Transcendental stalker on the intertubes. OR youtube's timeline is guided by operation OverLord

[Kharakov]

I'm thinking that e is transcendental because it contains every prime, by definition. Of course, since I said this before I watched the video, maybe I'm wrong. I'm always at least half right.

Pi is transcendental because it de-transcendentalizes e (e's function), which means it has every prime encoded in it, which means it is transcendental (it's really completely natural, but don't say that to a mathematician).

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This also means that the nested root \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+}}}} \, \dots\) has every prime encoded in it... except maybe 2 is the infinitesimalestiest (thus deepest) prime in this encoding. I conclude this because of the function to generate pi from the nested root, which multiplies 2^k (k being the number of nestings) times 2-nested root (which means 2 is the deep prime that you need to multiply tons of... or the prime that the pi needs the most....

\(\pi = \, \lim_{k\to\infty} \, 2^k \, \times \, \sqrt { 2 \, - \, \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ \,\dots }}}}}\)

 

[steve_bank]

Looks like infinite recursion. You would have to start at an infinite term and recursively work up to the top square root.

 

[Kharakov]

Well, the infinite recursion you mention is sort of a Julia of sorts. Looks sort of like the backwards iteration formula for the set.

I'm wondering if there is a way to apply this to the Mset (Mandelbrot)? I remember thinking about it before, but not sure if I looked into it. I'm thinking there are multiple roots, so one has to select the correct root path, etc.

One might be able to calculate the infinite depth, exact border (transcendental maybe) coordinates of the Mset if one had a way of reverse iteration that found the "right" root path.