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

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  • Опубликовано: 27 дек 2018
  • 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

Комментарии • 764

  • @Nondas8552
    @Nondas8552 5 лет назад +3307

    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
      @blackpenredpen  5 лет назад +596

      Yup!! : )

    • @leif1075
      @leif1075 5 лет назад +37

      @@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..

    •  4 года назад +8

      @@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
      @johnny_eth 4 года назад +67

      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
      @19divide53 2 года назад +9

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

  • @badmanjones179
    @badmanjones179 5 лет назад +1002

    black pen red suit

  • @alanturingtesla
    @alanturingtesla 5 лет назад +987

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

    • @tricky778
      @tricky778 5 лет назад +63

      @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
      @arnavanand8037 4 года назад +60

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

    • @GynxShinx
      @GynxShinx 4 года назад +18

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

    • @arnavanand8037
      @arnavanand8037 4 года назад +20

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

    • @oracle7858
      @oracle7858 4 года назад +4

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

  • @moumous87
    @moumous87 4 года назад +49

    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

  • @radiotv624
    @radiotv624 5 лет назад +565

    This is fascinating, I love Ramanujan

    • @kingbeauregard
      @kingbeauregard 5 лет назад +32

      Me too, especially the chicken-flavored stuff.

    • @davidrheault7896
      @davidrheault7896 5 лет назад +13

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

    • @radiotv624
      @radiotv624 5 лет назад +2

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

    • @davidrheault7896
      @davidrheault7896 5 лет назад +2

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

    • @indicgamer2907
      @indicgamer2907 5 лет назад +23

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

  • @themeeman
    @themeeman 5 лет назад +417

    Looking sharp

  • @saxbend
    @saxbend 5 лет назад +375

    Mathologer wants a word.

    • @NateROCKS112
      @NateROCKS112 5 лет назад +92

      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
      @angelmendez-rivera351 5 лет назад +105

      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
      @Lolwutdesu9000 5 лет назад +4

      @@angelmendez-rivera351 this is a brilliant post.

    • @literatedouchebag
      @literatedouchebag 5 лет назад +7

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

    • @angelmendez-rivera351
      @angelmendez-rivera351 5 лет назад +8

      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.

  • @benjaminbrat3922
    @benjaminbrat3922 5 лет назад +38

    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!

  • @dayzimlich
    @dayzimlich 5 лет назад +77

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

  • @angelmendez-rivera351
    @angelmendez-rivera351 5 лет назад +226

    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
      @user-ft2vp5yw6p 5 лет назад +7

      Wow, just wow. Amazing

    • @DarthRaven9000
      @DarthRaven9000 5 лет назад +15

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

    • @DanNguyen-oc3xr
      @DanNguyen-oc3xr 5 лет назад +3

      "Naturally...counterintuitive." Damn it.

    • @suprafluid3661
      @suprafluid3661 5 лет назад +1

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

    • @abustefano8225
      @abustefano8225 4 года назад

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

  • @YourPhysicsSimulator
    @YourPhysicsSimulator 5 лет назад +98

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

  • @spudhead169
    @spudhead169 2 года назад +80

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

    • @rykehuss3435
      @rykehuss3435 5 месяцев назад

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

  • @elthomaso10
    @elthomaso10 5 лет назад +14

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

  • @biswadeepghosh5568
    @biswadeepghosh5568 5 лет назад +18

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

  • @nekososu
    @nekososu 5 лет назад +88

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

    • @abhiruppaul5601
      @abhiruppaul5601 3 года назад +2

      correct bro

    • @arnoldo-probjeto3111
      @arnoldo-probjeto3111 2 года назад

      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
      @19divide53 2 года назад +1

      @@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).

  • @fCauneau
    @fCauneau 5 лет назад +3

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

  • @anindyabiswas1551
    @anindyabiswas1551 5 лет назад +1

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

  • @martind2520
    @martind2520 5 лет назад +83

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

    • @MrLecancre
      @MrLecancre 5 лет назад +3

      Try with cv riemann series

  • @nicolasgoubin
    @nicolasgoubin 5 лет назад

    Whaaaaaaah so HANDSOME ! Merry Christmas & Happy New Year YAAAAAAAY

  • @beatoriche7301
    @beatoriche7301 5 лет назад +17

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

  • @kostagerosky534
    @kostagerosky534 5 лет назад +1

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

  • @NoNTr1v1aL
    @NoNTr1v1aL 5 лет назад +2

    Thank you Ramanujan, very cool!

