"one way to use visual arguments to get at the proposed result" - I do appreciate the careful use of language here that doesn't overpromise that you've demonstrated a complete and airtight rigorous mathematical proof.
Abel and/or Cauchy Sums work well as an alternative to Σ_/infty. It does lead to to weird results, but I remember seeing some of those results and being surprised at how intuitive it is. E.g. The one so casually mentioned in the first Numberphile - 1/12 video, where Σ(-1)^n = 1/2 In some sense it just feels like the answer should be kinda between 0 and 1, since the parity of /infty is a philosophical discussion for another day.
@@lazenby5793 Because one of the basic properties of infinity is that if you subtract a finite amount from it, it stays the same. It doesn't get smaller and it doesn't get bigger either.
The Analytic continuation of the Riemann zeta function gives the value of -1/12 for s = -1. This does not mean that the summation of all natural numbers is -1/12 - the whole point of analytic continuation is that you are extending the function to the domain where the original function is not defined.
Precisely. I cant understand why people seem obsessed with ‘proving’ this result. In fact, I think these type of videos may do a lot of damage - it takes advanced mathematical knowledge to understand the context in which this result is correct (the meromorphically continued zeta function). Most people do not have nearly enough mathematical training to appreciate this context and will end up thinking that the sum of the natural numbers is a negative fraction. I blame Numberphile’s infamous video that started all of this… .😢
@@ralphhebgen7067 Yep, although I loved Mathologer's video picking apart Numberphile's video, so I would highly recommend that if you haven't watched it already!
Before watching the video: There's no way he can convince me but let's see what he has to come up with After watching the video: YOU DIVERGENT SON OF A-
@@MathVisualProofs it's a great video and you do emphasize that this only works in a certain context, but I really struggle to accept the fact that you move around the A^2 sum terms for one specific reason : in classical real analysis, you can make a semi-divergent series equal anything you like. With A^2 being 'even less convergent' than a semi-divergent series, I struggle to accept this visual argument, in the sense that we could probably make a visual argument to make it equal a lot of other results It's still a great video, don't get me wrong !
@@guillaume5313he has done it cleanly without “getting any value he likes” - he has correctly calculated the “residues” of the divergent sums. I used this same argument to defend my physics thesis in 1990. i abstracted from -1/12 to solving divergent integrals. This math was necessary to solve the Schrödinger equation for a proton orbiting INSIDE a helium nucleus as a model for the unstable isotope Li 5. The solutions to the “divergent” equation gave the decay times and radii of the resonant state that existed for the few femtoseconds thus thing held together before the coulomb force overcame the strong force and ejected the proton.
@@idjles that's great but it's besides my point The "getting any value he wants" part refers to a proven theorem in mathematics that is presentes in this video : ruclips.net/video/U0w0f0PDdPA/видео.htmlsi=al9GeWj4KE8cHkxo
@@rr.studios Zero and infinite are the only numbers that makes sense to Russell’s paradox. A set containing zero sets in of its self is zero which means the nothing sets also contains the same. nothing. Same for infinity.
@binbots but does the set of all sets contain the sets of all sets that contains all sets that contains all sets that contains all sets that contains all sets that doesn't contain itself.
You also get -1/12 when integrating the equation y = x(x+1)/2 evaluating over [-1, 0]. That's the formula for summing the natural numbers, but using a continuous curve rather than the discrete values in the normal N summation. I don't think that's a coincidence, but I don't know how to prove otherwise. (You can see for yourself by using a query of "integrate x(x+1)/2 from -1 to 0" in wolfram alpha.)
6:37-6:46 Are you allowed to do this? My maths is rusty but I vaguely recall that changing the order of terms in an infinite summation changes the value it limits to. (Or maybe that's me digging into the whole reason this maths is weird 😂😂)
You are correct. It's even possible to get any conditionally convergent infinite series to sum to *any* finite number by changing the order of the series. These series aren't conditionally convergent (they're divergent, which is the point), but the principle is the same. He would have got a different answer if he'd changed the order in a different way, and there are an infinite number of possible reorderings. I appreciate this video as a clever and creative magic trick, but I don't like the fact that he's left some people thinking that the magic is real.
By shifting he cut square into 2 triangles, moving 2nd one off screen and ignoring it for the rest of calculations. By shifting it more, you will get enen bigger part of square to ignore.
@@kicorse Thanks for the confirmation! I do think it's fair to say that he worded 9:42-9:46 very deliberately to avoid the implication that this is a concrete proof, but thanks for the explanation 😊 (And I thought that you could make them reach any number too, that's so cool that you can!! 🤩😂)
@@kicorseyeah. Since on video, shifting the columns is akin to ignoring future element alignments in the sequence, it can be manipulated by hiding elements in choice. One example I found to find out that the sum 1+2+3+4+5+..... Amounts to -1/6 and not -1/12 was also by manipulating the sequences in addition in some inherently flawed way. I will try to explain it here. Let S=1+2+3+4+5+...... Therefore S=1+3+5+7+9+..... +2+4+6+8+10+..... S=2(1+2+3+4+5+6+....) +1+3+5+... S=2S +1+3+5+7+.... So -S=1+3+5+7...... Now, we can do a similar trick. -S=1+3+5+7+9+11.. -S=0+1+3+5+7+9..... ------------------------------------- -2S=1+4+8+12+16+20... -2S=1+4(1+2+3+4+...) -2S=1+4S -6S=1 Therefore, S=-1/6. Yeah, realigning of elements of a sequence is a free way to modify the final answer, with no definite proof against it.
Using the same logic at the end you can argue it equals whatever number you want. When you are dividing an infinite number into different groups, you are always going to end up with equal groups.
@@valtersanches3124 Yea, so why do so many youtubers pretend it always equals that same number? The infinite hotel problem is a good way to think about it as well.
@@MegaLokopo It's not about "youtubers pretending" anything. It's actually used in theoretical physics (the number of dimensions needed by a particular string theory is an example of this result being applied). It's not about "summing" integers anymore, more like associating a number to a formal infinite sum, and there are multiple perfectly valid ways of doing this consistently. Convergence is just one of them, and it doesn't always hold, that's all. This particular result can be made quite natural with the use of regularization, Terence Tao wrote about it (what he calls smoothed sums). The way we evaluate a sum as a limit of partial sums is just one choice of "cutoff function" that happens to produce discretization artifacts that blow up at infinity. There are more sensible choices. The guy showed in what he calls "smoothed asymptotics" that the -1/12 is just the finite part of an asymptotic expansion of the regularized sum: S=-1/12+CN^2+O(1/N), with C vanishing for a particular class of cutoff functions. That's how you get rid of infinity. Again, very typical in quantum physics, where getting rid of infinite non-physical quantities is absolutely needed and can be made rigorous.
@@MegaLokopo Because they don't understand math, same as this guy. Even I could see the gaping holes in this logic and I have no formal mathematical background. I'm still under the impression that university doesn't necessarily teach you anything. You're under such immense pressure that all you have to time do is rote memorization. That's not understanding, however. That's not to say that bright people don't learn at the same time, just to say that just because someone has a qualification, does not mean they have any understanding of what they're parroting.
@@MegaLokopo I know, it is so dumb; they use a pseudo-justification to get 1 - 1 + 1 - 1... to equal 1/2, but the limit of that does not equal that; it can equal 0, or 1/3, or .0343423, or literally anything. But really all we need to know the proof is nonsense is that 1 + 2 + 3 + 4... clearly diverges and thus does not equal any finite number; we can also arbitrarily 'define' it to equal anything we want
This reminds me of the (in)famous Numberphile video on this topic, except that you actually give some (admittedly handwavy) justification to say Grandi's series sums to ½. Subscribed!
Yes, this is similar to the numberphile video (linked) and has some of the same problems, but I wanted to see the argument visually. Thanks for watching!
5:32 Not sure if this is allowed. Visually, it seems to converge to 1. Graphically however, it's undefined, as both limit from left is different from limit on right. If you want to do that, you have to make sure that the diagonal shifting from the outer side also converge to that same point.
This was freakin awesome and deserves more views! This is now my favorite way of trying to explain this to someone - especially those who don't understand math so well
Umm ik people know this this through Numberphile and BIgmathguy btw For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined. It's bit controversial topic for long time .. 😅
@@MathVisualProofs You do realize that a divergent sum can literally equal anything, yes? And that this isn't a proof of anything? You just defined 1 - 1 + 1 - 1... to equal 1/2, but one can make it equal any value one wants, because it diverges. In this way, you literally just defined 1 + 2 + 3... to be -1/12
@@pyropulseIXXI "this isn't a proof of anything". No shit. It's a visual proof. And he's not saying 1+2+3+... = -1/12. He's giving a visual interpretation of common arguments. Obviously the sum diverges. But that doesn't mean that the result is completely useless.
1:01 When You do S - N. You add 2 for subtracting 1 from S, similarly, add 4 for removing 3 and so on.. which means S - N = S + x where x = 1×(number of total terms) Which is not equal to 4S because the number of terms are still same in both S and N so the operation S - N can not increase it by additional threefold
A=1/2 is completely wrong. It's a divergent series with infinitive numbers of+1 and -1. You can make it equal to anything by carefully grouping the terms. e.g. to make it equal to 3 group terms like this (1+1+1+1-1)+(1+1+1+1-1)+...
let me ask a question: why is your standard for summable series given by limits of partial sums? can you prove that this is the only sensible way to think about infinite summation? another remark: the exact same logic can be used to say that 1 - ½ + ⅓ - ¼ + ... is "completely wrong;" this is precisely the content of riemann's rearrangement theorem.
The sequence: 1-1/2+1/3-1/4+1/5-1/6+1/7-1/8+... This can be rearranged to make any number. It still has a value of ln(2). Ask any mathematician. Any at all. Even ask fifty. They will say it has a value, or be indifferent due to not caring about this branch of math
@@MichaelDarrow-tr1mn there is a difference between absolutely convergent and conditionally convergent. Consider series 1-1+1/2-2/2+1/3-1/3... it has value 0 if you don't rearrange terms, and if you are allowed to rearrange you can make any value. A series is absolutely convergent if it is convergent even after making all negative terms positive. If that series diverges then the original series is called conditionally convergent and it won't have any fixed value.
