Timeline for solving $f(f(x))=g(x)$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S Dec 7, 2019 at 20:37 | history | edited | ARG | CC BY-SA 4.0 |
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| S Dec 7, 2019 at 20:37 | history | suggested | Kenta Suzuki | CC BY-SA 4.0 |
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| Dec 7, 2019 at 20:23 | review | Suggested edits | |||
| S Dec 7, 2019 at 20:37 | |||||
| Aug 8, 2014 at 1:52 | comment | added | Michael Albanese | Some details about the counterexample in Sergei Ivanov's comment have been discussed in this MSE answer. | |
| Dec 1, 2010 at 3:49 | comment | added | Anixx | It is defined on the whole real line, it is just non-real at some arguments. From the comment by Kevin it is not evident that he asks about the solution f:R-->R | |
| Nov 6, 2010 at 0:22 | comment | added | Sergei Ivanov | Anixx, your function is defined only on an interval, not the entire real line. | |
| Nov 5, 2010 at 4:43 | comment | added | Anixx | Sergei, this function $f(x)=2 \cos \left(\sqrt{2} \arccos \left(\frac{x-1}{2}\right)\right)+1$ is a half-iterate of $g(z)=x^2-2x$ | |
| Mar 11, 2010 at 15:23 | comment | added | Kevin Buzzard | Yes indeed. OK, no further questions m'lud! | |
| Mar 11, 2010 at 7:56 | comment | added | Sergei Ivanov | Sure, for example let $g(x)=x^2-2x$. | |
| Mar 10, 2010 at 22:52 | comment | added | Kevin Buzzard | Probably the trick for C using g(z)=quadratic can be beefed up to give an example of a continuous g:R-->R for which no f (continuous or not) exists? I just noticed that we didn't yet deal with this case. | |
| Mar 9, 2010 at 18:40 | comment | added | Kevin Buzzard | I had just logged on to this site to edit my quetsion and add Q4: what about g:C-->C continuous but no restriction on f, and I see you've answered it before I even asked it! Talk about efficient ;-) Thanks for your most excellent answers Sergei. | |
| Mar 9, 2010 at 17:43 | comment | added | Sergei Ivanov | A counter-example for C: let g be a quadratic polynomial such that $g(z)-z$ has distinct roots, e.g. $g(z)=z^2$. Then there are four solutions of $g(g(z))=z$: two roots of g and two points a,b such that g(a)=b and g(b)=a. But if g(z)=f(f(z)), then the point z=f(a) must be another solution to g(g(z))=z. This works for non-continuous f as well. | |
| Mar 9, 2010 at 17:02 | comment | added | Kevin Buzzard | In fact, as you will well know, your answer to Q1 generalises to the situation where R is replaced by any set with at least 2 elements. | |
| Mar 9, 2010 at 17:01 | vote | accept | Kevin Buzzard | ||
| Mar 9, 2010 at 17:01 | comment | added | Kevin Buzzard | Very nice indeed. I suspected it would all be easy. So now we know that there's a chance that there's no function R-->R with f(f(x))=cos(x). Can you prove this? [this was the original question; see the link above] | |
| Mar 9, 2010 at 16:59 | comment | added | Gabriel Benamy | Okay, so that excludes the possibility of anywhere-decreasing functions having a functional square root in the reals. But what about complex functions? | |
| Mar 9, 2010 at 16:48 | history | answered | Sergei Ivanov | CC BY-SA 2.5 |