So it seems to me, that finding real and complex extensions to tetration (finite) is kind of symmetric to finding real and complex extensions to selfroots and superroots which are related to infinite tetration.
By extension I mean self root is usually made of 2 discrete parts: number x, its inverse 1/x and than exponentiation x^(1/x).
it is possible to extend it to 4, 8, 16...2^n discrete parts easily by repeated application. How to make it consist of non-integer number of parts? Real number? Complex?
Second superroot can be extended to 3,4,5 etc easily but not by simple iteration of f(x) =x^x as it will miss 3,5,7 and all odd superroots. First superroot is also not obvious. By looking at x^x^x = y where x is 3rd superroot, x^x^x^x =y, where now x is 4th superroot etc it is clear that odd superroots exist. How to extend this to fractional or real or complex number of superroots?
By analogy not by logic odd self roots would be: 3rd = x^(1/x) ^x =y, 5th x^(1/x)^x^(1/x)^x=y etc.
But what is then: x^((1/x)^x)? The change in grouping gives another value for 3rd self root- there are 2 of them.
For 5th self root, there are probably more different values of y by changing grouping.
Can it be done by looking into iterations?
if f(x)=x^(1/x), than f(f(x)) = (x^1/x))^(1/(x^(1/x));
if f(x) =x^x, than f(f(x) = (x^x)^(x^x):
Application of brackets or grouping of operations seems not to be trivial at all in these cases. Obviously, ((x^x)^x)^x is not equal to (x^x)^(x^x)=x^(x^x)^x= x^x^(x^x) which is not equal to x^(x^(x^x)), the last one being the biggest if x>1. So there are at least 3 4th superroots , 2 3rd superroots , 1 2nd superroot for any y >1. Interestingly, if we try with x=3, smallest of 4th superroots, 3^3^3^3 equals biggest of 3rd supperroots:
3^3^3^3 = 3^(3^3)= 7,6256E+12
The same is not true neither for 2, nor for 4 , nor probably any other real number which is not 3.
2^2^2^2=256 > 2^(2^2)=16
4^4^4^4=3,40282E+38 < 4^(4^4) = 1,3408E+154
Most of the effects of bracketing can be clearly seen by plotting self and superroots in polar coordinates, but what about analyzing them?
Ivars
By extension I mean self root is usually made of 2 discrete parts: number x, its inverse 1/x and than exponentiation x^(1/x).
it is possible to extend it to 4, 8, 16...2^n discrete parts easily by repeated application. How to make it consist of non-integer number of parts? Real number? Complex?
Second superroot can be extended to 3,4,5 etc easily but not by simple iteration of f(x) =x^x as it will miss 3,5,7 and all odd superroots. First superroot is also not obvious. By looking at x^x^x = y where x is 3rd superroot, x^x^x^x =y, where now x is 4th superroot etc it is clear that odd superroots exist. How to extend this to fractional or real or complex number of superroots?
By analogy not by logic odd self roots would be: 3rd = x^(1/x) ^x =y, 5th x^(1/x)^x^(1/x)^x=y etc.
But what is then: x^((1/x)^x)? The change in grouping gives another value for 3rd self root- there are 2 of them.
For 5th self root, there are probably more different values of y by changing grouping.
Can it be done by looking into iterations?
if f(x)=x^(1/x), than f(f(x)) = (x^1/x))^(1/(x^(1/x));
if f(x) =x^x, than f(f(x) = (x^x)^(x^x):
Application of brackets or grouping of operations seems not to be trivial at all in these cases. Obviously, ((x^x)^x)^x is not equal to (x^x)^(x^x)=x^(x^x)^x= x^x^(x^x) which is not equal to x^(x^(x^x)), the last one being the biggest if x>1. So there are at least 3 4th superroots , 2 3rd superroots , 1 2nd superroot for any y >1. Interestingly, if we try with x=3, smallest of 4th superroots, 3^3^3^3 equals biggest of 3rd supperroots:
3^3^3^3 = 3^(3^3)= 7,6256E+12
The same is not true neither for 2, nor for 4 , nor probably any other real number which is not 3.
2^2^2^2=256 > 2^(2^2)=16
4^4^4^4=3,40282E+38 < 4^(4^4) = 1,3408E+154
Most of the effects of bracketing can be clearly seen by plotting self and superroots in polar coordinates, but what about analyzing them?
Ivars