So... at 17:45, You say that is never converges to any number, as it wanders around and never approaches any number? But why? Are there multiple accumulation points? Why it doesn't converge?? Then, (at 18:35) after talking for most of the video about these powers of primes summed together with primes themselves, we instead just jump straight to something else, just reciprocals of primes (so without these powers?? Were these only for that "intuition"? What was the purpose to talk about them for so long?), ignoring what we were doing for most of the video, to "measure that in slightly different way" and we get that it... suddenly converges to 0.9959..? But didn't Euler proved that sum of reciprocals of primes diverge to infinity? What exactly are we doing here? Are we taking these reciprocals with different signs, depending on the team that they belong to? Why it was just "hand-waved" and neglected, wasn't this supposed to be the main conclusion from the video? What exactly converges to that number: 0.9959.. ? This all was slightly chaotic, with the last fact being just barely "mentioned" in a matter of seconds at the end of the video, without much of explanation, so I would be glad, if someone could add up to what he said and explain it slightly better or fill in the holes (like that divergence at 17:45) :-) Thanks!
3 is also the oddest prime in that it's the only prime divisible by 3. AND it's actually odd. Also 5 is the oddest prime, in that it's the only prime divisible by 5. Oh, but we have a name for divisibility by 2, so it appears more special, when really, all the primes have the same generalized property
@@tolkienfan1972 There's only one instance of two consecutive primes though; 2 followed by 3. There are multiple instances, possibly infinitely many, of two primes being separated by two, known as twin primes. This, among other things, make 2 particularly unique. Multiples of two being easy to identify & think about certainly helps, but isn't the only thing that makes 2 odd.
Grant is so good at pulling abstract concepts down to earth. He's always asking the question "Ok, so this is definitely true, but how could I have thought about it to have guessed that before proving it?" It really gets at the heart of what makes math so fun!
There are very large stretches of primes where 1mod4 is in the lead: 17,624,031 primes in a row from prime 18,536,693,261 to prime 18,953,583,223 87,113,527 primes in a row from prime 1,491,324,485,141 to prime 1,493,766,257,567 89,473,585 primes in a row from prime 1,488,610,998,109 to prime 1,491,118,967,167 But then there is also a very long stretch from 19 billion (19,033,524,533) to 1488 billion (1,488,478,427,089) where 1mod4 is never in the lead. Edit: Found a much longer stretch of 1mod4 in the lead: 2,998,321,421 consecutive primes from prime 9,193,124,324,353 to prime 9,282,636,959,627
@@jonasba2764 Technically, maybe it'd just be easier to define numbers divisibly by two as odd and otherwise as even? Linguistically, the alternative - to redefine the adjective 'odd' as unremarkable and 'even' as bumpy - would only upset all English speakers, forever. It'd have to go to the UN or something.
@@LemoUtan Hmm, so then 2 would be the only odd prime, making it linguistically odd as well. I like your idea, now we just gotta convince everyone to make the switch :)
@@jonasba2764 Related is the theorem that there are no boring numbers. Because some numbers are interesting, like 1, or 2, and the first number to be boring instead of interesting - well, that's interesting, isn't it?
Grant really has something about him. He's the type of guy you want on your team.
Рік тому+11
Yeah, he has such a positive vibe, and he's amazing at explaining and analyzing things. I cannot imagine a single situation where a personality like this would be detrimental.
i just spoke to him and he said he would like to use that proof (which is apparently a pretty simple manipulation) to segue into a video about the prime number theorem and the riemann zeta function :)
A multiplicative series (like this one) can be factored in a clever way using prime numbers. The magic phrase you want to google for that is 'Euler product'. And log works nicely with products since log(ab)=log(a)+log(b). Then you write the logs themselves as a power series and the new series 3B1B gave pops out.
As a team 1 prime, I can confirm that we do live in a team 3 dictatorship, and that the composites are ready to help us revolt after grant mistreating them.
Thank you for clarifying the meaning of "almost" in this case, Grant. Probably and statistics have "almost certain" as a concept, which basically means certain. My own control field has a concept "almost strictly positive real", which are systems that can be made into SPR systems, where the strictly just means "not including zero" and the positive real mostly means "invertible". None of these are like the "almost" that humans mean, its fuzziness is a feature, outside of math.
Ok so I don't know if you meant to say "Probably" instead of 'Probability' in your comment, but it made me laugh. Thank you, I really needed that today
I loved watching numberphile back in my undergrad days. Even though I studied neuroscience, learning about topics in math such as imaginary numbers through this channel really helped strengthen my logic. Glad you guys are back!
It surprises me sometimes when 2 is singled out as a prime. Understandably, our definition of even vs odd is etched in our minds, but is it just because we gave it a name. We just separated the numbers into 2 buckets: 0mod2 and 1mod2. Or maybe we just defined the buckets as numbers with a factor of 2 and those without. If we had a name for numbers with factors of 3, would 3 be an odd prime as well?
I think it's also because 2 is small. In this case it needs to get singled out because 2 is the only prime that's 2 mod 4 since all numbers that are 2 mod 4 greater than 2 will have 2 as a divisor and so will not be prime.
I think what makes it feel special intuitively is that n mod 2 = 0 and n mod 2 != 0 are the same size, which is a unique property. Exactly how meaningful that is I don't know but I think it's what we latch on to
Another reason 2 is "odd" is that because it's so low, there are no natural numbers lesser than 2 that *aren't* divisors of it. When we mentally check whether a certain number is prime, we go over the numbers up to it making sure none of them divide our number. But the only numbers up to 2 are 1 and 2, both of which *do* divide 2 evenly, so it feels like it's only a prime number because there's no numbers to disqualify it with.
