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

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Gerry Myerson

Posts: 4,919
Registered: 12/6/04
Re: Help with finding zeros
Posted: Sep 13, 2010 6:45 PM
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In article
<793328e3-f7c4-4695-a034-bf08625a306e@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)



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