Talk:Z function

arg &zeta;
Riemann-Siegel theta function says $&theta;(x) = arg &zeta;(&frac12; + t i)$, so $arg (e&minus; i &theta;(x) &zeta;(&frac12; + t i)) = 0$. If $Im Z(x) = 0$ as well, we get $&theta;(x) = 0$ or $&theta;(x) = &pi;/2$. How is that possible?

What &Omega; means
Z times &Omega; does not tend to 0, this is true; but it does not imply that $sup {Z(<VAR >t</VAR >): <VAR >t</VAR > < <VAR >x</VAR >}$ is increasing, and that meaning of the symbol &Omega; is nonstandard. --Yecril (talk) 08:23, 7 October 2008 (UTC)


 * Quite: usually, an &Omega; 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)