Proving Grant's Little Theorem | A Surprising Geometric Result 

@Sebastian Henkins I can't really point to a single thing, it's something that developed/formed over time and unseen trial and error. I feel like I'm pulling from hundreds of influences from random sources when it comes to the visuals. I did a lot of web-design in high school in the form of templates for message boards as a member of the design team, and I feel like a lot of my visuals are rooted there when it comes to an overall aesthetic. That was probably influenced by a lot of websites I came across at the time. I think my biggest influence for presentation style is the channel "Because Science". It's hard to say. Sorry I can't help you remember what movie/game I'm reminding you of, I'm not really sure what it could be. I've used some Dr. Strange style effects in the past, but I don't know what matches the overall look. 😅

HELP FLAMMABLE MATHS PLEASE! HIS CHANNEL HAS BEEN HACKED

I only did it because I see strong potential in his channel. He'll be a big success someday, calling it now.

XD

Zach Star is already breathing down 3b1b’s neck lol. How I remember if it is 3 blue or 1 blue or whichever is first is to remember that I have seen people with blue eyes in the past with a tiny little stripe of brown in it. And usually the brown bit is the smallest, hence 3 blue 1 brown. Reverse ordered by area. Edit: To be more precise. Zack Star’s production rate is insane while keeping high quality content. I am not too familiar with this channel yet, but I will definitely keep my eyes open.

@@EpicMathTime isn't he a succes now? i mean, many people see him as the best math channel

@@Cat-yz1tk That's the joke.

And L'Hopital's rule.

Well, technically you can't because the closed formula you wrote for the geometric series holds for |z| 1, it explodes to infinity, and for z=1, it is just n.

I think you should rethink n=1. You're right that there aren't any lengths, but your conclusion is not correct. Awesome comment though!

You're right. The theorem still valid for n=1 if we apply the formula, even though there aren't any lengths here (which previously make me feel nonsense before I eventually use the theorem formula)

@@KienPS257 Right, there are no lengths, so this is an empty product, which is 1. This is the trivial case.

This is a good observation, thank you!

Thanks!

still recording everything on a 2.5 year old android phone 😅

I don't get it, is everyone in the comment being sarcastic or what? I mean, Grant has more than twice as many subscribers as Jon...

@@nanigopalsaha2408 They are just playing on the same joke in the video (calling him a small youtuber, young man, forgetting his name, etc). He's arguably the largest math-instructional-dedicated channel on the entire site, so the sarcasm is hard to miss.

@@EpicMathTime Oh. HEHE😃

@@nanigopalsaha2408 also, 3b1b has about 230 times as many subs, not 2.

@@brandonklein1 He said more than twice not twice, so he is correct.

big daddy energy

The proof is the same more or less, you just include the scalar factor r. The Limits and L'Hôpital's rule give n*r^(n-1).

I'm not sure the situation can be described very well if the points can be anywhere. We can immediately put bounds - it's between 0 and 2^(n-1), and I think it can be arbitrarily close to either of those. What else do you think can be said?

Epic Math Time Oh when I said “I am not sure it was mentioned...” I meant that as an apology if I missed it, not as a criticism for you not mentioning it. Sorry if it came across as criticism. That was not the intention. Thanks for the answer though!

@@roygalaasen No, I didn't take your comment that way at all. I'm actually inquiring, what else can be said? You mentioned n^n, where does that arise in the situation you're describing?

Epic Math Time ok, I have been thinking about this for a bit and honestly don’t think it would have any applications other than curiosity. The way I see this is as an extension to allow for arbitrary positioned points. Let us say three points placed in a way that they are not rotatable related to each other in any way. That is if you rotate the points chosen, and align one point to another at a time for all permutations, you would not have more than that one point touching anyhow you did the rotation. I was thinking you could add n^n points AT MOST to sort this problem to get to an arrangement where you would get all the points equidistant. You take note of the number of points added, then you get the new integer point count the way you presented in the video proof, subtract the number of points that you added and voila you have generalised the formula. I am thinking one way to make equidistant points from arbitrary points is to just do what I mentioned earlier: match each points with each of the other points one at a time. Then for each match you mark the newly created points. I imagine in some configurations you will have overlapping points. Thus less than n^n points in some cases. If anything is unclear, I am happy to explain further. I just tried to jot this down quickly while no one is talking in my head since I tried to write this yesterday, and it is dang hard to write while you are having a conversation. All in all this is probably just a silly though experiment. Edit: I said “add” n^n points, but what I mean is you go from n points to n^n points at most as said in my first post.

