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:
\( f(x)= \ln(x^{(1/x)}) \) 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:
When resolution is increased (step decreased) more and more shifts in phase happen in the region which seems to converge to approximately 0,6529204....
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
\( f(x)_n = \ln(f(x)_{n-1}) \text{ if } f(x)_{n-1}>0 \)
\( f(x)_n = \ln(-f(x)_{n-1}) \text{ if } f(x)_{n-1} <0 \)
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:
\( f(x)= \ln(x^{(1/x)}) \) 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:
When resolution is increased (step decreased) more and more shifts in phase happen in the region which seems to converge to approximately 0,6529204....
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
