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Revitalizing an old idea : estimated fake sexp'(x) = F3(x)
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Revitalizing an old idea : estimated fake sexp'(x).

It's an old idea but I want to put some attention to it again.

Perhaps with our improved skills and understanding this might lead us somewhere.

As most members and frequent readers know so-called fake function theory was developed by myself and sheldon to 

1) create a real entire function

2) that is an asymptotic of a given function for positive reals

3) that has all its taylor or maclauren coefficients positive or at least non-negative.

There are some extra conditions such as strictly rising for positive real x imput for the given function and such.
But basicly that is what fake function theory is about.

We call such created functions - or the attempts - fake functions.

We also considered the cases were the given function already satisfied 1 -- 3 and how our algorithms created " fake ones ".

Many variants occured , such as replacing sums with analogue integrals and stuff but that is not important here.

We got good results for fake exp^[1/2](x).
 
So far the intro to the fake function theory part.

We also discussed base change constants here.

And we discussed the " expontential factorial " function ; the analogue of tetration like 2^3^4^... , the analogue of the factorial 1*2*3*4*... .

We are also familiar with telescoping sums.

And we even discussed converging infinite sums (over natural index n )  of the n th iteration of a function.

We also know the derivative of exp^[n](x) = exp^[n](x) * exp^[n-1](x) * exp^[n-2](x)*...
And how close this is related to the derivative of sexp(x) ; d sexp(x)/dx = sexp(x) * d sexp(x-1)/dx.


Is this all related to an old idea of myself and partially others ??

Yes certainly.

Lets estimate the derivative of sexp(x) but without using fake function theory tools directly.


Impossible ?

Well far from , if you accept brute estimates.

So we want something faster than any fixed amount of exp iterations.

But slower than sexp(x^2) or sexp(x)^2 or so.

The exact speed of the function is something to investigate and discuss.

But a logical attempt is this :

This function is entire ...

F3(x) = (1 + exp(x)/2^3) (1 + 2^3 exp^[2](x)/ 2^3^4 ) (1 + 2^3^4 exp^[3](x)/ 2^3^4^5 ) ...

and it converges fast.

It is also faster than any exp(x)^[k](x) for fixed k.

And all the coefficients are positive.

Also notice the pseudotelescoping product.

This leads to the asymptotic conjecture :

Using big-O notation :


For 1 < x 

integral from 0 to x   F3(t) dt = F4(x) =  O ( fake sexp(x + c) ).

For some integration constant , and some constant c.

How about that ?

***

Next idea 



lim v to oo ;

ln^[v] F4(x + v) = ??

Which also looks very familiar.


Regards

tommy1729

Tom Marcel Raes
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