Timeline for How to solve $f(f(x)) = \cos(x)$?
Current License: CC BY-SA 4.0
8 events
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| Mar 5, 2022 at 15:24 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Math Jaxed
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| Jun 25, 2017 at 4:46 | comment | added | Joel David Hamkins | The solution in my answer only works on an interval, and this is an important difference. | |
| Jun 25, 2017 at 4:34 | comment | added | Ovi | Sorry if I'm misunderstanding this: You said "When g is continuous, then this function f will be continuous also", but the answer with the most upvotes claims that there is no continous solution. | |
| Aug 24, 2011 at 21:50 | comment | added | Joel David Hamkins | Anixx, I could give you a plot, but I think you can get the idea without one. (Or you could make a plot.) The function $f$ is the straight line $y=x+z$ on the interval $[a,b]$, with $z$ constant. This maps the interval $[a,b]$ to $[a+z,b+z]$. On this interval, the function $f$ looks exactly like $g$ does on $[a,b]$, but translated by $z$. Thus, the function $f$ applied once moves you from $[a,b]$ to $[a+z,b+z]$, and applied again, gives you the result of $g$. So $f\circ f=g$ on the interval $[a,b]$, as desired. | |
| Aug 23, 2011 at 23:44 | comment | added | Anixx | Can you please show us a plot of a solution for cosine on interval say $[-\pi/2,\pi/2]$? | |
| Mar 9, 2010 at 15:43 | comment | added | Kevin Buzzard | I've just asked an explicit question about this: mathoverflow.net/questions/17614/solving-ffxgx | |
| Mar 9, 2010 at 15:33 | comment | added | Kevin Buzzard | OK but I don't see how to generalise this trick so that it works for all x in the reals. | |
| Mar 9, 2010 at 15:10 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |