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- Newsgroups: sci.math
- Path: sparky!uunet!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!mp.cs.niu.edu!rusin
- From: rusin@mp.cs.niu.edu (David Rusin)
- Subject: Re: A tarnished Gaussian sum?
- Message-ID: <1992Sep10.193634.18529@mp.cs.niu.edu>
- Organization: Northern Illinois University
- References: <1992Sep10.151807.25693@tdb.uu.se>
- Date: Thu, 10 Sep 1992 19:36:34 GMT
- Lines: 30
-
- In article <1992Sep10.151807.25693@tdb.uu.se> matsa@tdb.uu.se (Mats Andersson) writes:
- >
- >Let p be a prime and w a pth root of unity. Does there always exist a k
- >such that
- >
- > (1-w)(1-w^2)...(1-w^((p-1)/2)) = w^k * Gaussian sum ?
-
- Let Z be the left side. Then if Z* is its complex conjugate,
-
- Z*=(1-w^(-1))...(1-w^((p+1)/2))=(1-w^((p+1)/2)...(1-w^(p-1)),
-
- so Z.Z* is the product of all conjugates of (1-w); that is, you let
- x=1 in the polynomial \Prod (x-w^j) = 1+x+...+x^(p-1): Z.Z*=p.
- Thus certainly Z is \sqrt(p) times a complex number of norm 1.
-
- On the other hand, (1-w^(2j)) = w^j . (w^(-j)-w^j) and similarly
- (1-w^(2j+1))=(w^p-w^(2j+1))=w^p.(1-w^(2j+1-p)) can be factored in the
- form w^something.(w^a-w^(-a)). So if you collect all the factors in
- Z you get a big power of w times a product of terms (w^a-w^(-a)),
- all of which are purely imaginary. So Z is of the form
- (power of w ) . (i ^ (p-1)/2)) . real.
-
- So up to powers of w this product Z is just i^((p-1)/2) . sqrt(p).
-
- dave rusin@math.niu.edu
-
- PS - yes, I think there is a way to decide on the signs in the square root
- but I forget what it is.
-
-
-