Sum of Natural Numbers (second proof and extra footage)

MAIN VIDEO IS AT: • ASTOUNDING: 1 + 2 + 3 ...
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Ed Copeland and Tony Padilla are physicists at the University of Nottingham.
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[@JustOneAsbesto]

Ed's voice is the auditory equivalent of getting a hug from a feather pillow.

[@U014B]

One that isn't overly-warm, even.

[@billybishop5323]

Post op Physio for shoulder decompression

[@moregasmthepowerful2959]

And a pillow that makes you learn things.

[@jayquelin]

most apt statement i have ever read

[@SujitVital]

Nice

[@ravenlord4]

Someone in the US government is trying this with the national debt.

[@emilcatar4620]

Unless the debt is infinite,the result will be a big number,never -1/12

[@ravenlord4]

@@emilcatar4620 r/woosh

[@SumitKumar-go9fz]

@@ravenlord4 r/whooosh

[@ccv3956]

@@emilcatar4620 it's not -1/12 even if its infinite this video is just bunch of bs

[@georgewashingtonballs5603]

@@ccv3956 it is. Atleast in real maths it's fake

[@ethanmartin2781]

"1+2+3+4+5... = ∞?" "Yeah that makes sense" "So.... am I right??" "No."

[@michaelempeigne3519]

No. Ethan

[@NITINKUMAR-ss2ji]

Ethan Martin tell me how the sum become -1/2

[@jerry1378]

it didnt

[@oscarmarreroengstrom6115]

Every lecture in history of math

[@vagabond7199]

0:25 Yeah. I like that part.

[@LakeNipissing]

Professor has the most calming voice ever to listen to him describing this proof.

[@manueldelrio7147]

He does, doesn't he?

[@junkdubious]

The Bob Ross of numbers.

[@darkdelphin834]

That'd be really annoying at school. Excitement and mathematics hold hands together. Imagine hearing that monotonous voice at school. It'd make you sleepy and eventually you'd start to get bored of maths as a subject

[@darkdelphin834]

But really calm voice indeed xd

[@MHeymann]

Says only converges for |x| < 1 Proceeds to set x to -1

[@TheGarfield1337]

Well that was his trick. He only said the geometric sum converges for x

[@subarnasubedi7938]

Well he did that because thats the whole concept of analytical continuation setting the value of domain beyond its range and playing with the mathematics its almost like setting sqaure root of -1 to i .

[@aweebthatlovesmath4220]

Wasnt that |x|

[@jamesknapp64]

its the idea of extending summation of divergent series to give meaning to them. This is called Abel summation

[@puneetbajaj786]

​@@subarnasubedi7938 can't i write 1 + 2+ 3+......... = 1+1+1+1... = 1/(1-1) and this becomes infinite, why are we doing this, we can get all sorts of answers by these manipulations 🧐

[@mathunt1130]

Ed's comedy timing is beyond perfect: Ed: So what do you think the sum is? Brady: Well, I would think that the sum would tend to infinity. Ed: Yeah, makes sense doesn't it? Brady: Was I right? Ed: No!

[@Peter_1986]

An ordinary sum WILL approach infinity, though.

[@hikarifathan5143]

Mat Hunt wow @_@

[@O-Kyklop]

If the sum is infinite, the sum will be =1. You can never reach infinity with finite steps.

[@O-Kyklop]

Laurelindo *An ordinary sum WILL approach infinity, though.* By how much then?

[@josephcambron7060]

Bullsh!t, bullsh!t, bullsh!t!!!!!!!

[@hunternewberry5860]

If this is fundamental to string theory no wonder they're having a hard time working it out..

[@AJMansfield1]

You really aren't allowed to arbitrarily rearrange a sequence to get a sum unless the series is absolutely convergent. Otherwise you can use such manipulations to prove that a nonconvergent series is equal to any value you chose.

[@Tom_Het]

Anson Mansfield try and stop me

[@andyjabez9780]

Addition is commutative.

[@marloucabalquinto7514]

Euler: Reality can be whatever I want.

[@gulgaffel]

@@andyjabez9780 Yes normally, but with infinite sums it isn't unless the sum is absolutely convergent

[@ahmadrahimisudin8364]

That's what I've been thinking too. What is the reasoning behind shifting the series one term to the right.

[@matthieu9792]

I'm glad to live in a world where numbers stop at 2^64-1

[@jon_collins]

I'm looking for a bit more precision...

[@XB10001]

@@jon_collins what a negative comment! 😁

[@randolphbusch7777]

Laughs in python.

[@amaanali9525]

LAUGHS IN YO MOMMA LANGUAGE BINARY AHAHAHHA

[@m4inline]

Where's your point?

[@DaniloJAC]

I found this: Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to infinite divergent series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined. ( I still like to think that you can't add all integer number up to infinity)

[@Ohmriginal722]

...I need more paper Brady... Get me ma damn paper Brady...

[@angermanager2116]

As long as 1+1 is 2, If summation is adding, the summation of the series 1,2,3... will always tend to infinity and never have a finite sum. But special "summation" techniques can be applied to ASSIGN values to the series which are useful in physics.

[@SarahStarmer]

I agree "useful" rather than true.

[@oddlyspecificmath]

Thanks

[@executive]

And that's the disclaimer they should have included in the original video, instead of misleading people to farm views.

[@erikdietrich2678]

What's the difference between "assigning" a value to a summation using a set of mathematical steps, and a summation "equalling" that value? Isn't that the same thing? Imaginary numbers were invented to be used to "assign" values to the roots of polynomials that otherwise don't have real values. I don't see how this is different. Numbers don't have to "exist" in order to be useful in describing phenomena.

[@Jomczx]

"That would tend towards infinity" "Yeah, that makes sense doesn't it" "Am I right?" "No" I don't know why thats so funny to me xD

[@SquashBox]

Ed Copeland's voice is so soothing. Math with his voice is so beautiful. I love it. :)

[@camerongrant7721]

This is a very misleading video. What they don't tell you is that this is a divergent series and has no sum. It's clearly unbounded. However, using a different method of summation, Ramanujan summation in particular, we can assign a number to some divergent series and this happens to be one of them.

[@losboston]

Riemann put a complex number into the function. What happens if you put in a quaternion?

[@Dan1elAndrade]

Cam GG You say it like it is no big deal. Obviously it is not equal to -1/12 whatever that means. However the idea that you can assign a negative fraction to an infinite sum that goes to infinity it's mind-bending. It don't matter how you look at it.

[@MrCmon113]

Daniel Andrade Wrong. They claimed it to be the (normal) sum. Would you excuse it if a student made the same mistakes?

