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This Result Keeps Me Up At Night 

BriTheMathGuy
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What is the sum of all the natural numbers?! Of course the answer is obvious...right? Many have actually pointed to the result that the sum of natural numbers is negative 1/12 !
We're going to do a 1+2+3+...=-1/12 proof , then a false sum of all natural numbers proof. What can we make of this?!
You might want to check out Grandi's series! • I Just Can't Believe w...
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
1+2+3+4+5+6+n=-1/12 proof
#math #brithemathguy #sumofallnaturalnumbers

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12 апр 2022

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Комментарии : 1,9 тыс.   
@BriTheMathGuy
@BriTheMathGuy 2 года назад
Head to squarespace.com/brithemathguy to save 10% off your first purchase of a website or domain using code BRITHEMATHGUY
@TerryPlays
@TerryPlays 2 года назад
Video is 20 mins ago and you posted this 8 days ago. How?
@BambinaSaldana
@BambinaSaldana 2 года назад
Hey, Bri, have you heard?
@BambinaSaldana
@BambinaSaldana 2 года назад
@@TerryPlays Either it was unlisted or it was available only to members.
@TerryPlays
@TerryPlays 2 года назад
@@BambinaSaldana ok
@crystalkalem9289
@crystalkalem9289 8 месяцев назад
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.
@erikm8373
@erikm8373 2 года назад
Average complex math theory fan vs. Average "It clearly isn't, just look at it" enjoyer
@eterty8335
@eterty8335 2 года назад
After such rough comments on why the video is misleading/incorrect/incomplete, I really needed this comment lol
@thewall4069
@thewall4069 2 года назад
Based
@rjeverett6223
@rjeverett6223 2 года назад
me, an average “it clearly isn’t, just look at it” enjoyer
@catalintimofti1117
@catalintimofti1117 2 года назад
me a complex number theory fan(complex numbers apear in physics)
@spoopyscaryskelebones3846
@spoopyscaryskelebones3846 2 года назад
@@catalintimofti1117 they’re stupid (complex my ass)
@MaxxTosh
@MaxxTosh 2 года назад
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
@MagruderSpoots 2 года назад
@@CellarDoor-rt8tt Also because of its relation to quantum mechanics.
@MasterHigure
@MasterHigure 2 года назад
@@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
@nintendoswitchfan4953 2 года назад
It is -1/12
@Who_is_Emmy
@Who_is_Emmy 2 года назад
Mathtubers? First time I heard that term
@oni8337
@oni8337 2 года назад
@@nintendoswitchfan4953 cope
@Konzon
@Konzon 2 года назад
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
@lf9177 Год назад
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
@jamescollier3 Год назад
@@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
@emilemerten6535 9 месяцев назад
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
@fatmonkey4306 8 месяцев назад
@@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
@kyahua3 6 месяцев назад
@@lf9177 but you are assuming S is infinity, we start out with no assumptions about S to find its value
@RedRing-tech
@RedRing-tech Год назад
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.
@sheldonj.plankton163
@sheldonj.plankton163 2 года назад
Subtracting Infinity is not a valid operation, so everytime you subtract S you rule out that S could be infinity
@simongross3122
@simongross3122 2 года назад
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
@FermionPhysics 2 года назад
@@simongross3122 it’s also used in Casimir effect. Physicists have a knack for using mathematically ill-defined operations…
@simongross3122
@simongross3122 2 года назад
@@FermionPhysics Like cancelling infinities. That's another one that has me scratching my head.
@FermionPhysics
@FermionPhysics 2 года назад
@@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
@simongross3122 2 года назад
@@FermionPhysics Thank you, I'd appreciate that.
@kasuha
@kasuha 2 года назад
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
@fahdal-sebaey3322 2 года назад
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
@DTLRR 8 месяцев назад
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
@kyzer422 8 месяцев назад
Thanks for this comment, this is a simple way to show some of the problems with these "proofs".
@lory3771
@lory3771 7 месяцев назад
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
@theapexsurvivor9538 7 месяцев назад
​@@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.
