The Longest Standing Mathematical Conjecture That Was Actually False
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- Опубликовано: 7 сен 2024
- Looking into one of the longest running conjectures that was disproved 😮
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Link to the paper that disproves the conjecture: www.researchga...
Just for clarity: the supremum of a set is its least upper bound and the infimum is its greatest lower bound :) sorry for any confusion!
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Just for clarity: the supremum of a set is its least upper bound and the infimum is its greatest lower bound :) I also missed the minus sign for the last limit-sorry for any confusion! I filmed this video at 5am to make sure I had a video out for you all this weekend so apologies for the slip up ☺️
Also, a bit of mathematical notation: it should say that limsup (M(n)/sqrt(n)) > 1.06 and the same for infimum. I wrote it that way to make it easier for people new to the field to understand what the limits were in this sense! I understand it’s not “appropriate” mathematically speaking but for a less-technical audience, I thought it would make them understand a bit easier :)
The conjecture was named after Franz Mertens - so it should be Mertens’ not Merten’s :)
Some more information on lim sup and lim inf:
en.m.wikipedia.org/wiki/Limit_inferior_and_limit_superior
Link to the paper that disproves it: www.researchgate.net/publication/2355126_Disproof_of_the_Mertens_Conjecture
^ I’ll be making a follow up video on the proof of this paper! 👀
For those asking for a counter example: Pintz (1987) subsequently showed that at least one counterexample to the conjecture occurs for n
More precisely, limsup_{n -> infinity} M(n)/sqrt(n) > 1.06 is really shorthand for lim_{n -> infinity} (sup_{m>n} M(m)/sqrt(m)) > 1.06. One reason for considering the limsup rather than just the sup is that if sup_n M(n)/sqrt(n) > 1.06 then there might be only finitely many counterexamples to the conjecture. But if limsup_{n -> infinity} M(n)/sqrt(n) > 1.06 then there must be infinitely many counterexamples.
@@JohnDoe-ti2np all comments everywhere, on every platform, should display LaTeX notation...just sayin...
A question: I understand that these new bounds give room for a counter-example (of the Mertens conjecture) to exist. I will agree that it probably does. Is it a certainty that an actual counter-example exists ? or only a possibility ? in which case the bounds might never be approached and the Mertens conjecture still factually be true ?
@@yomguioim If there were no actual counter-example, then there would exist a smaller upper bound than 1.06*sqrt(n) (it would have to be
Thanks for the correction, I was wondering that the lim sup would be rather infinity...
It is also worth mentioning that Merten's conjecture is actually a strengthening of Riemann's conjecture, and there is a phrasing of Riemann's conjecture which is very similar to Merten's conjecture.
By the way, some time ago, I was also looking at the Four Color Theorem, and there was also a conjectured strengthening that stayed quite some time -- Tait's conjecture, disproved after 62 years (and there it really takes just a single counterexample).
Maybe an odd thing to praise, but I am genuinely impressed by your hand-writing. It is fast while still being quite legible and lines stay level.
my little eye spy a grid behind the writing 😂
@@lukahutinski9075 that has never stopped me from writing like an idiot
Unless I'm mistaken, the (Stieltjes) conjecture that M(n) = O(√n) has not been disproven. it only means that M(n)/√n is bounded by some pair of numbers. Mertens' disproven conjecture would set those numbers to ±1.
Euler's sum of powers conjecture stood unrefuted for even longer
Yeah that stood for over 200 years.
Basically Euler conjectured that if a sum of Nth poweres is a perfect Nth Power then there are at least N terms
Example
3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 which is 6^3
Thus a sum of 3 third powers is a perfect 3rd power
Well a computer found a counter-example of a sum of Four 5th powers is a perfect 5th power. All the numbers are in the thousands
One step ahead of you… that video is coming soon 👀
when
I'll have to prove this mathematically 👀
@@jamesknapp64 Reminds me of Fermat's last theorem, but a way stronger claim (and unlike Fermat's, false).
