Some complex value tetration:
If I have understood right (if not, tell me please) , basic idea to be able to have some analytics relations is to make \( ln(a)^z = 1 \) or
\( z*ln(a) =1 \) and if we have chosen complex or real \( z \) then we can find \( a \) such that :
\( ln(a) =1/z , b= a^{(1/a)} \)
So every time
\( a^{(1/a)} [4] {1/ln(a)} = 1 \)
E.g.
lower fixed point \( a=e^-I=cos(1) -I*sin(1)=0.5403023-I*0.8414098 \)
we get the base by \( b=(e^-I)^{(e^I)} \)
if \( z=I \) then
\( (e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}((e^{-I})*(1-\ln(e^{-I})^I) + \ln(e^{-I})^I \exp_b^{\circ n}(1)) \)
but :
\( \ln(e^{-I})=-I \)
\( \ln(e^{-I})^I=I*\ln(e^{-I})=I*-I=1 \)
\( (e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}((e^{-I})*(1-1) + 1* \exp_b^{\circ n}(1)) \)
\( (e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}(\exp_b^{\circ n}(1)) \)
this seems to equal 1. So with Imaginary height:
\( (e^{-I})^{(e^I)}[4] I= 1 \)
Few more interesting outcome :
\( I^{1/I} = e^{\pi/2}[4] {-I*2/\pi}=1 \)
\( e^{1/e}[4] 1 =1 \)
Which seems wrong
.So where all others, then...
Or actually a=e and base \( b=e^{1/e} \) seems to be rather unique point in selfroots as it is the only one for a>1 which has only 1 value, so h(e^(1/e))=e and that is the only value ( all other points for a>1 has 2 values for h(a^(1/a))), because 2 different a on different sides of a=e have the same selfroot values).
So if we speculate a little to continue with 2 values as a demand for \( h(a^{1/a}) a>=1 \) , we may propose that:
\( e^{1/e}[4] 1 =1 \)
\( e^{1/e}[4] 1 =e^{1/e} \)
\( h( e^{1/e}) =e \)
\( h( e^{1/e}) =1 \)
\( h(1)=1 \)
\( h(1)=\infty \)
Or, if \( ln(e) = 1+I*2\pi*k \)
\( e^{1/e}[4] 1 =e^{1/e} \)
\( e^{1/e}[4] {1/(1+2\pi*I*k)} =1 \)
\( 1^1[4] \infty =1 \)
\( 1^1[4] {1/(2\pi*I*k)} =1 \)
But it was good tex training...
Ivars
If I have understood right (if not, tell me please) , basic idea to be able to have some analytics relations is to make \( ln(a)^z = 1 \) or
\( z*ln(a) =1 \) and if we have chosen complex or real \( z \) then we can find \( a \) such that :
\( ln(a) =1/z , b= a^{(1/a)} \)
So every time
\( a^{(1/a)} [4] {1/ln(a)} = 1 \)
E.g.
lower fixed point \( a=e^-I=cos(1) -I*sin(1)=0.5403023-I*0.8414098 \)
we get the base by \( b=(e^-I)^{(e^I)} \)
if \( z=I \) then
\( (e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}((e^{-I})*(1-\ln(e^{-I})^I) + \ln(e^{-I})^I \exp_b^{\circ n}(1)) \)
but :
\( \ln(e^{-I})=-I \)
\( \ln(e^{-I})^I=I*\ln(e^{-I})=I*-I=1 \)
\( (e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}((e^{-I})*(1-1) + 1* \exp_b^{\circ n}(1)) \)
\( (e^{-I})^{(e^I)}[4]I =\exp_b^{\circ I}(1)=\lim_{n\to\infty} \log_b^{\circ n}(\exp_b^{\circ n}(1)) \)
this seems to equal 1. So with Imaginary height:
\( (e^{-I})^{(e^I)}[4] I= 1 \)
Few more interesting outcome :
\( I^{1/I} = e^{\pi/2}[4] {-I*2/\pi}=1 \)
\( e^{1/e}[4] 1 =1 \)
Which seems wrong
.So where all others, then...Or actually a=e and base \( b=e^{1/e} \) seems to be rather unique point in selfroots as it is the only one for a>1 which has only 1 value, so h(e^(1/e))=e and that is the only value ( all other points for a>1 has 2 values for h(a^(1/a))), because 2 different a on different sides of a=e have the same selfroot values).
So if we speculate a little to continue with 2 values as a demand for \( h(a^{1/a}) a>=1 \) , we may propose that:
\( e^{1/e}[4] 1 =1 \)
\( e^{1/e}[4] 1 =e^{1/e} \)
\( h( e^{1/e}) =e \)
\( h( e^{1/e}) =1 \)
\( h(1)=1 \)
\( h(1)=\infty \)
Or, if \( ln(e) = 1+I*2\pi*k \)
\( e^{1/e}[4] 1 =e^{1/e} \)
\( e^{1/e}[4] {1/(1+2\pi*I*k)} =1 \)
\( 1^1[4] \infty =1 \)
\( 1^1[4] {1/(2\pi*I*k)} =1 \)
But it was good tex training...
Ivars