Also, ln(π/4) is negative, which corresponds to the 3's, so that gives the sense that when you tend to infinity, the 3's are the ultimate winner, even if they occasionally lose the lead.
@@josephs.webster6484 Sure, I am just a bit surprised that the OP attributes the win to one term, out of an infinity of them. That requires a better motivation imho.
You misunderstood how it works. The point is that there's a preference for the + bucket, but it doesn't diverge, so there must be a tendency for the - bucket to contain bigger numbers. It doesn't matter what number the series converges to.
chi(n) oscillates between 0, +1, 0, -1, so another way of writing it is sin(n pi/2). So sum( sin(n pi/2) / n ) tends to pi/4. Now, I asked myself what the continuous version of this sum would be. Turns out int( sin(n pi/2) / n ) over the positive reals is pi/2. Twice as much!
Cuz in our 3d space here.... xyz, right? The sections these planes create if considering the 0 axis coordinate, sphere, or the reason fir circle in 2d, the outer bound of sphere only need radian. And that radian also defines perfect circle, at any relative angle. It's the meridian of sphere
I find it interesting that not only will team 1 always win at least some of the time forever, it will always win by larger and larger amounts as you keep looking to higher numbers in order to keep the value at 0.005%. So eventually team 1 will lead for 20 billion primes in a row and that 20 billion will still only be 0.005%
Not necessarily. It could lead for 5 primes in a row then lose the lead for 2 primes in a row then take the lead again and repeat that 20 billion times
I've been waiting since November for this video, and whether it's true or not, I'm blaming Grant's amazing visuals for slowing down the production process. What a great collab. Thank you both.
These are fascinating videos. I don't consider myself a math person but I find these videos are just as compelling as an engrossing novel you just can't put down or a mesmerizing painting you can't stop staring at.
I think another reason you should have some intuition that there are "more" 4n+3 primes is that (4n+3)*(4m+3) is always some number (4k+1). However, the converse is not true. (4n+1) * (4m+1) is also 1 more than a multiple of 4. There are just more ways to get one more than a multiple of 4 than three more by multiplying these numbers, and this allows 4n+3 primes to stay in the lead for longer as their products "take up" more space in the numbers 4n+1, and less composite numbers can break into 4n+3 territory.
4:45 my simplest explanation for why is that prime squares can be 1 mod 4, but not 3 mod 4 that accounts for the difference since powers of primes are considered to be partially prime for theorems regarding the prime number theorem (see the Van Mangoldt function for example)
My initial thought was well, any even power of an odd number will be of the form 4n+1, so since primes occur in the first place they can kind of by definition, there's less "room" for primes of the form 4n+1 than for 4n+3
Always great to have Grant teaming up with Brady. I gotta say though, as a number theorist myself, the "weirdest possible way to take a logarithm" does such an injustice to the actual process happening behind the scenes. In reality, it's a completely natural way to express a logarithm. Granted, you have to understand the notion of a completely multiplicative function, an Euler product, and Taylor series, but I don't think these are concepts that would be challenging to the typical numberphile viewer. At the very least, the rigorous calculation (or enough details to demonstrate it) should be included for those familiar with these topics. That critique aside, I really liked the intuition given here.
Classical mathematician shortcut. Just remember this short, simple, absolutly uninuitive, 20 step instruction and then its totally easy to work out :-) Still the fact, that team 3 never gets in the lead is verry nice. Those problems nomaly have the tendency to go to 0 or 100% or infinety, or 50:50. Something you would expect. Thx for sharing this. great vid.
Ha. Well, we think it look complicated, but that's because our normal logarithm algorithm is to hit a button on a calculator or look it up in a log table. The _actual_ other log algs are not so easy!
Crucial point: ln(pi/4) is negative (since pi/4 < 1). That's the intuition for why there are more 3 mod 4 primes than 1 mod 4; even after putting all the even powers in the positive bucket when half of them "should" have been in the negative bucket, the total sum is still negative.
It doesn’t matter what the value of ln(π/4) is, so long as the series in question converges. The idea grant spoke about is that the sum of the positives go to infinity, and the sum of the negatives go to -infinity; however, added together they converge to a value. This is the big point - for every positive value we need a negative to stop the sum going to infinity. Why should there be more 3 mod 4? Well that’s because there are more positives as you get lots more from prime powers, which means you’re left with basically only the primes that are 3 mod 4 doing the negative business. Which means more primes 3 mod 4 to stop the sum going to infinity. This is overly verbose but I hope it explains that you don’t care what the sum converges to, just that it does and we have the positives sum to infinity and the negatives sum to -infinity
@@benp2291 The only existence here of the sum is not enough in the argument, the value is also important. The sum of reciprocal of squares of prime also converge, so for the team 3 lead intuitive 'argument' to work what matters is the fact that essentially: sum of reciprocal of squares of primes (+ some more) > ln(pi/4) + sum of reciprocal of (some) cubes of primes (+ some more) + the 'some' can be computed by looking at the chi function.
9:35 - What's shown on the screen is not "reducing it by one third", though. It's reducing it *_to_* one third (of its original value). Reducing it *_by_* one third would mean subtracting 1/3, and reducing it *_by_* one third of its original value would mean multiplying it by 2/3.
