tag:blogger.com,1999:blog-5531528752382322417.post8975125640171597957..comments2019-09-12T08:19:13.430-07:00Comments on Data Bonanza: Why the sum of all natural numbers is not negative one twelfth Iwonahttp://www.blogger.com/profile/12119249286687307739noreply@blogger.comBlogger17125tag:blogger.com,1999:blog-5531528752382322417.post-58579978205054496712017-11-23T15:29:12.733-08:002017-11-23T15:29:12.733-08:00The falsehood of the notion that the sum of positi...The falsehood of the notion that the sum of positive integers is -(1/12) is so incredibly, staggeringly obvious, that I have to wonder if anyone would be arguing with the OP if she weren't a woman.the_green_bastardhttps://www.blogger.com/profile/15037910832804906693noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-27099540720328265212015-09-08T17:35:41.812-07:002015-09-08T17:35:41.812-07:00Hi, I saw the Riemmann zeta approach shown in this...Hi, I saw the Riemmann zeta approach shown in this link. https://www.youtube.com/watch?v=E-d9mgo8FGk<br /><br />It looks logical except that I've seen in other math sources that a sum of a geometric series is bound for IxI <1. The fellow in this video says that simply x<1, which naturally allows him later to use x= -1 in the derived formulas giving rise to the weird result of -1/12. However, if the right condition is IxI < 1 then x can only take values between -1 and 1 but not including -1 nor 1. Any thoughts about who is actually right?Anonymoushttps://www.blogger.com/profile/04514626117868201083noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-22867108861415412752015-09-08T17:32:31.885-07:002015-09-08T17:32:31.885-07:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/04514626117868201083noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-24757686500919996212015-02-06T20:43:24.670-08:002015-02-06T20:43:24.670-08:00Yes 0+0+0+0+0+0..... = 1Yes 0+0+0+0+0+0..... = 1Venkatesanhttps://www.blogger.com/profile/07003761613261649806noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-23441826451979806672015-02-06T20:42:08.666-08:002015-02-06T20:42:08.666-08:00Yes 0+0+0+0+0+0..... = 1Yes 0+0+0+0+0+0..... = 1Venkatesanhttps://www.blogger.com/profile/07003761613261649806noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-68445128241853469682014-08-01T10:53:15.746-07:002014-08-01T10:53:15.746-07:00Iwona, one cannot insert additional terms (yes, ev...Iwona, one cannot insert additional terms (yes, even zeroes) into a divergent series.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-29837729255568549352014-06-17T23:31:38.818-07:002014-06-17T23:31:38.818-07:00Thanks, Tom. Here is a simple example of how you c...Thanks, Tom. Here is a simple example of how you can use their logic to derive an inconsistent result: Let's take the first series S1 = 1 - 1 + 1 - 1 + 1 - 1 ... = 1/2. If this is an actual infinite sum, then we can replace each (1 - 1) with (1 - 1 + 0) without changing the result (since adding zero doesn't change anything, right?) This means that S1 = 1 - 1 + 0 + 1 - 1 + 0 + 1 - 1 + 0 ... = 1/2. But if you now try to apply the "expected value" logic to the new version of S1, you conclude that if you stop the series at position 3k + 1, then the sum is 1, if you stop it at position 3k + 2, then the sum is 0, and if you stop it at position 3k, then it is also 0. Averaging these three possible outcomes gives 1/3 and 1/3 <> 1/2, hence inconsistent result.Iwonahttps://www.blogger.com/profile/12119249286687307739noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-29159186258946468922014-06-17T22:12:59.066-07:002014-06-17T22:12:59.066-07:00Your argument would be more convincing if you coul...Your argument would be more convincing if you could provide an example where you use their logic to derive an inconsistent result. Anyway, thanks for sharing your thoughts on the matter.Tomhttps://www.blogger.com/profile/18416151831957145084noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-34314361556545904482014-05-21T09:54:45.968-07:002014-05-21T09:54:45.968-07:00There is no need to publish a paper, since it is a...There is no need to publish a paper, since it is a well-accepted fact that divergent series have no well-defined sum, I am not saying anything controversial here. See for example: http://mathworld.wolfram.com/DivergentSeries.html.<br /><br />The problem with the derivation in the first Numberphile video is that using the same "logic" I could derive also other results. A mathematical framework which allows one to arrive at different conclusions is not very useful. I was just trying to explain where the inconsistency lies in an approachable way.