A Prime Surprise (Mertens Conjecture) - Numberphile
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Күн бұрынDr Holly Krieger discusses Merterns' Conjecture.
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More links & stuff in full description below ↓↓↓
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Dr Holly Krieger is the Corfield Lecturer at the University of Cambridge and a Fellow at Murray Edwards College: www.dpmms.cam.ac.uk/~hk439/
On Twitter: / hollykrieger
PAPER: Disproof of the Mertens Conjecture: www.dtc.umn.edu/~odlyzko/doc/a...
More on Mertens Conjecture: mathworld.wolfram.com/MertensC...
Useful OEIS sequences:
oeis.org/A002321
oeis.org/A084237
Holly on the Numberphile Podcast: • Champaign Mathematicia...
The Riemann Hypothesis: • Riemann Hypothesis - N...
Mathematical correction from Dr Krieger: The first counterexample to Mertens conjecture must happen for some number no worse than about 10^(10^40), but the actual expectation is that the first counterexample is around 10^(10^23).
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@numberphile 4 жыл бұрын
More videos with Holly: bit.ly/HollyKrieger Holly on the Numberphile Podcast: kzfaq.info/get/bejne/h9OWhLSrz9bIoIE.html
@AgentM124 4 жыл бұрын
I didn't catch, but did they know anything about if it breaks the √n in the pos or the negative?
@warb635 4 жыл бұрын
Fyi: I see "Mertern's Conjecture" instead of "Merten's Conjecture" in the KZfaq description of this video.
@RobinDSaunders 4 жыл бұрын
@@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.
@rob6129 4 жыл бұрын
Could you do a video about how this problem relates to the Riemann Hypothesis? I find the interconnections of mathematics to be really interesting
@aesthetic1950 4 жыл бұрын
@@rob6129 Second that.
@StefanReich 4 жыл бұрын
5:02 "Living around zero but in a really complicated way"... Wow, you just described my bank account
@GaneshNayak 4 жыл бұрын
Lol
@stuffofmaking 4 жыл бұрын
Stefan not so Reich
@Oblivion1407 4 жыл бұрын
But you can be certain that at one point your money amount blows outside of the boundaries, it might be negative though.
@jwink7795 4 жыл бұрын
HOLLLLAAAAA
@TheTrueAltoClef 4 жыл бұрын
@@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?
@rad858 4 жыл бұрын
The square root of 10^(10^40) is about 10^(10^39.7). Freaky how it's so much smaller but barely looks any different
@5astelija75 4 жыл бұрын
Why isn't it 10^((10^40)*0.5) ? How does this math thing even work
@ReconFX 4 жыл бұрын
@@5astelija75 It is. 10^40 = 10*...*10 40 times, so 0.5*10^40 = 0.5*10*10^39 = 5*10^39. Since 10^0.7 is roughly equal to 5 we can also write this as 10^0.7*10^39 or simply 10^39.7
@fraserkennedy5497 4 жыл бұрын
To compare them properly - what power of 10 will give you 0.5? Take log (base 10) 0.5
@rad858 4 жыл бұрын
@@5astelija75 10^39.7 = 5.01187...x 10^39, so you're right. In fact 10^(5.01 x 10^39) is 10^37 orders of magnitude larger than 10^(5 x 10^39). Big numbers are bizarre. What does the word "about" even mean any more
@5astelija75 4 жыл бұрын
wow ok my mind is officially blown. restarting....
@fwiffo 4 жыл бұрын
"If this was true, it would imply the Riemann hypothesis!" "It's false." *uncomfortable digestive noises*
@fllthdcrb 4 жыл бұрын
Just remember: the inverse of an implication is not equivalent to the original implication, i.e., the antecedent (Mertens conjecture) being false does not imply the consequent (Riemann hypothesis) being false. The Riemann hypothesis could still be true; we just don't get any help from here.
@Gooberpatrol66 4 жыл бұрын
*jazz music stops*
@angelmendez-rivera351 4 жыл бұрын
Daniel Dawson I think the OP knows this. The discomfort comes from the fact that this doesn't help us with it
@leif1075 4 жыл бұрын
@@fllthdcrb why would you call it the inverse of an implication?..is that the right term?..wouldnt the negation or opposite be more correct..inverse is more like 3 vs 1 over 3 or reciprocal in math..think using that word is unclear..just saying..
@fllthdcrb 4 жыл бұрын
@@leif1075 It's the _logical_ inverse. Totally different from a reciprocal (multiplicative inverse, which is part of algebra, not logic), and also totally different from a negation: Original implication: P → Q ⇔ ¬P ∨ Q Inverse: ¬P → ¬Q ⇔ P ∨ ¬Q ⇎ ¬P ∨ Q Negation: ¬(P → Q) ⇔ ¬(¬P ∨ Q) ⇔ P ∧ ¬Q If you're not familiar with the symbols, P and Q are statements, ∧ means "and", ∨ means "or", ¬ means "not", → is implication, and ⇔ is equivalence. The first equivalence on each line is by the definition of implication, and the last one on the third line is applying one of De Morgan's laws. Anyway, what I was getting at is, the inverse is not equivalent to the original implication, which you can see above. One implication using the same statements that is equivalent is the contrapositive: ¬Q → ¬P ⇔ Q ∨ ¬P ⇔ ¬P ∨ Q ⇔ P → Q.
