Keep making videos, you are talented, you will get plenty of views after a while and help many people.
@maalls
6 күн бұрын
My mind is blown 🎉
@knowscope
8 ай бұрын
great video, ty
@jonathanlevy9635
9 ай бұрын
7:04 correct me if I'm wrong but I think there's a slight mistake with those examples. you can't get a vertex connected to an edge in Spermer lemma. otherwise you could have had a GBB triangle with two reds on his edges that prevent one turquoise edge and gives a counter example
@concavecuboid3253
9 ай бұрын
We are not actually proving Sperner's lemma, we are proving a slightly different result. We omitted some details, but because every line can have at most two colors on it, the overall statement still holds (I think, but it's been a while).
@dmitryvolovich4357
9 ай бұрын
8:15 Why is there a last argument? What if there are infinitely many such arguments, and no last one between them?
@dmitryvolovich4357
9 ай бұрын
And actually, if you look at a continuous function like x*sin(1/x), there are infinitely many points where it is 0 in any interval that contains 0
@concavecuboid3253
9 ай бұрын
The interval is closed. If there are infinitely many such arguments, then the sequence of points that all evaluate to the same value must have a limit point (and the problem you discuss is only an issue when ) but by continuity, the function evaluated at the limit point must also evaluate to that same value. xsin(1/x) is only continuous on [0, 1] if you define f(0)=0.
@ottolehikoinen6193
9 ай бұрын
This brings to mind that the ITCZ behavior is inherently chaotic and this might be one of the reasons how El Ninos are developed and to where the worse weather hits sometime afterwards. As the Northern and Southern hemisphere are more land or more ocean they have different style of responses that may delay or hasten the next cycle of ENSO. Thanks.
@SmileyMPV
9 ай бұрын
19:12 Isn’t it possible that xr*yb is equal to yr*yb? I don’t immediately see a fix for this issue.
@ianweckhorst3200
9 ай бұрын
With the double pendulum, I believe the time the two differ is when they hang together in the air, if you find an example like that in other chaotic systems, you could get very predictable behavior, although would likely be pretty boring
@RickyMud
9 ай бұрын
Tasteful animation
@trsarathi
10 ай бұрын
Simple things can create chaos. Chaos can not be simple.
@gabitheancient7664
10 ай бұрын
have I got tricked into learning real analysis? you mf good video
@concavecuboid3253
10 ай бұрын
lol glad you liked it
@Jaylooker
10 ай бұрын
Sharkovskii’s theorem at 15:24 motivates Bott periodicity being 2-periodic in stable homotopy theory. There are other periodic “v_n” maps of homotopy groups to consider as well.
@smiley_1000
10 ай бұрын
I think it would be interesting to consider the intersection I of all the I_k's. Since I is the intersection of a chain of compact sets, it's nonempty. Also we can see that f(I) is a subset of I. Consider x in I. Then for any k, x is in I_(k+1) and therefore f(x) is in I_k. Therefore, f(x) is in I, so we have that f(I) is a subset of I. I would also guess that we even have f(I) = I, but I'm unable to prove it.
@concavecuboid3253
10 ай бұрын
f(I)=I because f(I_k)=I_{k-1}. f(I) is the intersection of all f(I_k) which is the intersection of all l_{k-1} which is precisely I.
@smiley_1000
10 ай бұрын
@@concavecuboid3253 Why is f(I) the intersection of all the f(I_k)?
@concavecuboid3253
10 ай бұрын
Suppose x in I. Then, for all k, x in I_k, so f(x) is in I_k-1. Therefore, f(x) is in I. Suppose x not in I. Then, there is a k so that x is not in I_k. So f^-1(x) is disjoint from I_k+1. Therefore, f^{-1}(x) is disjoint from I and x is not in f(I). I think that I just consists of a single point with period 1.
@smiley_1000
10 ай бұрын
@@concavecuboid3253 I don't see what you're proving here. In the first part, you've proven the following claim: "Given x in I, f(x) is in I". This means that f(I) is a subset of I. In the second part, you've proven the following claim: "Given x not in I, x is not in f(I)". By contrapositive, this is equivalent to the claim "Given x in f(I), x is in I". This means that f(I) is a subset of I. So you've just proven the same claim two times. I'm not sure myself of how to prove that applying f to I will not make it smaller. Also, I don't think that I generally consists of a single point only. For example, if f is the identity function on some subinterval of [b, c], then our construction of the I_k's will stabilize, so we will keep getting the same subinterval, meaning that I will also be that subinterval instead of just a single point.
