Timeline for Are there any functions $f$ beyond trivial examples where $\int f(x +f(x + f(x +\dotsb)))\,dx$ $= F(x+F(x+F(x +\dotsb))) + C$ for some function $F$?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Jan 9 at 0:02 | vote | accept | gdoug | ||
| Dec 21, 2023 at 16:08 | comment | added | Zeta | Then you should consider to accept the corresponding answer. | |
| Dec 18, 2023 at 23:04 | comment | added | gdoug | Yes! My original intent was to find any such construction, and I think the way this has progressed satisfies that. | |
| Dec 16, 2023 at 17:23 | answer | added | Pietro Majer | timeline score: 2 | |
| Dec 16, 2023 at 14:36 | comment | added | Steven Landsburg | @FredHucht has turned this into a quite interesting question. It would be nice if you revealed whether that's the question you intended. | |
| Dec 16, 2023 at 10:54 | history | edited | gmvh |
Added top-level tags
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| Dec 15, 2023 at 13:20 | history | edited | Max Lonysa Muller |
Added a relevant tag
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| Dec 15, 2023 at 11:12 | answer | added | Fred Hucht | timeline score: 10 | |
| Dec 14, 2023 at 23:19 | comment | added | Fred Hucht | I think one can construct higher order solutions using a power series ansatz for $\phi(x)$ and $f(x)$. I found a linear and a quadratic solution in $\phi$, but I have to go to bed now :-). | |
| Dec 14, 2023 at 22:30 | comment | added | Fred Hucht | One possible not so trivial solution is $\phi(x)=f(x)=c$, $\psi(x)=c x$, $F(x)=c x/(1+c)$. | |
| Dec 14, 2023 at 21:36 | comment | added | Sidharth Ghoshal | Namely instead of picking any $\psi$ you must find a $\psi$ s.t. $\psi’(T^{-1}) (T^{-1})’ = \psi’(Q^{-1})$ which is a highly non linear equation since $T,Q$ depend on $\psi$ too, still unless something crazy is happening, this probably has some non trivial solutions. And once we find one such $\psi$ everything else falls into place | |
| Dec 14, 2023 at 21:32 | comment | added | Sidharth Ghoshal | If we assume we can take function inverses easily. Then if $T = x + \psi(x)$ then $\psi(T^{-1}) = F$ and similarly if $Q = x + \psi’(x)$ then $\psi’(Q^{-1}) = f$. So given a choice of $\psi$ you can formally speaking find the $F,f,\phi$ using this procedure. The situation gets a lot harder when adding the $F’ = f$ constraint | |
| Dec 14, 2023 at 21:22 | comment | added | Sidharth Ghoshal | With @LSpice’s nice idea we basically want to find an $f,F,\phi,\psi$ such that we have some subset of $S \subseteq \mathbb{C}$ where for every $x \in S$ $f(x+\phi(x)) = \phi(x)$ and $F(x+\psi(x)) = \psi(x)$ and $(\psi)’ = \phi$ seeing that it’s 3 equations and 4 unknown functions this should probably be possible. Adding the condition $F’ = f$ gives us 4 equations and 4 unknowns so perhaps with some luck that’s also tractable. | |
| Dec 14, 2023 at 21:11 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 2 characters in body; edited title
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| Dec 14, 2023 at 1:55 | review | Close votes | |||
| Dec 22, 2023 at 3:05 | |||||
| Dec 14, 2023 at 0:14 | comment | added | LSpice | @No-one, re, presumably a solution $\phi$ of the functional equation $\phi(x) = f(x + \phi(x))$? (But I don't know to what extent such a solution exists, or is unique.) | |
| Dec 14, 2023 at 0:13 | history | edited | LSpice | CC BY-SA 4.0 |
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| Dec 14, 2023 at 0:10 | comment | added | No-one | What do you mean by that infinite composition of functions? | |
| S Dec 13, 2023 at 23:50 | review | First questions | |||
| Dec 14, 2023 at 0:41 | |||||
| S Dec 13, 2023 at 23:50 | history | asked | gdoug | CC BY-SA 4.0 |