rumor result tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 11/14/2011, 09:35 PM 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 sheldonison Long Time Fellow 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 tommy1729minor 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 sheldonison Long Time Fellow 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. - ShelThe third approximation is very good. This is taking the iterated logarithm three times of the superfunction of 2sinh, and then recentering around zero. $\text{tommysexp}(z)=\log^{[3]}(\text{superfunction}_{\text{2sinh}}(z+k+3))$. 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 tommy1729 Ultimate Fellow 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