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 Mizugadro, pentation, Book Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 01/14/2015, 01:12 AM (01/13/2015, 09:02 PM)sheldonison Wrote: (01/13/2015, 06:52 PM)Kouznetsov Wrote: ? How can I see that the precision is better? Have you calculated the residual at the substitution into the transfer equation? Or comparison with the explicitly–holomorphic Taylor expansion?There is a Taylor series in that post#10, for Pent(x-1); you could compare it to results you have.Ah, now I understand. The truncated Taylor series, by construction, is holomorphic; so, we can substitute it into the transfer equation and see the agreement. Let the truncated Taylor series be called F; let the transfer function be called T. Could you plot the map of agreement $ A(z)= - \lg \left( \frac {|T(F(z-1))-F(z)|} {|T(F(z-1))|+|F(z)|} \right)$ ? This agreement indicates, with how many decimal digits the transfer equation is satisfied. Quote:When I rerun in higher precision today, the Taylor series coefficients I first posted are accurate to approximately 21 decimal digits; that Pentation result used a 32 decimal digits Tetration implementation for its base. The pentation pari-gp program hasn't been updated in over three years ... I would probably clean it up if I were writing it today, and also write more math equations (instead of just pari-gp code). The most recent version I have (roughly the same as post#13) works fine, and can be run in arbitrary precision and gives pent(-0.5)=0.4910543386356481974128179471452718984517, which gives an error term vs that very first post of pent(-0.5)=-7E-22. How wide is area of agreement with 21 decimal figures? Is it possible to implement your algorithm in C++ and/or in Mathematica? Could you submit your result to some mathematical journal? Then, it may be easier to cite it. « Next Oldest | Next Newest »