Exponential sum

A finite sum of complex exponentials, such as $S(\theta )={\sum }_n{a}_{n}e(n\theta )$ with $e(t)=\exp (2\pi it)$, is an exponential sum. Cancellation in an exponential sum between its terms can make it much smaller than the sum of their absolute values.

Taking a finite difference lowers the degree of a polynomial phase. For $S(\theta )={\sum }_{x=1}^{n}e(\theta P(x))$, expand $|S{|}^2$ by shifts $h$ and apply the Cauchy-Schwarz inequality to those shift sums. This gives $|S{|}^4\le 2n{\sum }_{h,l}{\sum }_{x\in I_{h,l}}e(\theta {\Delta }_h{\Delta }_{l}P(x))$, where $I_{h,l}$ enforces that $x,x+h,x+l,x+h+l$ lie in $[1,n]$. Although individual terms can be complex, the entire sum is real and nonnegative because it equals a sum of squares. For $P(x)=x^3$, the phase is $3hl(2x+h+l)$.

For a real degree-$k$ polynomial with leading coefficient $\theta$, $k\ge 2$, and coprime integers $a,q$ satisfying $|\theta -a/q|\le q^{-2}$, the displayed bound applies to its exponential sum over $1\le n\le N$. Repeated Weyl differencing reduces the phase to slope $k!\theta h_1\cdots h_{k-1}$. A geometric-sum estimate, the subpower bound for the divisor function and the reciprocal fractional-part sum near a rational complete the estimate. The arbitrary positive exponent $\eta$ absorbs divisor bounds and logarithms.

If $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$, the displayed estimate holds. Twice applying Weyl differencing bounds $|S{|}^4$ by $O(n^3+n{\sum }_{0<|h|,|l|<n}\min (n,\Vert 6\theta hl{\Vert }^{-1}))$. Group by $6hl$ and use the subpower bound for the divisor function. The reciprocal fractional-part sum near a rational bounds the resulting sum by $O(\log (2q)(n^3/q+n^2+n+q))$. Take fourth roots, absorbing logarithms into $n^{\eta }$ when $q\le n^3$; for larger $q$ the trivial bound suffices. Leiden lecture notes, Section 7.2 develop this cubic differencing estimate.

For $1\le j\le k$ and every $\eta >0$, the displayed estimate bounds a mean value of a polynomial exponential sum. For cubes its eighth-moment bound is $O_{\eta }(n^{5+\eta })$. It complements pointwise Weyl differencing estimates: a ninth moment over a region is bounded by its supremum there times this eighth moment. Leiden lecture notes, Section 8.1 give the cubic formulation.

Write $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $I_j={\int }_0^1|S{|}^j$. Character orthogonality gives $I_2=n$. For $I_4$, fix the difference $D=x^3-y^3$. If $D=0$, there are $n^2$ choices for the two equal pairs. If $D\ne 0$, any solution $z^3-w^3=D$ has $z-w\mid D$; fixing this divisor gives a quadratic equation for $w$, hence at most two choices. The subpower bound for the divisor function yields $I_4{\ll }_{\eta }{n}^{2+\eta }$.

The twice-differenced inequality in Weyl differencing reads $|S{|}^4\le 2n{\sum }_v{c}_{v}e(\theta v)$, with $c_v$ counting triples $(h,l,x)$ for which $v=3hl(2x+h+l)$ and all four shifted arguments are in $[1,n]$. Thus $c_0\ll n^2$, and $c_v{\ll }_{\eta }{n}^{\eta }$ for $v\ne 0$: the product $hl$ divides $v$, and the remaining equation determines $x$. The Fourier coefficients $r(v)$ of $|S{|}^4$ are nonnegative counts, with $r(0)=I_4$ and ${\sum }_{v}r(v)=n^4$. Multiply the inequality by $|S{|}^4$ and integrate. It follows that $I_8\ll n(n^2{I}_4+n^{\eta }{n}^4){\ll }_{\eta }{n}^{5+\eta }$, after choosing the divisor-bound exponents small enough.

For $a\ne 0$ in the finite field ${\mathbb{F}}_p$, put $d=\gcd (k,p-1)$. The multiplicative group of a finite field is cyclic, and character orthogonality expresses the number of $k$th roots of $y\ne 0$ as ${\sum }_{{\chi }^d=1}\chi (y)$. The trivial character contributes $-1$, cancelling the $x=0$ term. Each of the remaining $d-1$ Gauss sums of Dirichlet characters has modulus $\sqrt{p}$, by the prime-field character Gauss sum identity. Consequently $|G_{a,p}|\le (d-1)/\sqrt{p}\le k/\sqrt{p}$.

