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 Initial values for hyper operations bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 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: 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. « Next Oldest | Next Newest »

 Messages In This Thread Initial values for hyper operations - by bo198214 - 03/14/2008, 10:44 AM RE: Initial values for hyper operations - by GFR - 03/19/2008, 11:03 AM

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