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@TerryPlays

2 жыл бұрын

Video is 20 mins ago and you posted this 8 days ago. How?

@BambinaSaldana

2 жыл бұрын

Hey, Bri, have you heard?

@BambinaSaldana

2 жыл бұрын

@@TerryPlays Either it was unlisted or it was available only to members.

@TerryPlays

2 жыл бұрын

@@BambinaSaldana ok

@crystalkalem9289

7 ай бұрын

Not gonna lie, the answer is wrong. The reason the answer is wrong is because it doesn't use the correct definitions of the terms it uses. The problem is that you do a sequence of + and - to INFINITY! infinity - infinity is still infinity. infinity contains all other forms of infinity , such as infinity+1 and infinity+infinity repeat to infinity. If you minus infinity from infinity you still get infinity or 0, even if you minus 2 infinity from infinity, the answer is still infinity or 0.

Average complex math theory fan vs. Average "It clearly isn't, just look at it" enjoyer

@eterty8335

2 жыл бұрын

After such rough comments on why the video is misleading/incorrect/incomplete, I really needed this comment lol

@thewall4069

2 жыл бұрын

Based

@rjeverett6223

2 жыл бұрын

me, an average “it clearly isn’t, just look at it” enjoyer

@catalintimofti1117

2 жыл бұрын

me a complex number theory fan(complex numbers apear in physics)

@spoopyscaryskelebones3846

2 жыл бұрын

@@catalintimofti1117 they’re stupid (complex my ass)

I’m so happy with the context you provided and ended on “no the sum of all natural numbers is not -1/12.” Too many mathtubers don’t do that!

@MagruderSpoots

2 жыл бұрын

@@CellarDoor-rt8tt Also because of its relation to quantum mechanics.

@MasterHigure

2 жыл бұрын

@@CellarDoor-rt8tt The Riemann Zeta function isn't the ONLY reason. There are many reasons to associate this series with the number -1/12. Quantum mechanics (the Casimir effect uses this -1/12 in calculations and has been, IIRC, confirmed experimentally), Ramanujan summation, the standard naive series manipulation argument by way of "1+1-1+...=1/2" and "1-2+3-4+...=-1/4", and many more. It's just that conventional summation isn't one of them.

@nintendoswitchfan4953

2 жыл бұрын

It is -1/12

@Who_is_Emmy

2 жыл бұрын

Mathtubers? First time I heard that term

@oni8337

2 жыл бұрын

@@nintendoswitchfan4953 cope

The problem isn't using the distributive law, it's grouping the series so that you are changing the "speed" at which the series grows.

@lf9177

Жыл бұрын

There is another issue: he performs algebraic operations on infinity (for example S-9S = -8S), when in fact 9*inf=inf and inf-inf is undefined, so inf-9*inf is also undefined. The "mistake" is at the very beginning: treating infinity as a number (S)! INFINITY IS NOT A NUMBER! You cannot perform "normal" calculations on infinity.

@jamescollier3

Жыл бұрын

@@lf9177 exactly. I think you are correct. if 1+2+...= infinity, then you can't say 1+2+... = X, then use that to do math, as infinity is not a number or constant.

@emilemerten6535

8 ай бұрын

So say for example the speed it converges is the speed of light so would there be a limit if there is a limit in physics?

@fatmonkey4306

7 ай бұрын

@@emilemerten6535 well the speed of light is a physical limit by reality, numbers have no physical limit. There are rules for it to work, but there is no cap. Any rate at which you think a sequence could reach infinite quickly, I could make one go faster by just applying ^n or whatever

@kyahua3

5 ай бұрын

@@lf9177 but you are assuming S is infinity, we start out with no assumptions about S to find its value

Just found this video, and you instantly got a subscription from me. This irks me so much, and you succinctly described why and gave me a good counter example why it's flawed. Thank you.

Subtracting Infinity is not a valid operation, so everytime you subtract S you rule out that S could be infinity

@simongross3122

2 жыл бұрын

I like that argument. It's also pretty easy to prove that partial sums of S to n terms tend to infinity as n tends to infinity. Apparently this stupid result shows up in a textbook on string theory. What keeps me up at night is that people believe it.

@FermionPhysics

2 жыл бұрын

@@simongross3122 it’s also used in Casimir effect. Physicists have a knack for using mathematically ill-defined operations…

@simongross3122

2 жыл бұрын

@@FermionPhysics Like cancelling infinities. That's another one that has me scratching my head.

@FermionPhysics

2 жыл бұрын

@@simongross3122 the renormalization thing is much less absurd (if that is what you’re referring to). It actually makes some sense if explained in a less hand wavy way…I might make a video on it sometime.

@simongross3122

2 жыл бұрын

@@FermionPhysics Thank you, I'd appreciate that.

The main catch is assuming A is a number. While all the rules work for convergent series, in this case if we calculate A - A = 1 - (1 - 1) + (1 - 1) - (1 - 1) + (1 - 1) (with similar shift as we did with B), we get 0 = 1 and we just broke the math.

@fahdal-sebaey3322

2 жыл бұрын

Yes! This right here.. A is the catch.. A is in itself non-convergent.. the alternating sum of -1 and +1 is the geometric series sum of (-1)^n which is very famously divergent because geometric series only converge if the base is absolutely smaller than 1.

@DTLRR

7 ай бұрын

Hmm that's correct besides if you take something from an infinite series, it just doesn't remain the same as in the case of series A. There is something that's been subtracted and we can't just ignore it no matter how small it is. Furthermore what does this -1/12 imply? Does this show that the series is convergent? Perhaps it might have some uses but it's still a non conventional mathematics that doesn't apply accordingly to the regular mathematics

@kyzer422

7 ай бұрын

Thanks for this comment, this is a simple way to show some of the problems with these "proofs".

@lory3771

6 ай бұрын

My question was how can we be sure that 1 = 2A, when we can very easily add another A for the same result to get 1 = 3A, and again for 1 = 4A, and so on.

@theapexsurvivor9538

6 ай бұрын

​@@lory3771simple, 1=A*X where X is any positive integer, as A is equal to both 0 and 1, thus 2A is the same as 0+0, 1+0, and 1+1. So as long as we can be certain that 1=1+0(X-1) we can be certain that 1=A*X for whatever valid X we choose.

I am pointing at the culprit as being the "shifting the infinite series" and then adding them to infinity. And that's because at any step, you are counting 1 element of the first infinite series while completely disregarding the same order element of the other infinite series, which the further out you are, the bigger that number gets.

@ruce9269

2 жыл бұрын

Yes, which also means you are infinitely away from the true answer: infinity.

@evanfox3136

2 жыл бұрын

That step is fine, you are just adding a zero to the beginning of the series which cannot change its value.

@9core

2 жыл бұрын

​@@ruce9269 there is no value as it doesnt converge, but the limit is infinity

@kinyutaka

2 жыл бұрын

@@ruce9269 approximately.

its not a paradox its a mistake

@sxnchou

7 ай бұрын

yeah the indian who “proved” it was a fraud

@sakshamraj1566

12 күн бұрын

That is what a paradox is !!!!

@louisvollert1420

8 күн бұрын

@sakshamraj1566 i dont think so. I consider a Paradoxon as a problem which is solved logical but the solution isnt logical. But in thos Case the way of solving the Problem is the mistake . Because 1-A=A is only correct If one A ends with +1 anda so A=1 and one A ends with -1 and so A=0 . The way used for B is based on the same phenomen. The result for a + as last sign is not the same as if the last sign would be a - . Even the length is Not defined and so you Just Take an average of the results what is a good way ti simplify the Problem but technically Not absolutely correct the way its written. And so even the S=-1/12 is Not a paradox but based on the simplifications the only result which could come out . I would declare It this way: S IS the sum of all Numbers until n So S=Sum(x=1,n)=(n+1)*(n/2) But I havent studied Maths at University so I might be wrong

Brian, this is probably my favorite video you've done. Outstanding, clear explanations. When Numberphile made a video on this result, I commented that it was at odds with what I'd learned in real analysis, and the other commenters chimed in with, "I can't believe your arrogance, contradicting the work of high-level mathematicians." Well, the professors at UW Madison ARE high-level mathematicians! Thanks for showing the elasticity of this approach to computing divergent infinite... just super fun and informative.

