Thanks, Dr. Jantzen!
Sunday, April 13, 2014 at 5:14 PM
Bernd Jantzen said...
I totally agree. And I refer to my previous comments above, where I have made similar statements.
Please note that the discussion about this topic actually went on in two new blog posts by the same author:
[titled "Correction: Does 1+2+3+4+ . . . =-1/12? Absolutely Not! (I think)"]
[titled "If Not -1/12, What Does 1+2+3+4+... Equal? We Give You the Answer "]
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.
Sunday, April 13, 2014 at 3:43 PM
PS: please read the 3rd paragraph at books.google.de/books?id=n8Mmbjtco78C&pg=PA73
Sunday, April 13, 2014 at 1:10 PM
I'm afraid you're greatly confused.
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.
Also, n appeal to authority is a fallacy.
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.
3. You're engaging in trickery by juggling different definition of sums.
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.
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).
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.
But this doesn't justify the bad math, sorry.
Sunday, April 13, 2014 at 10:19 AM
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.
Sunday, April 13, 2014 at 10:07 AM
Imre Fabian said...
For me, not a mathematician, this is a perfect example for that you can prove anything with infinity mathematics.
In my view, infinity does not exist (in the real world), nor does 0.
Infinity = anything / 0
0 = anything / infinity.
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).
Wednesday, March 5, 2014 at 4:57 AM
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.
Tuesday, March 4, 2014 at 10:28 AM
Wondrful exchange guys. Thanks for the knowledge you've been sharing with us, and for the effort that takes.
Monday, February 3, 2014 at 5:22 PM
Buzz Skyline said...
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).
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.
Tuesday, January 28, 2014 at 5:36 PM
Bernd Jantzen said...
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.
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?
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
1^(-s) + 2^(-s) + 3^(-s) + 4^(-s) + ...
= 1/2 + integral from 1 to infinity of dx * x^(-s) + small remainder,
which is valid, as the series on the left-hand side, for s>1.
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:
integral = 1/(s-1) * ( 1 - infinity^(1-s) )
For s>1, the contribution from the upper boundary is infinity raised to a negative power, i.e. zero, and we and up with:
integral = 1/(s-1) > 0 for s>1.
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.
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:
integral = 1/(s-1) = -1/(1-s) < 0 for s<1.
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.
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.
Tuesday, January 28, 2014 at 4:17 PM
Bernd Jantzen said...
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.
Tuesday, January 28, 2014 at 3:10 PM
Buzz Skyline said...
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.
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.
Tuesday, January 28, 2014 at 2:46 PM
Bernd Jantzen said...
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.)
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.
Tuesday, January 28, 2014 at 2:32 PM
Buzz Skyline said...
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?
Tuesday, January 28, 2014 at 2:04 PM
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.
Tuesday, January 28, 2014 at 1:47 PM
Bernd Jantzen said...
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.
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.
Tuesday, January 28, 2014 at 1:42 PM
Keep in mind, I'm saying
If QED is true
and it relies on 1+2+3+4…= -1/12
then 1+2+3+4…= -1/12 must be true.
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.
Tuesday, January 28, 2014 at 1:37 PM
Ramanujan (and Euler) wrote
Ramanujan's exact words are ". . . the sum of an infinite number of terms of the series
1+2+3+4…= -1/12 . . ."
I am only repeating what he said.
See for yourself here http://books.google.com/books?id=Of5G0r6DQiEC&pg=PA53&dq=gratified#v=onepage&q&f=false
Are you saying he is wrong or that "=" means something other than equals?
Tuesday, January 28, 2014 at 1:29 PM
Davide Castelvecchi said...
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.
Tuesday, January 28, 2014 at 1:09 PM
I think the point is that no one is saying they converge. It's the same as for the series 1-1+1-1+1-1+...
It doesn't converge, but if you include all the terms to infinity (not the limit as you go to infinity, but the value when infinite terms are included), it equals 1/2.
