Відео

Bro stopped the timer that many times to muscle memory the place of the stop button?!!

Wow, i just realized now that youre the really big blind solver, this is such a cool channel pivot, love this math/geometry stuff

15:18 the profile picture

Nice video. It's related to Centrifuge problem (Numberphile has a video with Hannah Fry explaining it).

I hit the like button too early in the video because now we're talking about building polynomials and I'm even more excited then before but don't have any way to easily express it; and thus have been forced to use the treacherous comment function.

tip: set playback speed to 1.5x

7:10 off center day ruined 😔 literally unwatch able. (Great video actually❤)

Go hyperlegible!!

1:16 You mean digon.

0:01 My guess: the points are evenly spaced in the sense that they consist of several sets of points, each of which is rotationally symmetrical. This also means that if these points had mass, the center of mass would be in the middle of the circle. I'm assuming this based on a video about balancing 7 tubes in a 12 slot centrifuge

Just to nitpick (or because I like edge cases) I think Phi_1(z) = z - 1 should be interpreted as the empty collection, i.e. not having any points at all. Since in this case z = 1, so so it's a positive and a negative point at position 1 cancelling each other out. (Btw, Phi_1 doesn't actually matter for the proof, since we're only building collections of ≥2 points. Really, the base case could just be the primes.)

Well y₀ for the final answer seems to only be because the process as defined is only countable. But that does not seem necessary. Suppose I say that for every time t, in [0, 1), I will add a digon at angle πt. This builds the whole uncountable circle. We need to update our definitions of convergence, though, to account for multiple infinities. It seems easy enough if we just want to change it into positive and negative cardinalities for the weights, disallowing when a point appears in the same cardinality of positive and negative steps. But if we want something more complicated like measure theory, I don’t know enough to design definitions for it.

i take issue with the 1-point build, its like saying the sine of infinity is 0 because thats where it starts

Another commenter stated "A buildable set must always have all points be partnered up with others such that the partner is a rational distance from the original." Why impose this arbitrary limit? If we allow non-rational rotation of polygon's points, and allow the removal of balanced, polygon-based subsets of points, wouldn't all centered sets of points be buildable using a potentially infinite number of shapes and point nullification?

What's the name of your cat? 😺

This is so underrated

It seems like it would be possible to generalize this construction to uncountably-buildable sets by allowing uncountably many steps. Since the convergence of a point does not depend on the order of the steps, just the cardinality of the sets of positive and negative steps the point occurs in, this should not cause problems. Using this definition, I believe every set of points where each point has at most countable multiplicity (positive or negative) is uncountably-buildable (which I will use here in a sense that includes finitely- and countably-buildable). By partitioning the set into rational subcollections as in the previous video, you can use the technique from this video to build each one and take the union of their sets of steps, which cannot overlap on any points. This result cannot be extended to the case where points can have uncountable multiplicities. For instance, the point set that contains just 1 with uncountable multiplicity is not uncountably-buildable since by the pigeonhole principle, at least one other point a rational angle away from 1 would also need uncountable multiplicity.

Vigintillion un- dou- tre- quad- quin- sext- sep- oct- non- dec- btw ten douvigintillion is googol

There are still irrational numbers that when summed they can be rational. Does that change how you define the I_p collection?

I heard the cat. Meow.

10:33 and maybe an 11-gon (shows 12-gon)

24:06 I think Answer is 3 points, because we can distribute points in such way that them will be in one half of circle

I dont think the reasoning at 6:42 works generally, because we could have more points at every step so it would not converge. For example for every point expect first make the negative version of it and 2 new points this way number points is strictly increasing so it cant converge to the first point.

I did some study of base 1+i last fall. If you allow for p-adic values, the "infinite repeating ones" (...1111111) is equivalent to "i" and you can start from there and get a second interlocking spiral, rotated by 90 degrees, that completes the plane.

Keep in mind that Lego is *not* a term for a a type of product, that would be Klemmbausteine, and since are not using the product Lego you must not to refer to it as such, otherwise Lego will sue you. If you want more information on legal protection and Klemmbausteine you should check out the Held der Steine Inh. Thomas Panke

I don’t think I understand how it’s possible that an infinite sequence of steps that don’t change the average of all the points can result in something that has a different average. If I have an average score of 0 and (don’t) turn in infinite tests that each scored 0 points, how does my grade go up? I just don’t get how you can construct the single point even after infinitely many steps. If there are infinitely many steps than you’ve never actually completed the construction, and if we assume you’re at the end of the infinite construction after a series of infinite steps that all increased the number of points while maintaining a zero average, then there are infinitely many points spread around the circle. Just because one of those points was there the whole time doesn’t mean any of the following points don’t exist. Also why can’t you just define the empty collection of points as being centered? Logically speaking if there are no points on the circle then the average is at the center. I know mathematically there are divide by zero issues when calculating the average of nothing but if a finitely buildable construction is centered then adding and removing a polygon as a 2 step construction must be centered…

10:31 I don’t think there’s any need to check whether you’d hit a point you’ve already used in this way. You could probably just get away with using the next prime every step.

