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So if you have f_0 = cos(n) and then f_1 = cos(f_0) and so on to get f_x = cos(f_(x-1)), and then let n be any real number, f_x converges to approximately
If you know anything that might relate to this new irrational number please respond to this post!Edit:
I found out what this is, so don't clog the post with what you think it means, its meaning has already been revealed.
Last edited by Βεν Γ. Κυθισ (2018-10-10 10:11:05)
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1+1=|e^(π×i)-1|
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Hi Βεν,
Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to the fixed point of the cosine function, i.e. the solution to . (I think the two different answers come from using degrees versus radians rather than real vs imaginary.) There are ways of calculating this in terms of the Lambert W function or some nice infinite sums of Bessel functions I think.You can prove that a solution to the above equation exists via Brouwer's fixed point theorem (and probably the contraction mapping theorem too).
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Hi Βεν,
Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to the fixed point of the cosine function,
That makes so much sense! I also just tried
and it converged to its fixed point, 0. Tan has a fixed point but it is not attractive; sinh becomes a vertical line on 0 which means there is no attractive point, 0; cosh has no fixed point but it converges to ∞; 0 is tanh's attractive fixed point; cot has no attractive point due to it being tan with the x axis shifted; sec is related to tan because it shares half of its discontinuous points with tan suggesting it has no attractive point, and it does have no attractive point; csc is sec with the x axis shifted so it has no fixed point.Last edited by Βεν Γ. Κυθισ (2018-10-10 11:41:26)
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1+1=|e^(π×i)-1|
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