Derivative of exp^[1/2] at the fixed point?

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RE: Derivative of exp^[1/2] at the fixed point? - sheldonison - 01/01/2016

(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.