Proving Grant's Little Theorem | A Surprising Geometric Result

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Epic Math Time

Epic Math Time

Күн бұрын

Пікірлер: 208
@EpicMathTime
@EpicMathTime 4 жыл бұрын
Big thanks to Dashlane for sponsoring this video! I've used Dashlane for a very long time (it came pre-installed on my Sager laptop) and it is mega convenient. It's one of those things that I can't imagine going without. Use our link bit.ly/2FuimEs to get your 30 day free Premium version, and remember that by supporting our sponsor, you're supporting the channel too!
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@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. 😅
@johubify
@johubify 4 жыл бұрын
HELP FLAMMABLE MATHS PLEASE! HIS CHANNEL HAS BEEN HACKED
@zachstar
@zachstar 4 жыл бұрын
Love seeing Jon help out small creators like 3b1b. Great video man! Oh and first!
@EpicMathTime
@EpicMathTime 4 жыл бұрын
I only did it because I see strong potential in his channel. He'll be a big success someday, calling it now.
@isaacstamper7798
@isaacstamper7798 4 жыл бұрын
XD
@roygalaasen
@roygalaasen 4 жыл бұрын
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.
@Cat-yz1tk
@Cat-yz1tk 4 жыл бұрын
@@EpicMathTime isn't he a succes now? i mean, many people see him as the best math channel
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@Cat-yz1tk That's the joke.
@cinvhetin8054
@cinvhetin8054 4 жыл бұрын
This is such an elegant proof, I love it. Also, I might go and check out this "two red three green" guy you mentioned!
@upliftingcommunity2465
@upliftingcommunity2465 4 жыл бұрын
What I love about this proof is how it branches across so many areas! It uses complex numbers, Euler’s formula, derivatives, limits, algebra, geometry. Super cool
@Jack_Callcott_AU
@Jack_Callcott_AU 3 жыл бұрын
And L'Hopital's rule.
@chemistro9440
@chemistro9440 4 жыл бұрын
I think you might literally be the most underrated/underwatched/undersubscribed channel on youtube
@fiNitEarth
@fiNitEarth 4 жыл бұрын
I freaking LOVE this video! I'm gonna use this as a prime example when my class mates when they're starting to complain again about the "uselessness of maths". You first have to convert your problem into a arithmetic expression and THEN use those tools given to you..
@KienPS257
@KienPS257 4 жыл бұрын
1) The theorem is true for n=2 and n=3. For n=1, there will be only 1 point on the circle, so there isn't any length here (or the length would just be 0). Therefore, the theorem is valid for all n greater or equal to 2. 2) On my guess, if the radius of the circle is r (r >= 1), then the product of those lengths will be n*r^(n-1) (with n >= 2) Proving this is similar to your proof. Drawing the circle on the complex plane, with the center is at the origin, the length between the center and any point on it is r*e^((2*k*pi*i)/n) for k runs from 1 to n (see 7:20). Then shift the circle by r, the product of the lengths will be |r*e^((2*k*pi*i)/n) - r| with k starts from 1 to n-1 (not counting the point sits on the origin), and we want this to be equal to n*r^(n-1). By factoring r out, we will get r^(n-1) appear on the left-hand side, and canceling that factor on both sides, we will eventually back to the formula that you've already proved in the video. 3) First, the length of any 2 points A, B sits on the circle (O is the center) is 2*sin(theta*pi), where theta is the angle between OA and OB, measured in radian. So if I have n points evenly-spaced on the circle, the length between the point you choose to any other points on it would be 2*sin(k*pi/n) (k runs from 1 to n-1), and this result is algebraic (You can find this fact on the internet) for any k and n.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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!
@KienPS257
@KienPS257 4 жыл бұрын
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)
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@KienPS257 Right, there are no lengths, so this is an empty product, which is 1. This is the trivial case.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
Sorry for sounding like Chucky Finster in some parts, I got mega sick and had to wait a while to finish recording this. 😅
@jonahdembe
@jonahdembe 4 жыл бұрын
How am I just now finding this channel? This was so good.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
Thanks!
@GhizLob
@GhizLob 4 жыл бұрын
Great video. Just wanted to add that you can actually avoid having to deal with the modulus (up to a +- sign) as the product is already a real number: taking the complex conjugate will only invert the order of the roots of unity.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
This is a good observation, thank you!
