the most famous Ramanujan sum 1+2+3+...=-1/12

What is the sum of all the natural numbers? Isn't it just infinity or is it really -1/12? Here we will see 1+2+3+... "will be" -1/12 if you use the Ramanujan summation, which is a way to assign a value to a divergent series.
Check out Bose Integral: • Zeta function in terms...
Check out NOT -1/12, • Not -1/12
Sum of 1/n^2 by a Gucci Integral: • a spectacular solution...
Sum of n^2, • how Ramanujan did 1^2+...
Ramanujan Summation: en.wikipedia.org/wiki/Ramanuj...
Ramanujan summation in detail: hal.univ-cotedazur.fr/hal-011...
The sum of all natural numbers:
by 3b1b: • But what is the Rieman...
by Numberphile: • ASTOUNDING: 1 + 2 + 3 ...
by Mathologer: • Ramanujan: Making sens...
🛍 Shop my math t-shirt & hoodies: amzn.to/3qBeuw6
💪 Get my math notes by becoming a patron: / blackpenredpen
#math #mathforfun #blackpenredpen #calculus #ramanujan

[@Nondas8552]

When you have to prove that the sum of all natural numbers equals -1/12 at 10 pm and go on a date at 11 pm

[@blackpenredpen]

Yup!! : )

[@leif1075]

@@blackpenredpen isnt it only true if you include inaginary numbers like negative one square root..otherwise it is impossible and incorrect! Because the sum,always increases..

@@leif1075 If you taking the sum of all counting numbers from one to infinity then your sum is already increasing continuously. What controversy on this integral sum!!

[@johnny_eth]

It's not equals. The sum is divergent. The Ramanujan summation is a transform that assigns values to divergent series. Kind like how a Fourier transform gives you the domains of frequencies and phases of a periodic signal but it NOT equal to the signal.

[@19divide53]

@@johnny_eth The equal in OP's comment is obviously in a Ramanujan summation sense

[@badmanjones179]

black pen red suit

[@David-ww2sg]

*tom scott has joined the chat*

[@zwz.zdenek]

Still YAY!

[@gabbarisback6052]

😂😂😂

[@mukeshkumar_manigilla]

Legends thought

[@alanturingtesla]

Square root of -1‚ now convergent-divergent series. In a few years, I expect dividing by 0.

[@tricky778]

@Alan Turing, which size of 0? a big 0 or a little 0? I'm sure if you divide by a big 0 you get a smaller result than a little 0

[@arnavanand8037]

I feel like for that we need to redefine division. Just like there's gamma function for factorials

[@GynxShinx]

lim(x->0+) 1/x->infinity If you're too lazy to write the whole thing then 1/0+=infinity

[@arnavanand8037]

@@GynxShinx I think everyone knows the limit already. But people feel unsure about the actual answer

[@oracle7858]

Eagle Shows Down 1/0+ is infinity but what is 1/0 🤔

[@radiotv624]

This is fascinating, I love Ramanujan

[@kingbeauregard]

Me too, especially the chicken-flavored stuff.

[@davidrheault7896]

The monster is back again...i know it from physics waves in vacuum. -1/12 casimir effect

[@radiotv624]

David Rheault That’s right! What is/was your major

[@davidrheault7896]

@@radiotv624 I did a specialisation in physics It is above major.

[@indicgamer2907]

Ramanujan's work opened a new world of mathematics that astrophysicist use to study black hole , time travel , free energy quantum tunneling

[@moumous87]

2:00 you got to love this guy for putting the R on top of the = and for actually showing what the heck is this "Ramanujan summation" thing. Thank you

[@themeeman]

Looking sharp

[@blackpenredpen]

Clingfilm Productions Thank you!!!!

[@dayzimlich]

One of your best videos yet - keep up the great positive attitude in 2019!

