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Algebra simultaneous equations Question

  1. Oct 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##g_k = 2cos(k/2)## and ##z=e^{ip(N+1)}## where N is an integer.

    There are two simultaneous equations:

    ##E^2 = (g_k + e^{ip})(g_k + e^{-ip}) = 1 + g_k^2 + 2g_k cos(p) ## [1]

    ##(1+z^2)E^2 = (g_k + e^{-ip})^2 z^2 + (g_k + e^{ip})^2##[2]

    Eliminate ##E^2## to show that:

    ##sin[pN] + g_k sin[p(N+1)] = 0##[3]

    2. Relevant equations


    3. The attempt at a solution

    I've tried substitution of [1] into [2] and then equating the real parts/imaginary parts of the equation. Also have tried equating coefficients of terms like ##e^{i2p(N+1)}## but couldn't get anywhere with it.

    Subsituting [1] into [2] and expanding all complex exponentials into sines and cosines, equating real parts gives me:

    ##1 + cos(2p(N+1)) = cos(2pN) + cos(2p) + 2sin(2p(N+1))g_k sin(p)##

    And equating imaginary parts gives me:

    ##sin(2p(N+1)) = sin(2pN) - 2g_k sin(p)cos(2p(N+1)) + sin(2p) + 2g_k sin(p)##

    But I can't seem to get equation [3] from these. Also, equating the coefficients of ##e^{i2p}## I get:

    ## e^{i2pn} = -2i e^{i2pN} g_k sin(p) + 1##

    But again I can't rearrange it to equation [3] after equating real/imaginary parts. Does anyone have any insight as to how to do this?
     
  2. jcsd
  3. Oct 21, 2016 #2

    Simon Bridge

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    Looking at: ##E^2 = 1+g_k^2 + 2g_k\cos p##

    Looks like the cosine rule in trig doesn't it? Is there a clue by considering the matter geometrically?
    Solving for ##E^2## in [2] and subbing into [1] gives:
    $$\frac{(g_k+e^{-ip})^2z^2 + (g_k+e^{ip})^2}{1+z^2} = 1+g_k^2 + 2g_k\cos p$$... this looks like a combination of real and imaginary terms on the LHS and only reals on the RHS. Is this how you are thinking? The imaginary terms have to cancel out?

    Note: put a backslash before the trig function to get it to format correctly like ##\sin \theta## instead of ##sin\theta##.
    You may want to adopt a shorthand like ##s=\sin p## and ##S=\sin(N+1)p## to ease writing.
     
  4. Oct 27, 2016 #3
    This may be completely wrong, but I think there's a quick workaround to the problem. Even in the highly unlikely case that the answer is correct, the reasoning is probably wrong regardless.

    Let A = [tex](g_k + e^{ip})[/tex]
    Let B = [tex](g_k + e^{-ip})[/tex]

    Therefore [tex]E^2 = AB[/tex]

    If [tex](1 + z^2)E^2 = (g_k + e^{-ip})^2 z^2 + (g_k + e^{ip})^2 [/tex]

    then [tex]AB + ABz^2 = B^2z^2 + A^2[/tex]

    Just looking at both sides of the equation, it appears that

    [tex] AB = A^2 [/tex]
    [tex]ABz^2 = B^2z^2[/tex]

    As such, [tex]A = B[/tex]
    Therefore, [tex]e^{ip} = e^{-ip}[/tex]

    which means [tex]p = 0[/tex]

    With that in mind [tex]sin(0) + cos(k/2)*sin(0) = 0[/tex]
     
  5. Oct 27, 2016 #4

    haruspex

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    It was looking useful up to that point, but your next step seems to be treating the equation in A, B and z as true for all z with constants A, B. There is no basis for that.

    Anyway, it is just as well that you went wrong. This is a homework forum. The idea is to provide hints and point out errors, not lay out complete solutions.
     
  6. Oct 27, 2016 #5
    I knew it was too good to be true - way too simple. Back to the long form of calculation.

    By the way, what's the right answer? Does the angle p = 0 or is it irrelevant? Unless that's the key to the problem, which would not be allowed until after the original poster solved it.
     
    Last edited: Oct 27, 2016
  7. Oct 27, 2016 #6

    haruspex

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    As I posted, your working down to there looked good. Use it to express z in terms of A and B. Simplify, then substitute in what A, B and z represent.
    The problem is a bit strange... k plays no role. gk is just any real in the range -2 to +2, and it turns out that even that is not necessary. Maybe it is just inherited from some original context.
     
  8. Oct 27, 2016 #7
    Thanks haruspex, good thoughts! With some algebra, the equation simplifies even further to:

    [tex]AB - A^2 = B^2z^2 - ABz^2[/tex]
    [tex]A(B-A) = Bz^2(B - A)[/tex]

    At this point, even if I solve it, the original poster must show the answer first.

    I wouldn't be able to divide by (B - A) if A = B, since that would be division by zero. In the event A did not equal B, then [tex]A = Bz^2[/tex]
     
  9. Oct 27, 2016 #8

    haruspex

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    Right, but in that case p=0 and the equation to be proved is trivially true anyway.
     
  10. Oct 27, 2016 #9

    Mark44

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    The OP hasn't chimed in since last Friday, so let's hold off any more discussion until he comes back...
     
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