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[entire exp^0.5] The half logaritm.
#1
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

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#2
The weierstrass product expansion also seems mysterious ?

regards

tommy1729
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