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Searching for an asymptotic to exp[0.5]
#14
(05/10/2014, 11:56 PM)tommy1729 Wrote: ...
Hence the new conjecture is 1/(n ln(n)) ! as Taylor coefficients.

I used my new code algorithm, from post#9 in math equations earlier, to generate coefficients for ultra large values of n; these are unscaled values of a_n; f(x) using a_n Taylor series is on over approximation of exp^{0.5}, see post #9 and/or below for scaling equation.





If the conjecture is correct, then a_n ~= n ln(n)^2, since ln(n!) ~= n ln(n)-n. But the pattern does not work, since k from the third column grows arbitrarily large, instead of converging to k=2.


these are minimum possible values k_n, scaling with f2 very slightly increases k_n

Code:
n                  ln(a_n)                  k_n   ln(a_n) = -n(ln(n))^k_n
10.0000000        -43.23747618988           1.755474309098
100.000000        -1702.383415564           1.856110652219
10000.0000        -808235.9469405           1.978208308724
100000000.        -51574683648.64           2.143700628920
1.00000000E16     -4.843045366425E19        2.352700385134
1.0000000E32      -7.278717641299E36        2.603697260942
1.0000000E64      -1.897211261800E70        2.895281429488
1.0000000E128     -9.271029181788E135       3.226308077405
1.0000000E256     -9.170982024460E265       3.595948495976
1.0000000E512     -1.983493098320E524       4.003740240479
1.0000000E1024    -1.014825960853E1039      4.449613625772
1.0000000E2048    -1.333046642964E2066      4.933883260692
1.0000000E4096    -4.898918992448E4117      5.457211597134
1.0000000E8192    -5.514831365014E8217      6.020558126848
1.0000000E16384   -2.093094173164E16414     6.625126235356
1.0000000E32768   -2.964298724943E32803     7.272315222911
1.0000000E65536   -1.743728846523E65577     7.963680815351
1.0000000E131072  -4.770569826246E131119    8.700904681531
1.0000000E262144  -6.838009537646E262198    9.485771998591
1.0000000E524288  -5.820676972960E524350   10.32015555316
1.0000000E1048576 -3.356313289241E1048647  11.20600483435

For comparison, for one of our "fake" half exp functions, if we try a_n=1/(4n)! for n=1E1048576, we get k=1.0943, so this is a very good test for detecting "fake" half exp functions, where ultimately k=1. For the conjectured a_n=1/(n ln(n))!, we would expect k=2.000000386, for the last term. So it would be nice to understand k for our entire function which is conjectured to converge to a half exponential growth. For an earlier conjecture, a_n=(n^2)!, k_n for the last term would be 164282.152, so that too gives a very different value.

(05/12/2014, 11:35 PM)tommy1729 Wrote: Let a(x) and b(x) be entire functions that are asymptotic to exp^[0.5](x) for x > -1.
Also a(n) = b(n) for integer n.

Since a(n) and b(n) are entire and grow slower than exp , Carlsons theorem applies and

a(x) - b(x) = 0

!!!

So we can make a newton series.

I wasn't focused at all on uniqueness, just the fact that I wanted the asymptotic function to be entire. I don't claim that the asymptotic is equal to exp^0.5(x) anywhere, but I conjecture that the ratio, for the second function, which has been scaled approaches 1. So would Carlson's theorem allow us to generate an entire function that exactly equals exp^0.5(x) at the whole numbers, and also be entire? I don't know if a pure Newton series would have all positive derivatives, or whether or not that matters, or when an infinite Newton series would be entire.... It looks like Carlson's theorem requires that you start with an entire function, so I'm not sure it helps to turn a non-entire half iterate into an asymptotic entire function, and the existing conjectured asymptotic entire function already has a Taylor series, so we don't need Newton.

using a_n from above

using h_n from above, scaling is conjectured to allow the ratio to approach arbitrarily close to 1, and would be approximately




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Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by sheldonison - 05/13/2014, 04:23 AM

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