Ernie Parker (and others) in 1960 disproved Euler's conjecture from 1782 that there do not exist two mutually orthogonal latin squares of order 4n + 2 for every n. Bose, R. C.; Shrikhande, S. S.; Parker, E. T. (1960), "Further results on the construction of mutually orthogonal Latin squares and the falsity of Euler's conjecture", Canadian Journal of Mathematics, 12: 189-203, doi:10.4153/CJM-1960-016-5, MR 0122729
Making a video on this soon 👀
@@EllieSleightholm I knew Ernie when I was at U or IL. He was very poorly teated by the long term chair of the Math dept (who was fired in the late 70s for misspending federal funds). Ernie was kept at one of the lowest salaries (for his rank) and given a higher teaching load (than anyone at his rank). Yet he was one of the best known people in that department at that time.
Might also be interesting to mention Perron's formula. It's used a lot in analytic number theory to find bounds for these kinds of sums. Roughly speaking, (look up the integral that appears on the right-hand side of Perron's formula) you calculate the residues of the integrand at all its poles, and you use those to 'slide' the line of integration to the left until the integrand is suitably bounded. The pole(s) with the greatest real component will contribute the most to your bound for the sum. In this case, the integrand has a 1/ζ(s) term, so its poles are the Riemann zeta zeros. Every pole on the Re(s) = 1/2 line contributes something on the order of x^(1/2), while a pole with real part σ > 1/2 would contribute something on the order of x^σ. Summing up infinitely many things on the order of x^(1/2) we might get something slightly greater than x^(1/2) (maybe x^(1/2)*log(x) or something like that), in other words it's okay for the Riemann hypothesis to be true while |M(x)| < x^(1/2) is false. However, if you can prove |M(x)| < x^(1/2 + ε) for every ε > 0, then that's equivalent to the Riemann hypothesis.
6:45 unless I'm very mistaken, u(6)=1. 6 is square-free and has 2 prime factors (2 & 3), an even number.
I correct it at 7:26 if you watch the full video :))
Contrary to other comments, the upper bound of a counterexample is now around 10^(4.4 • 10²⁹), with a lower bound of 10¹⁶ (Rozmarynowycz, John; Kim, Seungki (2023)).
In other words, if I ever have to use the Merten's function in any practical engineering application, I'm safe in assuming it's between +/- sqrt(n).
@@kevinvanhorn2193Yes, but typically theorems are used to prove other theorems. In other words, a large counterexample in theorem A might lead to a small counterexample in theorem B.
can you explain mu(6) please?
the prime factors of 6 are 2 and 3. so it has 2 prime factors. i'm not seeing any squares.
so i'd have guessed mu(6) = 1. what am i missing?
nevermind. it was corrected a few seconds later :)
I think you're correct. I looked through the comments seeing if anyone agrees with me!
I also looked for this in the comments, thanks!
I corrected this in the video :)
Those two inequalities are written wrong. First of all, I really think we are talking about operators limsup and liminf; in fact, as written, the limit as n --> ∞ of the sup of M(n) doesn't make much sense, because the sup of M(n) doesn't depend on n. Also, on the right we have 1.06 sqrt(n), a function of n, which itself goes towards ∞ with n, so it can't be lower to the term on the left for every n.
I find it highly interesting when mathamatics is used in programming to controll the outside world like machinery and reading sensors plus calibration.
I built a interface for the Raspberry Pi to read a water pressure sensor and in the Python program a formula was written taking into account the values of the electronic components like the resistors, fixed reference voltage, including a voltage divider, from that any change to one of the electronic component values it would show up in the readout of the program.
A small example of what can be done and when the maths formula matches the real world values of sensors and electronic components makes one want to explore or achieve a much higher level of interesting projects.
The best we know is that the conjecture is disproved for at least one integer smaller than exp(3.21x10^64). The existence of bounds for M(n)/sqrt(n) implies the Riemann conjecture (Good luck!)
Mentioned this in the pinned comment! Can’t wait to see further advancement in this field 😅
@@EllieSleightholm The reference is "J. Pintz, An effective disproof of the Mertens conjecture , Astérisque, nos 147-148, 1987, p. 325-333". The author, János Pintz, is a specialist in analytic number theory.
@@EllieSleightholm After checking again, the first integer that disproves the conjecture is between 10^14 and exp(1.59x10^40), as shown here : "T. Kotnik et Herman te Riele, The Mertens Conjecture Revisited, dans Proceedings of the 7th Algorithmic Number Theory Symposium, coll. Lecture Notes in Computer Science (no 4 076), 2006, p. 156-167". Te Riele and Odlysko were the ones who disproved the conjecture.
As Hurst (in 2016) calculated M(n) up to 10^16, it seems that the lower bound for a counterexample should be 10^16.
