# Higher Order Trigonometric Function

Once in a time, I had to work with functions that have the following Taylor series expansion: $$t_m(x)=1-\frac{x^m}{m!}+\frac{x^{2m}}{(2m)!}+\cdots =\sum_{k=0}^\infty \frac{(-1)^k x^{km}}{(km)!}.$$ Plugging in $m=2$ this is obviously the Taylor series expansion for $\cos(x)$. Now I found the following nice formula to get a closed formula for this functions, which is (proven here): $$t_m(x)=\frac{1}{m}\sum_{k=0}^{m-1} \exp( e^{i\frac{2k+1}{m}\pi}x )$$ and again it is obvious that for $m=2$ I'll get $\displaystyle\frac{e^{i\pi x}+e^{-i\pi x}}{2}=\cos(x)$. It is also easy to see that $\displaystyle\frac{d(t_m)^m}{dx^m}=-t_m$.

And now I have 2 questions:

1. Do these functions have a name and any application? One possible use would be in solving $m$th order differential equation over $\mathbb{R}$.

2. When I ask Wolfram for the roots, if $m=4$, I get $x_n=\frac{2\pi n + \pi}{\sqrt{2}}$ (and also $i\cdot x_n$). Asking for other $m\neq2,4$, I (so far) just get numerical values. Are there closed formulas for the roots in all case of $m$. Do they have a geometric interpretation?

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What you have at hand is known as Mittag-Leffer function $t_m(x) = E_{m}(-x^m)$.
To the best of my knowledge roots are hard to come by in closed form. For instance, for $m=3$, $$t_3(x) = \frac{1}{3} \mathrm{e}^{-x} + \frac{2}{3} \mathrm{e}^{x/2} \cos\left(\frac{\sqrt{3}}{2} x\right)$$ Its roots are solutions of $$\exp\left(\frac{3}{2} x\right) \cos\left(\frac{\sqrt{3}}{2} x\right) = -\frac{1}{2}$$ which is a transcendental equation. It has infinitely many real solutions, all of which are positive: $$x_{n, n \geqslant 0} = \frac{\pi}{\sqrt{3}} \left(2n+1\right) + \frac{(-1)^n}{\sqrt{3}} \exp\left(-\frac{\sqrt{3}}{2} \pi (2n+1) \right) + \mathcal{o}\left(\mathrm{e}^{-\pi \sqrt{3} n}\right)$$