Two questions: 1. What is the LARGEST LEAD "Team 3" has? Is there an upper limit? 2. Who the heck comes up with these things? "Gee, I wonder what it would look like if you did this weird thing with prime numbers...."
That's a very interesting heuristic way of explaining the bias. Still though, all leads are temporary, and the asymptotic density of each of the residue classes is 1/2.
Fascinating video! I don't know if it's already been mentioned somewhere in the comments, but the fact that ln(π/4) < 0 lends even more credence to the theory being true on principle alone, as it requires even more tiny negative fractions to fill the bucket and tip the scales that direction.
Seriously, these are the kinds of friends I'd love to have. In my dream world, I live in a neighborhood where all my neighbors are guys like this - and their wives are doctors or something like that.
This makes sports rivalries quite interesting, in a way. If you just showed me the graph and asked me who's "better" I would say team 3 without thinking, but if you asked me at some point in time and I only were to take current form and recent results into consideration, it'd practically be a coin flip. If a "season" started at the beginning of the series and ended at a random point, team 3 would be 99.5% likely to win the title. If it started at a random point and ended at a random point, team 1 would win about half the time. Either this is mind-boggling or I'm completely wrong. I really can't tell, because I'm mind-boggled.
19:26 - "Does it ever get to the point where it'll be in the lead forever?" No. Because if it did, then the percentage of time that it's in the lead would be strictly increasing and approach 100%.
It seems clear that a math prodigy could spend an entire lifetime dissecting, testing, and understanding primes --- and never exhaust all the ways that they astound and confound us.
Brady's tortoise and hare comment at the end is interesting given that the two concepts meet in cryptography where you are modulo some prime and both its chi value and cycle length (found with Floyd or Brent's tortoise and hare method) are important.
My first intuition was: 2 numbers multiplied that are both 1 mod 4, result in another 1 mod 4 number. 2 numbers multiplied that are both 3 mod 4, result in a number 1 mod 4. Only if a 1 mod 4 and a 3 mod 4 number is multiplied, we get a 3 mod 4. So I thought if we think of the Sieve of Eratosthenes, there would be a higher chance that 1 mod 4 numbers get eliminated. But didn't think this through further...
That does not mean anything. 1 number being 1 mod 4 and the other 3 mod 4 is as likely than having both 1 mod 4 or both 3 mod 4 (if you flip a coin, you have 50% to not get the same result) You have 4 possivilities : both number you multiply are 1 mod 4 (the result is 1 mod 4), the first one is 1 mod 4 and the 2nd one is 3 mod 4 (the result is 3 mod 4), the first one is 3 mod 4 and the 2nd one is 1 mod 4 (the result is 3 mod 4) and both are 3 mod 4 (the result is 1 mod 4). So half of the time it will be 1 mod 4 and half of the time it will be 3 mod 4
@@quentind1924 well, yes but no. My first post is a bit misleading. The three sets don't have 1/3 chance each. But its also not a coin flip here. Lets say we have m primes 1 mod 4 already in the list, and n primes 3 mod 4. Then there are m(m-1)/2 + m = m(m+1)/2 combinations of 1 mod 4 primes ( m(m-1)/2 pairs different numbers and then there are m self-combinations (square of the primes) Same logic there are n(n+1)/2 combinations of 3 mod 4 primes Last, there are m*n combinations of different primes. m(m+1)/2+n(n+1)/2>(m^2)/2+(n^2)/2>=m*n So there are always more pairs that eliminate 1 mod 4 numbers than pairs that eliminate 3 mod 4 numbers.
In one video, you manage to touch on the prime numbers, pi, e (via ln), and the most mind-boggling way to calculate a natural logarithm ever seen. mind blown
The way the difference was plotted as a function with integer values being positive and occasionally negative gives a really nice way to think about this. The question with the answer of roughly 99.5 % (if I understood it correctly, but I think I did) is equivalent to asking what part of its values are positive. If the ratio of positive values to negative values was 1:1, the 3s would be in the lead half of the time - but that's not the case. But it gives me an idea for another, similar question. How does the integral of this function behave? Both from 0 to some finite bound X, and on the infinite scale (0 to infinity) - that one I think is most probably diverging given the heavy bias towards positive values, but it's not impossible for it to be finite-valued given this fact (meaning that you can have a function that's positive 99.5 % of the time and yet has integral equal to 0 - the negative areas just have to be deep enough to compensate for their rarity).
"Integral" of a discrete function? Do you mean a Σ sum, or can this be thought of as an actual integral in some [no pun intended] way? Anyway, I agree that it probably diverges, and the _lim n→∞_ is gonna be like -1/12.
Chebychev biases. Also, it seems like the twin primes conjecture should be tied to this somehow (as any pair of twin primes [ignoring the triplets 3, 5, 7] would contribute one to each of 1 mod 4 and 3 mod 4).
I find this topic analogous to entropy. How for most of the time everything is barreling towards chaos and disorder, we find these small bursts where order and complexity emerge, like in our own pocket of complexity we all live in.
I just want to say that, your videos always be able regain my interest in math when I feel burnout, it also help me to see problems from different perspective which is important for a math lover Wish you guys all the best!!
3:32 I think it's more of a bias in our language: we have a word for numbers that are divisible by 2 (even) and another word for numbers that are not (odd). Prime numbers have also a bias towards throdd numbers (3 is the only threeven number), 5 is the only quinary number, ...
