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 Interpolating an infinite sequence ? JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 06/14/2022, 12:55 AM (This post was last modified: 06/14/2022, 01:04 AM by JmsNxn.) 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, $$h(z+n)W(z+n) = 0 + 0 + 0....\frac{W(z+n) a_n}{W'(n)z}....+0+0\\\$$ I'm sorry, I refuse to believe you've never seen this before Tommy! This is a common consequence of Weierstrass/Mittlag Leffler theory. We are setting all the other terms of the sum to zero, and the only ones which evaluate to a value are those with simple poles.... Come On, you know this!!!! tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 06/15/2022, 11:56 PM i tried \exp but that did not work. guess i just had to refresh the page. tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 06/16/2022, 05:09 PM (06/14/2022, 12:55 AM)JmsNxn Wrote: No problem tommy. The tex is failing because you are writing Exp instead of \exp, not sure why it's formatting like that. I apologize I thought the result was obvious. The function: $$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 JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 06/17/2022, 10:26 PM (This post was last modified: 06/17/2022, 10:28 PM by JmsNxn.) 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)$$ the derivative of the recipricol gamma. And voila, it interpolates tetration. It's hot garbage though. Probably not monotone, probably extra chaotic. Would be a bitch to even attempt at showing that it satisfies tetration equation; where spoiler alert, it doesn't. I was just explaining that you can interpolate infinitely many different ways. And it's always possible. That's all I meant. Wasn't advocating for actually using it, lol. tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 06/17/2022, 10:41 PM (06/17/2022, 10:26 PM)JmsNxn Wrote: Oh, Tommy I was not advocating to use this as a method of interpolating. I'm just saying, IT IS A METHOD OF INTERPOLATING. It's absolutely garbage, if you ask me. It contains next to zero functional data. But there is always a solution for this. For example, let's use $$\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. ok then we agree « Next Oldest | Next Newest »

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