@@AgentM124 the MathWorld article linked in the video description mentions that the bound's eventually broken in both directions, but not much else, so I'd guess it isn't known which direction is broken first.
I'm being serious, this is the most interesting thing I've ever learned. How have I not been aware that it breaks the sqrt barrier. I have a math degree and live in math, thank you Numberphile!!
Dr Krieger is easily in my top three Numberphile experts. Simple explanations of complex problems that usually tend to be the type of math I’m interested in. Great video!
Excellent description of the Mertens Conjecture, and the counter-example found. Many thanks for the link to the mathematical paper disproving the Conjecture.
I love when they have these amazing women on! I've watched for a while, and this is one of the few inspiringly awesome channels who feature these women! I've introduced this to a lot of my friends who were on the fence on whether they should go into math/science or not (both boys and girls) and this channel really tipped that favor for most of them! I love it when a channel is both entertaining and also inspiring. Love the channel guys! Keep it up!
The 10^(10^40) is only an upper bound on the first occurrence, there are no proven lower bound except the limit it has been tested to: 10^16. So it could be small enough to write down, though probably not very likely with such a large upper bound.
It's conceivable that the number is way smaller than the upper bound. Remember, it's still possible that the solution to the Graham's number problem is 13.
I asked for a video just like this a long time ago: a video on a problem where it looked very likely that it was true based on computation or all known examples, but it was eventually proven false for some huge number. I really enjoyed watching this :)
Very nice explanation, but I do want to pick some nits. There *are* huge numbers we can write down, for example 10^10^40 +1. What you mean is that one cannot write it down in decimal notation which was not designed to handle really big numbers. I assume there is no constant C such that C * sqrt(n) bounds the Mertens function? Anyway, thanks for this very nice explanation.
My thought was that the sum would approach infinity as n approached infinity. I figured that it would start out well-balanced, but that as the primes became more sparse, the frequency of odd-factored numbers would drop as well.
I'm pretty sure that oscillates arbitrarily far from 0 if given long enough, it's just that it also comes back to 0 infinitely often. To put it another way, choose an n, then the sum will eventually be greater than n or less -n, and will necessarily eventually come back to 0 after reaching that value.
But it's not just primes that have mu(x) = -1, though. Any number with three, or five etc, different prime factors also counts, and as you get to large enough numbers these begin to dominate over the primes.
@@alexpotts6520 -- I know. What I'm surprised by is that the frequency of odd-factored non-primes increases at the same rate that the frequency of primes decreases.
Wait but if there are 10^80 atoms wouldn't it be possible to represent numbers up to 2^(10^80) using binary (eg. Divide universe into grid spaces, atom =1, no atom = 0)? And then 2^(10^80) = e^(ln(2)*10^80) is definitely > than 10^(10^40) ° e^(ln(10)*10^40)... So the number CAN be represented! It would take on the order of 10^40 or 10^41 atoms to do it...
You kept talking about it escaping the bounds of the square root and showing it doing it on the positive side, but didn't mention if the paper that disproved the conjecture indicated if it broke the bounds on the positive or negative side. I'm guessing it was the negative side because the primes "thin out" when numbers get that large.
The paper's results imply that the conjecture is broken infinitely many times on each side, though we don't know which side happens first. The paper in the video description focuses more on the positive side, however, because they get stronger results from the positive side. I can't speak for why the artist chose to show it break on the positive side, but I imagine it was because the paper focused more on the positive side. That's just a guess, though.
@@samgraf7496 Both, according to the paper (first lines of page 3). M(n) goes above sqrt(n) and below -sqrt(n) for some values of n (infinitely many times). Don't know which one occurs first.
of course you can write the number, just not linearly. If you flash the digits one at a time on screen eventually you will have described the number (if you have long enough..)
1/4 of all positive integers have repeating 2's in their prime factorization, 1/9 have repeating 3's, 1/25 have repeating 5's., etc. So, the probability that a given positive integer has no repeating prime factors is: (1 - 1/4)(1 - 1/9)(1 - 1/25) ... ((1 - 1/p^2) ... for all primes p. As I know from another video, this is equal to 6 / π^2. For the integers with square-free factorization, half will have an odd number of factors, and half will have an even number. Therefore, it should be possible to approximate pi by, for example, counting the number of +1's in Holly's formula and doing a little math. That's not entirely relevent to this video, but I think it's a fun idea. :)
Analogous to Skewes's number, as I'm sure others have pointed out (Skewes's number being an upper bound on the point where the prime counting function surpasses the logarithmic integral function). The original version was much bigger than the number in this video; the current best value is a lot smaller. Even the first one is much, much smaller than things like Grahams number etc..
