Talk:Bombieri–Vinogradov theorem: Difference between revisions
Surprisingly vanishing ''A'' |
→Surprisingly vanishing ''A'': Typo'ed "RHS" for "LHS" |
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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). |
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). |
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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 |
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. |
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-- [[User:Fuzzyeric|Fuzzyeric]] 05:28, 27 December 2006 (UTC) |
-- [[User:Fuzzyeric|Fuzzyeric]] 05:28, 27 December 2006 (UTC) |
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Revision as of 05:29, 27 December 2006
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)
- No. It is A. I. Vinogradov. Mon4 00:09, 30 June 2006 (UTC)
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.