@aokkishi5064
@aokkishi5064 4 жыл бұрын
Thats really cool, just a small detail: Instead of taking the limit you could have used the a^n - 1 formula and you'll get a sum and bim you finished haha
@awabqureshi814
@awabqureshi814 4 жыл бұрын
Thanks for shouting out a small math KZbinr!
@nanigopalsaha2408
@nanigopalsaha2408 4 жыл бұрын
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...
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@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.
@nanigopalsaha2408
@nanigopalsaha2408 4 жыл бұрын
@@EpicMathTime Oh. HEHE😃
@brandonklein1
@brandonklein1 4 жыл бұрын
@@nanigopalsaha2408 also, 3b1b has about 230 times as many subs, not 2.
@RanEncounter
@RanEncounter 4 жыл бұрын
@@brandonklein1 He said more than twice not twice, so he is correct.
@Jack_Callcott_AU
@Jack_Callcott_AU 3 жыл бұрын
This is one of the best maths videos I have seen. Bloody fantastic.
@MichaelRothwell1
@MichaelRothwell1 4 жыл бұрын
A nice proof and a great video. You could avoid limits and l'Hopital by using (z^n-1)/(z-1) = 1+z+z^2+z^3+...+z^(n-1) [sum of a geometric series] and then setting z=1 on right gives n.
@kazebaret
@kazebaret 4 жыл бұрын
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.
@camrouxbg
@camrouxbg 3 жыл бұрын
I like how carefully you go through this. It is nice and methodical, showing each step. Great for high school students and beyond. Nice work.
@Lance.2451
@Lance.2451 4 жыл бұрын
Wow the step up in production is looking amazing lately
@EpicMathTime
@EpicMathTime 4 жыл бұрын
still recording everything on a 2.5 year old android phone 😅
@toaj868
@toaj868 4 жыл бұрын
Verification for n=3. The points form an equilateral triangle. The centroid and circumcenter of this are the centre of the circle. The centroid divides the median/altitude (they are the same in an equilateral triangle) into a 2:1 ratio. The radius is thus 2/3 of the altitude. Therefore the altitude has length 3/2. The length of altitude=(sqrt(3)/2)*side. Therefore the length of each side is (3/2)*(2/sqrt(3))=sqrt(3). The product we want is the product of two of the sides=(sqrt(3))^2=3.
@ranjitsarkar3126
@ranjitsarkar3126 3 жыл бұрын
3b1b. You know, one of those SMALL youtubers with potential
@hassanalabdullah8814
@hassanalabdullah8814 4 жыл бұрын
The editing in this channel is insane
@perappelgren948
@perappelgren948 4 жыл бұрын
Love that magic buzz at 12:53 when taking limits!
@edwardus12
@edwardus12 4 жыл бұрын
I regret every moment of not knowing about this chanel, you're awesome
@berserker8884
@berserker8884 4 жыл бұрын
Got to the final expression myself pretty easily, but I would never try using the limit without you mentioning it. Maybe because I haven't had a complex analysis course yet... That comes in a few weeks tho so hypeee!
@braedenlarson9122
@braedenlarson9122 4 жыл бұрын
I taught myself about the roots of unity again for like 4 hrs yesterday just to understand this proof... honestly thank you for reminding me I forgot it! :)
@itaycohen7619
@itaycohen7619 4 жыл бұрын
Came here from flammable maths , and i gotta say this channel is remarkable , your vids are awesome bro , so underrated.
@toaj868
@toaj868 4 жыл бұрын
Solution to 2.] The unit circle is similar to a circle of radius r where the scaling factor is r. The length of each line has thus been scaled by r. So their product will be 2*r^(n-1) (since each line is scaled by r and there are n-1 lines).
@kristinacollins
@kristinacollins 3 жыл бұрын
If there has to be an ad interrupting all my math videos, I appreciate that at least that ad is Joe Rogen staring in wonderment that some infinities are bigger than others.
@mishikookropiridze
@mishikookropiridze 4 жыл бұрын
Nice Proof
@mickeeyyy
@mickeeyyy 4 жыл бұрын
Heck yeah! Papa Jon is back! 👏
@NonTwinBrothers
@NonTwinBrothers Жыл бұрын
Additionally I've been told after stretching the Y axis of the unit circle and points by sqrt(5) the product becomes N*Fib(N) How cool is that? :)
@gustavfranklin
@gustavfranklin 4 жыл бұрын
That's a really nice proof!
