The sum of all natural numbers = -1/12

Discussion in 'Physics & Math' started by Magical Realist, Feb 6, 2014.

  1. user41 Registered Member

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    14
    how are we supposed to picture such things? how can the sum be negative??
     
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  3. RJBeery Natural Philosopher Valued Senior Member

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    Dinosaur, while I agree that this series conflicts with common sense, the equation itself apparently has far-reaching consequences which are actually USEFUL in certain calculations. When this happens I'd say we can no longer dismiss the equation as "magic".
     
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  5. Dinosaur Rational Skeptic Valued Senior Member

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    CptBork: Your Post #20 seems to be missing some justification for redefining the integral. I assume that it was not entirely arbitrary.

    The following relates to the series: 1 -1 +1 -1 +1 -1 . . . . .

    James R: From your Post #17:
    From my Post #18:
    From your Post #19:
    I consider my Post #18 to be valid as is without need for further justification, but I will try to provide a bit.

    Since last term in an infinite series cannot be determined, the sum of the series is nebulous at best. However that does not refute the following from my Post #18, which makes no claim for the actual sum of an infinite number of terms:
    I do not claim that it is possible to determine whether the last term is odd or even. I only claim that the sum must be either one or zero depending on the parity of the number of terms remaining after truncation. I believe the following answers are correct.

    If you truncate after google terms, the sum is zero.

    If you truncate after (google + 1) terms, the sum is one.​

    The same can be said relating to googleplex terms or any other even number.

    For those not familiar with the terms google & googleplex:

    Google is 1 followed by 100 zeros (10[sup]100[/sup]).
    Googleplex is one followed by google zeros (10[sup]google[/sup]).​
     
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  7. RJBeery Natural Philosopher Valued Senior Member

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    Dinosaur, it isn't fair to argue with James R on this topic. Analytic Continuation is a well-established branch of mathematics; you might as well be arguing against the use of irrational numbers (which would be a logically plausible yet futile thing to do).
     
  8. phyti Registered Senior Member

    Messages:
    732
    It isn't convergent or divergent. Consider it a superpostion of states 0 and 1. It is (0 and 1) until you truncate it, when it becomes (0 or 1). Thus the flipped coin example.
     
  9. user41 Registered Member

    Messages:
    14
    this method works only for infinite series...trying to do it for a finite set of numbers, it failed miserably.
    i wonder why adding to infinity makes any difference since we are assuming the sum to be 1/2...
    and, if you solve it using the answer to the infinite series (1-1+1....) as 0, the final answer you get is S2 = 0 (since S2 is half S1, which is 0) and the difference between S3 and S2 becomes equal to 4 times S3 and since S2 is already 0, S3 equals 4 times itself, which probably fits better with the idea of infinity.
    while what i did above might not make sense, there is no reason for the sum to come out negative. i just dont seem to understand what is happening here.
     
  10. RJBeery Natural Philosopher Valued Senior Member

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    4,222
    Take any one of these infinite series and smooth them out; the summation value that you get at infinity is the smooth curve's Y-axis intersection when X = 0. This is why (1-1+1-1+1...) = .5 and why (1+2+3+4+5...) = -1/12.

    Now, as to WHY this is what happens is an intriguing mystery.
     
