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Topic:
Help with finding zeros
Replies:
7
Last Post:
Sep 14, 2010 3:35 PM




Re: Help with finding zeros
Posted:
Sep 13, 2010 6:45 PM


In article <793328e3f7c44695a034bf08625a306e@q16g2000prf.googlegroups.com>, Robert Adams <robert.adams@analog.com> wrote:
> On Sep 13, 10:01 am, "Rod" <rodrodrod...@hotmail.com> wrote: > > "Greg Neill" <gneil...@MOVEsympatico.ca> wrote in message > > > > news:BUpjo.55715$iw4.28826@unlimited.newshosting.com... > > > > > > > > > > > > > Robert Adams wrote: > > >> I could use a little help trying to solve the following engineering > > >> problem (no, I am not a student looking for homework help!) > > > > >> I need to find the values of w which cause the following summation to > > >> go to zero > > > > >> sum (n = 1 to N) of exp(i*w*ln(n)) > > > > >> where i is the usual sqrt(1) > > > > >> For a given N, there are infintely many solutions due to the > > >> periodicity of exp(i ...) > > >> How do I find these solutions? > > > > >> Thanks for any pointers. > > > > >> Bob Adams > > > > > Have you looked into expanding the exponentials into > > > trig form? You'll then have a sum of cos and > > > imaginary sin terms which will each have to sum to > > > zero separately. > > > > That's my approach as well > > Wouldn't w have to be complex to achieve both series being zero. > > > > Is this not a truncated zeta function? Hide quoted text  > > > >  Show quoted text  > > Yes, this is related to the zeta function. > > I was hoping to to find a solution where w is real, but this may not > be possible as you point out
I think that in the case N = 3 it's easy to prove that w can't be real. In the case N = 2, w = (2 k + 1) pi / log 2 for any integer k.
 Gerry Myerson (gerry@maths.mq.edi.ai) (i > u for email)



