sqrt thingy at MSE

[tommy1729]

I got inspired by this nested root topic at MSE

https://math.stackexchange.com/questions...tx2-0?rq=1

regards

tommy1729

[marcokrt]

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.

[bo198214]

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 \(1<x\le\eta\). 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 \(\eta\).

[JmsNxn]

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.