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Beautiful video. if we are able to regularise the most basic summation sum(n) without throwing away infinity, like this video, I guess very soon we would be able to regularise every other infinite sum with some combination of above weighting functions and some other known techniques. May be we misunderstood infinity till now. can someone do it for 1+1+1+1+1+..... ? may be we get a finite sum for this too. another thing that struck me is that all of this is not for a continuum set like R but for N , and Quantum Theory is about denying the continuum at various places. May be there is a subtle but a fundamental reason why such results find their shadows in bits and pieces in Quantum Theory. Something like you always find a circle somewhere whenever there is pi in some formula.
For some reason, the constant refocusing on this video made me slightly motion sick. PLEASE either stop down the aperture or add a separate overhead PaperCam!
but then you can choose a regulating function at results in the -1/12 being canceled with with some C +1/12. This shows that the series is divergent as you can get any result you like. I would also point out that all the infinitely small numbers are known as infinitesimals and are the basis of how we perform calculus in the first place, so you cannot simple ignore an infinite amount of infinitesimals. This goes along the same lines of you cannot zap an infinite amount of zeroes in an infinite series without changing the value.
Infinity plus 1 is NOT Infinity. Because if it was, then 1 has to be treated as 0 in an infinity equation. There is something called an infinitesimal unit. That unit can represent 1. And that unit is as far from 1 as infinity is from the integer 1. It means that 1 could even represent 1 google, if we count in Google units. -1/12 is only believed by those that are too lazy to use common sense. With all respect to Ramanujan.
One thing i love about numberphile is how passionate these very intelligent people are about such an awesome topic. Very nice change from the constant bombardment of low level nonsense. Thank you numberphile
Sqrt(-1) isn't intuitive either. But it's now well understood, and it wasn't well understood for hundreds of years, and yet it was used since Bernouilli. We are with inifinite series at the same place Bernouilli was when he was using sqrt(-1) without really knowing if that was legitimate.
@@InXLsisDeo No we're not. Infinite series are well understood and the imbecil in the video needs to give it up. The of whole numbers diverges. Always has, always will.
@@InXLsisDeo But there is no solution in the reals. No real times itself yields a negative product. Complex numbers, problem solved. This question can not be solved that way, because it obviously blows up to infinity. So what does Tony say that infinity + 1 equals? Less than infinity?
I was more or less mathematically illiterate and despised everything mathematics, and then I watched this video back in the day and it really intrigued me. Now, a few years later I am a grad student in pure mathematics, and it all started with watching these videos, particularly the one about -1/2! You can say what you want about the rigor of these computations, but for me, this is what started my love for mathematics!
That is great, thanks for sharing it. The -1/12 videos also rekindled my interest in maths, as did the superb video of hackerdashery on the p vs. np problem
That's awesome! I still do that hacky "proof" in my junior-level high school math classes sometimes just to rope students in in pursuit of exactly that response.
If you plot the graph f(N) = N(N+1)/2, it is a parabola that intersect X axis at points 0 and -1. The area bounded by parabola between 0 and -1 is exactly -1/12
@@L9MN4sTCUk ... An integral is not a partial sum but it is the limit of a partial sum as n goes to infinity. int[a,b] f(x) dx = lim [n to inf] sum [i=1 to n] f((b-a)/n*i)*(b-a)/n). It's the Riemann sum of the area of rectangles f(x)*Δx as Δx goes to zero (and hence n, the number of rectangles, goes to infinity)
I figured that was just a coincidence. But I looked at the sum of cubes, fifth powers, 7th powers and 9th powers, and it always worked. Note, the 5th and 9th powers have extra roots outside [-1, 0], but I still only integrated from -1 to 0 to get the Riemann zeta result. The graphs are really nice, I wish I could post them (with a zoomed in vertical axis). But search Faulhaber polynomials if you care to check yourself.
I remember feeling unsatisfied with the way -1/12 was justified with the zeta function. That's just one arbitrary function, who's to say if you used a different function you wouldn't get a different finite value, you know? This is much more convincing!
Whatever the method of regularization, if it allows shifting, adding, etc., its value must be -1/12. That is what the original video showed. There are many such regularization methods, some of which have physical realizations. But even if they don't have a physical realization, the math still exists.
@@vezokpiraka O(1/N) is something that converges to zero as N goes to infinity "by definition". The expression CN^2 - (1/12) + O(1/N) is telling us than *any* analytic continuation that gives us "1+2+3+4+..." at a non-singular point *must* have value -1/12 at that point.
I was an undergraduate when the infamous -1/12 video came out and now I’m close to finish a PhD in arithmetic geometry. This made me feel so nostalgic.
@@RingxWorld I asked my calc professor about that video at the time. Turns out his PhD advisor is the same as Terry Tao’s, and he gave a talk on the dept colloquium about the summability methods and how to make sense of this (which is half of the topic of this video). This was such a flashback 😂
@@bjshnog Yep. This whole thing could have been avoided if they had not been weaselly with terms. They're using "equals" in a novel sense, when they should have said that "this particular function is interesting. The value of this function of this series is -1/12", to which nobody would have objected, but instead they change the game whenever it's convenient to them, such as moving entities, spacing out entities, etc., which would not matter if this were basic algebra, and most importantly, pretending that this function is "equals". Go ahead and add as many positive integers as you want, the amount never decreases. So clearly this is not about a series "equalling" as sum. Maddening.
@@bjshnog I don’t think you understand the argument he was making. With certain choices of weighting functions, you get some value C*N^2 - 1/12 where C is determined by the chosen weighting function. If you choose a specific weighting function which gives C equal 0 (like the cos function described in the video), that function converges on -1/12 as N grows very large. However, you seemed to have missed his other point, which is as N grows very large, the sum becomes the standard 1+2+3… where each term is equally weighted. This is because the weighting function is defined as a function that begins at 1, and ends at 0. This means the weighting function equals 1 at w(0). Each terms weight in the series corresponds to w(n_i/N). This means that as N goes to infinity, each term tends to w(0) which equals 1. So as N goes to infinity, the sum equals -1/12, and each term becomes weighted as 1, giving the result: 1+2+3…=-1/12 There are no tangentially related concepts, he very clearly showed the same result just using a different method.
For those wondering why e^(-n/N)*cos(n/N) is so elegant, it's the fact that in C*N^2, C is the Mellin transform of the regulator function, which basically amounts to integrating x*e^(-x)*cos(x) from 0 to infinity, and it ends up being 0.
I'm not sure I see that beauty you speak about. This regulating function is a periodic function (which binds any value between two maxima) times a negative exponential, which makes the two maxima converge. It's not surprising at all it results in a finite value. It amounts to doing +1-1+(0,5)+(-0,5)+(0,2)+(-0,2).... Which eventually converges to 0. These kinds of functions are used everywhere in NMR spectroscopy to regulate noise in the output signals.
Lot more gray hairs on Tony. Cannot believe we have been growing up with this man for more than a decade. On a side note, any more of them big numbers?
my exact thought when I saw the thumbnail. Been heavy watching numberphile 10-12 years ago just when they were starting out, but then abruptly stopped following besides casual video here and there, and was kind of shocked and sad to realise so much changed
Professor Padilla is a brilliant physicist and mathematician and incredibly skilled at explaining his thinking to us, even when ,like here, it gets into the realm of wonder. Thank you, Brady, for bringing him to us.
I feel like the weighing functions should be strictly positive for this example. In the cos example, we are not summing everything anymore. Especially when N->Inf, you don't even know which numbers are summed which are subtracted (I like to think they are added and subtracted at the same time). I also feel like for weighing functions to make sense, they should be strictly decreasing from 1 to 0. In other words, 1>=w(n)>=0 and dw(n)/dn=0 .
@@volkeru2718 While you are correct, there is a caveat. The sum is infinite and cos(n/N) is not. They calculate the result of that infinite sum and only then set N to infinity. If they start with sum of N numbers and then looked at the limiting case N->Inf, I think your assertion would be valid. In other words, the initial sum has a period of T, which repeats infinitely within the sum already. Increasing its period to infinity voids the initial solution because It (probably) requires infinite repeat of that period (I assume this is correct. Else, we could just repeat the entire thing with sum up to N, then do N goes to infinity.). I am however wrong in my other assertions. We are not increasing frequency but decreasing it. However, the assumption that cos cycles infinite times within the sum remains even when N->Inf.
Thanks! I indeed overlooked the infinite sum. It is a pity that they did not even mention that the weighing function gets negative and why they are allowed to do so... Or maybe, as before, they do not ask these questions... They simply do it, as before when treating the sum of a non converging row as if it was just a normal number 😬
I am still reflecting about this one. And I Excel-led it... Guess I am much more a financial analyst / controller than a mathematician by now. 😅 However currently I tend to say that if they first proved that the result is proportional to C*N^2-1/12+epsilon independent of the weighing function then they are thereafter free to choose any weighing function, even a negative one. And indeed it is amazing that the results using the function shown in the video converge to -1/12 though nothing in the function itself hints in this direction. However interpreting the result as the sum of all integers would clearly be wrong. Allmost all weighing functions will result in a C ≠ 0 and therefore diverge to infinity which is the intuitive result.
@@volkeru2718 Lol my instinct after watching this video was to do the exact same thing--create an excel spreadsheet that approximates taking each component to infinity. Must be the engineer in me. 😆 Unfortunately, Excel limits any sequence to being about 10^6 items long, but that still gives you some leeway to play around. What's really interesting is if you try to apply this to other, established, convergent series. The ones I've tried it with so far still worked perfectly fine.
I remember getting upset at the original -1/12 video. I enjoyed this one much more and it was the first I'd heard of these regulators as a way of studying series! Also, 10 years since the original? I feel very old.
Ten years ago I was very unhappy with my career and watched the original video. I was so intrigued and inspired, I quit my job and went back to school for a math degree to better understand this result. I've never regretted this decision and seeing this video feels like everything has come full circle.
@@sbnwncI did a double major in math and comp. sci. after I retired from the USAF. Then, 17 years later, I completed a master's degree in math. Lots of computers, programming etc in all my careers I have done. Your comment was pretty close
Your videos were one of the things that really got me into mathematics, and now I'm in graduate school for my physics PhD. Looking in the comments, I'm not alone in this either! Thanks for ten years of accessible math education -- even if some of it isnt entirely rigorous :p This sum regulation method is fascinating... I wouldnt buy putting an equals sign anywhere, but it's still extremely compelling that -1/12 is just as important here as it is via analytic continuation. Edit: That connection to physics -- wow! It makes me think that there must be some universally "preferred" method of renormalizing sums, some way that's "natural" in both mathematics and physics for reasons that arent entirely clear. I'm very curious why certain regularizations produce that behavior but not others...
You can actually set this up in Excel by using a "SEQUENCE" function to create the series and then wrapping a simple "SUM" function around it. If you set n to be sufficiently large compared to N, it does converge on -0.08333. So, the infinite sum if you're using that weighting function does actually equal -1/12. No weird math tricks or anything. Just changing the way you weight it.
Me too, that's one of the reasons I like Brady's videos so much. Once he groks it, he throws curveballs at it. Sometimes people miss things that he sees pretty quickly.
@@YEC999 Almost the entirety of maths is developed this way, bro. You reverse engineer an answer *and then* check to see if it breaks stuff. Then you slowly prove that the answer couldn't be any other. Quite the opposite of a fraud. It's the most sure-fire way to prove things.
