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].
\( \exp^{[1/2]}(x) = \theta(x) \int_{\0}^{\infty} \((z (\2sinh^{[-1]}(z) - 1) )!)^{-1} x^{z}\,dz \)
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].
\( \exp^{[1/2]}(x) = \theta(x) \int_{\0}^{\infty} \((z (\2sinh^{[-1]}(z) - 1) )!)^{-1} x^{z}\,dz \)
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