10/26/2022, 08:20 PM

(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.