@cold-blooded49
@cold-blooded49 Год назад
I really enjoyed your video ❤️And now I will also drive my friends crazy in school in this way you told😉
@Infernal07
@Infernal07 2 года назад
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
@ruce9269 2 года назад
Yes, which also means you are infinitely away from the true answer: infinity.
@evanfox3136
@evanfox3136 2 года назад
That step is fine, you are just adding a zero to the beginning of the series which cannot change its value.
@9core
@9core 2 года назад
​@@ruce9269 there is no value as it doesnt converge, but the limit is infinity
@kinyutaka
@kinyutaka 2 года назад
@@ruce9269 approximately.
@rez188
@rez188 5 дней назад
​@@evanfox3136 You could also add 0 to any other position and use that to create whatever result you want
@MasterHigure
@MasterHigure 2 года назад
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
@user-xh9pu2wj6b 2 года назад
"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
@MasterHigure 2 года назад
@@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
@DinsAFK 2 года назад
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
@user-xh9pu2wj6b 2 года назад
@@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
@DinsAFK 2 года назад
@@user-xh9pu2wj6b those methods ignore divergence laws then and I really don't see any point in following them
@GlorifiedTruth
@GlorifiedTruth 2 года назад
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
@BriTheMathGuy 2 года назад
Really glad you enjoyed it! Thanks for watching and have a great day!
2 года назад
There are higher level mathematicians than UW Madison professors I believe. But the question about this series is more philosophical that mathematical.
2 года назад
I should add though that I'm a fan of public universities and in particular of UW Madison.
@GlorifiedTruth
@GlorifiedTruth 2 года назад
@🙂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 года назад
@Glorified Truth There is indeed very great mathematicians at UW Madison and has been before. Like Georgia Benkart.
@michaelcamp2870
@michaelcamp2870 Год назад
This was really interesting! I’ve never realized just how unruly infinite sums can be.
@sulfursw9286
@sulfursw9286 2 года назад
Actually, thank you very much for this clarification about axioms. I didn't realize that distributiveness could work differently with infinite series.
@leonro
@leonro Год назад
It's more about divergent series, as it still works about the same on infinite series that converge.
@kodirovsshik
@kodirovsshik 2 года назад
This is the best explanation of this phenomenon I've ever seen and probably will ever see. Great job
@BriTheMathGuy
@BriTheMathGuy 2 года назад
Glad you enjoyed it! Have a great day.
@alistairkentucky-david9344
@alistairkentucky-david9344 2 года назад
Mathologer has a fantastic video on this topic as well. Well worth watching.
@aravindswami6243
@aravindswami6243 6 месяцев назад
can you name the mathematician who had given this result??
@Wutheheooooo
@Wutheheooooo 3 месяца назад
@@aravindswami6243 Ramanujan
@dukenukem9770
@dukenukem9770 2 года назад
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.
@dannypipewrench533
@dannypipewrench533 2 года назад
Fascinating. I like this. Looking forward to learning calculus.
@peckychicken
@peckychicken 2 года назад
I love how he explains this to us like we understand right away
@leonthethird7494
@leonthethird7494 2 года назад
I mean, you should
@dqrksun
@dqrksun 2 года назад
Thanks for putting a smile on my face for showing the Riemann functional equation 😊
@beketyermek6853
@beketyermek6853 2 года назад
Brian, youre the best! Thanks for the video😍
@BriTheMathGuy
@BriTheMathGuy 2 года назад
You're the best!
@JSG-YT
@JSG-YT 6 дней назад
A key way that has helped my understanding of these ramanujan summations is that one should not allow themselves to "combine" terms into one term. As long as you do that, results tend to be fairly consistent
@nigelmokfilms7294
@nigelmokfilms7294 4 дня назад
This actually shows us we’re not even half way close to solving math.
@absence9443
@absence9443 2 года назад
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
@ChristAliveForevermore 2 года назад
Wonderful exposition of the 'why' for this result and others like it.
@kostasv6980
@kostasv6980 2 года назад
This is the best explanation on this topic that I've seen on youtube. Keep up the good work!!
@Geryboy666
@Geryboy666 Месяц назад
thanks for clearing that up.
@akiroclimbs2300
@akiroclimbs2300 8 месяцев назад
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.
@MsBombastik
@MsBombastik 2 года назад
you got my sub after this video and i was not a fan of some other video. Great job.
@BriTheMathGuy
@BriTheMathGuy 2 года назад
Welcome aboard!
@mathsexperts4901
@mathsexperts4901 2 года назад
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
@galaxyyy3427 2 года назад
3:55.
@AbelShields
@AbelShields 2 года назад
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
@rykehuss3435 2 года назад
Infinity - infinity = infinity, not undefined. Check out Hilbert's hotel paradox
@AbelShields
@AbelShields 2 года назад
@@rykehuss3435 that's just wrong, you can't just compare infinities like that, it's undefined. Infinity minus infinity can be anything.
@rykehuss3435
@rykehuss3435 2 года назад
@@AbelShields Hilbert's hotel paradox proves infinity - infinity = infinity. Go read about it.
@G8tr1522
@G8tr1522 4 месяца назад
this one only fucks me up because of its connection to the reimann hypothesis
@user-el3hr7jt4u
@user-el3hr7jt4u 9 месяцев назад
And that's why there are special rules to operating with series depending if they're converging or diverging
@dorian4387
@dorian4387 2 года назад
Ramanujan summation is a beautiful tool, anytime I see -1/12 I can't help but think about it
@monika.alt197
@monika.alt197 2 года назад
Ikr! Same
@FermionPhysics
@FermionPhysics 2 года назад
It’s nonsense
@spektator5418
@spektator5418 2 года назад
@@FermionPhysics ur mom
@createyourownfuture5410
@createyourownfuture5410 2 года назад
@@FermionPhysics shît up
@angelmendez-rivera351
@angelmendez-rivera351 2 года назад
@@FermionPhysics It isn't nonsense.
@Kurayami_mujou
@Kurayami_mujou 2 года назад
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
@adrianbozdog9702 2 года назад
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.
@upamanyu2317
@upamanyu2317 5 месяцев назад
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.
@dipayan8789
@dipayan8789 Год назад
You opened my eyes.
@ibrahimhaneef2074
@ibrahimhaneef2074 2 года назад
This is it, finally i find someone, brilliantly more educated than me, who agrees that this just doesn't seem possible. Thankyou brian
@noahvanhove6004
@noahvanhove6004 2 года назад
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 ❤️💪🏻
@RSLT
@RSLT Месяц назад
LOVED IT 😍 🥰 😍 🥰 😍
@foogod4237
@foogod4237 2 года назад
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
@scubasteve6175 2 года назад
Thank fuck someone said this
@mathsloverprashant9109
@mathsloverprashant9109 2 года назад
Huge respect to great mathematicians Sir Ramanujan
@mathsloverprashant9109
@mathsloverprashant9109 2 года назад
*mathematician
@maalikserebryakov
@maalikserebryakov Год назад
U are 100% an indian lol
@12345ngb
@12345ngb Год назад
Almost every Indian is proud of him. But this theorem that the sum of positive numbers ends up being negative is BS.
@CriticSimon
@CriticSimon 4 месяца назад
@@12345ngb Exactly! This sum cannot equal -1/12. Ramanujan was wrong hehe
@SP7_Pro
@SP7_Pro День назад
​@@CriticSimoneven though it's hard to believe, ramanujan proved it using simple math.
@seanmorrison3744
@seanmorrison3744 2 года назад
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 < p < 1, (1 - p^N)/(1 - p) = 1/(1 - p) in the limit as N -> 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 года назад
It doesn't even have to be valid. Z(-1) and the series discussed are litteraly the same and Z(-1)=-1/12.
@vascomanteigas9433
@vascomanteigas9433 Год назад
It is the direct result of the Unicity of Analytical Continuation.
@sciencefordreamers2115
@sciencefordreamers2115 Год назад
You are great!
@stevesmith9447
@stevesmith9447 4 месяца назад
An excellent illustration of the pointlessness of wondering about divergent infinite series.
@omer7895
@omer7895 2 года назад
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
@TheDannyAwesome
@TheDannyAwesome 2 года назад
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...