  • @basedoppenheimer1497
    @basedoppenheimer1497 Год назад

    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.

  • @SebastienPatriote
    @SebastienPatriote 5 лет назад +10

    This just went next level!!

  • @jzanimates2352
    @jzanimates2352 5 лет назад +27

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

  • @SartajKhan-jg3nz
    @SartajKhan-jg3nz 5 лет назад +1

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

  • @mitchelvalentino1569
    @mitchelvalentino1569 5 лет назад +4

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

  • @adamhrankowski1298
    @adamhrankowski1298 5 лет назад

    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
    @archithtelukunta4599 5 лет назад

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

  • @egalvaoepg
    @egalvaoepg 5 лет назад

    Excelent video, as always. Congrats!

  • @angelmendez-rivera351
    @angelmendez-rivera351 5 лет назад +12

    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
      @submarino006 Год назад +6

      I like your funny words, magic man

  • @ToHoUSA1
    @ToHoUSA1 5 лет назад +3

    ive been waiting for this

  • @sanatankaushik456
    @sanatankaushik456 7 месяцев назад +2

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

  • @yoavcarmel1245
    @yoavcarmel1245 5 лет назад

    very nice suit, and great video as usual!

  • @lautamn9096
    @lautamn9096 5 лет назад +5

    Happy new year bprp!

  • @debabrataadhikari4879
    @debabrataadhikari4879 5 лет назад

    Out of all yt channel this is the best explanation

  • @matthiashannesson7239
    @matthiashannesson7239 5 лет назад +26

    -1/12th?

  • @fmakofmako
    @fmakofmako 5 лет назад +4

    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.

  • @anweshaguha7366
    @anweshaguha7366 5 лет назад +2

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

  • @angelmendez-rivera351
    @angelmendez-rivera351 5 лет назад

    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.

  • @mulongoduncan9308
    @mulongoduncan9308 2 года назад +1

    Thank you. Your derivation makes much sense.

  • @harshitkumar4760
    @harshitkumar4760 5 лет назад +2

    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.

  • @scarbotheblacksheep9520
    @scarbotheblacksheep9520 5 лет назад +3

    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?

  • @TheNachoesuncapo
    @TheNachoesuncapo 5 лет назад +1

    looking good steve!

  • @jjeherrera
    @jjeherrera 5 лет назад +9

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

    • @science-y9209
      @science-y9209 3 года назад +1

      Ramanujan is dead but he was a genius..

  • @ianprado1488
    @ianprado1488 5 лет назад +7

    How many markers do you buy a month?

  • @EduardoHerrera-fr6bd
    @EduardoHerrera-fr6bd 5 лет назад

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

  • @wolframalpha8634
    @wolframalpha8634 5 лет назад +1

    Happy new year bprp

  • @maccook9951
    @maccook9951 5 лет назад +2

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

  • @stevenjohnson9466
    @stevenjohnson9466 5 лет назад +2

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

  • @tensoescalar1
    @tensoescalar1 3 месяца назад

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

  • @matteoserafini7704
    @matteoserafini7704 2 года назад

    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

  • @dppid083wk7
    @dppid083wk7 5 лет назад +1

    Love ramanujans work

  • @federicopagano6590
    @federicopagano6590 5 лет назад +44

    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
      @blackpenredpen  5 лет назад +36

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

    • @federicopagano6590
      @federicopagano6590 5 лет назад +3

      blackpenredpen OK got the point diverges lol

    • @angelmendez-rivera351
      @angelmendez-rivera351 5 лет назад +2

      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
      @shoobadoo123 5 лет назад +2

      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!