@Felixkeeg
@Felixkeeg 4 жыл бұрын
Checking this channel because of your uncle. Staying here for the math and 90s cgi background
@Achrononmaster
@Achrononmaster 4 жыл бұрын
If you change radius r you get lengths: n*r^(n-1) --- for the products. The sums of the vectors (z_k-r) do not change and still sum to ±n alternating. Here's a check for r=2...5, for n:1 thru 4 do ( print("n=",n), for r:2 thru 4 do (P:product(abs(r*exp(2*%pi*k*%i/n)-r),k,1,n-1), print("product for r=",r," is: ",float(realpart(P))), S:sum(exp(2*%pi*k*%i/n)-1,k,1,n-1), print("sum for r=",r," is: ",float(realpart(S)))) ); Output: n= 1 product for r= 2 is: 1.0 sum for r= 2 is: 0.0 product for r= 3 is: 1.0 sum for r= 3 is: 0.0 product for r= 4 is: 1.0 sum for r= 4 is: 0.0 product for r= 5 is: 1.0 sum for r= 5 is: 0.0 n= 2 product for r= 2 is: 4.0 sum for r= 2 is: - 2.0 product for r= 3 is: 6.0 sum for r= 3 is: - 2.0 product for r= 4 is: 8.0 sum for r= 4 is: - 2.0 product for r= 5 is: 10.0 sum for r= 5 is: - 2.0 n= 3 product for r= 2 is: 12.0 sum for r= 2 is: - 3.0 product for r= 3 is: 27.0 sum for r= 3 is: - 3.0 product for r= 4 is: 48.0 sum for r= 4 is: - 3.0 product for r= 5 is: 75.0 sum for r= 5 is: - 3.0 n= 4 product for r= 2 is: 32.0 sum for r= 2 is: - 4.0 product for r= 3 is: 108.0 sum for r= 3 is: - 4.0 product for r= 4 is: 256.0 sum for r= 4 is: - 4.0 product for r= 5 is: 500.0 sum for r= 5 is: - 4.0
@Achrononmaster
@Achrononmaster 4 жыл бұрын
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).
@roygalaasen
@roygalaasen 4 жыл бұрын
I am not sure it was mentioned at all, but I am assuming that all the points are equally spaced around the circle, or does the same hold if you pick points on the circle at random? I would imagine this falling apart without this requirement, but I might be wrong? Edit: from the formulas it does seem like a requirement. Edit2: I also see a way to get around this, but that would potentially make n points become up to n^n points in worst case.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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?
@roygalaasen
@roygalaasen 4 жыл бұрын
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!
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@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?
@roygalaasen
@roygalaasen 4 жыл бұрын
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.
@destroctiveblade843
@destroctiveblade843 4 жыл бұрын
@@roygalaasen if you want to distribute the points arbitrarely you can just add an arbitrary phase shift
@nocrates0408
@nocrates0408 4 жыл бұрын
Wooow what a quick way to get the hyp's for the 4 quadrants
@matron9936
@matron9936 4 жыл бұрын
That’s very cool!
@ChrisSutherlandPhys
@ChrisSutherlandPhys 4 жыл бұрын
Lookin like such a boss in that arm chair 👑
@EpicMathTime
@EpicMathTime 4 жыл бұрын
big daddy energy
@SakisStrigas
@SakisStrigas 4 жыл бұрын
Great as always my friend!
@Invalid571
@Invalid571 4 жыл бұрын
Excellent video & proof! 👏 👏 ☺ Also: subscribed Edit: lateralus outro -> biiig pp energy
@vtvtify
@vtvtify 4 жыл бұрын
2. suppose we stretch the plane by a factor of k, namely, every length in the plane whose value was I, would be k * I. Well, when applying such a transformation the radius of our original circle becomes k, and every distance out of the n-1 distances in our product is multiplied by k, which implies their total product is the original one, n, times k^(n-1) In conclusion, for a circle of radius k, the desired product is: n*k^(n-1) ⬛
@cycklist
@cycklist 4 жыл бұрын
Wonderful. Thank you.