@@EllieSleightholmis the mobius function any relation to the mobius strip?😂😂😂😂
@@user-lu6yg3vk9z They are named after the same person. I don't know of any other link between the two.
There's a famous book called Counterexamples in Topology by Lynn Steen which is full of false conjectures in the mathematical field of topology that seem intuitively plausible, along with a counterexample that disproves it.
Conjectures are normally understood to be unproven statements that are (or once were) widely believed to be true by experts in the area under discussion. I'd be surprised if most, or even all that many, of the statements for which that book provides counterexamples meet that description.
A small point, but useful for a mathematician: the plural of hypothesis is hypotheses, in which the last syllable sounds like 'seas'.
Actually, the supremum of the empty set is minus infinity (12:05) and the infimum of the empty set is plus infinity (12:47). I understand these were a typo! I loved your video, btw! Very interesting, indeed!
The empty set is bounded above though.
Any idea why there is an asymmetry between the constants in the lim sup and the lim inf that were found in the proof? This might suggest that, in a suitable sense, there are slightly more non-squares with an even number of prime factors than non-squares with an odd number of prime factors. However, it might just be an artefact of the method used in the proof.
Great video! Small correction tho at the end (13:05). The limit as n grows towards infinity of the infimum of M(n) is smaller than `- sqrt(n)` (missed the minus sign eventhough your statement "M(n) is smaller than sqrt(n)" is technically true as well)
This suggests in the real world there should be a counterexample. Are they still searching for a counterexample, or is it a case of having an existence proof, and not needing to find an actual example after that?
I am curious: were there a lot of theorems that were conditionals on the Merten conjecture?
No. Mathematicians rarely rely on unproven conjectures. The only unproven conjecture mathematicians use (and still it's a red flag) is the Riemann Hypothesis.
@@MK-13337 "Rely", of course not, if the word is strictly defined. You do not affirm that something is true based on a conjecture in mathematics at least.
Well there were hundreds of papers starting with "Assuming Taniyama-Shimura" is true.
My own colleague's dad has a a conjecture called the "Agoh" conjecture. A German mathematician(Kellner) has written a paper on proving that it is equivalent to Giuga's conjecture in number theory.
So mathematicians like to establish links and they will do so. In fact it may help them prove or disprove a conjecture in the future.
@@jceepf Well yes, sometimes people prove equivalence or that one theorem follows from another. If M(n)/sqrt(n) has a bound then that implies the Riemann hypothesis and by transitivity it implies everything built on top of the RH.
@@MK-13337 As you said, the Riemann hypothesis is also special. It has "practical" applications in theoritical physics for example. Therefore it makes a lot of sense to derive things based on its truth in "applied" math.
But it is true that for many conjectures, mathematicians build bridges only, which should not be viewed as "proofs" and they become useless the minute the conjecture is proven false. (Not obvious however in practical applications....physics does not necessarily demand perfect proofs.)
Coming from probability, the conjecture feels wrong. Looking at the graph it looks suspiciocly like a (rescaled) Brownian motion, which almost surely has sqrt(n log(log n)) asymptotics. That will eventually overcome the conjectured sqrt(n) bounds.
Assuming that the Möbius function is in some sense inpredictable, so that next values of it are seemingly independent of the previous ones, I feel like sqrt(n log(log n)) conjecture seems more natural. Of course, making sense of what is meant by "seemimgly independent" is hard, as usual in number theory.
I haven't touched this in a long time. I did not know that 6 out of pi-squared integers are square free. I feel like it has to be some crazy complex analysis proof maybe? I started trying to understand a proof of the prime number theorem once and didn't get very far.
I remember reading somewhere that for lots of conjectures the fact that it's only disproven by a very large number isn't necessarily meaningful because people tend to think in terms of "smaller" numbers, relatively speaking. i'm not claiming to be an expert in this field but I am curious what is the significance for a conjecture that fails for 10^164 versus a conjecture that first fails for 10^3. is there a fundamental reason "why" a conjecture may only fail for a large number or is it arbitrary?
Fermat conjectured that all integers of the form 1 + 2^(n^n) were prime. That lasted until 1732 when the Fermat number with n=5 was proven to be composite, by Euler, nonetheless. Now it's conjectured that it is only true for n=0,1,2,3 and 4. But, of course, it would only take one counterexample to disprove that conjecture as well.
