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- Newsgroups: sci.math
- Path: sparky!uunet!secapl!Cookie!frank
- From: frank@Cookie.secapl.com (Frank Adams)
- Subject: Re: An Interesting Function
- Message-ID: <1992Nov12.195839.76492@Cookie.secapl.com>
- Date: Thu, 12 Nov 1992 19:58:39 GMT
- References: <1992Oct29.001255.35683@Cookie.secapl.com> <BxKEDt.sL@unx.sas.com>
- Organization: Security APL, Inc.
- Lines: 27
-
- In article <BxKEDt.sL@unx.sas.com> sasrdt@shewhart.unx.sas.com (Randall D. Tobias) writes:
- >In article <1992Oct29.001255.35683@Cookie.secapl.com>, frank@Cookie.secapl.com (Frank Adams) writes:
- >|> Consider the function W(n) = sum(0<i<n,mod(n,i)). The problem is to find an
- >|> asymptotic expression for W(n).
- >|>
- >|> It is pretty clear that W(n) = O(n^2). In fact, we can write
- >|> W(n)=c n^2 + o(n^2). The question is, what is c? I know the answer, and
- >|> will post it later if no one gets it.
- >|>
- >|> I think in fact W(n) = c n^2 + O(n), but I can't prove it. It seems that
- >|> W(n) /= c n^2 + d n + o(n); the O(n) stuff "wanders around".
- >
- >Well?
-
- The key is to look at the summation from the "back end". Between n/2 and
- n, the values go in arithmetic sequence from just under n/2 to 0. Between
- n/3 and n/2, they go from n/3 to 0. This works well enough back to
- sqrt(n); and the terms up to sqrt(n) sum to at most O(n). So
-
- W(n) ~ SUM((k,1,sqrt(n)),(n/k - n/(k+1))*n/(2(k+1)))
- ~ n^2 * (SUM(k,1/(k*(k+1)))-SUM(k>0,1/(k+1)^2))/2
- = n^2 * (1 - (pi^2/6 - 1))/2
- = (1 - pi^2/12) * n^2
-
- So c = pi^2/12.
-
- (A more precise analysis is required to get an error estimate.)
-