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 \( a\neq 0 \) for [3] and [4] is anyway satisfied. If we consider exponentiation and tetration on the real numbers we must anyway impose that \( a>0 \) for [3] and \( a>1 \) 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, \( k\ge 0 \). 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 \( n\ge 3 \).
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 \( a\neq 0 \) for [3] and [4] is anyway satisfied. If we consider exponentiation and tetration on the real numbers we must anyway impose that \( a>0 \) for [3] and \( a>1 \) 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, \( k\ge 0 \). 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 \( n\ge 3 \).
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.
