# Tetration Forum

Full Version: Discussion of TPID 6
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(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.
(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
(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".
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.