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I'm working my way through this paper (https://www.math.ubc.ca/~gerg/teaching/613-Winter2011/LargeSieveBombieriVinogradov.pdf) and am not quite seeing how one goes from (Lemma 3 page 3):

Note: $\lambda_{r}$'s are positive distinct real numbers and the z denote complex numbers.

$\sum_{s} \sum_{t \neq s} \frac{|z_{s}|^{2} + |z_{t}|^{2}}{(\lambda_{s} - \lambda_{t})^{2}} = 2 \sum_{s} |z_{s}|^{2} \sum_{t \neq s} \frac{1}{(\lambda_{s} - \lambda_{t})^{2}} $

The paper states that it is because it is due to the pair $(i,j)$ occuring once as $ s= i, t =j$ and once as $s=j, t=i$

I've tried plugging these in these values into the first sum but i don't see how the result follows.

Thank you.

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Does this seem more clear? $$\sum_s \sum_{t \neq s} \frac{|z_s|^2 + |z_t|^2}{(\lambda_s - \lambda_t)^2} = \sum_{(s,t)\\s \neq t} \frac{|z_s|^2 + |z_t|^2}{(\lambda_s - \lambda_t)^2}$$ $$= \sum_{(s,t)\\s \neq t} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2} + \sum_{(s,t)\\s \neq t} \frac{|z_t|^2}{(\lambda_s - \lambda_t)^2} = 2 \sum_{(s,t)\\s \neq t} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2}$$ $$= 2 \sum_s \sum_{t \neq s} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2} = 2 \sum_s |z_s|^2 \sum_{t \neq s} \frac{1}{(\lambda_s - \lambda_t)^2}.$$

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  • $\begingroup$ Thanks, i just had a couple of questions. 1): Does interchanging the index on the summation i.e. $ s \neq t $ and $t \neq s $ matter? 2): I'm not following how we get the first 2 if $ s \neq t$ ? $\endgroup$ – TKD Oct 8 '17 at 9:35
  • $\begingroup$ @TDK - Are you having trouble with the first equality? That summing over all possible $s$ and all possible $t$ such that $t \neq s$ is identical to summing over all possible pairs $(s,t)$ such that $s \neq t$? $\endgroup$ – John M Oct 10 '17 at 0:02
  • $\begingroup$ The step from $\sum_{(s,t)\\s \neq t} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2} + \sum_{(s,t)\\s \neq t} \frac{|z_t|^2}{(\lambda_s - \lambda_t)^2} = 2 \sum_{(s,t)\\s \neq t} \frac{|z_s|^2}{(\lambda_s - \lambda_t)^2}$ I'm not following where the "2" comes from if $ s \ne t $ $\endgroup$ – TKD Oct 10 '17 at 19:55
  • $\begingroup$ The two sums on the left hand side are identical. Just switch the labels $s$ and $t$ for one of those sums. $\endgroup$ – John M Oct 13 '17 at 21:14

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