Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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**Βεν Γ. Κυθισ****Member**- Registered: 2018-10-09
- Posts: 21

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 theYou 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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**Βεν Γ. Κυθισ****Member**- Registered: 2018-10-09
- Posts: 21

zetafunc wrote:

Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to thefixed pointof 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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