08/13/2022, 08:48 AM
I got inspired by this nested root topic at MSE
https://math.stackexchange.com/questions...tx20?rq=1
regards
tommy1729
https://math.stackexchange.com/questions...tx20?rq=1
regards
tommy1729
sqrt thingy at MSE

08/13/2022, 08:48 AM
I got inspired by this nested root topic at MSE
https://math.stackexchange.com/questions...tx20?rq=1 regards tommy1729
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 , but I cannot find the exact value... it would be interesting since it compares (asymptotically) a unitary increment on hyper2 to a unitary increment on hyper4.
Let G(n) be a generic reverseconcatenated 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).
08/13/2022, 08:12 PM
(08/13/2022, 06:51 PM)marcokrt Wrote: Very interesting, thank you! 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 \(1<x\le\eta\). Also that it is strictly increasing for each m. Hence there is an M, such that \(f_M(\eta)<1\) and can not be a solution for all \(x\le \eta\) and \(m\ge M\). On the other hand for each \(x_1>\eta\) 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 \(\eta\). (08/13/2022, 08:48 AM)tommy1729 Wrote: I got inspired by this nested root topic at MSE 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: $$ f_n(xq_n) = g_n(x)\\ $$ 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. 
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