  11. rpenner Fully Wired Valued Senior Member

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    4,833
    Here's another (substantially similar) way to think about this from Luboš Motl
    \(S = 1 - 2 + 3 - 4 + 5 - 6 + \dots = \sum_{k=1}^{\infty} (-1)^{\tiny k+1} k \\ Z = 1 + 2 + 3 + 4 + 5 + 6 + \dots = \sum_{k=1}^{\infty} k S_n = \sum_{k=1}^{n} (-1)^{\tiny k+1} k = \frac{1 - (-1)^n(2 n + 1) }{4} Z_n = \sum_{k=1}^{n} k = \frac{(n+1)n}{2} Z_{2n} - 4 Z_n = S_{2n} S'_n = \sum_{k=1}^{n} (-1)^{\tiny k+1} k e^{-k \epsilon} = \frac{e^{(n+1) \epsilon} - (-1)^n n - (-1)^n (n+1) e^{\epsilon}}{ e^{(n+1) \epsilon} } \times \frac{ e^{\epsilon} }{ \left( e^{\epsilon} + 1 \right)^2 } Z'_n = \sum_{k=1}^{n} k e^{-k \epsilon} = \frac{e^{(n+1) \epsilon} + n - (n+1) e^{\epsilon}}{ e^{(n+1) \epsilon} } \times \frac{ e^{\epsilon} }{ \left( e^{\epsilon} - 1 \right)^2 } \lim_{\epsilon\to 0} S'_n = S_n \lim_{\epsilon\to 0} Z'_n = Z_n \lim_{\epsilon\to 0} \left( Z'_{2n} - 4 Z'_n \right) = S_{2n} = \lim_{\epsilon\to 0} S'_{2n} S' = \sum_{k=1}^{\infty} (-1)^{\tiny k+1} k e^{-k \epsilon} = \frac{e^{\epsilon}}{\left( e^{\epsilon} + 1 \right)^2} = \left( \frac{1}{2 \, \cosh \, \frac{\epsilon}{2} \right)^2 = \frac{1}{4} - \frac{1}{16} \epsilon^2 + \frac{1}{96} \epsilon^4 + O(\epsilon^6) \quad ; \quad \epsilon \neq 0 Z' = \sum_{k=1}^{\infty} k e^{-k \epsilon} = \frac{e^{\epsilon}}{\left( e^{\epsilon} - 1 \right)^2} = \left( \frac{1}{2 \, \sinh \, \frac{\epsilon}{2} \right)^2 = \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{1}{240} \epsilon^2 - \frac{1}{6048} \epsilon^4 + O(\epsilon^6) \quad ; \quad \epsilon \neq 0 Z' _{\epsilon \to x} - 4 Z' _{\epsilon \to 2 x} = \frac{e^x}{\left(e^x - 1 \right)^2} - 4 \frac{e^{2x}}{\left(e^{2x} - 1 \right)^2} = \frac{e^x}{\left(e^x + 1 \right)^2} = S' _{\epsilon \to x} S = \lim_{\epsilon\to 0} S' = \frac{1}{4} \quad ; \quad \textrm{this may be regarded as a trick involving improper manipulation of double limits} Z = \lim_{\epsilon\to 0} Z' = -\frac{1}{12} + \not\infty \quad ; \quad \textrm{ignoring the infinity is the weird part here} Z - 4 Z = S \)
     
  12. exchemist Valued Senior Member

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    12,451
    Yes, thank you. That is exactly what I started to try to do but then got stuck (my maths is 40yrs old). And there is an improper step in both S and Z , I see.
     
  13. CptBork Valued Senior Member

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    6,460
    I don't know what the original reasoning was for choosing that particular definition, but if the infinite series \(\sum n\) were to converge, then it would have to converge to the same quantity calculated after taking the limit. As a correction to my Post #20, I believe that the integral problem I posted was actually supposed to be something like \(\int_{-\infty}^\infty sinx\ \mathrm{d}x\), because I think the one I posted doesn't converge even after applying the altered definition.

    In any case, one of the rarely mentioned principles in physics is that tiny changes in the conditions of the problem will lead to correspondingly tiny changes in the solution. I think that principle applies even in the case of chaos theory and subjects like climate prediction, provided the changes in conditions are made sufficiently small. So the schemes I'm posting really just alter the initial problem and let it form an increasingly accurate set of approximations to the real problem, under the assumption that the calculable solutions in each case will become closer and closer to the real, physical solution to the original problem where the required calculations weren't well-defined.

