Timeline for How to solve $f(f(x)) = \cos(x)$?
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22 events
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| Jul 18, 2020 at 19:30 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
Fixed typo.
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| Aug 18, 2019 at 19:41 | comment | added | TLW | @Semola - for large factorials use Stirling's approximation. | |
| Jan 23, 2019 at 22:15 | comment | added | Semola | @Anixx Are you aware of any other similar formula, possibly with faster convergence? Computing large factorials is a bit of a problem numerically... | |
| Jun 9, 2014 at 16:35 | history | edited | Anixx | CC BY-SA 3.0 |
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| S Feb 19, 2014 at 17:57 | history | suggested | Cole Tobin | CC BY-SA 3.0 |
moving image to Stack Exchange Imgr
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| Feb 19, 2014 at 17:56 | review | Suggested edits | |||
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| S Feb 11, 2014 at 18:17 | history | suggested | Cole Tobin | CC BY-SA 3.0 |
Removed cruft
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| Feb 11, 2014 at 18:10 | review | Suggested edits | |||
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| Aug 10, 2011 at 12:48 | comment | added | Anixx | It is also possible to take $f^{[0]}(x)=\arcsin(\sin(x))$ to increase the rate of convergence, this is completely valid because sine has multi-branched inverse function, and as such, its zeroth iterate is also multi-branched (on the interval [0,π/2] this change will not affect the result). But this will lead to loss of differentiability of the partial sum in π/2+kπ. | |
| Jul 30, 2011 at 21:54 | comment | added | Anixx | This is like Taylor series. To get a function you do not necessary have to make the series around zero point, you can take any point if you know derivatives of higher orders there (and we can find the differences of higher order not only in 0 bus in 1 as well). | |
| Jul 30, 2011 at 21:26 | comment | added | Anixx | One can take the sum not from k=0 but from k=1. The limit will be the same, and the coefficient of x will be zero. But this will also converge very slowly. | |
| Jul 30, 2011 at 20:22 | comment | added | Will Jagy | A frustrating aspect of this construction is $f^{[0]}(x) = x.$ The infinite sum of all the coefficients of $x$ is 0. This must be true, $\sin x$ is periodic. The partial sum of a finite number of terms gives a nonzero coefficient for $x.$ This nonzero coefficient is necessary for uniform convergence on, say, the interval $(0,\pi - \varepsilon), \; \varepsilon > 0.$ On the other hand, by the time we have, say, $x > 2 \pi,$ the nonzero coefficient on $x$ itself (where all the other terms are periodic functions of $x$) makes matters worse, the partial summed function increases without bound. | |
| Jul 30, 2011 at 6:47 | comment | added | Anixx | Newton series is the following series: $$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$ It is equivalent to the above formula (just expand the deltas). Wikipedia: en.wikipedia.org/wiki/Finite_difference#Newton_series | |
| Jul 30, 2011 at 6:19 | comment | added | Ron Maimon | This is a truly great answer. But you use"Newton series" and "superfunction" in a way that will be misleading. By the "Newton series" you mean expansion of (1+x)^(1/2) in powers of x, not "Newton's iteration method", which is how some people read it. By "Superfunction" you mean an abstract iteration operator which when applied to functions composes them with other functions. This formula is a little hard to make precise. | |
| Jul 30, 2011 at 3:22 | history | edited | Anixx | CC BY-SA 3.0 |
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| Jul 27, 2011 at 18:34 | history | edited | Anixx | CC BY-SA 3.0 |
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| Jul 27, 2011 at 18:29 | history | edited | Anixx | CC BY-SA 3.0 |
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| Jul 25, 2011 at 11:23 | history | edited | Anixx | CC BY-SA 3.0 |
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| Jul 25, 2011 at 9:24 | history | edited | Anixx | CC BY-SA 3.0 |
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| Nov 3, 2010 at 21:59 | history | edited | Anixx | CC BY-SA 2.5 |
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| Nov 3, 2010 at 21:07 | history | edited | Anixx | CC BY-SA 2.5 |
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| Nov 3, 2010 at 20:58 | history | answered | Anixx | CC BY-SA 2.5 |