@@Voshchronos If you think to multiply whole number with a cos function is the most "sure-fire way" to prove that the sum of whole numbers=-1/12. than i would never believe anything scientific you say.Literally if i knew your name i would not believe anything you say. You should really play with a cos function before you make claims like this... Another thing that you don't understand: To manipulate to reverse engineer a weighing fucntion that it outputs what you want to have is a brutal fraud. It is like the definition of fraud: You use an manipulative function to get the result you wanted. The equivalent in economics is balance sheet manipulation like Enron or Bernie Madoff where people went 30 years to jail for.That is all that is: a manipulative weighing function to get to a sum that you wanted to have. in Case of Enron: lower debt than it would be. In case of this: lower sum to get to -1/12 And no nothing was invented in maths with manipulated weighing functions. No Complex numbers (they came through a very tanngible problem) no negative numbers no reals no nothing..
@@YEC999 I mean, I am not going to claim to understand the technicalities but it seemed very clear to me what he was saying. It wasn't that no other equation worked, it's just the particular one used showed it more clearly. All these "smoothing the curve" equations have a -1/12th in them, and the -1/12th becomes more prominent as they all get infinitesimally close to the sudden drop from 1 to 0.
10 years ago, because of your videos, I got really into this -1/12 thing. I kept researching it for quite a while. But this new video is just amazing; what a fascinating result!
I always hated mathematics, my brain just couldn't handle how numbers work, but I can appreciate people like Dr. Padilla who have devoted their lives to trying to explain things that were previously unexplainable. I admire his humility in saying "I don't want to understand infinity, I want to understand the journey of why we would want to get there in the first place". It may be just numbers, but it's an incredibly profound way of thinking and applying them to help explain how the very fabric of our universe works.
Watching your videos for many years I sometimes regret I left academia. So much to still be discovered. Thanks for making these intriguing videos that even I can keep track with.
I watched the first -1/12 video when I was in 7th grade, and I am now a junior in university. I remember in elementary math was my favorite subject and when I reached sixth grade I had a horrible math teacher which made me dislike math. Thanks to you guys for making that video as it absolutely rekindled my love for math when I was in seventh grade.
I'm completely math illiterate, I understand the individual concepts, but I can't put them together even when explained. I still felt the profoundness. It's like watching a foreign movie. I don't know what's going on, but the effects and action is awesome 😊
Terrence Tao’s formulation is so elegant and beautiful. The first video that was published 10 years ago had a lot of debate around it. This video probably won’t because it’s so convincing. One of the best math videos I have seen
Thanks! I love your channel, and this particular video is one of my all-time favorites. It explained a complex topic in a clear and compelling way. Amazing.
But then ... integers _do_ make for a rather sharp cutoff. As far as making it -1/12, I would agree with Tom Petty that the weighting is the hardest part.
Great video! I would love for it to become public and viewed by more people as this really helps explain where the "-1/12"th-ness of the sum is. To use Ed Frenkel's terminology, this is how to extract the gold nugget from the infinite dirt!
The best way my math professor described infinity is that it's not a number, but more like a direction. It's like traveling east, you never reach the end because you can always go more east. So, treating infinity like a number doesn't make sense since there's always more east.
Exactly. I think what we're seeing is that there might be different ways of travelling east (one step at a time or two steps at a time for a simple example), and there might be ways of quantifying those different destinations other than "they blow up to east".
I appreciate Tony’s passion for the subject! It’s always awesome to see someone apply their passion and continue to push it to new edges and in new ways!
Love your content, presented in a very accessible way. Coming from St. Helens, hearing someone who comes from the Northwest providing such great insights into mathematics feels comforting and even more relatable (and hopefully inspiring for kids growing up in the region). Thanks!
I personally think this is genuinely a momentous find. The fact that infinities can be numerically represented and define behavior of systems without having to be aggressively gotten rid of can possibly reconstitute many of the existing conjectures and hypotheses that are stuck due to infinities... not all, but even if one, that'd be huge.
Regularisation has been used for about a century... and there are a number of experimentally verified results that 'use' it. The novel intuition is that there may be a 'best' or 'preferred' way to regularise certain types of expressions.
He selected a weighting function that, for any value of N, goes NEGATIVE for half of the values of n. That's not something "between 1 and zero" as he claims and totally denaturalizes the addition of NATRUAL numbers which are NEVER negative.
I loved the "smells of String Theory" line around 13:30, so much of cutting edge mathematics is about feeling out into those intuitive spaces where we don't have models for, in order to get a clearer picture of what we're trying to model.
To be fair, String Theory is been shown to be pretty much a hoax as well. Michio Kaku is basically a Charlatan. Dr. Angela Collier (@acollierastro) has an excellent video about it.
I am absolutely entranced with this linking of such a bizarre result from number theory and the physical workings of our universe. Tony Padilla is such a treasure.
It makes me wonder if it is that our descriptions are wrong since infinities pop up that we have to throw out or is that since we can, in a sense, wrangle the infinity to finite using regularization, that nature at the quantum level is infinite but has a way of regulating itself back to finite.
@@hmx10011Well you are right in at least one way: in quantum physics there are infinities that are reduced back to finite. The values for any property of a particle is a superposition of all the values this property could have, and the possibilities are indeed infinite, yet they all add up in a way to a single usable physical value. So is this process more generalized that we are expecting or is it inherently only tied to possibilities ? Hard to tell
Processes in nature are computationally bounded in the same way that your computer does not have enough time in the universe to calculate all the digits of pi
Indeed, the optimist in me is already looking forward to a numberphile mention at a future nobel ceremony, this seems like it could be a potientialy fundamental insight once properly understood, was left very excited after watching this.
It really does. Full circle on a topic that kicked off controversy, excitement, and many peoples' math careers. But this time, it comes back with new insights. Amazing 😍🤩
You can express a partial sum of an infinite series in terms of the sum of the infinite series.... WHAT? I had to rewatch 3:30 to 5:30 again and then pause on this to ruminate how amazing what is being expressed here. Excellent videos. This feels like you all are mass producing the discovery of one person down for the rest of us to understand. Just as important, to me, than the discovery itself. Thank you.
Oh my how the time has passed. 10 years ago I was in high school, watching these videos with astonishment and a genuine feel of adventure and love for all these awesome discoveries. A decade and two degrees later, and nothing has changed, except maybe for greying Professor Padilla, but still as enthusiastic as ever. Thank you, sincerely.
I love that his takeaway was "it got people interested in mathematics, so I'm glad we did it." It's something that resonates with me so much. Sometimes in education and outreach, it's not about teaching the right answer, but getting people interested.
Astounding result. It really does seem like something quite fundamental has been uncovered, and it comes out of such a simple question that seemed unanswerable for the longest time. I wonder if this is what it felt like when people started taking imaginary numbers seriously and realised they gave real, verifiable answers to questions. It seems super weird, but if the maths gives you useful answers then you just have to follow it.
that is the result of Terrence Tao I believe. First time I kinda understand something he wrote, and you can really see a glimpse of his genius with this.
@@herbie_the_hillbillie_goat"the *sum* diverges". You didn't heard (or understood) the whole point of the video. You wouldn't have said "it was solved since". Not it wasn't "solved". And you need to understand clearly why by reflecting some more time on the topic.
What a delightful video! Finally an honest way to do renormalization, and great to see more of the amazing work that Terry's been doing since we were grad school room mates :)
Using the cos function to "smooth" the cutoff point seems a much more physical way of doing this. Any RF engineer will tell you there's no such thing as a pure step function; that rising edge is always a collection of ever increasing frequencies so that it just looks like a step.
Modular arithmetic (which is one easy way to describe integer underflow/overflow) is distinctly different from zeta regularization (which is what they are doing here). One does not directly inform understanding of the other. You can think of zeta regularization like squishing a function so that the annoying parts don't cause you problems and then taking the answer for the squished function instead of the original function. Like, what is the "value" of sin^2(x) at infinity? That depends entirely on where infinity "stops". (Which doesn't make sense, right?) Since the sin wave never converges, it doesn't have a value in the limit as x approaches infinity. Regularization allows us to assign a knowingly false, but still useful number to the nonsense equality. sin^2(x) does NOT equal 0.5, but there is a way of thinking about the function where you can wave your hand and say "for all intents and purposes, it's 1/2". Very useful for physicists that don't know why a problematic infinity can be ignored but are happy to ignore them, but extremely unhelpful to a mathematician trying to better describe mathematical systems. "I don't know why, but it's useful even if it's not necessarily true" seems to be the hallmark of modern maths and I am not a fan. (I'm looking at you Axiom of Choice...)
My problem is the statement "I am removing the regulator because I'm taking N to infinity". The weights are still there even as N goes to infinity. With W(n,N) = exp(-n/N)cos(n/N), even though for a fix n, W(n,N)--->1 as N--> infty, those weights are still N. For example W(2N,N) ~ -0.056, no matter how large N is. That means, even taking N to infinity ("removing the regulator"), the weighted sum always includes the term -0.056*2*N. It seems to me that these weights being considered are too general. The weighted sums they yield don't have much to do with the original sum 1+2+...+n+...
The fact that the form CN^2 - 1/12 + o(1) is independent of weight, still says there's something very special about -1/12. Wheter you think of it as "the" sum or 'the finite residue' is splitting hairs for me at that point.
@@johngalmann9579 Seeing this form makes me wonder if the result comes from an assumption that a 'C' that evaluates to zero can 'overpower' an N^2 that goes to infinity. Similar to how 0^0 has different limits when approached from different directions, 0 x infinity has different values depending how the 0 and infinity are arrived at. If I take a limit of a sum of 0x1 + 0x2 + 0x3 etc it is easy to say that all terms are zero, then extend to N->infinity, but if I take the sum of N first and then multiply, will I get a different answer?
It does change them. At 3:40 he tells you quite open. That he did everything he could to get to the -1/12 he is clearly not a scientist. He wanted that result.
@@YEC999The -1/12 is always there, as -1/12+CN^2. If you pick your weighting such that C=0, you'll get -1/12, but even if you don't, the -1/12 is still there. It's always the constant term of the asymptotic expansion.
Wow, thanks for an excellent revisit to -1/12. This was just beautiful! Pure partial and infinite sums, to end up where we all know we would end up... love it!❤
This how math moves forwards - craziness for a long time and denial and then a little piece of wisdom from a genius like Terry, Euler or Ramanujan and we sheeple get a little further.
This could be told like a story by H.P.Lovecraft. The protagonist thinks the -1/12 is not a real answer, it is just an analytic continuation, a trick. But it haunts him, he sees it in his dreams more and more often, then he gets visions during the day. He can no longer ignore it, he has to scratch the itch, searching deeper. And he finds it. The answer. -1/12 was real all along, he can no longer protect his mind by pretending that it is just a trick. And he goes insane. Oh good old infinity. We cannot understand you. But how sweet can ignorance be if you face the monster, the old god of chaos. Twisting, shouting, breaking what we meager humas tought to be the well ordered little world we inhabit.
Agree with the others, this is one of my favorite Numberphile viedos in a long while (and a good redemption of Dr Padilla :))) ). It's fun and interesting and it communicates a lot of high-level ideas without being too bogged down by details. It reminds me of being in undergrad and having much smarter people than me explain deep ideas over lunch, or during "office hours".
The sum of strictly positive integers from 1 to N, where N goes to infinite, is of course infinite. 1+2+3=6. Adding more strictly positive integers to this sum will clearly increase the sum. QED.
@@karljohanhaarberg6706This very case is a counterexample. 1+2+3...=-1/12 is infinitely many positive integers adding to a negative value. Another example would be 1+1+1...=-1/2
My layman understanding of the need for a weighting function is that you only see the sharp cutoff at N if you stop the partial sum at N, but you're interested in what happens when N goes to infinity so you will never see a cutoff: it's infinitely "far away".