@contaantiga5397
@contaantiga5397 4 жыл бұрын
Thanks to your chanel now I can do the two things I like the most at the same time. Learn math, and listen to Tool
@eliyasne9695
@eliyasne9695 4 жыл бұрын
I like the proof, well done. 👍
@lorenzodeiaco8902
@lorenzodeiaco8902 4 жыл бұрын
this is amazing, AWESOME video!!!!
@samhollins2076
@samhollins2076 4 жыл бұрын
sooo... you're playing a "character" now? in a spaceship traveling through abstract math space? And how was your DMT experience?😂 not hating at all, it's an awesome idea! keep going man!
@josejavierotazu4788
@josejavierotazu4788 4 жыл бұрын
The fact that everyone in the comment section understood the proof except for me make me feel stupid, I guess this is how my friends feel when I start talking about math and physics😆
@tatjanagobold2810
@tatjanagobold2810 4 жыл бұрын
Oh is that tool in the background music?
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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.
@tatjanagobold2810
@tatjanagobold2810 4 жыл бұрын
@@EpicMathTime stoked to hear that! This band is my jam 😍 and their songs can really be influential if you think about them deeply
@AV-ws2rz
@AV-ws2rz 4 жыл бұрын
@@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?
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@AV-ws2rz Let's do it brah
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@Ethan Beyak Not any one song specifically, and I don't consider the fibonacci use in Lateralus as important as its actual message (that I personally get from it) or its lyrics. My favorite from the album is Parabola, it almost seems too sacred for me to use it in my videos lol. Man, I really am the most stereotypical tool fan of all time. 😂
@mustafaa3370
@mustafaa3370 4 жыл бұрын
You should make a whole video on the beauty that is Dummit and Foote
@EpicMathTime
@EpicMathTime 4 жыл бұрын
Dummit & Foote is my #1 favourite textbook, I could definitely grab it and make a few series from it.
@hydraslair4723
@hydraslair4723 4 жыл бұрын
@@EpicMathTime please make a video on the algebra of tensors from D&F! My favourite section of the book!
@CoughSyrup
@CoughSyrup 4 жыл бұрын
Frickin' Derek, man... Screw that guy.
@Entropize1
@Entropize1 4 жыл бұрын
Nice! What's your academic background?
@TechToppers
@TechToppers 3 жыл бұрын
Isn't there any proof by similarity of triangles?
@Aj-ch5kz
@Aj-ch5kz 4 жыл бұрын
Are you writing everying in reverse like its mirror image for us to see!! That's REALLY impressive.
@kvandermeersch5753
@kvandermeersch5753 4 жыл бұрын
He just mirrors it when editing... Is this bait?
@EpicMathTime
@EpicMathTime 4 жыл бұрын
I flip it but the spaceship and holograms are real
@kvandermeersch5753
@kvandermeersch5753 4 жыл бұрын
Epic Math Time Dang. I want one of those glowing π-shirts
@plaustrarius
@plaustrarius 4 жыл бұрын
I'm about it good stuff man
@Achrononmaster
@Achrononmaster 4 жыл бұрын
You missed the best associated results!!! If you sum the vectors sum(exp(2*%pi*k*%i/n)-1,k,1,n-1) (so without abs value) you also get "n" or to be exact, "-n". If you take the products of (z_k-1) without abs values you get ±n alternating. There's an exception for n=1 where sum(...)=0. Here's a line in Maxima to check it for n=1 to 12: for n:1 thru 12 do (P:product(exp(2*%pi*k*%i/n)-1,k,1,n-1), print([float(realpart(P)), float(imagpart(P))]), S:sum(exp(2*%pi*k*%i/n)-1,k,1,n-1), print( [float(realpart(S)), float(imagpart(S))]) );
@isaacstamper7798
@isaacstamper7798 4 жыл бұрын
Please upload more
@infinitedegreesoffreedom584
@infinitedegreesoffreedom584 4 жыл бұрын
My god this channel is so great...
@thatdude_93
@thatdude_93 4 жыл бұрын
I hear Rosetta stoned, I like
@thatdude_93
@thatdude_93 4 жыл бұрын
Btw you could have also factored out z-1 from z^n-1 and just prove it algebraically, without limits
@berserker8884
@berserker8884 4 жыл бұрын
@@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....