@BriTheMathGuy

2 жыл бұрын

Really glad you enjoyed it! Thanks for watching and have a great day!

2 жыл бұрын

There are higher level mathematicians than UW Madison professors I believe. But the question about this series is more philosophical that mathematical.

2 жыл бұрын

I should add though that I'm a fan of public universities and in particular of UW Madison.

@GlorifiedTruth

2 жыл бұрын

@🙂And I wasn't implying that their professors trump all others, just that I when I was claiming "but that isn't what I was taught," I wasn't referring to some goofball's blog or some random Tik Tok video.

2 жыл бұрын

@Glorified Truth There is indeed very great mathematicians at UW Madison and has been before. Like Georgia Benkart.

So glad to see a video about -1/12 that doesn't just stop at -1/12, but keeps going and demonstrates that no, it is not indisputably the case that the sum of all the natural numbers is -1/12. Yes, the series of natural numbers is closely linked to -1/12, but the conventional sum is not that connection.

@user-xh9pu2wj6b

2 жыл бұрын

"it is not indisputably the case that the sum of all the natural numbers is -1/12" it is tho. Every single regularization method that will give you a finite value for this sum gives you exactly -1/12.

@MasterHigure

2 жыл бұрын

@@user-xh9pu2wj6b But those regularization methods aren't what sum conventionally means. That's the entire issue here. Many popmath sources (including Numberphile, mind you) say more or less uncritically that the sum is -1/12, and that just isn't true.

@DinsAFK

2 жыл бұрын

But it isn't at all. You can't use cesaros sum like that. People like to just be like "hey, sum(-1^n) from n=0 to infinity is = 1/2, but that is the cesaros sum which is a separate arithmetic. It's more like saying csum(sum(-1^n)) instead. For instance, we can say if a = b, and a = 2, then b = 2, but we can't say that if a^2 = b^2 and a = 2 that b = 2. It goes against fundamental laws of mathematics. This cesaros sum shit just annoys me because when I integrate cosine from 0 to infinity, it's undefined, however this would suggest it has a solution.

@user-xh9pu2wj6b

2 жыл бұрын

@@DinsAFK (-1)^n can be summed up to 1/2 using pretty much any other method aside from the classic one tho, it's not just Cesaro.

@DinsAFK

2 жыл бұрын

@@user-xh9pu2wj6b those methods ignore divergence laws then and I really don't see any point in following them

I really enjoyed your video ❤️And now I will also drive my friends crazy in school in this way you told😉

This is the best explanation of this phenomenon I've ever seen and probably will ever see. Great job

@BriTheMathGuy

2 жыл бұрын

Glad you enjoyed it! Have a great day.

@alistairkentucky-david9344

2 жыл бұрын

Mathologer has a fantastic video on this topic as well. Well worth watching.

@aravindswami6243

5 ай бұрын

can you name the mathematician who had given this result??

@Wutheheooooo

2 ай бұрын

@@aravindswami6243 Ramanujan

When I do my taxes later this week, I am going to use the analytic extension of my gross annual income to assess my tax burden.

I love how he explains this to us like we understand right away

@leonthethird7494

2 жыл бұрын

I mean, you should

Witnessed something great here.Someone had to say these.I am glad that the 'someone' is one of my favourite KZreadrs! You keep things neat & clean,everytime.

@BriTheMathGuy

2 жыл бұрын

You're the best!

The ruleset is sometimes not as unclear as presented here, both diverging and alternating sums don't have a general sum that can be simplified without implementing supersums or regarding a generalized function, e.g. Riemann-Zeta. Thereby neither shifting a sum "left" or "right" completely breaks the initial result, nor can the distributive law be utilized for an infinite sum whose identity isn't convergent. Just wanted to slightly highlight it more, bc if we were to allow all forms of algebraic operations on infinite sums, then we'd not only get proofs for -1 = 1 or N = R, but also break mathematical operations by applying mathematical operations. The scorn of the goddess of infinity is eternal, infinite even.

@ChristAliveForevermore

2 жыл бұрын

Wonderful exposition of the 'why' for this result and others like it.

Thanks for putting a smile on my face for showing the Riemann functional equation 😊

I guess one of the issues that I've always had with these sums is that it looks like a simple proof by contradiction. Where the base assumption we're making is that we can assign a value to a divergent series at all. It also appears that depending on how you work with your divergent series you can derive other values that are not necessarily equal to each other. Sounds quite similar to how if you're not careful you may assume that 1/0=1=2=..., which is obviously a contradiction because 1/0 is undefined. However, people are free to define their own 'rules of the game' so to speak.

@adrianbozdog9702

2 жыл бұрын

yep that's what it looked like to me, If I say that 4x= 5x and then divide the entire thing by x then 4 = 5. Since you don't know what the divergent series are and you are doing something to estimate the result no shit it winds up wonky.

This is the best explanation on this topic that I've seen on youtube. Keep up the good work!!

Brian, youre the best! Thanks for the video😍

@BriTheMathGuy

2 жыл бұрын

You're the best!

Ramanujan summation is a beautiful tool, anytime I see -1/12 I can't help but think about it

@monika.alt197

2 жыл бұрын

Ikr! Same

@FermionPhysics

2 жыл бұрын

It’s nonsense

@spektator5418

2 жыл бұрын

@@FermionPhysics ur mom

@createyourownfuture5410

2 жыл бұрын

@@FermionPhysics shît up

@angelmendez-rivera351

2 жыл бұрын

@@FermionPhysics It isn't nonsense.

Actually, thank you very much for this clarification about axioms. I didn't realize that distributiveness could work differently with infinite series.

@leonro

Жыл бұрын

It's more about divergent series, as it still works about the same on infinite series that converge.

I tried showing this result to my Calculus teacher but couldn’t remember the whole process of getting to that -1/12. Now this result really does keep me up at night.

Infinity - Infinity is undefined, when you subtract you didn't consider about the last terms probably +infinity or -infinity. Then we can't do it.

@galaxyyy3427

2 жыл бұрын

3:55.

@AbelShields

2 жыл бұрын

Ahh, it's only undefined when we don't know anything about the infinities. Lim (x->inf) (x+1)-x is always 1, even though in the limit it is inf-inf

@rykehuss3435

Жыл бұрын

Infinity - infinity = infinity, not undefined. Check out Hilbert's hotel paradox

@AbelShields

Жыл бұрын

@@rykehuss3435 that's just wrong, you can't just compare infinities like that, it's undefined. Infinity minus infinity can be anything.

@rykehuss3435

Жыл бұрын

@@AbelShields Hilbert's hotel paradox proves infinity - infinity = infinity. Go read about it.

Great and easy to understand video. I saw this on Numberphile a few years ago with the professional math guys who got it wrong. Don’t ask physics guys to do simple math. I disagreed as they didn’t take into account convergent vs. divergent. Comments weren’t kind. Mathologer did two very nice videos on it also.

@coursmaths138

2 жыл бұрын

Numberphile didn't got it wrong. FYI the exact calculation Numberphile made, was made by Ramanujan. So, it's not a problem of "physics guys" or whatever....and the videos of Mathologer were not accurate on all points... There's no need to "take in account" convergent vs divergent series, because such calculation can be made with properties satisfied with convergent AND divergent series!

@csb178

2 жыл бұрын

@@coursmaths138 I went back looked at the videos again. On Numberphile (which I enjoy) physicist Tony Padilla obviously doesn’t understand what is happening and even says, “you have do the mathematical ‘hocus focus’ to see it.” (kzread.info/dash/bejne/qWF9mLqNhr2-p9o.html 6:48) I’m not saying they aren’t good physicists, but obviously there are mathematically principals escaping them here very specially in what you can and cannot do with convergent and divergent series. He refers to Konrad Knopp’s book and string theory. In a referenced video in the description they discuss the Riemann Zeta function which is not a sum but a value at -1. Mathologer discusses Ramanujan’s note’s and infinite sums and how to sum them. Numberphile did not discuss this at all. Mathologer video (kzread.info/dash/bejne/i6l9q8yFopnchaQ.html) I think is much on this topic and in greater detail on why the infinite sum of is not -1/12 and why it just isn’t mathematical “hocus pocus”. He discusses why this is the case, where in the Riemann Zeta function the and hypothesis there is a value. How Ramanujan fits into this (not in any sense via a proof or acceptance, but an initial note). kzread.info/dash/bejne/i6l9q8yFopnchaQ.html And I’d be interested to know where you think Mathologer got it wrong. You mention videos plural, however I was referring this one singular video on the infinite sum equalling -1/12, so please confine yourself to that video, unless you’d like t open a different discuss on another topic. You certainly do need to take into account divergent and convergent series or it’s math versus hocus pocus. Can you please show in detail how such a calculation can be made otherwise as you say, “because such calculation can be made with properties satisfied with convergent AND divergent series!” Thanks for your engagement even if you disagree.