[@ezrasingh2486]

Cameron Grant The reason why you can't simply sum over the natural numbers is because the complexity of addition is optimistically O(n) and the infinite sum grows faster than linear time. This is why it is difficult to literally concieve a value like -1/12 however, analytic operations are abstract and have no time complexity. This is what I believed allowed Riemann to circumvent the divergence at s=1 he looped around the complex plane. An operation like this has no algorithmic reasoning therefore a solution can only be made via analytical operations. Essentially, there does not exist an algorithmic approach to resolve these class of problems. Q.E.D

[@user-xz1yr8on6v]

wrong , even under other summation methods, this series is still divergent

[@atomicmrpelly]

20:19 - Ghostbusters walking past the window?

[@currentlyspeakingbmwmusic1793]

WTF XD

[@caitanialuiz]

Indeed

[@abraribnmizanlabib6028]

Lol

[@SImonK1996]

The fact that an unlisted maths video on KZbin can overwhelm the view counter gives me some faith in humanity...

[@JoshiaSeam]

Hi five! People all around the world love interesting things. It is this curiosity that had driven us to amazing things. Don't let the loud and disgusting actions of mankind elude the quite and amazing things that people do daily. Captain Enthusiastic For Humanity flies away.

[@MrCmon113]

Well that video presents absolutely shitty mathematics. It is not just wrong, but much worse than that.

[@samb443]

but its not even wrong

[@YoungColCol]

But 1 + 2x + 3x^2 + 4x^3 +... = 1(1-x)^2 is only valid for |x|

[@ferojshaikh2880]

you can see the binomial expansion and can see that it is true

[@avdheshgupta22]

Yes. I cannot rely on this proof. You cant put -1 in that equation

[@itchyknows]

but you can put x->-1, this proof shows that 1 + 2 + 3 + 4 + ... ->-1/12

[@redd8551]

matt de leeuw first of all: you cant say that without proofing it that easily seccondly: infinity is greater than -1/12 so this proof would be irrelevant

[@itchyknows]

Daryl can you speak clearer, because I'm uncertain as to what your saying. all I was saying is that if you take the right hand limit as x -> -1 that it works for the proof to show that sum sums to -1/12, this isn't really a proof that's required, because it's common knowledge for anyone who's taken at least high school math

[@user-cd5kp2qd5l]

The thing is that the series DOES NOT approach -1/12, it IS -1/12, this is why you can't get it by doing a limit or a summation, your result not only will be imprecise but ultimately incorrect, it only makes sense as a whole. For every other application, INCLUDING when it approaches infinity this series does not equal -1/12 and behaves "normally". Another way to see it: no function nor series is continuous at infinity, and we know that in non continuous "points" the limit and the value often don't match, these calculations allow you to find not the limit, but the actual value at infinity.

[@jacobm5167]

This video has been completely discredited. I believe even these guys discredited this video. In no way shape or form does 1+2+3+...=-1/12. Infinite sums such as 1+2+3+... are thoroughly discussed in a 2nd semester calculus class in a chapter on sequences and series. 1+2+3+... is an example of a "divergent series." It's an undefined expression something like 1/0.

[@angelogandolfo4174]

Oh my goodness. I’ve been watching these, thinking “this guy reminds me of a professor I had at uni, who taught me vector calculus, further calculus, and more; and he took a 4-person study group every week that I was in.” Then I look at my uni records, and.... he was called Ed Copeland!!! I can’t believe it! Mr Copeland, we had great times in Brighton, UK!! BIG UP THE UNIVERSITY OF SUSSEX!!

[@ethanwilliamson782]

I hope he got fired after this abomination of a “proof”

[@snoxh2187]

@@ethanwilliamson782 only because u cant understand it?

[@joshuaum473]

@@snoxh2187 it's clearly wrong, nothing to "understand" here

[@manvith1878]

This theory was proved by Indian mathematician Srinivasa Ramanujan.

Somewhere in the process of adding/substracting infinite sums the equal sign does not mean "equal" anymore. All this demonstration is about getting the Zeta function "extendable" to -1. The trick is that they keep using the sign and word "equal", while they shouldn't (that's why physics is physics... always making nice asumptions but using bad mathematical language). The Zeta function with -1/12 added isn't "equal" to the infinite sum anymore, it is more than that (a new mathematical object). Let's take an example. On both sides of a river there are two similar roads. But you can't get a car above the river until you build a bridge. But the road with the bridge is not the sum of the two previous roads anymore, it is a new road. And while before it would have been absurd to say that you can be driving a car above that river, now you can even stop on this bridge, and keep the car parked here. Brady parked his car at -1 (the river) on the Zeta function, but before that, he made a mathematical bridge appear that "fits" between the two roads so that his car does not sink ;)

[@daminkon246]

that example is just really bad

[@oddlyspecificmath]

This makes a certain kind of sense to me; thanks.

[@GlobalWarmingSkeptic]

I think that mathematicians over-sensationalize this particular sum because the -1/12 result is not based on traditional sums. You have to treat the number in a certain way and relate it to other functions in order to get this. The problem is that this isn't communicated in this video or by any of the others that talk about this sum. It is mathematically interesting, but it doesn't mean you can literally add up all the natural numbers and come up with a negative fraction.

[@golfer435]

People this isn't a finite sum, it is infinite. The second you put this in a finite realm, it ceases to be true.

[@evilkillerwhale7078]

+Logan Hollis You're bad at math.

[@NebulusVoid]

+evilkillerwhale 2+2=5

[@divxxx]

It doesn't need to be true if it's useful.

[@sagantucker2097]

I find it really interesting that 1 - 2 + 3 - 4 + 5 -... = 1/4 which ( i realized since I just came from some videos on it) is the point at which real numbers begin to "escape" from the mandelbrot set or whatever. Makes me wonder, where does -1/12 fall on the mandelbrot set? I'm certainly no mathematician but its fun to think about these things.

[@jamesthorn9992]

People misunderstand the point of this. It is not saying that the sum of all the natural integers is equal to minus one twelfth , it is beautiful mathematical experiment to see if anything meaningful and consistent can come from viewing diverging series in a different way. As it happens it is consistent, he is using a solid mathematical framework witch requires bending the rules of existing frameworks for it to work. It is not wrong. Once upon a time people would have thought that negative number break the fundamental laws of arithmetic but every educated person fully understands that the theory of negative numbers makes something of quantities less than zero in a useful and consistent way without contradictions.

