Posts: 1,372
Threads: 336
Joined: Feb 2009
hi all
i dont have time to explain or post alot.
but i managed to get some partial results about tetration i believe.
i seem to have gotten a similar equation as andrew's slog ( the infinite matrix ) for my tommysexp(tommyslog(x)+k) in base e.
also this seems to relate and possibly answer many uniqueness and existance questions that seemed somewhat unrelated and unattackable before.
im still working on it but it seems promising and i might even solve some of those hard matrix questions.
in a sense i also seem to have found a trick to attach a fixpoint to an infinite matrix equation. ( yes this is vague and informal , i dont have time to explain fully )
regards
tommy1729
Posts: 641
Threads: 22
Joined: Oct 2008
11/15/2011, 09:22 AM
(This post was last modified: 11/17/2011, 10:35 PM by sheldonison.)
(11/14/2011, 09:35 PM)tommy1729 Wrote: hi all
i dont have time to explain or post alot.
but i managed to get some partial results about tetration i believe.
i seem to have gotten a similar equation as andrew's slog ( the infinite matrix ) for my tommysexp(tommyslog(x)+k) in base e.
also this seems to relate and possibly answer many uniqueness and existance questions that seemed somewhat unrelated and unattackable before.
im still working on it but it seems promising and i might even solve some of those hard matrix questions.
in a sense i also seem to have found a trick to attach a fixpoint to an infinite matrix equation. ( yes this is vague and informal , i dont have time to explain fully )
regards
tommy1729 minor edit
I spent some time analyzing the derivatives of the basechange function, and the work can also be applied to tommysexp. I didn't post the results because they're complicated, and I didn't think they were good enough to provide a proof of nowhere analytic, but the results were very interesting. Here's a summary. The derivatives of the basechange/tommysexp functions act somewhat like a Russian stacking doll. For tommysexp, centered at x=0, with tommysexp(0)=1, the first 5 million or so derivatives are consistent with a radius of convergence of ~0.46, and appear to be well behaved. But then somewhere between the 5 millionth and the 6 millionth derivative, the radius of convergence abruptly changes to ~0.035. Then, again abruptly around the sexp(4)th derivative, the radius of convergence changes again, to approximately 0.0000000058. Near the sexp(5)th derivative, the radius of convergence abruptly changes to approximately 1/sexp(4). This goes on forever, with the radius of convergence getting arbitrarily small, but only for superexponentially huge derivatives.
 Shel
Posts: 641
Threads: 22
Joined: Oct 2008
11/15/2011, 04:48 PM
(This post was last modified: 11/17/2011, 10:50 PM by sheldonison.)
minor edit. I dug out my basechange delta approximation program. For the base change function, the crossover frequency for three versus four iterated logarithms of the superfunction of (exp(z)1) was at approximately the 2,700,000th taylor series term. For Tommysexp, it should be 2x that frequency.
(11/15/2011, 09:22 AM)sheldonison Wrote: I spent some time analyzing the derivatives of the basechange function, and the work can also be applied to tommysexp. I didn't post the results because they're complicated, and I didn't think they were good enough to provide a proof of nowhere analytic, but the results were very interesting. Here's a summary. The derivatives of the basechange/tommysexp functions act somewhat like a Russian stacking doll. For tommysexp, centered at x=0, with tommysexp(0)=1, the first 5 million or so derivatives are consistent with a radius of convergence of ~0.46, and appear to be well behaved. But then somewhere between the 5 millionth and the 6 millionth derivative, the radius of convergence abruptly changes to ~0.035. Then, again abruptly around the sexp(4)th derivative, the radius of convergence changes again, to approximately 0.0000000058. Near the sexp(5)th derivative, the radius of convergence abruptly changes to approximately 1/sexp(4). This goes on forever, with the radius of convergence getting arbitrarily small, but only for superexponentially huge derivatives.
 Shel The third approximation is very good. This is taking the iterated logarithm three times of the superfunction of 2sinh, and then recentering around zero.
.
If instead, you use four iterated logarithms, the first 5 million taylor [/color]series terms are virtually unchanged, to an accuracy of more than a million of digits. But eventually, the extra logarithm abruptly takes over the taylor series, and the radius of convergence switches. Here is k, and the first 30 terms of the taylor series of Tommysexp, with a radius of convergence of ~0.46. I should probably post the detailed work I did, which was really on the basechange function's taylor series at different points on the real axis, showing equations for how the taylor series changes with different numbers of logarithms, and showing the breakpoints in the taylor series caused by the iterated logarithms, before I forget it.
