03/17/2008, 10:54 AM

I would suggest to move the discussion about the "proper" zeration to the zeration thread. For this discussion it makes anyway not that much difference which zeration we choose:

because the condition for [3] and [4] is anyway satisfied. If we consider exponentiation and tetration on the real numbers we must anyway impose that for [3] and for [4].

I called the values a[n]0 initial values because they, together with the rule a[n+1](x+1)=a[n](a[n+1]x), determine the operation (at least on the natural numbers for the right operand) like an initial condition would do for a differential/difference equation.

Thatswhy I also dont consider the values 1[n]x, 0[n]x here.

And the values a[n]1 follow then from a[n]0 by a[n]1=a[n-1](a[n]0).

It was already shown in the zeration thread that if we obey the law a[n+1](b+1)=a[n](a[n+1]b) then it must a[-k]b=b+1, . No room for a[-1]a=a[0]2=3 or a[0]a=a+2.

If we now backwards search for the initial condition that makes

a[-k]b=b+1, given that a[-k-1]b=b+1, then we see that it is:

a[-k]0=1. Quite in accordance with a[n]0=1 for .

The difference why [-k] is an increment but [1] is addition, though both are the super operation of the increment, lies only in the initial condition:

a[-k]0=1 vs. a[1]0=a.

The only thing is that a[-k]0=1 does not fit in my rules for chosing the initial value.

GFR Wrote:a[0]0 = a + 1, for a >< 0

a[1]0 = a

a[2]0 = 0

a[3]0 = 1, for a >< 0

a[4]0 = 1, for a >< 0

because the condition for [3] and [4] is anyway satisfied. If we consider exponentiation and tetration on the real numbers we must anyway impose that for [3] and for [4].

I called the values a[n]0 initial values because they, together with the rule a[n+1](x+1)=a[n](a[n+1]x), determine the operation (at least on the natural numbers for the right operand) like an initial condition would do for a differential/difference equation.

Thatswhy I also dont consider the values 1[n]x, 0[n]x here.

And the values a[n]1 follow then from a[n]0 by a[n]1=a[n-1](a[n]0).

Quote:So, no possibility to have: a[-1]a = a[0]2 ?

Schade ...! Aber, warum nicht.

It was already shown in the zeration thread that if we obey the law a[n+1](b+1)=a[n](a[n+1]b) then it must a[-k]b=b+1, . No room for a[-1]a=a[0]2=3 or a[0]a=a+2.

If we now backwards search for the initial condition that makes

a[-k]b=b+1, given that a[-k-1]b=b+1, then we see that it is:

a[-k]0=1. Quite in accordance with a[n]0=1 for .

The difference why [-k] is an increment but [1] is addition, though both are the super operation of the increment, lies only in the initial condition:

a[-k]0=1 vs. a[1]0=a.

The only thing is that a[-k]0=1 does not fit in my rules for chosing the initial value.