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Interpolating an infinite sequence ?
#11
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!!!!
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#12
i tried \exp but that did not work.

guess i just had to refresh the page.
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#13
(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
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#14
Oh, Tommy

I was not advocating to use this as a method of interpolating. I'm just saying, IT IS A METHOD OF INTERPOLATING. It's absolutely garbage, if you ask me. It contains next to zero functional data. But there is always a solution for this. For example, let's use \(\beta(z)\), base \(2\) arbitrary multiplier \(\lambda\). You can even make a faster growing \(\beta\) I don't really care.

$$
F(z) = \frac{\beta(2z)}{\Gamma(-z)} \sum_{n=0}^\infty \frac{2 \uparrow \uparrow n}{\beta'(2n)A(2n)(z-n)}
$$

for \(A(2n)\) the derivative of the recipricol gamma.

And voila, it interpolates tetration. It's hot garbage though. Probably not monotone, probably extra chaotic. Would be a bitch to even attempt at showing that it satisfies tetration equation; where spoiler alert, it doesn't.

I was just explaining that you can interpolate infinitely many different ways. And it's always possible. That's all I meant. Wasn't advocating for actually using it, lol. Smile
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#15
(06/17/2022, 10:26 PM)JmsNxn Wrote: Oh, Tommy

I was not advocating to use this as a method of interpolating. I'm just saying, IT IS A METHOD OF INTERPOLATING. It's absolutely garbage, if you ask me. It contains next to zero functional data. But there is always a solution for this. For example, let's use \(\beta(z)\), base \(2\) arbitrary multiplier \(\lambda\). You can even make a faster growing \(\beta\) I don't really care.

$$
F(z) = \frac{\beta(2z)}{\Gamma(-z)} \sum_{n=0}^\infty \frac{2 \uparrow \uparrow n}{\beta'(2n)A(2n)(z-n)}
$$

for \(A(2n)\) the derivative of the recipricol gamma.

And voila, it interpolates tetration. It's hot garbage though. Probably not monotone, probably extra chaotic. Would be a bitch to even attempt at showing that it satisfies tetration equation; where spoiler alert, it doesn't.

I was just explaining that you can interpolate infinitely many different ways. And it's always possible. That's all I meant. Wasn't advocating for actually using it, lol. Smile

ok then we agree Smile
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