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 Doubts on the domains of Nixon's method. JmsNxn Long Time Fellow Posts: 398 Threads: 80 Joined: Dec 2010 03/02/2021, 10:43 PM Hey MPHlee, Honestly I was just probing what might happen. I wasn't standing by what I wrote absolutely. I suggest reading the construction of $C^\infty$ hyper-operations, which I just posted. I was trying to visualize how we might generalize this to more exotic scenarios, and I was largely hypothesizing (guessing?) what might happen. That is, guessing what type of structure we may be able to develop--I think you're right though. What I wrote is definitely not correct, but it was more of an exercise in thought. After working through hyper-operators, I think I can more clearly say what I can say. But still, I honestly do not know.  I chose the space rather arbitrarily as, let's say, "I think it might look something like this." Perhaps a better way to say it is, to list the requirements of what we need in the process of the proof. The first thing we need is good control over our function $f$ such that, $ \Phi(s) = \Omega_{j=1}^\infty e^{s-j}f(z) \bullet z\\$ Converges.  This will be $C^k$ if $f$ is. Further, this should be pretty simple to derive convergence. All we really need from this is that $f(\mathbb{R}^+)\subset \mathbb{R}^+$ is defined. So that the infinite composition converges. Second thing we need is for $f^{-1}$ to be well behaved, which is why I said diffeomorphism. I said that loosely, and you are correct. diffeomorphism is probably too strong, I simply meant it as a differentiable isomorphism. I chose $\mathbb{R}^+$ because it's convenient (and yes that would require it fixes zero). This is by no means necessary. However, $f'(x) > 0$ is needed, or at least, where $f'(x) = 0$ is easily controlled, not too sure. I'd just stick to non-zero derivative tbh. All we really need is for iterates $f^{-1}$ to be defined well enough, and for $f^{-n} \Phi(s+n)$ to be a meaningful thing. Again, this is very malleable. In the construction of the hyper-operators I use either a function $f^{-1} : (\alpha,\infty) \to \mathbb{R}$ bijectively, or $f^{-1} : \mathbb{R} \to (\alpha,\infty)$ bijectively. So we have a good amount of freedom here. If I were being honest, we just want $f^{-n} \Phi(s+n)$ to be a meaningful thing. The next thing we need is good control as a lipschitz condition of $f^{-1}$. We want, $ |f^{-1}(x) - f^{-1}(y)| \le g(min(x,y)) |x-y|\\$ Where as $x,y \to \infty$ the value $g(min(x,y)) \le \lambda < 1$. This allows us to apply Banach's fixed point theorem very cleanly and efficiently. Again, this may not be necessary, but for a quick simple proof, yes this is good. This is why I said that exponential growth, or something like it, would be good. It would mean that $f^{-1}$ looks something like the logarithm. Again, I did this in the construction of the hyper-operators, where $\text{slog}$ and its higher order equivalents, have a lipschitz condition AT LEAST as good as the logarithm. Which is when $g(x) = \frac{1}{x}$. Which makes things pretty quick and not very messy. But again, not really necessary. As long as $f^{-1}$ is decently behaved and has some kind of lipschitz condition, we are all green. Now all of this is enough for us to show that, $ F = \lim_{n\to\infty} f^{-n} \Phi(s+n)\\$ Converges to a CONTINUOUS function such that $F(x+1) = f(F(x))$. We will know that $\Phi$ is $C^k$ if $f$ is, but deriving that $F$ is $C^k$ is something I am not sure how to do yet. I managed to do it for hyper-operations, but it required a lot of nice things to happen. Mainly that, $ \Lambda_n(s) = f^{-n}\Phi(s+n) - \Phi(s)$ Had some pretty convenient differential properties. Namely that $\Lambda^{(j)}_n(s)$ looks like $\Lambda_1^{(j)}(s)$ for large values of $s$. This is a very very crucial step in the proof to derive $C^\infty$, and I'm not sure how to guarantee this as  a condition on $f$. It happened conveniently with $\phi$ and $\tau$ and its converging sequence $\tau_n$; which look like $s$ for large $s$ and similarly its derivatives looked like $s^{(j)}$ for large $s$. Luckily, something similar happens with hyper-operators. Although I am not certain, I do believe that the missing ingredient will be something of the form $g(x) \le \frac{1}{x}$. Or that, the lipschitz constant of the inverse function $f^{-1}$ looks like the logarithmic lipschitz constant. This would mean, we want $f$ to grow faster, or like the exponential $e^x$ and that its inverse $f^{-1}$ is slow like the logarithm.  Again though, there's probably more to it. I think something like $f^{(j)}(x) > 0$ for all $j$ may come into play at some point. I cannot say for certain. But this isn't necessary, because I never showed $e \uparrow^n x$ satisfied this--but for large $x$ this will be true (again though, I don't use this.) Again, don't take everything I'm saying to heart. I'm still trying to get a feel for the general case. I also have to clean up the hyper-operators a bit--and in doing so, hopefully be better at identifying what actually allowed for them to be $C^\infty$. The proof seems to be pretty hardwired to the super-fast growth of hyper-operators, and the excessively slow growth of their inverses. When I get the time, I might run through a more elementary example, and see if we can construct a super function to something simple like, $ f(x) = x^3 +1\\$ My suspicions are the method should still work in this case, but I may have to modify the approach; or at least fine-tune it. Especially when it comes to trying to make the super-function $C^\infty$. I don't foresee any problems with constructing a continuous super-function--it's the $C^\infty$ part that has me scratching my head. Regards, James. PS: I'll leave the spaces and category; or what sets these things live in, and how the conjugacy classes are structured to you, lol. I'll just focus on trying to create a reasonable set of criteria on $f$ such that we can get a super function $F$ which is as differentiable as $f$. My guess is that it's going to be really tricky to nail down a single criteria. However, what I just laid out above is the gist of how the hyper-operators were constructed. « Next Oldest | Next Newest »

 Messages In This Thread Doubts on the domains of Nixon's method. - by MphLee - 02/27/2021, 11:03 AM RE: Doubts on the domains of Nixon's method. - by JmsNxn - 03/02/2021, 10:43 PM

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