Talk:Z function

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arg ζ

Riemann-Siegel theta function says $θ(x) = arg ζ(½ + t i)$ , so $arg (e - i θ(x) ζ(½ + t i)) = 0$ . If $Im Z(x) = 0$ as well, we get $θ(x) = 0$ or $θ(x) = π/2$ . How is that possible?

What Ω means

Z times Ω does not tend to 0, this is true; but it does not imply that $sup {Z(t): t < x}$ is increasing, and that meaning of the symbol Ω is nonstandard. --Yecril (talk) 08:23, 7 October 2008 (UTC)[reply]

Quite: usually, an Ω theorem is the negation of the corresponding little-o form -as you say! Also, a product is not involved - it's a quotient of the absolute values - usually! Hair Commodore (talk) 17:24, 10 January 2009 (UTC)[reply]