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Here is a link on the internet: https://www.dropbox.com/s/su3uak2a057yrqv/YitangZhang.pdf

Can someone teach me how to use trivial estimation to reach (6.1) on page 24? Namely, how to impose $(d,P_0)<D_1$ and replace $\theta$ by $\Lambda$ with acceptable O-term?

Thanks a lot!

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up vote 8 down vote accepted

Let me remind some of notations for readers: $$\mathcal{L}=\log x,$$ $$k_0=3.5\times 10^6, \varpi=\frac{1}{1168},$$ $$D_0=\exp(\mathcal{L}^{\frac{1}{k_0}}), P_0=\prod_{p < D_0}p,$$ $$D_1=x^{\varpi},P=\prod_{p < D_1}p,$$

$$D=x^{\frac{1}{4}+\varpi}, D_2 = x^{\frac{1}{2}-\epsilon}$$

Here, the effect of replacing $\theta$ by $\Lambda$ produces an error of $$O(x^{\frac{1}{2}}\mathcal{L}^B),$$ for some positive $B$. So, this change is negligible compared to $O(x\mathcal{L}^{-A})$.

For imposing $(d,P_0)< D_1$, we claim that the following sum is also negligible compared to $O(x\mathcal{L}^{-A})$: $$ \sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)| $$ Note that the trivial bound for $|\Delta(\Lambda,d,c)|$ is: $$ |\Delta(\Lambda,d,c)|=O(\frac{x\mathcal{L}}{d})$$

Since we have $(d,P_0)\geq D_1$, the number $w(d)$ of distinct prime divisors of $d$, must satisfy $$ w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})}.$$

The sum is bounded by: $$ x\mathcal{L} \sum_{\substack{{ d {<} D^2} \\\ {w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})} } }} \frac{\tau_{k_0}(d)}{d }$$

Standard argument now applies, and the sum above is bounded by: $$ \frac{1}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}\sum_{d < D^2} \frac{2^{w(d)}\tau_{k_0}(d)}{d}=O(\frac{\mathcal{L}^B }{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}),$$ for some positive constant $B$.

Combining all together, we obtain an upper bound: $$ \sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)| =O(\frac{x\mathcal{L}^{B+1}}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}} })$$

Hence, our claim follows.

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It sounds nice, and I just remark that the sum over $c\in C_i(d)$ gives a $\tau_{k_0}(d)$ which can be handled by your method similarly. Thanks! –  ericc May 29 '13 at 3:58
    
@ericc: Yeah, I missed that. The extra $\tau_{k_0}(d)$ will give extra powers of $\mathcal{L}$ on the numerator. But, my argument still works as you noticed. –  i707107 May 29 '13 at 4:50
    
@ericc: Edited. Implemented your correction, and simplified a little. I did not need to have $d=d_0 k$, since what I needed is $d$ having many prime factors. –  i707107 May 29 '13 at 18:41
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A typographical correction: the character represented by "$\bar w$" is actually $\varpi$ (\varpi). –  Noam D. Elkies Jun 5 '13 at 1:34
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I changed $\bar{w}$ to $\varpi$ in this post. –  GH from MO Jun 6 '13 at 22:38
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