tag:blogger.com,1999:blog-35187314.post525565253866621317..comments2020-07-04T04:08:56.100-04:00Comments on Physics Buzz: Does 1+2+3+4+ . . . =-1/12? APS Webmasterhttp://www.blogger.com/profile/05951833208918853453noreply@blogger.comBlogger45125tag:blogger.com,1999:blog-35187314.post-85189854655391475122018-01-12T05:17:34.909-05:002018-01-12T05:17:34.909-05:00What if the analytic continuation is real, and we ...What if the analytic continuation is real, and we simple-minded apes initially failed to notice anything but the banana-shaped set of natural numbers sticking out?David Turnerhttps://www.blogger.com/profile/14738491011019732316noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-18752057901255622192017-01-04T09:09:54.733-05:002017-01-04T09:09:54.733-05:00 Can you explain this regularization? Can you explain this regularization?Anonymoushttps://www.blogger.com/profile/04383553420379472039noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-81679663800909700842016-12-07T14:04:24.416-05:002016-12-07T14:04:24.416-05:00Please please can someone point out the flaw in my...Please please can someone point out the flaw in my reasoning. This theory is completely WRONG.<br /><br />If I take the series S = 1 + 2 + 3 + 4 + ....<br />and the series 4S = 4 + 8 + 12 + 16 + ...<br /><br />And subtract the 2, I cannot just simply shift and then pretend everything works out ok. Why shift the bottom row with intervals that line up numerically with the top row? Why not just shift it all an infinite amount and end up with:<br />S - 4S = (1 + 2 + 3 + 4 + 5 + 6) - 4?<br /><br />The amount of shift is completely arbitrary and is just done for convenience sake. It completely disregards the rate of numerical increase.<br /><br />We all know that a function f(x) = 2x [x limit to infinity] is larger than the function f(x) = x [x limit to infinity]. It's absolutely important that the numbers in each position are subtracted with the corresponding position.<br /><br />Please tell me why I am wrong, or tell me how some string theory is actually based off of this??Denzilhttps://www.blogger.com/profile/09121459653381361628noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-38652497355791115192016-05-02T03:37:32.611-04:002016-05-02T03:37:32.611-04:00Thanks for bringing some clarity into this horribl...Thanks for bringing some clarity into this horrible mess!Giuseppehttps://www.blogger.com/profile/16141971215901955013noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-88537064858996300812015-11-25T16:33:55.927-05:002015-11-25T16:33:55.927-05:00https://m4t3m4t1k4.wordpress.com/2015/11/25/genera...https://m4t3m4t1k4.wordpress.com/2015/11/25/general-method-for-summing-divergent-series-using-mathematica-and-a-comparison-to-other-summation-methods/Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-15986042189644291142015-02-14T16:10:50.440-05:002015-02-14T16:10:50.440-05:00Problems of summation divergent series is solved! ...Problems of summation divergent series is solved! https://m4t3m4t1k4.wordpress.com/2015/02/14/general-method-for-summing-divergent-series-determination-of-limits-of-divergent-sequences-and-functions-in-singular-points-v2/Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-43775175501973454602014-04-13T17:14:59.474-04:002014-04-13T17:14:59.474-04:00Thanks, Dr. Jantzen!Thanks, Dr. Jantzen!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-50507312790682056562014-04-13T15:43:38.244-04:002014-04-13T15:43:38.244-04:00I totally agree. And I refer to my previous commen...I totally agree. And I refer to my previous comments above, where I have made similar statements.<br /><br />Please note that the discussion about this topic actually went on in two new blog posts by the same author:<br /><br />http://physicsbuzz.physicscentral.com/2014/01/redux-does-1234-112-absolutely-not.html<br />[titled "Correction: Does 1+2+3+4+ . . . =-1/12? Absolutely Not! (I think)"]<br /><br />and<br /><br />http://physicsbuzz.physicscentral.com/2014/02/so-what-does-1234-equal-we-give-you_11.html<br />[titled "If Not -1/12, What Does 1+2+3+4+... Equal? We Give You the Answer "]<br /><br />You may see from these blog posts (and their comments) that the author changed his mind several times about this topic. Unfortunately, he did not link his follow-up posts from this first one.Anonymoushttps://www.blogger.com/profile/08953408454915030105noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-7996727387995873472014-04-13T13:10:49.873-04:002014-04-13T13:10:49.873-04:00PS: please read the 3rd paragraph at books.google....PS: please read the 3rd paragraph at books.google.de/books?id=n8Mmbjtco78C&pg=PA73Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-41167545209500613762014-04-13T10:19:38.941-04:002014-04-13T10:19:38.941-04:00I'm afraid you're greatly confused.
