[seqfan] Re: tan(n). Was: Into subtleties of musical information
veikko at nordem.fi
Wed May 27 09:08:56 CEST 2015
My point is that instead of random search one could simply accept my conjecture for the moment and focus to proving it (probably indirectly), as a separate project.
As for the potential proof, it should be remembered that also the number of levels of nesting is without limit. A quick intuitive playing with the potential of finding a case where nesting would not lead to 0 or 1 suggests a fictive loop where floor(tan(n)) returns to its value at some earlier level of nesting.
As an example, between 0…10^5 to reach the limiting value (0) the deepest level required is 8 for n=18116.
The approach is straight-forward: 18116, -168, -14, -8, 6, -1, -2, 2, -3, 0
israel at math.ubc.ca kirjoitti 26.5.2015 kello 23.48:
> Yes, it's known that tan(n) is irrational, in fact it's transcendental for all algebraic n, by Lindemann's theorem.
> On May 26 2015, Antti Karttunen wrote:
>> But it seems that at least up to 1108341089274117551 = A249836(13)
>> there is none at the positive side. I assume that tan(n) is irrational
>> for all integers > 0, thus for function floor(tan(n)) to have a fixed
>> point k, we have to have tan(k) > k, right?
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