A Sum Of Cosines = Mobius Function
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From: Leroy Quet on 5 Jul 2010 10:09 This is in all probability a well-known result, since it was very easily found. But in case it is not well known, I post it here to help publicize it, since the result certainly deserves to be well-known. Let m be any positive integer. Then: sum{1<=k<=m, GCD(m,k)=1} cos(2pi*k/m) = mu(m), where the sum is over all those positive integers k coprime to m and <= m, and where mu(m) is the Mobius function of m. (mu(m) = 0 if m is not squarefree; mu(m) = (-1)^(number primes dividing m) if m is squarefree.) This result can be generalized slightly. Let n be any positive integer, given m, where n divides m and where GCD(n,m/n) = 1. Then: sum{1<=k<=m, GCD(m,k)=1} cos(2pi*k/n) = mu(n)*phi(m/n), where phi(k) is the number of positive integers <= k and coprime to k. Sorry about the simplicity of this result, but I am wasting away mathematically now that I have gotten older, as many of us do. Thanks, Leroy Quet
From: George Jefferson on 5 Jul 2010 13:45 "Leroy Quet" <qqquet(a)mindspring.com> wrote in message news:bfc0fdb8-6194-42fb-a9a3-84f6e9a31244(a)c10g2000yqi.googlegroups.com... > This is in all probability a well-known result, since it was very > easily found. But in case it is not well known, I post it here to help > publicize it, since the result certainly deserves to be well-known. > > > Let m be any positive integer. Then: > > sum{1<=k<=m, GCD(m,k)=1} cos(2pi*k/m) = mu(m), > > where the sum is over all those positive integers k coprime to m and > <= m, and where mu(m) is the Mobius function of m. (mu(m) = 0 if m is > not squarefree; mu(m) = (-1)^(number primes dividing m) if m is > squarefree.) > > This result can be generalized slightly. > > Let n be any positive integer, given m, where n divides m and where > GCD(n,m/n) = 1. Then: > > sum{1<=k<=m, GCD(m,k)=1} cos(2pi*k/n) = mu(n)*phi(m/n), > > where phi(k) is the number of positive integers <= k and coprime to k. > > Sorry about the simplicity of this result, but I am wasting away > mathematically now that I have gotten older, as many of us do. > > Thanks, > Leroy Quet > http://en.wikipedia.org/wiki/M%C3%B6bius_function under properties and applications
From: Leroy Quet on 5 Jul 2010 14:49 George Jefferson wrote: > > http://en.wikipedia.org/wiki/M%C3%B6bius_function > > under properties and applications Ah, I figured the result must have been well know. I am just glad I didn't make a mistake in figuring it out! Because I have been rather error-prone lately. Thanks, Leroy Quet
From: rancid moth on 6 Jul 2010 18:03 "Leroy Quet" <qqquet(a)mindspring.com> wrote in message news:16231fe7-4aeb-4961-9751-7250cda5991f(a)b35g2000yqi.googlegroups.com... > > George Jefferson wrote: > >> >> http://en.wikipedia.org/wiki/M%C3%B6bius_function >> >> under properties and applications > > > Ah, I figured the result must have been well know. I am just glad I > didn't make a mistake in figuring it out! Because I have been rather > error-prone lately. > > Thanks, > Leroy Quet > here is something quite nice that involves a sum of cosines _and_ a sum of mobius functions and a sum of bernoulli polynomials. sum{n=0..inf} x^(2n)/B_(2n) = 1 + 1/2 sum(k=1,oo) Mu(k)*(1-cos(2*pi*x/k)) I thought that was rather pretty because the sum on the left seems to converge very slowly, while the one on the right does so rather quickly. I worked this out answering this thread... http://groups.google.com.au/group/sci.math/browse_thread/thread/20738a6fa22eaca8/89dc79dc511b8a31?hl=en&lnk=gst&q=why+does+this+function+do+what+it+does#89dc79dc511b8a31
From: OwlHoot on 9 Jul 2010 10:03
On Jul 5, 3:09 pm, Leroy Quet <qqq...(a)mindspring.com> wrote: > This is in all probability a well-known result, since it was very > easily found. But in case it is not well known, I post it here to help > publicize it, since the result certainly deserves to be well-known. > > Let m be any positive integer. Then: > > sum{1<=k<=m, GCD(m,k)=1} cos(2pi*k/m) = mu(m), > > where the sum is over all those positive integers k coprime to m and > <= m, and where mu(m) is the Mobius function of m. (mu(m) = 0 if m is > not squarefree; mu(m) = (-1)^(number primes dividing m) if m is > squarefree.) > > This result can be generalized slightly. > > Let n be any positive integer, given m, where n divides m and where > GCD(n,m/n) = 1. Then: > > sum{1<=k<=m, GCD(m,k)=1} cos(2pi*k/n) = mu(n)*phi(m/n), > > where phi(k) is the number of positive integers <= k and coprime to k. > > Sorry about the simplicity of this result, but I am wasting away > mathematically now that I have gotten older, as many of us do. > > Thanks, > Leroy Quet Hey Leroy, check out this paper: http://www.man.poznan.pl/cmst/2008/v_14_1/cmst_47-54a.pdf In particular, look at that graph of Fig 2 on Page 5 of 8. If you can prove that it stays an obvious cosine function, as the associated formulae suggest, with non-increasing amplitude and no new trends creeping for larger n to throw it out, then you'll have established quite a significant result (and won a million smackers into the bargain ;-) With your expertise and practice in summing recondite series, you have a better chance than most, even most professionals! Good luck! Cheers John Ramsden ( jhnrmsdn(a)yahooo.co.uk ) remove letter to reply |