Iterated tangent

  • #1
CRGreathouse
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Let [itex]\tan_1x=\tan x[/itex] and [itex]\tan_{k+1}x=\tan\tan_kx[/itex].

It's fairly clear that the sequence [itex](\lfloor\tan_n1\rfloor)[/itex] = http://www.research.att.com/~njas/sequences/A000319 [Broken] is chaotic, in the sense that it can diverge from [itex](\lfloor\tan_n(1+\varepsilon)\rfloor)[/itex] even for small [itex]\varepsilon[/itex].

1. Any thoughts on how to calculate members of this sequence efficiently? The loss of precision at each step makes this very difficult to calculate; also, the tangent is difficult to compute compared to more elementary functions. For comparison's sake, 4000 decimal digits of precision are needed to calculate a(2,000,000), which takes perhaps a week in naive implementation on Pari.
2. Does any proof idea spring to mind regarding the idea that for all integers k, a(n) = k for some n? This seems natural in light of the chaotic nature of the problem... but I have nothing in mind.
3. Can anyone thing of a good explanation for the 'probability' that a random member of the sequence will be equal to a given number k? Assuming some kind of equidistribution I thought it was reasonable to consider
[tex]\frac{\tan^{-1}(k+1)-\tan^{-1}(k)}{\pi}[/tex]
but this seems to vastly overestimate the chances for large k. To wit, it would predict that about 91 out of every million randomly selected elements would be in the range 190-200, but in the first 3 million or so elements I haven't found one.
 
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Answers and Replies

  • #2
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There are specific algorithms to calculate trigonometric functions used in calculators. But even the Taylor series doesn't look so bad especially if you use hard coded Bernoulli numbers.
 

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