This is interesting, square root grows quite slow for huge numbers, so really you could be adding or subtracting any epsilons without knowing if it escapes the parabola, not just the 1s.
The sequence of -1, 0, and 1 makes me think of balanced ternary and mod 3... and the upper bound has me thinking of something akin to Graham's number, which is threes all the way up. I'm also seeing similarities to ternary combinatorics, matrix indices, and so much more. Maybe I'm grasping at straws here...? 🤷
A number that big can be made possible by thinking about connections between particles rather than about particles, because the number of the former is the factorial of the number of the latter.
Now some natural questions come to mind: 1)How many of these numbers such that M(n) > sqrt(n) exist? Is it known if they are maybe infinitely many? 2)Is there another boundary function f(n) larger than sqrt(n) but smaller than n such that M(n) < f(n) for all natural n's?
People keep talking about Amy Adams or even Nicole Kidman, but I always think I'm getting stuff explained by Hanna Alström/Princess Tilde from Kingsman
If experiment was the only avenue available we would conclude with evidence far stronger than needed to convict in court or accept as theory in science that the conjecture is for all practical purposes true. And we'd be wrong. Is there a moral here?
Great video. My hunch though it that it's totally not true that all the atoms in the universe couldn't be arranged to represent the number. Totally true that there are not enough to be used to make ink to write the number in tiny digits. But... say 10^80 baryons - in the universe. Each can be arranged in space to an accuracy of a plank length - 10^-35 meters, across 10^18 meters (diameter of the observable universe) in three dimensions. In other words if the position in space of a particle is used to represent a number then we have = 10^35+18+3 = 10^56 possible values for that 'digit'. So we have 10^80 digits in base 10^56 and 10^56^80 is the maximum number that can be expressed. This is comfortably larger than 10^10^40. Except more than one particle cannot meaningfully be in the same location (there's no where to stack them) and that messes things up.
Can we do anything with those numbers with repeated prime factors? Eg. could we mix the results of the function (-1,0,1) that 0 goes to even or odd number of factors and check how the function then behaves. Does itjust go to infinity or also jumps around 0?
Interesting. We could instead sum µ(rad(n)), the number of distinct prime factors of n. Or sum µ(sqf(n)), where sqf(n) is n's square-free part: the product of all n's prime factors p where p^2 does not divide n, so µ(sqf(n)) is the number of distinct non-repeating prime factors of n. Or sum µ(n/hsqf(n)) where sqf(n) is n's highest factor which is square -- I don't know if that is more pertinent.
@@rosiefay7283 I though of something similar. The issue is that all composite numbers are repeated prime factors so it won't hover around 0. If you do it with coprime numbers and primes it for sure won't hover around 0.
A conjecture is a kind of problem. So do they have solutions and logically derived from other proofs and conditions. FM modulators for high ratio dividend from huge resonance. Root is a split of halftime.
Why do we toss out the numbers with repeats in their prime factorization? How does it behave if we count them (either as 8 = 3 (as in 3 2s) or 8 = 1 (as in "this number only has one prime factor))?
What exactly do you mean by not being able to write it down? By writing 10^10^40 you write the instructions to get to one specific number, same is true for the decimal system writing "12" is the instruction for writing xxxxxxxxxxxx. So if you're lucky and the number that breaks the pattern happens to be a very special number, you could in fact write it down right? Of course not in a unary system but in some other system?
Dr Krieger did not really say what this means for Riemann's hypothesis, since it has been disproved. In other words, is this "dis-proof" a path to disproving the Riemann hypothesis?
another question do we don't know if the number n where Mu(n) is greater the square root of n but is it more likely to be a prime number itself or a compound number
why aren't repeated prime factors counted towards the total sum? I mean, 40 has4, and 20 has three prime factors? So why are both of those counted as 0?
Cool video. I hope you will do a video explaining how it is proven that Merten's Conjecture is not true, even if it is only a superficial explanation of a complicated proof.