Divergent series are the new "imaginary" numbers. We must accept them and start learning how to work with them, what can or cant be done, instead of ignoring or avoiding it. Great video, congrats
They are not the "new imaginaries" infinite series have been well studied for over a century and a handful of results been proved with careful definitions. Accepting these results in such a naive way only leads to an incomplete picture of what they really meant in the first place and in consequence making people think that the sum of natural numbers is a some negative fraction
The problem being that once you assign an imaginary root "i," it only works in one way, and you can figure out exactly how that one way works. Different attempts at assigning values to divergent series don't work the same way. A bunch of different methods of assigning value to "1 - 1 + 1 - 1 + 1 - 1 + ..." all agree on "1/2," but more complicated series will get different results when you try to assign a value to the same one in different ways.
Lmao what? People already know how they work for hundreds of years. You can literally make a divergent sum equal anything you want. They aren't the same as imaginary numbers at all. This proof is false because the limit of 1 - 1 + 1 - 1... does not equal 1 - 1 + 1 - 1... All he did was arbitrarily define 1 - 1 + 1 - 1 to equal 1/2 and used a pseudo-justification to do so With this same 'stupid' logic, you can set 1 - 1 + 1 - 1... = 1/3 and it would be just as valid as 0.
These aren't proofs; any divergent sum can be made to equal litearlly any number; he just arbitrarily choose numbers to get to -1/12. Ironically, he can just skip the proof and just arbitrarily define 1 + 2 + 3.. = -1/12
The first bad math was trusting the limit from one side. The second was assuming you could slide the columns of the square grid into the infinite triangle without altering that sum. Pretty sure there's a flaw in the S-N=4S argument too, but not as sure, maybe I'm just sleepy
(6:43) shenanigans! That shifting trick is like reordering the terms of the sum, as used in other Σn = -1/12 videos. That means you have a completely different series of partial sums which you're taking a limit of. It's like if you have f(a, b) = a ^ b, which is undefined when a and b are zero, but you can take the limit; but you get a different limit depending on what direction (a, b) approaches (0, 0) from. The limits are not interchangeable.
If instead of shifting the columns like this, we take the diagonals (which is essentially the same thing), we get the same thing… And like, if the terms were coefficients of a power series, then the diagonals would be the coefficients of the same power. So, I think it would be the most natural way of converting the product of two series into a single series? Like, if I have two formal series (sum_n a_n) and (sum_n b_n) I think the product should be the formal series (sum_n (sum_{k=0}^n a_k b_{n-k} ) So, like, \sum_{n=0}^\infty (-1)^n x^n is (1/(x+1)) (for |x|
Ah, I think I maybe see where I made my error? Or, kind of? First of all, I had an error in my power series for (x+1)^{-2}, it should have been The series \sum_{n=0}^\infty (-1)^{n} (n+1) x^n . If you substitute in -x for x in that, you obtain \sum_{n=0}^\infty (n+1) x^n for |x|
In some sense, 1/2 begins to describe some kind of ratio regarding the number of negative and positive numbers in A. Whatever the definition of that ratio is, I can see the argument following from there. But that is not saying that the sum is -1/12. Rather some aspect of that sum can be numerically described.
1/2 doesn't describe anything; it is an arbitrary assigning. 1 - 1 + 1 - 1... can literally be made to equal anything; 0, 0.324234, 1/3, 1/17... literally anything. It is arbitrary. Which means it equals nothing, because it diverges
I think this video actually shows that if the Grandi's series is convergent, the sum of all natural number also is. Some still get mad when we talk about those series convergence, but it is just a minor mathematical artifact. Chose your axioms and you can decide if the series diverges or converges. However the real question is not if the sum of natural numbers is convergent, but why -1/12 pops up in so many different ways. Do the mathematical operations that we perform have a underlying equivalence and we are always just getting the same "wrong" result? Or is there a -1/12 hidden behind the infinity? Great video nevertheless!
"I think this video actually shows that if the Grandi's series is convergent, the sum of natural numbers also is." This isn't true. Grandi's series coverages in the Abel sense, but 1+2+3+... does not.
@@martinepstein9826 Abel summation does not prove conventional convergence. If a series is convergent any mathematical operation of the series can be replaced by its converging value. If Grandi's series converges, the alternating sum of natural number converges because it can be obtained from the square of Grandi's series. The positive sum can then be obtained from the alternating series if it is convergent. In conventional axioms, 1-1+1-... is undefined, but the video shows a valid link between the series.
he arbitrarily defined 1 - 1 + 1 - 1... to equal 1/2, but it can equal anything you want. In this way, I can skip the unnecessary nonsense and litearlly just assign -1/12 to 1 + 2 + 3.... and the proof is done via 'definition.' This is how nonsensical it is; he defined 1 - 1 + 1 - 1... to equal 1/2 to get his 'proof,' so it is a proof by definition; skip all the nonsense and just define 1 + 2 + 3... directly and get -1/12
@@pyropulseIXXI You forgot about the fact that this is a visual proof. Visual proofs are not meant to be formal proofs. They just give a visual interpretation of something that is already proven.
It's interesting: the alternating sum 1 - 1 + 1 - 1 + 1 - 1 + 1 -..... doesn't converge as a sequence of partial sums. But I guess the idea here is that it does converge as a sequence of infinite series 1 - r^2 + r^3 - r^4 .... if you let r approach 1 from the left. It's like, depending on the order / approach of trying to make the series converge, it either does or doesn't converge. I'm not sure I'm articulating this well.
@marcevanstein This is Abel summation, which can also be used to define 1 - 2 + 3 - 4 + ... as ¼. (I used to be somewhat disheartened whenever I found out that my latest 'discovery' was already known. Now, I'm at least somewhat comforted by the realization that it was significant enough to be written down and remembered.)
-1/12. "Can it be demonstrated on the Cartesian plane?" The sum of all positive Integers up to N is traditionally represented by the equation f(n) = (n*(n+1))/2. Plot the curve represented by this quadratic and one will observe a parabola (open upward towards infinity) which slightly crosses below the x-axis at points A =(-1,0) and B=(0,0). The integral area below the x-axis between points A and B = -1/12 or -0.08333. (Also, Euler's zeta(-1)). One can visualize a finite part below the x- axis and an infinite part above the same. The validity of this construct can be argued in a similar manner as to how we explain the real and imaginary parts of complex numbers.
With regards to your last statement, are you saying the finite part below the x-axis is akin to the real part of a complex number, and the infinite part above it is the imaginary part? Or is it the other way around?
@@Fluoman_ f(n) = (n*(n+1))/2 will output finite sums of positive integers from (1 to n). An infinite sum of (-1/12) is absurd. I find the integral under the x-axis a curiosity. Personally, it helps me make sense of a perplexing topic. I propose no real relationship.
@@MathVisualProofs I hope you are just playing this up but actually realize how absurd this argument is. It seems people in the comments are taking this way too seriously, when you literally just arbitrarily defined 1 - 1 + 1... to be 1/2 I consider this to be spreading misinformation, because the sum of the naturals does not equal -1/12, ever. One can 'make' a divergent series equal any number they want, so it is just an arbitrary assigning. So just because there are infinites that exist, and in some situations, it would be nice to have that infinity not exist, that doesn't mean the sum to -1/12 is useful. Why not make it sum to -2? What if I need it to do that? It would be just as useful
A divergent series is, by definition, divergent. With a proper rearrangement, we can obtain any value for ( S ), ( N ), and ( A ) in your examples. The point is that these series do not converge to any specific number.
I know; this is funny, because he could've skipped all the pseudo-justification and just assigned -1/12 to the infinite sum of the natural numbers; that is literally what he did anyway, just in a round-about way
@@pyropulseIXXI There are a lot of people happy to share their negative opinion in these comments but you must be the happiest! The point of -1/12 is not that it's unique. We can obviously get whatever we want by manipulating the divergent infinite sum. The point is that we get this particular result through analytic continuation of the Riemann Zeta function. This video is trying to give a visual intuition of that fact. I assume you know this, since you seem to have a mathematical background, so I don't understand what your whole comment crusade is about. Please, enlighten me
Just for fun. As we made infine half triangle from square, we can make square from triange of S just bring every segment to upper limit. So now Infinity ^ 2 = S = -1/12 So infinity = √ -1/12 = i/2√3 😂😂😂
this is the most contentious math topic... fun video, i like having a visual intuition for something that is not correct. i think it is interesting that you can do seemingly normal adjustments to the objects and get a bogus result, like in the shifting triangles and rearranging terms
Imagine getting more money everyday, but you end up having a debt of 0.50$ at the end of your life. How can you lose money by earning when you didn't even spend it 😭
That one is how you multiply two infinite series to get the product. But of course it’s suspicious for divergent series (works perfectly well for formal power series). And the Abel sum of that second series is 1/4. The very last one is the most suspicious to me and not really justifiable but just there for fun :)
Is this the equivalent of mathematical sophistry? (Edit: your video is magnificent. Seriously well done and clear. I was just referring to the -1/12) :)
Nice. Really good job. Obviously there are some big hand waving, non rigorous arguments made here. It is essentially a version of the numberphile proof, which has some technical problems. But all this opens up some really interesting topics around infinite sums and convergence.
This is exactly the reason why real analysis needed to be formalized using proper definitions lol... Is quite interesting for sure but definitely misleading.
I agree with your point that this demonstrates why it is necessary to formalize things carefully. However, I don’t know about “misleading”? It suggests that there may be a reason to assign the value (-1/12) to the series, and, there *is* a reason to associate it with the value (-1/12) for some purposes. It just isn’t the sum in the usual sense (as it diverges). I guess it could be misleading in the sense that one mistakenly might take it to indicate that (-1/12) *is the sum* of the series? But, in any case, yes, does show importance of formalizing.
My weakness has been advanced mathematics and it started from calculus and binomial theorem. My hold on normal maths is excellent, I was too lazy to learn it.
The biggest problem is the possible invalid use of the equals sign (=). I guess you can define 'equals' as anything you want. However, IMO, equals means "exactly the same as" and when a non-convergent series of numbers is presented it is simply invalid to put an equals sign anywhere near it.