  • @jorgecanales4864
    @jorgecanales4864 4 года назад +9

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

  • @djsmeguk
    @djsmeguk 5 лет назад +64

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

    • @angelmendez-rivera351
      @angelmendez-rivera351 5 лет назад +4

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

    • @nathanisbored
      @nathanisbored 5 лет назад

      i dont think its derived, its just a definition

    • @angelmendez-rivera351
      @angelmendez-rivera351 5 лет назад +5

      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
      @angelmendez-rivera351 5 лет назад +5

      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
      @sl.murarikumar9675 5 лет назад

      ruclips.net/video/odn6weKzk1Y/видео.html
      Ohhhhhhh Nooooooo
      Ramanujan questions answer is right or wrong
      Very confusing
      -1/12. OR. -1/8 🤔🤔🤔

  • @victoryforvictims3522
    @victoryforvictims3522 2 года назад

    Excellent explanation. It all makes sense now.

  • @cagnusmarlsen4050
    @cagnusmarlsen4050 4 года назад +1

    I love confusing my math teacher watching your videos :D.

  • @QuippersUnited
    @QuippersUnited 5 лет назад

    Quite the dapper jacket, Mr. Pen.

  • @somnathdash4428
    @somnathdash4428 5 лет назад +1

    You got addicted to India!!! Love from India

  • @integrando1847
    @integrando1847 3 года назад +1

    very interesting that maths say the same thing in a lot of ways, this summ appears in quantum mechanics, infinite series, complex variable, etc..

  • @sohamkumar2346
    @sohamkumar2346 3 месяца назад +1

    Love from India 🫡 Proud of Ramanujan

  • @AbbeyRoad69147
    @AbbeyRoad69147 5 лет назад +1

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

  • @JoshuaHillerup
    @JoshuaHillerup 5 лет назад +1

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

  • @rajatkhandelwal7276
    @rajatkhandelwal7276 5 лет назад

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

  • @Sush
    @Sush 5 лет назад

    Oooh looking fancy today!!

  • @user-zz3gt9wo4t
    @user-zz3gt9wo4t 5 лет назад

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

  • @kamidhunkumar3021
    @kamidhunkumar3021 5 лет назад

    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.

  • @SupratheekSreedharaNagaVenkata
    @SupratheekSreedharaNagaVenkata 5 месяцев назад

    @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

  • @blazedinfernape886
    @blazedinfernape886 5 лет назад +71

    The views should be -1/12

    • @kishanthakor971
      @kishanthakor971 4 года назад +1

      😂😂yaaa

    • @gabbarisback6052
      @gabbarisback6052 3 года назад

      😂😂

    • @shivamchouhan5077
      @shivamchouhan5077 3 года назад +1

      🤣

    •  3 года назад +1

      your comment is at 69 likes already, so I'm using *this* comment as a like button instead

    • @pardeepgarg2640
      @pardeepgarg2640 2 года назад

      @ I ruined that 69 likes 😈😈

  • @raiedahmednishat8883
    @raiedahmednishat8883 5 лет назад

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

  • @ssaamil
    @ssaamil 2 года назад

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

  • @late7245
    @late7245 5 лет назад +80

    1-1+1-1+........=1/2 confirmed

  • @deathwildrift3225
    @deathwildrift3225 5 лет назад

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

  • @khaledelhoushi8834
    @khaledelhoushi8834 9 месяцев назад

    Very fascinating video.

  • @Gezraf
    @Gezraf 6 месяцев назад

    the "2" you wrote in 7:03 is the most beautiful 2 ive seen

  • @peterheerens3093
    @peterheerens3093 2 года назад

    It always amazes me how many way there are to prove 1+2+3+4+5+...... = -1/12. It is counterintuitive, but you cannot argue with the prove

  • @olegt962
    @olegt962 5 лет назад

    In fact we can say that if a serie converges conditionally or diverges, by Riemann's theorem there exists an arrangement of serie such that converges for any k real we want.

  • @binitkumarsingh8296
    @binitkumarsingh8296 5 лет назад

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

  • @tylertorsiello8450
    @tylertorsiello8450 3 года назад

    this guy is my spirit animal

  • @alainrogez8485
    @alainrogez8485 5 лет назад

    One word : brilliant!

  • @batmanthedarkestknight
    @batmanthedarkestknight 9 месяцев назад +1

    "I didn't make this up " made me laugh so much

  • @adamwalker8777
    @adamwalker8777 2 года назад +2

    If I take 1,2,3... coins from every inhabitant of the multiverse, then I will remain in debt.