[@LfeinYT]

This demonstration is a classic bait and switch. It begins by claiming exactly that the result is the sum of all the natural positive integers, but it is actually about a different relationship altogether. I believe Copeland and Padilla are correct that there is an important relationship between the sum of the positive integers and -1/12. However, they clearly claim at the start that the result they will calculate is the arithmetical sum of the positive integers: 1 + 2 + 3 + 4 . . . (It's certainly not nearly as astonishing that there is some relationship between the sum of the integers and -1/12) In making precisely the claim "that the sum of all the natural [positive] integers is minus one-twelfth," they do make fools of themselves and insult the intelligence of their viewers, and for this they owe us an apology. If you Google this topic you'll quickly come across a comment by a real mathematician--which I am not--that the first video is "seriously misleading."

[@orenzeshani]

Well you can't bend the rules. It's mathematics. Either you stick to the rules or it doesn't mean anything at all.

[@Arkayjiya]

I don't like this way of putting things. Changing the rules implies that there is a "standard". There is not, every consistent set of rules is equally valid. But even if I don't like the way you formulated it, I agree that in a video that is supposed to touch a large public they should have made the fact that they were not using the most common set of rules clearer. They didn't "cheat" or do anything "wrong" except in the way of pedagogy.

[@bennattj]

Aesahethr I have a very real problem with people stating that they "changed the rules"...they "didn't" (notice the quotation marks). Here is the bottom line, if you take a geometric series: x + x^2 + x^3 + ..., and you evaluate it at -9/10, -99/100, -999/1000, etc. it approaches 1/2. Therefore _as you approach -1_ the geometric sum approaches +1/2. The problem is that 1) the geometric sum of 1 - 1 + 1 - 1 + 1... is _obviously not_ 1/2 (it's obviously indeterminate) and 2) the above reasoning (that as you get closer to -1 the sum approaches +1/2) requires more and more terms to converge until you get to the point of x = -1 where it requires an _infinite_ amount of terms to converge...

[@orenzeshani]

Jared Bennatt Not true because it only approaches from one side. Now here is the point: Maybe it is possible, in certain fields of Mathematics, to have equivalent set of rules, but what you can't do is to use a rule from one set in another set in which it is "broken" and this is what is going on in this movie. You can't take the diffrential of a function if you just waved off convergence..

[@lraoux]

"But 1 + 2x + 3x^2 + 4x^3 +... = 1(1-x)^2 is only valid for |x|

[@pitreason]

Ed: for this abs(x) must be less than 1 ... few moments later Let x = -1 Wtf???

[@viannedemirel]

pitreason -1

[@pitreason]

@@viannedemirel yeah but abs(-1)=1

[@ViratKohli-jj3wj]

N word

[@chriswebster24]

@@ViratKohli-jj3wjAfrican-American is the proper term.

[@Yupppi]

Some people were quite upset that the material in these extras wasn't mentioned in the original video. I guess editing cut some key definitions.

[@OmegaRainbow]

I just received the brown papers from this video - thanks Brady :D As soon as I get back home (I'm abroad for work) I'm going to frame them, hang them on a wall and send you a pic ;)

[@numberphile]

Look forward to seeing the pics!!!

[@hermokuolio01]

+Numberphile it's been over 2 years, have you seen any pictures?

[@elave16]

its over two years an two months!! still waiting? :P

[@OmegaRainbow]

elave16 for what?

[@FreddyMercry]

For picture confirmation!

[@tomselleck8204]

as i understand it, to put it in layman terms. if you added all the numbers together the resulting integer is so huge its dense and it sucks in on itself into another dimension. it then poops out gummi bears. joy

[@sammed92]

lol

[@connorhorman]

2:51, you also need the sum rule of derivatives (f+g)'(x)=f'(x)+g'(x), on top of the power rule f(x)=x^n (n is constant with respect to x), f'(x) = nx^(n-1).

[@panga131]

What amazing (and unexpected) results. Great video by the way, and the Dutch angle makes the video even cooler haha

[@smallfry4973]

"Yeah, that makes sense" "Am I right?" "No."

[@iamthereaImaster]

1:47 shouldn't the range of x be -1

[@jamesknapp64]

In the strictness sense, no you can't. But at the same time then the sum 1 + 2 + 3 + 4 + ... doesn't work either. To give meaning to these diveragent sums you can do things. This is what is called Abel summation; in the fullest rigor you take a limit as x -> -1 which gives you the sum in question. Then you do other trick, such as "analytic continuation" with the zeta function thus you can "uniquely" define the sum 1 + 2 + 3 + 4 + ... = -1/12

[@julianwood6625]

You're right - the range of x has to be -1 < x < 1 for the summation to converge. End of.

[@hansbauer760]

3:43 "OMG, did he actually just set x to -1???" lololol

[@Nylspider]

What

[@enesb7412]

The more people argue, the more laws of math they violate. He also gives wrong information about zeta function. (such as using the given representation for s=-1) He also assigns values to alternating sums and changes orders of the numbers like it is very acceptable. If that proof is valid, 1 = 0, and numbers mean nothing.

[@miloweising9781]

By extending the zeta function through analytic continuation to negative integer values you forfeit the ability to write that the sum of positive natural numbers equals zeta(-1) and therefore the sum of positive integers in no way equals -1/12. It is an absolutely divergent sum. The methods in which are used in this video to prove this result are simply incorrect. This video is extremely misleading and condescending towards its audience.

[@AstAMoore]

Loved this. But Brady, you desperately need a parfocal zoom lens.

[@numberphile]

I need lost of stuff…. if you want the unedited footage as "extras" there is a price to be paid! :)

[@MrLemonsChannel]

+Numberphile lots*

[@EvolBob1]

+Numberphile lol - love to see the video of you guys working out who gets what, from the KZbin payment for the views. I bet the maths is extreme.

[@labibbidabibbadum]

The person in the pink parka at 16:44 gets £0.5772156649

[@thesinglemathnerd]

labibbidabibbadum lol

[@ioneldragomirescu4179]

The formula 1 + x + x^2 + x^3 + x^4 + ... = 1 / (1 - x) work only for -1 < x < 1. So, x cannot be -1. Ionel

[@jamma246]

Ionel Dragomirescu Yes, but the value approaches 1/2 as you let x approach -1. Since in the framework of all of this stuff everything varies continuously, that is enough.

[@lizokitten2]

Colin Smith That's also what upper division math professors tell you, because the series does not converge, because the sequence does not converge, which is easy to prove. If they want to assign number values to divergent sequences, fine, but that's not mind-blowing.

[@jamma246]

lizokitten2 I didn't say anything about the Taylor series. Anyway, they both converge to the same answer from the left as you approach x=-1. This fact is sufficient by what we know about the continuity of the Riemann Zeta function etc., as I said before.

[@lizokitten2]

The series you're referring to is the Taylor series.

