# Tetration Forum

Full Version: tetration limit ??
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(04/14/2011, 08:17 PM)JmsNxn Wrote: [ -> ]
(04/29/2009, 01:08 PM)tommy1729 Wrote: [ -> ]so , what are the answers ?

using logarithmic semi operators (lol)
{q : 0 <= q <= 1 q E R}, if S(x) is the identity function, q:ln(x) = exp^[-q](x):

$\lim_{h\to\0}\,(S(1-q)\, \{-q\}\, q:ln(h))\,\{2-q\}\,\frac{1}{h} = q:ln(e)$

euh ... does this still relate to the first few posts by me and bo ?

we are looking for functions !
Sure it is, it's a limit that's related to tetration, though a slight variant of what your original inquisition was. It's related to half-iterates of e, and is a generalization of the original limit in question $\lim_{n\to\infty} (1 + \frac{1}{n})^n = e$I was just posting it because it popped into my head as I was reading through your thread.

But, I get your point. I wasn't really contributing to your question. It's just that I couldn't think of anything to contribute from that angle, but I still read the whole thread and wanted to post something. Didn't mean to appear off-topic and shrewd
(04/02/2009, 09:56 PM)bo198214 Wrote: [ -> ]Perhaps then you should start with the simpler case of the double iterate. And look what a suitable function f you would find that:

$\lim_{n\to\infty} (1+f(n))^{(1+f(n))^{n}} = Q$
I dont see what useful function f that could be.

we reduce to (1+1/f(n))^((1+1/f(n))^n) = Q

in essense we only need to understand the relation between n and f(n).

further switch f(n) and n to get

(1+1/n)^((1+1/n)^f(n)) = Q

ln(1+1/n) * (1+1/n)^f(n) = ln(Q)

replace ln(Q) by Q

f(n) = ln(Q/ln(1+1/n)) / ln(1+1/n)

done.

but lim n-> oo sexp_(1+f(n))[slog_(1+f(n))[n] + 1/2] = n + C

0 < C

seems harder and not so related at first.

worse , it might have problems stating it like above ... because our n needs to be after the second fixpoint and our superfuntions need to be defined at their second fixpoint ... which " evaporates " at oo as n goes to oo.

and hence our superfunctions become valid and defined > q_n where lim q_n = oo !!

if f(n) does not grow to fast this might be ok , but on the other hand to arrive at C at our RHS f(n) seems to need some fast growing rate.

so f(n) is strongly restricted and C must be unique and existance is just assumed.

i do not know anything efficient to compute f(n) apart from numerical *curve-fitting* upper and lower bounds as described above.

or another example , actually the original OP rewritten :

lim n-> oo sexp_(1+f(n))[slog_(1+f(n))[n] + 1/2] = C

0 < C

now we must take the first fixpoint approaching 1 .. or the second ??

it seems easiest to take the first fixpoint , if we take the second we have the same problem of the " evaporating ' fixpoint as above.

on the other hand , we dont know the radiuses for bases 1+f(n) expanded at their first or second fixpoint.

again , its hard to find f(n) and C despite they are probably strongly restriced - even unique -.

another idea that might make sense is that there exists a function g(n) such that

but lim n-> oo sexp_(1+f(n))[slog_(1+f(n))[n] + 1/2] = g(n)

0 < g(n) < n

and that g(n) gets closer and closer towards the end of the radius of one of the fixpoint expansions as n grows.

and that might be inconsistant with the other equations/ideas above.

so many questions.

regards

tommy1729
nice proof, however as you already said, it doesnt seem to help with your original question.
I noticed post 7 and post 33 resemble the idea of generalized golden numbers.
In particular solving polynomials with a parameter.

I think there is more hidden in this.

Regards

Tommy1729
I think fake function theory Will help with these limits.

Though that seems nontrivial.

Regards

Tommy1729
Let n be a positive integer going to +oo.

lim [e^{1/e} + 1/n]^^[(10 n)^{1/2} + n^{A(n)} + C + o(1)] - n = 0.

