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 extension of the Ackermann function to operators less than addition JmsNxn Long Time Fellow Posts: 568 Threads: 95 Joined: Dec 2010 11/06/2011, 01:31 AM (This post was last modified: 11/06/2011, 01:35 AM by JmsNxn.) Well the result is surprisingly simple, and is derived from a single defining axiom. I feel if there are any qualms with the proof it comes from the axiom itself. we assume for now that all variables can be extended to the complex domain $a, b, \sigma \in \text{C}$ and thus, the Ackermann series of operators are defined by the single property: $\text{(I):}\,\,a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\sigma + 1}\,\,b) = a\,\,\bigtriangleup_{\sigma + 1} (b+1)$ starting the Ackermann function with addition at 1 gives us: $a\,\,\bigtriangleup_1\,\,b = a + b$ therefore, if we plug in from our defining axiom (I) and try to solve for when sigma is zero we get: $a\,\,\bigtriangleup_0\,\,(a\,\,\bigtriangleup_1\,\,b) = a\,\,\bigtriangleup_1\,\, (b + 1)$ $\text{(II):} a\,\,\bigtriangleup_0\,\,(a + b) = a + b + 1$ with this, we try to solve for the more general: $a\,\,\bigtriangleup_0\,\,b = a\,\,\bigtriangleup_0\,\,(a + (b - a))$ which by (II) gives $\text{(III):}\,\,a\,\,\bigtriangleup_0\,\,b = a + b - a + 1 = b + 1$ this was derived from only a single defining axiom and no other assumptions were made. But that's not it, we can continue this sequence and try to solve for when sigma is negative one. Again, we'll start with the defining axiom: $a\,\,\bigtriangleup_{-1}\,\,(a\,\,\bigtriangleup_{0}\,\,b) = a\,\,\bigtriangleup_{0}\,\,(b+1)$ of course, by (III) we have the simple formula: $a\,\,\bigtriangleup_{-1}\,\,(b+1) = b+2$ which of course, gives the fantastic equation: $a\,\,\bigtriangleup_{-1}\,\,b = b + 1$ now, by induction, we can make the complete argument: $k \in \mathbb{Z}; k \le 0$ $a\,\,\bigtriangleup_k\,\,b = b + 1$ Or basically, we have the result that every operator less than addition which is equal to an integer is the equivalent to successorship. Q.E.D. Now if we add the additional property that sigma plus one be an iteration count of sigma, we run into many problems. And I will shortly argue here: The second axiom which need not be in play is the axiom of iteration given as $n \in \mathbb{Z}; n > 0$ $\text{(IV):}\,\,a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_{\sigma}\,\,(a_2\,\,\bigtriangleup_{\sigma} \,\, (a_3 \,\, \bigtriangleup_{\sigma}\,\,...\,\,(a_{n-1}\,\, \bigtriangleup_\sigma \,\,a_n)$ with this axiom we instantly have a contradiction with the first axiom when we extend operators less than addition: by (III) $a\,\,\bigtriangleup_{0}\,\,a = a + 1$ by(IV) $a\,\,\bigtriangleup_{0}\,\,a = a + 2$ therefore if we allow the axiom of iteration we cannot extend the Ackermann function to operators less than addition. Furthermore, if we allow the axiom of iteration we also have the cheap result: $x \in \mathbb{R}; x \ge 2$ $a\,\,\bigtriangleup_x\,\,1 = a$ this causes the function $f(\sigma) = a\,\,\bigtriangleup_\sigma\,\,1$ to no longer possibly be analytic. we can however accept a modified axiom of iteration which is written as: $n \in \mathbb{Z}; n > 1$ $\text{(V):}\,\,a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_{\sigma}\,\,(a_2\,\,\bigtriangleup_{\sigma} \,\, (a_3 \,\, \bigtriangleup_{\sigma}\,\,...\,\,(a_{n-1}\,\, \bigtriangleup_\sigma \,\,a_n)$ this allows for different identities at non-integer values of sigma, however, this still disallows sigma to be extended to values less than addition. So, I leave it open ended, hoping someone else has some comment. Do we use axioms (I) and (IV), or (I) and (V), or, as I prefer, (I) alone? « Next Oldest | Next Newest »

 Messages In This Thread extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 01:31 AM RE: extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 04:56 PM RE: extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 08:06 PM

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