[@blackpenredpen]

dayzimlich thanks!!! : )

[@benjaminbrat3922]

Thank you very much for this much-needed relativism when talking about divergent series. You could go even further by presenting several alternative summations (Cesàro, VP, Lambert, Borel, etc), I know Cesàro to be quite easy to grasp, and this would contribute bringing diverging sums enough breathing room for actual exploring, instead of bantering. Happy New Year!

[@angelmendez-rivera351]

The discrepancy between saying that the series diverges and the sum is -/12 is that they are not even talking about the same thing. The fact that it diverges is a statement about asymptotic behavior and sequences, whereas the latter is a statement about arithmetic and infinite sets. Calculus is a theory about sets and functions. When we deal with summations in calculus, we never truly deal with an infinite summation, so to speak, although we do call them this out of bad tradition (just like how calling imaginary numbers imaginary is bad tradition). In calculus, what we do deal with instead is sequences of partial sums. Why? Because this tells us about the algorithm of after one number after another. If I start adding the natural numbers, in a specific order, and form a sequence for every step, then what is the behavior as I increase the number of steps? The behavior is that this sequence simply becomes infinite, every number I get is larger and larger on several orders. Although we tend to associate this with adding infinitely many terms, this is not what is truly happening and is merely an informality that happens in the calculus classroom, since explaining the real details behind standard analysis is complicated and outside the scope of the syllabus. In a sense, using the limit of the sequence of partial sums is already a way of assigning values to infinite series. It already is “a summation method”, but it is an error to call it a summation in the arithmetic sense because it obviously is not, since it has very different properties. This bears no contradiction with what Ramanujan postulated because Ramanujan is not talking about sequences and their limits. What Ramanujan is trying to do instead is imitate more closely the properties of summation as given by number arithmetic. Naturally, since we are adding elements of infinite sets, the results we produce are counterintuitive and outside of what induction can allow us to prove. Keep in mind, though, that using Ramanujan summation is not the only way to arrive at this result. Abel summation and Borel summation are both more intuitive than Ramanujan summation, and they also give -/12 as a result. Also, here is an interesting article by Terrence Tao that justifies this non-classical result in a very convincing way. terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/#zeta-s One thing I find interesting is that the integral from -1 to 0 of x^2/2 + x/2 is 1/6 - 1/4 = 2/12 - 3/12 = -1/12. It is interesting because this integral is the area between the x-intercepts of the function f(x) = x^2/2 + x/2 = x/2•(x + 1), which is equal to the nth partial sum of the sequence of natural numbers for x = n.

[@user-ft2vp5yw6p]

Wow, just wow. Amazing

[@DarthRaven9000]

Very well written. This clarifies what seems contradictory when the topic is usually discussed without a proper perspective. Thank you.

[@DanNguyen-oc3xr]

"Naturally...counterintuitive." Damn it.

[@suprafluid3661]

@@DanNguyen-oc3xr But i was going to say that hmmm.. 😣

[@abustefano8225]

Mmmm... Something wrong here. According to Abel-Olana formula en.m.wikipedia.org/wiki/Abel%E2%80%93Plana_formula the series diverges. Please elucidate

[@saxbend]

Mathologer wants a word.

[@NateROCKS112]

Mathologer actually hinted at what is going on in this video. He showed that the area under the x-axis of f(n) = n(n+1)/2 is -1/12, which can be interpreted as a definite integral from -1 to 0 of n(n+1)/2. n(n+1)/2 is a graph of this infinite series (as n→∞) since it's the sum of all natural numbers up until n. Also, BPRP used the proper notation with the Ramanujan summation, and did not claim that the natural numbers "summed" to -1/12, just that they could be assigned to it.

