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 extension of the Ackermann function to operators less than addition JmsNxn Long Time Fellow Posts: 628 Threads: 102 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? JmsNxn Long Time Fellow Posts: 628 Threads: 102 Joined: Dec 2010 11/06/2011, 04:56 PM (This post was last modified: 11/06/2011, 08:10 PM by JmsNxn.) furthermore, if we define the identity function as: $a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a$ since S(1) = 0, and S(2) = 1. If we want S(x) to be analytic and continuous (which we do), by the mean value theorem there is an operator who's identity is 0.5. Let's call this q. $1 < q < 2; q \in \R$ $a\,\,\bigtriangleup_q\,\,0.5 = a$ doing some simple manipulations and we get $a\,\,\bigtriangleup_{q-1}\,\,a = a\,\,\bigtriangleup_q\,\,1.5$ this is by the first axiom. Given that. We can let a = 2, and get the contradictory result: $2\,\,\bigtriangleup_{q-1}\,\,2 = 2\,\,\bigtriangleup_{q}\,\,1.5$ if we want: $2\,\,\bigtriangleup_\sigma\,\,2 = 4$ we arrive at a contradiction or a result that is definitely not desired. Either $2\,\,\bigtriangleup_{\sigma}\,\,2 \neq 4$ or $2\,\,\bigtriangleup_{q}\,\, 1.5 = 2\,\,\bigtriangleup_{q}\,\,2 = 4$ this would imply our function at q has the following behaviour (if we simply reapply the first axiom recusively): $r \ge 1; r \in \mathbb{Z}$ $2\,\,\bigtriangleup_{q}\,\,r + 0.5 = 2\,\,\bigtriangleup_{q}\,\, (r + 1)$ this result can be extended to any value of sigma: $2\,\,\bigtriangleup_{\sigma}\,\,S(\sigma) = 2$ $2\,\,\bigtriangleup_{\sigma - 1}\,\,2 = 2\,\,\bigtriangleup_{\sigma}\,\,1 + S(\sigma)\,\, = 4 = 2\,\,\bigtriangleup_{\sigma}\,\,2$ giving: $2\,\,\bigtriangleup_{\sigma}\,\,r + S(\sigma) = 2\,\,\bigtriangleup_{\sigma}\,\, r + 1$ therefore we have three options 1. the identity function is not analytic 2. $2\,\,\bigtriangleup_\sigma\,\,2 \neq 4$ 3. or in general $f(x) = 2\,\,\bigtriangleup_{\sigma}\,\,x$ is not a smooth monotonic increasing function unless sigma is an integer. If we let 3 then we also get the result that operators extended less than addition are impossible. In this we imply $2\,\,\bigtriangleup_0\,\,2 = 4$ but we already know $2\,\,\bigtriangleup_0\,\,2 = 2 + 1 = 3$ The first option, that the identity function be not analytic isn't desirable at all. Furthermore, it's more than it just not be analytic, but it implies the only results possible are 0 and 1. And that it only takes 0 at addition and everywhere else we have 1. In conclusion, if we want to have the Ackermann function extended to the complex domain everywhere and $f(x) = 2\,\,\bigtriangleup_\sigma\,\,x$ be monotonically increasing everywhere we're going to have to lose the aesthetic property: $2\,\,\bigtriangleup_\sigma\,\,2 = 4$ JmsNxn Long Time Fellow Posts: 628 Threads: 102 Joined: Dec 2010 11/06/2011, 08:06 PM I realize now we have to create a second axiom in order that the series of operators become the true Ackermann function. Consider the possibility that: $a\,\,\bigtriangleup_2\,\,2=a$ we still have the result that $a + (a\,\,\bigtriangleup_2\,\,2) = a\,\, \bigtriangleup_2\,\, (2 + 1) = a\,\,\bigtriangleup_2\,\,3$ the only difference is that $S(2) = 2$ and in return we get $a\,\,\bigtriangleup_2\,\,b = a \cdot (b-1)$ So in order to get the true Ackermann function we must make the second assertion: $n\,\,\in\,\,\mathbb{Z};n \ge 2$ $S(n) = 1$ This is actually the equivalent to the iteration axiom: $a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_\sigma\,\,(a_2\,\,\bigtriangleup_\sigma\,\,(a_3\,\, \bigtriangleup_\sigma \,\,...(a_{n-1}\,\,\bigtriangleup_\sigma\,\,a_n)$ though only true for integer values of sigma greater than or equal to two. Therefore we define the Ackermann function from axioms: $\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b$ such that $a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\sigma + 1}\,\,b) = a\,\,\bigtriangleup_\sigma\,\,(b+1)$ $a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a$ $n\,\,\in\,\,\mathbb{Z}\,\,;\,\,n \ge 2$ $S(n) = 1$ $a\,\,\bigtriangleup_1\,\,b = a + b$ Does anyone see any modifications necessary? « Next Oldest | Next Newest »

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