Could you link to the paper with the actual proof? I was hoping for a bit more details in your video. Thanks
Ask wikipedia. Mertens_conjecture page has links to newer papers.
Thanks. Way to obvious.😊
It’s in the pinned comment and description of the video :) I’ll also be making a follow up video on the proof 👀
This video was aimed more at people new to the field of mathematics so will include more mathematical detail in the follow up video 😅
Not a mathematician. But am I wrong in assuming that the only way to find a counter example would be to do the calculations for every value of n up to the point where the conjecture fails? In other words, if the conjecture fails at 10^100, then you are doing 10^100 calculations to find the counter example?
If you derive a counter example in some other way than brute force checking increasing numbers, you only need to compute the function at that value, not the value of the function at every n before it.
Off the top of my head, I don't have an instance of a counter example computed by anything besides brute force.
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Edit: Mertens' conjecture here wasn't disproven via counter example, so the following isn't really relevant.
Mertens' conjecture only needs to be evaluated at the counter example. You don't need to re-compute Mertens' function and test Mertens' conjecture for all natural numbers up to your counter example. You just need to compute the sum of μ(k) from 1 to n one time; and show that that sum is not between ±sqrt(n).
---
If you can produce a counter example in another way than brute force checking, you don't need to check all the numbers before it. With Mertens' function, when computing M(n+1), you only need M(n) + μ(n+1). You don't need to re-compute all of the sums for every new n+1. If you have derived an _n_ that breaks the conjecture some other way, you can skip the sqrt and comparison part of the conjecture because it's sufficient to demonstrate that this _n_ breaks it, not that this _n_ is the smallest one that breaks it.
Anyway, in this case, I'm not sure how much computation skipping the sqrt and comparison at each step saves you when compared to the cost of calculating μ(k). (Also, with numbers as large as I am guessing the counter example is, I'm not sure the relative cost of sqrt vs squaring when you're dealing with arbitrarily large integers.)
I've only just started the video, so I'm not sure how the counter example was found, and how "expensive" computing μ is compared to the sqrt and compare.
I’m glad that you noticed u(6) = 1; that was bugging me! But now I’m reading more about these functions, from making sure I understood the definition of u(n), so that’s a positive. :)
I love your style - the presentation, pacing, and voice. Relaxed and informative. 😊
right, I was like "did they really not bother checking if it was true after n=5?" haha
😂😂 I mentioned this in the pinned comment but I filmed this at 5am so my brain wasn’t fully working when filming this 😂 really wanted to get a video out this weekend for you all!
Thank you so much, I really appreciate that ☺️
With such peculiar constants, 1.06 and -1.009, I'm curious as to whether these are exact values.
Probably not. If there are exact values for which those inequalities hold, it would imply Riemann's Hypothesis, and we don't know if Riemann's Hypothesis is true.
Just a clean Mathematics video.I like your clear cut explanations.
To be clear, we still don't know if M(x) = O(√x). We just know that 1.06*√x is an lower bound for lim(x→∞) sup M(x).
Thanks. This reminded me of an old TV show I recorded on VHS back in the 80s: "A Mathematical Mystery Tour" [NOVA]. It said the Mertens conjecture was proven to fail before 10^10^70. It mentioned that this is larger than the number of atoms in the [visible] universe. I transfered the show to DVD years ago. It's one of my favorites, & I still go back & watch it. It's available free online. Well worth it. tavi.
Great video!
Sorry to point out one more typo: in the title at the start, the apostrophe is in the wrong place, because the conjecture is named after Franz Mertens. So MERTENS' CONJECTURE woukd be the correct title.
Imagine if P=NP or navies stokes is smooth.
Very interesting, thanks for sharing!
A random walk is expected to be ~√n from the start after n steps. So if μ(n) were simply random, you'd expect a√n to be look like a bound (for some a) for many terms, but not infinitely many. In other words, the conjecture was no more predictive than random guessing.
Looking at this conjecture from a statistics perspective, it seems almost purposefully engineered to produce an contradiction at an extremely high value. Square-free numbers have a fixed density at 6/pi^2 (~60%). So this is very close to asking whether the sum of the first n non-zero terms of the Mobius series is bounded by ±sqrt(n)*pi^2/6. The Mobius function isn't completely random, as there are closed-form 1-to-1 mappings from the -1's to the 1's. But the log difference of the values generating these pairings diverges very slowly, making it act more and more like noise for very high n.