Does this mean that for p>3 the sum of the reciprocals of the team 3 primes is the same as the sum of the reciprocals of the team 1 primes?, I.e. if you calculated the 'pseudo-alternating' sum 1/5-1/7-1/11+1/13+1/17-1/19-1/23 ... it'd converge to zero?
The theorem at 20:28 was not immediately clear to me, until I realized that 1/ln(x) . sum(1/x) converges to 1. So if you leave out the terms where team 1 wins, the sum is a little less.
Play the same game flipping a coin for some large number of flips. Team H and team T. Whoever starts out leading, will tend to hold the lead for the duration of the flips. Both will have about the same number of wins, but at any given moment in time, whoever started out leading will have a higher number of wins. And this is random, unlike the primes race.
i'm not sure i fully understand the way grant explained this. the fact that the even powers of primes always land in the positive bucket in my head doesn't necessiate that there ought to be more items in the negative bucket because the sum of all the inverses of the even powers of primes might just converge to some small number (
Being aware of what seems like a natural imbalance, have there been any deeper looks into the commonalities between the threshold points? Given the appearance of pi, plus the actual results data, it feels like there must be some periodic pattern on a spiral that guarantees a minority batch within some max interval, and whose duration is commensurate with magnitude of n. Anyway, I'm a first time commenter, but have watched and enjoyed both of your guys videos for a long time, keep up the great work!
I think this argument is not sound. Sum of reciprocals of powers of natural numbers larger than 1, even without dividing and with counting 6th power as a square and a cube etc. is finite (and equal to 1). So in fact the powers don’t contribute anything to the convergence, is all about the difference of reciprocals of 4k+1 and 4k+3 being small.
See all three videos in this series - Grant's Prime Pattern Trilogy: bit.ly/PrimePatternTrilogy
I KEEP TELLING YOU TO FIND THE ALLSPARK BEFORE PRIME DOES!!!
very interesting
So... at 17:45, You say that is never converges to any number, as it wanders around and never approaches any number? But why? Are there multiple accumulation points? Why it doesn't converge??
Then, (at 18:35) after talking for most of the video about these powers of primes summed together with primes themselves, we instead just jump straight to something else, just reciprocals of primes (so without these powers?? Were these only for that "intuition"? What was the purpose to talk about them for so long?), ignoring what we were doing for most of the video, to "measure that in slightly different way" and we get that it... suddenly converges to 0.9959..?
But didn't Euler proved that sum of reciprocals of primes diverge to infinity? What exactly are we doing here? Are we taking these reciprocals with different signs, depending on the team that they belong to?
Why it was just "hand-waved" and neglected, wasn't this supposed to be the main conclusion from the video? What exactly converges to that number: 0.9959.. ?
This all was slightly chaotic, with the last fact being just barely "mentioned" in a matter of seconds at the end of the video, without much of explanation, so I would be glad, if someone could add up to what he said and explain it slightly better or fill in the holes (like that divergence at 17:45) :-)
Thanks!
Why isn't this video not available in 3blue1brown ?
i don't think numbers should be called, 'prime' because it implies the others are not prime. How about indivisble or non-composite?
"2 is disquialified. It's the ODDEST prime as it's the only one that is EVEN". I really love that statement 🤩
3 is also the oddest prime in that it's the only prime divisible by 3. AND it's actually odd. Also 5 is the oddest prime, in that it's the only prime divisible by 5. Oh, but we have a name for divisibility by 2, so it appears more special, when really, all the primes have the same generalized property
@@tolkienfan1972 I feel like you missed the joke there
@@Tankem nope. I got it. My comment is an aside
@@tolkienfan1972 There's only one instance of two consecutive primes though; 2 followed by 3. There are multiple instances, possibly infinitely many, of two primes being separated by two, known as twin primes. This, among other things, make 2 particularly unique. Multiples of two being easy to identify & think about certainly helps, but isn't the only thing that makes 2 odd.
Just a thought: in base 12 divisibility by 3 would also have a name, and corresponding last digits: 0, 3, 6, 9.
Grant is so good at pulling abstract concepts down to earth. He's always asking the question "Ok, so this is definitely true, but how could I have thought about it to have guessed that before proving it?" It really gets at the heart of what makes math so fun!
There are very large stretches of primes where 1mod4 is in the lead:
17,624,031 primes in a row from prime 18,536,693,261 to prime 18,953,583,223
87,113,527 primes in a row from prime 1,491,324,485,141 to prime 1,493,766,257,567
89,473,585 primes in a row from prime 1,488,610,998,109 to prime 1,491,118,967,167
But then there is also a very long stretch from 19 billion (19,033,524,533) to 1488 billion (1,488,478,427,089) where 1mod4 is never in the lead.
Edit: Found a much longer stretch of 1mod4 in the lead:
2,998,321,421 consecutive primes from prime 9,193,124,324,353 to prime 9,282,636,959,627
How do we know this?
Thanks for sharing. I was wondering what the streaks looked like at somewhat higher numbers.
This suggests a conjecture about whether for any natural k, there is a stretch of length k where team 1 has the lead
@@samp-w7439 presumably someone has checked using a computer
@@tonygruff maybe not for every natural number but i wonder if they get arbitrarily long
0:40 "It's the oddest prime. [...] It's the only one that is even." What a peculiar sentence.
