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 Repeated Differentiation Leading to Tetrationally Fast Growth Catullus Fellow Posts: 204 Threads: 46 Joined: Jun 2022   07/13/2022, 02:45 AM (This post was last modified: 07/22/2022, 09:43 AM by Catullus.) Does anyone know of a function $\dpi{110}f(k)$, such that the xth derivative of $\dpi{110}f(k)$ grows like $\dpi{110} a\uparrow\uparrow x$, for a real a, and for a fixed k? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus Daniel Fellow Posts: 183 Threads: 57 Joined: Aug 2007 07/13/2022, 08:43 AM (07/13/2022, 02:45 AM)Catullus Wrote: Does anyone know of a function $\dpi{110}f(k)$, such that the xth derivative of $\dpi{110}f(k)$ 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 $a\in\mathbb R$, $\sum_{n=0}^\infty ^na x^n$ Daniel MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 07/13/2022, 10:32 AM (This post was last modified: 07/13/2022, 10:54 AM by MphLee.) (07/13/2022, 08:43 AM)Daniel Wrote: (07/13/2022, 02:45 AM)Catullus Wrote: Does anyone know of a function $\dpi{110}f(k)$, such that the xth derivative of $\dpi{110}f(k)$ 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 $$\frac{h'_{n+1}(x+1)}{h'_{n+1}(x)}=h'_{n}(h_{n+1}(x))$$ MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 912 Threads: 111 Joined: Dec 2010 07/13/2022, 07:00 PM (This post was last modified: 07/13/2022, 07:02 PM by JmsNxn.) (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 $\dpi{110}f(k)$, such that the xth derivative of $\dpi{110}f(k)$ 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: $$\lim_{n\to\infty} \left(\frac{^n a}{n!}\right)^{1/n} = \infty\\$$ Catullus Fellow Posts: 204 Threads: 46 Joined: Jun 2022   07/14/2022, 08:37 AM (This post was last modified: 07/16/2022, 10:46 PM by Catullus.) (07/13/2022, 07:00 PM)JmsNxn Wrote: 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: $$\lim_{n\to\infty} \left(\frac{^n a}{n!}\right)^{1/n} = \infty\\$$ How many times real differentiable do you think it could be? Do you have any more guesses about that? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 Joined: Feb 2009 07/16/2022, 07:26 AM (This post was last modified: 07/16/2022, 07:30 AM by tommy1729.) (07/14/2022, 08:37 AM)Catullus Wrote: (07/13/2022, 07:00 PM)JmsNxn Wrote: 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: $$\lim_{n\to\infty} \left(\frac{^n a}{n!}\right)^{1/n} = \infty\\$$How many times differentiable do you think it could be? Do you have any more guesses about that? 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 « Next Oldest | Next Newest »

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