06/15/2014, 06:22 PM
(This post was last modified: 06/15/2014, 06:37 PM by sheldonison.)
(03/28/2014, 12:04 AM)tommy1729 Wrote: Notice 0 < b^z < oo for finite complex z !
hence since b^z is unbounded to make f(z,1) b^z bounded we need f(z,1) going to 0 near the imaginary limits +/- oo i.
But f(z,1) b^z needs to be entire. And hence (by the above) f(z,1) cannot be entire.
Thus if f(z,1) b^z needs to be entire and f(z,1) goes to oo (f(z,1) goes to oo somewhere because its not entire ) ,we MUST conclude we need points z1 where b^z1 are 0.
But b^z is NEVER 0 in any strip.
I viewed this proof only applied to b^z; that any other solution of b^(z+theta(z)) would be unbounded on the strip, and that's how I interpreted the proof. I didn't view this as a proof relating to Tetration, and I didn't think the OP viewed this as applying to Kneser's tetraton solution, a non-entire function, since tet(z) and sexp(z) are never mentioned in the post. But for b^z, you can multiply by a 1-cyclic function and still have a solution to b^z, so I interpreted the OP as intending this post to only apply to b^z. You can't multiply tet(z) by a 1-cyclic function and get an alternative solution to tet(z). But maybe I'm missing something.
- Sheldon