@yellowblueua
@yellowblueua Год назад
Keeps you up? I fell asleep while watching this already!
@saurindragoswami468
@saurindragoswami468 День назад
1:47 The only logical argument is when you're doing the cheap trick of moving B to the right by 1 digit, you're agreeing that right now Im adding the 1 to the -2, the +3 to the -2 and at the end of the string I will come back and add the last remaining digit of the lower B to the first 1 of upper B. But since the string never ends you're never doing that despite it being a requirement, it can't be ignored just because the string is infinite. You CAN NOT leave the 1 like that, because some day you'd have to add something to the 1 but you're never doing that and saying hey it keeps on forever! So B ≠ 1/4 No.
@DuangJai99
@DuangJai99 2 года назад
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
@X22GJP 2 года назад
So?
@owenaspinall2046
@owenaspinall2046 2 года назад
I gauss that would work (pun fully intended)
@dilnasheenzafar4751
@dilnasheenzafar4751 8 месяцев назад
its not a paradox its a mistake
@sxnchou
@sxnchou 8 месяцев назад
yeah the indian who “proved” it was a fraud
@sakshamraj1566
@sakshamraj1566 Месяц назад
That is what a paradox is !!!!
@louisvollert1420
@louisvollert1420 Месяц назад
@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
@chillxkill...
@chillxkill... 15 дней назад
well a paradox is not something that can be proven till now obviously but it can be applied to prove others things .
@altairyanyan1570
@altairyanyan1570 12 дней назад
It is not. This is used as a regularization method in Physics. There is a paper using this to explain zero-point energy. There are also other methods giving meaningful finite values to divergent quantities. An example is the analytic continuation of the Mellin transform or finite-part integration. It might seem weird to Mathematicians, but these kinds of methods were found to be useful in physics with EXPERIMENTAL evidence. People used to mock the notion of imaginary numbers, and then boom: complex analysis.
@sjoerdstougie
@sjoerdstougie 2 года назад
most videos take 20 minutes of lead-up and dont explain it this detailed, wow thanks
@Snoozefor10minutes
@Snoozefor10minutes 2 месяца назад
Thanks to Ramanujan for giving us this wonderful series
@csb178
@csb178 2 года назад
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
@coursmaths138 2 года назад
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
@csb178 2 года назад
@@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.” (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-w-I6XTVZXww.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 (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YuIIjLr6vUA.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). ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YuIIjLr6vUA.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
@coursmaths138 2 года назад
@@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 (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-jcKRGpMiVTw.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
@coursmaths138 2 года назад
@@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
@csb178 2 года назад
@@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. 🙂
@aryatripathi2486
@aryatripathi2486 8 месяцев назад
Help, we watched this today in furthermaths and it came in my yt front page! yt be stalking me ;-;
@vitovitovito3693
@vitovitovito3693 4 месяца назад
Amazing
@riskyabigael_0960
@riskyabigael_0960 Год назад
I love how we know that the answer it's clearly incorrect but we agree that the work did make sense.
@wx39
@wx39 Год назад
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
@findmebro Месяц назад
damn
@prim16
@prim16 2 года назад
Thanks for not making the same blunder Numberphile did, I was worried going into this
@ketsuppi
@ketsuppi Год назад
I saw a clip of this on tiktok today. Google is really listening and watching.
@Haunted.01
@Haunted.01 Год назад
Thanks to Ramanujan for all of this. May his soul rest in peace 😔.
@Occ881
@Occ881 4 месяца назад
What is the application of his discoveries?
@fireworxz
@fireworxz 4 месяца назад
@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
@CriticSimon 4 месяца назад
This sum cannot equal -1/12. Ramanujan was wrong hehe
@ProfAmeen08
@ProfAmeen08 2 года назад
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).
@lavieag6172
@lavieag6172 7 месяцев назад
wow!!! how do you even get to think about this?
@Bedfford
@Bedfford Год назад
I can't believe this topic is a thing. Many videos on YT explaining the issue.
