A Prime Surprise (Mertens Conjecture) - Numberphile

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Dr Holly Krieger discusses Merterns' Conjecture.
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More links & stuff in full description below ↓↓↓
More videos with Holly (playlist): bit.ly/HollyKrieger
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).
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumberphile
We are also supported by Science Sandbox, a Simons Foundation initiative dedicated to engaging everyone with the process of science. www.simonsfoundation.org/outr...
And support from Math For America - www.mathforamerica.org/
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Animation by Pete McPartlan
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16 июн 2024

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Комментарии : 1,5 тыс.

@numberphile 4 года назад

More videos with Holly: bit.ly/HollyKrieger Holly on the Numberphile Podcast: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-QmfQQzjpdpM.html

@AgentM124 4 года назад

I didn't catch, but did they know anything about if it breaks the √n in the pos or the negative?

@warb635 4 года назад

Fyi: I see "Mertern's Conjecture" instead of "Merten's Conjecture" in the RU-vid description of this video.

@RobinDSaunders 4 года назад

@@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 года назад

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 года назад

@@rob6129 Second that.

@StefanReich 4 года назад

5:02 "Living around zero but in a really complicated way"... Wow, you just described my bank account

@GaneshNayak 4 года назад

Lol

@stuffofmaking 4 года назад

Stefan not so Reich

@Oblivion1407 4 года назад

But you can be certain that at one point your money amount blows outside of the boundaries, it might be negative though.

@jwink7795 4 года назад

HOLLLLAAAAA

@TheTrueAltoClef 4 года назад

@@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 года назад

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 года назад

Why isn't it 10^((10^40)*0.5) ? How does this math thing even work

@ReconFX 4 года назад

@@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 года назад

To compare them properly - what power of 10 will give you 0.5? Take log (base 10) 0.5

@rad858 4 года назад

@@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 года назад

wow ok my mind is officially blown. restarting....

@fwiffo 4 года назад

"If this was true, it would imply the Riemann hypothesis!" "It's false." *uncomfortable digestive noises*

@fllthdcrb 4 года назад

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 года назад

*jazz music stops*

@angelmendez-rivera351 4 года назад

Daniel Dawson I think the OP knows this. The discomfort comes from the fact that this doesn't help us with it

@leif1075 4 года назад

@@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 года назад

@@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 года назад

Dr. Holly one of the best.

@spicemasterii6775 4 года назад

Can Ali Tomruk Or Isla Fisher

@pH7oslo 4 года назад

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 года назад

@@mostlynothing8130 Pugsley Addams will play me in that movie.

@codycast 4 года назад

pH7oslo “I’ve never heard of famous people. Aren’t I edgy and cool?”

@fgc_rewind 4 года назад

dont you have world records on CTR?

@tatjoni 4 года назад

Holly's laugh makes my heart smile!

@aaaaanomaly 3 года назад

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 года назад

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 года назад

Just read the paper, link is in the description. But first, in order to understand it, go get a PhD in mathematics.

@yeoman588 4 года назад

@@w00tehpwn Ideally I'd like an explanation that _doesn't_ require a PhD to understand. 😅

@lukesteeves1291 4 года назад

@@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 года назад

@@w00tehpwn But first, we need to talk about parallel universes

@effuah 4 года назад

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.

@JNCressey 4 года назад

2:08 "lets forget about zero and start with two. " *sad one noises

@dlevi67 4 года назад

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 года назад

math people seem to forget 1 is a prime, its only factors are 1 and itself, 1.

@Eliseo_M_P 4 года назад

@@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 года назад

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 года назад

no primes are numbers with 2 factors

@nosuchthing8 4 года назад

One of the best channels on RU-vid

@numberphile 4 года назад

thanks!!

@EdbertWeisly 3 года назад

@@numberphile Numberphile Deserve have 314,159,265,358,979,323 SUBSCRIBERS

@aradhya_purohit 3 года назад

@@EdbertWeisly or 2,718,281,828,459,045,235,360?

@EdbertWeisly 3 года назад

@@aradhya_purohit sure

@OMGclueless 4 года назад

8:37 "That's my new favorite number." Why do I get the feeling Brady says this a lot?

@omikronweapon 4 года назад

only because he does

@YtseFrobozz 4 года назад

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 года назад

@@YtseFrobozz No, he doesn't pick them uniformly.

@TavartDukod 3 года назад

@@ShankarSivarajan I mean there's literally no way to pick natural numbers uniformly because of sigma-additivity of probability.

@ucantSQ 4 месяца назад

It's the name of the show.

@apollion888 4 года назад

Holly is my favorite Numberphile speaker. I am delighted to see her again, and it's primes too!

