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(06/12/2022, 10:05 PM)JmsNxn Wrote: (06/12/2022, 01:50 PM)tommy1729 Wrote: Dear James your 2 last posts are inspirational but I cannot accept them at the moment - maybe later -.

Here is why : it is not shown to be analytic in x,s,y and at the same time satisfy the superfunction property where x <s+1> y is in a way the super of x <s> y.

I am not convinced that they can be united with x <s> f(x) = x for some analytic f(x) and that that f(x) is almost free to choose.

( the integers have to match with the defintions of x<0>y and x<1>y etc ofcourse * unless you let those be free to choose as well )

Not trying to be annoying

regards

tommy1729

You have misinterpreted the notation.

There is no solution \(x <s> f(x) = x\)... At least, not in the purview of this solution. By construction I'm assuming \(x > e\) and \(y > e\), and there's no value \(x <s> y = x\) for these values.

You asked if there is a value \(\varphi\) such that \(x [s]_{\varphi} y = x\). Which there is. This is just:

\[

x[s]_{\varphi} y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\

\]

By which the answer is \(\varphi = - y\) for the equation you asked to solve. This would not happen when we solve for the actual semi-operator, as there is no solution...

I think you're mixing things up. Remember that \(x [s] y\) is absolutely analytic in all variables. It just equals \(\exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y\right)\)--which is absolutely analytic in all variables... The question is whether \(x <s> y\) can be found in a neighborhood of \(x [s] y\) and be analytic. I understand I haven't shown this yet. But you're mixing things up.

I know this thread is a mess, and very disorganized. I plan on writing up much more fluidly the observations, but I'm waiting until I have something concrete.

oh yes.

then we agree.

but then the question becomes prove that there is NO solution : quote : " There is no solution \(x <s> f(x) = x\)... ".

For me that is cruxial to the actual hyperoperator.

regards

tommy1729

Posts: 1,214

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(06/12/2022, 11:13 PM)tommy1729 Wrote: (06/12/2022, 10:05 PM)JmsNxn Wrote: (06/12/2022, 01:50 PM)tommy1729 Wrote: Dear James your 2 last posts are inspirational but I cannot accept them at the moment - maybe later -.

Here is why : it is not shown to be analytic in x,s,y and at the same time satisfy the superfunction property where x <s+1> y is in a way the super of x <s> y.

I am not convinced that they can be united with x <s> f(x) = x for some analytic f(x) and that that f(x) is almost free to choose.

( the integers have to match with the defintions of x<0>y and x<1>y etc ofcourse * unless you let those be free to choose as well )

Not trying to be annoying

regards

tommy1729

You have misinterpreted the notation.

There is no solution \(x <s> f(x) = x\)... At least, not in the purview of this solution. By construction I'm assuming \(x > e\) and \(y > e\), and there's no value \(x <s> y = x\) for these values.

You asked if there is a value \(\varphi\) such that \(x [s]_{\varphi} y = x\). Which there is. This is just:

\[

x[s]_{\varphi} y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\

\]

By which the answer is \(\varphi = - y\) for the equation you asked to solve. This would not happen when we solve for the actual semi-operator, as there is no solution...

I think you're mixing things up. Remember that \(x [s] y\) is absolutely analytic in all variables. It just equals \(\exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y\right)\)--which is absolutely analytic in all variables... The question is whether \(x <s> y\) can be found in a neighborhood of \(x [s] y\) and be analytic. I understand I haven't shown this yet. But you're mixing things up.

I know this thread is a mess, and very disorganized. I plan on writing up much more fluidly the observations, but I'm waiting until I have something concrete.

oh yes.

then we agree.

but then the question becomes prove that there is NO solution : quote : " There is no solution \(x <s> f(x) = x\)... ".

For me that is cruxial to the actual hyperoperator.

regards

tommy1729

Oh yes, this would follow from monotone in \(s\). I'd need to show that \(x <s> y\) is monotone in \(s\) then we're good. Because there's no solution at \(s=0\) or at \(s=1\) and at \(s=2\). That's because this looks like \(x +y = x\) and \(x\cdot y = x\) and \(x^y = x\), while \(x,y > e\). But yes, not a proof, need monotone.