Tricki

## Revision of Getting rid of nasty cutoffs from Thu, 02/04/2009 - 11:15

### Quick description

When faced with an unpleasant looking cutoff condition in a quantity one wishes to estimate it, try expanding that condition using Fourier analysis.

### Prerequisites

Basic harmonic analysis

### Example 1

The classic example of this technique is the Pólya-Vinogradov theorem. Let be a prime, and write for the Legendre symbol: thus if is a quadratic residue modulo and is otherwise. The Pólya-Vinogradov theorem states that if then

What this means is that, provided , about 50 percent of the numbers less than are quadratic residues modulo .

There is a particular value of for which the result is very easy: when , one is just counting the number of quadratic residues and nonresidues modulo (excluding zero), and it is well-known that there are . Thus

The only obstacle lying between this easy result and the Pólya-Vinogradov theorem is a \emph{cutoff} hiding in the sum . To see this more clearly let us write the quantity we are to bound as

The notation means a function which equals on the set and is zero elsewhere. Now we apply the trick under discussion: we claim that the cutoff may be expanded as a Fourier series

where . To see this one uses (discrete) Fourier analysis on . The Fourier coefficients

$\hat{f}(r) = \frac{1}{p}\sum_{x = 0}^{p-1} f(x) e^{-2\pi i xr/p}$

of are just geometric series, the common ratio of series one must sum for being . An easy exercise shows that , where is the size of when reduced to lie between and . The claim now follows from the Fourier inversion formula.

Applying this decomposition of and the triangle inequality, our task now follows if we can establish that [{math]] \sum_{x = 1}^{p-1} e^{2\pi i xr/p} (x | p) = O(\sqrt{p}) [{\math]] for all . We have now \emph{completed} the sum to the whole range , though we have paid the price of introducing the exponential .

It turns out that this is not a huge price to pay: these new sums are \emph{gauss sums} and it is possible to show that they are . We leave the details as an exercise (the solution to which may be found in any number of places).

### Example 2

This trick is ubiquitous in analytic number theory. It is very important, for example, in the treatment of sums using the so-called \emph{method of bilinear forms}; see B.~Green's article Three topics in additive prime number theory, Chapter 2, for more information.

### General discussion

Sometimes this principle is described as all cutoffs are the same, or as completing exponential sums. It can lead to extra logarithmic factors as in the Pólya-Vinogradov inequality mentioned above; in this example reducing the size of these factors is a major unsolved problem.