<br /><br />Now the Riemann zeta approach is a whole different story, since it actually does provide a consistent framework under which you get the same results no matter how you approach the problem and I am not arguing its usefulness. But this framework is not operating on natural numbers anymore, but an abstract analytic continuation of the numbers domain. Results you get in this abstract domain do not have to carry over to the natural numbers domain, so there is no conflict here. When you talk about real-world natural numbers, the divergent series sum is still infinity or simply undefined.Iwonahttps://www.blogger.com/profile/12119249286687307739noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-25064754631252596892014-05-20T20:31:19.116-07:002014-05-20T20:31:19.116-07:00Blah. Everyone, the author fails to address the tw...Blah. Everyone, the author fails to address the two other derivations in the other video. One using the Reimann Zeta Function and the other simply using the summation of squares formula defined for x<1. <br /><br />Here is the video: https://www.youtube.com/watch?v=E-d9mgo8FGk&src_vid=w-I6XTVZXww&feature=iv&annotation_id=annotation_2226441133<br /><br />I highly doubt a huge proof in the mathematical community could be foiled by a single blog. If she was serious she would be publishing a paper, not writing a blog. She would get so much fame for disproving this. Anonymoushttps://www.blogger.com/profile/17043970834687364941noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-8371215136868557632014-05-20T20:06:30.998-07:002014-05-20T20:06:30.998-07:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/17043970834687364941noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-11756249361184521332014-05-20T19:45:11.398-07:002014-05-20T19:45:11.398-07:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/16139731099688372430noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-55917581833291360402014-05-05T03:38:58.948-07:002014-05-05T03:38:58.948-07:00You didn't disprove anything. Merely pointed o...You didn't disprove anything. Merely pointed out some arguments which they didn't explain in detail. The rules in this case make perfect sense and are proven by physics (see the casimir effect) they are just not intuitive to us.Anonymoushttps://www.blogger.com/profile/09790266295289870175noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-10035613589048864542014-03-22T15:46:51.479-07:002014-03-22T15:46:51.479-07:00There's nothing wrong with S1 = 1/2. If you do...There's nothing wrong with S1 = 1/2. If you don't like the explanation using cesaro sums you can do it equivalently by 1 - s1 = s1.<br /><br />Both sides of the equation are termwise equal and with an equivalently infinite number of terms. It's a genuine equality.<br /><br />You don't have to assume anything there to get S1 = 1/2. Just that infinite really means what it says.Anonymoushttps://www.blogger.com/profile/09155331833751310670noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-21023828399581886142014-02-10T06:54:06.672-08:002014-02-10T06:54:06.672-08:00Excellent and Concise explanation . When we try to...Excellent and Concise explanation . When we try to define something that is undefined by the rules of the game we end up with paradox like this . Anand K. Ghuryehttps://www.blogger.com/profile/00355011717915074437noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-45540511527581309972014-02-08T16:10:18.330-08:002014-02-08T16:10:18.330-08:00This assumes that X is a well-defined value in the...This assumes that X is a well-defined value in the first place. If X were well-defined then your argument would show that it must be equal to 1/2. But there is no reason to assume that the sum exists - and if you think of the sum as an infinite series, X does *not* exist, because the series diverges.)Michael Weisshttps://www.blogger.com/profile/04541032124336933501noreply@blogger.comtag:blogger.com,1999:blog-5531528752382322417.post-30571230920219706012014-01-22T15:40:31.333-08:002014-01-22T15:40:31.333-08:00If the series starting with +1 is S, then the ser...If the series starting with +1 is S, then the series starting with -1 can be expressed both as -1 + S and as -S, and the ... in both cases mean "infinite terms follow".<br /><br />if -1 + S and -S are the same and have the same sum X, then -1+X = -X, so X=1/2 <br /><br />Anonymousnoreply@blogger.com