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- Path: sparky!uunet!ogicse!das-news.harvard.edu!husc-news.harvard.edu!ramanujan!elkies
- From: elkies@ramanujan.harvard.edu (Noam Elkies)
- Newsgroups: sci.math
- Subject: Re: real exponential sums
- Message-ID: <1992Oct13.113705.16330@husc3.harvard.edu>
- Date: 13 Oct 92 15:37:04 GMT
- Article-I.D.: husc3.1992Oct13.113705.16330
- References: <1behulINNm4e@uwm.edu>
- Distribution: world
- Organization: Harvard Math Department
- Lines: 24
- Nntp-Posting-Host: ramanujan.harvard.edu
-
- In article <1behulINNm4e@uwm.edu> litow@csd4.csd.uwm.edu
- (bruce e litow) writes:
- >Let a_k = exp(k/N) where 0 <= k < N and let S1 and S2 be two sums
- >over two disjoint subsets of the a_k. By Lindemann's Thm,
- >S1 - S2 can never vanish but is there a decent lower bound on
- >|S1 - S2| ? In particular, 1/N^O(1) ?
-
- That's too much to hope for. There are 2^N subsums in all, each of
- absolute value <=N. However you put 2^N points in a radius-N circle,
- some pair must be at distance at most O(N/2^(N/2)), which decreases
- faster than 1/N^O(1). [If all distances had been >=2r, you'd have
- a packing of 2^N circles of radius r in a circle of radius N+r, whence
- 2^N*r^2 >= (N+r)^2; now solve for r.] Now eliminate the roots of
- unity occurring in both sums to get two disjoint subsums S1, S2 at
- the same small distance.
-
- This bound can be reduced a bit further (by a factor of sqrt(N)) by
- noting that a positive proportion of the 2^N sums must have absolute
- value O(sqrt(N)) [Central Limit Theorem and all that]. Actually
- finding such S1 and S2 when N is at all large (say N=100) is of
- course a much harder problem...
-
- --Noam D. Elkies (elkies@zariski.harvard.edu)
- Dept. of Mathematics, Harvard University
-