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UDC 511.9
MATHEMATICS
A. A. KARATSUBA
ASYMPTOTIC FORMULAS
FOR A CERTAIN CLASS OF TRIGONOMETRIC SUMS
(Presented by Academician I. M. Vinogradov on 22 X 1965)
In this paper an asymptotic formula is obtained for a certain class of trigonometric sums, making it possible to judge the behavior of the modulus of such sums as the interval of summation changes.
We shall use the following notation: \(n, m, t, P, a\) are integers, \(n \geqslant 20\); \(r\) is a real number, \(1 \leqslant r \leqslant 0.1n\); \(t \geqslant n\); \(p\) is a prime number, \((a,p)=1\); \(P \gg 1\);
\[ \delta_n(x)= \begin{cases} 1, & \text{if } x \equiv 0 \pmod n,\\ 0, & \text{if } x \not\equiv 0 \pmod n. \end{cases} \]
Theorem 1. Let \(q=p^t,\ P=q^{1/r}\). If
\[
P=a_0p^s+a_1p^{s-1}+\cdots+a_{s-1}p+a_s,\quad
1\leqslant a_0\leqslant p-1,\quad 0\leqslant a_\nu\leqslant p-1,\quad \nu\geqslant 1,
\]
is the \(p\)-adic expansion of the number \(P\), then the asymptotic formula
\[
S=\sum_{x=1}^{P}\exp 2\pi i\,\frac{ax^n}{q}
=Aa_0p^{s-\alpha}+a_0p^{s-\alpha+\beta}+O\!\left(P^{1-\gamma/r^2}\right),
\]
holds, where the quantities \(A,\alpha,\beta\) are determined by the equalities
\[
A=\delta_n(t-1)\sum_{x=1}^{p-1}\exp 2\pi i\,\frac{ax^n}{p},
\qquad
\alpha=\left[\frac{t}{n}\right]+1,
\qquad
\beta=\delta_n(t).
\]
The formula has a particularly simple form when \(P=p^s\) and \(t\equiv 0\pmod n\). In this case we obtain:
\[
a_0=1,\quad a_\nu=0,\quad \nu=1,2,\ldots,s;\qquad
\delta_n(t-1)=0,\quad A=0,\quad \delta_n(t)=1;
\]
\[
P^r=p^{sr}=p^t,\qquad s=t/r;\qquad t/ns=r/n;
\]
\[
S=P^{1-r/n}+O\!\left(P^{1-\gamma/r^2}\right).
\]
For the proof of the theorem we shall need the following lemma.
Lemma. Let \(q=p^m,\ m\geqslant n+1,\ (a_1,p)=\cdots=(a_{n+1},p)=1\);
\[
P^r=q,\qquad 1\leqslant r\leqslant 0.5n;\qquad
f(x)=a_1x+a_2px^2+\cdots+a_{n+1}p^nx^{n+1}.
\]
Consider the trigonometric sum
\[
S_L=\sum_{x=1}^{P}\exp 2\pi i\,\frac{f(x)}{q}.
\]
Then the estimate
\[
|S_L|\leqslant CP^{1-\gamma_1/r^2},
\]
holds, where \(C\) and \(\gamma_1>0\) are absolute constants.
The proof of this assertion is contained in (1), p. 239.
Proof of Theorem 1. Consider the sum
\[
S'=\sum_{x=1}^{p^m}\exp 2\pi i\,\frac{a(x+bp^m)^n}{p^t},
\tag{1}
\]
where \((a,p)=1,\ r=t/m,\ 1\le r\le 1/4n\), and obtain an asymptotic formula for it.
Let \(t=t_1n+t_2,\ 0\le t_2\le n-1\). We split the sum (1) into \(t_1-\delta_n(t)+1\) sums, collecting together the terms in \(x\) not divisible by \(p\), divisible by \(p\) but not by \(p^2\), divisible by \(p^2\) but not by \(p^3\), and so on. According to this decomposition we obtain the equality
\[ S'=p^{m-t_1+\delta_n(t)-1}+\sum_{\nu=0}^{t_1-\delta_n(t)} S'_\nu, \]
where
\[ S'_\nu= \sum_{\substack{x=1\\(x,p)=1}}^{p^{m-\nu}} \exp 2\pi i\, \frac{a(x+bp^{m-\nu})^n}{p^{t-\nu n}}, \qquad \nu=0,1,\ldots,t_1-\delta_n(t). \]
We split the sum over \(\nu\) into two:
a) \(\quad 0\le \nu\le \nu_0'=\left[\dfrac{t}{n-1}\left(1-\dfrac1r\right)\right]\);
b) \(\quad \nu_0'+1\le \nu\le t_1-\delta_n(t)\).
Consider case a). In the sum \(S'_\nu\) we make a linear change of the summation variable of the form
\[ x=py+z,\qquad 1\le z\le p-1,\qquad 0\le y\le p^{m-\nu-1}-1. \]
Then we have the equality
\[
x^n=(py+z+bp^{m-\nu})^n=a_0+a_1py+\cdots+a_np^ny^n,
\qquad (a_\nu,p)=1,
\]
\[
\nu=0,1,2,\ldots,n.
