be conceived as other than representations of something physical? and that anything physical cannot be accurately represented on a 2D plane because physical things occur in 3D? So there will never be a match? If this is absurd please ignore. (Also since I’m here Gauss was definitely a chad because he was teaching his father calculus at the age of four - though I would not go so far as to claim for him as being a chad that he was at the time -26.)

@pausesmaths3086 Жыл бұрын

@@markhughes7927 Vectors can be thought in 2, 3 , ..., 141, ... 5 468, ..., any finite, dimensions, and even in infinite dimension. The definition of a vector is that it is an element of a vector space. No dimensions involved first.

@francescaerreia8859 Жыл бұрын

I think I understood everything you said because I think every individual sentence made sense to me as you stepped through, but I still don’t see where you proved this here. If I lost you, I don’t even know where.

@FishSticker Жыл бұрын

Your proof technically doesn’t account for terms other than the leading one and the constant, it’s a slight modification to get there. besides that, great video

@stimulantdaimamld2099 Жыл бұрын

great

@stimulantdaimamld2099 Жыл бұрын

wastage of time.

@annaclarafenyo8185 Жыл бұрын

Not quite the best proof along these lines, because it doesn't translate to a good algorithm. You are using an "inside-outside" argument in the proof to argue that 0 is "inside" the big circle, but "outside" the small circle which means that as you deform the big circle to the small one, zero must cross at some point. The rigorous version of this is called the "Jordan Curve Theorem", and if you look inside any proof of the Jordan Curve Theorem, you'll see it uses a winding-number argument. Points in the inside of the curve are those where the curve makes a winding number of one (or minus one, depending on orientation of the curve), while points on the outside have a winding number of zero. You can use the winding-number argument directly to prove the theorem without going through the intermediate step of the Jordan curve theorem. Just notice that the function f(z)=z^n has a winding number of n around the biggest circles, and it can only have a nonzero winding number around a zero of the polynomial. Since winding number is additive when you chop a curve in two, one of the two halves has a nonzero winding. So you just keep chopping up the curve until you get a nonzero winding number around an arbitrarily small region, and that's converging to a zero of the polynomial. This is constructive up to minor Brouwer issues (you need to bound the topmost coefficient away from zero), and IMO this is the "book proof" of the theorem. It is essentially due to Gauss, although he didn't quite formulate it in terms of winding number, the ideas are analogous enough to give him credit.

@patato5555 Жыл бұрын

The FTA only states an every polynomial over the complex numbers has at least one root. This implies the statement you made.

@ingiford175 Жыл бұрын

Depends on the version of the FTA you read, But once you got one, you can get the rest. I have seen it stated with one solution (one I prefer) or that it has n solutions (which can be proved once you got the one solution).

@deeppurrple Жыл бұрын

Some day you should come out with a better visual presentation.

@zevfarkas5120 Жыл бұрын

Rehearsal might help...

@momeme2925 Жыл бұрын

I agree with you. Check out my latest video I think it's a big jump in presentation

@AndresFirte Жыл бұрын

What happens if C happens to be 0?

@michaelguenther7105 Жыл бұрын

Then you know that one root is x=0, divide your polynomial by x to get another polynomial of degree one less. You can keep doing this until either c for the new polynomial is nonzero, or show that all roots are zero (which would mean your starting polynomial was a*x^n = 0).

@GeodesicBruh Жыл бұрын

Nice memes. Gauss Chad.

@kparag01 Жыл бұрын

Are you indian?

@reidflemingworldstoughestm1394 Жыл бұрын

Mathesar? Is that you?

@soniahazy4880 Жыл бұрын

🎨✨⛲️💎🦋🪷🙏🌈🫧

@cacostaangulo Жыл бұрын

This is an explanation, not a proof

@MuffinsAPlenty Жыл бұрын

It's a sketch of a proof.

@zeeshanalamkhan2320 3 жыл бұрын

First

@momeme2925 3 жыл бұрын

No, I was first

@agrajyadav2951 Жыл бұрын

@@momeme2925 -.-

@rainerausdemspring3584 Жыл бұрын

A polynomial is not a function. These are completely different things.

@MuffinsAPlenty Жыл бұрын

Polynomials absolutely are functions.

@rainerausdemspring3584 Жыл бұрын

@@MuffinsAPlenty Take any book about algebra and study the chapter about polynomials over rings (or fields),

@MuffinsAPlenty Жыл бұрын

@@rainerausdemspring3584 I know what you're trying to say, but you're saying it wrong. Polynomials are functions. Given any ring R and any polynomial ring S over R and any element f in S, then f defines a function from R to R. Yes, polynomials are more than functions, but they're still functions. It would be like saying that a field extension E/F isn't an F-vector space because it's actually an F-algebra. It's a technically wrong statement. Field extensions _are_ vector spaces over the ground field, even if they are also more than that. It's really quite unbecoming of a mathematician (math student? math enthusiast?) to harp on technical details which are irrelevant to a given presentation. Polynomials aren't _just_ functions, but does the fact that the video described them as functions take away from the presentation at all? No, it doesn't. It also doesn't mean that polynomials _aren't_ functions. Also, I would love to see a proof of the Fundamental Theorem of Algebra which doesn't use the fact that polynomials are functions. Surely, you must know of one if you find it important to go out to videos explaining the Fundamental Theorem and making a big deal about how polynomials aren't functions.

@bobh6728 Жыл бұрын

A polynomial function is sometimes just called a polynomial. So I think both of you are correct. A polynomial like x+2 is an expression, but it is also a function, since for every value of x it gives you a unique value. It just doesn’t give a name to those values like f(x) or y, but they are the codomain of a function. A polynomial equation like x+3=x^2+1 is not because it is not producing an output based on values of x. It is saying x=2, but it doesn’t map 2 to anything because there is no mapping. So polynomial is an adjective describing what kind of expression, or what kind of function, or what kind of equation. But the shortened form of just saying polynomial, would have a meaning depending on the context.

@vazn4143 Жыл бұрын

@@rainerausdemspring3584 Polynomials and polynomials functions are isomorphic (and the isomorphism is quite obvious), hence they're the same from an algebraic perspective.