Worse than that. He selected a weighting function that, for any value of N, goes NEGATIVE for half of the values of n. That's not something "between 1 and zero" as he claims and totally denaturalizes the addition of NATRUAL numbers which are NEVER negative.
@@dexterPL ... I copied it only 2 or 3 times as responses to other comments where it was relevant. In some of these places it triggered some constructive discussion. I also wrote a long comment myself (not in response to another comment) explaining my position more formally and in more detail. You don't need to like what I do and I don't need to care whether you like what I do or not. I will not make a yt video.
@@adb012 I don't know why you're so concerned about negative numbers. Using just e^(-n/N) as the "weighting function" produces a series where _not one single term_ of the series is a natural number. So I find it a bit odd that you are outraged that he uses a regulator which allows for negative terms but aren't equally outraged by the fact that other regulators produce series which have no natural numbers at all. But maybe this point has already been made in another discussion. I don't know where that discussion is, though.
I honestly think, that this was the way, that Ramanujan approached these problems. That was his talent, to be open minded to techniques. I'm not surprised, that Terence Tao was the one to decode parts of this secret...
@@idjles May well be, for sure. Sadly he died too young. But the questions, he asked, about the nature of mathematics, are similar to this stuff. But he was in mathematical spheres, that were never documented, but they gave him some answers, that we, over a hundred productive years later, find hard to understand!
@@idjles actually, if you check the paper, you'll see that Ramanujan precisely used this kind of trick a lot. It is indeed pretty much what he did. The insane thing is, that he did so without having to be told that this is even a thing by a Fields Medal winning mathematician. He just kinda did it all auto didact style with barely any support until he got discovered.
10:17 this weighting function does not go between 1 and 0 as stated, as cos(n/N) will have negative values for π/2 +2kπ < n/N < 3π/2 + 2kπ. (integer k) the ne^(-n/N) factor is always positive, so it oscillates.
I think it is exactly the purpose of this setup, it is a weight function that allows a smooth landing in infinity without having to throw away a CN². The interesting part is that the landing is exactly on -1/12. The reason why breaks my brain but I find it fascinating!
@@j.r.8176 Unless I got it wrong, I believe that n is not smaller than N., therefore it will not be equal to 1. N is an arbitrary cutting point in the sum and n will go past that number. Cos will then oscilate and the sum will be -1/12
Without a doubt the most important topic ever covered on Numberphile. People probably laughed the first time someone said “let’s just call SQRT(-1) i and see what happens”. Now, instead of just ignoring the fact that -1/12 keeps popping up, a few courageous mathematicians are seeing where the clues lead. Kudos to Dr. Padilla and others for boldly going where no mathematicians have gone before.
Adding the cosine function into the summation completely changes it, you now have a summation of positive and negative numbers. The original and your proposed are as different as summing 1+2+3 ...N and summing 1 - 2 + 3 - 4 + 5 - 6 ....N, in each case with N going to infinity. The first goes to N(N+1)/2 as you said. The second toggles, going to -N/2 for N even, and (N+1)/2 for N odd.
_"The original and your proposed are as different as summing 1+2+3 ...N and summing 1 - 2 + 3 - 4 + 5 - 6 ....N"_ Not quite. The method using the weighting function has some desirable properties. Padilla has stated these in his paper on this subject (which is more scientific and less obscurantistic than this video). He cites G. H. Hardy: _"Hardy has argued that any new summation method should satisfy three properties: regularity, linearity and stability._ _Regularity states that a summation method yields the known results for convergent series obtained using partial sums. [...]"_ Your above method would violate that _regularity_ requirement, other than the weighting sum method of the video. Hardy and Ramanujan were at that time discussing suitable methods to define a "sum-like" functional for sequences which are divergent in the usual sense.
I'd hazard that arriving at -1/12 tells us more about these weighting functions than it does about the concept of infinity. So in a way we're still "sweeping infinity under the rug."
I thought that as well, but I think the idea is, that as N goes to infinity, these weighting functions all go to w(n/N)=1, so you're back at the original infinite sum. The only thing that wasn't really clear to me, is if there are weighting functions that go to a different finite constant. He says he reverse-engineered the cos(n/N) one, so is it possible to reverse-engineer a w that makes the sum go to eg. -1/42?
@@ComaVN the thing is, you're not "back at the original sum". fundamentally, infinite sums are not sums at all, they are limits of sequences. the original sum is a limit of N(N+1)/2. and that's not because we just chose to "cut it off", it's because that's the way the idea is formally defined. and that same definition is used even in these other regularization methods. when you introduce the weights, you're taking some limit of infinite sums (which are themselves limits of sequences). you're changing the sequence under the limit, so you're changing the result. and yes, by the way, you could get any result you want. any of the methods that will give you -1/12 are basically just asking "what if we treated these sums as some other object". because the idea of "adding to infinity" is meaningless without a formal definition, we're allowed to explore alternatives like this. you just have to be careful not to conflate the different definitions.
That's a wonderful way to look at series!! For anyone curious about the formula at 7:54, this can be gotten nicely using generating functions! For simplicity of notation, let A = e^(-1/N). Let G(x) be sum from n=0 to infinity of A^nx^n. Then that inner term (the summand) can be written as (Ax)^n, so G(x) is a geometric series equal to 1/(1-Ax). Taking the derivative of the series G(x) gives a sum from n=1 to infinity of nA(Ax)^(n-1). So xG'(x) has summand xnA(Ax)^(n-1) = n(Ax)^n and setting x = 1 gives our desired sum. So taking the derivative of 1/(1-Ax) gives A/(1-Ax)^2, and setting x = 1 gives A/(1-A)^2, which they rearranged as A/(A-1)^2 because the square cancels out the negative anyway. That second one with cosine...I'd have to think on!
Actually, it looked so similar to the last problem and it made me think of that nice trick where you use Euler's identity e^(ix) = cos(x) + isin(x) to get info about a sum multiplied by cosine and sine at the same time. Then taking the real part gives the cosine sum. Let's try! In this case, we let B = e^((i-1)/N), so that Sum(B^n) = Sum(e^(-n/N)cos(n/N)) + iSum(e^(-n/N)sin(n/N)). Then writing G(x) = Sum((Bx)^n) gives two infinite sums C(x) = Sum(e^(-n/N)cos(n/N)) and S(x) = Sum(e^(-n/N)sin(n/N)), so that G(x) = C(x) + iS(x). So our desired sum is again xG'(x) at x = 1, which is equal to B/(1-B)^2. Writing bar(z) for the complex conjugate of a complex number z, the real part of z is (z + bar(z))/2. Since bar(B/(1-B)^2) = bar(B)/(1-bar(B))^2, we can use this to calculate the real part C(1). And we immediately get S(1) as well!
Great video!! It truly felt like this "coincidence" of weighting functions, unique pathways to the minefield of the infinite and particle physics is telling us something. Like a jab of the prospector pick, hitting straight into the gilded veins flowing from the inner most reality and up into our realm.
Yep. And this is all undoubtedly connected to the p-adics in some way. We just don’t know how. This is the profound mathematical and scientific mystery of our time.
@@4984christian It is easy to represent and manipulate these sums in -adic notation. For example, the powers of tens sum mentioned in the video is just ….1111 in the 10-adics. Now I know 10 is not prime but that is the general idea. These infinite sums in the -adics act a lot like repeating decimals do in the ordinary reals (technically the reals are a subset of each n-adic number system.)
@@LookToWindward So you could represent the infinite series over the integers by a lot of infinite "power-series" over some p? Like the powerseries over p = 10 would yield 11111111...?
@@4984christian yes but the dots are on the left. Adic numbers can extend infinitely to the left just like the reals can extend infinitely to the right.
Great video! This video gives another beautiful argument why -1/12 is closely related to the series 1 + 2 + 3 + ... I think the problem with the first video is that it claimed that 1 + 2 + 3 + ... = -1/12 using an invalid proof. The value of an infinite series is defined in litterature as the limit of partial sums. This video basically changes the series to make it converge by adding weights and then compute it using partial sums. Mathematics is based on definitions so you cannot say 1 + 2 + 3 + ... = -1/12 without specifying what this statement means.
So when you introduce the cut off weight function with a cos, then the signs of numbers that you will sum can be + and - like the sign of the cos and, thus consequence numbers can eat each other when u will sum them. For example: if the cut of factor for x = cos(x*pi/2)*(1/x) wich even doesn't use N (but who cares? in the end you will never reach the value N for it will we infinite) and you take this weightfunction on each x of the natural numbers and then the sum or those, you get the sum of this sequence 0,+1,0,-1,... what obviously doesn't go infinite but is either 0, 1 or -1.
New ideas always start off sounding like this. Calculus (as Newton and Leibniz knew it) sounded even more like a horoscope for over a hundred years; people like George Berkeley in 1734 criticized the whole subject as being nothing but the study of "ghosts of departed quantities." It wasn't until the 1800s that mathematicians like Weierstrass and Dedekind came around in the to formalize what was going on. You can't expect a new mathematical idea to sound like it came straight from a textbook the moment people start thinking about it.
@@japanada11 Anyone that believes the natural numbers actually sums to -1/12 is a fool. Period. Believe whatever fantasies you wish. The rest of us will be doing real math while you twiddle your thumbs with neo-sacred geometry.
What Tony says in the end is actually a request to *redefine* infinite sums, because the very definition (being the limit of a finite sum upto N, where N goes to infinity) has built this sharp cutoff in. Like there's a Cesaro sum etc, we will be having the Padilla Sum.
the "limit of partial sums" definition is much more ad hoc than it might seem because it's essentially mathematical induction used outside of it's domain of applicability. i.e. induction only works for a countably finite number of steps, whereas here we're DEFINING the value of the sum such that it keeps the induction valid past infinity. though ad hoc, this could potentially be a reasonable approach. but it turns out there are - arguably - better ways of assigning unique values to divergent sums, as shown in the video.
@@toxicore1190 Sure but if you want to go "past" a limit ordinal (like the infinity of natural numbers), the proposition has to be proved separately at the "limit ordinal"th step, it's truth doesn't follow via the usual induction principle.
@@toxicore1190 when assigning the value infinity to divergent sums whose finite partial sums blow up, no such proof is involved, instead it's just a definition.
I was screaming that this is exactly the basis for why renormalization works in QFT.(once I saw the power series come out of the sum) But I had no idea that only the regulators with convergent answers preserve the symmetry. I'm reading this paper for sure.
As a layperson, I feel like the "weighting" was unduly glossed over. I would have loved to see what happens to the actual numbers in the series when you apply some of those complex weighting functions.
@@DonnieX6 one of the properties the regulator function has is that in the limit x→0 it is 1. for x=n/N and N→∞, that means it's going to be 1 too, once you take that limit.
I kinda understand the idea that "chopping off" an infinite sum is effectively removing its most important property (that of the infinitely large tail), so after this operation, the "single value answer" is very far away from the original concept. I'm just confused on how the gentler limiters are applied without similarly shaping the result. Especially in physics etc, I would imagine that modifying the sum in any way effectively breaks away from the physical grounds it is based on. When applying these limiters, what properties of the infinite sum are you preserving to make the answer meaningful?
