Let f(z) = sum z^n/(2^n!)

I really wonder how fast f^[a](z) grows ...

Im fascinated by my own function

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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