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Solving two equations

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi everybody, I am trying to find the solution to this system of equations for a bigger physics problem:

    h+k+l=2n+1 with n=0,1,2,3...
    ei(π/2)(h+k+l)+ei(π/2)(3h+3k+l)+ei(π/2)(3h+k+3l)+ei(π/2)(h+3k+3l)=0

    2. Relevant equations

    None

    3. The attempt at a solution

    Ihave tried to solve it by taking the common factor ei(π/2) out and trying to solve
    e(h+k+l)+e(3h+3k+l)+e(3h+k+3l)+e(h+3k+3l)=0 by using h+k+l=0; then, I arrive to 1+e2k+2h+e2h+2l+e2k+2l=0.

    From here I don't know how to keep going, so if anybody could point me what to do next it wolud be very helpfull.

    Thanks for reading.
     
    Last edited by a moderator: Oct 8, 2015
  2. jcsd
  3. Oct 8, 2015 #2

    haruspex

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    That is a factor of ei(π/2)+x, not of ei(π/2)x.
     
  4. Oct 8, 2015 #3

    RUber

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    To make these equal to zero, you need to enforce conditions on the periodic functions--both real and imaginary.
    ##\cos( \frac{ \pi(h+k+l)}{2} ) + \cos( \frac{ \pi(3h+3k+l)}{2} )+\cos( \frac{ \pi(3h+k+3l)}{2} )+\cos( \frac{ \pi(h+3k+3l)}{2} ) =0##
    and
    ##\sin( \frac{ \pi(h+k+l)}{2} ) + \sin( \frac{ \pi(3h+3k+l)}{2} )+\sin( \frac{ \pi(3h+k+3l)}{2} )+\sin( \frac{ \pi(h+3k+3l)}{2} ) =0.##

    From these, you should be able to say something about the cases where this must be true.
     
  5. Oct 8, 2015 #4

    RUber

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    Think of these as :
    ##\cos( \frac{ \pi(h+k+l)}{2} ) + \cos( \frac{ \pi(h+k+l)}{2} +(h+k)\pi )+\cos( \frac{ \pi(h+k+l)}{2}+ (h+l)\pi )+\cos( \frac{ \pi(h+k+l)}{2}+(k+l)\pi ) =0##
    and
    ##\sin( \frac{ \pi(h+k+l)}{2} ) + \sin( \frac{ \pi(h+k+l)}{2} +(h+k)\pi )+\sin( \frac{ \pi(h+k+l)}{2}+ (h+l)\pi )+\sin( \frac{ \pi(h+k+l)}{2}+(k+l)\pi ) =0.##
     
  6. Oct 8, 2015 #5

    Ray Vickson

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    Please do not use all bold fonts for your solution; it looks like you are yelling at us!
    Mod note: Removed excess bold font

    Anyway, if you let ##m = 2n + 1##, and if you note that ##h+k+l = m## implies
    [tex] \frac{i \pi}{2} (3h+3k+l) = \frac{i \pi}{2} (3h + 3k + 3l - 2l) = \frac{i \pi}{2} 3m - i \pi l, [/tex]
    and so forth, you get a much simpler problem, especially if you multiply through by ##e^{-3 m \pi i/2}##.

    I would suggest you look at some of the initial cases ##m = 1, 3, 5, \ldots## separately, until you have gained sufficient insight into the nature of the solution for general odd ##m##.
     
    Last edited by a moderator: Oct 8, 2015
  7. Oct 11, 2015 #6
    Thank you very much for your anwsers, guess I will be able to proper solve it.
     
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