 Shel
Code: k= 0.067838366070752225
a0= 1.0000000000000000
a1= 1.0914653607683951
a2= 0.27333490639412095
a3= 0.21521847924246615
a4= 0.065271503768026496
a5= 0.039165656430892704
a6= 0.017152131406819940
a7= 0.011705880632484224
a8= 0.0047195886155923871
a9= 0.0012366758967825547
a10= 0.0022628815033596365
a11= 0.0032155986809687338
a12= 0.0082015427101427038
a13= 0.0021877770703785160
a14= 0.047737214601532077
a15= 0.11724756374916340
a16= 0.078884922073312641
a17= 0.88303830799631482
a18= 0.61016681587894246
a19= 5.1047079727741415
a20= 7.8161210634700306
a21= 28.647958035475997
a22= 60.048152189155451
a23= 173.31480125153131
a24= 382.32333460877720
a25= 1156.1406836504859
a26= 2075.6510351695731
a27= 8124.1576973283600
a28= 8589.2177234097241
a29= 56026.718096127404
Posts: 1,372
Threads: 336
Joined: Feb 2009
11/15/2011, 08:40 PM
(This post was last modified: 11/15/2011, 08:48 PM by tommy1729.)
Dear Sheldon
do you have the coefficients for tommysexp(tommyslog(x) + k) expanded at x = 0 with respect to x and some k : 0<k<1/2 and tommyslog(0)=0 ?
or tommysexp(1)=1
thanks
Posts: 641
Threads: 22
Joined: Oct 2008
(11/15/2011, 08:40 PM)tommy1729 Wrote: Dear Sheldon
do you have the coefficients for tommysexp(tommyslog(x) + k) expanded at x = 0 with respect to x and some k : 0<k<1/2 and tommyslog(0)=0 ?
or tommysexp(1)=1
thanks These are the first 30 terms for the taylor series for tommyslog(z) at z=1 and z=0, where tommyslog(z) is the inverse of tommysexp(z). I previously posted the taylor series for tommysexp(0)=1
Code: Taylor series for tommyslog(z), inverse of tommysexp
centered around z=1 where tommysexp(0)=1, tommyslog(1)=0
a0= 0.0000000000
a1= 0.9161994837
a2= 0.2102157372
a3= 0.05518392127
a4= 0.07650306010
a5= 0.02195729221
a6= 0.01723953659
a7= 0.01858002983
a8= 0.001674698879
a9= 0.008103278578
a10= 0.005531080198
a11= 0.001448690788
a12= 0.004865277068
a13= 0.01190384981
a14= 0.002255103164
a15= 0.07929305008
a16= 0.1797516453
a17= 0.04543643270
a18= 1.037251071
a19= 1.429908490
a20= 3.373553499
a21= 11.26149187
a22= 5.788279805
a23= 60.82409187
a24= 12.40696731
a25= 289.5467632
a26= 176.2348389
a27= 1331.084516
a28= 1061.765848
a29= 6150.410748
Taylor series for tommyslog(z), inverse of tommysexp
centered around z=0 where tommysexp(1)=0, tommyslog(0)=1,
a0= 1.000000000
a1= 0.9161994837
a2= 0.2478840046
a3= 0.1126997445
a4= 0.09072369004
a5= 0.01714998778
a6= 0.03540651071
a7= 0.003672300096
a8= 0.01892382429
a9= 0.0002980121835
a10= 0.01423518108
a11= 0.005494586590
a12= 0.007771395743
a13= 0.003971299138
a14= 0.006583409223
a15= 0.01295612958
a16= 0.07528917759
a17= 0.03705601856
a18= 0.5394289338
a19= 0.07994465159
a20= 3.644502979
a21= 1.240563331
a22= 24.48611084
a23= 4.182310661
a24= 163.5000785
a25= 48.75658767
a26= 1038.709146
a27= 1064.101937
a28= 5809.792386
a29= 12259.53117
For completeness, taylor series for tommysexp(z),
centered around z=1, where tommysexp(1)=0
a0= 0.000000000
a1= 1.091465361
a2= 0.3223134105
a3= 0.3503025671
a4= 0.2361619604
a5= 0.2014245880
a6= 0.1608281229
a7= 0.1430181261
a8= 0.1236438971
a9= 0.1089356872
a10= 0.1008682201
a11= 0.09506845024
a12= 0.09556770389
a13= 0.09154481719
a14= 0.03710079064
a15= 0.08963659000
a16= 0.01735261878
a17= 0.8895096731
a18= 1.563219154
a19= 3.642024787
a20= 12.03222965
a21= 16.78748395
a22= 80.80580588
a23= 92.01432366
a24= 500.6034989
a25= 650.7982979
a26= 2898.542045
a27= 5237.364750
a28= 14929.88995
a29= 41726.98836
