10/21/2022, 08:16 PM
(10/21/2022, 07:23 PM)bo198214 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(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 seem complicated, but let's take an example. To calculate \({^{-0.5}}10\), we take \({^2}10\) (m=2), we iterate f a lot of times (n times), wa substract \(\lambda\) which is 0, we multiply by \({\frac{1}{\ln(10)}^{0.5} }\) (k=0.5), we add \(\lambda\) (still 0) we iterate g the same number of times as f to cancel it out (n times), so here we have an approximation of \({^{1.5} }10\), so we iterate \(\log_{10}\) twice (p=2) to get an approximation of \({^{-0.5}}10\).
It is not clear to me how you choose the values m and p. Above you just chose m=p=2, but you could have taken any other natural number (say m=p=5), and I guess the result would be different.
Or do you take another limit with \(m=p\to \infty\)?

