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Exp(i infinity) = 0 ?

  1. Sep 9, 2005 #1
    :grumpy: :yuck:
    Hello there,

    please tell me whether [tex]e^{iS} = 0[/tex] when [tex]S = +\infty[/tex]

    S is the action of a particle [tex] = \int L dt[/tex]
    e.g. when the particle goes to infinity and comes back, kinetic energy blows up,
    then S blows up.
     
  2. jcsd
  3. Sep 9, 2005 #2

    LeonhardEuler

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    Gold Member

    It does'nt:
    [tex]e^{ix}=\cos{x}+i\sin{x}[/tex]
    [tex]\lim_{x\rightarrow \infty}e^{ix}=\lim_{x\rightarrow \infty}\cos{x}+i\lim_{x\rightarrow \infty}\sin{x}[/tex]
    But niether the real nor imaginary parts of the limit exist; as x approaches infinity they oscilate between -1 and 1.
     
  4. Sep 9, 2005 #3
    That's what my math teacher told me,
    but my physics teacher told me entirely different thing.
     
  5. Sep 9, 2005 #4

    Hurkyl

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    Gold Member

    I would suspect both your math teacher and physics teacher are right!

    The map x → e^(ix) is, indeed, discontinuous at +∞, and thus one cannot continuously extend this map to have a value at +∞. (continuous extension is the process that is generally used to justify statements like arctan +∞ = π / 2 or x/x = 1).


    However...


    As is often the case, there is probably some related concept that is practical to use here, and whether he knows it or not, it's what your physics professor meant.
     
  6. Sep 9, 2005 #5
    Do you mean, for computation purpose, we can define the limit to be zero
    it can't be proven?
     
  7. Sep 12, 2005 #6
    :eek:
    this is no college level!
    Feymann Path Integral!
     
  8. Sep 13, 2005 #7

    College=university. Don't get freaked out.
     
  9. Sep 13, 2005 #8
    Feynmann Path Integral is at graduate level :rofl:
     
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