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 Holomorphic semi operators, using the beta method tommy1729 Ultimate Fellow     Posts: 1,625 Threads: 364 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: 767 Threads: 104 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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