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 the inverse ackerman functions JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 04/14/2011, 08:48 PM (This post was last modified: 04/14/2011, 08:54 PM by JmsNxn.) Every one here has heard of the ackermann function; we'll define it as: $\{x,\, y,\, u\, \epsilon\, R^{+}\}$ $A(x,\, y,\, u)\, =\, x\, \{u\}\, y$ I'm interested in itss inverses though. There would be three inverses therefore, the first one with relation to x: $R(x,\, y,\, u)\, =\, x\, \}u\{\, y$ which means R for roots, and }u{ is taken to mean: }0{ is subtraction, }1{ division, }2{ roots, }3{ super roots, etc etc and then are next inverse is taken with relation to y $L(x,\, y,\, u)\, =\, u;\log_x(y)$ where 0;log_x(y) = y-x, 1;log_x(y) = y/x, and 2;log_x(y) = $\log_x(y)$ and now, the final inverse, and definitely the most interesting inverse is the inverse taken with relation to u $I(x, y, u) = ?$ if $A(x, y, I(x, y, u)) = u)$ this question is the equivalent of asking $3\,\{u\}\, 4\, =\, 9$ what is u? it's the act of rearranging an equation to solve for u. It might even seem intuitively impossible. I've only found one solution to this problem... and that's by using logarithmic semi operators, though I is only defined over domain [0, 2] Does anyone know if anybody else has ever looked into this inverse function? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 04/14/2011, 10:57 PM When there is a (local) unique inverse , you might use the so-called " series reversion " for its taylor series at a fixpoint. Or you might try to express ackermann in terms of sexp slog etc but im not sure its possible. maybe this has occured in older threads. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 04/14/2011, 11:04 PM (This post was last modified: 04/14/2011, 11:06 PM by JmsNxn.) (04/14/2011, 10:57 PM)tommy1729 Wrote: When there is a (local) unique inverse , you might use the so-called " series reversion " for its taylor series at a fixpoint. Or you might try to express ackermann in terms of sexp slog etc but im not sure its possible. maybe this has occured in older threads. you can express the domain [0, 2] using sexp slog algos, it's piece wise though. but I'll show you what happens; 0 <= q <= 1 $x\, \{q\}\, y\, =\, -q:ln(q:ln(x)\, +\, q:ln(y))\, = sexp(slog(sexp(slog(x)-q)+sexp(slog(y)-q))+q)$ solve for q, it's practically impossible; same deal for [1, 2] Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 09/18/2016, 11:02 AM (04/14/2011, 08:48 PM)JmsNxn Wrote: $3\,\{u\}\, 4\, =\, 9$ what is u? Well, James, according to PARI/gp u is about 1.5737, because 3[1.5737]4 = ~9 But I do not know why. I need a taylor-series formula or a recursive formula to get know why. Xorter Unizo « Next Oldest | Next Newest »

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