1. Ju...I'm afraid you're greatly confused.<br /><br />1. Just because Euler derived something doesn't make it true. Euler was known for "breaking the rules" all too often - that was surely part of his genial nature - but it doesn't mean that all of his results and proofs are correct, in fact his handling of infinite diverging sums is not strictly mathematically correct.<br /><br />Also, n appeal to authority is a fallacy.<br /><br />2. It's true that 1+2+3... results in the so-called Ramanujan sum of -1/12. But a Ramanujan sum is not at all the same as *a* sum in a traditional sense. Look it up.<br /><br />3. You're engaging in trickery by juggling different definition of sums.<br />If we're talking about traditional sums, then no: divergent infinite series cannot be summed up and -1/12 is certainly not a sum of any such series.<br /><br />Now, we can extend the definition of "summation" to cover the assignment of finite numbers to infinite divergent series. But in such a case it *must* be clear that we're talking about something entirely different than traditional summation. Because 1+2+3...=-1/12 is only mindblowing when we're thinking of traditional sums (in that case it also happens to be incorrect).<br /><br />4. Now, certainly it must be admitted that the persistence with which -1/12 (and other numbers) pops up when all those different kinds of operations are performed (legally or not), as well as the fact that it is applicable to the real world does mean that it's not merely a fluke and that there is indeed a deep connection between these things on a fundamental level.<br /><br />But this doesn't justify the bad math, sorry.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-56470937402428360552014-04-13T10:07:57.312-04:002014-04-13T10:07:57.312-04:00It is a trick for a simple reason - zeta function ...It is a trick for a simple reason - zeta function is only defined for s>1, whereas to perform the trick of "summing up" the natural numbers if set s=-1. This may be a useful trick, but it's still a trick.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-68914565453418790822014-03-05T04:57:43.189-05:002014-03-05T04:57:43.189-05:00For me, not a mathematician, this is a perfect exa...For me, not a mathematician, this is a perfect example for that you can prove anything with infinity mathematics.<br /><br />In my view, infinity does not exist (in the real world), nor does 0.<br /><br />Infinity = anything / 0 <br /><br />so<br /><br />0 = anything / infinity.<br /><br />Physical argument: the smallest thing (measure) is the planck dimension (planck lenght, planck time etc.) so there cannot be an infinite number of since the beginning of time (the big bang).Anonymoushttps://www.blogger.com/profile/09258302203873963536noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-42628361941269904622014-03-04T10:28:15.604-05:002014-03-04T10:28:15.604-05:00Definition error. The sum of n to infinity diverge...Definition error. The sum of n to infinity diverges, goes really big. The "regularization" converges to -1/12. The technique is proven, duh, it would be odd if they were using the regularized sum without a proof. Its like saying 2+2=10. Very misleading unless you realize that the equation has been manipulated by being turned into 2, base ten, +2, base ten, equals 10 base 2. Although the regularization is legitimate, you can't say the sum adds to -1/12. You have to say the sum can be regularized through the Riemann zeta function and through a variety of other methods to equal -1/12. Although for all intents and purposes no ones checking the addend up to infinity to make sure that it equals what we say it does so feel free to say it equals -1/12, but make sure add in that regularized part if your talking with any self respecting mathematician or physicist.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-35187314.post-91413733609274480362014-02-03T17:22:06.455-05:002014-02-03T17:22:06.455-05:00Wondrful exchange guys. Thanks for the knowledge...Wondrful exchange guys. Thanks for the knowledge you've been sharing with us, and for the effort that takes.Uniquityhttp://www.swiftgrass.com/uniquity/noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-8746138229436430602014-01-28T17:36:50.868-05:002014-01-28T17:36:50.868-05:00Bernd, my boss happened to stop by and say basical...Bernd, my boss happened to stop by and say basically the same thing (he's a physicist who has a lot of experience with quantum field theory) and illustrate it using the Casimir Effect. I have to admit, I'm coming around to see what you're saying. It's easier to understand with a white board handy. In any case, I'll have to think about it for a while before I have anything much to say (or ask).<br /><br />Along the way, I happened to find a relatively simple way to evaluate zeta functions that I'd never seen before, for odd, negative values of n. I guess it doesn't matter, really, but it happens to make calculating some Bernoulli numbers (B_n, for n even) relatively quick and easy. I have no idea whether that's useful, interesting, or new. It entertained me though.