Not a full answer, but just vaguely intuitively, in physics, continuity matters a lot. I mean, modern (classical Netwonian) physics is literally built on functional equations (the Lagrangian). It's pretty much what calculus was invented for! And that is all about continuous motion and what not. So at least intuitively, it kinda makes sense, that doing things "smoothly" preserves more properties than doing it with a butcher's ax. One very easy example that comes to mind is to approximate a circle on a square grid. If you only follow the actual grid horizontally/vertically, you can approximate a circle's *area* arbitrarily well, while keeping the *circumference* completely constant at every step. (A "circle" of radius 1 drawn this way will converge to an area of π, but a circumference of 8 rather than 2π) As soon as you introduce even the slightest bit of smoothing though (by allowing arbitrary diagonals to better approximate the circumference), you suddenly converge to the correct area *and* circumference. And all you did there was jump one level of smoothness, from C(0) to C(1) continuous. In the proposed method, you jump from non-continuous to infinitely continuous (because you are using Schwartz functions) I think this is kind of a similar situation to that.
I think the short answer is that when you first apply the limiters, it does affect the "value" that the series converges to, but if you look at their design, they all converge to weights of "1" being applied to every term in the sum as N (your partial sum cutoff) goes to infinity. So in the limit, the result is the same as the full unweighted sum itself. It kind of reminds me of approaching troublesome values in a function from different directions and getting different answers from every choice of approach path. The variation of weight functions approach is nice because it replaces our brutish and discontinuous first intuition of weight functions with a smoother alternative. The big shining revelation here is that this huge class of approaches involving continuous weight functions all agree when it comes to their first order approximation term of -1/12. Even better, those that apparently respect the symmetries of QFT have no higher order terms and converge exactly to -1/12. It makes physical sense that the way nature would opt to regularize its infinite potential processes on the quantum level respects the symmetries underlying the mechanics. Maybe something about the logic of our universe's laws allows infinite things to occur as long as the way those things are counted up give us finite, physically realizable outcomes. Buried in the thick of this particular series' infinite tendencies lies a finite portion that can be allowed to show itself if things are counted the right way. It's always there, even if the series diverges, it's just being covered up by infinity in those cases. Really makes me think of residues in complex analysis. Your function can blow up to infinity at certain points, but if you integrate around those points (which in complex analysis equates to zero-summing the nicely behaved parts of your function) you can extract finite values that sort of "stick out" from the parts of them that aren't behaving so nicely.
The smoothed asymptote parabola of the partial sums of the series 1 + 2 + 3 + 4 + ⋯ has a y-intercept of −1/12, as shown nicely on Wikipedia, for visual thinkers. That is *the* intuitive justification for "most people in the street" (17:45-18:15), and that's Level 1, if you will. Level 2 is: I'm a visual thinker, too, so I took this in and tried to fit that very same parabola, from scratch, using those very same partial sums, and the y-intercept came to not -1/12 but -1/8, so then I tried to contact the wiki image creator as to their method, and have not heard back. This deserves a more focused investigation for Numberphile's 2034 short & sweet conclusive return to the topic, or much sooner, please! :)
I remember back when I was obssesed about both the Numberphile video about -1/12, and Death Battle's video about Goku vs Superman. Ten years later and they both revisited the same subjects
I don't feel like this weighting is fair. He draws the step weighting first (hard transition from 1 to 0) and then pretends that e^(n/N) is a smoothed-out version of that transition. It is not. He draws the graph as if its starts at 1 horizontally for n=0 and curves down when n approaches N and then it curves up again as n goes from N to infinity. Almost nothing of that is true. It odes start from 1 for n=0, but it already has a very marked negative slope regardless of the value of N (d(e^(-n/N))/dn = -1/N * e^(-n/N) which, for n=0, is -1/N), it never ever approaches horizontal at any point before n=N, and it never ever curves down (the second derivative is always positive so it is always concave up). So it is constantly diminishing and the rate of diminishing is greatest at n=0, at the beginning, not as n approaches N), by the time n is approaching N, tit already lost 63% of its weight. And, again, is not that it loses a significant part of that weight as it approaches N, most of the weight is lost in the beginning where the derivative has its minimum (i.e max negative slope). So with that kind of weighting it is not surprising that you turn an initially diverging series in a converging one. Except that even with all that you don't. You still gen N^2-1/12 which as N approaches infinite it is still infinite - 1/12 which is infinite. But wait there is more: Then he adds another weighting term: cos(n/N). He describes it as "this still goes between one and zero with some transitioning around N, it does a bit of wiggling around". That is profoundly deceptive, and I can only thing that it is intentionally deceptive. Why didn't he attempt to do a graph of this weighting function? Did he just forget to mention that, for any value of N, that weighting factor is NEGATIVE for HALF of the values of n?????? Little detail, hu? So it doesn't go between one and zero. It OVERSHOOTS zero and goes NEGATIVE... HALF OF THE TIME!!!!! So ware not adding NATURAL numbers with a weighting anymore. -1,000,000 is not a natural number. I wonder what would happen if you take |cos(n/N)| instead. I bet it doesn't converge.
cos of (n/N) can only range from cos(0)=1 till cos (1) which is 0.999.... it is in fact almost always 1. Halfway of N at 0.5 it is 0.99996 so i really don't understand what he wants with this.
@@YEC999 .... No. n doesn't stop at N. As you make n tend to infinity, n will take values of 2N, 10N, 9999999999N, tending to infinity times N. The argument inside of cos will be from cos(0) to cos(tending to infinity), so the cos will cycle between -1 and 1 as the cos function does.
@@adb012 You are right, i am not a mathematician, thought it would Stop at N!! But it really goes into the negativ!! i saw this yesterday later when i drew the function. But one thing: I don't know if you mistyped, but it should be e^(-n/N)...
You're missing the point. If you pick a weighting that is a smoothed version of the step, you get -1/12+CN^2. If you smooth it out more or less, you still get the same -1/12+CN^2. If you create your own weighting function, you'll once again get -1/12+CN^2. That -1/12 is always there; it's just that, if you carefully pick your weighting function such that C=0, it's more apparent. That weighting function isn't *why* the sum is said to be -1/12, it's just a particular strange example of a weighting function that makes the connection clearer. Try using a weighting function of f(x)=(1-erf(k(x-1)))/2 for some large k. The larger k gets, the closer this gets to the step function. For any k, the weighted sum will asymptotically approach -1/12+CN^2 for appropriate C. As k gets larger, it approaches more slowly, but it still approaches it. The -1/12 is always there.
Also newly uploaded today about -1/12: ruclips.net/video/FmLIGN8ZGdw/видео.html
And see the full -1/12 playlist at: ruclips.net/p/PLt5AfwLFPxWK2zCU-4X1iuuu5m8hf6L1B
-1/12 shield sticker and tee: numberphile.creator-spring.com/listing/1-12-shield-numberphile
Beautiful video.
if we are able to regularise the most basic summation sum(n) without throwing away infinity, like this video, I guess very soon we would be able to regularise every other infinite sum with some combination of above weighting functions and some other known techniques. May be we misunderstood infinity till now. can someone do it for 1+1+1+1+1+..... ? may be we get a finite sum for this too.
another thing that struck me is that all of this is not for a continuum set like R but for N , and Quantum Theory is about denying the continuum at various places. May be there is a subtle but a fundamental reason why such results find their shadows in bits and pieces in Quantum Theory. Something like you always find a circle somewhere whenever there is pi in some formula.
For some reason, the constant refocusing on this video made me slightly motion sick. PLEASE either stop down the aperture or add a separate overhead PaperCam!
but then you can choose a regulating function at results in the -1/12 being canceled with with some C +1/12. This shows that the series is divergent as you can get any result you like. I would also point out that all the infinitely small numbers are known as infinitesimals and are the basis of how we perform calculus in the first place, so you cannot simple ignore an infinite amount of infinitesimals. This goes along the same lines of you cannot zap an infinite amount of zeroes in an infinite series without changing the value.
Infinity plus 1 is NOT Infinity. Because if it was, then 1 has to be treated as 0 in an infinity equation. There is something called an infinitesimal unit. That unit can represent 1. And that unit is as far from 1 as infinity is from the integer 1. It means that 1 could even represent 1 google, if we count in Google units.
-1/12 is only believed by those that are too lazy to use common sense. With all respect to Ramanujan.
There are 12 fundamental particles that interact with the higgs field that causes them to slow down from the speed of light providing the minus sign🤷
"I saw a tweet that made me so mad that I disproved it and wrote a paper about it" is the best way to do research
Based
"I saw a tweet that made me so mad" is always true
now that tweeter is called "X" what are tweets called?
someone write a paper about that please
@Paulo_Dirac No one calls it X, we all still call it Twitter and tweets
@@supersuede91 Xitter
xeets
For what it's worth, the original -1/12 video is the reason I went on to study maths and physics in uni and here I am now 😂
Doubt
@@co2metal 🤷♂️
So -1/12 didn't protect you from infinities, quite the opposite.
And now you go to work for corps/state/mill and start removing ppl with fake saiens.
@@Intellllect Some must make the sacrifice of fighting infinity to save the rest of us from it.
Good lord have i really been watchng your videos for 10 years or more? Time sure flies by.
i had the same thought.
@@dagudelo88Likewise!
same!
I've been watching these videos for -1/12 years. That's negative one month.
It’s been 12 years… I can’t imagine
One thing i love about numberphile is how passionate these very intelligent people are about such an awesome topic. Very nice change from the constant bombardment of low level nonsense. Thank you numberphile
well said.
You can genuinely feel Tony going "I told you so!" to everyone by the way he's talking. Man's been brooding for literally a decade
Sqrt(-1) isn't intuitive either. But it's now well understood, and it wasn't well understood for hundreds of years, and yet it was used since Bernouilli. We are with inifinite series at the same place Bernouilli was when he was using sqrt(-1) without really knowing if that was legitimate.
@@InXLsisDeo No we're not. Infinite series are well understood and the imbecil in the video needs to give it up. The of whole numbers diverges. Always has, always will.
You do realise he didn't tell anyone anything. This entire video is just avoiding the simple fact that he's wrong, with extra steps.
@@InXLsisDeo
But there is no solution in the reals. No real times itself yields a negative product. Complex numbers, problem solved. This question can not be solved that way, because it obviously blows up to infinity.
So what does Tony say that infinity + 1 equals? Less than infinity?
Maybe that's part of why I'm so put off by this one.
I was more or less mathematically illiterate and despised everything mathematics, and then I watched this video back in the day and it really intrigued me. Now, a few years later I am a grad student in pure mathematics, and it all started with watching these videos, particularly the one about -1/2! You can say what you want about the rigor of these computations, but for me, this is what started my love for mathematics!
haha that's a great story!
That is great, thanks for sharing it. The -1/12 videos also rekindled my interest in maths, as did the superb video of hackerdashery on the p vs. np problem
That's awesome! I still do that hacky "proof" in my junior-level high school math classes sometimes just to rope students in in pursuit of exactly that response.
I subscribed to numbphile 10 years ago because of -1/12 video
Wow, that's a story! Hahaha. What a surprise.
If you plot the graph f(N) = N(N+1)/2, it is a parabola that intersect X axis at points 0 and -1. The area bounded by parabola between 0 and -1 is exactly -1/12
My dad showed me that just after the first -1/12 video!
@@L9MN4sTCUk ... An integral is not a partial sum but it is the limit of a partial sum as n goes to infinity. int[a,b] f(x) dx = lim [n to inf] sum [i=1 to n] f((b-a)/n*i)*(b-a)/n). It's the Riemann sum of the area of rectangles f(x)*Δx as Δx goes to zero (and hence n, the number of rectangles, goes to infinity)
I figured that was just a coincidence. But I looked at the sum of cubes, fifth powers, 7th powers and 9th powers, and it always worked. Note, the 5th and 9th powers have extra roots outside [-1, 0], but I still only integrated from -1 to 0 to get the Riemann zeta result.