We don't know whether the _first_ breakthrough point is in the positive or negative direction. But the work done actually implies that there are infinitely many breakthrough points in the positive direction and infinitely many breakthrough points in the negative direction.
In this specific proof they don't work with massive numbers at all. Their strategy was to find a link to a different field and work there. Their results in this different field are then used to disprove Merten's conjecture because they imply that M(x) becomes at least as big as 1.06x².
Someone needs to start a anime about the Riemann Hypothesis vs the entire mathematical community for the century. That Hypothesis just keeps dodging and weaving everything that's thrown at it.
For whatever reason I just want to know if it breaks this rule on the negative side or positive side first, and also does it eventually break it on both sides.
i just love numbers that we can only describe using mathematical statements rather than explicit constructions. like Rayo's Number: "The smallest number bigger than any finite number named by an expression in the language of first-order set theory with a googol symbols or less." or that famous proof about describing natural numbers: Claim: Every natural number can be described in 20 English words or less "Proof": Assume this is not true. Then there is some first natural number n that cannot be described in 20 English words or less. Then n is "the smallest natural number that cannot be described in 20 English words or less." This is 14 words, contradicting the assumption that n could not be described in 20 words or less. Therefore, every natural number can be described with 20 English words or less
Do the gaps between 0 crosses get arbitrarily large? Additionally, is there another number x >1 for which merten(n) never exceeds n^x? Something like 0.6?
Given the numbers at 8:22 doesn’t this mean there are more atoms in the universe than digits in this number? It looks to me as though you could take 10^40 atoms to write each number digit of the number 10^10^40
Cancelling out anything that has any duplicate prime factor seems a bit arbitrary. What happens if you go for -1 for an odd number of unique prime factors (eg 8) and 1 for unique even prime factors (eg 12)?!
That mu with the long tail was so pretty I just had to go do an image search for "math cat mu". Awww, so cute. The Internet was obviously built for cats.
I would suggest calling it the Mertens Number, and represent it with the Greek letter Mu. Question, is the Mertens number bigger or smaller than TREE(3) and/or Graham's Number?
The video is incorrect, I think. There is no proof that the first counterexample is "too big". We know that it exists and is smaller than 10^(10^40), which is indeed a too big finite number (too big to be encoded on any imaginable computer), but "smaller than" means that the first counterexample may well be MUCH smaller than that.
It is not 10^400. Let's look at an example. 2^3^4 is not the same as 2^12. It also doesn't equal 8^4. You actually work your way down. 3^4 is 81, so you have 2^81, which is a humongous number in the septillions. So if we apply this to 10^10^40, we see that it's actually the same as saying 10^1000000000000000000000000000000000000000. Totally different level of big
Clearly u(n) can never be greater than n or less than -n. Is there any known bound that is tighter than this? Also how does this have anything to do with the Riemann hypothesis?
More videos with Holly: bit.ly/HollyKrieger
Holly on the Numberphile Podcast: th-cam.com/video/QmfQQzjpdpM/w-d-xo.html
I didn't catch, but did they know anything about if it breaks the √n in the pos or the negative?
Fyi: I see "Mertern's Conjecture" instead of "Merten's Conjecture" in the TH-cam description of this video.
@@AgentM124 the MathWorld article linked in the video description mentions that the bound's eventually broken in both directions, but not much else, so I'd guess it isn't known which direction is broken first.
Could you do a video about how this problem relates to the Riemann Hypothesis? I find the interconnections of mathematics to be really interesting
@@rob6129 Second that.
5:02 "Living around zero but in a really complicated way"... Wow, you just described my bank account
Lol
Stefan not so Reich
But you can be certain that at one point your money amount blows outside of the boundaries, it might be negative though.
HOLLLLAAAAA
@@Oblivion1407 What if my bank account is 5+3i? Would that be money that works not just across space, but also through time and/or dimensions?
I'm being serious, this is the most interesting thing I've ever learned. How have I not been aware that it breaks the sqrt barrier. I have a math degree and live in math, thank you Numberphile!!
sarcasm, right?
@@thishandleistaken1011 Many of us have math degrees and PhDs and didn't know it, that's the point, it's a new discovery.
Gregory Fenn by new you mean 1985
It breaks the Squarrier
@@gregoryfenn1462 how did so many people manage to get PhDs without knowing they got them? /s
Dr Krieger is easily in my top three Numberphile experts. Simple explanations of complex problems that usually tend to be the type of math I’m interested in. Great video!
