10/28/2022, 07:11 PM
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:
\[ {^{r}}b = \lim_{n\to\infty} \log_b^{n}(g^{r}(\exp_b^{n}(1))) = \lim_{n\to\infty} \log_b^{n}(g^{r}({^n}b)) \]
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.
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.