    In ever physics calculation I've ever seen, you don't simply ignore the infinities, even though some mathematical methods may do so for expediency. You keep track of the infinities and what sorts of series sums or integrals are used to represent them, and they all cancel off to leave behind a final answer. In cases where the infinities don't automatically cancel, you're always left with at least one parameter which you then pick so as to cancel off the infinity and leave behind an answer which matches with physical measurements, which then fixes those parameters for every other calculation you do (this is how QFT renormalization works).

    There was never a case where we had to assume \(\sum n = \infty - \frac{1}{12}\) and then simply ignore the infinity, it always cancels off in physical calculations.
     
  14. Dinosaur Rational Skeptic Valued Senior Member

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    4,885
    The sum of all natural numbers = -1/12

    I do not understand why a result like the above is not viewed as a reductio ad absurdum proof that there is a logical error in the statements leading to such a conclusion.

    The so called proofs remind me of a favorite phrase often used by a classmate:
     
  15. CptBork Valued Senior Member

    Messages:
    6,460
    Again, none of the proofs above actually make that statement. They make statements like what the sum would equal if the conditions for the series to be analytically continued were satisfied. Since those conditions aren't actually satisfied, this method can't be used to perform the sum and derive the correct answer, since the correct answer is larger than any finite number. It does, however, provide us with convenient shortcuts in certain circumstances, ultimately yielding the same result as more rigorous methods even though certain illegal/invalid operations were performed along the way, as if the errors cancelled each other out.

    When you solve problems in physics such as what I described, you don't have to do any fudging, you only have to assume that when you get something involving a difference of two infinities, nature resolves the issue by defining the result as the limit of a sequence of problems involving the subtraction of finite quantities, with those finite quantities jointly being allowed to approach the infinite quantities of the original problem.
     
  16. user41 Registered Member

    Messages:
    14
    dinosaur's #31 seems to make sense. there must be some invalid statement in the proof.
    one of my friends pointed out that the step in which 4+8+12+... is reduced to 4 times the equation we wanted to solve could be wrong, since both the equations go on to infinity. now, when u subtract one infinity from another, can you get some other multiple of infinity?
     
  17. Farsight

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    3,492
    Bah.

    Woo.
     
  18. RJBeery Natural Philosopher Valued Senior Member

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    4,222
    Code:
    S1 = 1 + 2 + 3 + 4 + 5... = -1/12
    
       1 + 2 + 3 + 4 + 5 + ...  (S1)
    -  0 + 1 + 2 + 3 + 4 + ...  (S1)
    ------------------------
       1 + 1 + 2 + 3 + 4 + 5 + ...  (1 + S1)
    
    Therefore -1/12 - (-1/12) = 11/12

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  19. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    Code:
    s1 = 1 + 2 + 3 + 4 + 5... = -1/12
    
       1 + 2 + 3 + 4 + 5 + ...  (s1)
    -  0 + 1 + 2 + 3 + 4 + ...  (s1)
    ------------------------
       1 + 1 + [color=red]1 + 1 + 1 + 1 +[/color] ...  
    
    I don't think your method shows that.
     
  20. RJBeery Natural Philosopher Valued Senior Member

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    4,222
    Hah! Friends don't let friends do math before coffee...

    It's interesting to note, however, the above implies that 1 + 1 + 1 + 1 ... = 0 which also happens to coincide with the y-axis intercept of the smooth graph (as they do with the sum of natural numbers as shown below for -1/12)

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  21. rpenner Fully Wired Valued Senior Member