N=5 is the Minimum number of Points for a non buildable collection brause N=1 does Not Work AS Seen in the Video N=2 call the Points A=(1,0) (roate if requierd) given It is centerd B has to bee (-1,0) thus it is constructeble N=3 Let A =(1,0 ) -> B=(bx.by) C=(cx cy) With by+cy=0 and bx +by=1 -> by= -cy =y Given B,C are on the circle WE get that bx= +-sqrt(1-y^2) =x and the Same for cx Obviusly cx hav to have the Same sign in oder to add Not add to Zero and x=-1/2 else they would Not add to one -> B=(-1/2, sqrt(3/4)) similar for C Thus all cases are perfekt triangels this buildable. N=4 similar results in only perfekt rectangels. All of them are buildable As a Set of two balanced points. And N=5 is possible as shown by the example in the Video.

Alright, here's my reasoning as to why uncountable collections are not infinitely-buildable: 1. an uncountable set cannot be mapped to a countable set 2. For any infinitely-buildable collection, we can map each point to the step it was first added 3. There are always a countable number of steps since they are integers and the integers are countable 4. Since an uncountable set cannot be mapped to a countable set, and we can map all infinetly-buildable collections to a countable set, all infinitely-buildable collections must be countable

I love this idea, it's nice to see you decided to make a second part going more in depth on it!

The reason they're not symmetrical is because you're not using all of the points.

Yes. There's a non-bailable centered collection. Consider putting two vertical lines πr/4 from the centre where r is the radius. The dots from intersection are centered, but not buildable

Oh I am so eager to investigate centered collections on the surface of a sphere! Trivially all of the properties of centered collection in 2D generalize to 3D, but I don't have any intuition about what more interesting configurations might exist!

This channel is gold

galois theory, done in a way a highschooler can understand. amazingly well done!

So we have seen that the roots of cyclotomic polynomials can get you any rational/buildable collection. if we allow ourselves to use the roots of any (integer) polynomial can we then make any centered collection? Or are there even transcendental collections??

I love this ! Allowing for negative points seems like it makes this a vector space over R/2piR ? or R/4piR ? idk

I'm having trouble with "remove the triangle of points". Removing is not the same as adding, and can skew the results.

nice song

This is so amazing, it let me wondering if there is such a way to make all the contructible shapes instead of regular polygons, maybe that could give some insight into transcendental numbers.

It's all about how to balance a centrifuge.

3 points can only be balanced im almost sure. as for 4, i suspect not, and it probably has something to do with not being able to find an equation for a polynomial of degree 5. you need to be able to "split" the points of either a line or a triangle in a way that i dont think you can do

Those collections you chose study are neat, the questions you posed quite interesting, and your investigation intellectually satisfying.

I'll wait and watch this channel boom in 2 years

as a programmer it kinda stuck out to me that the "negative" points or polygons could be nicely replaced by a XOR (addition modulo 2) operation, so when two points match, 1 XOR 1 = 0

There’s actually many irrational collections that are buildable. For instance, two lines, one rotated 1/sqrt(5) of the circle away from the other, form a clearly buildable, yet irrational collection. This works because for any p and I sub p you can pick, its average contains the opposite point, and so works out to be the center.

1 isn't possible cause it's not centred 2 isn't possible because any deviation from the opposite positions isn't centred 3 isnt' possible because any deviation from the equilateral triangle isn't centred 4 may have the necessary degrees of freedom, but I cannot produce a counter-example, starting from an arbitrary 3 point collection that doesn't contain a pair in opposition generates a balancing point off the circle. I don't believe 4 points is possible either but I can't prove it. There may be a pathological singular counter example I cannot think of. 5 is the first arrangement where there is definitively the necessary degrees of freedom and you can generate an infinite number of counter examples. Start with a point at (0,1) and then for every point (x,y) there's another point (-x,y) for every point (p,q) where q<0 along with (-p,q) there exists a pair of points (r,s), (-r,s) that balances out the collection, x is always balanced by virtue of each point having it's mirror in the y axis (or lying on the y axis) while y can be balanced because of the degrees of freedom offered to the pair of points (r,s), (-r,s)

Wait a second ... THIS IS VERY COOL!!!

bless this channel, did you find this yourself?

This is so interesting!