(06/14/2022, 12:55 AM)JmsNxn Wrote: No problem tommy. The tex is failing because you are writing Exp instead of \exp, not sure why it's formatting like that.

I apologize I thought the result was obvious.

The function:

$$

h(z) = \sum_{n=0}^\infty \frac{a_n}{W'(n)(z-n)}\\

$$

Has a simple pole at \(z=n\). Similarly, \(W(z)\) has a simple zero at \(z=n\). Thereby:

$$

W(z+n)h(z+n) = a_n + Qz\\

$$

I mean this result from complex analysis.

$$

\frac{W(z+n) a_n}{W'(n)z} = a_n \,\,\text{as}\,\,z\to 0\\

$$

All the other coefficients of the sum defining \(h\) disappear because they are finite and \(W(n)\) is zero...

So,

$$

h(z+n)W(z+n) = 0 + 0 + 0....\frac{W(z+n) a_n}{W'(n)z}....+0+0\\\

$$

I'm sorry, I refuse to believe you've never seen this before Tommy! This is a common consequence of Weierstrass/Mittlag Leffler theory.

We are setting all the other terms of the sum to zero, and the only ones which evaluate to a value are those with simple poles....

Come On, you know this!!!!

Yeah that is trivial by using l'hopital if the sequence $$ a_n $$ converges.

let f(n) = 2^^n.

and let n < x < n+1.

Now we want the function f(x) to still converge.

And we want f(n) < f(x) < f(n+1).

Now letting W ' (n) be equal to 2^^n makes it converge but feels a bit weird ...

I mean that feels like interpolation of the constant function 1 then.

I might as well interpolate 1/ 2^^n and then take the inverse again.

Or interpolate ( 2^^n + 2 )/(2^^n) and compute 2^^x from that.

Maybe Im being silly.

But it does not feel like " the " interpolation.

Let me reconsider with a W that is entire and has zero's at all integers without multiplicity.

W(z) = exp(h(z)) sin(2pi z)

W ' (z) = exp(h(z)) sin(2 pi z) h ' (z) + cos( 2pi z) exp(h(z)) 2 pi.

thus

W ' (n) = 2 pi exp(h(n)).

So - if am not mistaken - to make the sequence a_n = 2^^n converge we need W ' (n) to grow at a rate about 2^^ n as well.

So we set up the equation :

W ' (n) = 2 pi exp (h(n)) = 2^^n

and we notice that h(n) must grow at about 2^^(n-1).

Thus basicly the interpolation problem shifted from finding f(x) s.t. f(n) = 2^^n to finding an interpolation function h(x) s.t. h(n) = ln(2^^n / 2 pi).

This seems a self-reference fractal like problem.

because by the method above

finding an interpolation function h(x) s.t. h(n) = ln(2^^n / 2 pi) requires finding an interpolation U(x) s.t. U(n) = ln( ln(2^^n / 2 pi) / 2 pi ).

That is the fractal/convergeance issue ( assuming i made no silly error ).

than there is still the case :

And we want f(n) < f(x) < f(n+1).

Funny thing ; fake function theory gives an entire function approximation of 2^^n without fractal or convergeance issues. Although it requires a tetration solution ofcourse ...

I could use fake function theory to approximate h(x) as ln( 2^^x / 2pi ) but that would nothing new , approximation and perhaps " cheating ".

Maybe Im wrong or confused.

Or maybe I should have used a different W from the other product theorem. But that might give a similar problem.

Also , not sure what you mean by : Weierstrass/Mittlag Leffler theory.

I know some theorems of these guys such as the product expansion of entire function from Weierstrass.

( that seem to relate weakly )

Maybe you mean some theorems instead of " theory "

***

I also wonder if you consider the presented method you gave as better or inferiour to the fractional calculus methods ?

I guess it depends on whether you want 1 interpolated as constant 1 or not.

***

Ofcourse I had the idea of

lim n to oo

sexp(m) = ln^[n] interpolate e^^(m+n)

to arrive at sexp(x).

...as you might have guessed.

***

Thank you for you time discussing this.

regards

tommy1729