Excellent description of the Mertens Conjecture, and the counter-example found. Many thanks for the link to the mathematical paper disproving the Conjecture.
It is intuitively clear (for a physicist) why the magnitude of M(n) should be about sqrt(n): it is similar to a random walk on the number line.
Taking it to be related to a random walk would imply the conjecture is false, by the law of the iterated logarithm (even after accounting for the 0s).
Is it possible that the first failure of Riemann's Hypothesis is a number as big as this one?
Yes, unfortunately.
It could be even bigger
it almost definitely is. no failures have ever been found
@@leofisher1280 We probably haven't looked quite that far, but yes, we've looked very far
You take that back
I'm thinking of a number between 1 and Tree(3).
we're gonna need a bigger universe!
Best comment
I love when they have these amazing women on!
I've watched for a while, and this is one of the few inspiringly awesome channels who feature these women!
I've introduced this to a lot of my friends who were on the fence on whether they should go into math/science or not (both boys and girls) and this channel really tipped that favor for most of them!
I love it when a channel is both entertaining and also inspiring.
Love the channel guys! Keep it up!
my typical numberphile viewing experience:
start of the video: **yawn**
end of the video: **screams geometrically**
then don't watch - dork
The 10^(10^40) is only an upper bound on the first occurrence, there are no proven lower bound except the limit it has been tested to: 10^16.
So it could be small enough to write down, though probably not very likely with such a large upper bound.
It's conceivable that the number is way smaller than the upper bound. Remember, it's still possible that the solution to the Graham's number problem is 13.
Just imagine working with numbers so large. That you would need more mass of ink than there is mass in the universe to write them down
That problem then is unsolvable.
I think Ron Graham can relate to that
@@RB-jl8sm not really. Math is filled with numbers you can describe and not write down. E.g. all irrational numbers
@@BrosBrothersLP i see, so we dont need to imagine it because it exists anyway and hence it is not too interesting.
That was an unfunny joke.
I love Dr. holly's Laugh 😇
I asked for a video just like this a long time ago: a video on a problem where it looked very likely that it was true based on computation or all known examples, but it was eventually proven false for some huge number. I really enjoyed watching this :)
Very nice explanation, but I do want to pick some nits. There *are* huge numbers we can write down, for example 10^10^40 +1. What you mean is that one cannot write it down in decimal notation which was not designed to handle really big numbers. I assume there is no constant C such that C * sqrt(n) bounds the Mertens function? Anyway, thanks for this very nice explanation.
6:50 STONKS
My thought was that the sum would approach infinity as n approached infinity. I figured that it would start out well-balanced, but that as the primes became more sparse, the frequency of odd-factored numbers would drop as well.
I'm pretty sure that oscillates arbitrarily far from 0 if given long enough, it's just that it also comes back to 0 infinitely often. To put it another way, choose an n, then the sum will eventually be greater than n or less -n, and will necessarily eventually come back to 0 after reaching that value.
But it's not just primes that have mu(x) = -1, though. Any number with three, or five etc, different prime factors also counts, and as you get to large enough numbers these begin to dominate over the primes.
@@alexpotts6520 -- I know. What I'm surprised by is that the frequency of odd-factored non-primes increases at the same rate that the frequency of primes decreases.
Wait but if there are 10^80 atoms wouldn't it be possible to represent numbers up to 2^(10^80) using binary (eg. Divide universe into grid spaces, atom =1, no atom = 0)?
And then 2^(10^80) = e^(ln(2)*10^80) is definitely > than 10^(10^40) ° e^(ln(10)*10^40)...
So the number CAN be represented! It would take on the order of 10^40 or 10^41 atoms to do it...
Hmm, I wonder if the mu function is breaks the square root bound on the positive or the negative side at the first counter-example.
You kept talking about it escaping the bounds of the square root and showing it doing it on the positive side, but didn't mention if the paper that disproved the conjecture indicated if it broke the bounds on the positive or negative side. I'm guessing it was the negative side because the primes "thin out" when numbers get that large.
The paper's results imply that the conjecture is broken infinitely many times on each side, though we don't know which side happens first. The paper in the video description focuses more on the positive side, however, because they get stronger results from the positive side.
