06/12/2022, 12:02 AM
maybe you considered this before or maybe not.
But I am fascinated by taking the inverse super of x^3 repeatedly.
In fact it might result in a deeper understanding of the superfunction operator , fractional superfunction operators etc.
We seem to approach some kind of pattern or fixpoint.
so we start
x^3
( x^(1/3) + 1 ) ^3
...
We seem to be getting closer to the identity or successor function.
The asymptotics for large x are fun.
x^3
x + O( x^(2/3) )
x + O( x^(1/3) )
...
See also :
https://math.stackexchange.com/questions...v-n-sqrt-7
Maybe worth some attention.
A type of koenigs function for this repeated operator might be a nice result ...
It is late so maybe i missed something ...
time to sleep.
regards
tommy1729
But I am fascinated by taking the inverse super of x^3 repeatedly.
In fact it might result in a deeper understanding of the superfunction operator , fractional superfunction operators etc.
We seem to approach some kind of pattern or fixpoint.
so we start
x^3
( x^(1/3) + 1 ) ^3
...
We seem to be getting closer to the identity or successor function.
The asymptotics for large x are fun.
x^3
x + O( x^(2/3) )
x + O( x^(1/3) )
...
See also :
https://math.stackexchange.com/questions...v-n-sqrt-7
Maybe worth some attention.
A type of koenigs function for this repeated operator might be a nice result ...
It is late so maybe i missed something ...
time to sleep.
regards
tommy1729