The graphs are really nice, I wish I could post them (with a zoomed in vertical axis). But search Faulhaber polynomials if you care to check yourself.
Area can't be negative
@@B3Band In math yes, area can absolutely be negative.
C N² - 1/12 + O(1/N) such that C happens to be 0 makes way more sense than just a blanket -1/12. This is very cool
I remember feeling unsatisfied with the way -1/12 was justified with the zeta function. That's just one arbitrary function, who's to say if you used a different function you wouldn't get a different finite value, you know? This is much more convincing!
@@khaduopha2640Analytic continuation is kind of weird. There is only one correct continuation for any function
Whatever the method of regularization, if it allows shifting, adding, etc., its value must be -1/12.
That is what the original video showed.
There are many such regularization methods, some of which have physical realizations.
But even if they don't have a physical realization, the math still exists.
But what about the O(1/N). How do you get rid of it? Or what does its existence imply?
@@vezokpiraka O(1/N) is something that converges to zero as N goes to infinity "by definition". The expression CN^2 - (1/12) + O(1/N) is telling us than *any* analytic continuation that gives us "1+2+3+4+..." at a non-singular point *must* have value -1/12 at that point.
I was an undergraduate when the infamous -1/12 video came out and now I’m close to finish a PhD in arithmetic geometry. This made me feel so nostalgic.
We played that video in my calc 2 class when we got to infinite series. The professor basically said nah it's divergent vid is wrong. 💀
@@RingxWorld I asked my calc professor about that video at the time. Turns out his PhD advisor is the same as Terry Tao’s, and he gave a talk on the dept colloquium about the summability methods and how to make sense of this (which is half of the topic of this video). This was such a flashback 😂
@@RingxWorld yeah that's what it is simply wrong...
Finally, the long-awaited -1/12 redemption arc.
@@bjshnog You think you have average IQ but really its room temperature. Ignorance is bliss!
"falsehood"@@bjshnog
@@bjshnogyou are firmly in the middle of the bell curve meme
@@bjshnog Yep. This whole thing could have been avoided if they had not been weaselly with terms. They're using "equals" in a novel sense, when they should have said that "this particular function is interesting. The value of this function of this series is -1/12", to which nobody would have objected, but instead they change the game whenever it's convenient to them, such as moving entities, spacing out entities, etc., which would not matter if this were basic algebra, and most importantly, pretending that this function is "equals". Go ahead and add as many positive integers as you want, the amount never decreases. So clearly this is not about a series "equalling" as sum.
Maddening.
@@bjshnog I don’t think you understand the argument he was making. With certain choices of weighting functions, you get some value C*N^2 - 1/12 where C is determined by the chosen weighting function. If you choose a specific weighting function which gives C equal 0 (like the cos function described in the video), that function converges on -1/12 as N grows very large. However, you seemed to have missed his other point, which is as N grows very large, the sum becomes the standard 1+2+3… where each term is equally weighted. This is because the weighting function is defined as a function that begins at 1, and ends at 0. This means the weighting function equals 1 at w(0). Each terms weight in the series corresponds to w(n_i/N). This means that as N goes to infinity, each term tends to w(0) which equals 1. So as N goes to infinity, the sum equals -1/12, and each term becomes weighted as 1, giving the result: 1+2+3…=-1/12
There are no tangentially related concepts, he very clearly showed the same result just using a different method.
For those wondering why e^(-n/N)*cos(n/N) is so elegant, it's the fact that in C*N^2, C is the Mellin transform of the regulator function, which basically amounts to integrating x*e^(-x)*cos(x) from 0 to infinity, and it ends up being 0.
Beautiful!
Thanks!
YOu explained it very well, now i only need a maths phd to understand it:)
I'm not sure I see that beauty you speak about. This regulating function is a periodic function (which binds any value between two maxima) times a negative exponential, which makes the two maxima converge. It's not surprising at all it results in a finite value. It amounts to doing +1-1+(0,5)+(-0,5)+(0,2)+(-0,2).... Which eventually converges to 0. These kinds of functions are used everywhere in NMR spectroscopy to regulate noise in the output signals.
I have no idea what my entry point to begin understanding that is.
Lot more gray hairs on Tony. Cannot believe we have been growing up with this man for more than a decade.
On a side note, any more of them big numbers?
Besides his age?
Somewhere between Grahams number and Tree(3)?@@ashleylawrence2110
my exact thought when I saw the thumbnail. Been heavy watching numberphile 10-12 years ago just when they were starting out, but then abruptly stopped following besides casual video here and there, and was kind of shocked and sad to realise so much changed
Rude hahaha it’s graceful.
Also I think he’s got kids.
I have not seen Tony so excited for years. I like it. Good luck on his way
Professor Padilla is a brilliant physicist and mathematician and incredibly skilled at explaining his thinking to us, even when ,like here, it gets into the realm of wonder. Thank you, Brady, for bringing him to us.
I feel like the weighing functions should be strictly positive for this example. In the cos example, we are not summing everything anymore. Especially when N->Inf, you don't even know which numbers are summed which are subtracted (I like to think they are added and subtracted at the same time).
I also feel like for weighing functions to make sense, they should be strictly decreasing from 1 to 0. In other words, 1>=w(n)>=0 and dw(n)/dn=0 .
I was thinking exactly the same till I looked more closely: n/N is always a positive number below 1 and therefore cos(n/N) is also always positive 😊
@@volkeru2718 While you are correct, there is a caveat. The sum is infinite and cos(n/N) is not. They calculate the result of that infinite sum and only then set N to infinity. If they start with sum of N numbers and then looked at the limiting case N->Inf, I think your assertion would be valid.
In other words, the initial sum has a period of T, which repeats infinitely within the sum already. Increasing its period to infinity voids the initial solution because It (probably) requires infinite repeat of that period (I assume this is correct. Else, we could just repeat the entire thing with sum up to N, then do N goes to infinity.).
I am however wrong in my other assertions. We are not increasing frequency but decreasing it. However, the assumption that cos cycles infinite times within the sum remains even when N->Inf.
Thanks! I indeed overlooked the infinite sum. It is a pity that they did not even mention that the weighing function gets negative and why they are allowed to do so... Or maybe, as before, they do not ask these questions... They simply do it, as before when treating the sum of a non converging row as if it was just a normal number 😬
I am still reflecting about this one. And I Excel-led it... Guess I am much more a financial analyst / controller than a mathematician by now. 😅 However currently I tend to say that if they first proved that the result is proportional to C*N^2-1/12+epsilon independent of the weighing function then they are thereafter free to choose any weighing function, even a negative one. And indeed it is amazing that the results using the function shown in the video converge to -1/12 though nothing in the function itself hints in this direction. However interpreting the result as the sum of all integers would clearly be wrong. Allmost all weighing functions will result in a C ≠ 0 and therefore diverge to infinity which is the intuitive result.
@@volkeru2718 Lol my instinct after watching this video was to do the exact same thing--create an excel spreadsheet that approximates taking each component to infinity. Must be the engineer in me. 😆 Unfortunately, Excel limits any sequence to being about 10^6 items long, but that still gives you some leeway to play around. What's really interesting is if you try to apply this to other, established, convergent series. The ones I've tried it with so far still worked perfectly fine.
Best Numberphile video I have seen in a while. Thanks Brady.
That video was 10 years ago?? Feel like the biggest mathematical mystery here is where that time went.
I hear that.
No. It was like three Tuesdays ago, MAX.
I remember getting upset at the original -1/12 video. I enjoyed this one much more and it was the first I'd heard of these regulators as a way of studying series! Also, 10 years since the original? I feel very old.
Hey! Lovely seeing you here :)
Ten years ago I was very unhappy with my career and watched the original video. I was so intrigued and inspired, I quit my job and went back to school for a math degree to better understand this result. I've never regretted this decision and seeing this video feels like everything has come full circle.
What do you do for work now?
@@sbnwncfishes for likes on RUclips...
@@sbnwncnever ask a math major that question
@@oscar278 Is it always computer programming? Lol
@@sbnwncI did a double major in math and comp. sci. after I retired from the USAF. Then, 17 years later, I completed a master's degree in math. Lots of computers, programming etc in all my careers I have done. Your comment was pretty close
Your videos were one of the things that really got me into mathematics, and now I'm in graduate school for my physics PhD. Looking in the comments, I'm not alone in this either! Thanks for ten years of accessible math education -- even if some of it isnt entirely rigorous :p
This sum regulation method is fascinating... I wouldnt buy putting an equals sign anywhere, but it's still extremely compelling that -1/12 is just as important here as it is via analytic continuation.
Edit: That connection to physics -- wow! It makes me think that there must be some universally "preferred" method of renormalizing sums, some way that's "natural" in both mathematics and physics for reasons that arent entirely clear. I'm very curious why certain regularizations produce that behavior but not others...
You can actually set this up in Excel by using a "SEQUENCE" function to create the series and then wrapping a simple "SUM" function around it. If you set n to be sufficiently large compared to N, it does converge on -0.08333. So, the infinite sum if you're using that weighting function does actually equal -1/12. No weird math tricks or anything. Just changing the way you weight it.
Man I love how Brady just goes in with the tough questions and points. Great video!
Me too, that's one of the reasons I like Brady's videos so much. Once he groks it, he throws curveballs at it. Sometimes people miss things that he sees pretty quickly.
But the answeer was. "Of course i reverse engineerd it..." = fraud.
@@YEC999 Almost the entirety of maths is developed this way, bro. You reverse engineer an answer *and then* check to see if it breaks stuff. Then you slowly prove that the answer couldn't be any other. Quite the opposite of a fraud. It's the most sure-fire way to prove things.
@@Voshchronos If you think to multiply whole number with a cos function is the most "sure-fire way" to prove that the sum of whole numbers=-1/12. than i would never believe anything scientific you say.Literally if i knew your name i would not believe anything you say. You should really play with a cos function before you make claims like this...
Another thing that you don't understand: To manipulate to reverse engineer a weighing fucntion that it outputs what you want to have is a brutal fraud. It is like the definition of fraud: You use an manipulative function to get the result you wanted. The equivalent in economics is balance sheet manipulation like Enron or Bernie Madoff where people went 30 years to jail for.That is all that is: a manipulative weighing function to get to a sum that you wanted to have. in Case of Enron: lower debt than it would be. In case of this: lower sum to get to -1/12
And no nothing was invented in maths with manipulated weighing functions. No Complex numbers (they came through a very tanngible problem) no negative numbers no reals no nothing..
@@YEC999 I mean, I am not going to claim to understand the technicalities but it seemed very clear to me what he was saying.
It wasn't that no other equation worked, it's just the particular one used showed it more clearly.
All these "smoothing the curve" equations have a -1/12th in them, and the -1/12th becomes more prominent as they all get infinitesimally close to the sudden drop from 1 to 0.
10 years ago, because of your videos, I got really into this -1/12 thing. I kept researching it for quite a while. But this new video is just amazing; what a fascinating result!
I always hated mathematics, my brain just couldn't handle how numbers work, but I can appreciate people like Dr. Padilla who have devoted their lives to trying to explain things that were previously unexplainable. I admire his humility in saying "I don't want to understand infinity, I want to understand the journey of why we would want to get there in the first place". It may be just numbers, but it's an incredibly profound way of thinking and applying them to help explain how the very fabric of our universe works.
Watching your videos for many years I sometimes regret I left academia. So much to still be discovered. Thanks for making these intriguing videos that even I can keep track with.