@nadiyayasmeen3928
@nadiyayasmeen3928 4 жыл бұрын
Amazing
@okoyoso
@okoyoso 4 жыл бұрын
proof. ACCESS DENIED BOI
@ahmedamraniakdi2143
@ahmedamraniakdi2143 4 жыл бұрын
What if you have as many points as the lenght of the circumference, Will you get 2*pi*r as the result of the product?
@EpicMathTime
@EpicMathTime 4 жыл бұрын
The length of the circumference is 2pi, so I'm not quite sure what you're asking.
@benjaminbrady2385
@benjaminbrady2385 4 жыл бұрын
Flammable Maths had his channel hacked. I was wondering if you could make a video about that to help him out or something
@Higgsinophysics
@Higgsinophysics 4 жыл бұрын
damn it's epic math time - but stronger
@EpicMathTime
@EpicMathTime 4 жыл бұрын
I have become more powerful than I ever imagined possible.
@Intrebute
@Intrebute 4 жыл бұрын
Every step of the proof feels justified to me, except one. What justifies taking limits? I get that every other algebraic step is in itself meaningful. "Since this equation holds, this other equation holds, thus this other equation holds". I don't get why suddenly putting everything in limits is meaningful. I'm sure it is, it just doesn't feel like it is. "This equation holds for all values but z = 1, so something something limit towards 1"?
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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.
@gdsfish3214
@gdsfish3214 4 жыл бұрын
@@EpicMathTime its a R E M O V A B L E S I N G U L A R I T Y
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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.
@non-inertialobserver946
@non-inertialobserver946 4 жыл бұрын
All Dereks gonna be like wtf bruh
@kornrade3967
@kornrade3967 4 жыл бұрын
I would have expected to keep the proof in the field of complex numbers, not go to limits... z^n-1 = = (z-1)*SUM(z^k) [for k = 0..n-1] (algebraic decomposition) z^n-1 = PROD(z-exp(j*2*k*pi/n)) [for k = 0..n-1] (factor product) ------------- SUM(z^k) = PROD(z-exp(j*2*k*pi/n)) [for k = 1..n-1] = [notation] PROD(Ek) [for k = 1..n-1] ABS(SUM(z^k)) [for k = 0..n-1] = ABS(PROD(z-Ek)) [for k = 1..n-1] = PROD(ABS(z-Ek)) [for k = 1..n-1] ------------- For z = 1 => 1+1+1+...+1 = n = PROD(ABS(1-Ek)) for k = 1..n-1
@upliftingcommunity2465
@upliftingcommunity2465 4 жыл бұрын
I was wondering if he was joking 😂 I was like.... that dude has 2.5 million subscribers or something ridiculous... Also Geaux Tigers!! I’m getting a minor in math major in Econ at LSU right now!
@nanigopalsaha2408
@nanigopalsaha2408 4 жыл бұрын
This is some real good effort, but your proof is effectively the same as Grant's proof. Not saying that you stole it or something, and the video was really great!
@samhollins2076
@samhollins2076 4 жыл бұрын
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.
@chemistro9440
@chemistro9440 4 жыл бұрын
@@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. 😂
@nanigopalsaha2408
@nanigopalsaha2408 4 жыл бұрын
@@samhollins2076 Stupid me, I should have written effectively.
@nanigopalsaha2408
@nanigopalsaha2408 4 жыл бұрын
@@chemistro9440 He DOES give a proof
@samhollins2076
@samhollins2076 4 жыл бұрын
@@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.
@tomctutor
@tomctutor 3 жыл бұрын
Bloody π-Demon from Alpha Centauri got to have a go at our 3B1B friend. This guy comes along wants to prove it with all sorts of complex wizardry; I mean if 3B1B says it is so then it is so, good enough for me! _Don't worry guys i've just subd to this dude and will be watching his shenanigans closely_ ﭢ.
@EpicMathTime
@EpicMathTime 3 жыл бұрын
π-daemon*
@tomctutor
@tomctutor 3 жыл бұрын
@@EpicMathTime Hope you dont mind my satire, I do like your maths vidz.
@dorukhan8707
@dorukhan8707 4 жыл бұрын
I am stuck at 9:32 , probably because I miss some knowledge, can anyone explain what happens afterwards, the thing with raising nth power equaling 1. Edit: I understood the whole other proccess. The video was neat man really thx for it :D
@EssentialsOfMath
@EssentialsOfMath 4 жыл бұрын
Noticed the beanie, are you from Louisiana?
