Timeline for Cancellation in a very rapidly oscillating exponential sum
Current License: CC BY-SA 4.0
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| Apr 30, 2023 at 15:34 | comment | added | Terry Tao | Your particular phase function $f(n) := e^{2\pi i T/n}$ does obey a somewhat unusual equation $f(n) = f(pn)^p$ in lieu of multiplicativity, but it is not clear to me how to exploit this equation to get any further cancellation. | |
| Apr 30, 2023 at 15:30 | comment | added | Terry Tao | ...the function $n \mapsto e^{2\pi i T \log n}$ is of course multiplicative, and so one could in principle use various flavors of multiplicative number theory (elementary, pretentious, complex-analytic) to try to analyze the $\log n$ sum in particular, but this basically boils down to requesting more information about the Riemann zeta function (particularly near the right edge $\{1+it\}$ of the critical strip), so in the absence of powerful hypotheses such as RH this is somewhat circular, since the best control we have on zeta near this edge comes from the Vinogradov estimate. | |
| Apr 30, 2023 at 15:25 | comment | added | Terry Tao | For $\log n$ there are potentially other ways to estimate this sum using other information on the Riemann zeta function, for instance if one assumes RH, and there are various techniques (e.g., moment method, large sieve type methods, etc.) that permit one to obtain estimates for almost all $T$ rather than all $T$, but I don't know of any unconditional method that can go beyond the Vinogradov range uniformly in $T$ for any non-trivial exponential sum (excluding degenerate cases such as geometric series). | |
| Apr 30, 2023 at 13:14 | comment | added | Random | One more question: as you've mentioned here and in the blog, both methods essentially approximate $1/n$ by its Taylor expansion. Are there other methods that work specifically for $1/n$ or $\log n$ and not for generic smooth functions $f$ (that maybe give less optimal bounds than Vinogradov's method)? The reason I ask is that for $k > \log x$ (that is, $T > e^{\log(x)^2}$) we cannot approximate $T/n$ with its Taylor expansion, so another idea must be used to get past this. | |
| Apr 29, 2023 at 20:44 | comment | added | Terry Tao | For the sum $\sum_{x \leq n \leq 2x} e^{2\pi i T/n}$, yes; but for our specific application we were actually concerned with the variant sum $\sum_{x \leq p \leq 2x} e^{2\pi i T/p}$ over primes, to which your upper bound construction doesn't apply. (That said, currently the only way to control the sum over primes is to first control the sum over natural numbers, so some other method would have to be used if we were to ever get close to that conjecture.) | |
| Apr 29, 2023 at 19:59 | vote | accept | Random | ||
| Apr 29, 2023 at 19:59 | comment | added | Random | Thanks, this is exactly what I was looking for. Regarding your second edit, I think the upper bound I describe in the question shows that contrary to what is conjectured in footnote $2$, the upper bound on $M, N$ cannot be relaxed all the way to $O(\text{exp} (P^c))$ right? | |
| Apr 29, 2023 at 19:33 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Apr 29, 2023 at 16:18 | comment | added | Terry Tao | All of the exponential sum methods mentioned here can handle sums of the form $\sum_{x \leq n < 2x} e^{2\pi i f(n)}$ where $f$ is a smooth function obeying bounds such as $\alpha^{-j^2} F \leq \frac{t^j}{j!} |f^{(j)}(t)| \leq \alpha^{j^2} F$ for some constants $\alpha, F$ and all $t \in [x,2x]$, and some suitable range of $j$. Both $t \mapsto T/t$ and $t \mapsto T \log t$ obey estimates of this form (with $\alpha \asymp 1$ and $F$ equal to $T/x$ in the former case and $T$ in the latter case). | |
| Apr 29, 2023 at 16:16 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Apr 29, 2023 at 15:55 | comment | added | mathworker21 | $+1$, thanks! Why is $\sum_{x \le n < 2x} e^{2\pi i T \log n}$ "extremely similar"? | |
| Apr 29, 2023 at 15:49 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Apr 29, 2023 at 15:43 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Apr 29, 2023 at 15:38 | history | answered | Terry Tao | CC BY-SA 4.0 |