Haha that's a classic joke number theorists make. It really is quite odd to be even as a prime!
@@jonasba2764 Technically, maybe it'd just be easier to define numbers divisibly by two as odd and otherwise as even? Linguistically, the alternative - to redefine the adjective 'odd' as unremarkable and 'even' as bumpy - would only upset all English speakers, forever. It'd have to go to the UN or something.
@@LemoUtan Hmm, so then 2 would be the only odd prime, making it linguistically odd as well. I like your idea, now we just gotta convince everyone to make the switch :)
@@jonasba2764 Related is the theorem that there are no boring numbers. Because some numbers are interesting, like 1, or 2, and the first number to be boring instead of interesting - well, that's interesting, isn't it?
@@jonasba2764 Yup. Shouldn't take too long.
Grant really has something about him. He's the type of guy you want on your team.
Yeah, he has such a positive vibe, and he's amazing at explaining and analyzing things. I cannot imagine a single situation where a personality like this would be detrimental.
We need a 3B1B video on why that natural log trick works
He also showed this in his 7th lockdown-math video (at 7:12), but also with no explanation
i just spoke to him and he said he would like to use that proof (which is apparently a pretty simple manipulation) to segue into a video about the prime number theorem and the riemann zeta function :)
@@trobin What an insane flex - I just spoke to Grant Sanderson
A multiplicative series (like this one) can be factored in a clever way using prime numbers. The magic phrase you want to google for that is 'Euler product'. And log works nicely with products since log(ab)=log(a)+log(b). Then you write the logs themselves as a power series and the new series 3B1B gave pops out.
Let me know when anything related to this comes out
As a team 1 prime, I can confirm that we do live in a team 3 dictatorship, and that the composites are ready to help us revolt after grant mistreating them.
Perhaps you could vote to leave.
@@robertpearce8394 Primexit?
@@robertpearce8394 We tried that a few years back, they blocked all the exits and still are blocking them.
@@robertpearce8394 It's a prime time to leave.
😂😂😂
I have no joke been waiting for this video for 3 and a half months. So excited to watch this later today after rewatching the first 2 vids
Thank you for clarifying the meaning of "almost" in this case, Grant.
Probably and statistics have "almost certain" as a concept, which basically means certain. My own control field has a concept "almost strictly positive real", which are systems that can be made into SPR systems, where the strictly just means "not including zero" and the positive real mostly means "invertible".
None of these are like the "almost" that humans mean, its fuzziness is a feature, outside of math.
Ok so I don't know if you meant to say "Probably" instead of 'Probability' in your comment, but it made me laugh.
Thank you, I really needed that today
"Do you know someone who's good at making mathematical visuals?" - love it
I can't help but feel itchy by this missed chance, in the "teams" plot, to have made 3 "blue", 1 "brown" ...
you're absolutely right!
Grant Sanderson's (aka 3Blue1Brown) voice is just so soothing. I can listen to whatever this guy has to say.
Colab between 3b1b and Hannah Fry? 👀👂
I loved watching numberphile back in my undergrad days. Even though I studied neuroscience, learning about topics in math such as imaginary numbers through this channel really helped strengthen my logic. Glad you guys are back!
Back? They never left!
After months of waiting, right when I already start to believe the part 3 does not actually exist, here it comes ...
It surprises me sometimes when 2 is singled out as a prime. Understandably, our definition of even vs odd is etched in our minds, but is it just because we gave it a name. We just separated the numbers into 2 buckets: 0mod2 and 1mod2. Or maybe we just defined the buckets as numbers with a factor of 2 and those without. If we had a name for numbers with factors of 3, would 3 be an odd prime as well?
Yeah, sometimes you look at primes mod 6, and you have to single out both 2 and 3 as the special cases
I think it's also because 2 is small. In this case it needs to get singled out because 2 is the only prime that's 2 mod 4 since all numbers that are 2 mod 4 greater than 2 will have 2 as a divisor and so will not be prime.
I think what makes it feel special intuitively is that n mod 2 = 0 and n mod 2 != 0 are the same size, which is a unique property. Exactly how meaningful that is I don't know but I think it's what we latch on to
Another reason 2 is "odd" is that because it's so low, there are no natural numbers lesser than 2 that *aren't* divisors of it. When we mentally check whether a certain number is prime, we go over the numbers up to it making sure none of them divide our number. But the only numbers up to 2 are 1 and 2, both of which *do* divide 2 evenly, so it feels like it's only a prime number because there's no numbers to disqualify it with.
2 has a lot of other annoying properties. For example F_2 (the field with 2 elements) is a counterexample to pretty much everything
Also, ln(π/4) is negative, which corresponds to the 3's, so that gives the sense that when you tend to infinity, the 3's are the ultimate winner, even if they occasionally lose the lead.
Hmm, but that constant is tiny if not negligible compared to all the terms that have even powers of primes...
@@josephs.webster6484 Sure, I am just a bit surprised that the OP attributes the win to one term, out of an infinity of them. That requires a better motivation imho.
But that’s from adding and subtracting the actual values of the reciprocals of the primes, not from adding or subtracting 1 for each prime.
You misunderstood how it works. The point is that there's a preference for the + bucket, but it doesn't diverge, so there must be a tendency for the - bucket to contain bigger numbers. It doesn't matter what number the series converges to.
Hearing Grant saying susses makes my life complete
chi(n) oscillates between 0, +1, 0, -1, so another way of writing it is sin(n pi/2). So sum( sin(n pi/2) / n ) tends to pi/4.
