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 the infinite operator, is there any research into this? JmsNxn Long Time Fellow Posts: 312 Threads: 69 Joined: Dec 2010 07/09/2011, 08:45 PM (This post was last modified: 07/09/2011, 08:50 PM by JmsNxn.) The infinite operator is simply defined as: $a\,\,\bigtriangleup_{0}\,\,b = a + b\\ \\a\,\,\bigtriangleup_{1}\,\,b = a\cdot b\\ \\a\,\,\bigtriangleup_{2}\,\,b = a^b\\ ...\\ \lim_{n\to\infty} a\,\,\bigtriangleup_{n}\,\,b = a\,\,\bigtriangleup_{\infty}\,\,b$ I know that for negative integers values of b (and zero) it returns to successorship, i.e $a\,\, \bigtriangleup_{\infty}\,\, 0 = 1\\ \\a\,\, \bigtriangleup_{\infty}\,\, -1 = 0\\ \\a\,\, \bigtriangleup_{\infty}\,\, -2 = -1\\ ...$ but we also have an interesting formula, since the euler constants and the eta constants converge to 2 and 4, and since infinite iteration of eta constant x with hyper operator x results in euler constant x we have this very odd formula: $2\,\, \bigtriangleup_{\infty} \,\,(2\,\, \bigtriangleup_{\infty} \,\,(2\,\, \bigtriangleup_{\infty} \,\,(2... \,\, = 4$ but we also know that: $2\,\,\bigtriangleup_{\infty}\,\,2 = 4$ and that $2 = 4\,\,\bigtriangledown_{\infty}\,\,4 = 2$ if this is $\bigtriangledown_{\infty}$ is the root inverse of the infinite operator. we find that $f(x) = 2 \bigtriangleup_{\infty} x$ equals 4 at both 2 and 4. This means, if we want the infinite operator to be analytic and continuous, we have at least one zero in the second derivative. But we'd probably want two so that f(x) is increasing towards infinity. Also, I was wondering, since it's completely possible to take the superfunction of the infinite operator, how would we denote that? Since it can't be $\infty + 1$ should we make it something like $\infty + k$ where k is just another unit but denotes a seperate branch of superfunctions on top of the infinite operator? $g(x) =2\,\,\bigtriangleup_{\infty+k}\,\,x$ $g(x) = 4$ for all $x\ge2$ This means we have a flatline for the infinite operators superfunction. Interesting. tommy1729 Ultimate Fellow Posts: 1,386 Threads: 339 Joined: Feb 2009 07/13/2011, 12:14 PM in general we consider analytic as intresting , rather than local flatlines ... you could consider your flatline as a fixpoint and that takes away some magic ... tommy1729 JmsNxn Long Time Fellow Posts: 312 Threads: 69 Joined: Dec 2010 07/15/2011, 02:23 AM (07/13/2011, 12:14 PM)tommy1729 Wrote: in general we consider analytic as intresting , rather than local flatlines ... you could consider your flatline as a fixpoint and that takes away some magic ... tommy1729 Oh, so you're proposing $g(x) =2\,\,\bigtriangleup_{\infty+k}\,\,x$ $g(n) = 4$ for all $n\ge2 \,\,n \in N$ Yes, you are quite right, that is much more interesting. « Next Oldest | Next Newest »

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