[@angelmendez-rivera351]

NateROCKS112 Mathologer did show this, but he equally also spoke against any association between adding the natural numbers and obtaining the result -/12 via analytic methods. The problem lies in how poorly and inconsistently the technical terminology of calculus is being used, and how poorly the word “summation” is interpreted. The problem is that, nowadays, when dealing with sequences of partial sums, we tend to associate them to summations themselves, without much justification, even though these summations are not sequences, but numbers, and the summations are furthermore the result from a set of rules of arithmetic operations on a field, whereas statements about these sequences are statements about vector spaces and linear functionals. When we see the sentence 2 + 2 = 4, this statement is provable. This has nothing to do with sequences. It is a statement that is the consequence of the rules of addition and basic arithmetic. When we see that f(x) = x^2, and we calculate f(1) + f(2) + ••• + f(100), this is also a statement about arithmetic, and summation notation is merely a notation to abbreviate this result: the operation is ultimately still addition. I see no reason why 1 + 1/4 + 1/9 + ••• = π^2/6 should be treated any different. In fact, to prove this identity, we typically just use algebra and trigonometry, not much from calculus itself though. I see no reason why 1 + 2 + 3 + ••• = -/12 should be treated any different either. Caught hated the concept of infinity. He was, so to speak, an ultra finitist. This is why he invented the concept of a limit, and whenever we wanted to speak of adding infinitely many numbers together, he rejected the idea, and instead proposed to talk about the limit of the sequence of partial sums. For some practical purposes, and for his specific theoretical, both mathematical and philosophical purposes, this definition works just fine. Nowadays, we treat it is a method to assign value to infinite sums, but strictly speaking this method does not give you the sum itself. In fact, as I already said, Cauchy would have said such sums do not exist, and the expressions are nonsensical, precisely because he did not believe in the notion of infinity. So associating actually adding infinitely many terms with these limits is conceptually incorrect, and an equivocation of bounded behavior with infinite quantity, which should not happen. What the limit of the sequence of partial sums tell us is information about the asymptotic behavior of some function, and this function represents an algorithm. In the case of the natural numbers, what this limit does is answer the question, “What happens when, at every step of some process, I add the next natural number to the total I already have?” It tells us that this process will result in a number which at every step is larger than it was before, and at an increasing rate. I am not succeeding in getting closer to some value when I do this process. This is the proper meaning of what divergence is for this case. This tells us nothing about ACTUALLY adding the natural numbers, which intuitively and arithmetically should have nothing to do with sequences (because if we defined addition by sequences, it could never be commutative or associative, but we know addition is a commutative operator for fields). In fact, it tells us nothing about adding infinitely many numbers in general in the first place. We make the association because we want to and because it is practical in some contexts, but again, they strictly are not the same thing, especially when you understand the mathematical logic behind the axioms and definitions. Obviously, mathematicians tend to understand this, but when students learn calculus, they do not learn any of the rigorous details behind the definitions or theorems involved, so naturally they get confused.

[@Lolwutdesu9000]

@@angelmendez-rivera351 this is a brilliant post.

[@literatedouchebag]

@@angelmendez-rivera351 i couldn't have said it any better. Amazing post my dude

[@angelmendez-rivera351]

