Sum of Natural Numbers (second proof and extra footage)

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  • Опубліковано 10 січ 2015
  • MAIN VIDEO IS AT: • ASTOUNDING: 1 + 2 + 3 ...
    More links & stuff in full description below ↓↓↓
    Ed Copeland and Tony Padilla are physicists at the University of Nottingham.
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    Videos by Brady Haran
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  • Наука та технологія

КОМЕНТАРІ • 3,3 тис.

  • @JustOneAsbesto
    @JustOneAsbesto 7 років тому +1724

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

    • @U014B
      @U014B 7 років тому +40

      One that isn't overly-warm, even.

    • @billybishop5323
      @billybishop5323 7 років тому +3

      Post op Physio for shoulder decompression

    • @moregasmthepowerful2959
      @moregasmthepowerful2959 7 років тому +13

      And a pillow that makes you learn things.

    • @jayquelin
      @jayquelin 7 років тому +8

      most apt statement i have ever read

    • @SujitVital
      @SujitVital 7 років тому +2

      Nice

  • @ravenlord4
    @ravenlord4 4 роки тому +755

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

    • @emilcatar4620
      @emilcatar4620 3 роки тому +20

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

    • @ravenlord4
      @ravenlord4 3 роки тому +19

      @@emilcatar4620 r/woosh

    • @SumitKumar-go9fz
      @SumitKumar-go9fz 3 роки тому +5

      @@ravenlord4 r/whooosh

    • @ccv3956
      @ccv3956 3 роки тому +9

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

    • @georgewashingtonballs5603
      @georgewashingtonballs5603 3 роки тому +1

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

  • @ethanmartin2781
    @ethanmartin2781 7 років тому +2984

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

  • @LakeNipissing
    @LakeNipissing 8 років тому +591

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

    • @manueldelrio7147
      @manueldelrio7147 8 років тому +3

      He does, doesn't he?

    • @junkdubious
      @junkdubious 7 років тому +11

      The Bob Ross of numbers.

    • @darkdelphin834
      @darkdelphin834 4 роки тому +6

      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
      @darkdelphin834 4 роки тому +2

      But really calm voice indeed xd

  • @MHeymann
    @MHeymann 5 років тому +235

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

    • @TheGarfield1337
      @TheGarfield1337 5 років тому +35

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

    • @subarnasubedi7938
      @subarnasubedi7938 2 роки тому +16

      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
      @aweebthatlovesmath4220 2 роки тому +1

      Wasnt that |x|

    • @jamesknapp64
      @jamesknapp64 2 роки тому

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

    • @puneetbajaj786
      @puneetbajaj786 Місяць тому

      ​@@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
    @mathunt1130 10 років тому +653

    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
      @Peter_1986 6 років тому +7

      An ordinary sum WILL approach infinity, though.

    • @hikarifathan5143
      @hikarifathan5143 6 років тому +1

      Mat Hunt wow
      @_@

    • @O-Kyklop
      @O-Kyklop 6 років тому +1

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

    • @O-Kyklop
      @O-Kyklop 6 років тому +1

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

    • @josephcambron7060
      @josephcambron7060 Рік тому

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

  • @hunternewberry5860
    @hunternewberry5860 6 років тому +189

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

  • @AJMansfield1
    @AJMansfield1 9 років тому +583

    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
      @Tom_Het 8 років тому +95

      Anson Mansfield try and stop me

    • @andyjabez9780
      @andyjabez9780 4 роки тому +30

      Addition is commutative.

    • @marloucabalquinto7514
      @marloucabalquinto7514 4 роки тому +86

      Euler: Reality can be whatever I want.

    • @gulgaffel
      @gulgaffel 4 роки тому +100

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

    • @ahmadrahimisudin8364
      @ahmadrahimisudin8364 4 роки тому +26

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

  • @matthieu9792
    @matthieu9792 9 років тому +175

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

    • @jon_collins
      @jon_collins 3 роки тому +16

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

    • @XB10001
      @XB10001 3 роки тому +6

      @@jon_collins what a negative comment! 😁

    • @randolphbusch7777
      @randolphbusch7777 3 роки тому +2

      Laughs in python.

