possible tetration extension part 3
(10/22/2022, 09:29 PM)bo198214 Wrote:
(10/22/2022, 04:42 PM)Shanghai46 Wrote: Technically I don't think so

Come on, be a bit more elaborative! It should be in your interest to make your finding more understandable!

Let me reiterate: you write
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

This formula is not defining \({^r}x\) in a unique way (apart from the typo "k any real integer number" - what you mean is "k any real number"), because for different choices of m and p we get different results. Say x=2 and r=0.9; when we choose
m=p=2 then \({^{0.9}}2\approx\) 1.8482158
m=p=3 then \({^{0.9}}2\approx\) 1.8802187
m=p=4 then \({^{0.9}}2\approx\) 1.8802194
(which btw seems to be an error in your given value 0.9109247189 which is not \({^{0.9}}2\) but \({^{-0.1}}2\) because \(1={^0}2 < {^{0.9}}2 < {^1}2 =2 \) )

So to make it a proper definition (that does not depend on the choice of m and p), I suggested to take the limit \(m=p\to\infty\) which would be just the execution of what you described in text form in the beginning. So I am wondering a bit why you reject this formula.
For simplicity lets just assume that \(0<r<1\) if that helps to accept the formula.

The thing is that you can technically do k time the function Log_x(a+1) for any k, like -0.5 times, etc. It's just more precise the less k is. And since tetration is defined recursively, we can take the inverse function enough times to get to the value we want. Also, the more we apply f ang g, the more it's precise. And the more the number a is, the more Log_x(a+1) will approach Log_x(a). So a and (the number of times you apply f and g) should go to infinity, but the amount of time you take the inverse of tetration depends on the precision and the desired value.
I think we need to distinguish "best ways to numerically calculate a value" and "mathematically defined by a limit"!

For example the limit 
$$ g^t(x) = f^{-t}(x)=\lim_{n\to\infty}g^n((f^n(x)-\tau)/\lambda^t)+\tau) $$
is not more precise or less precise depending whether t is big or small, it is by definition maximum precise (if the limit *exists*).
Only when you want to calculate numerically - where you can't go all the way up to infinity with n - you can think how to improve the precision of your result. Maybe you split t into integer and fractional part \(t = p + \varrho\), \(0 \le \varrho < 1\) and \(p\in\mathbb{Z}\) and then
$$ g^{p+\varrho}(x) = g^p(\lim_{n\to\infty}g^{n}((f^n(x)-\tau)/\lambda^\varrho)+\tau)) $$
but that is just a numerical consideration.

So lets for now \(g^t\) be already defined in a suitable way for all \(t\in\mathbb{R}\). Then we want to obtain \({^r}b\), first for \(0<r<1\). According to your idea we apply a lot of \(\exp_b\) then we apply  a fractional iteration of \(g\) or a negative fractional iteration of \(f\) - because for high \(y\) the difference between \(f(y)=\log_b(y+1)\) and \(\log_b(y)\) vanishes, and then we go back with \(\log_b\). And putting this in one formula is:
$$ {^{r}}b = \lim_{n\to\infty} \log_b^{n}(g^{r}(\exp_b^{n}(1))) = \lim_{n\to\infty} \log_b^{n}(g^{r}({^n}b)) $$

Quote:but the amount of time you take the inverse of tetration depends on the precision and the desired value.

The part with the precision is exactly expressed with \(\lim_{n\to\infty}\) - the higher you go with n the more precise is the result.

And whether for arbitrary \(r\in\mathbb{R}\) you split again \( r = m - p + \varrho \), \(0\le \varrho<1\) and set
$$ {^{m-p+\varrho}}b = \log_b^{p} \lim_{n\to\infty} \log_b^{n}(g^{\varrho}(\exp_b^n({^{m}}b))) $$
is more a numerical consideration. At the heart of the formula is the limit with \(n\to\infty\) which did not occur in the formulas you gave.

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