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The super 0th root and a new rule of tetration?
#5
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):
Code:
szrt(a,h)={
x=a;
for(i=0,h,sexpinit(x);x=sexp(1/h-1);x=a^(1/x));
return(x);
}
But somewhy, fatou.gp does not work at this point.
Could anyone give me an advice how have it worked?
Thank you.
Xorter Unizo
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Messages In This Thread
RE: The super 0th root and a new rule of tetration? - by Xorter - 11/29/2017, 11:53 AM

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