@@roygalaasen if you want to distribute the points arbitrarely you can just add an arbitrary phase shift

The length of the circumference is 2pi, so I'm not quite sure what you're asking.

The limit's existence means, by definition, that the limit is the same regardless of the direction we approach the limit. If you believe l'hopital for real numbers, you believe it for complex numbers too. It doesn't change anything.

@@EpicMathTime that's what I meant when I said that I could avoid the problem by using real numbers, but my question was if we were for example approching an imaginary number and not a real number, would it still work ?

@@destroctiveblade843 Yes, l'hopital's rule is innately true of _calculus_ . What I mean is just that l'hopital's rule is true on any smooth manifold, which is essentially "any structure where calculus concepts can be well defined." The proof of l'hopital's rule makes no reference to what kind of "numbers" are being considered. It's not relevant, it's something it can't see. As an analogy, you learn in linear algebra that any two bases of a vector space have the same cardinality. If one asks, "is this true for complex numbers" the answer is yes, because linear algebra doesn't care what's "in" the vector space, it can't see that, it's an extraneous detail. It's really kind of like asking: is this true for vector spaces written in blue ink! It's kind of like that. The theorems of _calculus itself_ aren't about real numbers or complex numbers at all. They are about something far more general (smooth/differential manifolds). I hope this makes sense. We get misled by learning elementary calculus before really being exposed to more modern math, so we often don't realize that the calculus we learned also has a modern and more general treatment. Think about a theorem you've learned in abstract algebra, we probably wouldn't think to ask if a certain theorem about groups also holds for complex numbers, because right from the beginning, groups and their theorems are treated generally/abstractly without ever referencing what the things "in the group" are (because it's an extraneous detail, irrelevant to _groups_ ).

@@EpicMathTime makes sence, idk about differential manifolds, I would guess it has to do with diffenrential functions, but thanks for the explanation

@@destroctiveblade843 Well, think of it this way. The core concepts of calculus are mainly an idea of "distance" and the idea of "limits." A really rough version of what I'm saying is that the results of calculus are true on _anything_ that has notions of distance and limits. Those things are called differentiable manifolds. An example of a differentiable manifold is a sphere, another is a torus. An infinite line and infinite plane are also differentiable manifolds. The results of calculus are true on any of these "surfaces", it doesn't matter what we call the points on these surfaces. It's an irrelevant detail. Calculus works on real numbers and complex numbers because topologically, they are differentiable manifolds (real number _line_ , complex _plane_ ) it works because of the overall shape of the _whole set_ (its topology), it doesn't care what the individual points are, it doesn't know what real numbers are complex numbers are.

Tool played a part in my decision to major in math, so I will always pay homage to them. I'll explain in detail one day.

@@EpicMathTime stoked to hear that! This band is my jam 😍 and their songs can really be influential if you think about them deeply

@@EpicMathTime my man, I myself am somewhat of a 'late bloomer' in terms of mathematical endeavours, but a passionate educator nonetheless, and a fellow Tool fan. Let's keep up the good work, a'ight yes?

@@AV-ws2rz Let's do it brah

@@EpicMathTime Lateralus and their use of the Fibonacci sequence?

Dummit & Foote is my #1 favourite textbook, I could definitely grab it and make a few series from it.

@@EpicMathTime please make a video on the algebra of tensors from D&F! My favourite section of the book!

He just mirrors it when editing... Is this bait?

I flip it but the spaceship and holograms are real

Epic Math Time Dang. I want one of those glowing π-shirts

I'm not going to say that _can't_ be done, but I think the argument will be much more difficult. Furthermore, the connection with the roots of unity is really hard to ignore. I do suspect that there are nice "collapsings" that happen in the measurements. In the n=4 case we get a clear view of that "collapse" in the form of (sqrt(2))^2. I have a hunch that similar things happen in the form of conjugate multiplication, which happens for points that are across the circle horizontally as opposed to vertically, perhaps? It might be worth exploring.

π-daemon*

@@EpicMathTime Hope you dont mind my satire, I do like your maths vidz.