@CriticSimon
@CriticSimon 4 месяца назад
I know. This sum cannot equal -1/12. Ramanujan was wrong hehe
@nickharland9207
@nickharland9207 2 года назад
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 года назад
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
@lorenzobarbato4558 2 года назад
@ 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 года назад
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
@ethanbottomley-mason8447 8 месяцев назад
@ 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.
8 месяцев назад
@@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.
@nucleushyena8330
@nucleushyena8330 2 года назад
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
@coursmaths138 2 года назад
This only prove that the limit of partial sums doesn't exist. Not that it has no "actual solution".
@garethreynolds557
@garethreynolds557 Год назад
@@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
@MrTrollo2 Год назад
@@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.
@vincenzofranchelli2201
@vincenzofranchelli2201 5 дней назад
Calling A 1/2 is like calling every single square root 0. You cant just average the answers
@monochr0m
@monochr0m 4 месяца назад
thanks for actually showing *how* the maths breaks down when you just assume things that don't hold
@caliburnleaf9323
@caliburnleaf9323 2 года назад
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_
@_sayan_roy_ Год назад
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.
@RSLT
@RSLT 2 года назад
Good Job! just a minor correction. Riemann himself(and many others ) analytically continued zeta function to the entire complex plane except one.
@absolutezero6190
@absolutezero6190 2 года назад
Didn’t they say that in the video?
@RSLT
@RSLT 2 года назад
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
@absolutezero6190 2 года назад
@@RSLT oh I see what you’re saying
@PhrenicDemand
@PhrenicDemand 4 месяца назад
This reminded me of how rhythm turns into pitch at a higher speed. Funny how it turned into 1/12 just like equal temperament
@HerzhaTheShapeshifter
@HerzhaTheShapeshifter 2 года назад
Mind blown
@janseta5162
@janseta5162 2 года назад
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
@angelmendez-rivera351 2 года назад
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 года назад
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
@angelmendez-rivera351 2 года назад
@ The harmonic series is regularized to γ, so that defeats your point.
2 года назад
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
@angelmendez-rivera351 2 года назад
@ 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.
@scottk3292
@scottk3292 2 года назад
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
@kanavbansal3760 Год назад
Yep, that was also my question while watching the video
@ahmethakanuysal9324
@ahmethakanuysal9324 4 месяца назад
I think the same This is the main problem
@frysauce8653
@frysauce8653 2 года назад
This makes sense to me, that doesnt happen very often
@Perhaps-.
@Perhaps-. 4 месяца назад
I want to cry.
@yottawatt
@yottawatt 2 года назад
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
@X22GJP 2 года назад
infinity + 1 is a different "infinity" to the that which you added 1 to. There are an infinite number of different infinities.
@THEREALOGDUCK
@THEREALOGDUCK 2 года назад
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
@testmicrowavedinner4438 2 года назад
Its a summation of an infinite series. In calculus, it is undefined
@ajshowdown8033
@ajshowdown8033 8 месяцев назад
To me its like an excuse to not being able to find out whether its gonna land on 1 or 0
@archananethi9233
@archananethi9233 7 месяцев назад
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…
@destroctiveblade843
@destroctiveblade843 8 месяцев назад
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.
@zyxziangamerx8065
@zyxziangamerx8065 10 месяцев назад
Hats off to Ramanujan Sir
@michaelbujaki2462
@michaelbujaki2462 2 года назад
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
@carvalhorosolen Год назад
What you mean? that the sum is indeed -1/12 or that it is not? 🤔
@MrWin-pj4sg
@MrWin-pj4sg 2 года назад
I think a key issue is at 1:30. I don't think that is something you can just do like that.
@manthansagar200
@manthansagar200 Год назад
You can
@adityagoyal7110
@adityagoyal7110 6 месяцев назад
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
@rubixpuzzlechamp
@rubixpuzzlechamp Год назад
Partial sum formula: y = x(x+1)/2, area under this curve from -1 to 0 is -1/12
5 месяцев назад
It would be nice if you mentioned the relationship between Ramanujan's result and the Casimir effect in physics.