<br />Buzz Skylinehttps://www.blogger.com/profile/04255849304022062681noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-13691122301712747772014-01-28T16:17:30.724-05:002014-01-28T16:17:30.724-05:00Buzz, let me show you one way to better understand...Buzz, let me show you one way to better understand, at least in a qualitative way, how such divergent sums of positive numbers can be related to finite and sometimes even negative numbers.<br /><br />Look at the physics case of the Casimir effect (http://en.wikipedia.org/wiki/Casimir_force#Derivation_of_Casimir_effect_assuming_zeta-regularization): Even when properly regularized by using the parameter s, how can a sum (and integral) of energy contributions, which are each individually positive, end up in a negative total Casimir energy at the end?<br /><br />In order to explain this, let me first approximate the series representation of the zeta function by an integral which is easier to handle. Using the Euler-Maclaurin-Formula (http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula) in its 0th-order approximation, we can approximate<br /><br />1^(-s) + 2^(-s) + 3^(-s) + 4^(-s) + ...<br />= 1/2 + integral from 1 to infinity of dx * x^(-s) + small remainder,<br /><br />which is valid, as the series on the left-hand side, for s>1.<br /><br />Now, as the integrand x^(-s) is positive all over the integration domain from 1 to infinity, one would expect this integral to be positive as well. And in the convergent case s>1, this is as expected. Using the primitive of the integrand, -1/(s-1) * x^(1-s), one gets:<br /><br />integral = 1/(s-1) * ( 1 - infinity^(1-s) )<br /><br />For s>1, the contribution from the upper boundary is infinity raised to a negative power, i.e. zero, and we and up with:<br /><br />integral = 1/(s-1) > 0 for s>1.<br /><br />But what happens in the divergent case, when s<1? Then the contribution from the upper boundary is divergent: infinity raised to a positive power. So the integral should be positive and infinite, as one would also expect from summing all natural numbers (the above series with s = -1) or all cubes of the natural numbers (series with s = -3). However, we are using a kind of regularization here which is based on analytic regularization. As the result 1/(s-1) for our integral is valid all over the complex half-plane where the real part of s is larger than 1, we analytically continue this result to all values of s (except s=1). Practically this means that the contribution from the upper boundary of the integral, 1/(s-1) * infinity^(1-s), is omitted although it diverges.<br /><br />So our analytically regularized integral is still given by minus the primitive of the integrand at the lower boundery, which is now a negative contribution:<br /><br />integral = 1/(s-1) = -1/(1-s) < 0 for s<1.<br /><br />Summarizing, the negative and finite result is obtained by dropping the positive and infinite contribution from the upper boundary of the integral. This prescription to drop such an infinite contribution arises from the analytic continuation in the parameter s.<br /><br />In physics (example of the Casimir effect) such a prescription makes sense because constant energy shifts (even if they are infinite) are irrelevant for the Casimir force.<br />Anonymoushttps://www.blogger.com/profile/08953408454915030105noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-22518492921911051762014-01-28T15:10:33.565-05:002014-01-28T15:10:33.565-05:00No conspiracy. But some people tend to oversimplif...No conspiracy. But some people tend to oversimplify things when they present them to non-experts. Then they are presenting them in a way which is actually not correct, but behind the wrong picture is a true meaning which though would have been more complicated to explain in a correct way. That's what the Numberphile people have done in my opinion. I always try to avoid this, but sometimes it is hard.<br />Anonymoushttps://www.blogger.com/profile/08953408454915030105noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-79668831419490509282014-01-28T14:46:05.944-05:002014-01-28T14:46:05.944-05:00Well, finite, and *correct* I assume, is what your...Well, finite, and *correct* I assume, is what your after. I'm sure there are many ways to deal with infinities. I imagine the ways that give you precise answers are fewer.<br /><br />So your proposal is that the answer is more along the lines of a conspiracy of physicists who are telling us this stuff to impress. It's possible, I suppose, but hard to believe.Buzz Skylinehttps://www.blogger.com/profile/04255849304022062681noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-82995985573454942652014-01-28T14:32:56.090-05:002014-01-28T14:32:56.090-05:00I have also done a lot of QED calculations, and I ...I have also done a lot of QED calculations, and I have never needed to sum divergent series without a proper regularization. Probably the physicists you are speaking about just wrote down such "identities" with divergent series as a very simple picture of what they did, omitting all notion of regularization for sake of simplicity? Maybe they even wanted to impress you by showing you what kind of "crazy things" they are doing? (Things which aren't crazy at all, once you properly define your regularization procedure.)