The theory of exponent pairs provides bounds for $$\sum_{N<n<2N} e(f(n)),$$ where f behaves like a monomial. Precise formulations of this are in Graham and Kolesnik (GK) which seems to be what is cited in the literature when one wants to apply the A and B processes.

The problem is that for a function such as $f(n) = n^{3/2}$, it does not satisfy equation 3.3.3 in GK. Nevertheless, one should typically be able to apply the A and B processes.

I am asking for a reference that I can cite which allows one to apply the A and B processes to such functions.


2 Answers 2


See Montgomery's book, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, specifically the examples on p.55. He considers variants of your sum, namely $\sum_{n \leq N} e((n/3)^{3/2})$, and $\sum_{n \leq N} e((2n/3)^{3/2})$, and shows that they have very different asymptotic behavior. The former is asymptotically $c N^{3/4}$ while the latter is $O(N^{1/4})$. These examples illustrate why the hypothesis (3.3.3) in [GK] is assumed.

  • $\begingroup$ Thanks for your response! But, for instance if we consider $e(tn^{3/2})$ and t is transcendental, one should always be able to apply the A and B processes, with some basic assumptions avoiding the case where the derivatives of the resulting amplitude function become too small. Do you know if this is the case? $\endgroup$ Oct 4, 2017 at 16:58
  • $\begingroup$ You may certainly apply Poisson summation or the Weyl/van der Corput differencing method to the exponential sum, which you can analyze in an ad-hoc way. The point of the two examples above is that you cannot automatically determine the behavior of the exponential integral only using derivative bounds on the phase function. $\endgroup$
    – Matt Young
    Oct 4, 2017 at 17:22

Applying the A^2B process to n^3/2 will get you to a linear function, for which certainly this method does not work.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.