I managed to plot in Excel first 6 integer iterations of f(x)=x^(1/x) for 1<=x<=10; I took only positive real values for f and its iterates.
What is obvious, for every x>e there is 1<x<e which has the same selfroot; most notably, 2^(1/2)=4^(1/4); but for e.g. x=3:
3^(1/3)= 1,442249...= 2,47805..^(1/2,47805..) meaning from h(x^(1/x))= x
h(1,442249..) = 2,47805.. if x<e
h(1,442249..)= 3 if x>e
So for any 1=<z=x^(1/x)<e^(1/e) there are 2 real values of infinite tetration.
Only for x=e there is exactly 1 value, which is also strange and requires further attention.
Since it seems that 1^(1/1) = lim n^(1/n) = infinity^(1/infinity) = 1 ,even for 1 there are 2 values
h (1) = 1 if x<e ,
h(1)= infinity if x>e;
So I understand that for x<1 there may be more interesting things. Also, going backwards with negative integer iterations might be instructive.
Ivars
IntegerIteratesSelfroot.bmp (Size: 198.75 KB / Downloads: 1,262)
Full range x>0 leads to a graph whihc looks like step function at x=1, continuing iterations to infinity will lead to step function? for x>0, I do not understand what happens with point x=e and 2 values of \( f^{on} (x) \) at x>1 and n-> infinity-do they all converge into one value =1? Andrew had somewhere in a thread a nice picture where these iterations where continued to negative.
IntegerPositiveIterationsSeflrootmono.bmp (Size: 40.49 KB / Downloads: 1,207)
What is obvious, for every x>e there is 1<x<e which has the same selfroot; most notably, 2^(1/2)=4^(1/4); but for e.g. x=3:
3^(1/3)= 1,442249...= 2,47805..^(1/2,47805..) meaning from h(x^(1/x))= x
h(1,442249..) = 2,47805.. if x<e
h(1,442249..)= 3 if x>e
So for any 1=<z=x^(1/x)<e^(1/e) there are 2 real values of infinite tetration.
Only for x=e there is exactly 1 value, which is also strange and requires further attention.
Since it seems that 1^(1/1) = lim n^(1/n) = infinity^(1/infinity) = 1 ,even for 1 there are 2 values
h (1) = 1 if x<e ,
h(1)= infinity if x>e;
So I understand that for x<1 there may be more interesting things. Also, going backwards with negative integer iterations might be instructive.
Ivars
IntegerIteratesSelfroot.bmp (Size: 198.75 KB / Downloads: 1,262)
Full range x>0 leads to a graph whihc looks like step function at x=1, continuing iterations to infinity will lead to step function? for x>0, I do not understand what happens with point x=e and 2 values of \( f^{on} (x) \) at x>1 and n-> infinity-do they all converge into one value =1? Andrew had somewhere in a thread a nice picture where these iterations where continued to negative.
IntegerPositiveIterationsSeflrootmono.bmp (Size: 40.49 KB / Downloads: 1,207)