Does anyone know of a function , such that grows like , for a real a, and for a fixed k?
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Repeated Differentiation Leading to Tetrationally Fast Growth

Does anyone know of a function , such that grows like , for a real a, and for a fixed k?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
07/13/2022, 08:43 AM
(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 \( a\in\mathbb R \), \( \sum_{n=0}^\infty ^na x^n \)
Daniel
(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? Hi Daniel, I was thinking the same but then... what the kth derivatives of nhyperexponentiation 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 \(\frac{h'_{n+1}(x+1)}{h'_{n+1}(x)}=h'_{n}(h_{n+1}(x))\) MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)\) S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\) (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? No, Daniel's right, there is no such functionat 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: \[ \lim_{n\to\infty} \left(\frac{^n a}{n!}\right)^{1/n} = \infty\\ \] (07/13/2022, 07:00 PM)JmsNxn Wrote: No, Daniel's right, there is no such functionat least no such analytic function. I doubt there's even a smooth function.How many times real differentiable do you think it could be? Do you have any more guesses about that?
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ (07/14/2022, 08:37 AM)Catullus Wrote:(07/13/2022, 07:00 PM)JmsNxn Wrote: No, Daniel's right, there is no such functionat least no such analytic function. I doubt there's even a smooth function.How many times differentiable do you think it could be? Radius zero implies not defined outsite its expansion point. So nothing to take the derivative from. unless you have another nontaylor definition that does compute values ofcourse. regards tommy1729 ps : a taylor expansion expanded at another point will not work since your derivatives diverges faster than exponential. to see this : if f(x) has radius zero and derivatives diverges faster than exponential then f( c x ) has the same problem. Plz think about it before you reply. regards tommy1729 
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