If you approach it by thinking in terms of convergence, as you point out, there's nothing you can do with zeta(-1), zeta (-3), or 1-1+1-1+1 . . . And yet, using the values that Euler, Reimann, and Ramanujan calculated for those things give you meaningful answers to real, testable theories.
I can't see a way to rectify those two things - either the equations and the theories are true, in some sense anyway, or they're each wrong in ways that precisely (to 12 or more decimal places) correct the errors the other one produces. The second option seems much harder to believe than the first.
Tuesday, January 28, 2014 at 12:49 PM
Davide Castelvecchi said...
Analytical continuation is not a way to make divergent series converge. It just isn't.
Analytical functions are by definition those that admit a power
expansion at every point. That means that *locally* you can write them
as the sum of their Taylor expansion. (And no, not all functions are
analytic, even if they admit infinitely many derivatives: sometimes
you can write the Taylor expansion but that does not equal the
function you started with. But I digress.)
Being analytic does not mean that the function is the same as its
Taylor expansion at a point. The Taylor expansion, in particular, may
start diverging at points where the function is perfectly well
defined. Taylor expansions have a well-defined convergence radius in
the complex plane, and that radius may be infinite (as it is for the
case of e^z, say) but it is often finite.
So, if you start with a given Taylor series, there can be a function
that the Taylor series was the expansion of (at a given point), and
that function may be well-defined on a larger domain than the Taylor
series was. But that doesn't mean that you are making the original
Taylor series converge. It still diverges: it's just that it only
represents the function _within_ its radius of convergence.
If you want to play with an example, take f(z)=1/(z-1). f(z) is
perfectly well defined at any complex number z, as long as it's not 1.
Now take its Taylor expansion at z=0, meaning
f(0) + f'(0)z + 1/2f"(0)z^2 + 1/6 f'''(0)z^3 + ...
Its radius of convergence is 1, which means that the Taylor series
*diverges* for any |z|>1. The fact that f(z) is well defined for *all*
|z|>1 doesn't magically make the series converge there. It's the
*function* that is well defined there, not the Taylor series. If you
had started with the Taylor series, and then realized that it was the
Taylor series of 1/(z-1), then you could say that f(z) was an
analytical continuation of the Taylor series. But again, doing that
doesn't make the series any more convergent than it was before. This
is stuff that has been figured out once and for all 200 years ago.
About 1+2+3... Evelyn Lamb has already said everything that there is to say: "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."
End of story.
Tuesday, January 28, 2014 at 12:10 PM
The twelfth root of 2 is how we divide the musical scale, each note is 1/12th of the octave, wonder if there is a connection. Its often assumed that all music strictly cultural but the evidence does not support it, for example tone/tempo of voice carries emotional information you can understand even if you don't know the language and have no other clues, play a film in a foreign language while blindfolded and you can still tell what the characters are feeling by tone/tempo of voice alone. The variances in tempo and pitch that convey this information are indistinguishable from musical variances, we speak with music only some of which carries abstract associative meaning but most of which carries universally understood emotional state information. This suggests that the emotional content of music is conveyed by adherence to universal rules that are just as solid as the rules of physics with the 12th root of two the simplest generator of the frequencies for both tempo and tone.
Sunday, January 26, 2014 at 4:04 PM
Bernd Jantzen said...
So, what did I mean by saying before "This 'identity' as it stands here is simply wrong. But it can be useful for a calculation if you properly regularize the divergent series and apply analytic continuation."?
The "identity" 1 + 2 + 3 + ... = -1/12 is wrong, because the summation of a divergent series does not have a value and cannot be equal to a finite number.
However, the following statement is correct:
"The Riemann-zeta-regularized summation of 1 + 2 + 3 + ... yields -1/12."
If you mean a specific regularization for your divergent series, then you have to name it. Simply writing "1 + 2 + 3 + ..." implies standard summation by adding consecutively more and more terms, and this does not yield a finite value here.
What "Riemann-zeta-regularized" means is that you replace the original divergent series by the new series 1^(-s) + 2^(-s) + 3^(-s) + ... which converges for s>1. (I am now omitting "real part" for simplicity.) This new series is summed, it yields zeta(s). Then this function zeta(s) is analytically continued to negative values of s, where the series itself has no meaningful value any more. The result of these steps is zeta(-1) = -1/12. Mark: zeta(-1) is not equal to the original divergent series, but it follows from it through the regularization steps described here.
