Discussion of TPID 6 - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Discussion of TPID 6 (/showthread.php?tid=526) Discussion of TPID 6 - JJacquelin - 10/22/2010 (10/07/2009, 12:03 AM)andydude Wrote: Conjecture $\lim_{n\to\infty} f(n) = e^{1/e}$ where $f(n) = x$ such that ${}^{n}x = n$ Discussion To evaluate f at real numbers, an extension of tetration is required, but to evaluate f at positive integers, only real-valued exponentiation is needed. Thus the sequence given by the solutions of the equations $x = 1$ $x^x = 2$ $x^{x^x} = 3$ $x^{x^{x^x}} = 4$ and so on... is the sequence under discussion. The conjecture is that the limit of this sequence is $e^{1/e}$, also known as eta ($\eta$). Numerical evidence indicates that this is true, as the solution for x in ${}^{1000}x = 1000$ is approximately 1.44. I think that the conjecture is false. First, the numerical computation have to be carried out with much more precision. The solution for x in ${}^{1000}x = 1000$ is approximately 1.44467831224667 which is higher than e^(1/e) The solution for x in ${}^{10000}x = 10000$ is approximately 1.4446796588047 which is higher than e^(1/e) As n increases, x increasses very slowly. But, in any case, x is higher than e^(1/e) = 1.44466786100977 Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of ${}^{n}x$ is e , for n tending to infinity. So, the limit isn't = n , as expected. RE: Limit of self-super-roots is e^1/e. TPID 6 - sheldonison - 10/22/2010 (10/22/2010, 11:27 AM)JJacquelin Wrote: .... As n increases, x increasses very slowly. But, in any case, x is higher than e^(1/e) = 1.44466786100977 Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of ${}^{n}x$ is e , for n tending to infinity. So, the limit isn't = n , as expected.I think that's a good starting point. For x=e^(1/e), the $\lim_{n \to \infty}\text{sexp}_\eta(n)=e$ Another limit that I think holds is that the slog(e) gets arbitrarily large as the base approaches eta from above. Note that for these bases with B>eta, sexp(z) grows super exponentially when z gets big enough. $\lim_{b \to \eta+}\text{slog}_b(e)=\infty$ Now lets pick 10000. Solve for base b>eta $\text{slog}_b(e)=10000$. We know there is another number n>10000, for which $^n b=n$, because we know that super exponential growth will eventually set in, as n grows past 10000, and that $\lim_{n \to \infty}^n b=\infty$. Then, for some number n>10000, $^n b=n$. I actually have a hunch that somewhere around n=20000 or so that superexponential growth finally kicks in. I guess what I'm trying to get at is that we can probably prove that for $^{n}b=e$, solving for b as n grows arbitrarily large, b approaches eta+. For each particular base b, there is another larger number, call it "m>n", for which Andrew's equation holds. $^{m}b=m$. And that might be a pretty good step in proving Andrew's lemma. - Sheldon RE: Limit of self-super-roots is e^1/e. TPID 6 - nuninho1980 - 10/23/2010 (10/22/2010, 11:27 AM)JJacquelin Wrote: I think that the conjecture is false. First, the numerical computation have to be carried out with much more precision. The solution for x in ${}^{1000}x = 1000$ is approximately 1.44467831224667 which is higher than e^(1/e) The solution for x in ${}^{10000}x = 10000$ is approximately 1.4446796588047 which is higher than e^(1/e) As n increases, x increasses very slowly. But, in any case, x is higher than e^(1/e) = 1.44466786100977 Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of ${}^{n}x$ is e , for n tending to infinity. So, the limit isn't = n , as expected.The solution for x in ${}^{1000}x = 1000$ is approximately 1.44467831224667 -> is not correct! but yes - ${}^{1000}x = 1000 \Rightarrow$ x=1.44467829141456 The solution for x in ${}^{10000}x = 10000$ is approximately 1.4446796588047 -> is not correct but yes - ${}^{10,000}x = 10,000 \Rightarrow$ x=1.4446679658595034 you mistake!?!! eheheh! lool ${}^{100,000}x = 100,000 \Rightarrow$x=1.444667862058778534938 therefore, the conjecture is NOT false! I calculated the numbers corrects by program "pari/gp". RE: Discussion of TPID 6 - bo198214 - 10/24/2010 Hey guys, 1. please dont post discussion in the open problems survey! Its reserved for problems exclusively. 2. The conjecture is already proven: By me here By tommy here. I update the stati of the problems.