@smiley_1000
10 ай бұрын
@@concavecuboid3253 I think I figured out how to prove that I is a subset of f(I). Take x in I. Then, for all k, x is in I_k, meaning that there is y_k in I_(k+1) with f(y_k) = x. Now, (y_k)_k is a bounded sequence which contains a convergent subsequence by the Bolzano-Weierstrass theorem. Let y be the limit of that convergent subsequence. Then we have by continuity of f that f(y) = x. Additionally, since for any k, the terms of the sequence are contained in I_k from some point on, and I_k is a closed set, y is also contained in I_k. Therefore, y is contained in I. Therefore, for an arbitrary x in I, we have found y in I with f(y) = x. Therefore, I is a subset of f(I). Since we have already proven that f(I) is a subset of I, we can conclude that f(I) = I.
@bscutajar
10 ай бұрын
Ah yes, loud music and a mumbling voice. Perfect
@pierreabbat6157
10 ай бұрын
Sharkovsky's initials are АНШ in Russian, but ОМШ in Ukrainian.
@johnchessant3012
10 ай бұрын
Great video!
@federook78
10 ай бұрын
∀ r ∈ script, r = v 😂
@user-pr6ed3ri2k
10 ай бұрын
0:42 collatz and tent things
@user-pr6ed3ri2k
10 ай бұрын
binary shift chaot
@user-pr6ed3ri2k
10 ай бұрын
2:17 insane functioj
@user-pr6ed3ri2k
10 ай бұрын
5:23 intermediate val
@user-pr6ed3ri2k
10 ай бұрын
5:34 baking
@gametimewitharyan6665
10 ай бұрын
Such an intensive video, good work
@superawesomename5027
11 ай бұрын
Amazing video! Keep up the good work!
@juancristi376
11 ай бұрын
Nice video!
@spacelem
11 ай бұрын
Ugh, I remember learning "period 3 implies chaos" during my maths degree in the early 2000s, then promptly blanked on it during the exam!
@hvok99
11 ай бұрын
I am dying 😂 "'period 3 implies chaos' is basically mathematical click bait"
@blacklistnr1
11 ай бұрын
17:30 quite insane that Dee Dee's "2 steps forward, 1 step back" song from Dexter's Lab is at the heart of a chaos theorem
@TheOneMaddin
11 ай бұрын
This was very entertaining and non-trivial. Not the same old stuff that all the other math KZitemrs endlessly regurgitate (uh... oh... quaternions bruh, ... and ... and Fourier trafos dude).
@TheOneMaddin
11 ай бұрын
To the question: "why no other valuations?". Because there are no other valuations by some famous theorem whose name I forgot :D
@alexf2008
3 ай бұрын
Ostrowski
@TheOneMaddin
3 ай бұрын
@@alexf2008 I was thinking of Hadwiger. Is Ostrowski a generalization?
@TheOneMaddin
11 ай бұрын
The background music at times sounds like the Gravity Falls intro :D
@blacklistnr1
11 ай бұрын
0:30 "Continuous system" **shows discrete animation of discrete simulation** 😂
@concavecuboid3253
11 ай бұрын
increase your framerate to infinite ;)
@filipbanek1514
11 ай бұрын
great video, i’m leaving a comment for the algorithm
@ChaoteLab
10 ай бұрын
Im replying for the algorithm. Thx for your repliable comment.
@hammadhassan9125
10 ай бұрын
top lad!
@asailijhijr
9 ай бұрын
Engagement for The Engagement God.
@mightymoe333_
11 ай бұрын
Excellent video! I have not encountered Sharkovskii's theorem before. Can you explain why the logistic map for r=1+sqrt(8) does not appear to have higher order periods or chaos? Is the additional assumption of an unstable 3-cycle necessary?
@japanada11
11 ай бұрын
From the Wikipedia article on Sharkovskii's theorem: "Sharkovskii's theorem does not state that there are stable cycles of those periods, just that there are cycles of those periods. For systems such as the logistic map, the bifurcation diagram shows a range of parameter values for which apparently the only cycle has period 3. In fact, there must be cycles of all periods there, but they are not stable and therefore not visible on the computer-generated picture."
@concavecuboid3253
11 ай бұрын
If you set f(x)=rx(1-x) and graph y=f^5(x) and y=x, you will see that these lines intersect (apart from where f(x)=x). This means that there is a point with period 5. But the period 5 is unstable, so does not appear in the bifurcation diagram for the logistic map. The same thing is true for all other periods, which is why the bifurcation diagram only has 3 points on it.
@octosaurinvasion
11 ай бұрын
comment
@aweebthatlovesmath4220
10 ай бұрын
Replay
@bashirabdel-fattah9499
11 ай бұрын
This is a very interesting video, and it's a shame that it doesn't have more views...
