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3Blue1Brown
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Comments by "Anonymous" (@Anonymous-df8it) on "3Blue1Brown" channel.
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Do the same thing but instead of the bit, you have -log base 6 of p(x).
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@AlienValkyrie No! I meant -log base 10 of 1/p(x)
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@bhaskarpandey8586 LOL! hospital rule!
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@NewWesternFront ???
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@NewWesternFront ???
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@jmkyarrow Black paper will eliminate the cost!
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@angelorondini5835 He would probably add flipbooks to the textbook. However, like he monetizes his content on Youtube, he would probably sell them separately (see the connection!)!
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@user-zp5yu6pg6r 'stale' is actually an anagram of 'salet' so you chose an optimal answer!
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I love how you crossed out the "1,000,000" and instead put "1,048,576". 🤣🤣🤣
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Why isn't this pinned? I laughed so hard at it!
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@elreyrey7537 3b1b, please can you make the video already? I have been waiting for 9 months and got silence. Please!!!!!!!!!!!!!!!!!!
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What! The best Wordle answer is actually salet? Also, what is the shortest game of Absurdle?
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@Nebulisuzer Should I delete my initial reply and this reply?
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LOL.
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@cheshire1 No. I meant log base 10 of p(x)
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@cheshire1 No no no. I meant -log base 10 of p(x)
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@3blue1brown If 2's complement is just p adic numbers, can we represent fractions without floating point?
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1/((1/x)+(1/y))=1/((y/xy)+(x/xy))=1/((y+x)/xy)=xy/(x+y) You're welcome!
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Lol! For a challenge, see if you can figure out what this is: (-b+-/2\b/2\-4ac)/(2a). Good luck!
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@mjtsquared Which shut down because PBS said so! That's one Parker infinite series you got there!
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@ontoverse Can you add more definitions to remove uncountable infinity?
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@BudgieInWA Right...
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This needs to be pinned!
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The 'unreasonable' results are perfectly explained in the 2-adic numbers!
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3b1b, do math, not meth!
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@Sean Francis Waters Lancaster And in 5d 1 in 32.
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Well 1/3 in 2-adic numbers would be ...1010101011. Because ...1010101011*3=1 if you do regular multiplication in binary!
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@tjejojyj Very slow! iilii
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5:46. Actually, there are infinitely many solutions to the unsolved problem. There is only so much energy you can cram into the 10^80 elementary particles inside the observable universe. Any further out and it would take longer than the existence of the universe for the particles outside that bound to have an effect. But, then, you can substitute final radius as initial radius forever. And thus, we have proven that there is only a finite amount of energy to deal with for any finite time (the reason is, in a finite radius, there is only a finite volume despite what the TARDIS would have you believe, and you just can't cram an infinite amount of energy over a finite volume). There cannot be a state of infinite kinetic energy because you would run out of energy in other forms before you get to the undesired state. Thus, the answer you were looking for is yes!
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He should make ASMR...
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I read e^x as ex lol!
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No. It was actually 3.08 minutes
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That sure is a Parker idea to follow!
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Pls do intuition for matrices and vectors with non-real inputs. Also, what do things like non-real determinants and such even mean?
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Why does he do powers of 2 instead of powers of 10 like a normal person would?
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Except it requires 1! So you can plug the equation into the equation: (1/2+1/4+1/8...)/2+(1/2+1/4+1/8...)/4+(1/2+1/4+1/8...)/8...=1. And then you can plug THAT equation into itself, and so on forever...
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@sodiboo Like absurdle, but you still only have 6 guesses!
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👍👍👍
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The real question is why he explained e^ipi in 3.08 minutes instead
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This needs to be pinned!
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Yeah, in general the probability is 2^-n in n dimensions!
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@paulzapodeanu9407 LOL!
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liar
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No it wouldn't. Because the polar co-ordinate for e^(a+bi) is (b, e^a).
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Welp. I guess Wolfram Alpha has rounding errors!
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Which is a bad thing apparently, because he hates people studying art!
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