@edghe119 4 жыл бұрын
Dr. Holly one of the best.
@spicemasterii6775 4 жыл бұрын
Can Ali Tomruk Or Isla Fisher
@pH7oslo 4 жыл бұрын
I know who Dr. Holly Krieger is, but I have no idea who Amy Adams or Isla Fisher are.. I'm perfectly happy with that.
@christosvoskresye 4 жыл бұрын
@@mostlynothing8130 Pugsley Addams will play me in that movie.
@codycast 4 жыл бұрын
pH7oslo “I’ve never heard of famous people. Aren’t I edgy and cool?”
@fgc_rewind 4 жыл бұрын
dont you have world records on CTR?
@tatjoni 4 жыл бұрын
Holly's laugh makes my heart smile!
@aaaaanomaly 3 жыл бұрын
The Pólya conjecture is similar, but you count DISTINCT prime factors (e.g. 10 and 20 both have two factors), and the conjecture is that the running total never goes above 0. It's true for a while, but it eventually fails, although at a more reasonable number: 906,150,257.
@yeoman588 4 жыл бұрын
I really want to know how it was proven that this number exists and breaks out of the parabola, since it is so big that we can never know what it actually is.
@w00tehpwn 4 жыл бұрын
Just read the paper, link is in the description. But first, in order to understand it, go get a PhD in mathematics.
@yeoman588 4 жыл бұрын
@@w00tehpwn Ideally I'd like an explanation that _doesn't_ require a PhD to understand. 😅
@lukesteeves1291 4 жыл бұрын
@@yeoman588 Not really an explanation but from a quick look at the paper, here's the idea: There's a thing in math called the limsup - if you know what a limit and a supremum are, then it's the limit as x goes to infinity of sup f(x). If you don't know what those are, it asks what's the highest value y of the function f so that no matter how big x is, there's a bigger x_0 with f(x_0) really close to that value y. In other words even when x is big, the function keeps wandering back to that maximum value. The paper looked at limsup M(x)/sqrt(x), the ratio between the sums of mu values in the video and the square root of x, and they found that limsup M(x)/sqrt(x)>1.06. In other words, M(x)>sqrt(x) infinitely often for large x. To show this limit, they needed to use a lot of computation with complex integrals which look yucky and I ain't gonna try to understand them :p . But that's math for ya!
@WheelDragon 4 жыл бұрын
@@w00tehpwn But first, we need to talk about parallel universes
@effuah 4 жыл бұрын
Short version: find another function, which bounds the biggest values of M (involves Zeros of the Riemann ζ function) and then approximate this function. This is possible by knowing a lot (a few thousands) of zeros of ζ and it is a "nicer" function. Since this gets large enough sometimes, the conjecture is disproven.
@nosuchthing8 4 жыл бұрын
One of the best channels on KZfaq
@numberphile 4 жыл бұрын
thanks!!
@EdbertWeisly 3 жыл бұрын
@@numberphile Numberphile Deserve have 314,159,265,358,979,323 SUBSCRIBERS
@aradhya_purohit 2 жыл бұрын
@@EdbertWeisly or 2,718,281,828,459,045,235,360?
@EdbertWeisly 2 жыл бұрын
@@aradhya_purohit sure
@OMGclueless 4 жыл бұрын
8:37 "That's my new favorite number." Why do I get the feeling Brady says this a lot?
@omikronweapon 4 жыл бұрын
only because he does
@YtseFrobozz 4 жыл бұрын
Because there's an infinite number of numbers from which to pick a new favorite number, so the probability of any particular number being Brady's favorite number is zero.
@ShankarSivarajan 4 жыл бұрын
@@YtseFrobozz No, he doesn't pick them uniformly.
@TavartDukod 3 жыл бұрын
@@ShankarSivarajan I mean there's literally no way to pick natural numbers uniformly because of sigma-additivity of probability.
@ucantSQ 3 ай бұрын
It's the name of the show.
@JNCressey 4 жыл бұрын
2:08 "lets forget about zero and start with two. " *sad one noises
@dlevi67 4 жыл бұрын
I thought much the same, but in a less funny way, then I think I worked it out: 1 has exactly 0 prime factors, so it has an even number of them.
@pulsefel9210 4 жыл бұрын
math people seem to forget 1 is a prime, its only factors are 1 and itself, 1.
@Eliseo_M_P 4 жыл бұрын
@@pulsefel9210 One is not a prime. A prime number has exactly one factor, not including itself. 1 has zero factors, not including itself.
@pulsefel9210 4 жыл бұрын
no primes can only be factored by multiplying itself by 1, so 1 fits since you cant multiply anything to get 1 except 1.
@PickleRickkkkkkk 4 жыл бұрын
no primes are numbers with 2 factors
@dhoyt902 4 жыл бұрын
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!!
