Talk:Bombieri–Vinogradov theorem

The right Vinogradov?

The article says A. I. Vinogradov, who I have never heard of. Could it be the analytic number theorist I. M. Vinogradov instead? 165.189.91.148 15:47, 29 June 2006 (UTC)[reply]

No. It is A. I. Vinogradov. Mon4 00:09, 30 June 2006 (UTC)[reply]

Surprisingly vanishing A

After establishing that A is a positive real, it's surprising that A doesn't appear in the first equation. Since it does not, the second equation should collapse (since nothing blocks taking the limit A --> 0).

It's also unclear what role y takes in the maximum over y<=x. Specifically, for x on the RHS to be the same x on the LHS, x must be free on the LHS. This implies that y is being bound by the maximum and we should have psi(y;q,a) and y/phi(q). The semantics, corrected this way, would the be : let x be the value of y such that "max over a s.t. 1<=a<=q and (a,q)=1, ..." is maximized and the value of this subexpression be that maximum. So then x is bound to the value producing the maximum and that maximum value of the subexpression is passed out to the sum.

-- Fuzzyeric 05:28, 27 December 2006 (UTC)[reply]

A is an absolute constant: it is not allowed to take limits of such things. Beware of 'implied constants' that are actually dependent on A. Charles Matthews 11:23, 27 December 2006 (UTC)[reply]

There should be at least some mentioning of A somewhere in the text (by someone more knowledgeable of the topic) explaining whether the statement is true for any A or any positive real A or one specific A known as the Bombieri-Vinogradov constant with value approximately equal to blah blah or some other possibility I haven't even thought of. As the article is written now the appearance of A is even more surprising than its vanishing.Octonion (talk) 11:39, 30 October 2015 (UTC)[reply]