<br /><br />As I told you before, physicists tend to be pragmatic. So they might even work with "identities" like 1+2+3+4+... = -1/12 because they know that, in principle, it can also be done properly and mathematically. Anyway, theories in physics do not depend on such "identities" being true, they just depend on being regularizable and renormalizable such that their predictions are finite in the end.<br />Anonymoushttps://www.blogger.com/profile/08953408454915030105noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-47321575756651130922014-01-28T14:04:50.331-05:002014-01-28T14:04:50.331-05:00By saying that the theories rely on the equation, ...By saying that the theories rely on the equation, I'm saying that, as I understand it, the theories would make different predictions if the "sum of the infinite series" (as Ramanujan says) equaled something other than -1/12. If you can do without them, they why to physicists who do QED calculations tell me they need them?Buzz Skylinehttps://www.blogger.com/profile/04255849304022062681noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-90502024012652819002014-01-28T13:47:53.477-05:002014-01-28T13:47:53.477-05:00This is proof of my Theory of Dark Numbers. This t...This is proof of my Theory of Dark Numbers. This theory postulates that Dark Numbers exist (but yet unobserved) to tame the large expansion of currently observable whole numbers 1+2+3+4+... so that the infinite series is equal to -1/12 instead of infinity. Tesla alluded to using Dark Numbers when he built his transport ray machine. According to my theory, the number line contains 4.9% ordinary numbers, 26.8% dark Robinson hyper-reals and 68.3% dark numbers.uncountablehttps://www.blogger.com/profile/12056612904448942712noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-25754670831713559532014-01-28T13:42:39.139-05:002014-01-28T13:42:39.139-05:00Thanks, Davide, for your comments. I really like E...Thanks, Davide, for your comments. I really like Evelyn Lamb's statement which you cited: You may assign -1/12 to the divergent series, but it is misleading to call this the sum of the series.<br /><br />And, Buzz (are you the same as "Buzz Skyline" or a different person?), it is sad when the only remaining argument for a mathematical equation like 1+2+3+4+… = -1/12, is the correctness of theories in physics (like QED). As a physicist I tell you, there is no meaningful theory in physics which relies on "identities" like 1+2+3+4+… = -1/12. Whenever infinities and divergences are encountered in physics, they are properly regularized (and eventually renormalized, especially in QED). And that is also what Euler, Ramanujan & co. did: They attached a meaning to divergent series by choosing a certain regularization.<br /> Anonymoushttps://www.blogger.com/profile/08953408454915030105noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-42657913491216235352014-01-28T13:37:39.589-05:002014-01-28T13:37:39.589-05:00Keep in mind, I'm saying
If QED is true
and i...Keep in mind, I'm saying<br /><br />If QED is true<br />and it relies on 1+2+3+4…= -1/12<br />then 1+2+3+4…= -1/12 must be true. <br /><br />I have no reason to want 1+2+3+4…= -1/12 to be true. I am happy with it being infinite. But then what's going on with QED? I don't accept that luck can lead to such precise predictions.Buzzhttps://www.blogger.com/profile/01468651273398333954noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-73012465915443221082014-01-28T13:29:49.875-05:002014-01-28T13:29:49.875-05:00Ramanujan (and Euler) wrote
1+2+3+4…= -1/12
Ram...Ramanujan (and Euler) wrote <br /><br />1+2+3+4…= -1/12<br /><br />Ramanujan's exact words are ". . . the sum of an infinite number of terms of the series <br />1+2+3+4…= -1/12 . . ."<br /><br />I am only repeating what he said.<br /><br />See for yourself here http://books.google.com/books?id=Of5G0r6DQiEC&pg=PA53&dq=gratified#v=onepage&q&f=false<br /><br />Are you saying he is wrong or that "=" means something other than equals?<br />Buzzhttps://www.blogger.com/profile/01468651273398333954noreply@blogger.comtag:blogger.com,1999:blog-35187314.post-62806883175371735382014-01-28T13:09:05.280-05:002014-01-28T13:09:05.280-05:00I am sorry Physics Buzz, but you are talking nonse...I am sorry Physics Buzz, but you are talking nonsense. The statement "include all the terms to infinity" is devoid of meaning. And as Evelyn Lamb explained, you are misrepresenting what Ramanujan & co. did. I don't think there is anything more I can say at this point other than suggesting that you read the sentence "There is a meaningful way to associate the number -1/12 to the series 1+2+3+4…, but in my opinion, it is misleading to call it the sum of the series" again and again. Perhaps it would have been advisable to do so before blogging about this on a website that's supposed to be educational. Davide Castelvecchinoreply@blogger.com