Math Encounters - Massive Numbers
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- čas přidán 11. 09. 2024
- Every kid at some point wonders, "How much is a kajillion? Or a bazillion?" Even adults may have the impression that math is mostly about numbers and finding ways to deal with them, whatever their size. Join Po-Shen Loh, Carnegie Mellon math professor and national lead coach of the USA International Math Olympiad team, as he gives the inside scoop on whether and when numbers more gigantic than you've ever dreamed of really do turn up in the world of mathematics - and get a glimpse into the little-known drama of competitive math contests, as well.
what a delightful human being
STEM education needs more like him
Terrific teacher. CMU hit a home run getting him.
Astounding. Thanks for putting these online for free. Very inspiring!
Truly fantastic. I watched this all in one sitting and thoroughly enjoyed it.
Very amazing videos in this Math Encounters line. I like a good puzzle and the Olympiad question is definitly one.
1:08:50 the feeling of satisfaction when he says so I actually can make a tower of 2's whose height is that number there.
I was wondering if the strategy demonstrated for the IMO'10 problem was optimal. i'd love to think/know about the proof of bounds for the same.
Maharshi Ray yes i understand what your saying
the swap represents a recursive call am I incorrect?
This is so much fun!
The volume is too low . turned up volume on chromebook. Please correct.
I think I've watched this solution like 6 or 7 times. It's just mind boggling how big that number is.
this ith thooo good
isn't 2^2^2^2 = 256??
Yes, however that is not what he is calculating.
Think about it this way, you are squaring 2 and getting an answer (which is 4) and then squaring that number and getting another answer (16) and again to get 256. Which is correct.
He is squaring 2 by a 2 that is being squared by a 2. So for example. He is doing (2) to the power of (2^2) which is (2^4) which is 16. then is he putting that to the power of 16^2 which is 256 so its 2^256 which is the answer he got.
Hope this makes sense.
Oh dear I'm lost! 2^2 is 4 and 2^2^2 is (2^2)^2 = 4^2 = 16 and surely 2^2^2^2 =(2^2^2)^2 = 16^2 which is 256.
Is there some ambiguity in writing 2^2^2... ?
CockofDootie Thank you very much.
but it is still not clear, so what he did is 2^2 then 2^4 then 2^16??
Got it! 2^2^2^2 is not well-defined; it could mean
(2^2^2)^2 = 16^2 = 256 or
2^(2^2^2) = 2^16 = 65536.
It's really rather silly for an eminent mathematician to have written down an ambiguous expression! (note that 2^2^2 is not ambiguous as 2^(2^2) and (2^2)^2 have the same value 16)