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 using sinh(x) ? sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 11/20/2012, 12:01 AM (This post was last modified: 11/21/2012, 04:04 PM by sheldonison.) (11/18/2012, 11:41 PM)tommy1729 Wrote: tommysexp(z) and sheldontommysexp(z) agree on the reals (converge speed may differ though ). Now tommysexp(z) seems to be only defined for the real z so it requires analytic continuation ( assuming it is possible ). However sheldontommysexp(z) IS periodic and might be analytic in that period. However to go beyond that period we will need analytic continuation because the C property forbids nonreal periodicity. ( again assuming it is possible ). $\text{tommysexp}_3(z)=\log^{[3]}(\text{2sinh}^{[z]}(\exp^{[3]}(1)))$ This is tommysexp generated with three iterated logarithms. This function is reasonably well behaved at the origin, with a radius of convegence of approximately 0.457. For reference, the taylor series coefficients are listed below. $\text{tommysexp}_4(z)=\log^{[4]}(\text{2sinh}^{[z]}(\exp^{[4]}(1)))$ But the next step, for n=4, is not so well behaved in the complex plane. The radius of convergence in the complex plane is about 0.03469, and by my count, there are 587507 singularities within a radius of 0.035. And its not just a matter of extending the function beyond the wall of singularities, because the function misbehaves inside the wall of singularities too. Here is where singularities occur, due to the log(0). $\text{2sinh}^{[z]}=n\pi i$ $\text{2sinh}^{[z+1]}=0$ The other interesting thing about the singularities is how quickly they fall off. At 99.999% of the singularity radius, $|\text{tommysexp}_4(z)-\text{tommysexp}_3(z)|<10^{-50}$. Its almost as if one can totally ignore the effect of the fourth iteration. And in fact the 4th iteration has no measurable effect on any taylor series coefficient one would use in normal computation. But, somewhere around the 1352620th taylor series term there is an abrupt transition. The $\text{tommysexp}_4(z)$ function starts behaving radically differently than the $\text{tommysexp}_3(z)$, as the taylor series terms finally become dominated by the wall of nearby singularities. By the way, estimating the 1.35 millionth taylor series coefficent for a function is a very delicate calculation involving algorithms to make approximations cauchy integrals at carefully picked radius's of $s(z)-\log(e^{s(z)}-e^{-s(z)})=\sum_{n\to\infty}-e^{-2n s(z)}$, where $s(z)=\text{2sinh}^{[z]}$. After picking an appropriate approximation radius, I can estimate the log of the desired taylor series coefficient. It helps that usually the different values of "n" in the sum can be treated independently. I have a pari-gp program, but haven't posted the details, which are complicated. The algorithm works for both the 2sinh(z) superfunction, as well as the exp(z-1) superfunction. One final thought. $\text{tommysexp}_3(z)$ has a radius of convergence of ~0.457 which can be seen in the taylor series coefficients below. It is not as well behaved as sexp(z) in the complex plane. The interesting thing is that $\text{tommysexp}_4(z)$ has the same first 40 taylor series coefficients listed below, accurate to millions of decimal digits. This would also be true of the first million or so terms, until the transition. So, in addition to misbehaving at the 1.35 millionth taylor series term, $\text{tommysexp}_4(z)$ still has all smaller taylor series coefficients dominated by the singularities in $\text{tommysexp}_3(z)$. $\text{tommysexp}_5(z)$ would have a radius of convergence of less than 10^-7, that would effect uncountably large taylor series terms due to a wall of uncountably many nearby singularities. There's nothing particularly special about doing the approximations for tommysexp(0), and similar misbehavior would be expected for any value of tommysexp(z). I could post the methods I used for these approximations which might lead to a rigorous proof, but the general problem is complicated. You need to show the taylor series terms eventually grow faster than any radius of convergence. The difficulty is coming up with a rigorous language to express appropriate approximations of superexponentially large taylor series terms. Even so, I am convinced that tommysexp(z) is nowhere analytic, even though it is infinitely differentiable at the real axis. - Sheldon added pretty graph showing accuracy of tommysexp3 taylor series terms, and switchover to tommmysexp4 taking over near 1.35 millionth taylor series term     Code:Taylor series coefficients for tommysexp(0) a0=   1.00000000000 a1=   1.09146536077 a2=   0.273334906394 a3=   0.215218479242 a4=   0.0652715037680 a5=   0.0391656564309 a6=   0.0171521314068 a7=   0.0117058806325 a8=   0.00471958861559 a9=   0.00123667589678 a10= -0.00226288150336 a11=  0.00321559868096 a12= -0.00820154271015 a13=  0.00218777707040 a14=  0.0477372146016 a15= -0.117247563749 a16= -0.0788849220733 a17=  0.883038307996 a18= -0.610166815881 a19= -5.10470797278 a20=  7.81612106347 a21=  28.6479580355 a22= -60.0481521891 a23= -173.314801252 a24=  382.323334609 a25=  1156.14068365 a26= -2075.65103517 a27= -8124.15769733 a28=  8589.21772341 a29=  56026.7180961 a30= -10973.8915911 a31= -353646.843999 a32= -264336.726813 a33=  1880963.01782 a34=  3669078.97541 a35= -7012765.98983 a36= -30701470.8277 a37=  1548864.00646 a38=  184151139.583 a39=  254566511.140 « Next Oldest | Next Newest »

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