05/11/2014, 06:09 PM

If f is entire and grows not faster than exp(|z|) on the complex plane, it either takes all values or it is exp(az+b)+c.

Therefore since f takes all values, ln(f) has at least 1 logaritmic singularity.

Now let f be our beloved entire approximation of exp^[1/2].

Then it follows exp^[1/2] and exp^[-1/2] cannot both be entire.

In fact exp^[-1/2] is never entire because it is independant of the entirehood of exp^[1/2].

( a logarithm of a nonentire function is also nonentire ! )

Ofcourse this does not yet rule out an entire exp^[-1/2] if we allow a "fake" log.

However Im worried about how good of an approximation a fake log would give us.

The story is getting " complex ".

regards

tommy1729

Therefore since f takes all values, ln(f) has at least 1 logaritmic singularity.

Now let f be our beloved entire approximation of exp^[1/2].

Then it follows exp^[1/2] and exp^[-1/2] cannot both be entire.

In fact exp^[-1/2] is never entire because it is independant of the entirehood of exp^[1/2].

( a logarithm of a nonentire function is also nonentire ! )

Ofcourse this does not yet rule out an entire exp^[-1/2] if we allow a "fake" log.

However Im worried about how good of an approximation a fake log would give us.

The story is getting " complex ".

regards

tommy1729