Need a hard math problem

Discussion in 'Physics & Math' started by rian.wrenn, Sep 7, 2007.

  1. §outh§tar is feeling caustic Registered Senior Member

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    Ouch!
     
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  3. Absane Rocket Surgeon Valued Senior Member

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  5. D H Some other guy Valued Senior Member

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  7. rian.wrenn Registered Member

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    lim
    N!1
    N Xn=0
    an this is what i say to my teacher???
     
  8. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Hmmm. I don't think so. This idea of mathematically continuing a function outside it's radius of convergence is a mathematically valid thing to do, unlike dividing by zero to prove 1=2.

    We do these things in physics all of the time. For example, Euler's gamma function is

    \(\Gamma(n) = (n-1)!, n\in\mathbb{Z}\).

    In quantum field theory, we end up with answers like \(\Gamma(-2)\). So how do you take a factorial of a negative number??? -3! = ? The answer is to analytically continue the gamma function, and regulate it.

    Tha amazing thing is that doing this mathematical trickery gets you answers that are more accurate than any other theory man has ever written down.

    someone asked about the zeta functions use---in string theory, this result is necessary for vacuum energy cancellations.
     
  9. iceaura Valued Senior Member

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    Still might want to clarify the indices of summation, back there, to show that you aren't dividing by zero - unless I'm reading everything wrong, somehow ?
     
  10. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    yeah you're right.

    I'm home in Texas and have been more concerned with fishing and drinking beer than zeta functions

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  11. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Oh well.

    I can't edit the post anymore.

    The zeta function should be defined as

    \(\zeta(s) \equiv \sum_{s=1}^{\infty} \frac{1}{n^s}\)

    One can show (by analytic continuation), that

    \(\zeta(-1) = \sum_{s=1}^{\infty} n = -\frac{1}{12}\).
     
  12. D H Some other guy Valued Senior Member

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    One can show that the analytic continuation of the zeta function is indeed -1/12 at -1. This does not mean the series evaluates to -1/12 at -1. The sum of a set of elements, all of which are positive, is never negative. The series diverges at -1, end of story.

    Given a function f(z) with a limited domain, the analytic continuation of f(z) is some other function F(z) with a domain larger than that of f(z) and such that F(z)=f(z) everywhere f(z) is defined. This does not mean f(z) magically been given a broader domain.

    Physicists are just too damn loose when they use math.
     
  13. Tom2 Registered Senior Member

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    D H, that was exactly my point. Here was the original quote:

    The middle member of the equality shouldn't be there. In truth, I didn't know that \(\zeta(-1)=\frac{-1}{12}\) by analytic continuation, but I'm perfectly happy to believe you on that one. But if \(\zeta(-1)\) does equal \(\frac{-1}{12}\), then it does not equal the indicated sum. My point is that complex analysis simply does not and can not overturn the results of real analysis.

    Right, but when you analytically continue \(\Gamma(n)\) for \(n\) not in \(\mathbb{Z}\), then obviously the expression for computing \(\Gamma(n)\) that is valid in \(\mathbb{Z}\) is no longer applicable.
     
  14. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    I'll drink to that

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  15. devire Registered Member

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    i'm not sure if this has been said before, but i think you can only reduce the second part of the riemann zeta function, i.e. the part that is taken over the gamma function, to that series when the real part of 's' is greater then one.

    i think it also says that the riemann zeta function can only be defined that way when the real part of 's' is greater than 1 on that mathworld.wolfram.com article on the riemann zeta function.

    actually, you could say that it can be defined that way when the real part of 's' is equal to one, with imaginary part of zero, though, i think, since that series would give you infinity when the real part of 's' equals 1 and the imaginary part of 's' equals 0.
     
  16. rian.wrenn Registered Member

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    dam you ppl r smart
     
  17. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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  18. iceaura Valued Senior Member

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    The story does not end there.

    The continuation would be: " the series diverges, therefore its sum is undefined". There is a big difference between "undefined" and "impossible". The difference is that the possibility of defining a sum for the series exists. This has been accomplished.

    So the statement "the sum of a set of elements, all of which are positive, is never negative" needs attention - it appears to be an illegitimate extrapolation of finite-conditioned intuition to manipulations of infinities, with the key an overlooking of some problems with the word "never".

    More hints might be drawn from geometrical language - the "point at infinity", the "circle at infinity", etc.

    Or modular arithmetic, in which sums of positive numbers are negative. Such might be better grounds for extrapolation and intuition than a directed number line, in some situations involving infinities.
     
  19. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    I had forgotten about the Casimir effect, and the fact that the zeta function pops up there, too.
     
  20. [a-5] Sex machine, coin operated. Registered Senior Member

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    Math is difficult.
     
  21. temur man of no words Registered Senior Member

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    think there is no problem since it is not really the sum, but some extension of a function defined by certain sum we are talking about here.
     
  22. one_raven God is a Chinese Whisper Valued Senior Member

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    Do you think you deserve the credit if someone else gives you the equation?
     
  23. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Read Lubos's blog entry. The sum pops up in physical problems, and can (in essence) be measured. An example of this is in quantum field theory, where we get expressions like

    \(\Gamma(-2) = \frac{1}{\epsilon} + finite + \mathcal{O}(\epsilon)\)

    for epsilon small. You substract the infinite part, neglect the small part, and use the finite part. Doing so gives you physical results which are more accurate than any physical results ever measured (for example, precision QED processes).

    So in some sense, adding all of the positive integers and getting a negative fraction is experimentally verified

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    Just like taking a negative integer factorial

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