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Infinite tetration and superroot of infinitesimal
If we apply Mother Law of hyperoperations to infinite tetration (and with a=i^(1/i)=e^(pi/2) than:

i^(1/i)+(i^(1/i)*oo) = i^(1/i)*(oo+1)

(i^(1/i))*(i^(1/i)^oo) = i^(1/i)^(oo+1)

(i^(1/i))^(i^(1/i)#oo)) =i= i^(1/i)#(oo+1)

Tetration is the first operation where something qualitatively new happens.

so (i^(1/i))^i=i=(i^(1/i))#(oo+1)

This allows(?) to apply formula recursively, so that by denoting oo'=oo+1:

(i^(1/i))^(i^(1/i))#oo')) =i= (i^(1/i))#(oo'+1)

If this is continued infinitely, then

(i^(1/i))^((i^(1/i))#(oo+ oo-1) )) =i= (i^(1/i))#(2*oo)

This again can be repeated, so going via 3 oo, 4oo etc we get:

(i^(1/i))^((i^(1/i))#(oo*oo-1)) )) =i= (i^(1/i))#(oo*oo)=(i^(1/i))#(oo^2)

It does not end there as well, as by repeating (oo^2+1, oo^2+2...) we get to oo^2+oo, then oo^2+oo^2= 2*oo^2 then n*oo^2 then oo*oo^3= oo^3
(i^(1/i))^(i^(1/i)#(oo^3-1)) )) =i= (i^(1/i))#(oo^3) and continuing this:
(i^(1/i))^(i^(1/i)#(oo^oo-1)) )) =i= (i^(1/i))#(oo^oo) and it still goes on, so as a point to have a look at this and may be stop:

(i^(1/i))^(i^(1/i)#(oo[oo]oo-1)) )) =i= (i^(1/i))#(oo[oo]oo)

What could be done next? Is this a full circle now or is there a way to add 1 still to oo[oo]oo? or next operation is 0[0]0?

(i^(1/i))^(i^(1/i)#(oo[oo]oo)) )) =(i^(1/i))^((i^(1/i))#(0[0]oo)-1) =i= i^(1/i)#(0[0]0)?

What it shows is that starting from tetration, i^(1/i) = (e^pi/2) = 4,81...ignores all operations, returning the same i at infinite application of any hyperoperation. Except for (1/i)^i returning 1/i I do not think there are ANY (or not MANY) other numbers able to stay the same during so fast hyperoperations. The smallest deviation from i^(1/i) or (1/i)^i will lead to immedeate amplifying of it via hyperfast operations away from itself .


Ivars
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RE: Infinite tetration and superroot of infinitesimal - by Ivars - 04/05/2008, 06:42 PM

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