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I Infinity question

  1. Mar 24, 2017 #1
    If there is no upper limit on t, can you find a t such that: [itex] e^{iat} = e^{ia_0}[/itex], [itex] e^{ibt} = e^{ib_0}[/itex] and [itex] e^{ibct} = e^{ic_0}[/itex] at the same time?

    No matter what a,b and c is, though given a != b , a!=c, b!=c and a!= 0, b!= 0, c!=0

    Or maybe rather:
    [itex]at=a_0 +k_12\pi[/itex], [itex]bt=b_0 +k_22\pi[/itex] and [itex]ct=c_0 +k_32\pi[/itex], where the k's are integers

    I think it seems reasonable that you can, or at least come arbitrarily close to the equations being satisfied... But don't know how to prove it, or if I am right... Any pointers?
     
  2. jcsd
  3. Mar 24, 2017 #2
    Are you really talking factorials up there, or were you trying to type "not equals" (≠)?
     
  4. Mar 24, 2017 #3

    mfb

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    Staff: Mentor

    Not in general.

    Consider a=2, b=1, a0=0 and b0 = 0.1. Clearly t=kπ solves the first equation, but does not solve or even approximate the second one.

    If you require that a,b,c do not have a pair which have a common multiple, there should be approximations to arbitrary precision. In general there won't be an exact solution here either.
     
  5. Mar 24, 2017 #4
    Not equals, mjc123 ;)

    Thanks mfb! Forgot to mention, I assumed a is not a multiple of b etc.
     
  6. Mar 24, 2017 #5

    mfb

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    Staff: Mentor

    That is not general enough.
    a=2 and b=3 still lead to a common multiple of 6, and the same result.

    If we don't have that case, in general all you get is an approximation.
     
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