I can't speak for why the artist chose to show it break on the positive side, but I imagine it was because the paper focused more on the positive side. That's just a guess, though.
Proposal for the name of this number: Mertens downfall.
Edit: i really like all proposed alternatives in the replies!
Mertens' Bane ;-)
Odlyzko's Number named after the author who proved it
Mertens' Folly.
Merten's Conjecture First Counterexample (There could be more than one).
MCFC
@@samgraf7496 Both, according to the paper (first lines of page 3).
M(n) goes above sqrt(n) and below -sqrt(n) for some values of n (infinitely many times). Don't know which one occurs first.
of course you can write the number, just not linearly. If you flash the digits one at a time on screen eventually you will have described the number (if you have long enough..)
1/4 of all positive integers have repeating 2's in their prime factorization, 1/9 have repeating 3's, 1/25 have repeating 5's., etc. So, the probability that a given positive integer has no repeating prime factors is:
(1 - 1/4)(1 - 1/9)(1 - 1/25) ... ((1 - 1/p^2) ... for all primes p.
As I know from another video, this is equal to 6 / π^2. For the integers with square-free factorization, half will have an odd number of factors, and half will have an even number. Therefore, it should be possible to approximate pi by, for example, counting the number of +1's in Holly's formula and doing a little math. That's not entirely relevent to this video, but I think it's a fun idea. :)
Love Dr Holly !!!
Square root is more parabolic so cross should be possible for any wave. The concept is parabolic is parallel as you approach higher numbers.
I like to believe that graphing irrational numbers a specific way will produce a visual pattern. Graphs like this or the recaman sequence etc..
That part at 6:00 actually blows my mind
It's time for another Dr. Holly video. I have spoken.
Fantastic! I'm studying number theory in my undergraduate course but I hadn't found this one. Classic.
The second best Dr. Krieger.
the Riemann Hypothesis is really everywhere.
Analogous to Skewes's number, as I'm sure others have pointed out (Skewes's number being an upper bound on the point where the prime counting function surpasses the logarithmic integral function). The original version was much bigger than the number in this video; the current best value is a lot smaller. Even the first one is much, much smaller than things like Grahams number etc..
One of the finest explanations I've ever seen.
This is interesting, square root grows quite slow for huge numbers, so really you could be adding or subtracting any epsilons without knowing if it escapes the parabola, not just the 1s.
The sequence of -1, 0, and 1 makes me think of balanced ternary and mod 3... and the upper bound has me thinking of something akin to Graham's number, which is threes all the way up. I'm also seeing similarities to ternary combinatorics, matrix indices, and so much more. Maybe I'm grasping at straws here...? 🤷
A number that big can be made possible by thinking about connections between particles rather than about particles, because the number of the former is the factorial of the number of the latter.
Now some natural questions come to mind:
1)How many of these numbers such that M(n) > sqrt(n) exist? Is it known if they are maybe infinitely many?
2)Is there another boundary function f(n) larger than sqrt(n) but smaller than n such that M(n) < f(n) for all natural n's?
People keep talking about Amy Adams or even Nicole Kidman, but I always think I'm getting stuff explained by Hanna Alström/Princess Tilde from Kingsman
0:00 If your date says this, congratulations.
If experiment was the only avenue available we would conclude with evidence far stronger than needed to convict in court or accept as theory in science that the conjecture is for all practical purposes true.
And we'd be wrong.
Is there a moral here?
PifflePrattle I think it’s fair to question the utility of a conjecture whose truth depends on a quantity we could never compute.
Great video. My hunch though it that it's totally not true that all the atoms in the universe couldn't be arranged to represent the number. Totally true that there are not enough to be used to make ink to write the number in tiny digits. But... say 10^80 baryons - in the universe. Each can be arranged in space to an accuracy of a plank length - 10^-35 meters, across 10^18 meters (diameter of the observable universe) in three dimensions. In other words if the position in space of a particle is used to represent a number then we have = 10^35+18+3 = 10^56 possible values for that 'digit'. So we have 10^80 digits in base 10^56 and 10^56^80 is the maximum number that can be expressed. This is comfortably larger than 10^10^40. Except more than one particle cannot meaningfully be in the same location (there's no where to stack them) and that messes things up.
