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As we all know, there is a thing about polynomial interpolation when using equally-spaced nodes on certain functions called Runge's phenomenon. However, there are a lot of functions that do not have this problem (those whose radius of absolute convergence is infinite).

Given the function $f(x) ={}^x2$, meaning $2$ tetrated to $x$, where tetration means the hyperoperation of the exponential:

$$ {}^42 = 2[4]4 = 2\upuparrows4 = 2^{2^{2^{2}}} = 2^{2^{4}} = 2^{16} = 65536 $$

We know its values for integers $x = k \geq-1$. Therefore, we should be able to construct an infinite Lagrange interpolating "polynomial" as follows:

$$ p_\infty(x) = \sum_{j=-1}^{\infty}{}^{j}2 \prod_{\substack{i=-1\\i\neq j}}^{\infty} \frac{x-i}{j-i} $$

The question is simple: does this series converge uniformly/absolutely for all $x\in(-2,+\infty)$? If not, which is the largest interval of uniform/absolute convergence?

Myungheon Lee
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Rosario Martínez de Meticedio
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  • True, my bad. Let me correct it. – Rosario Martínez de Meticedio Apr 13 '25 at 23:43
  • Why invent a new notation when $\uparrow\uparrow$ already exists? – Greg Martin Apr 13 '25 at 23:44
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    What makes you think the infinite products converge to anything? That certainly doesn't represent a "power series, since the sum for $f(0)=a_0$ in the expansion as terms that don't go to zero. It might be a function, but his doesn't give us a power series. – Thomas Andrews Apr 13 '25 at 23:45
  • Knuth arrows are not commonly used when working with tetration, the same way we do not use $\uparrow$ for the exponential (only when typewriting, where we use the circumflex accent as a way to denote the Knuth arrow). You can use it tho, I have no problem. It just looks cleaner without it imo. – Rosario Martínez de Meticedio Apr 13 '25 at 23:47
  • You can construct an interpolating function using Mittag–Leffler's theorem as described in this answer. – Greg Martin Apr 13 '25 at 23:47
  • @ThomasAndrews If you plug in $x=0$, all of the terms in the infinite product will cancel out except for the one where $j=0$, yeilding $^02 = \ ^02 = 1$, which is trivial. – Rosario Martínez de Meticedio Apr 13 '25 at 23:50
  • @GregMartin The notations used are not new nor invented here. – jjagmath Apr 13 '25 at 23:53
  • Well, I don't see how this defines a power series. I didn't say it doesn't have a value at $0,$ just that it's first coefficient doesn't have an obvious value from the terms given. If you expand each term of the sum as a function, do you have a power series? What is the constant there? You can easily show this function has a value at integers $\geq -1.$ If it was somehow computable from this form, But not that it is computed by a power series available directly from this form. If it was, you'd definitely get no value for $x<-1,$ so you wouldn't have a radius of convergence more than $1.$ – Thomas Andrews Apr 14 '25 at 00:15
  • @ThomasAndrews I made a mistake there and said power series when it is just a series. The question has already been updated – Rosario Martínez de Meticedio Apr 14 '25 at 00:17
  • you could have avoided me typing so much by confessing your error here. It is hard to monitor the changes in the question while writing comments. – Thomas Andrews Apr 14 '25 at 00:20
  • In any event, I'm pretty sure when $x$ is not an integer $\geq-1$ the products all converge to $0.$ – Thomas Andrews Apr 14 '25 at 00:21
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    We have $$\frac{x-i}{j-i}=1+\frac{x-j}{j-i}$$ and for the product not to converge to zero, you need $\sum_{i\neq j}\frac{x-j}{j-i}$ to converge. This is because all but finitely many values of $i$ give a negative value. – Thomas Andrews Apr 14 '25 at 00:31
  • If I am correct, then there is no non-trivial intervals on which it converges. – Thomas Andrews Apr 14 '25 at 00:37
  • Technically, I guess this means it is defined everywhere, but the value is $0$ at all real numbers other than integers $x\geq-1.$ – Thomas Andrews Apr 14 '25 at 02:34
  • The key is that there is no such thing as an infinite polynomial. – Thomas Andrews Apr 14 '25 at 03:19

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