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 Generalized recursive operators bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/07/2008, 06:58 PM andydude Wrote:I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:  \begin{align} \lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\ \lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\ \lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\ \lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6 \end{blign} ... I suppose you could see this from the integer versions of these operators, but I think the continuous (or if not continuous, mostly real-valued) versions make it easier to see. First, Andrew, these are really fascinating findings. andydude Wrote:$\lim_{N\rightarrow\infty} (a \begin{tabular}{|c|}\hline N \\\hline\end{tabular} b) = b+1$ for all $a>1, b<0$ meaning, in the limit, all hyper-operators return to the successor operation, like the circle of life... I dare a proof by induction which only needs the integer operations. Proposition: If we have a sequence of operations [n] on the natural numbers (>0) that satisfy b[n+1]1=b, b[n+1](x+1)=b[n](b[n+1]x) for n$\ge$ 1, then we can extend the domain of the right operand of [n] to integer k with k$\ge$ 3-n and the only way to do so still satisfying the above conditions and injectivity of the functions f(x)=b[n]x is by b[n](-k)=-k+1 for 0$\le$ k$\le$ n-3. Proof: We prove by induction over k that b[n](-k)=-k+1 for all n$\ge$ k+3. Induction Start k=0: b[n]1=b=b[n+1]1=b[n](b[n+1]0), for n$\ge$2, by injectivity follows 1=b[n+1]0 for n+1$\ge$3=0+3 Induction Step k=k+1: by induction assumption for n$\ge$k+3 : b[n](-k)=-k+1=b[n+1](-k)=b[n](b[n+1]-(k+1)) by injectivity: -k = b[n+1]-(k+1) which is the induction assertion: -(k+1)+1=b[n+1]-(k+1) for n+1$\ge$k+1+3 « Next Oldest | Next Newest »

 Messages In This Thread Generalized recursive operators - by Whiteknox - 11/23/2007, 06:42 AM RE: Generalized recursive operators - by bo198214 - 11/23/2007, 08:41 AM RE: Generalized recursive operators - by Whiteknox - 11/23/2007, 03:57 PM RE: Generalized recursive operators - by andydude - 11/25/2007, 01:02 AM RE: Generalized recursive operators - by andydude - 11/29/2007, 04:45 AM RE: Generalized recursive operators - by andydude - 11/29/2007, 05:55 AM RE: Generalized recursive operators - by andydude - 11/29/2007, 06:20 AM RE: Generalized recursive operators - by Gottfried - 11/29/2007, 08:14 AM RE: Generalized recursive operators - by andydude - 11/30/2007, 06:12 PM RE: Generalized recursive operators - by andydude - 11/30/2007, 09:18 PM RE: Generalized recursive operators - by bo198214 - 03/07/2008, 06:58 PM RE: Generalized recursive operators - by Ivars - 02/02/2008, 10:11 PM RE: Generalized recursive operators - by Whiteknox - 12/01/2007, 04:59 AM RE: Generalized recursive operators - by Ivars - 02/03/2008, 10:41 AM RE: Generalized recursive operators - by andydude - 02/11/2008, 09:47 PM RE: Generalized recursive operators - by Ivars - 02/14/2008, 06:05 PM RE: Generalized recursive operators - by GFR - 02/03/2008, 04:12 PM RE: Generalized recursive operators - by Ivars - 02/03/2008, 08:48 PM RE: Generalized recursive operators - by GFR - 02/06/2008, 02:44 PM RE: Generalized recursive operators - by Ivars - 02/06/2008, 02:56 PM RE: Generalized recursive operators - by Ivars - 02/06/2008, 03:43 PM RE: Generalized recursive operators - by GFR - 03/10/2008, 09:53 PM RE: Generalized recursive operators - by GFR - 03/11/2008, 10:24 AM RE: Generalized recursive operators - by bo198214 - 03/11/2008, 10:53 AM RE: Generalized recursive operators - by GFR - 03/12/2008, 12:13 AM RE: Generalized recursive operators - by GFR - 03/13/2008, 06:41 PM RE: Generalized recursive operators - by Stan - 04/04/2011, 11:52 PM

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