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$$\sum_{i=0}^n\cos(\omega_0\cdot2^{i/12})$$ When we put successively $i=0,4,3$ we have a major chord. But what is an analytic expression for this limited sum?

Chris Culter
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Franck
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    The sum given at the beginning of this question is $$ 1 + \cos(\omega_0\cdot 2^{1/12}) + \cos(\omega_0\cdot 2^{2/12}) + \cos(\omega_0\cdot 2^{3/12}) + \cos(\omega_0\cdot 2^{4/12}) + \cdots + \cos(\omega_0\cdot 2^{n/12}) $$ The sum $$\cos(\omega_0\cdot 2^{0/12}) + \cos(\omega_0\cdot 2^{3/12}) + \cos(\omega_0\cdot 2^{7/12}) $$ is not an instance of that. Might it be that instead of $\displaystyle \sum_{i=0}^n$ you meant $\displaystyle \sum_{i,\in, I}$ where $I$ is some specified set of integers? $\vphantom{\frac{\displaystyle\sum} {}} \qquad$ – Michael Hardy Nov 16 '17 at 20:57
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    In addition to Michael Hardy's comment, did you mean $i=0,4,7$ for a major chord? – Chris Culter Nov 16 '17 at 20:59
  • Usually, we do it the other way around! We start with a musical tone, then decompose it into a linear combination of trig functions, since that is much simpler to do calculations with. – GEdgar Nov 16 '17 at 21:17
  • I agree with your remarks due to the fact I met some problems to use the math editor. The first answer I hope, brings about the major chord formula , which in turn treats the simple problem of summing three trigonometric lines. – Franck Nov 17 '17 at 17:12

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