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