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:
\( 2\,\,\bigtriangleup_\sigma\,\,2 = 4 \)
\( a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a \)
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:
\( 2\,\,\bigtriangleup_\sigma\,\,2 = 4 \)