@1ab1
Жыл бұрын
A subtle yet important detail is the that the triangles in the triangulation must meet edge to edge (the lemma isn't correct if this condition is dropped). The proofs themselves still remain correct but it is perhaps mislieading looking at the pentagon graph which its green-green-red triangle shares an edge with 2 triangles. Otherwise, really a masterpiece of a video from all perspectives!!! Definitly at least in my top 5 #some2 videos :)
@SmileyMPV
9 ай бұрын
6:09 assuming that lines cannot contain more than two colors Took me a while to catch this myself as he glossed over it bit this crucial remark makes the theorem correct
@JohnDlugosz
Жыл бұрын
12:34 you state that the property v(x)-0 <==> x=0 but you just said that v(p)=0 for all primes" 2,3,5,... are not zero. If I recall correctly from earlier, you only range x from 0 to 1. So, that would mean that the property applies in the domain of this usage. If that's right, you should review that point here rather than just handwaving over it. If that's wrong, it really needs explaining!
@concavecuboid3253
Жыл бұрын
I am not sure what you mean. Could you give me the time stamp where I say that "v(p)=0 for all primes" 2,3,5,... are not zero."? If p and q are different primes, than v_p(q)=1 because q is divisible by no powers of p, and p^(-0)=1.
@accountname1047
Жыл бұрын
Sperner's Lemma used in Monsky is in some sense is a discrete fixed point theorem
@taranknutson175
Жыл бұрын
Awesome
@WannesMalfait
Жыл бұрын
Great video. This is the first time I've seen this technique with valuations used. The resulting solution is very elegant! If you're planning on making more videos, I would suggest to lower the volume of the music. I found it a bit too loud and distracting during the video. Because the same music is played throughout the whole video, the structure of the video falls apart a bit. I would suggest only playing the music at transitions between segments or to use different music for each of the segments. That way the visual structure and the audio structure work together instead of against each other. Hope to see more from you!
@durian7551
Жыл бұрын
Wording at 7:42 confused me for a while. It says "sum of turquoise segments" when it means "number of exterior turquoise segments + 2 * number of interior turquoise segments". I got confused thinking of the following counterexample: G - B | \ | B - B The number of turquoise segments is 3 but the number of colorful triangle is 0, clearly not congruent mod 2.
@concavecuboid3253
Жыл бұрын
I should have been clearer about that. What's being counted is not actually the number of turquoise line segments, (which is 3 in your example) but the sum over all triangles of the number of turquoise segments on the outside of that triangle (which is 4 because the diagonal is on two triangles, so is counted twice).
@amaarquadri
Жыл бұрын
What a crazy proof! It's surprising that such a simple question has such a complicated answer.
@Fishelstix
Жыл бұрын
This proof is absolutely wild! Sperner's also plays a key role in Nash's Theorem (I guess Brouwer's Fixed Point Theorem in general). Crazy to see it connect like this with p-adics! Good job dude, thanks for sharing this proof with the world.
@drdca8263
Жыл бұрын
This is quite the ride. Very nice. I think I’ve typically seen the non-archimidean property here be called the “ultrametric inequality” iirc?
@SmileyMPV
Жыл бұрын
What an interesting problem and crazy proof, thanks for sharing!
@WhiterockFTP
Жыл бұрын
very cool video! :) - the music is a bit too loud for my taste
@JM-us3fr
Жыл бұрын
I think modern sources define the p-adic valuation as the n at 12:10 , while your definition is what is called a p-adic _absolute value_ .
@concavecuboid3253
Жыл бұрын
Sorry about that and thank you for catching it. I should definitely have referred to them p-adic norms or p-adic absolute values instead.
@JM-us3fr
Жыл бұрын
@@concavecuboid3253 No problem, there’s actually a correspondence between valuations and non-Archimedean absolute values (up to equivalence).
@NoNTr1v1aL
Жыл бұрын
Absolutely amazing video! Subscribed.
@concavecuboid3253
Жыл бұрын
Thanks!
@rohanvanklinken3029
Жыл бұрын
Fascinating! I would never have guessed that there was only one known way to prove that result. I love proofs that bring in maths that seems to be completely unrelated to the problem at first glance :)
@hujackus
Жыл бұрын
When this summer is over, we will all be p-addicts.
@uroscolovic1357
Жыл бұрын
Example in Sperner's lemma is not triangulation.
@concavecuboid3253
Жыл бұрын
This is a fair point. I think the ambiguity lies in calling Sperner's Lemma a result from graph theory, which was incorrect since the exact location of vertices does matter. I am instead using triangulation to mean a partition of a polygon into triangles in a geometric sense which means the result should still hold.
@AexisRai
Жыл бұрын
@@concavecuboid3253 what does "three different colored vertices" mean when you have a triangle with four vertices on it, as you do in the Sperner example?
@concavecuboid3253
Жыл бұрын
We only care about the 3 vertices on the triangle itself. The colors of the vertices on the edges don't matter.
@sachs6
Жыл бұрын
@@concavecuboid3253 divide each side of a triangle in 3 with 2 false vertices, colour all actual vertices red and, in each side, one false vertex green and one blue. The triangle is not colorful, yet it has an odd number of cyan segments. I think this lemma should respect graph theoretical triangles.
Пікірлер