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Continuously Iterating Modular Arithmetic
#11
My idea would be to use Fourier series to approximate [Image: png.image?\dpi%7B110%7D%20mod(k,x)], and then somehow find iterations of the Fourier approximations, such that near the fixed points and n-cycles of the Fourier approximations (If any.) they would approach continuously iterating mx+b, and are as holomorphic as possible.
But, I do not know how to do that.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
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Sincerely: Catullus
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#12
Idk man, if your having such a strong reaction maybe I don't remember correctly.
But since you're having such a strong reaction maybe it would be interesting to go back to my 2020 notes where I was playing with this idea.

I vaguely remember that it was not so hard to extend \(F_a(n)=a\,{\rm mod}\, n\) from \(n\) integer to \(x\) real. And \(F_a(x)\) had to do with dividing stuff modulo the integers, so in the group of complex roots of units, or something like that. Like the graph of the thing was just a normal hyperbola of the form \(k/x\) but displayed on a quotient of the plane by a 2-lattice, i.e. a torus. In other words it was just division but topologically wrapped up.

Maybe the Fields' medal-part of it was that integer points of the hyperbole were important and maybe there is something hard in finding rational/integer points of algebraic curves... it may be that is critical point.

But I don't remember... need to get back to it.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#13
(07/19/2022, 10:03 AM)MphLee Wrote: Idk man, if your having such a strong reaction maybe I don't remember correctly.
But since you're having such a strong reaction maybe it would be interesting to go back to my 2020 notes where I was playing with this idea.

I vaguely remember that it was not so hard to extend \(F_a(n)=a\,{\rm mod}\, n\) from \(n\) integer to \(x\) real. And \(F_a(x)\) had to do with dividing stuff modulo the integers, so in the group of complex roots of units, or something like that. Like the graph of the thing was just a normal hyperbola of the form \(k/x\) but displayed on a quotient of the plane by a 2-lattice, i.e. a torus. In other words it was just division but topologically wrapped up.

Oh yes, I apologize. No, what you are saying here is perfectly possible.

Quote:Maybe the Fields' medal-part of it was that integer points of the hyperbole were important and maybe there is something hard in finding rational/integer points of algebraic curves... it may be that is critical point.

But I don't remember... need to get back to it.


It's the finding the rational/integer solutions that are ridiculously hard. Number theory is all about asking the simplest questions and having 100 pages to almost prove it but only for specific cases, lol.

Ya, solving these iterates in the natural/integer case/describing the equivalence classes. No, that shit is so hard.
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#14
I see but when talking about iterating it, I understood it was just matter of first extending \(F_a(x)\) from \(\mathbb Z\) to the reals, and then looking for \(F_a^t\).
I was just saying that I was able to do something in that direction, just this.

Anyways, at this point I'd like to know, in your opinion, how exactly having a fully continuous iteration of \(F_a\) or even analytic or holomorphic, assume we get one, can help with hard number theory problems.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#15
Well it wouldn't, I had misunderstood you is all.

I thought you were talking about iterating modulo operators in the integers, that's the really hard part. That's the only part that would be crazy. Also, I'd doubt it'd help with many hard number theory problems--it would just be a very hard number theory problem, lol. I think our wires just got crossed is all.

But solving things like:

$$
p \mod x \mod y \mod x \mod y = n\\
$$

For large numbers and large primes would be crazy hard. Interesting. Not sure it would prove anything significant. It'd just be really hard, lol. When I said they'd probably give you a fields medal, it's just a joke that one of my profs would always say. If a problem was just unreasonably hard, it was always--they'd probably give you a fields medal if you solved this, lol.
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#16
haha We need a Fields' medal for solving non-integer ranks... but I doubt they will give us one if we  also can't be convincing that goodstein maps are relevant in mathematics... and not just a weird gadget...
Or maybe a new Einstein in 2103 will use it as a basis for the Theory of everything.

Btw... I'm trained in graphic design... and I have a crush for math typesetting... so I feel bad looking at that thing without parentheses and in italic...
pls... if you care for my health, fix it xD

PS: is that meant to be
$$p \mod (x \mod (y \mod (x \mod y))) = n\\$$?

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#17
(07/22/2022, 01:58 AM)MphLee Wrote: haha We need a Fields' medal for solving non-integer ranks... but I doubt they will give us one if we  also can't be convincing that goodstein maps are relevant in mathematics... and not just a weird gadget...
Or maybe a new Einstein in 2103 will use it as a basis for the Theory of everything.

Btw... I'm trained in graphic design... and I have a crush for math typesetting... so I feel bad looking at that thing without parentheses and in italic...
pls... if you care for my health, fix it xD

that's as good as I'm making it, too lazy. Get bo to give you mod powers and then you can fix it, lmao.
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#18
I actually have powers to edit post, but I'm not gonna do it, and for two good reasons: it is outside my mandate, the reason bo gave me them, secondly... I fear that I'll go back adding latex to every single Tommy's post xd wasting the rest of my life on it just to satisfy my obsession hahah.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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