Timeline for How to solve $f(f(x)) = \cos(x)$?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2023 at 5:52 | history | edited | Bjorn Poonen | CC BY-SA 4.0 |
I believe that the construction was not correct on the grand orbit of the fixed point, so I proposed a correction.
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| Mar 5, 2022 at 15:15 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing
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| Apr 28, 2020 at 13:06 | comment | added | Lasse Rempe | Often the term "grand orbit" is used for the equivalence class you describe, to differentiate from the usual (forward) orbit, and the backward orbit. | |
| May 29, 2012 at 3:10 | comment | added | Anixx | See my new answer. | |
| Jul 30, 2011 at 22:49 | comment | added | Harry Gindi | I think that "orbit" is actually the correct term. Pairs $(X,g)$ of a set $X$ and endofunction $f$ are precisely the sets equipped with an $\mathbf{N}^+$-action for the additive monoid $\mathbf{N}^+$ of natural numbers. Then the partitions you noted are the orbits for the action. | |
| Apr 30, 2010 at 10:20 | comment | added | Homology | since $f$ is continuous, if it has no fixed point you have e.g. $f(x)>x$ for all $x$, and thus $g(x)>x$ also. | |
| Mar 9, 2010 at 21:09 | comment | added | Sergei Ivanov | @Ben: then there will be two fixed points of $f\circ f$. | |
| Mar 9, 2010 at 21:06 | comment | added | Ben Weiss | Im confused as to why x_0 must be a fixed point of f. Can't it be an involution (order two) point of f? | |
| Mar 9, 2010 at 21:04 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
fixed typos
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| Mar 9, 2010 at 20:41 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
elaborated
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| Mar 9, 2010 at 19:50 | comment | added | Sergei Ivanov | Actually I found that I overlooked the exceptional orbit containing 1, so I might be wrong. I'll update the answer once I sort it out. I meant to use the sufficient condition that you have an even number of (or infinitely many) orbits of every type, where "orbit" is an equivalence class of the equivalence relation generated by $x\sim \cos x$ and "type" of the orbit is that of a set equipped with a map to itself. | |
| Mar 9, 2010 at 19:20 | comment | added | Kevin Buzzard | Sergei: how can you see there exists a discontinuous solution? what formal properties of cos(x) do you need to come to this conclusion? | |
| Mar 9, 2010 at 18:55 | history | answered | Sergei Ivanov | CC BY-SA 2.5 |