@coursmaths138

2 жыл бұрын

@@csb178 ​ Thanks for your precise answer and your respectful tone 😊. I totally agree with you on the fact that " _Tony P doesn't understand what is happening_ " . And i know you didn't question their physicists' habilities. But you basically said (1st comment; if i understood well) that " _their math was wrong, because they're physicists_ " and that was the point i wanted to invalidate. This point is invalid, because the calculation wasn't made by them, but by the mathematicians Ramanujan (last part, on his note) and Euler (the 2 first series). So, their misunderstanding (although true) of the subject isn't the cause of the weirdness of that computation. Because, it is not *from* them. Yes, they do not understand what they're doing; but they just did _exactly_ what was already made before them. And if someone just made the same (write all these calculations as they already were in the previous litterature (mainly Euler and Ramanujan)), the result would have been the same: a weird calculation we don't really understand (and it wouldn't have nothing to do with that person being a physicist, a nobel prize, a mathematician, a field medal, a plumber, a soccer player or anything else...). " _the Riemann Zeta function which is not a sum but a value at -1_ " It is not a sum in itself. But, the value of -1/12 (as well as other values of zeta) can be obtained by using a summation operator (ie a "sum") acting on a subspace of sequences, and constructed with the zeta function. It is called the Dirichlet regularization (improperly known as " Zeta regularization"). And studying the properties of this summation method enlighten why the weird calculation using "sum properties" works well (as unexpected). I mentionned "videos" because your 1st comment mentionned "two very nice videos on it". I thought you talked about the one with the link you posted (in your second answer) and another on Ramanujan summation (kzread.info/dash/bejne/nJd_tKmpfcy8hNo.html). But, yes we can focus on the 1st one. What was wrong about Mathologer's video then? I think his video is very well supported, and yet paradoxically very bad. Because he missed the main point of all of this. And his explanations, though clear and (almost everywhere) mathematically right, didn't really clarify the subtlety of the topic. And you can see it in almost *every* (i said almost *every* ) comment refering to his video (including yours). So far, i have just met one single guy (english and others languages included), over all the hundreds of comments i read (litteraly), that understand well the subtlety of the topic. So clearly, something went wrong...and you'll soon get what and why. Mathologer's clearly right about saying that the limit of partial sum of the sequence (1;2;3;4;....) is obvisoully not -1/12. And he is also right about saying that answering this, would have make you fail your exam (5:35). But here's the main problem: every time he talks about the limit of partial sum, he says "sum". Why is it a problem? Because a limit of partial sum *isn't* a sum! No, it is not a sum. If you think the opposite, you're confusing the vocabulary and the mathematical meaning of math objets. Yes, we call the "limit of partial sum" the *sum of the serie* (the choice of word is questionable). But it is only a choice or *words*. And though this choice seems legitimate, it made forget that we're dealing with and operation other sequences that is not a sum. A sum is an operator other finite sequences (2-lenghts sequences to be exact, but can be extended and generalized over n-lenght sequences because of associative property). And the limit of partial sum clearly do not correspond to this definition. Moreover, limit of partial sums just doesn't have all the properties of sums: commutativity (Riemann Theorem), associativity, inversion with derivative and integral sign....and so on.... But, it is true that the limit of partial sums is an extension that has some properties in common with "sums". And that's why, i think the choice to call it "sum" isn't that "bad". But the problem is that almost every student fail to understand (ou memorize) that *this is not a sum* , and that it's often dangerous to think of it as it is (though it is often convenient). The problem is so bad that many teachers and students claim (with vigor and confidence) that *we cannot do anything* with divergent series. If the absurdity of this sentence doesn't shock you, let me explain: We said you have an operation other sequences (the limit of partial sums). Naturally, like all functions/operators, this function (the limit of partial sums) has a domain. And this domain is exactly the set of convergent series (by definition). And, of course putting a divergent serie (by definition a sequence not in the domain) inside a limit of partial sum is absurd. But it is just *one particular* function. The fact that some sequences doesn't fit with that function, obviously *doesn't mean* that we cannot do *anything* with these sequences at all! It's like saying we cannot do anything with the numbers -1 or -2, because log(-1) or log (-2) doesn't exists (or doesn't have a sense in the reals). It's obviously false and totally absurd....many functions can take -1 as an input... And it is the same for divergent series. Many operators can take divergent series as inputs. But, maybe you're asking "yes it's great, but it is not a sum". I would say yes; but (again) it's the same with the limit of partial sums. And we chose a function which has many properties (though not all) of actual sums and choose to view it as a kind of "infinite sum". That's why the limit of partial sums is called "sum" (when it exists). But again an infinite sum doesn't exists (mathematically). If we want to talk about "infinite sums" (that's some intuitive concept), we have to make a *choice* (with an actual definition of an operation we will see as corresponding to the intuitive notion). We make a *choice* . Thus, it is arbitrary. Because, the recipe "limit of partial sum" isn't the only one to generalize the notion of sum (in the math sense; ie for finitely many numbers). There are many other... And, knowing that we made *a choice* to generalize the actual sum, we can also make *another choice* to generalize the "infinite sum", by choosing some operator that can act on the sequences that weren't summable in respect with the limit of partial sum: such an operator is called a divergent serie operator (en.wikipedia.org/wiki/Divergent_series).

@coursmaths138

2 жыл бұрын

@@csb178 The argument of the exam is pretty weak (5:35). It's basically saying that "if you write something (the sum of 1+2+3...) that is not what we *chose* to call infinite sums (meaning the limit of partial sums), you're wrong". The same could've be said about complex numbers in elementary school or earlier (when it wasn't accepted). Write something as aparently stupid as x²+1=0 is false in all elementary school exams and was viewed as extremely suspicious during a long time....but does it makes it wrong mathematically? Or course not. And that's one other missing point in his video. It surely talks about super sums later, but all the begining (and the video in general) is assuming that Numberphile calculation is (or must be) about the limit of partial sum (8:48 - 9:05), and then is "nonsense" (cause the limits of partial sums don't exist). But *the calculation absolutely doesn't say that* . The only thing that the calculation says, is that this "sum" is an operation with some algebraic properties, and that these properties imply values for 1-1+1-1+1-...., 1-2+3-4+5-.... and 1+2+3+4+5+.... (1/2, 1/4 and -1/12). Nothing else. So, it *can't be wrong* , unless you prove such an operation cannot exist (and not only that it doesn't fit with the particular operation "limit of partial sum") ....but here's the thing: we can prove that such a function exist. And he is wrong on an other point (17:40): there exists a supersum on all 3 series (indeed over all series, with the axiom of choice; and only on a "big" set of sequences (including many divergent series) if you don't use AC). He is also wrong saying that "they don't have a supersum" (19:49; 20:30). The last serie just doesn't fit with *one particular type* of supersum (he is talking about Cesaro). But the fact that *this particular supersum* doesn't work on this serie, doesn't mean it does'nt have a supersum at all. Moreover, his (false) reasonning tends to make believe that all supersums are only about Cesaro (and Holder) and averaging (you can see in the link above that it is obviously false). And in 22:18, he is saying that inserting infinitely many 0 in the "sum" can change the value, and it's true. But in *that* timing, with *that* insertion of 0, the sum is unchanged. And his argument was that the insertion in the numberphile calculation was illegitimate, except....it is totally legitimate. That particular insertion of 0 is legitimate and can be justified with (almost) only the 3 basic properties of supersums (contrary to his claim in 22:42). Again, the calculation doesn't say that *any 0-insertion is okay* , but just use this one particular 0-insertion. At 23:20, he claims that the 3 properties (mentioned earlier in the video) are contradictory with 1+2+3+4+.... and it's true (in fact only the first two properties are sufficient to exhibit the contradiction). But, by saying that, he also claims that 1+2+3+4+.... *doesn't have* a supersum; which again is false. 1+2+3+....can't be supersumed with these 3 properties, but again it doesn't mean it cannot be supersumed with other supersums *at all* . And actually, it can (cf Dirichlet summation). At 25:40, he says that the zeta function is the "genuine, real, actual, connection between 1+2+3+....and -1/12" meaning that because supersums fail on this serie (according to what he said just before; again even if it is false ). And though, zeta function is clearly one aspect of that connection, it is not the only one. More: the connection between 1+2+3+...... and -1/12 can be made *without* any use of zeta function (and complexe analysis), in at least 2 different ways: one analytic made by Tao (terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/) and one with purely algebraic methods (sorry don't have the courage to develop, maybe an other day 😘). So no, zeta is not *the* connection between the serie and -1/12 (as many comments say in all video about this theme), but just an aspect of it. What is strange is that he put the link of Tao's article in the description. But this article specifically says that it is possible to give an interpretation of these computations _by purely real-variable methods, without recourse to complex analysis methods such as analytic continuation, thus giving an “elementary” interpretation of these sums that only requires undergraduate calculus_ . And then, he is confusing supersuming and analytic continuing. Eta isn't extended by supersums, but by complex analysis methods, and then corresponds to some supersums values. And in 35:20, the Numberphile computation isn't nonsense because there's a coherent set of rules which allow all the calculations made, and imply the values 1/2, 1/4 and -1/12. And to answer your question, no we don't have to check wether the series are convergent or divergent, because the calculation is legitimate *if* such an operator (on these series) exists. So, we just have to *prove that it actually exists* , and then it immediatly follows that all the calculations are mathematically justified. And guess what? Yes, you can prove it 😉. For now, i'm not writing the details, but there's an excellent article on divergent series (arxiv.org/abs/0705.1578 ) which can help you to understand how we can construct this operator (left as an exercice to the reader 😝; maybe i'll post the solution if you can't find).

@csb178

2 жыл бұрын

@@coursmaths138 Thank you for your quick and courteous reply. I think you saw an insult where more of tongue and check humor was meant. I would say math and physics are certainly overlapping disciplines, but each still have their own nuances and specialties that the other does not necessarily deal with. My point was not to invalidate the physicist’s math, but rather for the physicist to stay more in the physicist’s proper lane rather than in the mathematician’s lane, and they do as I say overlap much. My point here was in Numberphile bringing in the physicists’ credence as professionals and experts particularly as physicists in order to explain an especially nuanced part of math that is not part of physics (although can be used in physics) and that they certainly did not fully understand. Numberphile is watched by many, especially young aspiring mathematicians and the explanation was permitted and accepted as true but only via “hocus pocus” and not through the rigor for which math is done. I also do think it also goes beyond semantics of simply what we want to (or chose) to call infinite sums. I do not think this is a language problem or an epistemological problem. In this instance, I think it is purely an understanding or comprehension problem. In elementary school I was never told x2+1=0 was false. Not being able to explain why it is not false does not make it false even if the explanation is unknown. We are both beyond that. And it wouldn’t not have satisfied the rigorous mind especially from an equation for graph point of view. However, I do think if it were explained as wrong, even if not understoof, there are not answers in the real numbers, that would have been acceptable. I just haven’t been taught imaginary numbers and that there exists it was just not taught to me until later so I could be taught and understand that ι is the √-1 Their reference to Ramanujan also I think was deceptive because it was not for his summation of divergent series but his note on it as an oddity, with no supplement, further confusing the matter for viewers. I appreciate your in-depth discussion (the arxiv link does not work, it gives a 404 error) and also your polite tone and interaction. Here is my biggest problem with Numberphile’s video on this. And I will give you much credit because I think you hit the nail on the head, you have a very in-depth knowledge of mathematics, and I would say myself to a degree. Numberphile usually presents a range of math from easy to difficult in a manner that most people can comprehend in order to further the desire to study and appreciate math. They also use well-known and established professors to do that. In this video, under 8 minutes, they took and presented what in their own words is A POSITIVE INFINITE SUM OF REAL NUMBERS (not verbatim) and told their audience that it equals -1/12. At this level to present this as they have in the time they did at the very least THEY MUST discuss divergent and convergent series as Mathologer did. Mathologer had to do it in a very longer way in order to discuss the 7 minute and 50 second video which was not sufficient. Numberphile did not talk about imaginary numbers. I brought up the Riemann Zeta function and hypothesis because I know these properties. Now, let me ask you, do you think anyone was thinking about Dirichlet regularization in this video? Or even understands it (I’m not saying they can’t understand it, but they currently did not, I would say. They talk about String Theory to back up -1/12 but not Dirichlet regularization). I can see your issues with Mathologer and comments, perhaps even my own. However, to me it is very clear Numberphile set-up an 8 minute video talking about the infinite sum of real number equaling -1/12. I listened and watched the video, I saw the equations they wrote and referenced (math is universal, there was no word choice, and the equations they chose did not use limits, etc.) it is presented as simply the sum of all the real positive to infinity. Presented as such, and that is how I, and almost universally everyone else, interpret that particular Numberphile video, I cannot accept it. And on those terms, I think Mathologer did an objectively better job explaining how the sum of all the real positive numbers to infinity cannot equal -1/12. Notwithstanding higher math and the wonderful work you did, and that I agree there is a place for imaginary numbers, axioms (not particularly in math I would say generally), but this video was not the place for it. In the end, I will be glad to say it was poorly explained on their end. I do not think, even with your explanation, that your explanation is what THEY meant. 🙂

This was really interesting! I’ve never realized just how unruly infinite sums can be.

_You will get the S when you fix this damn door_ ~Maguire the Bully

you got my sub after this video and i was not a fan of some other video. Great job.

@BriTheMathGuy

2 жыл бұрын

Welcome aboard!

Huge respect to great mathematicians Sir Ramanujan

@mathsloverprashant9109

2 жыл бұрын

*mathematician

@maalikserebryakov

Жыл бұрын

U are 100% an indian lol

@12345ngb

Жыл бұрын

Almost every Indian is proud of him. But this theorem that the sum of positive numbers ends up being negative is BS.

@CriticSimon

3 ай бұрын

@@12345ngb Exactly! This sum cannot equal -1/12. Ramanujan was wrong hehe

thanks for clearing that up.

I think it’s because 1-1+1-1… is never actually 1/2, it averages out that way but the result is always either 0 or 1

@testmicrowavedinner4438

2 жыл бұрын

Its a summation of an infinite series. In calculus, it is undefined

@ajshowdown8033

7 ай бұрын

To me its like an excuse to not being able to find out whether its gonna land on 1 or 0

@archananethi9233

6 ай бұрын

Yes you are right indeed, in that case, since we have two results(0,1) for a single question, we would take the average of the result which would be 0+1/2 = 1/2…

One of these sums involves the partial sum of an alternating series which is conditionally convergent at best. In a conditional convergent sum, the order of addition and subtraction DOES matter. So 1-1+2 = 2 + 1 - 1 for a finite set of numbers, but not for infinite

The various other details about applicability of the distributive law and such are really important and I'm glad you went over that as well, but regarding the original proof, the point at which I immediately said "whoa, hold on there" (and everybody else with some critical thinking should too, IMHO) was the whole "2B = A" addition trick. Because you're shifting the terms over a space, the resulting sum must logically have _one more term_ than the original ones did, which is not being accounted for anywhere. This is actually the "last" term of the second B sequence, which will be "dangling off the end" somewhere out there infinitely far down the way. If this is actually taken into account, then 2B does not equal A. 2B actually equals "A + (the final term of B)". However, there arguably _is no final term of B,_ (B is an infinite sequence, after all) so this effectively becomes "2B = A + (undefined)". That is, it's not an actual value you can substitute into other equations in the first place. This is all a great lesson in the pitfalls of blindly applying mathematical operations without understanding their limitations or ramifications, though..

@scubasteve6175

2 жыл бұрын

Thank fuck someone said this

Fascinating. I like this. Looking forward to learning calculus.

