Derivative of exp^[1/2] at the fixed point? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 01/01/2016, 03:58 PM (This post was last modified: 01/01/2016, 04:12 PM by sheldonison.) (12/31/2015, 01:25 PM)tommy1729 Wrote: The 5 th derivative of $\approx \left(4.44695n+1.05794ni\right)^{z-L}$ is equal to of $\approx \left(4.44695n+1.05794ni\right)^{z-L} ln(4.44 n + 1.05 n i)^5$ ?? No singularity ? Im sure you make sense , but it is not clear what you are doing to me. Regards Tommy1729 I apologize for the typos, which I corrected. The correct equation is (z-L)^p, where (z-L) is being raised to a complex power. $h_k(z) \approx h(z) + c \cdot (z-L)^{(4.44695+1.05794i )}\;\;\;\;$ p ~= 4.44695+1.05794i is the pseudo period of sexp The fifth derivative has the real part of the power term negative, so the value is no longer defined at L, just like $z^{-0.553}=\frac{1}{z^{0.553}}$ is not continuous at z=0. But the first four derivatives are defined and equal to zero at L. - Sheldon « Next Oldest | Next Newest »

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