Tetration extension for bases between 1 and eta

[bo198214]

If we set \( f(x+1)=b^{f(x)} \) which is to say
\( \lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) ) \)
then reduce it to
\( \lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b) \)
Then just strait up plug in infinity for k
we get \( log_{b} {}^\infty b = {}^\infty b \) which is the same as \( {}^\infty b = {}^\infty b \)

Slowly, slowly. The first line is what you want to show. You show from the first line something true, but you can also show something true starting from something wrong; so thats not sufficient. Also it seems as if you confuse limit equality with sequence equality.

Lets have a look at the inverse function \( g=f^{-1} \),
\( g(x)=\lim_{k\to\infty} \frac{\exp_b^{\circ k}(x)-{^k b}}{{^k b}-({^{k-1} b})} \) it should satisfy
\( g(b^x)=g(x)+1 \).

Then lets compute
\( g(b^x)-g(x)=
\lim_{k\to\infty}\frac{\exp_b^{\circ k+1}(x)-{^k b}}{{^k b}-({^{k-1} b})} - \frac{\exp_b^{\circ k}(x)-{^k b}}{{^k b}-({^{k-1} b})}
=\lim_{k\to\infty} \frac{\exp_b^{\circ k+1}(x) - \exp_b^{\circ k}(x)}{{^k b}-({^{k-1} b})}
\)

Take for example \( x=1 \) then the right side converges to the derivative of \( b^x \) at the fixed point; and not to 1 as it should be.

This is the reason why the formula is only valid for functions that have derivative 1 at the fixed point, e.g. \( e^{x/e} \), i.e. \( b=e^{1/e} \).

This is really weird because if i do \( \lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \) for \( b= sqrt2 \)
i get about 1.558 which is substantially larger then \( sqrt2 \)

Can anyone else confirm that \( \lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558 \) for \( b= sqrt2 \) ?

Try the same with \( b=e^{1/e} \) and it will work; but for no other base; except you use the modified formula I described before.