@palahnuk1 2 года назад

you need to get out more

@solandge36 4 года назад

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 года назад

I love how excited Dr. Krieger is about this one! excellent stuff thank you!!!

@dhoyt902 4 года назад

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 года назад

sarcasm, right?

@gregoryfenn1462 4 года назад

@@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 года назад

Gregory Fenn by new you mean 1985

@BauerMonty 4 года назад

It breaks the Squarrier

@TavartDukod 3 года назад

@@gregoryfenn1462 how did so many people manage to get PhDs without knowing they got them? /s

@jppagetoo 4 года назад

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 года назад

@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 года назад

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 RU-vid 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 года назад

@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 года назад

@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 года назад

That's all fine and dandy but what on earth is that design on that tele

@robertschlesinger1342 4 года назад

Excellent description of the Mertens Conjecture, and the counter-example found. Many thanks for the link to the mathematical paper disproving the Conjecture.

@gl1500ctv 4 года назад

"You go to the gym to get in shape but what about your brain?" Uh, I come here.

@revenevan11 4 года назад

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!

@ClevorBelmont 4 года назад

I never REALLY understand any of her videos but I always instant click. Dr Holly is a legend.

@marklemoine1634 3 года назад

Always a pleasure to see Dr. Krieger featured on Numberphile!

@DanTheStripe 4 года назад

Surely you've got to name it something awesome like Mertens' Nemesis if it's the first number to break the rule?

@PhilBagels 4 года назад

The Anti-Mertens Number.

@lapiscarrot3557 4 года назад

A Mertensplex, perhaps?

@badmanjones179 4 года назад

brb writing this down on my math rock song name ideas list

@gregnixon1296 4 года назад

Merten's Foil? How's that?

@stapler942 4 года назад

Mertens' Bane.

@fghjghjfhgjfhgj 4 года назад

More videos with Holly! Please more Holly!

@jamirimaj6880 3 года назад

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.

@unvergebeneid 4 года назад

Really beautiful example for the power of proof over both intuition and brute force.

@ZachGatesHere 4 года назад

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 года назад

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 года назад

but in this case zeta has a deep connection with primes and this Merten thing is built on the primes itself

@gregoryfenn1462 4 года назад

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 года назад

@@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 года назад

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".

@grapheist612 4 года назад

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 :)

@kevwang0712 4 года назад

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 года назад

Therefore the number should be called "Rie...maaaaan!"

@SuperSpruce 4 года назад

No, it should be called the reeeeeeeee-mann

@TimothyGowers0 4 года назад

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 года назад

Still a great video though!

@simplebutpowerful 4 года назад

@@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).

@senororlando2 4 года назад

Love Holly’s guest spots

@JonathonV 4 года назад

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!

@jcantonelli1 3 года назад

Incredible, love this video - the graph looks very similar to a Brownian motion with no drift.

@jrbleau 2 года назад

Makes me think of a random walk.

@MinorCirrus 4 года назад

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 года назад

This is a matter of definition only.

@rad858 4 года назад

The usual mathematical reason: because it's more interesting that way

@AlexJones-ue1ll 4 года назад

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 года назад

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 года назад

@@sambachhuber9419 I see now. Thanks!

@11pupona 4 года назад

The connection with the RH is very interesting and I worked on that in an expository paper for my number theory course.

@shtfeu 4 года назад

I love when Nicole Kidman explains maths to me.

@anglo2255 4 года назад

Ha ha , I totally see that now

@robertheikkila4045 4 года назад

You mean Amy Adams?

@Wecoc1 4 года назад

ok mister killjoy

@ditzfough 4 года назад

Holly is alot prettier than nicole kidman.

@chinmaybhoir6955 4 года назад

Robert Heikkilä exactly my thought

@smudgepost 4 года назад

I came for Numberphile and charting /prediction techniques and also got a glorious redhead - Thanks!

@benterrell9139 4 года назад

Fantastic! I'm studying number theory in my undergraduate course but I hadn't found this one. Classic.

@CharlesPanigeo 3 года назад

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.

@uladzislaushulha1994 4 года назад

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 года назад

Uladzislau Shulha We are on the same page

@alsorew 4 года назад

So, you UNclicked “like” second time you clicked it, then.

@fakestory1753 4 года назад

That really depends on defining "click like" as an event or command

@xCorvus7x 4 года назад

@@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 года назад

But you didn't post a comment to your own comment in which you saw both "prime" and "Dr.Holly"

@phlogchamp 3 года назад

Dr. Holly, Cliff, and Matt are the three best Numberphiles on this channel.

@nripendrakrdeb1327 2 года назад

Dr. Holly and this video,just beauty all around 🤩

@sandman7955 4 года назад

Love Dr Holly !!!

@tomelifeisjustonebig 4 года назад

More Dr Krieger please!