Can we do anything with those numbers with repeated prime factors? Eg. could we mix the results of the function (-1,0,1) that 0 goes to even or odd number of factors and check how the function then behaves. Does itjust go to infinity or also jumps around 0?
Interesting. We could instead sum µ(rad(n)), the number of distinct prime factors of n. Or sum µ(sqf(n)), where sqf(n) is n's square-free part: the product of all n's prime factors p where p^2 does not divide n, so µ(sqf(n)) is the number of distinct non-repeating prime factors of n. Or sum µ(n/hsqf(n)) where sqf(n) is n's highest factor which is square -- I don't know if that is more pertinent.
@@rosiefay7283 I though of something similar. The issue is that all composite numbers are repeated prime factors so it won't hover around 0. If you do it with coprime numbers and primes it for sure won't hover around 0.
07:33 Reiman hypo jumpscare 😅
Why was root-n thought to be the bounding function for mu-n. And since it's not that, is mu-n bounded by something else?
Well, obviously it is bounded by n. I don't know if it is bounded by something that is not in Omega(n) though.
A conjecture is a kind of problem. So do they have solutions and logically derived from other proofs and conditions. FM modulators for high ratio dividend from huge resonance. Root is a split of halftime.
We should definitely name that number...
I vote for Mertens Bane
Why do we toss out the numbers with repeats in their prime factorization? How does it behave if we count them (either as 8 = 3 (as in 3 2s) or 8 = 1 (as in "this number only has one prime factor))?
What exactly do you mean by not being able to write it down? By writing 10^10^40 you write the instructions to get to one specific number, same is true for the decimal system writing "12" is the instruction for writing xxxxxxxxxxxx. So if you're lucky and the number that breaks the pattern happens to be a very special number, you could in fact write it down right? Of course not in a unary system but in some other system?
That looks like technical analysis of the prime numbers.
Well it says the lower bound is 10^14 so it could break the barrier at reasonable number.
6:17 It could still be wrote down in a some kind of time-crystal
Why does 10 have 2 primes (2, 5) excluding 10 itself, but 5 has 1 prime (5)?
Because 10 isn't a prime
Got it. "Prime" factors ;)
Please do more on this! How is it related to Riemann?
What is fractal dimension of the curve? It can be interesting.
Dr Krieger did not really say what this means for Riemann's hypothesis, since it has been disproved. In other words, is this "dis-proof" a path to disproving the Riemann hypothesis?
another question do we don't know if the number n where Mu(n) is greater the square root of n but is it more likely to be a prime number itself or a compound number
why aren't repeated prime factors counted towards the total sum? I mean, 40 has4, and 20 has three prime factors? So why are both of those counted as 0?
wait waitwait, there is letter "meow"? ;)
"mu", the Greek m
Cool video. I hope you will do a video explaining how it is proven that Merten's Conjecture is not true, even if it is only a superficial explanation of a complicated proof.
Is it known whether the "breakthrough" point is in the positive or negative direction?
We don't know whether the _first_ breakthrough point is in the positive or negative direction. But the work done actually implies that there are infinitely many breakthrough points in the positive direction and infinitely many breakthrough points in the negative direction.
I'd love to see a video explaining roughly how you work with numbers in a proof that are so massive you can't describe them.
In this specific proof they don't work with massive numbers at all.
Their strategy was to find a link to a different field and work there.
Their results in this different field are then used to disprove Merten's conjecture because they imply that M(x) becomes at least as big as 1.06x².
Did I miss it, or do they not tell us if it breaks +ve, -ve or we don’t know?
Someone needs to start a anime about the Riemann Hypothesis vs the entire mathematical community for the century. That Hypothesis just keeps dodging and weaving everything that's thrown at it.
If you're looking for something like this, I'd recommend Simon Singh's book on Fermat's Last Theorem.
What was the basis for choosing the square root as the limit anyway?
Yea viewers cant believe their eyes when they see Dr Holly Krieger on Numberphile
Bottom line is, was there any reason to put square root as the limit in the first place?
My immediate curiosity is why that conjecture? What was it about square root?
Doctor Krieger ❤️❤️ mmmm
I find it difficult to concentrate on the subject sometimes...
@@Kris.G yep...two very big distractions
The slope of that square root curve gets closer and closer to zero. I wonder if that's the main reason it breaks out.