VeryEvilPettingZoo “But it’s the only definition that makes any conceptual wrong.” And that’s obviously wrong, as mathematicians from the 17th century have proven with their understanding of divergent series, which literally preceded any understanding of limits of sequences. The fact that surreal numbers and transfinite addition exists, without need to use limits and all, further proves this point. You must know any mathematics beyond those of your calculus II textbook if you think it’s the only conceptually sensical definition. But the other reason you are wrong is because there is no such a thing as “the ordinary conceptual understanding” of something in mathematics. That doesn’t exist. Mathematics is strictly and exclusively about what axioms you work with. And depending on the axioms, something is true, or it isn’t. And here is the catch: there is no standard set of axioms that is used in every field of research. Even within the same field of research, if we are having a conversation and I decide to get a different level of rigor, even there we’ve already changed the scope of the axioms and of the theory. Because of this, there is no ordinary understanding of anything: any understanding of anything just comes from axioms and that’s that. Math has nothing to do with intuition, and because of this, nothing is ordinary or conventional. That is what people refuse to understand. The only reason some things seem conventional to some people is because they’ve been limited to only specific math courses or they were exposed to something first for the longest time before being taught something they personally would consider “non classical”. If you had been raised with a calculus course learning about Ramanujan summation first and then being exposed to it continuously to work with it in engineering for the rest of your life, then you would perceive that to be the only conceptually ordinary sense to add infinitely many things. But that too would be an illusion. If there were such a thing as an ordinary conceptual understanding, then it wouldn’t be necessary to learn maths in such a way that in every new course, everything you learned previously is a lie and that there is another way to look at things. And look: mathematicians said the exact same thing about complex numbers centuries ago. The fact that people continuously make these claims and centuries later are always disproven is an obvious indication to the fact that such ideas about understanding are subjective and merely illusions, artifacts of tradition, not real in any mathematical or logical sense. “...of what it would mean to add up the terms of that series “forever”.” There is no such a thing as adding things up “forever” in mathematics. I don’t care if you put it in quotation marks or not: that word should not be in that sentence, not even in a remote, metaphorical sense. You’re projecting feelings and intuition that don’t exist in mathematics into the subject. Adding infinitely many numbers does not have anything to do with time. If I add one by one, then yes it would take me a very long time to finish the process, mechanically speaking, but processes don’t define anything in mathematics. If we can agree that a sum of infinitely many numbers can have a value, then that has nothing to do with time. Talking about time only furthers the misconception about what infinite sums represent notionally. If I have a way of adding infinitely many numbers and knowing what that is, then in principle there is no reason I shouldn’t be able to do this immediately, in less than a picosecond, or faster. And if time was of relevance, then it would be impossible to talk about infinity in the first place, so the concepts of convergence and divergence would make no sense whatsoever. “Conceptually, it’s indisputable that adding up the series “indefinitely” drives the sum up towards infinity, not -1.” And once again you made the mistake I pointed out in my comment. It’s like you just missed my point altogether. If you have a sequence defined by steps, where in every step, you add the next natural number, then yes, you get a process by which this number will increase to infinity as more and steps are performed, arbitrarily. And unfortunately for your argument, that’s NOT what the sum itself is. Summation is an operator, not a process. Nothing in mathematics in any field is a process. The only field in mathematics that talks about things related to processes may be some subfield of discrete mathematics concerning computing power and what not. But those things don’t define operations, nor should they. A mathematical identity and the process you get by it computationally are strictly unrelated. Perhaps children have a strong association between operations and processes, because it’s the only exposure they’ve ever had of operations, since they cannot really understand the abstract essence of what operations are, but that’s it. If I can add all the powers of 2 at once, then there is no “driving the sum” anywhere because there is no process, there is no sequence. Any extension of an operation should capture this abstract essence of what the operation is rather than any false non/existing notion of a procedure. You can choose to create a mathematics which is based on procedures, maybe define something call procedure theory. But that wouldn’t be arithmetic of real numbers anymore, that would be, well, procedure theory. And worse for the argument is that this concept of limit to infinity depends strictly on the type of infinity you are choosing to use as a boundary condition on the real line. Calculus on the projective line looks very different. Which only proves my point further. I will watch the video, though I hardly doubt they’ll say anything fruitful I haven’t already addressed, unless they end up agreeing with me. I say this as I’ve been in over 50 different discussions about the subject.

[@fCauneau]

That's what I call a Christmas gift ! Thanks !!! And Merry Xmas to you and your readers !!

[@biswadeepghosh5568]

Great presentation, felt so good since this concept is related to the work of Ramanujan, many thanks to help me understand this concept.

[@YourPhysicsSimulator]

So elegant... I'm talking about you, though.

[@blackpenredpen]

Thanks!!

[@spudhead169]

Thank you, both Numberphile and Mathologer didn't really explain this properly, that it's a "non-canon" summation.

[@rykehuss3435]

Nice way of saying its just made up by Ramanujan to goof around with

[@anindyabiswas1551]

I've waited a long time for this video. Thank you very much.