If it were pure noise you'd expect that the sequence would eventually fall outside of k*±sqrt(n) for the first time at median value n=e^(a+bk^2), where a and b are constants I haven't calculated. (The key part of this result [e^(k^2)] is a result of Chebyshev's Theorem. The constant b has to do with the sampling frequency at which a partial sum is sufficiently uncorrelated with a previous partial sum.)
This obviously gets very big very fast for large k. k effectively starts out very large due to the early correlation of +1's and -1's in the series, and slowly converges to pi^2/6 as the series tends toward infinity. The expectation would be that the first exception to Mertens conjecture would occur at the intersection between that descending value and k(n) from the prior equation. That would be an incredibly large number, but a fairly predictable one. From that point forward, I'd expect to see another interval with contradictions at what converges to an average of exp(b*pi^4/36) times the size of the previous interval.
The 0 axis line pattern in the graph reminds me of something that was frequently used in advertising and marketing for decades. Either that, or it’s one of the many things Fractint could generate that wasn’t strictly fractal art 😅
Was there any reason whatsoever for Mertens to believe this conjecture was true in the first place, other than just looking at graphs and such? And what implications does this have?
Interesting result. The paper you link, "Disproof of the Mertens Conjecture" by A.M. Odlyzko and H.J.J te Riele, 2013, claims that Merten's conjecture is linked to the Riemann Hypothesis. That's pretty cool. In fact, the paper mentions that the Riemann Hypothesis is equivalent to |M(x)| = O(x^(0.5 + epsilon)) for all epsilon greater than zero.
It’s such an interesting read isn’t it!
@@EllieSleightholm Ah well, I can only really read most serious papers at a superficial level. But I do enjoy poking around and seeing what I can glean 🙂
This is the first time I've heard of this conjecture!
may I suggest to use the Runge-Kutta iteration = possible look for an eigen vector in an inner product space
hi, i wanted to know if you major was just mathematics or if you majored in mathematics with a concentration in something specific
Euler's sum of powers conjecture I think.
Very nice job. It would be interesting to know three more points of history. First, when was the Merten's conjecture presented, and when was it figured to be likely true.
Also, what other conjectures fell when Merten's conjecture was shown to be false?
And is anyone actually working to find an exception?
It’s a bit disappointing as I did not really learn anything outside of some definition of some functions. Can u talk about the intuition behind why this thing should only grow like sqrt(n) and why it can appear so random?
A question regarding pronunciation: when I first encountered the word 'infimum', I figured it was similar to minimum, and like you, I said "IN-fimum". I was quite surprised to find when I looked it up in a dictionary (I think it was the Oxford dictionary) that the pronunciation given was "in-FYE-mum". Now, I have NEVER heard a mathematician say "in-FYE-mum". In the USA I have the impression that most mathematicians say "in-FEE-mum", which rhymes with supremum. Have you heard either of these pronunciations?
At 12:30, shouldn’t infimum be “greatest lower bound”, or am I mistaken?
Aww. I was hoping you'd show some BIG NUMBERS (you know how we love big numbers!) that have been tried and failed to falsify the conjecture. Then maybe a delicious, even hand-wavy BIG NUMBER where this new disproof suggests we could expect to find a counter-example.
Is the supremum a *hard* upper bound? in the context of this conjecture, is it the case that there must be *some* integers n for which root(n) < mu(n)
Very interesting conjecture... Thanks for your presentation.
But how do they show that the region between sqrt(n) and 1.06 sqrt(m) is populated? Or something similar for - sqrt(n)? Without that, it's just proving a looser bound than the conjecture has.
They are proving the least upper bound and the greatest lower bound exceeds +/- sqrt(n). If sqrt(n) really was a bound, then the numbers provided wouldn't be the least/greatest
@silver6054 does the least upper bound mean that there MUST exist some n for which M(n) is equal to 1.06 root n?
@@tomgrunwald8166 Yes, if there were no n for which M(n) > 1.06 sqrt(n) then the *least* upper bound of M(n)/sqrt(n) would be 1.06. The word "least" is important here. If the region between sqrt(n) and 1.06 sqrt(n) were empty then it would be possible to find a *smaller* upper bound than 1.06.
