# Tetration Forum

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I have been playing with iterations of
$f(x)_n = \ln(f(x)_{n-1}) \text{ if } f(x)_{n-1}>0$
$f(x)_n = \ln(-f(x)_{n-1}) \text{ if } f(x)_{n-1} <0$

And of course I tried $f(x) = \ln(x^x)$ and $f(x)= \ln(x^{(1/x)})$

$ln(x^x)$ is not very interesting, it converges to 2 values ${1/e}; {-1/e}$ depending on integer iteration number. Here are picture of 200 iterations:
[attachment=283]

$f(x)= \ln(x^{(1/x)})$ is more interesting. This iteration converges to 4 values for each x, cyclicaly,and their dependance on n is shifting depending the region x is in the interval ]0:1[. The convergence values are :

$e; 1/e ; -e;-1/e$

Here is what happens:

[attachment=284]

[attachment=287]

When resolution is increased (step decreased) more and more shifts in phase happen in the region which seems to converge to approximately 0,6529204....

[attachment=285]

[attachment=286]

This number has properties:

$(0,6529204^{(1/0,6529204)})^{(1/0,6529204)}={1/e}$

${1/0,6529204.}=1,531580266$

$(({1/0,6529204})^{(1/0,6529204)})^{(1/0,6529204)}= e$

So its 2nd selfroot is $1/e$, while its reciprocal 1,531580266.. is 3rd superroot of e.

It has also following properties, at least approximately numerically, so it might be wrong, but interesting:

if we denote it $A=0.6529204..$, than:

$(((A^A)^A)^A)^..n times = \exp^{(-(A^{(n+1)})}$

$(({1/A})^A)^A)^A)^...n times = \exp^{(A^{(n+1)})}$

So:
$\ln(({1/A})^{(1/A)})^{(1/A)}))=1$
$\ln(({1/A})^{(1/A)})=A$
$\ln ({1/A}) =A^2$
$\ln (({1/A})^A)=A^3$
$\ln(({1/A})^A)^A) = A^4$
........
$\ln(A^{(1/A)}=-A$
$\ln(A) = -A^2$
$\ln(A^A) = -A^3$
$\ln((A^A)^A)=-A^4$
$\ln(((A^A)^A)^A)=-A^5$
..........

if this is so, what happens if instead of integer n we take x, so that:

maybe:

$\ln( A[4Left]x) = - A^{(x+1)}$
$\ln((1/A)[4Left]x)= A^{(x+1)}$

I hope I used [4Left] correctly. So:

$\ln(A[4Left]e] = -A^{(e+1)}=-0,2049265$ and

$A[4Left]e=0,81470717$

If accuracy is not enough, so it is numerically only 3-4 digits, perhaps going for n-th superroot of e would improve situation?

Probably this is old knowledge.

Ivars