(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, such that the xth derivative of
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:
\[
\lim_{n\to\infty} \left(\frac{^n a}{n!}\right)^{1/n} = \infty\\
\]

