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On Fri, 29 Jul 2005 21:57:16 EDT, Eldes <ilnippon@xxxxxxxxx> wrote:
>In other words, the proof is rather of functional analytical type:
>
>Let f(s) be complex-valued, analytic (i.e., holomorphic, <=> continuous,)
>everywhere except for s = 1, f(0 + it) \not= 0 for any positive t, and satisfy
>
>|f(Re(s) + it)| = |g(s)||f(1 - Re(s) + it)|,
>
>where g(s) is nonzero everywhere. Then f(s) can have zeros only at Re(s) =
>1/2.
>
>In other words, if you could show a counterexample to the theorem above, then
>my proof will fall on some part.
If I didn't drop any minus signs then f(s) = (s - i)(s - (1+i))
is a counterexample, with g(s) = 1.
>Eldes
>
>----------------------------------------------------------
>This is the way I do math, miss!
************************
David C. Ullrich
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