@thishandleistaken1011 4 жыл бұрын
sarcasm, right?
@gregoryfenn1462 4 жыл бұрын
@@thishandleistaken1011 Many of us have math degrees and PhDs and didn't know it, that's the point, it's a new discovery.
@b3z3jm3nny 4 жыл бұрын
Gregory Fenn by new you mean 1985
@BauerMonty 4 жыл бұрын
It breaks the Squarrier
@TavartDukod 3 жыл бұрын
@@gregoryfenn1462 how did so many people manage to get PhDs without knowing they got them? /s
@jppagetoo 4 жыл бұрын
I love numbers. It's probably why in the middle of my college career I switched from engineering to math (and got a degree in it). I hated engineering but I loved all the math I was doing. I loved number theory but at that time it wasn't an area of math I could study, it was just a single class. Numberphile is so cool.
@kdawg3484 4 жыл бұрын
@Paul O'Reilly As a chemical engineer, I had to take calculus all the way through differential equations, then Engineering Math which is specific extensions of those areas, and also probability and statistics. Differential equations are pretty much the heart of engineering, because most things that matter in engineering problems revolve around change. Check out 3blue1brown's growing series on DEs to get a sense for this. All chemical engineering classes revolve around DEs. ChemEs, more that any other discipline are familiar with the Navier-Stokes equations (see the Numberphile playlist on these.) Now, as an engineer, I don't directly solve DEs...well, basically ever. But that's because most of that work has already been done or is underpinning the tools and equations and software we use on a daily basis. Engineering is about optimization, not exact answers. Not that there's anything wrong with that; nearly everything manmade you see right now in front of your eyes wherever you are is the result of engineers. We have to fold all considerations together to come up with an optimal design, because there's no perfect design. But math still underpins it all, because math is the language of physics, chemistry, biology, economics, and much more, and those are all the legs that the engineer's table is built on.
@theowleyes07 4 жыл бұрын
I am a Medical Student here but I love to Play with Numbers because it helps me to relax. I was trying to make Cross Product of Vectors easy as it was a nightmare in 12th Standard Maths and Physics I self discovered mu ijk in different form later I saw a video from Andrew Dawtson in KZfaq to confirm if anything like that is there or Not. Yep it is there it was called something but it was epsilon ijk made me happy. Small ideas for fun
@christianbarnay2499 4 жыл бұрын
@Dr Deuteron Engineering is the science of understanding the high complexity of exact mathematical equations and approximating them with much more simple equations that are practical to compute in reasonable time and still very close to the hard real stuff.
@lincolnsand5127 4 жыл бұрын
@Dr Deuteron Electrical Engineering involves fourier series, differential equations, linear algebra, and more. Control engineering (a sub field) is almost entirely math that involves lots of calculus.
@sohamsengupta6470 4 жыл бұрын
That's all fine and dandy but what on earth is that design on that tele
@DanTheStripe 4 жыл бұрын
Surely you've got to name it something awesome like Mertens' Nemesis if it's the first number to break the rule?
@PhilBagels 4 жыл бұрын
The Anti-Mertens Number.
@lapiscarrot3557 4 жыл бұрын
A Mertensplex, perhaps?
@badmanjones179 4 жыл бұрын
brb writing this down on my math rock song name ideas list
@gregnixon1296 4 жыл бұрын
Merten's Foil? How's that?
@stapler942 4 жыл бұрын
Mertens' Bane.
@MinorCirrus 4 жыл бұрын
Dr Krieger, also known as Mathematician Amy Adams. Also, perhaps I missed something, but why exactly does the function ignore (attribute zero) numbers with repeated prime factors?
@nahidhkurdi6740 4 жыл бұрын
This is a matter of definition only.
@rad858 4 жыл бұрын
The usual mathematical reason: because it's more interesting that way
@AlexJones-ue1ll 4 жыл бұрын
What value would you assign to it then? 1, 0 and -1 are already taken. Plus or minus 1/2? And which one when?
@sambachhuber9419 4 жыл бұрын
By doing this you make sure that this is multiplicative, where if m has no common divisors to n and we let f(x) denote the function, then f(n*m)=f(n)*f(m). This might still sound a little arbitrary, but the function in this form pops up pretty naturally in a number of places like the mobius inversion formula.
@MinorCirrus 4 жыл бұрын
@@sambachhuber9419 I see now. Thanks!
@gl1500ctv 4 жыл бұрын
"You go to the gym to get in shape but what about your brain?" Uh, I come here.
@apollion888 4 жыл бұрын
Holly is my favorite Numberphile speaker. I am delighted to see her again, and it's primes too!
@palahnuk1 2 жыл бұрын
you need to get out more
@kevwang0712 4 жыл бұрын
With that link to the Riemann Hypothesis, I can almost hear the collective groan in the world of maths when this was disproven
@unvergebeneid 4 жыл бұрын
Therefore the number should be called "Rie...maaaaan!"
