We must notice that

lim h->infinity a^^(1/h) = 1

and

lim h->infinity super 1/h-th root of a = infinity,

but

lim h->infinity super 1/h-th root of a^^(1/h) = a

So for the most of non-trivial functions it gives interesting results.

Yes, as you said lim->infinity log( f(x+1/h) )/log( f(x) ) approaches to 1, but the whole result by the super root will not approach to only 1 or only infinity.

The remaining question is that: How can we calculate lim h->infinity super 1/h-th root of f(x,h)?

Update:

I have an interesting formula for super roots like this:

super Nth root of A = (A^(1/(x^^(N-1))))^oInfinity

So we are looking for the fixed point of the formula above. It also shows that:

super 1st root of A = A

and

lim h->0 super h-th root of A = infinity

Update:

I have a code, too for Super Zeroth RooT of a number (h->infinity):

But somewhy, fatou.gp does not work at this point.

Could anyone give me an advice how have it worked?

Thank you.

lim h->infinity a^^(1/h) = 1

and

lim h->infinity super 1/h-th root of a = infinity,

but

lim h->infinity super 1/h-th root of a^^(1/h) = a

So for the most of non-trivial functions it gives interesting results.

Yes, as you said lim->infinity log( f(x+1/h) )/log( f(x) ) approaches to 1, but the whole result by the super root will not approach to only 1 or only infinity.

The remaining question is that: How can we calculate lim h->infinity super 1/h-th root of f(x,h)?

Update:

I have an interesting formula for super roots like this:

super Nth root of A = (A^(1/(x^^(N-1))))^oInfinity

So we are looking for the fixed point of the formula above. It also shows that:

super 1st root of A = A

and

lim h->0 super h-th root of A = infinity

Update:

I have a code, too for Super Zeroth RooT of a number (h->infinity):

Code:

`szrt(a,h)={`

x=a;

for(i=0,h,sexpinit(x);x=sexp(1/h-1);x=a^(1/x));

return(x);

}

Could anyone give me an advice how have it worked?

Thank you.

Xorter Unizo