I watched the first -1/12 video when I was in 7th grade, and I am now a junior in university. I remember in elementary math was my favorite subject and when I reached sixth grade I had a horrible math teacher which made me dislike math. Thanks to you guys for making that video as it absolutely rekindled my love for math when I was in seventh grade.
Wow can't believe it's been 10 years since that video
Right!!
And they're still wrong.
I feel like I've watched something incredibly profound. What a great discovery! I hope there's something to it.
I'm completely math illiterate, I understand the individual concepts, but I can't put them together even when explained. I still felt the profoundness.
It's like watching a foreign movie. I don't know what's going on, but the effects and action is awesome 😊
Terrence Tao’s formulation is so elegant and beautiful. The first video that was published 10 years ago had a lot of debate around it. This video probably won’t because it’s so convincing. One of the best math videos I have seen
Thanks! I love your channel, and this particular video is one of my all-time favorites. It explained a complex topic in a clear and compelling way. Amazing.
But then ... integers _do_ make for a rather sharp cutoff. As far as making it -1/12, I would agree with Tom Petty that the weighting is the hardest part.
Yeah, but that may only be our "idealized" perception of integers. Perhaps there's a better way to grasp that which involves weightings.
@@radoskan I see it not as a way of _perceiving_ integers, but of extending them for injection into the squishy world of ℝeal numbers.
@@radoskanit’s not idealized or a perception, it’s just the definition
Is this a joke? Let's redefine integers as squishy real numbers?
@@CalifornianViking You're right - I said it wrong. I should have said real _world_ - as in "protecting us from infinity"
Great video! I would love for it to become public and viewed by more people as this really helps explain where the "-1/12"th-ness of the sum is. To use Ed Frenkel's terminology, this is how to extract the gold nugget from the infinite dirt!
No this is the dirt. Infinity is just OK.
The best way my math professor described infinity is that it's not a number, but more like a direction. It's like traveling east, you never reach the end because you can always go more east. So, treating infinity like a number doesn't make sense since there's always more east.
Wow I love this
Exactly. I think what we're seeing is that there might be different ways of travelling east (one step at a time or two steps at a time for a simple example), and there might be ways of quantifying those different destinations other than "they blow up to east".
You can only go east as far as the international date line. From there, there's only north, south, west, up, and down.
@@andrewj22 love when people are pedantic about a metaphor, that doesn’t deride the point at all
@@Varooooooom Of course the point stands. It's just a bad analogy.
I appreciate Tony’s passion for the subject! It’s always awesome to see someone apply their passion and continue to push it to new edges and in new ways!
Love your content, presented in a very accessible way. Coming from St. Helens, hearing someone who comes from the Northwest providing such great insights into mathematics feels comforting and even more relatable (and hopefully inspiring for kids growing up in the region). Thanks!
I personally think this is genuinely a momentous find. The fact that infinities can be numerically represented and define behavior of systems without having to be aggressively gotten rid of can possibly reconstitute many of the existing conjectures and hypotheses that are stuck due to infinities... not all, but even if one, that'd be huge.
Regularisation has been used for about a century... and there are a number of experimentally verified results that 'use' it. The novel intuition is that there may be a 'best' or 'preferred' way to regularise certain types of expressions.
but isn't this basically the gist of many of the videos on the topic 10 years ago?
This is pretty much the point of another video, "-1/12 is a Gold Nugget"
The Chonotop conjecture
He selected a weighting function that, for any value of N, goes NEGATIVE for half of the values of n. That's not something "between 1 and zero" as he claims and totally denaturalizes the addition of NATRUAL numbers which are NEVER negative.
I loved the "smells of String Theory" line around 13:30, so much of cutting edge mathematics is about feeling out into those intuitive spaces where we don't have models for, in order to get a clearer picture of what we're trying to model.
To be fair, String Theory is been shown to be pretty much a hoax as well. Michio Kaku is basically a Charlatan. Dr. Angela Collier (@acollierastro) has an excellent video about it.
I know, it blew my mind. More so than Eulers identity which was the big thing that blew my mind in math
I am absolutely entranced with this linking of such a bizarre result from number theory and the physical workings of our universe. Tony Padilla is such a treasure.
It makes me wonder if it is that our descriptions are wrong since infinities pop up that we have to throw out or is that since we can, in a sense, wrangle the infinity to finite using regularization, that nature at the quantum level is infinite but has a way of regulating itself back to finite.
@@hmx10011Well you are right in at least one way: in quantum physics there are infinities that are reduced back to finite. The values for any property of a particle is a superposition of all the values this property could have, and the possibilities are indeed infinite, yet they all add up in a way to a single usable physical value.
So is this process more generalized that we are expecting or is it inherently only tied to possibilities ? Hard to tell
The separation between mathematics and reality is the distinction between the computable and the infinite/continuous
Processes in nature are computationally bounded in the same way that your computer does not have enough time in the universe to calculate all the digits of pi
@@discotecc A circle doesnt need to compute the value of pi to have it's diameter as the exact value.
Your argument is ridiculously wrong
Well, I agree that our intuition isn't necessarily right, but neither is doing something BECAUSE it gives the answer you want it to give you.
It’s an endless debate. People will still be arguing about about this 1 month ago.
The debate ended in the original video where he pointed into a textbook that didn’t use an equals sign
@@soonahero 1 month ago means -1/12th year, implying infinite years.
very nice
This feels like the most significant and profound numberphile video ever made.
New knowledge! Very exciting.
Or least! Who knows!
Indeed, the optimist in me is already looking forward to a numberphile mention at a future nobel ceremony, this seems like it could be a potientialy fundamental insight once properly understood, was left very excited after watching this.
It really does. Full circle on a topic that kicked off controversy, excitement, and many peoples' math careers. But this time, it comes back with new insights. Amazing 😍🤩
Yeah, but string theory....
This is astounding! Relating a childhood fascination of mine with a current intrest of mine. What a delight! I'm going to read this paper
What a Valentine's day treat XD
this is so insightful and to see how Terrance approached the problem with regulator functions and Tony's way of explaining it
You can express a partial sum of an infinite series in terms of the sum of the infinite series.... WHAT? I had to rewatch 3:30 to 5:30 again and then pause on this to ruminate how amazing what is being expressed here. Excellent videos. This feels like you all are mass producing the discovery of one person down for the rest of us to understand. Just as important, to me, than the discovery itself. Thank you.
This is the most exciting science news I've heard for a long time!
Do we need to start a gofundme to buy Tony a new phone charger
Came here to say this!
I saw that too!!😬😬
That charger does not do a smooth transition.
Oh my how the time has passed. 10 years ago I was in high school, watching these videos with astonishment and a genuine feel of adventure and love for all these awesome discoveries. A decade and two degrees later, and nothing has changed, except maybe for greying Professor Padilla, but still as enthusiastic as ever.
Thank you, sincerely.
"That's what you're really doing here," is gaslighting.
I love that his takeaway was "it got people interested in mathematics, so I'm glad we did it." It's something that resonates with me so much. Sometimes in education and outreach, it's not about teaching the right answer, but getting people interested.
I love Tony’s passion in how he presents things. And a bit of string theory thrown in for good measure!!
Astounding result. It really does seem like something quite fundamental has been uncovered, and it comes out of such a simple question that seemed unanswerable for the longest time. I wonder if this is what it felt like when people started taking imaginary numbers seriously and realised they gave real, verifiable answers to questions. It seems super weird, but if the maths gives you useful answers then you just have to follow it.
"I wonder if this is what it felt like when"
No, this is _exactly_ what it felt like
that is the result of Terrence Tao I believe. First time I kinda understand something he wrote, and you can really see a glimpse of his genius with this.
This unanswerable question was solved in the 16th century. The sum diverges. Sorry, but It is not and will never be -1/12.
@@herbie_the_hillbillie_goat"the *sum* diverges". You didn't heard (or understood) the whole point of the video.
You wouldn't have said "it was solved since". Not it wasn't "solved". And you need to understand clearly why by reflecting some more time on the topic.
18:48 Infinity didn't drive Cantor mad. Other mathematicians who failed to recognize his findings and simply mocked and bullied him did.
What a delightful video! Finally an honest way to do renormalization, and great to see more of the amazing work that Terry's been doing since we were grad school room mates :)
You were roommates with Terrence Tao??
Using the cos function to "smooth" the cutoff point seems a much more physical way of doing this. Any RF engineer will tell you there's no such thing as a pure step function; that rising edge is always a collection of ever increasing frequencies so that it just looks like a step.
Ask the Computerphile guys to explain. They'll tell you it's common or garden integer overflow to sum positives and end up with a negative.
Modular arithmetic (which is one easy way to describe integer underflow/overflow) is distinctly different from zeta regularization (which is what they are doing here).
One does not directly inform understanding of the other.
You can think of zeta regularization like squishing a function so that the annoying parts don't cause you problems and then taking the answer for the squished function instead of the original function. Like, what is the "value" of sin^2(x) at infinity?
That depends entirely on where infinity "stops". (Which doesn't make sense, right?)
Since the sin wave never converges, it doesn't have a value in the limit as x approaches infinity. Regularization allows us to assign a knowingly false, but still useful number to the nonsense equality.
sin^2(x) does NOT equal 0.5, but there is a way of thinking about the function where you can wave your hand and say "for all intents and purposes, it's 1/2".
Very useful for physicists that don't know why a problematic infinity can be ignored but are happy to ignore them, but extremely unhelpful to a mathematician trying to better describe mathematical systems.
"I don't know why, but it's useful even if it's not necessarily true" seems to be the hallmark of modern maths and I am not a fan. (I'm looking at you Axiom of Choice...)
My problem is the statement "I am removing the regulator because I'm taking N to infinity". The weights are still there even as N goes to infinity. With W(n,N) = exp(-n/N)cos(n/N), even though for a fix n, W(n,N)--->1 as N--> infty, those weights are still N. For example W(2N,N) ~ -0.056, no matter how large N is. That means, even taking N to infinity ("removing the regulator"), the weighted sum always includes the term -0.056*2*N. It seems to me that these weights being considered are too general. The weighted sums they yield don't have much to do with the original sum 1+2+...+n+...
The fact that the form
CN^2 - 1/12 + o(1) is independent of weight, still says there's something very special about -1/12. Wheter you think of it as "the" sum or 'the finite residue' is splitting hairs for me at that point.
@@johngalmann9579 Seeing this form makes me wonder if the result comes from an assumption that a 'C' that evaluates to zero can 'overpower' an N^2 that goes to infinity. Similar to how 0^0 has different limits when approached from different directions, 0 x infinity has different values depending how the 0 and infinity are arrived at. If I take a limit of a sum of 0x1 + 0x2 + 0x3 etc it is easy to say that all terms are zero, then extend to N->infinity, but if I take the sum of N first and then multiply, will I get a different answer?
It does change them. At 3:40 he tells you quite open. That he did everything he could to get to the -1/12 he is clearly not a scientist. He wanted that result.
The summation function is from n=0 to n=N. You do not add the terms for n>N because the summation function explicitly tells you to stop at n=N.
@@YEC999The -1/12 is always there, as -1/12+CN^2. If you pick your weighting such that C=0, you'll get -1/12, but even if you don't, the -1/12 is still there. It's always the constant term of the asymptotic expansion.
Awesome to see the joy and excitement in the eyes of Tony when he talks about this. ❤
Even though i enjoy what I'm studying this video honestly made me regret not studying maths or physics
It's never too late!
@@numberphile If only I could still pay my rent while being back at school.