@@tomgrunwald8166 No, just that for any smaller bound there exists an n for which M(n) is bigger (so in particular there exists n s.t. M(n) > sqrt(n). But there doesn't have to be n such that M(n) = least upper bound.
@@silver6054 You didn't really answer the question. How do they prove this without giving an example?
12:28
inf is the greatest lower bound, not the greatest upper bound ;)
beautiful content, thanks
This is what Bertrand Russell had to say about mathematics: ''Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true.''
I was expecting a specific counterexample -- like M(k) > sqrt(k) for a specific counterexample value of k? Does the proof only establish that such values must exist, without giving a single specific example of one?
A very clear and concise explanation. Thank you!
Thank you very much for sharing your intelligence and education filled videos !! Scientific and correct !! Outstanding !!
Greetings from California … wishing you and folks good health , success and happiness !! Much Love ✌️😎💕
On my first day of mf probability class, the instructor asked what is the probability of the sun rising the next day given it has risen 10 million days before.
The article proving the result is avaliable online for those who are interested. It is amazing how many results for number theory are still unsolved.
It’s also in the description and the pinned comment for those interested!
The axiom of choice is obviously false, but maybe the axiom of determinancy is true. I think one or the other but not both are consistent with set theory.
6:55 Isn't mu(6) = 1?
6 has two prime factors: 2 and 3.
I corrected it later in the video if you watch the full thing :)
I really want to see an odd perfect number. Especially if it's right around the largest twin prime.
And what does the Mertens Funktion signify? What is it about?
It has a direct link to Riemann hypothesis. Namely, Riemann hypothesis is equivalent to M(n) growing asymptotycaly slower than n^(1/2 + epsilon) for any epsilon > 0. Thus, Merten's conjecture would've proven Riemann hypothesis .
@@tetraedri_1834 Thank you very much.
Just saying: it's the Mertens function, named after Franz Mertens. So it's Mertens's, not Merten's, conjecture.
en.wikipedia.org/wiki/Mertens_function
en.wikipedia.org/wiki/Franz_Mertens
yep already corrected this in my pinned comment
Is there a table somewhere of long-standing conjectures that were widely assumed to be true, along with their status (proven false after N years, proven true after N years, still open after N years)? This would be useful for getting a feel for how reliable "everybody assumes it is true" has proven to be.
In 1985, Andrew Odlyzko and Herman te Riele proved this....However, the Monty-Hall paradox is wrong
Not sure what a square number is. I am guessing it is one of the numbers that is another numgpber squared,.....4,9,16,25,36, .....
What tablet are you using to write?
I fail to understan what möbius' function is good for without summing it.
What software do you use to write on the screen? I've seen others use it too.
I would like to know also so that I can do synchronous teaching online.
Very interesting video, thank you!
Thank you
How consequential was this conjecture? Were there any somewhat significant pieces of maths that relied on the conjecture?
It's not entirely wrong. It'd be interesting to know at which number exactly the line moves off the graph in either field. I don't even know if there would be real number we calculate that could. It's just when approaching infinity. Which you know, it's not exactly that far off the graph in either direction, so it could just be moot.
But basically, what she's saying, is you see that horse shoe thing? That's a quadratic function. And the line, according to the conjecture, will stay within it. Which, when you calculate it and approach infinity, it slightly deviates off of it in either direction. Which, infinity can be weird like that.
Like, generally, it's wrong... the conjecture is disproven, but in a moot way. I don't know if we could ever have a number in real terms, that wouldn't obey that law. It'd be an interesting thing for mathematicians to find, is the exact point where the function goes off the graph (if they even can, that's also a question to ponder), but seeing how slightly it deviates off of it, I'd say you'd probably be looking at some absurdly huge numbers. Or maybe not? But finding that would be interesting, because all the numbers up to that point, would indeed obey the rules of the conjecture. So finding that point, you'd find the place where the conjecture is false, and would have a set of numbers that obey the conjecture.
I couldn't do that. I'm just an amateur, and can barely do long division. But, I understand the shape.
@@BKNeifert Some proofs rely on prior conjectures. Like a lot of proofs rely on the Riemann Hypothesis. If RH is ever proven, hundreds more theorems will finally have a solid proof. Or if it is disproved, decades of mathematical advances would be shattered, as alluded to in the video. The idea behind RH can also be drawn on a graph, but the importance of it is far greater.
