05/21/2008, 06:24 PM
Inspired by Kouznetsov's consideration I investigated a bit more in this direction and found interesting results:
1.
Let F be holomorphic on the right half plane, and let F(z+1)=zF(z),
F(1)=1, further let F be bounded on the strip 1<=Re(z)<2, then F is
the gamma function.
2.
\( \exp_b \) is also determined as the only function that satisfies \( b^{x+1}=bb^x \), \( b^1=b \) and is bounded on the strip 1<=R(z)<2 (or 0<=R(z)<1).
3. I started a thread in sci.math.research and it turned out that the same criterion also makes the Fibonacci function
\( F(z)=\frac{\phi^z - (1-\phi)^z}{\sqrt{5}} \), \( \phi=\frac{1+sqrt(5)}{2} \) the unique extension of \( F(n+1)=F(n)+F(n-1) \), \( F(0)=0 \), \( F(1)=1 \).
So I really would guess that this criterion (which is a slightly weaker demand than Kouznetsov's criterion) also implies uniqueness for tetration, at least for base \( b>e^{1/e} \).
Maybe its not even difficult to prove.
1.
Let F be holomorphic on the right half plane, and let F(z+1)=zF(z),
F(1)=1, further let F be bounded on the strip 1<=Re(z)<2, then F is
the gamma function.
2.
\( \exp_b \) is also determined as the only function that satisfies \( b^{x+1}=bb^x \), \( b^1=b \) and is bounded on the strip 1<=R(z)<2 (or 0<=R(z)<1).
3. I started a thread in sci.math.research and it turned out that the same criterion also makes the Fibonacci function
\( F(z)=\frac{\phi^z - (1-\phi)^z}{\sqrt{5}} \), \( \phi=\frac{1+sqrt(5)}{2} \) the unique extension of \( F(n+1)=F(n)+F(n-1) \), \( F(0)=0 \), \( F(1)=1 \).
So I really would guess that this criterion (which is a slightly weaker demand than Kouznetsov's criterion) also implies uniqueness for tetration, at least for base \( b>e^{1/e} \).
Maybe its not even difficult to prove.