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Trig functions satisfy $\cos^2t+\sin^2t=1$, which is an expression of the Pythagorean theorem. Hyperbolic trig functions satisfy $\cosh^2t-\sinh^2t=1$ which may perhaps be viewed as a generalization. The trig functions parametrize the circle $x^2+y^2=1$ and the hyperbolic trig functions parametrize the "generalized circle" $x^2-y^2=1.$

The generalized hyperbolic functions $F^\alpha_{n,r}(t)$ are defined by a power seres

$$ F^\alpha_{n,r}(t)=\sum_{k=0}^\infty\frac{\alpha^k}{(nk+r)!}t^{nk+r}. $$ These functions satisfy the differential equation

$$\left(\frac{d}{dt}\right)^nF^\alpha_{n,r}(t)=\alpha F^\alpha_{n,r}(t)$$

and

$$ \frac{d}{dt}F^\alpha_{n,r}(t)=\begin{cases} F^\alpha_{n,r}(t) & 0<r\leq n-1\\\ \alpha F^\alpha_{n,n-1}(t) & r=0. \end{cases} $$

So they are eigenfunctions of $n$th order differentiation; generalizations of trig and hyperbolic trig functions, and an alternate basis for the solution space to $e^t,e^{\omega_n t},e^{\omega_n^2 t},\dotsc,e^{\omega_n^{n-1} t},$ for $\omega_n$ a primitive $n$th root of unity. For $n=2,\alpha=1$ we recover the classical hyperbolic trig functions, and for $n=2,\alpha=i$ we get the classical trig functions. Or alternatively for $n=4,\alpha=1,$ we get a linear combination of trig and hyperbolic trig.

These functions share many properties with trig and hyperbolic trig functions, such as a generalized Euler identity, and a generalized "angle sum" formula, as seen in this paper by Jens Schwaiger.

Question: Is there also a polynomial relation among the generalized hyperbolic functions, which generalizes the pythagorean relation satisfied by the trig and hyperbolic trig functions? Do these functions parametrize a conic section or a higher degree curve?

I believe since these functions are, up to a change of basis, equivalent to the exponentials $e^t,e^{\omega_n t},e^{\omega_n^2 t},\dotsc,e^{\omega_n^{n-1} t},$ for $\omega_n$ a primitive $n$th root of unity, therefore I believe it is equivalent to asking for expressing for $e^{\omega_n^k t}$ as a polynomial in $e^t$. Somehow a cyclotomic version of a Chebyshev polynomial. Or if not as straightforward as $e^{\omega_n^k t}$ as a function of $e^t$, then at least a polynomial relating all the $e^{\omega_n^k t}$.

Edit: Since the sum of the roots of unity is zero, we have $$e^t\cdot e^{\omega_n t}\cdot e^{\omega^2_n t}\cdot\dotsb e^{\omega^{n-1}_n t}=1.$$ Then using the generalized Euler identity $$e^t=\sum_{r=0}^{n-1}F_{n,r}^1(t)$$ and the easy fact that $F_{n,r}^1(\omega t)=\omega^r F_{n,r}^1(t),$ we have

$$\left(F_{n,0}^1(t)+F_{n,1}^1(t)+\dotsb+F_{n,n-1}^1(t)\right)\cdot\\\left(F_{n,0}^1(t)+\omega F_{n,1}^1(t)+\dotsb+\omega^{n-1}F_{n,n-1}^1(t)\right)\cdot\dotsb\cdot\left(F_{n,0}^1(t)+\omega^{n-1}F_{n,1}^1(t)+\dotsb+(\omega^{n-1})^{n-1}F_{n,n-1}^1(t)\right)=1.$$

So that is a polynomial relation among the functions. The answer to my first question is "yes". In the $n=2$ and $n=4$ cases it does reproduce the Pythagorean identity. In the $n=3$ case we get the equation

$$x^3+y^3+z^3-3xyz=1.$$

So that is my $n=3$ candidate for what we might call a generalized Pythagorean theorem.

Note that this equation defines a hypersurface. The curve parametrized by the generalized hyperbolic functions lies on this hypersurface. To answer the second question (what curve do these functions parametrize?), presumably there are more polynomial relations, $n-1$ of them, to define a curve. I tried to use Vieta's formulas for the roots of unity to find some more polynomial relations, but they all seemed to give the same result as above. How do I find the $n-1$ polynomials that $\left(F_{n,0}^1(t),F_{n,1}^1(t),\dotsc,F_{n,n-1}^1(t)\right)$ is a parametrization of the intersection of?

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  • $\begingroup$ Can you write $e^{ix}$ as a polynomial in $e^x$? $\endgroup$ – Gerry Myerson Feb 9 '18 at 2:05
  • $\begingroup$ @GerryMyerson Hmmm obviously not. Right, thanks. I guess the Chebyshev analogy was not the right direction. $\endgroup$ – ziggurism Feb 9 '18 at 3:07

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