So the importance of a conjecture isn't always obvious at first glance. The graphs are sometimes just the easier bit to communicate, relating to deeper mathematical ideas. If the idea breaks, it breaks, even if it happens at a number greater than has been calculated.
@@dfw-k6z K.
@@dfw-k6z K. So, like the Riemann Hypothesis is that primes will be organized in a specific way that can be predicted by a formula? That's funny. That's a little less knowable than this, I'd think.
I mean, obviously, because we know this. Lol.
@@dfw-k6z NVM I watched a bad video, and so consulted my bookshelf. No clue what the Riemann hypothesis is. Probably never will.
If you say the concept is what matters, I'll believe you. I'm still learning the basics of geometry. But, I understood that the squiggle line is between the quadratic function. That I do see. Everything else, I'm a little clueless on, and would like to remain that way. And I also know when you approach a limit, in infinity, it can make things very strange.
But how did they prove those new limits??
What do you think of Roger Penrose's belief that mathematicians can look at a conjecture and say "Yeah, probably right" or "No, probably wrong" with good accuracy? He claims this as proof that consciousness cannot be computational.
ehhh. this is just heuristics. if people had 100% accuracy for this I'd believe him but this is a stretch
Suppose we had a list of all the conjectures that Penrose thinks is "probably right". Now suppose you were paid $X every day, but if one of those conjectures was ever disproved you would be immediately executed. What is the smallest value for X that you would you accept the deal?
That makes no sense. How does that prove consciousness cannot be computational?
@@Galinaceo0 there are theorems that cannot be proven by any computational process. since we can determine whether they are true or not, we must do so by a process that cannot be computational. there's a recent PBS Space Time episode on it, about quantum processes in the brain.
@@Valdagast Theorems are things that can be proven, by definition. What there can be are statements that are true despite being unprovable. You say "we can determine whether they are true or not". I don't know why you say that. I don't know any way to determine whether a statement is true or not other than trying to prove or disprove it. Intuitively thinking something is true does not mean it is true. There are many cases of conjectures which people thought to be true that turned out to be false.
I thought 1 was an odd number? In my mind - an even number is any number divisible by two with a remainder of zero.
So shouldn’t u(n=1)= -1 ?
1 is an odd number. But it has an even number of prime divisors. (it has 0 prime divisors, which is an even number)
@@Galinaceo0 Thanks! That's the detail I didn't quite understand. Appreciate your response
Is Merten's Function supposed to (or was conjectured to) behave like a pseudo-random choice between -1, 0 and 1? Moebius and Merten's functions both seemed to be pulled out of thin air here.
The Möbius mu function comes from the coefficients of the terms in the reciprocal of the zeta function. (1-2^x)(1-3^x)(1-5^x)(1-7^x)(1-11^x)... = 1 - 2^x - 3^x - 5^x + 6^x - 7^x + 10^x - ... = mu(1)1^x + mu(2)2^x + mu(3)3^x + mu(4)4^x + ...
(This expression is for 1/zeta(-x), as IMO the expression for 1/zeta(x) would've had minus signs in the exponents and been even less readable.)
The Mertens function? AFAIK, it's just a running total of the values of the Möbius function.
On our first date and on our last date, the two plates were squared. On our second date and our infinite data, there were two slices of bread given. On our first movie night, and never ending movie, gone but not forgotten. Maybe you get the picture. Art vs science.
I may have missed it, any impact of this finding on other theories that now are wrong?
Square free could be said as no even powers of its prime factors
6:44 crazy that they never checked up to 6 - you would've thought it would break at a higher value :p
I have no clue if you check your comments but what is. 6/2(2+1) Now the only reason I’m asking is because I saw a RUclips short and many people said 1 and many others said 9 personally I say 9 but I decide to go around a bunch of math professional people on RUclips and put this in their comments
What am I missing? If I conjecture that a sequence is bounded for n\in\N by \pm C*f(N) and someone comes along and says, "no way buddy, that bound should be 1.0006*C.f(n)"... and neither of us has any examples..... ??? I expect the truth is that the original conjecture wasn't shown to be disproven until an actual n\in\N was found for which `\sqrt{n}` counterexample was found?
That depends on _why_ the "someone" asserts that there is such a bound.
Why does mu(6) = -1? It is square free and has two prime factors. Should it equal 1?
Edit: I just saw that you caught it yourself shortly after making that mistake 😊
I see. Amazed. Thank you.