[@nekososu]

or you should say the zeta(-1) is -1/12

[@abhiruppaul5601]

correct bro

[@arnoldo-probjeto3111]

Not correct bro. In fact, ACzeta(-1)= -1/12, where ACzeta is the analytic continuation of zeta function, that is NOT the zeta function.

[@19divide53]

@@arnoldo-probjeto3111 But we call the analytic continuation of zeta function by ζ(s) as well. A bit of abuse of notation but I think that's the convention. In fact, in Riemann's 1859 paper he did denote the analytic continuation of zeta function (sum of n^(-s) over all positive integers n) ζ(s), which you can see by looking at the paper online. There's no reason to say ζ(s) may only be the sum of n^(-s) over all positive integers n. When I'm doing some other problem unrelated to the Riemann Zeta function, I could define a function, say, s^4sin(3s)-cos^5(s^2)+6/s-9/s^7, and call that ζ(s). That's just as valid as calling my function f(s) or g(s).

[@SebastienPatriote]

This just went next level!!

[@nicolasgoubin]

Whaaaaaaah so HANDSOME ! Merry Christmas & Happy New Year YAAAAAAAY

[@NoNTr1v1aL]

Thank you Ramanujan, very cool!

[@elthomaso10]

This is hands down the most terrifying video title I've ever seen.

[@kostagerosky534]

This video was so enjoyable, you explain things so nice!

[@SartajKhan-jg3nz]

Everything in the video is🔥🔥🔥. The integral, the result and yes...you!

[@basedoppenheimer1497]

Bro I don't even know what to say. This is mind blowing in such ways you feel like you got transported to another dimension.

[@angelmendez-rivera351]

UPDATE: I found that if one takes the polynomial in n of degree s + 1 that gives you the sum 1^s + 2^s + ••• + n^s, then if we integrate the same polynomial in x with respect to x from -1 to 0, one obtains the same value one would obtain if one applied Ramanujan summation instead. The key is in using Faulhaber’s formula, finding the antiderivative of it in x, where n = x, then set x = -1, and the resulting formula is equal to -B(s + 1)/(s + 1), where B(s) is the s-th Bernoulli number. This gives the same result as Ramanujan summation, which also gives the analytic continuation of ζ(s). I’m mentioning it because I thought it was interesting. So this generalizes the result from my first comment.

[@submarino006]

I like your funny words, magic man

[@ToHoUSA1]

ive been waiting for this

[@blackpenredpen]

: ))))

[@mitchelvalentino1569]

Yes!!! I’ve been waiting for this exact content. Very well explained. Thank you!!

[@egalvaoepg]

Excelent video, as always. Congrats!

[@adamhrankowski1298]

Thumbed up without even watching. Loved Mathologer's vid. Have this stuff on a T-shirt. Really looking forward to what BPRP has to offer. #Yay

[@archithtelukunta4599]

Just few days ago I was just thinking whether u would ever upload a video on this series and here it is😃

[@fmakofmako]

Would it be possible for you to do a video on analytic continuation or smoothed sums? Both have relevance to this video and the riemann zeta function.

[@beatoriche7301]

Great video! If I may ask, how do you actually derive this formula?

[@martind2520]

This is awesome. Can you do a video proving that Ramanujan Summation is consistent with regular summation for non-divergent series?

[@MrLecancre]

Try with cv riemann series

[@jzanimates2352]

You should make a collab video with 3blue1brown or numberphile!

[@europeankid98]

YES

[@mulongoduncan9308]

Thank you. Your derivation makes much sense.

[@JoshuaHillerup]

Is there a different term for when the limit goes off to say positive infinity, versus when the limit has multiple finite values?

[@lautamn9096]

Happy new year bprp!

[@blackpenredpen]

Thank you!! You too!!

[@federicopagano6590]

we have 2 options depending on which class we are in. But wich one is the correct answer if u are alone no classroom and u have to answer

[@blackpenredpen]

That will be depending if I want to make my life easy or not.