I’ve literally had lectures taught by this guy and this still doesn’t make much sense to me
This is incredible. You should really consider doing this as a full release video (not unlisted).
It will be - just when we release two videos at once we keep one unlisted for a few hours.
@@numberphile oh no. It smells of math-drama now.
The best kind of drama! 🤗
"Lemme give you another example of another choice you could make"
Best words to hear in math video 😊
Wow, thanks for an excellent revisit to -1/12. This was just beautiful! Pure partial and infinite sums, to end up where we all know we would end up... love it!❤
-1/12 is the error code you receive when part of your formula tries to treat an infinity as a number.
This how math moves forwards - craziness for a long time and denial and then a little piece of wisdom from a genius like Terry, Euler or Ramanujan and we sheeple get a little further.
This could be told like a story by H.P.Lovecraft. The protagonist thinks the -1/12 is not a real answer, it is just an analytic continuation, a trick. But it haunts him, he sees it in his dreams more and more often, then he gets visions during the day. He can no longer ignore it, he has to scratch the itch, searching deeper. And he finds it. The answer. -1/12 was real all along, he can no longer protect his mind by pretending that it is just a trick. And he goes insane. Oh good old infinity. We cannot understand you. But how sweet can ignorance be if you face the monster, the old god of chaos. Twisting, shouting, breaking what we meager humas tought to be the well ordered little world we inhabit.
You looked so hard into the distance for the monster, only to see it was right behind you!
Agree with the others, this is one of my favorite Numberphile viedos in a long while (and a good redemption of Dr Padilla :))) ). It's fun and interesting and it communicates a lot of high-level ideas without being too bogged down by details. It reminds me of being in undergrad and having much smarter people than me explain deep ideas over lunch, or during "office hours".
The sum of strictly positive integers from 1 to N, where N goes to infinite, is of course infinite. 1+2+3=6. Adding more strictly positive integers to this sum will clearly increase the sum. QED.
Only for finitely many integers.
@@RaRa-eu9mw No, a sum of strictly positive integers cannot be negative, be it finite or infinite.
@@karljohanhaarberg6706This very case is a counterexample. 1+2+3...=-1/12 is infinitely many positive integers adding to a negative value. Another example would be 1+1+1...=-1/2
« Infinity has made Cantor mad … »
Nicely said !
My layman understanding of the need for a weighting function is that you only see the sharp cutoff at N if you stop the partial sum at N, but you're interested in what happens when N goes to infinity so you will never see a cutoff: it's infinitely "far away".
Worse than that. He selected a weighting function that, for any value of N, goes NEGATIVE for half of the values of n. That's not something "between 1 and zero" as he claims and totally denaturalizes the addition of NATRUAL numbers which are NEVER negative.
@@adb012 you copy this comment like 20 times already. Make a yt video proving this is wrong, don't create spam
@@dexterPL ... I copied it only 2 or 3 times as responses to other comments where it was relevant. In some of these places it triggered some constructive discussion. I also wrote a long comment myself (not in response to another comment) explaining my position more formally and in more detail. You don't need to like what I do and I don't need to care whether you like what I do or not. I will not make a yt video.
Hey, maybe you were the guy from the tweet he talked about @1:08 ;-) @@adb012
@@adb012 I don't know why you're so concerned about negative numbers. Using just e^(-n/N) as the "weighting function" produces a series where _not one single term_ of the series is a natural number. So I find it a bit odd that you are outraged that he uses a regulator which allows for negative terms but aren't equally outraged by the fact that other regulators produce series which have no natural numbers at all.
But maybe this point has already been made in another discussion. I don't know where that discussion is, though.
I honestly think, that this was the way, that Ramanujan approached these problems. That was his talent, to be open minded to techniques. I'm not surprised, that Terence Tao was the one to decode parts of this secret...
No, Ramanujan was way beyond this kind of stuff.
@@idjles May well be, for sure. Sadly he died too young. But the questions, he asked, about the nature of mathematics, are similar to this stuff. But he was in mathematical spheres, that were never documented, but they gave him some answers, that we, over a hundred productive years later, find hard to understand!
@@idjles actually, if you check the paper, you'll see that Ramanujan precisely used this kind of trick a lot. It is indeed pretty much what he did.
The insane thing is, that he did so without having to be told that this is even a thing by a Fields Medal winning mathematician. He just kinda did it all auto didact style with barely any support until he got discovered.
10:17 this weighting function does not go between 1 and 0 as stated, as cos(n/N) will have negative values for π/2 +2kπ < n/N < 3π/2 + 2kπ. (integer k)
the ne^(-n/N) factor is always positive, so it oscillates.
I think it is exactly the purpose of this setup, it is a weight function that allows a smooth landing in infinity without having to throw away a CN². The interesting part is that the landing is exactly on -1/12. The reason why breaks my brain but I find it fascinating!
Yes but as N tends to infinity e^(-n/N)cos(n/N) tends to e(0)cos(0) = 1 meaning you are multiplying every term by 1 which is trivial.
@@j.r.8176 Unless I got it wrong, I believe that n is not smaller than N., therefore it will not be equal to 1. N is an arbitrary cutting point in the sum and n will go past that number. Cos will then oscilate and the sum will be -1/12
@@j.r.8176 n is not a finite constant.
@@BigDBrian So what if it's not? I'm talking about N not n. Also when did I say n was a constant?
Without a doubt the most important topic ever covered on Numberphile. People probably laughed the first time someone said “let’s just call SQRT(-1) i and see what happens”. Now, instead of just ignoring the fact that -1/12 keeps popping up, a few courageous mathematicians are seeing where the clues lead. Kudos to Dr. Padilla and others for boldly going where no mathematicians have gone before.
Glad to see a few commenters picking up on this. I agree completely, and hope to live long enough to see what the -1/12 is really all about.
Adding the cosine function into the summation completely changes it, you now have a summation of positive and negative numbers. The original and your proposed are as different as summing 1+2+3 ...N and summing 1 - 2 + 3 - 4 + 5 - 6 ....N, in each case with N going to infinity. The first goes to N(N+1)/2 as you said. The second toggles, going to -N/2 for N even, and (N+1)/2 for N odd.
_"The original and your proposed are as different as summing 1+2+3 ...N and summing 1 - 2 + 3 - 4 + 5 - 6 ....N"_
Not quite. The method using the weighting function has some desirable properties. Padilla has stated these in his paper on this subject (which is more scientific and less obscurantistic than this video). He cites G. H. Hardy:
_"Hardy has argued that any new summation method should satisfy three properties: regularity, linearity and stability._
_Regularity states that a summation method yields the known results for convergent series obtained using partial sums. [...]"_
Your above method would violate that _regularity_ requirement, other than the weighting sum method of the video.
Hardy and Ramanujan were at that time discussing suitable methods to define a "sum-like" functional for sequences which are divergent in the usual sense.
I'd hazard that arriving at -1/12 tells us more about these weighting functions than it does about the concept of infinity. So in a way we're still "sweeping infinity under the rug."
Yes but now we are doing it with a Roomba instead of a broom.
That would be my conclusion too. Introducing a weighting function of anything other than 1 or 0 is just sleight of hand, ie a deception
I thought that as well, but I think the idea is, that as N goes to infinity, these weighting functions all go to w(n/N)=1, so you're back at the original infinite sum. The only thing that wasn't really clear to me, is if there are weighting functions that go to a different finite constant. He says he reverse-engineered the cos(n/N) one, so is it possible to reverse-engineer a w that makes the sum go to eg. -1/42?
@@ComaVN the thing is, you're not "back at the original sum". fundamentally, infinite sums are not sums at all, they are limits of sequences. the original sum is a limit of N(N+1)/2. and that's not because we just chose to "cut it off", it's because that's the way the idea is formally defined. and that same definition is used even in these other regularization methods.
when you introduce the weights, you're taking some limit of infinite sums (which are themselves limits of sequences). you're changing the sequence under the limit, so you're changing the result. and yes, by the way, you could get any result you want. any of the methods that will give you -1/12 are basically just asking "what if we treated these sums as some other object". because the idea of "adding to infinity" is meaningless without a formal definition, we're allowed to explore alternatives like this. you just have to be careful not to conflate the different definitions.
@@ComaVN No either C=0 which makes the limit -1/12 or C≠0 and makes the limit blow up. There are no other finite limit only -1/12.
I understood this about as well as the original -1/12 video.
but the math in the first video was HORRIBLY wrong. You can't attach a fixed value to a infinite series that is not convergened.
I feel like I'm in a Numberphile speakeasy
?
@@codycastthis video was previously unlisted
That's a wonderful way to look at series!!
For anyone curious about the formula at 7:54, this can be gotten nicely using generating functions! For simplicity of notation, let A = e^(-1/N).
Let G(x) be sum from n=0 to infinity of A^nx^n. Then that inner term (the summand) can be written as (Ax)^n, so G(x) is a geometric series equal to 1/(1-Ax). Taking the derivative of the series G(x) gives a sum from n=1 to infinity of nA(Ax)^(n-1). So xG'(x) has summand xnA(Ax)^(n-1) = n(Ax)^n and setting x = 1 gives our desired sum.
So taking the derivative of 1/(1-Ax) gives A/(1-Ax)^2, and setting x = 1 gives A/(1-A)^2, which they rearranged as A/(A-1)^2 because the square cancels out the negative anyway.
That second one with cosine...I'd have to think on!
Actually, it looked so similar to the last problem and it made me think of that nice trick where you use Euler's identity e^(ix) = cos(x) + isin(x) to get info about a sum multiplied by cosine and sine at the same time. Then taking the real part gives the cosine sum. Let's try!
In this case, we let B = e^((i-1)/N), so that Sum(B^n) = Sum(e^(-n/N)cos(n/N)) + iSum(e^(-n/N)sin(n/N)).
Then writing G(x) = Sum((Bx)^n) gives two infinite sums C(x) = Sum(e^(-n/N)cos(n/N)) and S(x) = Sum(e^(-n/N)sin(n/N)), so that G(x) = C(x) + iS(x).
So our desired sum is again xG'(x) at x = 1, which is equal to B/(1-B)^2. Writing bar(z) for the complex conjugate of a complex number z, the real part of z is (z + bar(z))/2. Since bar(B/(1-B)^2) = bar(B)/(1-bar(B))^2, we can use this to calculate the real part C(1). And we immediately get S(1) as well!
i find this explanation so much more insightful than any other treatment of this topic ive seen before!
Great video!! It truly felt like this "coincidence" of weighting functions, unique pathways to the minefield of the infinite and particle physics is telling us something. Like a jab of the prospector pick, hitting straight into the gilded veins flowing from the inner most reality and up into our realm.
Yep. And this is all undoubtedly connected to the p-adics in some way. We just don’t know how. This is the profound mathematical and scientific mystery of our time.
@@LookToWindwardhow are the p-adics involved in this? Please explain a little more.
@@4984christian It is easy to represent and manipulate these sums in -adic notation. For example, the powers of tens sum mentioned in the video is just ….1111 in the 10-adics. Now I know 10 is not prime but that is the general idea. These infinite sums in the -adics act a lot like repeating decimals do in the ordinary reals (technically the reals are a subset of each n-adic number system.)
@@LookToWindward So you could represent the infinite series over the integers by a lot of infinite "power-series" over some p? Like the powerseries over p = 10 would yield 11111111...?
@@4984christian yes but the dots are on the left. Adic numbers can extend infinitely to the left just like the reals can extend infinitely to the right.
