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Hello I was wondering if anyone knows how the following was derived:

$$\Phi_{{n}} \left( x \right) =\prod _{k=1}^{n} \left( {x}^{{\it \gcd} \left( k,n \right) }-1 \right) ^{\cos \left( {\frac {2\pi \,k}{n}} \right) }$$

I know that there are a lot of other related expressions like the Euler totient function, the gcd itself, the Mobius function, the modulo function that all have expressions that contain the real part of ${{\rm e}^{{\frac {2\pi \,ik}{n}}}}$ or ${{\rm e}^{{\frac {-2\pi \,ik}{n}}}}$ in a finite summation or product with another arithmetic function being a part of its summand formula, and I can see this is related to the roots of unity, but each of these formulas have unique characteristics and I have not been able to find what I can only assume to be a generalized method for finding them.

For the totient we have

$$\varphi \left( n \right) =\sum _{k=1}^{n}\gcd \left( k,n \right) { {\rm e}^{-{\frac {2\pi \,ik}{n}}}}. $$

And I am aware that this is a discrete Fourier transform, but as you can see from comparison of the two, this is not the generalized method that I am assuming to exist.

the_fox
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Adam Ledger
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The Wikipedia article on cyclotomic polynomials states:

The Möbius inversion formula allows the expression of $\;\Phi_n(x)\;$ as an explicit rational fraction: $\;\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)},\;$ where $\mu$ is the Möbius function. The Fourier transform of functions of the greatest common divisor together with the Möbius inversion formula gives: $\;\Phi_n(x) = \prod_{k=1}^n \left(x^{\gcd(k,n)}-1\right)^{\cos(2\pi k/n)}.$

The inversion formula means that the cyclotomic polynomials are uniquely determined by $\;x^n-1 = \prod_{d|n}\Phi_d(x).\;$ That is, if for all $n$ $\;x^n-1 = \prod_{d|n}c_d(x)\;$ for some sequence $\;\{c_n(x)\}\;$ of polynomials, then for all $n$ $\;c_n(x)=\Phi_n(x)\;$ follows. Now define $$\;c_d(x) := \prod_{k=1}^d \left(x^{\gcd(k,d)}-1\right)^{\cos(2\pi k/d)}.$$ We can partition the product as $$c_d(x) = \prod_{t|d} \left( \prod_{m=1\; \land\; t=\gcd(m,d)}^d (x^t-1)^{\cos(2\pi mt/d)} \right) = \prod_{t|d} (x^t-1)^{a(d,t)} = \prod_{t|d} (x^t-1)^{\mu(d/t)} $$ where $\;a(n,t):=\sum_{m=1\; \land\; t=\gcd(m,n)}^n\cos(\frac{2\pi m}n) = \mu(n/t)$ if $n$ is divisible by $t$ and $0$ otherwise.

Now $\;x^n-1 = \prod_{d|n}c_d(x)\;$ by a property of the Möbius function and we are done.

Somos
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  • I don't quite understand how we can then define $c_d({x})$ as you describe from the conclusion drawn from the inversion formula – Adam Ledger May 02 '18 at 09:45
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    @user548331 I can understand why you might think that. It looks like we are starting with what we want to prove instead of deriving it from a simple starting point using Fourier transform. Mathematics doesn't always work that. way. – Somos May 02 '18 at 10:41
  • Really what I have asked is to show how Mobius inversion applies to any abelian group, hence allowing the derivation of formula from the gcd such as we have for the Euler totient as a summation, and the $n^{th}$ cyclotomic polynomial as a product formula. – Adam Ledger May 02 '18 at 15:39
  • "The generalized method" I claimed to have not been able to find at the time being Mobius inversion, I just wasn't reading carefully enough as per usual – Adam Ledger May 02 '18 at 15:41
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    @user548331 Ok, I didn't quite grasp what you were looking for. It is just elementary linear algebra. Summing over divisors is a linear map and using Mobius inversion is just the inverse map. Any abelain group is just a $Z$-module. – Somos May 02 '18 at 16:02