    • @amaanali9525
      @amaanali9525 3 роки тому

      LAUGHS IN YO MOMMA LANGUAGE BINARY AHAHAHHA

    • @m4inline
      @m4inline 2 роки тому +1

      Where's your point?

  • @DaniloJAC
    @DaniloJAC 9 років тому +68

    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
    @Ohmriginal722 8 років тому +270

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

  • @angermanager2116
    @angermanager2116 4 роки тому +49

    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
      @SarahStarmer Рік тому +11

      I agree "useful" rather than true.

    • @oddlyspecificmath
      @oddlyspecificmath Рік тому +1

      Thanks

    • @executive
      @executive 7 місяців тому +4

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

    • @erikdietrich2678
      @erikdietrich2678 3 місяці тому +3

      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
    @Jomczx 10 років тому +117

    "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
    @SquashBox 10 років тому +27

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

  • @camerongrant7721
    @camerongrant7721 7 років тому +194

    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
      @losboston 7 років тому +10

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

    • @Dan1elAndrade
      @Dan1elAndrade 7 років тому +22

      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
      @MrCmon113 6 років тому +23

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

    • @ezrasingh2486
      @ezrasingh2486 5 років тому +12

      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
      @user-xz1yr8on6v 5 років тому +4

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

  • @atomicmrpelly
    @atomicmrpelly 4 роки тому +59

    20:19 - Ghostbusters walking past the window?

  • @SImonK1996
    @SImonK1996 10 років тому +79

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

    • @JoshiaSeam
      @JoshiaSeam 10 років тому +9

      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
      @MrCmon113 6 років тому

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

    • @samb443
      @samb443 5 років тому

      but its not even wrong

  • @YoungColCol
    @YoungColCol 10 років тому +205

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

    • @ferojshaikh2880
      @ferojshaikh2880 7 років тому +5

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

    • @avdheshgupta22
      @avdheshgupta22 7 років тому +12

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

    • @itchyknows
      @itchyknows 7 років тому +2

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

    • @redd8551
      @redd8551 7 років тому +1

      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
      @itchyknows 7 років тому

      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
    @user-cd5kp2qd5l 4 роки тому +7

    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
      @jacobm5167 Рік тому +2

      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
    @angelogandolfo4174 3 роки тому +95

    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
      @ethanwilliamson782 3 роки тому +19

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

    • @snoxh2187
      @snoxh2187 2 роки тому +23

      @@ethanwilliamson782 only because u cant understand it?

    • @joshuaum473
      @joshuaum473 2 роки тому +13

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

    • @manvith1878
      @manvith1878 Рік тому +2

      This theory was proved by Indian mathematician Srinivasa Ramanujan.

  •  10 років тому +58

    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 ;)

  • @GlobalWarmingSkeptic
    @GlobalWarmingSkeptic 10 років тому +35

    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
    @golfer435 8 років тому +13

    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
      @evilkillerwhale7078 8 років тому +4

      +Logan Hollis You're bad at math.

    • @NebulusVoid
      @NebulusVoid 8 років тому +1

      +evilkillerwhale 2+2=5

    • @divxxx
      @divxxx 3 місяці тому +1

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

  • @sagantucker2097
    @sagantucker2097 7 років тому +30

    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
    @jamesthorn9992 9 років тому +356

    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
      @LfeinYT 9 років тому +43

      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
      @orenzeshani 9 років тому +12

      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
      @Arkayjiya 9 років тому +5

      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
      @bennattj 9 років тому +3

      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
      @orenzeshani 9 років тому +2

      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
    @lraoux 10 років тому +32

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

  • @pitreason
    @pitreason 6 років тому +39

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

  • @Yupppi
    @Yupppi Рік тому +4

    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
    @OmegaRainbow 10 років тому +270

    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
      @numberphile  10 років тому +88

      Look forward to seeing the pics!!!

    • @hermokuolio01
      @hermokuolio01 8 років тому +60

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

    • @elave16
      @elave16 8 років тому +7

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

    • @OmegaRainbow
      @OmegaRainbow 8 років тому +4

      elave16 for what?

