+- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Iterations of ln(mod(ln(x^(1/x))) and number with property Left a[4]n=exp (-(a^(n+1)) (/showthread.php?tid=139) Iterations of ln(mod(ln(x^(1/x))) and number with property Left a[4]n=exp (-(a^(n+1)) - Ivars - 03/30/2008 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