Do you think the reason that 1-1+1-1 etc comes out to 1/2, is because the "last digit" has to be either a 1 or a -1, and since it starts with 1 and repeats -1+1 then the sum would either be a 0 or a 1, so the average is just 1/2? Is that essentially why it gives that number?
As a programmer, the idea of adding up positive numbers and ending up with a negative result isn't that weird. It happens all the time, and we call it integer overflow.
@@tomkerruish2982 I feel like you people don't understand anything when you say things like this and think the result is quite insane. It is just turning the LHS of the decimal to 'behave' similar to how the RHS of the decimal behaves in 'normal' numbers (have the LHS have infinite digits) So it is litearlly just a different 'notation' for numbers. The summation of 1 + 2 +4 +8 + 16 +... never equals -1. It is like getting up in arms about 0.333... = 1/3 The p-adic numbers are not the same as real numbers.
@pyropulseIXXI I believe you misunderstood my post. I never claimed the result is 'insane'; I was simply pointing out a fact about the 2-adic numbers (which I never claimed are identical with the reals). In the topology of the 2-adics, the sequence of partial sums 1, 3, 7, 15, 31, ... does indeed approach -1 as a limit. I was never 'up in arms' about this result; I simply thought the original poster might be interested in this nugget of mathematical truth. And yes, this video uses reasoning that could be called questionable if one were feeling exceptionally generous. It would've been better to introduce the idea of regularization; however, that wouldn't've been very visual.
Me: So, if I add like a few thousand consecutive integers, from one up, I will get a negative outcome? Math person: No, you won’t. Me: Okay, but if I add like the next one million consecutive integers, at some point the result will turn negative? Math person: No, it won’t Me: Well, but if I add like a few billion more, maybe? Math person: Nope, still not negative Me: I see. And a few trillion more will not help either? Math person: Correct. Still a positive sum. Me: Hmmm. I’m getting worried. Since the numbers we’re adding are getting bigger and bigger, we’re only getting further away from a negative outcome. Math person: You’re right Me: But maybe after adding quadrillion times quadrillion times quadrillion integers, and then a few more? Math person: No. Still not negative Me: Wait, but at what point will the sum finally ‘magically’ flip from positive to negative? Math person: Only at infinity Me: And how many numbers do I need to add before I arrive at infinity? Math person: You won’t arrive there. You have to keep adding numbers forever Me: Forever? Then when will I arrive at infinity? Math person: Never Me: I get it: the sum of positive integers NEVER adds up to a negative number. That’s all I wanted to know.
Notice: Midwits are swarming the comment section. Yes, we all know the series doesn’t converge, and it is stated clearly in the video from the beginning
"I know it's divergent, but I don't care as long as I get the answers I want" Has the same energy as "I know the experiment shows the Earth is round, but I don't care as long as I wanted to believe it is flat"
According to the Abel summmation and Cesaro summation processes, that is a value you could assign to the divergent series :) But I agree it doesn't make much sense from the real analysis point of view.
6:40 "we can take each column and shift it down. In particular we shift column i down i spots." Why can we do this? Why doesn't it change the value? I don't understand this and frankly I don't believe it.
He was computing the product of 2 series. In standard calculus, we would use the cauchy development technique to do so to regroup terms of the same degree which is the same thing as what he did by "shifting column i down i spots". I still don’t understand how that works tho. Maybe lower degree term tends to have more importance than higher degree ones which would make sense if it is supposed to converge
Umm ik people know this this through Numberphile and BIgmathguy btw For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined. It's bit controversial topic for long time .. 😅
This kinda reminds me to that "You can't shoot a turtle with an arrow and a bow" line, because both of them make 100% sense when breaking down, but still have next to zero logic to them
the sum of all integers diverges, it is not -1/12. This should be made clear. One can re-arrange elements on infinite sums only if they converge but not if they diverge.
All comments like this miss the point of extending mathematical ideas. Limits of partial sums is only one way (also the standard and most common) of assigning values to infinite sums. In that context, it diverges. Using other summation methods, which is a perfectly valid thing to do, it converges. Also, you can rearrange terms as much as you want. The question is what might happen afterwards and that may depend on the summation method being used.
the very first thing I say in the video is that the series diverges using standard real analysis techniques. There are real regularization techniques that say -1/12 makes some sense in other contexts, so this is just a visual way to get to that answer.
@@MathVisualProofs sure, there are regularization techniques but rearranging the terms like you do at 6:40 is illegal under all of them. This proof technique is invalid and unsalvageable. That it comes to -1/12 is just sleight-of-hand.
@@MathVisualProofs right, I skipped through parts of the video while I should have watched the whole thing. Nonstandard approaches of infinite sums can lead to different outcomes. What we want from math, I would think, is consistency. Sum of all integers have a connection to -1/12, like with the analytical continuation of the Riemann-Zeta-function. Some particular approaches of nonstandard summation can also leave to have the sum of all integers being -1/12 but there is a lot of handwaving there, or having to use Abel's summation like in that video, which in my point of view is not really the sum, just a particular way to compute averages instead of getting to the real answer which is a divergence. I am aware that I am a part-time math hobbyist at best, my opinion is not important.
I'll still never get this. If you're calculating the sum of aleph-null, it's divergent. It's addition. Why do all those other calculations? They have no business adding aleph-null. It's transforming something without need.
All is well until the bit from 5:40 to 6:00. But however you dress it up, Abel summation or not, you cannot assign the value 1/2 to the divergent series 1 - 1 + 1 - 1 ... Everything thereafter is built on a fallacy. You might as well base the argument on dividing by zero.
7:25 eh, yeah, logically A^2 _should_ be 1/4, but as a visual proof that checkerboard looks just as much like 1/2 as the alternating sum A did, so just as S - N = 4S, A^2 should equal A 🤔
ok didn't know that this result -1/12 is actually in the plane of complex numbers... which means that -1/12 is not a correct representation, the correct representation would be i^{2}/12, but then there is no mystery is then, is there?
Another reasonable explanation is simply that A = 0, and N = 0. Because it's also very sensible to be able to group these infinite, alternating sums into groups that equal 0, and it doesn't look like you're completely trying to game the system while doing so. And then, A^2 = N ; because 0^2 = 0, which still checks out. Then, S - N = S still, because N = 0. And all the fuzzy math breaks down there. Even if you staunchly refuse that A and N are zero and they must be some small finite number at least, at 8:00 the visualization is STILL wrong. Just because you take two pyramids of blocks, invert half the blocks in one pyramid, and then add up all the positive blocks with positive and negate the others, what you've created is a shape that LOOKS like it's 4x S, but it's not really the case, because you haven't been keeping track of how fast your infinities are growing. From another perspective, it's much more obvious that what REMAINS of S, you've DOUBLED. And then, keeping track of the pyramid rows, you have HALVED the number of rows in your pyramid. So, HMM. Double row lengths, but half the row amount ... looks like a simple case of 2x * 1/2 to me, which is just 1, so S - N still = S, which again makes sense if N = 0. Alternatively, yet another way to look at it is, if S is absolutely infinite and unbounded, and we subtract zero or a small finite number from it, you have not affected S in any significant way whatsoever.
Oh man, so you didn't like the visualization of the alternating geometric series from the Viewpoints group? Or the visualization of the Cauchy product of two formal series? I definitely agree that the last part (the trickiest in the argument because it has the most problems) is just using boxes instead of numbers :)
I agree until you used the triangle shape by moving and infinite matrix into a diagonal form (I will say it has a 45º angle down from the horizontal, for simplification). It was arbitrary and you could had arranged it into an infinitely different arbitrary angles, since we are talking about an infinite matrix. You'd have to justify why you picked up the 45 angle.
The fact that the sum of all positive integers can be made to look like -1/12, is proof....Proof that if you overthink anything hard enough, you can attain from even the most commonplace of things, absurdity.
The fallacy comes from the fact that nobody proved that S is not infinite, and infact it IS infite, so while this is true: S - N = 4S, N=1/4 This is also true: S = inf and if we substitute in: inf - 1/4 = 4*inf inf = inf Which proves that S = inf
Good visual proof, but this still is no better than pluging in values into the formula to arrive at the solution, that is essentially what we have done here. The problem is, the alternating sums are convergent series whereas the sum of natural numbers is divergent for which we cannot apply the method presented here
This is the first time I've felt able to put an objection to this result into words. I understand that your video is focused on putting a visual argument to the result, not necessarily primarily advocating for the result itself. And it was a fascinating approach. Anyway, here's my problem: When looking for the sum of (-r)^n, assuming that the sum is finite, a limit gets taken as r approaches 1 from below. That is fine. But my objection is then taking the limit to be the value of the series when r=1. In calculus class, when limits are first introduced, we are *very* careful to make sure that we all understand that just because the left- and right-side limits of a function or expression are equal, that does *not* automatically imply that f(c)=L. I've seen this argument here, and I've seen other videos that just want to "average the partial sums" of the 1-1+1-1+... alternating unit series, and none of that feels sufficient to give the geometric series a value of 1/2 when a=1 and r=-1. Just a few statements prior, we had established that the 1/(1+r) convergence general formula was only valid for 0
Thanks! The thing to look at for more justification is "Abel summation" (there are a couple of things called this). For convergent series, the Abel sum will agree with the limit of the sequence of partial sums, but there are other series that are "Abel summable" that give different sums. If you do the Abel summation on the divergent series 1-1+1-1+... you do get 1/2, but again, this is a different type of summation and does not mean that the series really equals that value. Instead, it suggests that if you want to assign a value to that series, then 1/2 is perhaps the best choice.