  • @kimothefungenuis
    @kimothefungenuis 5 лет назад +6

    Better than numberphile

  • @youtubecommentsguy9805
    @youtubecommentsguy9805 Месяц назад +1

    That whole video was the kind of vibe it was in high school when you cannot follow a single step what the math teacher is teaching you understand nothing but don‘t dare to ask because you feel like he does a good job and is a genuine friendly guy so you don‘t want to interrupt him.

  • @muterchak
    @muterchak Год назад

    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? :)

  • @GenesisRussell-jt2rp
    @GenesisRussell-jt2rp 5 лет назад

    you need more views!

  • @hamsterdam1942
    @hamsterdam1942 5 лет назад

    Ze best video about this freaky number!

  • @lalitverma5818
    @lalitverma5818 5 лет назад

    Ohh great respected Ramanujam was best mathmetician of india

  • @coursmaths138
    @coursmaths138 2 года назад

    Once upon a time, there was a group of French people who had always lived only in Paris. They had seen nothing else. One day, they had the idea to formalize what a human could be. So they looked at the humans directly at their fingertips, and listed different properties to try to characterize them. Since they had living examples of humans satisfying all these properties, it was clear that their properties characterized the humans of Paris. Which meant, in their minds, humans at all (having seen nothing else). One of these properties was "A human necessarily likes cheese". All humans in Paris satisfied it. But a far-sighted scientist thought that this property did not necessarily seem to be attached to the intrinsic notion of human itself, and that the choice of this property to characterize a human seemed arbitrary. So he tried a new list of properties by removing this property ("like cheese"). He unrolled the implications, and found with surprise that nothing in his calculations opposed the existence of such "objects" (humans not liking cheese). However, the shocked Parisians claimed loud and clear that it was "absurd", that one had never seen "someone who didn't like cheese", and "that these people couldn't exist". It was also heard that "they couldn't really be human". The idea was proposed to burn the scientist at the stake. To save his life, he had to flee his hometown and settle elsewhere.
    Time passed, and our scientist finally met humans in distant lands who did not like cheese. So he had found an example of existing objects, satisfying the properties he had posed. He returned to his city and presented them to the Parisians. The skepticism was palpable. They looked at them very fixedly. Then, after several days of observation, they finally had to face the facts: there were real humans who did not like cheese. Although this property seemed intuitive and natural to them, the facts seemed to show that their intuition (probably conditioned by their environment) was wrong. It now seemed very unreasonable to redefine the notion of human, to exclude those who did not like cheese. Because in them, everything worked almost like the humans they were used to. Time passed, and it was finally accepted that these individuals were indeed humans. They settled down and eventually settled in Paris. With even more time, it turned out that these humans even made an important civic contribution (through their work, knowledge and skills, ...) to this new, and richer, beautiful city of Paris. 🙃

  • @divergentmaths
    @divergentmaths 4 года назад

    If you are interested to learn more about divergent series and want to understand why and how 1+2+3+4+5+6+... = -1/12,
    I recommend the online course “Introduction to Divergent Series of Integers” on the Thinkific online learning platform.

  • @Gideon_Judges6
    @Gideon_Judges6 5 лет назад +1

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

  • @akk92278
    @akk92278 3 года назад +1

    Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined. ...

  • @chandankar5032
    @chandankar5032 5 лет назад +1

    Its a request to make video on euler mclaurin summation. No one has made it in RUclips yet. Please🙏

  • @sugarfrosted2005
    @sugarfrosted2005 5 лет назад +3

    Fun with analytic continuation!

  • @abhijeetdas6279
    @abhijeetdas6279 5 лет назад

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

  • @mikejackson19828
    @mikejackson19828 3 года назад +1

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

  • @nicolarossi8728
    @nicolarossi8728 5 лет назад

    You are the best!

  • @stratonikisporcia8630
    @stratonikisporcia8630 10 месяцев назад +1

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

  • @stanislasnicolau2412
    @stanislasnicolau2412 5 лет назад +1

    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
      @keoviwath 3 года назад

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