Timeline for How to solve $f(f(x)) = \cos(x)$?
Current License: CC BY-SA 3.0
15 events
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| Oct 18, 2016 at 18:54 | comment | added | user78249 | @NoahSnyder I've found this same function independently written much differently. I have a proof around here somewhere, but it definitely works for $z$ in the immediate basin about the geometrically attracting fixed point on the real positive line. Every expression I've found once it hits the julia set just turns to bleh though--diverges as the terms $\phi(k,z)$ grow super exponentially with $k$. As does this one. | |
| Dec 24, 2014 at 11:36 | history | undeleted | Anixx | ||
| Dec 24, 2014 at 11:35 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 31, 2012 at 10:42 | history | deleted | Anixx | ||
| May 30, 2012 at 2:42 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 29, 2012 at 5:43 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 29, 2012 at 5:18 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 29, 2012 at 5:13 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 29, 2012 at 3:59 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 29, 2012 at 3:58 | comment | added | Noah Snyder | Are you claiming that your formula works for all complex numbers z? | |
| May 29, 2012 at 3:56 | comment | added | Anixx | The difference in sign in negative half-periods is due the fact that arccos is not the true inverse function but just only one branch. So the first iterate becomes $\cos(\arccos(x))$. One have to use multi-valued arccos to get the correct sign. | |
| May 29, 2012 at 3:53 | comment | added | Anixx | The function is complex. I added description to the plot. | |
| May 29, 2012 at 3:52 | history | edited | Anixx | CC BY-SA 3.0 |
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| May 29, 2012 at 3:37 | comment | added | Noah Snyder | I can't follow this answer... What's the meaning of the two colors in the plots? Is this function real or complex? And doesn't "follow quite well, at least at positive half periods" mean that it doesn't actually equal cos? | |
| May 29, 2012 at 3:09 | history | answered | Anixx | CC BY-SA 3.0 |