(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.
....
\( f = \sum_{n = 1}^{\infty}x^{n}{(2n)!} = \cosh(\sqrt{x})-1\approx \frac{1}{2}\exp(\sqrt{x}) \)
....
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
\( f(x) = \sum_{n = 1}^{\infty} a_n x^n \; < \; \exp^{0.5}(x) \)
\( \forall {x \gt 0 } \; \; a_n x^n \; < \; \exp^{0.5}(x) \)
\( \forall {x \gt 0 } \; \; \log(a_n x^n) \; < \; \log(\exp^{0.5}(x)) \)
\( \forall {x \gt 0 } \; \; \log(a_n) + n\log(x) \; < \; \log(\exp^{0.5}(x)) \)
\( \forall {x \gt 0 } \; \; \log(a_n) \; < \; \log(\exp^{0.5}(x)) - n\log(x) \)
For log(exp^0.5(x)), we can substitute exp^0.5(log(x)).
\( \forall {x \gt 0 } \; \; \log(a_n) \; < \; \exp^{0.5}(\log(x)) - n\log(x) \)
Next, replace x with log(x), and change the range from x>0, to all x
\( \forall {x } \; \; \log(a_n) \; < \; \exp^{0.5}(x) - nx \)
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.
\( \frac{d}{dx} \exp^{0.5}(x) - nx = -n + \frac{d}{dx} \exp^{0.5}(x)=0 \)
At the minimum, the derivative will be equal 0, so defining \( \text{dexphalf}(x)=\frac{d}{dx} \exp^{0.5}(x) \)
Now define
\( h_n = \text{dexphalf}^{-1}(n) \)
\( \log(a_n) \; < \; \exp^{0.5}(h_n) - n h_n \)
\( a_n \; < \; \exp(\exp^{0.5}(h_n) - n h_n)) \)
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".
\( a_n = \exp(\exp^{0.5}(h_n) - n h_n) \)
Here is the first entire asymptotic approximation of halfx, where \( a_0=\exp^{0.5}(0) \)
\( f_1(x) = \sum_{n=0}^{\infty} a_n x^n \)
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 \( \exp(h_n) \), and compare to exp^0.5(exp(h_n)). Empirically, the scale factor is approximately 1/sqrt(n).
\( b_n=a_n\frac{f_1(\exp(h_n))}{\exp^{0.5}(\exp(h_n))} \)
Then a much better entire approximation of exp^0.5(x) is:
\( f_2(x) = \sum_{n=0}^{\infty} b_n x^n \)
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.
\( h_n = \text{dexphalf}^{-1}(n) \)
\( a_n = \exp(\exp^{0.5}(h_n) - n h_n) \)
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.
\( m(n) = \frac{n \times \exp(\exp^{0.5}(h_n))}{\exp(h_n)} \)
\( \log(a_m) = \exp(\exp^{0.5}(h_n))\times(-n+1) \)
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