@nevokrien95
@nevokrien95 3 жыл бұрын
Its sooooo easy to prove with complex polynomials
@MasterHigure
@MasterHigure 4 жыл бұрын
Limits (and in particular l'Hopital) is overkill. You have an equality of polynomials, and you can divide on both sides by common factors of polynomials and keep equality everywhere without worrying about the roots of those factors making the operation invalid.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
I'm not sure I follow how dividing by more factors would yield the result.
@MasterHigure
@MasterHigure 4 жыл бұрын
@@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).
@mattermonkey5204
@mattermonkey5204 4 жыл бұрын
@@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.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@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.
@mattermonkey5204
@mattermonkey5204 4 жыл бұрын
@@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.
@destroctiveblade843
@destroctiveblade843 4 жыл бұрын
Again excellent choice of music, rosetta stoned, fear inoculum, invincible are all some of my favourite tool songs. Anyway I wanted to ask if the l'hopital's rule still works with complex numbers, ik I could use real numbers here and avoid the problem but I am just curious
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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.
@destroctiveblade843
@destroctiveblade843 4 жыл бұрын
@@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 ?
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@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_ ).
@destroctiveblade843
@destroctiveblade843 4 жыл бұрын
@@EpicMathTime makes sence, idk about differential manifolds, I would guess it has to do with diffenrential functions, but thanks for the explanation
@EpicMathTime
@EpicMathTime 4 жыл бұрын
@@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.
@TopCuber
@TopCuber 4 жыл бұрын
You wouldn't have needed L'Hopital or limits if you used the geometric series expansion for (z^n -1)/(z-1)
@DRACOBUCIO
@DRACOBUCIO 4 жыл бұрын
Cool, man. Now I'm a suscriber :D .
@duckymomo7935
@duckymomo7935 4 жыл бұрын
Really good proof But I was hoping to be able to use Euclidean geometry
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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.
@i.i
@i.i 3 жыл бұрын
Ben 10 watch have no charge 😂
@arekkrolak6320
@arekkrolak6320 3 жыл бұрын
"small time math youtuber"? are you referring to 3 blue 1 brown or am i hallucinating? :) this is _the_ math channel on KZbin :)
@paradoxicallyexcellent5138
@paradoxicallyexcellent5138 4 жыл бұрын
A lot of high school students are gonna lose you at 10:15. In my experience they wouldn't be able to use the Fundamental Theorem of Algebra without a lot of guidance.
@EpicMathTime
@EpicMathTime 4 жыл бұрын
As someone who has taught several precalculus classes, I wholely agree with you! 😂
@theemptyset715
@theemptyset715 4 жыл бұрын
I took a double take, and I’m a high schooler going on to be a math major!
@Joe-bb4yi
@Joe-bb4yi 4 жыл бұрын
F for Derek
@olanmills64
@olanmills64 3 жыл бұрын
Outset Island 😊
@Safwan.Hossain
@Safwan.Hossain 4 жыл бұрын
5/7 would prove again
@nikolaalfredi3025
@nikolaalfredi3025 4 жыл бұрын
Hey, I love it....
@CousinoMacul
@CousinoMacul 4 жыл бұрын
Also, (z^n-1)/(z-1) = z^(n-1)+z^(n-2)+ ... +z+1 Now plug in z=1 and you get n
@EpicMathTime
@EpicMathTime 4 жыл бұрын
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.
@CousinoMacul
@CousinoMacul 4 жыл бұрын
@@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) ;-)
@ProfOmarMath
@ProfOmarMath 4 жыл бұрын
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.
@hybmnzz2658
@hybmnzz2658 3 жыл бұрын
It should be apparent for those that can spit the zeroes of polynomials.
@ARBB1
@ARBB1 4 жыл бұрын
This video feels like a Quake 3 level kek
@GeodesicBruh
@GeodesicBruh 4 жыл бұрын
Poor derrick
@Adam-wm4ys
@Adam-wm4ys 4 жыл бұрын
love u
@japalocoturbo
@japalocoturbo 4 жыл бұрын
Thank god I chose engineering... But it's fun to watch.
@trololollolololololl
@trololollolololololl 4 жыл бұрын
👀
@aashsyed1277
@aashsyed1277 3 жыл бұрын
WHO IS DEREK
@mantratrambadia3063
@mantratrambadia3063 4 жыл бұрын
I think grant already proved this.