@@MathVisualProofs Thank you. Took me the better part of the day to find what looks like the right wiki hole. But I think the theorem is here: en.wikipedia.org/wiki/Abel%27s_theorem The Remarks section even talks about this 1-1+1-1+... series specifically. In the Theorem section, it says, "Suppose that the sum from 0 to infinity of a_k converges". Does this mean that convergence of the infinite series of coefficients is a required prerequisite to application of the theorem? Or does "Suppose" mean that "we'll assume for now that the series converges, and if we can find a good plausible value for it through other principles, then great for us"? The Remarks section, about the alternating unit series, it says "the claims of the theorem may fail", and it seems like that's when the series of a_k diverges (violating the "Suppose" bit?) because of oscillating partial sums, rather than because of partial sums heading to infinity. Furthermore, Alexandre Eremenko at Purdue has this proof on his university website: www.math.purdue.edu/~eremenko/dvi/abel.pdf Eremenko seems to regard convergence of the series of coefficients as a requirement for applying Abel's theorem to a power series. He uses "Let" to establish the converging series of coefficients in the theorem statement, rather than the "Suppose" found on Wikipedia. To me, "Let" reads like instantiation, like "If you want to apply this theorem, you'll need a thing that already has these well-established properties before you even think about getting a result." He also uses the alternating unit series as his last Exercise, but if I'm reading it correctly, he states that the theorem doesn't apply to this series. So, what's the most mathematically correct way to read this theorem in order to sort out prerequisite conditions and consequent results? So sorry to keep bothering you in RUclips comments, as if I'm at your office hours or something. I don't have enough graduate degrees to credibly argue with Math RUclipsrs. That's probably why I need so much help getting on board with this particular result.
@@MathVisualProofs You can litearlly make 1 - 1 + ...1 to sum to anything; I can create one that makes it sum to 1/3 and call it the 'pyro summation.' It doesn't suggest that 1/2 is the best choice at all. It suggests that if you do it that way, then it equals 1/2. There is no 'best choice' This is just spreading massive misunderstandings of mathematics to math ignorant viewers
@@emmeeemm So the idea there is that if the series is convergent, then the abel summation will give the right sum. The idea after that is that you can take the consequence of that theorem to somehow assign a value to an infinite series (even though the series diverges). Typically, if you want to try to assign a value to a divergent series (which you really shouldn't do), then you want to use one of the techniques that will agree with convergent sums. As you found in the notes, you can assign an Abel sum to 1-1+1-1+... and 1-2+3-4+... and you get 1/2 and 1/4, respectively. It turns out you cannot do the Abel sum of 1+2+3+4+..., and that's why, to do it like in the video, I had to resort to the arithmetic idea (which isn't really a good idea). There are other regularization techniques that give 1+2+3+4+... = -1/12, so this one isn't as misleading as it seems. Though again, it is really a weird world when you start trying to assign values to divergent series, and the rules should be made more clear than I had time for in this video.
@@MathVisualProofs "[...], and the rules should be made more clear than I had time for in this video." Respectfully chap, the time that you have now set aside to excuse your video's non-thoroughness in this comments section has probably eclipsed the time it would have taken you to write this explanation into your script. This is a quite a shame, because from what I am reading in this comment section, you possess a great deal of understanding and insight into the topic. Your video lives a bit short of that. (but not by much, as others in this comment thread would like to suggest).
9:22 But what does this even mean?... And also, remember, in order to arrive at this conclusion, you had SUBTRACTED N from S - as in S-N - so that isn't adding all of the positive integers!
numbers with different infinite numbers of digits are not equal. 111....111 is not 222....2222. In base2, 111....1111_2 = -1. in base10, 999...999_10 = -1. It happens to be that 1+2+3+4+... = 1111....1111_13. ie: in base13, it's all 1s. S = 1 + 13S = -1/12. Once you are at peaces with T = 1 + 2 T, that it solves for -1, and expands as powers of 2; you can no longer object to "a sum of positive numbers being negative"; because there is no way to make a contradiction by iteratively performing this sum. It's the same when you get a negative fraction. -1/12 is just a thing that if you multiply it times 12, and add 1, you get zero. And there are objects with an infinite number of digits that act like this. Saying that any sum is "equal to" infinity is nonsense. Because firstly, infinity isn't a value. And it's more that if you try to do the sum iteratively, it never stops growing. But if the final sum is all 9 digits, that's completely different from if the final sum is all 5 digits: X = 5 + 10 X Y = 9 + 10 Y Two different "infinite values". But they straightforwardly solve for finite sums as well: X = 5555.....5555 = -5/9 Y = 9999.....9999 = -9/9 The sum for powers of 2 summing to -1 is very very easily done algebraically. It's more difficult with the sum of integers because these steps that he did visually; you can use completely finite arithmetic with recursion to prove it. It requires a differentiation to produce 1 + -2 + 3 + -4 + ..., but you can do it with no references at all to infinity. If you have a sum in mind, if you know S and TailS, then you can solve for the finite formula SumS. If you only known SumS, you can use S to solve for TailS: // S isn't the real SumS. (S-TailS[n])=SumS[n] // n isn't "infinite", it's just the number of terms expanded that you want, and it can always be higher S = SumS[n] + TailS[n] = SumS[n+1] + TailS[n+1] SumS[n] = (n(n+1)/2) ... // discover that S is constrained to be -1/12 ... // solve for TailS[n]
The digression from 2:00 was not very useful. Look at the columns of the third picture, they are just the middle pictures. Each column starts alternatively with white box and black box. Using the limit of r approaching 1, you already showed that the middle picture is 1/2 of the starting box. So the first column of the third picture is half of white box, and the second column of the third picture is half of black box, and third column of the third picture is again half of the white box. So the entire third picture is again half of the middle picture.
7:10 what if you shift the layers down twice as much? wouldn't you get 1-1+0+0+1-1..., and removing the zeros you get 1-1+1-1..., which is supposed to equal 1/2? Wouldn't that mean 1/4=1/2?
This is an interesting note for sure. But the checkerboard-to-triangular array is the standard, natural way to multiply two formal power series, so it is the normal way to interpret the infinite product here... yes, still a stretch to do this with divergent series :)
Actually if you don’t remove the zeros you can get 1/4, because the partial sums are 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0… and if you take the mean you get 1/4. This is Cesàro summation, a way to assign a value to infinite sums such as this (it can also be applied to 1−1+1−1…). It’s quite wierd that adding or removing zeros can change the result though.
"one way to use visual arguments to get at the proposed result" - I do appreciate the careful use of language here that doesn't overpromise that you've demonstrated a complete and airtight rigorous mathematical proof.
Yes. This one is a stretch, and I know some people really get up in arms about visual proofs 😀
He still did have to use Abel summation
Which is less than airtight
Abel and/or Cauchy Sums work well as an alternative to Σ_/infty. It does lead to to weird results, but I remember seeing some of those results and being surprised at how intuitive it is.
E.g. The one so casually mentioned in the first Numberphile - 1/12 video, where
Σ(-1)^n = 1/2
In some sense it just feels like the answer should be kinda between 0 and 1, since the parity of /infty is a philosophical discussion for another day.
@@TheFabledSCP7000and doesn’t work for the actual last sum :)
Cept that the Ramanujan sum assigns a number to an infinite series and convergence doesn't so it's correct since it's the only closed function defined
This man can do visual proof of anything
Hah! This one is a stretch :)
Now i want a visual proof of why i can't get a girlfriend 🥺🥺🥺 /j
We should make him proof the Millennium Prize Problems
Make a visual proof that 1=2
@theromanticist8023 you can't prove this statement because it's a proof of itself.
9:14 S - N = 4S
I like how this implies that you have an infinite sum, you subtract a finite amount from it, and the sum gets bigger 😂😂😂
Well
infinite - finite = infinite
and
4*infinite = infinite
So
infinite = infinite
Which is true
@@Neo-vz8nh That's........................not how it works 😅
Doesn’t it just imply that the sum’s magnitude gets bigger, not its value? Why is that problematic?
@@feynstein1004 then how it works? Because with limits, this is how works.
@@lazenby5793 Because one of the basic properties of infinity is that if you subtract a finite amount from it, it stays the same. It doesn't get smaller and it doesn't get bigger either.
The Analytic continuation of the Riemann zeta function gives the value of -1/12 for s = -1. This does not mean that the summation of all natural numbers is -1/12 - the whole point of analytic continuation is that you are extending the function to the domain where the original function is not defined.
That's exactly what it is 👍!
Thx for this precise clarification!
🙂👻
THANK YOU
Precisely. I cant understand why people seem obsessed with ‘proving’ this result. In fact, I think these type of videos may do a lot of damage - it takes advanced mathematical knowledge to understand the context in which this result is correct (the meromorphically continued zeta function). Most people do not have nearly enough mathematical training to appreciate this context and will end up thinking that the sum of the natural numbers is a negative fraction. I blame Numberphile’s infamous video that started all of this… .😢
@@ralphhebgen7067 Yep, although I loved Mathologer's video picking apart Numberphile's video, so I would highly recommend that if you haven't watched it already!
@@abhinavanand8618 I have watched it years ago and agree completely. It’s excellent!
Before watching the video:
There's no way he can convince me but let's see what he has to come up with
After watching the video:
YOU DIVERGENT SON OF A-
Haha! Still a stretch :)
@@MathVisualProofs it's a great video and you do emphasize that this only works in a certain context, but I really struggle to accept the fact that you move around the A^2 sum terms for one specific reason : in classical real analysis, you can make a semi-divergent series equal anything you like.
With A^2 being 'even less convergent' than a semi-divergent series, I struggle to accept this visual argument, in the sense that we could probably make a visual argument to make it equal a lot of other results
It's still a great video, don't get me wrong !
love the pun
@@guillaume5313he has done it cleanly without “getting any value he likes” - he has correctly calculated the “residues” of the divergent sums. I used this same argument to defend my physics thesis in 1990. i abstracted from -1/12 to solving divergent integrals. This math was necessary to solve the Schrödinger equation for a proton orbiting INSIDE a helium nucleus as a model for the unstable isotope Li 5. The solutions to the “divergent” equation gave the decay times and radii of the resonant state that existed for the few femtoseconds thus thing held together before the coulomb force overcame the strong force and ejected the proton.
@@idjles that's great but it's besides my point
The "getting any value he wants" part refers to a proven theorem in mathematics that is presentes in this video : ruclips.net/video/U0w0f0PDdPA/видео.htmlsi=al9GeWj4KE8cHkxo
Math is just counting an infinite amount zeros.
Tell that to Set theory
@@rr.studios Zero and infinite are the only numbers that makes sense to Russell’s paradox. A set containing zero sets in of its self is zero which means the nothing sets also contains the same. nothing. Same for infinity.