    • @FreddyMercry
      @FreddyMercry 8 років тому +5

      For picture confirmation!

  • @tomselleck8204
    @tomselleck8204 10 років тому +27

    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

  • @connorhorman
    @connorhorman 4 роки тому +2

    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
    @panga131 7 років тому +1

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

  • @smallfry4973
    @smallfry4973 10 років тому +17

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

  • @iamthereaImaster
    @iamthereaImaster 10 років тому +21

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

    • @jamesknapp64
      @jamesknapp64 2 роки тому

      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
      @julianwood6625 7 місяців тому

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

  • @hansbauer760
    @hansbauer760 7 років тому +31

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

    • @Nylspider
      @Nylspider 4 роки тому

      What

    • @enesb7412
      @enesb7412 3 роки тому +4

      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
    @miloweising9781 8 років тому +6

    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
    @AstAMoore 10 років тому +53

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

    • @numberphile
      @numberphile  10 років тому +49

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

    • @MrLemonsChannel
      @MrLemonsChannel 8 років тому +6

      +Numberphile lots*

    • @EvolBob1
      @EvolBob1 8 років тому +10

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

    • @labibbidabibbadum
      @labibbidabibbadum 7 років тому +3

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

    • @thesinglemathnerd
      @thesinglemathnerd 7 років тому

      labibbidabibbadum lol

  • @ioneldragomirescu4179
    @ioneldragomirescu4179 9 років тому +24

    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
      @jamma246 9 років тому +3

      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
      @lizokitten2 9 років тому +2

      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
      @jamma246 9 років тому

      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
      @lizokitten2 9 років тому

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

    • @lizokitten2
      @lizokitten2 9 років тому +1

      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
    @acm-gs6bl Місяць тому +1

    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
    @MgtowRubicon 2 роки тому +2

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

  • @frepi
    @frepi 10 років тому +39

    Beautiful handwriting!

  • @manmanman784
    @manmanman784 10 років тому +41

    finally some more difficult stuff :) thanks numberphile!

  • @ChrisBandyJazz
    @ChrisBandyJazz 7 років тому +295

    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
      @ChrisBandyJazz 7 років тому +5

      Thank you!

    • @renatomagretti7824
      @renatomagretti7824 7 років тому +71

      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
      @connorsmith3282 6 років тому +62

      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
      @NitinChauhan-vh2yk 6 років тому +20

      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
      @iankretschmer559 6 років тому +14

      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
    @MrCmon113 8 років тому +37

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

    • @evilkillerwhale7078
      @evilkillerwhale7078 8 років тому +2

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

    • @MrCmon113
      @MrCmon113 8 років тому +2

      *****
      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
      @evilkillerwhale7078 8 років тому +7

      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
      @MrCmon113 8 років тому +1

      *****
      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
      @evilkillerwhale7078 8 років тому +2

      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
    @tamasdemjen4242 9 років тому +174

    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
      @MuffinsAPlenty 9 років тому +15

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

    • @Spix_Weltschmerz-Pucket
      @Spix_Weltschmerz-Pucket 9 років тому +2

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

    • @bharathp666
      @bharathp666 9 років тому +1

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

    • @MuffinsAPlenty
      @MuffinsAPlenty 9 років тому +2

      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
      @roblaquiere8220 9 років тому

      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
    @Poldovico 9 років тому +14

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

  • @diabl2master
    @diabl2master 8 років тому +61

    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
      @orbitalteapot21 8 років тому +1

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

    • @diabl2master
      @diabl2master 8 років тому +2

      Muhammad Abdullah i think you misunderstand me

    • @muddafiggagagamel1820
      @muddafiggagagamel1820 8 років тому +13

      +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
      @CharcoalBlasterdog 8 років тому +13

      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
      @TheAusrali 8 років тому

      thats exactly what i said on another video!!

