Timeline for Using Vinogradov's theorem for finding prime solutions to a linear equation (an exercise from Vaughan's book)
Current License: CC BY-SA 3.0
13 events
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| Jan 22, 2012 at 1:20 | history | edited | Timothy Foo | CC BY-SA 3.0 |
added term
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| Jan 21, 2012 at 19:48 | comment | added | Tal H | I tried to use your idea for "case 2" to get the general case. Does this seem true to you? | |
| Jan 21, 2012 at 1:20 | history | edited | Timothy Foo | CC BY-SA 3.0 |
formatting; Post Made Community Wiki
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| Jan 21, 2012 at 1:17 | comment | added | Timothy Foo | You are welcome. Some special cases have been added. Let me know if you have more ideas too. | |
| Jan 21, 2012 at 1:14 | history | edited | Timothy Foo | CC BY-SA 3.0 |
latex
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| Jan 21, 2012 at 1:08 | history | edited | Timothy Foo | CC BY-SA 3.0 |
latex, phrasing
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| Jan 20, 2012 at 13:22 | comment | added | Tal H | Thank you for a very detailed answer. I think the result you mentioned about $u_i(y)\ll \min (N,\frac {1}{||b_iy||})$ can be used to obtain: $$\int_\frac {Q}{N} ^\frac {1}{2b} u_1(y)u_2(y)u_3(y)e(-b_4y)dy\ll (b_1b_2b_3)^{-1} (\frac{Q}{N})^2$$since if $y\in (\frac{Q}{N},\frac{1}{2b})$, then $$y\leq \frac{1}{2b}\Rightarrow by\leq \frac {1}{2} \Rightarrow b_iy\leq \frac {1}{2}$$which gives $||b_iy||=|b_iy|$ and therefore $u_i(y)\ll \frac {1}{|b_iy|}$. But, like you said, this is the easy part of the interval. Would be happy to know if you have more ideas. | |
| Jan 20, 2012 at 12:44 | history | edited | Timothy Foo | CC BY-SA 3.0 |
added references
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| Jan 20, 2012 at 9:00 | history | edited | Timothy Foo | CC BY-SA 3.0 |
added stuff
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| Jan 20, 2012 at 8:03 | history | edited | Timothy Foo | CC BY-SA 3.0 |
removed part
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| Jan 20, 2012 at 7:29 | history | edited | Timothy Foo | CC BY-SA 3.0 |
phrasing
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| Jan 20, 2012 at 7:20 | history | edited | Timothy Foo | CC BY-SA 3.0 |
latex
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| Jan 20, 2012 at 7:10 | history | answered | Timothy Foo | CC BY-SA 3.0 |