Timeline for Periodical functions with $f^{(n)} = f$, but $f^{(k)} \neq f$ for $k\in \{1,\ldots,n-1\}$ [closed]
Current License: CC BY-SA 4.0
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| Nov 25, 2020 at 18:33 | history | closed |
Emil Jeřábek abx Wojowu user44191 ARG |
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| Nov 25, 2020 at 15:09 | vote | accept | Dominic van der Zypen | ||
| Nov 25, 2020 at 9:16 | comment | added | Jack L. | @abx: it appears you are/would be right if the OP is asking for periodic functions; I had taken “periodical” to be derivative-wise—that is, the derivative is periodic with period $n$—rather than $f$ itself being a periodic function. My comment above and answer below had been in this regard; perhaps the OP could clarify this as a re-edit to the question, and I will remove my answer accordingly if it’s as you have understood it. | |
| Nov 25, 2020 at 8:59 | comment | added | abx | The only possibility are $n=4$. The functions $f:\mathbb{R}\rightarrow \mathbb{C}$ satisfying $f^{(n)}(t)=f(t)$ form a vector space with basis $(e^{\zeta t})$, where $\zeta $ runs through the $n$-th roots of 1. The function $e^{\zeta t})$ is periodic of period $2\pi i/\zeta $. Since you want the period to be real, you must have $\zeta =\pm i$, giving $n=4$, $f(t)=a\cos t+b\sin t$. | |
| Nov 25, 2020 at 8:55 | review | Close votes | |||
| Nov 25, 2020 at 18:33 | |||||
| Nov 25, 2020 at 8:38 | answer | added | Jack L. | timeline score: 3 | |
| Nov 25, 2020 at 8:17 | comment | added | Dominic van der Zypen | Oh - that's right, thanks @JackL.! Will remove the question - or do you want to post this as an answer with a few additional remarks? | |
| Nov 25, 2020 at 8:15 | comment | added | Jack L. | Isn’t this equivalent to solving for particular cases of the differential equation $y^{(n)}=y$, so that, generically, the functions you are looking for could be determined from the general solution $$f=\sum_{j=0}^{n-1}c_k\exp(\zeta^jx)\,,$$ where $\zeta=\exp(2\pi i/n)$ for coefficients $c_k$ that make $f$ real-valued? | |
| Nov 25, 2020 at 8:02 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |