Hey Konstantin, I am pleased to see you writing your first article here.
I really honor this, knowing that you nearly had no previous knowledge of the english language.
Ok, that was perhaps a wrong usage of the word "postulate". I meant that this evaluation rule does not follow from the initial rule, but is rather an interesting suggestion for an operation that satisfies the initial rule.
But to be compatible with the definition of tetration (and also Ackermann function)
a^a = a[4]2
a^(a^a) = a[4]3
a^(a^(a^a)) = a[4]4
(as a note to Gianfranco: I really like the ASCII notation a [n] b, which was I think introduced by you)
the bracketing must be to the right. So
ao(ao(aoa))=a+4
ao(aoa)=a+3
aoa=a+2
a = a+1 ???
One notices that the last equation, though being the logical progression of the previous lines, is not possible; while for all higher operation we have:
a+a=a*2
a=a*1
a*a=a^2
a=a^1
a^a=a[4]2
a=a[4]1
No, thats not a strict consequence. The following is a strict consequence:
We start with the axiom
ao(ao(ao......))) = a+n for integer n>1 (and a being real or complex)
then we conclude
ao(a+n)=a+n+1
If we assume a and b to be an integer, then we can write for b>a+1
aob=ao(a+(b-a))=a+(b-a)+1=b+1.
So what to do then about real number arguments? For tetration we have a similar situation:
(1) a^(a[4]n)=a[4](n+1)
(2) a[4]1=a
and we just demanded that the above rule (1) also shall be valid for real (or even complex) n, and that the function f(x)=b[4]x shall be analytic (or at least continuous). For Zeration we probably would look for an operation (on the reals) that satisfies
(1') ao(a+x)=a+x+1 for each real x>1
(2') aoa=a+2
So we dont need any continuity or analyticity to see that
aob=b+1 for b>a+1 (a,b real). If we would now demand that f(x)=aox is analytic, then it would follow that f(x) must be x+1 for all complex x (because an analytic function is already determined by being defined on a set which is dense at one point, and here it is even defined for all x>a+1). So this would contradict f(x)=x+2 for a=x. The next try would be to keep f(x) at least continuous or infinitely differentiable, but I see no canonic definition that one would chose for f(x)=aox on the yet undefined interval (a,a+1].
Can you btw show me, how you derived the commutativity of zeration?
I really honor this, knowing that you nearly had no previous knowledge of the english language.
KAR Wrote:bo198214 Wrote:But see Gianfranco, the postulate
aob = max{a,b}+1 for a!=b
aob = a+2=b+2 for a=b
does not follow from your initial rule
aoa=a+2
It not a postulate!
Ok, that was perhaps a wrong usage of the word "postulate". I meant that this evaluation rule does not follow from the initial rule, but is rather an interesting suggestion for an operation that satisfies the initial rule.
Quote:aoa=a+2
aoaoa=a+3=(aoa)oa=(a+2)oa, a+2>a
aoaoaoa=a+4=((aoa)oa)oa=(a+3)oa, a+3>a
......
ao...oa=a+n, a+(n-1)>a if n>1.
But to be compatible with the definition of tetration (and also Ackermann function)
a^a = a[4]2
a^(a^a) = a[4]3
a^(a^(a^a)) = a[4]4
(as a note to Gianfranco: I really like the ASCII notation a [n] b, which was I think introduced by you)
the bracketing must be to the right. So
ao(ao(aoa))=a+4
ao(aoa)=a+3
aoa=a+2
a = a+1 ???
One notices that the last equation, though being the logical progression of the previous lines, is not possible; while for all higher operation we have:
a+a=a*2
a=a*1
a*a=a^2
a=a^1
a^a=a[4]2
a=a[4]1
Quote:From definition Zeration for "a" it is possible to write for b>a:
boa=b+1 and boa=a+2=b+2 if a=b.
This base definition
The rule of evaluation Zeration is a corollary from extended base definition, but not a postulate!
No, thats not a strict consequence. The following is a strict consequence:
We start with the axiom
ao(ao(ao......))) = a+n for integer n>1 (and a being real or complex)
then we conclude
ao(a+n)=a+n+1
If we assume a and b to be an integer, then we can write for b>a+1
aob=ao(a+(b-a))=a+(b-a)+1=b+1.
So what to do then about real number arguments? For tetration we have a similar situation:
(1) a^(a[4]n)=a[4](n+1)
(2) a[4]1=a
and we just demanded that the above rule (1) also shall be valid for real (or even complex) n, and that the function f(x)=b[4]x shall be analytic (or at least continuous). For Zeration we probably would look for an operation (on the reals) that satisfies
(1') ao(a+x)=a+x+1 for each real x>1
(2') aoa=a+2
So we dont need any continuity or analyticity to see that
aob=b+1 for b>a+1 (a,b real). If we would now demand that f(x)=aox is analytic, then it would follow that f(x) must be x+1 for all complex x (because an analytic function is already determined by being defined on a set which is dense at one point, and here it is even defined for all x>a+1). So this would contradict f(x)=x+2 for a=x. The next try would be to keep f(x) at least continuous or infinitely differentiable, but I see no canonic definition that one would chose for f(x)=aox on the yet undefined interval (a,a+1].
Can you btw show me, how you derived the commutativity of zeration?