@binbots but does the set of all sets contain the sets of all sets that contains all sets that contains all sets that contains all sets that contains all sets that doesn't contain itself.
Well yes but no
Zero is actually the alternative form of infinity ... 🤔
You also get -1/12 when integrating the equation y = x(x+1)/2 evaluating over [-1, 0]. That's the formula for summing the natural numbers, but using a continuous curve rather than the discrete values in the normal N summation. I don't think that's a coincidence, but I don't know how to prove otherwise.
(You can see for yourself by using a query of "integrate x(x+1)/2 from -1 to 0" in wolfram alpha.)
6:37-6:46 Are you allowed to do this? My maths is rusty but I vaguely recall that changing the order of terms in an infinite summation changes the value it limits to.
(Or maybe that's me digging into the whole reason this maths is weird 😂😂)
You are correct. It's even possible to get any conditionally convergent infinite series to sum to *any* finite number by changing the order of the series.
These series aren't conditionally convergent (they're divergent, which is the point), but the principle is the same. He would have got a different answer if he'd changed the order in a different way, and there are an infinite number of possible reorderings.
I appreciate this video as a clever and creative magic trick, but I don't like the fact that he's left some people thinking that the magic is real.
Very good! You're correct, this is an illegal maneuver that invalidates the entire proof. This video is wrong.
By shifting he cut square into 2 triangles, moving 2nd one off screen and ignoring it for the rest of calculations. By shifting it more, you will get enen bigger part of square to ignore.
@@kicorse Thanks for the confirmation! I do think it's fair to say that he worded 9:42-9:46 very deliberately to avoid the implication that this is a concrete proof, but thanks for the explanation 😊 (And I thought that you could make them reach any number too, that's so cool that you can!! 🤩😂)
@@kicorseyeah. Since on video, shifting the columns is akin to ignoring future element alignments in the sequence, it can be manipulated by hiding elements in choice. One example I found to find out that the sum 1+2+3+4+5+..... Amounts to -1/6 and not -1/12 was also by manipulating the sequences in addition in some inherently flawed way. I will try to explain it here.
Let S=1+2+3+4+5+......
Therefore S=1+3+5+7+9+..... +2+4+6+8+10+.....
S=2(1+2+3+4+5+6+....) +1+3+5+...
S=2S +1+3+5+7+....
So -S=1+3+5+7......
Now, we can do a similar trick.
-S=1+3+5+7+9+11..
-S=0+1+3+5+7+9.....
-------------------------------------
-2S=1+4+8+12+16+20...
-2S=1+4(1+2+3+4+...)
-2S=1+4S
-6S=1
Therefore, S=-1/6.
Yeah, realigning of elements of a sequence is a free way to modify the final answer, with no definite proof against it.
Using the same logic at the end you can argue it equals whatever number you want. When you are dividing an infinite number into different groups, you are always going to end up with equal groups.
There's a theorem from Riemann that proves exactly that. Divergent sums can be rearranged to add up to any arbitrary number
@@valtersanches3124 Yea, so why do so many youtubers pretend it always equals that same number? The infinite hotel problem is a good way to think about it as well.
@@MegaLokopo It's not about "youtubers pretending" anything. It's actually used in theoretical physics (the number of dimensions needed by a particular string theory is an example of this result being applied). It's not about "summing" integers anymore, more like associating a number to a formal infinite sum, and there are multiple perfectly valid ways of doing this consistently. Convergence is just one of them, and it doesn't always hold, that's all.
This particular result can be made quite natural with the use of regularization, Terence Tao wrote about it (what he calls smoothed sums). The way we evaluate a sum as a limit of partial sums is just one choice of "cutoff function" that happens to produce discretization artifacts that blow up at infinity. There are more sensible choices.
The guy showed in what he calls "smoothed asymptotics" that the -1/12 is just the finite part of an asymptotic expansion of the regularized sum: S=-1/12+CN^2+O(1/N), with C vanishing for a particular class of cutoff functions. That's how you get rid of infinity.
Again, very typical in quantum physics, where getting rid of infinite non-physical quantities is absolutely needed and can be made rigorous.
@@MegaLokopo Because they don't understand math, same as this guy. Even I could see the gaping holes in this logic and I have no formal mathematical background. I'm still under the impression that university doesn't necessarily teach you anything. You're under such immense pressure that all you have to time do is rote memorization. That's not understanding, however. That's not to say that bright people don't learn at the same time, just to say that just because someone has a qualification, does not mean they have any understanding of what they're parroting.
@@MegaLokopo I know, it is so dumb; they use a pseudo-justification to get 1 - 1 + 1 - 1... to equal 1/2, but the limit of that does not equal that; it can equal 0, or 1/3, or .0343423, or literally anything.
But really all we need to know the proof is nonsense is that 1 + 2 + 3 + 4... clearly diverges and thus does not equal any finite number; we can also arbitrarily 'define' it to equal anything we want
I had no idea how you'd pull it off but this is beautifully and elegantly done! Incredible work!
Thanks! Still a stretch, but I am glad you liked it :)
He pulled it off by arbitrarily setting 1 - 1 + 1 - 1... to 1/2; the limit of 1 - 1 + 1 - 1... does not equal 1 - 1 + 1 - 1...
Now I wonder if there is a construction that would make us identify 12S as 1 hollow square.
😀 hah!
This reminds me of the (in)famous Numberphile video on this topic, except that you actually give some (admittedly handwavy) justification to say Grandi's series sums to ½. Subscribed!
Yes, this is similar to the numberphile video (linked) and has some of the same problems, but I wanted to see the argument visually. Thanks for watching!
5:32
Not sure if this is allowed. Visually, it seems to converge to 1. Graphically however, it's undefined, as both limit from left is different from limit on right.
If you want to do that, you have to make sure that the diagonal shifting from the outer side also converge to that same point.
He's approaching the limit from below ;)
It's funny when people play around divergent sums. Triggers me everytime man.
This was freakin awesome and deserves more views! This is now my favorite way of trying to explain this to someone - especially those who don't understand math so well
so it's just the numberphile argument but with more squares
Yes. It is just an attempt at a visual of the standard argument.
Umm ik people know this this through Numberphile and BIgmathguy btw For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.
It's bit controversial topic for long time .. 😅
The more squares the better, right?
@@MathVisualProofs You do realize that a divergent sum can literally equal anything, yes? And that this isn't a proof of anything? You just defined 1 - 1 + 1 - 1... to equal 1/2, but one can make it equal any value one wants, because it diverges.
In this way, you literally just defined 1 + 2 + 3... to be -1/12
@@pyropulseIXXI "this isn't a proof of anything". No shit. It's a visual proof. And he's not saying 1+2+3+... = -1/12. He's giving a visual interpretation of common arguments.
Obviously the sum diverges. But that doesn't mean that the result is completely useless.
1:01 When You do S - N. You add 2 for subtracting 1 from S, similarly, add 4 for removing 3 and so on.. which means S - N = S + x where x = 1×(number of total terms)
Which is not equal to 4S because the number of terms are still same in both S and N so the operation S - N can not increase it by additional threefold
"If I can do whatever I want, I can obtain whatever I want"
6:37 alternatively, without any shifting, you can just turn your head to the left 45 degrees
For sure! I almost just rotated instead :)
A=1/2 is completely wrong. It's a divergent series with infinitive numbers of+1 and -1. You can make it equal to anything by carefully grouping the terms. e.g. to make it equal to 3 group terms like this (1+1+1+1-1)+(1+1+1+1-1)+...
let me ask a question: why is your standard for summable series given by limits of partial sums? can you prove that this is the only sensible way to think about infinite summation?
another remark: the exact same logic can be used to say that 1 - ½ + ⅓ - ¼ + ... is "completely wrong;" this is precisely the content of riemann's rearrangement theorem.
okay, but consider: he didn't rearrange them
@@MichaelDarrow-tr1mn if a series yields different outcomes based on arrangements then we can't assign a value it.
The sequence: 1-1/2+1/3-1/4+1/5-1/6+1/7-1/8+...
This can be rearranged to make any number.
It still has a value of ln(2). Ask any mathematician. Any at all. Even ask fifty. They will say it has a value, or be indifferent due to not caring about this branch of math
@@MichaelDarrow-tr1mn there is a difference between absolutely convergent and conditionally convergent. Consider series 1-1+1/2-2/2+1/3-1/3... it has value 0 if you don't rearrange terms, and if you are allowed to rearrange you can make any value.
A series is absolutely convergent if it is convergent even after making all negative terms positive. If that series diverges then the original series is called conditionally convergent and it won't have any fixed value.
Divergent series are the new "imaginary" numbers. We must accept them and start learning how to work with them, what can or cant be done, instead of ignoring or avoiding it. Great video, congrats
Clifford algebras and Spinors seem like the modern "imaginary numbers" to me. Geometric algebras are fascinating in general.
For me, divergent series relate more to negative numbers before they existed.
They are not the "new imaginaries" infinite series have been well studied for over a century and a handful of results been proved with careful definitions. Accepting these results in such a naive way only leads to an incomplete picture of what they really meant in the first place and in consequence making people think that the sum of natural numbers is a some negative fraction
The problem being that once you assign an imaginary root "i," it only works in one way, and you can figure out exactly how that one way works.
Different attempts at assigning values to divergent series don't work the same way. A bunch of different methods of assigning value to "1 - 1 + 1 - 1 + 1 - 1 + ..." all agree on "1/2," but more complicated series will get different results when you try to assign a value to the same one in different ways.
Lmao what? People already know how they work for hundreds of years. You can literally make a divergent sum equal anything you want. They aren't the same as imaginary numbers at all.
This proof is false because the limit of 1 - 1 + 1 - 1... does not equal 1 - 1 + 1 - 1... All he did was arbitrarily define 1 - 1 + 1 - 1 to equal 1/2 and used a pseudo-justification to do so
With this same 'stupid' logic, you can set 1 - 1 + 1 - 1... = 1/3 and it would be just as valid as 0.
i've seen the algebraic version of this proof (the one that uses S, A, and N), but seeing it adapted into a visual form is a delight!
