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 The super 0th root and a new rule of tetration? Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 11/29/2017, 11:53 AM (This post was last modified: 11/30/2017, 04:33 PM by Xorter.) 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 « Next Oldest | Next Newest »

 Messages In This Thread The super 0th root and a new rule of tetration? - by Xorter - 11/27/2017, 01:44 PM RE: The super 0th root and a new rule of tetration? - by Xorter - 11/27/2017, 06:33 PM RE: The super 0th root and a new rule of tetration? - by sheldonison - 11/28/2017, 04:15 PM RE: The super 0th root and a new rule of tetration? - by Xorter - 11/28/2017, 05:45 PM RE: The super 0th root and a new rule of tetration? - by Xorter - 11/29/2017, 11:53 AM

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