@bearbrown2597
@bearbrown2597 2 года назад
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
@rykehuss3435 2 года назад
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
@biblebot3947 Год назад
@@rykehuss3435 it approaches -infinity
@tayfunsaral5024
@tayfunsaral5024 Год назад
1 - A doesnt equal A
@KwassaKwassa_
@KwassaKwassa_ 22 дня назад
Doesn't equal a what
@drewm3807
@drewm3807 2 года назад
For analytic continuation, the analogy I like to use is a tree that is 10 feet tall and 10 years old. A year ago, it was 9 feet tall. Next year, it's 11 feet tall, etc. How tall was it 20 years ago? -10 feet tall. Is negative height a real property now? Can trees actually have negative height? No, it's an abstraction.
@ranjittyagi9354
@ranjittyagi9354 6 месяцев назад
6:36 is where it forces divergence and gives a wrong answer otherwise.
@angelmendez-rivera351
@angelmendez-rivera351 2 года назад
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
@BriTheMathGuy 2 года назад
Thanks for the great insights Angel! Hope you're doing well.
@kazedcat
@kazedcat 2 года назад
I did not read your blog post so it seems you forgot the issue of being brief.
@user-kw3vg4jn8u
@user-kw3vg4jn8u 2 года назад
k
@user-up7nb6id1f
@user-up7nb6id1f 2 года назад
@@kazedcat never said that learn to read bozo
@D-a-MiR
@D-a-MiR 2 года назад
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
@coursmaths138 2 года назад
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...
@tryingtomakeanamebelike7245
I don’t think you get to pick and choose from which rows you subtract from. That’s like me subtracting 10 from one, but shuffling it over to get it to equal 0
@astreanightfang2622
@astreanightfang2622 2 года назад
i feel this is as math shenanagins as getting 1 to = 2, but just dealing with infinites, of a type. which sucks because we do have to deal with infinity, in a weird way, at least due to pi
@fuzzybanana0123
@fuzzybanana0123 2 года назад
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
@user-lh5hl4sv8z 2 года назад
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
@fuzzybanana0123 2 года назад
@@user-lh5hl4sv8z I literally explained it in my original comment.
@mertaliyigit3288
@mertaliyigit3288 2 года назад
Its because series doesnt converge. This is no different than finding 0=1 by dividing by zero
@georgecantu856
@georgecantu856 2 года назад
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
@kazedcat 2 года назад
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.
@deyomash
@deyomash 2 года назад
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
@simongross3122 Год назад
That's an odd result :)
@m.h.6470
@m.h.6470 2 года назад
You are only allowed to calculate with infinite series, if the series converges, if it diverges (meaning it goes to plus or minus infinity), it is not allowed in normal calculus (because it will break it)
@LarryPeteet
@LarryPeteet Год назад
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.
@yees_7568
@yees_7568 2 года назад
1:33 B + B is 2B me: 2B… or not 2B?
@pratyush_kumar_sahoo
@pratyush_kumar_sahoo 6 месяцев назад
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
@tolaras1571
@tolaras1571 2 года назад
Imagine receiving ever increasing money ( 1€ the first minute 2€ the second...) how will you be in debt
@stagelights_
@stagelights_ Год назад
this feels like those "proofs" that say 1=2 that do something like divide by 0 but in a way that isn't initially obvious.
@AlexeiFilippov
@AlexeiFilippov 2 года назад
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
@mbrusyda9437 2 года назад
And where'd you get that S is infinity
@rykehuss3435
@rykehuss3435 2 года назад
@@mbrusyda9437 1+2+3... = infinity. From there.
@rykehuss3435
@rykehuss3435 2 года назад
Hilbert's hotel paradox proves that infinity - infinity = infinity
@mbrusyda9437
@mbrusyda9437 2 года назад
@@rykehuss3435 that explains nothing...
@rykehuss3435
@rykehuss3435 2 года назад
@@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
@IS-py3dk
@IS-py3dk 2 года назад
IDK why these types of problems are so mysterious that i am always awake till 3 am at night thinking about these 😆
@websparrow
@websparrow Год назад
S by itself adds to infinity…but you are relating it to other factors introducing different properties.
@real-timelabel-freeimaging4653
@real-timelabel-freeimaging4653 2 года назад
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.
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