Discussion of TPID 6

[sheldonison]

....
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