  • @Integralsouls
    @Integralsouls 4 роки тому +17

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

    • @douggwyn9656
      @douggwyn9656 4 роки тому

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

    • @Integralsouls
      @Integralsouls 3 роки тому +3

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

  • @jhobbz45
    @jhobbz45 10 років тому +89

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

    • @Anytus2007
      @Anytus2007 10 років тому +20

      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
      @jhobbz45 10 років тому +2

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

    • @Anytus2007
      @Anytus2007 10 років тому +4

      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
      @jhobbz45 10 років тому +1

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

    • @Anytus2007
      @Anytus2007 10 років тому +31

      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
    @TheophilosPorter 10 років тому +16

    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
    @DrDress 8 років тому +91

    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
      @SensibleSine 8 років тому +20

      +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
      @DrDress 8 років тому +3

      +halofan313
      What an d thing to say...

    • @lakshbk
      @lakshbk 8 років тому +1

      +halofan313 I find ur explanation really convincing

    • @diptoneelde836
      @diptoneelde836 6 років тому

      Yeah that makes sense

    • @ezrasingh2486
      @ezrasingh2486 5 років тому

      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
    @grapheist612 7 років тому +7

    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
      @adhamm5503 3 роки тому +1

      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
      @oooBASTIooo Рік тому +6

      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
      @KarlDeux 10 місяців тому +1

      @@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
      @xavierstanton8146 8 місяців тому

      ​@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
      @KarlDeux 8 місяців тому

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

  • @ge2719
    @ge2719 10 років тому +43

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

    • @DrankisDank
      @DrankisDank 10 років тому +4

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

    • @Vackraetraed
      @Vackraetraed 10 років тому

      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
      @diabolic42090 10 років тому

      Exactly what I was wondering. Can someone explain this?

    • @ge2719
      @ge2719 10 років тому

      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
      @HebaruSan 10 років тому

      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
    @synchronizerman 10 років тому +39

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

    • @BillySkceuk
      @BillySkceuk 10 років тому +1

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

    • @Dextomus
      @Dextomus 10 років тому +1

      Was thinking the same thing

    • @ZipplyZane
      @ZipplyZane 10 років тому +1

      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
      @COHEMUACOS 10 років тому

      en.wikipedia.org/wiki/Geometric_series

    • @synchronizerman
      @synchronizerman 10 років тому

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

  • @Rjsipad
    @Rjsipad 8 років тому +12

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

    • @Falcon0408
      @Falcon0408 8 років тому +9

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

    • @Rjsipad
      @Rjsipad 8 років тому

      +OneSixteenMike same im curious

    • @christianbouwense4702
      @christianbouwense4702 8 років тому

      +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
      @crazymuthaphukr 8 років тому

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

    • @Rjsipad
      @Rjsipad 8 років тому +1

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

  • @remimahmoud7672
    @remimahmoud7672 8 років тому +1

    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
    @freethebrain 10 років тому +44

    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
      @MrCmon113 6 років тому +3

      The maths in that video are BS though.

  • @jordankay9082
    @jordankay9082 9 років тому +16

    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
    @shawniscoolerthanyou 8 років тому +7

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

  • @marcosrodriguez2496
    @marcosrodriguez2496 8 років тому +3

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

  • @Wizarth
    @Wizarth 10 років тому +18

    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
      @Anytus2007 10 років тому +2

      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
    @ASKaPHYSICIST 10 років тому +26

    "strictly true for x

  • @topilinkala1594
    @topilinkala1594 5 років тому +1

    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
    @AkiraUema 8 років тому +44

    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
      @isithardtobestupid7568 8 років тому +1

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

    • @ala4sox02
      @ala4sox02 8 років тому +8

      +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
      @isithardtobestupid7568 8 років тому +2

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

    • @s4archie
      @s4archie 8 років тому +6

      +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
      @s4archie 8 років тому +8

      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.

  • @DanielEstrada
    @DanielEstrada 10 років тому +8

    John Baez makes a surprise appearance at 15:50!

  • @LegionOnomaMoi
    @LegionOnomaMoi 10 років тому +4

    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 UA-cam 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).

  • @PC_Simo
    @PC_Simo Рік тому +2

    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
    @johndoyle2347 5 місяців тому

    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.

  • @LordSwordbreak
    @LordSwordbreak 10 років тому +10

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

  • @juanmanuelmunoz5645
    @juanmanuelmunoz5645 9 років тому +84

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

    • @longevitee
      @longevitee 6 років тому

      Juan Manuel Muñoz Bless you.

