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What to make out of series like 1+2+3+ = -1/12 ?

  1. Jul 16, 2009 #1
    Sometimes, analytic continuation of some functions (like zeta) gives bizarre summation values for series like the one in the title or for 1+1+1+.... = -1/2 etc... and they are also used quite often in physics.
    Are mathematicians really comfortable with such values for such series? Whats the general understanding about such appearance?
     
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  3. Jul 16, 2009 #2

    CompuChip

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    In mathematics, these series do not converge, hence statements like "diverging series" = "finite number" are nonsense. Even with series which are only absolutely divergent the problem arises (for example, the series 1 - 1 + 1 - 1 + ... can be given any value depending on the order in which the terms are summed).

    In physics, however, series like 1 + 2 + 3 + ... turn up in answers which are supposed to give physically sensible (i.e. finite) quantities. Instead of using this sum, people look at the function
    [tex]\zeta(s) = \sum_{k = 1}^\infty \frac{1}{k^s}[/tex]
    which is defined for values of s > 1. One then continues this function analytically to all of the complex plane (with exception of s = 1). Obviously then the function is no longer given by the above power series, although theoretical physicists may pretend it is and assign to
    [tex]\sum_{k = 1}^\infty \frac{1}{k^{-1}} = \sum_{k = 1}^\infty k[/tex]
    the value of [itex]\zeta(-1) = -1/12[/itex].
    Why this should make sense and why it should be particularly this value is, AFAIK, not really clear - although it seems to work well enough to use it.
     
  4. Jul 16, 2009 #3

    Hurkyl

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    Zeta regularization is a well-defined (partial) operation you can apply to sequences of real numbers. There is nothing mathematically wrong with it.

    Is it fair to call zeta regularization a kind of infinite sum? Yes, because (I believe) it has the properties we'd expect of such a thing.

    Does zeta regularization have anything to do with the "limit of partial sums" operation one learns in introductory calculus? Not really. (However, both being kinds of infinite sums, probably give the same answer in many interesting cases)

    The only "problem" that arises here is when people misinterpret such summations and think that we're saying the limit of partial sums of 1/n converges to -1/12.

    Incidentally, I believe this method originated in number theory, not physics.
     
  5. Jul 16, 2009 #4
    To just write it down and use it blindly is not comfortable for mathematicians.

    To say what you mean, then compute it (as, for example, [tex]\zeta(-1) = -1/12[/tex]) is fine with mathematicians. The warning is: there may be OTHER analytic functions that suggest in the same way that [tex]1+2+3+\dots[/tex] has a value, but that value may not be [tex]-1/12[/tex]. So that mythical physicist mentioned (in order to get the approval of the mathematicians) would have to say why he uses the zeta function and not something else.
     
  6. Jul 17, 2009 #5
    I'm surprised. Don't you think the series 1+1+1+.. should end up in infinity? How can it have a value -1/12?
    May i know where such a result is used? In any particular field of physics?
     
  7. Jul 17, 2009 #6

    CompuChip

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    In the first part below, the prime is used to denote a different function, so it's not a derivative. I tried using a tilde at first but it was invisible in the generated image, so it was impossible to distinguish [itex]\zeta[/itex] ($\zeta$) from [itex]\tilde\zeta[/itex] ($\tilde\zeta$).

    Yes, it does diverge (i.e. "end up in infinity"). It doesn't have the value -1/12. As I explained above, there exists at least one function which can be defined through a summation like
    [tex]\zeta(n) = \sum_{k = 1}^\infty \frac{1}{k^n}[/tex]
    on some domain and then extended to a larger domain which - technically - makes it a different function [itex]\zeta'(n)[/itex]. This does not mean that on the larger domain, it can still be expanded in such a formal sum. So "a priori", the equality [itex]\zeta(n) = \zeta'(n)[/itex] only holds on the domain of the original function, but in general takes different values outside that (for example, [itex]\zeta'[/itex] may have finite values when the series expansion defining [itex]\zeta[/itex] does not). Statements like "1 + 1 + .... = -1/12" arise from sloppy interpretation of such functions, where people (implicitly) do not distinguish between [itex]\zeta[/itex] and [itex]\zeta'[/itex].

    In theoretical physics this is used in (closely related) techniques called "regularization" and "renormalization". In modern-day quantum field theories, infinities often arise where physical (i.e. finite) quantities are expected. As a simplified example, consider obtaining for some mass or energy scale the expression
    [tex]\exp\left[ \hbar \sum_{k = 1}^\infty (2k) \right][/tex].
    By replacing the infinite sum by the finite value [itex]2 \zeta(-1)[/itex] (which is, technically, the analytic continuation of the zeta function defined by the sum, in the sense of my earlier story, because the latter is not defined for x = -1) theoretical physicists can make sense of such a value. The argumentation is that the "bare" value may be infinite, because of infinite quantum corrections; however when we actually measure the quantity there will be quantum screening effects which will give us a finite outcome of the experiment (cf. bare and screened electron charge, for example).