@SuperSpruce 4 жыл бұрын
No, it should be called the reeeeeeeee-mann
@TimothyGowers0 4 жыл бұрын
Actually that part was slightly misleading, as a weaker conjecture, not disproved by the counterexample to Mertens's conjecture, suffices for the Riemann hypothesis. (Basically the sum doesn't have to be smaller than the square root -- there's a bit of extra elbow room.) So the counterexample is very interesting, but not a tragedy for number theorists.
@TimothyGowers0 4 жыл бұрын
Still a great video though!
@simplebutpowerful 4 жыл бұрын
@@TimothyGowers0 If a weaker conjecture would have also proven Riemann's, then it's not misleading to say Merten's conjecture would have proven Riemann's. So, at the end of the day, Merten's Outlaw is still a disappointment (though neat to discover).
@solandge36 4 жыл бұрын
When the content is soo good, I put the effort to watch every second of the ad that you so considerately put at the end of the video.... Cheers!
@plaustrarius 4 жыл бұрын
I love how excited Dr. Krieger is about this one! excellent stuff thank you!!!
@ZachGatesHere 4 жыл бұрын
It's always crazy when one conjecture just happens to tie in another one when they seem to have nothing to do with one another.
@timbeaton5045 4 жыл бұрын
Well, that's what seem to happen in Mathematics, quite a lot. Think elliptical functions and modular forms. On the surface, no relation, until it was all tied up with Wiles proof (with a bit of help from others, of course) of FLT. Or go see 3blue1brown's video on colliding masses being a neat algorithm relating to Pi. Happens all over the place! That's all part of the fun!
@hamiltonianpathondodecahed5236 4 жыл бұрын
but in this case zeta has a deep connection with primes and this Merten thing is built on the primes itself
@gregoryfenn1462 4 жыл бұрын
Even weirder is how this conjecture if true would have proven the Riemann Hypothesis, which we really really think and hope is true. So the falsity of this is perhaps a little alarming and surprising.
@slsalkin 4 жыл бұрын
@@homelessrobot "Imply" in the sense of logical implication, so A implies B means that if A is true, B is true. Not a colloquial English sense of "suggests" or "hints".
@christianbarnay2499 4 жыл бұрын
This is not the case here. The Mertens and Riemann conjectures are twins. The original question is whether there is an identifiable pattern in the distribution of primes among natural numbers. One research path led to creating the Zeta function and formulating the Riemann hypothesis about its zeroes. Another research path led to studying the prime decomposition of all numbers and formulating the Mertens conjecture. Both are just different attempts at answering the same initial question: "Is there a a way to instantly check if a number is prime or not".
@fghjghjfhgjfhgj 4 жыл бұрын
More videos with Holly! Please more Holly!
@shtfeu 4 жыл бұрын
I love when Nicole Kidman explains maths to me.
@anglo2255 4 жыл бұрын
Ha ha , I totally see that now
@robertheikkila4045 4 жыл бұрын
You mean Amy Adams?
@Wecoc1 4 жыл бұрын
ok mister killjoy
@ditzfough 4 жыл бұрын
Holly is alot prettier than nicole kidman.
@chinmaybhoir6955 4 жыл бұрын
Robert Heikkilä exactly my thought
@robertschlesinger1342 4 жыл бұрын
Excellent description of the Mertens Conjecture, and the counter-example found. Many thanks for the link to the mathematical paper disproving the Conjecture.
@Tehom1 4 жыл бұрын
7:25 "Does it just break away from that square root limit or does it blast past it?" Geometrically or arithmetically? Arithmetically, it blasts past it. Geometrically it goes at least 6% past it and probably more than that, all according to the paper you linked.
@danielroder830 4 жыл бұрын
I wonder if the wiggles after that big number break away in both directions even further and further or if it calms down somewhere near TREE(3) or whatever.
@dlevi67 4 жыл бұрын
@@danielroder830 From what was said in the video, it doesn't break away very far, though what "far" means when dealing with numbers of that size is debatable (and almost certainly not intuitive - work out what the square root of 10^10^40 is, and I think you may be surprised). FWIW, 10^10^40 is nowhere near TREE(3). It's not even anywhere near 3↑↑↑3.
@danielroder830 4 жыл бұрын
@@dlevi67 AFTER that number, it breaks away at 10^10^40 and after that, i wonder what happens after that.
@dlevi67 4 жыл бұрын
@@danielroder830 What the paper says is that it continues to grow at a geometric rate that is about 6% bigger than the square root. It doesn't break away "suddenly" at 10^10^40; it just grows faster than the square root for large values of n.
@GerSHAK 4 жыл бұрын
+
@R2Cv1 4 жыл бұрын
One question not addressed is, DOES IT BREAK AWAY UPWARDS OR DOWNWARDS??
@fnors2 4 жыл бұрын
From what I gather from other people who read the paper : both, infinitely many times. As for the first break? No idea.
@ubertoaster99 4 жыл бұрын
Picture showed upwards.
@snbeast9545 4 жыл бұрын
@@ubertoaster99 There was a large disclamer that the picture was an artistic rendition.
@ubertoaster99 4 жыл бұрын
@@snbeast9545 The artist knew what they were doing. I'd bet on positive :)
@elltwo8393 4 жыл бұрын
If you’re referring to boundedness, then the authors of the paper say they think it’s not unlikely the limsup is infinite.