These aren't proofs; any divergent sum can be made to equal litearlly any number; he just arbitrarily choose numbers to get to -1/12. Ironically, he can just skip the proof and just arbitrarily define 1 + 2 + 3.. = -1/12
I love these types of videos so much!! They help me understand math a lot more!! 💗
The first bad math was trusting the limit from one side. The second was assuming you could slide the columns of the square grid into the infinite triangle without altering that sum. Pretty sure there's a flaw in the S-N=4S argument too, but not as sure, maybe I'm just sleepy
(6:43) shenanigans!
That shifting trick is like reordering the terms of the sum, as used in other Σn = -1/12 videos. That means you have a completely different series of partial sums which you're taking a limit of. It's like if you have f(a, b) = a ^ b, which is undefined when a and b are zero, but you can take the limit; but you get a different limit depending on what direction (a, b) approaches (0, 0) from. The limits are not interchangeable.
If instead of shifting the columns like this, we take the diagonals (which is essentially the same thing), we get the same thing…
And like, if the terms were coefficients of a power series, then the diagonals would be the coefficients of the same power.
So, I think it would be the most natural way of converting the product of two series into a single series?
Like, if I have two formal series (sum_n a_n) and (sum_n b_n)
I think the product should be the formal series (sum_n (sum_{k=0}^n a_k b_{n-k} )
So, like, \sum_{n=0}^\infty (-1)^n x^n is (1/(x+1)) (for |x|
Ah, I think I maybe see where I made my error?
Or, kind of?
First of all, I had an error in my power series for (x+1)^{-2}, it should have been
The series \sum_{n=0}^\infty (-1)^{n} (n+1) x^n .
If you substitute in -x for x in that, you obtain
\sum_{n=0}^\infty (n+1) x^n
for |x|
it is just convenient for 0^0 to be 1.
In some sense, 1/2 begins to describe some kind of ratio regarding the number of negative and positive numbers in A. Whatever the definition of that ratio is, I can see the argument following from there. But that is not saying that the sum is -1/12. Rather some aspect of that sum can be numerically described.
Exactly. This video is nonsense.
"converges to", "divergence from"
1/2 doesn't describe anything; it is an arbitrary assigning. 1 - 1 + 1 - 1... can literally be made to equal anything; 0, 0.324234, 1/3, 1/17... literally anything. It is arbitrary. Which means it equals nothing, because it diverges
I think this video actually shows that if the Grandi's series is convergent, the sum of all natural number also is. Some still get mad when we talk about those series convergence, but it is just a minor mathematical artifact. Chose your axioms and you can decide if the series diverges or converges.
However the real question is not if the sum of natural numbers is convergent, but why -1/12 pops up in so many different ways. Do the mathematical operations that we perform have a underlying equivalence and we are always just getting the same "wrong" result? Or is there a -1/12 hidden behind the infinity?
Great video nevertheless!
Thanks!
"I think this video actually shows that if the Grandi's series is convergent, the sum of natural numbers also is."
This isn't true. Grandi's series coverages in the Abel sense, but 1+2+3+... does not.
@@martinepstein9826 Abel summation does not prove conventional convergence. If a series is convergent any mathematical operation of the series can be replaced by its converging value. If Grandi's series converges, the alternating sum of natural number converges because it can be obtained from the square of Grandi's series. The positive sum can then be obtained from the alternating series if it is convergent.
In conventional axioms, 1-1+1-... is undefined, but the video shows a valid link between the series.
he arbitrarily defined 1 - 1 + 1 - 1... to equal 1/2, but it can equal anything you want. In this way, I can skip the unnecessary nonsense and litearlly just assign -1/12 to 1 + 2 + 3.... and the proof is done via 'definition.'
This is how nonsensical it is; he defined 1 - 1 + 1 - 1... to equal 1/2 to get his 'proof,' so it is a proof by definition; skip all the nonsense and just define 1 + 2 + 3... directly and get -1/12
@@pyropulseIXXI You forgot about the fact that this is a visual proof. Visual proofs are not meant to be formal proofs. They just give a visual interpretation of something that is already proven.
It's interesting: the alternating sum 1 - 1 + 1 - 1 + 1 - 1 + 1 -..... doesn't converge as a sequence of partial sums. But I guess the idea here is that it does converge as a sequence of infinite series 1 - r^2 + r^3 - r^4 .... if you let r approach 1 from the left. It's like, depending on the order / approach of trying to make the series converge, it either does or doesn't converge. I'm not sure I'm articulating this well.
@marcevanstein This is Abel summation, which can also be used to define 1 - 2 + 3 - 4 + ... as ¼.
(I used to be somewhat disheartened whenever I found out that my latest 'discovery' was already known. Now, I'm at least somewhat comforted by the realization that it was significant enough to be written down and remembered.)
-1/12. "Can it be demonstrated on the Cartesian plane?"
The sum of all positive Integers up to N is traditionally represented by the equation f(n) = (n*(n+1))/2.
Plot the curve represented by this quadratic and one will observe a parabola (open upward towards infinity) which slightly crosses below the x-axis
at points A =(-1,0) and B=(0,0). The integral area below the x-axis between points A and B = -1/12 or -0.08333.
(Also, Euler's zeta(-1)).
One can visualize a finite part below the x- axis and an infinite part above the same. The validity of this construct can be argued in a similar manner as to how we explain the real and imaginary parts of complex numbers.
With regards to your last statement, are you saying the finite part below the x-axis is akin to the real part of a complex number, and the infinite part above it is the imaginary part? Or is it the other way around?
@@erawanpencil It was only an analogy. Maybe a better metaphor would be "a finite part below the x-axis and an unimaginable infinite part above".
Ok, sure, but how is that supposed to relate to the "series" of all positive integers?
@@Fluoman_ f(n) = (n*(n+1))/2 will output finite sums of positive integers from (1 to n). An infinite sum of (-1/12) is absurd. I find the integral under the x-axis a curiosity. Personally, it helps me make sense of a perplexing topic. I propose no real relationship.
This might be your most insightful video yet.
Thanks! I think this one is a stretch, but I know people love this sum 👍
@@MathVisualProofs I hope you are just playing this up but actually realize how absurd this argument is. It seems people in the comments are taking this way too seriously, when you literally just arbitrarily defined 1 - 1 + 1... to be 1/2
I consider this to be spreading misinformation, because the sum of the naturals does not equal -1/12, ever. One can 'make' a divergent series equal any number they want, so it is just an arbitrary assigning.
So just because there are infinites that exist, and in some situations, it would be nice to have that infinity not exist, that doesn't mean the sum to -1/12 is useful. Why not make it sum to -2? What if I need it to do that? It would be just as useful
A divergent series is, by definition, divergent. With a proper rearrangement, we can obtain any value for ( S ), ( N ), and ( A ) in your examples. The point is that these series do not converge to any specific number.
I know; this is funny, because he could've skipped all the pseudo-justification and just assigned -1/12 to the infinite sum of the natural numbers; that is literally what he did anyway, just in a round-about way
@@pyropulseIXXI There are a lot of people happy to share their negative opinion in these comments but you must be the happiest! The point of -1/12 is not that it's unique. We can obviously get whatever we want by manipulating the divergent infinite sum. The point is that we get this particular result through analytic continuation of the Riemann Zeta function.
This video is trying to give a visual intuition of that fact. I assume you know this, since you seem to have a mathematical background, so I don't understand what your whole comment crusade is about. Please, enlighten me
Just for fun.
As we made infine half triangle from square, we can make square from triange of S just bring every segment to upper limit. So now
Infinity ^ 2 = S = -1/12
So infinity = √ -1/12 = i/2√3
😂😂😂
brilliant! XD
Now do it again using visualisations for ζ(s) regularisation!
All jokes aside, very fun visual proof, I definitely enjoyed the ladder limit!
👍😀
Clever
Very clever
It infuriates me that it's just that simple but yet somehow so complicated
I swear if this comes up in my college life
:)
@@MathVisualProofs OH YOU ...
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
You better make more great videos so we won't be behind in college :|
this is the most contentious math topic... fun video, i like having a visual intuition for something that is not correct. i think it is interesting that you can do seemingly normal adjustments to the objects and get a bogus result, like in the shifting triangles and rearranging terms
This was very elegant indeed. So satisfying in fact.
OK, now do a visual proof of something true only in dimension 26!
y=x, y=1-rx lines construction is actually very insightful. Thanks! The rest of the video is just crazy, you can't do this with divergent series.
just got this recommended after watching his biography. Ramanujan was such a Genius.
but isn't the limit not the real value? a limit is just a limit?
Imagine getting more money everyday, but you end up having a debt of 0.50$ at the end of your life. How can you lose money by earning when you didn't even spend it 😭
In the end, you demonstrated that the sum is based on a limit, and that's why both it makes sense and is not real. Cool!
Excellent visualisation for 1-1+1-1+...=1/2. Still not convinced that 1-2+3-4+...=1/4 though
That one is how you multiply two infinite series to get the product. But of course it’s suspicious for divergent series (works perfectly well for formal power series). And the Abel sum of that second series is 1/4. The very last one is the most suspicious to me and not really justifiable but just there for fun :)
Is this the equivalent of mathematical sophistry?
(Edit: your video is magnificent. Seriously well done and clear. I was just referring to the -1/12) :)
😀👍
Nice. Really good job. Obviously there are some big hand waving, non rigorous arguments made here. It is essentially a version of the numberphile proof, which has some technical problems. But all this opens up some really interesting topics around infinite sums and convergence.
Please give credits of this beautiful yet amazing series to the greatest Indian Mathematician Srinivasan Ramanujan 🗿🗿🗿🗿🗿
This is exactly the reason why real analysis needed to be formalized using proper definitions lol... Is quite interesting for sure but definitely misleading.
I agree with your point that this demonstrates why it is necessary to formalize things carefully.
However, I don’t know about “misleading”? It suggests that there may be a reason to assign the value (-1/12) to the series, and, there *is* a reason to associate it with the value (-1/12) for some purposes.
It just isn’t the sum in the usual sense (as it diverges).
I guess it could be misleading in the sense that one mistakenly might take it to indicate that (-1/12) *is the sum* of the series?