[@lizokitten2]

The functions they are using are not analytic for the values they are plugging in. That's the problem. Some people here are saying that they are simply creating a different, consistent system by ignoring where functions are analytic, which is interesting, but if that's the point of these videos, then what they're doing is misleading, not mind-blowing.

[@acm-gs6bl]

the problem with the zeta function argument is that the function only works with the condition that s>1 or the series won’t converge. the analytic continuation of the zeta function is ζ(s)=(2^s)(π^(s-1))(sin(πs/2)) Γ(1−s)ζ(1− s). notably not the same as the original function. plugging s=-1 into this will give us -1/12, however because this is not the zeta function, simply a continuation of it, we can’t expect the same result from the original zeta function.

[@MgtowRubicon]

In a mathematical "proof", the first erroneous argument nullifies (neither refutes or supports) all subsequent arguments, and thus the conclusion is unsupported.

[@frepi]

Beautiful handwriting!

[@manmanman784]

finally some more difficult stuff :) thanks numberphile!

[@ChrisBandyJazz]

My Calculus 2 textbook says that answers like these are "absurd" because the sequence diverges. Don't get me wrong, I love concepts like this and am all for the sum of the natural numbers being -1/12...but what justifies the addition of infinite series?

[@ChrisBandyJazz]

Thank you!

[@renatomagretti7824]

i dont think you would get a propper answer in a calculus 2 book (specially for this kind of math)... you should look for complex functios and analytic continuation. And i think there you may satisfy your doubt. Im sorry for my English, im from Argentina and its 2 o'clock in the morning... 😁

[@connorsmith3282]

Chris Bandy I think your misinterpreting your calculus book. They're probalbly talking about the limit of the series, this isn't the limit, its assigning a value to a divergent infinite sum.

[@NitinChauhan-vh2yk]

Divergence is not really caught in high school mathematics just as complex numbers are not caught in junior school where possibility of square root of -1 is termed absurd. :) There are ways to get sums like this. Each divergent sum like this also has a type and based on that type there are ways to get to the sum. Watch the video on Grandi series (1-1+1-1+1-1.....). There too, the sum (called Cesaro sum) of the divergent series comes out to be 1/2.

[@iankretschmer559]

Or ist just comes out as basic garbage, because the series 1-1+1-1 has no limit, If you where to apply the Theorem of Limits, you need first to proof that for every epsilon around said limit, you get a certain n, after which the series does not leave an intervall of epsilon around such limit. Which you cannot for the sum of 1-1+1. So if no limit exist, the sum cannot have fixed value. Only if there is proof of a sum being convergent, you can than apply summation of know, convergent sums to determine its value.

[@MrCmon113]

I guess "you can meaningfully associate -1/12 to an infinite sum on natural numbers" doesn't get you enough clicks...

[@evilkillerwhale7078]

+Taxtro It's correct. I'm not sure why you're whining.

[@MrCmon113]

***** No it's not correct to give a sum for an infinite series. When you treat infinite as a number, you can "prove" all kinds of nonsense. Every mathematician is well aware of that.

[@evilkillerwhale7078]

Taxtro I've got my mathematics degree (as well as an aerospace engineering degree). I work with mathematicians at MIT, from Cambridge, and from all over the world. You're wrong. The Riemann Hypothesis has been proven. Sum of the natural numbers has been proved from the Riemann Hypothesis. It's well understood, well agreed upon, and has been repeatedly rigorously proven.

[@MrCmon113]

***** Well then you guys are a very peculiar group. I'm not saying that - 1/12 has no significance, I'm saying that you cannot equate such an infinite series. Otherwise you can "prove" all kinds of other stuff like 0=16.

[@evilkillerwhale7078]

Taxtro That's not true. They didn't equate anything. He used a right hand differential to use a completely valid theorem at its boundary condition. Things like "0=16" come from a divide by 0 error. There WAS no divide by 0 here. That's why it's so beautiful. And in the Zeta proof, there's not even a 0, much less one to be divided by, required.

[@tamasdemjen4242]

The way I look at it is that generalization is a tool we use to give meaning to concepts that cannot be described otherwise. Consider x^2=-1. You could say that it can't be solved, because there's no such real number. Or you could imagine such a number, and call it i = sqrt(-1). As it turns out this imaginary concept solves a lot of physical problems, such as describing a rotating vector, and it opens up new ideas, such as frequency domain analysis, which has a real physical analogy. For example, you can sharpen your photos, thanks to the introduction of a new concept that is supposed to be imaginary and not real. There's no such number as infinite. Something is either convergent, when it clearly tends to get ever closer to a real number, or it's divergent. Just like sqrt(-1) didn't exist before, 1+2+3+...+inf doesn't exist, either. However, a genius called Ramanujan from India had invented a special summation, a way of assigning a value to divergent series. It may be terribly confusing, but if it allows us to describe something in the real world, then who's to stop us from doing it? It's certainly done in a logical way, by generalizing something that was meaningful before. For example, the factorial is easy to define for integer numbers, but what about real numbers, or complex numbers? Well, who's stopping us from defining a function that we call Gamma, where Gamma(n) = (n-1)! if n is positive integer? All of a sudden we can solve probability theory problems that we could never handle without it. We just introduced a new concept that solves something real.

[@MuffinsAPlenty]

One of the best comments I've ever seen on a KZbin video. You truly "get it" when it comes to mathematics.

[@Spix_Weltschmerz-Pucket]

***** I agree with muffins. This is by far the best comment I have ever seen in a youtube video. Thank you :)

[@bharathp666]

***** You sir, made my day! Thank you for the best insight. You would make a great teacher. :-)

[@MuffinsAPlenty]

Rob Laquiere Generalization is usually applied to definitions (e.g., we can generalize "product" from taking two inputs to taking any finite - or in some cases countably infinitely many - number of inputs). Generalization is not applied to properties, such as the commutative property. Properties of generalizations of definitions must be proven for each generalization of the definition. Also, no one will ever "prove Ramanujan wrong" about Ramanujan summation. What could happen is that someone could come up with a different generalization of the sum of an infinite series which is more useful. But if this happens, it won't prove Ramanujan summation to be wrong; rather, it would simply be a more useful generalization.