\]
For the modulus of the sum \(S'_\nu\) we obtain the estimate:
\[ |S'_\nu|\le p \left| \sum_{y=0}^{p^{m-\nu-1}-1} \exp 2\pi i\, \frac{a_1y+a_2py^2+\cdots+a_np^{n-1}y^n}{p^{t-\nu n-1}} \right|. \]
Defining the quantity \(r_\nu\) from the equality
\[ (p^{m-\nu-1})^{r_\nu}=p^{t-\nu n-1}, \]
we obtain
\[ r_\nu=\frac{t-\nu n-1}{m-\nu-1} = r\,\frac{t-\nu n-1}{t-\nu r-r}, \qquad 0\le \nu\le \nu_0'. \]
It is easy to show that \(1\le r_\nu\le 2r,\ \nu=0,1,\ldots,\nu_0'\). Applying the estimate of the lemma, after simple computations we arrive at the estimate
\[ |S'_\nu|=O\left(P^{1-\gamma/r^2}\right). \]
Consider case b). We have
\[ S'_\nu= \sum_{\substack{x=1\\(x,p)=1}}^{p^{m-\nu}} \exp 2\pi i\, \frac{a(x+bp^{m-\nu})^n}{p^{t-\nu n}}, \qquad \nu_0'+1\le \nu\le t_1-\delta_n(t). \]
For the selected values of \(\nu\) the inequality holds:
\[ m-\nu\ge t-\nu n,\qquad \nu=\nu_0'+1,\ldots,t_1-\delta_n(t). \]
Therefore, using the property of a complete trigonometric sum, we obtain
\[ S'_\nu=p^{m-\nu-t+\nu n} \sum_{\substack{x=1\\(x,p)=1}}^{p^{t-\nu n}} \exp 2\pi i\,\frac{ax^n}{p^{t-\nu n}}, \qquad \nu_0'+1\le \nu\le t_1-\delta_n(t). \]
If \(\delta_n(t)=0\), then
\[ l=t-\nu n\ge t-t_1n=t_2\ge 1. \]
If \(t_2 \ge 2\), then \(l \ge 2\) and
\[ \sum_{\substack{x=1\\(x,p)=1}}^{p^l}\exp 2\pi i\,\frac{a x^n}{p^l}=0. \]
If, however, \(t_2=1\), then \(S_\nu'=0\) for all \(\nu=\nu_0'+1,\ldots,t_1-1\), and
\[ S_{t_1}'=p^{m-t_1-1}\sum_{x=1}^{p-1}\exp 2\pi i\,\frac{a x^n}{p}. \]
If \(\delta_n(t)=1\), then \(t_2=0\),
\[ t=nt_1,\qquad \nu_0'+1\le \nu\le t_1-1;\qquad t-\nu n\ge n\ge 2; \]
\[ S_\nu'=0,\qquad \nu=\nu_0'+1,\ldots,t_1-1. \]
Consequently,
\[ S'=p^{m-t_1-1}\delta_n(t-1)\sum_{x=1}^{p-1}\exp 2\pi i\,\frac{a x^n}{p} +p^{m-t_1-1+\delta_n(t)}+O\left(P^{1-\gamma/r^2}\right). \]
- The equality is obvious
\[ S=\sum_{\nu=0}^{s-1}\sum_{m=1}^{a_\nu p-a_\nu+1} \left( \sum_{x=1}^{p^{s-\nu}-1}\exp 2\pi i\, \frac{a\left(x+b_m p^{s-\nu-1}\right)^n}{p^t} \right)+O(p), \]
where \(b_m\) are certain integers.
For \(\nu=0,1,\ldots,\nu_0=[s-1-4t/n]\), for each of the inner sums there holds the asymptotic formula which we derived above. If \(\nu>\nu_0\), then, estimating each of the remaining sums trivially, we obtain a remainder term of order \(O(P^{1-\gamma/r^2})\). Thus the theorem is completely proved.
Corollary. For the sum \(S\) the inequalities hold
\[ S\ll \begin{cases} P^{1-\gamma r/n}, & \text{if } 1\le r\le \sqrt[3]{n},\\ P^{1-\gamma/r^2}, & \text{if } \sqrt[3]{n}\le r\le 0.1n. \end{cases} \]
The proof follows from Theorem 1 and simple calculations. Applying the estimates obtained, we get the assertion (see \((2)\)).
Theorem 2. Consider the congruence
\[ x_1^n+\ldots+x_k^n\equiv y_1^n+\ldots+y_k^n \pmod {p^t}, \]
\[ 1\le x_\nu,y_\nu\le P,\qquad \nu=1,2,\ldots,k. \]
For the number \(I\) of solutions of this congruence, the following asymptotic formula holds for \(k\ge c_1 n\):
\[ I=\psi(p,k)P^{2k}p^{-t}+O\left(P^{2k-1}p^{-t}\right), \]
where
\[ \psi(p,k)=\sum_{\nu=0}^{\infty}p^{-2k\nu} \sum_{\substack{a=1\\(a,p)=1}}^{p^\nu} \left| \sum_{x=1}^{p^\nu}\exp 2\pi i\,\frac{a x^n}{p^\nu} \right|^{2k}, \]
and the constant in \(O\) depends only on \(k\) and \(t\).
This result cannot be improved, since it is easy to prove the inequality
\[ I\gg P^{2k}p^{-t}p^{\,n-2k}. \]
Theorem 1 can be extended to a broader class of trigonometric sums.
Moscow State University
named after M. V. Lomonosov
Received
18 X 1965
REFERENCES
- A. A. Karatsuba, Izv. Akad. Nauk SSSR, Ser. Mat., 28, 1, 237 (1964).
- A. A. Karatsuba, Vestn. Mosk. Univ., Ser. 1, 1, 38 (1962).