Where C is a constant.

Conjecture : lim A(n) = 1/e.

regards

tommy1729
No reaction ???

Regards

Tommy1729
(05/28/2015, 11:32 PM)tommy1729 Wrote: [ -> ]Let n be a positive integer going to +oo.

lim [e^{1/e} + 1/n]^^[(10 n)^{1/2} + n^{A(n)} + C + o(1)] - n = 0.

Where C is a constant.

Conjecture : lim A(n) = 1/e.

Its a curious equation. I viewed it from a different angle: What is the slog_{1/e+1/n}(n)? But I couldn't figure out why you were interested in slog(n) as opposed to say, slog(e^e) or something like that that made more sense to me. e^e is the cusp of where this tetration function takes off, and the function starts growing superexponentially. But the (1/n) means it might take 1 or 2 more iterations to reach (1/n), Or if n is hyperexponentially large = sexp(4.5), then 3 extra iterations. But most of the time is spent getting to e^e. And that equation is dominated by approximately real(Pseudo period)-2. And you included an O(1) term in your equation anyway, which implies C isn't an exact constant.

So then my counter conjecture would be that lim sexp_{1/e+1/n)(real(Period)-2)=constant, and that constant seems to be about 388 as n goes to infinity. But that seemed to be a very different equation than the one you had in mind, so I thought it would be off topic, so I didn't mention it. But yeah, I have equations for the pseudo period, which I posted below.

Then there is your approximation itself. slog_{1/e+1/n}(n) = (10n)^{1/2} + n^{A(n)} + C. Can you explain why you think this is the right approach or equation? It doesn't seem to match the approximation I have for real(pseudo_period)-2...

The equations for the fixed point and Period are approximately as follows. One can see that the resulting period has a sqrt term, but not sqrt(10n).
$\eta=\exp(1/e)\;\;k=\ln(\ln(\eta+\frac{1}{n}))+1\approx \frac{e}{\eta \cdot n} \approx \frac{1.8816}{n}\;\;\;$

Now we have switched it to a problem of iterating $z \mapsto \exp(z)-1+k\;$. In the limit, the fixed point goes to zero. This iteration mapping has a simpler Taylor series for the fixed point L, from which we can generate the Pseudo Period.

$x=\sqrt{-2k}\;\;\; L = x - \frac{x^2}{6}+\frac{x^3}{36}+...\;\;\approx \sqrt{ \frac{-2e}{\eta \cdot n}}\;\;\;\text{period}=\frac{2\pi i}{L}\;\;\$ this is the period at the fixed point.

$\Re(\text{period}) = \Re(\frac{2\pi i}{L}) \approx 2\pi \sqrt{\frac{\eta \cdot n}{2e}} \approx \sqrt{10.49n}\;\;\;$ this is close to sqrt(10n). But I don't understand your n^A(n)~=n^(1/e) term; [(10 n)^{1/2} + n^{A(n)}].

Anyway, my counter-conjecture is that
$\lim_{n \to \infty} \text{sexp}_{(\eta+1/n)}\left[\Re(\text{period})-2\right] =k \;\;\; k\approx 388.787398293917704779$, where the period is from the equation above.

The correct middle term is probably $\text{slog}_e(\frac{n}{e}-1)\;\;$ Note that $\text{slog}_e(n/e-1)=\text{slog}_e(\eta^{\eta^n})-2\;\;$ So then, we have the following conjectured equation, where I'm pretty sure a 1-cyclic theta is required as n gets arbitrarily large, $\theta(\text{slog}_e(n/e-1))$, whose predicted amplitude is probably about +/-0.002

$\lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[2\pi\sqrt{\frac{\eta\cdot n}{2e}} + \text{slog}_e(\frac{n}{e}-1) + C + \mathcal{O}(\theta) \right] -n = 0\;\;\;\;C \approx -2 - \text{slog}_e(388.7874/e-1)$

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