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    4,833
    \(C = 1 - 1 + 1 - 1 + 1 - 1 + \dots = \sum_{k=1}^{\infty} (-1)^{\tiny k+1} \\ K = 1 + 1 + 1 + 1 + 1 + 1 + \dots = \sum_{k=1}^{\infty} 1 C_n = \sum_{k=1}^{n} (-1)^{\tiny k+1} = \frac{1 - (-1)^n }{2} K_n = \sum_{k=1}^{n} 1 = n K_{2n} - 2 K_n = C_{2n} C'_n = \sum_{k=1}^{n} (-1)^{\tiny k+1} e^{-k \epsilon} = \frac{1 - (-1)^n e^{-n \epsilon}}{ e^{\epsilon} + 1 } K'_n = \sum_{k=1}^{n} e^{-k \epsilon} = \frac{1 - e^{-n \epsilon}}{e^{\epsilon} - 1 } \lim_{\epsilon\to 0} C'_n = C_n \lim_{\epsilon\to 0} K'_n = K_n \lim_{\epsilon\to 0} \left( K'_{2n} - 2 K'_n \right) = C_{2n} = \lim_{\epsilon\to 0} S'_{2n} C' = \sum_{k=1}^{\infty} (-1)^{\tiny k+1} e^{-k \epsilon} = \frac{1}{e^{\epsilon} + 1} = \frac{1 - \tanh \frac{\epsilon}{2}}{2} = \frac{1}{2} - \frac{1}{4} \epsilon + \frac{1}{48} \epsilon^3 - \frac{1}{480} \epsilon^5 + O(\epsilon^6) \quad ; \quad \epsilon \neq 0 K' = \sum_{k=1}^{\infty} e^{-k \epsilon} = \frac{1}{e^{\epsilon} - 1} = \frac{\coth \frac{\epsilon}{2} \, - 1}{2} = \frac{1}{\epsilon} - \frac{1}{2} + \frac{1}{12} \epsilon - \frac{1}{720} \epsilon^3 + \frac{1}{30240} \epsilon^5 + O(\epsilon^6) \quad ; \quad \epsilon \neq 0 K' _{\epsilon \to x} - 2 K' _{\epsilon \to 2 x} = \frac{1}{e^x - 1} - \frac{2}{e^{2x} - 1} = \frac{1}{e^x + 1} = C' _{\epsilon \to x} C = \lim_{\epsilon\to 0} C' = \frac{1}{2} \quad ; \quad \textrm{this may be regarded as a trick involving improper manipulation of double limits} K = \lim_{\epsilon\to 0} K' = -\frac{1}{2} + \not\infty \quad ; \quad \textrm{ignoring the infinity is the weird part here} K - 2 K = C \)

    It's neat that this technique winds up being identical to zeta regularization of these divergent sums.

    \(\zeta(0) = - \frac{1}{2} =^? K = \sum_{k=1}^{\infty} k^0 \\ \zeta(-1) = - \frac{1}{12} =^? Z = \sum_{k=1}^{\infty} k^1 \\ \zeta(-2) = 0 =^? \sum_{k=1}^{\infty} k^2 \\ \zeta(-3) = \frac{1}{120} =^? \sum_{k=1}^{\infty} k^3\)
     
    Last edited: Feb 13, 2014
  22. RJBeery Natural Philosopher Valued Senior Member

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    4,222
    I know you're using Lobl's method here, do you agree with it? Are you saying that you believe {1 + 1 + 1 + 1 + 1 + 1....} = 1/2?

    Hell, I could do this: 1 + 1 + 1 + 1 + 1 + 1 ... = 1 + (1 + 1) + (1 + 1 + 1) + (1 + 1 + 1 + 1) + ... = 1 + 2 + 3 + 4 + ... = -1/12!
     
  23. rpenner Fully Wired Valued Senior Member

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    4,833
    No -- I'm saying the divergent series can be regularized in a number of ways that give the same finite value (-1/2).
    Regularizing a divergent series is transforming it into which is in some sense very different (i.e. the result is a number) while some properties are preserved.

    Example: \(\lim_{\epsilon\to 0} K'_n = \lim_{\epsilon\to 0} \frac{1 - e^{-n \epsilon}}{e^{\epsilon} - 1} = \lim_{\epsilon\to 0} n - \frac{n(n+1)}{2} \epsilon + O(\epsilon^2) = n = K_n\)

    http://en.wikipedia.org/wiki/Divergent_series

    Grouping of divergent sequences is not particularly compelling.
     

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