This is the power of the person who knew infinity. "Shrinibas Ramanujan"

Hi I'm french and I love maths I can't do maths in English in my high school so I do maths in English with your videos and I understand all you say thank you so much for these ❤️💪🏻

An interesting way to get the initial result for Grandi's series is to use S = 1 + p + p^2 + p^3 + ... + p^{N-1} = (1 - p^N)/(1 - p), which is pretty straightforward to derive geometrically, and setting p = -1. When 0 infinity, but when p > 1, the typically quoted sum formula of 1/(1 - p) actually gives the result of S' = S + p^N/(1 - p). This means it's no longer an infinite sum as N -> infinity, but the output is still "correct"; evaluating 1/(1 - p) at p=-1 gives 1/2 for any value of N, because the p^N/(1 - p) term in S' ensures that the result is always 1/2 without any magic or faith required. We can also get 1 + 2 + 4 + 8 + ... = -1 for p = 2, which is another famous result, and again, the p^N/(1 - p) term ensures that this is true for any value of N. Similarly, plugging in the manipulations in your video gives -1/12 for K = 1 + 2 + 3 + ..., but you end up with the unfortunate situation that K + (-1)^N/(-12) = -1/12. The issue seems to be that the manipulations required to get from Grandi's series to 1 + 2 + 3 + ... are invalid.

2 жыл бұрын

It doesn't even have to be valid. Z(-1) and the series discussed are litteraly the same and Z(-1)=-1/12.

@vascomanteigas9433

Жыл бұрын

It is the direct result of the Unicity of Analytical Continuation.

Ive been trying to understand this mind boggling summation and all other youtubers were just showing the proof and thats all. Yours is the best video ive seen about this phenomenon so far.

thanks for actually showing *how* the maths breaks down when you just assume things that don't hold

I'm curious as to whether there exists some alternate metric on the natural numbers under which this result converges, or at least a partial proof...

This is it, finally i find someone, brilliantly more educated than me, who agrees that this just doesn't seem possible. Thankyou brian

Thanks to Ramanujan for giving us this wonderful series

Man proving being capable of finding 1/0

Overall a great video. However, there is a bit of an error. The Riemann Zeta function is not defined to be the sum, but instead the analytic continuation of the sum and thus zeta(-1)=-1/12 is true. Whereas the video says it's defined to be the sum, and the analytic continuation of the sum at -1 is -1/12.

2 жыл бұрын

The Riemann Zeta function may not be defined as the series, but zeta(-1) coincides with -- is the same thing as -- the series. So the sum is -1/12. As for the value -- the (part of) mathematics connecting to this part determines it numerically. This is valid because these parts has other -- from this matter independent -- connections.

@lorenzobarbato4558

2 жыл бұрын

@ No, the Riemann zeta function coincides with the series only in the region of the complex plane where the real part is greater than 1. Otherwise, that series does not make sense. Usually mathematicians call Riemann zeta function the analytic continuation. ζ(-1) is NOT the series evaluated in -1, but the value that the analytic continuation assumes in -1 (and cannot be expressed by the usual series that defines the function in the domain I referred to previously)

2 жыл бұрын

Lorenzo Barbato The Riemann zeta function is much older than Riemann but had another name of course. The treatment for complex values are due to Riemann and followers. The important thing is that the series has also a multiplicative definition. Euler was investigating the structure rather than particular values. The modern view of the zeta function may not easily correspond to the series but the original zeta function does. So this is a matter of interpretation. The important thing is that the value is -1/12. A value which is determined by methods that ultimately arose from the studies of the structure. Analytic continuation didn't come out of nothing.

@ethanbottomley-mason8447

7 ай бұрын

@ No, the Riemann zeta function is not the same thing as the series, it is a meromorphic continuation of the series using its functional equation. It does not evaluate to the same value when Re(z) = 0. This gives 1/(1-x) when |x|=1, the series diverges. This means that 1/(1-x) is a meromorphic continuation of the series. A continuation does not mean that you can now evaluate the series outside its original domain, it is something completely new.

7 ай бұрын

@@ethanbottomley-mason8447 No. It is not something completely new, The functional equation is not arbitrary and can be obtained from both "starting points". Via Abel, Malmsten, Riemann, Mittag-Leffler, Weil for example. But you are right that considered only in a narrowly defined series setting it is a a new thing. Of course the very goal of generalising is to unify.

Easiest way to add consecutive numbers starting with 1 is that if the number ends in an even number, just divide that number by 2 and multiply by the next number. If the last number ends in an odd number, divide the next number by 2 and multiply by the last number. Example: 1+2+3+4+5+6+7+8= 8/2=4, 4*9=36. 1+2+3+4+5+6+7+8+9=, 10/2=5, 5*9=45

@X22GJP

2 жыл бұрын

So?

@owenaspinall2046

Жыл бұрын

I gauss that would work (pun fully intended)

And that's why there are special rules to operating with series depending if they're converging or diverging

LOVED IT 😍 🥰 😍 🥰 😍

My opinion is: S = 1+9S & S-0.25 = 4S could be equal to two solutions: one is the algebraic usual way -1/12 & -1/8, two is the way that says S could be ∞ (infinity).

1:33 B + B is 2B me: 2B… or not 2B?

There is a french youtuber who did a video about this showing that these rules that we use for convergent sums can be extended to a sub set of divergent sums without contrdictions. This sum isn't one of them. For examples we can still keep using this rule for divergent geometric series as long as the common ratio isn't 1. can be very interesting for those who speak french and have some understanding of linear algebra.

Partial sum formula: y = x(x+1)/2, area under this curve from -1 to 0 is -1/12

05:45 You could also recursively define it. S = 1 + 9 ( 1 + 2 + 3 + ...) = 1 + 9S = 1 + 9 (1 + 9S) = … You could say S = 1 + 9S then 1 = -8S and then S = -1/8 Or you could continue with the recursive definition to say S = 1 + 9(1 + 9S) 1 = S - 9(1 + 9S) = S - 9 - 81S = -80S - 9 10 = 80S S = 1/8

@findmebro

6 сағат бұрын

damn

Given by legendary Ramanujan sir and credit by this man 😅 Legendary person in the history of mathematics who shook the entire West Great Indian and Mathematician SIR RAMANUJAN ❤

You opened my eyes.

Ramanujan sir great mathematician from India 🇮🇳

@YTMRCODER

6 ай бұрын

Yes

@CriticSimon

3 ай бұрын

This sum cannot equal -1/12. Ramanujan was wrong hehe

Good Job! just a minor correction. Riemann himself(and many others ) analytically continued zeta function to the entire complex plane except one.

@absolutezero6190

2 жыл бұрын

Didn’t they say that in the video?

@RSLT

2 жыл бұрын

time stamp 7:37 says "traditionally" riemann's zeta function.. only defined Re(s) > 1. Riemann himself extended the Riemann zeta function to the entire complex plane.

@absolutezero6190

2 жыл бұрын

@@RSLT oh I see what you’re saying

Nice video. This has been bothering me for a while too. Numberphile and Mathologer also cover it in similar detail. But I think I finally figured out why this is flat bad math. It goes wrong before you invoke the associative law. It goes wrong right at the start when you treat S as a number. Strictly speaking, it is not a number. It’s infinity from the beginning. You can prove all kinds of nonsense by treating infinity as a number. For example, infinity minus one is still infinity. Therefore S-1 = S or 1=0. This whole game is like dividing by zero or saying that 0/0 = 1.

@rykehuss3435

Жыл бұрын

Exactly. 1+2+3... = infinity, and infinity is not a number or a sum of numbers

@FrankGBoston

Жыл бұрын

Yes. The distributive and associative properties of the integer, real and imaginary numbers work for infinite sums, if the sums are convergent. Otherwise, one can rearrange the terms to get anything. IE one can write (without any mathematically defined meaning) that ∞ = 1 + ∞, then rearrange to ∞ - ∞ = 1, then 0 = 1. When something diverges (ie is not a number), you can derive anything.

@user-uh9bo2im1h

10 ай бұрын

Okay so according to you infinite is another form of number but we can’t use basic operations on it. For example let’s consider x = sqrt(x+sqrt(x…….. Well then x must be x = sqrt(1+x) x^2 = 1+x x = -(-1/2)+ or - sqrt((-1/2)^2) Hence why x is both 0 and 1 so those infinite sums are just useless as a whole

Thanks to Ramanujan for all of this. May his soul rest in peace 😔.

@Occ881

3 ай бұрын

What is the application of his discoveries?