[@federicopagano6590]

blackpenredpen OK got the point diverges lol

[@angelmendez-rivera351]

Federico Espil He did not say that. Obviously you want him to say that, though, because you already are prejudiced and already believed beforehand the sum diverges. I wrote a comment addressing this.

[@shoobadoo123]

When applied in the real world, -1/12 is the answer. This appears in physics, string theory specifically. You can google to learn more there’s a lot of info!

[@debabrataadhikari4879]

Out of all yt channel this is the best explanation

[@dppid083wk7]

Love ramanujans work

[@yoavcarmel1245]

very nice suit, and great video as usual!

[@matthiashannesson7239]

-1/12th?

[@djsmeguk]

What would be interesting is how the ramanujan summation formula is derived.. does it concur with convergent series as well?

[@angelmendez-rivera351]

djsmeguk I believe the answer is yes, but I’m not 100% certain.

[@nathanisbored]

i dont think its derived, its just a definition

[@angelmendez-rivera351]

nathanisbored It is inspired by some derivation. The definition only plays in a role in the domain of functions to which it applies. It is a function of functions.

[@angelmendez-rivera351]

Markus Steiner “I think it's neither derived nor it's just arbitrary.” Be more specific with what is it that you mean when using the word “derived”. This may not have a derivation from the axioms, but it has a rigorous foundation on time-scale calculus, which is a theory of which the calculus of sequences is a special case. It also has equally rigorous foundation in complex functional analysis and set theory. It uses the projective extension of the complex plane, known as the topological Riemann sphere. Also, it is not arbitrary, because it can be shown that, given any argument from heuristics, the continuation entailed by the heuristics is formalized with this method presented in the video. If any continuation to summation should exist so that divergent series could be evaluated, then this must be the one. Namely, had the method of convergence of the sequence of partial sum never existed, the field axioms of addition would heuristically entail these results. “The thing is that it also has to work for not divergent series.” It does. In his following video, which sums all the square numbers of the natural numbers, I wrote in the comments section that it is easy to prove that this series sums every convergent geometric series to its correct value, and it appears as though it does so with series involving trigonometric functions as well, as well as for the sum of all 1/n^2 for n in N. “So there might be more possible formulas which achieve this with different results for divergent series.” There are. Borel summation exists, as well as Abel summation and stronger linear methods. Wheel algebra and transfinite set theory can also derive these results independently. “Choosing one of these possibilities is in fact arbitrary.” No, it is not. If two different theories derive the same conclusions, then this entails the theories are, at the very least, consistent, but it could be the case that one is a sub-theory of the other, or even more radically, that given some super-theory containing both theories, that they are equivalent theories. Being able to prove the law of cosines using either Classical Greek geometry, or else using a Hamiltonian algebra of 4-ions does not mean that choosing either theory as the foundation for a method to obtain results is arbitrary. That is not how math works, obviously, nor has it ever been this way.

[@sl.murarikumar9675]

kzbin.info/www/bejne/pZXRZ6qbgN-eZ7s Ohhhhhhh Nooooooo Ramanujan questions answer is right or wrong Very confusing -1/12. OR. -1/8 🤔🤔🤔

[@AbbeyRoad69147]

Please can you do a video on the R. summation. Thanks, I'd like to see how it is derived.

[@EduardoHerrera-fr6bd]

Finally! I was specting for this including if I had to go for 2 weeks :c 7:13 Balls Integral.

[@sharathgowtham2157]

Lmfao!!!!!!!!!!!!!

[@ahmedbashandy4784]

Is there any application of ramanujan sum? Some thing like using complex number in schrodinger equation.

[@victoryforvictims3522]

Excellent explanation. It all makes sense now.

[@infernocaptures8739]

Is it possible to assign a value to Σ(n^n) with this method?

[@scarbotheblacksheep9520]

If "diverges" and -1/12 are correct depending on how you think of it, could the domains of these be different? Like, is the Ramanujan summation including transfinite numbers in its sum, or something like that?