@brucerosner3547 4 жыл бұрын
Is it possible that the first failure of Riemann's Hypothesis is a number as big as this one?
@romajimamulo 4 жыл бұрын
Yes, unfortunately. It could be even bigger
@leofisher1280 4 жыл бұрын
it almost definitely is. no failures have ever been found
@romajimamulo 4 жыл бұрын
@@leofisher1280 We probably haven't looked quite that far, but yes, we've looked very far
@TheTortuga58 4 жыл бұрын
You take that back
@christosvoskresye 4 жыл бұрын
I'm thinking of a number between 1 and Tree(3).
@unvergebeneid 4 жыл бұрын
Really beautiful example for the power of proof over both intuition and brute force.
@senororlando2 4 жыл бұрын
Love Holly’s guest spots
@B4der 4 жыл бұрын
Great video! But 10^(10^40) has 10^40 decimal digits. So if we have 10^80 atoms in Universe, we have about 10^40 atoms for each digit in that number. So we CAN write it down if we really wanted
@avz1865 3 ай бұрын
In fact seems like there are more than 10^40 atoms in a star. So we just need to use one.
@CharlesPanigeo 3 жыл бұрын
The Mobius function was featured heavily in my number theory course in university. It's a rather interesting function because of the Mobius inversion formula.
@marklemoine1634 3 жыл бұрын
Always a pleasure to see Dr. Krieger featured on Numberphile!
@ClevorBelmont 4 жыл бұрын
I never REALLY understand any of her videos but I always instant click. Dr Holly is a legend.
@maitland1007 4 жыл бұрын
I'd love to see some kind of description of how the proof was done if that's at all possible. Also, do we know if it breaks the bound pisitively or negatively? Thanks for another great video!
@jcantonelli1 3 жыл бұрын
Incredible, love this video - the graph looks very similar to a Brownian motion with no drift.
@jrbleau 2 жыл бұрын
Makes me think of a random walk.
@grapheist612 4 жыл бұрын
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 :)
@jamirimaj6880 3 жыл бұрын
That is so amazing. This is like a much lower example of the Graham's number, in which you have a higher bound but don't know exactly the value of the number you're pinpointing to.
@mrnicomedes 4 жыл бұрын
I feel like this is a glaringly unaddressed question, though it may take a few mathematical detours to answer (I have no idea): Why did Martens conjecture that the function was bounded by sqrt(n)? From the data we were shown, the function doesn't even seem to approach sqrt(n). Why not n^(1/3) or ... anything else? Very curious!
@rodrigorodders7173 2 жыл бұрын
It’s not quite sqrt (n) it’s most likely it’s n^1/2+epsilon
@WaluigiisthekingASmith Жыл бұрын
Iirc it's because sqrt(n) is how max(random walk) grows.
@revenevan11 4 жыл бұрын
Dr Holly Krieger is my favorite presenter on numberphile! All the videos with her explaining the mandelbrot set are incredibly mindblowing and inspiring to me! This video was a wonderful treat to start my morning with, thanks!
@11pupona 4 жыл бұрын
The connection with the RH is very interesting and I worked on that in an expository paper for my number theory course.
@douglasbrinkman5937 4 жыл бұрын
we're gonna need a bigger universe!
@akf2000 4 ай бұрын
Best comment
@JonathonV 4 жыл бұрын
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!
@NoIce33 4 жыл бұрын
Thinking about Skewes' number, there really seems to be something about primes that makes them break our seemingly natural expectations if we dig really deep, i.e. look stupidly far.
@nripendrakrdeb1327 2 жыл бұрын
Dr. Holly and this video,just beauty all around 🤩
@nitinjain1605 Жыл бұрын
I love Dr. holly's Laugh 😇
@mueezadam8438 4 жыл бұрын
my typical numberphile viewing experience: start of the video: **yawn** end of the video: **screams geometrically**
@palahnuk1 2 жыл бұрын
then don't watch - dork
@Sylocat 4 жыл бұрын
I was legit heartbroken when I found out this had wrecked a chance to finally prove the Riemann hypothesis.
@michaelbauers8800 4 жыл бұрын
I think that was the most interesting point made in the video. I didn't know that was under consideration.
@psps6623 2 жыл бұрын
Loved your work in "Arrival"
@_kopcsi_ 4 жыл бұрын
I am curious if all of this is related to the “law of large numbers” in some way. because there [if we consider the additive inverse (difference), and not the multiplicative inverse (quotient) of the numbers of positive and negativ outcomes] we have similar zigzag behaviour (naturally from the randomness), which zigzag is also mainly inside a parabola [+-sqrt(N), where N is the number of trials].
@sandman7955 4 жыл бұрын
Love Dr Holly !!!
@vizart2045 2 жыл бұрын
Brilliant video. It has to be said that not all hope is lost proving the Riemann hypothesis through some statement on these numbers. Just tweak the conjecture a little.
@smudgepost 4 жыл бұрын
I came for Numberphile and charting /prediction techniques and also got a glorious redhead - Thanks!