@fireworxz

3 ай бұрын

@Occ881 Until very recently, his formula was used as the basis for the fastest approximation of pi in computers. Also, his number partition formulas are used to study black holes. i suggest to read works by Ken Ono, who explains Ramanujan's contributions in detail

@CriticSimon

3 ай бұрын

This sum cannot equal -1/12. Ramanujan was wrong hehe

For people who want to know, this is also called the Ramanujan Paradox

It pisses me off. It is like making the rules what ever you want and just playing around with laws and numbers. In the B part, where the terms are shifted before getting added, what was that??

@EAOO

7 ай бұрын

Imagine instead of "shifting" it, you add zero at the beginning. You are just adding 0, and not changing anything so it isnt an issuem the issue arises from uses the distributive property on an infinite series.

At 1:54, you're leaving out a term in the second B. If you include the -6 on the bottom, then you have to include the +7 on top, which requires the matching +7 added to the bottom... so this trick of moving the numbers when adding uses an inequal number of terms on the top and bottom. So then why couldn't you rearrange the numbers as you please and create any value that you like? I think this violates some basic rules of algebra. In order to remove the -6 from the right, you need to add it to the left, so your result would be 2B + 6 = 1 -1 +1 -1.. Then if you instead stopped at 200, you would get 2B +200 = 1 -1 +1... I think the real issue here is that it's dangerous to do concrete algebraic operations on abstractions such as infinite series.

@kanavbansal3760

Жыл бұрын

Yep, that was also my question while watching the video

@ahmethakanuysal9324

3 ай бұрын

I think the same This is the main problem

Eh, I have issues with the way this was explained. For one, there is this inherent assumption that it only makes sense to talk about series in the context of convergence, all without even an attempt for an explanation for why that must be the case. I think that is flaw. Also, the counterargument presented in the video, regarding the distributive property, does not work. The video, at one point, explained that if we take terms of 1 + 2 + 3 + ••• by groups of 3s, we end up getting 1 + 9 + 18 + 27 + •••. But the issue is that this is not allowed by the distributive property. Instead, it assumes the associative property, which no one claims holds for series, not even for convergent ones (see the Riemann rearrangement theorem). The series 1 + 0 + 0 + 0 + ••• = 1 + (- 1 + 1) + (- 1 + 1) + ••• = 1 and 0 + 0 + 0 + ••• = (1 - 1) + (1 - 1) + (1 - 1) + ••• = 0 are both convergent, and so this demonstrates already how associativity does not work with convergent series. The series are different, and so no one expects divergent series to be the same under a regrouping like that. And both of the series mentioned above are different from the series 1 - 1 + 1 - 1 + •••, which is divergent. It has nothing to do with the distributive property failing. And on that note, that leads to me the biggest issue with the video, which is the treatment of series as a kind of summation of numbers. To be fair, the actual mathematical terminology we use does make it sound like we are just adding numbers together. But this is inaccurate, and that is revealed by the fact that we talk about convergence and divergence of series in terms of their partial sum sequence. Series are just that: sequences. We can add sequences, multiply them, and we can apply operations to them that will output a real number. Evaluation of a series is just that: applying some combination of operations on a sequence. Addition is involved in some of the operations, but describing the entirety of the operations as just being "summation" is misleading at best, and it leads to the kind of misunderstanding that led to the infamous Numberphile video in the first place. And this actually explains why the Riemann rearrangement theorem is true, and why "associativity" does not work for series: because in reality, we are not merely adding quantities, we are applying some kind of transformation to a sequence, and while addition is somewhat involved, there are also other things involved exclusive to sequences, that have nothing to do with adding numbers, and that is what makes this whole "associativity" issue not work. It does not invalidate a rigorous treatment of divergent series, which does exist, it just means we have to be more careful with our understanding of series. That is also why I find the insistence on focusing on the limit of partial sums to be a bad argument to use in this context: it is self-defeating. It misses the point of how series work, and it fails to appropriately bridge the gap between how series work, and how they are portrayed. But, the video does a good job at explaining the role analytic continuation plays in all of this, and the video does clarify that, ultimately, this is a matter of context. I just wish that explanation was better supplemented by providing a conceptual understanding of what a series really is, and why it is a conceptual mistake to just think of these manipulations as summation, and of these properties as analogous to distributivity and associativity.

@BriTheMathGuy

2 жыл бұрын

Thanks for the great insights Angel! Hope you're doing well.

@kazedcat

2 жыл бұрын

I did not read your blog post so it seems you forgot the issue of being brief.

@user-kw3vg4jn8u

2 жыл бұрын

k

@user-up7nb6id1f

2 жыл бұрын

@@kazedcat never said that learn to read bozo

Hats off to Ramanujan Sir

Help, we watched this today in furthermaths and it came in my yt front page! yt be stalking me ;-;

We can also use divergence/convergence tests to test whether the series actually come to a point, and we find that “1 + 2 + 3 + …” is divergent by the nth term test, and we find “1 - 1 + 1 - …” and “1 - 2 + 3 - …” are divergent by the alternating series test. Thus, you can’t find an actual solution for these

@coursmaths138

2 жыл бұрын

This only prove that the limit of partial sums doesn't exist. Not that it has no "actual solution".

@garethreynolds557

Жыл бұрын

@@coursmaths138 Ok but then you have to define what an "solution" would actually be. The epsilon definition of convergent sequences (along with the extension of this definition to series in terms of its partial sums) is what people generally are referring to when using the equals sign next to a series. The -1/12 comes from the analytic continuation of the Riemann Zeta function evaluated at -1, which is not the same as saying that 1+2+3+... actually converges to -1/12.

@MrTrollo2

Жыл бұрын

@@garethreynolds557 i mean, people do use the -1/12, as in the often mentionend casimir effect. So, in a way physics confirm the result experimentally. It seems like "sums" aren't defined clearly enough to concluce that the natural numbers do not converge at -1/12 or go to infinity depending on where you use it.

I know you can't just treat the end of the infinite series 1+2+3+4... as an actual number but lets say its k (even though it doesn't actual exist). So when he shifted over all those values and took the sum the very "last" number k would have no pair to be added to. This would mean it also is being added to 0 just like the initial 1 on the row above (because of the shift). Taking the limit, k = infinity, the sum is therefore equal to infinity +1 which is also infinity. it would be like saying which is larger -1+0+1+2+3+4... or 1+2+3+4....? Both are approaching infinity. I know there is a high chance I am violating some sort of rule in math but I am no educated mathematician. I understand limits but when they are being used in this way my confidence in my logic is very low. But if someone notices an obvious flaw let me know because i like to know if I made a valid point or learn where i got it wrong.

@X22GJP

2 жыл бұрын

infinity + 1 is a different "infinity" to the that which you added 1 to. There are an infinite number of different infinities.

wow!!! how do you even get to think about this?

Keeps you up? I fell asleep while watching this already!

I'm in University studying Engineering. Part of my degree is learning about sequences and series. What I just saw here was the mathematical equivalent of snake oil.

@carvalhorosolen

Жыл бұрын

What you mean? that the sum is indeed -1/12 or that it is not? 🤔

This is a good video that touches on a subject of critical importance to the real world: Perspective. Math can tell us any number of things about the physical world, but if you don't understand the meaning behind the mathematical operations you apply, then your result lacks meaning too. Certainly, the number "-1/12" can tell us something meaningful about the world, but only when you apply it in the specific set of circumstances that led to that result. For example, the "+1 -1 to infinity" series yielding a result of 1/2 makes perfect sense if you think of it as solving for an average. Infinity is itself not any particular number, even though we often like to think of it as one. Here, infinity is best represented not as one really big number, but as a distribution of many really big numbers. An infinite number of infinities, if you will. If you take the average of this function when evaluated at n = "really big number" and n = "really big number + 1," you get 1/2. If you continue adding pairs in this way, you will continue to get 1/2. If you stop after an infinite number of pairs and then add just half a pair more... you *still* get 1/2, because you're now measuring the limit approaching a single infinity rather than infinite infinities. So it makes sense that unintuitive answers like the sum of natural numbers being -1/12 would get used in fields like quantum mechanics. Quantum mechanics deal in probability distributions all the time, so being able to represent where a particle is "on average" is meaningful to them. This sum is derived not through conventional means, but as an extension of a probability function describing some other quantity.