[@stratonikisporcia8630]

Does it have anything to do with the p-adics such as the 10-adic form of -1/12 = ...3,333.25

[@ianprado1488]

How many markers do you buy a month?

[@tensoescalar1]

This is a very impressive video, thank you very much, greetings from Mexico.

[@TheNachoesuncapo]

looking good steve!

[@sohamkumar2346]

Love from India 🫡 Proud of Ramanujan

[@matteoserafini7704]

For 1/X you just notice that the function in odd, so in a "simnetrical" interval the area in the positive part of the interval is the same of the area in the negative part of the interval, but with the opposite sign. So the sum of the two is zero

[@benjaminparra4672]

What about the sum of all the factorials of the natural numbers?, I couldn't do it because the double integral (using the pi function).

[@elyseepasteur6163]

Please will you tell me the meaning of the R and its role to make more clear? Or give me the link of its video please 101.

[@sanatankaushik456]

This is the power of the "Man who knew Infinity"!

[@alainrogez8485]

One word : brilliant!

[@raiedahmednishat8883]

You even dresses up for it. I love you man. あなたおあうしている

[@angelmendez-rivera351]

Another way in which these results make sense is in wheel theory. The field of complex number can be extended by first equipping it with an involution operator / that satisfies certain properties. Then we introduce to new elements to the complex number, /0, and 0/0. Every element of this new set has a well-defined involution: for all z, /z is well defined. Whenever z^(-1), which is the reciprocal of z, or multiplicative inverse of z, exists, z^(-1) = /z, but even for z with no multiplicative inverse, /z is well-defined. Namely, /0 = ψ and 0/0 = θ, with the properties that /ψ = 0 and /θ = θ. Also, ψ + ψ = ψ - ψ = θ, ψ + a = ψ for any complex number a, 0ψ = θ, and aψ = ψ for any complex nonzero a. Furthermore, θ + z = yθ = θ for all wheel numbers z & y (by wheel numbers, I mean complex numbers plus the two new elements we added to the field, which is now a wheel). There are some identities I have not given concerning / for all these, though right now I cannot remember them all. But there is a Wikipedia article you can find if you simply search Wheel algebra on google. We can give this wheel the wheel topology. With this wheel topology, 0 and ψ are boundaries of this space. As such, ψ somewhat plays the role of infinity in this space, with a few key differences, since now this is the only infinity regardless of the direction.

[@harshitkumar4760]

It means that negative numbers are greater than positive and smaller too, it means its a cycle which repeats positive after negative and again positive. Think it this way as we increase the angle in tangent function its value reaches to infinity and then negative! And then again positive.

[@tomaszzywicki8446]

Can you use this formula for convergent series?

[@holyshit922]

Integral Int(u/(exp(u)-1),u=0..infinity) can be converted to Basel problem ?

[@user-zz3gt9wo4t]

Very nice video, thank you for that, but can you please make video about this integral summation what does it mean and why?

[@wolframalpha8634]

Happy new year bprp

[@tylertorsiello8450]

this guy is my spirit animal

[@kamidhunkumar3021]

Integration of 1/x from -1 to 1 seems to be elementary. The definite integration is introduced as the area under curve and it is evident that the area ' enclosed '(l mean symmetric with respect to origin) by 1/x in the 1st and 3rd quadrants cancel each other.

[@maccook9951]

At about 3 minutes in I had to actually check this wasn't uploaded April 1st

[@anweshaguha7366]

Okay, so I'm in love with that "Hmm" before you start solving a tough integral, I mean tough for me😊

[@stanislasnicolau2412]

Hi! Can you give me a link or an explanation of how we demonstrate that integral from -1 to 1 of 1/X dx can be equal to 0? Thx a lot

[@keoviwath]

To get that answer, I think we should use the fact that the integral from -a to a of an odd function is zero.