@tomelifeisjustonebig 4 жыл бұрын
More Dr Krieger please!
@averagesongcontestan 4 жыл бұрын
How do we know that the Mertens Conjecture is not true? What gives us the 10^(10^40)?
@averagesongcontestan 4 жыл бұрын
@@yareyaredaze9450 I would have wished for a small remark in the video itself. I'm far from being able to understand the proof after investing a few minutes to skim through the paper, but a few seconds in the video saying how it was achieved would have been greatly appreciated.
@nahidhkurdi6740 4 жыл бұрын
Odlyzko and a co-worker managed to prove that the limit of the supremum of Mertens function as x goes to infinity is greater than 1.06 which disproves Mertens conjecture. Their proof used extensive computations on the roots of the zeta function from which that number emerged.
@sykes1024 4 жыл бұрын
Keep in mind that 10^(10^40) is only a bound, not the exact number.
@8Clips 4 жыл бұрын
Imagine asking for a proof of this, when there's a proof in the description of the video. Unfortunately if you don't understand it, you simply don't understand it. It's not really possible to sum up a 32 page proof for a youtube comment section. It involves a lot of complex mathematics and the more someone explains to you, the more questions you would have.
@surfclimbcycle 2 жыл бұрын
I'm curious to know, whether it's known (or not): after continuing past the value that fails Mertens, does Fmu(n) remain greater than +/- root(n) for a significant series of n, or does it quickly return to values that satisfy Mertens?
@benterrell9139 4 жыл бұрын
Fantastic! I'm studying number theory in my undergraduate course but I hadn't found this one. Classic.
@j7m7f 4 жыл бұрын
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?
@rosiefay7283 4 жыл бұрын
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.
@gocrazy432 4 жыл бұрын
@@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.
@toolatetocolonize 4 жыл бұрын
If Dr. Holly Krieger were my maths professor I would be in college right now
@irwNd2 4 жыл бұрын
Why does everyone always blame their previous teacher on their own failure?
@hkr667 4 жыл бұрын
@@irwNd2 No one does, the only person to say so is you.
@tom_jasper2647 3 жыл бұрын
One of the finest explanations I've ever seen.
@adamel-sawaf4045 4 жыл бұрын
So do we know that this function will eventually break the upper bound of the square root limit and not the lower one (the positive and not the negative one)?
@r-prime 2 жыл бұрын
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...
@quocanhnguyen7275 4 жыл бұрын
I love your performance in Arrival!
@kaustubhnamjoshi4133 4 жыл бұрын
I came to the comment section just to see someone write this comment!
@ActuarialNinja 4 жыл бұрын
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.
@pqRachel 4 жыл бұрын
Do we know if the first rule breaking number breaks the boundary on the positive or negative side?
@viktornikolic6931 4 жыл бұрын
When i see notification my heart goes +1-1+1-1+1-1
@ze_rubenator 4 жыл бұрын
_It goes bom-bodi-bom-bodi bom-bodi-bom-bodi bom-bodi-bom-bodi bom_ _Goodness gracious me_
@andrewtan881 4 жыл бұрын
So your heart shrinks to half its size
@AbirInsights 4 жыл бұрын
@@andrewtan881 Nah its 0
@hamiltonianpathondodecahed5236 4 жыл бұрын
@@andrewtan881 his sum evaluates to zero as he has considered only finitely many (6 to be precise) terms
@andrewtan881 4 жыл бұрын
Fair enough
@uladzislaushulha1994 4 жыл бұрын
I'm a simple person: I see primes - I click like I see Dr. Holly - I click like . . . I see Primes and Dr. Holly - I post a comment
@leonhardeuler9839 4 жыл бұрын
Uladzislau Shulha We are on the same page
@alsorew 4 жыл бұрын
So, you UNclicked “like” second time you clicked it, then.
@fakestory1753 4 жыл бұрын
That really depends on defining "click like" as an event or command
@xCorvus7x 4 жыл бұрын
@@alsorew No, the second like overflows into the comment section, since both likes and comments matter to the algorithm (or so I gather).
@ganeshprasad9851 4 жыл бұрын
But you didn't post a comment to your own comment in which you saw both "prime" and "Dr.Holly"
@TumTum21x 4 жыл бұрын
Holly is the best! Thanks for this one guys :)
@joshuazelinsky5213 4 жыл бұрын
Note that there are a variety of statements involving Merten''s Function which are equivalent to the Riemann Hypothesis. One of the easier to state and prove ones is that RH is equivalent to there existing a constant C such that for sufficiently large x, |M(x)| < x^(1/2) e^(C (log x)/ log log x) . In fact, RH is equivalent to the even weaker statement that for any eps>0, we have |M(x)| < C_eps x^{eps} where C_eps is allowed to depend on epsilon. This is also a connected reason for actually believing RH. In particular, imagine you have a function made by randomly flipping a fair coin, where you add 1 every time you get a heads and subtract one every time you get tails, and we'll call the sum after n flips f(n). Then it turns out that with probability 1, one has |f(n)| < C_eps n^{eps} . So in a certain sense we expect the Riemann Hypothesis to hold with probability one. One other note: This is in a certain sense also connected to why it makes sense that the Riemann Hypothesis should tell us interesting things about primes. The function mu(n) shows up in a lot of circumstances where we need to do inclusion-exclusion arguments involving primes. Saying that M(x) is small essentially amounts to saying that when doing inclusion-exclusion arguments with primes, our inclusions and exclusions should roughly cancel.