@_sayan_roy_

Жыл бұрын

You see, I thought about the analogy of collapse of a wave function and intuitive average of 1/2 as a sum of 1-1+1... as well before, this same feeling can't be extended to -1/12 which is straightaway negative. But i understand that it can exude similar mental gymnastics if we try harder.

You are great!

Amazing

My explanation for this is not that distributive law does not work here. I think it does, but the problem is in the step of converting "S = 1 + 9S" to "-8S = 1". You effectively subtract infinity from another infinity, that's indefinite, i.e. it can be any number (or infinity as well).

@mbrusyda9437

2 жыл бұрын

And where'd you get that S is infinity

@rykehuss3435

Жыл бұрын

@@mbrusyda9437 1+2+3... = infinity. From there.

@rykehuss3435

Жыл бұрын

Hilbert's hotel paradox proves that infinity - infinity = infinity

@mbrusyda9437

Жыл бұрын

@@rykehuss3435 that explains nothing...

@rykehuss3435

Жыл бұрын

@@mbrusyda9437 Then you are really stupid. You asked where he got that S = infinity, and I just literally showed you very clearly where from. Try using your brain

Love it how infinity can give us all sorts of interesting behavior since we can't quantify it. While there are many arguments against why this doesn't work, videos like this (Not Bri's, but every video about this) never point out the obvious. At the step where you say S = 1 + 9(1 + 2 + 3 + ....), what you have in the parenthesis isn't technically another copy of S. Yes, it looks like it, but you set S equal to 1 + 9(....., so technically the parenthesis isn't S, it's missing that 1 at the beginning. This is the problem with trying to quantify infinity. I think each S on either side is a different infinity. Hence why we use things like limits. L'Hospital's rule, etc. Simply, infinity isn't enough. So setting these two infinities equal (when in reality they are not equal) gives you weird issues like this. However, this makes me wonder about your (and the usual method in other videos) about deriving x^x^x^....., you just set it to be x^y, since everything after the first x^ looks just like the original function. But it's not, it's always missing that first x^, so technically it's not the same. I guess doing this works though because you're not looking at it's behavior long term but rather a point? I don't know, now I'm starting to let my mind wander. Point is the two S's you are using are ever so slightly not the same infinity, giving weird results.

@user-lh5hl4sv8z

2 жыл бұрын

How do you measure sizes of infinity and if they’re really “bigger” or not? Why can’t it be smaller, have no size, be of same size, or there is no “size” scalar when working with infinities? Can you even ascertain their “ordinality” is by a factor different with respect to ordinals?

@fuzzybanana0123

2 жыл бұрын

@@user-lh5hl4sv8z I literally explained it in my original comment.

@mertaliyigit3288

2 жыл бұрын

Its because series doesnt converge. This is no different than finding 0=1 by dividing by zero

@georgecantu856

2 жыл бұрын

The S on both sides of the equation are the same. If they are not then I challenge you to find the difference between them. They both start at 1 and end at infinity. The reason why weird issues pop up is not because we are setting two infinities equal to each other, it is a result of treating a divergent series as a convergent series. And for the y = x^x^x^... = x^y has a missing x; well, we are not missing the first x because x is a variable for a number. Its like saying y = 2^2^2^... = 2^y is incorrect. We are not missing a first 2, it is already included. In short be careful with infinity.

@kazedcat

2 жыл бұрын

You compare the sizes of infinity by doing a bijection. If two infinities can be map into 1 to 1 corespondence then they are equal. For example all even numbers can be map to natural numbers. Every number n in the set of all natural numbers has a counterpart 2n in the set of all even numbers therefore the set of all natural numbers is equal to the set of all even numbers.

The series A is indeterminant, it doesn't equal 1/2 since it doesn't have a solution in the first place.

i know whats going on here, the set of s gets so big that it encounters an integer overflow and defaults to -1/12!

@napalm5

8 ай бұрын

12 factorial?

@dAni-ik1hv

8 ай бұрын

@@napalm5 nah -1/12 factorial

@ROForeverMan

7 ай бұрын

Best answer!

the way I always say it is "1+2+3+4+5... doesn't equal anything, it's a divergent series thus does not converge, but if we assume it does converge to a finite value, you get -1/12"

@angelmendez-rivera351

2 жыл бұрын

But that is a categorical mistake. Equality and convergence are not the same thing, so saying the series is not equal to anything because the partial sums do not converge is a mistake.

2 жыл бұрын

Convergence IN THE LIMIT and equality is the same thing. That what it is what is made/used for. There are infinite series that are truly divergent such as the harmonic series. The actual series has the numerical value -1/12 and the non-numerical value infinity.

@angelmendez-rivera351

2 жыл бұрын

@ The harmonic series is regularized to γ, so that defeats your point.

2 жыл бұрын

You can regularize any infinite series or infinite product. In particular you can show that the series sum to -1/12. The regularized value for the harmonic series is the difference between the harmonic series and the natural logarithm at infinity.

@angelmendez-rivera351

2 жыл бұрын

@ Both are intimately related by the Euler-Maclaurin formula, which is what Ramanujan summation is defined in terms of and is formally based on. Essentially, if we have some function f : R -> R, we can restrict it to f* : N -> R, such that we can ask about the summation of f*, the integral of f, and then analyze the difference between the two. This is what leads to Ramanujan summation and regularization.

I think a key issue is at 1:30. I don't think that is something you can just do like that.

@manthansagar200

Жыл бұрын

You can

@adityagoyal7110

5 ай бұрын

It's better and easy if you do A-B A-B = (1-1+1-1....) - (1-2+3-4+5......) Now on opening brackets A-B = 1-1+1-1...... -1+2-3+4-5.... Now on making pairs (such that one number from first bracket and second no. from 2nd no.) A-B = (1-1)+(-1+2)+(1-3)+(-1+4)+(1-5)...... A-B = 0 +1 -2 +3 -4........... A-B = 1-2+3-4+5..........., Notice that 1-2+3-4+5.... = B So A-B = B A = 2B 1/2 = 2B (becoz A = 1/2, already proved above) 1/4 = B

this one only fucks me up because of its connection to the reimann hypothesis

most videos take 20 minutes of lead-up and dont explain it this detailed, wow thanks

I am glad how ramanujan ( indian mathematician and discoverer of this series) discovered this while being in 12 class and blew the worlds mind, true genius

Also, if you double all the values in the set (1+2+3+... to infinity), you make the value (or sum) of the set smaller!

@rykehuss3435

Жыл бұрын

And if you keep doubling the values infinitely, the sum of all natural numbers is 0. Which again just proves how silly this is and has nothing to do with logic

@biblebot3947

Жыл бұрын

@@rykehuss3435 it approaches -infinity

An excellent illustration of the pointlessness of wondering about divergent infinite series.

I like th part which shows so nicely that S = -1/8 in the one layout, simply showing, if you run around with infinities, be careful, and as quite ofthen -1/12 is a results, does not mean it is correct... it is simply a convergent error which appears.... I was wondering, if it may be possible to group instead of the 3 numbers together, in other groups with more or less numbers, and larger and minor groups to basically generate any number you like.

IDK why these types of problems are so mysterious that i am always awake till 3 am at night thinking about these 😆

I think the best analogy are complex numbers. They’re a continuation of real numbers and taking the root of a negative number makes as much sense as the sum of all natural numbers being negative

2 жыл бұрын

Complex numbers makes sense and the result that the sum of all numbers is -1/12 too. Numbers are not just like bricks lining up.

@peterrussell6950

2 жыл бұрын

@ I am unsure what you mean by numbers are not just bricks

@peterrussell6950

2 жыл бұрын

We (as collective humans) used to say it made no sense to do 1-2. 1 apple minus 2 apples? That makes no sense! At one point we saw that it was advantageous to extend our number system to get some extra nice features. It's the same thing with imaginary numbers. We used to say that the square root of a negative number wasn't a thing. But we've since then extended our number system again because it turned out to be wildly advantageous to. They're regretfully dubbed imaginary not because they're actually some abstract thing that only pure mathematicians with little time on their hands use, but because "imaginary" was used early on to describe them, because at the time they weren't widely accepted. They are equally as valid and make just as much sense as the real number line. As for the sun of the natural numbers thing, yes you are right that makes no sense except in specifically defined circumstances, which is almost never

2 жыл бұрын

Numbers have no size other than relative to other numbers. No direction, weight or anything. They are abstract. The have value/size only relative eachother. In applied mathematics such as physics number asign values in the same way as written text. Mathematics is a language. It can not be contradictory. The -1/12 result is frequently used in physics. It may well turn out that there is a simpler reason for that unrelated to this question. That doesn't invalidate the mathematics though. Note that I don't say that infinity is wrong but I say -1/12 is right. Infinite series may have more than one sun without being contradictory. Pardon my lousy English. I might be wrong but I don't think so. Greetings to all readers.