[@roblaquiere8220]

MuffinsAPlenty Can you please define "generalization" so that I can be sure we are on the same page? When I first started to learn mathematics, it appeared to me that the basic laws of Algebra were in fact generalizations of Algebra. Just like x*y = y*x is the general form of the law of communicability of multiplication. But just like I stated in my post, this generalization is wrong in certain fields of mathematics! If that is not what you consider a generalization, then I need you to clarify your definition of "mathematical generalization". The properties of mathematics are generalized statements about how the math works out (properties = generalizations). No properties have been proven for all possible cases, as some families of equations are infinite in size (infinity cases cannot be checked). Therefore generalizations are not proven for all cases simply because they yield sensible results for some cases. Like in my example, the generalized form of the law of communicability of multiplication breaks down when vectors are multiplied. It is therefore possible that Ramanujan's generalization here is also wrong (as some generalizations have been proven wrong in the past). For the case of divergent series, the definition of divergence contradicts the result they got. To me, checking the results empirically is more sensible then trying to convince ourselves it's true. As a physicist, I tend to check results to make physical sense first, as that is indeed the goal of mathematics; to count real things and to achieve empirical results. Lastly, as I already stated, the generalization may very well be true... But I need to see this verified in reality (with tangible things), until then, it's just numbers written on a page in a particular way (not proof of anything really).

[@Poldovico]

Ok seriously. What the shit? Maths are broken, I want a refund.

[@diabl2master]

I think this type of result should be referred to more as "there is a function f so that f(1+2+3+4+...)=-1/12". It's not a summation in the conventional sense, but we could invent some kind of "divergent sum space" (of which 1+2+3+4+... is an element) which is the domain of this function and the co-domain is the reals, or the complex numbers, or other sets.

[@orbitalteapot21]

actually it is f (-1)= 1+2+3+4+5+6...=-1/12

[@diabl2master]

Muhammad Abdullah i think you misunderstand me

[@muddafiggagagamel1820]

+Davy Ker I think you are right. What is bugging me is that this whole process only works if you assume that the limit of the alternating series 1-1+1-1+1.... etc is (1/2) altough it is clear that this series has no limit. And if you watch the Video where he tries show us that the limit of this series is (1/2) you can point out some major flaws in his proof.

[@CharcoalBlasterdog]

The limit of that series is not 1/2, but rather the cesaro convergence of the series is 1/2 --- by taking the limit of the average of the partial sums, you can in a way assign a finite value to an otherwise "normally" non-convergent series. So theyre kind of not giving you the whole picture as they dont say "1-1+1... is not equal to 1/2 in the traditional sense of summation, but we can provide an alternate description based on some characteristic of the series, and say the series is equal to that description."

[@TheAusrali]

thats exactly what i said on another video!!

[@Integralsouls]

"critical to get the 26 dimentions in string theoriy". few seconds after:*proceeds to explain how to differentiate*

[@douggwyn9656]

Not only that, but without the -1/12 kludge one does get a solution, namely D=2.

[@Integralsouls]

@@douggwyn9656 fucc u Doug, nobody asked ....

[@jhobbz45]

Please make a video explaining how a series that diverges can equal some value that's not infinity.

[@Anytus2007]

Strictly speaking, the series diverges. It does not have a value. However, we can regard the geometric series as a function (this is the whole 1/(1-x) bit) which does converge for some values of x. Then we analytically continue (a process in complex analysis that gives us the unique function that covers a larger domain but agrees with our original function on the smaller domain) this function to other values. We take whatever value our analytic continuation gives and treat the series as having that value. But strictly speaking all the equals signs here aren't really equals signs in the same way that 2+2=4

[@jhobbz45]

Thanks Anytus2007! So 1+2+3+4+5..... doesn't really equal -1/12?

[@Anytus2007]

Not really. We can treat the series as having a value of -1/12 in many calculations, but this is why in the Polchinski book that they show, Polchinski uses '->' instead of '='.

[@jhobbz45]

So it is tending towards -1/12 but does not really equal -1/12?

[@Anytus2007]

It doesn't even tend toward -1/12. It tends toward +infinity. Again, the series does not converge. Its just that there is a consistent way of assigning this series a value and that value happens to be -1/12.

[@TheophilosPorter]

But you can't just remove a divergence! It tells you that there is no limit, no one answer it tends toward! So how is that valid? The obvious way to approach 1+2+3+4+...+infinity is infinity plus any finite number, an infinite number of times--which will continue to be infinity. I just can't accept this one.

[@DrDress]

13:08 If you press 1+2+3+4+5 ect for ever you'd never press the equal sign now would you? If you did that that means that you'd hve stopped

[@SensibleSine]

+DrDress but as you kept typing 1+2+3+4+... to infinity, with time... lots of time you'd get so bored and go so crazy that you would catch yourself typing -1/12 over and over again.. then realizing that you're now headed towards negative infinity. so then you would try to recompense by typing +1+2+3+4.... all over again and just consider giving up and when you finally did you would find out you gave up at the exact same point you began to start typing -1/12 (the point you lost your mind) and the calculator would read -1/12 . then you'd realize you where still typing. and that number would just haunt you all the way. for eternity..... its the devil....

[@DrDress]

+halofan313 What an d thing to say...

[@lakshbk]

+halofan313 I find ur explanation really convincing

[@diptoneelde836]

Yeah that makes sense

[@ezrasingh2486]

DrDress The reason why you can't simply sum over the natural numbers is because the complexity of addition is optimistically O(n) and the infinite sum grows faster than linear time. This is why it is difficult to literally concieve a value like -1/12 however, analytic operations are abstract and have no time complexity. This is what I believed allowed Riemann to circumvent the divergence at s=1 he looped around the complex plane. An operation like this has no algorithmic reasoning therefore a solution can only be made via analytical operations. Essentially, there does not exist an algorithmic approach to resolve these class of problems. Q.E.D

[@grapheist612]

I am a 6th year PhD student in Physics and Electrical Engineering, but I double-majored in Physics and Mathematics as an undergraduate, just to give some background on my expertise (or, more likely, lack thereof). A few thoughts: I notice that both Dr. Copeland and Dr. Padilla are physicists, according to the video description. My math professors held up their noses at physicists for their cavalier approach to mathematics, particularly for switching sums and integrals and their handling of infinite series like these without regard for rules and the applicability of formulae. I would be very interested in hearing what a strict mathematician had to say about this result, particularly because we simply "accepted" the insertion of x=-1 into a formula with applicability for |x|

[@adhamm5503]

And what about now? still accepting it? I mean after 4 years from your comment.! did you get a mathematician opinion on that? I really would like to know :) Regards

[@oooBASTIooo]

It is very simple: What they present here is wrong. Even reordering convergent series leads to arbitrary results, as Riemann showed. Plus, there is a proper definition of convergence of series, which is that the sequence of partial sums converges, hence their statement: 1+2+3+...=-1/12 is simply wrong. Additionally, the Zeta function is not defined for numbers whose real part is smaller than 1. What actually happens is that you can use analytic continuation to extend the zeta function to numbers with real values smaller than 1 and for this continuation, the value at -1 is -1/12. But this has not much to do anymore with 1+2+3+...