Great video! This video gives another beautiful argument why -1/12 is closely related to the series 1 + 2 + 3 + ... I think the problem with the first video is that it claimed that 1 + 2 + 3 + ... = -1/12 using an invalid proof. The value of an infinite series is defined in litterature as the limit of partial sums. This video basically changes the series to make it converge by adding weights and then compute it using partial sums. Mathematics is based on definitions so you cannot say 1 + 2 + 3 + ... = -1/12 without specifying what this statement means.
I really like this weighting function explanation.
So when you introduce the cut off weight function with a cos, then the signs of numbers that you will sum can be + and - like the sign of the cos and, thus consequence numbers can eat each other when u will sum them. For example: if the cut of factor for x = cos(x*pi/2)*(1/x) wich even doesn't use N (but who cares? in the end you will never reach the value N for it will we infinite) and you take this weightfunction on each x of the natural numbers and then the sum or those, you get the sum of this sequence 0,+1,0,-1,... what obviously doesn't go infinite but is either 0, 1 or -1.
It’s just the best youtube video of all time and i need to put this comment to be part of this wonderful work
seeing the phone cord at 0:22 worsened my asthma and shortened my life by 30 minutes
There is something very metaphysical about this episode.
That's a flaw, not a feature.
I personally don't want math videos to feel or sound like a horoscope.
New ideas always start off sounding like this. Calculus (as Newton and Leibniz knew it) sounded even more like a horoscope for over a hundred years; people like George Berkeley in 1734 criticized the whole subject as being nothing but the study of "ghosts of departed quantities." It wasn't until the 1800s that mathematicians like Weierstrass and Dedekind came around in the to formalize what was going on. You can't expect a new mathematical idea to sound like it came straight from a textbook the moment people start thinking about it.
@@japanada11 Anyone that believes the natural numbers actually sums to -1/12 is a fool.
Period.
Believe whatever fantasies you wish. The rest of us will be doing real math while you twiddle your thumbs with neo-sacred geometry.
It's because they crammed metaphysics into the video without any reason to do so, and in a very philosophically unsound manner.
What Tony says in the end is actually a request to *redefine* infinite sums, because the very definition (being the limit of a finite sum upto N, where N goes to infinity) has built this sharp cutoff in. Like there's a Cesaro sum etc, we will be having the Padilla Sum.
the "limit of partial sums" definition is much more ad hoc than it might seem because it's essentially mathematical induction used outside of it's domain of applicability. i.e. induction only works for a countably finite number of steps, whereas here we're DEFINING the value of the sum such that it keeps the induction valid past infinity. though ad hoc, this could potentially be a reasonable approach. but it turns out there are - arguably - better ways of assigning unique values to divergent sums, as shown in the video.
@@surrendherify induction is not limited to finite ordinals
@@toxicore1190 Sure but if you want to go "past" a limit ordinal (like the infinity of natural numbers), the proposition has to be proved separately at the "limit ordinal"th step, it's truth doesn't follow via the usual induction principle.
@@toxicore1190 when assigning the value infinity to divergent sums whose finite partial sums blow up, no such proof is involved, instead it's just a definition.
@@surrendherify Since when is induction limited to a finite number of steps?
The -1/12 thing is a product of assuming an uniform continuos space in an quantised discontinuous Spacetime I'm sure of it, like a technicality
I was screaming that this is exactly the basis for why renormalization works in QFT.(once I saw the power series come out of the sum) But I had no idea that only the regulators with convergent answers preserve the symmetry. I'm reading this paper for sure.
I like Numberphile. In other places, I sometimes don't understand the explanation. Here I don't understand the question, much refreshing.
As a layperson, I feel like the "weighting" was unduly glossed over. I would have loved to see what happens to the actual numbers in the series when you apply some of those complex weighting functions.
Kinda the point is, that, "in the limit", nothing happens with them:
n/N = 0 for 0
Thank you, but that leaves me even more confused. If that is the case, why weight at all?@@Kram1032
@@Kram1032 n/N is not defined as 0 here ( for 0
@@DonnieX6 one of the properties the regulator function has is that in the limit x→0 it is 1.
for x=n/N and N→∞, that means it's going to be 1 too, once you take that limit.
I kinda understand the idea that "chopping off" an infinite sum is effectively removing its most important property (that of the infinitely large tail), so after this operation, the "single value answer" is very far away from the original concept. I'm just confused on how the gentler limiters are applied without similarly shaping the result. Especially in physics etc, I would imagine that modifying the sum in any way effectively breaks away from the physical grounds it is based on. When applying these limiters, what properties of the infinite sum are you preserving to make the answer meaningful?
Not a full answer, but just vaguely intuitively, in physics, continuity matters a lot. I mean, modern (classical Netwonian) physics is literally built on functional equations (the Lagrangian). It's pretty much what calculus was invented for!
And that is all about continuous motion and what not.
So at least intuitively, it kinda makes sense, that doing things "smoothly" preserves more properties than doing it with a butcher's ax.
One very easy example that comes to mind is to approximate a circle on a square grid.
If you only follow the actual grid horizontally/vertically, you can approximate a circle's *area* arbitrarily well, while keeping the *circumference* completely constant at every step. (A "circle" of radius 1 drawn this way will converge to an area of π, but a circumference of 8 rather than 2π)
As soon as you introduce even the slightest bit of smoothing though (by allowing arbitrary diagonals to better approximate the circumference), you suddenly converge to the correct area *and* circumference. And all you did there was jump one level of smoothness, from C(0) to C(1) continuous.
In the proposed method, you jump from non-continuous to infinitely continuous (because you are using Schwartz functions)
I think this is kind of a similar situation to that.
I think the short answer is that when you first apply the limiters, it does affect the "value" that the series converges to, but if you look at their design, they all converge to weights of "1" being applied to every term in the sum as N (your partial sum cutoff) goes to infinity. So in the limit, the result is the same as the full unweighted sum itself.
It kind of reminds me of approaching troublesome values in a function from different directions and getting different answers from every choice of approach path. The variation of weight functions approach is nice because it replaces our brutish and discontinuous first intuition of weight functions with a smoother alternative. The big shining revelation here is that this huge class of approaches involving continuous weight functions all agree when it comes to their first order approximation term of -1/12. Even better, those that apparently respect the symmetries of QFT have no higher order terms and converge exactly to -1/12.
It makes physical sense that the way nature would opt to regularize its infinite potential processes on the quantum level respects the symmetries underlying the mechanics. Maybe something about the logic of our universe's laws allows infinite things to occur as long as the way those things are counted up give us finite, physically realizable outcomes. Buried in the thick of this particular series' infinite tendencies lies a finite portion that can be allowed to show itself if things are counted the right way. It's always there, even if the series diverges, it's just being covered up by infinity in those cases.
Really makes me think of residues in complex analysis. Your function can blow up to infinity at certain points, but if you integrate around those points (which in complex analysis equates to zero-summing the nicely behaved parts of your function) you can extract finite values that sort of "stick out" from the parts of them that aren't behaving so nicely.
Thank you for fighting the hate and discovering this exciting connection to sting theory as a result!
Always a pleasure to see the professor on the channel!
Thanks! Incredibly cool stuff!
Thank you
The smoothed asymptote parabola of the partial sums of the series 1 + 2 + 3 + 4 + ⋯ has a y-intercept of −1/12, as shown nicely on Wikipedia, for visual thinkers. That is *the* intuitive justification for "most people in the street" (17:45-18:15), and that's Level 1, if you will. Level 2 is: I'm a visual thinker, too, so I took this in and tried to fit that very same parabola, from scratch, using those very same partial sums, and the y-intercept came to not -1/12 but -1/8, so then I tried to contact the wiki image creator as to their method, and have not heard back. This deserves a more focused investigation for Numberphile's 2034 short & sweet conclusive return to the topic, or much sooner, please! :)
What's a visual thinker? Someone with functioning eyes and a functioning brain?
When someone takes an otherwise very simple idea and overly complicates it, you can be sure they are pulling the wool over your eyes.
Definetly one of your most interesting videos in recent memory!
This could turn out to be the first step in something big after all
{وَأَحۡصَىٰ كُلَّ شَیۡءٍ عَدَدَۢا }
[سُورَةُ الجِنِّ: ٢٨]
I remember back when I was obssesed about both the Numberphile video about -1/12, and Death Battle's video about Goku vs Superman.
Ten years later and they both revisited the same subjects
I don't feel like this weighting is fair. He draws the step weighting first (hard transition from 1 to 0) and then pretends that e^(n/N) is a smoothed-out version of that transition. It is not. He draws the graph as if its starts at 1 horizontally for n=0 and curves down when n approaches N and then it curves up again as n goes from N to infinity. Almost nothing of that is true. It odes start from 1 for n=0, but it already has a very marked negative slope regardless of the value of N (d(e^(-n/N))/dn = -1/N * e^(-n/N) which, for n=0, is -1/N), it never ever approaches horizontal at any point before n=N, and it never ever curves down (the second derivative is always positive so it is always concave up). So it is constantly diminishing and the rate of diminishing is greatest at n=0, at the beginning, not as n approaches N), by the time n is approaching N, tit already lost 63% of its weight. And, again, is not that it loses a significant part of that weight as it approaches N, most of the weight is lost in the beginning where the derivative has its minimum (i.e max negative slope).
So with that kind of weighting it is not surprising that you turn an initially diverging series in a converging one. Except that even with all that you don't. You still gen N^2-1/12 which as N approaches infinite it is still infinite - 1/12 which is infinite.
But wait there is more: Then he adds another weighting term: cos(n/N). He describes it as "this still goes between one and zero with some transitioning around N, it does a bit of wiggling around". That is profoundly deceptive, and I can only thing that it is intentionally deceptive. Why didn't he attempt to do a graph of this weighting function? Did he just forget to mention that, for any value of N, that weighting factor is NEGATIVE for HALF of the values of n?????? Little detail, hu? So it doesn't go between one and zero. It OVERSHOOTS zero and goes NEGATIVE... HALF OF THE TIME!!!!! So ware not adding NATURAL numbers with a weighting anymore. -1,000,000 is not a natural number. I wonder what would happen if you take |cos(n/N)| instead. I bet it doesn't converge.
cos of (n/N) can only range from cos(0)=1 till cos (1) which is 0.999.... it is in fact almost always 1. Halfway of N at 0.5 it is 0.99996 so i really don't understand what he wants with this.
@@YEC999 .... No. n doesn't stop at N. As you make n tend to infinity, n will take values of 2N, 10N, 9999999999N, tending to infinity times N. The argument inside of cos will be from cos(0) to cos(tending to infinity), so the cos will cycle between -1 and 1 as the cos function does.
@@adb012 You are right, i am not a mathematician, thought it would Stop at N!! But it really goes into the negativ!! i saw this yesterday later when i drew the function.
But one thing: I don't know if you mistyped, but it should be e^(-n/N)...
You're missing the point. If you pick a weighting that is a smoothed version of the step, you get -1/12+CN^2. If you smooth it out more or less, you still get the same -1/12+CN^2. If you create your own weighting function, you'll once again get -1/12+CN^2. That -1/12 is always there; it's just that, if you carefully pick your weighting function such that C=0, it's more apparent. That weighting function isn't *why* the sum is said to be -1/12, it's just a particular strange example of a weighting function that makes the connection clearer.
Try using a weighting function of f(x)=(1-erf(k(x-1)))/2 for some large k. The larger k gets, the closer this gets to the step function. For any k, the weighted sum will asymptotically approach -1/12+CN^2 for appropriate C. As k gets larger, it approaches more slowly, but it still approaches it. The -1/12 is always there.