05/05/2014, 06:18 PM
OMG. FOund a very nice theorem in complex analysis. I found it referenced in this paper.
http://algo.inria.fr/seminars/sem01-02/delabaere2.pdf
[4] Boas, Jr. (Ralph Philip). – Entire functions. – Academic Press, New York, 1954, x+276p.
It says that:
"In fact, if a and b are two [holomorphic] functions that are of exponential type α < π, if a(n) = b(n) for all
n ≥ 1, then a = b, due to a theorem by Carlson [4]."
THEREfore if:
\( |\psi(z)|,|\phi(z)| < C e^{\alpha|\Im(z)| + \rho|\Re(z)|} \) for \( 0 \le \alpha < \pi/2 \) and \( 0 \le \rho < \pi \)
We know \( \psi(n) = \phi(n) \) implies \( \psi = \phi \)!!!!!
This is a nice uniqueness result.
http://algo.inria.fr/seminars/sem01-02/delabaere2.pdf
[4] Boas, Jr. (Ralph Philip). – Entire functions. – Academic Press, New York, 1954, x+276p.
It says that:
"In fact, if a and b are two [holomorphic] functions that are of exponential type α < π, if a(n) = b(n) for all
n ≥ 1, then a = b, due to a theorem by Carlson [4]."
THEREfore if:
\( |\psi(z)|,|\phi(z)| < C e^{\alpha|\Im(z)| + \rho|\Re(z)|} \) for \( 0 \le \alpha < \pi/2 \) and \( 0 \le \rho < \pi \)
We know \( \psi(n) = \phi(n) \) implies \( \psi = \phi \)!!!!!
This is a nice uniqueness result.
