(05/11/2009, 09:12 PM)BenStandeven Wrote: Actually, \( \cos(2^x) = \cosh(i 2^x) = \cosh(2^{x + \frac{\pi i}{2 \ln 2}}) \), so they are translations of each other, albeit along the imaginary axis instead of the real axis.
Well observed! So this is not even the worst non-uniqueness.
This is generally the case, that there are these two regular super-functions at a fixed point. And one is the other translated by some imaginary value (up to real translations along \( x \)) which is half of the period of both functions.