    I must admit, again, that although I understand the ideas of regularization and renormalization and see how they are useful, I still fail to see why choosing specifically this function - in the above case, the Riemann zeta function - is "correct" (as in: yielding the correct measurable value).
     
  8. Jul 17, 2009 #7
    i do not know how tis works but it is really admirable technique, is just my favourite 'regularization scheme' for physics and think (in words of ELizalde) theat ' Zeta regularization is the regularization of the future , and should be taught at University'
     
  9. Jul 17, 2009 #8
    And this finite outcome matches the one theoretically found using the zeta function! So nature too prefers zeta like the physicists?! :-O
     
  10. Jul 17, 2009 #9
    @CompuChip: I would call analytical extension more of a 'rearrangement' within the series definition of a function than a totally different function altogether. Also, wasn't the equivalence of this extended function to the original series established by number theorists? So, is it really a sloppy usage?

    @all: I clearly understand that such a series can never converge to any finite value, leave alone a negative number. But, as Hurkyl put it, the fact that the limits of certain partial sums (like 1+2+3+...) matches the series definition of a continued zeta function which, in turn, has a finite negative value at that point is not only interesting but very bizarre. And, as sganesh88 said, the fact that such a usage is verified to a very high precession in QED experiments is an indirect validation by nature. This should be giving sleepless nights, but I get a feeling of general complacency towards this among mathematicians. Why is that so?
     
  11. Jul 23, 2009 #10
    Sorry for this bad behavior of bumping up my thread, but this thing is driving me crazy and I dont understand why it is not driving number theorists (or mathematicians, in general) the same way.

    I have read writings on this (age old) problem from Euler to Hardy, but I dont find them convincing. The only way I can pacify myself is how CompuChip took it - that is, analytic continuation changes the function altogether (which is not the common interpretation and hence can cause more confusion in lots of other areas that use analytic continuation as a legitimate 'non-changing' technique). Even with that, as CompuChip said, there is no explanation why only this function is preferred by nature.
     
  12. Jul 23, 2009 #11

    Hurkyl

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  13. Jul 23, 2009 #12

    CRGreathouse

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    Have you taken complex analysis? The special nature of analytic functions (in this case, the uniqueness of the analytic continuation) would be one good reason that this doesn't bother others.
     
  14. Jul 23, 2009 #13
    @Hurkyl: Sorry, I didnt get the sarcasm (or a genuine question/pointer?). I dont find anything wrong with generating functions per se. It is the analytic continuations that give results like 1+1+1+... = -1/2 that is bothering me. Maybe one reason I didnt get a full clarification from your earlier post is because of your usages like 'not really', 'kind of', 'probably' etc in your point that addresses the main source of my confusion - that is, how is it that both partial sums and regularised zeta coincides in certain cases and gives bizarre results like above.

    @CRGreathouse:
    I have read complex analysis and have come across uniqueness of analytic continuations. But I dont understand how that settles my confusion. If you are implying that the original function and the continued function are two different functions that happen to have the same values in some (the original) domain, then where in the continuation process are we changing the function as such? Arent we just using the holomorphic=analytic equivalence and shifting the 'center' of the laurent series as we move around the entire plane?
     
  15. Aug 13, 2009 #14
    but there are no problems ?, for example

    if i define [tex] F(s)= \sum_{n=1}^{\infty} (a(n))^{-s} [/tex] and [tex] G(s)= \sum_{n=1}^{\infty} a(n).n^{-s} [/tex]

    then [tex] \sum_{n=1}^{\infty} a(n)[/tex] can take TWO regularized values

    F(1) or G(1) but F(s) and G(s) are two completely different functions
     
  16. Aug 13, 2009 #15

    CompuChip

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    I think you meant F(1) or G(0), or want to put -s+1 in the exponent of G(s).
     
  17. Aug 13, 2009 #16
    right Compuchip i made the mistake

    i meant F(-1) or G(0) in both cases the results would be different since F(s) or G(s) would not be equal.

    as a real example [tex] \sum_{n=0}^{\infty}(n+b) = \zeta (-1,b) [/tex] is different from

    [tex] \zeta(-1)+b\zeta(0) [/tex] which you could assume by linearity of the series, although if 'b' is an integer then both method give similar results.
     
  18. Sep 2, 2009 #17
    @ zetafunction

    I dont think you can assume linearity of divergent series, but I could be wrong. If I am right, that leaves with only one value for your last example.
    On the other hand, I would be interested to know if you arrived at F(s) and G(s) through continuation of (two different) functions you had in mind or if you picked them up randomly.
     
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