    • @embedded_
      @embedded_ 6 років тому

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

    • @MrCmon113
      @MrCmon113 6 років тому +1

      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
      @DeathBringer769 5 років тому

      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.

  • @user-qd2nd6hi8j
    @user-qd2nd6hi8j Рік тому

    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
    @5oulcrusher 3 роки тому +2

    "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."

  • @grantkohler7612
    @grantkohler7612 10 років тому +11

    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?

  • @CabooseNor
    @CabooseNor 10 років тому +33

    It's so fascinating to listen to Ed Copeland!

  • @somni6756
    @somni6756 7 років тому

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

  • @yichern4351
    @yichern4351 7 років тому +2

    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

  • @arthurthegreat216
    @arthurthegreat216 10 років тому +82

    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
      @paultang5479 6 років тому +2

      ...

    • @nyxgraal6949
      @nyxgraal6949 6 років тому

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

    • @bow3i
      @bow3i 5 років тому +1

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

    • @bhobba
      @bhobba 5 років тому

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

  • @Pantopam
    @Pantopam 10 років тому +4

    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
      @MrCmon113 6 років тому +1

      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.

  • @herbie6721
    @herbie6721 4 роки тому +3

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

  • @saeedbawazeer9636
    @saeedbawazeer9636 7 років тому +3

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

  • @andrew_cunningham
    @andrew_cunningham 9 років тому +19

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

  • @magicstix0r
    @magicstix0r 10 років тому +457

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

  • @sethwhite4378
    @sethwhite4378 5 років тому +1

    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
    @DocBree13 4 роки тому +2

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

  • @xMagno313x
    @xMagno313x 10 років тому +9

    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?

  • @pete275
    @pete275 10 років тому +25

    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
      @MrCmon113 6 років тому +1

      That is completely unrelated.

  • @trevordennison9115
    @trevordennison9115 7 років тому +2

    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
    @ItsEverythingElse 8 років тому +2

    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 :)

  • @-dazz-
    @-dazz- 9 років тому +14

    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
      @HWEWSWEW 4 роки тому +16

      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
      @amanmavi4118 3 роки тому +4

      @@HWEWSWEW finally now I can sleep

  • @mynameisawesomeman
    @mynameisawesomeman 10 років тому +11

    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
      @zacharychaney9489 10 років тому

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

    • @KutuluMike
      @KutuluMike 10 років тому +1

      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
      @zacharychaney9489 10 років тому

      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.

  • @user-br6ry8bu5b
    @user-br6ry8bu5b 7 років тому +8

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

  • @JazzyBoiZ
    @JazzyBoiZ 9 місяців тому +4

    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

  • @Harlequin314159
    @Harlequin314159 10 років тому +25

    Much better explanation than the first video. Thank you!

  • @dlbattle100
    @dlbattle100 10 років тому +14

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

  • @Jacobswe
    @Jacobswe 8 років тому

    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
    @oooBASTIooo Рік тому +4

    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.

  • @mandydax
    @mandydax 10 років тому +5

    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.

  • @Mandragara
    @Mandragara 9 років тому +59

    How are the comments months older than the video lol Numberphile

    • @MahmoudYahyaoui
      @MahmoudYahyaoui 9 років тому

      Update

    • @WalterKingstone
      @WalterKingstone 9 років тому +17

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

    • @SunOfCinder
      @SunOfCinder 9 років тому +7

      Pure Mathematics!

    • @WalterKingstone
      @WalterKingstone 9 років тому +1

      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
      @Oners82 9 років тому

      Magic!

  • @timelsen2236
    @timelsen2236 Рік тому

    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
    @subarnasubedi7938 7 років тому +37

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

    • @enesulasdincer7049
      @enesulasdincer7049 4 роки тому

      i am agree with you.

    • @igoranisimov6549
      @igoranisimov6549 2 роки тому +1

      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
      @Invisifly2 2 роки тому +2

      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
      @igoranisimov6549 2 роки тому

      @@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
      @GBloxers 2 роки тому

      @@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?