(05/10/2014, 11:58 PM)tommy1729 Wrote: [ -> ] (05/10/2014, 11:48 PM)sheldonison Wrote: [ -> ].....

Wow -- both of these functions grow exactly exponentially, where growth is defined as slog(f^n)/n.

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Nice to see you agree.

But I kinda said all those things recently.

Agreed; and I'm sure I probably realized that before posting, my reply was also only a few minutes after your reply

Anyway sometimes I need time to think it all through. Now I need to switch from numerical approximations mode to theoretical mode, the end goal is to come up with an equation like the one you conjectured, which I am not yet able to do.

(05/10/2014, 11:58 PM)tommy1729 Wrote: [ -> ]Hence the new conjecture is 1/(n ln(n)) ! as Taylor coefficients.

We desire an asymptotic under-approximation of the half exp(x) with all positive Taylor series coefficients at x=0, based on the Kneser half exp(x) function, which is not entire. So we need

For log(exp^0.5(x)), we can substitute exp^0.5(log(x)).

Next, replace x with log(x), and change the range from x>0, to all x

Conjecture, for each a_n, there is a particular value of x that most limits a_n. In other words, we conjecture that for each value of n, exp^0.5(x)-nx has one minimum value, where the derivative is zero.

At the minimum, the derivative will be equal 0, so defining

Now define

This is the first step of my generation of an entire approximation of f(x), and is probably the most important, by replacing the "<" with equal".

Here is the first entire asymptotic approximation of halfx, where

With this approximation f(x) will be an entire over approximation of the exp^0.5(x). To get a better approximation, use the same values from above, and scale the a_n values to generate b_n. Formally, we can get a better approximation by scaling, where we evaluate f1(x) at

, and compare to exp^0.5(exp(h_n)). Empirically, the scale factor is approximately 1/sqrt(n).

Then a much better entire approximation of exp^0.5(x) is:

For values of x on the order 1E10, the ratio of f_2(x) to exp^0.5(x) is accurate to about 1 part in 10000. I think the ratio of f_2(x) to exp^0.5(x) goes to exactly 1 as x goes to infinity. By the way, this f_2(x) is a little bit more accurate than the results I posted earlier in this thread, which used rough approximations for the minimum of a_n, instead of derivatives. One can also scale twice (or more times though the lower derivatives probably start oscillating); scaling twice, I get accuracy of 0.2 parts per million for f_3(1E10)/exp^0.5(1E10). By scaling three times, I get an accuracy of 1 part per billion, for f_4(1E10)/exp^0.5(1E10).

I hope the next step is to come up with a recursive equation, given any particular value of a_n, to come up with another a_n, for another much larger value of n. Specifically, can we come up with an approximation for how fast the factorial in the denominator grows? Specifically, given n, I can generate a_m, where h_n is the location used to evaluate a_n, from above. Starting with n, and a_n, the first two equations just use a_n from above.

Skipping a lot of algebra, this is what I got to. I can include the algebra later. Here, we are generating a fractional Taylor series coefficient m, from the value n above, which would exactly match the equation above.

I haven't been able to do anything useful in terms of limiting behavior (yet), but I only got this recursive relationship cleaned up a few minutes ago, this morning; I expect the relationship will be there! If I apply these equations to n=6, I get h_6=5.546380530883 and a_6=0.0000001055905600243, and m=700791.2, and log(a_m)= -149682094.7, which is correct. Then a_m = 1/9906980.030456! which is 1/(m*14.137)!, which actually matches Tommy's conjecture reasonably well since log(m)=13.4599.

Hopefully, the equations I just posted don't have too many typos.... If there are typos, I can say that I have a working pari-gp program that implements these mathematical equations (without typos).

The goal is to use this recursive relationship to prove something about f(n) where a_n = 1/f(n)!. Also, the second set of equations allows generating accurate recursive coefficients for much much larger values of n than the original equation. This could be used to check Tommy's conjecture....

- Sheldon