For $S(a)={\sum }_{u}e(a{u}^k/p)$, character orthogonality counts representations of $x$ by $p^{-1}{\sum }_{a}S(a{)}^{3}e(-ax/p)$. Its zero-frequency term is $p^2$. The power Gauss sum over a prime field bounds the other terms in total by $k^3{p}^{3/2}$, strictly less than $p^2$ when $p>k^6$. Every residue class therefore has a representation.

A quadratic exponential sum has a polynomial phase of degree two. Its multiplicative derivative at lag $h$ has the linear phase $2\alpha hn$ up to a constant factor. This degree reduction lets the Van der Corput inequality for finite scalar sequences reduce its size to estimates for finite geometric series.

For coprime integers $a,q$ with $1\le q\le Q$ and $|\alpha -a/q|\le 1/(qQ)$, expand the squared quadratic exponential sum by its difference $h$. The exponential geometric sum bound gives $|S{|}^2\ll N+{\sum }_{h\le N}\min (N+1,\Vert 2\alpha h{\Vert }^{-1})$. In each block of length at most $q/4$, the rotations $2\alpha h$ are separated by at least $1/(2q)$: the reduced denominator of $2a/q$ is at least $q/2$, and the approximation error between two block points is at most $1/(2q)$. The separated reciprocal-distance sum bounds each block by $O(N+q\log (2N))$. There are $O(N/q+1)$ blocks, giving the displayed inequality. For bounded $q$ the trivial estimate supplies the same conclusion.

By orthogonality of integer Fourier modes, this integral counts pairs of $k$-tuples with the same first $r$ power sums. Diagonal pairs give $J_{k,r}(Z)\gg Z^k$. There are $O_{k,r}(Z^{r(r+1)/2})$ possible moment vectors; the Cauchy-Schwarz inequality gives $J_{k,r}(Z)\gg Z^{2k-r(r+1)/2}$. Upper bounds measure how much arithmetic coincidence remains beyond these necessary contributions, and enter the Vinogradov mean-value method for a bilinear exponential sum.

Expand moments of the inner sum and group equal power-sum differences. Their multiplicities are bounded by $J_{k,r}(Z)$ through the Cauchy-Schwarz inequality. Two applications of the Holder inequality yield a bound for $|U{|}^{4k^2}$ containing $Z^{8k^2-4k}{J}_{k,r}(Z{)}^2$ and a product of short geometric-sum bounds over the moment differences. Thus a sharp mean-value estimate combines with rational approximation or spacing of the coefficients to prove cancellation. When all coefficients are integers, $U=\lfloor Z{\rfloor }^2$, so mean-value estimates alone cannot force cancellation.

Translating an integer interval by $xy$ changes a sum of unit-modulus terms by at most $2xy$. Averaging gives the displayed formula. For $n\asymp N$, $Z=N^{2/5}$, expand $-t\log (1+xy/n)$ to degree $r=\lfloor 5.01\log t/\log N\rfloor$. Its remainder is at most $t{N}^{-(r+1)/5}<t^{-1/500}$. This reduces the logarithmic exponential sum to bilinear polynomial sums while keeping explicit boundary and approximation errors.

For a real $C^1$ phase with continuous monotone derivative and $|f^{\prime }|\le \delta <1$, the sum-integral discrepancy has the displayed uniform bound. Periodization and the Dirichlet-Jordan convergence theorem express it as symmetric nonzero Fourier modes. Integration by parts gives endpoint terms and reciprocal-derivative variations. Monotonicity bounds their total variation by ${\sum }_{h\ne 0}2\delta /(h^2-{\delta }^2)=O((1-\delta {)}^{-1})$. Pairing the leading $1/h$ endpoint terms reduces them to the uniformly bounded sine series. The exclusion of integer nonzero frequencies is essential.

Cancellation occurs when different complex numbers in an exponential sum have directions that reduce the absolute value of their sum. For example, the orthogonality of roots of unity makes ${\sum }_{a=0}^{p-1}e(a/p)=0$ for every integer $p>1$, despite all terms having absolute value one.