@KienPS257
@KienPS257 4 жыл бұрын
Yep. But for this proof, I found it somehow easier to extend the theorem with circle have a radius of r.
@JacopoBerzeatti10
@JacopoBerzeatti10 4 жыл бұрын
Sugoi
@ranjitsarkar3126
@ranjitsarkar3126 3 жыл бұрын
Where does this guy live? In heaven or the world of forms??
@CrittingOut
@CrittingOut 4 жыл бұрын
Hypee
@loukafortin6225
@loukafortin6225 4 жыл бұрын
Magnus Carlsen 2.0
@EpicMathTime
@EpicMathTime 4 жыл бұрын
😂 my trig students last year totally agreed with you. One of them brought it up before class and the rest promptly looked up pictures.
@chair547
@chair547 4 жыл бұрын
Me: can we get papa flammy Mom: no, we have papa flammy at home Papa flammy at home:
@roberto_mella
@roberto_mella 4 жыл бұрын
@petargameplay2998
@petargameplay2998 4 жыл бұрын
:O
@evanev7
@evanev7 4 жыл бұрын
You can't apply l'Hopital's rule here. The limit is just the definition of the derivative of z^n at 1, so using l'Hopital just gives back the exact same limit. rEEEEEEEEEEEEEEEEEEEEE BAD ANALYSIS REEEEEEEEEEEEEeeeeeeeeeeeeeeeee (cool vid tho)
@ivarangquist9184
@ivarangquist9184 4 жыл бұрын
6:34 "This is definitely the natural way to look at this problem" I'm not really sure about that.
@ivarangquist9184
@ivarangquist9184 4 жыл бұрын
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.
@MrYesman43
@MrYesman43 4 жыл бұрын
Ivar Ängquist When he says natural I think he means the way that will lead to the simplest proof
@cykkm
@cykkm 2 жыл бұрын
Y U NO ANXENT GREEK? _Hypoteni_ is even more wrong than _octopi!_ Octopuses are lucky to be so flexible, and the only thing that saves the poor creature from broken spine when he's inflected so horribly is the fact that he doesn't have one. But if you inflect _hypotenuse_ so carelessly, it will just snap and shatter apart, making your proof null and void! _Octopi_ would make sense if _octopus_ were Romans, but they're in fact Greek, so that when they gather a-many, they're _octopodēs!_ But _hypoteni???_ From singular masculine _hypotenus???_ No waaaay!!! _Hypotenuse_ snuck into English directly from Greek via a 16c. book loosely based on a sloppy translation of Euclides' _Elements¹._ The word is a present tense active feminine participle _hypoteinūsa,_ lit. “stretching under,” but meaning “subtending” in context: Euclides used η ὑποτεινουσα πλευρα, ≈ _hüpoteinūsa pleura_ (with _ei_ like in “stage;” _eu_ like _au_ in “cow,” only _eu;_ _ū_ like _oo-o-o_ in “boo-o-om!” and _ü_ like _ew_ in “new”, only shorter and more fronted, or in German “über”), meaning “the subtending side”, or, in the original context, “the side subtending the right angle.” The translation lost the “side,” and used _hypotenuse_ alone in the same sense we use the word to-day. And the plural nominative of ὑποτεινούσα is... 🥁 drum roll please...🥁 ὑποτεινούσαι, _hüpoteinūsai._ Now, you can't, just can't disagree that “hewpotaynooosye” sounds cooler by a factor of about a million than “hypoteni!” 😎 Oh, and I love the name “Grant's little theorem!” This kinda fell through in most top comment threads. __________ ¹ In a book printed in 1571 under a curiously precise title: “A geometrical Practise, named Pantometria, diuided into three Bookes, Longimetra, Planimetra, and Stereometria, containing Rules manifolde for mensuration of all lines, Superficies and Solides: with sundry straunge conclusions both by instrument and without, and also by Perspectiue glasses, to set forth the true description of exact plat of an whole Region: framed by Leonard Digges Gentleman, lately finished by Thomas Digges, his sonne. Who hathe also thereunto adioyned a Mathematicall treatise of the fiue regulare Platonicall bodies, and their Metamorphosis or transformation into fiue other eqiulater uniforme solides Geometricall, of his owne inuention, hitherto not mentioned of by any Geometricians.”
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