Repeated Differentiation Leading to Tetrationally Fast Growth
#4
(07/13/2022, 10:32 AM)MphLee Wrote:
(07/13/2022, 08:43 AM)Daniel Wrote:
(07/13/2022, 02:45 AM)Catullus Wrote: Does anyone know of a function [Image: gif.image?\dpi%7B110%7Df(k)], such that the xth derivative of [Image: gif.image?\dpi%7B110%7Df(k)] grows tetrationally fast, for a fixed k?

First I should mention that the function you are looking for would be wildly divergent. While I don't know of any publication of such a function, it's construction would be straight forward.

For a given \( a\in\mathbb R \),
\( \sum_{n=0}^\infty ^na x^n \)

Hi Daniel, I was thinking the same but then... what the kth derivatives of n-hyperexponentiation would look like?
Let \(h_n\) be hyperexponentiation of rank \(n\) of base \(b\) fixed. What about
\[{\mathcal H}_{n,x}(k)=\frac{d^k}{d^kx}h_n(x)\]

Does exist an \(n\) s.t. \(h_4(k)\leq{\mathcal H}_{n,x}(k)\)?

I wonder if this identity is telling us something \(\frac{h'_{n+1}(x+1)}{h'_{n+1}(x)}=h'_{n}(h_{n+1}(x))\)

No, Daniel's right, there is no such function--at least no such analytic function. I doubt there's even a smooth function.

Every analytic function must satisfy:

\[
\lim_{n\to\infty} \left(\frac{f^{(n)}(k)}{n!}\right)^{1/n} \le 1
\]

And it's safe to say:

\[
\lim_{n\to\infty} \left(\frac{^n a}{n!}\right)^{1/n} = \infty\\
\]
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Messages In This Thread
RE: Repeated Differentiation Leading to Tetrationally Fast Growth - by JmsNxn - 07/13/2022, 07:00 PM

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