Can roots of unity save the day?

Geometry Level 4

k = 1 2 cos 8 ( k π 5 ) , k = 1 3 cos 8 ( k π 7 ) , k = 1 4 cos 8 ( k π 9 ) , k = 1 5 cos 8 ( k π 11 ) , k = 1 6 cos 8 ( k π 13 ) , \sum_{k=1}^2 \cos^8\left(\frac{k \pi}{5}\right), \sum_{k=1}^3 \cos^8\left(\frac{k \pi}{7}\right), \sum_{k=1}^4 \cos^8\left(\frac{k \pi}{9}\right), \sum_{k=1}^5 \cos^8\left(\frac{k \pi}{11}\right), \sum_{k=1}^6 \cos^8\left(\frac{k \pi}{13}\right), \ldots

Above shows a sequence of numbers, what type of sequence of progression do they follow?

Bonus : Find its common difference and/or common ratio.

Arithmetic Progression Geometric Progression Harmonic Progression Arithmetic Geometric Progression

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1 solution

Kushal Dey
Mar 26, 2020

Well actually to be quite honest, I am pretty much confused myself. However, I will share my way of solving this problem. We can express cos^8(x) as, a1+a2cos(2x)+a3cos(4x)+a4cos(6x)+ a5cos(8x). The values of a(i)'s don't matter because they are constants anyways. Now, let S(n) denote the sum of cos^8(2r*pi/n), where r denotes the integers between 1 and (n-1)/2, and n is an odd integer. Therefore, S(n) equals the sum of multiple angles of 2pi/n, all of which sums up to be constant and independent of n.( If you are confused about the last statement, you may google up "n-th roots of unity".) This means that S(n) is a constant. Hence we can say it is an AP with common difference of 0. But it also can be a GP with common ratio of 1. I may be wrong in my last statement, but S(n) is definitely constant series.

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