Proving the Cosine Sum Identity for a Discrete Interval

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Homework Help Overview

The discussion revolves around proving the cosine sum identity for a discrete interval, specifically examining the expression \(\frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})\) and its behavior based on the congruence of \(m\) with respect to \(M\). The context involves splitting the interval \(-\pi\) to \(\pi\) into \(M\) equal parts and analyzing the midpoints.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the accuracy of the statement regarding the cosine sum identity and question the definition of \(x_j\). There are attempts to clarify the relationship between \(x_j\) and the midpoints \(y_K\), as well as considerations of the implications of different values of \(M\) on the identity.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided insights into the geometric series and roots of unity, while others are questioning the assumptions regarding the midpoints and the implications of specific values of \(M\). There is no explicit consensus yet.

Contextual Notes

Participants note the need for clarity on the definition of \(x_j\) and its relationship to the midpoints \(y_K\). There is also a mention of potential restrictions on \(M\) that could affect the validity of the identity being discussed.

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


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

midpoints are [tex]y_K[/tex]
 


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.
 


matlabber said:

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?

matlabber said:
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.
 
Last edited:


matlabber said:
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.
 

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