Now, I asked myself what the continuous version of this sum would be. Turns out int( sin(n pi/2) / n ) over the positive reals is pi/2. Twice as much!
Cuz in our 3d space here.... xyz, right? The sections these planes create if considering the 0 axis coordinate, sphere, or the reason fir circle in 2d, the outer bound of sphere only need radian. And that radian also defines perfect circle, at any relative angle. It's the meridian of sphere
YES, FINALLY! I waited so long for this.
I find it interesting that not only will team 1 always win at least some of the time forever, it will always win by larger and larger amounts as you keep looking to higher numbers in order to keep the value at 0.005%. So eventually team 1 will lead for 20 billion primes in a row and that 20 billion will still only be 0.005%
Not necessarily. It could lead for 5 primes in a row then lose the lead for 2 primes in a row then take the lead again and repeat that 20 billion times
11:16
One small thing to note is the "logarithm" in this sense only works for a completely multiplicative function, like χ(n) or 1/n^2.
Is there a name for this type of logarithm, so I can go look it up later?
I think he did cover that a bit earlier, at 7:23
I've been waiting for a new upload between NP and Grant for way too long! I'm glad it happened!!!! Woohooooooo
I've been waiting since November for this video, and whether it's true or not, I'm blaming Grant's amazing visuals for slowing down the production process. What a great collab. Thank you both.
A collaboration between 2 of the most valuable channels on yt
I like that prime number you snuck in there 😀
Gotta love these two during Uni times. They were better lecturers and conceptualized the ideas better than most professors.
Essence of Calculus was easily better than even Gilbert Strang I'd say.
@@well_said7846id respectfully disagree
It was fun to watch Grant tripping around on the word 'almost'. Mathematicians' language really is different than normal people.
These are fascinating videos. I don't consider myself a math person but I find these videos are just as compelling as an engrossing novel you just can't put down or a mesmerizing painting you can't stop staring at.
The log "shortcut" reminds me of leap seconds. Basically just some way to include some terms to get closer to a desired result.
When these 2 do a collab, you know it’s gonna be good
You mean, when these oddest primes do a collab!
I think another reason you should have some intuition that there are "more" 4n+3 primes is that (4n+3)*(4m+3) is always some number (4k+1). However, the converse is not true. (4n+1) * (4m+1) is also 1 more than a multiple of 4. There are just more ways to get one more than a multiple of 4 than three more by multiplying these numbers, and this allows 4n+3 primes to stay in the lead for longer as their products "take up" more space in the numbers 4n+1, and less composite numbers can break into 4n+3 territory.
4:45 my simplest explanation for why is that prime squares can be 1 mod 4, but not 3 mod 4 that accounts for the difference since powers of primes are considered to be partially prime for theorems regarding the prime number theorem (see the Van Mangoldt function for example)
Just listening to Grant speaking is also soothing
Thanks
I am curious about the "distance" away from 0 for Team3-Team1. Is that bounded? Or does it increase for both and just balances out?
My initial thought was well, any even power of an odd number will be of the form 4n+1, so since primes occur in the first place they can kind of by definition, there's less "room" for primes of the form 4n+1 than for 4n+3
Hey man been watching for years, and thank you so much for all the knowledge you provide to us. It's truly inspiring.
i thought this was never coming😭😭..i assumed ur footage was corrupted or something..so glad to see it finally posted ♥♥
Please make a video on Disarium numbers, and finding out the 20th possible disarium number 😀
Always great to have Grant teaming up with Brady. I gotta say though, as a number theorist myself, the "weirdest possible way to take a logarithm" does such an injustice to the actual process happening behind the scenes. In reality, it's a completely natural way to express a logarithm. Granted, you have to understand the notion of a completely multiplicative function, an Euler product, and Taylor series, but I don't think these are concepts that would be challenging to the typical numberphile viewer. At the very least, the rigorous calculation (or enough details to demonstrate it) should be included for those familiar with these topics. That critique aside, I really liked the intuition given here.
"Completely natural way to express a logarithm" - I see what you did there.
@@ps.2 😂 didn't even catch that when I wrote the comment.
For some reason, I chuckled when he said "2 is the oddest prime"
5 is. Because it's not half base 10
I think I'm in love with 3B1B ! You're amazing, Grant! Keep it up!
Great to see Grant again!~
Classical mathematician shortcut. Just remember this short, simple, absolutly uninuitive, 20 step instruction and then its totally easy to work out :-) Still the fact, that team 3 never gets in the lead is verry nice. Those problems nomaly have the tendency to go to 0 or 100% or infinety, or 50:50. Something you would expect. Thx for sharing this. great vid.
Ha. Well, we think it look complicated, but that's because our normal logarithm algorithm is to hit a button on a calculator or look it up in a log table. The _actual_ other log algs are not so easy!
19:26 "No!" means "What have I just been explaining to you for five minutes my dude!" 😄
This was a much better waste of my time than many other videos on UA-cam. Thank you.
This is probably the most involved and difficult to grasp of Grant's videos ever, I'm gonna need multiple rewatches
Crucial point: ln(pi/4) is negative (since pi/4 < 1). That's the intuition for why there are more 3 mod 4 primes than 1 mod 4; even after putting all the even powers in the positive bucket when half of them "should" have been in the negative bucket, the total sum is still negative.
