Representation of 0 as $\sum^{N}_{k=-N} \epsilon_k k$

Representation of 0 as $\sum^{N}_{k=-N} \epsilon_k k$

Citation for published version (APA):

van Lint, J. H. (1967). Representation of 0 as $\sum^{N}_{k=-N} \epsilon_k k$. Proceedings of the American Mathematical Society, 18(1), 182-184. https://doi.org/10.1090/S0002-9939-1967-0205964-8

DOI:

10.1090/S0002-9939-1967-0205964-8 Document status and date:

Published: 01/01/1967

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182 J. H. VAN LINT [February 6. D. Mumford, The topology of normal singularities of an algebraic surface and a cri- terion for simplicity, Inst. Hautes Etudes Sci. Publ. Math. No. 9, Paris, 1961.

7. H. Seifert and W. Threlfall, Lehrbuch der Topologie, Chelsea, New York, 1947. 8. N. Steenrod, The topology of fibre bundles, Princeton Univ. Press, Princeton,

N. J., 1951.

9. B. L. van der Waerden, Topologische Begriindung des Kalkuls der abzdhlenden Geometrie, Math. Ann. 102 (1929), 337-362.

PRINCETON UNIVERSITY AND BRANDEIS UNIVERSITY

REPRESENTATION OF 0 AS Ek= -N Ekk

J. H. VAN LINT

Abstract. If Ec are independent identically distributed random vari- ables with values 0 and 1, each with probability 2 then

P(

E

k=--N 7r

1. Introduction. Recently P. Erdos asked the following question (oral communication'). If A(N) denotes the number of representa- tions of 0 in the form Zk=-NEkk, where Ek=O or 1 for -N<k?N then determine the asymptotic behavior of A (N). We shall prove that

3 1/2

(1) ~~~~~A(Nf t) 22Nf+11V-3/2.

Another way of formulating this result is the following. Let Ek be

independent random variables identically distributed with values 0

and 1, each with probability 2. Then

(2) ( P

E

k=-N 7r

The referee has pointed out that (2) can be expected from the Linde- berg theorem as follows. If N is large

EkI=

mally distributed with mean zero and variance N(N+1)(2N+1)/12. The right hand side of (2) is the probability density at the origin of

Received by the editors June 27, 1966. 'See also [I].

i967] REPRESENTATION OF 0 AS FK_Ek 183

the normal distribution with the same mean and variance.

2. Proof of (1). The number A(N) is the constant term in the ex- pansion of In--N (1 +x1). Hence

1 N _dz 22N+2 /2N

(3) A(N) =

-

II

J R

2ri c k=-N z i k=l

(here C is the unit circle in the complex plane).

Now first we estimate the integrand for values of x not near to the origin.

N N N

II _Cos2kx = 11 _(1- sin2 kx) < exp -E sin2kx

k=1 k1=1 F N sin Nx cos (N + l)x- (4) = exp[--+2 2 sin x 7r 7r = O(e-NI14) for -?< x <2- 2N

-2

Next we remark that cos2 x<e-x2 for 0 x _?r/2. Therefore

r

fiI2N

1ff

[-

O k==l O k=1 (5)

1)f(2N

N(N

+

+ 1)

exp

L-

-d6x

0f3

Now for O<x<N-413 we have

N N N

II COS2 kx - T e-k2 2 II { 1 + O(k4x4) I k=1 k=1 k-1 N ][I e-k2X2 exp{ O(N5N-1613) k=l exp

{-

2

Using the fact that

x

N-43 N()N3 )

184 J. H. VAN LINT

and the result (5) we find

1

r/2N N 1

(6) T C cos2 kxdx (3)1/21f-8 /2

O kl 2

Now from (6), (4) and (3) we find / 1/2

A(N) (+)"22N+1N-8I2.

This completes the proof.

3. Acknowledgment. The author is indebted to B. F. Logan for a valuable suggestion which led to (4).

REFERENCE

1. A. SErkdzy and E. Szemer6di, Uber ein Problem von Erdes und Moser, Acta. Arith. 11 (1965), 205-208.