But, in any case, yes, does show importance of formalizing.
Man, are you Ramanujan? This is awesome
My weakness has been advanced mathematics and it started from calculus and binomial theorem. My hold on normal maths is excellent, I was too lazy to learn it.
This breaks with the common sense that integers in simple operations with other integers makes other integers, not fractions(excluding division)
great pictorial representation..
still I believe dealing with infinity is spooky
For sure. Perhaps not the best to do this, but I thought it would be fun to “see” this argument :)
Another stonkingly fantastic video. Love it.
Glad you enjoyed it
This was not only very intuitive but also very inspiring! You are amazing! ❤
Thanks! Glad you liked it.
@@thepro4805 This one is even more inspiring ruclips.net/video/MPuZDumAzDE/видео.htmlsi=8AmPDurysv1Tg13a
Very cool. Some neat tricks here but diagonalizing the infinite chess board is my favorite..
That’s essentially the same way you find the sum of two convergent infinite series :) just weird to use it here when the series doesn’t converge …
The biggest problem is the possible invalid use of the equals sign (=). I guess you can define 'equals' as anything you want. However, IMO, equals means "exactly the same as" and when a non-convergent series of numbers is presented it is simply invalid to put an equals sign anywhere near it.
Yeah, perhaps better to use “=“ :)
Do you think the reason that 1-1+1-1 etc comes out to 1/2, is because the "last digit" has to be either a 1 or a -1, and since it starts with 1 and repeats -1+1 then the sum would either be a 0 or a 1, so the average is just 1/2? Is that essentially why it gives that number?
Simply beautiful! But it's still difficult to comprehend the meaning 😅
Agree! You need to work through much more advanced ideas to get a real feel for this.
* For certain definitions of equals.
Maths is such that, the same animation, could also be used to prove countless other conjectures. Fractals within fractals...
As a programmer, the idea of adding up positive numbers and ending up with a negative result isn't that weird.
It happens all the time, and we call it integer overflow.
Check out 2-adic numbers, where 1+2+4+8+16+... = -1. (You'll likely be directed to p-adic numbers, where p can be any (specific) prime.)
@@tomkerruish2982 I feel like you people don't understand anything when you say things like this and think the result is quite insane.
It is just turning the LHS of the decimal to 'behave' similar to how the RHS of the decimal behaves in 'normal' numbers (have the LHS have infinite digits)
So it is litearlly just a different 'notation' for numbers. The summation of 1 + 2 +4 +8 + 16 +... never equals -1. It is like getting up in arms about 0.333... = 1/3
The p-adic numbers are not the same as real numbers.
@pyropulseIXXI I believe you misunderstood my post. I never claimed the result is 'insane'; I was simply pointing out a fact about the 2-adic numbers (which I never claimed are identical with the reals).
In the topology of the 2-adics, the sequence of partial sums 1, 3, 7, 15, 31, ... does indeed approach -1 as a limit.
I was never 'up in arms' about this result; I simply thought the original poster might be interested in this nugget of mathematical truth.
And yes, this video uses reasoning that could be called questionable if one were feeling exceptionally generous. It would've been better to introduce the idea of regularization; however, that wouldn't've been very visual.
Integer plus integer producing double is weird indeed.
This, as the original video , asumes that you can move the front part of an infinite divergent series and slide it, which it seems is where it breaks
Me: So, if I add like a few thousand consecutive integers, from one up, I will get a negative outcome?
Math person: No, you won’t.
Me: Okay, but if I add like the next one million consecutive integers, at some point the result will turn negative?
Math person: No, it won’t
Me: Well, but if I add like a few billion more, maybe?
Math person: Nope, still not negative
Me: I see. And a few trillion more will not help either?
Math person: Correct. Still a positive sum.
Me: Hmmm. I’m getting worried. Since the numbers we’re adding are getting bigger and bigger, we’re only getting further away from a negative outcome.
Math person: You’re right
Me: But maybe after adding quadrillion times quadrillion times quadrillion integers, and then a few more?
Math person: No. Still not negative
Me: Wait, but at what point will the sum finally ‘magically’ flip from positive to negative?
Math person: Only at infinity
Me: And how many numbers do I need to add before I arrive at infinity?
Math person: You won’t arrive there. You have to keep adding numbers forever
Me: Forever? Then when will I arrive at infinity?
Math person: Never
Me: I get it: the sum of positive integers NEVER adds up to a negative number. That’s all I wanted to know.
Any idea how to do something similar for the sum of the squares being 0, i.e. the smallest trivial root of the riemann zeta function?
Notice: Midwits are swarming the comment section. Yes, we all know the series doesn’t converge, and it is stated clearly in the video from the beginning
Thank you! I realize this will be contentious but just like the visualizations that go with this (problematic) derivation :)
"I know it's divergent, but I don't care as long as I get the answers I want"
Has the same energy as
"I know the experiment shows the Earth is round, but I don't care as long as I wanted to believe it is flat"
Brilliant video
But there is no way that 1-1+1... =1/2
According to the Abel summmation and Cesaro summation processes, that is a value you could assign to the divergent series :) But I agree it doesn't make much sense from the real analysis point of view.
Not only no, but no to the no. Approaching a limit is not the same as an equality. The limit of A is 1/2, but it is not equal to 1/2.
Insane and irrational! Doesn't make sense at all!
Thank you for visualization❤
😀
6:40 "we can take each column and shift it down. In particular we shift column i down i spots." Why can we do this? Why doesn't it change the value? I don't understand this and frankly I don't believe it.
He was computing the product of 2 series. In standard calculus, we would use the cauchy development technique to do so to regroup terms of the same degree which is the same thing as what he did by "shifting column i down i spots". I still don’t understand how that works tho. Maybe lower degree term tends to have more importance than higher degree ones which would make sense if it is supposed to converge
Umm ik people know this this through Numberphile and BIgmathguy btw For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.
It's bit controversial topic for long time .. 😅
This kinda reminds me to that "You can't shoot a turtle with an arrow and a bow" line, because both of them make 100% sense when breaking down, but still have next to zero logic to them
the sum of all integers diverges, it is not -1/12. This should be made clear. One can re-arrange elements on infinite sums only if they converge but not if they diverge.
right??? and the whole proof hinges at the fact that 1 - 1 + 1 - 1 + ... = 1/2, which it can't be, since it diverges
All comments like this miss the point of extending mathematical ideas. Limits of partial sums is only one way (also the standard and most common) of assigning values to infinite sums. In that context, it diverges. Using other summation methods, which is a perfectly valid thing to do, it converges.
Also, you can rearrange terms as much as you want. The question is what might happen afterwards and that may depend on the summation method being used.
the very first thing I say in the video is that the series diverges using standard real analysis techniques. There are real regularization techniques that say -1/12 makes some sense in other contexts, so this is just a visual way to get to that answer.
@@MathVisualProofs sure, there are regularization techniques but rearranging the terms like you do at 6:40 is illegal under all of them. This proof technique is invalid and unsalvageable. That it comes to -1/12 is just sleight-of-hand.
@@MathVisualProofs right, I skipped through parts of the video while I should have watched the whole thing.
Nonstandard approaches of infinite sums can lead to different outcomes. What we want from math, I would think, is consistency. Sum of all integers have a connection to -1/12, like with the analytical continuation of the Riemann-Zeta-function. Some particular approaches of nonstandard summation can also leave to have the sum of all integers being -1/12 but there is a lot of handwaving there, or having to use Abel's summation like in that video, which in my point of view is not really the sum, just a particular way to compute averages instead of getting to the real answer which is a divergence.
I am aware that I am a part-time math hobbyist at best, my opinion is not important.
The checkerboard already shows that the square of A is supposed to be 1/2, therefore A isnt 1/2 but rather the squareroot of it
I bet this guy could draw 7 red strictly perpendicular lines with green and transparent ink
I'll still never get this. If you're calculating the sum of aleph-null, it's divergent. It's addition. Why do all those other calculations? They have no business adding aleph-null. It's transforming something without need.
Some math beginners when you say "your answer to the problem is wrong":
Issue is assuming there is a limit 0.999... which is less than 1. 0.999... = 1 exactly. Any argument after this point is therefore invalid.
All is well until the bit from 5:40 to 6:00. But however you dress it up, Abel summation or not, you cannot assign the value 1/2 to the divergent series 1 - 1 + 1 - 1 ... Everything thereafter is built on a fallacy. You might as well base the argument on dividing by zero.
You can assign it the value 1/2, you just shouldn't use "=”. Doesn't matter though, because what happens at 6:40 is invalid too.
7:25 eh, yeah, logically A^2 _should_ be 1/4, but as a visual proof that checkerboard looks just as much like 1/2 as the alternating sum A did, so just as S - N = 4S, A^2 should equal A 🤔
“Give me a twelfth of a dollar and I’ll give you 1+2+3… dollars ”
Isn't A undefined
A=1-1+1-1+1-1+1+...
A=(1-1)+(1-1)+(1-1)+...
A=0+0+0+0+0+0+0+...
Since multiplication is repeated additions,
A= 0 × infinity
A= undefined ??
ok didn't know that this result -1/12 is actually in the plane of complex numbers... which means that -1/12 is not a correct representation, the correct representation would be i^{2}/12, but then there is no mystery is then, is there?
Conclusion : messing up with the weird concept of infinity introduces errors in calculus.
Another reasonable explanation is simply that A = 0, and N = 0. Because it's also very sensible to be able to group these infinite, alternating sums into groups that equal 0, and it doesn't look like you're completely trying to game the system while doing so. And then, A^2 = N ; because 0^2 = 0, which still checks out.
Then, S - N = S still, because N = 0. And all the fuzzy math breaks down there.
Even if you staunchly refuse that A and N are zero and they must be some small finite number at least, at 8:00 the visualization is STILL wrong. Just because you take two pyramids of blocks, invert half the blocks in one pyramid, and then add up all the positive blocks with positive and negate the others, what you've created is a shape that LOOKS like it's 4x S, but it's not really the case, because you haven't been keeping track of how fast your infinities are growing.