@tom_jasper2647 3 года назад

One of the finest explanations I've ever seen.

@nitinjain1605 Год назад

I love Dr. holly's Laugh 😇

@NoIce33 4 года назад

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.

@Sylocat 4 года назад

I was legit heartbroken when I found out this had wrecked a chance to finally prove the Riemann hypothesis.

@michaelbauers8800 4 года назад

I think that was the most interesting point made in the video. I didn't know that was under consideration.

@balaalalaslk 4 года назад

So glad we learned about TREE(3) because it makes everything look like a rounding error to 0 even this number here.

@johubify 4 года назад

I like these Number analysis videos the most

@douglasbrinkman5937 4 года назад

we're gonna need a bigger universe!

@akf2000 4 месяца назад

Best comment

@toolatetocolonize 4 года назад

If Dr. Holly Krieger were my maths professor I would be in college right now

@irwNd2 4 года назад

Why does everyone always blame their previous teacher on their own failure?

@hkr667 4 года назад

@@irwNd2 No one does, the only person to say so is you.

@ceythehun83 4 года назад

Great videos guys. μ is the coefficient of friction in my little world. your μ is waaay better, Dr Kelly you are the best!

@psps6623 2 года назад

Loved your work in "Arrival"

@Tehom1 4 года назад

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 года назад

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 года назад

@@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 года назад

@@dlevi67 AFTER that number, it breaks away at 10^10^40 and after that, i wonder what happens after that.

@dlevi67 4 года назад

@@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 года назад

@viktornikolic6931 4 года назад

When i see notification my heart goes +1-1+1-1+1-1

@ze_rubenator 4 года назад

_It goes bom-bodi-bom-bodi bom-bodi-bom-bodi bom-bodi-bom-bodi bom_ _Goodness gracious me_

@andrewtan881 4 года назад

So your heart shrinks to half its size

@AbirInsights 4 года назад

@@andrewtan881 Nah its 0

@hamiltonianpathondodecahed5236 4 года назад

@@andrewtan881 his sum evaluates to zero as he has considered only finitely many (6 to be precise) terms

@andrewtan881 4 года назад

Fair enough

@TumTum21x 4 года назад

Holly is the best! Thanks for this one guys :)

@call_me_stan5887 4 года назад

You do awesome job finding great people to explain things on your channel!

@B4der 4 года назад

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 месяца назад

In fact seems like there are more than 10^40 atoms in a star. So we just need to use one.

@maitland1007 4 года назад

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!

@vizart2045 2 года назад

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.

@LotsOfS 4 года назад

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.

@R2Cv1 4 года назад

One question not addressed is, DOES IT BREAK AWAY UPWARDS OR DOWNWARDS??

@fnors2 4 года назад

From what I gather from other people who read the paper : both, infinitely many times. As for the first break? No idea.

@ubertoaster99 4 года назад

Picture showed upwards.

@snbeast9545 4 года назад

@@ubertoaster99 There was a large disclamer that the picture was an artistic rendition.

@ubertoaster99 4 года назад

@@snbeast9545 The artist knew what they were doing. I'd bet on positive :)

@elltwo8393 4 года назад

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 года назад

Is it possible that the first failure of Riemann's Hypothesis is a number as big as this one?

@romajimamulo 4 года назад

Yes, unfortunately. It could be even bigger

@leofisher1280 4 года назад

it almost definitely is. no failures have ever been found

@romajimamulo 4 года назад

@@leofisher1280 We probably haven't looked quite that far, but yes, we've looked very far

@TheTortuga58 4 года назад

You take that back

@christosvoskresye 4 года назад

I'm thinking of a number between 1 and Tree(3).

@joshuazelinsky5213 4 года назад

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.

@waarschijn 4 года назад

Three statements in decreasing order of strength: |M(x)| < x^½ |M(x)| < Cx^½ for some constant C |M(x)| < Cx^{½+ε} for some constant C(ε), for all ε > 0 The first is the (false) Mertens conjecture. The third is equivalent to the Riemann Hypothesis. It has to do with the fact that 1/ζ(s) = Σ μ(n)/n^s. Mertens' function M can be approximated as a complex integral of 1/ζ just to the right of the critical line, hence the ½+ε. But this is only valid if ζ doesn't have any zeros with real part > ½.

@KnightsOfTheMemeTable 4 года назад

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!

@BrosBrothersLP 4 года назад

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 года назад

That problem then is unsolvable.

@TheAlps36 4 года назад

I think Ron Graham can relate to that

@BrosBrothersLP 4 года назад

@@RB-jl8sm not really. Math is filled with numbers you can describe and not write down. E.g. all irrational numbers

@RB-jl8sm 4 года назад

@@BrosBrothersLP i see, so we dont need to imagine it because it exists anyway and hence it is not too interesting.

