In an attempt to generalize $\sin(x)$ and $\cos(x)$—but just a curiosity—I found some functions with fairly interesting properties. The idea was to extent the definition of $\sin$ and $\cos$ as unique solutions to $y''+y=0$ (with $y(0)=0$ and $y'(0)=1$ in the former and $y(0)=1$ and $y'(0)=0$ in the latter case), as well as their relationship to each other through differentiation.
Define the "first" $n$'th root of unity as $\xi _{n}=\exp(2 \pi i / n)$. I define $X_{m}^{n}$ as $$X_{m}^{n}(x)=\frac{\xi_{2n}^{m}}{n}\sum_{i=0}^{n-1} \xi_{n}^{im} \exp\left ( \xi_{n}^{i}\xi_{2n}x \right )$$ I know the $n$'th roots can be simplified, but I find it easier to conceptualize in this way.
For example, with $n=2$ we get: $$X_{0}^{2}(x)=\frac{1}{2}\left ( e^{ix} + e^{-ix} \right )=\cos(x)$$ $$X_{1}^{2}(x)=\frac{i}{2}\left ( e^{ix} - e^{-ix} \right )=-\sin(x)$$
Properties
Note that $\frac{d}{dx}X_{m}^{n}(x)=X_{m+1}^{n}(x)$ for $m$ in $\left \{ 0, 1...n-1 \right \}$, and $\frac{d}{dx}X_{n-1}^{n}(x)=-X_{0}^{n}(x)$. If you continue differentiating you'll enter an "endless cycle".
You'll also find that $X_m^{n}$ is the unique solution to $y^{ \left ( n \right )}+y=0$ with $y^{\left ( i \right )}(0)=\begin{cases} 1 & \text{ if } i= m \text{ mod } 2n\\ -1 & \text{ if } i= m+n \text{ mod } 2n\\ 0 & \text{ otherwise } \end{cases}$
Question
The last thing I know about these functions is that they admit an invariant in the same way that $cos(x)^{2}+sin(x)^{2}=1$, and remarkably in a simple way. I'm looking for help in finding a simple expression for this invariant. I have found a method for generating this invariant but I find the algebra much too overwhelming.
Below I show how to find the invariant for $n=3$. You should probably write it out or it'll just look like gibberish. We have:
$$X_{0}^{3}(x)=\frac{1}{3} \left ( e^{\xi_6 x} + e^{\xi_6 \xi_3 x} + e^{\xi_6 \xi_3^{2} x} \right )$$ $$X_{1}^{3}(x)=\frac{\xi_6}{3} \left ( e^{\xi_6 x} + \xi_3 e^{\xi_6 \xi_3 x} + \xi_3^{2} e^{\xi_6 \xi_3^{2} x} \right )$$ $$X_{2}^{3}(x)=\frac{\xi_6^{2}}{3} \left ( e^{\xi_6 x} + \xi_3^{2} e^{\xi_6 \xi_3 x} + \xi_3 e^{\xi_6 \xi_3^{2} x} \right )$$
Next note $$e^{\xi_{6} x} = X_{0}^{3}(x) + \xi_{6}^{5} X_{1}^{3}(x) + \xi_{6}^{4} X_{2}^{3}(x)$$ Substituting $x\rightarrow \xi_{3}x$ and $x\rightarrow \xi_{3}^{2}x$ in the above and multiplying the resulting expressions, we find:
$$e^{\xi_{6} \left ( 1 + \xi_{3} + \xi_{3}^{2} \right ) x} = e^{\xi_{6} 0} = 1 = \left ( X_{0}^{3}(x) + \xi_{6}^{5} X_{1}^{3}(x) + \xi_{6}^{4} X_{2}^{3}(x) \right )\left ( X_{0}^{3}(\xi_{3} x) + \xi_{6}^{5} X_{1}^{3}(\xi_{3} x) + \xi_{6}^{4} X_{2}^{3}(\xi_{3} x) \right )\left ( X_{0}^{3}(\xi_{3}^{2} x) + \xi_{6}^{5} X_{1}^{3}(\xi_{3}^{2} x) + \xi_{6}^{4} X_{2}^{3}(\xi_{3}^{2} x) \right )$$
This may not look helpful at all, but you can check that $X_{m}^{3}(\xi_{3}^{i} x)=X_{m}^{3} \xi_{3}^{-im} (x)$. Again, try it with a few cases. We then get
$$1=\left ( X_{0}^{3}(x) + \xi_{6}^{5} X_{1}^{3}(x) + \xi_{6}^{4} X_{2}^{3}(x) \right ) \left ( X_{0}^{3}(x) + \xi_{6}^{3} X_{1}^{3}(x) + X_{2}^{3}(x) \right ) \left ( X_{0}^{3}(x) + \xi_{6} X_{1}^{3}(x) + \xi_{6}^{2} X_{2}^{3}(x) \right )$$
You can also tell at this point that it is constant by complex conjugating on the RHS, seeing that we have a real holomorphic function which is therefore constant. This we can multiply out (by hand) to get $$(X_{0}^{3})^{3}-(X_{1}^{3})^{3}+(X_{2}^{3})^{3}+3X_{0}^{3}X_{1}^{3}X_{2}^{3}=1$$
I graphed this with Python and it seems to hold true very well. If you decide to plot these functions yourself, because these functions taken individually blow up in an exponential way, you'll find that beyond a certain point this expression seems to blow up due to floating point (im)precision.
I find it absolutely amazing that such a simple expression exists when we know that multiplying (complicated expressions (three times!)) tends to complicate more than simplify.
The question is, in the case of $n=4$ and beyond, what is the invariant? For every $n$ I expect there to be a real $n$-th degree polynomial invariant expression in these functions. What are the coefficients? I would also like to know if there's more to these functions than I've written so far, or if there is another way to look at them?
I have found the general formula for the product (the one I used with $n=3$), in which sense I am really asking how you would simplify
$$1=\prod_{i=0}^{n} \sum_{j=0}^{n} \xi_{2n}^{-j\left ( 2i + 1 \right )}X_{j}^{n}$$
(For this I don't think you'll need any special properties of $X_{j}^{n}$)
EDIT
I found that if we name $(X_{0}^{4}(x),...,X_{3}^{4}(x)) = (a,b,c,d)$ then $$1 = a^{4}+b^{4}+c^{4}+d^{4} +2a^{2}c^{2}+2b^{2}d^{2}+4a^{2}bd+4acd^{2}-4bc^{2}d-4ab^{2}C$$
For $n=5$...
$$1 = a^{5} + 5 a^{3} b e + 5 a^{3} c d - 5 a^{2} b^{2} d - 5 a^{2} b c^{2} + 5 a^{2} c e^{2} + 5 a^{2} d^{2} e + 5 a b^{3} c + 5 a b^{2} e^{2} - 5 a b c d e - 5 a b d^{3} - 5 a c^{3} e + 5 a c^{2} d^{2} + 5 a d e^{3} - b^{5} - 5 b^{3} d e + 5 b^{2} c^{2} e + 5 b^{2} c d^{2} - 5 b c^{3} d + 5 b c e^{3} - 5 b d^{2} e^{2} + c^{5} - 5 c^{2} d e^{2} + 5 c d^{3} e - d^{5} + e^{5}$$
At this point I think it's fair to say that maybe I'm looking in the wrong place.