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+--- Thread: Multiple exp^[1/2](z) by same sexp ? (/showthread.php?tid=854)
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RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 05/06/2014
(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
RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 05/06/2014
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
RE: Multiple exp^[1/2](z) by same sexp ? - tommy1729 - 05/06/2014
I edited the previous post.