Btw you could have also factored out z-1 from z^n-1 and just prove it algebraically, without limits

@@thatdude_93 You are right. I was actually THAT close to do that, but in my mind I was too quick and came to the conclusion that I would probably just end up with the original product that I want to know what it is and get nowhere... If I'd actually stop and think for a second I would have solved the damn problem....

The thing to recognize is that since the right hand side is continuous, taking limits will make the desired "goal" expression appear. As for why this is justified, well, yes, it's what you said at the end. If two functions agree everywhere except for a single point, then certainly the limit as we tend to that point is the same for those expressions. They are the same for every point relevant to the limit. Graphically, these functions would be identical, except for a point sized hole at 1 for the left expression. If we take a limit as we tend to 1, what happens _at_ 1 is no concern to us. It happens to be the same thing for continuous functions (by definition) but generally, limits are concerned with what happens _around_ 1, and the functions are totally identical _around_ 1.

@@EpicMathTime its a R E M O V A B L E S I N G U L A R I T Y

Also, just to elaborate further, if f(x) and g(x) disagree on _any finite number_ of input values, but agree elsewhere, all of their limits are always equal. This can be traced back to simple ideas about sequences. If I have a sequence, and I change a finite number of its entries, it still converges to the same value.

I'm not sure I follow how dividing by more factors would yield the result.

@@EpicMathTime What I mean is, (z^n-1)/(z-1) represents a polynomial. This polynomial is valid for z=1 (even though it doesn't look that way at first glance). We don't need limits and l'Hopital to make it work (although it is, admittedly, a fancy and quick way to find the value we're after, in case we've forgotten the standard form of that polynomial).

@@EpicMathTime (z^n-1)/(z-1) is actually z^n-1 + z^n-2 + z^n-3 + ... + 1. Subbing 1 into here is no problem, and yields n.

@@mattermonkey5204 That's the route 3blue1brown went, but I just went for l'hopitale right away. I proved this before watching his justifications. I'm glad that our approaches had some differences, because otherwise it might have not been worth making the video.

@@EpicMathTime Yeah fair enough. I think I've come across this exact limit before, tough in a different context, and did it with l'hopital's rule, and honestly I feel like this way is probably easier, since you don't have to notice polynomial divisibility stuff.

I have become more powerful than I ever imagined possible.

I don't know which 3b1b proof you watched but the proofs look completely independently written to me. They both phrased the problem with complex polynomials, but that's what literally anyone would do.

@@samhollins2076 3Blue1Brown doesn't even give a step by step proof of this in the video because it wasn't the main point of the video. He gives an outline of a proof, and Epic Math Time's proof doesn't even follow the outline. 😂

@@samhollins2076 Stupid me, I should have written effectively.

@@chemistro9440 He DOES give a proof

@@nanigopalsaha2408 I think if you asked any random mathematician to prove this, their proof wouldn't be any more different from 3b1b's proof than emt's is. Anyone capable of proving this would recognize the roots of unity right away, it's immediate. So implying that there was any kind of "stealing" is kind of ridiculous. If you ask two people on different sides of the world to prove that R isn't homeomorphic to R^2, their proofs will be "effectively" the same. Why wouldn't they be "effectively" the same? There's generally set reasons why something is true and they are both going to reference those reasons.

As someone who has taught several precalculus classes, I wholely agree with you! 😂

I took a double take, and I’m a high schooler going on to be a math major!

Yes, that's what 3blue1brown mentioned in his video. I don't think this is immediately apparent, though? I wrote this proof before watching his own justifications, so I honestly haven't gone through the algebra on that yet.

@@EpicMathTime Maybe it would be a little more apparent (at least to anyone who's taken calculus) if you rewrote it as (1-z^n)/(1-z) ;-)

If you think about it as akin to summing a geometric series, it can become apparent. Another way is to think of z^n-1 as a polynomial with z-1 as a factor by the Remainder Theorem. Nice proof by the way.

It should be apparent for those that can spit the zeroes of polynomials.

Introducing complex numbers to a question that doesn't include them to begin with to make the calculations more straight forward is a very human and "Unnatural "thing to do.

Ivar Ängquist When he says natural I think he means the way that will lead to the simplest proof

Yep. But for this proof, I found it somehow easier to extend the theorem with circle have a radius of r.

😂 my trig students last year totally agreed with you. One of them brought it up before class and the rest promptly looked up pictures.