[@akshatsharma9882]

I love this guy.

[@epicawesomenessnucleartrin1585]

Can you integrate floor(x) from 1 to infinity? :)

[@SupratheekSreedharaNagaVenkata]

@blackpenredpen 10:54 u can just say integral 1/x from -1 to 1 is zero bcoz 1/x is odd function so areas are equal in magnitude from -1 to 0 , 0 to 1 but opposite in sign

[@MrBoubource]

What if you chose some other function, like integer part of x, or x + sin(pi x), its gonna produce a different value right ?

[@soccerbels7947]

Maybe its easier way

[@jjeherrera]

Nice as always. Still, it would be interesting to understand what is Ramanujan doing that helps him to make the analytic continuation.

[@science-y9209]

Ramanujan is dead but he was a genius..

[@Gideon_Judges6]

So could this "R equals" be replaced with a triple equals, meaning is defined to be equal?

[@abhijeetdas6279]

Please make a video on the Navier Stokes theorem problem proposed by the Clay mathematical institute

[@hamsterdam1942]

Ze best video about this freaky number!

[@rajatkhandelwal7276]

This video is so so so so so so so so much interesting plz make more videos like this

[@binitkumarsingh8296]

Finally i got the perfect method for that.thnx to u

[@jorgecanales4864]

Si lo tuvieras tus videos con sub titulos en español, serian aun mas geniales, eres un genio

[@Podzhagitel]

has aprendido inglés?

[@muterchak]

The formula for the partial sum of the series 1+2+3+... of N terms is S(N)=N*(N+1)/2. If we represent it as a function of X: f(x) = x*(x+1)/2 and plot this function, then the area under the graph between the points of intersection with the X axis will be just -1/12. Why is that? :)

[@nicolarossi8728]

You are the best!

[@shavishvilitornike]

I have a question: why did you select formula f(x)=x ? Why did not you select some other firmula, which would have different value at 0?

[@khaledelhoushi8834]

Very fascinating video.

[@user-cn7kf8te3v]

The last two examples are appropriate!

[@Ben-wv7ht]

if we take f(x) = 1/x^2 we get f(0)/2+zeta(2)=0 , but as f(0)/2 approaches +oo , would we be able to say zeta(2)= -oo ? ofc I know it's false x) just curious

[@somnathdash4428]

You got addicted to India!!! Love from India

[@dappermink]

1. Does this sum definitions gives the same result as the usual sum for any classical convergent series? 2. Where does it come from?

[@nandakumartulip5279]

1.yes or so I heard 2.given by ramanujan

[@dappermink]

@@nandakumartulip5279 I don't remember the video now actually but thanks!

[@ytrichardsenior]

The numberphile video explains that this answer '-1/12' is used in some fairly complex string theory mathematics. This is what fascinates me. We all know that negative, and imaginary numbers are not 'real'. We can use negative numbers in order to avoid having to 'scale' our calculations. We can use imaginary numbers in order to solve problems that result in 'real' numbers. Is there any way that we can use these 'golden parts' of divergent series as part of a larger calculation in order to somehow get back into reality, in the same way that we can with complex numbers etc.?

[@QuippersUnited]

Quite the dapper jacket, Mr. Pen.

[@sugarfrosted2005]

Fun with analytic continuation!

[@Sush]

Oooh looking fancy today!!

[@stasfilippov5916]

so sum of the positive (and only positive) values is a negative value?

[@siddhantchess]

this is so counter intuitive that its beautiful

[@stevenjohnson9466]

this dude loves simple maths you can see it in his smile :)

[@stefanotorelli3688]

But if I rember well, in every non convergent series we can rearrange the terms so the serie convergens to a given numbere... do I remember well? didn't I?

[@ssaamil]

-1/12 is my favourite number. I love this.

[@deathwildrift3225]

Uhuuu. The video is very nice, you is excellent. :)

[@mikejackson19828]

Can you make the sum of the natural numbers to be whatever you want it to be?