[@KarlDeux]

@@adhamm5503 Well the natural numbers with the addition is a group, meaning if you sum natural numbers you can only get a natural number. Moreover a positive one if all the numbers you add are positive. If you find something else, it's either you are wrong, or what you say is addition is in fact something else (which is what Ramanujan meant).

[@xavierstanton8146]

​@KarlDeux A bit of a correction, the set of natural numbers is not a group under addition; there's no inverse elements at all! However, the set of natural numbers including 0 is what's called a monoid. That is, it's a set equipped with a binary operation that satisfies all the group axioms except for the inverse axiom. Specifically, this tells us that adding two positive integers yields another positive integer.

[@KarlDeux]

@@xavierstanton8146I did not want to have the fish drowned but thx to have done this for me.

[@ge2719]

how have you put -1 into the formula for sum to infinity? the limits of that formula are -1< x

[@DrankisDank]

Well technically, the limits of the formula for the sum of a geometric series are not -1

[@Vackraetraed]

I was wondering the same. I was taught that the sum he uses is only defined for the absolute value of x not x. As in its defined in -1 < x < 1 as you said.

[@diabolic42090]

Exactly what I was wondering. Can someone explain this?

[@ge2719]

After doing a bit of reading ( and I do mean only a bit) I think its not supposed to be be "equal" to in the usual sense but is instead just meant as a property of the sequence and how the sum of it tends to infinity in a different way than how the sum of other sequences tend to infinity.

[@HebaruSan]

Around 17:45, Guy #2 admits that it's a "trick" called "analytic continuation", which apparently involves finding a convenient formula for a function that's valid over a certain range and then arbitrarily extending the range over which you can use it. So apparently it's OK to break basic rules of math as long as you give it a fancy name when you do it. :)

[@synchronizerman]

I wish that you wouldn't "accept it." I want to see the entire thought-process, though I may be in the minority.

[@BillySkceuk]

I think they have done that proof in an earlier video.

[@Dextomus]

Was thinking the same thing

[@ZipplyZane]

Assuming you are talking about the geometric series (1 + x + x^2 + x^3 + ...), I found it online here: mathcentral.uregina.ca/QQ/database/QQ.09.00/carter1.html

[@COHEMUACOS]

en.wikipedia.org/wiki/Geometric_series

[@synchronizerman]

I know what a generic geometric sequence is, but thanks.

[@Rjsipad]

4:10 this is incorrect. for the sum of an infinite geometric series to be 1/(1-x), |x|

[@Falcon0408]

+xxBIGBIRDxx it's the Taylor series expansion, nothing to do with what you have said

[@Rjsipad]

+OneSixteenMike same im curious

[@christianbouwense4702]

+xxBIGBIRDxx he took the derivative of 1/(1-x), he had already established that the sum of an infinite geometric series is a/(1-r) [which in this case is 1/(1-x)]

[@crazymuthaphukr]

+OneSixteenMike It's basic calculus. Look it up and prove it to yourself. Don't be an idiot.

[@Rjsipad]

+crazymuthaphukr theres always that one guy that decides to take it personally... congrats crazymutha

[@remimahmoud7672]

Hello Professor, nice video, but I would like to find a proof of this formula which uses the analytic prolongation of the Riemann function, because the proves with the manipulation of divergent series such as finites sums are not "mathematicaly correct", aren't they ? Could someone advice me some links or videos to go further on this subject ? Thank you

[@freethebrain]

I used to harbor a huge dislike for Maths when I was in high school and I simply feel the need to say that through the years this dislike has turned around 180 degrees and I found the maths in this video simply beautiful. The Universe is so stunning.

[@MrCmon113]

The maths in that video are BS though.

[@jordankay9082]

Thank you for the second explanation! Analytic continuation... I can buy that because it's not saying that this is actually valid, but that it is a 'what if' situation (just like imaginary numbers, where we said 'what if the square root of -1 existed?', and it has been a HUGE help in physics). Really wasn't happy with the first part, at 6:35, when we just started multiplying bases together because they shared a common exponent, but kept watching hoping for someone to make-up for that.

[@shawniscoolerthanyou]

"Yeah, that makes sense doesn't it..." "Am I right?" "No."

[@marcosrodriguez2496]

"this result is critical to getting the 26 dimensions of sting theory to pop out" made me smile.

[@Wizarth]

I would like to hear more examples in quantum field theory where this property plays out. String theory has a bit of a reputation for being nice in theory, but unprovable (from what people who have researched it more of told me). Quantum mechanics is something we're getting experimental results out of, and I'd love to hear about one (or more) which were predicted by a theory built on this sum.

[@Anytus2007]

As a physicist, I have to ask, why do you need to see such things? This result is a mathematical fact that can be made totally rigorous provided you have the years of training to talk about analytic continuation and zeta functions. Adding physics is just muddying the waters.

[@ASKaPHYSICIST]

"strictly true for x

[@topilinkala1594]

I have HP 50g where I can write the summation notation from 1 to infinity and it gives infinity as the answer. It also has Riemann's zeta function which give -1/12 at -1. Feynman in a lecture said that the mathematical calculations involve sums that go to infinity but the physical measurement shows finite answers and so the only way to compensate is to take the sum of all positive integers to be -1/12. Meaning that what you see is real but what you think is fantasy. Infinity is fantasy and -1/12 is real.

[@AkiraUema]

Numberphile, I've got a question. It might even be a silly one. At 1:45, Ed says that the sum of a geometric series equals to 1/(1-x) strictly when x < 1. However, in some textbooks, we see abs(x) < 1 instead of x < 1, which would render that formula not defined for x = -1 also. It would give as domain, the open interval -1 < x < 1. Why is it that makes it possible to consider all values x

[@isithardtobestupid7568]

+Akira Uema He did say that we will push the boundaries. Anyways, onstead of -1 you can think of -0.999999999999999999999999999999999999999999999999

[@ala4sox02]

+IsItHardToBeStupid? And in that case, the sum would be convergent. Eventually, the fraction ^ n would counteract the coefficient and it would approach a finite number. Ignoring those boundaries makes it divergent and not calculable. The Riemman Zeta function is a nice trick once you've ignored the very first restriction that was placed on your equation.

[@isithardtobestupid7568]

ala4sox02 and it would converge to something very close to 1/4.

[@s4archie]

+IsItHardToBeStupid? That doesn't make any difference. And no, the sum does not suddenly converge. Euler is notorious for doing things that are completely unjustified, such as putting (x=-1) into a formula that is not valid there.

