05/15/2014, 06:15 PM
(This post was last modified: 05/15/2014, 10:42 PM by sheldonison.)

(05/14/2014, 11:42 PM)tommy1729 Wrote: Btw sheldon edited his posts a few times, which makes them better naturally.

I've gone through at least 4 approximation versions, for the coefficients of an entire asymptotic half exponential function. Each one is better, more accurate, and more robust than the one before it. Sometimes, the next version shows up before I'm done documenting the previous version...

Quote:( I do not yet understand where his second derivative comes from , is this the same logic that goes from a_n to b_n ?? Also his integrals need more study and/or explaination. Forgive me if im a bit behind )The shows up in the denominator of the Gaussian approximation, which is version III, and is very accurate, and is robustly shown (some details left out) to be an approximation of version IV (also in post#16), which involves integrals, and converges to the Kneser half iterate extremely extremely well, accurate to 28 decimal digits for half iterates O(1E12). Gaussian can be shown to always give a slight over approximation for the a_n (compared to the Integral). Multiplying Gaussian by any constant<1 will eventually give an under approximation for all a_n bigger than some particular value of n, assuming the Taylor series for Kneser half exp is eventually well behaved for large enough hn(n), (needs a definition). So that might rigorously address "stuff < a_n".

...But a good boundary means

stuff < a_n < stuff

and the stuff < a_n is not adressed yet ...

Quote:More importantly , we have encountered h_n before on this forum !h_n first shows up in the version II approximation, in post #9 of this thread. I'm unaware of h_n before that...

I do not know exactly where and how but I remember it.

That might be usefull.

Quote:ln[1/2](n) VS h_n

....

Hence 50 % of the tommy sheldon conjecture has been proven, more specific :

1/a_n < O( exp(n * (ln^[1/2](n))^(1+o(1)) ) )

Tommy, empirical tests for your conjecture look remarkably good for this last equation, for n=10^1000000, this approximation is accurate to 99.8%, though only 60% accurate for n=100000000, so convergence is a little slow, and it appears to always be an under-approximation.

So, we have this nice asymptotic approximation, f=exp^0.5(x). Lets take the best one, version IV. What would f(f(z))/exp(z) look like???? Its an entire function, since f is entire, and since exp(z) has no zeros. The answer is really neat, and counter intuitive! I'll have to post some complex plane graphs tomorrow.

- Sheldon