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 [entire exp^0.5] The half logaritm. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/11/2014, 06:10 PM The weierstrass product expansion also seems mysterious ? regards tommy1729 « Next Oldest | Next Newest »

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