[@s4archie]

But that is *not* the part where this "proof" falls apart. The problem is where the -2.2^(-s) line is split up. Instead of -2(2^(-s)+4^(-s)+...) it becomes -2(0+2^(-s)+0+4^(-s)+0+...) which could have a different value. For example 1-1+1-1+1...=1/2, but 1+0-1+1+0-1+1...=2/3.

[@LegionOnomaMoi]

At about 12:50, Dr. Copeland refers to "removing the divergence". The problem is that the only way to do this is to not use infinite series. Infinite sums diverge or converge. About a minute later, he asserts "you've got to deal with divergent numbers- divergences very carefully", but the way he "deals" with infinite series wouldn't be considered "careful" even in Euler's time, as Wanner & Hairer note in Analysis by its History. Of another divergent infinite series that Euler wrote equaled a finite number, they state "its mathematical rigor was poor even by 18th century standards", So why are we repeating mistakes made in the 1700s? Even a defender of Euler's ridiculous equations, Sir Roger Penrose, admits "..,the rigorous mathematical treatment of series did not come about until the late 18th and early 19th century...Moreover, according to this rigorous treatment [Euler's equations] would be officially classified as 'nonsense'." While a rigorous treatment of the zeta function (or any other topic in complex function theory) is obviously beyond the scope of a KZbin clip like this, that doesn't entitle one to make patently false claims while ignoring one's own advice just to be able to make a mathematical claim that's seems mind-blowing (to the mathematician, it's only "mind-blowing" for its fundamentally flawed basis).

[@LordSwordbreak]

It looks kinda like the student in the background at 20:20 is walking by with a light-saber in his/her hand...

[@PC_Simo]

That ”1/(1-x)²” -formula is gonna spell trouble, if you let x = 1, which is, what it has to be to give the series: ”1 + 2 + 3 + 4 + …”. So, if you apply that formula that ”x = 1/(1-x)²” to the series: ”1 + 2 + 3 + 4 + …”, you’ll find that this particular series equals 1/(1-1)² = 1/0² = 1/0 = **ERROR!** 🤔🤯

[@johndoyle2347]

Willans' Formula for primes: 2 to the n part = vertical asymptote. 1/n part = vertical tangent. Factorial part = vertical line. These tensors from differential calculus determine singularities in stable matter as represented as primes.

[@juanmanuelmunoz5645]

Enormous misconception about Taylor power series: The theorem says that it can be done for -1

[@longevitee]

Juan Manuel Muñoz Bless you.

[@embedded_]

Thanks Great Gauss , that there are adequate people on the Earth.

[@MrCmon113]

Dismissing is is the only correct thing. If you try to "prove" stuff like Numberphile you'll make a total fool out of yourself at any university.

[@DeathBringer769]

Tons of people dismiss things out of hand without the proper knowledge, understanding/education or experience to do so in an intellectually honest way... It's unfortunate, but it shouldn't surprise you. People make their minds up about all kinds of stuff they actually have NO clue about, all the time. Many time them will get very angry with you if you even try to insinuate possibly ignorance about any possible topic, even though we ALL possess some level of remaining ignorance no matter how well educated we are, and we should never stop striving to learn everyday... I wish more people followed through with that though.

[@grantkohler7612]

So if instead of setting x at -1, you set it at +0.73(it's still less than 1), would you still get -1/12, or am I breaking math again?

[@user-qd2nd6hi8j]

As i saw in other video - the sum 1+2+3+...=inf (divergent). We can imagine summation as the spiral, that gets bigger and bigger with each turn, but the center of the spiral located at point -1/12. For convergent sums - their spiral twist to a point of convergense Divergent sum - spiral from a point, convergent - spiral to the point

[@5oulcrusher]

"It's called an analytic continuation. It's taking you from one regime where it clearly looks like it's divergent and moving it into a regime where it's better defined."

[@CabooseNor]

It's so fascinating to listen to Ed Copeland!

[@arthurthegreat216]

Brady, please ask the professor the following: In the beginning of the vid, the prof offers to prove the convergence formula for geometric series (a+ax+ax^2+...ax^n= a/(1-x), in our example a=1, so it's just 1/(1-x), as the prof stated) and you said we should just accept it. If he were to prove it, one thing would become very obvious; the convergence formula only holds for -1

[@paultang5479]

...

[@nyxgraal6949]

hes not claiming that its true,he just says that it is the thing that euler did

[@bow3i]

He plugs it into a differentiated formula to which the limits no longer hold true, and it's a hypothesis that euler used.

[@bhobba]

Look into Borel integral summation. You can rigorously write it in a form where instead of it being convergent at |x|

[@somni6756]

I love Numberphile, but why am I rewatching this at a quarter of midnight?!!

[@yichern4351]

The professor at the beginning sounds like the brilliant master mind behind a plan to take over the world revealing his plan to the main protagonist

[@Pantopam]

To those who see the problem where he just plugs s=-1, it kinda is wrong but what I think he secretly does is so called Abel summation, where you look at the limit when x approaches -1 it actually gives 1/4 so it's actually true. Although in this video just plugging x=-1 doesn't work but with Abel summation it gives 1/4. Google Abel summation for more information

[@MrCmon113]

pantopam You can always justify any BS in retrospect. Mathematics is about being clear and defining what you are doing precisely. The video is just wrong.

[@magicstix0r]

This sum is clearly god's middle finger to mathematicians.

[@Sainterinho]

THIS IS WROOOOOONG :D ... so do not feel special you snowflake

[@adityashankar5267]

😂😂true

[@user-uc3ks5sk4g]

This comment is soooo underrated.

[@informationparadox387]

I just Love this comment!😍🤣

[@herbie6721]

"It gets a little hairy but we'll do it together, we'll go slowly" oh professor, you're so silly.

[@saeedbawazeer9636]

I just can't accept that the sum of positive numbers could have a negative result

[@andrew_cunningham]

Abandon all common sense, ye who enter here. This is realm of logic alone.

[@xMagno313x]

I guess the thing that makes this the most confusing is that I assume that if you add an infinite amount of numbers the answer would be infinity. And if the answer is -1/12, then what happens if I add the"next" term in the sequence, as the sequence can never end?

