01/03/2013, 12:02 AM

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

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