How long till companies ask "write a python script to graph the mertin function?" During job interviews?
n days, where n is the smallest counterexample to Merten's conjecture
Shouldn't mu(6)=1? It's square free and has an even number of factors.
Should've expected you to correct that.
I did if you watch the whole video :)
Oh, it is a petty that you didn´t show HOW it was proofed and/or showed an example. Moreover, albeit that Merten was incorrect, it is amazing how close he was. So he must have understood the tendency very well.
Yes I can imagine the details of this proof are a nightmare and it would be intetesting to know when those bounds are broken given that it does not occur if n
It's so weird to write μ(n=1) or μ(n=4) instead of μ(1) or μ(4)
Is it a british thing ? I'm French and we would never write that.
I’d ordinarily write μ(1) or μ(4) but given this video was aimed more at people new to the field, I wanted everything to be clear :)
But… why? You didn’t go into where those numbers (1.06 and -1.009) came from, or even a hint. I get that the proof can be complex but surely it’s not so bad that you can’t give a hint of the reasoning?
This video was aimed more as an ‘intro’ for a less technical audience (I’m trying to make maths more accessible). I’m making a follow up video on this topic and will cover more details there.
@@EllieSleightholm If you had a second channel for more technical discussion, I'd love to subscribe to that. 🙂
Omg you’re so frickin right. Ellie is never wrong
😂😂 I wish
@@EllieSleightholmwhat was the class average on the test in you’re math classes? 40% average
Merten's conjecture is only partially false, it is still true that the Merten's function is bounded by Order(sqrt(n)), IE there is no other power or log or exponent or whatever of n that is better. a good follow on talk would be a description of Odlyzko and Riele's proof 🙂 !!!
What does Order(sqrt(n)) mean? O(sqrt(n))? That would imply Riemann Hypothesis, so we don't know if it's true or not.
@@Galinaceo0 Odlyzko and Riele's proof showed 1.06 x sqrt(n), that's O(sqrt(n)), so I don't know what you're trying to say, because it seems you're saying Odlyzko and Riele's proof proves Riemann ???
@@MostlyIC Their proof didn't show that it is bounded above by 1.06*sqrt(n). It only showed that if there is an upper bound, then it is at least as big as 1.06*sqrt(n).
@@Galinaceo0 I've always been a bit confused by the terms supremum / "least upper bound", and infinum / "greatest lower bound". in my mind if 1.06 is the "least upper bound" then there's no number smaller than 1.06 that is still a bound, and it being a bound means there are no "n" that makes mertens(n) > 1.06 * sqrt(n). but you're saying that's not relevant, you're saying they didn't show that a bound even exists, only that if it does then its at least 1.06, I believe you, but I didn't get that impression from the video, I guess I'll have to re-watch it 😞 !!!
@@Galinaceo0 thanks for clearing up my confusion 🙂 !!!
Great video.
Very well explained.
Instant sub. 😊
I think what this says is that the Merton function does not behave as a random walk
A random walk is not bounded by sqrt(n). It's expected distance from it's starting point is equal to sqrt(n), meaning half the time it will be outside that bound.
This proof means that M(n) behaves slightly more like a random walk than was expected.
happy to watch this video
excited to watch the next video
good job
All my homies randomly start thinking about Reimann Hypothesis
All I know is that was a quadratic function, and you had a squiggle line in-between it. So, I guess the thing is, that the squiggle line goes outside of the quadratic function, when you approach infinity.
Good job 👍
I'm missing a sketch of a proof.
It's difficult to give even a sketch. The Mertens conjecture was long suspected to be false because it would imply a variety of other statements that seemed unlikely. Various people searched computationally for counterexamples to these other consequences of the Mertens conjecture, but their algorithms and computers were not powerful enough. The breakthrough by Odlyzko and te Riele came because they figured out how to use a powerful lattice-basis reduction algorithm of Lenstra-Lenstra-Lovasz to find a counterexample to one of these consequences of the Mertens conjecture.
@@JohnDoe-ti2np Thanks a lot! Interesting. Sketching this in a generally comprehensible way would probably end up in a longer video.
Inf is the greatest LOWER band, not upper 🙄
See pinned comment :))
why is mu(6) = -1 ? its prime factors are 2, 3, not repeated. Shouldn't mu(6) = +1? (see 6:44)
Whoop, you fix this later. Nevermind.