2 жыл бұрын

Complex numbers are awsome. Consider as an example (2+2i)^3 = -16+16i. Here the absolute value of (2+2i)^3 = (2sqrt(2))^3 = 16sqr(2) is equal to the absolute value of -16+16i. But (2+2i)^(3+3i) has not the absolute value (2sqrt(2))^(3sqrt(2)). It's unfortunately a bit more complicated.

Thanks for not making the same blunder Numberphile did, I was worried going into this

A is either 1 or 0 but is undefined in my mind. Working with Undefined (vs Unknown) could go many places but means nothing especially since including another undefined thing like Infinity.

I also think it is impossible for the sum of all positive integer numbers from 1 to infinity to end up being equal to some minus and non-integer value But let's suppose that 1+2+3+4+... = -1/12. We know the formula of arithmetic progression: S(n) = (2a1 + d(n-1)) / 2 * n = (2 + n - 1) / 2 * n = n(n+1) / 2 So n(n+1)/2 = -1/12 n*n + n = -1/6 n*n + n + 1/6 = 0 Discriminant = 1 - 4 * 1 * 1/6 = 1/3 n1,2 = -1 +- sqrt(1/3) / 2 But isn't n supposed to be infinity (or tend to inf)? Or at least some *positibe* number? What am I doing wrong? inf + inf = inf inf * inf = inf I don't see any mistakes

@coursmaths138

2 жыл бұрын

Your mistake is to assume that "sum" is a finite sum. Which is not. So the formula of arithmetic progression doesn't apply. We're dealing with a totally different object here...

My favorite thing is from string theory where I was asked to compute the sum of all ODD numbers. sum_n>0 = sum odd + sum_n>0 (2n) so you split all numbers in even and odd. But then , sum odd = -1/12 - sum_n>0 (2n) = -1/12 - 2 sum_n>0 = -1/12 - 2*(-1/12) = -1/12 + 2/12 = +1/12 lol. So even if the odds are clearly a subset of all integers, their sum is "bigger" than all integers. Peace

@simongross3122

Жыл бұрын

That's an odd result :)

Clearly a super-advanced example of an integer overflow error. We live in a simulation. Sleep well.

It would be nice if you mentioned the relationship between Ramanujan's result and the Casimir effect in physics.

*"I applied the distributive law to a divergent series and proved that infinity equals -1/12. CONSEQUENCES!"* - BriTheMathGuy (2022), channeling his inner Key & Peele.

@user-xh9pu2wj6b

2 жыл бұрын

Not really, he applied a pretty crude regularization to this sum and got a correct answer. You can try another regularization, like exponential one, for example, and get the exact same result.

@rykehuss3435

Жыл бұрын

@@user-xh9pu2wj6b And what if I apply this regularization and got a correct answer. S = 1+1+1+1... there fore S = 1+S, which can be reduced to 0 = 1. I just proved with math that 0 = 1. This is why you cant treat infinity as a number. S = 1+2+3+4... = infinity. Not a number, and definitely not -1/12.

@user-xh9pu2wj6b

Жыл бұрын

@@rykehuss3435 "S = 1+1+1+1... there fore S = 1+S" prove that. You can't even add a zero to the beginning of an infinite sum and expect it to be regularized in the same way always. Indexing matters in some cases. "This is why you cant treat infinity as a number" nobody does that here tho. "1+2+3+4... = infinity" only under weak summation methods. "Not a number, and definitely not -1/12." you've been proven wrong and yet you insist on this like your life depends on this. Why?

@rykehuss3435

Жыл бұрын

@@user-xh9pu2wj6b Here's proof that S = 1+S. You cannot say where in that line of +1's you cant fit another one. Therefore S = 1+S. Also S= 1+1+1+1+S. And with that logic S = S+S. Because infinity + infinity = infinity. Also infinity +1 = infinity. Proven. "This is why you cant treat infinity as a number" nobody does that here tho. " Everyone who thinks that 1+2+3+4... = -1/12 is treating infinity as a number. Including yourself. And you area dead wrong if you do that. Ask any real mathematician. "1+2+3+4... = infinity" only under weak summation methods." I dont care if its weak or strong, and neither does mathematical logic. You cant sum all natural numbers and say its some number S. Its infinity, and nothing else. And infinity is not a number. ""Not a number, and definitely not -1/12." you've been proven wrong and yet you insist on this like your life depends on this. Why?" I just proved you wrong pretty easily. Maybe you should start learning some math and logic before you speak next time.

@user-xh9pu2wj6b

Жыл бұрын

@@rykehuss3435 "You cannot say where in that line of +1's you cant fit another one. " I can say that and you just proved why you indeed can't just add new members in front of a series and expect it to have the same value. Can't you really see how you contradict yourself? It's amazing, honestly. "Everyone who thinks that 1+2+3+4... = -1/12 is treating infinity as a number" wrong. They just treat equals sign differently. In case of infinite series, even convirgent ones, it doesn't mean that you literally add up all the members of the series one by one, if you didn't know. It just means you used some method to assign a value to this entire series. "I dont care if its weak or strong" once again you show that you don't have any idea about the things you try to disprove so hard. It's really odd behavior, honestly. It's like if you heard about Riemann integration and then learned that there's extensions to it, like Lebesgue integration and started screaming about it being wrong because it allows one to integrate functions that aren't Riemann-integrable. "and neither does mathematical logic" wrong, it cares a lot about it. Though here we don't exactly discuss logic, we discuss series and their regularization. Logic comes from the way we structure proofs from axioms we pick. "You cant sum all natural numbers and say its some number S" I can tho. You just don't wanna learn how to do it and why it's useful. "I just proved you wrong pretty easily" you've proven *yourself* wrong very easily. I said that you can't just add random stuff to the beginning of a series and expect it to be regularized in the same way under all methods. You then added random stuff in the beginning and then showed that it gives you a different result, proving my claim. Then you perform some exceptional mental gymnastics and conclude that I was wrong. It's really incredible how you managed that. Technically your proof goes in the wrong direction even earlier. You assumed that a value assigned to a series directly corresponds to adding all the numbers in it using regular summation rules(or in other words S represents a whole series as a sum of its members as an arithmetic expression and can be interchanged with a value assigned to it). That's wrong even for converging series and you should know that.

My logic here is that a set containing all positive numbers can never equal a negative number. Since subtracting a number is technically just adding a negative, that set containing all positive numbers must necessarily be only growing. Even for a set containing all real postive numbers you are still only ever adding more thus it can never be less that the first number of that set.

@JavaScripting64

7 ай бұрын

Yeah just cleary wrong on it’s face

@gamelovergold

5 ай бұрын

I given my comment logically and common sense should prevail. -1/12 looks like a wizadary rather than mathematics. Sum of all positive nos upto infinity . How can it be negative. For example person 1 has 1 apple person 2 has 2 apple person 3 has 3 apple and so on and person who is at infinity has infinite apple and all there sum is -1/12 apple. How it becomes Negative apple. Proof and theorms are good but common sense must prevail unless you are willing to believe water is milk.

As an extensively experienced math nerd, frankly, I've never believed -1/12 is the actual convergence. That proof is FILLED with assumptions.

I admit, I was so ready to chastise you based on the first half, but the second half made this one of the best explanations on KZread about infinite series and that little bogus trick made famous by Numberphile & co that represents Ramanujan sum as a real function and that algorithm as a valid way of calculating sum of a series no matter their convergence or divergence.

The equations A and B are not convergent series. You cannot do these operations like shifting etc. this proof isn’t the right proof.

1:23 it's illegal, I mean it.

@TheWorldsLargestOven

7 ай бұрын

Truen't

Imagine receiving ever increasing money ( 1€ the first minute 2€ the second...) how will you be in debt