Proving the Cosine Sum Identity for a Discrete Interval

[matlabber]

"Proof Beyond Scope of Course"

Homework Statement

[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}

Homework Equations

The Attempt at a Solution

is statement truthfully accurate? how to show this?

 

[HallsofIvy]

Your formula makes no sense without a statement of what the "[itex]x_j[/itex]" are.

 

[matlabber]

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

midpoints are [tex]y_K[/tex]

 

[Muphrid]

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.

 

[gabbagabbahey]

Homework Statement

[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}is statement truthfully accurate? how to show this?

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

midpoints are [tex]y_K[/tex]

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.

 

[dimension10]

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

midpoints are [tex]y_K[/tex]

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