Holomorphic semi operators, using the beta method tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 06/12/2022, 11:13 PM (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 y is in a way the super of x y. I am not convinced that they can be united with x 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 $x f(x) = x$... At least, not in the purview of this solution. By construction I'm assuming $x > e$ and $y > e$, and there's no value $x y = x$ 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 $x y$ 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 $x f(x) = x$... ". For me that is cruxial to the actual hyperoperator.  regards tommy1729 JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 06/13/2022, 08:33 PM (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 y is in a way the super of x y. I am not convinced that they can be united with x 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 $x f(x) = x$... At least, not in the purview of this solution. By construction I'm assuming $x > e$ and $y > e$, and there's no value $x y = x$ 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 $x y$ 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 $x f(x) = x$... ". 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 $x y$ 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. « Next Oldest | Next Newest »

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