Superfunctions in continu sum equations - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Superfunctions in continu sum equations (/showthread.php?tid=768) Superfunctions in continu sum equations - tommy1729 - 01/03/2013 Superfunctions in continu sum equations As the title says I will express Superfunctions in an equation involving the continu sum. First I need to say this post is in the context of real-differentiable functions. Also we avoid fixpoint issues or assume there is only 1 or 0 on the real line and/or 2 conjugate on the complex plane. Second I Must say that we need to find a real analytic function g(x) and it is not totally clear or proven what g(x) is and how many g(x) exist ; in other words uniqueness issues. I assume many g(x) exist and I think you will agree because there are many superfunctions. However I think it makes sense to say there are as many g(x) as analytic solutions to the superfunction. ( in the sense of a continu bijection ). Intuitively the equation seems logical to me. Lets take the exponential as example ( but this post/equation is thus in a more general setting ) Let CS ... ds be the notation for Continuum Sum with respect to s. ( inspired by integrals ... because we can express this in terms of integrals ! ) Let x > 0 consider an invertible realvalued f(x) such that f(x) = g(x) + g(exp(x)) + g(exp^[2](x)) + ... +g(exp^[oo](x)) Now assume f(x) always converges. This implies that g(x) goes to zero for large values of x. Let T(x) be the functional inverse of f(x). Now notice f(x) - f(exp(x)) = g(x) Such equation looks familar... If g(x) was GIVEN. But here it is not given yet. However if we continue ; f(x) - f(exp^[2](x)) = g(x) + g(exp(x)) f(x) - f(exp^[s](x)) = CS g(exp^[s-1](x)) d(s-1) - f(exp^[s](x)) = CS g(exp^[s-1](x)) d(s-1) - f(x) f(exp^[s](x)) = - CS g(exp^[s-1](x)) d(s-1) + f(x) exp^[s](x) = T ( - CS g(exp^[s-1](x)) d(s-1) + f(x) ) exp^[s](1) = T ( - CS g(exp^[s-1](1)) d(s-1) + f(1) ) Which seems to completely express the superfunction in terms of familiar calculus once we rewrite CS in terms of integrals and try to solve for g(x). Its almost like a differential equation. I think it might even be solvable by brute force iterations of truncations. For instance by taylor series. I was also intruiged by the idea of derivative ; we know the CP ( continuum product ) is the derivative of sexp and this seems related yet different. Thing is the CP is always(!) the derivative of ANY sexp so setting up that equation seems useless. I hope to do better with this one. But it seems tempting to differentiate (with respect to s) that last equation. Btw the equation(s) to turn a CS into an integral can be easily found , for instance on wiki or this forum. (This forum also contains a limit form that is well explained (and equivalent) but imho harder to compute.) Regards tommy1729