I tried to do analytically one case:
lower fixed point \( a=\Omega=0.567143.. \)
we get the base by \( b={1/e}=\Omega^{1/\Omega} \)
if \( z={1/\Omega} \) then
\( {1/e}[4]{1/\Omega}=\exp_{1/e}^{\circ 1/\Omega}(1)=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*(1-\ln(\Omega)^{1/\Omega}) + \ln(\Omega)^{1/\Omega }\exp_{1/e}^{\circ n}(1)) \)
but :
\( \ln(\Omega)=-\Omega \)
\( \ln(\Omega)^{1/\Omega }=({1/\Omega })*\ln(\Omega)={-\Omega/\Omega }=-1 \)
\( {1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*(1-(-1)) -1*\exp_{1/e}^{\circ n}(1)) \)
\( {1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*2 -1*\exp_{1/e}^{\circ n}(1)) \)
At this point I do not know what to do with limits -how do You apply n to iterating logarithms, and at the same time to iteration of exponentiation of 1/e inside it? If we could move limit inside, than:
\( \lim_{n\to\infty} \exp_{1/e}^{\circ n}(1))=h({1/e)}=\Omega \)
And \( 2*\Omega-\Omega=\Omega \)
\( \log_{1/e}(\Omega)= \Omega \)
for all n, so
\( {1/e}[4]{1/\Omega}=\Omega \) which equals
\( {\Omega^{1/\Omega}[4]{1/\Omega}=\Omega \)
I know it is not complex but real...
Must be a mistake with the limits, but looks nice still...
If someone could explain me on this example how I should have proceeded after the place where I got to limit taking inside iterated logarithm, I promise never to make the same mistake.
If we take the same but:
\( z={-1/\Omega} \)
Then I have to go through the whole procedure again.
In this case, instead of \( 2*\Omega \) we will have 0, and + before tower as:
\( \ln(\Omega)^{-1/\Omega }=(-{1/\Omega })*\ln(\Omega)={-\Omega/-\Omega }=1 \)
\( {1/e}[4]{-1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(1*\exp_{1/e}^{\circ n}(1)) \)
this seems to equal 1. So with negative height:
\( {1/e}[4]{-1/\Omega}= 1 \)
\( {\Omega^{1/\Omega}[4]{-1/\Omega}=1 \)
Ivars
lower fixed point \( a=\Omega=0.567143.. \)
we get the base by \( b={1/e}=\Omega^{1/\Omega} \)
if \( z={1/\Omega} \) then
\( {1/e}[4]{1/\Omega}=\exp_{1/e}^{\circ 1/\Omega}(1)=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*(1-\ln(\Omega)^{1/\Omega}) + \ln(\Omega)^{1/\Omega }\exp_{1/e}^{\circ n}(1)) \)
but :
\( \ln(\Omega)=-\Omega \)
\( \ln(\Omega)^{1/\Omega }=({1/\Omega })*\ln(\Omega)={-\Omega/\Omega }=-1 \)
\( {1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*(1-(-1)) -1*\exp_{1/e}^{\circ n}(1)) \)
\( {1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*2 -1*\exp_{1/e}^{\circ n}(1)) \)
At this point I do not know what to do with limits -how do You apply n to iterating logarithms, and at the same time to iteration of exponentiation of 1/e inside it? If we could move limit inside, than:
\( \lim_{n\to\infty} \exp_{1/e}^{\circ n}(1))=h({1/e)}=\Omega \)
And \( 2*\Omega-\Omega=\Omega \)
\( \log_{1/e}(\Omega)= \Omega \)
for all n, so
\( {1/e}[4]{1/\Omega}=\Omega \) which equals
\( {\Omega^{1/\Omega}[4]{1/\Omega}=\Omega \)
I know it is not complex but real...
Must be a mistake with the limits, but looks nice still...
If someone could explain me on this example how I should have proceeded after the place where I got to limit taking inside iterated logarithm, I promise never to make the same mistake.
If we take the same but:
\( z={-1/\Omega} \)
Then I have to go through the whole procedure again.
In this case, instead of \( 2*\Omega \) we will have 0, and + before tower as:
\( \ln(\Omega)^{-1/\Omega }=(-{1/\Omega })*\ln(\Omega)={-\Omega/-\Omega }=1 \)
\( {1/e}[4]{-1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(1*\exp_{1/e}^{\circ n}(1)) \)
this seems to equal 1. So with negative height:
\( {1/e}[4]{-1/\Omega}= 1 \)
\( {\Omega^{1/\Omega}[4]{-1/\Omega}=1 \)
Ivars
