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By splitting the Taylor series of an exponential $e^x$ into even and odd terms we obtain the hyperbolic trigonometric functions, $\cosh$ and $\sinh$, as discussed e.g. here.

What about splitting the series in different ways? In particular, we can think of the even-odd splitting as separating the indices in the different cosets of $\mathbb Z_2$. What if we do the same for other $\mathbb Z_n$?

For example, for $\mathbb Z_3$ we split into terms of the form $3n,3n+1, 3n+2$: $$e^x=\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!} + \sum_{n=0}^\infty \frac{x^{3n+1}}{(3n+1)!} + \sum_{n=0}^\infty \frac{x^{3n+2}}{(3n+2)!}.$$ Do the corresponding terms correspond to well-known functions? Are there geometrical interpretations similar to the ones for the hyperbolic functions?

To be clear, I'm asking about possible interpretations/formulae for the general $n$ case, the $\mathbb Z_3$ was just an example to clarify what I meant.

glS
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2 Answers2

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You're referring to a multisection of a series. In your case, where you split it up into three functions $E_0(x),E_1(x),E_2(x)$, you get the system: $$ \begin{cases} E_0(x)+E_1(x) + E_2(x) = e^x\\ E_0(0)=1;\qquad E_0'(0)=0\\ E_0'(x) = E_1(x);\qquad E_1'(x)= E_2(x);\qquad E_2'(x)=E_0(x) \end{cases} $$In this case, you get $E_0(x) = \frac{1}{3}\left(e^x+2 e^{-x/2}\cos\left(\frac{\sqrt{3}}{2} x\right)\right)$.

Integrand
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  • how did you solve the system and find $E_0(x)$ with this approach though? – glS Apr 20 '20 at 08:57
  • You can solve the homogeneous system $E_0 + E_0' +E_0'' =0$ using the characteristic polynomial. Then you can use variation of parameters to modify your solution to include the non-homogenity of $e^x$. TBH this isn't really my area of expertise, so someone else can probably give a better answer. – Integrand Apr 20 '20 at 18:59
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$$\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}=\frac{e^x}{3}+\frac{2}{3} e^{-x/2} \cos \left(\frac{\sqrt{3} }{2}x\right)$$ $$\sum_{n=0}^\infty \frac{x^{3n+1}}{(3n+1)!}=\frac{e^x}{3}-\frac{2}{3} e^{-x/2} \sin \left(\frac{\pi }{6}-\frac{\sqrt{3} }{2}x\right)$$ $$\sum_{n=0}^\infty \frac{x^{3n+2}}{(3n+2)!}=\frac{e^x}{3}-\frac{2}{3} e^{-x/2} \sin \left(\frac{\pi }{6}+\frac{\sqrt{3} }{2}x\right)$$

Claude Leibovici
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  • what about $\mathbb Z_n$? Can general formulae be derived? – glS Apr 19 '20 at 09:00
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    @glS. Be sure I tried ! I am still trying. Cheers – Claude Leibovici Apr 19 '20 at 09:16
  • So going from one of the answer in the linked question, I think I found some sort of solution: $$\sum_{k=0}^{n-1}\omega_n^{-kt} e^{\omega_n^k x}=\sum_{\ell\in[t]_n} \frac{x^\ell}{\ell!},$$ where $\omega_n\equiv e^{2\pi i/n}$ and $[t]_n$ is the set of positive integers of the form $4j+t$ for some $j$. I don't know if there is a nice way to write the sum of exponential in terms of real exponentials and trigonometric functions in general though – glS Apr 20 '20 at 08:55