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Proof Beyond Scope of Course

  1. Jun 17, 2012 #1
    "Proof Beyond Scope of Course"

    1. The problem statement, all variables and given/known data

    [tex] \frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})=[/tex]\begin{cases} 1, & \ m \equiv 0\pmod{M}\\ 0, & \text{else} \end{cases}

    2. Relevant equations



    3. The attempt at a solution

    is statement truthfully accurate? how to show this?
     
  2. jcsd
  3. Jun 17, 2012 #2

    HallsofIvy

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    Re: "Proof Beyond Scope of Course"

    Your formula makes no sense with out a statement of what the "[itex]x_j[/itex]" are.
     
  4. Jun 18, 2012 #3
    Re: "Proof Beyond Scope of Course"

    interval [tex]-\pi:\pi[/tex] split into M equal intervals.

    midpoints are [tex]y_K[/tex]
     
  5. Jun 18, 2012 #4
    Re: "Proof Beyond Scope of Course"

    This is a basic problem about geometric series and the roots of unity.

    Let [itex]Z_M[/itex] be an Mth root of unity. What you're considering is this:

    [tex]1 + Z_M + Z_M^2 + \ldots{} + Z_M^{M-1}[/tex]

    But, if you multiply through by [itex]z_M[/itex], you get

    [tex]Z_M + Z_M^2 + \ldots{} + Z_M^{M-1} + Z_M^M[/tex]

    But [itex]Z_M^M = 1[/itex], and you get the same sum all over again. No matter how many times you multiply by [itex]Z_M[/itex], the answer is the same. Since [itex]Z_M[/itex] is itself neither unity nor zero, the only possible solution is that the sum's value is zero.

    You can also confirm this by looking at the closed form result of the geometric series:

    [tex]1 + Z_M + Z_M^2 + \ldots{} + Z_M^{M-1} = \frac{1-Z_M^M}{1-Z_M}[/tex]

    Again, the numerator must be zero as [itex]Z_M[/itex] is an Mth root of unity.

    Edit: ah, I realize the problem is slightly more complicated than this. Well, this handles the case [itex]m=1[/itex] well, at any rate.
     
  6. Jun 18, 2012 #5

    gabbagabbahey

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    Re: "Proof Beyond Scope of Course"

    Do you mean that the [itex]x_j[/itex] in your first post are these midpoints [itex]y_k[/itex]? If so, then surely there must be given some restriction on [itex]M[/itex]? If we allow [itex]M=1[/itex] for example, then we have a single midpoint [itex]x_1 = 0[/itex] and [itex] \frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})= 1[/itex] regardless of what the value of [itex]m[/itex] is.

    Edit: nevermind, just realized that there is a congruence relationship, not an an m=0.
     
    Last edited: Jun 18, 2012
  7. Jun 18, 2012 #6
    Re: "Proof Beyond Scope of Course"

    So are you saying that the ##y_K=x_j##? In that case, try representing ##x_j## in terms of j.
     
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