For a physics calculation, it is only safe to apply such Riemann-zeta regularization if the regulator s is introduced at a point in the calculation where everything is still mathematically well-defined, i.e. before arriving at a divergent series. Otherwise there are caveats such that the calculation might lead to right or wrong results, you never know in general, and manipulating divergent expressions is a dangerous thing to do.
By the way, look at the Wikipedia page for "1 + 2 + 3 + 4 + ⋯" (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF). This page says that the series "can be manipulated to yield a number of mathematically interesting results", that it is "'summed' by zeta function regularization" ('summed' even in quotation marks), or that its "Ramanujan summation [...] is also -1/12". These formulations are different from stating plainly "1 + 2 + 3 + ... = -1/12", which is clearly avoided in the Wikipedia article.
Sunday, January 26, 2014 at 7:06 AM
Bernd Jantzen said...
Concerning the Riemann zeta function, I already answered this in my reply to Anonymous from January 24 at 3:02 AM.
In short: The Riemann zeta function zeta(s) is an analytic function defined for s in the whole complex except s=1 where zeta(s) has a pole. In my reply from January 24 at 3:02 AM above, I linked the PDF file of a book where you can find various definitions of zeta(s) which are also valid for negative s. However, the series representation zeta(s) = 1^(-s) + 2^(-s) + 3^(-s) + ... is only valid if the real part of s is larger than 1 (so for s>1 on the real axis). That means the series can only be used to define zeta(s) for Re(s)>1, but other definitions are available for Re(s)<1 which may be derived from the series representation via analytic continuation.
The Wikipedia page on the Riemann zeta function does not state zeta(-3) = 1^3 + 2^3 + 3^3 + ... Instead, it says in the introduction: the "Riemann zeta function [...] is a function of a complex variable s that analytically continues the sum of the infinite series [...], which converges when the real part of s is greater than 1." That is exactly what I explained above. And also the "Definition" section of the Wikipedia article restricts the given series definition to Re(s)>1, stating that the "infinite series defines zeta(s) in this case". Nowhere in the Wikipedia page do you find "zeta(-3) = 1^3 + 2^3 + 3^3 + ..." or "zeta(-1) = 1 + 2 + 3 + ...", which would be wrong.
The other Wikipedia page on the Casimir effect uses the Riemann zeta function in order to sum 1^(3-s) + 2^(3-s) + 3^(3-s) + ... = zeta(s-3), which is valid for real parts of s larger than 4. This zeta(s-3) is then identified with the analytic continuation of the zeta function which is valid also for negative values of s (in contrast to the series!). So, while intermediary steps of the calculation where valid only for Re(s)>4, the limit for s->0 of the final result is taken in the end. Such a procedure is often applied when using regularization, and it is justified by knowing that the calculated quantity is an analytic function of s.
If I find more time later, I will comment on what I meant by an identity being "simply wrong and yet useful".
Sunday, January 26, 2014 at 4:29 AM
Buzz Skyline said...
I confess, I don't understand what you mean by "This "identity" as it stands here is simply wrong. But it can be useful for a calculation if you properly regularize the divergent series and apply analytic continuation." How can it be simply wrong, and yet useful?
Also, as far as I can tell, the the Wikipedia piece is using the zeta function as defined here http://en.wikipedia.org/wiki/Riemann_zeta_function. So zeta (-3)=1^3+2^3+3^3+ . . .
In an earlier line, they write sum[abs(n)^(3-s)], then take the limit as s goes to zero, leaving sum[abs(n)^(3)] which looks like the usual zeta function with the argument -3.
So I still believe I'm right in saying that the Wikipedia includes the equation 1^3+2^3+3^3 +4^3+ . . . = 1/120. It may or may not be a correct equation, but it's what the page says.
Sunday, January 26, 2014 at 2:07 AM