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 Multiple exp^[1/2](z) by same sexp ? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/06/2014, 09:12 PM (05/06/2014, 09:14 AM)sheldonison Wrote: All three of these half exponential functions have positive Taylor series coefficients up to around x^14 or x^15th. This has been known to me for years. Many of these half exponential solutions ( not just the 3 you mentioned ) have this strange property. And why cant we easily find one that has all these coefficients positive ? This has troubled me for a long time. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/06/2014, 09:59 PM (This post was last modified: 05/06/2014, 10:57 PM by tommy1729.) Let f(z) = sum z^n/(2^n!) I really wonder how fast f^[a](z) grows ... Im fascinated by my own function --- A few more remarks on these slow growing functions. Let g(z) be a real entire nonpolynomial function. If the " growth (g(z)) = 0 " where growth is as defined by sheldon , then that g(z) is very close to its truncated Taylor series ; a polynomial. Since g(z) is close to a polynomial this has some important implications : It is UNLIKELY that g(z) has ONLY 1 fixpoint AND that 1 fixpoint has multiplicity 1 and derivative y with 0 < y < 1. It follows that the fractal for g(z) is PROBABLY very similar to that of a polynomial. Similar case for a conjugate fixpoint pair. Many fixpoints are also a complication when considering superfunctions ... Yet for the investigation of comparing g^[n] to sexp(n) ( or growth rate ) these superfunctions of g(z) are intresting and maybe needed ... Another question is : I wrote UNLIKELY AND PROBABLY. Can we be more specific ? We seem to have " growth = 0 " => implications. Does the inverse : implications => " growth = 0 " make sense ? For instance : if we have a fractal resembling that of a polynomial , can we conclude the function has growth = 0 ?? This also brings me to the next idea : If we want a fractal of a transcendental entire function with finite growth , do we Always end up in a fractal resembling a polynomial or one resembling exp ? ( This is somewhat suggested by the above. Note that the fractal of exp is the same as that of exp(exp) and that of exp^[1/2] ! ) If the answer to that is NO , then there are PROBABLY two types of slow functions : the ones who have polynomial-like fractals and those who dont. ( where the ones who dont, grow faster than the others ) ( edited ) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/06/2014, 10:55 PM I edited the previous post. « Next Oldest | Next Newest »

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