@mynewaccount2361 4 года назад

That was an unfunny joke.

@lyrimetacurl0 3 года назад

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.

@zyxzevn 4 года назад

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.

@Pageleplays 4 года назад

Can we please hold on for a minute an apprechiate that the video wasn‘t stretched to 10 minutes 🙏🏽❤️

@CosmiaNebula 4 года назад

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 года назад

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).

@AbiGail-ok7fc 4 года назад

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).

@ig2d 4 года назад

Dr Holly Kreiger + numberphile = happy days

@mueezadam8438 4 года назад

my typical numberphile viewing experience: start of the video: **yawn** end of the video: **screams geometrically**

@palahnuk1 2 года назад

then don't watch - dork

@gobdovan 4 года назад

6:50 STONKS

@supermaster2012 3 года назад

I will never not like a video with Dr Krieger.

@_PsychoFish_ 4 года назад

5:07 describes dating really well: complicated, a lot of cancellation and I never was too big xD

@diagorasofmelos4345 4 года назад

Damn it, Brady! Now I'm forced to take a study break.

@quocanhnguyen7275 4 года назад

I love your performance in Arrival!

@kaustubhnamjoshi4133 4 года назад

I came to the comment section just to see someone write this comment!

@kevingil1817 4 года назад

Excited for the part 2 for this video. Like, how?

@solandge36 4 года назад

More podcasts with holly!!

@mrnicomedes 4 года назад

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 года назад

It’s not quite sqrt (n) it’s most likely it’s n^1/2+epsilon

@WaluigiisthekingASmith Год назад

Iirc it's because sqrt(n) is how max(random walk) grows.

@professortrog7742 4 года назад

Proposal for the name of this number: Mertens downfall. Edit: i really like all proposed alternatives in the replies!

@fiddlinmacx 4 года назад

Mertens' Bane ;-)

@jsraadt 4 года назад

Odlyzko's Number named after the author who proved it

@NickMunch 4 года назад

Mertens' Folly.

@jessstuart7495 4 года назад

Merten's Conjecture First Counterexample (There could be more than one). MCFC

@GuzmanTierno 4 года назад

@@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.

@conexant51 4 года назад

You know there's a number that breaks a conjecture. You know it's larger than a certain amount. This certain amount is so large you can't put it on paper cos there are not enough paper nor ink nor computer memory, or even enough atoms in the universe to ever show this number visually. Mind blowing!

@violetasuklevska9074 4 года назад

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.

@j7m7f 4 года назад

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 года назад

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 года назад

@@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.

@Tzelemel 4 года назад

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 года назад

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.

@_kopcsi_ 4 года назад

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].

@aianvigare1158 4 года назад

Pondered this one as a kid thanks for explaining it!

@RussellFlowers 4 года назад

Call it "Merten's Bane"

@faokie 4 года назад

It's a big guy

@ghostsdefeated4078 4 года назад

@@faokie For you

@averagesongcontestan 4 года назад

How do we know that the Mertens Conjecture is not true? What gives us the 10^(10^40)?

@averagesongcontestan 4 года назад

@@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 года назад

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 года назад

Keep in mind that 10^(10^40) is only a bound, not the exact number.

@8Clips 4 года назад

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.

@TheChondriac 4 года назад

"Let's forget about zero I don't wanna even think about it" haha love it

@Einyen 4 года назад

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 года назад

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.

@r-prime 2 года назад

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...

@DaveSalwinski 4 года назад

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 года назад

both things happen infinitely often

@Czeckie 4 года назад

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 года назад

Awesome! Thanks!

@gamesandstuff7724 4 года назад

I saw "Mathematics of Life" on the shelf in the corner. Great book I think.

@RowBowLP 4 года назад

Awesome! Thank you so much :)

@UnimatrixOne 4 года назад

Dr. Holly Krieger the beauty of mathematics! :) 7:15 ❤️

@darklink1113 4 года назад

They should name it Holly's Number. Or Merten's outlaw

@Simpson17866 4 года назад

You win :)

@gizatsby 4 года назад

Merten's outlaw is the best one I've seen

@DoctorShaunB 4 года назад

Upvote Merten's Outlaw

@christosvoskresye 4 года назад

Or they could name it TWO (all caps).

@gregoryfenn1462 4 года назад

@@christosvoskresye ...? why would they do that?

@ActuarialNinja 4 года назад

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.

@sam111880 4 года назад

Nice explaination clear and concise , ya that merten conjuncture is a bummer ... It's similar to the Hardy prime conjecture on the bounds on prime number function it eventually fails to with what they term skews number. It be nice though to find bigo or littleo of merten conjuncture though 😉