[entire exp^0.5] The half logaritm. - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: [entire exp^0.5] The half logaritm. (/showthread.php?tid=864) [entire exp^0.5] The half logaritm. - tommy1729 - 05/11/2014 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 RE: [entire exp^0.5] The half logaritm. - tommy1729 - 05/11/2014 The weierstrass product expansion also seems mysterious ? regards tommy1729