The solution is out! So why do colliding blocks compute pi?
Added 2019-01-20 16:40:48 +0000 UTCFriends,
The solution is out! Well, the first solution at any rate, I plan on doing one more video after this outlining Galperin’s original approach, which is arguably more beautiful. I chose to show the one above first since I think it’s more conducive to a “discover fiction” style narrative, hopefully making the viewer feel they could have discovered it themselves.
I kind of like this format of splitting what would usually be one video into a short opener showcasing a phenomenon and a second more in-depth part 2. So far, the response to the initial video in this last week has been exceedingly positive. One interesting metric here is not just views, but the like-to-view ratio, which is much higher on the first video here than for most. Of course, I think this has much more to do with the topic than the runtime or format, but I do think there’s something to be said for people having a better experience if they’re left wanting more (so long as more actually comes).
My thanks, as always, for your continuing support. I hope you enjoy!
-Grant
Comments
You might notice that I sprinkled in some examples with powers of 4, just so that curious viewers like you might check. 16 gives 12 collisions, 1100. 64 gives 25 collisions, 11001. You might guess what 256 gives :)
3blue1brown
2019-01-25 17:04:24 +0000 UTCabsolutely!
Daniel Armesto
2019-01-24 21:07:03 +0000 UTCthe process to compute π with collisions required that the big block is a power of ten. The π spit out by the process is also in base 10. Could be some relation between the base of the big block and the base of π produced. For example if the bigger block is a n-th power of 2, (counting collisions in base 2) can be produced 11.00100 10000 11111 .... (π base 2) ?
2019-01-24 15:11:01 +0000 UTCFeels like a missed opportunity not to solve the inequality equation and show n=floor(10^d * pi) (approximately).
2019-01-24 04:34:33 +0000 UTCSo was Buckminister Fuller mistaken? "mother nature does not know how to calculate pi, she doesn't have time to":?)
2019-01-22 04:55:35 +0000 UTCWhat's interesting, and this is mentioned in the Galperin paper, is that this will actually break if at some point a sufficiently large number of digits in pi are all 9's. I can't remember exactly what it is, but it's something like if at any point, among the first 2n digits of pi, the second n of them are all 9's. This certainly seems unlikely! But it's shockingly hard to prove.
3blue1brown
2019-01-21 18:32:11 +0000 UTCWell... I tried to find if arctan is ever far enough from x to last digit to be incorrect... which resulted in inequality which is definitely always false (basically it was 1 minus very small number (for x close to 0), has to be less than 0) P.S. in O notation I suppose the inequality was: 1 - O(x) < 0 which is always false for x -> 0
Timur Sultanov
2019-01-21 04:12:02 +0000 UTCGlad you enjoyed it!
3blue1brown
2019-01-21 03:03:48 +0000 UTCwow, great video Grant! I was very excited for this one and I'm actually really excited for the next one. I was introduced to this phenomenon by Numberphile but I was left asking "but why though?" so when I saw the first video about how you're gonna cover this, I got very excited because you're always able to break these things down in an elegant way, like no other.
kendall
2019-01-20 23:46:33 +0000 UTCI loved that moment too! Forgetting things can be the greatest source of creative insight.
3blue1brown
2019-01-20 22:57:30 +0000 UTCThat's great. Any opportunity to get them thinking in terms of phase space is worth doing.
Gabe
2019-01-20 19:45:32 +0000 UTCThere's a moment in STEM cell's video (linked from Grant's in the description) that I particularly love. He lacks the inscribed angle theorem, so he has to invent it on his own. At one point he makes a lovely argument with words and a picture (no symbols, or none that matter anyway) that reminds me of the kind of arguments that Lockhart espouses in Measurement. It happens around 10:45 <a href="https://youtu.be/ils7GZqp_iE?t=642" rel="nofollow noopener" target="_blank">https://youtu.be/ils7GZqp_iE?t=642</a> For more about this kind of mathematical argument, see Lockhart's Lament: <a href="https://www.maa.org/external_archive/devlin/LockhartsLament.pdf" rel="nofollow noopener" target="_blank">https://www.maa.org/external_archive/devlin/LockhartsLament.pdf</a> Further edit: rereading Lockhart's Lament I came across this quote (after he elicits a nice piece of reasoning from a student): "So a great project and a beautiful piece of mathematics. I’m not sure who was more proud, the student or myself. This is exactly the kind of experience I want my students to have." In other words, well done Grant!
John Rauser
2019-01-20 18:55:26 +0000 UTC
