Iterated Tangent Sequence: Calculating & Proving Solutions

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In summary, the conversation focuses on the chaotic nature of the sequence (\lfloor\tan_n1\rfloor) and the difficulties in efficiently calculating members of the sequence due to loss of precision and the complexity of the tangent function. There is also a question about the probability of a random member of the sequence being equal to a given number, which is estimated using an equation but may not accurately predict the chances for large numbers. Finally, there is a mention of specific algorithms and the use of hard coded Bernoulli numbers to calculate trigonometric functions.
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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 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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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.
 

FAQ: Iterated Tangent Sequence: Calculating & Proving Solutions

What is an iterated tangent sequence?

An iterated tangent sequence is a mathematical concept that involves repeatedly applying the tangent function to a given starting value. The resulting values form a sequence, and the goal is to find a solution or prove that a solution does not exist for a given equation.

How is an iterated tangent sequence used to calculate solutions?

The iterated tangent sequence can be used to calculate solutions by starting with a given value and repeatedly applying the tangent function. If the sequence converges to a single value, that value is considered a solution to the equation. If the sequence diverges, then there is no solution.

What is the importance of proving solutions in an iterated tangent sequence?

Proving solutions in an iterated tangent sequence is important because it provides mathematical certainty that a solution exists or does not exist for a given equation. This can help validate the accuracy of calculations and provide a more thorough understanding of the equation.

What are some common applications of iterated tangent sequences?

Iterated tangent sequences have various applications in mathematics, including finding roots of equations, solving optimization problems, and modeling complex systems. They are also used in physics, engineering, and other scientific fields to analyze and solve problems.

How can I use computer software to solve iterated tangent sequences?

There are various computer software programs available that can help solve iterated tangent sequences. These programs use algorithms and numerical methods to calculate solutions and can provide visual representations of the sequence. Some popular options include Mathematica, MATLAB, and Wolfram Alpha.

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