For n colours, if you're guaranteed to have m colour-focused sequences of length 2 within a block of length b, that forces a new colour into the focus (or else there's a length 3 sequence) then by taking 2*((n^b)+1)) blocks of length b, you're guaranteed to have two duplicate blocks, each with the same m colour focused sequences of length 2 and same new colour in their focuses, within the first half, which means you also have the common focus of the cross-sequences formed by taking the first from the first block and second from the second block somewhere within the sequence, and that is also the third element in the arithmetic progression formed by the two beads of the new colour. So that shared focus is the focus of m+1 colour-focused sequences of length 2 within a much longer block. Since 2n+1 beads guarantees a repeat within the first n+1 beads, giving a single colour focused sequence of length 2 within the 2n+1 block, if you don't find a length 3 sequence early, you'll eventually find n colour-focused sequences of length 2 which forces there to be a length 3 sequence.
My first thought was "That upper bound is way too large to be practical, but I guess it does the trick to prove your point." A Graham's number kind of theorem. (It's not infinite, so at least there's that).
Alan Becker’s released A new video and I am absolutely discombobulated by the ending, with my main query being, “Why is Mr. Hat there?” (Not saying anything more for the sake of spoilers.)…
This takes me back a quarter of a century. I picked this module because Imre Leader was the lecturer, even though it was completely unrelated to all my other modules on geometry/topology. I really enjoyed it - particularly as I got to switch off from the other modules and think a different way. Knowing about ultrafilters actually came in useful once in my own research, when I was dealing with categories of infinite products of certain rings. On the other hand I have not thought about Van der Waerden's theorem since I was a student. The clear and engaging exposition in this video brings it all back to me - as good as Imre Leader himself.
Yes, any number when written out as a decimal will either terminate in an infinite number of zeroes, or have an infinite number of digits with no last non-zero digit. Either way, there must be some digit that repeats, evenly spaced, any given number of times within that infinite sequence (probably also appearing in between those evenly spaced occurrences). This is just cases where n=10 and k is whatever.
@@rmsgrey ah right that makes sense since given an infinite amount of digits there would just have to be patterns somewhere, however do you know if there is a limit to how big a pattern can be so for example in π could there be a part which goes 999999999999 or would it be impossible for 12+ 9's to repeat eachother?
@@lucidreconalt3229 That's a different question. If there is an infinite amount of digits, there will be arithmetic progression of *some* digit, but that doesn't mean it will be the digit 9. Just as in the video, there will be an arithmetic progression of some colour, but it may be that red only appears on position 10^k. In that case, there will be no arithmetic progression of red (of length greater than 2), just blue. Take for instance the number 4/11, with is .36363636... Lots of arithmetic progressions, but none with the digit 9. And none with length 1. Now, the if sequence 999999999999 appear in the digits of pi occurs as often as any sequence of a string of digits of the same length appears, then pi is normal, as that normal means that any sequence of digits occurs as frequent as any other sequence of digits of the same length. But while we know that most numbers are normal, for most numbers, we do not know whether they are. Including pi. It's expected to be normal, but not proven.
I like how you made a pretty complicated argument really intuitive. I learned something new today :) I also like how you quite naturally show that even for small cases you get these stupidly huge numbers. You gotta love the fact that one can make arbitrarily large finite numbers :D But when you discuss a theorem with a foreign name in a video, please make sure you look up how to pronounce the name. The IPA pronunciation is on Wikipedia. en.wikipedia.org/wiki/Bartel_Leendert_van_der_Waerden . The "ae" in Waerden is an old spelling for the "aa" sound.
![Why is this "Fundamental" to Arithmetic? #SoMEpi [5th]](http://i.ytimg.com/vi/46E0-XJuAXs/mqdefault.jpg)
Thanks for a fascinating and very well presented proof. I love the fact that there is a place on TH-cam for content like this.
This is presented so well
Most professors have awful communication skills.
Great video, Rebekah! 🎉
YO, ANIMATION VS GEOMETRY JUST DROPPED!!!
For n colours, if you're guaranteed to have m colour-focused sequences of length 2 within a block of length b, that forces a new colour into the focus (or else there's a length 3 sequence) then by taking 2*((n^b)+1)) blocks of length b, you're guaranteed to have two duplicate blocks, each with the same m colour focused sequences of length 2 and same new colour in their focuses, within the first half, which means you also have the common focus of the cross-sequences formed by taking the first from the first block and second from the second block somewhere within the sequence, and that is also the third element in the arithmetic progression formed by the two beads of the new colour. So that shared focus is the focus of m+1 colour-focused sequences of length 2 within a much longer block.
