sqrt thingy at MSE tommy1729 Ultimate Fellow Posts: 1,708 Threads: 376 Joined: Feb 2009 08/13/2022, 08:48 AM I got inspired by this nested root topic at MSE https://math.stackexchange.com/questions...tx2-0?rq=1 regards tommy1729 marcokrt Junior Fellow Posts: 12 Threads: 3 Joined: Dec 2011 08/13/2022, 06:51 PM (This post was last modified: 08/13/2022, 06:56 PM by marcokrt. Edit Reason: Adding LaTeX tags ) Very interesting, thank you! Recently, I posted another (and maybe easier) question on the same line that I would like to be answered: convergence value of this series. I suspect that the constant above could be very close to $e^{\frac{1}{e}}$, but I cannot find the exact value... it would be interesting since it compares (asymptotically) a unitary increment on hyper-2 to a unitary increment on hyper-4. Let G(n) be a generic reverse-concatenated sequence. If G(1)≠{2, 3, 7}, [G(n)^^G(n)](mod 10^d)≡[G(n+1)^^G(n+1)](mod 10^d), ∀n∈N\{0} (La strana coda della serie n^n^...^n, 60). bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/13/2022, 08:12 PM (08/13/2022, 06:51 PM)marcokrt Wrote: Very interesting, thank you! Recently, I posted another (and maybe easier) question on the same line that I would like to be answered: convergence value of this series. I suspect that the constant above could be very close to $e^{\frac{1}{e}}$, but I cannot find the exact value... it would be interesting since it compares (asymptotically) a unitary increment on hyper-2 to a unitary increment on hyper-4. Its not an exact proof, but we can take for granted that $$f_m(x):=^{m+1}x-^mx\to\infty$$ for $$m\to\infty$$ if $$x>e^{\frac{1}{e}}=:\eta$$ and $$f_m(x)\to 0$$ for $$1\eta$$ there is another M such that $$f_m(x)>1$$ for all $$x\ge x_1$$ and $$m\ge M$$. So for each $$x_2>\eta$$ there is an m such that the solution of $$f_m(x)=1$$ is in the open interval $$(\eta,x_2)$$. So the limit is $$\eta$$. JmsNxn Ultimate Fellow Posts: 994 Threads: 117 Joined: Dec 2010 08/14/2022, 05:44 AM (This post was last modified: 08/14/2022, 06:55 AM by JmsNxn.) (08/13/2022, 08:48 AM)tommy1729 Wrote: I got inspired by this nested root topic at MSE https://math.stackexchange.com/questions...tx2-0?rq=1 regards tommy1729 Fuck, that's a good thread. I really don't think it's as hard as they're making it though. For the simply reason, you can find constants: $$f_n(x-q_n) = g_n(x)\\$$ Which centers the solution for $$\Re(x) > 0$$. Then we can check that, for $$x_n \approx 0$$, that: $$\sum_{n=0}^\infty |g_n(x_n)| < \infty\\$$ Because: $$\sum_{n=0}^\infty |g'_n(x)|< \infty$$ This tells us that $$g_n(x_n)$$ converges; which casts a net of values about $$0$$. It won't be a neighborhood. But instead it'll mean that: The function $$g(x)$$ is real analytic, and probably holomorphic for $$\Re(x) > 0$$. I think the fact no one in this thread is mentioning that, is something crucial to the problem. I could go into detail on why this converges, but it would require me explaining about 3 years worth of infinite compositions. « Next Oldest | Next Newest »

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