From another perspective, it's much more obvious that what REMAINS of S, you've DOUBLED. And then, keeping track of the pyramid rows, you have HALVED the number of rows in your pyramid. So, HMM. Double row lengths, but half the row amount ... looks like a simple case of 2x * 1/2 to me, which is just 1, so S - N still = S, which again makes sense if N = 0. Alternatively, yet another way to look at it is, if S is absolutely infinite and unbounded, and we subtract zero or a small finite number from it, you have not affected S in any significant way whatsoever.
This is just normal proof. He just used boxes in place of numbers, nothing visual about the proof.
Oh man, so you didn't like the visualization of the alternating geometric series from the Viewpoints group? Or the visualization of the Cauchy product of two formal series? I definitely agree that the last part (the trickiest in the argument because it has the most problems) is just using boxes instead of numbers :)
I agree until you used the triangle shape by moving and infinite matrix into a diagonal form (I will say it has a 45º angle down from the horizontal, for simplification). It was arbitrary and you could had arranged it into an infinitely different arbitrary angles, since we are talking about an infinite matrix. You'd have to justify why you picked up the 45 angle.
This video is the proof that all the sufferings are created by us humans
The fact that the sum of all positive integers can be made to look like -1/12, is proof....Proof that if you overthink anything hard enough, you can attain from even the most commonplace of things, absurdity.
The fallacy comes from the fact that nobody proved that S is not infinite, and infact it IS infite, so while this is true:
S - N = 4S, N=1/4
This is also true:
S = inf
and if we substitute in:
inf - 1/4 = 4*inf
inf = inf
Which proves that S = inf
Because that is exactly the point. Is means that if this série was convergent it Will convergent to -1/12
It’s difficult to prove something visually (or in any way) if it’s incorrect
Good visual proof, but this still is no better than pluging in values into the formula to arrive at the solution, that is essentially what we have done here. The problem is, the alternating sums are convergent series whereas the sum of natural numbers is divergent for which we cannot apply the method presented here
This is the first time I've felt able to put an objection to this result into words. I understand that your video is focused on putting a visual argument to the result, not necessarily primarily advocating for the result itself. And it was a fascinating approach. Anyway, here's my problem:
When looking for the sum of (-r)^n, assuming that the sum is finite, a limit gets taken as r approaches 1 from below. That is fine. But my objection is then taking the limit to be the value of the series when r=1. In calculus class, when limits are first introduced, we are *very* careful to make sure that we all understand that just because the left- and right-side limits of a function or expression are equal, that does *not* automatically imply that f(c)=L. I've seen this argument here, and I've seen other videos that just want to "average the partial sums" of the 1-1+1-1+... alternating unit series, and none of that feels sufficient to give the geometric series a value of 1/2 when a=1 and r=-1. Just a few statements prior, we had established that the 1/(1+r) convergence general formula was only valid for 0
Thanks! The thing to look at for more justification is "Abel summation" (there are a couple of things called this). For convergent series, the Abel sum will agree with the limit of the sequence of partial sums, but there are other series that are "Abel summable" that give different sums. If you do the Abel summation on the divergent series 1-1+1-1+... you do get 1/2, but again, this is a different type of summation and does not mean that the series really equals that value. Instead, it suggests that if you want to assign a value to that series, then 1/2 is perhaps the best choice.
@@MathVisualProofs Thank you. Took me the better part of the day to find what looks like the right wiki hole. But I think the theorem is here: en.wikipedia.org/wiki/Abel%27s_theorem
The Remarks section even talks about this 1-1+1-1+... series specifically. In the Theorem section, it says, "Suppose that the sum from 0 to infinity of a_k converges". Does this mean that convergence of the infinite series of coefficients is a required prerequisite to application of the theorem? Or does "Suppose" mean that "we'll assume for now that the series converges, and if we can find a good plausible value for it through other principles, then great for us"? The Remarks section, about the alternating unit series, it says "the claims of the theorem may fail", and it seems like that's when the series of a_k diverges (violating the "Suppose" bit?) because of oscillating partial sums, rather than because of partial sums heading to infinity.
Furthermore, Alexandre Eremenko at Purdue has this proof on his university website: www.math.purdue.edu/~eremenko/dvi/abel.pdf
Eremenko seems to regard convergence of the series of coefficients as a requirement for applying Abel's theorem to a power series. He uses "Let" to establish the converging series of coefficients in the theorem statement, rather than the "Suppose" found on Wikipedia. To me, "Let" reads like instantiation, like "If you want to apply this theorem, you'll need a thing that already has these well-established properties before you even think about getting a result." He also uses the alternating unit series as his last Exercise, but if I'm reading it correctly, he states that the theorem doesn't apply to this series.
So, what's the most mathematically correct way to read this theorem in order to sort out prerequisite conditions and consequent results?
So sorry to keep bothering you in RUclips comments, as if I'm at your office hours or something. I don't have enough graduate degrees to credibly argue with Math RUclipsrs. That's probably why I need so much help getting on board with this particular result.
@@MathVisualProofs You can litearlly make 1 - 1 + ...1 to sum to anything; I can create one that makes it sum to 1/3 and call it the 'pyro summation.' It doesn't suggest that 1/2 is the best choice at all. It suggests that if you do it that way, then it equals 1/2. There is no 'best choice'
This is just spreading massive misunderstandings of mathematics to math ignorant viewers
@@emmeeemm So the idea there is that if the series is convergent, then the abel summation will give the right sum. The idea after that is that you can take the consequence of that theorem to somehow assign a value to an infinite series (even though the series diverges). Typically, if you want to try to assign a value to a divergent series (which you really shouldn't do), then you want to use one of the techniques that will agree with convergent sums. As you found in the notes, you can assign an Abel sum to 1-1+1-1+... and 1-2+3-4+... and you get 1/2 and 1/4, respectively. It turns out you cannot do the Abel sum of 1+2+3+4+..., and that's why, to do it like in the video, I had to resort to the arithmetic idea (which isn't really a good idea). There are other regularization techniques that give 1+2+3+4+... = -1/12, so this one isn't as misleading as it seems. Though again, it is really a weird world when you start trying to assign values to divergent series, and the rules should be made more clear than I had time for in this video.
@@MathVisualProofs "[...], and the rules should be made more clear than I had time for in this video."
Respectfully chap, the time that you have now set aside to excuse your video's non-thoroughness in this comments section has probably eclipsed the time it would have taken you to write this explanation into your script.
This is a quite a shame, because from what I am reading in this comment section, you possess a great deal of understanding and insight into the topic.
Your video lives a bit short of that. (but not by much, as others in this comment thread would like to suggest).
9:22 But what does this even mean?...
And also, remember, in order to arrive at this conclusion, you had SUBTRACTED N from S - as in S-N - so that isn't adding all of the positive integers!
Absolute bolleeks, trickery, infinity should not be juggled about.
how do i get the reference of the banach-tarski paradox from this tho
edit: i assumed that it is one of the case of the paradox (S-N=4S)
numbers with different infinite numbers of digits are not equal. 111....111 is not 222....2222. In base2, 111....1111_2 = -1. in base10, 999...999_10 = -1. It happens to be that 1+2+3+4+... = 1111....1111_13. ie: in base13, it's all 1s. S = 1 + 13S = -1/12.
Once you are at peaces with T = 1 + 2 T, that it solves for -1, and expands as powers of 2; you can no longer object to "a sum of positive numbers being negative"; because there is no way to make a contradiction by iteratively performing this sum. It's the same when you get a negative fraction. -1/12 is just a thing that if you multiply it times 12, and add 1, you get zero. And there are objects with an infinite number of digits that act like this. Saying that any sum is "equal to" infinity is nonsense. Because firstly, infinity isn't a value. And it's more that if you try to do the sum iteratively, it never stops growing. But if the final sum is all 9 digits, that's completely different from if the final sum is all 5 digits:
X = 5 + 10 X
Y = 9 + 10 Y
Two different "infinite values". But they straightforwardly solve for finite sums as well:
X = 5555.....5555 = -5/9
Y = 9999.....9999 = -9/9
The sum for powers of 2 summing to -1 is very very easily done algebraically. It's more difficult with the sum of integers because these steps that he did visually; you can use completely finite arithmetic with recursion to prove it. It requires a differentiation to produce 1 + -2 + 3 + -4 + ..., but you can do it with no references at all to infinity. If you have a sum in mind, if you know S and TailS, then you can solve for the finite formula SumS.
If you only known SumS, you can use S to solve for TailS:
// S isn't the real SumS. (S-TailS[n])=SumS[n]
// n isn't "infinite", it's just the number of terms expanded that you want, and it can always be higher
S = SumS[n] + TailS[n]
= SumS[n+1] + TailS[n+1]
SumS[n] = (n(n+1)/2)
... // discover that S is constrained to be -1/12
... // solve for TailS[n]
The limiting sum of a geometric series, like the one in this video, only converges for |r|
Imagine this guy’s wife cheats on him and he comes out with a visual proof of her cheating
It was fine uptil 5:40 . The statement is true for lim r -> 1-, but undefined without the limit. So A ≠ 1/2.
The digression from 2:00 was not very useful. Look at the columns of the third picture, they are just the middle pictures. Each column starts alternatively with white box and black box.
Using the limit of r approaching 1, you already showed that the middle picture is 1/2 of the starting box.
So the first column of the third picture is half of white box, and the second column of the third picture is half of black box, and third column of the third picture is again half of the white box.
So the entire third picture is again half of the middle picture.
I'm fan of this! Psychodelic cool!
7:10 what if you shift the layers down twice as much? wouldn't you get 1-1+0+0+1-1..., and removing the zeros you get 1-1+1-1..., which is supposed to equal 1/2? Wouldn't that mean 1/4=1/2?
This is an interesting note for sure. But the checkerboard-to-triangular array is the standard, natural way to multiply two formal power series, so it is the normal way to interpret the infinite product here... yes, still a stretch to do this with divergent series :)
Actually if you don’t remove the zeros you can get 1/4, because the partial sums are 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0… and if you take the mean you get 1/4. This is Cesàro summation, a way to assign a value to infinite sums such as this (it can also be applied to 1−1+1−1…). It’s quite wierd that adding or removing zeros can change the result though.