It doesn’t matter what the value of ln(π/4) is, so long as the series in question converges. The idea grant spoke about is that the sum of the positives go to infinity, and the sum of the negatives go to -infinity; however, added together they converge to a value. This is the big point - for every positive value we need a negative to stop the sum going to infinity. Why should there be more 3 mod 4? Well that’s because there are more positives as you get lots more from prime powers, which means you’re left with basically only the primes that are 3 mod 4 doing the negative business. Which means more primes 3 mod 4 to stop the sum going to infinity.
This is overly verbose but I hope it explains that you don’t care what the sum converges to, just that it does and we have the positives sum to infinity and the negatives sum to -infinity
@@benp2291 The only existence here of the sum is not enough in the argument, the value is also important.
The sum of reciprocal of squares of prime also converge, so for the team 3 lead intuitive 'argument' to work what matters is the fact that essentially:
sum of reciprocal of squares of primes (+ some more) > ln(pi/4) + sum of reciprocal of (some) cubes of primes (+ some more) +
the 'some' can be computed by looking at the chi function.
We’re talking about natural log here, so ln(π/4) is negative ∵ π/4 < e, even though π/4 < 1 is true.
9:35 - What's shown on the screen is not "reducing it by one third", though. It's reducing it *_to_* one third (of its original value). Reducing it *_by_* one third would mean subtracting 1/3, and reducing it *_by_* one third of its original value would mean multiplying it by 2/3.
Just a few days ago I watched the first two and was wondering where part 3 was. What a great payoff.
Two questions:
1. What is the LARGEST LEAD "Team 3" has? Is there an upper limit?
2. Who the heck comes up with these things? "Gee, I wonder what it would look like if you did this weird thing with prime numbers...."
Im happy to finally know what 1mod4 and 3mod4 means
That's a very interesting heuristic way of explaining the bias. Still though, all leads are temporary, and the asymptotic density of each of the residue classes is 1/2.
Fascinating video! I don't know if it's already been mentioned somewhere in the comments, but the fact that ln(π/4) < 0 lends even more credence to the theory being true on principle alone, as it requires even more tiny negative fractions to fill the bucket and tip the scales that direction.
Seriously, these are the kinds of friends I'd love to have. In my dream world, I live in a neighborhood where all my neighbors are guys like this - and their wives are doctors or something like that.
This makes sports rivalries quite interesting, in a way. If you just showed me the graph and asked me who's "better" I would say team 3 without thinking, but if you asked me at some point in time and I only were to take current form and recent results into consideration, it'd practically be a coin flip.
If a "season" started at the beginning of the series and ended at a random point, team 3 would be 99.5% likely to win the title. If it started at a random point and ended at a random point, team 1 would win about half the time.
Either this is mind-boggling or I'm completely wrong. I really can't tell, because I'm mind-boggled.
19:26 - "Does it ever get to the point where it'll be in the lead forever?"
No. Because if it did, then the percentage of time that it's in the lead would be strictly increasing and approach 100%.
Grant did a video explaning the series converging to pi. It's an old 3B1B video
Is there a common name for this logarithm technique for those who might want to look for more info?
Nice talk about Chebyshev’s bias with it mentioned against Dirichlet’s theorem of arithmetic progression while using Dirichlet’s series.
you videos are responsible for a math teacher saying i "know a lot about primes"
This is the most bizarre mathematical result I’ve heard of in a while.
It seems clear that a math prodigy could spend an entire lifetime dissecting, testing, and understanding primes --- and never exhaust all the ways that they astound and confound us.
Brady's tortoise and hare comment at the end is interesting given that the two concepts meet in cryptography where you are modulo some prime and both its chi value and cycle length (found with Floyd or Brent's tortoise and hare method) are important.
Bugs me that 1 mod 4 was blue and 3 mod 4 wasn't, we all know *3* is blue, 1 is _brown_
Loved that, and love Grant's work. He is one of the best explainers of difficult stuff I have ever come across on YT.
We are waiting for a great video on 3Blue1Brown after getting 5 million subscribers!🥰
My first intuition was:
2 numbers multiplied that are both 1 mod 4, result in another 1 mod 4 number.
2 numbers multiplied that are both 3 mod 4, result in a number 1 mod 4.
Only if a 1 mod 4 and a 3 mod 4 number is multiplied, we get a 3 mod 4.
So I thought if we think of the Sieve of Eratosthenes, there would be a higher chance that 1 mod 4 numbers get eliminated. But didn't think this through further...
That does not mean anything. 1 number being 1 mod 4 and the other 3 mod 4 is as likely than having both 1 mod 4 or both 3 mod 4 (if you flip a coin, you have 50% to not get the same result)
You have 4 possivilities : both number you multiply are 1 mod 4 (the result is 1 mod 4), the first one is 1 mod 4 and the 2nd one is 3 mod 4 (the result is 3 mod 4), the first one is 3 mod 4 and the 2nd one is 1 mod 4 (the result is 3 mod 4) and both are 3 mod 4 (the result is 1 mod 4). So half of the time it will be 1 mod 4 and half of the time it will be 3 mod 4
@@quentind1924 well, yes but no. My first post is a bit misleading. The three sets don't have 1/3 chance each. But its also not a coin flip here. Lets say we have m primes 1 mod 4 already in the list, and n primes 3 mod 4.