@BrosBrothersLP 4 жыл бұрын
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
@RB-jl8sm 4 жыл бұрын
That problem then is unsolvable.
@TheAlps36 4 жыл бұрын
I think Ron Graham can relate to that
@BrosBrothersLP 4 жыл бұрын
@@RB-jl8sm not really. Math is filled with numbers you can describe and not write down. E.g. all irrational numbers
@RB-jl8sm 4 жыл бұрын
@@BrosBrothersLP i see, so we dont need to imagine it because it exists anyway and hence it is not too interesting.
@mynewaccount2361 4 жыл бұрын
That was an unfunny joke.
@DaveSalwinski 4 жыл бұрын
So we know there exists a number n where |M(n)|>\sqrt{n}, but do we know whether M(n) is positive or negative there? I mean does the jagged graph break out of the bounding square root curves from the top or the bottom?
@Czeckie 4 жыл бұрын
both things happen infinitely often
@Czeckie 4 жыл бұрын
actually, the state of the art result is that it happens infinitely often that M(n)>1.82*\sqrt{n} and M(n)< -1.83*\sqrt{n}
@DaveSalwinski 4 жыл бұрын
Awesome! Thanks!
@whatitmeans 2 жыл бұрын
If you think of the Mertens function as one realization of a random walk, for BIG numbers the sum will behave as a Brownian motion, so it should be bounded by its modulus of continuity... like Upper/Lower bound f(n) = +/- sqrt[2*n*sqrt(pi^2+{log(log(n+1)+1)}^2)] right?
@ceythehun83 4 жыл бұрын
Great videos guys. μ is the coefficient of friction in my little world. your μ is waaay better, Dr Kelly you are the best!
@Pageleplays 4 жыл бұрын
Can we please hold on for a minute an apprechiate that the video wasn‘t stretched to 10 minutes 🙏🏽❤️
@gobdovan 4 жыл бұрын
6:50 STONKS
@phlogchamp 3 жыл бұрын
Dr. Holly, Cliff, and Matt are the three best Numberphiles on this channel.
@lyrimetacurl0 3 жыл бұрын
They predicted the "two horse conjecture" (from Cabinet of Mathematical Curiosities) to break at a 300 digit number but it breaks at a 9 digit number. And that is a similar problem, just number is prime factors odd=-1 and even=+1, it only first goes positive around 900 million but when it does it smashes through, to some extent, before falling back in.
@diagorasofmelos4345 4 жыл бұрын
Damn it, Brady! Now I'm forced to take a study break.
@RussellFlowers 4 жыл бұрын
Call it "Merten's Bane"
@faokie 4 жыл бұрын
It's a big guy
@ghostsdefeated4078 4 жыл бұрын
@@faokie For you
@LotsOfS 4 жыл бұрын
Love how it put a banana instead of the number 3, it not only broke the pattern on the screen, but all the other patterns videos like these have planted over a lifetime.
@zyxzevn 4 жыл бұрын
Another repeating-pattern problem is with primenumbers of 2^N-1 where N is a prime. It works until you have a large N. The pattern is still used to find extremely large primes.
@professortrog7742 4 жыл бұрын
Proposal for the name of this number: Mertens downfall. Edit: i really like all proposed alternatives in the replies!
@fiddlinmacx 4 жыл бұрын
Mertens' Bane ;-)
@jsraadt 4 жыл бұрын
Odlyzko's Number named after the author who proved it
@NickMunch 4 жыл бұрын
Mertens' Folly.
@jessstuart7495 4 жыл бұрын
Merten's Conjecture First Counterexample (There could be more than one). MCFC
@GuzmanTierno 4 жыл бұрын
@@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.
@zerid0 4 жыл бұрын
This reminds me a lot of random walks and Brownian motion. Do we have any results for C*√n where C is any constant? What about C*n^(0.5 + Ɛ ) where Ɛ is positive and as small as we want?
@christosvoskresye 4 жыл бұрын
I'm sure that's where the idea came from. On the other hand, look at that graph: it looks more structured than I would expect for flips of a fair coin. Of course, even a random walk has some chance of exceeding those bounds at some point.
@MagruderSpoots 4 жыл бұрын
@@christosvoskresye Given enough time, random flips of a three sided coin will produce this graph infinitely many times.
@thishandleistaken1011 4 жыл бұрын
@@MagruderSpoots That would produce any graph though.
