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.
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
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.
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.
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.
@@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.
@@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.
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?
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."
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.
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.
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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
great video, i’m leaving a comment for the algorithm
Im replying for the algorithm. Thx for your repliable comment.
top lad!
Engagement for The Engagement God.
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.
Ugh, I remember learning "period 3 implies chaos" during my maths degree in the early 2000s, then promptly blanked on it during the exam!
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
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.
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.
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.
@@concavecuboid3253 Why is f(I) the intersection of all the f(I_k)?
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.
@@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.
@@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.
I am dying 😂 "'period 3 implies chaos' is basically mathematical click bait"
My mind is blown 🎉
Great video!
This is a very interesting video, and it's a shame that it doesn't have more views...
Such an intensive video, good work
have I got tricked into learning real analysis? you mf
good video
lol glad you liked it
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?
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."
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.
Amazing video! Keep up the good work!
Tasteful animation
8:15 Why is there a last argument? What if there are infinitely many such arguments, and no last one between them?
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
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.
Nice video!
Simple things can create chaos. Chaos can not be simple.
comment
Replay
0:42 collatz and tent things
binary shift chaot
2:17 insane functioj
5:23 intermediate val
5:34 baking
Sharkovsky's initials are АНШ in Russian, but ОМШ in Ukrainian.
0:30 "Continuous system" **shows discrete animation of discrete simulation** 😂
increase your framerate to infinite ;)
∀ r ∈ script, r = v 😂