Discretized complex exponential

[cepheid]

I'm reading that if you have a complex exponential exp(iω₀n) where n is in the set of integers, then unlike for the case of a continuous independent variable, the set of complex exponentials that is harmonically-related to this one is finite. I.e. there is only a finite number of distinct complex exponentials that have frequencies that are integer multiples of the fundamental frequency ω₀. This is because if the frequency differs from the fundamental by 2π: then:

exp[i(ω₀+2π)n] = exp(i2πn)exp(iω₀n) = exp(iω₀n).

Then the book I'm reading gives an example figure in which a progression of discrete cosines is plotted, and pairs of them with different frequencies look identical. There is one with frequency π/8 that looks exactly the same as another with frequency 15π/8. But I would have thought that the next identical one should have frequency 17π/8, since that differs from the original by 2π. At first I thought it was a typo in the book, but when I tried plotting them myself, I found that the sequence with frequency 15π/8 is indeed identical to the one with frequency π/8. I cannot understand why this is.

EDIT: I think it might be because for cosine, if:

ω₀ = π/8

then

15π/8 = π/8 + 16π/8 - 2π/8

= ω₀ + 2π - 2ω₀

= -ω₀ + 2π

= -(ω₀ - 2π)

and a cosine with a frequency given by the above is ALSO the same as cos(ω₀n). Is THAT the reason? If so, then I have figured it out, and you can delete this thread.

 

[SteamKing]

It's because of Euler's formula:

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[cepheid]

It's because of Euler's formula:

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I'm well aware of Euler's formula. My question was why is cos[(π/8)n] = cos[(15π/8)n] when the aforementioned property of discrete complex exponentials (and by extension cosines) suggests that frequencies that differ by two pi should produce identical sequences? But I think I figured it out (in the edited portion of my post). Thanks anyway.