Then there are
m(m-1)/2 + m = m(m+1)/2 combinations of 1 mod 4 primes ( m(m-1)/2 pairs different numbers and then there are m self-combinations (square of the primes)
Same logic there are n(n+1)/2 combinations of 3 mod 4 primes
Last, there are m*n combinations of different primes.
m(m+1)/2+n(n+1)/2>(m^2)/2+(n^2)/2>=m*n
So there are always more pairs that eliminate 1 mod 4 numbers than pairs that eliminate 3 mod 4 numbers.
Always good to see Grant on numberplate 👍
on the rare occasions when team 1 is in the lead, does the length of time it spends in the lead increase on average?
Also, ln(pi/4) is negative, which also tends to tell us, that team 3 should always be ahead by a little bit.
I'm going back and forth on the fractional odds of getting a quarter of pi.
In one video, you manage to touch on the prime numbers, pi, e (via ln), and the most mind-boggling way to calculate a natural logarithm ever seen. mind blown
We love the crossover and collab!!!!!!
The way the difference was plotted as a function with integer values being positive and occasionally negative gives a really nice way to think about this. The question with the answer of roughly 99.5 % (if I understood it correctly, but I think I did) is equivalent to asking what part of its values are positive. If the ratio of positive values to negative values was 1:1, the 3s would be in the lead half of the time - but that's not the case. But it gives me an idea for another, similar question. How does the integral of this function behave? Both from 0 to some finite bound X, and on the infinite scale (0 to infinity) - that one I think is most probably diverging given the heavy bias towards positive values, but it's not impossible for it to be finite-valued given this fact (meaning that you can have a function that's positive 99.5 % of the time and yet has integral equal to 0 - the negative areas just have to be deep enough to compensate for their rarity).
"Integral" of a discrete function? Do you mean a Σ sum, or can this be thought of as an actual integral in some [no pun intended] way?
Anyway, I agree that it probably diverges, and the _lim n→∞_ is gonna be like -1/12.
Chebychev biases. Also, it seems like the twin primes conjecture should be tied to this somehow (as any pair of twin primes [ignoring the triplets 3, 5, 7] would contribute one to each of 1 mod 4 and 3 mod 4).
The wait is over! And it was worth it :)
I find this topic analogous to entropy. How for most of the time everything is barreling towards chaos and disorder, we find these small bursts where order and complexity emerge, like in our own pocket of complexity we all live in.
What about 1 mod 6 and 5 mod 6?
mod6=1 never leads mod6=5 through the first 10,000 primes (that is, through prime 104729).
I did not think i would like this video that much. But as is always the case with numberphile, i really enjoyed it.
I didn't quite notice we were talking about primes yet, and it took me a moment to work out why there were only two categories of number mod 4... :)
What would an alternating sum of the reciprocals of the primes be?
I love this guy, the best math tutor
I just want to say that, your videos always be able regain my interest in math when I feel burnout, it also help me to see problems from different perspective which is important for a math lover
Wish you guys all the best!!
3:32 I think it's more of a bias in our language: we have a word for numbers that are divisible by 2 (even) and another word for numbers that are not (odd). Prime numbers have also a bias towards throdd numbers (3 is the only threeven number), 5 is the only quinary number, ...
Aww, I should have read this before commenting. Seems like I had heard the term "throdd" before but just forgot about it.
Here's an incredibly intelligent and educated man who pronounces "often" ... _"ofen"_ .
I no longer feel so all alone.
3B1B! Love this guy as a guest and his channel!
Edit: this was somehow inspirational. Great video 👍
Does this mean that for p>3 the sum of the reciprocals of the team 3 primes is the same as the sum of the reciprocals of the team 1 primes?, I.e. if you calculated the 'pseudo-alternating' sum
1/5-1/7-1/11+1/13+1/17-1/19-1/23 ... it'd converge to zero?
The theorem at 20:28 was not immediately clear to me, until I realized that 1/ln(x) . sum(1/x) converges to 1. So if you leave out the terms where team 1 wins, the sum is a little less.
Brady's questions are on point.
There's some beauty in this pattern!
Play the same game flipping a coin for some large number of flips. Team H and team T. Whoever starts out leading, will tend to hold the lead for the duration of the flips. Both will have about the same number of wins, but at any given moment in time, whoever started out leading will have a higher number of wins. And this is random, unlike the primes race.
You know the video is gonna be good when it has Brady in it
I was beginning to think that part 3 of this would turn into a Valve situation.
i'm not sure i fully understand the way grant explained this. the fact that the even powers of primes always land in the positive bucket in my head doesn't necessiate that there ought to be more items in the negative bucket because the sum of all the inverses of the even powers of primes might just converge to some small number (
Indeed I think that the exact value of the number has importance, however ln(pi/4)
Love you guys doing these videos.
Does this have any relationship/connection to the twin prime conjecture?
Finally, the trilogy is complete
Being aware of what seems like a natural imbalance, have there been any deeper looks into the commonalities between the threshold points? Given the appearance of pi, plus the actual results data, it feels like there must be some periodic pattern on a spiral that guarantees a minority batch within some max interval, and whose duration is commensurate with magnitude of n. Anyway, I'm a first time commenter, but have watched and enjoyed both of your guys videos for a long time, keep up the great work!
I think this argument is not sound. Sum of reciprocals of powers of natural numbers larger than 1, even without dividing and with counting 6th power as a square and a cube etc. is finite (and equal to 1). So in fact the powers don’t contribute anything to the convergence, is all about the difference of reciprocals of 4k+1 and 4k+3 being small.