UDC 511.9
MATHEMATICS
Submitted 1966-01-01 | RussiaRxiv: ru-196601.82053 | Translated from Russian

Full Text

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,

\[ S'=p^{m-t_1-1}\delta_n(t-1)\sum_{x=1}^{p-1}\exp 2\pi i\,\frac{a x^n}{p} +p^{m-t_1-1+\delta_n(t)}+O\left(P^{1-\gamma/r^2}\right). \]

  1. 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

  1. A. A. Karatsuba, Izv. Akad. Nauk SSSR, Ser. Mat., 28, 1, 237 (1964).
  2. A. A. Karatsuba, Vestn. Mosk. Univ., Ser. 1, 1, 38 (1962).

Submission history

UDC 511.9