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Mizugadro, pentation, Book - Printable Version +- Tetration Forum (https://tetrationforum.org) +-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3) +--- Thread: Mizugadro, pentation, Book (/showthread.php?tid=949) |
RE: Mizugadro, pentation, Book - MphLee - 02/17/2015 (02/10/2015, 02:40 PM)Kouznetsov Wrote: I think, you are doing in the correct way. I hope, you made no errors.I have to notice that I didn't deduced it but just observed the progression of your formulas on your book. Anyways I tried to substitute YOUR solution \( \displaystyle F(z) = L+\sum_{n=1}^{\infty} a_n {\varepsilon^n} \) inside the transfer equation \( F(z+1)=\exp_b(F(x)) \)... but my deduction has some holes, I need some algebraic manipulation in order to come at your final form: In fact developing \( \exp_b(F(x)) \) I get \( \displaystyle \exp_b(F(x))= \exp_b( L+\sum_{n=1}^{\infty} a_n {\varepsilon^n} )=\sum_{i=0}^{\infty}{\ln(b)^i\over i!}\left ( L+\sum_{n=1}^{\infty} a_n {\varepsilon^n} \right)^i \) that looks very different from the series you show in your book at formula (6.4) \( \displaystyle \exp_b(F(x))=L+\sum_{i=0}^{\infty}{T^{[i]}(L)\over i!}\left ( L+\sum_{n=1}^{\infty} a_n {\varepsilon^n} \right)^i=L+\sum_{i=0}^{\infty}{\ln(b)^i b^L\over i!}\left (\sum_{n=1}^{\infty} a_n {\varepsilon^n} \right)^i \) (6.4) So the question is how one can go from \( \displaystyle\sum_{i=0}^{\infty}{\ln(b)^i\over i!}\left ( L+\sum_{n=1}^{\infty} a_n {\varepsilon^n} \right)^i \) TO \( L+\sum_{i=0}^{\infty}{\ln(b)^i b^L\over i!}\left (\sum_{n=1}^{\infty} a_n {\varepsilon^n} \right)^i \)? And how you can manipulate the series \( L+\sum_{i=0}^{\infty}{\ln(b)^i b^L\over i!}\left (\sum_{n=1}^{\infty} a_n {\varepsilon^n} \right)^i \) and obtain a series of the form \( L+\sum_{n=1}^{\infty} \tau_n {\varepsilon^n} \)? In other words Quote:Do you have any software at your computer to check your deduction? No I don't have powerful softwood and I don't even know how C++ works (I'd need a massive amount of time to learn it) ------------------ Other topic: With your supefunction formalism/metods/techniques(or theory or how you call it) is possible to find solution to factorial-like problems where there are involved families of different functions instead of only one function iterated many times? Consider this general problem Given a family of functions \( f_0,f_1,f_2,... \) we define a function \( H \) such that it satisfies this equations \( H(0)=\beta \) \( H(n+1)=f_n(H(n)) \) Is it possible to find an unique extensions of \(H\) to the real/complex numbers? This is a very general problem: if\( f_n(x)=n\cdot x \) and \( \beta=1 \) then \( H(n)=n! \) and its extension is the gamma function but we can still achieve uniqueness for other sequences \( f_0,f_1,f_2,... \) of functions? Thank you again for the kindness! RE: Mizugadro, pentation, Book - sheldonison - 03/02/2015 How many sample points are required for the Kouznetsov Cauchy integral method to get double precision (51 bits=16 decimal digits) accuracy? I seem to remember it was 1000-2000 sample points; also, how many iterations? Do you have any estimates on how fast the number of sample points required grows? For example, what it would take to get twice that precision, 32 decimal digits? |