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I have encountered such an exponential sum $$\sum_{m\leqslant M}\bigg|\sum_{n\leqslant N}a_ne(\alpha mn)\bigg|^{2k},$$ where $a_n$'s are arbitrarily coefficients bounded by 1.

I don't know how to estimate such an exponential sum, and I don't know whether Vinogradov's method works. I want to know whether there are certain references regarding its upper bound.

PS:

If $\alpha=p/q$ $((p,q)=1)$ is a rational number, I can deduce an estimate such as $$\ll(M+q)(N^{2k}/q+1)q^\varepsilon.$$

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    $\begingroup$ Using the large sieve, I come up with $(M+q)(N+q) N^{2k-1}/q$ for a bound. The diagonal term alone seems to give $MN^{2k-1}$ which is not consistent with your stated bound in case $M > q > N$. Can you sketch your argument? Also, what ranges of variables are you interested in? $\endgroup$
    – Matt Young
    May 21, 2011 at 4:52

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To Prof. Young,

The function of "comments" has the limitation on the number of words, so I write my ideas here.

Sorry, there is a serious mistake in my upper bound. In fact my argument yields $N^{2k-1}(N/q+1)(M+q\log q)$ with is just the same with yours up to a log facor. Here is the detail.

$$THE~~SUM=\mathop{\sum\sum}_{n_1,\cdots,n_{2k}\leqslant N}\sum_{m\leqslant M}e\left(\frac{pm((n_1+\cdots+n_k)-(n_{k+1}+\cdots+n_{2k}))}{q}\right) $$

$$\ll\mathop{\sum\sum}_{n_1,\cdots,n_{2k}\leqslant N}\min\left(M,\left\|\frac{p((n_1+\cdots+n_k)-(n_{k+1}+\cdots+n_{2k}))}{q}\right\|^{-1}\right)$$

$$\ll\sum_{a\bmod q}\min\left(M,\left\|\frac{q}{a}\right\|^{-1}\right)\mathop{\sum\sum}_{\substack{n_1,\cdots,n_{2k}\leqslant N\\(n_1+\cdots+n_k)-(n_{k+1}+\cdots+n_{2k})\equiv \overline{p}a(\bmod q)}}1.$$

Since the summations over $n_i$ are bounded by $N^{2k-1}(N/q+1)$ for each $a\bmod q$, we can conclude the estimate $$\ll N^{2k-1}(N/q+1)\sum_{a\bmod q}\min\left(M,\left\|\frac{q}{a}\right\|^{-1}\right)\ll N^{2k-1}(N/q+1)(M+q\log q).$$

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  • $\begingroup$ BTW: I am just interested in the range $M,N\leq q$ or $M,N\leq q^{1/2}$. $\endgroup$
    – arithboy
    May 21, 2011 at 10:44
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    $\begingroup$ Thanks for the clarification. You can also just create a sum of the form $\sum_{m \leq M} |\sum_{l \ll N} c_l e(\alpha l m)|^2$ where $c_l = \sum_{n_1 + \dots + n_k = l} a_{n_1} \dots a_{n_k}$. Then you can use the large sieve to bound the sum using bounds on $\sum_{l \ll N} |c_l|^2$. $\endgroup$
    – Matt Young
    May 22, 2011 at 3:51

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