(10/08/2008, 04:22 PM)bo198214 Wrote: We know that the recurrence for \( b>0 \)
(1) \( f(0)=1 \)
(2) \( f(z+1)=b{f(z)} \)
has \( f(z)=b^z \) as the only entire solution that is bounded on the strip \( S=\{z: 0<\Re(z)\le 1\} \).
The image of \( S \) under \( b^z \) is an annulus for \( b>0 \) and so bounded. We know that for complex \( b\not\ge 0 \) the function \( f(z)=b^z=\exp((\log(z)+2\pi i k)z) \) is not bounded on \( S \) (the image is kinda infinite spiral) for any \( k \). The question remains whether
Conjecture
There is no entire solution \( f \) that satisfies (1) and (2) and is bounded on \( S \) for complex \( b\not\ge 0 \).
the general solution is b^(z + theta(z)) where theta (z) is an entire 1 periodic function.
since theta(z) must reach to all complex values in its range and period , then so does t(b,z) = b^theta(z) apart from 0.
since b^z is never zero and does not approach it , this implies t(b,z) b^z is also unbounded.
Afterall bounded and not approaching 0 times a function that is unbounded = unbounded.
conjecture update = b^(z + theta(z)) reaches all nonzero complex numbers.
for more details search for TPID 4.
regards
tommy1729