[@sethwhite4378]

A long-winded explanation, but as short-and-sweet as I can make it. (**SUMMARY AT THE END) (-1/12) isn't "equal" to the zeta function evaluated for (-1), because the zeta function is undefined at (-1). HOWEVER, plugging in (-1) to the analytical continuation of the zeta function produces an output of (-1/12). This is only possible because the zeta function, which is only defined for SOME numbers (in the complex plane), has one specific analytical continuation into the rest of the complex plane (--into the part where the original function is undefined). Since it has one specific analytical continuation -- there is only one possible output for any and every input -- the analytical continuation can be represented as another function. This new function isn't exactly "defined" at (-1) either, but within it there is ANOTHER layer of "continuation", in this case a continuation of an infinite sum beyond its normal defined inputs (specifically, a Cesàro summation). So, the analytical continuation of the zeta function can be represented as another function, and if you plug (-1) into this new function, using the Cesàro summation for the infinite sum within said function, it turns out you get an output of (-1/12). SO... (-1/12) is the output of the extension of a function within the extension of another function which represents a series, a series which depends on a variable for which an input of (-1) produces the sum of all natural numbers; HOWEVER, the actual function ITSELF (not its extension) is UNDEFINED at (-1), because the series it represents (the sum of an infinite set of numbers) DIVERGES when the variable within it is set to (-1) (i.e., when it becomes the sum of all natural numbers) . **ERGO, the sum of all natural numbers does not “equal” (-1/12), but rather corresponds to (-1/12), according to a function which extends beyond the realms of the very numbers we’re adding together.

[@DocBree13]

He offered to demonstrate how he could “push” the |x|

[@pete275]

when he started talking about the strings and their harminics I realized that the standard music scale has 12 semitones, kinda freaked me out

[@MrCmon113]

That is completely unrelated.

[@-dazz-]

Can somebody riddle me this? At 1:45 that result is only valid if abs(x) < 1 But then at 3:40 ... "let x = -1" Doesn't that break it all? I know I must be missing something obvious here

[@HWEWSWEW]

dazz This is a method in complex analysis called analytic continuation in order to extend the domain of functions. Without that context, yes it breaks everything, but within this context it is okay. Obviously, 1+2+3... is undefined but in order to assign it a value, this framework is used. In the same way that no number squares will equal -1, we create a new framework to allow us to find roots to polynomials with no roots.

[@amanmavi4118]

@@HWEWSWEW finally now I can sleep

[@trevordennison9115]

Doesn't 2^(-s)*1^(-s)= 2^(-2s)? Wouldn't the two negative exponents be added to form the -2s exponent? If I am wrong please elaborate.

[@ItsEverythingElse]

I don't know if this was covered in any of the -1/12 videos but what is the practical aspect of this value? Is it used for other mathematical theories or proofs or what? Maybe you could do another video on that :)

[@mynameisawesomeman]

This proof is still incorrect. The faulty part is where he says that 1-2+3-4+5.. =1/4 based on the power series 1+2x+3x^2+...=1/(1-x)^2. If x = -1, then the LHS does not exist. In fact in the limit the power series oscillates around approx. 0 for say x = -.999.. But saying that for x=-1, that the RHS = LHS = 1/4 is totally preposterous, and is equivalent to dividing by zero. In fact the limit DOES NOT EXIST at x = -1 and saying that this is 1/4 is just pure lying!!

[@zacharychaney9489]

Watch the main video, they explain perfectly why that function will equal 1/4, using algebra.

[@KutuluMike]

The left hand side doesn't exist under the simple rules of limits taught in precalculus (the series' partial sums diverge, so it has no limit). Fortunately, people like Euler and Cesaro and Borel figured out how to sum divergent series hundreds of years ago, thus generalizing things like the power series limit for cases that had previously been undefined.

[@zacharychaney9489]

It may exist, however, when the function is expanded to the complex plane, as the actual [formal] proof of this involves the reiman-zeta function where s is a complex number.

[@Harlequin314159]

Much better explanation than the first video. Thank you!

[@user-br6ry8bu5b]

It is wrong !Because: 1+x+x^2+...=1/(1-x) , |x|

[@ThaiNguyen-wp6by]

True !!

[@JazzyBoiZ]

Id be curious to see Numberphile's response to debunking videos regarding this proof. If they are aware I feel like it would be interesting to see how they address the critiques or maybe even admit being wrong

[@dlbattle100]

Does this work in any number base, or would you get a different answer if you did it in, say, base 8?

[@mandydax]

This is actually making me a bit angry. The sum of any two integers is an integer and the sum of any two positive numbers is a positive number. How an infinite series of positive integers can be equal to a negative fraction is beyond me. Tony's explanation starts with something that doesn't really have an answer or has two answers, so lets average them, and then builds on that as if the average were really the answer. How is it not undefined like 1/0? It's mad, and it's maddening.

[@Jacobswe]

I really like the video, and the explanation. I don't like the final annotation (or whatever you call it) that is meant to send you to a discussion about physics between Tony and Ed. Today it just sends me to the Sixty symbols page (which is amazing), but not what I was after at this instance.

[@oooBASTIooo]

No idea why this video is still up. It is easily the numberphile video with the most wrong statements I have ever seen. Maybe someone should have told the guy that an implication with a false assumption is true, regardless of the truth of the conclusion. You can make a video about this, but then do it properly and introduce analytic continuation.

[@DanielEstrada]

John Baez makes a surprise appearance at 15:50!

[@Mandragara]

How are the comments months older than the video lol Numberphile

[@MahmoudYahyaoui]

Update

[@WalterKingstone]

I think because the video has been unlisted for about a year.

[@SunOfCinder]

Pure Mathematics!

[@WalterKingstone]

Haniff Din Yeah, unlisted as in uploaded, but only watchable from a link. So you probably watched the proper -1/12 video, then clicked on a link from that to watch this.

[@Oners82]

Magic!

[@timelsen2236]

S2=1/4 can be done directly as a 4 term partial sum average on even term sums and odd term sums giving 1/4[ ( n-n)+(-m+m+1) ] =1/4 as a variety average of types.

[@subarnasubedi7938]

Firstly he stated that 1/1-x is true for x

[@enesulasdincer7049]

i am agree with you.

[@igoranisimov6549]

That's actually how Romanian came up with his "prove" because he was not formally educated in math. This concept is similar to imaginary numbers. Square roots of negative numbers do not exist but we pretend that they do. For sanity, sum of natural numbers is not equal to -1/12, it is only a concept.

[@Invisifly2]

It's like saying √-1 is i. It's breaking the rules in a clever way to do what you want anyway, and yet, you get real results out of it that actually work and can be experimentally verified (and have been).

[@igoranisimov6549]

@@Invisifly2 yes, it is called ramanujan summation in case of the sum, not a conventional number. Imaginary number is another convention, and cannot be experimentally verified.

[@GBloxers]

@@Invisifly2 how have they been "experimentally verified?" can you show me an instance where the sum of natural numbers = -1/12 has been useful for describing things in the real world?