@jacksonpercy8044 4 жыл бұрын
@Mark W That got me thinking, what's the probability that a series of coin flips would exactly match that graph until it breaks out of bounds at 10^10^40? I wonder if that number is comparable to Graham's number
@sebastianjost 4 жыл бұрын
@@jacksonpercy8044 that probability is quite easy to work out. Let's say you want to copy this graph up to a point n (could be 10^10^40 or anything else) Then the probability that tossing a 3 sided coin/dice would produce the same graph up to that n is exactly (⅓)ⁿ. That's simply because for every 0≤k≤n you know exactly what the value should be so you have a chance of ⅓ that your coin matches that number
@jamma246 4 жыл бұрын
Obviously |mu(n)| is at most n=n^1 for any n. What instead of staying between + or - sqrt(n), we changed the conjecture to: there is some epsilon for which |mu(n)| < n^{1-epsilon} So it stays within some power of n (smaller than 1) of 0? Would that still be enough to prove the Riemann Hypothesis? I'd be surprised if we needed the power to be 1/2, but could be wrong?
@katzen3314 4 жыл бұрын
Couldn't we potentially write the number down though? Using the lower limit, in base 10 there would be 10^40 digits, which allows for 10^40 atoms per digit if we utilise the entire universe, which is actually quite generous.
@UnimatrixOne 4 жыл бұрын
Dr. Holly Krieger the beauty of mathematics! :) 7:15 ❤️
@CosmiaNebula 4 жыл бұрын
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.
@michaelrobertson714 4 жыл бұрын
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).
@johubify 4 жыл бұрын
I like these Number analysis videos the most
@acaryadasa 4 жыл бұрын
Question. Is it just a single number that is outside the square root boundary and the immediate previous and immediate after number are within the boundary or is it more than one?
@darklink1113 4 жыл бұрын
They should name it Holly's Number. Or Merten's outlaw
@Simpson17866 4 жыл бұрын
You win :)
@gizatsby 4 жыл бұрын
Merten's outlaw is the best one I've seen
@DoctorShaunB 4 жыл бұрын
Upvote Merten's Outlaw
@christosvoskresye 4 жыл бұрын
Or they could name it TWO (all caps).
@gregoryfenn1462 4 жыл бұрын
@@christosvoskresye ...? why would they do that?
@KnightsOfTheMemeTable 4 жыл бұрын
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!
@Einyen 4 жыл бұрын
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.
@alexpotts6520 4 жыл бұрын
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.
@balaalalaslk 4 жыл бұрын
So glad we learned about TREE(3) because it makes everything look like a rounding error to 0 even this number here.
@kelemnamare2535 4 жыл бұрын
When you want to comment first but you have to watch the whole video. I will edit this later anyways
@supermaster2012 4 жыл бұрын
If Holly had been my professor in collete I would have actually attended Maths classes.
@hamiltonianpathondodecahed5236 4 жыл бұрын
lol ~just replied to get notified for future comments~
@ammonkunzler3948 4 жыл бұрын
@@andermium ikr?
@eayllon1 4 жыл бұрын
I would have gone to office hours
@michaelbauers8800 4 жыл бұрын
I think I attended most math classes. Because you miss one math class, and you are seriously behind in many cases. It's a pyramid scam! :)
@MichaelDeHaven 4 жыл бұрын
Do we know if it breaks thru on the positive or negative side? The video seemed to indicate positive, but was that just an example?
@AbiGail-ok7fc 4 жыл бұрын
The number of atoms in the observable universe is estimated to be around 10^80 (give or take some factors of 10). Assuming it takes less than 10^40 atoms to write down a digit, that's more than enough to write down the number 10^(10^40) (it's "only" 10^40 digits, and 10^40 atoms to write down each of 10^40 digits needs 10^80 atoms).
@christosvoskresye 4 жыл бұрын
I wish he had asked if the sum exceeds the bounds infinitely many times.
@christosvoskresye 4 жыл бұрын
@Gnomicality Maybe. It seems clear that the failure of the Mertens Conjecture does not DISPROVE the Riemann Hypothesis, or that would have been the big story.
@dlevi67 4 жыл бұрын
@@christosvoskresye The Riemann hypothesis is "equivalent" to Mertens' function growing with exponent (1/2 + ε), rather than exponent 1/2 (which was Mertens' conjecture). So invalidating Mertens' conjecture is frustratingly close to disproving the RH, but only close... (as far as I know, it's not even known if there is a lower bound on the value of ε).
@felixmerz6229 2 жыл бұрын
It does.
@Tzelemel 4 жыл бұрын
I must admit: i got slightly distracted when I saw that sequence and went to OEIS to see if I could listen to Merten's Function. And I can.
@miklov 4 жыл бұрын
Cool. I was thinking that I wanted to hear it too but on OEIS I could only find it as a sequence of notes rather than a sequence of pressure samples as I was hoping for.
@WylliamJudd 3 жыл бұрын
What happens you use a different rule for repeated prime factors, such as counting odd and even total numbers, or odd and even unique prime factors?
@omikronweapon 4 жыл бұрын
For once Brady DIDNT ask the same question I wanted to know. Is it a single instance, or does it continue outside the perimeter for a bit? Though, I suppose, the way they talk about it implies it's just ONE integer at that range. Are there (infinitely) more?
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