Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Superfunctions in continu sum equations
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.)



Possibly Related Threads...
Thread Author Replies Views Last Post
  Moving between Abel's and Schroeder's Functional Equations Daniel 1 706 01/16/2020, 10:08 PM
Last Post: sheldonison
  Taylor polynomial. System of equations for the coefficients. marraco 17 19,081 08/23/2016, 11:25 AM
Last Post: Gottfried
  Finding continu iteration cycles. tommy1729 2 2,939 03/05/2016, 10:12 PM
Last Post: tommy1729
  Totient equations tommy1729 0 2,048 05/08/2015, 11:20 PM
Last Post: tommy1729
  Bundle equations for bases > 2 tommy1729 0 2,086 04/18/2015, 12:24 PM
Last Post: tommy1729
  Grzegorczyk hierarchy vs Iterated differential equations? MphLee 0 2,307 01/03/2015, 11:02 PM
Last Post: MphLee
  A system of functional equations for slog(x) ? tommy1729 3 5,147 07/28/2014, 09:16 PM
Last Post: tommy1729
  partial invariant equations ? tommy1729 0 2,090 03/16/2013, 12:32 AM
Last Post: tommy1729
  tetration base conversion, and sexp/slog limit equations sheldonison 44 62,768 02/27/2013, 07:05 PM
Last Post: sheldonison
  superfunctions of eta converge towards each other sheldonison 13 18,283 12/05/2012, 12:22 AM
Last Post: sheldonison

Users browsing this thread: 1 Guest(s)