6:40 that kind of looks like the graph of (plus or minus) x^(sin(x))
For whatever reason I just want to know if it breaks this rule on the negative side or positive side first, and also does it eventually break it on both sides.
Don’t know which is first but am told it breaks it on both sides.
More podcasts with holly!!
Interesting. So do we know any other bound then? When sqr(n) is not enough, perhaps just n? Or will it eventually reach any number?
O(n^(0.5+epsilon))
Epsilon being the smallest number greater than zero
And O() being “big O notation”
@@debblez I consider that a yes; it will reach any number.
I propose calling it: Holly's Number.
So how would you go from 10^(10^40) to Graham´s number?
i just love numbers that we can only describe using mathematical statements rather than explicit constructions. like Rayo's Number: "The smallest number bigger than any finite number named by an expression in the language of first-order set theory with a googol symbols or less." or that famous proof about describing natural numbers:
Claim: Every natural number can be described in 20 English words or less
"Proof": Assume this is not true. Then there is some first natural number n that cannot be described in 20 English words or less. Then n is "the smallest natural number that cannot be described in 20 English words or less." This is 14 words, contradicting the assumption that n could not be described in 20 words or less. Therefore, every natural number can be described with 20 English words or less
Do the gaps between 0 crosses get arbitrarily large?
Additionally, is there another number x >1 for which merten(n) never exceeds n^x? Something like 0.6?
Given the numbers at 8:22 doesn’t this mean there are more atoms in the universe than digits in this number?
It looks to me as though you could take 10^40 atoms to write each number digit of the number 10^10^40
Thats a lotta atoms
Is it known if this number generates a positive or negative divergence?
Cancelling out anything that has any duplicate prime factor seems a bit arbitrary. What happens if you go for -1 for an odd number of unique prime factors (eg 8) and 1 for unique even prime factors (eg 12)?!
I'd like to see the graph wrapped in a spiral, and see if any periodicity.
I just sat silently for five minutes wrapping my head around that.
That's why I love math.
That mu with the long tail was so pretty I just had to go do an image search for "math cat mu". Awww, so cute. The Internet was obviously built for cats.
it sounds kinda lame that numbers with repeating factors get assigned a 0...
I would suggest calling it the Mertens Number, and represent it with the Greek letter Mu.
Question, is the Mertens number bigger or smaller than TREE(3) and/or Graham's Number?
Pretty sure Mertens Outlaw (as I like to call it) is puny compared to either of those
So would the sum of all those numbers not deviate too much from 0? Is there a way to / is there proof to that?
Sooo... How many prime factors does that number have?
6:45 Am I the only one who sees this function similar to the stock market figures?
Do we know if it breaks through first on the positive side or the negative side?
We do not know that, no. But the work done in the paper proves that it breaks through infinitely many times on each side.
So is the first counterexample positive or negative?
I find this all super fascinating but as with many theoretical mathematical problems I have to ask: WHY????
What is the purpose of this?
'We know it's there' - where?
0:20 weeeee, that banana is my favorite number
The video is incorrect, I think. There is no proof that the first counterexample is "too big". We know that it exists and is smaller than 10^(10^40), which is indeed a too big finite number (too big to be encoded on any imaginable computer), but "smaller than" means that the first counterexample may well be MUCH smaller than that.
Pondered this one as a kid thanks for explaining it!
Wait, doesn't 10^10^40 equal 10^400 ? And if so, isn't that just a 1 followed by 400 zeros? And if so, why can't that number be written down?
It is not 10^400. Let's look at an example. 2^3^4 is not the same as 2^12. It also doesn't equal 8^4. You actually work your way down. 3^4 is 81, so you have 2^81, which is a humongous number in the septillions.
So if we apply this to 10^10^40, we see that it's actually the same as saying 10^1000000000000000000000000000000000000000. Totally different level of big
Does the big number break the negative bound or the positive bound of the square root function?
Clearly u(n) can never be greater than n or less than -n. Is there any known bound that is tighter than this?
Also how does this have anything to do with the Riemann hypothesis?
It seems like the sqrt(n) just barely fails. I'd imagine sqrt(n)*log(n) is more than sufficient but I obviously couldn't prove it.
Why does 1 give a positive?
Does it imply that Riemann's hypothesis stays true up to 10^10^40?
Riemann's hypothesis is about whether complex roots of the Zeta function all have real part 1/2, so your question is nonsense.