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The super 0th root and a new rule of tetration?
#1
Hi, everyone!

I have been working on the so-called iterated analysis. And I have found an interesting formula in that I am not so sure, so I share it with you to discuss it.
So, what if I say:
lim h->infinity super 1/h-th root of ( (f^^N)-th root of f^^N ) = lim h->infinity (super 1/h-th root of ((f)-th root of f))^^N
which is equivalent to
super 0th root of ( (f^^N)-th root of f^^N ) = (super 0th root of ((f)-th root of f))^^N
1st question: How can we calculate the super 0th root of a number or a function?
Any idea?
Xorter Unizo
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#2
Okay, I revised the formula, and I guess my intuation mislead me, maybe.
But I have a good news, too.
Maybe this formula will bring the trueth:
lim h->infinity (log(f(x+1/h)^^N)/log(f(x)^^N))^^h = lim h->infinity ((log(f(x+1/h))/log(f(x)))^^h)^^N
Do you think it can be correct?
Xorter Unizo
Reply
#3
(11/27/2017, 06:33 PM)Xorter Wrote: Okay, I revised the formula, and I guess my intuation mislead me, maybe.
But I have a good news, too.
Maybe this formula will bring the trueth:
lim h->infinity (log(f(x+1/h)^^N)/log(f(x)^^N))^^h = lim h->infinity ((log(f(x+1/h))/log(f(x)))^^h)^^N
Do you think it can be correct?

I can't understand the formula, but one question is why have f(x) as opposed to x?  Are you trying to iterate f(x)???  

Also one assumes you are only interested in this limit for exp(-e)<=a<=exp(1/e), 
 

otherwise it is not defined.  if  exp(-e)<=a<=exp(1/e), then it is the real attracting fixed point of a^L=L.  Is this a correct understanding of your intentions?

update: 
Also, the 1/h terms in your equation drop out as h->infinity.  Then you have:
                (log(x^^N)/log(x^^N))^^h = ((log(x)/log(x))^^h)^^N

The ^^h is interpreted as the attracting fixed point.  The attracting fixed point of a^^infty as a approaches 1 also approaches a: 
fixed(a^^infity) ~=  a + (a-1)^2 + O(a-1)^3,
So the numerators equal the denominators and we are left with 1 = 1
- Sheldon
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#4
(11/28/2017, 04:15 PM)sheldonison Wrote: I can't understand the formula, but one question is why have f(x) as opposed to x?  Are you trying to iterate f(x)???  

Also one assumes you are only interested in this limit for exp(-e)<=a<=exp(1/e), 
 
otherwise it is not defined.  if  exp(-e)<=a<=exp(1/e), then it is the real attracting fixed point of a^L=L.  Is this a correct understanding of your intentions?

Well, first of all, f(x) is any function, it can be x^2+3, cos(x), x, ... etc.
No, I have not tried iterate f(x) (functionally like f^oN), because these formulas are based on the iteration of addition (multiplication, exponentiation,...etc.), but I could try it, why not?
In fact it is not the lim h->infinity a^^h, because that is equals to W(-ln(a))/-ln(a). I have been looking for lim h->0 super h-th root of a.
You know, x^^(1/y) is not super y-th root of x, right?
This is my big problem, but I believe it might be our advantage.
Xorter Unizo
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#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
Reply


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