Since 2n+1 beads guarantees a repeat within the first n+1 beads, giving a single colour focused sequence of length 2 within the 2n+1 block, if you don't find a length 3 sequence early, you'll eventually find n colour-focused sequences of length 2 which forces there to be a length 3 sequence.
very interesting video, thanks!
You can find more videos made by interns working with Tom on the designated playlist here: th-cam.com/play/PLMCRxGutHqfn8YkbswsI3HDOaVQDUTZMp.html
Animation vs geometry reaction pls
Animation vs Geometry!!! Please
Are you going to react to Animation vs Geometry
Hey Tom you should react to animations VS Geometry!
My first thought was "That upper bound is way too large to be practical, but I guess it does the trick to prove your point."
A Graham's number kind of theorem. (It's not infinite, so at least there's that).
that was well presented and extremely interesting
Hey Tom, could you try the hsc maths extension 2 exam it’s Australias hardest highschool maths paper
Please react to Animation VS Geometry, they dropped another banger
Yo react to Animation VS Geometry pls continue this series
React to animation vs geometry! It came out today!
This has the looks of data compression. I wonder if there is a data compression algorithm that already uses something like this.
More than likely. It might not have a general name but private companies higher people with a background in optimization for this sort of thing.
Alan Becker’s released A new video and I am absolutely discombobulated by the ending, with my main query being, “Why is Mr. Hat there?” (Not saying anything more for the sake of spoilers.)…
This takes me back a quarter of a century. I picked this module because Imre Leader was the lecturer, even though it was completely unrelated to all my other modules on geometry/topology. I really enjoyed it - particularly as I got to switch off from the other modules and think a different way. Knowing about ultrafilters actually came in useful once in my own research, when I was dealing with categories of infinite products of certain rings. On the other hand I have not thought about Van der Waerden's theorem since I was a student. The clear and engaging exposition in this video brings it all back to me - as good as Imre Leader himself.
This gives a nice intuition into why Ramsey theory produces such crazily large numbers.
Waerden is pronounced as warden.
But what is warden pronounced as?
@@backwashjoe7864 with an a, and not as weirden.
Alan Becker posted "Animation vs. Geometry" to his channel, should be right up your alley
Fantastic job!
Can you react to Alan Becker animation vs geometry
Does this mean there has to be a monochromatic arithmetic progression somewhere in π?
Also maybe for primes aswell?
Yes, any number when written out as a decimal will either terminate in an infinite number of zeroes, or have an infinite number of digits with no last non-zero digit. Either way, there must be some digit that repeats, evenly spaced, any given number of times within that infinite sequence (probably also appearing in between those evenly spaced occurrences). This is just cases where n=10 and k is whatever.
@@rmsgrey ah right that makes sense since given an infinite amount of digits there would just have to be patterns somewhere, however do you know if there is a limit to how big a pattern can be so for example in π could there be a part which goes 999999999999 or would it be impossible for 12+ 9's to repeat eachother?
@@lucidreconalt3229 That's a different question. If there is an infinite amount of digits, there will be arithmetic progression of *some* digit, but that doesn't mean it will be the digit 9. Just as in the video, there will be an arithmetic progression of some colour, but it may be that red only appears on position 10^k. In that case, there will be no arithmetic progression of red (of length greater than 2), just blue.
Take for instance the number 4/11, with is .36363636... Lots of arithmetic progressions, but none with the digit 9. And none with length 1.
Now, the if sequence 999999999999 appear in the digits of pi occurs as often as any sequence of a string of digits of the same length appears, then pi is normal, as that normal means that any sequence of digits occurs as frequent as any other sequence of digits of the same length. But while we know that most numbers are normal, for most numbers, we do not know whether they are. Including pi. It's expected to be normal, but not proven.
I like how you made a pretty complicated argument really intuitive. I learned something new today :) I also like how you quite naturally show that even for small cases you get these stupidly huge numbers. You gotta love the fact that one can make arbitrarily large finite numbers :D
But when you discuss a theorem with a foreign name in a video, please make sure you look up how to pronounce the name. The IPA pronunciation is on Wikipedia. en.wikipedia.org/wiki/Bartel_Leendert_van_der_Waerden . The "ae" in Waerden is an old spelling for the "aa" sound.