I could improve the results up to iteration-height h ~ 2^22 which means values of b^^4194304 (which is very near the fixpoint... 
)
Let's denote b^^h as T(h), t the lower fixpoint, and ln(t) = u
Then with the limit-function
\( \hspace{48} d(h) = u - \ln\( \frac { T(2*h)} { T( h ) } \)^{\frac1h} \)
we seem to get approximation down to zero with increasing height h.
Moreover, if we use the ratio of two such limits in steps of h = 2^k , or even more general in h = c^k with some exponential step-base c :
\( \hspace{48} d( k,c ) = u - \ln\( \frac { T(c^{k+1})} { T( c^k ) } \)^{\frac1{c^k}} \)
or even better for computation
\( \hspace{48} d( k,c ) = u - \ln\(\( T(c^{k+1}-1) - T(c^k-1)\)*\ln(b)\)^{\frac1{c^k}} \)
and then
\( \hspace{48} q(k,c\) = \frac{ d(k-1,c) }{ d( k,c ) } \)
then we seem to arrive at the following
\( \text{Conjecture :} \\
\\
\hspace{48} \lim_{k\to\infty} q( k , c ) \to c \)
Heuristic:
The extremely high iteration-heights can be achieved in regular tetration when we use the schroeder-function: with this the powertower-iteration need not be done by h-times exponentiation but can be found by raising u to the h'th power. Denote S_a(x) the schroeder-function for the fixpoint-shifted argument and S_b(x) the inverse such that
\( \hspace{48} T(h) = S_b\( u^h * S_a\( \frac1t-1\) \) *t + t \hspace{120} \) // regular tetration
Different T(h) can then be computed conveniently after the value of the schroederfunction for x=1, s = S_a(1/t-1), was saved in a constant s . Also the numerator in the second d-formula can be simplified, since the difference of two T()'s also cancel the numerically uneasy addition of t, and finally ln(b)=u/t where then also t cancels, so
\( \hspace{48} d( k,c ) = u- \ln \( \( S_b(s*u^{c*c^k-1}) - S_b(s*u^{c^k-1})\)*u\)^{\frac1{c^k}}
\)
[Update: don't know: am I the only one who get's the vague feeling that one of these formulae really sucks ? ]
Gottfried
)Let's denote b^^h as T(h), t the lower fixpoint, and ln(t) = u
Then with the limit-function
\( \hspace{48} d(h) = u - \ln\( \frac { T(2*h)} { T( h ) } \)^{\frac1h} \)
we seem to get approximation down to zero with increasing height h.
Moreover, if we use the ratio of two such limits in steps of h = 2^k , or even more general in h = c^k with some exponential step-base c :
\( \hspace{48} d( k,c ) = u - \ln\( \frac { T(c^{k+1})} { T( c^k ) } \)^{\frac1{c^k}} \)
or even better for computation
\( \hspace{48} d( k,c ) = u - \ln\(\( T(c^{k+1}-1) - T(c^k-1)\)*\ln(b)\)^{\frac1{c^k}} \)
and then
\( \hspace{48} q(k,c\) = \frac{ d(k-1,c) }{ d( k,c ) } \)
then we seem to arrive at the following
\( \text{Conjecture :} \\
\\
\hspace{48} \lim_{k\to\infty} q( k , c ) \to c \)
Heuristic:
Code:
.
(Pari/GP, float-prec 800 , regular tetration, base b=sqrt(2),t=fp0=2, u=ln(2)
[c=2 , k, q(k,c)]
% 99 = [2, 15, 1.9999760334888695166]
%100 = [2, 20, 1.9999992510378326004]
%101 = [2, 21, 1.9999996255188461821]
%102 = [2, 22, 1.9999998127594055616]
[c=3 , k, q(k,c)]
%103 = [3, 7, 2.9989230328032237024]
%104 = [3, 12, 2.9999955667153233567]
%105 = [3, 13, 2.9999985222372279067]
%106 = [3, 14, 2.9999995074122745008]
[c=5 , k, q(k,c)]
%111 = [5, 4, 4.9874581459928721151]
%112 = [5, 7, 4.9998994774776347240]
%113 = [5, 8, 4.9999798952530082216]
%114 = [5, 9, 4.9999959790409007914]The extremely high iteration-heights can be achieved in regular tetration when we use the schroeder-function: with this the powertower-iteration need not be done by h-times exponentiation but can be found by raising u to the h'th power. Denote S_a(x) the schroeder-function for the fixpoint-shifted argument and S_b(x) the inverse such that
\( \hspace{48} T(h) = S_b\( u^h * S_a\( \frac1t-1\) \) *t + t \hspace{120} \) // regular tetration
Different T(h) can then be computed conveniently after the value of the schroederfunction for x=1, s = S_a(1/t-1), was saved in a constant s . Also the numerator in the second d-formula can be simplified, since the difference of two T()'s also cancel the numerically uneasy addition of t, and finally ln(b)=u/t where then also t cancels, so
\( \hspace{48} d( k,c ) = u- \ln \( \( S_b(s*u^{c*c^k-1}) - S_b(s*u^{c^k-1})\)*u\)^{\frac1{c^k}}
\)
[Update: don't know: am I the only one who get's the vague feeling that one of these formulae really sucks ? ]
Gottfried
Gottfried Helms, Kassel