10/22/2022, 10:09 AM
(10/22/2022, 09:50 AM)Shanghai46 Wrote:(10/22/2022, 08:10 AM)bo198214 Wrote:(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.
(10/21/2022, 08:16 PM)Shanghai46 Wrote: Technically it wouldn't give exactly the same number, but the limit would be the same one, so it's not really changing anything, but the simpler the better
You described letting \(^mx\) go to a big number and then taking the fractional iterate of \(f\), but it does not reflect in your formula. So I guess what you mean with "limit" in your comment is the limit over \(m\), so the formula would look like:
\[{^r}x=\lim_{m\to\infty}\lim_{n\to\infty}({\log_x}^m( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))\]
Is that what you mean? (Computationally of course one would just look for a suitably big \(^mx\) ) Just in the clarification phase
What I mean is that the larger an number is, the closer Log_a(x+1) will be to Log_a(x).
Is that a "yes" to the double limit formula?